Strength of Materials, 1e 9788131759097, 8131759091, 9781299445154, 1299445152, 9788131768549, 8131768546, 9788131798898, 8131798895, 9789332503519, 9332503516, 9789332514829, 9332514828

Table of contents :
Cover......Page 1
Contents......Page 4
Preface......Page 14
About the Author......Page 16
Introduction......Page 18
Tensile and Compressive Stresses......Page 19
Shear Stress and Shear Strain......Page 21
Complementary Shear Stresses......Page 23
Stresses on an Inclined Plane......Page 24
Bars of Varying Cross-Sections......Page 25
Longitudinal Strain, Lateral Strain and Poisson’s Ratio......Page 27
Tapered Bar......Page 29
Tapered Flat......Page 32
Bars Subjected to Various Forces......Page 33
Extension in Bar Due to Self-weight......Page 35
Bar of Uniform Strength......Page 37
Volumetric Stress and Volumetric Strain......Page 39
Statically Indeterminate Problems......Page 40
Strain Energy and Resilience......Page 42
Sudden Load......Page 43
Impact Load......Page 45
Tensile Test on Mild Steel......Page 47
Stress Concentration......Page 50
Factor of Safety......Page 51
Key Points to Remember......Page 69
Multiple Choice Questions......Page 70
Practice Problems......Page 71
Special Problems......Page 76
Answers to Practice Problems......Page 77
Answers to Special Problems......Page 78
Chapter 2: Composite Bars and Temperature Stresses......Page 79
Stresses in a Composite Bar......Page 80
Composite System......Page 82
Bars of Different Lengths......Page 84
Bolt and Tube Assembly......Page 85
Temperature Stresses in a Single Bar......Page 88
Temperature Stresses in a Composite Bar......Page 89
Review Questions......Page 107
Practice Problems......Page 108
Special Problems......Page 111
Answers to Multiple Choice Questions......Page 112
Answers to Special Problems......Page 113
Introduction......Page 114
Stresses on an Inclined Plane (I)......Page 115
Stresses on Inclined Plane (II)......Page 117
Principal Stresses......Page 120
Graphical Solution......Page 123
Ellipse of Stresses......Page 130
Strain Components......Page 132
Strain Components on an Inclined Plane......Page 133
Mohr’s Strain Circle......Page 137
Strain on Inclined Plane......Page 139
Three-Dimensional Stresses......Page 140
Principal Strains in Terms of Principal Stresses......Page 144
Rosettes......Page 146
Key Points to Remember Syste......Page 159
Review Questions......Page 160
Practice Problems......Page 161
Special Problems......Page 163
Answers to Special Problems......Page 164
Young’s Modulus and Poisson’s Ratio......Page 165
Determination of Modulus of Rigidity......Page 167
Relation Between Young’s Modulus and Modulus of Rigidity......Page 169
Relation Between Young’s Modulus and Bulk Modulus......Page 171
Multiple Choice Questions......Page 179
Practice Problems......Page 180
Answers to Special Problems......Page 181
Introduction......Page 182
Thin Cylinder Subjected to Internal Pressure......Page 183
Thin Spherical Shell......Page 187
Cylindrical Shell with Hemispherical Ends......Page 189
Wire Winding of Thin Cylindrical Shells......Page 191
Pressure Vessel with a Double Curved Wall......Page 194
Conical Water Tank......Page 196
Key Points to Remember......Page 209
Review Questions......Page 210
Practice Problems......Page 211
Special Problems......Page 213
Answers to Multiple Choice Questions......Page 214
Answers to Special Problems......Page 215
Introduction......Page 216
Lame’s Equations......Page 217
Thick Cylindrical Shell Subjected to External Pressure......Page 222
Compound Cylinder......Page 225
Shrinkage Allowance......Page 229
Hub and Shaft Assembly......Page 231
Thick Spherical Shell......Page 233
Key Points to Remember......Page 252
Multiple Choice Questions......Page 253
Practice Problems......Page 254
Special Problems......Page 255
Answers to Multiple Choice Questions......Page 256
Answers to Special Problems......Page 257
Introduction......Page 258
Shear Force (Positive and Negative)......Page 259
Bending Moments (Positive and Negative)......Page 263
SF Diagrams of Beams/Cantilevers Carrying Point Loads......Page 266
SF Diagrams of Cantilevers and Beams with UDL......Page 269
SF Diagrams of Beam/Cantilevers with Applied Moment......Page 273
BM Diagrams of Cantilevers/Beams with Point Loads......Page 275
BM Diagrams of Cantilevers/Beams with UDL......Page 279
SF and BM Diagrams of Beam/Cantilever with Load Through a Crank......Page 283
Variable Loading on a Beam......Page 287
Relation Between Rate of Loading, SF and BM in a Beam......Page 290
Multiple Choice Questions......Page 308
Practice Problems......Page 309
Special Problems......Page 311
Answers to Exercises......Page 313
Answers to Special Problems......Page 314
Assumptions in Theory of Simple Bending......Page 315
Theory of Simple Bending......Page 317
Neutral Axis......Page 318
Moment of Resistance......Page 319
Symmetrical I-Section......Page 322
T-Section......Page 324
Channel Section......Page 325
Unequal I-Section......Page 326
Modulus of Rupture......Page 328
Built Up Sections......Page 329
Beams of Uniform Strength......Page 331
Composite Beams......Page 335
Reinforced Cement Concrete Beam......Page 339
RCC Beams (Rectangular Section)......Page 340
Stress Concentration in Bending......Page 343
Review Questions......Page 355
Multiple Choice Questions......Page 356
Practice Problems......Page 357
Special Problems......Page 358
Answers to Special Problems......Page 359
Introduction......Page 360
Any Section (With Variable Breadth)......Page 361
Shear Stress Distribution in a Rectangular Section of a Beam......Page 364
Shear Stress Distribution in a Circular Section of a Beam......Page 366
Curves of Principal Stresses in a Beam......Page 369
Directional Distribution of Shear Stresses......Page 370
Review Questions......Page 383
Practice Problems......Page 384
Special Problems......Page 386
Answers to Special Problems......Page 387
Eccentric Axial Thrust on a Column......Page 388
Load Eccentric to Both Axes (Rectangular Section)......Page 392
Core of Rectangular Section......Page 395
Core of Circular Section......Page 396
Core of Any Section......Page 397
Wind Pressure on Walls......Page 400
Wind Pressure on Chimney Shafts......Page 401
Key Points to Remember......Page 413
Multiple Choice Questions......Page 414
Practice Problems......Page 415
Special Problems......Page 417
Answers to Special Problems......Page 418
Relation Between Bending Moment and Curvature......Page 419
Sign Conventions......Page 421
Simply Supported Beam with a Central Point Load......Page 422
A Beam Carrying UDL with Simply Supported Ends......Page 424
A Cantilever with the Point Load at Free End......Page 427
A Cantilever with a UDL......Page 429
Macaulay’s Method......Page 431
Eccentric Load on a Beam......Page 438
Impact Loading of a Beam......Page 442
Propped Cantilevers......Page 444
Stepped Beam......Page 448
Slope and Deflection by Area Moment Method......Page 450
Conjugate Beam Method......Page 452
Deflection......Page 453
Slope and Deflection of Stepped Beams......Page 455
Key Points to Remember......Page 479
Multiple Choice Questions......Page 480
Practice Problems......Page 481
Answers to Exercises......Page 484
Answers to Special Problems......Page 485
Development of Shear Stress and Angular Twist in a Shaft Due to Twisting Moment......Page 486
Horse-Power Transmitted by a Shaft......Page 491
Shafts of Varying Diameters......Page 492
Stresses in a Shaft Subjected to Twisting Moment......Page 498
Shafts Subjected to T and M......Page 499
Torsional Resilience of a Shaft......Page 501
Stresses Developed in a Key......Page 503
Stress Concentration in Torsional Loading......Page 505
Key Points to Remember......Page 521
Multiple Choice Questions......Page 522
Practice Problems......Page 523
Special Problems......Page 525
Answers to Exercises......Page 526
Answers to Special Problems......Page 527
Introduction......Page 528
Wahl’s Factor......Page 529
Close-coiled Helical Spring Subjected to an Axial Load......Page 530
Closed-coiled Helical Spring Subjected to an Axial Moment......Page 532
Open-coiled Helical Spring......Page 533
Strain Energy Method......Page 535
Open-coiled Helical Spring Subjected to Axial Moment......Page 537
Open-coiled Helical Spring—Stresses Developed in Spring Wire......Page 539
Plane Spiral Spring......Page 541
Conical Spring......Page 543
Leaf Spring......Page 545
Cantilever Leaf Spring (Quarter Elliptic Spring)......Page 548
Key Points to Remember......Page 561
Multiple Choice Questions......Page 562
Practice Problems......Page 563
Answers to Exercises......Page 565
Answers to Special Problems......Page 566
Introduction......Page 567
Euler’s Theory of Buckling......Page 568
Equivalent Length......Page 571
Limitations of Euler’s Theory of Buckling......Page 572
Higher-order Differential Equation For......Page 573
Rankine Gordon Formula......Page 576
Johnson’s Parabolic Formula......Page 578
Eccentric Loading of Columns......Page 581
Professor Perry’s Approximate Formula......Page 583
Long Columns with Eccentricity in Geometry......Page 586
Professor Perry Robertson Formula......Page 590
Loading of Strut with Point Loada......Page 592
Strut with an Uniformly Distributed Lateral Load......Page 595
Energy Approach......Page 599
Key Points to Remember......Page 611
Multiple Choice Questions......Page 613
Practice Problems......Page 614
Answers to Practice Problems......Page 615
Introduction......Page 616
Maximum Principal Stress Theory (Rankine’s Theory)......Page 617
Maximum Shear Stress Theory (Tresca Theory)......Page 619
Maximum Principal Strain Theory (St. Venant’s Theory)......Page 622
Strain Energy Theory (Beltrami, Haigh Theory)......Page 624
Shear Strain Energy or Distortional Strain Energy Theory (Von Mises Theory)......Page 627
Mohr’s Theory of Failure......Page 630
Review Questions......Page 640
Practice Problems......Page 641
Special Problems......Page 642
Answers to Special Problems......Page 643
Castigliano’s First Theorem......Page 644
Strain Energy Due to Axial Force......Page 646
Strain Energy Due to Shear Stress......Page 648
Strain Energy Due to Bending......Page 651
Strain Energy Due to Twisting Moment......Page 654
Maxwell’s Reciprocal Theorem......Page 656
Principle of Virtual Forces Applied to Trusses......Page 659
Practice Problems......Page 674
Answers to Practice Problems......Page 677
Introduction......Page 678
Stresses in a Curved Bar......Page 679
Ah2 for a Rectangular Section......Page 683
Value of h 2 for Sections Made Up of Rectangular Strips......Page 686
Ah2 for a Trapezoidal Section......Page 691
Ah2 for a Circular Section......Page 693
Ring Subjected to a Diametral Load......Page 696
Chain Link Subjected to a Tensile Load......Page 701
Deflection of Curved Bar......Page 708
Deflection of a Chain Link......Page 711
Multiple Choice Questions......Page 720
Practice Problems......Page 721
Answers to Practice Problems......Page 723
Introduction......Page 724
Principal Axes......Page 726
Parallel Axes Theorem for Product of Inertia......Page 728
Determination of Principal Axes......Page 730
Moment of Inertia About Any Axis......Page 733
Stresses Due to Unsymmetrical Bending......Page 736
Deflection of Beams Due to Unsymmetrical Bending......Page 740
Shear Centre......Page 744
Key Points to Remember......Page 764
Multiple Choice Questions......Page 765
Practice Problems......Page 766
Answers to Multiple Choice Questions......Page 768
Answers to Practice Problems......Page 769
Introduction......Page 770
Stress Tensor......Page 771
Stress at a Point......Page 772
Plane Stress Condition......Page 775
Strain Tensor......Page 776
Deformations......Page 778
Generalized Hooke’s Law......Page 782
Elastic Constants K and G......Page 785
Equilibrium Equations......Page 787
Second-degree Polynomial......Page 791
A Beam Subjected to Pure Bending......Page 792
Practice Problems......Page 802
Answers to Multiple Choice Questions......Page 803
Answers to Practice Problems......Page 804
Introduction......Page 805
Materials Behaviour Under Static Tension......Page 806
Non-linear Behaviour......Page 807
Shear Modulus......Page 808
Ductility......Page 809
Behaviour of Materials Under Static Compression......Page 810
Behaviour of Materials Under Bending......Page 811
Mild Steel......Page 813
Residual Stresses......Page 814
Behaviour of Materials Under Torsion......Page 815
Behaviour of Materials Under Impact......Page 816
Hardness......Page 820
Brinell’s Number......Page 821
Rockwell Hardness Test......Page 822
Rebound Hardness......Page 823
Superficial Hardness Test......Page 824
Cyclic Stresses......Page 825
Factors Affecting Fatigue Life......Page 827
Case Hardening......Page 828
Creep......Page 829
Stress Relaxation......Page 830
Review Questions......Page 831
Answers to Multiple Choice Questions......Page 832
Tensile Test on a Specimen Using a Tensometer......Page 833
Shear Punching Test on Sheet Sample......Page 836
Tensile Test on a Sample Using UTM......Page 837
Double Shear Test on a Specimen Using UTM......Page 839
Compression Test on a Cast Iron Specimen Using UTM......Page 840
Deflection Test on a Bar Using UTM......Page 841
Compression Test on Brick......Page 842
Compression Test on a Wooden Sample......Page 844
Hardness Test—Brinell Hardness Number......Page 845
Hardness Test—Rockwell Hardness Number......Page 846
Hardness Test—Vickers Pyramid Number......Page 847
Izod Impact Test......Page 848
Torsion Test......Page 850
Stiffness of a Helical Spring......Page 852
Stiffness of a Leaf Spring......Page 854
Introduction......Page 856
Stainless Steel......Page 857
Cast Iron......Page 859
Aluminium, Magnesium, Titanium and Their Alloys......Page 860
Bronzes......Page 861
Babbits......Page 862
Thermoplastics......Page 863
Elastomers......Page 864
Glasses......Page 865
Polymorph of Carbon......Page 866
Review Questions......Page 867
Multiple Choice Questions......Page 868
Answers to Multiple Choice Questions......Page 869
Index......Page 870

Citation preview

Strength of Materials U. C. Jindal

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DEDICATION To my students

Copyright © 2012 Dorling Kindersley (India) Pvt. Ltd Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material present in this eBook at any time. ISBN 9788131759097 eISBN 97899332501232 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

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Contents

Preface

xiii

About the Author

xv

Chapter 1 Simple Stresses and Strains

1

Introduction 1 Tensile and Compressive Stresses 2 Shear Stress and Shear Strain 4 Complementary Shear Stresses 6 Stresses on an Inclined Plane 7 Bars of Varying Cross-sections 8 Longitudinal Strain, Lateral Strain and Poisson’s Ratio 10 Tapered Bar 12 Tapered Flat 15 Bars Subjected to Various Forces 16 Extension in Bar Due to Self-weight 18 Bar of Uniform Strength 20 Volumetric Stress and Volumetric Strain 22 Statically Indeterminate Problems 23 Strain Energy and Resilience 25 Sudden Load 26 Impact Load 28 Tensile Test on Mild Steel 30 Stress Concentration 33 Factor of Safety 34 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

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iv

Contents

Chapter 2 Composite Bars and Temperature Stresses

62

Introduction 63 Stresses in a Composite Bar 63 Composite System 65 Bars of Different Lengths 67 Bolt and Tube Assembly 68 Temperature Stresses in a Single Bar 71 Temperature Stresses in a Composite Bar 72 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 3 Principal Stresses and Strains

97

Introduction 97 Stresses on an Inclined Plane (I) 98 Stresses on Inclined Plane (II) 100 Principal Stresses 103 Practical Cases of Principal Planes 106 Graphical Solution 106 Ellipse of Stresses 113 Strain Components 115 Strain Components on an Inclined Plane 116 Mohr’s Strain Circle 120 Three-Dimensional Stresses 123 Principal Strains in Terms of Principal Stresses 127 Strain Gauge Rosettes 129 Rosettes 129 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 4 Elastic Constants

148

Introduction 148 Young’s Modulus and Poisson’s Ratio 149 Determination of Modulus of Rigidity 150 Relation Between Young’s Modulus and Modulus of Rigidity 152 Relation Between Young’s Modulus and Bulk Modulus 154 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

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Contents

Chapter 5 Thin Cylindrical and Spherical Shells

v

165

Introduction 165 Thin Cylinder Subjected to Internal Pressure 166 Thin Spherical Shell 170 Cylindrical Shell with Hemispherical Ends 172 Wire Winding of Thin Cylindrical Shells 174 Pressure Vessel with a Double Curved Wall 177 Conical Water Tank 179 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 6 Thick Shells

199

Introduction 199 Lame’s Equations 200 Thick Cylindrical Shell Subjected to External Pressure 205 Compound Cylinder 208 Shrinkage Allowance 212 Hub and Shaft Assembly 214 Thick Spherical Shell 216 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 7 Shear Force and Bending Moment Diagrams

241

Introduction 241 Different Types of Beams 242 Shear Force (Positive and Negative) 242 Bending Moments (Positive and Negative) 246 SF Diagrams of Beams/Cantilevers Carrying Point Loads 249 SF Diagrams of Cantilevers and Beams with UDL 252 SF Diagrams of Beam/Cantilevers with Applied Moment 256 BM Diagrams of Cantilevers/Beams with Point Loads 258 BM Diagrams of Cantilevers/Beams with UDL 262 SF and BM Diagrams of Beam/Cantilever with Load Through a Crank Variable Loading on a Beam 270 Relation Between Rate of Loading, SF and BM in a Beam 273 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

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vi

Contents

Chapter 8

Theory of Simple Bending

298

Introduction 298 Assumptions in Theory of Simple Bending 298 Theory of Simple Bending 300 Neutral Axis 301 Moment of Resistance 302 Perpendicular Axes and Parallel Axes Theorems 305 Symmetrical I-Section 305 T-Section 307 Channel Section 308 Unequal I-Section 309 Modulus of Rupture 311 Built Up Sections 312 Beams of Uniform Strength 314 Composite Beams 318 Reinforced Cement Concrete Beam 322 RCC Beams (Rectangular Section) 323 Stress Concentration in Bending 326 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 9

Shear Stresses in Beams

343

Introduction 343 Shear Stress Distribution 344 Shear Stress Distribution in a Rectangular Section of a Beam 347 Shear Stress Distribution in a Circular Section of a Beam 349 Curves of Principal Stresses in a Beam 352 Directional Distribution of Shear Stresses 353 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 10 Combined Bending and Direct Stresses Introduction 371 Eccentric Axial Thrust on a Column 371 Load Eccentric to Both Axes (Rectangular Section) Core of Rectangular Section 378 Core of Circular Section 379

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Contents

vii

Core of Any Section 380 Wind Pressure on Walls 383 Wind Pressure on Chimney Shafts 384 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 11 Deflection in Beams

402

Introduction 402 Relation Between Bending Moment and Curvature 402 Sign Conventions 404 Simply Supported Beam with a Central Point Load 405 A Beam Carrying UDL with Simply Supported Ends 407 A Cantilever with the Point Load at Free End 410 A Cantilever with a UDL 412 Macaulay’s Method 414 Eccentric Load on a Beam 421 Impact Loading of a Beam 425 Propped Cantilevers 427 Stepped Beam 431 Slope and Deflection by Area Moment Method 433 Conjugate Beam Method 435 Deflection 436 Slope and Deflection of Stepped Beams 438 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 12 Torsion Introduction 469 Development of Shear Stress and Angular Twist in a Shaft Due to Twisting Moment 469 Modulus of Rupture 474 Horse-Power Transmitted by a Shaft 474 Shafts of Varying Diameters 475 Compound Shaft 478 Stresses in a Shaft Subjected to Twisting Moment Shafts Subjected to T and M 482 Torsional Resilience of a Shaft 484 Stresses Developed in a Key 486

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Contents

Stress Concentration in Torsional Loading 488 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 13 Springs

511

Introduction 511 Helical Springs 512 Wahl’s Factor 512 Close-coiled Helical Spring Subjected to an Axial Load 513 Closed-coiled Helical Spring Subjected to an Axial Moment 515 Open-coiled Helical Spring 516 Open-coiled Helical Spring Subjected to Axial Moment 520 Open-coiled Helical Spring—Stresses Developed in Spring Wire 522 Plane Spiral Spring 524 Conical Spring 526 Leaf Spring 528 Cantilever Leaf Spring (Quarter Elliptic Spring) 531 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 14 Struts and Columns

550

Introduction 550 Euler’s Theory of Buckling 551 Equivalent Length 554 Limitations of Euler’s Theory of Buckling 555 Higher-order Differential Equation 556 Rankine Gordon Formula 559 Johnson’s Parabolic Formula 561 Eccentric Loading of Columns 564 Professor Perry’s Approximate Formula 566 Long Columns with Eccentricity in Geometry 569 Professor Perry Robertson Formula 573 Lateral Loading of Strut with Point Load 575 Strut with an Uniformly Distributed Lateral Load 578 Energy Approach 582 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems

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Contents

Chapter 15 Theories of Failure

ix

599

Introduction 599 Maximum Principal Stress Theory (Rankine’s Theory) 600 Maximum Shear Stress Theory (Tresca Theory) 602 Maximum Principal Strain Theory (St. Venant’s Theory) 605 Strain Energy Theory (Beltrami, Haigh Theory) 607 Shear Strain Energy or Distortional Strain Energy Theory (Von Mises Theory) 610 Mohr’s Theory of Failure 613 Key Points to Remember • Review Questions • Multiple Choice Questions • Practice Problems • Special Problems

Chapter 16 Strain Energy Methods

627

Introduction 627 Castigliano’s First Theorem 627 Strain Energy Due to Axial Force 629 Strain Energy Due to Shear Stress 631 Strain Energy Due to Bending 634 Strain Energy Due to Twisting Moment 637 Maxwell’s Reciprocal Theorem 639 Principle of Virtual Forces Applied to Trusses 642 Multiple Choice Questions • Practice Problems

Chapter 17 Bending of Curved Bars Introduction 661 Stresses in a Curved Bar 662 Ah2 for a Rectangular Section 666 Value of h 2 for Sections Made Up of Rectangular Strips Ah2 for a Trapezoidal Section 674 Ah2 for a Circular Section 676 Ring Subjected to a Diametral Load 679 Chain Link Subjected to a Tensile Load 684 Deflection of Curved Bar 691 Deflection of a Chain Link 694 Multiple Choice Questions • Practice Problems

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Contents

Chapter 18 Unsymmetrical Bending and Shear Centre

707

Introduction 707 Principal Axes 709 Parallel Axes Theorem for Product of Inertia 711 Determination of Principal Axes 713 Stresses Due to Unsymmetrical Bending 719 Deflection of Beams Due to Unsymmetrical Bending 723 Shear Centre 727 Key Points to Remember • Multiple Choice Questions • Practice Problems

Chapter 19 Three-Dimensional Stresses

753

Introduction 753 Stress Tensor 754 Stress at a Point 755 Plane Stress Condition 758 Strain Tensor 759 Deformations 761 Generalized Hooke’s Law 765 Elastic Constants K and G 768 Equilibrium Equations 770 Second-degree Polynomial 774 A Beam Subjected to Pure Bending 775 Multiple Choice Questions • Practice Problems

Chapter 20 Mechanical Properties Introduction 788 Materials Behaviour Under Static Tension 789 Behaviour of Materials Under Static Compression Behaviour of Materials Under Bending 794 Behaviour of Materials Under Torsion 798 Behaviour of Materials Under Impact 799 Hardness 803 Fatigue 808 Creep 812 Review Questions • Multiple Choice Questions

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Contents

Chapter 21 Material Testing

816

Tensile Test on a Specimen Using a Tensometer 816 Shear Punching Test on Sheet Sample 819 Tensile Test on a Sample Using UTM 820 Double Shear Test on a Specimen Using UTM 822 Compression Test on a Cast Iron Specimen Using UTM Deflection Test on a Bar Using UTM 824 Compression Test on Brick 825 Compression Test on a Wooden Sample 827 Hardness Test—Brinell Hardness Number 828 Hardness Test—Rockwell Hardness Number 829 Hardness Test—Vickers Pyramid Number 830 Izod Impact Test 831 Charpy Impact Test 833 Torsion Test 833 Stiffness of a Helical Spring 835 Stiffness of a Leaf Spring 837

Chapter 22 Engineering Materials Introduction 839 Carbon Steels 840 Alloy Steels 840 Cast Iron 842 Aluminium, Magnesium, Titanium and Their Alloys Copper and its Alloys 844 Nickel, Cobalt and Their Alloys 845 Babbits 845 Plastics 846 Rubber 847 Ceramic Materials 848 Glasses 848 Polymorph of Carbon 849 Review Questions • Multiple Choice Questions Index

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Preface

Strength of Materials deals with the study of the effect of forces and moments on the deformation of a body. The book has been designed in a simple and lucid manner. All topics follow a simple approach along with numerous solved and unsolved problems to explain the basics. To understand the concepts easily, lot many figures are provided in all the chapters. There are some intricacies in every chapter, therefore problems and exercises are framed in a fashion so that these intricacies are exposed and resolved. After every topic students are tempted to solve an exercise to gain confidence to have learnt the topic. The book contains chapters on advanced topics such as curved bars, unsymmetrical bending, shear center, three-dimensional stresses, theories of failure, mechanical properties, material testing and Airy’s stress functions. These topics are generally included in the second course of Strength of Materials in the syllabus of many universities. Learning about mechanical properties of materials is as important as learning about the effect of application of external forces and moments on a body made from some engineering material. So, two chapters are devoted on various mechanical properties and the tests to verify the mechanical properties of common engineering materials. I have taken care to weed out typographical error and error in calculations, however, I will be happy to receive feedback and suggestions for the improvement of the book. U. C. Jindal

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About the Author

U. C. Jindal is a former Professor and Head of the Department of Mechanical Engineering, Delhi College of Engineering. He completed his M.Tech. from Indian Institute of Technology Kanpur and did his Ph.D. on Experimental Stress Analysis from the University of Delhi. For the last 47 years, Dr Jindal has been involved in teaching, research and development activities in the mechanics group of subjects such as engineering mechanics, strength of materials, machine design, theory of machines and materials science. He is the author of nine books, and has also published numerous research papers in the field of stress analysis, material testing, stress concentrations, adhesives and composite materials in various national and international journals. Dr Jindal was awarded the Toshiba Anand Prize in 1978 for original research in Theory and Practice of Standardization. He is a life member of the Indian Society for Construction Materials and Structures, New Delhi.

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1 Simple Stresses and Strains CHAPTER OBJECTIVES In this chapter, we will learn about:  Various types of (a) stresses as tensile and compressive stresses, positive and negative shear stresses, complementary shear stresses, volumetric stress; (b) strains as longitudinal strain, lateral strain, shear strain; and (c) strain energy absorbed by a body during deformation.



Variation of stress in a stepped bar, uniformly tapered bar and uniformly tapered flat.



Statically indeterminate problems in which an additional equation of deformation is made for solution of unknowns.

Difference between gradual, sudden and impact loads.



Difference between resilience and toughness, i.e., strain energy absorbed by a body within the elastic limit and strain energy absorbed by a body until fracture.



Mild steel being the most commonly used engineering material, tensile test on mild steel to study mechanical properties as ultimate strength, yield strength, ductility, resilience and toughness.



 Elastic constants of a material as Young’s modulus, Bulk modulus, Shear modulus, Poisson’s ratio as deformation in a body depend on these elastic constants. 

Stresses developed in a bar due to its own weight.



Bar of uniform strength in which stress due to self-weight remains constant.

Introduction In strength of materials, we study the effect of forces and moments on the deformation of a body. Figure 1.1(a) shows a bar subjected to an axial compressive force P which is perpendicular to the section of the bar. This is known as direct force producing change in length of the bar. A shear force acting parallel to the top plane of the block is shown in Fig. 1.1(b); bottom plane of block is fixed. This type of force produces a change in the shape of the body, that is, a rectangular plane (abcd) of the body is converted into a parallelogram (a′b′cd). Shear force produces shear strain f in block. Bending moment

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2

Chapter 1

P

P a

b

a

C

b

g

φ Section

Length

M

M d (a) Axial compressive load

P

c

(b) Shear force

(c) Bending moment T

f

T

q

(d) Twisting moment

Figure 1.1 M causes angular rotation g in the bar as shown in Fig. 1.1(c). Then, in Fig. 1.1(d), a twisting moment acts on the bar, causing angular twist in the bar as defined by angle q. Stresses and strains in bars due to direct and shear forces, angular rotation and bending stresses in beams due to bending moment and angular twist and shear stresses in shaft due to twisting moment are studied in detail in various chapters of the book.

Tensile and Compressive Stresses Figure 1.2(a) shows a bar AB of length L and diameter d subjected to an axial tensile force P. Note that a force is applied along the axis of the bar and in a direction perpendicular to all cross-section. The effect of this tensile force (acting in a direction away from the plane) on bar is to extend its length and reduce its diameter as shown in Fig. 1.2(b). Note that the bar is of uniform section and on each cross-section the same force P acts producing the same stress on all the sections. P Axial stress in the bar, σ = Area of cross-section A

A P

d P

d (a)

d L L L > L d < d (b)

Figure 1.2

MTPL0259_Chapter 01.indd 2

B

d

Axial tensile force

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Simple Stresses and Strains

=

Axial strain in the bar,

ε=

P 4P = π 2 πd 2 d 4

3

Strain s

L ′ − L δ L Change in length = = L L Original length

d − d′ δ d Change in diameter =− = d d Original diameter The axial load which produces extension in the length of the bar is e, Strain termed as tensile load and the stress due to tensile load is termed as Hooke’s law tensile stress. As the load on the bar is gradually increased, the stress s and the Figure 1.3 strain e also gradually increase in the bar. The relationship between s and e is linear, that is, s a e (stress is proportional to strain). This relationship between the stress and the strain, that is, s a e, is known as Hooke’s law (given by Professor Hooke) as shown in Fig. 1.3. or s=Ee where E is the constant of proportionality. σ Axial tensile stress E= = ε Axial strain E is also called the Young’s modulus of elasticity (given by Professor Young) of the material. P d Similarly, let us consider a bar subjected to an axial compressive force P, acting along the axis of the bar L but perpendicular to the section of the bar. Due to (a) this compressive force, the axial length of the bar is L reduced from L to L′ while the diameter is increased d from d to d′ (Fig.1.4). P 4P Axial stress in the bar, σ=− =− 2 π 2 πd d L 4 L < L d > d L′ −L δL Axial strain in the bar, ε= =− Compressive force L L Lateral strain in the bar, ε l =

δ L = −ε × L , where dL is change in length Lateral strain in the bar, εl =

(b)

Figure 1.4

d′ − d δ d Change in diameter =+ = . d d Original diameter

Note that if the tensile stress is taken as a positive direct stress, then the compressive stress will be a negative direct stress. Moreover, we have learnt that if the axial strain is positive in the bar, then the lateral strain will be negative or vice versa.

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4

Chapter 1

Example 1.1 A steel bar of a diameter of 20 mm and a length of 400 mm is subjected to a tensile force of 40 kN. Determine (a) the tensile stress and (b) the axial strain developed in the bar if the Young’s modulus of steel E = 200 kN/mm2 (Fig.1.5). Solution Diameter of the bar, Cross-sectional area

d = 20 mm

400 mm

d = 20 mm A = (p /4 ) d 2

Figure 1.5

Example 1.1

π × 20 2 = 100π mm 2 4 P = +40 kN = 40,000 N =

Axial load,

40, 000 = 127.32 N /mm 2 = 127.23 MPa 100π (1 MPa = 1 × 106 N/106 mm2 = 1 N/mm2) Young’s modulus of steel E = 2,00,000 N/mm2 Axial strain in the bar, ε = σ /E (as per Hooke’s law) 127.32 = 0.636 × 10 −3 2, 00, 000 There are no units of strain, as strain is change in length per unit length. Tensile stress in the bar, σ =

Exercise 1.1 A 100-mm-long copper bar is subjected to a compressive force such that the stress developed in the bar is 50 MPa. If the diameter of the bar is 15 mm, what is the axial compressive force? If E for copper is 105 kN/mm2, what is the axial strain in the bar?

Shear Stress and Shear Strain A rectangular block is fixed at the bottom plane and a force F is applied on the top plane as shown in   Fig. 1.6(a). Equal and opposite reaction F develops on the bottom plane. Applied force F and reaction F constitute a couple, tending to rotate the body in a clockwise direction. This type of shear force is a positive shear force and the shear force per unit area of the plane on which it acts is a positive shear stress. Figure 1.6(b) shows another block on which the shear force F applied on the top surface and the reaction F from the bottom surface produce anticlockwise or counter clockwise moment (the arm of the couple is equal to the height of the block). This type of shear force that tries to rotate the body in the anticlockwise (ccw) direction is termed as a negative shear force producing a negative shear stress in the block. The shear angle f shown in the figure is very small for any metal within the elastic limit. AA′ = tan φ AD However, f is very small, much less than 1°; therefore, Shear strain =

tan φ = sin φ = φ Shear angle f can be termed as shear strain because the angle f is very small within the elastic limit of the material.

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Simple Stresses and Strains

5

y y

D D

A

C F

F A

D

C B

f

B

A

+

A

C

B

B

+

f

F

D

C

F

x

(a)

x (b)

Figure 1.6 (a) Positive shear force and (b) negative shear force Shear stress is proportional to shear strain, that is, taf t = Gf, where G is the constant of proportionality. τ Shear stress The shear modulus or the modulus of rigidity, G = = φ Shear strain Example 1.2 A rectangular block of aluminium of size 60 mm × 50 mm × 40 mm is subjected to a shear force of 45 kN on the top surface as shown in Fig. 1.7. What is the shear stress developed in the block? If G for material is 25 kN/mm2, what is the shear strain? The bottom surface of the block is fixed to the ground.

50

mm

45 kN F

60 mm 40 F

Solution F = 45 kN = 45,000 N Area of the top surface, A = 60 × 50 = 3,000 mm2 45, 000 Shear stress, τ= = −15 N /mm 2 3, 000 (A negative shear stress as it tends to rotate the body in anticlockwise direction.) Shear modulus, G = 25,000 N/mm2 τ −15 Shear strain, φ= = G 25, 000 = − 0.6 × 10 −3

Figure 1.7

Example 1.2

50 F 50 50

Figure 1.8

Example 1.2

Exercise 1.2 A 50-mm cubical brass block is subjected to a shear force of 50 kN on the top surface as shown in Fig 1.8. What is the shear stress developed in the block? If G for brass = 38 kN/mm2, what is the shear strain developed in the block?

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6

Chapter 1

Complementary Shear Stresses Whenever any external shear force applied on the body produces a shear stress +t, a shear stress of nature -t is developed in the body on perpendicular planes to maintain equilibrium. Consider a rectangular block of size l × b × h as shown in Fig. 1.9. The bottom face of the block top face,  is fixed to the ground. On the  a shear stress t is applied. Stress force on the top face = τ × b × l . Reaction at the bottom face = τ × b × l . Both the forces constitute a couple of arm h as shown in Fig. 1.9. Moment of the couple, C = t × l × b × h (cw) t × b × h

 t × b × h

t× ×b H

G t

E

h

t

F

h

t

t D

C t

A

C =t××b ×h

b



t× ×b AB = , EA = h, BC = b

B

Figure 1.9 Rectangular block subjected to shear stress To balance this couple, an anticlockwise couple of the same magnitude must act on the body. Say the shear stress developed on vertical faces is τ ′ as shown in the figure. Shear forces on the vertical faces = τ ′ × h × b Arm of couple = l Moment of the couple, C ′ = τ ′ × h × b × l However, couple C′ = C to maintain equilibrium or τ ′lbh = τ lbh G F Negative shear stress τ ′ = positive shear stress t m 6.4 kN 0m 4 τ ′ is the complementary shear stress to applied shear stress D (numerically) t, acting at an angle of 90°. C 40

Example 1.3 A 40-mm cube with its lower face fixed to the ground. A positive shear force of 6.4 kN acts on the top surface of cube. Draw the shear stress distribution on the vertical face of the cube (fig 1.10). Solution Shear force, F = 6.4 kN = 6,400 N Area of the top surface = 40 × 40 = 1,600 mm2

MTPL0259_Chapter 01.indd 6

40 mm

E 6.4 kN

A

Figure 1.10

B

Example 1.3

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Simple Stresses and Strains

τ = +6, 400/1, 600 = + 4 N/mm2 (as shown)

Shear stress,

7

4 N/mm2

Shear stress on the vertical face CBEF = t ′ = - 4 N/mm2 Shear stress distribution on vertical surface is t ′ = - 4 MPa (uniform) Shear stress tends to rotate the body in anticlockwise direction. Figure 1.11 shows the shear stress distribution on face ABCD. Students should learn how to draw complementary shear stress. Many a times in interviews, they fail to explain the concept of complementary shear stresses.

4

4 MPa

4

Figure 1.11

Shear stress distribution

Exercise 1.3 A rectangular block of dimensions 60 mm × 40 mm × 30 mm is fixed to the ground on face of size 40 mm × 30 mm. On the top face, a negative shear stress of -10 N/mm2 is applied. Draw the complementary shear stress distribution on a vertical surface of the block (Figure 1.12).

Stresses on an Inclined Plane Consider a bar of rectangular section bh as shown in Fig. 1.13. A plane abcd is inclined at an angle a with the axis of the bar. Axial load P applied on bar is acting at an angle a on the inclined plane. There are two components of this force P: Normal component, Pn = P sina (a tensile force; the arrow of the force is pointing away from the plane abcd) Tangential component, Pt = P × cosa (a positive shear force, tending to rotate the part I of the bar in the clockwise direction. Take a point A and consider the effect of Pt) Side ab =

−10 40 mm

t  = +10 MPa

t  = +10 MPa

60 mm

−10

Figure 1.12

h sin α

Exercise 1.3

Side ad = b Pn d b II

a P

Pt

h I

A

a

P

c

a b

Figure 1.13

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Forces on inclined plane

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8

Chapter 1

Cross-sectional area of the inclined plane, A = ad × ad =

hb sin α

Normal stress on the plane abcd,

σ=

Pn P sin 2 α = (a tensile stress) A bh

(1.1)

Shear stress on the plane abcd,

τ=

Pt P cos α sin α P sin 2α = =+ 2bh A bh

(1.2)

(a positive shear stress) Example 1.4 Consider Fig. 1.13. Take b = 50 mm, h = 40 mm, P = 40 kN and a = 30°. Determine the normal and shear stresses on the inclined plane. Solution Angle, Force,

a = 30°, cosa = 0.866, sina = 0.5 P = 40 kN = 40,000 N

Normal component,

Pn = Psina = 40,000 × 0.5 = 20,000 N

Tangential component,

Pt = Pcosa = 40,000 × 0.866 = 34,640 N h 40 ab = = = 80 mm sin α 0.5 ad = 50 mm

Side Side

Area of inclined section = 80 × 50 = 4,000 mm2 Normal stress on the inclined plane, σ n =

Shear stress on the inclined plane,

τs =

20, 000 = +5 N /mm 2 4, 000 34, 640 = +8.66 N /mm 2 4, 000

Exercise 1.4 A circular bar of a diameter of 40 mm is subjected to an axial force of 50 kN. A plane is inclined at an angle of 45° to the bar. What are the normal and shear stresses on the plane, if the axial force is compressive? [Hint: Inclined plane is an ellipse with major axis = d/sin a and minor axis = d. Area of ellipse = p /4 × major axis × major axis.]

Bars of Varying Cross-Sections Till now we have considered bars of uniform section throughout. Let us consider a stepped bar of diameters d1, d2 and d3 with axial lengths L1, L2 and L3, respectively, as shown in Fig. 1.14. The bar is subjected to an axial tensile force P as shown. Each section of the bar is subjected to the same force P, but the cross-sectional area of portions I, II and III are different. Minimum axial stress occurs in portion I and maximum axial stress occurs in portion III. Axial stresses in portions I, II and III, respectively, are

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Simple Stresses and Strains

9

I II

III

d1

P

d2

L1

L2

d3

P

L3

Figure 1.14 Bar of varying cross-sections

σI =

4P π d12

σ II =

4P π d22

σ III =

4P π d32

If E is the Young’s modulus of the material of the bar, then the strain in each portion can be obtained by using Hook’s law. Axial strains in 3 portions respectively are

εI =

σI 4P = E π Ed12

ε II =

σ II 4P = E π Ed22

ε III =

σ III 4P = E π Ed32

Total change in length of the bar

δ l = δ lI + δ lII + δ lIII = =

4PL1 4PL 2 4PL 3 + + π Ed12 π Ed22 π Ed32 4 P  L1 L2 L3  + + π E  d12 d 22 d32 

Example 1.5 Consider a bar in three steps, with diameters d1 = 20 mm, d2 = 15 mm and d3 = 10 mm and axial lengths L1 = 50 mm, L2 = 75 mm and L3 = 100 mm as shown in Fig 1.14. The bar is subjected to an axial tensile force of 8 kN. Determine the stress in three portions of the bar and the total change in length of the bar of E = 67,000 N/mm2.

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10

Chapter 1

Solution Stresses are

σI =

4P 4 × 8, 000 = = 25.46 N /mm 2 π d12 π × 20 2

σ II =

4P 4 × 8, 000 = = 45.27 N /mm 2 π d22 π × 152

σ III =

4P 4 × 8, 000 = = 101.86 N /mm 2 π d32 π × 10 2

E = 67,000 N/mm2 Total change in length, δ l =

4 P  L1 L2 L3  4 × 8, 000  50 75 100  + + = + +   π E  d12 d 22 d32  π × 67, 000  202 152 102 

= 0.152 (0.125 + 0.333 + 1) = 0.152 × 1.458 = 0.2216 mm Exercise 1.5 A circular stepped bar of diameters of 16, 12 and 8 mm and axial lengths of 40, 60 and 70 mm, respectively, is subjected to an axial compressive force of 4.8 kN. The bar is made of steel. If E for steel = 208 kN/mm2, determine the axial strains in three portions of the bar and the total change in the length of the bar.

Longitudinal Strain, Lateral Strain and Poisson’s Ratio There is an axial (or longitudinal) strain in the bar due to an external axial load, there is a lateral strain of opposite nature in the bar. This is due to Poisson’s effect. Many students may feel that if the length is increased, the diameter is decreased as the volume may remain constant. However, in reality, there is a change in volume within the elastic limit of the material. Figure 1.15 shows a bar of length L and diameter d subjected to an axial tensile force P. Length is increased to L′ and diameter is decreased to d ′. Note that the diameter of the bar is lateral (perpendicular) to its length. Therefore, the strain or change in diameter per unit diameter is termed as lateral strain. L P

d

d

P

L

Figure 1.15

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Simple Stresses and Strains

Axial (longitudinal) strain in the bar,

εa =

L ′ − L δL = L L

εl =

δd d′ − d =− d d

Since L′ > L, it is a positive strain. Lateral strain bar,

11

Since d′ < d, el is a negative strain. Ratio of lateral strain to longitudinal strain is

εl δd L = a negative ratio =− × d δL εa This is known as Poisson’s ratio, given by Professor Poisson, a French scientist. el = -vea, where v is Poisson’s ratio. Though the ratio is negative, the negative sign is not generally attached with this ratio. 4P σa = Axial stress in the bar, πd 2 σ Axial stress in the bar, ε a = a as per Hooke’s law, E where E is the Young’s modulus of the material. σ Lateral strain ε l = −vε a = −v a , E where v is Poisson’s ratio. For common engineering materials, the values of elastic constants (E and v) are given in Table 1.1. Figure 1.16 shows variation of axial stress sa and axial strain ea for a bar subjected to axial load. O to A is a straight line and only up to A, material obeys Hooke’s law. In strength of materials, all loadings are confined up to point A, such that the stress developed in the body is less than the elastic limit stress, se. If the bar is unloaded within the elastic limit, then the strain developed in the body is fully recovered. However, if the load is removed in the plastic region, then unloading curve is along BC, line BC is parallel to OA and residual strain equal to OC remains in the bar after unloading. Table 1.1

Elastic region se

Axial stress

Young’s modulus × 103 MPa, E Poisson’s ratio, v

MTPL0259_Chapter 01.indd 11

A

sa unloading

Plastic region O

C Axial strain, ea

Figure 1.16

Elastic constants

Material

B

Axial stress versus axial strain graph

Steel

Aluminium

Brass

Cast iron

Glass

Copper

200 – 210

67–70

100 –105

100 –105

–

100 –102

0.3

0.33

0.35

0.25

0.20

0.34

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12

Chapter 1 20 P

P 30

500 mm

Figure 1.17 Example 1.6 A copper bar of a rectangular section of 20 mm × 30 mm and a length of 500 mm is subjected to an axial compressive load of 60 kN. If E = 102 kN/mm2 and v for copper = 0.34, determine the changes in the length and the sides of the bar (Fig. 1.17). Solution Bar dimensions l = 500 mm b = 20 mm h = 30 mm Cross-sectional area = 20 × 30 = 600 mm2 60, 000 Compressive stress, σ a = − = −100 N/mm 2 600 E = 102 kN/mm2 = 102,000 MPa Axial strain,

εa =

σa 100 =− = −0.98 × 10 −3 102, 000 ε

Lateral strain, el = -vea = 0.34 × 0.98 × 10-3 = +0.333 × 10-3 Changes in dimensions, dl = ea × l = -0.98 × 10-3 × 500 = -0.49 mm Change in length, dh = el × h = +0.333 × 10-3 × 30 = +10 × 10-3 mm Changes in sides, db = ea × b = 0.333 × 103 × 20 = 6.66 × 10-3 mm. Exercise 1.6 A steel bar of square section 30 × 30 mm and a length of 600 mm is subjected to an axial tensile force of 135 kN. Determine the changes in dimensions of the bar. If E = 200 kN/mm2, v = 0.3.

Tapered Bar Consider a circular bar uniformly tapered from diameter d1 at one end and gradually increasing to diameter d2 at the other end over an axial length L as shown in Fig. 1.18. Take an elementary disc of diameter dx and length dx at a distance of x from end A. d − d1 = d1 + kx Diameter dx of elementary disc = d x = d1 + 2 L

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Simple Stresses and Strains

d1

P

13

d2

d x

P

A

B x

dx L

Figure 1.18 Circular tapered bar d2 − d1 L π π Ax = (d x ) 2 = (d1 + kx) 2 4 4 k=

where constant Cross-sectional area, Axial stress,

σx =

P 4P = Ax π (d1 + kx) 2

Axial strain,

εx =

σx 4P = E π E (d1 + kx) 2 4 P dx π E (d1 + kx) 2

Change in length over dx,

δ dx =

Total change in length,

δ L = ∫O

L

=

as Therefore, Putting the value of k =

MTPL0259_Chapter 01.indd 13

4 P dx π E (d1 + kx) 2

4 P (d1 + kx) −1 −k πE

L

O

=−

4P  1 1 −   π Ek  d1 + kL d1 

=−

4P  1 1  − π Ek  d 2 d1 

d1 + kL = d1 +

δL =

d2 − d1 × L = d2 L

4 P  1 1  4 P(d 2 − d1 ) − = π Ek  d1 d 2  π Ekd1d 2

d2 − d1 , we get L

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14

Chapter 1

δL = Total change in length = δ L =

4P (d2 − d1 ) L × π E (d1d2 ) (d2 − d1 )

4PL π Ed1d2

Example 1.7 A brass bar uniformly tapered from diameter 20 mm at one end to diameter 10 mm at the other end over an axial length 300 mm is subjected to an axial compressive load of 7.5 kN. If E = 100 kN/mm2 for brass, determine (Fig.1.19): (a) the maximum and minimum axial stresses in bar and (b) the total change in length of the bar. Solution Axial load, P = 7.5 kN. Maximum stress occurs at end A with minimum cross-sectional area: (a) σ max = −

4P 4 × 7.5 × 1, 000 =− = −95.94 N /mm 2 at end A π (10 )2 π × 100

(b) σ min = −

4P 4 × 7.5 × 1, 000 =− = −23.87 N /mm 2 at end B 2 π (20 ) π × 20

δL = − =−

4PL , as the load is compressive π Ed1d2

4 × 7.5 × 1, 000 × 300 = −0.143 mm π × 100 × 1, 000 × 10 × 20

10 mm f

B

A P

P

20 mm f

300 mm

Figure 1.19

Exercise 1.7 A 400-mm-long aluminium bar uniformly tapers from a diameter of 25 mm to a diameter of 15 mm. It is subjected to an axial tensile load such that stress at middle section is 60 MPa. (a) What is the load applied? (b) What is the total change in the length of the bar if E = 67,000 N/mm2? [Hint: At middle, the diameter is (25 + 15) / 2 = 20 mm]

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Simple Stresses and Strains

Tapered Flat

Breadth, bx = b +

t

c

Consider a flat of same thickness t throughout its length, tapering uniformly from a breadth B at one end to a breadth b at the other end over an axial length L. The flat is subjected to an axial force P as shown in Fig. 1.20. Thickness of elementary strip abc is t at a distance of x from end A.

P b

b bx A

B−b x = (b + kx), L

B −b where k = L Cross-sectional area of elementary strip, Ax = bxt

15

B

P

a

x dx

L Tapered flat

Figure 1.20

= (b + kx)t P P = Ax (b + kx)t

Stress at elementary strip,

σx =

Change in length over dx,

δ dx =

Total change in the length of flat,

δL = ∫

Pdx Et (b + kx) L 0

δL = = Putting the value of k,

δL =

Pdx Et (b + kx)

P [ln( B ) − ln(b)] Etk P B × ln    b Etk PL B ln   Et ( B − b )  b 

Example 1.8 A steel flat of an axial length of 600 mm and a uniform thickness of 10 mm has a uniformly tapered width of 40 mm at one end to 20 mm at the other end. If the axial force is P = 18 kN and E = 208 kN/mm2, then what is the change in axial length of flat? Solution Load,

P = 18,000 N

Breadth,

B = 40 mm b = 20 mm

Length, Thickness, Young’s modulus,

MTPL0259_Chapter 01.indd 15

L = 600 mm t = 10 mm E = 208,000 N/mm2

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16

Chapter 1

δL =

Change in length,

=

PL B ln   Et ( B − b )  b  18, 000 × 600  40  ln   208, 000 × 10 × ( 40 − 20 )  20 

= 0.2596 × ln 2.0 = 0.2596 × 0.693 = +0.18 mm Exercise 1.8 An aluminium flat of a thickness of 8 mm and an axial length of 500 mm has a tapered width of 15 mm tapering to 25 mm over the total length. It is subjected to an axial compressive force P, so that the total change in the length of flat does not exceed 0.25 mm. What is the magnitude of P, if E = 67,000 N/mm2 for aluminium?

Bars Subjected to Various Forces A stepped bar ABCD of diameters d1, d2 and d3 and axial lengths L1, L2 and L3 is subjected to axial forces P1, P2, P3 and P4 as shown in Fig. 1.21. If E is the Young’s modulus of the material, let us determine the change in the length of three portions as shown. The bar is in equilibrium. B P1

P3

A

D

d2

P2

d3 P4

d1

L1

L2

A P1

C

L3

B P1

I

B

C P4 + P2

II P3 + P1

C

D III

P2

P2

Figure 1.21

MTPL0259_Chapter 01.indd 16

Stepped bar subjected to various forces

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Simple Stresses and Strains

17

Therefore, for equilibrium forces P1 + P3 = P2 + P4. Let us consider three portions independently (for equilibrium). Portions I: P1 forces on one side P1 forces on the other side (for equilibrium) Portion II: Forces at end B, P3 - P1 Forces at end C, P4 - P2 Portion III: P2 forces at end C P2 forces at end D (for equilibrium) For equilibrium of portions II   Net force at end B is P3 + P1 - P1 = P3   Net force at end C is P4 + P2 - P2 = P4 Note that force in each portion is compressive. Stresses in three portions

σI =

4P1 π d12

σ II =

4( P1 + P3 ) 4 ( P4 + P2 ) = π d 22 π d 22

σ III =

4P2 π d32

Knowing the values of Young’s modulus E, strains and change in the lengths of three portions can be worked out. Example 1.9 Considering Fig. 1.21, let us take d1 = 20 mm, d2 = 14 mm, d3 = 10 mm, L1 = 100 mm, L2 = 140 mm, L3 = 150 mm. Forces are P1 = 30 kN, P2 = 16 kN, P3 = 20 kN, P4 = 34 kN. If E = 100 kN/ mm2, determine the change in length in each portion and the overall change in length. Solution P1 = 30 kN P1 + P3 = 50 kN P2 = 16 kN P2 + P4 = 50 kN Portion I

δ lI = − =−

4 × 30 × 1, 000 × 100 due to P1 π × 20 2 × E 9, 549 9, 549 =− E 100, 000

= -0.0955 mm Portion II

δ lII = −

4 × 50 × 1, 000 × 140 due to P1 + P3 π × 142 × 100, 000

= - 0.455 mm

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18

Chapter 1

δ lIII = −

Portion III

4 × 16, 000 × 150 due to P2 π × 10 2 × 100, 000

= -0.3056 mm Overall change in length = -0.0955 - 0.455 - 0.3056 = -0.8561 mm Exercise 1.9 A stepped circular steel bar of a length of 150 mm with diameters 20, 15 and 10 mm along lengths 40, 50 and 65 mm, respectively, subjected to various forces is shown in Fig. 1.22. If E = 200 kN/mm2, determine the change in its length. A 20 kN

B 20 f

C 15

15 f f

f 40 mm

45 mm

Figure 1.22

D

10 f f

15

10 kN

65 mm

Exercise 1.9

Extension in Bar Due to Self-weight Problem of stresses or strains exists in long hanging cables where the length of the cable becomes large; self-weight of the cable causes tensile stress in cable with increasing magnitude from top to bottom. A bar of a cross-sectional area of A and a length of L is suspended vertically with its upper end rigidly fixed in ceiling as shown in Fig 1.23. Say the weight density of the bar is w. Consider a section yy at a distance y from the lower end. wL

Weight of the lower portion abyy = yAw Cross-sectional area at yy = A A yw = wy (tensile) Normal stress of yy, σ y = A (weight is acting downwards on section yy) Strain in section yy, ey = wy/E, where E is the Young’s modulus of material.

A

L y

Change in length over small length dy of element wydy δ dy = E L

Total change in length, δ L = ∫

O

MTPL0259_Chapter 01.indd 18

wydy E

dy

y y

a

Figure 1.23

b

O

Bar under self-weight

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Simple Stresses and Strains

δL =

19

wL 2 A wL 2 = 2E 2A E

=

(wL A )L 2A E

=

W 2A E

where W is the total weight of the bar. Stress in the bar or cable due to self-weight gradually increases from bottom to top as shown in Fig. 1.23. Example 1.10 A stepped steel bar is suspended vertically. The diameter in the upper half portion is 10 mm, while the diameter in the lower half portion is 6 mm. What are the stresses due to self-weight in sections B and A as shown in Fig. 1.24. E = 200 kN/mm2. Weight density, w = 0.7644 N/mm3. What is the change in its length? Solution

A 10 mmf (A1)

1m

B

w = 0.7644 × 10-3 N/mm3

6 mmf

Cross-sectional area,

π A1 = × 10 2 = 78.54 mm 2 4

(A1)

Cross-sectional area,

π A2 = × 6 2 = 28.27 mm 2 4

1m

C

Figure 1.24

Example 1.10

Weight of upper portion, W1 = A × l × w = 78.54 × 1,000 × 0.7644 × 10-3 = 60.03 N Weight of lower portion, W2 = 28.27 × 1,000 × 0.7644 × 10-3 = 21.61 N

σ at A =

W 1 + W 2 60.03 + 21.61 = = 1.0395 N /mm 2 A1 78.54

σ at B =

W 2 21.61 = = 0.7644 N /mm 2 A2 28.27

Change in length W 2L 21.01 × 1000 = = 0.191 × 10 −2 mm 2 A2 E 2 × 28.27 × 200 × 103

at B,

δl =

at A,

δ l = 0.191 ×10 −2 + = 0.191 × 10 −2 +

W1 2 A1 E 60.03 × 1000 2 × 78.54 × 200 × 103

= 0.191×10-2 + 0.191 × 10-2 mm = 0.382 × 10-2 mm due to self-weight

MTPL0259_Chapter 01.indd 19

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20

Chapter 1

dl′ = change in length of upper part due to weight of lower part. =

W 2 1, 000 × A1 E

=

21.61 × 1, 000 78.54 × 200 × 103

= 137.57 × 10-3 mm = 0.13757 × 10-2 mm Over all change in length = (0.382 + 0.13757) × 10-2 = 0.5196 × 10-2 mm Exercise 1.10 A steel wire of a diameter of 6 mm is hanging freely over a length of 10 m. The weight density of the wire is 0.7644 × 10-3 N/mm3. What is the maximum stress developed in the wire and what is the change in its length due to self-weight?

Bar of Uniform Strength A bar of varying cross-sections such that under the action of axial load F applied at the lower end produces the same stress throughout its length L as shown in Fig. 1.25. The addition of stress due to selfweight is compensated by the increased area so that the stress developed remains the same. The area of bottom section is A1 and the area of top section is A2 if the self-weight of the bar is w F+W c

d A2

L dy

s g f

h e

y

e

A + dA g f A

s

A1 a

h

b F

Figure 1.25

thus

Bar of uniform strength F F +W = A1 A2

(1.3)

Reaction at the upper end of bar is F + W. Consider an elementary thin disc efgh with area at ef as A and area at gh as A + dA. If w is the weight density of the bar, the weight of elementary thin disc is wAdy

MTPL0259_Chapter 01.indd 20

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Simple Stresses and Strains

21

For the equilibrium of forces

σ A + wA d y = σ ( A + d A ) wA d y = σ d A dA w = dy A σ

(1.4)

Integrating both sides of Eq. (1.4), we get A2

∫

A1

ln

A2 A1

=

dA L w = ∫ dy A 0σ

w × L, A2 = A1 e wL / σ σ

or at any distance y from lower end A = A1 e wy /σ . Example 1.11 A 2.5-m-long vertical circular bar is subjected to a uniform stress of 0.4 N/mm2 throughout its length. The diameter at the bottom edge is 60 mm. What is the diameter at the top section if the weight density w = 0.07644 N/mm3? Solution Weight density, w = 7.644×10-5 N/mm3 Length of bar, L = 2,500 mm s = 0.4 N/mm2 wL 7.644 × 10 −5 × 2, 500 = 0.4 σ = 0.4775 e0.4775 = 1.612 A2 = A1 e0.4775 A2 = A1 × 1.612

π 2 π 2 d2 = d1 × 1.612 4 4 d2 = d12 × 1.612 = d1 1.612 = 60 1.612 = 60 × 1.2696 = 76.18 mm Diameter at the top = 76.18 mm. Exercise 1.11 A 2.0-m-long vertical circular bar is subjected to a uniform stress of 0.3 N/mm2 throughout its length. The diameter of the bottom is 50 mm. What is the diameter of the bar at the middle section if the weight density of the bar is 5.2 × 10-5 N/mm2?

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22

Chapter 1

Volumetric Stress and Volumetric Strain A stress which acts on a body equally from all the directions is known as volumetric stress; as an example hydrostatic pressure acts equally in all the directions. Consider a sphere of diameter D at a depth h (from free surface of the liquid) in a liquid of weight density, w (Fig. 1.26). p p

Volum. stress

pe, elastic limit pressure

h D

p p

O

p D

Figure 1.26

Vol. strain, e v

Volumetric stress and volumetric strain

Pressure on sphere, p = wh. As per Pascal’s Law of Fluid Mechanics, a liquid transmits pressure equally in all the directions. Say V = initial volume of sphere V′ = final volume of sphere under pressure V - V′ = dV, change in volume Volumetric strain,

eV = δV V

As the depth of the body in the liquid increases, the pressure also increases on the body. Volumetric strain is proportional to pressure that is, p a ev. p = Kev, where K is the constant of proportionality or Bulk modulus. Bulk modulus,

K=

p Volumetric stress = εV Volumetric strain

Until elastic limit pe, the pressure is linearly proportional to the volumetric strain. Example 1.12 A spherical ball of a material of a diameter of 160 mm goes down to a depth of 600 m in sea water. If the specific weight of sea water is 10.2 kN/m3 and the bulk modulus of the material of the ball is 160 kN/mm2, determine the change in volume of ball. Solution Specific weight, Depth, Pressure,

w = 10.2 kN/m3 h = 600 m p = wh = 10.2 × 600 = 6,120 kN/m2 = 6.12 N/mm2

Bulk modulus,

MTPL0259_Chapter 01.indd 22

K = 160 × 1,000 N/mm2

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Simple Stresses and Strains

εv =

Volumetric strain,

23

p 6.12 = = 3.825 × 10 −5 K 160 × 1, 000

π D3 π = × 1603 = 2.144 × 106 mm3 6 6

Initial volume,

V=

Now,

ev = 3.825 ×10-5

Change in volume dV = evV = 3.825 × 10-5 × 2.144 × 106 = 82 mm3 Exercise 1.12 An aluminium ball of a diameter of 150 mm is immersed in sea water to a depth of h m. The specific weight of sea water is 10.2 kN/m3. What is h if the pressure acting on ball is 4.8 N/ mm2? If K for aluminium is 66 kN/mm2, what is the change in volume of balls?

Statically Indeterminate Problems In a statically determinate problem, the equations of equilibrium of forces are sufficient to determine the unknown forces as reactions in a structure. However, in certain problems, the equations of equilibrium of forces are not sufficient to determine the unknown forces. These are called statically indeterminate problems and an additional equation involving the geometry of deformed structure is necessary to determine unknown forces. This type of problem is a statically indeterminate problem. This equation provides condition for structural compatibility. In Fig. 1.28, a bar held between two rigid supports is shown. A force P is applied to the bar. Reactions at ends are P1 and P2. Then, P = P1 + P2

(1.5)

This equation is not sufficient to determine the values of P1 and P2 independently.

Additional Equation A bar is held between two rigid supports; as shown in Fig.1.27 therefore, net deformation in the bar along its axis is zero. Due to reactions P1 and P2, portion I comes under tension and portion II comes under compression, as shown in Fig.1.28. If E is the Young’s modulus of this material, then P

E P2

P I

II

a

b

P1

P2

I

I

MTPL0259_Chapter 01.indd 23

Bar field between two rigid supports

P1 P1

L

Figure 1.27

P2

Figure 1.28

II

Two portions under axial loads

5/23/2012 10:46:16 AM

24

Chapter 1

P2 a (extension in portion I) AE Pb δ lII = 1 (contraction in portion II) AE where A is the cross-sectional area of the bar.

δ lI =

P2 a P1b − =0 AE AE P2 a = + P1 b

Then, or

(1.6)

b P1 (1.7) a Now, Eqs (1.5) and (1.6) can determine the reactions, P1 and P2. The second equation i.e. Eq. (1.6) is known as deformation compatibility equation. P2 =

Example 1.13 An aluminium bar of a diameter of 10 mm and a length of 400 mm is rigidly held at its upper end. It is subjected to an axial force P = 12 kN at section B as shown in Fig. 1.29. E for aluminium = 67 kN/mm2. Gap between the lower end and the floor is 0.2 mm. Determine the stresses in portions AB and BC of the bar.

P

A Aluminium bar

150 B

Solution dia − 10 mm Under the action of the force P, the portion AB of the bar 250 extends and the end C of the bar touches the floor. Then, P = 12kN reaction offered from floor provides a compressive force for portion BC of bar. C Say P1 = tensile force in portion AB d = gap P2 = compressive force in portion BC 0.2 mm P = P1 + P2 (1.8) Figure 1.29 P Extension in AB, δ lI = 1 × 150 AE P Contraction in BC, δ lII = 2 × 250 AE where A is the cross-sectional area of the bar and E is the Young’s modulus of the bar. Then, δ lI − δ lII = 0.2 mm gap (1.9) Area,

π × 10 2 = 78.54 mm 2 4 P1 × 150 P × 250 − 2 = 0. 2 78.54 × E 78.54 × E A=

150 P1 - 250 P2 = 0.2 × 78.54 × 67, 000 150 P1 - 250 P2 = 1,052,436 P1 = 1.6667 P2 + 7016.24

MTPL0259_Chapter 01.indd 24

(1.10)

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Simple Stresses and Strains

P1 + P2 = 12 kN = 12,000 or

(1.11)

1.6667 P2 + 7016.24 = 12,000 - P2

A

A1 = 1 cm2

2.6667 P2 = 4983.76

or

1m

P2 = 1868.9 N

B

P1 = 12,000 - 1888.9 = 10131.1 N A2 = 2 cm2

Stresses in portions AB and BC:

σI = +

P 2m

10131.1 = +129.0 N /mm 2 78.54

σ II = −

25

1868.9 = −23.8 N /mm 2 78.54

C

0.4 mm

Figure 1.30 Exercise 1.13 A 3-m-long stepped steel bar, shown in Fig. 1.30, with a cross-sectional area of the portion AB is 1 cm2 and that of the portion BC, 2 cm2 is subjected to a load of P = 25 kN. Gap between the bar and the ground floor is 0.4 mm. What are the stresses developed in the portions AB and BC of the bar, given that E = 200 GPa? (Note that 1 GPa = 10 9 N /m 2 = 103 N/mm2)

Strain Energy and Resilience When a body is subjected to an external force, stress and strain are produced in the body and the work done on the body is absorbed as strain energy in the body. If the external forces are removed then strain energy absorbed is fully recovered, and it is said to be resilience of the body. For the property of resilience, springs are used in mechanisms and machines. Figure 1.31 shows a graph between stress (s) and strain (e) on a body. se is the elastic limit stress. Within the elastic limit, the strain energy is fully recoverable. Within the elastic limit, say, stress is s and strain is e ; then 1 1 σ .σ σ 2 Strain energy per unit volume = σ .ε = = 2 2 E 2E σ   using Hooker’s law ε =  E Shaded area ose shows resilience up to the elastic limit, se. σ2 Strain energy absorbed = e × V , where V is the volume 2E of the body.

MTPL0259_Chapter 01.indd 25

Elastic region se

Plastic region

Stress s s

o

e Strain

Figure 1.31

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26

Chapter 1

σ2 σ e2 × V is known as proof resilience. Modulus of resilience e is proof resilience per unit volume. 2E 2E In case of t (shear stress) and f (shear strain), τ e2 2G where te is the shear stress at elastic limit and G is shear modulus. In case of volumetric stress, pe (at elastic limit), and volumetric strain, ev, Modulus of resilience =

2 Modulus of resilience = pe 2K where pe is hydrostatic pressure at elastic limit and K is the bulk modulus.

Example 1.14 A 40-mm cubical block is subjected to shear stress and it is observed that te = 240 N/mm2. If shear modulus G = 84 kN/mm2, determine (i) the modulus of resilience, (ii) the shear strain at elastic limit and (iii) the total strain energy absorbed at elastic limit.

te

40 f

40 mm

40

Solution 1 Figure 1.32 shows a 40-mm cube subjected to 2 a shear stress, te, of 240 N/mm . Volume of the block = 40 × 40 × 40 = 64 × 103 mm3 Figure 1.32 Shear modulus, G = 84,000 N/mm2 22 22 τ 240 = 342.86 × 10 −−33 Nmm/mm3 Modulus of resistance = ee = 2G 2 × 84 × 1, 000 240 ττee 240 857××10 10−−33 == == 22..857 84,,000 000 G 84 G Total strain energy absorbed at elastic limit, Ue = Proof resilience = 342.86 × 10 −3 × 64 × 103 = 21931.5 N mm = 21.93 Nm Shear strain at elastic limit, =

Exercise.1.14 A copper bar of a diameter of 20 mm and a length of 600 mm is subjected to an axial tensile load. If σ e = 215 N/mm 2 , determine 1. the resilience at 100 N/mm2, 2. the proof resilience and 3. the modulus of resilience, given E = 103 kN/mm2.

Sudden Load Till now, we have considered the application of gradually applied loads, where the load is gradually increased starting from zero magnitude and the stress produced in a material is the load/cross-sectional

MTPL0259_Chapter 01.indd 26

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Simple Stresses and Strains

area. However, if the whole of the magnitude of the load is applied suddenly on a body, then the stress produced is more than the stress produced by the same load when applied slowly and gradually. Say a heavy load is being lifted by a crane hook, slings are tightened on the load and slings pass through the throat of the crane hook. When wire rope moves up, loose slings become taut and the whole of the load acts suddenly on the hook and the wire rope. Consider that a load W is applied suddenly on a bar and sudden deformation produced in bar is d as shown in Fig. 1.33. Work done on the bar = Wd = strain energy absorbed by the bar However, the internal resistance of the body develops gradually.

27

D

W R A

B

d

W

C d

O

Figure 1.33

Graph between sudden load and deformation

So that (1/ 2) Rδ = W δ where internal resistance, R = 2W. Stress is developed as an internal resistance of the body per unit area. R 2W Stress developed in the bar = = , where A is the cross-sectional area. A A 2W = 2σ gradual. A This shows that the stress developed due to load applied suddenly is double the stress developed in the body due to the same load applied gradually.

σ sudden =

Example 1.15 Water under a pressure of 5 N/mm2 is suddenly admitted on a plunger of a diameter of 120 mm. Plunger is connected to a steel connecting rod of a length of 4 m and a diameter of 30 mm. Determine the amount of stress developed in the connecting rod and the deformation produced in the connecting rod. E for steel = 200 kN/mm2 (Fig. 1.34). Solution Water pressure, Plunger diameter, Area of plunger,

Force on plunger,

p = 5 N/mm2 D = 120 mm π π A = D 2 = × 120 2 4 4 = 11309.7 mm2 F = p × A = 5 × 11309.7

CR P

p

p = Plunger CR = Connecting rod

Figure 1.34

= 56548.5 N

π × 30 2 = 706.86 mm 2 4 L = 4 m = 4,000 mm

Cross-sectional area of connecting rod = Length of connecting rod,

Stress produced in connecting rod, σ sudden = 2F /A

MTPL0259_Chapter 01.indd 27

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28

Chapter 1

=

2 × 56, 548.5 706.86

= 160 N/mm2

σs ×L E 160 = × 4, 000 = 3.2 mm 2, 00, 000

Deformation in connecting rod, δ =

Exercise 1.15 A copper column of a diameter of 50 mm and a length of 500 mm is subjected to a load W = 20 kN applied suddenly as shown in Fig. 1.35. If E for copper = 105 kN/mm2, what is the stress developed in copper and what is the deformation produced?

Impact Load

W

The load which is applied with some velocity on a body is gap = 0 known as impact or shock load. The kinetic energy of the load is absorbed as strain energy in the body. This type of load is unsafe and must be avoided in normal applications. However, d − 50 mm the impact loads are used in industry to produce forged parts 500 mm and to drive piles in the ground for reinforcing the earth for heavy building structure. In other words, if a load is allowed to fall through a height, it will gain some velocity due to gravity and will strike the body with kinetic energy. Figure 1.36 shows a bar of a uniform cross-sectional area of A and a length Figure 1.35 of L with a collar at lower end and upper end fixed in ceiling. A weight W falling under gravity through height h strikes the collar producing an instantaneous elongation d1 in the bar. Loss of potential energy of weight = W ( h + δ i ) σ2 A Gain in strain energy of the bar = i × V 2E where σ i = instantaneous stress produced in the bar L W V = volume = A × L E = Young’s modulus σ h d i = i × L, where L is length of bar E σ2 di W (h + δ i ) = i × AL (1.12) Collar di 2E Using the principle of conservation of energy, Figure 1.36 Bar under σ L σ2 impact load Wh + W i = i × AL 2E E 2E Multiplying throughout by , we get AL

MTPL0259_Chapter 01.indd 28

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Simple Stresses and Strains

Wh

2 E WL 2 E + × σ i = σ i2 AL E AL

σ i2 −

or

29

2W W 2 Eh =0 σi − × A A L

(1.13)

Solving the quadratic equation for instantaneous stress σ i , 1  2W 4W 2 8W Eh  σi =  ± +  A2 AL  2 A =

W W 2 EA h ± 1+ A A WL

=

W W 2 EA h + 1+ A A WL

Note that negative sign is inadmissible, as the stress produced in the bar is tensile and the term in the under root is greater than 1. W 2 EAh  Instantaneous stress produced, σ i = 1 + 1 + A WL  As the load strikes the collar, the bar starts vibrating with some amplitude, but the vibrations die down due to air and material damping and finally the load comes to rest W 2W 1+ 1+ 0 = If distance h = 0, load will become sudden load, h = 0, σ s = A A σ Instantaneous elongation in the bar, δ i = i × L, where L is the length of the bar E

(

)

Example 1.16 A 50-N load falls through a height of 50 mm onto a steel bar of a diameter of 25 mm and a length of 400 mm, supported on the ground. If E = 200 GPa, what is the instantaneous stress developed in the bar and what is the change in its length? Solution The load falls on the bar producing a compressive stress in the bar and its length has instantaneous contraction. W = 50 N h = 50 mm d = 25 mm π 2 π A = d = × 252 = 490.87 mm 2 4 4 Length of the bar, L = 400 mm E = 2,00,000 N/mm2 2 EAh 2 × 2, 00, 000 × 490.87 × 50 = WL 50 × 400 = 4,90,870

MTPL0259_Chapter 01.indd 29

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30

Chapter 1

W 50 = = 0.10186 A 490.87 Putting these values in the expression,

(

σ i = 0.10186 1 + 1 + 4, 90, 870

)

= 0.10186 (1 + 700.62) = 71.47 N /mm 2 (compressive stress)

δi =

σi 71.47 ×L= × 400 = 0.143 mm (contraction) E 2, 00, 0000

Exercise 1.16 A copper bar of a diameter of 10 mm and a length of 100 cm is stressed by a weight of 120 N dropping freely through a height of 40 mm onto a collar provided at the end of the bar before commencing to stretch the bar. Find the instantaneous stress and the instantaneous elongation in the bar. Given E = 105 kN/mm2, sut of material = 260 N/mm2. Comment whether the bar will break.

Tensile Test on Mild Steel Mild steel is the most commonly used engineering material. Majority of engineering components and structures are made of mild steel and the material is easily available in the market. In material testing labs of engineering colleges, majority of the tests are performed on mild steel samples. A specimen of the shape shown in Fig. 1.37(a), circular section, with collars at the ends is fitted in the grips of a testing machine. A gradually and slowly increasing tensile load P is applied on the sample through a mechanism provided in the machine. Due to the increasing tensile load, the specimen is continuously stretched. Tensile load and extension in the specimen are continuously recorded by the operator. Nowadays, there are machines available in which an automatic graph between the load and the extension is obtained as shown in Fig. 1.37(b). Many important mechanical properties are obtained from this load extension graph as follows: 1. O to A is a straight line, P ∝δ L (proportional), that is, the material obeys Hooke’s law. Slope of this line provides information on the Young’s modulus of the material. 2. Point B, an elastic limit point, is very close to A, and many a times it is hardly noticed. If the load is removed from the sample at this point, then the residual strain in sample will be zero. 3. Point C, upper yield point of the material. The material has yielded, meaning that plastic strain is developed in the sample. 4. Yield strength of the material is defined at this point, that is, load at upper yield point/crosssectional area. 5. Resistance of the material falls from C to D, lower yield point. This is due to the presence of extra rows of carbon and nitrogen atoms in the material. 6. D to E is called yield plateau. In this portion average stress remains constant. 7. At E, resistance of the material again starts increasing and point E is termed as point of strain hardening.

MTPL0259_Chapter 01.indd 30

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Simple Stresses and Strains Collar

d

P

31

P

L (a) Specimen F

P

B

C A D

Ultimate load C at F

G, Breaking load

E

At F Neck

C - Upper yield point

Cup

Cone

D - Lower yield point Shear fracture O Extension

Recovered extension Tensile fracture

(b) Load extension diagram

(c) At necking

Figure 1.37 8. At point F, in any ductile material as mild stress, necking takes place, that is, in some weaker section of the bar, the diameter is heavily reduced. Further extension takes place in the vicinity of this neck. Point F gives the ultimate load and the ultimate load/original area of sample gives the ultimate tensile strength of material 9. Resistance of the material falls through the portion F to G and finally specimen breaks into two pieces, showing the cup and cone type of fracture, marked by two regions: (i) shear fracture in outer ring, marked by shining lines and (ii) straight tensile fracture in inner core marked by dull granular structure. Join the two broken pieces to find out (i) the final gauge length, L′ (ii) the diameter at the neck, d ′. L′ −L × 100 indicates ductility of the material Percentage elongation = L π 2 π 2 d − d′ d 2 − d ′2 4 Percentage reduction in area = 4 × 100 = × 100 2 π 2 d d 4 Percentage reduction in area provides the information on degree of cold working of the material. Two important information on (i) yield strength and (ii) percentage elongation decide the suitability of steel for structural applications. Up to the point F, ultimate load extension in sample is proportional to its length and beyond F up to breaking load extension is proportional to lateral dimension, that is, the diameter of the bar.

MTPL0259_Chapter 01.indd 31

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Chapter 1

32

Total extension = bL + c A dL = bL + c A , which is known as Barba’s Law, where b and c are Barba’s constants, L is gauge length and A is the cross-sectional area of specimen. Example 1.17 A round section of wrought iron of a diameter of 12.5 mm and a gauge length of 100 mm was tested in tension up to fracture and the following observations were made. (i) (ii) (iii) (iv) (v)

Yield load 29.6 kN Maximum load 44.8 kN Load at fracture 37 kN Diameter at neck 9.1 mm Total extension in sample 27.6 mm.

Determine (i) the yield strength, (ii) the ultimate tensile strength, (iii) the actual breaking strength and (iv) the percentage elongation. Solution Diameter, Gauge length, Diameter at neck,

d = 12.5 mm L = 100 mm d′ = 9.1 mm

π 2 π d = × 12.52 4 4 = 122.72 mm2

Original cross-sectional area,

A=

Area at neck,

A′ =

π 2 π d ′ = × 9.12 4 4

= 65.04 mm2 syp = 29.6 × 100 = 241.2 N/mm2 122.72

(i) Yield strength,

sut = 44.8 × 1, 000 = 365.06 N/mm2 122.72

(ii) Ultimate tensile strength,

(iii) Actual breaking strength,

sbk = 37 × 100 = 568.9 N/mm2 65.04 =

(iv) Percentage elongation,

27.6 × 100 = 27.6% 100

Exercise 1.17 A specimen of aluminium alloy of a diameter of 12 mm was tested under tension. Linear strain parallel to the applied load P was measured accurately with the help of strain gauges as follows: Pin kN

4

8

12

16

20

24

28

32

36

38

39.5

ea ×10

0.343

0.69

1.03

1.39

1.72

2.07

2.40

2.734

4.30

5.65

8.40

-3

MTPL0259_Chapter 01.indd 32

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Simple Stresses and Strains

33

Plot graph (P ) versus axial strain (ea) and determine Young’s modulus of elasticity and 0.1 per cent proof stress from the graph. To determine 0.1 per cent proof stress, a line from 0.001 × length is drawn parallel to initial slope of load-extension graph. This line intersects the graph providing load.

Stress Concentration We have studied stresses and strains in bars of uniform section throughout the axial length and we find that the stress is equal to the load/cross-sectional area, a uniform stress. However, if there is an abrupt change in the geometry of a component, then the stress distribution becomes non-uniform with larger values of stress at corners, fillets and holes. Consider a stepped bar of diameters D and d as shown in Fig. 1.38, subjected to 4P 4P axial load P, stress in portion I, that is, suddenly increases to in portion II at the corner section bb 2 πD πd 2 of the stepped bar. Increase in stress is concentrated along the circumferential edge bb, and stress distribution becomes non-uniform with smax at corner b, as shown in Fig. 1.38(b). However, at section cc, away from discontinuity stress distribution again becomes uniform. Due to stress concentration the stepped bar under load will try to break along corner bb. To avoid failure due to stress concentration, the cross-sectional area should be gradually reduced by providing a fillet radius as shown in Fig. 1.39. Effect of fillet radius R on stress concentration factor (SCF) in a bar shown in Fig. 1.39 is obtained through photoelasticity (an experimental technique). a

I

P

D

a II

b

c D

d

b

b P b

c

smax

c

smax

c

s = 4 P2 pd

a

a

Figure 1.38 Stepped bar under axial load Moreover, one can consider the effect of a hole in a plate as shown in Fig. 1.40(a). Stress distribution in the section abba becomes non-uniform, with maximum stress σmax at the edge of the hole. σ P SCF = max σ av =

Maximum stress , Average stress

s max

with average stress over a section with no effect of abrupt change. a

b

b

a

R = Fillet radius P

B

P R P

Figure 1.39

MTPL0259_Chapter 01.indd 33

Figure 1.40 (a)

5/23/2012 10:46:32 AM

34

Chapter 1 P

Detailed discussion on SCFs is beyond the scope of this book. An empirical relation of SCF for an elliptical hole in a plate [Fig. 1.40(b)] subjected to an axial load is 2b SCF = 1+ a where a = semi-minor axis and b = semi-major axis.

2a

Example 1.18 In a rectangular section bar, shown in Fig. 1.39, breadth B = 25 mm. What should be the minimum fillet radius R so that SCF does not exceed 1.65? Solution Table 1.2 provides values of SCF for various ratios of R/B i.e. fillet radius/breadth From Table 1.2 for SCF of 1.65, R = 0.143 B Fillet radius, R = 0.143 × B = 0.143 × 25 = 3.575 mm

2b

P

Figure 1.40(b)

Plate with elliptical hole

Exercise 1.18 In a sample made from a 40-mm-broad thin sheet, a fillet radius of 7 mm is provided. Find SCF for this flat. R [Hint: Plot a graph between and SCF using values given in Table 1.2 and find SCF.] B Table 1.2 R/B, Ratio

0.333

0.222

0.143

0.083

SCF

1.25

1.50

1.65

1.8

Factor of Safety A material is tested for its ultimate strength. While designing a machine component or any engineering component, we do not want that it should fail in service; therefore, it is designed based for a safe allowable stress. σ σ allowable = ut F OS where σut is the ultimate strength and FOS is the factor of safety. The necessity for an FOS or safe design depends upon the application of an engineering component such as (i) a wire rope in a crane carrying the load through crane hook and the wire rope should not break, as the breaking of the wire rope may cause the casualty of human beings those who are working with the crane; (ii) a structural bridge should not collapse during service, as the collapse of a bridge may cause extensive damage to the property and human lives; and (iii) a bed of machine damage may cause less problems; therefore, a smaller FOS can be taken in this type of applications. A higher FOS is desired for bridges and wire ropes. Then, we assume that the martial is homogeneous (having same physical properties throughout the volume used) and isotropic (having same elastic properties throughout); however, in reality, the material may contain internal defects such as holes, cracks and slag inclusions and as such no material is homogeneous.

MTPL0259_Chapter 01.indd 34

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Simple Stresses and Strains

35

Therefore, to account for all these factors, we take some suitable value of FOS depending upon (i) the type of load on the component such as gradual, sudden or shock load; (ii) the type of service conditions and (iii) the mechanical properties of the material such as strength, ductility and hardness. Example 1.19 Two 10-mm-thick plates are joined by a single rivet of a diameter of 20 mm. The ultimate shear strength of the material of the rivet = 460 N/mm2. Take FOS as 3 and determine the load which can be safely applied on the plates, as shown in Fig. 1.41. The plates are 50 mm wide. Solution Ultimate shear strength = 460 MPa Factor of safety = 3 Allowable shear stress,

τ all =

Rivet diameter, d = 20 mm

460 = 153.33 MPa 3

π 2 π 2 d = 20 = 100π mm 2 4 4 The rivet can break in shear along the section aa, and the rivet is in single shear. π Permissible load, P = d 2 × τ all 4 = 100p × 153.33 = 48,170 N A=

Cross-sectional area of rivet,

= 48.17 kN

P

a

a P

10 mm

b = 50 mm

Riveted joint

Figure 1.41

Exercise 1.19 The ultimate tensile strength of mild steel is 660 N/mm2. A tie bar of circular section is subjected to an axial load of 180 kN. Take FOS as 4 and determine the diameter of the tie bar. Problem 1.1 A hole of a diameter of 50 mm is to be punched in a 1.6-mm-thick mild steel sheet. If the ultimate shear stress of mild steel is 480 N/mm2, determine the force required to punch the hole.

MTPL0259_Chapter 01.indd 35

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36

Chapter 1

Solution Figure 1.42 shows a thin sheet of thickness t and punch of diameter d. When the hole is punched, shearing area = pdt. Ultimate shearing strength tus Thus shear force. P = pdt tus

P

d

Punch

Hole

= p × 50 × 1.6 × 480

t

= 120,637 N = 120.637 kN, force required to punch the hole. t

Problem 1.2 Two parts of a certain machine component are joined by a rivet of a diameter of 20 mm. Determine the shear and normal stresses in the rivet if the axial force P = 15 kN and the angle of joint is 60° to the axis of the load as shown in Fig. 1.43. Solution Rivet diameter, Cross-sectional area

d = 20 mm π = × 20 2 4 = 100p mm2

Figure 1.42

60°

P

P

P = 15,000 N

30°

Normal component, Pn = P cos 30° = 15,000 × 0.866 = 12,990 N Tangential component, Pt = P sin 30° = 15,000 × 0.5 = 7,500 N

Figure 1.43

Normal stress,

P = 1200 N

P 12, 990 σ= n = = 41.35 N /mm 2 100π A Shear stress,

τ=

Pt 7, 500 = = 23.87 N /mm 2 A 100π

Problem 1.3 A 1-m2 rigid square plate is supported over four legs A, B, C and D of equal length. A load of 1,200 N is applied on a square plate at point K, such that AF = 0.4 m and AG = 0.3 m as shown in Fig. 1.44. Determine the reactions in each leg. Solution Say the reactions in legs are RA, RB, RC and RD, respectively. Then, RA+ RB+ RC + RD =1,200 N (1.14)

MTPL0259_Chapter 01.indd 36

D

G

AF = 0.4 m

m

0.3

AG = 0.3 m

A 0.4

K

C

E

F 1m B

1m

Figure 1.44

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Simple Stresses and Strains

37

Plate ABCD is a rigid plate and will remain in one plane when the legs get contracted due to compressive force. There is unsymmetrical application of load. Diagonals AC and BD of plate will remain straight meeting at mid-point E. The deflection in legs is proportional to the compressive force in each. δ + δc δ B + δ D δE = A = 2 2

δ ∝ force (deflection is proportional to force) R A + RC = R B + R D =

1, 200 = 600 N 2

Therefore, RA + RC = 600 N Taking moments of forces about the edge AB of plate,

(1.15)

1, 200 × 0.3 = RC × 1 + R D × 1 RC + R D = 360 N

(1.16)

Similarly taking moments about the edge AD of plate, 1, 200 × 0.4 = R B × 1 + RC × 1 480 = RB + RC RB + RC = 480 N

(1.17)

RB - RD = 120 N

(1.18)

From Eqs (1.16) and (1.17),

However,

RB + RD = 600 N RB = 360 N

Therefore,

RD = 240 N RC = 480 - RB = 480 - 360 = 120 N

From Eq. (1.17),

RA - 600 - RC = 600 - 120 = 480 N Problem 1.4 In a bar of rectangular section, the width tapers from 25 mm to 15 mm, while thickness tapers from 12 mm to 8 mm over a length of 500 mm. The bar is subjected to a tensile force P and E = 200 GPa. Determine the magnitude of P if the change in its axial length is 0.2 mm (Fig.1.45). Solution Consider a section at a distance x from one end as shown in Fig. 1.45. 25 − 15 bx = 15 + x = 15 + 0.02 x Breadth, 500 12.8 x = 8 + 0.008x 500

Thickness,

tx = 8 +

Cross-sectional area,

Ax = (15 + 0.02 x)(8 + 0.008 x)

or

MTPL0259_Chapter 01.indd 37

= 0.02 × 0.008(750 + x)(1, 000 + x)

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38

Chapter 1

25 mm

15 P

P

12 8 dx X 500 mm

Figure 1.45 or Stress,

= 16 × 10 −5 (750 + x)(1, 000 + x)

σx =

P P = −5 Ax 16 × 10 (750 + x)(1, 000 + x)

P E × 16 × 10 (750 + x)(1, 000 + x) P Putting the value of E, ε x = 5 −5 2 × 10 × 16 × 10 (750 + x) (1, 000 + x) Strain,

εx =

−5

=

P 32 (750 + x ) (1, 000 + x )

=

P 1  1  − 32 × 250  750 + x 1, 000 + x 

Change in length over dx, δ dx =

Pdx  1 1  −  8, 000  750 + x 1, 000 + x  500

Over all change in length,

δl = ∫

0

MTPL0259_Chapter 01.indd 38

P  1 1  dx − 8, 000  750 + x 7, 000 + x 

=

P [ln (750 + x ) − ln (1, 000 + x )]500 0 8, 000

=

P  1, 250 1, 500  ln − ln   8, 000 750 1, 000 

=

P  1, 250 1, 000  ln × 8, 000  750 1, 500 

=

P × ln 1.1111 8, 000

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Simple Stresses and Strains

39

P × 0.10535 8, 000

0.2 mm = P=

0.2 × 8, 000 = 15187.47 N 0.10535

= 15.187 kN Problem 1.5 The cross-section of a bar is given by (144 + 0.01x 2 ) mm 2 , where x is its distance from one end in millimetres. If the length of the bar is 240 mm and E = 100 kN/mm2, find the change in length under a load of 4 kN. Solution Consider a small length dx at a distance of x from one end of the bar. L. Cross-sectional area, Ax = (144 + 0.01x 2 ) mm 2 P = 4,000 N

Load, Stress,

σx =

P 4, 000 = Ax (144 + 0.01x 2 )

Strain,

εx =

σx 4, 000 = E E (144 + 0.01x 2 )

Change in length over dx, δ dx =

4, 000dx E (144 + 0.01x 2 ) 240

Total change in length,

δl = ∫

0

240

= ∫

0

4, 000d x E (144 + 0.01x 2 ) 4, 000dx 1, 00, 000 (144 + 0.01x 2 ) 240

= 0.04 ∫

0

dx 0.01(14, 400 + x 2 ) 240

=4 =

1 x tan −1 120 120 0

4  −1 240  4 × 1.107  tan = 120  120  120

= 0.0369 mm Problem 1.6 Two portions of a member AB are glued along a plane making an angle a with the horizontal as shown in Fig. 1.46. The ultimate stress of glued joint is σ u = 17 MPa and τ u = 9 MPa. Determine the range of values of angle a, for which the FOS of the member is at least equal to three. Solution Allowable stress,

MTPL0259_Chapter 01.indd 39

σ allo =

17 = 5.667 MPa , in tension 3

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40

Chapter 1

P = 10 kN

9 = 3 MPa in shear 3 Cross-sectional area of inclined joint plane 25 300 = × 12 = cos α cos α Normal force = P cos a Tangential force = P sin a

τ allo =

Allowable stress,

P cos α ≤ 5.667 A P cos α ≤3 A P × cos α 2 ≤ 5.667 300 P sin α .cos α ≤3 300

Now

or

cos 2α ≤

A a

B

P 12 mm

25 mm

Figure 1.46

Problem 1.6

5.667 × 300 10, 000

cos α = 0.4123, α = 65.6° = 65° 36 ’ sin α cos α =

3 × 300 10, 000

sin 2α 900 = 2 10, 000 sin 2α = 0.18, 2α = 10.37°, α = 5°11′ (Range of α = 5°11′ − 65°36′ ) Problem 1.7 A circular bar ABC, shown in Fig. 1.47, is subjected to an axial compressive force P. Show that the displacement of the point B is one-third the vertical displacement of point A. Solution Say E is the Young’s modulus of the material, change in length,

δ lBC

4PL 2PL = = π E 2d × d E π d 2

δ lA B

4PL = Eπd 2

Total contraction =

δA =

MTPL0259_Chapter 01.indd 40

2PL 4PL 6PL + = 2 2 π Ed π Ed π Ed 2

6PL ,displacement at A ↓ π Ed 2

P A L

d B

L

2d

Figure 1.47

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Simple Stresses and Strains

41

2PL ,displacement at B ↓ π Ed 2 1 δB = δA 3

δB =

Problem 1.8 A round steel bar of a length of 400 mm and a diameter of 15 mm at the middle portion is subjected to an axial tensile force of 17 kN. What should be the diameter at the outer collars if the stress in the collar is not to exceed 40 MPa? What is the magnitude of ‘a’ if the total extension in the bar is not to exceed 0.17 mm? E = 200 GPa (Fig. 1.48).

P

15 mm

a

(400 - 2a)

D

P = 17 kN

a

Figure 1.48 Solution Stress in collar, s1 = 40 MPa Load, P = 17,000 π 2 Cross-sectional area = D 4 π 2 D × 40 = 17, 000 4 D = 23.26 mm (diameter of collar) Stress in middle portion, s2 =

17, 000 × 4 π × 152

= 96.20 MPa Overall change in length =

40 (400 − 2a ) × 96.2 = 0.17 mm × 2a + E E

80 a + 96.2 × 400 - 2a × 96.2 = 0.17 × 2,00,000 or

-112.4a = -38,480 + 34,000

a = 39.86 mm (length of each collar) Problem 1.9 A 2,500-N load is suspended at a bar attached to ropes of a cross-sectional area of 40 mm2 and lengths of 4 and 5 m, respectively, as shown in Fig. 1.49. The load is suspended in such a manner that the bar AB remains horizontal. Determine (i) stress in each rope, (ii) distance a and (iii) downward movement of bar. E = 210 kN/mm2

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42

Chapter 1

Solution Say tensions in two ropes are T1 and T2, respectively, as shown. Load W is suspended in such a manner that

δ l1 = δ l2

5m

T1 4, 000 T 2 × 5, 000 or × = , where A is A E AE cross-sectional area of ropes.

a

A

B W = 2500 N

T1 + T2 = 2,500 N

Now,

T2

T1

T1 = 1.25T2

or

200 mm

4m

Figure 1.49

1.25T2 + T2 = 2,500 N T2 = 1111.11 N T1 = 1388.89 N

σ1 =

Stress in rope,

1388.89 = 34.72 N /mm 2 40

Taking moments about point A of the bar 2,500 × a = T2 × 200 a=

1111.11 × 200 = 88.89 mm 2, 500

δ l1 = δ l2 =

T1 × 4, 000 1388.89 × 4, 000 = AE 40 × 210 × 1, 000

= 0.66 mm, downward movement of bar. Problem 1.10 A 1-m-long stepped steel bar is subjected to axial loads as shown in Fig. 1.50. The diameters of two steps are 20 and 10 mm, respectively. If E = 200 GPa, calculate the stresses in each portion AB, BC and CD. What is the total change in length of the bar? Solution Portion AB Compressive force = 20 kN D = 10 mm

Steel bar d = 10 mm

20 kN

30 kN A

B

10 kN

Figure 1.50

MTPL0259_Chapter 01.indd 42

D

C

0.2 0.4 m

0.6 m

Problem 1.10

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Simple Stresses and Strains

43

π × 20 2 = 100π mm 2 4 −20, 000 σ AB = = −63.66 N/mm 2 100π

Cross-sectional area =

d l AB =

−63.66 × 200 = −0.0636 mm 2, 00, 000

Portion BC Tensile force = +10 kN

σ BC = + δ lBC = + Portion CD Tensile force = 10 kN Cross-sectional area =

10, 000 = +31.83 N /mm 2 100π

31.83 × 200 = 0.0318 mm 2, 00, 000

π × 10 2 = 25π = 78.54 mm 2 4 10, 000 σ CD = = 127.324 N /mm 2 78.54 δ lCD =

127.324 × 600 2, 00, 000

= +0.382 mm Overall change in length,

dl = -0.0636 + 0.0318 + 0.382 = +0.3502 mm

Problem 1.11 A rod ABCD has two different cross-sectional areas of 200 mm2 and 100 mm2 as shown in Fig. 1.51. The rod is held between two rigid supports at its ends. The loads P each act at sections B and C of the rod. If P = 20 kN, determine the axial stresses in each section of the rod. a = 100 mm2 P1

A = 200 mm2

a = 100 mm2

P A

B 0.3 m

P1

P C 0.3 m

D 0.3 m

Figure 1.51 Problem 1.11 Solution Load P acts as a compressive force on rod BC and as a tensile force on rod AB, such that the compressive force on rod BC is (P - P1). Compatibility condition Extension in AB + contraction in BC + extension in CD = 0

MTPL0259_Chapter 01.indd 43

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44

Chapter 1

P1 0.3 (P − P1) 0.3 P1 × 0.3 × − × + =0 100 E 200 100 E E

or

0.6 P1 ×

or

0.3 P 0.3 + P1 = 0 2 2

0.75 P1 - 0.15 P = 0 P = 5 P1 P1 = 0.2 P = 0.5 × 20 = 4 N P - P1 = 20 - 4 = 16 kN Axial stresses P1 4, 000 = = +40 N /mm 2 100 100 P − P1 16 × 1, 000 =− =− = −80 N /mm 2 200 200

σ A B = σ CD = σ BC

Problem 1.12 A uniform concrete slab of total weight W is to be attached to hangers as shown in Fig. 1.52, so that the slab remains horizontal. Determine the ratio of cross-sectional areas of wires so that the slab remains in the level. Solution Say Ta and Ts are tensions in aluminium and steel wires Aa and As: For the bar to remain horizontal

D E

70 GPa Ta

Aa

A

T a 2, 000 70 × 1, 000 Aa A = × × = 0.5833 a (1.19) 200, 000 As T s 1, 200 As

L=2m

1.2 m

dla = dla T a × 1, 200 T s × 2, 000 = E a × Aa E s × As

As

B

200 GPa

Ts G

C

W 1.5

1m

2m 0.5 m

Figure 1.52

Problem1.12

Taking moments about G of slab Ta × 1 = Ts × 2 Ta = 2Ts, pushing this line in Eq. (1.19)

or

2T s A = 0.5833 a Ts As Aa = 3.428 As

Problem 1.13 Two bars A and B are subjected to the same axial load P. Bar A is circular tapered from diameter 2d to d over axial length L. The bar B is of uniform diameter d throughout the length of the bar. Compare the strain energy UA/UB of two bars (see Fig. 1.53). Material of both the bars is the same.

MTPL0259_Chapter 01.indd 44

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Simple Stresses and Strains A P

B

P 2d

d

d

P

L

45

P

L

Figure 1.53

Problem 1.13

Solution Say Young’s modulus = E Tapered bar A

Strain energy,

δ lA =

4PL 2PL = π E (2d )(d ) π Ed 2

UA =

1 P2 L Pδ l A = 2 π Ed 2

δ lB =

4PL π Ed 2

UB =

1 2P2 L Pδ lB = 2 π Ed 2

Uniform bar B

Strain energy,

UA 1 = UB 2 Problem 1.14 A steel wire rope of a diameter of 10 mm is used for hosting purposes during construction of a building. If a 50-m-long wire rope is hanging vertically and a 1-kN load is being lifted at the lower end of wire, determine the total extension of wire. Given the specific weight of wire rope = 0.06 N/cm3 and E = 200 kN/mm2. Solution Wire rope diameter, Cross-sectional area, Weight at lower end,

d = 10 mm

π π × d 2 = × 10 2 = 78.54 mm 2 4 4 W = 1,000 N A=

E = 200 × 100 N/mm2 L = 50 m Specific weight, Extension due to load W,

w = 0.06 × 10-3 N/mm3

δ l1 =

1, 000 × 50, 000 WL = AE 7, 800 × 2, 00, 000

= 3.183 mm

MTPL0259_Chapter 01.indd 45

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46

Chapter 1

Extension due to self-weight

δ l2 =

−3 wL2 0.06 × 10 (50, 000) = 2E 2 × 2, 00, 000

2

= 0.375 mm Total extension = dl1 +dl2 = 3.183 + 0.375 = 3.558 mm Problem 1.15 A load W is suspended from a hook of a crane and is being lowered at a speed of 1 m/s. At a particular instant when the length of the unwound chain is 8 m, the motion is suddenly arrested. The chain links are made of a 10-mm round steel bar, as shown in Fig. 1.54. Determine the maximum load that the chain can carry under this condition if the instantaneous stress produced in the chain is not to exceed 100 N/ mm2. E for steel = 210 GPa (neglect the effect of loss of PE of the weight) and g = 9.81 m/s2. Solution Diameter of chain section, d = 10 mm π Cross-sectional area, A = 2 × × 10 2 = 157.07 mm 2 4 (Note that there are two sections) Length L = 8 m V = 1 m/s

10 mm

Figure 1.54

Chain link

E = 2,10,000 N/mm2 KE lost by the weight =

W  V 2  W × 12 = g  2  2 × 9 .81 W × (1, 000 ) = 50.97 W N mm 2 × 9.81 × 1, 000 2

=

Strain energy absorbed by chain,

U= =

σ i2 × A × L 2E σ i2 × 157.07 × 8, 000 2 × 2,10, 000

= 2.992 σ i2 Neglecting the effect of PE loss by weight due to extension in chain, 2.992σ i2 = 50.97 × W Nmm

σ i2 = 100 N/mm 2 W =

MTPL0259_Chapter 01.indd 46

2.992 × 100 2 = 587 N 50.97

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Simple Stresses and Strains

Problem 1.16 A copper bar of a diameter of 20 mm and a length of 1.6 m is rigidly fixed to a bracket as shown in Fig. 1.55. The other end of the collar rests on a support. A mass of 20 kg moving horizontally along the bar at a velocity of 1.5 m/s is brought to rest by a collar at the other end. The bracket deflects horizontally by 0.05 mm/kN of the tensile force induced in the bar. Calculate the maximum tensile stress developed in the bar. E = 105 kN/mm2.

47

1.5 m/s

Bracket

Figure 1.55

Solution Mass,

m = 20 kg

Velocity,

v = 1.5 m/s KE =

1 2 1 mv = × 20 × 1.52 2 2

= 22.5 N m = 22.5 × 103 N mm Say stress developed in bar = si N/mm2

π 2 × (20 ) = 100π = 314.16 mm 2 4 Length of bar = 1.6 m = 1,600 mm Cross-sectional area,

A=

Force developed in bar,

P = σ i′ A = 314.16 σ i′

Deflection in bracket = 0.05 × 10 −3 mm/N

δ i = 0.05 × 10 −3 × P

Deflection,

= 0.05 × 10 −3 × 314.16 σ i′ = 0.015708 σ i′ . Work done by sudden load P, = Pdi 31.4.16 si × 0.015708 si = 4.935 si2 Strain energy in bar =

σ i′ 2 × AL σ i′ 2 × 314.16 × 1, 600 = 2E 2 × 105 × 100 = 2.3936 σ i′ 2

using the principle of conservation of energy. (2.3936 + 4.935) σ i′ 2 = 22.5 × 103 = 22, 500

σ i2 =

22, 500 = 3, 070 7.3286

σ i = 55.4 N/mm 2 Problem 1.17 A 150-mm-long vertical rod is rigidly secured at its upper end and a collar is provided at the lower end. A weight of 15 N is allowed to fall freely on the collar through a distance of 15 mm.

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48

Chapter 1

What is the maximum stress developed in the upper part if the length of the upper part is 90 mm. The diameters of the upper and lower parts are 16 and 8 mm, respectively. The bar is made of copper (Fig. 1.56). E =108 kN/mm2. (Neglect loss of PE due to extension in bar). Solution W = 15 N h = 15 mm Wh = 15 × 15 = 225 Nmm Say stress developed in the upper part is s1 and stress developed in the lower part is s2

σ 1A2 = σ 2 A2 for equilibrium π × 16 2 = 64π mm 2 4 π A2 = × 82 × 16π mm 2 4 A1 =

d1 = 16 mm

90 mm

σ 1 × 64π = σ 2 × 16π σ 2 = 4σ 1

(1.20) d2 = 8 mm

L1 = 90 mm

60 mm

L2 = 60 mm Volume,

V1 = 90 × 64π = 18095.5 mm 2

h

V 2 = 60 × 16π = 3015.93 mm 2 Loss of PE = gain in strain energy of the bar 225 = =

Figure 1.56

σ σ × V1 + ×V 2 2E 2E 2 1

2 2

σ 12 × 18095.5 σ 22 × 3015.43 + 2E 2E

225 × 2 E = 18095.5σ 12 + ( 4σ 1 ) × 3015.93 2

= σ 12 [18095.5 + 48254.8] = σ 12 × 66350.30

σ 12 =

450 × 108, 000 60350.3

= 732.476

σ 1 = 27.06 N /mm 2 σ 2 = 4σ 1 = 108.24 N /mm 2

MTPL0259_Chapter 01.indd 48

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Simple Stresses and Strains

49

Problem 1.18 Figure 1.57 shows rods ABC and CDE, hinged at points A and C. Rod ABC is connected to another rod BF at point B. Rod BF has a uniform rectangular section of 10 × 20 mm. Determine the magnitude of P of the forces, for which the normal stress in rod BF is 80 MPa. F

A

B

P

P

C

D

E

Hinged 0.2 m

0.2 m

0.15 m

0.15 m

Figure 1.57 P

P + 0.5P T A

C

B 0.2 m

C P = Rc 2 0.15

0.2 m

D

E P = RE 2 0.15

Figure 1.58 Solution Normal stress in rod BF, s = 80 MPa Cross-sectional area, A = 10 × 20 = 200 mm2 Tensile force in rod BF, T = 200 × 80 = 16,000 = 16 kN Beams ABC and CDE are hinged at C, as shown in figure 1.58 Consider beam CDE Moments at C, P × 0.15 = 0.3R E Reaction at E, RE = 0.5P Reaction at C, RC = P - 0.5P = 0.5P Beam ABC To balance reaction RC, force at C for beam ABC will be 0.5P + P = 1.5 P Taking moments about A, T × 0.2 = 1.5 × 0.4 0.6 T = = 3P 0.2 However, T = - 16 kN 16 P= = 5.333 kN 3

MTPL0259_Chapter 01.indd 49

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50

Chapter 1

Problem 1.19 Two 6-m-long steel rods AB and BC are connected at B as shown in Fig. 1.59. Determine the diameter of each rod if the working stress in the rod is limited to 100 MPa. Calculate the vertical displacement of point B. Given E = 200 GPa. C A

30°

30° 120°

6m

B B B

150 kN B

B 60° B

Figure 1.59 Solution Using Lame’s theorem, FBC FA B 150 kN = = sin 120° sin 120° sin 120° FAB = FBC = 150 kN

Area,

Stress = 100 MPa π 1, 50, 000 A= = 1, 500 mm 2 = d 2 100 4 4 × 1, 500 π = 43.20 mm

D iameter of bars A B or BC =

Extension in bars,

δ BC =

100 6 × 105 = 3 mm × 6, 000 = 2 × 105 E

Point B will come down to B′. Angle of 30° will change slightly. In the enlarged figure BB′ is vertical displacement, B′B″ (extension in bar BC). BB ′ =

B ′B ′′ 3 mm = cos 60° 0. 5

= 6 mm

MTPL0259_Chapter 01.indd 50

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Simple Stresses and Strains

51

Problem 1.20 The copper bars BE and CD each have a cross-sectional area of 60 mm2.. E for copper is 100 GPa. Determine the deflections at the points A, B and C due to the force of 1,500 N at C. Refer to Fig. 1.60. Solution Taking moments of the forces about point B, 1, 500 × 200 = FA D × 100 FAD = 3,000 N (tensile) (internal pulling at joint) For equilibrium FBE = 1, 500 + 3, 000 = 4, 500 N Extension in BE,

4, 500 300 × 60 1, 00, 000

δ BE =

= 0.225 mm Extension in AD,

3, 000 300 × 60 1, 00, 000

δ AD =

= 0.15 mm C

1500 N

200 mm E B 100 mm A 300 mm

Figure 1.60

D 300 mm

Problem 1.20

BG 0.225 = BG = 1.5AG A G 0.15

Distance,

BG + AG =100 mm

B

2.5AG = 100 mm AG = 40 mm BG = 60 mm Deflection, dc =

MTPL0259_Chapter 01.indd 51

dC

C

Deflections at B and A are shown in Fig. 1.61.

0.225 × (200 + 60 ) = 0.975 mm. 60

0.225 dBE

G dAD A 0.15 mm

Figure 1.61

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52

Chapter 1

Key Points to Remember  Tensile stress is normal to the plane and pointing away from the plane.  Compressive stress is normal to the plane and pointing towards the plane.  Shear stress is parallel to the plane and positive shear stress tends to rotate the body in the clockwise direction.  Negative shear stress tends to rotate the body in the anticlockwise direction.  Material obeys Hooke’s law within the elastic limit.  Whenever a shear stress is applied on a body, a negative shear stress develops on a perpendicular plane, this shear stress is known as complementary shear stress.  Young’s modulus, E =

N ormal stress Longitudinal strain

.

 A specimen is being tested under the axial tensile or compressive load. If the load is removed from the specimen within the elastic limit, then all the strains are recovered. However, if the load is removed in the plastic stage, then a residual strain remains in the specimen.  Tapered bar, change in length = 4PL . π EDd  Tapered flat, change in length, dl =

 B PL ln   . Et ( B − b)  b 

 Extension in bar due to self-weight, δ l =

wl 2 . 2E

 Bar of uniform strength, A2 = A1e wl / σ .  Bulk modulus, K =

p Volumetric stress . = ε v Volumetric strain

 To solve a statically indeterminate problem, an equation of compatibility is necessary.  Strain energy, U =

1 Pdl. 2

 Strain energy per unit volume =  Modulus of resilience =

σ2 . 2E

σ e2 . 2E

 Stress due to sudden load, ss = 2 × sg (gradual stress) =  Instantaneous stress developed in a bar, σ i =

2W . A

W 2 EAh  1+ 1+ A  WL 

.

 Mild steel is a ductile material. In the σ − ε curve, there is an upper yield point and a lower yield point. This type of yielding is called discontinuous yielding.  Barba’s law dl = bl + c A , where b and c are Barba’s constants.

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Simple Stresses and Strains

 Stress concentration factor at abrupt change in a dimension, SCF =  Factor of safety, F OS =

53

σ max . σ av

U ltimate strength . Allowable stress

Review Questions 1. Explain the following: Hooke’s law, elastic limit, resilience, toughness, ductility, index of cold working and Poisson’s ratio. 2. With the help of sketches, explain the difference between positive shear stress and negative shear stress. 3. Explain how the stress due to sudden load is two times the stress when the same load is gradually applied. 4. Make a sketch of load extension diagram of mild steel and mark important points on it. 5. Explain the difference between the following: (i) resilience and toughness (ii) upper yield point and lower yield point 6. 7. 8. 9. 10.

(iii) elastic stage and plastic stage σ Consider a bar subjected to an axial strain and show that volumetric strain is (1 − 2v) ,where s is axial stress, E v is Poisson’s ratio and E is Young’s modulus. Explain why the stresses developed in a rope due to its own weight are negligible. What is the effect of gauge length on percentage elongation of a bar? Differentiate between sudden load and impact load. Explain how complementary shear stresses are developed in a body.

Multiple Choice Questions 1. A bar is subjected to an axial tensile stress. If the volumetric strain in the bar is 0.44 times the axial strain, what is the Poisson’s ratio of the material? (a) 0.44

(b) 0.30

(c) 0.28 (d) None of these 2. On a plane, resultant stress is inclined at an angle of 30° with the plane. If the normal stress on the plane is 50 MPa, what is the shear stress on the plane? (a) 43.3 MPa

(b) 86.6 MPa

(c) 100 MPa (d) None of these 3. A bar is of square section a × a subjected to a tensile load P. On a plane inclined at 45° to the axis of the bar, normal stress will be (a) 2P/a2

(b) P/a2

(c) P/2a2 (d) P/4a2 4. A 100-mm-long steel bar is tested in tension, so that the change in the length is 0.05 mm. If E = 200 GPa, what is the stress developed in the bar?

MTPL0259_Chapter 01.indd 53

(a) 200 MPa

(b) 100 MPa

(c) 50 MPa (d) None of these 5. Two tie rods are connected through a pin of a cross-sectional area of 40 mm2. If the tie rods carry a tensile load of 10 kN, the shear stress in the pin is (a) 125 MPa

(b) 250 MPa

(c) 500 MPa

(d) None of these

6. A 10-m-long wire rope is suspended vertically from a pulley. The wire rope weighs 12 N/m in length. The cross-sectional area of the wire rope is 20 mm2. What is the maximum stress developed in the wire? (a) 1.2 N/mm2

(b) 2.4 N/mm2

(c) 3.0 N/mm2

(d) 6 N/mm2

7. A spherical ball of volume 1,000 cm3 is subjected to a hydrostatic pressure of 90 N/mm2, and bulk modulus of the material is 190 kN/mm2 What is the change in volume of the ball?

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54

Chapter 1 (a) 473 mm3

(b) 940 mm3

(a) Ductility

3

(c) 502 mm (d) None of these 8. A bar of a length of 100 mm and a cross-sectional area of 64 mm2 is tested under tension. The Barba’s constants for the material are b = 0.21 and c = 0.5. What is the percentage elongation in the bar? (a) 30

(b) 25

(a) Upper yield point (b) Breaking load (c) Ultimate load

(d) None of these

12. A circular tapered bar tapering from a diameter of 20 mm to 10 mm over a length of 1,000 mm is subjected to an axial force of 10 kN. If E = 105 N/mm2, what is the change in length of bar?

(c) 21 (d) 20 9. Poisson’s ratio of aluminium is (a) 0.30

(b) Strength

(c) Resilience (d) None of these 11. In s –e curve for mild steel, load at which point considerable extension occurs with decrease in resistance is known as

(b) 0.33

(c) 0.35 (d) None of these 10. In stress–strain curve, the area up to the elastic limit stress indicates which mechanical property?

10 mm π 2 (c) mm π

(b)

(a)

21 mm π

(d) None of these

Practice Problems 1. A hole of a diameterof 40 mm is to be punched in a 1-mm-thick mild stress sheet. If the ultimate shear stress of mild steel is 490 N/mm2, determine the force required to punch the hole. 2. Two parts of a certain machine component are joined by a rivet of a diameter of 25 mm. Determine the shear and normal stresses in the rivet if P = 20 kN and the angle of joint is 30° to the axis of the load, as shown in Fig. 1.62. 3. A rigid square plate 1 × 1 m is supported over four legs A, B, C and D. A load W is applied at point K as shown in Fig. 1.63. Determine the compressive force in each leg. If the cross-sectional area of each leg is 20 mm2 and W = 1,000 N, what are the stresses in each leg? W

D

0.4 m A

C 0.4 m

K

30° B P

1.0 m

P

Figure 1.62

Figure 1.63

4. A rectangular tapered steel bar tapering from a section 30 × 20 mm to 12 × 10 mm area over an axial length of 100 mm is shown in Fig. 1.64. It is subjected to an axial force P. If the maximum stress is not to exceed 100 MPa, what is the value of P? If E = 210 GPa, what is the change in the length of the bar?

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Simple Stresses and Strains

55

5. The cross-section of a bar is given by (169 + 0.01x 2 ) mm 2 , where x is its distance from one end of the bar in millimetres. If the length of the bar is 200 mm, find the change in length under a load of 5 kN and E = 2 × 105 N/mm2. 6. Two portions of a member AB are glued along a plane inclined at an angle a to the horizontal as shown in Fig. 1.65. Determine the range of angle a if the normal stress in joint is not to exceed 4.8 N/mm2 and the shear stress in joint is not to exceed 5.4 N/mm2. 7. A circular tapered bar ABC is shown in Fig. 1.66. It is made of copper. How much axial force P is required so that the total contraction in the bar is 0.1 mm and E = 1,05,000 N/mm2.

20

P

12

P

30 mm

x

10

dx 100 mm

Figure 1.64

9 kN

A A

a

B C

P

B

30 mm

15 mm

0.2 m

0.2 m

P

15 mm 15 mm

Figure 1.65

MTPL0259_Chapter 01.indd 55

Figure 1.66

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56

Chapter 1

P

P

d

40

40

L 300 mm

Figure 1.67 8. A round brass bar is shown in Fig. 1.67. It is subjected to a tensile force of 50 kN. What should be the diameter of the middle portion if the stress in the A middle portion is not to exceed 150 MPa? What should be the length of the C middle portion if the total extension in the bar is 0.30 mm? E for brass = 100 kN/mm2. 5m 4m 9. A load of 2,500 N is suspended through a pulley of radius 0.1 m by a rope of a cross-sectional area of 40 mm2. If E = 2 × 105 N /mm 2 , what is the stress in rope and what is the downward movement of load (Fig. 1.68)? W [Hint: Tension will be same throughout, T = ] • R 2 10. A 480-mm-long stepped copper bar is subjected to an axial force as shown in R = 0.1 m B Fig. 1.69. The diameter at two steps are 30 and 15 mm, respectively. Calculate the stresses in each portion of the bar. What is the change in the length of the 2500 N bar? E for copper = 105 GPa. [Hint: Portions AB and BC will be in compression and portion CD will be in Figure 1.68 tension.] 11. A 1-m-long bar is rigidly held between two supports as shown in Fig. 1.70. What is the value of P if the stress in portion BC is 40 MPa and what is the stress developed in portion AB?

D = 30 mm A

Copper bar d = 15 mm

B C

20 kN

A1 = 100 mm2

D P

30 kN

A2 = 50 mm2

B

10 kN

P

A

C

100 180 mm

300 mm

Figure 1.69

0.4 m

0.6 m

Figure 1.70

12. A rigid beam is pin supported at A by two wire hangers B and D as shown in Fig. 1.71. A hoisting crane located at C lifts a load of 1 kN as shown. Determine the axial stress and deformation in wire hangers made of steel. The diameter of wires is 3 mm and E = 208 GPa. 13. Compare the strain energy absorbed by two bars A and B as shown in Fig. 1.72 subjected to the same amount of axial force. The material of both the bars is the same. Bar A is stepped bar of diameters 2d and d over axial length L/2 each. Bar B is of uniform section of diameter d throughout its length L.

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Simple Stresses and Strains

1m A

B

57

1m C

D W

2m

1m

1m

dB

dD

Figure 1.71 A B P

2d

d

L 2

L 2

P

P

P d

I

Figure 1.72 V 14. A steel wire of a diameter of 5 mm is used for hoisting purposes during construction of a building. If a 100-m-long wire is hanging vertically and a load of 0.6 kN is being lifted at the lower end of the wire, determine the total extension in wire. Given the specific weight of steel = 0.07644 N/cm3 and E = 210 kN/mm2. Bracket Support 15. A load W is suspended from a crane hook and is being lowered at a speed of 1 m/s. At a particular instant, when the length of unwound chain is 6.5 m, the motion is suddenly arrested. Chain links are made of 8 mm round steel bar. Figure 1.73 Determine the maximum load that the chain can carry under these conditions, if the instantaneous stress produced in the chain is not to exceed 120 MPa. E for steel = 200 kN/mm2 and g = 9.81 m/s2. Neglect the effect of loss of PE of weight. 16. A steel bar of a diameter of 3 cm and a length of 2 m is rigidly fixed to a bracket as shown in Fig. 1.73. The other end of the bar with the collar rests on a support. A weight of 25 kg moving horizontally along the bar at a velocity of 3 m/s is brought to rest by a collar at the other end. The bracket deflects horizontally by 0.04 mm/kN of the force induced in the bar. Calculate the maximum tensile stress developed in the bar. E = 210 GPa and g = 9.80 m/s2. 17. A 1,200-mm-long vertical steel rod is rigidly secured at its upper end and a collar is provided at the lower end. A weight of 10 N is allowed to fall freely onto the collar through a height of 20 mm as shown in Fig. 1.74.

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58

Chapter 1

A1 = 100 mm2 800 mm F

Steel bar

P

A2 = 50 mm2

400 mm

A

B

5 kN

C

P D

10 N

E

20 mm 0.15 m

0.25 m

0.2 m

0.1 m

Figure 1.75

Figure 1.74

The cross-sectional areas of the upper and lower parts are 100 mm2 and 50 mm2, respectively. Determine the stress developed in the lower part, if E = 200 GPa. 18. Figure 1.75 shows rods ABC and CDE, hinged at A and C. Rod ABC is connected to another rod BF at point B as shown. Rod BF has a rectangular section of 15 mm × 10 mm. Determine the stress in rod BF if P = 5 kN. 19. Two 5-m-long copper rods AB and BC are connected at B as shown in Fig. 1.76. Determine the diameter of each rod if the stress in rod is not to exceed 120 MPa. What is the vertical movement of point B? E = 100,000 N/mm2. 20. The steel bars BE and AD each has a cross-sectional area of 5 mm × 16 mm. E for steel is 200 GPa. Determine the deflection at the points A, B and C. (Fig. 1.77)

C 1.6 kN C

A

240 mm E

45°

45°

B

5m

80 mm D A

B 320 mm

320 mm

100 kN

Figure 1.76

MTPL0259_Chapter 01.indd 58

Figure 1.77

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Simple Stresses and Strains

59

Special Problems 1. A circular bar of a cross-sectional area of A is subjected to a force P as shown in Fig. 1.78. If w is the weight density of the material and E is the Young’s modulus, what is the change in length of the bar? 2. A stepped bar ABCD has a length of 2.25 m with the cross-sectional areas of 4, 3 and 2 cm2, respectively, as shown in Fig. 1.79. If the normal stress of the largest magnitude in the stepped bar is not to exceed 80 MPa, determine the magnitude of force F. If E = 200 GPa, what is the displacement of point D? [Hint: Tensile forces 7F, 3F and F in AB, BC and CD.] 3. A plain concrete compression specimen is tested up to failure. The original sample has a diameter of 150 mm and a length of 250 mm when subjected to compressive loads. The specimen fails suddenly by material break down, since the specimen is too short for instability to occur. The force in kilonewtons and the deformation in millimetres are as follows, respectively:

a

P

b

Figure 1.78

(0.0, 0.0), (140, 0.065), (280, 0.10), (425, 0.20), (495, 0.25), (570, 0.325),(635, 0.40), (705, 0.65), (685, 0.80), (565, 0.90) Plot the stress versus strain curve for the concrete specimen. What is the ultimate stress of concrete? What is E for 30 per cent of the ultimate stress? [Hint: Find slope of s–e curve at 30 per cent of suc.]

4 cm2 A

B

3 cm2 C

2 cm2 D

4. A mild steel test piece of a gauge length of 200 mm is 2F F 4F marked off in a length of 20 mm and is tested up to destruction. The extension in each marking measured from one end is equal to 3.2, 3.6, 4.6, 5.0, 9.6, 8.4, 5.2, 4.4, 4.6 and 3.2 1.0 m 0.5 0.75 mm. By using gauge lengths of 80, 120, 160 and 200 mm, respectively, explain the effect of gauge length on the percentage elongation Figure 1.79 [Hint: Consider gauge length from the centre.] 5. A small light piston of 100 mm2 in area compresses oil in a rigid container of 16 litres capacity. When a weight of 37.5 N is applied gradually on the piston, its movement is observed to be 16.8 mm. Find the bulk modulus of oil (Fig. 1.80).

37.5 N a = 100 mm2 16.8 m

V = 16 litres

Figure 1.80

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60

Chapter 1

Answers to Exercises -8.835 kN, -0.476 × 10-3 Exercise 1.12: 470 m, dV = 128 mm3 2 -3 Exercise 1.13: σ 1 = +165 N/mm 2 , +20 N/mm , 0.526 × 10 s = -19.88 N/mm2, t = -19.88 N/mm2 σ 2 = −42.5 N/mm 2 3 3 -0.1147 × 10 , -0.2009 × 10 , -0.459 × 103, -0.0449 mm Exercise 1.14: 9.15 Nm, 42.297 Nm, -3 Exercise 1.6: dl = +0.45 mm, db = -6.75 × 10 mm, 0.2244 Nmm/mm3 where b is side, that is, b = 30 mm Exercise 1.15: 20.37 MPa, 0.097 mm Exercise 1.7: (a) 18.85 kN, (b) + 0.382 mm Exercise 1.16: 114.35 N/mm 2 , 1.0937 mm, bar will not break Exercise 1.8: -5.246 kN -2 -2 Exercise 1.9: +1.273 × 10 mm, +0.637 × 10 mm, 114.35 N/mm 2 , 1.0937 mm, bar will not break -4.138 × 10-2 mm, overall, -2.228 × 10-2 mm Exercise 1.17: (105 kN/mm2, 310 N/mm2). Exercise 1.10: (7.644 N/mm2, 0.191 mm) Exercise 1.18: 1.59 Exercise 1.19: 37.27 mm Exercise 1.11: dm = 54.52 mm Exercise 1.1: Exercise 1.2: Exercise 1.4: Exercise 1.5:

Answers to Multiple Choice Questions 1 2 3 4

(c) (b) (c) (b)

5 6 7 8

9 10 11 12

(b) (d) (a) (b)

(b) (c) (a) (c)

Answers to Practice Problems 1. 61.575 kN

11. P = 16 kN, +60 N/mm2

2. 20.37 N/mm 2 , 35.284 N/mm 2

12. 53.05, 106.1 MPa, 0.51mm, 0.255 mm

3. RA = 350 N, RB = 250, RC = 150, RD = 250 N

U = 0.75; σ A = −17.5 MPa , σ B = −12.5 MPa , σ C = −7.5 MPa , σ D =13. −12.5A MPa UB Pa , σ B = −12.5 MPa , σ C = −7.5 MPa , σ D = −12.5 MPa 14. 14.551 + 1.82 = 16.371 mm 4. P = 12 kN, δ l = 0.021 mm 15. 461.56 N 5. 0.019 mm 16. 69.4 N/mm2 6. 7°50′–69°44′ 7. 12.373k/N 8. 20.6 mm, L = 163.9 mm 9. 31.25 N/mm 2 , 0.727 mm

17. 44.72 N/mm2 18. 44.43 MPa, normal

19. 27.4 mm, 8.486 mm 10. −28.29 N/mm 2 , − 113.177 N/mm 2 , + 56.59 N/mm 2 , δ l = − 0.0485 mm   2 20. dAD = 0.096 mm, dBE = 0.128 mm , dC = 0.80 mm , − 113.177 N/mm 2 , + 56.59 N/mm 2 , δ l = − 0.0485 mm

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Simple Stresses and Strains

61

Answers to Special Problems 1.

w(a + b) 2 Pb + 2E AE

2. 4.571 kN, 0.486 mm

MTPL0259_Chapter 01.indd 61

4. Percentage elongation varies from 35.25 to 25.4 per cent 5. 3.57 × 103 N/mm3

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2 Composite Bars and Temperature Stresses CHAPTER OBJECTIVES In this chapter, we will learn about: 

A composite bar made of two bars of different materials rigidly fixed together so that both bars strain together under external load.



Since strains in the two bars are same, the stresses in the two bars depend on their Young’s modulus of elasticity. In other words, a stiffer bar will share major part of external load.



In a composite system the two bars of different materials are not joined together, but they support a third rigid bar. Load application on the third rigid bar is such that the change in length of both bars is the same, producing different stresses in two bars.



An assembly of two bars, placed coaxially, but having different lengths. Load is applied on the assembly such that both bars share the external load, depending upon Young’s modulus and the difference in their lengths.



A bolt and a tube assembly in which the bolt is tightened by a nut on the bolt and the tube gets compressed. Tensile force in the bolt is equal to the compressive force in the tube.

MTPL0259_Chapter 02.indd 62



A single bar fixed at ends is subjected to a temperature change, which causes strain and stress developed in the bar due to the prevention of free expansion or free contraction in the bar.



In a composite bar with two rigidly fixed components of different materials, stresses are developed in both the components as the temperature changes. As both the bars are rigidly fixed all along their lengths, both bars strain by the same amount causing one bar to get extended beyond its free expansion and the other bar to get compressed from its free expansion causing stresses of opposite nature developed in both bars. Yet equilibrium is maintained, that is, tensile force in one bar is equal to compressive force in another bar.



To circumvent the introduction of temperature stresses in reinforced cement concrete (RCC), cement concrete mix and mild steel bars having nearly same coefficient of thermal expansion are taken.

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Composite Bars and Temperature Stresses

63

Introduction There are a number of examples of composite bars used in industry such as (a) bimetallic strip, used for temperature control in refrigerators; (b) RCC columns/slabs in which cement concrete is reinforced with steel bars to share the tensile load, as concrete is a brittle material and can take up only compressive load and (c) filched beams used in house construction at hilly areas, in which a wooden beam is sandwiched between two steel flitches. In such composite bars and many others in engineering applications, stresses of different magnitude are developed due to external load. Similarly stresses of opposite nature are developed in the two components if the temperature of the composite bar is changed.

Stresses in a Composite Bar Consider a composite bar, a solid bar, of diameter d completely encased in a hollow tube of outer diameter D and inner diameter d, as shown in Fig. 2.1. Grips at ends are provided so that load can be applied. Say under a tensile force P, the composite bar of length L extends by dL. Young’s modulus of the tube = E1 Young’s modulus of the bar = E2 Change in length of the bar, dL = change in length of the tube, dL. Using Hooke’s law, s2 s × L = 1 × L (where s1 and s2 are stresses developed in the tube and the bar, respectively) E2 E1 s1 s 2 = E1 E2

or

s1 E1 = s 2 E2 p A1 = ( D2 − d 2 ) 4 π 2 A2 = d 4

or Area of cross-section of the tube, Area of cross-section of the bar,

(2.1)

F F D

D

P

d

d

L dL

Figure 2.1 Composite bar

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Chapter 2

64

Force shared by the tube, Force shared by the bar, But total force,

P1 = s1A1 P2 = s2A2 P = P1 + P2 = s1A1 + s2A2 = s1 A1 +

(2.2)

E s1 E2   A2 = s1  A1 + 2 A2  E   E1 1

(2.3)

s E  =  2 1  A1 + s 2 A2  E2 

or

E   = s 2  A2 + 1 A1  E2  

(2.4)

From these equations, stresses in the bar and the tube can be worked out. s s Change in length, δ L = 1 × L = 2 × L E1 E2 Example 2.1 A brass bar of 40 mm diameter is completely encased in a steel tube of outside diameter 60 mm so as to make a composite bar of length 400 mm. Assembly is subjected to a compressive load such that the stress developed in brass bar is 50 MPa. How much compressive load is applied on the composite bar? What is the stress developed in the steel tube? Given Es = 2Eb = 210 GPa. What is the change in length of the composite bar? Solution Diameter of the brass bar, Outside diameter of the steel tube, Area of cross-section of the brass bar, Area of cross-section of the steel tube, Stress in the brass bar, Load shared by the brass bar,

d = 40 mm D = 60 mm Ab =

π × 402 = 400p mm2 4

π (602 − 402 ) = 500p mm2 4 sb = 50 N/mm2 Pb = sb × Ab As =

= 50 × 400p N = 62,832 N = 62.832 kN E s s = s b × s = 50 × 2 = 100 MPa Eb

Stress in the steel tube,

Ps = s s × As = 100 × 500π N

Load shared by the steel tube,

= 1,57,080 N = 157.08 kN Total compressive load on the composite bar,

P = Ps + Pb

62.832 + 157.08 = 219.912 kN

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Composite Bars and Temperature Stresses

Change in length of the composite bar, δ L = =

65

sb × 400 Eb 50 × 400 105, 000

= 0.190 mm Exercise 2.1 A flat bar of steel of 24 mm wide and 6 mm thick is placed between two aluminium alloy flats 24 mm × 9 mm each so as to form a composite square section. The three flats are fastened together at their ends. An axial tensile load of 20 kN is applied to the composite bar. What are the stresses developed in steel and aluminium alloy? Es = 3Ea = 210 kN/mm2 [Hint: Aa = 2 × 9 × 24 = 432 mm2]

Composite System Two or more wires or bars of different materials attached to a rigid bar may be defined as a composite system. The load W is applied at such a location that the rigid bar remains horizontal; the change in length of two bars/wires is the same, as we have already learnt in the section as shown in Fig. 2.2 ‘Stresses in a Composite Bar’. Consider bars AB and CD of areas of cross-section A1 and A2, respectively. Bar AB is E1 and say Young’s modulus of bar CD is E2. Then Taking moments about B A

W × a = W2 × (a +b) Reaction,

Wa W2 = a+b W1 = W − W2 =

E1

Wb a+b

Stress in bar CD,

s2 =

Corresponding strains are

ε1 =

s1 W = 1 E1 A1 E1

ε2 =

s2 W = 2 E2 A2 E2

MTPL0259_Chapter 02.indd 65

a W1

B

W1 A1

s1 =

Therefore,

.

A1

A2

L

Stress in bar AB,

Changes in length, dL = e1L = e2L

C

W2 A2

W AE ε1 = 1 × 2 2 ε 2 A1 E1 W2

b D

E2 W2

dL

W

Figure 2.2

Composite system

(2.5)

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66

Chapter 2

s1 W1 A2 = × s 2 A1 W2

Ratio of stresses,

(2.6)

The bar will remain horizontal under such conditions. But if the position of the load suspended on the rigid bar is such that dL1 ≠ dL2, say dL1 > dL2, then the rigid bar will deflect from horizontal position making an angle a, such that  dL − dL2   Difference in changes in lengths  α = tan −1  1 = tan −1   Distance between two bars   a + b  Example 2.2 A rigid bar is attached to two wires, one of steel (3 mm f) and other of copper (2 mm f), of length 1 m each as shown in Fig. 2.3. A load of 300 N is suspended at the centre. The distance between the wires is 200 mm. If Es = 2Ecu = 208 kN/mm2, what is the angle of inclination of the rigid bar with the horizontal?

3 mm φ 1m

1m 200 mm

Steel

Solution A load of 300 N is applied at the centre, so reaction Ws = Wcu =

2 mm φ

Copper 100 Wcu

Ws A

B 300 N

300 = 150 N 2

Figure 2.3

Diameter of the copper wire = 2 mm Diameter of the steel wire = 3 mm Stress in the copper wire,

s cu =

150 × 4 = 47.75 N/mm2 π × 22

Stress in the steel wire,

ss =

150 × 4 = 21.22 N/mm2 π × 32

Change in length of the copper wire, δ Lcu =

47.75 × 1,000 = 0.459 mm 1,04,000

δ Ls =

21.22 × 1,000 = 0.102 mm 2,08,000

Change in length of the steel wire, Angle of inclination of the rigid bar,

α = tan −1

(δ Lcu − δ Ls ) a+b

 0.459 − 0.102  = tan −1    200  0.357  = tan −1  = tan −1 (1.785 × 10−3 )  200  a = 0.1°

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67

Exercise 2.2 A rigid bar is suspended from two wires, one of aluminium and other of copper, length of the wire is 1.2 m and diameter of each is 2.5 mm. A load of 500 N is suspended on the rigid bar such that the rigid bar remains horizontal. If the distance between the wires is 150 mm, determine the location of line of application of load. What are the stresses in each wire and by how much distance the rigid bar comes down? Given Es = 3Ea = 201 kN/mm2.

Bars of Different Lengths P

Let us consider two bars, one is solid of diameter d and length L and the other is hollow of diameters d2 and d1 and length L2, placed co-axially as shown in Fig. 2.4. An axial load P is applied such that it is shared by both hollow and solid bars, meaning that a contraction in the length L1 of the solid bar is more than the gap between the two bars. P = P1 + P2

Load,

(2.7)

d , gap d d1

L1

d2

= Load shared by the solid bar + load shared by the hollow bar

E2 L2

E1

d = gap = L1 − L2 Stresses in the bars,

σ1 =

P1 4P = 1 A1 π d 2

σ2 =

P2 4 P2 = A2 π ( d22 − d12 )

Figure 2.4

Contraction in the solid bar − gap = Contraction in the hollow bar 4 P1 L1 4 P2 L2 × − = gap π d 2 E1 π ( d22 − d12 ) E2

(2.8)

Knowing the various parameters, ratio of P1/P2 can be obtained from Eq. (2.8). Then, the magnitude of P1 and P2 can be determined from Eq. (2.7). Example 2.3 An aluminium bar of diameter 15 mm and length 200 mm is co-axially placed in a hollow copper bar of inner diameter 16 mm and outer diameter 20 mm, with length 199.9 mm. An axial compressive load of 30 kN is applied on the top of the solid bar as shown in Fig. 2.4. If E for copper is 105 kN/mm2 and E for aluminium is 67 kN/mm2, determine the stresses developed in both the bars. Solution Diameter of the solid bar, d = 15 mm Area of cross-section of the aluminium bar, A1 =

π × 152 = 176.7 mm2 4

Say, load shared by the aluminium bar = P1 Load shared by the copper tube = P2 P1 + P2 = 30,000 N Length, L1 = 200 mm E1 = 67,000 N/mm2

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Chapter 2

L2 = 199.9 mm

Length,

E2 = 1,05,000 N/mm2 Area of cross-section of the copper tube =

π (202 − 162 ) 4

= 36p = 113.1 mm2 P1 L1 P2 L2 × − = gap = 200 − 199.9 A1 E1 A2 E2 or

P1 × 200 P2 × 199.9 = 0.1 mm − 176.7 × 67, 000 113.1 × 1, 05, 000

P1 × 1.6893 × 10−5 − P2 × 1.6893 × 10−5 = 0.1 P

P1 − P2 = 5,940.7 N P1 = P2 + 5,940.7 N P1 + P2 = 30, 000 N P1 = 17970.36 N P2 = 12,029.63 N Stresses

160.04 mm

16

17970.36 = 101.7 N/mm2 (compressive ) 176.7

σ2 =

12029.63 = 106.36 N/mm2 (compressive ) 113.1

160 mm

20 30

σ1 =

Gap 0.04 mm

Steel

Copper

Figure 2.5

Exercise 2.3 A steel bar of diameter 16 mm and length 160 mm is placed co-axially in a copper tube of outer diameter 30 mm and inner diameter 20 mm and axial length 160.04 mm. A load of 36 kN is placed on the top of the tube shown in Fig. 2.5. Determine the stresses developed in the bar and the tube if Es = 200 GPa and Ecu = 105 GPa.

Bolt and Tube Assembly Consider a bolt and tube assembly, wherein a bolt passes co-axially through a tube as shown in Fig 2.6. Washers are fixed at the ends of the tube. A nut on the threaded part is screwed so as to tighten the assembly, producing a tensile stress in the bolt and a compressive stress in the tube. Say stress developed in the bolt = sb, a tensile stress Stress developed in the tube = st, a compressive stress Diameter of the bolt = d Outer and inner diameters of the tube = d2 and d1 as shown

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Washer

69

Washer Nut

Tube

Thread d1

d2

Bolt

d

L Bolt head

Bolt and tube assembly

Figure 2.6

Area of cross-section of the bolt =

π 2 d 4

Area of cross-section of the tube =

π 2 (d − d12 ) 4 2

Compressive force in the tube,

Pt = σ t ×

π 2 (d − a12 ) 4 2

Tensile force in the bolt,

Pb = σ b ×

π 2 d 4

We have not applied any external load, but these loads are developed due to tightening of the nut on the bolt. For equilibrium, Pt = Pb π π σ t ( d22 − d12 ) = σ b × d 2 (2.9) 4 4 If Eb = Young’s modulus of the bolt Et = Young’s modulus of the tube Strain in the tube,

εt =

σt Et

Strain in the bolt,

εb =

σb Eb

Contraction in the length of the tube,

δt =

σt ×L Et

Extension in the length of the bolt,

δb =

σb ×L Eb

Axial movement of the nut = extension in the length of the bolt + contraction in the length of the tube or (σ b /Et + σ b / Eb ) L = axial movement of the nut.

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Chapter 2

Note that numerical value of (σ t / Et ) × L is taken, otherwise st is a negative stress and sb is a positive stress. Example 2.4 A steel bolt of diameter 18 mm passes through a copper tube of inner diameter 24 mm and outer diameter 30 mm. Washers are placed on both the sides of the tube. The bolt has threads of pitch 2.4 mm at one end. Nut is turned on the bolt through 60° so as to tighten the assembly. Determine the stresses developed in the bolt and the tube. The length of the assembly is 40 cm. Es = 2Ecu = 210 kN/mm2. Solution Pitch of the threads, Angle of rotation of the nut,

p = 2.4 mm = 60° 2.4 × 60° = 0.4 mm = 360° π Ab = × 182 = 254.47 mm2 4 π At = × (302 − 242) = 254.47 mm2 4

Axial movement of the nut, Area of cross-section of the bolt, Area of cross-section of the tube, Say, stress in the bolt = sb, tensile Stress in the tube = st, compressive For equilibrium,

sb Ab = st At sb × 254.47 = st × 254.47 sb = st, both are equal in magnitude. Now,  σb σt   E + E  L = 0.4 mm b t

σ t  0.4  σb  2,10, 000 + 1, 05, 000  = 400 sb + 2st = 210 210 σb = = 70 N/mm2 (tensile) 3 st = 70 N/mm2 (compressive) Exercise 2.4 Figure 2.7 shows a steel bolt of diameter 20 mm and length 200 mm centrally passing through a copper tube of length 150 mm and outer diameter 40 mm and inside diameter 28 mm. The screw on the bolt has a pitch of 2 mm and initially the nut is just tight. If the nut is tightened through 45°, what are the stresses developed in the bolt and the nut? Es = 210 GPa, Ecu = 100 GPa [Hint: Note that the length of the bolt is 200 mm but the length of the tube is 150 mm, during the axial movement of the nut take this fact into account.]

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71

Thread

Tube

20 mm f

40 28 Bolt

150 mm Bolt and tube assembly

50

Figure 2.7

Temperature Stresses in a Single Bar

Fixed end

Fixed end a

d

If a bar is held between two rigid supports and its temperature is E raised, then a compressive stress is developed in the bar and if its temperature is reduced, then a tensile stress is developed in the bar due to the reason that its free thermal expansion is prevented by L rigid supports. a∆TL Let us consider a bar of diameter d and length L rigidly held between two supports as shown in Fig 2.8. Say, a is the coefficient of Figure 2.8 linear expansion of the bar and its temperature is raised by ∆T (°C). Free thermal expansion in the bar = a∆TL Final length = (L + a∆TL) But the final length remains the same, so the fixed ends exert compressive force on the bar and thus the length (L + a∆TL) is compressed to L. α∆TL α∆TL L= Compressive strain = L + α∆T L = α∆T Compressive stress = a∆TE (using Hooke’s law) sT = thermal stress = a∆TE π Area of cross-section of the bar = d 2 4

π Compressive force developed in the bar = (α ∆TE ) d 2. 4 Sometimes, this force becomes excessive and the bar buckles. Similarly, if the temperature of the bar is reduced, the tensile stress developed in the bar = a∆TE and if the thermal tensile stress is excessive, then the bar will break in tension. Example 2.5 A rail line is laid so that there is no stress in rails at 30°C. The rails are 30 m long and there is a clearance of 5 mm between the rails. If the temperature of the rail rises to 60°C, what is the stress developed in the rails? a for the rail material = 11.5 × 10−6/°C, E = 210 GPa

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Chapter 2

Solution Length of the rails = 30 m Coefficient of linear expansion, a = 11.5 × 10−6/°C Temperature rise, ∆T = 60 − 30 = 30°C Free expansion in the rails = 30,000 × 11.5 × 10−6 × 30 = 10.35 mm Clearance between the rails = 5 mm Free thermal expansion prevented by the rails = 10.35 − 5 = 5.35 mm

ε=

Strain developed in the rails,

5.35 = 0.1783 × 10−3 30, 000

E = 2,10,000 N/mm2 Compressive stress developed in the rails, eE = 0.1783 × 10−3 × 2,10,000 = 37.43 N/mm2. Exercise 2.5 A rail line is laid so that there is no stress in the rails at 24°C. Determine the stress developed in the rails if its temperature rises by 35°C. The rails are 30 m long and there is no clearance between the rails. E = 210 GPa and a = 11.5 × 10−6/°C. What will be the stress in rails if its temperature falls by 12°C?

Temperature Stresses in a Composite Bar A composite bar is made up of two bars of different materials perfectly joined together so that during temperature change both the bars expand or contract by the same amount. The coefficient of expansion of the two bars is different; therefore, thermal stresses are developed in both the bars. Consider a composite bar of different materials with coefficients of expansion and Young’s modulus of elasticity, as a1, E1 and a2, E2, respectively, as shown in Fig. 2.9. The temperature of the bar is raised by DT. Free expansion in bar 1 = a1∆TL Free expansion in bar 2 = a2 ∆TL a 1∆TL a 1,E 1

A1

a 2,E 2

A2 a 2∆TL L

∆L

Figure 2.9 Temperature stresses in a composite bar

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73

But, both the bars strain together by ∆L as shown such that ∆L < a1∆TL ∆L > a2∆TL In other words, (L + a1∆TL) of bar 1 gets compressed to (L + ∆L) and bar 2 gets further extended to L + ∆L from (L + a2∆TL) Consequently, a compressive stress is developed in bar 1 and a tensile stress is developed in bar 2. Compressive strain in bar 1,

α1 ∆TL − ∆L α1 ∆TL − ∆L  L + α1 ∆TL L

Tensile strain in bar 2,

∆L − α 2 ∆TL ∆L − α 2 ∆TL  L + α 2 ∆TL L

Compressive strain in bar 1 + tensile stress in bar 2 = a1∆T − a2∆T e1 + e2 = (a1 − a2)∆T

or

It s1T and s2T are the temperature stresses in bars, then

σ1T σ 2T + = (σ1 − σ 2)∆T E1 E2

(2.10)

There are two unknowns but, only one equation, so we have to make another equation. We have not applied any external force on the bar, but for equilibrium the compressive force in bar 1 is equal to the tensile force in bar 2. s1T A1 = s2T A2

or for equilibrium

(2.11)

From these two equations, the temperature stresses s1T and s2T can be calculated. It the temperature of the composite bar is reduced, then a tensile stress will be developed in bar 1 with a1 > a2 and a compressive stress will be developed in bar 2 with a2, thermal coefficient. In RCC, to avoid the development of thermal stresses due to temperature variation, the coefficient of thermal expansion of mild stress reinforcement is more or less the same as that of concrete. Example 2.6 A flat steel bar of 20 mm × 10 mm is fixed with aluminium flat of 20 mm × 10 mm so as to make a square section of 20 mm × 20 mm. The two bars are fastened together at their ends at a temperature of 26°C. Now the temperature of whole assembly is raised to 55°C. Find the stress in each bar. Es = 200 GPa, Ea = 70 GPa, as = 11.6 × 10−6/°C, aa = 23.2 × 10−6/°C. Solution Temperature rise, Area of cross-section of steel,

∆T = 55 − 26 = 29°C As = Aa, area of cross-section of aluminium

10 × 20 mm2 = 10 × 20 mm2 Say, the stresses developed are s1T in aluminium and s2T in steel. s1T × 200 = s2T × 200 s1T = s2T Moreover,

MTPL0259_Chapter 02.indd 73

(2.12)

σ1T σ 2T + = (σ1 − σ 2 )∆T E1 E2

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Chapter 2

σ 1T

+

70, 000

σ 2T = (23.2 − 11.6) × 10−6 × 29 20, 000

σ1T σ 2T + = 3.364 7 20 or

0.14285s1T + 0.05s2T = 3.364 s1T = s2T (numerically)

but

0.19285s1T = 3.364

So,

s1T = 17.443 N/mm (comp) in aluminium sT = 17.443 N/mm2 (tensile) in steel. 2

Exercise 2.6 A flat steel bar of 20 mm × 8 mm is placed between two copper bars of 20 mm × 6 mm each so as to form a composite bar of section of 20 mm × 20 mm. The three bars are fastened together at their ends when the temperature of each is 30°C. Now the temperature of the whole assembly is raised by 30°C. Determine the temperature stress in the steel and copper bars. Es = 2Ecu = 210 kN/mm2, as = 11 × 10−6/°C, acu = 18 × 10−6/°C. Problem 2.1 A weight of 250 kN is supported by a short concrete column of 25 cm × 25 cm in section strengthened by four steel bars near the corners of the cross-section. The diameter of each steel bar is 30 mm. Find the stresses in the steel and the concrete. Es = 15 Ec = 210 GPa. If the stress in the concrete is not to exceed 2.1 N/mm2, what area of the steel bar is required in order that the column may support a load of 350 kN? Solution Load, W = 250 kN Area of cross-section = 25 cm × 25 cm = 625 cm2 = 62,500 mm2 Diameter of steel bars, d = 30 mm π Area of cross-section of steel bars, As = 4 × × 302 = 2827.4 mm2 4 Area of cross-section of the concrete, Ac = 62,500 − 2,827.4 = 59,672.6 mm2 Load,

W = ss × 2,827.4 + sc × 59,672.6 = 2,50,000 N

ss = 15sc as Es /Ec = 15, where ss = stress in steel bars and sc = stress in the concrete. Putting this value, 15 × sc × 2,827.4 + sc × 59,672.6 = 2,50,000 sc (42,411 + 59,672.6) = 2,50,000 sc =

250, 000 = 2.45 N/mm2 1, 02, 083.6

ss = 2.45 × 15 = 36.75 N/mm2

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Stress in the concrete = 2.1 N/mm2 Stress in the steel = 15 × 2.1 = 31.5 N/mm2 Say, As = area of the steel (62,500 − As) = Ac, area of the concrete As × 31.5 + 2.1(62,500 − As) = 3,50,000 29.4As = 3,50,000 − 1,31,250 = 1,28,750 2,18, 750 As = = 7,440 mm2 (area of cross-section of steel bars). 29.4 Problem 2.2 Two aluminium strips are rigidly fixed to a steel strip of section 25 mm × 8 mm and 1 m long. The aluminium strips are 0.5 m long each with section 25 mm × 5 mm. The composite bar is subjected to a tensile force of 10 kN as shown in Fig. 2.10. Determine the deflection of point B. Es = 3EA = 210 kN/mm2. Solution AC is a composite bar and CB is a single steel bar of section 25 mm × 8 mm Portion CB P = 10,000 N A = 25 mm × 8 mm = 200 mm2 s= dlCB =

1, 000 = 50 N/mm2 200

σ 50 × 500 = × 50 = 0.1190 mm E 2,10, 000

Portion AC P = Ps + PA σ s′ Es = =3 σ a′ Ea As = 25 × 8 = 200 mm2 As = 25 × 2 × 5 = 250 mm2 Ps = ss × 200 PA = 250 × sA Aluminium strip

05 08 05

Steel strip

25 mm A

C B 500 mm

P 10 kN

500

Figure 2.10

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Chapter 2

10, 000 = σ s′ × 200 + 250σ ′A But, σ s′ = 3 × σ A′ as Es = 3EA Therefore,

10, 000 = 3σ ′A × 200 + 250σ ′A

σ ′A =

10, 000 = 11.764 N/mm2 850

σ s′ = 3 × σ ′A = 35.294 N / mm 2 δ lAC =

σ s′ 35.294 × 500 × 500 = = 0.084 mm 2,10,000 Es

Deflection of the point B, dB = dlAC + dlCB = 0.119 + 0.084 = 0.203 mm. Problem 2.3 Figure 2.11 shows a composite bar of steel and copper. The temperature of the bar is raised by 80°C. Determine the compressive force developed 0.5 mm in the bars after the rise in temperature and the change 1000 mm2 600 mm2 in length of the copper bar. The area of the copper bar is Cu Steel 600 mm2 and that of the steel bar is 1,000 mm2. Ecu = 1,05,000 N/mm2, Es = 2,10,000 N/mm2. as = 11 × 10−6/°C, acu = 18 × 10−6/°C, clearance between support and composite bar is 0.5 mm.

0.5 m

0.5 m

Solution Free expansion in bars due to temperature Figure 2.11 change = 18 × 10−6 × 80 × 500 + 11 × 10−6 × 80 × 500 = 1.16 mm Temperature rise = 80°C Clearance = 0.5 mm. Contraction in length of the bars = 1.16 − 0.5 = 0.66 mm Say, the compressive force developed is P Newton. Then, P 500 P 500 × + × 0.66 = 600 1, 05, 000 1, 000 2,10, 000 P = (7.937 × 10−6 + 2.38 × 10−6) 0.66 × 106 = 63,972 N P= 10.317 = 63.972 kN (compressive force developed) P 500 Change in length of the copper bar = acu × ∆T × 500 − × Acu Ecu = 18 ×10−6 × 80 × 500 − = 0.72 − 0.5077 = 0.2123 mm.

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Problem 2.4 Two steel rods each of 4 cm in diameter are joined end to end by means of a turn buckle as shown in Fig. 2.12. The other end of each rod is rigidly fixed and there is initially negligible tension in the rods. The effective length of each rod is 5.4 m. Calculate the increase in tension when the turn buckle is tightened by one-third of a turn. There are threads on each rod with a pitch of 4.2 mm. E = 208 kN/mm2. If a = 11.2 × 10−6/°C (coefficient of linear expansion of steel), what rise in temperature would nullify the increase in tension? Solution Length of the rod, L = 5.4 m Pitch of the threads = 4.2 mm. Movement of the nut = one-third of a turn Axial movement of the nut = Strain developed in the bar,

4 cm

Turn buckle

1 × 4.2 = 1.4 mm 3 e=

Figure 2.12

1.4 × 10−3 5.4 × 1, 000

e = 0.259 × 10−3 E = 2,08,000 N/mm2 Tensile stress developed in the steel bars, s = E:e = 0.2593 × 10−3 × 2,08,000 = 53.926 N/mm2

π (40)2 = 1, 256.6 mm2 4 T = sA = 53.926 × 1,256.6

A=

Area of cross-section, Increase in tension,

= 67,763.4 N = 67.763 kN Increase in temperature to nullify the stress Strain, e = ∆T∆ 0.259 × 10−3 = 11.2 × 10−6 × DT Change in temperature,

DT = 23.125°C.

Problem 2.5 A steel tie rod of diameter 20 mm is encased in a copper tube of external diameter 36 mm and internal diameter 24 mm with the help of washers and nuts. The nut on the tie rod is tightened so as to produce a tensile stress of 50 N/mm2 in steel rod. Now the combination is subjected to a temperature rise of 80°C. Determine the resultant stresses developed in the rod and the tube if as = 11 × 10−6/°C, acu = 18 × 10−6/°C. Es = 2Ecu = 210 GPa. Solution Diameter of the steel rod, d = 20 mm Outer diameter of the copper tube, d1 = 36 mm Inner diameter of the copper tube, d2 = 24 mm

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Chapter 2

Area of cross-section of the steel rod π As = × 202 = 100π mm2 4 Area of cross-section of the copper tube Ac =

π (362 − 242 ) = 180π mm2 4

Stresses due to tightening of the rod

σ s As = σ c Ac 50 × 100p = σ c ×180p

or

σ c = −27.777 N/mm2 (compressive) σ s = +50 N/mm2 ( tensile) Temperature stresses Say, σ ST and σ CT are the temperature stresses developed in the steel rod and in the copper tube

σ ST × 100p = σ CT × 180p σ ST = 1.8σ CT σ ST σ CT + = (α1 − α 2 ) ∆T ES EC σ ST σ CT + = (18 − 11) × 10−6 × 80 2,10, 000 1, 05, 000 sST + 2sCT = 117.6 But sST = 1.8sCT, putting this value 1.8sCT + 2 sCT = 117.6 117.6 = −30.95 N/mm2 (compressive) 3.8 sST = 1.8sCT = + 55.70 N/mm2 (tensile)

sCT =

Resultant Stresses sRS = +50 + 55.70 = 105.70 N/mm2 (tensile) sRC = −27.77 − 30.95 = −58.72 N/mm2 (compressive) Problem 2.6 Three vertical rods carry a tensile load of 80 kN. The area of cross-section of each rod is 400 mm2. Their temperature is raised by 60°C and the load is now so adjusted that they extend equally. Determine the load shared by each rod. The outer two rods are of steel and the middle one is of brass. ES = 2EB = 200 GPa, aS = 11 × 10−6/°C, aB = 18 × 10−6/°C.

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Solution In this problem, stresses due to temperature change and external load are combined. Due to temperature change or rise in temperature, there will be tensile stress in the steel rod and compressive stress in the brass rod. Due to an external load, the tensile stress in the steel is more than that in the brass, because ES > EB. Due to load adjustment, there is an equal deformation in three bars. Say total extension in each rod = d Extension in the steel rods due to load = d − as LT, Extension in the brass rod due to load = d − ab LT, Strain in the steel roads,

δ ES =  − α sT  L 

Strain in the brass rod,

δ  EB =  − αbT  L 

Area of cross-section of each rod, A = 400 mm2 Stress in the steel rod = ES ES Stress in the brass rod = EB EB δ  δ  Total load = 2  − α sT  × 400 Es +  − α sT  × 400 Eb = 80,000 L  L  Putting the value of ES and EB, δ  200 = 2  − α sT  × 200, 000 + L 

δ   − αbT  × 100 × 1, 000 L

200 δ  δ  = 4  − α s T  +  − α bT  L  L  100, 000 1 δ = 5 − 4 × α s T − α bT 500 L =

5δ − 4 × 11 × 10−6 × 60 − 18 × 10−6 × 60 L

1 δ = 5 − 2.64 × 10−3 − 1.08 × 103 500 L 2 × 10−3 = 5 5

δ − 3.72 × 10−3 L

δ = −5.72 × 10−3 L δ = −1.144 × 10−3 L

Now,

δ  −3 −6  − α sT  = 1.144 × 10 − 11 × 10 × 60 L = 1.144 × 10−3 − 0.66 × 10−3 = 0.484 × 10−3

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Chapter 2

δ  −3 −6  − αbT  = 1.144 × 10 − 18 × 10 × 60 L = 1.144 × 10−3 − 1.08 × 10−3 = 0.064 × 10−3 δ  Load on each steel rod =  − α sT  × 400 × 1, 00, 000  L = 0.484 × 10−3 × 400 × 2,00,000 = 38,720 N = 38.72 kN Load on the brass rod

δ  =  − αbT  × 400 × 1, 00 × 1, 000 L  = 0.064 × 10−3 × 400 × 1,00,000 = 2,560 N = 2.56 kN

Total load

= 2 × 38.72 + 2.56 = 77.44 + 2.56 = 80 kN

Problem 2.7 Three vertical wires, one of steel and two of aluminium, are suspended in the same vertical plane from a horizontal support. They are all of same length and same area of cross-section and carry a load by means of a rigid cross bar at their ends. The load is increased and the temperature is changed in such a way that the stress in each wire is increased by 15 N/mm2. Find the change in temperature. ES = 3EA = 210 kN/mm2, aS = 11 × 10−6/°C, aa = 22 × 10−6/°C Solution There will be a direct stress and a temperature stress in each wire. Force due to temperature stresses are balanced; therefore, Say A is area of cross-section of each wire Then,

ssA + sA2A = 3 × 15 × A ss +2sA = 45 N/mm2

Al

S

Al

ss and sA are direct stresses in the steel and in the aluminium.

σs E = s =3 σa Ea

Cross bar

Figure 2.13

3sA + 2sA = 45 sA =

45 = 9.0 N/mm2 5

sS = 3 × 9.0 = 27.0 N/mm2

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Composite Bars and Temperature Stresses

81

Say stresses in the steel and in the aluminium due to the temperature change are sST and sAT, respectively, then sS + sST = 15 sA + sAT = 15 as given in problem, so 27 + sST = 15 sST = −12 N/mm2 sAT = 15 − 9 = + 6 N/mm2 This shows that there is a reduction in temperature, as stress sST in the steel is compressive.

σ ST σ = AT = (22 − 11) × 10−6 × ∆T ES EA 12 6 + = 11 × 10−6 × ∆T 2,10, 000 70, 000 5.714 × 10−5 + 8.571 × 10−5 = 11 × 10−6 × DT 14.2854 × 10−5 = 11 × 10−6 × DT DT = −13°C (reduction in temperature) Problem 2.8 A rigid bar of negligible weight is supported by steel and aluminium bars as shown in Fig. 2.14. It carries a load W = 60 kN. Determine the temperature change that will cause the stress in the steel bar to be 50 N/mm2. aS = 11 × 10−6/°C, aA = 22 × 10−6/°C, ES = 3EA = 210 kN/mm2. Solution Due to the combined effect of load and temperature change, the final stress in the steel wire is 50 N/mm2 Area of cross-section, AS = 300 mm2 Load WS in the steel bar = 300 × 50 = 15,000 N = 15 kN Taking moments of forces about point A,

4m 500 mm2

2m Ws A

300 mm2

Al

Steel

WA

B

C

D

dD

60 kN 1m

1m

1m

WS × 1 − 2 × 60 + WA × 3 = 0 WA =

120 − 15 = 35 kN (putting the value of WS ) 3

Figure 2.14

35, 000 = 70 N/mm2 500 dD = 3dB, as ABCD is a rigid bar, using principle of similar triangles. Stress in the aluminium, σ A =

Now deflection,

δA =

70 × 4, 000 + α A × ∆T × 4, 000 = δ D 70, 000

= 4.0 + 22 × 10−6 × 4000DT = 4.0 + 0.088DT

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Chapter 2

50 × 2000 + as DT × 200 = 0.5 + 11 × 10−6 × DT × 200 2, 00, 000 = 0.5 + 0.022DT

ds =

dA = 3dS

But

4.0 + 0.088DT = 3(0.5 + 0.022DT) 2.5 = −0.022DT DT = −113.63°C (reduction in temperature) Problem 2.9 A steel bar and a brass bar are each secured to a rigid support and fastened together at their ends by a 20-mm diameter pin as shown in Fig. 2.15. The area of cross-section of the steel rod is 600 mm2 and that of the brass bar is 800 mm2. Determine the maximum normal stress induced in the brass bar by 40°C drop in temperature. ES = 2Eb = 200 GPa, aS = 11.2 × 10−6/°C, aB = 18.6 × 10−6/°C. What is the shear stress developed in the pin? Solution Decrease in temperature = 40°C Free contraction in length = L × 40 × 11.2 × 10−6 + L × 40 × 18.6 × 10−6

Brass 800 mm2

−6

= L(448 + 744) × 10

Steel 600 mm2

Pin L

= 1.192 × 10−3L mm Say tensile stress in the bar of brass, sBT Stress in the steel bar, sST For equilibrium, sBT × 800 = sST × 600 sBT = 0.75sST

L

Figure 2.15

(2.13)

Tensile stress will be developed in both bars because there is a reduction in the temperature. So

σ ST σ BT ×L+ × L = 1.1992 × 10−3 L 1, 00, 000 2, 00, 000 sBT + 0.5sST = 119.2 or

0.75sST + 0.5sST = 119.2 1.25sST = 119.2 sST = 95.36 N/mm2 sBT = 71.52 N/mm2

Maximum normal stress in the brass bar is 71.52 of N/mm2 tensile. Now Force,

PS = PB = 95.36 × 600 = 57,216 N

d = 20 mm π Area of cross-section, Apin = × 202 = 100p mm2 4 57, 216 Shear stress developed in the pin = 100π =182.12 N/mm2 Pin diameter,

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Problem 2.10 An aluminium link of square section 30 mm × 30 mm and a steel rod having dimensions shown in Fig. 2.16 at 25°C. The steel rod is cooled until it fits freely into the link. The temperature of the whole assembly rises to an ambient temperature of 25°C. Determine the normal stress in the aluminium link and in the steel rod. ES = 3EA = 210 kN/mm2,

Aluminium link 30 200 30

aS = 11 × 10−6/°C, aA= 22 × 10−6/°C Solution Gap in the link = 200 mm Length of the steel rod = 200.12 mm Extra length = 0.12 mm Reduction in temperature of the steel rod, ∆T =

83

30 200.12 mm

30

Figure 2.16

0.12 = 54.54°C 200 × 11 × 10−6

After the steel rod fits freely in the link, its temperature will rise to 25°C. Steel rod will exert pressure on aluminium link and in reaction aluminium link will compress the steel rod. For equilibrium, sST As = sAT As sST × 900 = sAT × 2 × 900 (Refer to Fig 2.16) So temperature stress in the steel = 2 × temperature stress in the aluminium Now extension in the aluminium link + contraction in the steel rod = 0.12 mm (free expansion during rise in temperature)

σ ST × L σ AT × L + = 0.12 Es Ea σ ST × 200 σ AT × 200 + = 0.12 2,10, 000 70, 000 sST × 9.524 × 10−4 + 28.57 × 10−4 × σ ST = 0.12 But, sST = 2sAT

2 × sAT × 9.524 + 28.57 × sAT = 1,200 sAT (19.048 + 28.57) = 1,200 sAT =

1, 200 = 25.2 N/mm2 (tensile) 47.618

sST = 50.4 N/mm2 (compressive) Problem 2.11 A circular copper bar of area 300 mm2 is fitted in a square steel frame of area of crosssection 300 mm2 as shown in Fig. 2.17. At a temperature of 25°C, there is clearance of 0.1 mm between the upper end of copper rod and the top of the frame. Determine the compressive force in the copper rod

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84

Chapter 2

if the temperature of the system is raised to 50°C. Neglect the bending of the frame and the bar. as = 11 × 10−6/°C, acu = 18 × 10−6/°C, Es = 2Ecu = 210 kN/mm2 Solution At temperature 25°C, clearance between the copper bar and the frame is 0.1 mm. At 50°C, say extension in the steel frame = δ mm. Total extension in the copper rod = d + 0.1 mm. As acu >as, the copper rod will tend to extend more than the steel frame, but the steel frame will prevent free expansion of the copper rod and in reaction, the copper rod will exert push on two vertical steel bars and steel bars will extend beyond their freely expanded length. ecu =

Strain in the copper rod,

α cu (50 − 25) × (1, 200 − 0.1) − (δ + 0.1) 1, 200 − 0.1

Taking 1,200 − 0.1; 1,200 ecu = =

18 × 10−6 × 25 × 1, 200 − δ − 0.1 1, 200 0.44 − δ 1, 200

(2.14)

Area of cross-section Acu = 300 mm2 Compressive force in the copper rod = ecu × Ecu × Acu =

0.44 − δ × 1, 05, 000 × 300 1, 200

= (0.44 − d ) × 26,250

(2.15)

Tensile force in two steel bars Fs = esEs2As es =

Where

δ − α s (50 − 25) × 1, 200 1, 200

=

δ − 11 × 10−6 × 25 × 1, 200 δ − 0.33 = 1, 200 1, 200 0.1 mm Steel frame Cu bar

120 mm

300

300 mm2

300

Copper bar

Figure 2.17

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85

 δ − 0.33  Fs =  × 2,10, 000 × 2 × 300  1, 200  = (d − 0.33) × 1,05,000

(2.16)

But for equilibrium, compressive force in one copper rod = tensile force in two steel bars (0.44 − d ) 26,250 = (d − 0.33) × 1,05,000 (0.44 − d ) 0.25 = (d − 0.33) 0.11 − 0.25d = d − 0.33 d=

Strain in the copper rod,

ecu = =

0.44 = 0.352 mm 1.25 0.44 − δ 0.44 − 0.352 = 1, 200 1, 200 0.088 1, 200

Compressive force in the copper rod = ecu × Acu × Ecu =

0.088 × 300 × 1, 05, 000 = 2,310 N 1, 200

= 2.31 kN Checking Fs = (d − 0.33)1,05,000 from Eq. (16) = (0.352 − 0.33) × 1,05,000 = 2,310 N = 2.31 kN Solution is correct. Problem 2.12 A rigid bar AOB is shown in Fig. 2.18 pinned at O and attached to two vertical rods of steel and aluminium as shown. Initially the rigid bar is horizontal and vertical rods are stress free. Determine the stress in the aluminium rod if the temperature of the steel rod is decreased by 40°C. Neglect the weight of bar AOB. α s = 11.7 × 10−6/°C, α a = 23 × 10−6 /°C, Es = 200 GPa, Ea = 700 GPa. Solution The steel rod will try to contract when temperature is reduced; its free contraction will be prevented by the aluminium rod connected at the other end. Say, sST = stress in the steel sAT = stress in the aluminium Ps = sST × 300 PA = sAT × 1,200

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86

Chapter 2

Al

300 mm2

S

1.2 m

0.9 m

1200 mm2 O

dA A

B

0.6 m B

1.0 m

dB

Figure 2.18 Taking moments of forces about O, PS × 0.6 = PA × 1.0 sST × 0.6 × 300 = sAT × 1,200 × 1.0 sST =

1, 200 sAT 180

sST = 6.667sAT Extension in the aluminium bar =

(2.17)

σ AT × 1,200 EA

Net contraction in the steel bar = α × 900 × 40 −

σ ST × 900 = δ A , Es

But dA = 0.6dB, because AOB is a rigid bar. dA = 11.7 × 10−6 × 900 × 40 −

σ ST × 900 200, 000

= 0.4212 − 4.5 × 10−3 × sST (of steel) dB =

σ AT × 1, 200 σ AT × 1, 200 = EA 70, 000

= 17.14 × 10−3 × sAT (of aluminium) But dA = 0.6dB (by similar triangles) So,

0.4212 − 4.5 × 10−3 × sST = 0.6 × 17.14 × 10−3sAT = 10.284 × 10−3sAT

But sST = 6.667sAT from Eq. (2.17) 0.4212 − 4.5 × 10−3 × 6.667 × sAT = 10.284 × 10−3sAT

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87

0.4212 = sAT (30 + 10.284) × 10−3 = 40.284sAT × 10−3 sAT =10.455 N/mm2 Tensile stress developed in the aluminium bar. Problem 2.13 A rigid steel plate is supported by three vertical posts of 2 m height each but accidentally the height of middle post is 0.5 mm less as shown in Fig. 2.19. The area of cross-section of each post is 200 mm × 200 mm. Determine the safe value of the load P if the allowable stress for concrete in compression is 16 N/mm2. Econ = 12 kN/mm2.

P

0.5 mm 200 × 200 mm

2.0 m

Solution sallowable = 16 N/mm2. Under the action of load P, there will be less strain or less stress in middle post II. So, stress in the outer posts = 16 N/mm2. Stress in the middle post = s N/mm2 Compatibility equation,

I

II

III

Figure 2.19

16 σ × 2, 000 − × 2, 000 = 0.5 Econ Econ 32,000 − 2,000 s = 12,000 × 0.5 = 6,000 2,000 s = 26,000 s = 13 N/mm2 Safe load,

P = 2 × 200 × 200 × 16 + 200 × 200 × 13 = 4 × 104(32 + 13) = 180 × 104 = 1,800 kN

A

Problem 2.14 Two steel rods BE and CD, each of 16 mm in diameter are connected through a link ABC as shown in Fig. 2.20. The end of the rod BE is single threaded with a pitch of 2.5 mm. After being snugly fitted, the nut at B is tightened by one full turn. Determine (a) the tension in rod CD and (b) the deflection at point C of the rigid member ABCE = 200 GPa. Solution Due to tightening of the nut on the bolt, tensile force in CD, say P1, and tensile force in BE, say P2, are developed, taking moments of forces about point A,

MTPL0259_Chapter 02.indd 87

400 mm P2

B E

200 mm C

P1

D 2m

2m

Figure 2.20

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88

Chapter 2

P2 × 400 = P1 × 600 P2 = 1.5P1 d = 16 mm π A = × 162 = 64π mm2 4

Diameter of rods, Area of cross-section Change in length of CD,

dlCD =

Change in length of BE,

dlBE =

But axial movement of the nut = δ lCD + =

(2.18)

P1 × 2, 000 AE P2 AE

× 2, 000

600 δ lBE 400

P1 P × 2, 000 + 1.5 × 2 × 2000 AE AE

Putting the values of A and B, =

P1 × 2, 000 P 1.5 × 2, 000 + 2 × 64p × 2, 00, 000 64p 2, 00, 000

2.5 = P1(4.9736 × 10−5) + P2 × 7.4604 × 10−5 2.5 × 105 = 4.9736P1 + 7.4604P2 Putting P2 = 1.5P1 2.5 × 105 = 4.9736P1 + 7.4604 × 1.5P1 = P1(4.9736 + 11.190) = 16.1636P1 P1 =

2.5 × 105 = 0.15466 × 105 16.1636

= 15.466 kN (tension in rod CD) P2 = 23.200 kN dBE =

P2 23.20 × 1, 000 × 2, 000 × 2, 000 = AE 64p × 2, 00, 000

= 1.15387 mm Deflection of point C will be equal to 1.5dBE = 1.7308 mm. Problem 2.15 A circular ring is suspended by three vertical bars A, B and C of different lengths. The upper rods of the bars are held at different levels. Bar A is 2 m long, 15 mm diameter and made of steel. Bar B is 1.6 m long 12 mm diameter and made of copper. Bar C is 1 m long and 18 mm diameter and made of aluminium. A load of 30 kN is hung on the ring. Calculate how much of this load is carried by each bar, if the circular ring remains horizontal after the application of load. Es= 2Ecu = 3 Ea = 210 GPa.

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89

Solution Say, the load shared by steel, copper and aluminium bars is PS, Pcu and PA respectively. Length of the steel bar, Ls = 2,000 mm Length of the copper bar, Lcu = 1,600 mm Length of the aluminium bar, LA = 1,000 mm π Area of cross of the steel bar, As = × 152 = 176.7 mm2 4 π Area of cross of the copper bar, Acu = × 122 = 113.1 mm2 4 π Area of cross of the aluminium bar, AA = × 182 = 254.5 mm2 4 Change in lengths,

But So,

δ Ls =

Ps Ls × As Es

δ Lcu =

Pcu Lcu × Au Ecu

δ La =

Pa La × Aa Ea

δ Ls = δ Lcu = δ La Ps Ls P L × = cu × cu As Es Acu Ecu Ps 176.7 1, 600 Es = × × = 1.25 × 2 = 2.5 Pcu 113.1 2, 000 Ecu

Similarly

Ps 1, 000 Es 176.7 E = × × = 0.347 × s = 1.0415 Pa 2, 000 Ea Ea 254 Pa =

Ps = 0.96 Ps 1.0415

Ps = 0.4 Ps 2.5 Ps + Pcu + Pa = 30 kN Pcu =

Ps + 0.4 Ps + 0.96 Ps = 30 kN 30 = 12.71 kN 2.36 Pcu = 5.0847 kN Ps =

Pa = 12.20 kN Total = 29.9963 kN Loads shared by the steel, copper and aluminium bars are 12.71, 5.085 and 12.20 kN, respectively.

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Chapter 2

Key Points to Remember  A composite bar made of two bars of areas of cross-section A1 and A2 with E1 and E2 as Young’s modulii of elasticity. If it is subjected to an external load W, then W = W1 + W2 W = A1s1 + A2s2, where

σ 1 E1 = σ2 E2

 In a composite system, a rigid bar is supported by two wires/bars of different materials, with E1 and E2 being Young’s modulii of elasticity. A load W is suspended on the rigid bar and the load can be applied at such a position that a rigid bar remains horizontal.  If a load is applied on two bars placed co-axially, then a compatibility equation has to be formed involving the gap between the axial lengths of two bars.  In a bolt and tube assembly, due to tightening of the nut on the bolt, tensile stress is developed in the bolt and compressive stress is developed in the tube. Then, sb Ab = st·At

(2.19)

 Extension in length of bolt + contraction in length of tube = axial movement of nut on bolt. If

a bar of length L, Young’s modulus E and the coefficient of thermal expansion a, then during change of temperature DT of bar, stress is developed in bar if it is held between two rigid supports. sT = aEDT

 A composite bar of length L is composed of two bars of area of cross-section and Young’s modulus A1, E1 and A2, E2, with their coefficient of thermal expansion being a1, a2, respectively, and s1T and s2T are temperature stresses developed in two bars. Then, s1T A1 = s2T A2

σ 1T σ 2T + = (α1 − α 2)∆T E1 E2

Review Questions 1. What is a composite bar? Under the action of an external load on the bar, state how stresses are developed in different components of the composite bar. 2. What is the difference between a composite system and a composite bar. Explain with the help of a bimetallic strip and a bolt and tube assembly. 3. Consider a composite bar with its temperature changed. Explain how tensile and compressive stresses are induced in two components of the composite bar. 4. Consider a bar rigidly held between two supports. Explain why the bar bends if its temperature is increased. 5. Consider a composite bar with external load and change in temperature. Explain the development of resultant stresses due to the direct load and the temperature change in two components of the composite bar.

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91

Multiple Choice Questions 1. A composite bar is made of steel and aluminium strips, with Aa = 3As, where Aa and As are areas of cross-section of aluminium and steel bars, respectively. Es/Ea = 3. Due to an external load, if the stress developed in the aluminium is 30 MPa, then what is the stress developed in the steel bar? (a) 10 Mpa (b) 30 MPa (c) 90 MPa (d) None of these 2. A copper bar of area of cross-section 200 mm2 is encased in a steel tube of area of cross-section 400 mm2. Due to an external load, the stress in copper bar is 10 MPa and the load on composite bar is P. What is the load shared by the steel bar? Es /Ecu = 2. (a) 0.5 P (b) 0. 6 P (c) 0.8 P (d) None of these 3. In a bolt and tube assembly, the pitch of thread on the bolt is 2.4 mm. If the nut is tightened by onequarter of a turn and a reduction in length of the tube is 0.4 mm, what is the increase in length of the bolt? (a) 0.2 mm (b) 0.4 mm (c) 0.6 mm (d) None of these 4. In a composite bar, bars having same cross-sectional area, one bar shares one-third of the total load. What is the ratio of E1/E2? (a) 3 (b) 2 (c) 1.5 (d) None of these 5. A bimetallic strip is made of two metals with equal areas of cross-section. Due to temperature change, the stress developed in one strip is −40

6.

7.

8.

9.

10.

N/mm2. What is the stress developed in another component of the composite bar? (a) −40 N/mm2 (b) +20 N/mm2 2 (c) +40 N/mm (d) None of these A wire of diameter 1 mm is held between two rigid supports. The temperature of the wire drops by 10°C, á = 12 × 10−6/°C, E = 2,00,000 N/mm2. What is the stress developed in the wire? (a) +24 N/mm2 (b) −24 N/mm2 2 (c) +12 N/mm (d) None of these What is the approximate ratio of Esteel/Ealuminium (a) 2.0 (b) 3.0 (c) 2.5 (d) None of these Three strips of same area of cross-section share a load of 5.5 kN. If their Young’s modulus are in the ratio of E1 = 2E2 = 3E3, then what is the load shared by the strip with Young’s modulus E1? (a) 3 kN (b) 3.5 kN (c) 2.5 kN (d) None of these A steel bar of diameter 20 mm is encased in a copper tube of outside diameter 30 mm. An external load produces a stress of 30 N/mm2 in the steel bar. What is the stress developed in the copper tube? Given Es/Ecu = 2. (a) 30 N/mm2 (b) 24 N/mm2 2 (c) 15 N/mm (d) None of these A steel rail track is laid by joining 30 m long rails end to end. At 30°C, there is no stress in the rails. At 50°C, what will be the stress in the rails if a = 11 × 10−6/°C and E = 200 × 103 N/mm2. (a) 88 MPa (c) 22 MPa

(b) 44 MPa (d) None of these

Practice Problems 1. A short hollow cast iron column of external diameter 25 cm and internal diameter 20 cm is filled with concrete. The column carries a total load of 350 kN. If ECI = 6 ECon, calculate the stress in cast iron and concrete. What must be the internal diameter of the cast iron column if a load of 420 kN is to be carried and stresses and external diameter of the column remain unchanged? 2. Two brass strips are rigidly fixed to a steel strip of section 20 mm × 6 mm and length 1.2 m. The brass strips are 0.6 m long each with section 20 mm × 4 mm. The composite bar is subjected to a tensile force of 12 kN as shows in Fig. 2.21. Determine the deflection of point B if ES = 2 EB = 200 GPa.

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Chapter 2

Brass strip 20 4 6 4

Steel A

C B 600

12 kN

600 mm

Figure 2.21 3. Figure 2.22 shows a composite bar of bronze and 0.6 mm aluminium. The temperature of the composite bar is raised by 100°C. Determine the compressive force 450 360 developed in the bars after the rise of temperature and B A the change in length of aluminium bar. The area of cross-section of bronze bar is 1,300 mm2 and of aluminium bar is 1,600 mm2Eb = 105 GPa, Ea = 70GPa, ab = 18 × 10−6/°C, aa = 23 × 10−6/°C. Aluminium Bronze A = 1600 mm2 A = 1300 mm2 4. Two steel rods each 2.5 cm in diameter are joined end to end by means of a turn buckle. The other end of Figure 2.22 each rod is rigidly fixed and there is initially a small tension in the rods. The effective length of each rod is 4 m. Calculate the increase in this tension, when the turn buckle is tightened one-quarter of a turn. There are threads on each rod with a pitch of 3 mm. E = 210 GPa. If a = 11 × 10−6/°C, what rise in temperature would nullify this increase in tension? 5. A steel rod of diameter 20 mm is encased in a copper tube of an external diameter 36 mm and an internal diameter 24 mm with the help of washers and nuts. The nut on the tie rod is tightened so as to produce a tensile stress of 40 N/mm2 in steel rod. The combination is subjected to a tensile load of 20 kN. Determine the resultant stresses in the steel rod and in the copper tube. Es = 2Ec = 210 GPa.

3m

6. Three vertical rods carry a tensile load of 100 kN. The area of cross-section of each rod is 500 mm2. Their Bronze temperature is raised by 50°C and the load is now so adjusted that they extend equally. Determine the load 400 mm2 Steel 800 mm2 shared by each rod. The outer two rods are of steel and 1.5 m the middle one is of aluminium. Es = 3EA = 201 GPa, WB Ws as = 11 × 10−6/°C, aA = 20 × 10−6/°C. 7. Three vertical wires, one of steel and two of copper, are A B C D suspended in the same vertical plane from a horizontal support. They are all of same length and same area of 1m 1.5 m 1.5 m cross-section and carry a load by means of a rigid cross bar at their ends. The load is increased and the temperature is changed in such a way that the stress in each wire W = 80 kN is increased by 10 N/mm2. Find the change in tempera−6 ture.ES = 2Ecu = 200 GPa, as = 11 × 10 /°C, acu = 18 × Figure 2.23 10−6/°C. 8. A rigid bar of negligible weight is supported by steel and bronze bars shown in Fig. 2.23 and carries a load W. If W = 80 kN, compute the temperature change that will cause the stress in each bar to be 60 N/mm2,

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Composite Bars and Temperature Stresses

9.

10.

11.

12.

93

as = 11 × 10−6/°C, aB = 19 × 10−6/°C, Es = 200 GPa, Al Steel EB = 83 GPa. A steel bar and an aluminium bar are each secured to a rigid support and fastened together at their free ends 500 mm2 by a 20-mm diameter pin as shown in Fig 2.24. The 1000 mm2 2 area of cross-section of steel bar is 500 mm and that of 2 aluminium bar is 1,000 mm . Determine the maximum L L normal stress induced in the aluminium bar by 40°C drop in temperature. Es = 3EA = 210 kN/mm2, as = 11 × 10−6/°C, aa = 22 × 10−6/°C. What will be the shear stress Figure 2.24 developed in the pin? A brass link of square section 40 mm × 40 mm and a Brass steel rod having dimensions shown in Fig. 2.25 at 20°C temperature. The steel rod is cooled until it fits freely 40 in the link. The temperature of the whole assembly rises to an ambient temperature of 20°C. Determine 250 mm the normal stresses in the steel rod and the final length −6 of the steel rod. Es = 2EB = 200 GPa, as = 11 × 10 /°C, 40 aB = 19 × 10−6/°C. 2 A circular aluminium rod of area 300 mm is fitted in a Steel rod square steel frame of area of cross-section 400 mm2 as 40 shown in Fig. 2.26. At a temperature of 25°C, there is 40 mmf a clearance of 0.05 mm between the upper end of rod 250.15 mm and the top of the frame. Determine the compressive force in the aluminium rod if the temperature of the system is raised to 50°C, neglecting the bending of the Figure 2.25 frame and the rod. Given aS = 11 × 10−6/°C, aa = 22 × 10−6/°C, ES = 3Ea = 210 kN/mm2. A rigid bar AOB is shown in Fig. 2.27 pinned at O and attached to two vertical rods of steel and brass as shown. Initially the rigid bar is horizontal and vertical rods are stress free. Determine the stresses in the brass bar and the steel bar if the temperature of the steel bar is developed 50°C. Neglect the weight of the bar. aS = 11.7 × 10−6/°C, aB = 22 × 10−6/°C, ES = 2EB = 210 GPa.

Brass

0.05 mm Steel frame Aluminium rod

Steel

1.0 m

0.8 m

1000 mm2

400 mm2 300 mm2

300

O

400 mm2 A

B 0.5 m

Figure 2.26

MTPL0259_Chapter 02.indd 93

1.0 m

Figure 2.27

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94

Chapter 2

13. A rigid steel plate is supported by three vertical posts of 2.5 m height each but accidentally the height of middle post is 1 mm less as shown in Fig. 2.28. The diameter of each post is 250 mm. Determine the safe value of the load P if the allowable stress for concrete in compression is 18 N/mm2 Econ = 12 kN/mm2. 14. Two steel rods BE and CD each have an area of cross-section 180 mm2. The ends of rods are single threaded with a pitch of 2.4 mm. After being snugly fitted, the nut at C is tightened one full turn. Determine (a) the tension in rod CD (Fig. 2.29) and (b) the deflection at point C of the rigid member ABCE = 200 GPa. P A 150 mm 1 mm

B

100 mm

d = 250 mm

2.5 m

D

C 2m

3m

Figure 2.28 Figure 2.29

Special Problems 1. A short post is made by welding steel plates into a square section and then filling inside with concrete. The side of square is 200 mm and the thickness t = 10 mm as shown in Fig. 2.30. The steel has an allowable stress of 140 N/mm2 and the concrete has an allowable stress of 12 N/mm2. Determine the allowable safe load on the post. EC = 20 GPa, Es = 200 GPa. Refer to Fig. 2.30. [Hint: ( Es Ec ) = 10, if sS = 140, then sC = 14 > 12, so sC = 12 N/mm2, sS = 120 N/mm2] 2. A steel tube of internal diameter 100 mm and 5 mm thick has inner and outer brass liners of same length and thickness. The composite section carries an axial thrust of 200 kN. Calculate the thrust shared by each tube and the amount by which it is shortened in length of 150 mm (Fig. 2.31). ES = 200 GPa, EB = 100 GPa.

Steel Concrete

10 mm 200 mm

Figure 2.30

π π π (110 2 − 100 2 )Ab = (100 2 − 90 2 ) + (120 2 − 110 2 )] 4 4 4 3. A structural steel beam of a bridge structure is 30 m long. A gap of 20 mm is allowed for the expansion at right end as shown in Fig. 2.32. The beam was placed in a position at a temperature of 20°C. Determine (a) the highest temperature at which the beam can touch before it bears against the adjacent abutment and (b) the stress in the beam if its temperature is raised to 90°C. E = 210 GPa, a = 12 × 10−6/°C. [Hint: As =

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Composite Bars and Temperature Stresses

95

Brass liner

Steel tube

120 mm f Brass liner

20 mm 30 m Beam 90 mm 100 mm 110 mm

Figure 2.31

Figure 2.32

4. A tapered bar of steel is rigidly held between two vertical supports as shown in Fig. 2.33. If the temperature of the bar is increased by 20°C,.what is the maximum stress developed in the bar? E = 200 GPa, a = 11.7 × 10−6/°C. 5. A copper wire of diameter 2 mm is stretched tightly between two supports, 1 m apart under an initial tension of 200 N. If the temperature of the wire drops by 10°C, what is the maximum stress in the wire? E = 105 GPa, a = 18 × 10−6/°C.

Steel 20 mm f

40 mm f

300 mm

Figure 2.33

Answers to Exercises Exercise 2.1: 69.45; 23.15 N/mm2 Exercise 2.2: 37.5 mm from steel wire, 76.37 N/mm2, 25.458 N/mm2, downward movement, dL = 0.456 mm Exercise 2.3: −33.45 MPa, −113.72 MPa

Exercise 2.4: 72.6 MPa, +148.13 MPa Exercise 2.5: −84.52 N/mm2, +28.98 N/mm2 Exercise 2.6: +18.9 MPa in steel, −12.6 MPa in copper

Answers to Multiple Choice Questions 1. 2. 3. 4.

(c) (c) (a) (b)

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5. 6. 7. 8.

(c) (a) (b) (a)

9. (c) 10. (b)

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96

Chapter 2

Answers to Practice Problems 1. 2.55 N/mm2, 15.28 N/mm2, internal diameter = 181.7 mm 2. 0.48 mm 3. P = 162.7 kN, dla = 0.382 mm (increase) 4. 39.375 kN; 17.05°C 5. ss = 40 + 33.5 = 73.5 N/mm2 (tensile), sc = −22.22 + 16.75 = −5.47 N/mm2(compressive); 6. Ps = 49.31 kN, PA = 1.38 kN 7. −7.14°C

8. DT = +20.9°C 9. 34.66 N/mm2 (tensile), shear stress in pin = 110.3 N/mm2 10. 6.2 N/mm2 (compressive), 250.66 mm 11. 4.3575 kN 12. sBT = 32.76 N/mm2; sST = 163.83 N/mm2 13. 2,365.9 kN 14. Tension in rod CD = 17.280 kN, dC = 1.935 mm

Answers to Special Problems 1. 1,236 kN = safe load 2. 45.24 kN, 100 kN, 54.76 kN, 0.0433 mm 3. 55.5 + 20 = 75.5°C; s = 36.4 N/mm2 (compressive)

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4. 93.6 N/mm2 (compressive) 5. 82.56 N/mm2

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3 Principal Stresses and Strains CHAPTER OBJECTIVES We will learn about principal stresses, planes of principal stresses and strains on principal planes. Practical examples of principal planes in actual components will also be discussed. In this chapter we will learn about principal stresses/strains. 

Determination of the normal and shear stresses on a plane inclined on two planes on which stresses are given.



Determination of the angle of the plane on which shear stress is zero.



Determination of the maximum and minimum principal stresses and the planes of such stresses.



Graphical method to determine principal stresses at a point. In many typical cases students are confused to locate the centre of Mohr’s stress circle. This concept is clearly discussed.



Typical cases of engineering components with principal planes.



Determination of principal strains on principal planes in terms of principal stresses.



Determination of principal strains with the help of strain gauges and then calculation of principal stresses with the help of principal strains.



With the help of ellipse of stresses, determination of normal and shear stresses on any plane inclined with a given principal plane.

Introduction The efficient design of a machine component is an essential feature of mechanical engineering and design is based on the knowledge of principal stresses at the critical section of a component. For example, a crankshaft in an engine is a very important part as it converts the translatory motion of the piston into the rotary motion of the crank and makes the power available in the form of torque to the engine shaft. So, the determination of principal stresses at desired section in the crank shaft is an important and useful exercise.

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98

Chapter 3

Knowing the principal stresses on the principal planes and using the proper theory of failure, an engineering component is designed. For any practical design problem, first of all principal strains are determined experimentally using strain gauges. Then from these principal strains the values of principal stresses are determined. Using Mohr’s stress circle and Mohr’s strain circle in graphical method, principal stresses and principal strains are determined. To draw the Mohr’s stress/strain circle correctly is a salient feature of this chapter. However, complex may be a state of stress at any point; there always exist a set of three orthogonal planes on which shear stresses are zero and there are only normal stresses on these planes. These planes are called principal planes and normal stresses on these planes are called principal stresses. In a two-dimensional case two principal stresses are determined and the third principal stress is taken as zero.

Stresses on an Inclined Plane (I) Consider a small element ABC of thickness t with planes BC and AC perpendicular to each other. Plane AB makes an angle q with plane BC (i.e., plane BCFE as a reference plane) as shown in Fig. 3.1; say, there are only normal stresses s1 on plane BC and s2 on plane AC. Let us determine normal and shear stresses on inclined plane AB due to stresses on planes BC and AC. Normal force on plane BC, P1 = s1BCt D

sq

A

q

tq

P2 s2

cos q

E F t

B

C

P2 sinq

s1

P1 sinq

q

Internal resistance

P1 cosq P1 = s1Bct

Figure 3.1 A triangular element with stresses on planes

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Principal Stresses and Strains

99

Components of P1 along the plane and perpendicular to the plane AB are P1sin q and P1cos q, respectively. Normal force on plane AC, P2 = s2ACt. Components of P2 along the plane and perpendicular to the plane AB are P2·cos q and P2 sin q, respectively. Stress is the internal resistance per unit area. So, internal resistances are equal and opposite to the applied forces, shown by dotted lines. Normal resisting force on plane AB, Pn = P1 cos q + P2 sin q. Shear resisting force on plane AB, Pt = P1 sin q − P2 cos q. (Note that P1 sin q is a positive shear force as it tends to rotate the element in clockwise direction and P2 cos q is a negative shear force as it tends to rotate the element in anticlockwise direction.) If sq and tq are normal and shear stresses developed on inclined plane AB, then ABsq t = P1 cos q + P2 sin q = s1BCt cos q + s2ACt sin q or

σ θ = σ1

But

BC AC cos θ + σ 2 sin θ AB AB

BC BC = cos θ , = sin θ AB AB sq = s1cos2 q + s2sin2 q = σ1

σθ =

(1 + cos 2θ ) (1 − cos 2θ ) + σ2 2 2

σ1 + σ 2  σ1 − σ 2  + cos 2θ  2  2

Similarly, tq = ABt = P1sin q − P2cos q = s1BCtsin q − s2ACtcos q

τ θ = σ1

BC AC × cos θ sin θ − σ 2 AB AB

= s1cos q sin q − s2sin q cos q =

MTPL0259_Chapter 03.indd 99

(σ1 − σ 2 ) sin 2θ 2

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100 Chapter 3

Example 3.1 Stresses at a point in a stressed body are shown in Fig. 3.2. On plane BC, s1 = +120 MPa and on plane AC, s2 = −60 MPa. Determine normal and shear stresses on inclined plane AB. Solution (Note that the thickness of the element is not shown in the figure; it can be taken as unity.)

sq

tq

s2 = -60MPa 30° B

C s1 = 120 MPa

q = 30°

Figure 3.2

s1 = +120 MPa, s2 = −60 MPa Normal stress, σθ = =

A

σ1 + σ 2 σ1 − σ 2 + cos 2θ 2 2

120 − 60 120 − ( −60 ) + cos 60° 2 2

80 MPa q = 40° B

= 30 + 90 × 0.5 = +75 MPa Shear stress,

A

C 150 MPa

Figure 3.3

Exercise 3.1

σ1 − σ 2 sin 2θ 2 120 − ( −60 ) = sin 60° 2 = +77.44 MPa (a positive shear stress)

τθ =

Exercise 3.1 Stresses at a point in a stressed body are shown in Fig. 3.3. Determine the normal and shear stresses on the inclined plane AB inclined at an angle of 40° to the plane BC.

Stresses on Inclined Plane (II) Consider a small triangular element of thickness t, with planes AB and BC perpendicular to each other and plane AB inclined at an angle q with plane BC (as reference plane). Stresses on plane BC are normal stress s1 and shear stress t and stresses on plane AC are normal stress s2 and shear stress t opposite to the shear stress on plane BC. Forces on Planes Normal force on plane BC, P1 = s1BCt Shear force on plane BC, F1 = tBCt Normal force on plane AC, P2 = s2 ACt Shear force on plane AC, F2 = tACt Total horizontal force, PH = (P2 + F1) Total vertical force, PV = (P1 + F2)

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101

Let us resolve these forces parallel and perpendicular to plane AB (Fig. 3.4). Normal component, Pn = (P1 + F2) cos q + (P2 + F1) sin q (represented by dotted lines, internal resistance is equal and opposite to the force applied on the element is in equilibrium). Tangential component, Pt = (P1 + F2) sin q − (P2 + F1) cos q (shown by dotted lines). Note that (P1 + F2) sin q is a positive shear force, as it tends to rotate the element in the clockwise direction. If sq and tq are normal and shear stresses on inclined plane, then Pn = sq ABt = (P1 + F2) cos q + (P2 + F1) sin q Putting the values of forces sq ABt = s1BCt cos q + tACt cos q + s2ACt sin q + tBCt sin q

σ θ = σ1

BC AC AC BC cos θ + τ cos θ + σ 2 × sin θ + τ sin θ AB AB AB AB

= s1 cos2 q + t sin q cos q + s2 sin2 q + t cos q sin q =

σ1 (1 + cos 2θ ) σ (1 − cos 2θ ) + 2τ sin θ cos θ + 2 2 2

=

σ1 + σ 2 σ1 − σ 2 + cos 2θ + τ sin 2θ 2 2

(3.1)

Similarly, Pt = tq ABt = s1BCt sin q + tACt sin q − s2ACt cos q − tBCt cos q sq

sq

A tq

(P 2 s2 t

E t

C

q

) sin + F2

q

P 2 + F1

t

B

(P 1

)co + F1

(P2 + F1)sinq

s1

q

(P1 + F2) cosq

(P1 + F2)

Figure 3.4

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102 Chapter 3

τ θ = σ1

BC AC AC BC sin θ + τ sin θ − σ 2 cos θ − τ cos θ AB AB AB AB

= s1 cos q sin q + t sin2 q − s2 sin q cos q − t cos2 q =

(σ1 − σ 2 ) sin 2θ − τ (cos2 θ − sin 2 θ ) 2

=

σ1 − σ 2 × sin 2θ − τ cos 2θ 2

(3.2) sq

Note that we have negative shear stress on reference plane. Expressions (1) and (2) are derived for the normal and shear stresses on the inclined plane. If the shear stress on reference plane BC becomes positive as shown in Fig. 3.5, these expressions are changed as follows:

τθ =

t s2

q B

σ + σ 2 σ1 − σ 2 σθ = 1 + cos 2θ − τ sin 2θ 2 2

A

tq

C

t

s1

Figure 3.5

σ1 − σ 2 sin 2θ + τ cos 2θ 2

A

(for the state of stresses shown in Fig. 3.5) Example 3.2 Stresses at a point are as shown in Fig. 3.6. On plane BC, s1 = 100 MPa, t = 30 MPa and on plane AC, s2 = −60 MPa, t = 30 MPa. Determine the normal and shear stresses on the inclined plane AB, q = 30°. At what angle of plane AB, the shear stress will become zero?

B

Solution s1 = +100 MPa, t = 30 MPa

t

s2 = –60

t

30° t = 30

C

s1 = 100 MPa

Figure 3.6

Example 3.2

s2 = −60 MPa, t = 30 MPa

σ1 + σ 2 σ − σ2 = 20 MPa , 1 = 80 MPa 2 2 q = 30°, cos 2q = cos 60° = 0.5, sin 2q = sin 60° = 0.866 σθ =

σ1 + σ 2 σ1 − σ 2 + cos 2θ + τ sin 2θ 2 2

(putting the values) = 20 + 80 × 0.5 + 30 × 0.866 = 20 + 40 + 25.98 = 85.98 MPa

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Principal Stresses and Strains

τθ =

103

σ1 − σ 2 sin 2θ − τ cos 2θ (putting the values) 2

= 80 × 0.866 − 30 × 0.5 = 69.28 − 15 = 54.28 MPa Plane of zero shear stress

τθ =

σ1 − σ sin 2θ − τ cos 2θ 2

= 80 sin 2q − 30cos 2q = 0 tan 2θ =

30 = 0.375 80

q = 10°16.5′

A

tan(2q + 180) = tan 2q

s2 = 80 MPa

so, q = 90 + 10°16.5′ = 100°16.5′

t

q B

C t = 40 MPa s1 = 120

So, angles are q1 = 100°16.5′, q2 = 100°16.5′

Figure 3.7

Exercise 3.2

Exercise 3.2 Stresses at a point are shown in Fig. 3.7. On plane BC, s1 = 120 MPa, t = 40 MPa and on plane AC, s2 = 80 MPa, t = 40 MPa. If q = 35°, what are the normal and shear stresses on the inclined plane AB?

Principal Stresses

sq

For the state of stress of an element shown in Fig. 3.8, we have derived expressions for normal and shear stresses on inclined plane AB (with plane BC as reference plane) as follows:

σθ =

σ1 + σ 2 σ1 − σ 2 + cos 2θ + τ sin 2θ 2 2

τθ =

σ1 − σ 2 sin 2θ − τ cos 2θ 2

tq

A s2 t

q B

t

C

s1

Figure 3.8

Now on a principal plane, the shear stress is zero; therefore,

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104 Chapter 3

τθ = 0 = tan 2θ =

or

σ1 − σ 2 sin 2θ − τ cos 2θ 2

2τ σ1 − σ 2

(3.3)

Moreover principal stresses are the maximum and minimum normal stresses at a point; therefore, to determine the maximum and minimum principal stresses, dσθ d  σ1 + σ 2 σ1 − σ 2  =0= + cos 2θ + τ sin 2θ    dθ dθ  2 2

σ1 − σ 2 ( −2 sin 2θ ) + τ × 2 cos 2θ = 0 2

or

tan 2θ =

or

2τ σ1 − σ 2

(3.4)

We get the same expression for angle tan 2q. So, to obtain the angles of principal planes with reference to plane BC as above, tan 2θ =

tan (2θ1 ) =

or

tan (2θ1 + 180) =

Moreover,

2τ σ1 − σ 2 2τ 1 2τ ; θ1 = tan −1 σ1 − σ 2 2 σ1 − σ 2 2τ σ1 − σ 2

q2 = q1 + 90°

where

tan 2θ =

So,

(3.5)

τ (σ1 − σ 2 ) / 2

As shown in Fig. 3.9 From the triangle sin 2θ = ±

cos 2θ = ±

MTPL0259_Chapter 03.indd 104

A

τ 2

 σ1 − σ 2  2  +τ  2 

(3.6)

(σ1 − σ 2 ) + τ2 2

(

)

2

+ t2

t

2q

(σ1 − σ 2 )/ 2 2

±

s1 – s2 2

B

(3.7)

( s 2s ) 1– 2

C

Figure 3.9

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Principal Stresses and Strains

105

Putting these values in expression for sq, we get values of principal stresses.

σ1 + σ 2 σ1 + σ 2 (σ1 − σ 2 )/ 2 + × + 2 2 (σ1 − σ 2 )2 2 +τ 2

p1 =

τ ×τ (σ1 − σ 2 )2 + τ2 2

2

p=

or maximum principal stress,

σ1 + σ 2  σ − σ2  +  1 + τ2  2  2

(3.8)

Taking negative values of sin 2q and cos 2q 2

p2 =

Minimum principal stress,

σ1 + σ 2  σ − σ2  −  1 + τ2  2  2

(3.9)

Adding Eqs. (3.8) and (3.9) p1 + p2 = σ1 + σ 2 , this is known as stress invariant. Sum of the normal stresses on any set of two perpendicular planes at a point is constant. Example 3.3 Stresses on an element at a point are shown in Fig. 3.10. Determine the principal stresses and the angle of principal planes with respect to plane BC. Solution On reference plane BC

A

s1 = +100 MPa

-50 MPa q

t = −40 MPa

B

On perpendicular plane AC s2 = −50 MPa, t = 40 MPa

t

40 40

C

100 MPa

Figure 3.10

σ1 + σ 2 100 − 50 = = 25 MPa 2 5 σ1 − σ 2 100 − ( −50) = = 75 MPa 2 2 2

 σ1 − σ 2  2 2 2   + τ = 75 + 40 = 7, 225 2  2

 σ1 − σ 2  2 2   + τ = 85 N/mm 2  2

Principal stresses, p1 =

σ1 + σ 2  σ − σ2  +  1 + τ2  2  2

= 25 + 85 = 110 MPa

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106 Chapter 3

2

p2 =

σ1 + σ 2  σ − σ2  −  1 + τ2  2  2

= 25 − 85 = −60 MPa p1 + p2 = 110 − 60 = 50 = σ1 + σ 2 = 100 − 50 = 50 MPa A

(a check on calculations) Principal angles

q2 = 104° – 2

q2

q1

1 1 40 τ = tan −1 θ1 = tan −1 2 75 (σ1 − σ 2 )/2 2

=

q1 = 14° – 2

D

B

E

C

Figure 3.11

1 1 × (tan −1 0.5333) = × (28° − 4′ ) 2 2

A

q1 = 14°2′ q2 = 90 + q1 = 104°2′

80 MPa B

Principal planes are inclined to reference plane BC at angles 14°2′ and 104°2′as shown in Fig. 3.11 by planes BD and DE.

40 40 MPa

C

80 MPa

Figure 3.12

Exercise 3.3 Normal and shear stresses on two perpendicular planes are as shown in Fig. 3.12 at a point in a stressed body. Determine the principal stresses at the point and the principal angles with respect to the reference plane BC.

Practical Cases of Principal Planes Various cases of principal planes are shown in Fig. 3.13. Figure 3.13(a) shows a bar under an axial tensile load P. All planes perpendicular to axis OO such as a–a and b–b are principal planes, but plane c–c (inclined to axes) is not a principal plane as it carries both normal and shear stresses. Figure 3.13(b) shows a column under an axial compressive load P. All planes such as a–a and b–b perpendicular to axis O–O are principal planes, but inclined plane c–c is not a principal plane. Similarly in the case of a tapered round bar subjected to an axial load, all planes perpendicular to axis O–O such as a–a and b–b are principal planes.

Graphical Solution A graphical method to determine principal stresses, principal planes, maximum shear stress, planes of maximum shear stress, and normal and shear stresses on any inclined plane at a point is given by Mr Mohr by drawing a circle known as Mohr’s stress circle. Following sign conventions are taken.

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Principal Stresses and Strains

107

P y b

C x

P

a

O

a O

P

a

b

b c

a

C b (a) A bar in tension

O

c

C b

P

a

O

O (b) A column in compression O

C

b

P

a

Figure 3.13 Tapered bar under compression (i) Since normal stress and shear stress are two perpendicular stresses, these two stresses can be represented along x–y coordinate system (ii) Normal stress is represented along abscissa (iii) Shear stress is represented along ordinate (iv) Tensile normal stress along positive x-direction (v) Compressive normal stress along negative x-direction (vi) Positive shear stress along positive y coordinate (vii) Negative shear stress along negative y coordinate (viii) Principal stresses are normal stresses (no shear stress on principal planes), so principal stresses are obtained along x-axis. Procedure (i) Figure 3.14 shows the state of stress at point BC on sq A a reference plane. tq Normal stress is +s1 Reference plane s2 Shear stress is -t t q On AC perpendicular to reference plane, B C t normal stress is +s2, say s2 < s1 s shear stress is +t 1 Our objective is to find principal stresses and prinFigure 3.14 cipal planes and normal and shear stresses are sq , and tq on inclined plane AB. (ii) Draw x–y coordinate with origin at O as shown in Fig. 3.15. (iii) To some suitable scale take OA = + σ1 , A B = −τ , so that the coordinates of point B are (σ1 , −τ ) .

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108 Chapter 3 + Shear stress

L K

(s2, + τ) D

tq (–) Normal stress O

sr

2q3 I

C

M

E

Mohr’s stress circle tensile

A H

2q

Normal stress

2q1 B

(s1, −τ)

N

OM = sq

OA = s1 AB = t OC = s2

(–) Shear stress

CD = D

Figure 3.15

(iv) To same scale take OC = + σ 2 (σ 2 < σ1 ), CD = +t, so that the coordinates of point D are (σ 2 , τ ). (v) Join B and D. Line BD intersects the x-axis at E, which is the centre of Mohr’s stress circle. (vi) From E as centre and radius R = EB = ED, draw a circle (called Mohr’s stress circle). The circle intersects the x-axis at points H and I; at these points shear stress is zero. (vii) OH = p1, maximum principal stress. OI = p2, minimum principal stress. (viii) Since on reference plane stresses are σ1 and −τ point B represents the state of stress of reference plane so radius EB is the reference line for principal stresses or any stress at any plane. (ix) From line BE draw ∠BEK = 2q, intersecting Mohr’s stress circle at point K. Coordinates of point K, that is, OM, MK are sq,tq, respectively, on the inclined plane. (x) Angle of principal plane. ∠BEH = 2q1

θ1 =

1 ∠BEH 2

∠BEI = 2q1 + 180 = q2 q2 = q1 + 90°

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Principal Stresses and Strains

109

Note that the angle between principal stresses is 90° but in Mohr’s stress circle points H and I represent principal stresses, angle HEI =180°, so in Mohr’ stress circle the angle between any two planes is represented by double the angle as ∠BEK = 2q for the inclined plane. (xi) EL and EN represent ±tmax, that is, the maximum shear stress at the point Proof of principal stresses Radius of the circle

R = EA2 + AB2 OA − OC σ1 − σ 2 = 2 2

EA =

AB2 = t2 (σ1 − σ 2 )2 + τ2 2

Radius,

R=

Principal stress,

p1 = OH + OE + EH = OE + Radius OE = OC + CE = σ 2 +

σ1 − σ 2 σ1 + σ 2 = 2 2

σ1 + σ 2 (σ1 − σ 2 )2 + + τ2 2 2

Therefore,

p1 =

Principal stress,

p2 = OI = OE − EI = OE − Radius =

σ1 + σ 2 (σ1 − σ 2 )2 − + τ 2, this has been proved earlier 2 2

tmax = EM = EN = ±Radius = ±

(σ1 − σ 2 )2 + τ2 2

Stress on inclined plane

σθ = OE + EM = =

From the triangle EAB,

sin 2q1 =

σ1 + σ 2 + R cos(2θ − 2θ1 ) 2

σ1 + σ 2 + R cos 2θ cos 2θ1 + R sin 2θ sin 2θ1 2

(σ − σ 2 ) AB τ EA = ; cos 2θ1 = 1 = Radius R Radius 2R

Putting these values we get,

σθ = =

MTPL0259_Chapter 03.indd 109

σ1 + σ 2  σ − σ2  τ + R cos 2θ ×  1 + R sin 2θ     R  2R  2 σ1 + σ 2 σ1 − σ 2 + cos 2θ + τ sin 2θ 2 2

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110 Chapter 3

tq = KM = R sin (2q − 2q1) = R sin 2q cos 2q1 − R cos 2q sin 2q1 Putting the values of sin 2q1 and cos 2q1

τθ = R sin 2θ ×

EA τ − R cos2θ R R

τ  σ − σ2  = R sin 2θ  1 − R cos 2θ  2R  R  σ − σ2  = 1 sin 2θ − τ cos 2θ as proved analytically.  2  Principal directions tan 2θ1 =

θ1 =

AB τ = EA (σ1 − σ 2 )/2 1 −1 2τ tan (σ1 − σ 2 ) 2

q2 = q1 + 90° Direction of planes carrying maximum shear stress tmax 2q3 = 2q1 + 90° = ∠BEL q3 = q1 + 45° q4 = q3 + 90° Note that tmax, −tmax are the complementary shear stresses; they act at an angle of 90° to each other Example 3.4 State of stress on two perpendicular planes AC and BC is given in Fig. 3.16. Draw Mohr’s stress circle, taking BC on reference plane, determine (i) Principal Stresses (ii) Principal angles (iii) Maximum shear stress (iv) Normal and shear stresses on inclined plane AB, with q =30°. Solution Draw x–y coordinates ox and oy (Fig. 3.17) To some suitable scale, Say, 20 MPa = 10 mm

MTPL0259_Chapter 03.indd 110

A

40 40 MPa

30° B

C

40 100

Figure 3.16

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Principal Stresses and Strains

111

y K, L tmax (82 MPa)

B (100, 40)

2 × 30 = 60°

(–52)

C I

O

2q1 = –30°

(M) E

(110 MPa) A

H

x

q1 = –15°

D (–40, –40)

KM = 82 MPa OM = 29 MPa

(–82 MPa)

Figure 3.17 Take OA = 100 MPa AB = 40 MPa (the shear stress on reference plane is positive) OC = -40 MPa CD = −40 MPa Join B and D points and the line BD intersects abscissa at E, that is, the centre of Mohr’s stress circle. With E as centre and radius of the circle equal to EB or ED, draw a circle, intersecting x-axis at H and I. Draw a vertical line from E, that is, LEN. OH = p1 (maximum principal stress) = 110 MPa OI = p2 (minimum principal stress) = −52 MPa. ∠AEB = 2q1 = −30°, q1 = −15° q2 = −15 + 90 = 75° tmax = LE = EN = 82 MPa, line EB is reference line.

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112 Chapter 3

Take angle BEK = 2q3 = 2 × 30 = 60°, the coordinates of point K (OM, MK) give the normal and shear stresses on inclined plane sq = OM = 30 MPa (note that points M and E coincide) tq = MK = 82 MPa Verification through analytical solution s1 = +100 MPa s2 = −40 MPa t = 40 MPa 2

p1 , p2 =

σ1 + σ 2  σ − σ2  ±  1 + τ2  2  2 2

=

100 − 40  100 + 40  2 ±   + 40  2 2

= 30 ± 80.6 p1 = 110.6 MPa p2 = −50.6 MPa There is some graphical error in solution. Principal angles tan 2θ1 = −

τ −40 = (σ1 − σ 2)/ 2 70

= −0.5714, 2q1 = −29.74° q1 = −14.87° 2

 100 + 40  2 τ max = ±   + 40 = ±80.6 MPa  2  Stresses on inclined plane

σθ =

σ1 + σ 2 σ1 − σ 2 + cos 2θ − τ sin 2θ 2 2

(Note that the shear stress on reference plane is positive.) sq = 30 + 70 × cos 60° − 40 × sin 60° = 30 + 35 − 40 × 866 = 65 − 34.64 = 30.36 MPa

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Principal Stresses and Strains

tq =

113

σ1 − σ 2 sin 2q + t sin 2q 2

= 70 × sin 60° + 40× cos 60° = 60.62 + 20 = 80.62 MPa Exercise 3.4 State of a stress at a point is shown in Fig. 3.18 in a stressed body. Planes BC and AC are perpendicular to each other. Draw a Mohr’s stress circle for the case and with reference to plane BC and determine (i) principal stresses (ii) principal angles (iii) maximum shear stresses (iv) angles of planes of maximum shear stresses, and (v) normal and shear stresses on the inclined plane.

A 40 q = 35° B

30

C

30

120 MPa

Figure 3.18

Exercise 3.4

Ellipse of Stresses Ellipse of principal stresses is used to determine normal and shear stresses on any plane inclined with a principal plane. Say at a particular point, the principal stresses are +p1 and +p2, YY is the principal plane for p1 = OA and XX is the principal plane for p2 = OB. From the origin O, draw two concentric circles of radii p1 and p2, respectively, as shown in Fig. 3.19. Say, plane yy′ is inclined with YY plane at an angle q, from O draw a perpendicular on yy plane, intersecting the two concentric circles at points C and D. From C draw a vertical line and from D draw a horizontal line (parallel to X–X axis), both the lines meeting at point P, and the point P lies on an ellipse with semi-major axis, p1, and semi-minor axis, p2. Join OP, which gives sr (resultant stress on inclined plane). From P draw a perpendicular PK on line OC. Then, OK = sq (normal stress on inclined plane yy) KP = tq (shear stress on inclined plane yy) sq = OK = OD + DK = p2 + DP cos q DP = DC cos q = (p1 − p2) cos q sq = p2 + (p1 − p2) cos2 q  1 + cos 2θ  = p2 + ( p1 − p2 )    2 = p2 + =

MTPL0259_Chapter 03.indd 113

p1 p2  p1 − p2  − +  cos 2θ 2 2  2 

p1 + p2  p1 − p2  + cos 2θ  2  2

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114 Chapter 3 Y

OA = p1

y

OB = p2

q Ellipse

q

D

P

q

X

C

K

B

O

A

y

X

Minor axis OB = h1

Y

Figure 3.19

tq = KP = DP sin q = DC cos q sin q = (p1 − p2 ) =

sin 2θ 2

(p1 − p2 ) sin 2θ 2

Example 3.5 At a point the major and minor principal stresses are +80 MPa and −40 MPa, respectively. With the help of ellipse of stresses, determine the normal and shear stresses on a plane inclined at an angle 30° with the plane of major principal stress. Solution Choose x–y coordinates. To some scale, draw two concentric circles of radii 80 and 40 MPa, respectively, from origin as shown in Fig. 3.20. YY is the plane of principal stress p1 and XX is the plane of principal stress p2. Take plane yy inclined at angle of 30° with the plane of major principal stress. On this plane yy, draw a perpendicular on both sides, intersecting outer circle at C and inner circle at D as shown (at D, in quadrant III, p2 is negative). Draw lines CP || YY and DP || XX.

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Principal Stresses and Strains

115

Y

y

C K

q sq

tq

p A

X D

+sq

O

X

P

B (−40)

OK = 50 MPa KP = 52 MPa y Y

Figure 3.20 Example 3.5

Then OP = sr (resultant stress on inclined plane) OK = sq = 50 MPa (to same scale) KP = tq = 52 MPa (to same scale) Exercise 3.5 At a point major and minor principal stresses are +90 and +50 MPa, respectively. With the help of ellipse of stresses, determine the normal and shear stresses on a plane inclined at an angle of 60° with the plane of major principal stress.

Strain Components In a general case there are normal and shear strains. Normal strains change the volume and shear strains change the shape of the body. Consider a thin elementary plate of dimensions Dx, Dy subjected to strains exy, eyy and γ xy /2. Total shear angle between x- and y-axes = a + b = gxy . Shear strain, gxy, (change of angle between x- and y-axes) is taken to be equally divided on both the sides.

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116 Chapter 3

Normal strain component,

εx x =

δ∆x ∆x

ε yy =

δ∆y ∆y

γ xy α + β Shear strain 2 = 2 as shown in Fig. 3.21. Graphical representation of plane strain of state is given in Fig. 3.22. e yy

y

g xy 2

g xy 2

B

C d∆y = eyy , ∆y

eyy

C

B

exx ∆y

e xx

exx

e xx

a

O

eyy

b A

g xy

A x

2

g xy 2

∆x

e yy

d ∆ x = exx . ∆x

Plane strain state

Figure 3.21 Strain components

Strain Components on an Inclined Plane In the previous section we have studied about the normal and shear strain components on an element. Let us determine the normal and shear strain components on any inclined plane with respect to a reference plane. Consider an element ABCD, subjected to strains exx, eyy and γ xy /2 as shown in Fig. 3.22(a). Under the action of strains, the element takes the final shape AB1C3D2 as shown. B is displaced to B1 and D is displaced to D1 due to normal strains exx and eyy. Rectangle ABCD changes to AB1C2D1. Now due to shear strain γ xy /2 , point D1 is displaced to D2 and C2 is displaced to C3 resulting in final shape AB1C3D2. BB1 CC1 As per definition of strain, exx = = AB AB eyy =

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DD1 C1C 2 = A D B1C1

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Principal Stresses and Strains

117

e yy g xy 2

D2

D1

K

q

C

D

C2 q

N

q

C3

P

g xy 2

C1 F

f

e xx

e xx M

a

90°

q

f

A

B1

B g xy 2

e yy (a)

q

N

C2

K

C3

P

q

q

C

q

C1

F

f

a A A (b)

Figure 3.22

Shear strains

(a) Strain components on a small element and (b) enlarged view

gxy =

C2C3 C2C3 = B1C2 B1C1 + C1C2

But

C1C2  1  Hint:  1  2   2  

  

A

A 50 50 MPa

B

C

100 MPa B

C

50

50 150 MPa

100

Figure 3.45

Figure 3.44

4. The state of stress at a point in a stressed body is shown in Fig. 3.45. Draw a Mohr’s stress circle and determine the principal stresses and principal angles with respect to plane BC. [Hint: Centre of the Mohr’s stress circle lies at origin of x–y coordinate] 5. A rectangular bar is subjected to three principal stresses p1, p2 and p3 on perpendicular planes. If E is Young’s modulus and n is Poisson’s ratio, show that the volumetric strain

δν 1 = ( p1 + p2 + p3 )(1 − 2ν ). E ν [Hint: Volumetric strain = ε p1 + ε p2 + ε p3] 6. At a point in a strained material, the resultant stress on a certain plane is 140 MPa having an angle of obliquity of 30° and the resultant stress on another plane is 210 MPa with an angle of obliquity of 15°. Both the stresses are tensile. Determine the principal stresses at the point. [Hint: Locate two points in x–y coordinates from an origin, taking OA = +140 at 30° to x-axis, OB = +210 MPa at 15° to x-axis. Draw a perpendicular bisector of AB to get the centre of Mohr’s stress circle.]

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Principal Stresses and Strains

147

Answers to Exercises Exercise 3.1: 121.07 MPa, +34.47 MPa Exercise 3.2: 144.42 MPa, +5.11 MPa Exercise 3.3: p1 = 120 MPa, p2 = 40 MPa; q1 = +45°, q2 = 135° or −45° Exercise 3.4: 130 MPa, 30 MPa; 18°26′, 108°26′; ±50 MPa; 63°26′, 153°26′; 121.87 MPa, 27.3 MPa Exercise 3.5: 60 MPa, 17.32 MPa Exercise 3.6: eq = 275 × 10−6, gq = -250 × 10−6

Exercise 3.7: ±250 m strain; 16°, 106°; 92 m strain, 155 m strain Exercise 3.8: 120 MPa, −80, +80 MPa; 26.6°, 63.40°, 90°; 33.33 MPa, 83.8 MPa Exercise 3.9: p1 = 257.14 MPa, p2 = 217.14 MPa Exercise 3.10: 188.57 MPa, 108.57 MPa Exercise 3.11: 365.7 m strain, −165.7 m strain; 35.07 MPa, −4.296 MPa Exercise 3.12: 230.55 m strain, −130.55 m strain, +42.07 MPa, −13.49 MPa

Answers to Multiple Choice Questions 1. (b) 2. (a) 3. (c) 4. (b)

5. (a) 6. (c) 7. (b) 8. (b)

9. (b) 10. (c)

Answers to Practice Problems 1. t = 200 MPa, p2 = −200 MPa, tmax = 250 MPa 2. 95.035 MPa, 152.28 MPa, −91.325 MPa, 12°27′, 102°27′ 3. p1 = 170 MPa, p2 = 70 MPa, ep1 = 0.745 × 10−3, ep2 = 0.095 × 10−3, major axis = 200.149 mm, minor axis = 200.019 mm 4. f = 30°, sr = 173.2 MPa

5. q = 26.56°, (s2 = 100); p1 = 220.7 MPa, p2 = 79.3 MPa 6. q = 98°; p1 = 130 MPa, p2 = 20 MPa 7. On AB, s1 = 180.8 MPa, t = +59.95; On BC, s2 = 155.83 MPa,t = -59.95 MPa; p1 = 229.545 MPa, p2 = 107.085 MPa

Answers to Special Problems 1. 100 MPa; sq = 160 MPa, tq = 60 MPa 2. 161.8 MPa, −61.8 MPa; q1 = 13.28°, q2 = 103.28° 3. τ < σ1σ 2

MTPL0259_Chapter 03.indd 147

4. p1 = +111.8 MPa, p2 = -111.8 MPa; q1 = -13.280, q2 = 76.720 6. 225 MPa, 73 MPa

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4 Elastic Constants CHAPTER OBJECTIVES A separate chapter is devoted only to elastic constants. In many books, the topic of elastic constants is discussed with principal stresses and strains. Even after studying for four years, students fumble for answers when they are asked for the nominal values of elastic constants of various materials. For the determination of deformation in a body, as changes in length, breadth, volume and shape, the knowledge of correct values of elastic constants is essential. In this chapter, we will learn about: 

Determination of Young’s modulus and Poisson’s ratio of a material.



Relationship among Young’s modulus, bulk modulus and Poisson’s ratio.



Determination of modulus of rigidity of a material.



Numerical problems involving elastic constants of various materials.



Relationship among Young’s modulus, modulus of rigidity and Poisson’s ratio.

Introduction For any structural component, rigidity is an important consideration and the rigidity of any component depends on the elastic constants of the material of the component. As an example, a tie member subjected to tensile force; stretch in tie member depends on Young’s modulus. A shaft transmitting power is subjected to angular twist due to torque transmitted which depends on shear modulus and for an efficient and safe design; the angular twist per unit length of shaft should be within the permissible limits. Similarly, a beam subjected to bending moment and the ends of beam suffer angular rotation, which again depends on Young’s modulus of the material. Experimental determination of basic elastic constants such as Young’s modulus, Poisson’s ratio and modulus of rigidity will be discussed later in the chapter. Nowadays, strain gauges are employed for accurate measurement of strains and accurate determination of elastic constants.

Young’s Modulus and Poisson’s Ratio Young’s modulus E and Poisson’s ratio v are the basic elastic constants as values of modulus of rigidity G and bulk modulus K can be analytically derived from the values of E and v, respectively. For accurate

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Elastic Constants

149

determination of E, v, one takes the help of electrical resistance strain gauges for the measurement of axial and lateral strains. We have already discussed about strain gauges in chapter 3. A flat sample of standard dimensions is made as shown in Fig. 4.1. Strain gauges are installed on sample in axial and lateral directions using adhesive. After proper curing, the strain gauges are connected in the desired electrical circuit. Sample is gripped in the wedge grips of a universal testing machine. As the load on sample increases both stress and strain increase gradually and simultaneously. Load per unit area is defined as stress. Axial and lateral strains with the increasing load are measured using strain indicators. Therefore, the readings of s, ea and ela, i.e., axial stress, axial strain and lateral strain, are recorded. Graphs between s and ea and between ela and ea are plotted as shown in Figs 4.2(a) and 4.2(b). Slope of curve s versus ea gives Young’s modulus of elasticity and slope of the curve ela versus ea gives the value of Poisson’s ratio of the material of the sample.

Strain gauge SG 1 Leads SG t

Gauge length

Figure 4.1

A s ea

0

ea (a)

Figure 4.2

ea (b)

(a) Axial stress versus axial strain and (b) lateral strain versus axial strain

Example 4.1 A mild steel bar of 10 mm in diameter and 100 mm in gauge length is tested under tension. A tensile force of 10 kN produces an axial strain = 610 ms, lateral strain = 180 ms. Determine E and v for mild steel. Solution Tensile force, Sample diameter,

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P = 10,000 N d = 10 mm

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150 Chapter 4

π × 102 = 25 π mm2 4 P 10,000 σ= = = 127.324 N/mm2 A 25π

Area of cross radius, A = Axial stress, Axial strain,

ea = 610 ms = 610 × 10−6

Young’s modulus,

E=

Lateral strain,

= 208.7 kN/mm2 ela = 180 ms

Poisson’s ratio,

σ 127.324 = ε a 610 × 10−6

εla 180 µs = ε a 610 µs = 0.295

v=

Exercise 4.1 A copper bar of 10 mm in diameter and 100 mm in gauge length is tested under tension. A tensile force of 10.2 kN produces an axial strain of 1237 ms. The diameter of bar is reduced by 0.0042 mm. Determine E and v for copper.

Determination of Modulus of Rigidity Modulus of rigidity of a material is conveniently determined by performing torsion test on a sample subjected to gradually increasing twisting moment. A specimen of standard dimensions and shape as shown in Fig. 4.3(a) is made. One end of the specimen is fitted in a fixed hub and the other end is fitted into a moveable hub of a torsion-testing machine. Through hand wheel, torque is applied on the movable hub, so specimen is twisted. At the fixed end, there is resisting torque, which is measured through a lever and balance arm mechanism. Angular twist in terms of degree is measured at the movable head of the machine. A gradually increasing torque is applied which produces gradually increasing angular twist in the sample. A graph is plotted between torques, T, and angular twist, q, as shown in Fig. 4.3(b). Torque T

Keyway

d L Torsion test specimen (a)

q Angular twist graph between torque and angular twist (b)

Figure 4.3

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Elastic Constants

151

Within the elastic limit, torque, T ∝ q, angular twist (T is proportional to q) T=

πd3 τ, 16

where d is the diameter of specimen t = maximum shear stress developed on the surface of the specimen

τ=

16T πd3

f = shear angle =

dθ 2L

where q = angular twist in radians L = length of the sample

τ shear stress 16T 2 L 32TL = = × = φ shear strain π d 3 dθ π d 4θ But τ /φ = modulus of rigidity, G =

32TL π d 4θ

Example 4.2 An aluminium specimen with gauge length of 200 mm and diameter of 25 mm is tested under torsion. A torque of 16.4 Nm produces an angular twist of 0.2° in the specimen. Determine the modulus of rigidity of aluminium. Solution Diameter, Gauge length, Torque, Angular twist,

d = 25 mm L = 200 mm T = 16.4 Nm = 16,400 Nmm

θ = 0.2° = 0.2

Modulus of rigidity, G = =

π π = radian 180 900

32TL π d 4θ 32 × 16400 × 200 × 900 π × 254 × π

= 24.5 × 103 N/mm2 Exercise 4.2 A steel specimen with gauge length of 200 mm and diameter of 25 mm is tested under torsion. A torque of 55.8 Nm produces an angular twist of 0.2° in the specimen; determine the modulus of rigidity of steel.

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152 Chapter 4

Relation Between Young’s Modulus and Modulus of Rigidity Consider a cubical block of dimensions Dx = Dy = Dz = a and Fig. 4.4(a) shows only the front view ABCD, top surface of block is subjected to a shear stress t as shown, then on sides CB and AD, there will be complementary shear stresses as shown in Fig. 4.4. On the bottom surface AB, shear stress t acts as shown in the Fig. 4.4(a). Due to the action of shear stresses, shear strain or shear angle f occurs in the block. Point D is shifted to D′ and point C is shifted to C′ so that ∠DAD′ = ∠CBC′ = f. Diagonal AC of the block is extended to AC′ and diagonal BD of the block is reduced to BD′. Note that shear angle f is a very small quantity and not too large as shown in the figure 4.4(a) to explain the concept. Similarly, angle a between two diagonals AC and AC′ is very small. If one draw CC ″ perpendicular to AC′; thus, AC = AC ″ (because a is very small). Extension in length of diagonal AC = C ″C′ ∠DCA = ∠CC′ A = 45° C ″ C ′ = CC ′ cos 45° =

AC =

Strain along

CC′ 2

C ″C ′ CC ′ CC ′ CC ′ = × = AC ″ 2a 2 2a

as AC ″ = 2a, where a is side of cube Shear strain,

φ=

CC ′ CC ′ = BC a

Longitudinal strain,

ε=

CC ′ 2a

p 1 = +t

p 2 = −t D

t

D

45°

C

Shear stress C

+t

45° C t

f

f

a

p 2 = −t

t

2 × 45 Normal stress

O

p 1 = +t A

p 1 = +t

t

B

Principal stress

p 2 = −t

a

−t

Stresses on a plane of cubical block

Mohr’s stress circle (b)

(a)

Figure 4.4

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Elastic Constants

153

CC ′ a or CC ′ = 2ae = af (4.1) or f = 2e Now draw a Mohr’s stress circle, for the case of a cube subjected to pure shear stress t, as shown in Fig. 4.4(b). Principal stresses are p1 = +t and p2 = −t If E = Young’s modulus, v = Poisson’s ratio p vp Strain along diagonal AC = 1 − 2 E E However, p1 = +t and p2 = −t

φ=

Shear strain,

τ vτ τ + = (1 + v) E E E τ ε = (1 + v) E 2τ φ = 2ε = (1 + v) E

AC =

Strain along diagonal or From Eq. (4.1)

φ=

However,

τ stear stress = (by using Hooke’s law) G modulus of rigidity

Therefore,

τ 2τ = (1 + v) G E

or

E = 2G (1 + v)

Young’s modulus, E = 2 × shear modulus (1+ Poisson’s ratio) Example 4.3 A copper bar specimen of 15 mm diameter and 150 mm gauge length when tested as a tensile test specimen, a force of 7.8 kN produces an extension of 0.066 mm. When the same specimen is tested under torsion, a twisting moment of 6.50 Nm produces an angular twist of 0.3°. Determine Young’s modulus, modulus of rigidity and Poisson’s ratio of copper. Solution Length of a copper bar,

L = 150 mm

Diameter,

d = 15 mm

Area of cross-section,

A=

Tensile force,

P = 7.8 kN = 7800 N

Tensile stress,

σ=

Extension in length, Axial strain,

MTPL0259_Chapter 04.indd 153

π 2 π d = × 152 4 4 = 176.7 mm2 P 7800 = = 44.14 N/mm2 A 176.7

d L = 0.066 mm

ε=

δ L 0.066 = = 0.44 × 10−3 L 150

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154 Chapter 4

Twisting moment,

σ 44.14 = × 103 0.44 ε = 100.32 kN/mm2 T = 6.50 Nm = 6500 Nm

Angular twist,

φ = 0.3° = 0.3 ×

E=

Young’s modulus,

π radian 180

32TL πd4 ×θ 32 × 6500 × 150 × 180 = = 37.46 kN/mm2 π × 154 × 0.30 × π E 100.32 v= −1 = −1 2 × 37.46 2G = 0.34

Modulus of rigidity, G =

Poisson’s ratio,

Exercise 4.3 A steel bar specimen of 15 mm diameter and 150 mm gauge length when tested as a tensile test specimen, a force of 15 kN produces an extension of 0.064 mm. When the same specimen is tested under torsion, a twisting moment of 6.70 Nm produces an angular twist of 0.15°. Determine Young’s modulus, modulus of rigidity and Poisson’s ratio of the material.

Relation Between Young’s Modulus and Bulk Modulus Bulk modulus is the ratio of volumetric stress by volumetric strain ( p/ev). Volumetric stress is of equal magnitude in all the directions; therefore, let us consider a rectangular block of dimensions L × B × H as shown in Fig. 4.5 subjected to hydrostatic pressure p in all the directions 1, 2 and 3 say E is Young’s modulus and v is the Poisson’s ratio of the material. Principal strains in three directions are as follows. 2 e1

p, e 2

p

p

p

p e3

H p

e1 1

O

3 L

p

B

Figure 4.5

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Elastic Constants

ε1 = =

155

p vp vp − − E E E p (1 − 2v) E p (1 − 2v) E

Similarly,

ε 2 = ε3 =

Volume

V=L×B×D V = LD ∂B + L B∂D + BD∂L ∂V ∂B ∂D ∂L = + + = ε1 + ε 2 + ε3 V B D L

Volumetric strain,

εv = ε1 + ε 2 + ε3 =

Therefore, volumetric strain,

εv =

or

E=3

However, Finally,

3 p (1 − 2v) E

3 p (1 − 2v) E ( p) (1 − 2v) εv

p = K bulk modulus εv E = 3K (1 − 2v)

Example 4.4 What change in volume would a 300-mm cube of steel suffer at a depth of 5 km in sea water? For steel E = 208 GPa, v = 0.3 and weight density of sea water, w = 1 × 10−5 N/mm3. Solution Young’s modulus,

E = 208 × 103 N/mm3 v = 0.3

Bulk modulus,

K= =

E 3(1 − 2v)

208 × 103 = 173.33 × 103 N/mm2 3(1 − 2 × 0.3)

Depth of cube in sea water, h = 5 km = 5 × 103 × 103 = 5 × 106 mm Weight density of sea water, w = 1 × 10−5 N/mm3 Pressure on cube,

p = w × h = 1 × 10−5 × 5 × 106 = 50 N/mm2

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156 Chapter 4

Volumetric strain, εv =

p 50 = K 173.33 × 103

= 0.2885 × 10−3 Original volume,

V = 3003 = 27 × 109 mm3 = 27 × 106 cm3 dV = Vev = 27 × 106 × 0.2885 × 10−3 = 7789.5 cm3 = 7789.5 cc

Exercise 4.4 What change in volume a brass copper of 100 mm diameter would suffer at a depth of 2 km in sea water?   Hint: For brass E = 100 kN/mm2   v = 0.32     Weight density of sea water, w = 1.02 × 103 kg/m3   3 −6  = 1.02 × 9.89 × 10 N/mm    = 10.0062 × 10−6 N/mm3   Problem 4.1 At a point in a strained material, the principal stresses are p1, p2 and p3 while principal strains are e1, e2 and e3, respectively. Show that principal stress p1 = 2G (Bev + e1) where G = shear v modulus, B = and ev = volumetric strain. 1 − 2v Solution Principal stresses are p1, p2 and p3 E and v are Young’s modulus and Poisson’s ratio respectively Principal strains p1 vp2 vp3 − − E E E p2 vp1 vp3 ε2 = − − E E E p3 vp2 vp1 ε3 = − − E E E

ε1 =

 p + p2 + p3  2v εv = ε1 + ε 2 + ε3 =  1  − E ( p1 + p2 + p3 )  E 1 − 2v ( p1 + p2 + p3 ), = E v B= 1 − 2v

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Elastic Constants

εv × B = =

ε1 = Bεv + ε1 = p1 =

157

v  1 − 2v   ( p1 + p2 + p3 )  (1 − 2v)  E  v ( p1 + p2 + p3 ) E p1 vp2 vp3 − − E E E p vp vp p v v v p1 + p2 + p3 + 1 − 2 − 3 = 1 (1 + v) E E E E E E E E ( Bεv + ε1 ) (1 + v)

(4.2)

Putting the value of E = 2G(1 + v) in Eq. (4.2) p1 =

2G (1 + v) ( Bεv + ε1 ) = 2G ( Bεv + ε1 ) (1 + v)

Problem 4.2 A round bar of 15 mm in diameter and 150 mm in length is tested in tension for the determination of Young’s modulus. It was observed that change in length was 33 times the change in its diameter. If E = 208 GPa, calculate the modulus of rigidity of the bar. What twisting moment is required to produce an angular twist of 0.15° in the bar? Solution Diameter of the bar, d = 15 mm Length of the bar, L = 150 mm Assume that dL is change in length and dd is change in diameter of the bar. However, dL = 33 × dd. v=

δd l × d δL

(4.3)

Poisson’s ratio,

v=

1 150 × = 0.30 33 15

(4.4)

Young’s modulus,

E = 208 GPa

Poisson’s ratio Putting the values in Eq. (4.3)

(1 + v) = 1 + 0.3 = 1.3 Shear modulus,

G=

E 208 = 2(1 + v) 2 × 1.3

= 80 GPa = 80 × 1000 N/mm2 Torque,

MTPL0259_Chapter 04.indd 157

T=

Gπ d 4θ 32 L

(4.5)

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158 Chapter 4

ε = 0.15° = 0.15 ×

π π = 180 1200

Putting the values in Eq. (4.5) Torque,

T=

80,000 × π × 154 × π = 6.939 Nmm 32 × 150 × 1200

= 6.939 Nmm Problem 4.3 The modulus of rigidity of a material is 39 kN/mm2. A 10 mm diameter rod of the material is subjected to an axial tensile force of 5 kN and the change in its diameter is observed to the 0.002 mm. Calculate the Poisson’s ratio and modulus of elasticity of the material. Solution Modular of rigidity, G = 39 kN/mm2 Diameter, d = 10 mm Tensile force, P = 5kN = 5000 N Tensile stress,

σ=

4 P 4 × 500 = = 63.66 N/mm π d 2 π × 102

Change in diameter, dd = 0.002 mm

δ d 0.002 = = 0.0002 d 10

Lateral strain,

εla =

Axial strain,

εa = εa =

Lateral strain

εla = vε a =

Shear modulus,

G=

σ 63.66 = E E E = 0.0002 2(1 + v)

E = 39,000 N/mm2 2(1 + v)

(4.6) (4.7)

From Eqs (4.6) and (4.7), 63.66 × v E × = 39,000 × 0.0002 E 2(1 + v) 31.83v = 7.8 1+ v or

31.83 v = 7.8 + 7.8 v

or

24.03 v = 7.8 7.8 = 0.324 24.03

Poisson’s ratio,

v=

Young’s modulus,

E = 2G(1 + v) 2 × 39(1 + 0.324) = 103.27 kN/mm2

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Elastic Constants

159

Problem 4.4 The determination of E and G for a particular metal gives values as 208 kN/mm2 and 80 kN/mm2, respectively; calculate Poisson’s ratio and bulk modulus. If both the moduli are liable to an error of ±1 per cent, find the maximum percentage of error in the derived value of Poisson’s ratio: Solution E = 208 kN/mm2 G = 80 kN/mm2 Poisson’s ratio,

v=

E 208 −1 = −1 2G 2 × 80

= 0.3 Bulk modulus,

K=

E 208 = 3(1 − 2v) 3(1 − 2 × 0.3)

= 177.33 kN/mm2 Poisson’s ratio, Emax = 208 × 1.01 kN/mm2, Emin = 208 × 0.99 kN/mm2 Gmax = 80 × 1.01 kN/mm2, Gmin = 80 × 0.99 kN/mm2 vmax =

208 × 0.99 − 1 = 0.326 2 × 80 × 1.01

vmm =

208 × 1.01 − 1 = 0.274 2 × 80 × 0.94

Error in the derivation of Poisson’s ratio, (0.326 − 0.3) to (0.274 − 0.3) 0.026 to −0.26 % error =

0.026 × 100 = ±8.66 per cent 0.3

Problem 4.5 A small piston of area 150 mm2 compresses oil in a rigid container of 25 litres capacity. When a weight of 90 N is gradually applied to the piston, its movement is observed to be 27 mm. Determine bulk modulus of oil. Solution Weight, W = 90 N Piston area, A = 150 mm2

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160 Chapter 4

W 90 = = 0.6 N/mm2 A 150 d = 27 mm

p=

Pressure developed, Movement of piston,

Change in volume of oil, dV = 27 × 150 = 4050 mm3 V = 20 l = 20 × 106 mm3 Volumetric strain,

εv =

4050 δV = = 202.5 × 10−6 εv 20 × 106

Bulk modulus,

K=

0.6 p = = 2963 N/mm2 εv 202.5 × 10−6

Problem 4.6 A copper bar of 200 mm long with sections 50 × 50 mm is subjected to stresses of +20 N/mm2 along the axis, and −40 and +40 N/mm2 along sides. The increase in volume was observed to be 183 mm3. Determine the value of Poisson’s ratio, modulus of rigidity and bulk modulus, if E = 105 GPa. Solution Principal stresses are +120, −40 and −40 N/mm2 Volumetric strain,

εv =

1 [( p1 + p2 + p3 ) − 2v ( p1 + p2 + p3 )] E

where E = Young’s modulus v = Poisson’s ratio

εv =

Volume,

1 [(120 − 40 + 40) − 2v (120 − 40 + 40)] 10,500

=

1 [120 − 240 v] 10,500

=

120(1 − 2v) δV = 1,05,000 V

V = 200 × 50 × 50 = 5 × 105 mm3 dV = 1 × 3 mm3

δV 1+ 3 = = 36.6 × 10−5 V 5 × 105 Therefore,

120(1 − 2v) = 36.6 × 10−5 1,05,000 1 − 2v = v=

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36.6 × 10−5 × 1,05,000 = 0.320 120 1 − 0.32 = 0.34 (Poisson’s ratio) 2

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Elastic Constants

Modulus of rigidity,

G=

1,05,000 E = = 39.18 × 103 N/mm2 2(1 + v) 2 × 1.34

Bulk modulus,

K=

1,05,000 E = = 1,09,375 N/mm2 = 109.375 kN/mm2 3(1 − 2v) 3 × 0.32

161

Problem 4.7 A bar of elastic material subjected to a compressive stress sx in an axial direction and its strains at right angles are reduced to half to those which normally would occur in an ordinary tension member. If E = 208 kN/mm2 and v = 0.3, what is modified modulus of elasticity in x-direction. Solution Note that all the stresses in x-, y- and z-directions will be negative, i.e., −sx, −sy , −sz

say Strains

εx = − εy = − εz = −

σx v + (σ y + σ z ) E E σy

+

E

vσ v (σ x + σ z ) = x E 2E

vσ σz v + (σ y + σ x ) = x E E 2E

As per the condition given in the problem Since, ey = ez; sy = sz =

So, −σ y E

+

vσ y

( −1 + v) or,

E

−σ y

=−

E

+

vσ v (σ x + σ y ) = x E 2E

vσ x 1v σ x + E 2 E

σy v σx =− E 2 E

v σ y (1 − v) = σ x 2 v 1 0.3 σy = × σx = × σx 2 (1 − v) 2(1 − 0.3) σ y = 0.2143σ x

σ z = 0.2143σ x Both are negative stresses, Strain,

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σx v σ v × 2 × 0.2143σ x + (2σ y ) = − x + E E E E σ 0.3 × 2 × 0.2143σ x =− x + E E

εx = −

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162 Chapter 4

=−

σ x σx + (0.1286) E E

σx ( −1 + 0.2186) E σ = x ( −0.8714) E σx E = ε x −0.8714 =

Em = −

208 = −238.7 kN/mm2, modified modulus of elasticity. 0.8714

Key Points to Remember  Young’s modulus of elasticity, E =

Axial stress Axial strain

Lateral strain Axial strain Shear stress  Modulus of rigidity, G = Shear strain Volumeteric stress  Bulk modulus, K = Volumeteric strain  E = 2(1 + v)G  Poisson’s ratio, v =

 E = 3K(1 − 2v)  E = 9 K G/(3K + G)

Review Questions 1. 2. 3. 4. 5.

Describe the procedure to determine Young’s modulus of elasticity and Poisson’s ratio of a material. Explain torsion test to determine the modulus of rigidity of a material. How is bulk modulus K of an oil determined? Explain. Derive relationship among E, G and v. Derive relationship among E, K and v.

Multiple Choice Questions 1. For a material, Poisson’s ratio is 0.3. What is the ratio of E/K? (a) 1.0 (b) 1.20 (c) 1.50 (d) None of these

MTPL0259_Chapter 04.indd 162

2. For a material with E = 208 GPa and v = 0.3, what is the modulus of rigidity? (a) 78 GPa (b) 79 GPa (c) 80 GPa (d) None of these

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Elastic Constants 3. For a certain material, Poisson’s ratio in elastic stage 0.35. What is Poisson’s ratio in the fully plastic region? (a) 0.3 (b) 0.38 (c) 0.50 (d) None of these 4. Which material has Poisson’s ratio equal to 0.33? (a) Mild steel (b) Phosphorus bronze (c) Wrought iron (d) Aluminium 5. For a material with E = 105 GPa and G = 39 GPa, what is Poisson’s ratio of the material? (a) 0.31 (b) 0.33 (c) 0.34 (d) 0.346 6. A body is subjected to hydrostatic pressure of 100 N/mm2, Young’s modulus of material is

163

100,000 N/mm2 and Poisson’s ratio is 1/ 3. What is the volumetric strain in the body? (a) 1000 ms (b) 750 ms (c) 660 ms (d) None of these 7. Principal strains at a point are +800 ms, +400 ms and −1200 ms. If Poisson’s ratio is 0.3, what is the volumetric strain? (a) 1280 ms (b) 800 ms (c) 400 ms (d) None of these 8. For a material with K = 170 GPa and v = 0.3, what is Young’s modulus of elasticity? (a) 170 GPa (b) 204 GPa (c) 221 GPa (d) None of these

Practice Problems 1. At a point in a strained material, principal stresses and principal strains are p1, p2, p3 and e1, e2, e3, respectively.

2.

3.

4.

5.

6.

7.

Ev Show that principal stress p1 is given by aev + 2Ge1, where α = , ev = volumetric strain, G is ( 1 + v )( 1 − 2v) shear modulus and v is Poisson’s ratio. A round bar 15 mm in diameter and 150 mm in length is tested in tension for the determination of Young’s modulus. It was observed that change in length was 30 times the change in its diameter. Calculate the modulus of rigidity of the bar if its E = 70 GPa. What twisting moment is required to produce an angular twist of 0.12° in the bar? The modulus of rigidity of a material is 24.5 kN/mm2. A 10-mm-diameter rod of the material is subjected to an axial tensile force of 5 kN and the change in its diameter is observed to be 0.0032 mm. Calculate the Poisson’s ratio and modulus of elasticity of the material. The determination of E and G for a metal gives values as 70 kN/mm2 and 26.3 kN/mm2, respectively. Calculate Poisson’s ratio and bulk modulus. If both the moduli are liable to an error of ±1%, find the maximum percentage of error in the derived value of poisson’s ratio. A small piston of area 200 mm2 compresses oil in a rigid container of 25 litre capacity. A weight of 100 N is gradually applied to the piston. If the bulk modulus of oil is 3000 N/mm2, by how much distance the weight will move down the piston. A steel bar 250 mm long of section 50 mm × 50 mm is subjected to stresses of +100 N/mm2 along the axis and −30 N/mm2 along the sides. The increase in volume was observed to be 50 mm3. Determine the values of Poisson’s ratio, modulus of rigidity and bulk modulus if E = 200 GPa. A piece of material 15 cm long by 3 cm × 3 cm square is in compression under a load of 135 kN. If the modulus of elasticity is 105 GPa and Poisson’s ratio is 0.32, find the alteration in length if all strain is prevented in lateral direction by the application of lateral pressures.

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164 Chapter 4

Special Problems 1. Express the value of Poisson’s ratio v in terms of modulus of rigidity G and bulk modulus, K 2. Derive a relationship between modulus of elasticity, E, modulus of rigidity, G, and bulk modulus, K. 3. A prismatic bar of circular cross-section of 30 mm diameter and axial length of 3 m is loaded by an axial tensile force of P = 80 kN. It is made of aluminium with E = 70 GPa and v = 1/ 3. Calculate the change in volume of the bar. σ    Hint: ev = E (1 - 2v), where σ is axial stress   

Answers to Exercises Exercise 4.1: 104.98 kN/mm2, 0.34 Exercise 4.2: 83.37 kN/mm2

Exercise 4.3: 198.8 kN/mm2, 77.24 N/mm2, 0.287 Exercise 4.4: 113 mm3

Answers to Multiple Choice Questions 4. (d) 5. (d) 6. (a)

1. (b) 2. (c) 3. (c)

7. (d) 8. (b)

Answers to Practice Problems 2. G = 26.25 GPa, T = 1.82 Nm 3. v = 0.327, E = 65 kN/mm2 4. 0.33, 68.62 kN/mm2, ±8.0 per cent

5. 20.83 mm 6. v = 0.3, G = 76.92 GPa, K = 166.67 GPa 7. −0.149 mm

Answers to Special Problems 1. Solution: v =

3K − 2G 6 K + 2G

2. Solution: E =

9 KG 3K + G

MTPL0259_Chapter 04.indd 164

3. Solution: 1143 mm3

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5 Thin Cylindrical and Spherical Shells CHAPTER OBJECTIVES Thin cylindrical and spherical shells are used mainly for storage of gas, petrol, liquid, chemicals, grains and so on. Some are subjected to internal/external pressures and the order of pressure is low (10–30 atmospheres). Their D/t ratio, that is, the ratio of diameter to wall thickness, is large, that is, D/t is greater than 20. Because in comparison to diameter, thickness is very small, so the variation of stresses along the thickness is taken negligible. In this chapter, we will learn about: 

How to determine hoop stress and axial stress in a thin shell subjected to internal pressure.



How to determine axial, circumferential and volumetric strains in shell subjected to internal pressures.



The hoop stress developed is double the axial stress in thin cylindrical shell. To reduce the effect of hoop stress; cylinder is wound with a wire under tension, with the purpose of reducing the effect of hoop stress.



A thin spherical shell is equally strong in two hoop directions; therefore,

generally the cylindrical shells are provided with hemispherical ends. 

Some pressure vessels have double curved wall, that is, at a particular element radii of curvature in two directions are r1 and r2 (radius of curvature). So pressure vessels with double curved wall will also be analysed.

 Conical tanks are available for storage of water. The hoop stress developed is proportional to the square of the depth of element from free water surface. Stress in conical water tanks will also be analyzed.

Introduction Thin pressure vessels are subjected to internal uniform pressure or hydrostatic pressure of low magnitude. Development of axial and hoop stresses in the wall thickness of the pressure vessel provides useful data for the designer. Stresses in thin shells are dependent on the D/t ratio, that is, the ratio of shell diameter to wall thickness.

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166 Chapter 5

Derivation of axial and hoop stresses, strains in cylindrical pressure vessel and derivation of hoop stresses, hoop strain and volumetric strain in both shells forms the text of this chapter. The effect of wire winding on the reduction of hoop stresses (which is greater than axial stress) will be analysed. Pressure vessels with the double curved surface of wall and the conical surface of wall will also be analysed. In such shells, equations for hoop stresses will be derived. Cylindrical pressure vessel can be conveniently used to determine experimentally E and n, that is, elastic constants of material of pressure vessel.

Thin Cylinder Subjected to Internal Pressure Consider a thin cylindrical shell of inner diameter D, length L, with wall thickness equal to t. The shell is full of liquid. The volume of the liquid inside the shell is [p /4 (D 2 L)] and the pressure of the liquid is atmospheric. Now, the additional volume of the liquid is pumped under pressure inside the shell; as a result, there is overall expansion in the shell, that is, expansion in both diameter and length. However, the expansion of liquid is partly prevented by the wall of the metallic cylinder, which offers equal and opposite reaction and compresses the liquid inside the cylinder ( Fig 5.1). Cylinder

r

r

r

r (a)

Figure 5.1

Liquid

t

(b)

Small element of cylinder subjected to internal pressure

Mathematically dV = additional volume of the liquid pumped inside. = δV1 + δV 2 = expansion in the volume of cylinder + contraction in the volume of the liquid. Since the liquid exerts pressure on cylinder wall, the length and diameter of the cylinder get increased introducing axial and circumferential stresses, in the cylinder. Figure 5.2 shows a small element of the cylinder subjected to internal pressure p; sc is the circumferential a hoop stress developed (along circumference) and sa is the axial stress developed (along the axes); at outer surface pa is atmospheric pressure, such that p >>pa. Axial stress Diameter of cylinder = D Wall thickness = t Axial stress = sa

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Thin Cylindrical and Spherical Shells

167

sC

pa

t sa

sa p sC

(Internal pressure)

p

Axis

Figure 5.2 Small element under internal pressure sa

p

sa Pa

Axis p

sa Resisting area

Figure 5.3

Axial bursting force

Axial bursting force on cylinder in axial direction,  Pa = p / 4( D 2 p )  acting on the end plates as shown in Fig. 5.3 Resisting area = pDt Axial stress developed,

σa =

Pa π D 2 p pD = = π Dt 4 π Dt 4t

(5.1)

Circumferential stress (sc) Consider a small portion along length dL and inner radius D/2, subjected to internal pressure p, Fig 5.4 Take a small element at an angle q from x-axis, subtending an angle dq at inner surface. Area of the small element Force on the small element, Vertical component of

D dθ dL 2 dP = D dq dL p 2 D dP = p sin q dq dL 2 =

D cos q dq dL 2 The horizontal component of the force is cancelled out when the force is integrated over the semicircular portion of the cylinder. Horizontal component of

MTPL0259_Chapter 05.indd 167

dP = p

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168 Chapter 5 y PD

Projected area of curved surface = Dd L

p

D 2

q dq

dL

q

t sC

x sC

Figure 5.4

Diametral bursting force π

Therefore, total diametral bursting force,

PD = ∫ p 0

= p =

D sin θ dθ dL 2

D dL( − cos θ )π0 2

pD dL × 2 = pD dL 2

(= p × projected area of the curved surface) Area of cross-section resisting the diametral bursting force = 2t dl Circumferential stress developed,

σC = =

PD 2t dL pD dL pD = 2t dL 2t

Strains Consider Fig. 5.2, sc, sa and p stresses are perpendicular to each other. Note that p acts as a compressive radial stress on inner surface of cylinder. Stresses sc and sa are much larger than p; Therefore in calculation of strains, the effect of internal pressure p is neglected. Say E is Young’s modulus and n is Poisson’s ratio of the material. Circumferential strain,

MTPL0259_Chapter 05.indd 168

εc =

σ c νσ a − E E

=

pD ν pD − 2tE 4tE

=

pD (2 − ν ) 4tE

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Thin Cylindrical and Spherical Shells

Axial strain,

Note that ec = circumferential strain ea = axial strain Change in the diameter of cylinder,

εa =

σ a νσ c − E E

=

pD ν pD − 4tE 2tE

=

pD (1 − 2ν ) 4tE

dD = ecD =

Change in the length of cylinder,

169

pD 2 (2 − ν ) 4tE

dL = eaL =

pDL (1 − 2ν ) 4tE

Volumetric strain Volume of cylinder [V = π /4 ( D2 L)], taking partial derivative ∂V =

π π 2 D ∂DL + D 2 ∂L 4 4

∂V ∂D ∂L =2 + = 2ε c + ε a V D L

or

= 2 × circumferential strain + axial strain = 2× ev =

pD pD (2 − ν ) + (1 − 2ν ) 4tE 4tE

pD pD ( 4 − 2ν + 1 − 2ν ) = (5 − 4ν ) 4tE 4tE

Change in volume Change in volume of cylinder,

dV1 = ev V pD π (5 − 4ν ) × D 2 L 4tE 4 p p π 2 δV 2 = × V = × D L K K 4 =

Contraction in volume of liquid,

where K is bulk modulus of liquid. Additional volume of liquid pumped inside the cylinder, δV = δV1 + δV 2. Note that if, in any problem, K for the liquid is not given that dV2, that is, reduction in volume of liquid can be taken as negligible.

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170 Chapter 5

Example 5.1 A thin cylindrical shell made of 5-mm-thick steel plate is filled with water under pressure of 3 N/mm2. The internal diameter of the cylinder is 200 mm and its length is 1.0 m. Determine the additional volume of the water pumped inside the cylinder to develop the required pressure. Given for steel E = 208 kN/mm2 and n = 0.3, and for water K = 2,200 N/mm2. Solution p = 3 N/mm2, E = 208 kN/mm2 D = 200 mm, n = 0.3 t = 5 mm L = 1.0 m Volume,

V =

π 2 π D L = × 200 2 × 1, 000 = π × 107 mm 3 4 4

Volumetric strain,

εv =

pD (5 − 4ν ), putting the values 4tE

=

3 × 200 [5 − 4 × 0.3] 4 × 5 × 2, 08, 000

= 0.548 × 10−3 Change in volume of the cylinder, dV1 = evV = 0.548 × 10−3 × p × 107 = 1.72 × 104 mm3

Change in volume of water, dV2 = =

p V , where p/K is volumetric strain on water K

3 × π × 107 = 4.284 × 10 4 mm 3 2, 200

Additional volume of water pumped in shell, dV = dV1 + dV2 = 1.72 × 104 + 4.284 × 104 mm3 = 6.004 × 104 mm3 = 60 cm3 = 60 cc of water. Exercise 5.1 A thin cylindrical shell made of 4-mm-thick copper plate is filled with oil under a pressure of 2.4 N/mm2. The internal diameter of the cylinder is 200 mm and its length is 800 mm. Determine the additional volume of oil pumped inside the cylinder so to develop the required pressure. Given E for copper = 104 kN/mm2, n for copper = 0.32, and K for oil = 2,800 N/mm2.

Thin Spherical Shell Consider a thin spherical shell of diameter D and wall thickness t subjected to internal fluid pressure p as shown in Fig. 5.5(a). The internal pressure p acts throughout the inner curved surface and the

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Thin Cylindrical and Spherical Shells

171

PD

pDt

p

sC s C − Hoop stress p 2 D Projected area = 4 (b)

Wall thickness, t (a)

Figure 5.5 Thin spherical shell diametral bursting force tends to break the thin shell in two halves, that is, two hemispherical halves, as shown in Fig. 5.5(b). Diametral bursting force, PD = p × projected area of curved surface = p×

π 2 D 4

(5.2)

Say sc = hoop stress developed in the wall of the shell. Area resisting the diametral bursting force = pDt

σ c π Dt = p ×

So

sc, hoop stress =

π 2 D 4

pD 4t

Strains Consider a small element of thin spherical shell subjected to internal radial stress p, due to which circumferential stress sc is developed in shell, as shown in Fig. 5.6. pa = atmosphere pressure on outer surface. If E and n are Young’s modulus and Poisson’s ratio of the material, respectively, then σ νσ σ Circumferential strain ε c = c − c = c (1 − ν ) E E E =

(5.3)

pa

sc

sc

pD (1 − ν ), putting value of sc 4tE

Internal pressure p >a for steel

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188 Chapter 5

σ c , = pD = p × 3, 000 = 125 p 4t 4×6 Volumetric strain in sphere, evs = volumetric strain in water, evw p ε vw = − + α w ( T −To ) K Hoop stress in shell,

ε vs =

3σ c (1 − v ) + 3 × α s (T − T o ) E

125 p  = 3 (1 − 0.3) + 11 × 10 −6 (60 − 20 )  E   125 p × 0.7  = 3 + 440 × 10 −6   2, 00, 000  =

2.1 × 125 p +1320 × 10 −6 2, 00, 000

= 1, 312.5 × 10 −6 p + 1, 320 × 10 −6

ε vw = −

(5.17)

p + 0.207 × 10 −3 (60 − 20 ) 2, 200

= −0.4545 p × 10 −3 + 8.28 × 10 −3

(5.18)

ε vs = ε vw

Equating

1312.5 p × 10 −6 + 1320 × 10 −6 = −0.4545 p × 10 −3 + 8.20 × 10 −3 1.3125 p × 10 −3 + 1.32 ×10 −3 = −0.4545 p × 10 −3 + 8.28 × 10 −3 (1.767 p ) × 10 −3 = 6.96 × 10 −3 Pressure,

p=

6.96 = 3.94 N/mm 2 1.767

The volume of water which escapes through the leak is simply the difference of free expansion of the water and the vessel, as after leakage there is no pressure remaining in the shell.

δV = (8.28 × 10 −3 − 1.32 × 10 −3 ) × = 6.96 × 10 −3 ×

p × (3, 000)3 6

p D3 6

p × 27 × 106 mm3 6 = 98.395 × 106 mm 3 = 6.96 ×

Problem 5.9 A brass hoop of an inside diameter of 40 cm and a thickness of 1 cm fits snugly at 180°C over a steel hoop which is 1.5 cm thick. Both the hoops are 5 cm wide. If the temperature drops to 20°C, determine the circumferential stress in each hoop and the radial pressure at common radius. For steel E = 200 GPa , α = 12 × 10 −6/ °C and for brass E = 100 GPa , α = 20 × 10 −6/ °C ( Fig 5.15).

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Thin Cylindrical and Spherical Shells

189

p Brass hoop p

p Steel hoop Steel hoop

185

mm

p

200

mm

370 mm 420 mm

Figure 5.15 Solution Brass hoop initial temperature = 180°C Final temperature = 20°C Reduction in temperature = 160°C As the brass hoop cools down, brass will try to contract, exerting pressure on steel hoop, say the radial pressure developed at common radius of 200 mm is p′ as shown. Due to p′, compressive hoop stress in steel hoop will be developed. At the same time steel hoop exerts outer pressure p′ on brass hoop and tensile hoop stress will be developed in brass hoop. In brass

σ cb =

p′ × 200 p ′ × 200 = = 20 p ′ (tensile) t 10

σ cs =

p ′ × 200 p ′ × 200 = = 13.33 p ′ (compressive) 15 t′

In steel

Total strain =

σ cd σ cs + = ∆T (α b − α s ) Eb Es

20 p ′ 13.33 p ′ + = 160(20 − 12) × 10 −6 2, 00, 000 1, 00, 000 10 p ′ + 13.3 p ′ = 160 × 8 × 10 −6 × 105 23.33p′ = 128 p′=

128 = 5.48 N/mm 2 23.33

σ cb = 20 × p ′ = 5.48 × 20 = 109.6 N /mm 2 (tensile) σ cs = 13.33 p ′ = 5.48 × 13.33 = −73.05 N/mm 2 (compressive)

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190 Chapter 5

Problem 5.10 A bronze sleeve of an internal diameter of 200 mm and a thickness of 5 mm is pressed over a steel liner of an external diameter of 200 mm and a thickness of 10 mm with a force fit allowance of 0.06 mm on diameter. Considering both the bronze sleeve and the steel liner as thin cylinder, determine (a) radial pressure at the common radius, ( b) hoop stresses in sleeve and liner, and (c) percentage of fit allowance met by the sleeve. Given for bronze E = 120 kN/mm 2 ,ν b = 0.33 and for steel E = 208 kN /mm 2 , vs = 0.3. Solution Figure 5.16 shows the bronze sleeve under an initial pressure p and the sleeve liner under an external pressure p. The internal pressure p is developed due to force fit between the bronze sleeve and the steel liner. p

D = 200 mm

p

p p

105

90

100 p

100 mm

p Bronze sleeve

Steel liner

Figure 5.16 (a) Stresses Bronze,

σ cb =

pD p × 200 = = 20 p (tensile) 2t b 2×5

Steel,

σ cs =

pD p × 200 = = 10 p (compressive) 2t s 2 × 10

Radial strain in bronze sleeve

ε cb = =

σ cb ν b p + Eb Eb 20 p + 0.33 p 20.33 p = Eb Eb

where σ cb is tensile and p is compressive Radial strain in steel liner

ε cs = Total radial clearance, Common radius, So,

MTPL0259_Chapter 05.indd 190

σ cs ν s p −10 p + 0.3 p −9.7 p + = = Es Es Es Es δ R = 0.06 mm R = 100 mm

 20.33 p 9.7 p  0.06 + =  E E s  100 b

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191

20.33 p 9.7 p 0.06 + = 1, 20, 000 2, 08, 000 100 16.941 p + 4.663 p =

105 × 0.06 = 60 100

21.604p = 60 Radial pressure,

p=

60 = 2.78 N /mm 2 at common surface 21.604

( b) Hoop stresses

σ cb = 20 p = 2.78 × 20 = 55.6 N/mm 2 σ cs = −10 p = −2.78 × 10 = −27.8 N/mm 2 (c) Radial clearance On bronze sleeve 20.33 p 20.33 × 2.78 × 100 = × 100 Eb 1, 20, 000 = 0.047 mm On steel liner 9.7 p × 100 9.7 = × 2.78 × 100 5 2, 08, 000 = 0.013 mm Percentage of fit allowance of sleeve

=

0.047 × 100 0.047 × 0.013

= 78.33 per cent Problem 5.11 A steel tyre of a thickness of 10 mm, a width of 80 mm and an inside diameter of 1,500 mm is heated and shrunk on to a steel wheel of a diameter of 1,501 mm. If the coefficient of static friction between the tyre and the wheel is 0.3, what twisting moment is required to twist the tyre relative to the wheel? Neglect the deformation of the wheel. E = 200 GPa. Solution Width of tyre, b = 80 mm Thickness of tyre, t = 10 mm Tyre diameter, D = 1,500 mm Wheel diameter, D′ = 1,501 mm Change in the inner diameter of tyre = 1,501 − 1,500 = 1 mm Circumferential or diametral strain,

εc =

1 1, 500

E = 2,00,000 N/mm2

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192 Chapter 5

σ c = εc × E =

Hoop stress in tyre,

1 × 20, 000 1, 500

= 133.33 N/mm2 =

pD 2t

where p is junction pressure between the tyre and the wheel pD = 133.33 2t

or,

p × 1, 500 = 133.33 2 × 10 p= Twisting moment Total radial force on tyre,

133.33 = 1.77 N /mm 2 75 R = b × p Dp = 80 × p × 1, 500 × 1.77 = 6,67,274 N µ = 0.3

Coefficient of friction, Force of friction,

F = µR = 0.3 × 667, 274 = 2,00,182 N T = F × radius of wheel

Twisting moment,

= 2,00,182 × 750 = 1,50,136,500 N mm = 150.136 kN m

Key Points to Remember  For a shell, D/t ratio is greater than 20, it is classified as a thin shell.  For a cylindrical thin shell is subjected to internal pressure, p. Hoop stress,

σc =

Axial stress,

σa =

Hoop strain,

εc =

Axial strain,

εa =

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pD 2t pD 4t pD 4tE pD 4tE

(2 − v ) (1 − v )

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Thin Cylindrical and Spherical Shells

εv =

Volumetric strain,

pD 4tE

193

(5 − 4 v )

 For a thin spherical shell subjected to internal pressure, p. pD

Hoop stress,

σc =

Diametral strain,

εc = εd =

Volumetric strain,

εv =

4t

3 pD 4tE

pD 4tE

(1 − v )

(1 − v )

 δV, addition of volume of liquid pumped inside the cylinder dV1 (increase in the volume of cylinder) + dV2 (decrease in the volume of liquid)  Due to wire winding, initial compressive hoop stress is developed in cylinder. Pressure bearing capacity of cylinder is increased.  In double curved wall of a shell under pressure σ1 σ 2 p  + = r1 r2 t  In a cylindrical shell, σ 1 = σ c , σ 2 = σ a , r2 = ∝ (infinity)  In a conical water tank, stresses

σ1 =

w ( Hy − y 2 )

σ2 = ×

t

×

tan α cosα

w tan α  2 2  Hy − y  2t cosα 3

Where a is semi-cone angle of tank, t is wall thickness, w is specific weight and H is depth of liquid.

Review Questions 1. What is a thin cylindrical/spherical shell? On what D/t ratio it is classified as a thin shell? 2. Derive the expressions for hoop and axial stresses developed in a thin cylindrical shell subjected to internal pressure p. 3. Take a small element of a thin spherical shell and show the stresses acting on this element. 4. Derive the expression for the volumetric strain of a thin spherical shell subjected to internal pressure p. 5. A cylinder of diameter D and wall thickness t is wound with a single layer of wire under tension sw. Derive the expression for hoop stress developed in cylindrical shell. 6. Derive the following expression for a double curved surface under pressure p.

σ1 r1

+

σ2 r2

=

p t

7. Derive the expression for maximum hoop stress developed in a conical water tank. 8. How the strain gauges mounted on an external surface of a thin shell subjected to an internal pressure can be used to measure E and n of the material of the shell.

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194 Chapter 5

Multiple Choice Questions 1. A thin cylindrical shell with D/t = 30 is subjected to an internal pressure of 3 N/mm2. What is the hoop stress developed in shell? (a) 90 MPa

(b) 45 MPa

(c) 22.5 MPa (d) None of these 2. Athin spherical shell of an inner diameter of 400 mm is subjected to an internal pressure of 2.5 N/mm2. If the hoop stress is not to exceed 100 MPa, what is the thickness of shell? (a) 2.5 mm

( b) 5 mm

(c) 10 mm (d) None of these 3. A thin cylindrical shell is made of steel with n = 0.3. It is subjected to an internal pressure p. What is the ratio of hoop strain to axial strain? (a) 1.0

(b) 2.0

(c) 3.0 (d) 4.25 4. A thin spherical shell is subjected to an internal pressure p. What is the ratio of volumetric strain to circumferential strain in shell? (a) 1.0

(b) 2.0

(c) 3.0 (d) None of these 5. A cylindrical tank of an inside diameter of 1 m and a height of 20 m is filled with water of a specific weight of 10 kN/m3. If the thickness of tank is 25 mm, then the maximum stress developed in wall of the tank is

(a) 4 N/mm2

(b) 2 N/mm2

(c) 1 N/mm2 (d) None of these 6. A thin cylindrical shell is made of steel with n = 0.30. It is subjected to an internal pressure. What is the ratio of volumetric strain to circumferential strain? (a) 2.0

(b) 2.235

(c) 9.5 (d) None of these 7. A closed pressure vessel of a length of 400 mm, a wall thickness of 5 mm and an internal diameter of 100 mm is subjected to an internal pressure of 8 N/mm2. The normal stress on an element of the cylinder on a plane 30° to the longitudinal axis will be (a) 140 MPa

(b) 70 MPa

(c) 77.32 MPa (d) None of these 8. A steam boiler of an internal diameter of 1.5 m is subjected to an internal pressure of 2 N/mm2. If the efficiency of the longitudinal joint is 80 per cent and the maximum tensile stress in the plate section is not to exceed 125 MPa, what is the thickness of the plate? (a) 6 mm

(b) 3 mm

(c) 15 mm

(d) None of these

Practice Problems 1. A steam boiler of an internal diameter of 1.5 m is subjected to an internal pressure of 1.2 N/mm2. What is the tension per linear metre of the longitudinal joint in the boiler shell? Calculate the thickness of the plate if the maximum tensile stress in the plate section is not to exceed 90 N/mm2, taking efficiency of longitudinal joint as 70 per cent. 2. A single strain gauge making an angle of 15° with the horizontal plane is used to determine the gauge pressure in a cylindrical tank with its axis vertical as shown in Fig. 5.17. The tank is 6 mm in thickness and 500 mm in diameter. It is made of steel with E = 200 GPa and n = 0.29. The strain gauge reading is 350 m strain. Determine the pressure inside the tank.

εa + εc εa − εc    Hint: εθ = 2 + 2 cos 2θ , θ =105°    3. The ends of a thin cylindrical shell are closed by flat plates. It is subjected to an internal fluid pressure of 3 N/mm2, but the ends of the cylinder are rigidly stayed and no axial movement is permitted. The diameter

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195

Axis c a 15° Horizontal

O

b

sC q = 105°

15°

sa

Figure 5.17 of cylinder is 200 mm and its length is 800 mm, while the wall thickness is 5 mm. Determine the change in the volume of cylinder, if E = 200 GPa and n = 0.3. pD    Hint: ε v = tE (1 − 0.2ν )  4. A thin-walled copper alloy pressure vessel of a diameter of 250 mm and a wall thickness of 5 mm is subjected to an internal pressure of 5 MPa. The strain gauges that are bonded on the surface in the hoop and axial directions give readings of 988 and 191 ms at full pressure, respectively. Determine E and n for the material. 5. A 20-cm-long copper tube with closed ends is 100 mm in diameter and has a wall thickness of 4 mm with no internal pressure. The tube just fits between the two rigid end walls. Calculate the axial and hoop stresses in the cylinder for an internal pressure of 3 N/mm2. E = 105 kN/mm2 and n = 0.33. 6. A 50-mm bore gun metal tube of a wall thickness of 1.25 mm is closely wound externally by a steel wire of a diameter of 0.5 mm. Determine the tension under which the wire must be wound on the tube if an internal pressure of 1.5 N/mm2 is required before the tube is subjected to a tensile stress in the circumferential direction. For gun metal E = 102 GPa and n = 0.35 and for steel E = 210 GPa 7. A 150-mm-long bronze tube closed at its ends is 80 mm in outer diameter and has a thickness of 3 mm. It fits without clearance in an 80-mm hole in a rigid block. The tube is then subjected to an internal pressure of 4 N/mm2. Take v = 1/3 and E = 83 kN/mm2, determine the tangential stress in the tube. pDi pDi p ′ × 80   Hint: σ a = 4t , σ c = 2t , σ c′ = 4 × t  1  [σ a − σ c ′ − σ a ′v] = 0 E 

    

8. A thin spherical shell of copper has an internal diameter of 2.5 m and a wall thickness of 5 mm and is just filled with water at 25°C and at atmospheric pressure. If the temperature of water and shell rises to 60°C, determine the volume of water that would escape if a small leak is developed at the top of the vessel. For copper E = 105 GPa, acu = 18 × 10-6/°C and n = 0.34 and for water K = 2, 200 N/mm 2 , α = 0.207 × 10 −3/ °C (for volumetric expansion). 9. A steel tube of an internal diameter of 150 mm and a wall thickness of 8 mm in a chemical plant is lined internally with a well-fitting copper sleeve of a wall thickness of 2 mm. If the composite tube is initially unstressed, calculate the hoop stresses set up assumed to be uniform throughout the wall thickness, in a unit

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196 Chapter 5 length of each part of the tube due to increase in temperature of 100°C. Neglect the temperature effect in axial direction. [Hint: compressive stress in copper, α c > tensile stress in steel, α s]

σ cs =

p ′ × 150 p ′ × 150 , σ cc = 2×8 2×2

σ cs = 4σ cs ; use

σ cs σ cs + = (α c − α s ) ∆T Es Ec

For steel,

E = 208 GPa , α = 11 × 10−6 / °C

For copper,

E = 104 GPa , α = 18 × 10−6 / °C

10. A copper sleeve of an internal diameter of 150 mm and a thickness of 2.5 mm is pressed over a steel liner of an external diameter of 150 mm and a thickness of 2.5 mm with a force fit allowance of 0.04 mm on diameter. Considering both the copper sleeve and the steel liner as thin cylinders, determine (a) radial pressure at common radius, ( b) hoop stresses in sleeve and liner (c) percentage of fit allowance met by the sleeve. For copper, E = 105 kN/mm2, nc = 0.34 For steel, E = 210 kN/mm2, ns = 0.3 11. A thin tyre is to be shrunk onto a rigid wheel of a diameter of 1 m. Determine the necessary internal diameter of the tyre if the maximum hoop stress in the tyre is 10 N/mm2. Also determine the least temperature to which the tyre must be heated above that of wheel before it could be slipped on. a for tyre = 11 × 10-6/°C

E = 204 GPa

100 δD    Hint: ε a = E = ∆T α = D = εc   

Special Problems 1. A closed cylindrical vessel made of steel plate of a thickness of 5 mm with plane ends carries fluid under a pressure of 4 N/mm2. The diameter of the cylinder is 250 mm and its length is 750 mm. Calculate the longitudinal and hoop stresses developed in the cylinder. What are the changes in the diameter, length and volume of the cylinder?

E = 210 GPa, n = 0.3 2. One method of determining Poisson’s ratio for a material is to subject a cylinder to internal pressure and to measure axial strain ε a and hoop strain εc on the outer surface of cylinder show that

ν=

ε c − 2ε a 2ε c − ε a

pD p    Hint: εc = 4tE (2 − ν ), ε a = 4tE (1 − 2ν )   3. A cylindrical tank of an inside diameter of 2 m and a height of 20 m is filled with water of a specific weight of 10,000 N/mm3. The material of the tank is structural steel with yield strength of 250 N/mm2. What is the minimum thickness required at the bottom of the steel tank if the efficiency of the longitudinal seams is 80 per cent. Take factor of safety (FOS) as 4.  250 pD   Hint: p = wH , FOS = 2tη   

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197

4. A thin spherical shell made of copper alloy is 300 mm in diameter and 1.5 mm in wall thickness. It is full of water at an atmospheric pressure. Find by how much the internal pressure will increase if 20 cc of water is pumped inside the shell. E = 100 GPa, n = 0.29 For water, K = 2,200 N/mm2 5. A pressurized steel cylinder tank has an inner radius of 600 mm and a thickness of 16 mm. The tank is subjected to a pressure p = 1,750 kPa and an axial force P = 125 kN. The butt weld seam forms an angle of 54° with the longitudinal axis of the tank. Determine (a) the normal stress perpendicular to weld and (b) the inplane shear stress parallel to weld. pD    Hint: p = 1, 750 kPa = 1.75 MPa , σ c = 2t  pD p σa = , σ a′ = , σ a − σ a′ , net stress in axial direction 4t πDt q = 54° with the plane of σ c 6. A spherical steel vessel is made of two hemispherical portions fitted together at flanges. The inner diameter of the sphere is 600 mm and the wall thickness is 6 mm. Assuming that the vessel is a homogeneous sphere, what is the maximum working pressure for an allowable tensile stress in a shell of 150 MPa. If 20 bolts of a diameter of 16 mm are used to hold flanges together, what is the tensile stress in bolts when the sphere is under full pressure? pD π   2  Hint: 150 = 4t , PD = 20 × 4 × 16 × σ  7. For a hydraulic test, a steel tube of an internal diameter of 80 mm, a wall thickness of 2 mm and a length of 1.2 m is fitted with end plugs and filled with oil at a pressure of 2 MPa. Determine the volume of oil leakage which would cause the pressure to fall to 1.5 MPa. For oil, K for oil = 2.8 GN/m2, For steel, E = 208 GPa, n for steel = 0.29 p    Hint: δV = δV1 + δV2 = εvV + K V for ∆p   

Answers to Exercises H ; 2 3H 3.331 N/mm2 at y = 4

Exercise 5.5: 4.441 N/mm2 , at y =

Exercise 5.1: 48.51 cc Exercise 5.2: 3mm, ec = 0.3888 × 10−3, dD = 0.1166 mm Exercise 5.3: t2 = 2 mm, 1.235 Exercise 5.4: σ cr = +37.864 N/mm2 ,

σ wr = 43.02 N/mm2

Answers to Multiple Choice Questions 1. (b) 2. (a) 3. (d)

MTPL0259_Chapter 05.indd 197

4. (c) 5. (a) 6. (b)

7. (b) 8. (c)

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198 Chapter 5

Answers to Practice Problems 1. 2. 3. 4. 5. 6. 7.

400 kN/linear metre, t = 14.3 mm p = 2.07 N/mm2 14.17 cc E = 105 kN/mm2, n = 0.34 37.5 N/mm2, 12.335 N/mm2 64.52 N/mm2 8.72 N/mm2

8. 43,810 cm3 9. σ cs =16.18 N/mm2 (tensile)

= 57.325 − 1.91(18.5) 10. (a) 1.24 N/mm2; (b) 37.2 N/mm2 (tensile), 37.2 N/mm2 (compressive); (c) 67.14% 11. 44.56°C, 999.51 mm

Answers to Special Problems 1. 50 MPa, 100 MPa, 0.101 mm, 0.0714 mm, 33.3 × 103 mm3 3. 4 mm 4. 0.93 MPa

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5. 42.46 MPa, 14.51 MPa 6. p = 12 N/mm2; 211 Mpa 7. δV = 1.63 cc

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6 Thick Shells CHAPTER OBJECTIVES For low pressures, thin pressure vessels are used where D/t ratio is greater than 20. For high internal pressure as in the case of tubes used for high pressure gauge and bulk compression gauge, tubes with considerable thickness are used where D/t R22 + R12

)

) =σ )

c min

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Thick Shells

Therefore, the maximum hoop stress occurs at inner radius. Axial stress, sa = −

(

pR22 (compressive) R22 − R12

)

207

sc

sr

sa

sa

as the pressure acts on the outer surface. Stresses on an element of a cylinder subjected to pressure p are shown in Fig. 6.6. Note that all the stresses sc, sr and sa are compressive. The maximum hoop stress (though compressive in this case) occurs at inner radius.

sc

Figure 6.6

Example 6.2 A thick cylinder of inner radius 85 mm and outer radius 120 mm is subjected to an external pressure of 50 N/mm2. Determine scmax and scmin and show variation of sc and sr along the thickness of the cylinder. Solution sr = p, at outer radius

sr

sr = 0, at inner radius scmax = −

2 pR 22 R 22 − R12

= −

2 × 50 × 120 2 120 2 − 852

= −

1, 440, 000 = −200.7 N /mm 2 7,175

scmin = − p × − = −p× −

120 mm = R 2

85 mm = R 1

− 200.7

sc

(−50 N/mm2)

−150.7 N/mm 2

Figure 6.7

R 22 + R12 120 2 + 852 = − p × R 22 − R12 120 2 − 852 14, 400 + 7, 225 12, 625 = −p× 14, 400 − 7, 225 7,175

= −p × 3.0139 = −150.7 N/mm2 Figure 6.7 shows the variation of sc and sr in the cylinder due to external pressure. Exercise 6.2 A thick cylinder with an external diameter of 240 mm and an internal diameter of D is subjected to an external pressure of 50 MPa. Determine the diameter D of the maximum hoop stress in the cylinder that is not to exceed 200 MPa.  2 pR22  Hint: σ −   c max R22 − R12  

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208 Chapter 6

Compound Cylinder To increase the pressure-bearing capacity of a cylinder and to decrease the variation in hoop stress developed in cylinder, another cylinder is force fitted or shrink fitted over the cylinder. This process is called compounding of cylinders. Therefore, for compounding, the inner diameter of the outer cylinder is slightly less than the outer diameter of the inner cylinder so as to provide the force fit allowance or shrinkage allowance. In shrinkage fitting, the outer cylinder is heated such that its inner radius R 3′′ expands to R 3′ and then the outer cylinder is slipped over the inner cylinder. When the outer cylinder cools, it exerts a radial pressure p′ on the outer surface of the inner cylinder. In reaction, the inner cylinder exerts a pressure p′ on the inner surface of the outer cylinder as shown in Fig. 6.8(c). The outer pressure p′ on the inner cylinder develops compressive hoop stresses in the inner cylinder. Similarly the inner pressure p′ on the outer cylinder develops tensile hoop stresses in the outer cylinder. The final junction radius between two cylinders is R3 such that R3 < R 3′ but R3 > R3′′. The difference R 3′′− R 3′ is termed as shrinkage allowance. Stresses due to junction pressure Inner cylinder

σ ciR 1 = − p′ ×

2R 3 2 (compressive) at R1 R 32 − R12

σ ciR 3 = − p ′ ×

R32 + R12 (compressive) at R3 R32 − R12

(expression derived earlier in case of cylinder with external pressure) Outer cylinder

σ coR 3 = + p ′ ×

R 2 2 + R 32 (tensile) at R3 R 2 2 − R 32

σ coR 2 = + p ′ ×

2R 32 (tensile) at R2 R 22 − R 32 R 3 < R 3 R 3 > R 3

R2

R 3 < R 3 R 3

p

R3

R 3

R1

R1

p

p

R2

p Inner cylinder (a)

Outer cylinder (b)

Compound cylinder (c)

Figure 6.8

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Thick Shells

209

Figure 6.9 shows the variation of hoop stresses developed in the inner and outer cylinders due to junction pressure. Compound cylinder Say the compound cylinder is now subjected to an internal pressure p. Say A and B are Lame’s constants, thus

or

p=

B −A R12

(6.13)

0=

B −A R 22

(6.14)

A=

B R 22

R2

R3 R1

scoR3 scoR2

sciR 3 sciR1

(

2 2 B B B R2 − R1 p= 2 − 2 = R1 R2 R12 R22

B=p

constant,

Inner cylinder

)

Outer cylinder

Figure 6.9

R12 R 22 R 22 − R12 R12 R 22 − R12

A= p

Stresses at R1, R2 and R3 due to internal pressure

σ cR1

(

R 22 + R12 R 22 pR12 =p 2 + =p 2 R 2 − R12 R 22 R12 R 2 − R12

σ cR3 = p

)

R12 R 22 R12 + p R 22 − R12 R 32 R12 − R12

(

)

(R R + R R ) =p R (R − R ) 2 1

2 2

2 3

σ cR 2 = 2 p

2 1

2 2

2 3

2 1

R12 R 22 − R12

Resultant stresses Inner cylinder

σ cR ′ 1 = σ cR1 + σ ci R1 =p

σ cR ′3=p

MTPL0259_Chapter 06.indd 209

(R

2 2

+ R12

R 22 − R12

) − p ′ × 2R R 32 − R

2 3 2 1

R12 R 22 + R12 R 32 R 32 + R12 − p × ′ R 32 − R12 R 32 R 22 − R12

(

)

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210 Chapter 6

Outer cylinder

σ cR ′′ 3 = σ cR3 + σ coR3 =

(

p R12 R 22 + R12 R 32 2 3

(

2 2

R R −R

2 1

)

) + p′ × R

2 2 2 2

+ R 32 R − R 32

σ cR ′′ 2 = σ cR 2 + σ coR 2 =

p ′ × 2R 32 2 pR12 + 2 2 2 R 2 − R1 R 2 − R 32

Figure 6.10 shows the variation of resultant hoop stresses developed due to junction pressure p′ and internal pressure p on the inner and outer cylinders.

R2

sCR1

sCR3

R1 R3

Figure 6.10 Resultant hoop stress in the inner and outer cylinders of a compound cylinder

Example 6.3 A compound cylinder is made by shrinking one cylinder over another such that the outer diameter is 200 mm, the inner diameter is 100 mm and the junction diameter is 150 mm. If the junction pressure developed between two cylinders is 10 N/mm2 and the internal pressure is 50 N/mm2, what are the hoop stresses at inner and outer radii of both the cylinders? Solution Radii R1 = 50 mm R2 = 100 mm R3 = 75 mm Pressure p′ = 10 N/mm2 p =50 N/mm2

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Thick Shells

211

Resultant hoop stresses Inner cylinder p(R 2 2 + R12 ) p’(2R 32 ) − (R 2 2 − R12 ) R 32 − R12

σ cR ′1= Putting the values

50(1002 + 502 ) 10(2 × 752 ) − 1002 − 502 752 − 502

=

= 83.33 − 36 = 47.33 MPa

σ cR ′3=

p(R12 R 2 2 + R12 R 32 ) (R 2 + R12 ) − p ′ × 32 2 2 2 R 3 ( R 2 − R1 ) (R 3 − R12 ) 50(50 2 × 100 2 + 50 2 × 752 ) 10 × (752 + 50 2 ) − 752 (100 2 − 50 2 ) 752 − 50 2

=

= 46.29 − 26 = 20.29 MPa Outer cylinder

σ cR ′′ 3 =

p(R12 R 2 2 + R12 R 32 ) p’ × (R 32 + R 2 2 ) + R 32 (R 2 2 − R12 ) R 2 2 − R 32

= 46.29 + 10 ×

(752 + 100 2 ) 100 2 − 752

= 46.29 + 10 ×

15, 625 = 46.29 + 35.71 4, 375

= 82.00 MPa

σ cR ′′ 2 = =

P ′ × 2 R32 2 pR12 + R 22 − R 12 R 22 − R 32 2 × 50 × 50 2 10 × 2 × 752 + 100 2 − 50 2 100 2 − 752

= 33.33 + 25.71 = 59.04 MPa Exercise 6.3 A steel cylinder of an outer diameter of 180 mm is shrunk on another cylinder of an inner diameter of 120 mm. The common diameter is 150 mm. If after shrinkage the radial pressure at the common surface is 6 N/mm2, determine the magnitude of internal pressure to which the compound cylinder can be subjected so that the maximum hoop tension in the inner and outer cylinders is equal. Then, t [σ ′ cR1 = σ ″ cR3 ]

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212 Chapter 6

Shrinkage Allowance In a compound cylinder at the junction radius due to junction pressure, tensile hoop stress in the outer cylinder and compressive hoop stress in the inner cylinder are developed. At common radius R3 Inner cylinder

σ ciR3 = − p ′ ×

R32 + R12 (compressive ) R3L − R12

Junction pressure, p′ (compressive) Radial or circumferential strain,

(

p ′ R32 × R12

ε ciR3 = −

(

R32 − R12

)

) × 1 + v p′ i

Ei

Ei

(6.15)

When vi and Ei are the Poisson’s ratio and the elastic modulus of the inner cylinder, respectively ( R3′ − R3 ) = ε ciR3 × R3 =−

 p ′ × R3  R32 + R12 − νi  2 2  Ei  R3 − R1 

(6.16)

Outer cylinder

σ coR3 = + p ′ ×

R22 + R32 − (tensile) R22 − R32

Junctions pressure, p′ (compressive)

ε coR3 = + p ′ ×

ν p′ ( R22 + R32 ) + 0 2 2 E0 ( R2 − R3 ) E0

(6.17)

where v0 and E0 are the Poisson’s ratio and the elastic modulus of the outer cylinder. ( R3 − R3′′ ) = + p ′ ×

 R3  R22 + R32 + ν0  2 2  E0  R2 − R3 

(6.18)

Shrinkage allowance ( R3′ − R3′′ ) from Eqs (16) and (18)

(

)

(

)

2 2  R 2 + R32 ν 0 1 R3 × R1 ν  = p′ R3  1 2 + + − i 2 2 2 2 E0 Ei R3 − R1 Ei   E0 R2 − R3 

(6.19)

If both the cylinders are of different materials, then the shrinkage allowance is calculated from Eq. (6.19). If Ei = E0 = E,ni = n0 = n, that is, both the cylinders are of same material. The shrinkage allowance on common radius R3 is

δ R3 =

MTPL0259_Chapter 06.indd 212

p ′R3  R22 + R32 R32 + R12  + E  R22 − R32 R32 − R12 

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Thick Shells

213

where R1, R3 and R2 are inner radius, junction radius and outer radius of compound cylinder. δ R3 1 Remember that (numerical sum of hoop stresses developed at the common radius of two = R3 E cylinders). Example 6.4 A compound cylinder is formed by shrinking one outer steel cylinder over bronze cylinder. The final dimensions are internal diameter 100 mm, external diameter 200 mm and junction diameter 160 mm, and the shrinkage pressure at the common surface is 12 N/mm2. Calculate the necessary difference in radii of two cylinders at the common surface. For steel, E0 = 200 GPa, ni = 0.3 For bronze, Ei = 100 GPa, n0 = 0.32 What is the maximum temperature through which the outer cylinder should be heated before it can be slipped on? For steel a = 11 × 10−6/°C Solution Junction pressure,

p′ = 12 N/mm2 R1 = 50 mm, R2 = 100 mm, R3 = 80 mm dR3 =

  p1 × R3  R22 + R32 p ′ × R3  R32 + R12 − νi  + + ν0  2 2 2 2   E1  R3 − R1 E R − R    2 0 3

Putting the values

δ R3 =

 12 × 80  1002 × 802  12 × 80  802 + 502 − 0.32 + + 0.3 2 2  100 × 1, 000  80 − 50  200, 000  1002 − 802 

= 9.6 × 10−3(2.282 − 0.320) + 4.8 × 10−3 (4.555 + 0.3) = 18.8 × 10−3 + 23.304 × 10−3 = 42.104 × 10−3 mm = 0.0421 mm a = 11 × 10−6/°C dR3 = R3 × aDT 0.0421 = 80 ×11×10−6 × DT DT = 47.84°C The outer cylinder should be heated by 47.84°C so that it can be slipped over inner cylinder. Exercise 6.4 A steel tube of an outside diameter of 220 mm is shrunk on another steel tube of an inside diameter of 140 mm. The diameter at junction is 180 mm after shrinking on. The shrinkage allowance provided on the radius of the inner tube is 0.08 mm. Determine (a) the junction pressure, (b) the hoop stresses at the inner and outer radii of the inner tube and (c) the hoop stress at the inner and outer radii of the outer tube if E = 210 kN/mm2.

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214 Chapter 6

Hub and Shaft Assembly In a power transmission system, the shaft is connected to the hub with the help of a key inserted in keyways cut on the shaft and in the hub. These keys severely reduce the strength of shaft and hub due to the keyways having sharp corners. To avoid keyways, the hub can be shrunk fitted on the shaft as in the case of compound cylinders. Due to shrink fitting the junction pressure p′ is developed between the shaft and the hub. Let as consider a shaft of radius R1 fitted in a hub of outer radius R2 and inner radius R1 as shown in Fig. 6.11. R2

Hub

R1

p p

Shaft

Junction pressure p

B Hub and shaft

Figure 6.11 Shaft Let as take Lame’s constants A and B for shaft. B σr = 2 − A Radial stress, r B Circumferential stress, σc = 2 + A r But material exists at centre, that is, at r = 0, and the stress cannot be infinite; therefore, constant B = 0. In shaft sr = − A = p′ (junction pressure) or A = −p′ sc = + A = − p′ While deriving Lame’s equations, we have considered sr as negative and sc as positive. So now sc and sr both are compressive and equal to junction pressure p′. p′ ν s p′ Radial strain in shaft = + − at radius R1 Es Es εes =

p′ (1 − ν s ) negative strain Es

Hub Due to junction pressure p′, the hoop stress developed

 R 2 + R12  σ ch = p ′  22 (tensile)  R 2 − R12  Radial strain in hub, ε ch =

MTPL0259_Chapter 06.indd 214

p ′  R22 + R12  ν h p ′ , positive strain + Eh  R22 − R12  Eh

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Thick Shells

215

Because p′ is compressive Shrinkage allowance at radius R1

δ R1 = ε cs R1 + ε ch R1 If both shaft and hub are of same material such that E s = E h = E ; ν s = ν h = ν , then  R 2 + R12  δ R1 = R1 p ′  22 +1  R 2 − R12  =

Shrinkage allowance on radius =

R1 p ′  R 22 + R12 + R 22 − R12   E  R 22 − R12

R1 p ′  2 R 22  E  R 22 − R12 

Example 6.5 A steel shaft of a diameter of 100 mm is driven into a bronze hub. The driving allowance provided is on diameter 1/1,000 of the shaft diameter. Determine the thickness of the hub, if the maximum bursting stress in the hub is limited to 80 N/mm2. Es = 200 GPa, ν s = 0.3

Given

EB = 100 GPa, ν B = 0.34

Solution R1 = 50 mm R 2 = to be determined

δ D1 δ R1 1 = = = 0.001 D1 R1 1, 000 Say junction pressure is p′ Hub

σ ch = p ′ ×

R2 2 + R12 = 80 MPa R2 2 − R12

ε ch =

100 0.34 p ′ + EB EB

(6.20)

ε cs =

p ′ 0.30 p ′ − Es Es

(6.21)

Shaft

But total strain = ε ch + ε cs =

MTPL0259_Chapter 06.indd 215

100 0.34 p ′ p ′ 0.3 p ′ + + − = 0.001 EB EB Es Es

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216 Chapter 6

As given in the problem putting the values of Es and EB 80 0.34 p ′ p′ 0.3 p ′ + + − = 0.001 1, 00, 000 1, 00, 000 2, 00, 000 2, 00, 000 Multiplying throughout by 1,00,000, we get 80 + 0.34 p ′ + 0.5 p ′ − 0.15 p ′ = 100 0.69 p ′ = 20 Junction pressure, Outer radius of hub

p′ =

20 = 28.98 N /mm 2 0.69

σ ch = p ′ ×

R 22 + R12 , taking R1 = 50 mm R 22 − R12

80 = 28.9 ×

R 2 2 + 2, 500 R 2 2 − 2, 500

R 2 + 2, 500 80 = 22 28.98 R 2 − 2, 500

(

)

2.76 R 22 − 2, 500 = R 22 + 2, 500 1.76 R 22 = 3.76 × 2, 500 R 2 = 5340.5 = 73.08 mm Thickness of hub

= R 2 − R1 = 73.08 − 50 = 23.08 mm

Exercise 6.5 A bronze liner of an outside diameter of 60 mm and an inside diameter of 39.94 mm is forced over a steel shaft of 40 mm in diameter. Determine (a) the radial pressure between the shaft and the liner and (b) the maximum hoop stress in the liner. For steel, For steel,

E = 208 GPa, v = 0.29 E = 125 GPa, v = 0.33   δ R1 p ′  R2 2 + R12  ν h p ′ p ′ = ε ch + ε cs =  Hint :  R 2 − R 2  + E + E (1 − ν s )  R E 1 h h s 2 1  

Thick Spherical Shell To determine radial and hoop stresses in a thick spherical shell subjected to an internal pressure, let us first learn about the radial and circumferential strains in such axi-symmetric cases. Consider a thin disc of inner radius R1 and outer radius R2 subjected to an internal pressure p. A small element abcd is subtending an angle δθ at the centre is considered as shown in Fig. 6.12.

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Thick Shells

sc

217

s r + dsr

d a

sr

c b sc

R2

d

R1

a dq

c b

p

O

r

dr

Thin disc

Figure 6.12 Radius oa = r Radius od = r + δ r before the application of internal pressure. After the application of internal pressure, say r → changes to r + u δ r → changes to δ r + δ u. 2π (r + u − r ) Circumferential strain in element, εc = 2π r u = (6.22) r δr + δu − δr Radial strain in element, εr = δr δu = (6.23) δr These strains are tensile if positive. On the element of a spherical shell, say stresses are σ r + δσ r at radius r + δ r Circumferential stress on element = σ c Bursting force P on the elementary shell (taking the projected area of hemispherical portion) π r 2σ r − π (r + δ r ) 2 (σ r + δσ r )

MTPL0259_Chapter 06.indd 217

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218 Chapter 6

Resisting force = σ c × 2π rδ r , where δ r is the radial thickness. For equilibrium

π r 2σ r − π (r + δ r )2 (σ r + δσ r ) = σ c × 2π rδ r 2σ r δ r − rδσ r = 2σ cδ r (neglecting the small quantities of higher order terms)

δσ r δr Three principal stresses at any point in the elementary shell are: (1) radial pressure, σ r (compressive), (2) hoop stress, σ c (tensile), and (3) hoop stress, σ c , on another plane (at right angle to the first plane). Radial strain at any point σ σ ε r = r + 2ν c (compressive) E E 2σ c = −2σ r − r

εr = −

or

σ σr − 2v c (tensile) E E

(6.24)

(6.25)

Circumferential strain at the point

εc = =

σc σ σ −v c +ν r E E E σc σ (1 − ν ) + r (tensile) E E

(6.26)

From initial Eqs. (22) and (23)

δ u du u εc = , εr = = r δ r dr δε d (r ε c ) = ε c + r c dr δr Substituting for ε r and ε c from Eqs. (25) and (26) εr =

−

(6.27)

σ σ σr σ σ  d σ − 2v c = c (1 − v ) + v r +  c (1 − v ) + v r  E E E E dr  E E

After simplification we get (1 + ν )σ r + (1 + ν )σ c + r(1 − ν ) From Eq. (6.24)

σ c = −σ r −

σc vd σ r +r =0 dr dr

r dσ r 2 dr

(6.28)

(6.29)

Differentiating Eq. (6.29) we get dσ c dσ dσ r  1  d 2σ = − r −  + 2r + dr dr 2  dr d r 

MTPL0259_Chapter 06.indd 218

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Thick Shells

219

Substituting for σ c and dσ c /dr in Eq. (6.24) and after multiplication we get r

d 2σ r dσ +4 r =0 dv2 dr

Let us put dσ r /dr = k , then we have dk + 4k = 0 dr d k 4d r + =0 k r ln k = −4 ln r + ln C1

r

So where lnC1 is a constant of integration

C1 r4 d σ r C1 = 4 dr r k=

or

(6.30)

Integrating Eq. (6.30), C1 + C 2 , where C2 is another constant of integration. 3r3 r dσ r σ c = −σ r − (from Eq. 6.29) 2 dr C C σ c = + 13 − C 2 − 13 (using Eq. 6.30) 3r 2r C σ c = 13 − C 2 6r

σr =

Now or

Let us put constants C1 = 6 B and C 2 = − A Then, 2B Radial stress, σ r = 3 − A (compressive) r B Hoop stress, σ c = 3 + A (tensile) r For a thick spherical shell subjected to an internal pressure p, the boundary conditions are σ r = p at r = R1 p=

2B −A R13

σ r = 0 at r = R 2

Constants,

MTPL0259_Chapter 06.indd 219

0=

2B 2B − A or A = 3 3 R2 R2

B=

p R13R 23 pR3 and A = 3 1 3 3 3 R 2 − R1 2 R 2 − R1

(

)

5/23/2012 11:13:12 AM

220 Chapter 6

Example 6.6 Calculate the thickness of a spherical shell of an inside diameter of 120 mm to withstand an internal pressure of 50 MPa, if the maximum permissible tensile stress in the shell is 120 MPa. Solution R1 = 60 mm

σ c max =

pR 23 pR13 + 3 3 2 R 23 − R13 R 2 − R1

(

)

Putting the values 120 = =

(

50R 23 50 × 603 + 3 3 2 R 23 − 1203 R 2 − 120

(

)

25R 23 50 × 603 + R 23 − 1203 R 23 − 1203

)

120 R23 − 1203 = 25 R23 + 50 × 603

or

95R 23 = 20, 736 × 10 4 − 1, 080 × 10 4 = 19, 656 × 10 4 R23 = 206.905 × 104 = 2.06905 × 106 R 2 = 127.42 mm Thickness of shell = 127.42 − 60 = 67.42 mm Exercise 6.6 Determine the maximum shear stress at the inner surface of a spherical shell, having a diameter ratio of 1.5 for an internal pressure of 7 N/mm2.   Hint: 

σ c max +   τ max = 2

p   an inner surface  

Problem 6.1 Two thick cylinders are of the same dimensions. The external diameter is 1.5 times the internal diameter. Cylinder A is subjected to an internal pressure p1, while cylinder B is subjected to an external pressure p2. Find the ratio of p1/p2 if the greatest hoop strain in both cylinders is the same. Poisson’s ratio = 0.3 Solution Cylinder A Inner radius = R1 Outer radius, R 2 = 1.5 R1 Internal pressure = p1 At inner radius, σ c = p1 × = p1 ×

MTPL0259_Chapter 06.indd 220

R 22 + R12 R 22 − R12 2.25 + 1 = 2.6 p1 2.25 − 1

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Thick Shells

σ a = p1

Axial stress,

221

R12 1 = p1 × = 0.8 p 2 R − R1 2.25 p 2 2

σ r = p1 (compressive) ε c′ =

Hoop strain,

σ c νσ a νσ r − + E E E

=

2.6 p1 0.3 × 0.8 p1 0.3 p1 − + E E E

=

2.66 p1 E

Cylinder B External pressure = p2

σ c = p2 ×

At inner radius,

2R 22 2 × 1.52 = p × 2 R 22 − R12 1.52 − 12

= 3.6 p2 (compressive)

σ a = p2 ×

Axial stress,

R 22 = 1.2 p2 (compressive) R 22 − R12

σr = 0 σ c vσ a 3.6 p2 0.3 × 1.8 p2 3.06 p2 − = − = E E E E E

Hoop strain,

ε c′′ =

However,

ε c′ = ε c′′ 2.66 p1 3.06 p2 = E E p1 3.06 = = 1.150 p2 2.66

Problem 6.2 A steel cylinder of an inside diameter of 800 mm and a length of 6 m is subjected to an internal pressure of 40 MPa. Determine the thickness of the cylinder if the maximum shear stress of cylinder is not to exceed 65 MPa. What is the increase in the volume of the cylinder? E = 200 GPa, n = 0.3. Solution Inner radius,

R1 = 400 mm p = 40 MPa

τ max = 65MPa =

MTPL0259_Chapter 06.indd 221

σ c max + p = 65MPa 2

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222 Chapter 6

σ c max = p

where

R 22 + R12 R 22 − R12

 1  R22 + R12 p + p = 65 2  R22 − R12 

So

p  R22 + 4002  + 1 = 65 2  R22 − 4002  40  R22 + 4002 + R22 − 4002   = 65 2  R22 − 4002

(

2 R22 = 3.25 R22 − 4002

or

2 2

1.25 R = 3.25 × 400 R 22 =

)

2

3.25 × 400 2 = 41.6 × 10 4 1.25

R 2 = 645 mm t = R 2 − R1 = 645 − 400 = 245 mm

Thickness of cylinder, At the inner radius

σ c max = p ×

R 22 + R12 6452 + 4002 = 40 × R 22 − R12 6452 − 4002

= 40 ×

σa = p

57.60 = 90 MPa 25.60

R12 400 2 = 40 × = 25 MPa, Axial stress 2 R − R1 6452 − 4002 2 2

At the inner surface

σ c = 90MPa σ a = 25MPa p = −40MPa E = 200, 000 N /mm 2 , ν = 0.3 Strains

Volumetric strain,

MTPL0259_Chapter 06.indd 222

εc =

90 0.3 × 25 0.3 × 40 94.5 − + = E E E E

εa =

25 0.3 × 90 0.3 × 40 10 − + = E E E E

ε v = 2ε c + ε a =

2 × 94.5 10 + E E

5/23/2012 11:13:19 AM

Thick Shells

= Volume of cylinder,

V =

223

199 199 = E 200, 000

π π × D 2 × L = (0.8)2 × 6 4 4

= 3.0159 m 3 Change in volume,

δV = ε v × V =

199 × 3.0159 200, 000

= 3 × 10 −3 m3 = 3, 000 cc Problem 6.3 Strain gauge are mounted on the outer surface of a thick cylinder with a diameter ratio of 2.5. The cylinder is subjected to an internal pressure of 150 MPa. The recorded strains are (1) longitudinal strain = 60 × 10 −6 and (2) circumferential strains = 241 × 10 −6 Determine E and n of the material. Solution Internal radius = R1 Outer radius = R 2 D2 R 2 = = 2.5 D1 R 2 p = 150 MPa Outer surface

σc = p ×

2R12 2 = p× 2 2 R 2 − R1 6.25 − 1

= 150 ×

σa = p ×

2 = 57.14 N /mm 2 525

R12 = 28.57 N /mm 2 R − R12 2 2

Strains

or

MTPL0259_Chapter 06.indd 223

εc =

57.14 ν × 28.57 − = 241 × 10 −6 E E

εa =

28.57 ν × 57.14 − = 60 × 10 −6 E E 57.14 − ν × 28.57 = 241 × 10 −6 × E

(6.31)

28.57 − ν × 57.14 = 60 × 10 −6 E

(6.32)

5/23/2012 11:13:22 AM

224 Chapter 6

Dividing Eq. (6.31) by Eq. (6.32) 57.14 − ν × 28.57 241 = 28.57 − ν × 57.14 60 2 − ν 241 = 1 − 2ν 60 120 − 60ν = 241 − 482 ν 422 ν = 121 121 ν= = 0.286 422

or

Poisson’s ratio, Putting the value of n is Eq. (6.31)

57.14 − 0.286 × 28.57 = 241 × 10 −6 × E 48.969 = 241 × 10 −6 E 48.969 E= × 10 −6 241

Young’s modulus, = 203,000 N/mm2 = 203 kN/mm2 Problem 6.4 A thick cylinder is subjected to both internal and external pressures. The internal diameter of the cylinder is 120 mm and the external diameter is 200 mm. If the maximum permissible stress in the cylinder is 25 N/mm2 and the external radial pressure is 5N/mm2, determine the intensity of internal fluid pressure (Fig. 6.13).

5

p R1 5

Solution R1 = 60 mm

p

R 2 = 100 mm

R2

Radial stress

σ r = 5 N /mm 2 at

5 N/mm2

r = R2

Figure 6.13

σ r = p (say ) at r = R1 as shown in Fig. 6.13 Using Lame’s equation, B −A r2 B 5= −A 100 2 B p= 2 −A 60

σr =

MTPL0259_Chapter 06.indd 224

(6.33) (6.34)

5/23/2012 11:13:24 AM

Thick Shells

From Eqs. (6.33) and (6.34)

p −5 =

225

 100 2 − 60 2  B B − = B 60 2 100 2  60 2 × 100 2 

p − 5 = B × 1.777 × 10 −4 B = ( p − 5)5, 625 A=

(6.35)

B − 5 from Eq. (6.33) 100 2

= ( p − 5) × 0.5625 − 5

(6.36)

Circumferential stress B +A r2 at r = R1

σc =

σ c max

B ( p − 5) × 5, 625 +A = + ( p − 5) × 0.5625 − 5 2 60 602 Putting the value of scmax, 25 = ( p − 5) (2.125) − 5

σ c max =

p=

25 + 5 + 5 = 14.117 + 5 = 19.117 2.125

= 19.117 N /mm 2

Internal pressure

Problem 6.5 A compound cylinder is formed by shrinking one cylinder over another. The outer diameter of the compound cylinder is 240 mm, the inner diameter is 100 mm and the diameter at the common surface is 170 mm. The junction pressure due to shrinkage is 5 N/mm2. If the compound cylinder is now subjected to an internal fluid pressure of 60 N/mm2, determine the maximum hoop tension in the cylinder. How much heavier a single cylinder of an internal diameter of 100 mm would be if it is subjected to the same internal pressure in order to withstand the same maximum hoop tension? Solution R1 = 50 mm, R3 = 85 mm, R 2 = 120 mm p = 60 N /mm 2 , p ′ = 5 N /mm 2 , Inner cylinder Resultant at R1,

σ c = 60 ×

R 32 R 22 + R12 p − 2 × ′ R 22 − R12 R 32 − R12

120 2 + 50 2 852 − 2 × 5 × 120 2 − 50 2 852 − 502 10 × 7, 225 = 85.2 − 4, 725

= 60 ×

= 85.2 − 15.29 = 69.91 N/mm 2

MTPL0259_Chapter 06.indd 225

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226 Chapter 6

Outer cylinder Resultant at R3,

 60 R 2 + R32 R2 R2 60 × R12  σc =  2 × 2 1 2 2 + 2 + p ′ × 22 2 R 2 − R32  R3 R1 − R 2 R 2 − R1  =

60 50 2 × 120 2 60 × 502 1202 + 852 × + +5× 2 2 2 2 2 85 120 − 50 120 − 50 1202 − 852 60 0.210 21, 625 = × 3025.2 + 60 × +5× 7, 225 1 7,175

= 25.12 + 12.6 + 15.07 = 52.79 N /mm 2 At inner radius of inner cylinder, σ c is maximum. Single cylinder p = 60 N /mm 2 R1 = 50 mm R 2′ = ?

σ c max = 69.91 69.91 = 60 × 1.165 =

R2′ 2 + 502 R2′ 2 − 502

R2′ 2 + 502 R2′ 2 − 502

1.165 R2′ 2 − 1.165 × 502 = R22 + 502 0.165 R22 = 5412.5 R2′ 2 = 32803 R2′ = 181.11 mm

(

)

(

A′ = π R2′ 2 − 502 = π 181.112 − 502

Single cylinder,

)

= π × 30303.5 Compound cylinder,

A = π (120 2 − 50 2 ) = π (11, 900 )

weight of single cylinder A′ π (30303.5) = = weight of compound cylinder A π × 11, 900 = 2.546 (154.6 per cent heavier) Problem 6.6 A compound cylinder has a bore of 120 mm, an outer diameter of 240 mm and the diameter at common surface is 180 mm. Determine the radial pressure at the common surface which must be provided by shrinkage fitting if the resultant maximum hoop stress in the inner cylinder under the superimposed internal pressure of 60 N/mm2 is to be 60 per cent of the value if the maximum hoop tension at the inner radius of overall cylinder subjected to an internal pressure of 50 N/mm2 (Fig. 6.14).

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Thick Shells

227

p

R3 50

p R3

p

50

p

Figure 6.14 Solution Inner radius, R1 = 60 mm Junction radius, R 3 = 90 mm Outer radius, R 2 = 120 mm Internal pressure, p = 60 N /mm 2 Considering overall cylinder

σ c max at R1 = p ×

R22 + R12 R22 + R12

= 60 ×

120 2 + 60 2 = 100 N /mm 2 120 2 − 60 2

Compound cylinder Inner cylinder subjected to junction pressure p′ and internal pressure p, with σ c max equal to 100 × 0.6 = 60 N /mm 2 as given in the problem. This stress occurs at inner radius R1 (Fig. 6.14).

σ r = 50 at r = R1 = 60 mm σ r = p ′ at r = R 3 = 90 mm

Moreover,

MTPL0259_Chapter 06.indd 227

50 =

B −A R12

(6.37)

p′ =

B −A R 32

(6.38)

σc =

B +A r2

5/23/2012 11:13:33 AM

228 Chapter 6

at r = R1 , σ c = 60 N /mm 2 as given B 60 = 2 + A R1 From Eqs. (6.37) and (39)

(6.39)

2 A = 60 − 50 = 10 A = 5 N /mm 2 From Eq. (6.37) 50 =

B −5 60 2

B = 60 2 × 55 B 602 × 55 − A = − A = 24.44 − 5 = 19.44 N /mm 2 (junction pressure due to shrinkage which R32 902 must be provided so as to achieve the required condition). p′ =

Problem 6.7 A thick cylinder of an internal diameter of 100 mm is subjected to an internal pressure of 20 N/mm2. If the allowable stress for cylinder is 110 N/mm2, determine the wall thickness of the cylinder. The cylinder is now strengthened by wire winding so that it can be safely subjected to an internal pressure of 25 N/mm2. Find the radial pressure exerted by wire winding. Solution Radius R1 = 50 mm R 2 = to be calculated Internal pressure,

p = 20 N /mm 2

σ c max = p ×

Allowable maximum stress,

R 22 + R12 R 22 − R12

110 = 20 × or

(

R 22 + 502 R 22 − 502

)

5.5 R22 − 502 = R22 + 502 2 2

4.5 R = 6.5 × 50 2 R 2 = 1.2018 × 50 = 60.1 mm Internal pressure

σ c max =

25 × 110 , due to increased p. 20

= 137.5 N /mm 2

σ c′′, reduction in σ c′ max due to wire winding = 137.5 − 110 = 27.5 N /mm 2

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5/23/2012 11:13:36 AM

Thick Shells

= p′ × 27.5 = p ′ ×

229

2R 22 , where p′ is junction pressure R − R12 2 2

2 × 60.12 7, 244 = p′ × 2 2 60.1 − 50 3, 612 − 2, 500

= p ′ × 6.496 Pressure exerted by wire winding,

p ′ = 4.233 N /mm 2 .

Problem 6.8 A compound cylinder consists of a steel cylinder of an internal diameter of 120 mm and an external diameter of 180 mm and a bronze liner of a thickness of 10 mm with an external diameter of 120 mm as shown in Fig. 6.15. Assuming liner to be a thin cylinder and that there is no stress in the compound cylinder due to shrink fitting, determine the maximum hoop stress in the liner due to an internal pressure of 40 N/mm2. Ignore the longitudinal stress and strain. Solution For compound cylinder R1 = 50 mm R 2 = 90 mm p = 40 N /mm 2 B −A r2 B 40 = 2 − A R1

σr =

0= 40 =

B B − A, or A = 2 2 R2 R2

40 × R12 R 22 40 × 502 × 902 B B − , B = = R12 R 22 R 22 − R12 902 − 502 p

Steel cylinder

p

p Bronze liner

p

60

p

Liner

50

Figure 6.15

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230 Chapter 6

B = 144, 643 N A=

B 144, 643 = = 17.86 N /mm 2 R 22 90 2

σ r at r = 50 mm, σ r = 40 N/mm 2 σ r at r = 60 mm, σ r =

B 144, 643 − A= − 17.86 602 602

= 40.18 − 17.86 = 22.32 N /mm 2 Liner p = 40 N/mm 2 , at D1 = 100 mm p ′ = 22.32 N/mm 2 , at D ′ = 120 mm

σc = =

40 × 100 22.32 × 120 − , where t = 10 mm 2×5 2×5 40 × 100 22.32 × 120 − = 200 − 133.92 = 66.08 N /mm 2 2 × 10 2 × 10

= hoop stress in liner. Problem 6.9 A steel cylinder has an internal diameter of 100 mm and an external diameter of 200 mm. Another cylinder of same steel is to be shrunk over the first cylinder so that shrinkage stresses just produce a condition of yield at the inner surface of each cylinder. Determine the necessary difference in diameters of the cylinder at the mating surface before shrinking and the required external diameter of the outer cylinder, assuming that yielding occurs according to the maximum shear stress criteria’s and that no axial stresses are set up due to shrinking. Yield stress in simple tension or compression = 270 N/ mm2 and E = 200 GPa. Solution In a simple tensile test on a bar, at yield point stress σ yp, the maximum shear stress occurs at ±45° to the axis of the bar and it is equal to 0.5 syp. This can be obtained by drawing a Mohr’s stress circle. Maximum shear stress in simple tension or compression σ yp 270 = = = 135 N/mm 2 2 2 Say, junction pressure is p′ R1 = inner radius of liner cylinder = 50 mm R 3 = junction radius = 100 mm R 2 = outer radius of outer cylinder = ? Inner cylinder 2R2 σ c max occurs at inner radius due to p′ at outer radius = − p ′ 2 3 2 R3 − R1 = − p′ ×

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2 × 1002 = −2.667 p ′ 1002 − 502

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231

σ r = 0, at inner radius of inner cylinder. σ c max −2.667 p ′ = = −1.333 p ′ 2 2 1.333 p ′ = 135 τ max =

So,

p ′ = 101.25 N /mm 2 Outer cylinder

σ c max = p ′ ×

R 22 + R 32 (tensile) R 22 − R 32

= 101.25 ×

R 22 + 100 2 R 22 − 100 2

Junction pressure or σ r = p ′ (compression)

σ c max + p ′ 2  R 2 + 1002 1 = 101.25 × 22 + 101.25 = 135 2 2 R2 − 100 

τ max =

or

R 22 + 100 2 = 1.6667 (on simplification) R 22 − 100 2

or

R 22 + 100 2 = 1.667 R 22 − 1.667 × 100 2 0.667 R 22 = 2.667 × 100 2 R 2 = 200 mm

Outer diameter, Shrinkage allowance

D2 = 400 mm

δ R3 = =

R3 × p ′  R22 + R32 R12 + R32  + E  R22 − R32 R32 − R12  100 × 101.25  1002 + 502 2002 + 1002  + 200, 000  1002 − 502 2002 − 1002 

= 0.050625 (3.333) = 0.1688 mm Shrinkage allowance on diameter, δ D3 = 2δ R 3 = 0.337 mm Problem 6.10 A bronze sleeve is pressed onto a steel shaft of 60 mm in diameter. The radial pressure between steel shaft and bronze sleeve is 14 N/mm2 and the hoop stress at the inner surface of the sleeve is 50 N/mm2. If an axial compressive force of 60 kN is now applied to the shaft, determine the change in radial pressure. E = 200 GPa , ν = 0.3 For steel, For bronze,

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E = 100 GPa , ν = 0.33

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232 Chapter 6

Solution Bronze sleeve Junction pressure, p ′ = 14 N /mm 2 Hoop stress at inner surface = +50 N /mm 2 Diameter of shaft = 60 mm Radius, R1 = 30 mm When compressive force is applied onto the shaft, axial compressive stress will be developed in the shaft, which will cause positive radial strains (lateral strain in shaft) putting internal pressure p″ on sleeve. In reaction, sleeve will exert p″ radial pressure on the shaft. Additional junction pressure p″ p ′′ Additional hoop stress in sleeve, σ c′′ = × 50 14 = 3.5714 p ′′ (tensile) Additional circumferential strain in sleeve σ ′′ ν p ′′ ε cs′′ = c + Eb Eb = Shaft Axial compressive stress,

σ a′′ =

3.5714 p ′′ 0.33 p ′′ 3.9014 p ′′ + = Eb Eb Eb 60 × 1, 000 π (30 )2

= 21.22 N /mm 2 (compressive) Additional stresses in shaft are p″, radial compressive p″, circumferential compressive σ α′′, axial compressive Circumferential strain in shaft − p ′′ ν s p′′ ν sσ a′′ − p ′′ + 0.3 p ′′ 0.3 × 21.22 6.366 − 0.7 p ′′ ε cs′′ = + = + + Es Es Es Es Es Es ε cb′′ = ε cs′′ Strain compatibility, 3.9014 p ′′ 6.366 − 0.7 p ′′ = Eb Es Es (3.9014 p ′′ + 0.7 p ′′ ) = 6.366 Eb 2 (3.9014 p ′′ + 0.7 p ′′ ) = 6.366 p ′′ =

3.183 = 0.69 N /mm 2 4.6014

Problem 6.11 A steel rod of a diameter of 50 mm is forced into a bronze sleeve of an outside diameter of 80 mm, thereby producing a tension of 40 N/mm2 at the outer surface of sleeve. Determine (a) the radial pressure between the bronze sleeve and the steel rod and (b) the rise in temperature which would eliminate the force fit.

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For steel, For bronze,

233

E = 210 GPa , ν = 0.25, α = 11.2 × 10 −6 / °C E = 114 GPa , ν = 0.33, α = 18 × 10 −6 / °C

Solution Bronze sleeve R1 = 25 mm R 2 = 40 mm Tension on outer surface, σ c =

2 R12 × p ′, where p′ is junction pressure R − R12 2 2

40 = p ′ ×

2 × 252 1, 250 = p′ × 402 − 252 975

40 Radial pressure between bronze sleeve and rod p′ = = 31.20 N/mm 2 1 . 282 At junction

σ c′ = p ′ ×

R 22 + R12 R 22 − R12

= 31.2 ×

ε c′ =

Circumferential strain,

=

40 2 × 252 = 71.2 N /mm 2 (tensile) 40 2 − 252

σ c′ ν p ′ + Eb Eb 71.2 + 0.33 × 31.2 81.496 = Eb Eb

Shaft

σ c′′ = − p (compressive) both are compressive σ r′′ = − p′ Circumferential stress,

ε c′ = =

p ′ ν s p ′ p ′ × (1 − 0.25) − = Es Es Es 31.2 × 0.75 23.4 = Es Es

Total numerical sum of circumferential strains

ε c = ε c′ + ε c′′ =

81.946 23.4 + 114, 000 210, 000

= (71.488 + 11.143) × 10 −6 = 82.631 × 10 −6

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234 Chapter 6

=

δ R1 shrinkage allowance = R1 shaft radius

α b > α s (on heating bronze sleeve will expand more than steel shaft.) (α b − α s )∆T = ε c (18 − 11.2 ) × 10 −6 × ∆T = 82.631 × 10 −6 ∆T =

Rise in temperature,

82.631 × 10 −5 = 121.56°C 6.8 × 10 −6

Problem 6.12 A steel plug of a diameter of 50 mm is forced into a copper ring of an external diameter of 80 mm and a width of 30 mm. From a strain gauge fixed on the outer surface of the ring in the circumferential direction, the strain is found to be 64 × 10 −6. Considering that the coefficient of friction between the mating surfaces is 0.25, determine the axial force required to push the plug out of the ring. For bronze, E = 100 × 103 N /mm 2 For steel, E = 200 × 103 N /mm 2 Solution Bronze ring R1 = 25 mm R 2 = 40 mm

ε c , = 64 × 10 −6

At outer surface,

σ c = 64 × 10 −6 × E B

At outer surface,

= 64 × 10 −6 × 100, 000 = 6.4 N /mm 2 Say p′ = junction pressure

σ c at R 2 = p ′ ×

2R12 2 × 252 = p′ × 2 2 R − R1 40 − 252 2 2

1, 250 975 6.4 × 975 p′ = = 4.99 N /mm 2 1, 250

6.4 = p ′ ×

Normal force on shaft, N = 2π R1 × B × p ′, where B is the width of ring N = 2π × 25 × 30 × 4.99 = 23, 515 N Tangential force, F = µ N = 0.25 × 23515 = 5878.7 N (along the length of shaft) Axial force required to push the plug out of the ring = 5878.7 N

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Key Points to Remember  In a thick shell, the wall thickness is significant and stresses vary along the thickness of the cylinder and cannot be assumed to be uniform.  In a thick cylindrical shell subjected to internal pressure, Lame’s equations can be used to determine radial and hoop stresses.

σr =

B − A (compressive) r

σc =

B + A (tensile) r2

Constants A and B are determined by using boundary conditions.  In a thick cylindrical shell with inner radius (R1) and outer radius (R2) subjected to internal pressure p

σ c max = p

At inner radius,

R22 + R12 R22 − R12

pR12 (constant tensile) R − R12  In a thick cylindrical shell with inner radius (R1) and outer radius (R2) subjected to external pressure p

σa =

Axial stress,

2 2

σ c max = − p

At inner radius,

2 R22 R − R12 2 2

R22 (compressive) R − R12  In a compound cylinder, the outer cylinder is shrink fitted over the inner cylinder. The junction pressure is developed which causes tensile hoop stress in the outer cylinder but compressive hoop stress in the inner cylinder. Constant axial stress,

σa = p

2 2

R1 = inner radius, R2 = outer radius and R3 = junction radius. If both cylinders are made of same material, that is, same E 2 2 2 2 δ R3 = shrinkage allowance = p ′ × R3  R2 + R3 + R3 + R1  2 2 2  E  R2 − R3 R3 − R12 

 In a hub and shaft assembly, junction pressure introduces tensile hoop stress in the hub but compressive hoop stress in the shaft. In shaft, σ c = − p ′, σ r = p ′ (junction pressure)  In a thick spherical shell subjected to internal pressure, Hoop stress, σ c =

B + A (tensile) r3

2B − A (compressive) r3 With the help of boundary conditions, the values of Lame’s constants can be determined. Radialstress, σ r =

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236 Chapter 6

Review Questions 1. In a thick cylinder subjected to internal pressure, explain the assumption that plane sections remain plane after the application of internal pressure. 2. Show the variation of σ c and σ r in a thick cylinder along its thickness due to external pressure p. 3. In a compound cylinder, in which one cylinder is shrink fitted over another cylinder, show the variation of hoop stresses in outer and inner cylinders along the wall thickness. 4. Derive expressions of shrinkage allowance in a hub and shaft assembly. 5. Explain the purpose of compounding two cylinders. 6. In an assembly of bronze sleeve and steel shaft, if the temperature of the assembly is raised, at a particular temperature, the junction pressure becomes zero, why?

Multiple Choice Questions 1. In a thick cylindrical shell with R2 = 2 R1 subjected to external pressure of 45 N/mm2, what is the maximum hoop stress developed in the cylinder? (a) 120 N/mm2

(b) 75 N/mm2

2

(c) 60 N/mm (d) None of these 2. In a compound cylinder at junction, the sum of circumferential strains in outer and inner cylinders is 120 × 10−6 . If the junction’s diameter is 200 mm, what is the shrinkage allowance? (a) 0.048 mm

(b) 0.024 mm

(c) 0.0024 mm (d) None of these 3. In a thick cylindrical shell, σ c max = 1.25 p , what is the ratio of R2/R1? (a) 1.5

(b) 2.0

(c) 3.0 (d) None of these 4. In a shaft and hub assembly with shaft diameter 50 mm, hub diameter 100 mm and junction pressure 30 N/mm2, what is the hoop stress in the shaft? (a) +50 N/mm2

(b) +30 N/mm2

2

(c) −30 N/mm (d) None of these 5. In a thick cylinder, R2/R1 = 2, the internal pressure is 60 N/mm2. What is the maximum shear stress at inner radius? (a) 20 N/mm2 2

(b) 30 N/mm2

(c) 80 N/mm (d) None of these 6. In a thick cylindrical shell, σ c max = 120, p = 50 N/mm2, what is the value of Lame’s constant A?

MTPL0259_Chapter 06.indd 236

(a) 30 MPa

(b) 35 MPa

(c) 70 MPa (d) None of these 7. In a compound cylinder, the hoop stresses developed at the junction in outer and inner cylinders are +84 MPa and −66 MPa. If E = 200 GPa, the junction radius is 100 mm, what is the shrinkage allowance as diameter? (a) 0.3 mm

(b) 0.15 mm

(c) 168 mm (d) None of these 8. The variation of hoop stress across the thickness of a thick cylinder is (a) Linear

(b) Uniform

(c) Parabolic (d) None of these 9. Purpose of compounding cylinder is (a) To increase pressure-bearing capacity of a single cylinder (b) To reduce the variation in hoop stress distribution (c) To increase the strength of the cylinder (d) All the above 10. A bronze sleeve of an outer diameter of 100 mm is forced over a solid steel shaft of a diameter of 60 mm. If the junction pressure is 32N/mm2, the hoop strain at outer radius of sleeve is given by (if E = 100 GPa) (a) 320 mstrain

(b) 360 mstrain

(c) 720 mstrain

(d) None of these

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237

Practice Problems 1. Two thick cylinders A and B are of the same dimensions. The external diameter is double the internal diameter. A is subjected to an internal pressure p1, while B is subjected to an external pressure p2. Find the ratio of p1 / p2 if the greatest hoop strain developed in both cylinders is the same. The Poisson’s ratio of the material is 0.3. 2. A steel cylinder of an inside diameter of 1 m and a length of 4 m is subjected to an internal pressure of 60 MPa. Determine the thickness of the cylinder if the maximum shear stress in the cylinder is not to exceed 80 MPa. What is the change in the volume of the cylinder if E = 210 GPa, and n = 0.3? 3. A pressure vessel of an internal diameter of 200 mm, an external diameter of 250 mm and a length of 1 m is tested under a hydraulic pressure of 20 N/mm2. Determine the change in its internal and external diameters if E = 208 kN/mm 2 and ν = 0.3. If a strain gauge is mounted on the surface of the pressure vessel in an axial direction, what will be the strain reading? 4. A thick cylinder is subjected to both internal and external pressures. The internal diameter of the cylinder is 150 mm and the external diameter is 200 mm. If the maximum permissible stress in the cylinder is 20 N/mm2 and the external radial pressure is 4 N/mm2, determine the intensity of internal radial pressure. 5. A compound cylinder is formed by shrinking one cylinder over another. The outer diameter of the compound cylinder is 200 mm, the inner diameter is 120 mm and the diameter at the common surface is 160 mm. The junction pressure developed due to shrinkage is 4 N/mm2. The compound cylinder is now subjected to an internal pressure of 50 N/mm2, determine the maximum hoop tension in the cylinder. How much heavier a single cylinder of an internal diameter of 120 mm would be if it is subjected to the same internal pressure in order to withstand the same maximum hoop stress. 6. A compound cylinder has a bore of 120 mm and an outer diameter of 200 mm and the diameter at the common surface is 160 mm. Determine the radial pressure at the common surface which must be provided by the shrinkage fitting, if the resultant maximum hoop stress in the inner cylinder under a superimposed internal pressure of 50 N/mm2 is to be half the value of the maximum hoop tension in the inner cylinder if this cylinder alone is subjected to an internal pressure of 50 N/mm2.  2 p ′R32  R32 + R12 1 R22 + R12 ; σ = 50 × −  Hint: σ c max = 50 × 2  c max R3 − R12 2 R22 − R12 R32 − R12   7. A thick cylinder of an internal diameter of 160 mm is subjected to an internal pressure of 5 N/mm2. If the allowable stress for cylinder is 25 N/mm2, determine the wall thickness of the cylinder. The cylinder is now strengthened by wire winding so that it can be safely subjected to an internal pressure of 8 N/mm2. Find the radial pressure exerted by wire winding. 8. A compound cylinder consists of a steel cylinder of an internal diameter of 180 mm and an external diameter of 250 mm and bronze liner of an external diameter of 180 mm and an internal diameter of 170 mm as shown in Figure 6.16. Assuming the liner to be thin cylinder and that there is no shrinkage stress in compound cylinder, determine the hoop stress developed in the liner due to an internal pressure of 50 N/mm2. Ignore the longitudinal stress and strain. For steel, E = 208 kN/mm2 , ν = 0.28 For bronze, E = 112 kN/mm2 , ν = 0.30 9. A steel tube has an internal diameter of 25 mm and an external diameter of 50 mm. Another tube of the same steel is to be shrunk over the first tube so that shrinkage stresses just produce a condition of yield at the inner surface of each tube. Determine the necessary difference in the diameter of the tubes at the mating surface before shrinking and the required external diameter of the outer tube. Assuming that yielding occurs according to the maximum shear stress condition and that no axial stresses are set up due to shrinkage. The yield stress in simple tension or compression is 414 N/mm2, E = 207 GPa.

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238 Chapter 6 Steel cylinder

Bronze liner

Figure 6.16

Compound cylinder

10. A steel sleeve is pressed onto a steel shaft of a diameter of 50 mm. The radial pressure between the steel shaft and the steel sleeve is 20 N/mm2 and the hoop stress at the inner surface of sleeve is 56 N/mm2. If a compressive force of 50 kN is now applied to the shaft, determine the change in radial pressure. E = 210 GPa, ν = 0.3. 11. A steel rod of a diameter of 60 mm is forced into an aluminium sleeve of an outside diameter of 80 mm, thereby producing a tension of 54 N/mm2 at the outer surface of sleeve. Determine (a) the radial pressure between the bronze sleeve and the steel rod and (b) the rise in temperature which would eliminate the force fit. For steel, E = 200 GPa , ν = 0.3, α = 11.2 × 10−6 / °C For aluminum, E = 68 GPa , ν = 0.33, α = 22 × 10−6 / °C 12. A steel plug of a diameter of 80 mm is forced into a steel ring of an external diameter of 120 mm and a width of 50 mm. From a strain gauge fixed on the outer surface of the ring in the circumferential direction, the strain is found to be 41 × 10 −6. Considering that the coefficient of friction between the mating surface is 0.2, determine the axial force required to push the plug out of the ring. E = 210 GPa. [Hint: Normal force, N = p ′ × 20 × 40 × 50, F = µ N]

Special Problems 1. A compound cylinder is made by shrinking a cylinder of an outer diameter of 200 mm over another cylinder of an inner diameter of 100 mm. If the numerical value of maximum hoop stress developed due to shrinkage fitting in both the cylinders is the same, find the junction diameter.  2 R32 R2 + R32  = p′ × 22  Hint: p′ × 2  2 R2 − R1 R2 − R32   2. A thick cylinder of internal diameter D and wall thickness t is subjected to an internal pressure of p. If the maximum hoop stress developed in the cylinder is 2.5p, determine the ratio of t/D. 3. A steel shaft of a diameter of 120 mm is force fitted in a steel hub of an external diameter of 200 mm, so that the radial pressure developed at the common surface is 12 MPa. If E = 200 GPa, determine the force fit allowance on the diameter. What is the maximum hoop stress developed in the hub?

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239

  2 R22    Hint: δ R1 = p ′R1  2   R2 − R12    4. A steel ring of internal radius r and external radius R is shrunk onto a solid steel shaft of radius r + dr. Prove  r 2  dr that the intensity of pressure at the mating surface is equal to 1 − 2  E , where E is the modulus of elasR  2r  ticity of steel. [Hint: use shrinkage formula for shaft + hub] 5. Two thick cylinders A and B are of same dimensions. The external diameter is double the internal diameter. Cylinder A is subjected to an internal pressure p1, while cylinder B is subjected to an external pressure p2. Find the ratio of pressure p1 to p2 if the greatest circumferential stress developed in both the cylinders is the same. 6. A steel cylinder of an internal diameter of 100 mm and an external diameter of 150 mm is strengthened by shrinking another cylinder onto it, the internal diameter of which before heating is 149.92 mm. Determine the outer diameter of the outer cylinder at the junction if the pressure at the junction after shrinking is 20 N/mm2. E = 210 GPa. [Hint: use expressions of shrinkage allowance are taking R1 = 50 mm and R3 = 75 mm] 7. A compound cylinder is made by shrinking a tube of an outer diameter of 150 mm over another tube of an inner diameter of 100 mm. Find the common diameter if the greatest hoop stress in the inner tube is numerically 0.7 times that of outer tube.   2 R32 R 2 + R32 = 0.7 × 32 × p′   Hint: p ′ × 2 2 2 R R R R − − 2 3 3 1   8. A thick cylinder of an internal diameter of 120 mm and an external diameter of 180 mm is used for a working pressure of 15 N/mm2. Because of external corrosion, the outer diameter of the cylinder is machined to 178 mm. Determine by how much the internal pressure is to be reduced so that the maximum hoop stress in the cylinder remains the same as before machining.  R22 + R12 R2′ 2 + R12  = × p ′  Hint: R2 = 90 mm, R ′2 = 89 mm 15 × 2  R2 − R12 R2′ 2 − R12  

Answers to Exercises Ecercise 6.1: 109.7, 56.9 ms

Exercise 6.5: (a ) 55.576 N/mm 2 , (b) 144.5 N/mm 2

Exercise 6.2: D = 169.7 mm

Exercise 6.6: 7.46 MPa

Exercise 6.3: p = 102.78 N/mm

2

Exercise 6.4: 20.485, −103.70, −83.22, +103.45, + 82.96 MPa.

Answers to Multiple Choice Questions 1. 2. 3. 4.

(a) (b) (c) (c)

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5. 6. 7. 8.

(c) (b) (b) (c)

9. (d) 10. (b)

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240 Chapter 6

Answers to Practice Problems 1.

p1 = 1.214 p2

2. t = 0.5 m, δV = 5, 206 cc 3. 0.1707 mm, 0.0727 mm, 68.375 µ strain = 68.375 × 10 −6 µ strain 4. 10.72 N/mm 2 2 5. σ c max = 100.71 N/mm at inner radius, single cylinder is 41.5 per cent heavier than the compound cylinder

6. p′ = 3.71 N/mm2 due to shrinkage 7. p′ = 2.5 N/mm 2 8. 130.74 N/mm2 9. R2 = 50 mm, D2 = 100 mm, δ D 3 = 0.125 mm 10. 2.01 N/mm2 11. (a ) 21 N/mm 2 , (b) Rise in temperature 118.4°C 12. 13.52 kN

Answers to Special Problems 1. 2. 3. 5.

R3 = 75.12 mm, D3 = 150.24 mm 0.2635 0.0225 mm, 25.5 MPa p1 = 1.6p2

MTPL0259_Chapter 06.indd 240

6. 212 mm 7. 129.92 mm 8. 0.373 N/mm2

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7 Shear Force and Bending Moment Diagrams CHAPTER OBJECTIVES Any structure is composed of beam and column members. Structure is subjected to external loads such as dead loads, wind loads and hydraulic loads. Axial load in any member produces tensile or compressive stresses and if load is compressive and structural member is long, it causes buckling in member resulting in the collapse of a member or structure as a whole. Transverse load (perpendicular to the axes of the member) produces bending moment and shear force in the sections of the member and at a critical section beam/cantilever member may fail due to excessive bending moment. Basic objective of the chapter is to make the readers learn as to how to calculate bending moment and shear force at any section of a beam and then to locate the section where bending moment is maximum and to design the structural member efficiently taking a factor of safety. Various cases of loads such as uniformly distributed, point or variable loads will be discussed.

Introduction Transverse loads on a member of a structure cause bending in the member, resulting in the development of bending moment and shear force in the section of the member. To locate the critical section, the distribution of bending moment and shear force along the length of the member (called beam or cantilever) are drawn. To draw these SF and BM diagrams, students should be conversant with the positive and negative bending moments and shear forces. We will consider the application of point loads, uniformly distributed load, variable load or a moment so as to develop the concepts of bending moment and shear force in a beam. Beam with hinged end, roller supported end, a simply supported end and a fixed end will be considered. Sign convention for bending moment and shear force will be taken. Procedure for the calculation of BM and SF will be discussed. In many beams, the section of maximum bending moment, the section of maximum shear force or a section acting as point of contraflexure are important for the safe design of the beam/cantilever member.

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242 Chapter 7

Different cases of loads and end conditions will be discussed to make the understanding of SF and BM diagrams easy.

Different Types of Beams A beam is a structural member, on which loads perpendicular to its axis are applied. These loads bend the beam and shear force and bending moment are induced in the section of the beam. There are various types of beams such as (a) simply supported beam, (b) cantilever, (c) continuous beam, (d) fixed beams and (e) beams, hinged at one end and roller supported at the other end. We will not discuss continuous beams and fixed beams in this chapter. Figure 7.1(a) shows a simply supported beam. The beam is simply supported on supports A and B. When transverse loads are applied on beam, it bends and lateral (axial) movement of ends is permitted. Figure 7.1(b) shows a cantilever; one end of the cantilever is free but the other end is fixed in wall. When a load is applied on a cantilever, the wall applies fixing moment and a reaction to maintain slope and defection at fixed end to be zero. Figure 7.1(c) shows a beam with one end hinged and the other end roller supported. Under the action of transverse loads, when beam bends, the axial movement of beam at roller support is permitted. Figure 7.1(d) shows a beam with an overhang on one side; L1 is the span length between hinged and roller support ends and L2 is overhang (as cantilever length).

Fixed end

Wall A A

L (a) Simply supported

B

B L (b) Cantilever

L1

Free end

A

B

Hinged end

Roller

L (c) Hinged and roller supported beam

L2

(d) Beam with overhang

Figure 7.1 Different types of beams

Shear Force (Positive and Negative) In the Chapter 1, we have studied that a shear force acts parallel to the plane of a body and it changes the shape of the body; for example, a rectangular element is changed to a parallelogram element. We have also learnt that a shear force, which tends to rotate the body in a clockwise direction, is a positive

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243

shear force. A shear force which tends to rotate the body in an anticlockwise direction is a negative shear force. Let us take a beam AB of length L and depth D subjected to a load W. The beam is supported at A and B. Support reactions are RA and RB respectively. Reactions RA + RB = W (load applied); for equilibrium under the transverse load W, the beam bends and takes the shape as shown in Fig. 7.2. Take a small element abcd in portion AC of beam. The force on the left plane ac is RA↑ and the force on the right plane bd is (W − RB)↓ or RA↓. The shape of the element changes to a′b′c′d′ as shown in the figure. These two forces RA↑ and RA↓ try to rotate the body in the clockwise direction as shown in the figure or these are positive shear forces or

a

b

W e

f

C

A c

d

c

d

RA

h

g g

h

e

f

L a

RA d

c

RB

D

RB h

g

+

+

(+ ve) SF

(− ve) SF

Figure 7.2

Depth

RB

b

RA

B

Simply supported beam with a point load

(i) Upward force on the left side of an element is a positive shear force. (ii) Downward force on the right side of an element is a positive shear force. Similarly, consider an element efgh in portion CB of the beam. The shape of the element changes to e ′f ′g ′h ′. On the left side of the element, net force is W − RA = RB↓ and on the right side of the element, net force is RB↑. These two forces RB↓ on the left side and RB↑ on the right side try to rotate the element in anticlockwise direction. This type of shear force is a negative shear force or (iii) Downward force on the left side of the element is a negative shear force. (iv) Upward force on the right side of the element is a negative shear force. Reader is advised to take shear force only on one side, that is, left side and remember that: (i) Upward force on the left side is positive. (ii) Downward force on the left side is negative. In this chapter, we will draw SF diagrams. Generally, the beam is represented by its axes only and depth is not shown.

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244 Chapter 7

Example 7.1 An 8-m-long beam AB is hinged at end A and roller supported at end B carries transverse loads of 4 kN at point C (2 m from A) and 6 kN at point D (5 m from A) as shown in Fig. 7.3. Determine support reactions and shear force at sections E, F and G of beam as shown in the figure. 4 kN E

A RA = 5.25 kN

6 kN F

C 2m

1

G D

3m 3m

B 3m

3m

RB = 4.75 kN

1m

Figure 7.3 Solution Taking moment of forces about end A 4 × 2(cw ) + 6 × 5(cw ) − RB × 8(ccw ) = 0 Reaction,

Reaction,

38 = 4.75 kN 8 RA + RB = 4 + 6 = 10 kN RB =

RA = 10 − 4.75 = 5.25 kN

Considering forces only on the left side of the section

+1.25 kN +5.25 kN

D A

C

2m

B −4.75 kN

3m 3m SF diagram

Point E Shear force,

FE = +RA = +5.25 kN

Point F Shear force,

FF = + RA − 4 = +5.25 − 4 = +1.25 kN

Point G Shear force,

FG = + RA − 4 − 6 = 5.25 − 4 − 6 = −4.75 kN

Figure 7.4

Note that shear force +5.25 kN remains constant between A and C. Shear force +1.25 kN remains constant between C and D. Shear force −4.75 kN remains constant between D and B. Figure 7.4 shows the shear force diagram of the beam. It is the general convention to hatch the boundary of shear force diagram as shown in the figure. Exercise 7.1 A 6-m-long beam AB is hinged at end A and roller supported at end B carries loads of 5 and 4 kN at parts C and D as shown in Fig. 7.5. Determine support reactions. Determine shear forces at E and F. Draw SF diagram.

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Shear Force and Bending Moment Diagrams 5 kN

245

4 kN

E

F C 1m

A

B

D 1m

2m

3m

1m

Figure 7.5 Exercise 7.1 Example 7.2 A 6-m-long beam AB carries uniformly distributed load of 6 kN/m over CD (2 m) and a concentrated load of 12 kN at E, as shown in Fig. 7.6. Calculate support reactions and determine shear force at points C, D and E. 12 kN 6 kN/m C

A

D

1m

2m

E

2m

B

1m

RA = 10 kN

RB = 14 kN

Figure 7.6 Solution udl (uniformly distributed load) on beam = 6 × 2 = 12 kN CG of the load acts a distance of 1 m from C or 2 m from A. Taking moments of the forces about end A 12 × 2 (cw ) + 12 × 5 (cw ) − 6 RB (ccw ) = 0 24 + 60 = 6 RB Reaction, Total load on beam

84 = 14 kN 6 = 6 × 2 + 12 = 24 kN

RB =

RA + RB = 24 kN Reaction,

RA = 24 − 14 = 10 kN

Shear force At section C, FC = +10 kN (upward force on the left side of section is positive). At point D, FD = +10 − 2 × 6 = −2 kN (downward force on the left side of the section is negative).

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246 Chapter 7

At point E, FE′ = +10 − 12 = −2 kN ( just on the left side of section E). FE′′ = +10 − 12 − 12 = −14 kN ( just on the right side of section E). At point E, shear force varies from −2 to −14 kN. Exercise 7.2 A 7-m-long beam AB is hinged at A and roller supported at D carries udl of 10 kN/m over AC = 3 m and a point load of 5 kN at D as shown in Fig. 7.7. Determine support reactions and calculate shear force at E, AE = 1.5 m, C and D.

5 kN 10 kN/m A

E

1.5

C

3m

B

D 2m

2m

Figure 7.7 Exercise 7.2

Bending Moments (Positive and Negative) There are two types of bending moments, that is, (1) Bending moment which produces concavity upward as shown in Fig. 7.8(a) is taken as positive bending moment. (2) Bending moment which produces convexity upwards as shown in Fig. 7.8(b) is taken as negative bending moment.

W A

W B A B

Left side

D

A

C A

W

B

C

D

RC

RB

Concavity upwards positive BM

Left side Convexity upwards negative BM

(a)

(b)

Figure 7.8

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Shear Force and Bending Moment Diagrams

247

Figure 7.8(a) shows a simply supported beam AB carrying a load W. Due to this load, the beam bends and takes the shape AC′B, showing concavity upwards. Figure 7.8(b) shows a simply supported beam ABCD, supported at B and C and carrying loads W each at both ends A and D, producing convexity upwards. From the figure, it is amply clear that: (i) Clockwise moment on the left side of the section produces positive bending moment. (ii) Counterclockwise moment on the right side of the section produces positive bending moment. Similarly, (iii) Counterclockwise moment on the left side of the section produces negative bending moment. (iv) Clockwise moment on the right side of the section is negative bending moment. Example 7.3 A 6-m-long beam ABCD carries concentrated loads of 6 and 3 kN at B and C as shown in Fig. 7.9. Calculate reactions and determine bending moment at sections B and C of the beam. 6 kN A

3 kN

B

2m

C

2m

D

2m

RD = 4 kN

RA = 5 kN

Figure 7.9 Solution Bending moment at ends A and D is zero, until and unless a moment is specifically applied at these simply supported, hinged or roller supported ends. Taking moments about end A 6 × 2(cw ) + 3 × 4(cw ) − RD × 6(ccw ) = 0 Reaction,

RD =

24 = 4 kN 6

Total load on beam = 6 + 3 = 9 kN Reaction,

RA = 9 − RD = 9 − 4 = 5 kN

Bending moment at B At section B of the beam, on the left side, there is only one force RA = 5 kN ↑. On the left side,

M = 5 × 2 = +10 kNm(cw).

On the right side of the section, there are two forces 3 kN ↓ at C and 4 kN ↑ at D. On the right side,

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M B = +4 × 4 − 3 × 2 = +10 kN(ccw ).

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248 Chapter 7

Note that clockwise moment on the left side of a section is positive and counterclockwise moment on the right side of a section is positive. Therefore, M B = bending moment at B = +10 kN m (producing concavity upwards). Now, we will consider only one side of the section, that is, only the left side to find out bending moment. Section C On the left side, there are two forces RA = 5 kN ↑ and 6 kN ↓. Bending moment at C Mc = +5 × 4(cw) − 6 × 2(ccw) = 20 − 12 = +8 kN m (producing concavity upwards). Finally, we can write bending moments as follows: MA = 0 M B = +10 kN m MC = +8 kN m MD = 0

Exercise 7.3 A 4-m-long beam AB is hinged at end A and roller supported at end B. A load of 10 kN is applied on the beam as shown in Fig 7.10. Determine support reactions. Find bending moments at D, C and E sections of the beam. Example 7.4 A 6-m-long beam AB is hinged at end A roller supported at end B carries udl of 6 kN/m over AC = 3 m and a point load of 12 kN at D as shown in Fig. 7.11. Determine support reactions and bending moments at C, D. Solution Note that beam is supported at ends carries transverse loads as shown in Fig. 7.11. The beam will obviously bend producing concavity upwards. Therefore, bending moment at any section of the beam is positive. Support reactions Total udl = 6 × 3 = 18 kN CG of udl lies at a distance of 1.5 m from A.

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10 kN

A

D 1m

C 1m

1m

Figure 7.10

B

E 1m

Exercise 7.3 12 kN

6 kN/m C

A

3m

D

2m

B

1m

RA = 15.5 kN

RB = 14.5 kN

Figure 7.11

Example 7.4

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Shear Force and Bending Moment Diagrams

249

Taking moments of forces about end A 6 × 3 × (1.5) + 12 × ( 2 + 3) − 6 RB = 0 27 + 60 = 6RB Reaction, RB = 14.5 kN Total vertical load on horizontal beam = 3 × 6 + 12 = 30 kN RB + RA = 30

Reactions,

RA = 30 − RB = 30 − 14.5 = 15.5 kN

Reaction,

Bending moment at C Taking forces only on the left side of the section C. Forces are FA = 15.5 kN ↑ and 18 kN ↓ at distance of 1.5 m for C (Note that C.G. of the udl lies at its centre.) Mc = 15.5 × 3 − 18 × 1.5 = 46.5 − 27 = 19.5 kN m For MD, the forces on the left side of the section D, RA = 15.5 kN↑ at a distance of 5 m from D and udl = 18 kN at a distance of 5 − 1.5 = 3.5 m from D. Bending moment, M D = +15.5 × 5 − 18 × 3.5 = 77.5 − 63 = +14.5 kN m Exercise 7.4 A 5-m-long beam is hinged at end A and roller supported at end B carries udl of 5 kN/m over CD = 2 m and a point load of 10 kN at D as shown in Fig. 7.12. Calculate support reactions and determine bending moments at C and D. [Hint: Beam is supported at ends, loads only produce bending showing concavity upwards and CG of udl lies at the centre of CD.] 10 kN 5 kN/m A

C 1m

B

D 2m

Figure 7.12

2m

Exercise 7.4

SF Diagrams of Beams/Cantilevers Carrying Point Loads Till now, we have studied about SF and BM at various sections of a beam. Now, we will consider only point loads on cantilevers/beams with or without overhang. Example 7.5 A 5-m-long cantilever AB free at end A and fixed at end B carries vertical loads of 6 kN at end A and 4 kN at end C as shown in Fig. 7.13. Draw shear force diagram for the ↑ cantilever.

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250 Chapter 7 x

6 kN

4 kN x

A

Y

Y C Y

Y

2m

B

3m (a)

RB = 10 kN

−6 kN (−)

−10 kN

(b)

Figure 7.13 SF diagram Solution There are two portions AC and CB of the cantilever. Consider portion AC, x = 0 to x = 2 m Take a section YY at a distance of x from A Shear force, Fx = −6 kN ( x = 0 to x = 2 m) FAC = −6 kN (note that downward force on the left side of the section is negative) Shear force remains constant from A to C Portion CB Take a section YY at a distance of x from A in portion CB, x = 2 to 5 m Shear force Fx = −4 − 6 = −10 kN (two forces on the left side of the portion) FCB = −10 kN, shear force remains constant from C to B Figure 7.13(b) shows the SF diagram of the cantilever hatching along boundary of SF diagram is done as a convention. Example 7.6 A 6-m-long beam AB is hinged at end A and roller supported at end D, which is 1 m from B. It carries loads of 10 kN at point C and 5 kN at end B as shown in Fig. 7.14(a). Draw the SF diagram of the beam. Solution Support reactions Taking moments about A 10 × 3(cw ) + 5 × 6(cw ) − RD × 5(ccw ) = 0 60 Reaction, RD = = 12 kN 5 Total load on beam = 10 + 5 = 15 kN RA + RD = 15 kN Reaction, RA = 15 − 12 = 3 kN

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Shear Force and Bending Moment Diagrams

251

x x x A

Y

10 kN

Y

Y

C

Y

3m

2m

RA = 3 kN

C

−7 kN

Y

B

1m

+5

+5

A

D

5 kN

RD = 12 kN

(a)

+3 kN

Y

D

B

SF diagram (b)

Figure 7.14

There are three portions AC, CD and DB in the beam. Let us consider these portions one by one to draw SF diagram. Taking forces only on the left side of the section, upward force is positive consider section YY in portion AC. Shear force,

Fx = + RA = + 3 kN = FAC (remains constant from A to C)

Portion CD Shear force,

Fx = + RA − 10 = +3 − 10 = −7 kN (remains constant from C to D)

Portion DB Shear face

Fx = RA − 10 + RD = 3 − 10 + 12 = +5 kN (remains constant from D to B)

Figure 7.14(b) shows SF diagram of beam. Exercise 7.5 A 7-m-long cantilever is free at end A and fixed at end B carries three loads as shown in Fig. 7.15. Determine support reaction and draw SF diagram of the cantilever. Exercise 7.6 A 6-m-long beam AB with overhang DB = 2 m is hinged at end A and roller supported at end D, as shown in Fig. 7.16. Determine support reactions and draw SF diagram of the beam.

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252 Chapter 7

6 kN

4 kN

2 kN

C

D

8 kN

2m

2m

Figure 7.15

D

A

B

A

B

C

3m

2m

Exercise 7.5

2 kN

Figure 7.16

2m

2m

Exercise 7.6

SF Diagrams of Cantilevers and Beams with UDL We have studied about SF diagrams of cantilever/beams subjected to point loads, where SF diagram consists rectangles or portions of constant shear force. Let us now consider cantilevers/beam with uniformly distributed loads. In the portion of the beam which conY x tains udl, shear force no longer remains constant but varw ies along the length of the beam. Consider a cantilever with udl as shown in Fig. 7.17 cantilever is fixed at end B. B (where w is rate of loading and wx is total load on the A Y L left side of section YY ) RB Total load on cantilever = wL↓ Reaction at end B (a) RB = wL↑ D Consider a section YY at a distance of x from end A (−) Shear force, Fx = − wx −wL At any distance x from face End A shear force, Fx = −wx (b) dF /dx = − w (minus the rate of loading) SF diagram = Slope of the line of SF diagram Fx = 0

at x = 0

L = −w 2

Figure 7.17

L at x = + 2

= −wL at x = L as shown in Fig. 7.17(b). Now consider a simply supported beam AB of length L, carrying uniformly distributed load of intensity w per unit length throughout its length as shown in Fig. 7.18. Total load on beam = wL. For reactions, taking moments about end A  L wL   (cw) − RB L(ccw) = 0  2 (note that CG of the load lies at a distance of L/2 from A) Reaction,

MTPL0259_Chapter 07.indd 252

RB =

wL ↑ (Fig 7.18) 2

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Shear Force and Bending Moment Diagrams

wL wL = ↑ 2 2 Consider a section YY of the beam at a distance x from end A. wL − wx (note that upward Shear force Fx = 2 force on the left side of the section is positive).

Y

RA = wL −

Reaction,

wL − wx 2 wL = at x = 0 2 =

=

wL 4

at x =

= 0 at x = =

x

w B

A wL 2

Y

RA L

RB = wL 2

(a) +wL 2

−wL 2

L 4

L 2

L 2

L 2 SF diagram

− wL 3L at x = 4 4

(b)

Figure 7.18

− wL = at x = L 2

Figure 7.18(b) shows the SF diagram of the beam changing from + Fx =

Now,

253

wL − wx 2

wL wL at A to − at B. 2 2

dFx = − w (minus the rate of loading). dx Example 7.7 A 10-m-long cantilever AB carries udl of w = 2 kN/m over AC = 3 m and a point load of 6 kN at D as shown is Fig. 7.19. Draw SF diagram of cantilever. Solution Total load on cantilever = 3 × 4 + 6 = 18 kN ↓

Reaction,

RB = 18 kN ↑

There are three portions AC, CD and DB in cantilever. Portion AC (x = 0 to 4 m) Shear force

Fx = − wx = −3 x = 0 at x = 0 = −6 kN at x = 2 m = −12 kN at x = 4 m

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254 Chapter 7

x

x

x

6 kN

Y x

3 kN/m

C

Y

A

Y

Y

D

Y

B Y RB = 18 kN

3m

4m

3m

(a)

A

−18 kN −12

C

D −18

B

SF diagram (b)

Figure 7.19 Portion CD (x = 4 to 7 m) Shear force Fx = −3 × 4 = −12 kN (remains constant from C to D) Portion DB ( x = 7 to 10 m ) Shear force Fx = −3 × 4 − 6 = −18 kN (remains constant from D to B) Figure 7.19(b) shows the SF diagram of the cantilever. Examples 7.8 A 5-m-long beam AB is hinged at A and roller supported at D as shown in Fig. 7.20. It carries a udl of 10 kN/m over AC = 2 m and a point load of 2 kN at B as shown in the figure. Draw SF diagram of the beam. Solution Support reactions Take moments about end A (note that CG of a udl lies at its centre) Total udl = 10 × 2 = 20 kN Distance of CG of udl from A = 1 m 20 × 1(cw ) + 2 × 5(cw ) = 4 RD (ccw )

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Shear Force and Bending Moment Diagrams x A

x Y

x

2 kN

10 kN/m

Y

Y

C

Y

RA

D

RB 2m

255

B Y 7.5 m

2m

B

1m

(a) +14.5 kN +2 kN

SF diagram (b)

−5.5

Figure 7.20

Reaction, Total load on beam Reaction,

RD = 7.5 kN ↑ = 2 × 10 + 2 = 22 kN RA = 22 − 7.5 = 14.5 kN

Shear force There are three portions in beam, that is, AC, CD and DB. Portion AC (x = 0 to 2 m) Shear force at a distance x from A

Fx = 14.5 − wx = 14.5 − 10 x (as w = 10 kN / m) = +14.5 kN at x = 0 = +4.5 kN at x = 1 m = −5.5 kN at x = 2 m

Portion CD (x = 2 to 4 m) Fx = 14.5 − 20 = −5.5 kN (remains constants from C to D) Portion DB (x = 4 to 5 m) Fx = RA − w × 2 − RD = 14.5 − 10 × 2 + 7.5 = +2 kN (remains constant from D to B). Exercise 7.7 A 5-m-long cantilever AC B is fixed at end B carries a point load of 10 kN at end A and a udl of 5 kN/m over CB = 2 m as shown in Fig. 7.21. Draw SF diagram of cantilever. Exercise 7.8 A 5-m-long beam AB is hinged at B and roller supported at C as shown in Fig. 7.22. Determine support reactions and draw SF diagram.

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256 Chapter 7 3 kN 8 kN/m

10

5 kN/m C

A

B

A

B

D C

2m 1m

2m

2m

5m

Figure 7.21

Figure 7.22

Exercise 7.7

Exercise 7.8

SF Diagrams of Beam/Cantilevers with Applied Moment Consider a cantilever AB of length L which is fixed at end B is subjected to a moment, M (cw) at A as shown Fig. 7.23. To maintain equilibrium of the cantilever, this moment will be balanced by a resisting moment M (ccw) at end B. As a result, there is no shear force on the cantilever. Now take the case of a beam AB, of length L, which is fixed at end B is subjected to a cw moment at the centre C as shown in Fig. 7.24(a). (Please note that if the moment applied is counterclockwise, then beam will rotate about hinged end A, because end B is roller supported and cannot provide reaction in vertically downward direction.) For support reactions, taking moments about A. M (cw ) − RB × L = 0 Reaction,

RB =

M ↑ L

M

M A

B

(ccw)

Resisting moment L

Figure 7.23

y RA = M L A

M

x

B Y

L

M

RB = +

M L

−

M L

(a)

No force is applied on the beam, therefore, to balM M ance the reaction RB = ↑ reaction, RA = ↓. In SF diagram L L other words RB and RA constitute a couple of equal (b) M and opposite forces + , with L = arm length. Figure 7.24 L M Couple due to reaction, M = ×L= M L = M (ccw) Shear force diagram Taking a section at a distance of x from A shear force remains constant from A to B, SF diagram is shown in Fig. 7.24(b). (−)

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Shear Force and Bending Moment Diagrams

257

Example 7.9 A 10-m-long beam AB is hinged at A roller supported at B is subjected to a clockwise moment of 10 N m at C and a point load of 10 kN at point D as shown in Figure 7.25(a). Determine support reactions and draw SF diagrams. x 10 kN x

Y

10 kN/m Y

A

D

C

Y

RA = 2 kN 3m

+2

B

B

(+) D

A

RB = 8 kN 4m

−4 kN

(−)

3m (b)

(a)

SF diagram

Figure 7.25

Example 7.9

Solution For support reactions, taking moments about end A, 10(cw ) + 10 × 7(cw ) − 10 RB = 0 Reaction, Reaction,

80 = 8 kN ↑ 10 RA = 10 − 8 = 2 kN

RB =

(Note: At this stage that 10 kN m is a moment and not a force, so in the calculation of shear force, this will not be taken into account after the determination of support reactions.) Shear force diagram There are two portions AD and DB of the beam. In portion AD Fx = +2 kN (remains constant from A to D) In portion DB Shear force, Fx = +2 − 10 = −8 kN, note that downward force on the left side of the section is negative. Figure 7.25(b) shows the SF diagram of the beams.

4 kN 6 kN/m D

A

B

C 2m

Figure 7.26

2m

1m

Exercise 7.9

Exercise 7.9 A 5-m-long beam AB is hinged at end A and roller supported at end D as shown in Fig. 7.26 is subjected to a clockwise moment of 6 kN m at C and a point load of 4 kN at B. Determine support reactions. Draw SF diagram of the beam.

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258 Chapter 7

BM Diagrams of Cantilevers/Beams with Point Loads Consider a cantilever AB of length L which is fixed at end B and carrying a point load of W at free end A. Reaction at B, RB = W ↑ fixing moment provided by wall at fixed end B, M B = WL(cw ) to balance WL(ccw) acting on the cantilever. Consider a section Y − Y at a distance of x from A, bending moment, M x = −Wx (we have already discussed that clockwise moment on the left side of a section is a positive bending moment and counterclockwise moment on the left side of the section is a negative bending moment). Therefore, M x = −Wx (equation of a straight line) =0 at x = 0 MB = WL W WL L =− at x = x y 2 2 A B = −WL at x = L y

Figure 7.27(b) shows bending moment diagram of the cantilever. Now, consider a beam AB of length L, which is hinged at end A and roller supported at end B, carrying a point load W at the centre C of the beam. For support reactions, taking moments about A

RB

L (a) (−)

−WL

L (b) (cw ) − RB × L(ccw ) = 0 BM diagram 3 W Reaction, RB = ↑ Figure 7.27 2 W W = ↑ (for equilibrium of the beam) Reaction, RA = W − RB = W − 2 2 There are two portions AC and CB in the beam. BM diagram x (Taking the convention that clockwise moment on the W left side of the section is positive) x L  Y Portion AC  x = 0 to  A   2 Y C Taking section YY at a distance x from A L L W×

Bending moment, =0 WL = 8 = Moreover,

WL 4

W Mx = + x 2 at x = 0 L at x = 4 at x =

L 2

dMx W =+ shear force in portion AC dx 2

MTPL0259_Chapter 07.indd 258

RA = W 2

2

2

Y

B

Y RB = W 2

(a) WL 4 A

C

B

BM diagram (b)

Figure 7.28

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Shear Force and Bending Moment Diagrams

259

L Portion CB  x = to L   2 There are two forces on the left side of the section. Bending moment,

Mx =

W L  x −W  x −   2 2

=

WL 4

at x =

L 2

=

WL 8

at x =

3L 4

=0

at x = L

dM x W = −W dx 2 =−

W (shear force in portion CB) 2

Figure 7.28(b) shows the bending moment diagram of the beam. Example 7.10 A 5-m-long cantilever AB is fixed at end B carries point loads as shown in Fig. 7.29(a). Determine support reaction and support moment. Draw the bending moment of the cantilever. Solutions There are two portions AC and CB of the cantilever. Portion AC ( x = 0 to 2 m ) Section YY at a distance x from A Bending moment, Mx = −6x

=0

at x = 0 m

= −6 kN m = −12 kN m

at x = 1 m at x = 2 m

Portion CB (x = 2 to 5 m) There are two forces on the left side of the section Bending moment, Mx = –6x − 4 (x − 2) = −12 kN m = −22 kN m = −32 kN m = −42 kN m

MTPL0259_Chapter 07.indd 259

at x = 2 m at x = 3 m at x = 4 m at x = 5 m

x 6 kN

4 kN

x

MB = Fixing moment

Y Y Y

C A

Y

3m

2m

B RB = 10 kN

(a) B

C

A

−12 −12 kN m

(−)

−42 kN m

Bending moment diagram (b)

Figure 7.29

Bending moment diagram

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260 Chapter 7

42 kN m is the fixing moment at fixed vend B. Reaction RB = 4 + 6 = 10 kN ↑, as shown in Fig. 7.29. Figure 7.29(b) shows the bending moment diagram of the cantilever. Example 7.11 A 7-m-long beam AB is hinged at end A and roller supported at point D carries point loads of 10 and 2 kN at C and B as shown in Fig. 7.30(a). Determine (a) Support reactions (b) Draw BM diagram (c) Location of point of contraflexure x x 10 kN 2 kN x

Y

A

Y C

Y

D P

2.5 m

2.5 m RA = 4.2 kN

B

Y 2.0 m RD = 7.8 kN

(a)

+10.5 D

A 4.31 m

C

P

B

−4.0 kNm BM diagram (b)

Figure 7.30 Solution For support reactions, taking moments about A 10 × 2.5(cw ) + 2 × 7(cw ) − 5 × RD (ccw ) = 0 39 = 7.8 kN 5 Total load on beam = 10 + 2 = 12 kN Reaction, RA = 12 − 7.8 = 4.2 kN There are three portions of beam, that is, AC, CD and DB. In each portion, a section YY is considered at a distance of x from end A. Reaction,

MTPL0259_Chapter 07.indd 260

RD =

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Shear Force and Bending Moment Diagrams

Portion AC Bending moment,

261

Mx = + 4.2x (equation of straight line) = 4.2 kN m at x = 1 m = 10.5 kN m at x = 2.5 m

Portion CD Bending moment,

Mx = + 4.2x − (x − 2.5) = 10.5 kN m at x = 2.5 m = 4.7 kN m at x = 3.5 m = −4 kN m at x = 5 m

Portion DB Bending moment,

Mx = 4.2 x − 10( x − 2.5) + 7.8( x − 5) = −4 kN m at x = 5 m = 25.2 − 35 + 7.8 = −2 kN m at x = 6 m = 4.2 × 7 − 10 × 4.5 + 7.8 × 2 = 29.4 − 45 + 15.6 = 0 at x = 7 m.

Figure 7.30(b) shows the bending moment diagram. In portion CD of the beam, bending moment changes sign from positive to negative at point P. The point P is called the point of contraflexure. Location of point of contraflexure In portion CD, M x = 4.2 x − 10( x − 2.5) = 0 or 4.2 x − 10 x + 25 = 0 5.8 x = 25 25 x= = 4.31 m 5.8 Point of contraflexure lies at a distance of 4.31 m from A. Exercise 7.10 A 4-m-long cantilever AB is fixed at end B carries point loads of 3 and 5 kN at A and C as shown in Fig. 7.31. Determine support reaction, support moment and draw BM diagram of cantilever. Exercise 7.11 A 6-m-long beam AB is hinged at end A and roller supported at point D carries 8 and 4 kN loads as shown in Fig. 7.32. Determine support reactions and draw BM diagram. Locate the point of contraflexure, if any.

3 kN

5 kN

A

C

B

8 kN

4 kN

A

B C

2m

2m

Figure 7.31 Exercise 7.10

MTPL0259_Chapter 07.indd 261

2m

D 2m

Figure 7.32

2m

Exercise 7.11

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262 Chapter 7

BM Diagrams of Cantilevers/Beams with UDL Consider a cantilever of AB of length L, fixed at end B, carrying udl of intensity w over the entire length. Total load on cantilever = wL Reactions, RB = wL ↑ Bending moment diagram Consider a section YY at a distance x from A, udl up to the section = wx CG of the load lies at x/2 from YY section. x

 x Bending moment at the section, M x = − wx    2

A

B

2

=−

wx 2

=−

wL2 8

wL2 =− 2

at x =

L 2

(a) A

B (−)

L at x = 2

wL (cw ) 2 Figure 7.33(b) shows the bending moment diagram of cantilever. Moreover, dM x /dx = − wx (this is the shear force at the section). Now take a beam AB of length L, which is hinged A at end A and roller supported at end B, carrying udl of intensity w throughout its length as shown in Fig. 7.34(a). RA = wL 2 For support reactions, taking moments about and A,

wL ↑ 2 Total load as beam = wL RB =

Reaction

RA = wL −

Reaction,

wL wL = ↑ 2 2

RB = wL

L

−wL2 2

BM diagram (b)

Fixing moment at B, fixed end =

L (cw ) − RB L (ccw ) = 0 2

Fixing cupid

Y

2

wL ×

MB

w

Y

Figure 7.33

x

Y

w B

Y L (a)

RB = wL 2

wL2 2 BM diagram (b)

Figure 7.34

Bending moment diagram Consider a section YY at a distance x from A (note that CG of a udl lies at its centre): Mx =

MTPL0259_Chapter 07.indd 262

wL  x x − wx   , that is, CG of load wx ↓ lies at x/2 from section YY  2 2

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Shear Force and Bending Moment Diagrams

Mx =

263

wLx wx 2 − 2 2

= 0, at x = 0 =

3 wL2 32

at x =

L 4

wL2 8

at x =

L 2

=+

3 wL2 32 =0

3L 4 at x = L

=

at x =

Figure 7.34(b) shows BM diagram of the beam. Note that maximum bending moment occurs at x = L / 2 , that is, at the centre of the beam. In the section ‘SF Diagrams of Cantilevers and Beams with udl’, we have learnt that at the centre of the beam loaded with udl, shear force is zero. Example 7.12 A 5-m-long cantilever AB is fixed at end B carries a udl of 6 kN/m over AC = 3 m as shown in Fig. 7.35(a). Determine support reactions. Draw BM diagram of the cantilever.

x x A

Y

Y

6 kN/m Y

2m

3m −3 m

B

Y

C

(a)

A

C

B

−6.75 kNm −27 kNm

−63 kNm

(b)

Figure 7.35 Solutions Supported reaction Fixing moment,

Bending moment diagram

RB = 6 × 3= 18 kN M B = 18(5 − 1.5) = 63 kN m(cw), note that CG of udl lies at its centre.

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264 Chapter 7

Bending moment diagram There are two portions AC and CB in cantilever Portion AC (x = 0 to 3 m)  x M x = − wx   (equation of a parabola)  2 =−

wx 2 x2 = −6 = −3 x 2 , substituting the value of w 2 2

=0 = −6.75 kN m = −27 kN m Portion CB (x = 3 to 0.5 m),

at x = 0 m at x = 1.5 m at x = 3 m

Mx= −18 (x −1.5) (equation of straight line) = −27 kN m at x = 3 m = −45 kN m at x = 4 m = −63 kN m at x = 5 m Figure 7.35(b) shows the BM diagram of cantilever. Example 7.13 A 5-m-long beam AB carries a udl of intensity 4 kN/m over AC = 3 m and a point load of 6 kN at C as shown in Fig. 7.36(a). Determine support reactions and draw BM diagram of beam.

x x

6 kN

4 kN/m

A

Y Y

RA = 10.8 kN

Y B

C

3m

Y 2m

RB = 7.2 kN

(a) 13.6

14.4 kNm

. 8.8 BM diagram (b)

Figure 7.36

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Shear Force and Bending Moment Diagrams

265

Solution For support reactions, taking moments about A, (note that CG of udl lies at its centre) 4 × 3 × 1.5(cw ) + 6 × 3(cw ) − 5 RB (ccw ) = 0 18 + 18 = 5 RB Reaction

RB = 7.2 kN

Total load an beam = 4 × 3 + 6 = 18 kN Reaction RA = 18 − 7.2 = 10.8 kN Bending moment diagram There are two portions of the beam, that is, AC, CB Portion AC (x = 0 to 3 m) Bending moment,

 x M x = + RA x − wx    2  x = 10.8 x − 4 x    2 = 10.8 x − 2 x 2 =0 = +8.8 kN m = +13.6 kN m = +14.4 kN m

Portion CB (x = 3 to 5 m) Bending moment,

at x = 0 m at x = 1 m at x = 2 m at x = 3 m

M x = RB x − 3w ( x − 1.5) − 6( x − 3) = 10.8 x − 3 × 4( x − 1.5) − 6( x − 3) = 10.8 x − 12( x − 1.5) − 6( x − 3) = 14.4 kN m

at x = 3 m

= +7.2 kN m

at x = 4 m

=0

at x = 5 m

Figure 7.36(b) shows the BM diagram of the cantilever. Exercise 7.12 A 4-m-long cantilever is fixed at end B carries a point load of 6 kN at A and a udl of intensity 6 kN/m on AC = 2 m as shown in Fig. 7.37. Determine support reaction, support moment and draw BM diagram. Exercise 7.13 A 5-m-long beam AB is hinged at end A roller supported at end B carries udl and a point load as shown in Fig. 7.38. Determine support reactions and draw BM diagram of the beam.

MTPL0259_Chapter 07.indd 265

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266 Chapter 7 5 kN 6 kN 6 kN/m 6k N/m

A

C

C

B

D

B

A 2m

2m

3m

Figure 7.37 Exercise 7.12

1m

Figure 7.38

1m

Exercise 7.13

SF and BM Diagrams of Beam/Cantilever with Load Through a Crank Let us consider a cantilever AB of length L, which is subjected to a load W, not on cantilever but at the end of a crank ADC, such that DC = a. This load W at C can be replaced by a couple Wa(cw) and a load W ↓ at the point A of the cantilever. Reaction RB = W ↑ at fixed end B and fixing couple at B will be (WL − Wa)cw. Equivalent diagram of loading for Fig 7.39(a) is shown in Fig 7.39(b). Similarly, consider a beam AB of length L which is subjected to a load W, at the end of crank CDE. The load W at E can be replaced by a couple Wa ↓ and a load W ↓ at the point C. Equivalent diagram of loading for Fig. 7.40(a) is shown in Fig. 7.40(b). For support reactions, taking moments about A. W

D A

W

(WL −Wa) cw

C

MB

W.a B

a

B

A L

L

(a)

RB = W (b)

Figure 7.39 W W

D

Wa

E A

C L

A a RA =

(

W Wa − 2 L

(

(a)

B

C L 2

L 2

RB = WL + Wa 2 L

(

(

(b)

Figure 7.40

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Shear Force and Bending Moment Diagrams

Wa(cw ) + W

267

L (cw ) − RB L = 0 2 RB =

W Wa + 2 L

Total load on beam = W↓ RA = W −

Reactions,

W Wa W Wa − = − L L 2 2

Example 7.14 A 5-m-long cantilever AB is fixed at end B. It carries a point load of 5 kN at E of crank ADE and a point load of 10 kN at C as shown in Fig. 7.41(a). Draw SF and BM diagrams of the cantilever. Where is the location of point of contraflexure?

5 kN D

A

E

10 kN

1m A

C

C

B

(−)

1m 3m

2m +5 kN m

(a) 5 kN 5 kN m

Y

A

Y

C

−40 kN m B

2m

3m

(−10)

10 kN Y Y

−15 kN

SF diagram (c)

A P

x x

B

−5 kN

BM diagram (d)

(b)

Figure 7.41 Example 7.14

Solution Load of 5 kN at E is replaced by a load of 5 kN and a 5 kN m couple (cw) at A and arm of the couple is 1 m. Equivalent loading diagram of cantilever is shown in Fig. 7.41(b). There are two portions of the cantilevers, that is, AC and CB. SF diagrams Portion AC (x = 0 to 3 m) Fx = −5 kN (remains constant from A to C) Portion CB (x = 3 to 5 m) Fx = −5 − 10 = −15 kN (remains constant from C to B)

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268 Chapter 7

BM diagrams (clockwise moment on the left side of the section is positive) Mx = 5 − 5x, = 5 kN m at x = 0 m = 0 at x = 1 m = −5 kN m = −10 kN m Portion CB

at x = 2 m at x = 3 m

M x = +5 − 5 ( x ) − 10 ( x − 3) = −10 kN m

at x = 3 m

= −25 kN m

at x = 4 m

= −40 kN m

at x = 5 m

Figure 7.41(d) shows the BM diagram of the cantilever. Point P (point of contraflexure) lies at a distance of 1 m from A as shown in Fig. 7.41.

5 kN

A

MTPL0259_Chapter 07.indd 268

1

5 kN

5 kN

RB = 3 kN

5 kN m 2

2

A

D

C

1 m RB = 7 kN

(b)

(+)

+3 kN m

−2

(−)

−7 kN m

SF diagram (c)

RB = 7 kN ↑ 11

Total load on beam = 5 + 5 = 10 kN

Portion CD,

2m

(a)

10 + 5 + 20 = 5 RB

RA = 10 − 7 = 3 kN ↑ Reaction, SF diagram (x = 0 to 2 m) Portion AC, FF + 3 kN (remain constant from A to C)

B

D 1

2

5 × 2(cw ) + 5(cw ) + 5 × 4(cw ) − 5 × RB (ccw ) = 0

Reaction,

F

C

Example 7.15 A 5-m-long beam AB carries a load 5 kN at end F of a crank CEF and another point load of 5 kN at D as shown in Fig. 7.42(a). Draw SF and BM diagrams of the beam. Solutions Load of 5 kN at F is replaced by a load of 5 kN and a couple 5 kN m(cw) at C and equivalent loading diagram is shown in Fig. 7.42(b). There are three portions of beam, that is, AC, CD and DB as shown in the figure. For support reactions, taking moments about point A:

5 kN

E

6 A

(+) BM diagram

+7 kN m

B

(d)

Figure 7.42

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Shear Force and Bending Moment Diagrams

269

F = +3 − 5 = 2 kN (remains constant from C to D) Portion DB, F = 3 − 5 − 5 = −7 kN (remains constant from D to B) BM diagram Fig. 7.42(c) shows the SF diagram Portion AC, Mx = + 3x (straight line equation) =0 at x = 0 = +3 kN m at x = 1 m = +6 kN m at x = 2 m Portions CD, Mx = 3x + 5 − 5(x − 2) = 11 kN m at x = 2 m = 9 kN m at x = 3 m = 7 kN m at x = 4 m Portions DB, Mx = 3x + 5 − 5(x − 2) − 5(x − 4) = 7 kN m =0

at x = 4 m at x = 5 m

Figure 7.42(d) shows the BM diagram of the beam. Exercise 7.14 A 4-m-long cantilever AB is fixed at end B carries a point load of 4 kN at end of crank ACD as shown in Fig. 7.43. What are reaction and bending moment at fixed end? Draw SF and BM diagrams for the cantilever. Exercise 7.15 A 5-m-long beam AB carries a point load 4 kN at the end of a crank CEF and on other point load of 6 kN at D as shown in Fig. 7.44. Determine support reaction and draw SF and BM diagrams of beam. Where lies the point of contraflexure?

4 kN 4 kN 6 kN

C D B

A

1m

A

4m

Figure 7.43 Exercise 7.14

MTPL0259_Chapter 07.indd 269

E

F C

1

1m

Figure 7.44

D

2m

B

1m

Exercise 7.15

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270 Chapter 7

Variable Loading on a Beam

x

wx

A beam AB of length L carries a linearly variable load of maximum intensity w over length L1, as shown in Fig. 7.45.

Y

Y A

wL Total load as shown = 1 2

x

CG of the load lies at a distance of 2 L1/3 from end A (Fig. 7.45). For support reaction, let us take moments about end A.

w Y

C

Y

B

L1 L

Figure 7.45

WL1  L1 × 2  (cw ) − RB × L(ccw ) = 0 2  3  Reaction,

Reaction at

RB =

A, RA =

wL1 L1 1 wL12 × = 6 6L L wL1 wL2 1 6wLL1 − 2wL12 − = 2 6L 6L =

3wLL1 − wL12 3L

(7.1)

At any section YY at a distance of x from end A (x varies from o to L1). x Shear force, Fx = RA − wx where wx is the rate of load at x 2 wx = Therefore,

w x (rate of loading) L1

Fx = RA −

(n varies from o to L1) Note that CG of the load up to x, that is, of wx

Bending moment at n,

wx 2 2 L1

(7.2)

x2 lies at a distance of x /3 from section YY. 2

M x = RA x − = RA x −

wx 2 x × 2 L1 3 wx3 6 L1

(7.3)

(x various from o to L1) In the portion CB

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Shear Force and Bending Moment Diagrams

wL12 2 wL12 M x = RA x − 2 Fx = RA −

Shear force, Bending moment,

271

(7.4) 2 L1    x − 3 

(7.5)

Example 7.16 A 5-m-long beam AB carries a triangular load of maximum intensity of w = 6 kN/m over AC = 3 m as shown in Fig. 7.46(a). Determine support reactions. Draw SF and BM diagrams.

x

y

6 kN/m

+5.4

wx A

B

−3.6 kN

Y RA = 5.4 kN

3m

2m

RB = 3.6 kN SF diagram (b)

(a) 8.133 7.2 5.07

M max

M max = 8.3656 kNm

BM diagram (c)

Figure 7.46 Solution Reactions 6×3 2 Total load on beam = = 9 kN, CG of this load at 3 × = 2 m from A. 2 3 Moments about A, 9 × 2(cw ) − RB × 5(ccw ) = 0 18 = 3.6 kN 5

Reaction,

RB =

Reaction,

RA = 9 − 3.6 = 5.4 kN

SF diagram (portion AC3 x = 0 to 3 m) Take a section YY at a distance x from A. Rate of loading,

MTPL0259_Chapter 07.indd 271

wx =

6× x = 2 x kN/m. 3

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272 Chapter 7

Total load up to

wx =

wx x 2 xx = . 2 2

= x2 Shear force

Fx = 5.4 − x 2 = 5.4 kN = 4.4 kN = 1.4 kN = −3.6 kN

at at at at

x=0 x =1m x=2m x=3m

Portion CB ( x = 3 to 5 m ) Fx = 5.4 − 9 = −3.6 kN (remains constant from C to B) Bending moment diagram Load up to x = x 2 10 kN CG load lies at x / 3 from reaction YY Bending moment,

x2 x 3 x3 = 5.4 x − 3 =0 = 5.07 kN m = 8.133 kN m = 7.2 kN m

M x = 5.4 x −

at x = 0 m at x = 1 m at x = 2 m at x = 3 m

Portion CB ( x = 3 to 5 m ) M x = 5.4 x − 9( x − 2), please note that total load is 9 kN and its CG is at a distance of 2 m from A. = 7.2 kN m at x = 3 m = 3.6 kN m at x = 4 m =0 at x = 5 m Figure 7.46(c) shows the BM diagram of the beam. Note that maximum bending moment occurs in a beam where shear force is zero. Shear force is zero in portions AC, or where 5.4 − x 2 = 0 x = 2.328 m, SF is zero M max = 5.4 x −

x3 3

2.3283 3 = 12.5712 − 4.2056 = 8.3656 kN m > 8.133 kN m = 5.4 × 2 − x = 2 m.

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Shear Force and Bending Moment Diagrams

273

Exercise 7.16 A beam AB of length L, which is hinged at end A and roller supported at end B, carries a linearly variable load with maximum intensity w per unit length at end B as shown in Fig 7.47. Determine support reactions. Draw SF and BM diagrams.

w A

B L

Figure 7.47

Exercise 7.16

Relation Between Rate of Loading, SF and BM in a Beam Till now, we have studied a number of cases of cantilevers and beams subjected to various types of loading and learnt that: dF = − w (rate of loading) dx dM = F (shear force) dx If M is constant, F is zero. Similarly, if F remains constant w = 0. Let us now derive relationship between these parameters of a beam. Consider a beam subjected to various loads W1, W2 and udl equal to w as shown in Fig. 7.48(a). Take a small element of beam of length dx at a distance x from end A. Shear force on the left side of element, F ↑ Shear force on the right side of element F + dF↓ Bending moment on the left side of element M(cw) Bending moment on the right side of element M + dM(ccw) Considering equilibrium of forces or or

W1

w A

dx (a) M

M + dM w F + dF

F

dF = − w (minus the rate of loading) dx

dx

The rate of change of shear force at a section is numerically equal to rate of loading. Now, taking moments about the right-hand end of the section.

MTPL0259_Chapter 07.indd 273

B x

F = F + dF + w dx dF = − wdx

M + dM = M + Fdx − wdx

W2

Figure 7.48(a)

dx 2

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274 Chapter 7

 dx 2  , neglecting high-order term = M + Fdx − w   2 

=

dM = F. The rate of change of bending moment at a section is equal to the shear force at the section. dx dF = − w = 0 (rate of loading) (i) Now, taking dx Alternatively, shear force F = a constant dM =F dx

Moreover,

∫ dM = ∫ Fd M = F x + e, where C is constant of integration dF = −w dx

(ii)

dF = − wdx F = − wx + C , where C1 is another constant of integration. dM = − wx + C1 dx

or

M=−

wx 2 + C1 x + C2 , where C2 is another constant of integration. 2

Example 7.17 A beam of length L which is simply supported at ends carries transverse load. SF diagram for the beam is shown in Fig. 7.48(b). Draw BM diagram.

+W 2 A

C

B −W 2

SF diagram

Solution

L 2

BM at A = 0 (free end)

L 2

BM at B = 0 (free end)

(a)

dM W =F=+ as shown. dx 2

WL 4

Portion AC

A

c

c

A

A

M = ∫ Fdx − ∫ +

L 2

W W W dx = ∫ + dx = x 2 2 2 o o =

MTPL0259_Chapter 07.indd 274

L 2

WL 8

at x =

C BM diagram

B

Figure 7.48(b)

L 4

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Shear Force and Bending Moment Diagrams

Mc =

WL 4

at x =

275

L 2

Portion CB B

B

L

−W W x =− 2 2

L  L − 2   

−W −W dx = ∫ dx 2 2 C o

M = ∫ Fd x = ∫ C

L

M B − MC = =

L 2

−WL 4

M B = MC −

WL WL WL = − =0 4 4 4

Exercise 7.17 A 6-m-long beam AB is simply supported at ends carries two loads on the beam. SF diagram of beam is shown in Fig. 7.49. Draw BM diagram of the beam.

+6 kN D

B

C

A

2m

−6 kN

2m

2m

Figure 7.49 Problem 7.1 A 4-m-long beam AB and a 6-m-long beam BC are hinged at their ends at B as shown in Fig. 7.50(a). Beam AB carries a load 6 kN at its centre and beam BC carries udl of intensity 6 kN/m over EC = 4 m. Draw BM diagram of the beam ABC. Solution Beam AB, hinged at both the ends, carries a load 6 kN at centre. Reactions, RA = RB = 3 kN ↑ each. For beam BEC, 3 kN ↓ load acts at end B. Figure 7.50(b) shows equivalent loading diagram. Beam AB M max =

6×4 = +6 kNm at centre 4

Beam BEC Taking moments about E 2 × 3 + RC × 4 = 6 × 4 × 2 (note that CG of udl lies at its centre)

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276 Chapter 7 6 kN 6 kN/m

B

A

C E

D

2 4m

2m

6 kN A

4m (a)

B 3 kN

D

Y

X

3 kN

6 kNm

3 kN B

E

C RC = 10.5 kN

Y

RE 2

4

(b)

RE = 16.5 9

+6 kN

7.5

4.5 E

A

B

C

P −6 kNm BM diagram

(c)

Figure 7.50 Reaction,

RC = 10.5 kN

Reaction,

RE = 6 + 24 − 10.5 = 16.5 kN

BM diagram Portion BE ( x = 0 at 2 m )

Portion EC ( x = 2 to 6 m )

M x = − 3x =0 = −3 kN m = −6 kN m

at x = 0 at x = 1 m at x = 2 m

6 ( x − 2)2, (CG of udl lies at its centre) 2 = −3 x + 16.5( x − 2) − 3( x − 2)2 = −9 + 16.5 − 3 = + 4.5 kN m at x = 3 m

M x = −3 x + 16.5 ( x − 2) −

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Shear Force and Bending Moment Diagrams

277

= −12 + 16.5 × 2 − 12 = +9 kN m at x = 4 m = −15 + 16.5 × 3 − 27 = +7.5 kN m at x = 5 m = −18 + 16.5( 4) − 3 × 16 = 0

at x = 6 m

Problem 7.2 A 6-m-long beam AB, which is roller supported at C hinged at B as shown in Fig. 7.51(a), carries a udl of 5 kN/m over ACD = 3 m and an inclined load of 10 kN at E, at angle of 60° as shown in the figure. Determine support reactions and draw SF and BM diagrams. 10 sin 60 = 8.66 kN 10 kN

5 kN/m A

C

Hinged

60

B

E 5

D

RBV = 8.43

RC = 15.23

1m

RBH = 5

1m

2m

2m (a)

10.23 0.23 −5 kN

−8.43 SF diagram

−8.43 kN

(b)

7.96

8.42 kNm

5.23

A

C −2.5 KNm

D

E

B

BM diagram (C)

Figure 7.51 Solution Vertical load 10 kN Vertical component Horizontal component

= 10 sin 60° = 8.66 kN = 10 cos 60° = 5 kN

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278 Chapter 7

RBM = 5 kN

Reaction at B, fixed end

udl = 5 × 3 = 15 kN CG of udl lies at its centre, that is, at 0.5 m from C or 1.5 m from A. For support reactions, taking moments about C: 15 × 0.5 + 8.66 × 4 = 5 RBV RBV = vertical component of reaction at B = 8.43 kN ↑ = 15 + 8.66 = 23.66 kN

Total vertical load on beam

RC = 23.66 − 8.43 = 15.23 kN ↑

Reaction, SF diagram Portion AC ( x = 0 to 1)

Fx = −5 x (equation of straight line) =0 at x = 0 = −5 kN at x = 1 m Portion CD ( x = 1 to 3 m ) Fx = −5 x + 15.23 kN = 10.23 at x = 1 m = 5.23 at x = 2 m = 0.23 at x = 2 m Portion DE ( x = 3 to 5 m ) Fx = −15 + 15.23 = +0.23 kN (remains constant) Portion EB ( x = 5 to 6 m ) Fx = −15 + 15.23 − 8.66 = −8.43 kN (remains constant from E to B) Bending moment diagram Portion AC ( x = 0 to 1 m ) M x = − wx

x wx 2 =− 2 2

= −2.5 x 2 =0 = −0.625 kN m = −2.5 kN m

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at x = 0 at x = 0.5 m at x = 1 m

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Shear Force and Bending Moment Diagrams

279

Portion CD ( x = 1 to 3 m ) M x = 2.5 x 2 + 15.23( x − 1) = −2.5 kN m = +5.23 kN m

at x = 1 m at = 2 m

= +7.96 kN m

at x = 3 m

Portion DE ( x = 3 to 5 m ) M x = −15 × ( x − 1.5) + 15.23( x − 1) = 7.96 kN m

at x = 3 m

= +8.16 kN m

at x = 4 m

= +8.42 kN m

at x = 5 m

Portion EB M x = −15 × ( x − 1.5) + 15.23( x − 1) − 8.6( x − 5) = +8.42 kN m

at x = 5 m

=0

at x = 6 m

Problem 7.3 A 6-m-long propped cantilever AB carries transverse loads as shown in Fig. 7.52(a). Determine support reaction and support moment and draw SF and BM diagrams. Solution Total vertical load on cantilever = −6 + 4 × 6 = 18 kN Reaction, RB = 18 kN ↑ Fixing moment at B

x x A 6 kN

y

2m

Portion CB ( x = 2 to 4 m ) Fx = 6 − w ( x − 2) = 6 − 6( x − 2) kN

MTPL0259_Chapter 07.indd 279

y

4m

A

MB = B RB = 18 kN

+6 kN

B

C SF diagram

−18 kN

(b)

= −12 kN m

Fx = +6 kN (remains constant from A to C)

C (a)

Applied, M B = 6 × 6 − 6 × 4 × 2 = 36 − 48 Fixing moment = +12 kN m (cw) to balance the applied moment. There are two portion of the cantilever that is AC and CB. Portion AC ( x = 0 to 2 m )

Y 6 kN/m

Y

12

15

P 5.236 m

−12 kNm

BM diagram (c)

Figure 7.52

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280 Chapter 7

= 6 kN =0 = −6 kN = −18 kN

at x = 2 m at x = 3 m at x = 4 m at x = 6 m

Figure 7.52(b) shows the SF diagram of the cantilever. BM diagrams Portion AC (x = 0 to 2 m) Mx = + 6x =0 = 6 kN m = 12 kN

at x = 0 at x = 1 m at x = 2 m

Portion CB ( x = 2 to 6 m )  x − 2 M x = 6 x – w ( x – 2)   2   x − 2 Note that udl up to x is w ( x − 2), and CG of the load lies at a distance of  from Y  2  section as shown in Fig. 7.53. 6 x Y M x = + 6 x − ( x − 2) 2 G (x − 2) 2 2 = 6 x − 3( x − 2)2 2m (x − 2) Y

= 12 kN = 15 kN = +12 kN m

at x = 2 m at n = 3 m at x = 4 m

= +3 kN = −12 kN m

at x = 5 m at x = 6 m

6 kN

Figure 7.53

Figure 7.52(c) shows the BM diagrams of the cantilever. Point of contraflexure lies in portion CB, where bending moment is zero 6 x − 3( x − 2)2 = 0 6 x − 3( x 2 − 4 x + 4) = 0 18 x − 3 x 2 − 12 = 0 x2 − 6 x + 4 = 0 x=

6 ± 36 − 16 6 + 20 = = 5.236 m 2 2

This type of cantilever is knows as a propped cantilever. Problem 7.4 A beam AB of length L is hinged at both the ends. It carries W at L/6 from A and −W at L/ 6 from B as shown in Fig. 7.54(a). Determine support reactions. Draw SF and BM diagrams of the beam.

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Shear Force and Bending Moment Diagrams

W C

A

D

L 6

L 3

2W 3

2W 3

B

E L 3

L 6 RB =

RB =

281

−W

(a)

B +2W 3

D

C

+2W 3

E

A

−W 3 SF diagram (b)

+WL 9

P A

E B

D

BM diagram (c)

−WL 9

Figure 7.54 Solution For support reactions taking moments about A WL 5 LW (ccw ) (cw ) − + RB L(cw ) = 0 6 6 RB =

Reaction,

2W ↓ 3

(as shown) +2W To balance this force reaction on beam, reaction, RA = ↑ 3 SF diagram L  Portion AD  x = 0 to  6 FAD = +

2W (constant from A to D) 3

L 5L   Portion DE  x = to   6 6

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282 Chapter 7

FDE = +

2W −W −W = 3 3

FEB = +

2W 2W −W +W = + 3 3

(constant from C to E) 5L   Portion EB  x = to L   6

(remains constant from E to B) Figure 7.54(b) shows the SF diagrams of the beam. BM diagram Portion AD 2W x (equation of a straight line) 3 =0 at x = 0

Mx = +

=

WL 9

at x =

L 6

5L  L  to Portion DE  x =   6 6 Mx =

2W L  x −W  x −   3 6

L 6 L =0 at x = 2 2W 5 L  5L L  = × −W  −  6 6  3 6 =

W 9

at x =

5WL 4WL 5WL 2WL − = − 9 6 9 3 WL  5L  =−  at x =  9  6 =

5L   to L Portion EB  x =  6

Mx =

2W L 5L    x −W  x −  +W  x −    3 6 6

= 0 at x = L Figure 7.54(c) shows the bending moment diagram of the beam. Note that the centre of the beam becomes the point of contraflexure.

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Shear Force and Bending Moment Diagrams

283

Problem 7.5 A 7-m-long beam AB which is supported at C and B on 5-m span is subjected to a point load of 4 kN at A, 10 kN m (ccw) moment at D and a udl = 10 kN/m over EB = 2 m. Determine support reactions and draw BM diagram only Fig 7.55(a). 4 kN

10 kN/m

10 kNm

A

C 2m

D 2m

B

E 1m

RB = 12.4kN

2m

Rc = 11.6 (a) 7.2

7.4 4.8

O −2.8 −8 kNm

BM diagram

Figure 7.55 Solution Taking moment about C 10 (ccw ) + 5 RB (ccw ) −10 × 2 × (5 − 1)(cw ) + 4 × 2(ccw ) = 0 10 + 5 RB − 80 + 8 = 0 62 RB = 5 = 12.4 kN Total load Reaction, Rc

= 4 + 20 = 24 kN = 24 − 12.4 = 11.6 kN

Bending moment diagram Portion AC (x = 0 to 2 m) Mx

= −4x =0

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at x = 0

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284 Chapter 7

= −4 kN m = −8 kN m

Portion CD (x = 2 to 4 m) Mx

at x = 1 m at x = 2 m

= −x + 11.6(x − 2) = −8 kN m = −0.4 kN m = +7.2 kN m

Portion DE (x = 2 to 4 m) Mx

at x = 2 m at x = 3 m at x = 4 m

= −4x + 11.6(x − 2) − 10 = −2.8 kN m at x = 4 m = +4.8 kN m at x = 5 m

Portion EB ( x = 5 to 7 m ) M x = −4 x + 11.6( x − 2) − 10 −

w ( x − 5)2 2

= −4 x + 11.6( x − 2) − 10 − 5 ( x − 5)2 = 4.8 kN m

at x = 5 m

+7.4 kN m

at x = 6 m

=0

at x = 7 m

Note that there are three points of contraflexure in the beam. Problem 7.6 SF diagram for a beam simply supported at B and C with overhang on both the sides is shown in Fig. 7.56(a). Draw BM diagram of the beam. Solution Beam is supported at B and C, therefore AB is overhang, 2 m and CD is overhang, 1 m long. At B vertical reaction RB = 5 + 5 = 10 kN ↑ At C vertical reaction RC = 5 + 5 = 10 kN ↑ Point load at D = 5 kN ↓ Rate of loading Between A and B w′ =

+5 kN 2

= 2.5 kN/m ↓ 5+5 4 = 2.5 kN/m.

Rate of loading between B and C, w″ =

checking for reaction Total load = 6 × 2.5 + 5 = 20 kN

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Shear Force and Bending Moment Diagrams +5

A

285

+5

B

C −5 kN

−5 4m

2

D

1m (a)

2.5 kN/m

5 kN B

A

C

RC = −10 kN

RB = 10 kN

2m

D

4m

1m

Loading diagram (b)

−5 kNm

−5 kNm BM diagram (c)

Figure 7.56 Moments about B 2.5 × 6 × 1 + 5 × 5 − RC × 4 = 0 40 = RC × 4 RC = 10 kN, reactions are correct load is correct. BM diagram Portion AB ( x = 0 to 2 m ) − w ′′x 2 −2.5 x 2 = = −1.25 x2 2 2 =0 at x = 0 m = −1.25 kN m at x = 1 m = −5 kN m at x = 2 m

Mx =

Portion BC ( x = 2 to 6 m ) Mn = −

MTPL0259_Chapter 07.indd 285

w ′′x 2 + 10( x − 2) 2

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286 Chapter 7

= −1.25 x 2 + 10( x − 2) = −5 kN m = −1.25 kN m =0 = −1.25 kN m = −5 kN m

at at at at at

x=2m x=3m x=4m x=5m x=6m

Portion CD ( x = 6 to 7 m ) M x = −15( x − 3) + 10( x − 2) + 10( x − 6) = −45 + 30 = −5 kN m at x = 6 m = −60 + 50 + 10 =0 at x = 7 m Problem 7.7 An 8-m-long timber beam which is 0.4 m2 in section floats horizontally in seawater. The weight of the timber is 640 kN/m3 and of water is 1,020 kg/m3. Two equal weights just sufficient to immerse the beam are placed on it, 2 m from each side. Calculate the value of weight (Fig 7.57). What is Mmax? Solution Area of cross-section = 0.4 × 0.4 = 0.16 m 2 Length of the beam = 8 m Volume of the beam = 8 × 0.16 = 1.28 m3 Weight of beam = 1.28 × 640 = 819.2 kg Weight of water displaced = 819.2 kg Now, the beam is to be completely immersed in seawater; the weight of water displaced = 1.28 × 1, 020 = 1, 305.6 kg Extra weight to be placed on the beam = 1,305.6 − 819.2 = 486.2 kg 486.2 Two equal weights W = 2 = 243.1 kg = 243.1 × 9.81 = 2,385.8 N 2.4 kN 2.4 kN @ 2.4 kN E B When the timber beam is wholly immersed, the A C D upward thrust of buoyancy will be 1,305.6 kg. 0.6 kN/m Net upward force 1,305.6 − 819.2 = 480.2 kg 4m 2m 2m @ 4.8 kN distributed over 8 m length. 4.8 udl in upward direction = = 0.6 kN/m 8 Figure 7.57 Problem 7.7

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Shear Force and Bending Moment Diagrams

287

Maximum bending moment on beam will occur at C or D. Maximum =

0.6 × 22 2

= 1.2 kN m. Problem 7.8 A 10-m-long floor joist is simply supported at each end and carries a distributed load of grains over the whole span loading distribution is represented by the equation w = A x 2 + Bx + c, where w is the load intensity in kN/m at a distance x along the beam. Load intensity is zero at each end and maximum value of 5 kN/m at the middle of the span. Derive equation of rate of loading. What is maximum rate of loading? Solution Rate of loading

w = A x 2 + Bx + c, w = 0 at x = 0 0=0+0+C

Constant C = 0

or

However,

Equation of rate of loading, Total load on beam,

w = 0 at x = 10 m 0 = A × 100 + 10 B B = −10 A w = 5 kN/m at x = 5 m 5 = 25 A + 5 B B = −10 A 5 = 25 A − 50 A A = −0.2 B = +2 w = −0.2 x 2 + 2 x

(7.6) (7.7)

10

w = ∫ ( −0.2 x 2 + 2 x ) dx 0

0.2 x3 2 x 2 = − + 3 2

10

0

0.2 × 1, 000 = 100 − 3 = 100 − 66.667 = 33.33 kN m Rate of loading at

x = 2.5 m, w = −0.2 × 2.52 + 2 × 2.5 = +3.75 kN/m at centre maximum rate of loading

Problem 7.9 A 7-m-long beam AB is hinged at A and uniformly supported over DE = 2 m as shown in Fig. 7.58(a). Determine reaction at A and udl reaction over DE. Draw BM diagram of the beam. Determine the position of the point of contraflexure.

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288 Chapter 7 1 kN 3 kN/m D A

E

C 2.3 kN

1.6 kN/m

B u.d. reacttion

RA 3m

1m

2m

1m

(a)

0.2

A

B

P

4.1 m

BM diagram (b)

Figure 7.58 Solution Total triangular load 3×3 = 4.5 kN 2 CG of the triangular load lies at 2 m for A. CG of udl reaction lies at 3 + 1 + 1 = 5 m for A. Taking moments about A =

4.5 × 2 + 1 × 7 = R ′ × 5 3.2 = 1.6 kN / m 2 Total vertical load as beam udl reaction =

= 4.5 + 1 = 5.5 kN ↓ Reaction,

RA = 5.5 − R′ = 5.5 − 3.2 = 2.3 kN

BM diagram There are four portions of the beam, that is, AC, CD, DE and EB.

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Shear Force and Bending Moment Diagrams

289

Portion AC (x = 0 to 3 m) M x = 2.3 x − wx where wx = ratio of loading at x =

x  x   2  3

3x =x 3

M x = 2.3 x − x = 2.3 x −

xx 23

x3 6

=0 = +2.134 kN m = +3.27 kN m = +2.4 kN m

at x = 0 m at x = 1 m at x = 2 m at x = 3 m

Portion CD (x = 3 to 4 m) M x = 2.3 x − 4.5( x − 2) = 2.4 kN m = 0.2 kN m

at x = 3 m at x = 4 m

Portion DE (x = 4 to 6 m) M x = 2.3 x − 4.5( x − 2) +

1.6 ( x − 4 )2 2

= 2.3 x − 4.5( x − 2) + 0.8( x − 4)2

Portion EB (x = 6 to 7 m)

= 0.2 kN m

at x = 4 m

= −1.2 kN m

at x = 5 m

= −1 kN

at x = 6 m

M x = 2.3 x − 4.5( x − 2) + 3.2( x − 5) = −1 kN

at x = 6 m

=0 Point of contraflexure lies in portion DE. M x = 2.3 x − 4.5( x − 2) + 0.8( x − 4)2 = 0 2.3 x − 4.5 x + 2 + 0.8( x 2 − 8 x + 16) = 0 2.3 x − 4.5 x + 9 + 0.8 x 2 − 6.4 x + 12.8 = 0 x = 4.1 m. (as shown in Fig 7.58(b) Problem 7.10 A frame DCBA is subjected to a vertical load W at end A. As shown in Fig. 7.59(a), draw free body diagrams of members AB, BC and CD of the frame.

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290 Chapter 7

2L C D L W

A

B L (a)

W WL

WL

WL W C

D W

2L W WL A

B

W B W

Figure 7.59 Solution Bar AB At end A, vertical load is W. To balance W ↓ at end W ↑ at B (as shown). To balance couple WL(ccw) a couple WL(cw) at end B. Bar AC To balance W at B, in bar AB, W ↓ at B for bar BC. To balance WL(cw) at B is bar AB, couple WL(ccw) at B in bar BC. To balance WL(ccw) at B, couple WL(cw) at C. To balance W ↓ at B, reaction W ↑ at C. Bar CD To balance W ↓ at C in BC, load W ↓ at C in bar CD. To balance WL(cw) at c in BC, WL(ccw) at C in CD. At end D moment due to W, 2WL(cw) − WL(ccw) = wL(ccw) To balance WL(ccw) at D, reaction couple WL(ccw) acts at D. To balance W ↓ at C reaction W ↑ acts at D.

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Shear Force and Bending Moment Diagrams

291

Key Points to Remember Sign conventions (a) On the left side of a section of a beam, upward force is positive shear force. (b) On the left side of a section of a beam, downward force is a negative shear force. (c) On the left side of a section, clockwise moment is a positive bending moment (tends to produce concavity upwards). (d) On the left side of a section of a beam, anticlockwise moment is a negative bending moment (tends to produce convexity upwards). (i) On a particular portion of a beam, with concentrated loads on both sides of the portion, shear force remains constant and bending moment changes linearly. (ii) On a particular portion of a beam with udl, shear force changes linearly and bending moment curve is a parabolic curve. (iii) Point of contraflexure in a beam is the point where bending moment changes sign. (iv) If F is shear force, w is rate of loading and M is bending moment, then, in any portion of a beam, dF dM = − w, = F. dx dx (v) In a beam, maximum bending occurs at a section where shear force changes sign or shear force is zero.

Review Questions 1. What do you understand by positive shear force and negative shear force? 2. What is the importance of drawing SF and BM diagrams? 3. What do you understand by bending moment producing concavity upwards and bending moment producing convexity upwards? 4. What is the point of contraflexure? 5. What is the relation between F and w (rate of loading) if F is shear force? 6. A beam is uniformly supported on a flat surface and concentrated loads are applied on the beam? What type of reaction is provided by the flat surface? 7. What is the purpose of making one support of beam as pin joint and the other end as roller support? 8. What is the purpose of fixing couple provided by the wall for fixed end of cantilever?

Multiple Choice Questions 1. A 5-m-long cantilever carries a load of 10 kN at free end and 10 kN at middle. What is the bending moment at the fixed end? (a) −25 kN m (b) −50 kN m (c) −75 kN m (d) None of these 2. A 10-m-long beam which is supported over 8 m span and having equal overhang on both the sides carries loads of 8 kN each at its ends and a load of 2 kN at its centre, the point of contraflexure lies at

MTPL0259_Chapter 07.indd 291

(a) the support (b) the section (c) 2 m from each end (d) None of these 3. An 8-m-long beam which is supported over a span of 6 m carries a point load of 20 kN at the centre of the span. What is Mmax in beam? (a) 80 kN m (b) 60 kN m (c) 40 kN m (d) 30 kN m 4. In a beam, the point of contraflexure is a point where

5/23/2012 11:11:22 AM

292 Chapter 7 (a) 20 kN m (b) 30 kN m (c) 40 kN m (d) None of these 8 A 10-m-long beam is supported over 6-m span with equal overhang on both the sides. It carries point loads of 40 kN each at its ends and a point load of 80 kN at the centre; if the points of contraflexure lie at a distance x from each end, the value of x is (a) 4 m (b) 3 m (c) 2 m (d) None of these 9 A 6-m-long cantilever carries a point load of 100 kN at the free end and another point load of W at the middle of its length. If the maximum BM in the cantilever is 900 kN m, the value of W is (a) 50 kN (b) 100 kN (c) 150 kN (d) 200 kN 10 An 8-m-long beam which is simply supported at its ends carries a point load of 800 N at distance of 3 m from one end. The B M under the load is (a) 4 kN (b) 1.6 kN m (c) 1.2 kN m (d) None of these

(a) shear force is maximum (b) shear force is zero (c) bending moment changes sign (d) bending moment is maximum 5. A beam carries transverse loads. SF and BM diagrams for the beam are drawn. In the portion of the beam, where SF remains zero, bending moment. (a) maximum (b) minimum (c) constant (d) none of these 6. A 10-m-long beam which is hinged at both the ends is subjected to a clockwise moment of 40 kN m at a distance of 3 m from one end. The SF at the centre of the beam is (a) 6 kN (b) 2 kN (c) 4 kN (d) None of these 7. A 10-m-long beam carries point loads. When SF diagram is drawn, there are two rectangles of 10 kN × 2 m side; one is starting from one end and above the base and the other stating from the other end and below the base line. The BM at the centre of the beam is

Practice Problems 1. A 6-m-long beam AB and an 8-m-long beam BC are hinged at end B. Loads on the beam are shown in Fig. 7.60. Draw BM diagram of beams AB and BC.

9 kN 15 kN A

F

B

D

C

E 3

4m

2

2

6m

Figure 7.60 2. A 6-m-long beam AB which is hinged at B and roller supported at C as shown in Fig. 7.61 carries point loads of 4 kN at A and 8 kN at D and inclined load of 10 kN at E. Determine support reactions and draw SF and BM diagrams of the beam.

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Shear Force and Bending Moment Diagrams 4 kN

8 kN

293

10 kN 45

A

C 1

D

2m

E 2m

B 1

Figure 7.61 3. A 5-m-long cantilever AB carries loads as shown in Fig. 7.62. It is propped at A by a force of 4 kN. Determine support reaction and support moment. Draw SF and BM diagrams of cantilever. Point of contraflexure lies at 4.76 m from A. 4. A 6-m-long beam AB is hinged at both the ends; C is the centre of the beam. From A to C, there is downward udl of 5 kN/m and from C to B, there is upward udl of 5 kN/m as shown in Fig. 7.63. Determine support reactions and draw SF and BM diagrams. 5. A 7-m-long beam AB carries point loads of 4 kN at each end. It is hinged at D and roller supported at C as shown in Fig. 7.64. A moment of 10 kN m(ccw) acts at point E as shown in the figure. Draw SF and BM diagrams of the beam.

5 kN/m A

B C

4 kN 2m

3m

Figure 7.62 5 kN/m C A

−5 kN/m

3m

B

3m

Figure 7.63 4 kN

4 kN 10 kNm C

A

1m

D

E

2m

3m

B

1m

Figure 7.64 6. Shearing force diagram for a 5-m-long beam simply supported at A and B is shown in Fig. 7.65. Draw loading diagram of the beam. What is BM at D? 7. A 1-m-long beam AB which is supported throughout its length carries a udl of 30 kN/m over middle 0.4 m length as shown in Fig. 7.66. Draw SF and BM diagrams of the beam. Hint [ w1 × 0.4 = w2 × 1] 8. An 8-m-long floor joist is simply supported at each end and carries a distributed load of grains over the whole span. Loading distributed is given by the equation w = Ax 2 + Bx + c when w is the load intensity in kN/m at a distance of x along the beam. Load intensity is zero at ends and maximum value of 4 kN/m at the middle of span. Draw SF and BM diagrams. What are reactions at ends and Mmax?

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294 Chapter 7

0.4 m

w1 = 30 kN/m

C

+4.5 kN

D

A

B

+3 kN

w2 1m

Figure 7.66 D

O

B

A

2 kN 6 kN/m

−1.5 kN

−2

D

E

C

A

B

−3.5 kN 2m

1.5

w

1.5 m

3

Figure 7.65

1m

3m

1m

Figure 7.67

9. An 8-m-long beam AB is hinged at A and uniformly supported over DE = 3 m as shown in Fig. 7.67. Determine reaction at A and uniformly distributed reaction at DE. What is bending moment at C and D of the beam? 10. A frame ABCD is subjected to a load of 2 kN ↑ at A. Draw free body diagram of bars AB, BC and CD (Fig. 7.68). D

2m C

2 kN

1m

A

B 1m

Figure 7.68

Special Problems 1. A 14-m-long beam AB has overhang of 2 m on both of sides. Loads of 10 kN each act at ends A and B and load of 40 kN acts at the centre of the beam. Draw SF and BM diagrams. Where lie the points of contraflexure (Fig. 7.69)?

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Shear Force and Bending Moment Diagrams 10 kN

40 kN

295

10 kN E

A

D

C 2m

5m

B

5m

2m

Figure 7.69 2. A beam loaded as shown in Fig. 7.70 consists of two segments ABC and CD joined by a frictionless hinge at C, at which bending moment is zero. Draw BM diagram and determine maximum bending moment. 10 kN/m B

C

A 1m

2m

Hinge D 1m

Figure 7.70 3. A 10-m-long beam which is hinged at its ends is subjected to clockwise couples of 60 and 80 kN m at distances of 3 and 7 m from the left and support as shown in Fig. 7.71. Draw SF and BM diagrams and determine the position of the point of contraflexure if any. 4. For the cantilever as shown in Fig. 7.72, determine SF (x) and BM (x) and reaction at the support and draw SF and BM diagrams, w = kx2 as shown in the figure.

60 kNm B

A

80 kNm C

D

E 3m

4m

3m

Figure 7.71

y

w = kx 2

A

B

x

x L

Figure 7.72

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296 Chapter 7

Answers to Exercises Exercise 7.1: RA = 4 kN , RB = 5 kN , FE = 4 kN , FF = −1 kN Exercise 7.2: RA = 25 kN , RB = 10 kN , FE = +10 kN , FC = −5 kN, FD = −5 kN / −10 kN Exercise 7.3: RA = RB = 5 kN ↑, M D = + kN m, M c = +10 kN m, M E = +5 kN m Exercise 7.4: RA = RB = 10 kN ↑, MC = +10 kN m, M D = 20 kN m Exercise 7.5: FAC = −6 kN , FCD = −10 kN , FDB = −12 kN , RB = 12 kN ↑ Exercise 7.6: RA = 3 kN , RD = 7 kN , FCD = +3 kN , FAC = −5 kN , FDB = 2 kN Exercise 7.7: R B = 20 kN; FAC = −10 kN , FCB = −10 − 5 x kN , FC = −10 kN , FB = −20 kN Exercise 7.8: RC = 15.75 kN , RB = 3.25 kN , FAC = −3 kN , FCD = −3 + 15.75 − 8( x − 1), FDB = −3.25 kN Exercise 7.9: RA = −2.5 kN ↓, RB = 6.5 kN ↑, FAD = −2.5 kN , FDB = +4 kN Exercise 7.10: RB = 8 kN ↑, M B = 22 kN m(cw ), M A = 0, MC = −6 kN m, M B = 22 kN m Exercise 7.11: RA = 2 kN , RD = 10 kN; M C = +4 kN m, M D = −8 kN m (Point of contraflexure lies at 2.667 m from A). Exercise 7.12: RB = 18 kN ↑, M B + 60 kN m(cw ) bending moment MA = 0. M1 = −9 kN m M2 = − 2.4 kN m M3 = − 42 kN m, MB = − 60 kN m] Exercise 7.13: RA = 13.6 kN , RB = 9.4 kN , M1 = +10.6 kN m, M 2 = +15.2 kN m, M c = 13.8 kN m, M D = + 9.4 kN m Exercise 7.14: RB = 4 kN ↑, fixing moment at B = 20 kN m ( cw ) FAB = −4 kN , FA = −4 kN m, FB = −20 kN m Exercise 7.15: RA = 4.4 kN, RB = 5.6 kN, FAC = +4.4 kN, FCD = +4 kN, FDB = − 5.6 kN, M A = 0, M C = 8.8 kN m, MC′ = +4.8 kN m, M D = +5.6 kN m, M B = 0 No point of contraflexure. Exercise 7.16: RA =

wL wL wL wx 2 wL2 wx3 , RB = , Fx = − , Mx = x− 6 3 6 2 L. 6 6L M max =

wL2 9 3

Exercise 7.17: M A = 0, MC = M D = +12 kN m, M B = 0

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Shear Force and Bending Moment Diagrams

297

Answers to Multiple Choice Questions 1. 2. 3. 4.

(c) (d) (d) (c)

5. 6. 7. 8.

9. (b) 10. (d)

(c) (c) (a) (a)

Answers to Practice Problems 1. M A = 0, M F = +12 kN m, M B = 0, M D = −12 kN m, M E = +16.5 kN m, MC = 0 2. Rc = 11.014 kN, RBV = 8.056 kN, RBH = 7.07 kN, FAC = −4 kN, FCD = 7.014 kN, FDE = −0.986 kN, FEB = −8.056 kN, MA = 0, MC = −4 kN m MD = 10.028 kN m, ME = +8.056 kN m 3. 2.5 kN m(cw), 11 kN ↑; FAC = 4 kN, FCB = +4 − 5(x − 2) kN MA = 0, MC = +8 kN, M4 = +6 kN m, M5 = −2.5 kN m 4. RA = 7.5 kN ↑, RB = 7.5 kN ↓, MA = MB = 0 M1.5 = +5.625 kN m, M4.5 = −5.625 kN m the centre of the beam becomes point of contraflexure 5. RC = 6 kN ↑, RD = 2 kN ↑, FAC = −4 kN, FCD = +2 kN, FDB = −2 kN, MA = 0, MC = −4 kN m, ME = 0

6. w = 0.75 kN/m from O to A, w = 1 kN/m from A to B, point load 5 kN at D MD = +4.125 kN m 7. w2 = 12 kN/m, Fx ( AC ) = + w2 x ( x − 0 to 3 m ) Fx (CD ) = + w2 x − w1 ( x − 0.3) ( x = 0.3 to 0.7 m ) x2 ( x = 0 to 0.3) 2 x 2 w1 (CB ) M x = + w2 − ( x − 0.3)2 2 2 Mmax = 0.9 kN m ( AC ) M x = + w2

2 8. w = −0.25 x + 2 x; RA = RB = 10.666 kN. M max at centre = 32.66 kN m

9. RA = 4.8 2 kN , w = 2.06 kN/m M c = +5.46 kN m, M D = 1.28 kN m

M E ′ = −10 kN m, MD = −4 kN m, MB = 0

Answers to Special Problems 1. Rc = RE = 30 kN; FAC = −10 kN, FCD = +20 kN, M c = −20 kN; M D = +80 kNm, Points of coutraflexure lie at 3 m from each end 2. Mmax = −12.5 kN m at D 3. 3, 4.29 and 7 m from A, points of contraflexure

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4. Fx = Fmax

kx3 kx 4 , Mx = 3 4 kL3 kL4 ; M max = = 3 4

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8 Theory of Simple Bending CHAPTER OBJECTIVE A relationship between the bending moment on a section of a beam and the normal stress developed in a particular layer of the beam section will be derived. Students will learn about the flexure formula, a relationship between bending moment, normal stress and second moment of area. Different sections of the beam such as square, rectangular, I, T, channel and circular will be considered and stress distribution along the depth of the section will be plotted. Stresses in built up sections and in composite beams will also be analysed. Students will also learn about the stresses developed in reinforced cement concrete (RCC) beam.

Introduction Stress in various sections of the beam will be determined by using flexure formula. Beams of uniform strength (with varying cross-section) will be discussed here. There are many applications of such beam as a tapered pole subjected to transverse loads. The problems related to the position of centroid of a beam section and single second moment of the area about centroidal axes are determined. Therefore, various sections as simple section and built up sections are considered and complicated sections are also be analysed. For efficient design of beam section, the section of the beam where maximum bending moment occurs is considered and dimensions of the section are designed accordingly. To save steel and to use available material as wood for beams in hilly areas, composite beam is most suitable. Stress distribution along the depth of the section will be determined in each case.

Assumptions in Theory of Simple Bending A relationship is developed between the bending moment on a section of a beam and the stress developed in the beam section. To achieve a simple relationship between bending moment and stress following assumptions are taken:

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Theory of Simple Bending

299

1. The beam is initially straight, that is, no initial curvature in the beam. 2. The material of the beam is homogeneous, that is, it possesses same physical properties throughout the volume. 3. The material of the beam is isotropic, that is, its elastic constants do not vary throughout its volume. 4. Young’s modulus of the material in tension Et is the same as Young’s modulus in compression EC. 5. The elastic limit of the material se is not exceeded, that is, stresses in the material at any section at any layer do not exceed se, in other words when bending moment is removed from the beam, it comes back to its original shape and original dimensions. 6. Each layer of the beam is independent to contract or to extend irrespective of the layers above or below it, that is, whole beam acts as a pack of cards or sheets. 7. The beam section is symmetrical about the plane of bending, as Fig. 8.1 shows plane of bending and section symmetrical about plane of bending.

b

y

a

b

Symmetrical about yy plane

a x Plane of bending

x

G

y

Figure 8.1 8. Transverse sections which are plane before bending remain plane after bending and only the direction of plane is changed. Figure 8.2(a) shows a plane abcd in a straight beam which has not bent. After the application of bending moment, plane abcd has changed its position to a′b′c′d ′, that is, it has rotated by an angle f as shown in Fig. 8.2(b). d

d a

a

f

c

c

f

90

b

b Plane before bending (a)

Plane after bending (b)

Figure 8.2

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300 Chapter 8

Theory of Simple Bending Consider that a small length dx of the beam is subjected to bending moment M as shown in Fig. 8.3(a). After the beam bends, the element dx also bends and the shape is changed to a′b′c′d ′ such that a′b′ < ab and c′d′ > cd, that is, the upper length of the element is decreased and the lower length of the element is increased. So in the upper layer ab, there is a compressive strain and in the lower layer cd, there is a tensile strain. Strain changes from negative to positive from top layer to bottom layer as shown in the figure. Therefore, when there is a continuous change in strain from negative to positive, there must be a layer in which the strain becomes zero. Layer ef = e′f ′, the final length, No change in length, so no strain, and also no stress. This layer as ef is known as neutral layer. Original condition ab = ef = cd a′b′ < ab, e′f ′ = ef, c′d ′ > cd

Finally,

Now consider a layer gh at a distance y from the neutral layer ef, the length of the layer gh is changed to g ′h′. The beam has bent and the layer forms a portion of circular segment, subtending at an angle q at the centre of curvature, Fig. 8.3(b). From the centre of curvature, upto neutral layer, R is known as radius of curvature gh = e f = Rq

Initial layer,

g ′h′ = (R + y)q

Finally,

c

q

M

M a

R

b

Compressive

a

e

f h

y g

y

b

e

f h

g c

C

yt N

T

A

d d

c

dx

(b)

(a)

Figure 8.3

MTPL0259_Chapter 08.indd 300

yc

Tension (c)

Element of beam under bending moment

5/23/2012 11:07:55 AM

Theory of Simple Bending

e=

Strain in the layer,

301

g ′ h′ − gh gh

( R + y ) θ − Rθ Rθ y = R =

Using Hooke’s law, strain, ε = So, or,

σ Stress = E Young ’s modulus

σ y = E R σ E = y R

(8.1)

E Plane of , but E and R are constants stress, sC R bending s a y (stress is proportional to distance, y). Stress is proportional to the distance of layer yc under consideration from the neutral layer. N Within the elastic limit, stress in any layer G is directly proportional to its distance from the C A yt neutral layer. Figure 8.3(c) shows a section of the beam, st showing that due to bending, some portion of the section above neutral layer comes under (a) (b) compression and other portion below the neutral layer comes under tension. Since strain in Figure 8.4 Linear stress distribution a layer is linearly proportional to its distance from neutral layer. Fig. 8.4 shows the linear distribution of bending stress in a general section of a beam. sc is the maximum compressive stress developed in one extreme layer and st is the maximum tensile stress in another extreme layer of the beam section. s=y

Neutral Axis Consider any section of the beam as shown in Fig. 8.5, stress in layer ab E or, s=y R Force in a small element of area dA, dP = s dA + yc yE Total force, on section, d P = σ d A = ∫ ∫ ∫− y R dA t +y

=

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E c ydA R −∫yt

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302 Chapter 8

We have not applied any force, but we have applied only moment M.

∫ dP = 0 ,

So

∫

b

+y

or

+ yc

or

sc

E c ydA = 0, R −∫yt

s

Pc

ydA = 0, or the first moment of area is zero.

– dA

a

h

N

− yt

dy y

b A

G

Pt

Note that the first moment of any area is zero only about st its centroidal axis, therefore, neutral axis of the beam passes through the centroid G of the section. Figure 8.5 NA = neutral axis, Moreover, Pc = net force on compression side of the h = arm of couple section. Pt = net force on tension side of section. Pc = Pt, both these equal and opposite forces constitute a couple of arm h as shown. Pc × h = Pt × h = moment applied on the beam section.

Moment of Resistance We have considered a small element of area dA at a distance of y from neutral layer neutral axis. Force on the area,

dP = s·dA =y

E dA R

Moment of force dP, about neutral layer dMr = ydP = Total moment of resistance,

Mr = ∫

E 2 y dA R

E 2 y dA R y

=

E c 2 y dA R −∫yt

yc

But,

∫

y 2 dA = I NA , second moment of area about the neutral axis,

− yt

or the moment of resistance, Mr = M = applied moment, or

M E = I NA R

(8.2)

From Eqs (1) and (2) M σ E = = − flexure formula I NA y R The reader should carefully note the significance of each term in the flexure formula, where M = applied bending moment INA = Ixx = moment of inertia of a section about the neutral axis

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Theory of Simple Bending

303

s = stress in a layer under consideration y = distance of layer from the neutral axis E = Young’s modulus of the material R = radius of curvature Please note that the term E / R will be used only when the bending moment remains constant and there is a constant radius of curvature. M σc σt Now = = , taking into account the bending stresses in the section of the beam I NA yc yc I I M = sc NA = σ c Z c if Z c = NA yc yc where Zc = section modulus of the beam in compression Zt = section modulus of the beam in tension For some of the common sections of the beam, Table 8.1 gives section modulus. Example 8.1 A steel strip, 60 mm wide and 20 mm thick, is bent into an arc of radius 30 m. What is the maximum stress developed in the steel strip (Fig. 8.6)? E = 2 × 105 N/mm2

Steep strip 10 mm

Solution E = 2 × 105 N/mm2 Radius of curvature, R = 30 m = 30 × 103 mm For maximum stress y = ±10 mm

σ= =

Table 8.1

N

A 60 mm = b

Figure 8.6

y ×E R

20 mm = t

Example 8.1

10 × 2 × 105 = 66.67 N /mm 2 30 × 103

Section modulus of different sections

Section

INA

y1

y2

Z1

Z2

Circular

πd 4 64

d 2

d 2

πd3 32

πd3 32

bd 3 12

d 2

d 2

bd 2 6

bd 2 6

bh3 36

h 3

2h 3

bh 2 12

bh 2 24

d b

Rectangular

d

h

Triangular b

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304 Chapter 8

Example 8.2 A beam is of circular section of diameter d mm. Bending moment on a section is 2 × 106 Nmm. What is the diameter d, if the maximum stress in the section does not exceed 50 N/mm2? Solution sc = st = 50 N/mm2 Zc = Zt =

πd3 32

M = sc Zc = st Zt 2 × 106 = 50 × d3 =

πd3 32

2 × 106 × 32 = 0.4074 × 106 π 50

Diameter, d = 74.13 mm. Example 8.3 A beam section in an isosceles triangle has base 40 mm and altitude 60 mm. Bending moment at the section is 0.4 kNm, which produces tension in the base. Draw stress distribution along the altitude of the section (Fig. 8.7). C

Solution Triangular section

yt = 40 mm

b = 40 mm h = 60 mm yt =

80 = 20 mm 3

yc = 60 − 20 = 40 mm I NA

bh3 40 × 603 = = = 24 × 104 mm 4 36 36

N yt = 20 mm D

Figure 8.7

A G

40 mm

E

Example 8.3

yc = 40 mm M = 0.4 kN m = 0.4 × 106 N/mm

σc =

M 0.4 × 106 40 × yc = × = 66.67 N/mm 2 I NA 24 × 104 1

yt = 20 mm

σt =

MTPL0259_Chapter 08.indd 304

0.4 × 106 × 20 M × yt = = 33.33 N/mm 2 I NA 24 × 106

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Theory of Simple Bending

305

Exercise 8.1 A brass strip, 80 mm wide and 30 mm thick, is bent into an arc of radius 60 m. What is the maximum stress developed in strip? If E = 1 × 105 N/mm2 Exercise 8.2 A beam of rectangular section b × d is subjected to a bending moment, M = 1 × 106 N mm at a particular section. If the maximum stress in the beam section is not to exceed 80 MPa, determine b and d. If d = 2b Exercise 8.3 The section of a beam is an equilateral triangle of side a = 60 mm. How much bending moment can be applied on the section? If smax = 60 N/mm2 h    Hint: h = 0.866a, CG lies at 3 from base 

Perpendicular Axes and Parallel Axes Theorems In Fig. 8.8, an area A is shown with the centroid located at G and, Gx and Gy are centroidal axes. Polar moment of inertia

Y

Y

IGG = Ixx + Iyy Area, A = Second moment of area about xx axis + Second hx moment of area about yy axis. This is called perpendicular axes theorem. G X X Parallel axes theorem hy Again consider Fig. 8.8, an area A has CG located at G X  X O centroidal axes XX and YY are passing through G. Axis x′x′ is parallel to x-x axis but at a vertical distance of hy shown. Y Y Moment of inertia, I x ′x ′ = I xx + Ahy2 Similarly axis y′y′ is parallel to yy axis but at a horizontal Figure 8.8 Area with centroidal distance of hx as shown. co-ordinates Moment of inertia, Iy′y′ = Iyy + Ahx2 The parallel axes theorem is extensively used in this chapter to determine the moment of inertia (second moment of area) about the neutral layer of a beam section.

Symmetrical I-Section Symmetrical I-section is most commonly used as structural members, utilized in building construction. Figure 8.9 shows a symmetrical section with overall breadth, B and depth, D. The thickness of the flange is tf and thickness of the web is tw. The dimension d is the depth of the web. Generally in a rolled section, tw < tf , the centroid of the section G lies at the centre of the web. Neutral axis or NA, or x-x axis passes through G as shown in the figure.

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306 Chapter 8 y

Moment of inertia, I NA

sc

BD 3 ( B − tw )d 3 = − 12 12

Rolled steel sections are available in the market, IS specification identifies I-section and gives values of B, D, Ixx, Iyy, Zx and Zy in tables. As an example, ISLB 400 stands for D = 400 mm, B = 165 mm (Indian standard), tf = 12.5 mm, tw = 8.0 mm, Ixx = 1,930.3 × 104 mm4, Iyy = 716.4 × 104 mm4, Zx = 965.3 × 103 mm3, Zy = 88.8 × 103 mm3. Stress distribution in I-section for the positive bending moment is shown in Fig. 8.9.

tf

N D

x G

A x



d d – 2

tw tf

y

d – 2

st

B I-Section

Example 8.4 A beam of I-section 300 × 140 mm Figure 8.9 Stress distribution in has a flange of 15 mm thickness and a web of 10 mm I-section thickness. Determine section modulus of I-section. What will be the maximum stress developed in I-section for a bending moment of 36 kNm. Solution D = 300 mm B = 140 mm tf = 15 mm tw = 10 mm ymax = Moment of inertia,

I NA =

300 = 150 mm 2

140 × 3003 (140 − 10) (300 − 2 × 15)3 − 12 12

= 315 × 106 − 213.2325 × 106 = 101.7675 × 106 mm4 Section modulus,

Z=

I NA 101.7675 × 106 = = 678.45 × 103 mm3 150 ymax

M = 36 kN m = 36 × 106 Nmm Maximum stress,

σ max =

M 36 × 106 = Z 678.45 × 103

= 53.06 N/mm2 Exercise 8.4 A beam of I-section 300 mm × 120 mm has a flange of 20 mm thickness and web of 10 mm thickness. What is its section modulus? What will be the maximum stress developed for a bending moment of 30 kNm

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Theory of Simple Bending

T-Section

y

Rolled steel sections are also available in T-section. These are used in applications such as door frames. A T-section is shown in Fig. 8.10 and the section is symmetrical about yy axis only. Let us determine the location of its controid along yy axis (about edge aa). d Bt f ( D − 0.5t f ) + dtw 2 y1 = Bt f + dtw

B

tf

y2

B(t f )3 12

Section modulus, Z1 =

X

Flange

X

N

G

A d D

y1 tw

where B = breadth of the section D = overall depth d = depth of the web tf = thickness of the flange tw = thickness of the web y2 = D − y1 NA passes through G, or axis xx is passing through G I NA = I xx =

307

Web a

y

a

Figure 8.10

2

+ Bt f ( y2 − 0.5t f ) 2 +

tw d 3 d  + tw d  y1 −  (Using parallel axes theorem)  12 2

I xx I , Z 2 = xx y1 y2

Example 8.5 Dimensions of a T-section are: B = 110 mm. D = 150 mm, tf = 20 mm and tw = 10 mm. Determine Z1, Z2. A cantilever 4 m long is of this T-section. Determine the point load W at the free end so that the maximum stress in the section does not exceed 90 MPa. Solution B = 110 mm, tf = 20 mm, D = 150 mm, d = 150 − 20 = 130 mm, tw = 10 mm y1 =

110 × 20 × (150 − 10) + 130 × 10(65) 110 × 20 + 130 × 10

y1 =

3, 08, 000 + 84, 500 3, 92, 500 = = 112.14 mm 2, 200 + 1, 300 3, 500

y2 = 150 − 112.14 = 33.86 mm, I NA =

110 × 203 10 × 1303 + 110 × 20 (37.86 − 10) 2 + + 1, 300 (112.14 − 65) 2 12 12 (Using parallel axes theorem)

= 73,333.33 + 17,07,595.12 + 18,30,833.33 + 28,88,833.48 = 65,00,595.26 mm4 = 6.5 × 106 mm4

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308 Chapter 8

Cantilever

120 mm

Length, L = 4 m = 4,000 mm Point load at free end = W N Maximum bending moment = 4,000 W N mm.

20 mm

smax = 90 MPa

200 mm

y1 = 112.14

12 mm

Maximum stress will occur at lower edge as y1 > y2. 90 4, 000 W = 112.14 6.5 × 106

Figure 8.11

Exercise 8.5

6

Point load,

W=

6.5 ×10 × 90 =1,304 N 112.14 × 4,000

= 1.30 kN Exercise 8.5 A T-section with dimensions, flange 120 mm × 20 mm and web 180 mm × 12 mm, is shown in Fig. 8.11. It is subjected to a positive bending moment of 5 kN m. What are the stresses developed at extreme edges of the section? [Hint: Flange in compression, web below NA in tension]

Channel Section A channel section of width B and depth D is shown in Fig. 8.12. This section is symmetrical about yy axis but it is not symmetrical about xx axis. Therefore, xx axis cannot be the plane of bending and only yy axis is the plane of bending. [If load is applied on a beam of channel section about xx axis then, unsymmetrical bending will take place in beam.] CG of the section bar at a distance y1 from load edge MN along yy axis, t  D −t Bt   + ( D − t )b × 2  t +   2  2  y1 = 2 t B + ( D − t ) × 2b 2

y2

+

y1

Bt 3 t  + Bt  y1 −   12 2

L

b

P

Flunges

G

y2 = D − y1 Moment of inertia, Ixx =

Y

b

X

D X

t

Web

2

2b   D −t × ( D − t )3 + 2b( D − t )  y1 −  t +   12 2   

M

Y

N

B 2

YY-Plane of bending

Figure 8.12

Using parallel axes theorem I I Section modulus, Z 1 = x x , Z 2 = x x y1 y2

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Theory of Simple Bending

309

Example 8.6 A channel section has D = 50 mm, B = 100 mm, b = 5 mm, t = 2 mm. Locate CG of channel section and determine Ixx. What is the maximum bending moment that can be applied on the beam section, if smax ≤ 60 N/mm2? Solution y1 = =

100 × 2 ×1 + 48 × 5 × 2 × (2 + 24) 200 + 2 × 48 × 5 200 + 480 × 26 200 + 12, 480 12, 680 = = = 18.65 mm 200 + 480 680 680

y2 = 50 − 18.65 = 31.35 mm. Moment of inertia, I xx =

2 × 5 × 483 100 × 23 + 200 (18.65 − 1) 2 + + 2 × 5 × 48(31.35 − 24) 2 12 12

= 66.67 + 62,304.5 + 92,160 + 480 (54.0225) = 66.67 + 62,304.5 + 92,160 + 25,930.8 = 180.462 mm4 = 180.462 × 103 mm4 Z1 =

I x x 180.462 × 103 = = 9.676 × 103 mm 3 18.65 y1

Z2 =

I x x 18.402 × 103 = = 5.756 × 103 mm 3 31.35 y2

Maximum stress, s = 60 N/mm Maximum bending moment, M = s × Z2 as Z2 < Z1 = 60 × 5.756 × 103 N mm = 345.36 × 103 N mm = 345.36 N m 2

10

10

D = 60 8 mm B

Figure 8.13

120 mm

Exercise 8.6

Exercise 8.6 Figure 8.13 shows a channel section of given dimensions: B = 120 mm and D = 60 mm. Locate G of the section and calculate Ixx. If a beam of the channel section is used, what is the maximum bending moment that can be applied and also the stress should not exceed 50 N/mm2.

Unequal I-Section This section is symmetrical only about one axis, YY, as shown in Fig. 8.14. There are some applications of unequal I-section as railway tracks are made of rolled unequal I-section. Dimensions of unequal I-section has flanges B1 × t1, B2 × t2 and web d × t3 as shown. Location of CG of section from edge NP y1 =

MTPL0259_Chapter 08.indd 309

B1t1 ( D − 0.5t1 ) + dt3 (t2 + 0.5d ) + B2t2 × 0.5t2 B1t1 + dt3 + B2t2

5/23/2012 11:08:12 AM

310 Chapter 8

y2 = D − y1 Moment of inertia, I xx =

B1 Y 3

3 11

td Bt + B1t1 ( y2 − 0.5t1 ) 2 + 3 12 12

L

2

Bt3 d   + td3  t2 + − y1  + 2 2 + B2 t2 ( y1 − 0.5t2 ) 2   2 12 Section modulus,

Z1 =

I xx I , Z 2 = xx y1 y2

Example 8.7 A CI beam of unequal I-section with top flange 150 mm × 10 mm, bottom flange 200 mm × 15 mm and web 275 mm × 10 mm is supported as a cantilever of length 3 m. What load can be applied at the free end of the cantilever if the tensile stress in the section is limited to 80 N/mm2, the top flange of the beam comes under tension?

t1

M a1

y2

a3

d G

D

x

X a2

y1

t3 t2

y

N

p

B2

Figure 8.14

Unequal I-section

Solution B1 = 150 mm, t1 = 10 mm B2 = 200 mm, t2 = 15 mm d = 275 mm, t3 = 10 mm D = 275 + 10 + 15 = 300 mm Location of G

Distance,

275   150 × 10 (300 − 5) + 10 × 275 15 +  + 200 × 15 × 7.5  2  y1 = 1, 500 + 2, 750 + 3, 000 =

4, 42, 500 + 4,19, 375 + 22, 500 8, 84, 375 = = 122 mm 7, 250 7, 250

y 2 = 300 − 122 = 178 mm I xx =

2 150 × 103 2753 × 10 275   + 1, 500(178 − 5)2 + + 2, 750 15 + − 122   12 12 2

200 × 153 + 3, 000(122 − 7.5) 2 12 = 12,500 + 44,893,500 + 17,330,729 + 2,558,187.5 + 50,250 + 39,330,750 +

= 104,181,916.5 mm4 = 104.182 × 106 mm4 Z1 = Z2 =

MTPL0259_Chapter 08.indd 310

104.182 × 106 = 8, 53, 950 mm3 122 104.182 × 106 = 5, 85, 292 mm3 178 Z2 < Z1

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Theory of Simple Bending

311

Allowable stress, s = 80 N/mm2, moreover upper flange is in tension. M = s Z2 = 80 × 5,85,292 = 4,68,23,360 Nmm = 46.823 kNm Cantilever, L = 3 m Maximum bending moment, WL = 3 W kNm W =

46.823 = 15.607 kN 3

Exercise 8.7 The cross-section of a cast iron column is an I-section with top flange 50 mm × 15 mm, web 255 mm × 12 mm and bottom flange 100 mm × 30 mm. The load being in the plane of the web, the upper portion of the section is in compression. If the allowable stresses are 60 N/mm2 in tension and 150 N/mm2 in compression, determine the moment of resistance of the section.

Modulus of Rupture M σ E = = is derived on the assumption that stress in a section does not exceed the proI XX y R portional limit stress. Moreover, strain in any layer is proportional to its distance from neutral layer. However, even though stress in any layer exceeds the proportional limit stress, the strain remains proportional to the distance of the layer from neutral layer. Then stress distribution along the depth of the beam becomes nonlinear and if the bending moment is applied, the stress in the extreme layer reaches the ultimate stress, and the beam is supposed to have failed. Say, Mult = ultimate bending moment determined experimentally. Then, 6 M ult Modulus of rupture = for a rectangular section beam, bd 2 where b = breadth of the rectangular section, d = depth of the rectangular section. The flexure formula is not valid if the stresses go beyond proportional limit stress. The theoretical value of the fracture stress obtained by flexure formula using ultimate bending moment is known as the modulus of rupture. Flexure formula

Example 8.8 A 150 mm × 150 mm pine wood beam was supported at the ends on a 4.5 m span and loaded at third point. The beam failed when a 8 kN load was placed at 1.5 m from each end. Find the modulus of rupture. Solution Beam is of square section, side = 150 mm = b Section modulus Z = =

MTPL0259_Chapter 08.indd 311

b3 6 1503 = 56.25 × 10 4 mm 3 6

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312 Chapter 8 8 kN

Figure 8.15 shows the loading diagram of the beam. Reaction, RA = RB = 8 kN Portion CD of the beam is under pure bending.

A

Mc = Md = 8 × 1.5 = 12 kN m.

D

B

C

Mult = 12 × 106 N mm Modulus of rupture, sm =

8 kN

1.5

1.5

1.5 m

RA

M ult 12 × 106 = Z 56.25 × 104

RB

Figure 8.15

Example 8.8

= 21.33 N/mm

2

Exercise 8.8 A wooden beam of square section 125 mm × 125 mm is supported over a span of 3 m. A central load W is continuously increased till it breaks at W = 10.85 kN. Determine the modulus of rupture of the beam.

Built Up Sections To increase section modulus so as to improve the capacity of a beam to withstand high bending moment, a compound section is built up by using standard rolled sections as I, T, channel and angle (L) section. To reduce the effect of stress concentration at inner corners and to eliminate sharp outer edges, the steel section of different shapes is rolled out by having fillet radius at inner corners and rounded outer edges. I, channel and L-section with rounded edges and fillet radius are shown in Fig. 8.16. Rounded edge

Rounded edge

(Fillet radius)

fillet

R Rounded edge L-section

Channel section

I-Section

Figure 8.16 Rolled steel sections Figure 8.17 shows built up sections: (1) a box section, consisting of four angle sections with four side plates and (2) two channels connected to two flange plates. Built up sections are made with the help of riveting or welding of plates with standard section. CG of a built up section is located and moments of inertia Ixx and Iyy are calculated. Example 8.9 Two channel sections, 300 mm × 100 mm are placed back to back and the flanges are joined by plates 200 mm × 10 mm (Fig. 8.18). The compound section forms a simply supported beam 2 m long and carries a uniformly distributed load of 100 kN/m run. Determine the maximum stress developed in the section. The properties of a channel section are area = 4,219 mm2, Ix ′x ′ = 60 × 66 × 104 mm4 and Iy ′y ′ = 346.9 × 104 mm4.

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Theory of Simple Bending y

y

G

G

313

Flange plate

Channel

x

x

x

x

y y (a)

Figure 8.17

(b)

(a) Box section 4-angle sections with side plates and (b) two channel sections with two flange plates built up section

Solution Section is symmetrical about xx axis, even after adding the flange plates, 200 mm × 10 mm, each on top and bottom of channel sections. Moment of inertia, Ixx = 2 Ix ′x ′ +2×

200 × 103 + 2 × 200 × 10 (150 + 5) 2 12

= 2 × 6, 066 × 10 4 + 3.333 × 10 4 + 4, 000 × 1552 = (12,132 + 3.333) × 10 4 + 9, 610 × 10 4 = (12,135.33 + 9, 610) × 104 = 21,745.33 × 104 mm4 ymax = 160 mm Z =

21, 745.33 × 10 4 = 135.9 × 10 4 mm 3 160

10 Channel 150

Simply supported beam l=2m w = 100 kN/m 2

x

wl 100 × 2 = = 50 kN m 8 8 = 50 × 106 N mm Mmax = smaxZ

Maximum bending moment =

σ max

MTPL0259_Chapter 08.indd 313

x

2

50 × 106 = = 36.79 N /mm 2 4 135.9 × 10

150

10 200

Figure 8.18

Example 8.9

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314 Chapter 8

Exercise 8.9 Two channel sections ISJC 100 placed back to back at a distance of 30 mm are joined by two plates, 120 mm × 15 mm at the top and bottom flanges. Determine the moment of inertia Ixx of the built up section. The properties of the channel section are: area = 741 mm2, depth = 100 mm, flange width = 45 mm and Ix ′x ′ = 123.8 × 104 mm4.

Beams of Uniform Strength

b

To achieve a beam of uniform strength for the purpose of saving in material cost of the beam, the term M/Z, that is, ratio of bending moment N and section modulus Z is to be maintained constant either (a) by varying the breadth of the beam or (b) by varying the depth of the beam. Maximum stress in a section under bending is developed at extreme edges of the section. This maximum skin stress is maintained constant. Figure 8.19 In the formula

A

d

Rectangular section

M σ = I y

σ=

or

M  y  M Bending moment ×y =M   = = I Z I Section modulus

For a rectangular section, Z = bd 2 / 6 , as shown in Fig. 8.19.

σ=

So,

M M  6 ×6 =   2 2  bd bd

Considering the example of a simple supported beam with a concentrated load at the centre, Fig. 8.20 shows the BM diagram in portion AB of the beam, the bending moment at distance x from end A. W Mx = x (varies linearly) X W 2 If breadth is also varied linearly from zero at A to bx at distance x, then W  x σ   σ = 2  2 bx d  

Substituting bx = B

σ=

MTPL0259_Chapter 08.indd 314

C L — 2

w — 2

Y

BM diagram

bx

x 2B ×2 = (x ) L L

W x L 6 1.5WL × × = 2 2B x d 2 Bd 2

w — 2

B L — 2

wL — 4

W x 2 = constant bx

or,

A

(8.3)

Figure 8.20

B

Beam with varying width

5/23/2012 11:08:19 AM

Theory of Simple Bending

A constant term s is the maximum tensile/compressive stress at extreme edges of a rectangular section. Similarly in the case of a cantilever, with a concentrated load at the free end, as shown in Fig. 8.21. Bending moment at a distance x from free end A,

W y

Fixed end B

= bx

If,

x A

y

L

Mx = Wx Section modulus

315

d2 6

WL

σ=

Wx 6 × bx d 2

bx =

x ×B L

σ=

Wx 6 6W L ×L × 2 = Bx d Bd 2

bx

B x

(8.4)

Figure 8.21

Cantilever with varying width

where B is width at fixed end, d depth remains constant throughout the length of the beam. Variation of Depth (Breadth remains constant) Simply supported beam Mx =

W x 2

d x2 , breadth B remains the same but depth varies as shown in Fig. 8.22. 6 L — Say depth at middle is d. Zx = B

2

M 3Wx W x6 σ= x = = 2 2 Bd x Bd x 2 Zx At the centre,

x=

L , dx = d 2

3WL 3Wx σ= = 2 Bd 2 Bd x2 or

or

MTPL0259_Chapter 08.indd 315

d=

3WL 2 3WL ,d = 2 Bσ 2 Bσ

dx =

3Wx 2 3Wx , dx = Bσ Bσ

d x2 2 x = d2 L

x A

B dx

Figure 8.22

d

C

Variation of depth

(8.5)

5/23/2012 11:08:22 AM

316 Chapter 8

dx = d

2x , L

x

B

A

= 0 at x = 0 d L at x = 4 2

=

W

L

L as shown in Fig. 8.22. 2 Taking the case of a cantilever with a point load at free end, the depth can be varied along the length as shown in Fig. 8.23.

dx

d

= d at x =

Figure 8.23 Variation of depth, breadth remains constant

dx x 2 = d L2 x2 ×d (8.5) L2 Cantilever with a uniformly distributed load Considering a cantilever of length L is subjected to uniformly distributed load w per unit length. Say s is the uniform extreme stress in a rectangular section beam. Bending moment at fixed end dx =

MB =

wL2 2

Bending moment at any section at a distance x =

wx 2 2

Consider depth d remains constant, bx = breadth at the section B = breadth at end Then,

or

wL 2 wx 2 6 6 × = × = σ which is a constant 2 BD bx D 2 2 2

x w

L2 x 2 = B bx

B

A L

x (a parabolic variation) L2 2

or

bx = B

= 0 at x = 0 B = 4 =Bx

MTPL0259_Chapter 08.indd 316

L x= 2 = L as shown in Fig. 8.24

x B d

bx

bx

B at B

Figure 8.24

D at x

Variable width of cantilever

5/23/2012 11:08:24 AM

Theory of Simple Bending

If breadth B remains constant and depth varies dx to D then,

x

w

wL 2 wx 2 6 6 × = × 2 BD Bd x2 2 2

A L

dx x = , a linear variation of depth (Fig 8.25). D L

or

317

dx D

Example 8.10 A beam of uniform strength and variable rectangular section is simply supported over a span of 4 m. It carries a uniformly distributed load of 8 kN/m. The uniform strength is 80 N/mm2. (a) Determine the depth at a distance of 1 m from one Figure 8.25 Variation of depth as breadth remains end if breadth remains the same. (b) Find the breadth at the centre constant of the span if depth is constant throughout the length of the beam and equal to 120 mm (Fig. 8.26). Solution (a) s = 80 N/mm2 Figure 8.26 shows a beam with 4 m span and w = 8 kN/m RA = RB = reactions

x

8 kN/m

A

B

C 2m

2m

16 kN

16 kN

= 16 kN 70.7

Bending moment of centre M max =

81.65

w L2 8 × 42 = 8 8 Figure 8.26

= 16 kN m = 16 × 106 kN m

Example 8.10

Breadth, B = 180 mm, constant 6M = σ = 80 BD 2 6 × 16 × 106 = 80 180 × D 2 D=

96 × 106 = 81.65 mm , 180 × 80

Bending moment at distance of 1 m for A, M 1 = 16 × 1 − Depth, d = ? So

6 × 12 × 106 = 80 180 × d 2 d=

MTPL0259_Chapter 08.indd 317

8 × 12 = 12 kN m = 12 × 106 N mm 2

72 × 106 = 70.7 mm 80 × 180

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318 Chapter 8

Variation of depth is shown in Fig. 8.26. (b) Depth is a constant, D = 120 mm Breadth is a variable At the centre, M = 16 × 106 N mm Say, breadth is B 6M 1 = 80 BD 2 6 × 16 × 106 = 80 B × 120 2 B=

or Breadth,

96 × 106 = 83.33 mm 120 2 × 80

Exercise 8.10 A cantilever of length 2 m carries a udl of 20 kN/m. The breadth of the section remains constant and is equal to 100 mm. Determine the depth of the section at the middle of the length of the cantilever and at fixed end if maximum stress remains the same throughout and is equal to 120 N/mm2.

Composite Beams A composite beam is made by joining two or more beams of different materials. As an example, a steel beam is flitched in between two wooden beams. Bending moment acting on the composite beam is shared by steel as well as wooden beams. Steel beam is stronger than the wood, it can sustain higher bending stresses and therefore can share more bending moment. Figure 8.27 shows a steel beam of breadth b and depth D is sandwiched between two wooden beams of breadth B and depth D each as shown. If the section is subjected to a bending moment M, Then, M = Ms + Mw Bending moment shared by the steel beam + bending moment shared by two wooden beams. Since it is a composite beam, strain at any Steel Wood ec distance from neutral layer is same both in steel beam as well as in wooden beam. D Say ec = maximum compressive strain in — 2 wooden beam. D = strain in steel beam. N A Similarly et in the maximum tensile strain in D — both steel and wooden beam 2 et in steel = ew in wood. Using Hooke’s law et

σ ts σ tw = Es Ew

MTPL0259_Chapter 08.indd 318

B

b

Figure 8.27

B

Composite beam

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Theory of Simple Bending

319

Say, ss = maximum stress developed in steel sw = maximum stress developed in wood

σs σw = Es Ew σs =

Es σw Ew

Es Young’s modulus of steel = =m Ew Young’s modulus of wood

If

= modular ratio ss = mss So,

MS = σS

bD 2 6

Mw = σw

BD 2 × 2, 6

(8.6)

Note that these are rectangular section wooden beams M = σs

bD 2 BD 2 + 2σ w 6 6

but, σ s = mσ w

(8.7)

σs (8.8) m ss, sw are extreme stresses in steel and wooden beams. Let us take another composite beam; the depth of steel section is less than the depth of wooden section. Figure 8.28 shows a steel beam of section b × d sandwiched between the wooden beams of section B × D each, neutral layer NL passes through the centre as shown. Say, strain in wood at point a is ew. Steel σw =

or

d Strain in the steel at point b, es = ew × D Strain in any layer is directly proportional to its distance from the neutral layer. Using Hooke’s law

b D — 2

= mσ w

MTPL0259_Chapter 08.indd 319

d — 2

N

L

d — 2

σs σw d = × Es Ew D E d σs = s σw Ew D

Wood

a

B

Figure 8.28

b

D

B

Composite beam

d , where m is modular ratio. D

5/23/2012 11:08:29 AM

320 Chapter 8

Example 8.11 Let us take B = 25 mm, b = 10 mm, d = 48 mm and D = 60 mm. If the maximum stress in steel and wood sections is not to exceed 120 and 10 MPa, respectively, calculate the safe bending E moment on the section. Ratio of s = 16 Ew Solution Let us take sw = 10 MPa

σs =

Es 10 × 48 d × σ w = 16 × 60 Ew D

= 128 MPa >120 MPa (allowable stress for steel) So we will take ss = 120 N/mm2 120 = 16 × σ w ×

σw =

48 60

120 = 9.375 MPa < 10 MPa (allowable stress for wood) 12.8

Now b = 10 mm, B = 50 mm M = Ms + Mw bd 2 BD 2 + σw2 6 6 2 48 60 2 = 120 × 10 × + 9.375 × 2 × 25 × 6 6

= σS

= 4,60,800 + 281,250 = 7,42,050 N mm = 742.05 N m There is another type of composite beam in which steel reinforcement is provided at top and bottom of wooden beam as shown in Fig. 8.29. The section can be converted into an equivalent steel section with top and bottom flange B × t B and also web of steel with width , where m is modular ratio, Es /Ew. m t

Steel

D — 2

Wood

N

A

≡

D

t B

t

D

A

N

B b=— m

t B

Figure 8.29 Equivalent steel section

MTPL0259_Chapter 08.indd 320

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Theory of Simple Bending

321

Section is symmetrical about the centre, NA passes through the centre of the web. B ( D + 2t )3  B  D3 −B−   12 m  12 D ymax = + t 2 Section modulus, Z = INA /ymax Section of the composite beam can also be converted into equivalent wooden section. In that case, steel plates are converted into equivalent wooden plates of width mB and thickness t remains the same. If, ss = allowable stress in steel M = allowable bending moment = ss Z, where Z is section modulus of equivalent steel section. I NA =

Example 8.12 A wooden beam of breadth 60 mm and depth 80 mm is reinforced with steel plates 40 mm × 10 mm at the top and 60 mm × 10 mm at the bottom. Modular ratio = 15. If allowable stress in steel is 105 N/mm2, determine the safe moment of the section of the beam as shown in Fig. 8.30. Solution Equivalent steel section Steel plate 40 mm × 10 mm at top, Web 80 mm × 4 mm, where 4 = 60/m = 60/15 Steel plate 60 mm × 10 mm at bottom Location of G

40 mm

40 mm

10 mm

10 mm

y2 = 56.82

80 mm

80 mm

N

A 4 mm

y1 = 43.18 B

10 mm 60 mm

10 mm 60 mm

Figure 8.30 Equivalent steel section

MTPL0259_Chapter 08.indd 321

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322 Chapter 8

y1 =

60 × 10 × 5 + 80 × 4 × (10 + 40 ) + 40 × 10 (90 + 5) 3, 000 + 16, 000 + 38, 000 = = 43.18 mm 600 + 320 + 400 1, 320 Wood Steel

y2 = 100 − 43.18 = 56.82 mm, Moment of inertia I NA =

60 × 103 4 × 803 + 600 (43.18 − 5) 2 + + 320 (50 − 43.18) 2 12 12 +

300

250

40 × 103 + 400 (56.82 − 5) 2 12

= 5,000 + 8,74,627.4 + 1,70,666.67 + 14,884 + 3,333.33 + 1,074,125 = 2,142,636.4 mm4 Maximum stress will occur at upper edge as y2 > y1 Z2 =

I NA = 37709.2 mm3 56.82

200 10

10

Figure 8.31 a b

Exercise 8.11 20

ss = 105 N/ mm2 300

Allowable bending moment, M = Z2ss = 37,709.2 × 105 = 3,959,465.7 N mm = 3.96 kN m

20 250

Figure 8.32

Exercise 8.12

Exercise 8.11 A wooden beam of dimension 300 mm × 200 mm is strengthened by two steel plates of 250 mm × 10 mm each as shown in Fig. 8.31. Determine the allowable bending moment if the allowable stress in steel is 150 MPa and in wood is 8 MPa. Es = 200 GPa, Ew = 8 GPa Exercise 8.12 A composite beam consists of a wooden beam 300 mm deep × 200 mm wide with two steel plates, 250 mm × 20 mm each on top and bottom faces as shown in Fig. 8.32. A bending moment of 30 kN m is applied on the section, determine the maximum stress developed in the steel and wood. Given

Es = 20 Ew

Reinforced Cement Concrete Beam Concrete is a common building material used in the construction of columns and beams which come under tension and compression due to bending moment acting on these members. While concrete has very useful strength in compression but it is very weak in tension and so minute cracks may develop in concrete due to tensile stresses. Therefore, tension side of the beam is reinforced with steel bars.

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Theory of Simple Bending

323

Concrete and steel make a very good composite material. During setting, concrete contracts and grips the steel reinforcement. Moreover, the coefficient of thermal expansion of steel and concrete for most common mix is 1:2:4, more or less is the same. Ratio 1:2:4 stands for 1 part of cement, 2 parts of sand and 4 parts of aggregate by volume. To develop a theory for RCC structure, following assumptions are made: 1. 2. 3. 4. 5.

Concrete is effective only in compression and stress in concrete on the tension side of the beam is zero. Transverse sections which are plane before bending remain plane after bending. Strain in any layer is proportional to its distance from neutral layer. Stress is proportional to strain in concrete also. There is a constant ratio between modulus of elasticity of steel and modulus of elasticity of concrete.

The allowable stresses for concrete and value of Young’s modulus of concrete, Ec depend upon the type of the concrete mix.

RCC Beams (Rectangular Section) Figure 8.33 shows a rectangular beam of breadth B, depth D (of reinforcement from compression face). Let H be the distance of neutral axis from the compression face, and maximum stresses developed in steel and concrete are ss and sc, respectively, as shown in Fig. 8.33. Strain in any longer is proportional to its distance from neutral layer. ec aH et a (D − H) εc H = εt D − H Using Hooke’s Law,

σ c Es H × = Ec σ s ( D − H ) σ s Es ( D − H ) = × σ c Ec H Compression face

ec

N

sc P 2H – 3

H D

(8.9)

A

p et

Figure 8.33

MTPL0259_Chapter 08.indd 323

ss

Stresses and strains in RCC beam

5/23/2012 11:08:34 AM

324 Chapter 8

where Es = Young’s modulus of steels Ec = Young’s modulus of concrete. ES = m, modular ratio Ec So,

σs D−H = m  H  σc

Consider that the beam is under the action of pure bending, and then resultant force P in steel is the same as resultant force P in concrete. or

P=

σc × Ac = σ s As 2

σc × BH = σ s As 2

(8.10)

where Ac = area of concrete above neutral layer AC = BH As = area of steel reinforcement Stress in concrete linearly varies along the depth H. σc = Mean stress 2 The resultant compressive force P in concrete and tensile force P in steel form a couple resisting the applied moment M. H  Arm of the couple =  D −  as is obvious from Fig. 8.33.  3 H σ H H    M = P  D −  = c BH  D −  = σ s As  D −     3 2 3 3

(8.11)

If the maximum allowable stresses in steel and concrete are given using the ratio of s s /s c , the value of H can be determined for the given dimension of the beam using Eq. (9). This is known as Economic section.

σs D−H = m  H  σc σ s  BH  = from Eq. (10) σ c  2 As  So,

m( D − H ) BH = H 2 As 2As m(D − H) = BH2

MTPL0259_Chapter 08.indd 324

(8.12)

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Theory of Simple Bending

325

The value of H can be determined from Eq. (12). For the given value of M, actual values of stresses ss and sc are determined. Example 8.13 A reinforced concrete beam is 200 mm wide and 400 mm deep. The maximum allowable stresses in steel and concrete are 120 and 7.5 N/mm2, respectively. What area of steel reinforcement is required if both the stresses are developed and steel reinforcement is 60 mm above the tension face. If modular ratio, m = 16, determine the moment of resistance of the beam. Solution

Breadth, Cover for steel = 60 mm Depth, H = 400 − 60 = 340 mm

ss = 120 N/mm2 sc = 7.5 N/mm2 B = 200 mm

σs (D − H ) =m H σc but, m = 16 120 D −H  = 16   H  7.5 H=D−H D 340 H = = = 170 mm 2 2

σ s BH 120 = = = 16 σ c 2 As 7.5 As = Moment of resistance, M =

BH 200 × 170 = = 1, 062.5 mm 2 32 32

σc H  BH  D −   2 3

600 mm Compression face

Pushing the value, 7.5 170   M = × 200 × 170  340 −   2 3 

H 300 mm A

N

= 1,27,500 (283.33) = 3,61,25,000 N mm = 36.125 kN m

Figure 8.34

Exercise 8.13

Exercise 8.13 The reinforced concrete beam of T-section, shown in Fig. 8.34, has maximum stresses of 5 N/mm2 in concrete and 100 N/mm2 in steel. The modular ratio of the concrete and steel is 16. Assume that the neutral axis lies within the full width of the concrete section, find the area of steel reinforcement and moment of resistance.

MTPL0259_Chapter 08.indd 325

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326 Chapter 8

Stress Concentration in Bending In Chapter 1, we have learnt about stress concentration due to abrupt change in section due to holes, fillets, grooves etc, in members subjected to axial loads. In this chapter, we will discuss the effect of fillets, grooves etc., in members subjected to bending moment on the stress distribution causing high stresses at discontinuity as a fillet and a groove. If a member is subjected to bending moment M, then stress,

σ=

My Bending moment × distance from nextal axis = ment of area about neutral axis I NA Second mom

If the flexure member has an abrupt change in geometry then stress change due to this abrupt change in geometry can be determined by

σ max = K

ymax , I NA

where K is the stress concentration factor, the values of K for flat and round members are given in Fig. 8.35(a)–(d). Figure 8.35(a) and (b) show stress concentration factors for flat bars in bending with a fillet radius and groove. Figure 8.35(c) and (d) show stress concentration factors for round bars with fillet and groove, respectively

3.0

3.0 r

2.8

B

2.6

2.8 b

2.4 D/d = 1.10 D/d = 1.25

K 2.0 1.8

D/d = 2.00

2.2 K

2.0 1.8

1.6

1.6

1.4

1.4

1.2

1.2

1.0

0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 r — b (a)

MTPL0259_Chapter 08.indd 326

b

2.6

2.4 2.2

B

1.0 0

D/d = 1.10 D/d = 1.25 D/d = 2.00

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 r — b (b)

5/23/2012 11:08:39 AM

Theory of Simple Bending 3.0

3.0

r

2.8

M

2.6 2.2

2.2

K 2.0

1.8

1.8

1.6

D/d = 2

1.4

1.1

1.2 1.0

1.03 0

0.1

r — d (c)

0.2

0.3

M

d

save = Mr – 32 M I pd 3

2.4

K 2.0

D

M

2.6

save = Mr – 32 M I pd 3

2.4

r

2.8

M

d

D

1.6

D/d = 2 1.1 1.03

1.4 1.2 1.0

327

0

0.1

r — d (d)

0.2

0.3

Figure 8.35 (a) and (b): flat sample, stress concentration factors for the fillet and groove; (c) and (d): for round sample, stress concentration factors for the fillet and groove. Example 8.14 Figure 8.36 shows a round bending member subjected to a bending moment. There is a groove of radius 1 mm in this member. Determine this magnitude of M if the maximum stress in the groove does not exceed 150 MPa. Solution Outer diameter, D = 22 mm Diameter at grooving d = 20 mm Grove reduces, r = 1 mm Rates r = 1 = 0.05 d 20 From the plot of stress concentration factors, Fig. 8.35(d) K = 2.2 smax = 150 s at outer radius

Bending moment,

r d

D

D = 22 mm d = 20 mm r = 1 mm groove radius

Figure 8.36

Example 8.14

150 = 68.18 N /mm 2 2.2 π d 2 π × 203 Z = section modulus = = 32 32 = 785.4 mm3 M = sZ =

= 68.18 × 785.4 = 53,550 N mm = 53.55 kN mm

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328 Chapter 8

Exercise 8.14 A steel flat is to have a groove cut into both edges as shown in Fig. 8.37. The full depth of the bar is 100 mm and its thickness is 10 mm. The depth where the grooves are cut is required to be 80 mm. The allowable bending stress is 120 MPa. If the radius of the groove is 10 mm, determine the maximum bending moment to which the beam can be subjected. r    Hint: d = 0.125, K = 1.85 Problem 8.1 A cast iron water pipe 400 mm outer diameter with 16 mm thickness is supported over a span of 10 m. Find the maximum stress in cast iron when the pipe is running full of water. Density of cast iron = 71.54 kN/m3 Density of water = 9.8 kN/m3

10 mm

M

D

d

100 mm

Solution Figure 8.38 shows a CI pipe full with water. Cast Iron Outer diameter = 0.4 m Inner diameter = 0.368 m Area of cross-section, Ac

M

r

Flat

Figure 8.37

Exercise 8.14

π (0.42 − 0.3682 ) 4 p = (0.16 − 0.135424) 4

Ac =

= 0.0193 m2

Water

Cast iron

368 mm 400 mm

Figure 8.38 Problem 8.1

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Theory of Simple Bending

329

Considering 1 m length, Weight of CI pipe per unit length = 0.0193 × 71.54 = 1.380 kN/m Water p Area of cross-section = × 0.3682 = 0.10636 m 2 4 Considering 1 m length Weight of water in pipe per unit length = 0.10636 × 98 = 1.0423 kN/m Total uniformly distributed load of water pipe, w = 1.38 + 1.0423 = 2.4223 kN/m Length of the pipe supported at ends = 10 m Maximum bending moment, M = =

wl 2 8 2.4223 × 102 = 30.27875 kN m 8

Cast iron pipe Moment of inertia, I =

π π ( D 4 − d 4 ) = (0.44 − 0.3684 ) = 3.564 × 10−4 m4 64 64

ymax = 0.2 m Maximum stress developed is CI =

M 30.27875 ×y= × 0.2 I 3.564 × 10 −4

= 16,991.4 kN/m2 = 16.99 N/mm2 Problem 8.2 Three steel tubes are welded together so as to make a beam of section as shown in Fig. 8.39. Outer diameter of each tube is 20 mm and inner diameter is 16 mm, centres of the tubes make an equilateral triangle of side 20 mm as shown. Determine INA of the section about neutral axes. How much bending moment can be applied on the section if stress due to bending is not to exceed 80 MPa. Solution Tube, centre diameter = 20 mm Inner diameter = 16 mm Area of cross-section of ends π A = (202 − 162 ) 4 = 36p mm2

MTPL0259_Chapter 08.indd 329

D

C Welding

h

N

A

G A

B

h — 3

tube, 20 mm

Figure 8.39

Problem 8.2

5/23/2012 11:08:43 AM

330 Chapter 8

Moment of inertia,

π (204 − 164 ) 64 π (1, 60, 000 − 65, 536) = 64 94, 464 × π = = 1, 476π mm 4 of each tube 64

I=

G of the sections lies at h/3 from side AB of triangle ABC h = 20sin 60° = 17.32 mm h = 5.773 mm 3 2h = 11.546 mm 3 I = 2 × 1,476p + 2 × 36p (5.773)2 + 1,476p + 36p (11.546)2 = 2,952p + 2,399.6p + 1,476p + 4,799.16p = p (4,428 + 7,198.76) = 11,626.76p = 36,526.5 mm4 Maximum stress will develop at point D: 2h + outer Radius 3 = 11.546 + 10 = 21.546 mm

y=

smx = 80 MPa M = σ mx = 80 ×

I y 36, 526.5 Nmm 21.546

= 1,35,622.5 N mm

sc

Allowable bending moment = 135.622 Nm Problem 8.3 A beam of rectangular section b × d is subjected to a bending moment M = 0.4 × 106 N mm, at a particular section shown in Fig. 8.40. The bending moment is a positive bending moment. Determine the force in the shaded area of the section.

20

N

a

b

Solution

A

10 mm

b = 20 mm d = 40 mm I NA

MTPL0259_Chapter 08.indd 330

5

bd 3 20 × 403 = = 12 12 = 10.667 × 104 mm4

c

d

5

st 10 20

Figure 8.40

Problem 8.3

5/23/2012 11:08:45 AM

Theory of Simple Bending

σ τ ,σ c = ± =±

331

My I NA 0.4 × 106 × 20 , note that y max = 20 mm 10.667 × 10 4

= +75 N/mm2 In the portion below NA, there is tensile stress because of the positive bending moment on the section. 75 × 5 σ ab = = 18.75 N/mm 2 20 75 × 15 σ cd = = 56.25 N/mm2 20 18.75 + 56.25 Average stress, sav = 2 = 37.5 N/mm2 Area abcd = 10 × 10 = 100 mm2 Force in the shaded area = 37.5 × 100 = 3,750 N = 3.75 kN Problem 8.4 A beam subjected to bending moment M is of T-section as shown in Fig. 8.41 with width 150 mm and depth 300 mm. Determine the thickness of the flange and the web if the flange is two times as thick as the web and maximum tensile stress is two times the maximum compressive stress. Solution Stress, s a y, distance from neutral layer st = 2sc or y1 = 2y2 150 mm

y1 + y2 = 300 mm sC

2y2 + y2 = 300 mm y2 = 100 mm

b

a

2t

y2

Flange

y1 = 200 mm That is, neutral axis passes through a layer at a distance of 200 mm from edge cd. Taking moment of areas about edge cd  300 − 2t  (300 − 2t )   × t + 150 × 2t (300 − t )  2  y1 = (300 − 2t )t + 150 × 2t

N

A 300 mm

y1 Web

c st

d t

2

200 =

0.5t (90, 000 + 4t − 1, 200t ) + 300t (300 − t ) 300t − 2t 2 + 300t

MTPL0259_Chapter 08.indd 331

Figure 8.41

Problem 8.4

5/23/2012 11:08:46 AM

332 Chapter 8

200 =

(45,000 + 2t 2 − 600t ) + 90,000 − 300t , dividing by t throughout 300 − 2t + 300 200 =

1, 35, 000 + 2t 2 − 900t 600 − 2t

200(600 − 2t) = 1,35,000 + 2t2 − 900t 1,20,000 − 400t = 1,35,000 + 2t2 − 900t 2t2 + 15,000 − 500t = 0 t2 − 250t + 7,500 = 0 t = 250 ±

250 2 − 4 × 7, 500 2

=

250 ± 62,500 − 30,000 2

=

250 − 180.28 2

= 34.85 mm Thickness of the web = 34.85 mm Thickness of the flange = 2 × 34.85 = 69.70 mm Problem 8.5 A floor has to carry a load of 12 kN/m2 (including its own weight). If the span of each joist is 4 m, calculate the spacing centre to centre between the joists. The breadth of each joist is 100 mm and depth is 240 mm and permissible stress due to bending is 10 N/mm2 (Fig. 8.42). Solution Length of each joist = 4 m Say the space between the joist = Z m Floor area per joist = 4 × Z = 4Z m2 Uniformly distributed load per metre length of joist, w = Z × 1 × 12 = 12 Z kN/m

Floor 240 mm Joist

Z

Z

Z 100 mm Joist

Figure 8.42

MTPL0259_Chapter 08.indd 332

Problem 8.5

5/23/2012 11:08:47 AM

Theory of Simple Bending

Maximum bending moment, Mmax =

333

wl 2 8

= 12 Z ×

42 = 24 Z 8

bd 2 100 × 240 2 = 6 6 = 0.96 × 106 mm3 Permissible stress, s = 10 N/mm2 Mmax = s.z = 10 × 0.96 × 106 N mm Section modulus, Z =

= 9.6 kN m = 24 Z Spacing, Z =

9.6 = 0.4 m = 400 mm 24

Problem 8.6 A girder of I-section is simply supported over a span of 8 m. It is subjected to a central concentrated load of 300 kN. The beam is strengthened wherever necessary by the addition of flange plate 15 mm thick. Find the length and depth of the flange plates such that the maximum stress due to bending does not exceed 90 N/ mm2, dimensions of I-section are shown in Fig. 8.43. Solution The section is symmetrical about both the axis, G of the section lies at centre of the web. y1 = y2 = 300 mm is shown. I xx =

240 × 6003 220 × 5503 − 12 12

= 106(4,320 − 3,050.2) = 1,269.8 × 106 mm4

Flange plate y

Beam carries a concentrated load of 300 kN at the centre of the span of 8 m. Mmax =

W L 300 × 8 = = 600 kN m = 600 × 106 N mm 4 4

Maximum stress, smax =

300 mm

M max 600 × 106 ×y= × 300 I xx 1,269.8 × 106

= 141.75 N/mm2 > 90 allowable stress. Therefore, Ixx has to be increased, thickness of flange plates = 15 mm Say the width of flange plates = b Moment of inertia, Ix ′x ′ = Ixx +

15 mm

25

G x, N

A, x

300 mm

2b × 153 + 2b × 15(300 + 7.5) 2 12

20 mm

25 y 240

15 mm

b

3

2b × 15 + 2b × 15(300 + 7.5) 2 12

MTPL0259_Chapter 08.indd 333

Figure 8.43

Problem 8.6

5/23/2012 11:08:49 AM

334 Chapter 8

= 1,269.8 × 106 + 562.5b + 28,36,687.5b

300 mm

x

= 1,269.8 × 10 + 28,37,525 × 10 b 6

6

A

Allowable stress = 90 N/mm2 y′ = 300 + 15 = 315 mm

4m

RB = 150 kN

Figure 8.44

or 1,269.8 + 2.837525b = 2,100 b=

4m

RA = 150 kN

600 × 106 × 315 90

1,269.8 × 106 + 2.837525b × 106 =

B

C

Problem 8.6

830.2 = 292.58 mm 2.837525

Taking only I-section Mx = bending moment corresponding to allowable stress of 90 N/mm2 =

90 × 1, 269.8 × 106 = 380.94 × 106 Nmm 300

= 380.94 kN m Beam carries a central load, so Reactions, RA = RB =

300 2

As shown in Fig. 8.44 Mx = 150x = 389.94 kN m x = 2.6 m Central length of the beam requiring attached flange plates (8 − 2.6 × 2) = 2.8 m Up to a length of 2.6 m from each end there is no necessity of additional flange plates. Problem 8.7 A compound beam for a crane runway is built up of a 150 × 80 rolled steel joist with a 100 × 45 rolled steel channel attached to the top flange. Calculate the position of neutral axis of the composite section and determine Ixx of the composite section. For I-section, area = 1,808 mm2, Ixx = 688.2 × 104 mm4, Iyy = 55.2 × 104 mm4, for the channel section, Area = 741 mm2, web thickness = 3.0 mm, Ixx = 123.8 × 104 mm4, Iyy = 14.9 × 104 mm4, distance of CG from the outer edge of the web = 14 mm. Solution Figure 8.45 shows the compound beam consisting of I-section and a channel section. Depth of the composite section = 150 + 3 mm = 153 mm (depth of I-section + web thickness of channel section)

MTPL0259_Chapter 08.indd 334

100 Channel section 14

y2 = 59.4 x

x

153 mm I section

a

y1 = 93.6

b 80

Figure 8.45

Problem 8.7

5/23/2012 11:08:50 AM

Theory of Simple Bending

y1 =

335

1, 808 × 75 + 741 × (153 − 14) 1, 35, 600 + 1, 02, 999 = 1, 808 + 741 2, 549

2, 38, 599 = 93.6 mm from lower edge ab 2, 549 y2 = 153 − 93.6 = 59.4 mm =

I x x = 688.2 × 10 4 + 1, 808(93.6 − 75)2 + 14.9 × 10 4 + 741(59.4 − 14)2 = 688.2 × 104 + 6,25,495.7 + 14.9 × 104 + 15,27,319.6 = 104(688.2 + 62.55 + 14.9 + 152.73) = 918.38 × 104 mm4 Problem 8.8 A flitched beam is made up of two timber pieces of 120 mm wide and 240 mm deep with a 20 mm thick × 160 mm wide steel plate firmly attached to them and placed in a groove as shown in Fig. 8.46. Calculate the moment of resistance of the composite section if the permissible bending stress in timber is 10 N/mm2, Es /Et = 20. Solution Modular ratio m = 20 Equivalent steel section width at edge, = ab

240 = 12 mm 20

width at edge, = cd 160 + width at edge, = ef

(240 − 160) = 164 mm 20

240 = 12 mm 20

12 Y a

b

Wood 155.95

180 mm

180 mm

c

d L

N 20

Steel 84.05

20

40

e

f

40

Y

160 240

12 164

(a)

Equivalent steel section (b)

Figure 8.46

MTPL0259_Chapter 08.indd 335

Problem 8.8

5/23/2012 11:08:52 AM

336 Chapter 8

Figure 8.46(b) shows an equivalent steel section. Let us calculate the distance of CG from edge ef (the section is symmetrical about yy axis) y1 = =

40 × 12 × 20 + 164 × 20( 40 + 10 ) + 180 × 12(60 + 90 ) 40 × 12 + 164 × 20 + 180 × 12 9, 600 + 1, 64, 000 + 3, 24, 000 4, 97, 600 = 480 + 3, 280 + 2,160 5, 920

= 84.05 mm y2 = 240 − 84.05 = 155.95 mm Moment of inertia, I=

164 × 203 12 × 403 + 12 × 40(84.05 − 20) 2 + + 3, 280(84.05 − 50) 2 12 12 +

12 × 1803 + 12 × 180(155.95 − 90) 2 12

= 64,000 + 480 × 4,102.4025 + 1,09,335.33 + 3,280 × 1,159.1 + 5,832,000 + 2,160 × 4,349.4025 = 64,000 + 19,69,153.2 + 1,09,333.33 + 38,02,840.2 + 58,32,000 + 93,94,709.4 = 21,172,036 mm4 = 21.172 × 106 mm4 ymax = 155.95 mm Z=

21.172 × 106 = 1, 35, 761.7 mm3 155.95

M = Zss = 1,35,761.7 × ss ss = mst = 10 × 20 = 200 N/mm2 M = 135,761.7 × 200 = 2,71,52,338.7 N mm = 27.15 kN m Problem 8.9 A brass bar of 150 mm diameter is completely encased in a steel tube of outer diameter 200 mm and inner diameter 150 mm, so as to make a composite beam. If the stress in the steel bar is not to exceed 120 N/mm2, how much bending moment can be applied on the composite beam (Es = 2EB). Solution Diameter of the brass beam, d = 150 mm Outer diameter of the steel beam, D = 200 mm If maximum stress in the brass beam at d/2 is sbc or sbt (in compression and tension Fig. 8.47).

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Theory of Simple Bending

337

ssc Steel

sbc 75

100

Brass

sbt sst

Figure 8.47

Problem 8.9

Stress in the steel beam at

d Es = × σ b = 2σ b 2 Eb

Stress in the steel beam at

D 100 , σ s = 2σ b × 2 75 = 2.66sb

σb =

σs 2.66

(8.13)

Section modulus Brass beam, Z b =

π × d 3 π × 1503 = = 331.34 × 103 mm3 32 32

Steel beam, Z s =

π ( D 4 − d 4 ) π (2004 − 1504 ) = D 32 32 200

=

π (16 × 108 − 5.0625 × 108 ) 32 × 200

=

10.9375 × 11 × 108 = 536.89 × 103 6, 400

M = Mb + Ms = sb × 331.34 × 103 + 2.66sb × 536.89 × 103 =

σs (868.23) × 103 = 326.40 × 103 s s 2.66

= 326.4 × 103 × 120, by substituting the value of ss = 39,168.27 × 103 N mm = 39.168 kN m

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338 Chapter 8

Key Points to Remember  Flexure formula M = E = σ = σ c = σ τ I NA R y yc yτ where M = bending moment at the section INA = moment of inertia of section about neutral axis E = Young’s modulus of the material R = radius of curvature of beam at section s = stress in a layer y = distance of layer from neutral axis sc = maximum stress in compression st = maximum stress in tension yc = distance of extreme layer in compression from neutral axis yt = distance of extreme layer in tension from neutral axis M = scZc = st Zt I I where, Zc = NA , Z t = NA yc yt where Zc and Zt are the section modulus in compression and in tension  Modulus of rupture =

σ Mult bd 2

 Mult = ultimate bending moment a beam b, d = breadth and depth of a section, respectively  In a beam of uniform strength ratio M/Z , that is, bending moment/section modulus is maintained constant  In flitched beam M = M1 + M2 resisting moment of beam of material (1) + resisting moment of beam of material (2) Equivalent section of a composite beam is made by considering modular ratio E1/E2

Review Questions 1. A section of a beam is subjected to bending moment, explain how stress distribution changes from negative to positive over depth of the section. 2. What is neutral layer? Why stress and strain are zero in the neutral layer? 3. Explain the formula M = sZ, where s is stress and Z is section modulus. 4. What is plane of bending? Why a section should be symmetrical about the plane of bending? 5. Take the case of channel section, explain symmetrical bending and unsymmetrical bending. 6. What is the most important assumption in a composite beam? 7. In simple bending, plane transverse sections remain plane after bending; explain this assumption with the help of simple sketch.

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Theory of Simple Bending

339

8. What is a beam of uniform strength, explain? 9. In RCC, why concrete section is assumed to carry zero tensile stress? 10. What do you understand by an equivalent section in a composite beam?

Multiple Choice Questions 1. A beam is of triangular section with base 10 mm and height 12 mm. what is its minimum section modulus? (a) 60 mm3

(b) 90 mm3

(c) 120 mm3 (d) None of these 2. A beam of rectangular section is subjected to bending moment of 14.4 N m. If b = 0.5d, where b is breadth and d is depth and maximum stress developed is 100 MPa, what is the breadth of the section? (a) 18 mm

(b) 12 mm

(c) 6 mm (d) None of these 3. A steel strip of breadth 50 mm and depth 10 mm is bent around a drum of radius 10 m. If E = 2 × 105 N/mm2, what is the maximum stress developed in steel strip? (a) 200 MPa

(b) 100 MPa

(c) 50 MPa (d) None of these 4. T-section beam has a width of 40 mm and a depth of 100 mm. CG of section is located at a distance of 25 mm from outer edge of flange. Stress developed at upper edge of flange in 75 MPa, what is the stress developed at lower edge of web? (a) 150 MPa

(b) 75 MPa

(c) 25 MPa (d) None of these 5. A beam is of I-section with flanges 200 mm × 10 mm and web 180 mm × 10 mm. Due to the bending moment applied on the beam section, maximum stress developed in the beam section is 100 MPa, what is the stress developed at inner edge of the flange? (a) 110 MPa

(b) 100 MPa

(c) 90 MPa

(d) None of these

6. An ms beam is subjected to a bending moment, such that a stress of 100 MPa is developed in a layer at a distance of 100 mm from the neutral layer. If E = 200 GPa, what is the radius of curvature of the beam?

MTPL0259_Chapter 08.indd 339

(a) 400 m

(b) 200 m

(c) 100 m

(d) None of these

7. A beam of square section (with side of square horizontal and vertical) is subjected to a bending moment which produces 60 N/mm2, maximum stress in the beam section. If the diagonals of the sections take vertical and horizontal directions, the bending moment remains the same, what is the maximum stress developed in the section? (a) 120 N/mm2 2

(c) 60 N/mm

(b) 90 N/mm2 (d) None of these

8. A cantilever of uniform strength s, having rectangular section of constant breadth b but variable depth d is subjected to a udl throughout its length. If the depth of the section is 150 mm at the fixed and what is the depth at the middle of the length of cantilever? (a) 150 mm

(b) 100 mm

(c) 75 mm (d) None of these 9. A beam of rectangular section of breadth 100 mm and depth 200 mm is subjected to a bending moment of 20 kN m. Stress developed at a distance of 100 mm from top face is (a) 30 MPa

(b) 15 MPa

(c) 7.5 MPa (d) None of these 10. A cantilever of uniform strength s having rectangular section of constant depth d but variable breadth b is subjected to a point load W at its free end. If the length of the cantilever is L, breadth of the cantilever at the middle of its length is (a)

WL σd 2

(c) 2W L2 σd

(b) 3WL σd 2 (d) None of these

5/23/2012 11:08:56 AM

340 Chapter 8

Practice Problems 1. A cast iron water pipe 500 mm bore and 20 mm thick is supported over a span of 8 m, find the maximum stress in the metal, when the pipe is running full. Density of cast iron = 71.54 kN/m3 Density of water = 9.8 kN/m3 2. Four steel tubes of outer diameter 18 mm and inner diameter 16 mm are welded together so that their centres make a square of side 18 mm. This combination is used as a beam over a length of 2 m. How much central load can be applied on the beam, if smax is not to exceed 60 MPa. WL    Hint: M max = 4    3. A beam is of triangular section with base b = 25 mm and altitude h = 36 mm. If the bending moment on the section is +81 N/m, what force is developed in breadth shaded area of Fig. 8.48? 4. A T-beam of depth 120 mm is used as a beam with simply supported ends so that the flange comes under tension. The thickness of the flange and web is 30 mm. The material of the beam can be subjected to 90 N/mm2 in compression and 30 N/mm2 in tension. It is desired to achieve a balanced design so that the largest possible bending stresses are reached simultaneously. Determine the width of the flange. Find how much concentrated load can be applied to the beam at its centre if its span length is 4 m. [Hint: y1 = 3y2] 5. A floor has to carry a load of 10 kN/m2 (including its own weight). If the span of each joist is 5 m, calculate the spacing centre between the joists. The breadth of each joist is 100 mm and depth is 300 mm, permissible stress due to bending is 8 N/mm2. 6. An I-section girder is simply supported over a span of 10 m, it is subjected to a udl of 9.6 kN/m. Properties of I-section are Zx = 965.3 × 103 mm3, depth 400 mm, and width 105 mm. It is strengthened by flange plates on both sides 20 mm thick so that maximum stress in the section is not to exceed 80 N/mm2. Determine the width of the flange plates. [Hint: Ixx = Zx × 200] 7. A compound beam for a crane runway is built up of a 150 mm × 50 mm rolled steel joist with a 100 mm × 50 mm rolled steel channel attached to the top flange. Calculate the position of NA of the section and determine moment of inertia Ixx of the composite section. For I-section, area = 901 mm2, Ixx = 322.1 × 104 mm4, Iyy = 9.2 × 104 and for the channel section, area = 1,002 mm2, web thickness = 4 mm, Ixx = 164.7 × 104 mm4, Iyy = 24.8 × 104 mm4. The distance of CG from outer edge of web = 16.2 mm. 30 mm 12 mm

36 mm

Wood

200 mm

100 25 mm

Figure 8.48 Practice Problem 3

MTPL0259_Chapter 08.indd 340

120

Figure 8.49

120

Practice Problem 8

5/23/2012 11:08:57 AM

Theory of Simple Bending

341

8. A steel beam, 30 mm wide × 200 mm deep, is fitted in to grooves cut in two wooden beams, 120 mm wide × 300 mm deep, as shown in Fig. 8.49. If the allowable stress in steel is 90 N/mm2 and modular ratio is 15, determine the allowable bending moment on the section of the beam. [Hint: make equivalent steel section] 9. An aluminium bar 120 mm in diameter is completely encased in a steel tube of 180 mm outer diameter and 120 mm inner diameter so as to make a composite beam. The composite beam is subjected to a bending moment of 12 kN/m. Determine maximum stress developed in each of the material given. E steel = 3E aluminium

Special Problems 1. A beam of I-section of moment of inertia 954 × 104 mm4 and depth 140 mm is freely supported at its ends. Over what span can a udl of 5 kN/m be carried if maximum stress is 60 N/mm2. What additional central load can be carried, when the maximum stress is 100 N/mm2. 2. A cantilever has a free length of 3 m. It is of T-section with a flange of 100 mm × 20 mm and a web of 200 mm × 10 mm, the flange being in tension. What load per metre run can be applied if the maximum tensile stress is 40 N/mm2? What is the maximum compressive stress? 3. Two 50 mm × 150 mm full-sized wooden planks are glued together so as to form a T-section as shown in Fig. 8.50. If a positive bending moment of 3 kN m is applied to such a beam around a horizontal axis, determine (a) stress at extreme fibres, (b) total tensile force below neutral axis and (c) total compression force above neutral axis. 4. A beam of T-section, 4 m long, carries a uniformly distributed load w per metre run throughout its length. The beam is simply supported at ends. The T-section is 200 mm × 100 mm × 12 mm. What is the maximum value of w so that the stress in the section does not exceed 60 N/mm2. [Hint: B = 100 mm, D = 200 mm, t = 12 mm] 5. A composite beam of the section shown in Fig. 8.51 is subjected to a bending moment M. If the maximum stress developed in steel is 80 N/mm2, what is the magnitude of M? Modular ratio is 15. 6. A steel tube of 8 mm bore and 1 mm wall thickness is fully charged with mercury and forms the part of an apparatus in a laboratory. The tube is 600 mm long and is supported over a span of 500 mm. What is the maximum stress in tube due to bending? Given r for steel = 76.44 × 10−6 N/mm3 r for mercury = 133.28 × 10−6 N/mm3

150 mm

50 mm

Wooden flanks

150 mm

50

Figure 8.50

Special Problem 3

t = 5mm 35

Steel

75 mm

Wood

150 mm

Figure 8.51

Special Problem 5

[Hint: Find total w = rate of loading]

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342 Chapter 8

Answers to Exercises Exercise 8.1: Exercise 8.2: Exercise 8.3: Exercise 8.4: Exercise 8.5: Exercise 8.6: Exercise 8.7: Exercise 8.8:

±25 N/mm2 b = 26.56 mm, d = 53.13 mm 0.405 kN m Z = 725.9 × 103 mm3, smax = 41.32 N/mm2 −16.6 MPa, +41.27 MPa 0.845 kN m 41.484 kN m 25 N/mm2

Exercise 8. 9: 1,444.6 × 104 mm4 Exercise 8.10: 70.7 mm, 141.4 mm Exercise 8.11: 52.85 kN m Exercise 8.12: ss = 17.93 N/mm2, sw = 0.791 N/mm2 [Hint: maximum stress in steel at a, maximum stress in wood at b] Exercise 8.13: 2,000 mm2, 51.1 kNm Exercise 8.14: M = 711 Nm

Answers to Multiple Choice Questions 1. 2. 3. 4.

5. 6. 7. 8.

(a) (c) (b) (d)

(c) (b) (d) (c)

9. (d) 10. (b)

Answers to Practice Problems 1. 2. 3. 4.

8.32 N/mm2 167 N 2,250 N 270 mm, 9.72 kN

5. 384 mm 6. b = 77.6 mm 7. y1 = 108 mm, y2 = 46 mm, Ixx = 534 × 104 mm4

8. 28.548 kN m 9. 7.75 N/mm2, 23.25 N/mm2

Answers to Special Problems 1. 3.62 m, 6.02 kN 2

2. 2.657 kN/m, 95.38 N/mm

MTPL0259_Chapter 08.indd 342

3. −4.32 N/mm2, +7.2 N/mm2 22.5 kN, 22.5 kN 4. 3.377 kN/m

5. 1.665 kN m 6. 4.586 N/mm2

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9 Shear Stresses in Beams CHAPTER OBJECTIVES Beams and cantilevers are subjected to transverse loads, producing bending moment and shear force in beam/cantilever section. In Chapter 8, stress distribution due to bending moment along the depth of the section is analysed showing maximum stresses in outer skin of the beam. In this chapter, shear stress distribution along the depth of the section will be studied. For some sections, shear stress at the neural axes of the section is maximum, while bending stress is zero at the section. Basic objective of this chapter is to make the student learn about shear stress at any layer of beam section which depends upon the first moment of the area above the section or below the section depending upon the compressive or tensile zone of the area.

Introduction In this chapter, we will study about the variation of shear stress in a beam section. As the bending stress is not uniform in beam section, it varies linearly. Similarly, the shear stress (due to shear force) is not uniform and varies making a parabolic curve. For various sections such as: rectangular and triangular, stress distribution along the depth of the section will be plotted. The shear stress developed is of small magnitude, not as large as bending stress due to bending moment, but it is of academic interest to learn about variation of shear stress along the depth. Shear force in a beam section occurs where there is variation in bending moment. Variation in the bending moment causes unbalanced force along the length of the section, which is balanced by the shear stress on a plane parallel to the axes. Complementary shear stresses are produced in transverse sections of a beam. Variation of complementary shear stress is plotted along the depth of transverse section. We will also discuss shear stress distribution in hollow sections as thin hollow rectangular, hollow triangular, and hollow circular sections.

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344 Chapter 9

Shear Stress Distribution Consider a beam of rectangular section with depth D and breadth B. Along the length, consider an elementary of length dx. On the left side of element the bending moment is M and on the right-hand side of the element the bending moment is M + dM as shown in the Fig. 9.1(a). dM = F, Since there is change in bending moment, there will be a shear force in the section as dx shear force. Stresses due to M are sc and st as extreme compressive and tensile stresses, respectively, and stresses due to M + dM are sc′ and st′, extreme stresses as shown in Fig. 9.1(b). Note that sc′ > sc as given in Fig. 9.1(b), similarly st′ > st. Due to difference in longitudinal stresses (bending stresses) on both the sides, there will be difference in resultant. Push (Fc′ − Fc) or pull (on the other side of neutral layer), which will be balanced by a horizontal force developed on the longitudinal plane of beam Fc′ − Fc = Q = tdxB, where dx and B are the dimensions of longitudinal plane and t is the shear stress on this plane. d sC9 > sC M + dM

M

FC

sC9

sC

D 2

YC

D 2

N

a

b

e

f

D 2

L L

N

Tension

FC9

Q

y L

g

h

(Comp) N

d

Yt B

dx

st

dx

st9

dx

st9 > st

Figure 9.1

Stresses due to variable bending moment

Any Section (With Variable Breadth) Now, consider a plane of length dx and breadth b, at a distance y from neutral layer as shown in Fig. 9.2(a). Beam is subjected to bending moment M on left side and M + dM on right side as shown in the figure. Consider a layer at a distance y from neutral layer as shown in Fig. 9.2(a). Thickness of layer is dy. M Stress on left side, s= y I NA M + dM y I NA INA = moment of inertia of section about neutral axis Stress as right side,

MTPL0259_Chapter 09.indd 344

σ′ =

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Shear Stresses in Beams

345

M + dM

M

b s

G

s9

e

yC

f

dy y

L

N

yt

f

y

g

N

Compression zone

A Tension zone

dx (a)

(b)

Figure 9.2

Section with variable width

δ P = σδ a = σ bd y

Forces,

δ P ′ = σ ′δ a = σ ′bd y Summing up all the unbalanced forces on elemental area from y to yc yc

δF =

dM

∫I y

ybdy

NA

yc = distance of extreme layer in compression δ F = τ bd x where t is shear stress developed in plane efgh

τ=

1 b I NA

yc

∫ y

dM ybdy dx

dM = F , shear force on section dx

τ=

F yc ∫ by d y bI NA y

yc

yc

y

y

∫ by d y = ∫ y d a = A y

where A = area of cross-section above the layer ef (shaded area) y = distance of CG of the area A from neutral layer FA y I NA b or shear stress, t at distance y from neutral layer F = shear force A = area of cross-section above the layer under consideration

Shear stress,

MTPL0259_Chapter 09.indd 345

τ=

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346 Chapter 9

y = distance of CG of the area above the layer INA = moment of inertia of the section about neutral axis b = breadth of the layer Example 9.1 A round beam of circular section of diameter D is simply supported at the ends and carries a point load W at its centre. Determine the magnitude of the shear stress along the plane passing through the neutral axes at a particular section, which lies at a distance of L/4 from left-hand side support where L is span length. Solution Figure 9.3(a) shows a beam simply supported at each end over a span length L caring a central load W and also shows a shear force diagram. F at

W L from A is + 2 4

W 2 Breadth of the layer, b = diameter at neutral layer, D π Area above the layer, A = D2 8 Therefore,

F=+

CG of area from NL,

y = Ay = I =

2D 3π

π D 2 2D D 3 × = 8 3π 12 πD4 64

W L 4

A

Y

Y B

C

L 2

RA

+

Y

G

RB

W 2

N

L

y=

2D 3p

B A

−

D

W 2

SF diagram (a)

(b)

Figure 9.3

MTPL0259_Chapter 09.indd 346

Example 9.1

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Shear Stresses in Beams

τ=

Shear stress,

347

FA y W D 3 1 64 = × × × Ib 2 12 D π D 4

8W (shear stress at neutral axis) 3π D 2 W W 4 2W = × = t = τ av = 2 2 Area 2 π D π D2

τ NA = Average shear stress,

τ 8W πD2 4 × = = 2 τ av 3π D 2W 3

4 τ NA = τ av (average shear stress) 3

Shear stress at neutral axis,

Shear Stress Distribution in a Rectangular Section of a Beam Section of the beam is rectangular with breadth B and depth D is subjected to shear force F. Consider D a layer ab of breadth B, at a distance y for neutral axis NA. Neutral axis lies at a distance of from 2 extreme ends as shown in Fig. 9.4(a)

A

G y

1.125 tav

a

b

y N

1.5 tav

A

1.125 tav

B (a)

Figure 9.4

MTPL0259_Chapter 09.indd 347

(b)

Shear stress distribution in a rectangular section

5/23/2012 11:06:51 AM

348 Chapter 9

Shear stress,

Breadth, b = B

t=

FAy I NAb

I NA =

BD 3 12

(9.1)

D  A = B  − y 2  1D Y D y  D y = y +  − y = + y − = +  4 2 2 2 4 2 2  D   D y B  D Ay = B  − y  ×  +  =  − y2  2   4 2 2  4 

Substituting the values in Eq. (9.1)

τ=F× =

τ NA =

 6F  D 2 − y2 BD 3  4 

3 F 2 BD

= =

at y = 0

6F  D 2 D 2  − BD 3  4 16 

9 F × 8 BD

=0

τ av = Therefore,

 12 B  D2 1 × − y2 × × 3 2  4 B  BD

at y =

at y =

D 4

D 2

F BD

τ max = τ NL = 1.5av τ = 1.125τ av

at y =

D 4

Figure 9.4(b) shows the shear stress distribution along the depth of rectangular section. Example 9.2 A wooden beam of rectangular section 15 cm × 30 cm is simply supported over a length of 4 m. It carries a udl of 4 kN/m throughout its length. What is the maximum shear stress developed in the beam section?

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349

w = 4 kN/m

L=4m D +8kN −8kN SF diagram

B

(a)

Figure 9.5

(b)

Simply supported beam with udl

Section L = length of beam = 4 m w = 4 kN/m RA = RB =

Reactions,

4×4 = 8 kN 2

Figure 9.5(a) shows the SF diagram of the beam. Fmax = maximum shear force = 8 kN (occurs at ends of beam) Breadth, B = 15 cm = 150 mm D = 30 cm = 300 mm

Depth, Shear stress at neutral axis is maximum.

τ NA =

1.5F 8, 000 × 1.5 = BD 150 × 300

= 0.266 N/mm2 Exercise 9.1 A beam of square section, side 40 mm, carries a central load of 10 kN over a span of 6 m. The beam is simply supported at its ends. What is the shear force at a distance of 1 m from one end? Draw the shear stress distribution along the depth of the section.

Shear Stress Distribution in a Circular Section of a Beam Consider a beam of circular section of diameter D is subjected to a shear force F at the section under consideration. Neutral layer passes through the centre of the circular section. Shear force at any layer ab

τ=

MTPL0259_Chapter 09.indd 349

FA y I NA b

(9.2)

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350 Chapter 9

Consider a layer at a distance y from neutral layer, Fig. 9.6, subtending an angle q at the centre from neutral layer. b

b

dy y

θ

N

D

θ

L

4 t 3 av

D

Figure 9.6

Shear stress distribution in a circular section y R y = R sin q dy = R cos q dq b = 2R cos q (as shown in Fig. 9.6)

sin θ = or

π /2

R

Ay = ∫ bdyy = ∫ (2 R cos θ ) ( R sin θ ) (R cos θ )dθ θ

θ

π /2

= 2 R 3 ∫ cos 2 θ a sin θ dθ = θ

π 2R3 2 × cos3 θ θ 3

π 2R   = 2 R 3 ∫  cos3 − cos3 θ  = − cos3 θ   2 3 3

as cos

π =0 2

Substituting the value in Eq. (9.2)

τ= =

τ= =

MTPL0259_Chapter 09.indd 350

F I NA

×

2R 3 1 cos3 θ × 3 2R cos θ

FR 2 cos 2 θ 3I NA

where, I NA =

as b = 2R cos q

πR 4 4

FR 2 cos 2 θ × 4 3 × πR 4 4 F cos 2 θ × 4 3π R 2

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Shear Stresses in Beams

=

4F at θ = 0, neutral layer 3π R 2

=

F π R2

=

2 F × 3 π R2

=

F 3π R 2

351

at θ = 30° at θ = 45°

at θ = 60°

= 0 at q = 90°

τ av =

F πR 2

4 τ = τ av , at θ = 0 3 t = tav,

at q = 30°

2 τ = τ av , at θ = 45° 3 1 τ = τ av , at θ = 60° 3 = 0,

at q = 90°

Shear stress distribution is symmetrical on the other side of the neutral layer. Figure 9.6 shows shear stress distribution along the depth D of the section. Note that the maximum shear stress occurs at neutral axis. Example 9.3 A beam is of circular section of diameter 80 mm. At a particular section SF is 40 kN. Draw the shear stress distribution along the depth of the section. Solution Diameter, D = 80 mm Radius, R = 40 mm Area of cross section, A = pR2 = 1,600p mm2 4F Shear stress, τ= × cos 2θ 3A F = τ av , average shear stress A =

40, 000 = 7.96 N /mm 2 1, 600π

4 τ = τ av × cos 2 θ 3

MTPL0259_Chapter 09.indd 351

D

θ N

A

Circular sections

4 t 3 av = 10.61 N/mm 2

Figure 9.7

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352 Chapter 9

4 4 = τ av = × 7.96 3 3 = 10.61 N/mm2, at q = 0° 2 = τ av at q = 45° 3 = 0 at q = 90° It is symmetrical on the upper side of the neutral axis NA. Figure 9.7 shows the shear stress distribution along the depth of the section. Exercise 9.2 A beam is of circular sections of diameter D mm. At a particular section of the beam, shear force is 10 kN. Determine the diameter D if the maximum shear stress at neutral layer is not to exceed 15 N/mm2.

Curves of Principal Stresses in a Beam While analysing the bending stress and shear stress in a beam, it is observed that at the extreme fibres, bending stress is maximum and shear stress is zero, so the principal stress direction is parallel to the axes of the beam or the principal plane is perpendicular to the axis at extreme fibres. Similarly at the neutral axes, bending stress is zero and shear stress is maximum, principal stress inclined at 45° to neutral axis. At any other layer, there is a combination of normal stress due to bending and shear stress due to transverse shear force and principal stresses and their directions can be worked out. Then, the curves for principal planes can be drawn. At any point along the curve, a principal plane is tangential to the curve. There are two sets of stress curves known as curves of principal stresses as shown in Fig. 9.8. Two sets of curves of principal stresses/principal planes possess following characteristics: (a) Tangent and normal to these curves give directions of principal stresses and principal planes. (b) Two curves intersect each other at right angles. (c) Curves cross the neutral axes at 45°. (d) Curves are normal to top and bottom surfaces. (e) The intensity of principal stress is maximum when it is parallel to the axes of the beam.

Compression

45 Neutral

90

90

Layer

Tension

Figure 9.8 Curves of principal planes

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353

Directional Distribution of Shear Stresses A shear stress is always accompanied by a complementary shear stress at right angles. Therefore, the directions of shear stresses on an element are either both towards the corner or both away from the corner to produce balancing couples. However, near a free boundary, the shear stress at any section acts in a direction parallel to the boundary as shown in Figs 9.9(a) and 9.9(b) for an I-section and a circularsection, respectively. Note that in a circular section, shear stress is parallel to the boundary and at the centre the shear stress is perpendicular to the boundary, so as to provide complementary shear stress to the shear stress along the boundary. In I-section, note that shear stress has to follow the boundary of flange and the web; therefore, the distribution of shear stress in I-section is as shown in Fig. 9.9(a).

(a) I-section

(b) Circular section

Figure 9.9 Directional distribution of shear stress Problem 9.1 A rectangular beam of depth d, breadth b and length L is supported at ends and carries a point load P at the middle of span. Show that the principal stress at a point in the central cross-section 3 PL  9d 2  1 + 1 + at a distance 0.25d from top is   8 bd 2  4 L2  Solution Simply supported beam with a central load P is shown in Fig. 9.10. INA =

Moment of inertia,

bd 3 12

Bending moment at central cross-section = y= Normal stress at

d 4

d from top edge or from NL, 4

σ=

MTPL0259_Chapter 09.indd 353

PL 4

M PL 12 d 3 PL ×y= × × = I NA 4 bd 3 4 4 bd 2

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354 Chapter 9

P

L 2

L 2

d

d 4

G a

d 4

+P 2 −P 2

c

b

N

SF digram

y

d

L

b Section

PL 4 8M digram

Figure 9.10 Problem 9.1

Shear stress, t A y of area abcd = b ×

d  3d  3bd 2  = 4 8  32

Breadth = b

τ=

FAy P , where F = at central section I NAb 2

=

P 3bd 2 12 1 × × 3× 2 32 b bd

=

9P 16bd

Stresses at a point in layer ab are shown in Fig. 9.11.

σ=

3 PL 4 bd 2

τ=

9P 16bd

σ 3 PL = 2 8 bd 2

MTPL0259_Chapter 09.indd 354

t

s

t

Figure 9.11

Stresses on an element

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Shear Stresses in Beams

355

2

=

σ σ +   +τ2  2 2

=

3 PL 9 P 212 81 P2 + + × 8 bd 2 64 b 2 d 4 256 b 2 d 2

=

3 PL 3 PL 9 d2 + 1 + 8 bd 2 8 bd 2 4 L2

=

3 PL 8 bd 2

 9 d2  1 + 1 +   4 L2  

Problem 9.2 A rolled steel section is shown in Fig. 9.12. It is subjected to a vertical force of 20 kN. Determine shear stress at points A, B and C of the section.

A

30

30

Solution Moment of inertia about neutral layer,

B

R

=

Principal stress

100 × 120 π R − 12 4 2

I NA =

4

π × 30 4 = 14.4 × 106 − 63.6 17 × 104 4 = 13.764 × 106 mm4 Shear stress, tA = 0 (on top edge) Point B b = 100 mm A y = 30 × 100(60 − 15) = 13.5 × 104

60

N C

A

= 14.4 × 106 −

Shear stress,

30

100

Figure 9.12

Problem 9.2

FAy 20, 000 × 13.5 × 104 τB = = I NAb 13.764 × 106 × 100 = 1.96 N/mm2 (at B)

Point C b = 100 − 60 = 40 mm π R 2  4R  A y = 60 × 100 × 30 −   ; note that CG of semi circular area lies at 2  3π   4R    from base diameter 3π  30 2  4 × 30  = 18 × 104 − π ×   2  3π  = 18 × 104 − 450 × 40 = 18 × 104 − 18 × 103 = 16.2 × 104 mm3 20, 000 × 16.2 × 10 4 Shear stress, τ c = = 5.88 N/mm2 (at C) 13.764 × 106 × 40

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356 Chapter 9

Problem 9.3 Two steel pipes of outer radius 20 mm and inner radius 18 mm are welded to a steel plate, 120 mm deep and 5 mm wide, so as to form a beam section as shown in the Fig. 9.13. How much shear force can be applied on this section if the shear stress in the weld does not exceed 10 N/mm2?

20 Pipe 18 weld

Solution Area of cross-section of pipe

y = 60

= π (20 2 − 182 )

120

= 238.76 mm

N

2

Moment of inertia I NA =

5 × 1203  π + 2  (204 − 184 ) + (60 + 20) 2 (238.76)  12  4

5 mm A

plate weld

π  = 72 × 10 4 + 2  (16 − 10.4976 ) × 10 4 + 152.806 × 10 4  4 

Pipe

= (72 + 314.2552) × 104 = 386.2552 × 104 mm4 Section above the weld A y = 238.76 (60 + 20) = 1.91 × 104 Breadth,

Figure 9.13

Problem 9.3

b = 5 mm

τ=

FA y F × 1.91 × 10 4 = I NA b 386.2552 × 10 4 × 5

t = 10 N/mm2 Shear force,

F=

386.2552 × 5 × 10 = 10,111.4 N 11.91

= 10.111 kN Problem 9.4 The section of a beam is a square with diagonal d. Diagonals are placed in vertical and horizontal positions as shown in Fig. 9.14. Determine the position of the layer at which transverse shear stress is maximum. What is the shear stress at neutral layer? Plot shear stress distribution for d = 80 mm and F = 20 kN. Solution Section of the beam is symmetrical about XX and YY axes Side =

d 2 4

Moment of inertia,

1  d  I NA =  ×  2  12 =

MTPL0259_Chapter 09.indd 356

d4 48

(9.1)

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Shear Stresses in Beams

357

Y Light shade a y

•

G

b

d 2

b

y

X, N

A, X

d 2

Y

Figure 9.12

Problem 9.4

Consider a layer ab at a distance y from N 1 d d y 2y d  y = y +  − y = y + − = +  3 2 6 3 3 6 Height of the triangle, Y ab

d  =  − y 2 

Base of the triangle

d  = 2  − y 2  1 d  d d  Area =  − y   − y  2 × =  − y   2 2   2 2

2

2

d   2y d  A y =  − y  + 2   3 6  Breadth, b = side,

d  ab = 2  − y  2  2

Transverse shear stress,

MTPL0259_Chapter 09.indd 357

d   2y d  F  − y  + 2   3 6  FA y τ= = × 48 I NA b d  d 4 × 2  − y 2 

5/23/2012 11:07:07 AM

358 Chapter 9

d   2y d  24 F  − y   + 2   3 6  = d4 =

dd 24 F 1   ×  2y +   − y 4  3 2  2 d

=

 8 F  d 2 dy + − 2 y2  4  2 d  4 

For shear stress to be maximum dτ d  = 0 =  − 4 y   dy 2 d y = or substituting the value 8 d2  8F  d 2 d d τ max = 4  + × − 2 ×  64  d  4 2 8 =

8F  d 2 d 2 d 2  + − d 4  4 16 32 

=

8F d 2 × 9 9 F × = × 2 32 4 d d4

Shear stress at neutral layer Breadth of the layer = d A=

d2 , y=0 4

Ay =

d2 d d3 × = 4 6 24

I NA =

d4 48

τ NL = F × =

d 3 48 1 × × 24 d 4 d

40

40 mm

2F d2

7.3 N/mm2

d = 80 mm F = 20 kN

τ max = 2.25 ×

6.25

Figure 9.15

Shear stress distribution

20, 000 = 7.03 N /mm 2 80 × 80

2, 000 = 6.25 N /mm 2 80 × 80 Figure 9.15 shows the shear stress distribution along diagonal of the section.

τ NL = 2 ×

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359

Problem 9.5 An extruded aluminium alloy section is of the shape and dimensions as shown in the figure. Vertical shear force on the section is 4 kN. Determine the dimension d if the average shear stress in the section is 8 N/mm2. What is the shear stress at the NA (neutral axes) (Fig. 9.16)? Solution Shear force, F = 4 kN Average shear stress, tav = 8 N/mm2 Area of cross-section =

4, 000 F orce = 8 Average shear stress

= 500 mm2  60 × 30 d d  = − ×  = 500 mm 2  2 2 2 2

d = 900 − 500 = 400 4 d d = 40 mm, = 20 mm 2

yc

d b

yt

Location of centroid C 20 900 × 10 − 400 × 3 yt = 900 − 400 9, 000 − 2, 666.67 = = 12.67 mm 500

e

a 30 mm

•G f

c

d 2

45 d 60 mm

Figure 9.16

Problem 9.5

y c = 30 − 12.67 = 17.33 mm Moment of inertia,

I NA =

60 × 303 2 + 900 (12.67 − 10 ) 36

2  40 × 203 20    − + 400 12.67 −    3    36

I NA = 45 × 103 + 6.416 × 103 − (8.889 × 103 + 14.4 × 103 ) = 51.416 × 103 − (23.289 × 103 ) = 28.127 × 103 mm 4 Area above centroidal layer or neutral layer = abc − edf dG = 20 − 12.67 = 7.33 mm ef = 2 × dG = 14.66 mm yc = 17.33 bc = 2 × 17.33 = 34.66 Ay =

(34.66) 2 17.33 14.662 7.33 × − × 4 3 4 3

= 1734.9 − 131.78 = 1603.13 mm 3

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360 Chapter 9

b = be + fc = 10 + 10 = 20 mm

Breadth, Shear stress at neutral layer,

τ NL =

Substituting the value,

τ NL =

FAy I NA b

4, 000 × 1, 603.13 28.127 × 103 × 20 = 11.4 N/mm2

Problem 9.6 A 300 mm × 120 mm RSJ of I-section with flanges 120 mm × 20 mm and web 260 mm × 10 mm is subjected to a shear force F. What percentage of shear force is carried by the web? What is average shear stress in the section and what are shear stresses at: (a) lower edge of flange, (b) upper edge of web and (c) at neutral layer? Solution Figure 9.17 shows I-section with flange 120 mm × 20 mm and web 260 mm × 10 mm. 3

Moment of inertia, I NA = 120 × 300 − 110 × 260 12 12

20 dy N

3

20

= 108.887 × 106 mm4

FA y τ= , b = 10 mm I NA b

y 10 mm

= (270 − 161.113) × 106

Shear stress at any distance y from neutral layer, in web

L

260 mm

120

Figure 9.17

Problem 9.6

INA = 108.887 × 106 mm4 130 − y   A y = 120 × 20(150 − 10) + 10(130 − y)  y +   2   2 y + 130 − y  = 336 × 103 + 10(130 − y)    2 = 336 × 103 + 5(130 − y)(130 + y) = 336 × 103 + 5(16,900 − y2)

τ=

F × [336 × 103 + 5(16, 900 − y 2 )] 108.887 × 106 × 10

Consider a small area of depth dy dA = 10 × dy dF = t·dA = t × 10 × dy dF ′ =

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F (336 × 103 + 5(16, 900 − y 2 )) dy 108.887 × 106

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F ′ = Total shear force shared by web 130, 130

= 2 ∫ dF ′ 0

2 F 130 3 2 = ∫ [336 × 10 + 5(16, 900 − y )] dy I 0 130

 2F  y3   3 = 336 × 10 y + 5 16, 900 y −   3  0 I   =

 2F  1303   3 336 × 10 × 130 + 5 16, 900 × 130 −  3   I  

2F [4, 368 × 104 + 5 (2,197, 000 − 732, 333.33)] I 2F = [4, 368 × 104 + 5(1, 464, 666.67)] I 2F = (4, 368 × 104 + 732.33 × 104 ) I 10, 200.667 × 10 4 = F , substituting the value of I I =

F′ =

10, 200.667 × 104 F 108.887 × 106

F ′ = 0.9368F or 93.68% of total shear force is carried by the web. tav = average shear stress = =

F F = total area 2 × 120 × 20 + 260 × 10

F = 1.35135F × 10 −4 7, 400

t1 at lower edge of flange A y = 120 × 20 × (150 − 10) = 3,36,000 b = 110 mm

τ1 =

FA y F × 336, 000 = Ib 108.887 × 106 × 110

= 2.805F × 10−5

τ1 3.08576 × 10 −5 F = = 0.207 τ av 1.35135 × 10 −4 F t2 at upper edge of web Ay = 120 × 20 × (150 − 10) = 3, 36, 000

b = 10 mm

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362 Chapter 9

τ2 =

FA y F × 33, 600 = I .b 108.887 × 106 × 10

= 3.0857610−4F

τ 2 3.08576 × 10 −4 F = = 2.283 τ av 1.35135 × 10 −4 F t3 at upper edge of I-section A y = 120 × 20 × 140 + 130 × 10 × 65 = 336,000 + 84,500 = 420,500 mm3 = 4.205 × 105, b = 10 mm t3 =

4.205 × 105 × F = 3.8618 × 10 −4 F 108.887 × 106 × 10

τ3 3.8618 × 10 −4 F = = 2.8578 τ av 1.35135 × 10 −4 F Problem 9.7 A beam is made up of five wooden planks each with a width of 100 mm and a depth of 40 mm is placed one above the other so as to from a section as shown in Fig. 9.18. The planks are tied together by means of a 16-mm-diameter steel bolt passing across the overall depth with tightened nuts. Find the maximum shear stress developed in bolt section. The bolts are spaced 120 mm centre to centre along the beam length. Neglect the self-weight of the beam and the loss of cross sectional area due to bolts. Shear force in the section is 10 kN.

Bolt head Wooden planks

40 16 mm bolt

40

40

a

a

N

L

a

a

40 40 mm Nut 100 mm

Figure 9.18 Problem 9.7

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Solution F = 10 kN = 10,000 N Neutral axes passes through the middle of depth as shown 100 × 2003 12 = 66.67 × 106 mm4

I NA =

Maximum shear stress between the planks will occur along section aa, 80 mm from top or from bottom. About neutral layer, A y = 100 × (80) (20 + 40) = 48 × 104 mm3 breadth,

b = 100 mm

Shear stress,

τ=

Spacing between the bolts Shear force per bolt Area of cross-section of each bolt Shear stress in bolt

FA y 10, 000 × 48 × 10 4 = I NA b 66.67 × 106 × 100

= 0.72 N/mm2 = 120 mm = 0.72 × 120 × 100 = 8,640 N π = × 16 2 = 64π mm 2 4 = 8, 640 64π = 42.97 N/mm2

Problem 9.8 A beam of box cross-section shown in Fig. 9.19 carries a vertical load at its mid span such that the maximum bending stress produced is 10 N/mm2. The beam is supported over a span of 3 m. Upper and lower flanges are connected to the webs with the help of screws. Determine spacing between the screws if each screw can transmit a shear force of 2.2 kN. Solution The section is symmetrical, so neutral axes will be at the centre of the depth as shown in the figure. 3

2

flange a

a

web N

L

a

a

MTPL0259_Chapter 09.indd 363

200

50 50

100

50

Box cross-section

200 × 300 100 × 200 − = (4.5 − 0.66) × 108 12 12 Figure 9.19 = 3.833 × 108 mm4 Maximum bending stress will occur at top and bottom y = 150 mm s = 10 N/mm2 10 σ M = I NA × = 3.833 × 108 × = 25.53 × 106 N mm 150 y I NA =

50

Problem 9.8

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364 Chapter 9

Span length, L = 3 m = 3,000 mm Maximum bending moment occur outer = W ×

WL 4

3, 000 = 25.53 × 106 4 W = 34.07 × 103 N F =

Shear force at the centre,

W = 17.035 × 103 N 2

Screws are connecting the flanges on webs along the plane aa. Shear stress along the plane aa A y = 200 × 50 (150 − 25) = 1.25 × 106 mm 3 Breadth,

b = 50 + 50 = 100 mm

τ aa =

FAy 17.035 × 103 × 1.25 × 106 = I NAb 3.833 × 108 × 100

= 0.555 N/mm2 Shear force per meter length of beam = 0.555 × 100 × 1,000 = 0.555 × 105 N = 55.5 × 103 N Say spacing between the screws Force transmitted by a screw There are two screws on each side Therefore,

= Sm = 2.2 kN

2 × 2.2 kN = 55.50 kN S S=

4.4 = 0.079 m 55.5

= 79 mm Problem 9.9 A steel joist ISLB 150 has following parameters, Ixx = 688.2 cm4, area = 18.08 cm2, depth = 150 mm, flange thickness = 6.8 mm and web thickness = 4.8 mm. At a particular section of beam, M = 8 kN m and F = 32 kN. Determine principal stresses at edges bb and cc (section) as shown in Fig. 9.20.

a b 75 mm

a c

c

35 x

N

A

x

75 mm 4.8 mm

Solution I NA = 688.2 × 10 4 mm 4 M = 8 × 106 Nmm

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6.8 mm

b

80

Figure 9.20

Problem 9.9

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365

Bending stresses

σ bb =

8 × 106 × (75 − 6.8) as y = (75 − 6.8) at bb 688.12 × 104

= 79.28 N/mm2

σ cc =

8 × 106 × 35 688.2 × 10 4

= 40.68 N/mm2 (at cc) Shear stress F = 32 kN = 35,000 N

Shear force, Edge bb

A y = 80 × 6.8 × (75 − 3.4 ) = 38950.4 mm 3 b = 80 mm

Breadth,

τ bb =

FA y 32, 000 × 38950.4 = I NA b 688.2 × 10 4 × 80

= 2.26 N/mm2 (along edge bb) Layer cc A y = (75 − 6.8) = 80 × 6.8 (75 − 3.4) + (71.6 − 35) 4.8 × (35 + 18.3) = 38,950.4 + 36.6 × 4.8 × 53.3 = 38,950.4 + 9363.74 = 48,314.14 mm3 b = 4.8 mm

τ cc =

FAy 32, 000 × 48314.4 = I NA b 688.2 × 104 × 4.8

= 46.8 N/mm2 (along edge cc) Principal stress, 2

σ σ  2 Pbb = bb +  bb  + τ bb  2  2 2

 79.28   79.28  = ±  + 2.262 = 39.64 ± 1, 571.33 + 5.108   2   2  = 39.64 ± 39.70 = 79.044 N/mm2, −0.06 N/mm2

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366 Chapter 9

2

pcc =

σ cc σ  +  cc  + τ cc2  2  2 2

 40.68   40.68  = ±  + 46.82  2   2  = 20.34 ± 413.716 + 2,190.24 = 20.34 ± 51.032 = 71.37 N/mm2, −30.69 N/mm2

Key Points to Remember  Shear stress in any layer at a distance y from neutral layer

τ=

FAy , where I NA b

F = shear force at section

A y = first moment of area above the layer (or below the layer as the case may be) about neutral axis. INA = second moment of area of the section about neutral axis. 4  In a circular section, τ max = τ av 3  In a rectangular section, τ max = 1.5 τ av  In a thin circular section,

tmax = 2tav, where tav is average shear stress

 In the case of I-section, most of the shear force F is shared by the web.  In the case of I-section, most of the bending moment M is shared by the flanges.  In the case of square, circular and rectangular, I-section maximum shear stress occurs at to the neutral layer. 9  In the case of a square section with one diagonal becoming neutral layer, maximum shear stress is × 8 d average shear stress and occurs at a distance of from the neutral layer, while d is diagonal of the section. 8  Near a free boundary, the shear stress on any section acts in a direction parallel to the boundary.

Review Questions 1. Explain how shear stress is developed in a section of a beam where there is change in bending moment? 2. Why the ratio of maximum shear stress/average shear stress is more in a thin circular section than in a solid circular section? 3. A rectangular section of a beam is subjected to a bending moment M and a shear force F. Why bending stresses are maximum at extreme layer while shear stress is zero at these layers? 4. In a triangular section of a beam, why shear stress due to shear force is not maximum at neutral axis? 5. In a circular section, why at the centre of section the direction of shear stress is perpendicular to boundary?

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367

Multiple Choice Questions 1. In a rectangular section, tav = 50 MPa, what is shear stress at neutral axis? (a) 50 MPa

(b) 75 MPa

(c) 90 MPa (d) None of these 2. A thin circular tube is subjected to a transverse stress shear force F. If maximum shear stress developed in the section is 80 MPa, what is average shear stress in beam section? (a) 60 MPa

(b) 50 MPa

(c) 40 MPa (d) None of these 3. A I-section is subjected to transverse shear force. At which layer maximum shear stress is developed?

6. A beam with a square section of 80 × 80 mm is simply supported at its ends. A load W is applied at the centre of the beam. If the maximum shear stress developed in beam section is 6 N/mm2, what is the magnitude of W ? (a) 2.56 kN (b) 25.6 kN (c) 51.2 kN (d) None of these 7. In a particular section of a beam, the maximum shear stress is double the average shear, what is the section of the beam?

(a) At neutral layer

(a) Rectangular section

(b) At top edge of flange

(b) Solid circular section

(c) At bottom edge of flange (d) None of these 4. A square section with side ‘a’ is used in a beam but the diagonals are placed in horizontal and vertical position. F is the shear force of a section, what is the maximum shear stress? (a) 1.5 F/a2

(b) 1.25 F/a2

(c) 1.125 F/a2 (d) None of these 5. A circular section of a beam with area 100 mm2 is subjected to a transverse shear force of 750 N. Magnitude of maximum shear stress developed in the section is (a) 10 N/mm2 2

(c) 7.5 N/mm

(b) 8.75 N/mm2 (d) None of these

(c) Square section with one diagonal vertical (d) None of these 8. Match the section and ratio of tmax/tmean Section

Ratio

A Rectangular

I 1.33

B Circular

II 1.125

C Square with vertical diagonal

III 2.0

D Thin tubular section

IV 1.5

(a) (b) (c) (d)

A B IV II I II IV I None of these

C III III II

D I IV III

Practice Problems 1. A 6-m-long beam of a rectangular section 40 mm wide × 60 mm deep is subjected to two loads of 2 kN each, at a distance of 2 m from each end. Calculate the principal stresses at a section under the load in layer 15 mm from the top. 2. A rolled steel section 60 mm × 40 mm is shown in Fig. 9.21. A transverse shear force of 50 kN is acting on this section. Determine shear stresses at points A, B and C. 3. A beam is made up by glueing four pieces of wood of size 50 mm × 80 mm to a 25 mm × 500 mm (deep) wooden beam. Allowable stress in the glued joint is 0.5 N/mm2. Determine the maximum allowable shear force on the section (Fig. 9.22). [Hint: Maximum shear stress in glued joint will occur along aa] 4. Section of beam is circular of diameter D with a square hole of diagonal and d ′ = 0.8 D as shown in Fig. 9.23. Shear force on the section is F. Compare the magnitude of transverse shear stress at neutral axis NA with the average shear stress.

MTPL0259_Chapter 09.indd 367

A 20 mm 20

B

C 20 mm

20

40

Figure 9.21

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368 Chapter 9

80

Glued joint a

a

Wooden beam

340

N a

A

a

80 d

50 50 125

D

Figure 9.22 Practice Problems 3

Figure 9.23

5. Show that for a thin tube acting as a beam subjected to a shear force F, the maximum shear stress due to F is two times the mean shear stress (Fig. 9.24). 2R   3  Hint: I NA = π tR , b = 2t , y = π  6. A beam of T-section with flange 150 mm × 20 mm and web 180 mm × 20 mm is subjected to a shear force F at a particular section. Determine how much percentage of shear force is shared by the web (Fig. 9.25). [Hint: Find shear force in the two portions of web, upper and lower from neutral axis] 7. A laminated wooden beam 180 mm × 180 mm is made from three planks of the size 60 mm × 180 mm glued together all along the length of the contact surfaces. The beam is simply supported over a span of 4 m. If the allowable shear stress in the glued joint is 1 N/mm2, find the maximum concentrated load which the beam can carry at its mid span, neglecting self-weight of the beam? 8. A beam of box with a cross-sectionof 400 mm × 300 mm as shown in Fig. 9.26 carries a vertical load at its mid span such that the maximum bending stress produced in beam section is 15 N/mm2. Beam is supported

R

O N

A

t

Figure 9.24

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Practice Problems 5

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Shear Stresses in Beams

369

150 20

180

20

Figure 9.25 over a span of 2.4 m. Upper and lower flanges are connected to the webs with the help of screws. Determine spacing between the screws as if the each screw can transmit 12.5 kN of shear force. 9. A T-section with a flange 100 mm × 15 mm and web 100 mm × 10 mm is subjected to a bending moment of 12 kN m producing tension in flange and a shear force of 16 kN. Determine the principal stresses at points, A, B and C as shown in Fig. 9.27. Note that point B lies on NA. 40

220

40 100 mm 100 mm

200 mm

A

15

N

A B

100 mm 300 mm

Figure 9.26 Practice Problems 8

100 mm 10 mm

C

Figure 9.27

Practice Problems 9

Special Problems 1. A beam section shown in Fig. 9.28 is subjected to a shear force of 10 kN. Determine the shear stresses at a, b and neutral layer. 2. The cross-sectionof a beam is a rhombus with diagonals 120 mm × 60 mm as shown in Fig. 9.29. If the vertical shear force at the section is 2.4 kN, determine the shear stresses along depth at an internal of 20 mm 3. A box section with a depth of 250 mm, a width of 150 mm and a wall thickness of 15 mm is used for a beam on simple supports. In a certain portion of the beam, there is a constant change in bending moment by 10 kN m per meter along the length of the beam. What is the maximum shear stress developed in the section? dMx    Hint: dx = Fx   

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370 Chapter 9

60 mm 20

20

20

a 30 b

60 mm

30 mm 60 mm

60 mm

Figure 9.28 Special Problems 1

Figure 9.29

Special Problems 2

Answers to Exercises Exercise 9.1: Fmax = 5 kN

Exercise 9.2: 33.64 mm

at y = 0, t = 4.6875 N/mm a y = , t = 3.515 N/mm2 4 a y= ,t=0 2

2

Answers to Multiple Choice Questions 1. (b) 2. (c) 3. (a)

7. (d) 8. (c)

4. (c) 5. (a) 6. (c)

Answers to Practice Problems

2 2 2 1. τ = 0.9375 N/mm , σ = 83.33 N/mm , p = 83.34 N/mm 6. 2 2 7. , σ = 83.33 N/mm , p = 83.34 N/mm 2 2 8. 2. tA = 0,tB = 27.9 N/mm , tc = 61.63 N/mm 9. 3. F = 5.8 kN

93.37% 48.6 kN 88.9 mm spacing 7.068 − 0.532, ±20.076, −35.64 N/mm2

4. tNA = 3.565tav

Answers to Special Problems 1. ta = 0, tb = 6.185 N/mm2, tNL = 3.33 N/mm2, INA = 87.3 × 104 mm4

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2. 0, 0.52, 0.74 and 0.66 N/mm2 3. 1.67 MPa

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10 Combined Bending and Direct Stresses CHAPTER OBJECTIVES The columns and struts are subjected to axial compressive load and if the load becomes eccentric; then, in addition to direct compressive stress, there will be a bending stress developed in the column/strut and these bending stresses are compressive and tensile as well. However many columns made of brick/stone masonry are incapable of withstanding tensile stress. 

Students will learn, through the chapter, to calculate the resultant stresses due to eccentric load and to mark the area of cross section, on which if the load is applied at any point there will not be any tensile stress anywhere in the section.



Similarly, frame of manufacturing machines are subjected to eccentric load and section of frame has to be designed so as to withstand the compressive and/or tensile stress.

Introduction In this chapter, we will take various sections of columns/struts such as circular, rectangular, I-section and hollow section subjected to eccentric loading causing a bending moment in addition to a compressive force on struts/columns. For each section, a core will be established, and if load is applied within the core, there will not be any tensile stress developed anywhere in column section. Wind loads on walls and chimneys cause bending moment on wall/chimney due to which tensile and compressive stresses are developed in section of wall/chimney. Depending upon the stress due to selfweight of the wall/chimney, resultant stresses at the base of wall/chimney will be determined.

Eccentric Axial Thrust on a Column A column of rectangular section B × D is shown in Fig. 10.1. G is the centroid of the section abcd. A vertical load P is applied at G along xx axis such that GG′ = e, eccentricity. If this load acts at the CG of the section, then a direct compressive stress is developed in section. Effect of the eccentric load is to produce bending moment, M = Pe, on the section producing tensile

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372 Chapter 10 P

P y

d G

x

G

GG= e P

x

e

a

C

e

G

G

b

y

P

. M

D

P.e

B

Figure 10.1

Column under eccentric load

and compressive stresses in the section. Along the edge ad, there will be maximum tensile stress due to bending and along edge bc, there will be maximum compressive stress due to bending. Say, Iyy = moment of inertia of section about yy axis =

DB3 12

(10.1)

Direct compressive load = P Direct stress,

σd =

Section modulus,

Zy =

P (compressive) BD I yy B /2

=

DB2 6

(10.2)

Maximum bending stresses developed in section,

σb = ±

MTPL0259_Chapter 10.indd 372

Pe 6 Pe =± Zy DB2

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Combined Bending and Direct Stresses

373

Resultant stress at edge bc =

P 6 Pe + (compressive) BD DB2

Resultant stress at edge ad P 6 Pe − BD DB2 If sb > sd, then resultant stress at edge ad will be tensile. Resultant stress distribution along xx axis is shown in the Fig. 10.2. =

sb

sd

sb + sd

sd a

sb − sd

b (Compressive) (Tensile)

(Compressive)

Resultant stress along xx axes of the column

Figure 10.2 Along edge bc sb + sd = compressive stress =

P 6 Pe + BD DB2

Along edge ad

σb − σd =

6 Pe P − = Tensile stress 2 BD DB

Note that if sb − sd = 0, no tensile stress is developed along edge ad Therefore,

6 Pe P < 2 BD DB

B B , eccentricity must be less than for no tensile stress to develop in section, when load is 6 6 applied along xx axis e
> 80 kmin 320 For mild steel column, the slenderness ratio should be greater than 80, so that Euler’s formula can be used to predict the buckling load of a column.

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556 Chapter 14

Example 14.4 For what length of a mild steel bar of 60 mm in diameter used as strut, the Euler’s theory is applicable, if the ultimate compressive strength is 0.33 kN/mm2 and E = 210 kN/mm2, when one end of the strut is hinged and the other end is fixed? Solution End conditions, one end hinged and the other end fixed Le = L/ 2

Equivalent length,

kmin = the minimum radius of gyration =

d 60 = = 15 mm 4 4

Le E >π kmin σc >π

2,10, 000 330

> π × 25.226 > 79.25 Le = kmin × 79.25 = 15 × 79.25 = 1,188.75 mm = Length of the column,

L 2

L = 2 Le = 1,188.75 × 2 = 1, 680 mm

Length should be greater than 1.68 m. Exercise 14.4 For what length of a cast iron column of 80 mm in diameter, the Euler’s theory is applicable, if σ c = 550 N/mm2 for CI and E = 102 kN/mm2, the column is hinged at both the ends?

Higher-order Differential Equation For any end conditions, the buckling load of a column/strut can be determined by a higher order differential equation as follows: d4 y d2 y P + k 2 2 = 0, where k 2 = 4 EI dx dx Solution of this differential equation is y = C1sin kx + C2 cos kx + C3 x + C4

(14.7)

Constants C1, C2, C3 and C4 are determined using end conditions.

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Struts and Columns

Example 14.5 Let us consider a column with both the ends fixed as shown in Fig. 14.5.

P B

Solution Column AB of length L is fixed at ends A and B. At the ends, both slope and deflection are zero, that is, the boundary conditions of the column are:

Fixed end

L

y

At end A, x = 0, y = 0, dy /dx = 0

x

At end B, x = L, y = 0, dy /dx = 0

Fixed end

A P

Differentiating Eq. (14.7) with respect to x, we get dy /dx = C1k cos kx − C2 k sin kx + C3

557

(14.8)

Figure 14.5

Taking x = 0, y = 0, x = L, y = 0, we get, 0 = C1sin 0 + C2 cos 0 + C3 x 0 + C4 0 = 0 + C2 + 0 + C4 0 = C1sin kL + C2 cos kL + C2 L + C4

(14.9) (14.10)

Taking dy /dx = 0 at x = 0, and x = L we get, 0 = C1k cos 0 − C2 k sin 0 + C3 0 = C1k − 0 + C3

(14.11)

0 = C1k cos kL − C2 sin kL + C3

(14.12)

Using Eqs (14.9), (14.10), (14.11) and (14.12), the following matrix is made: 0 1 0 1 sin kL cos kL L 1 =0 k 0 1 0 k cos kL − k sin kL 1 0 L 1 sin kL sin kL cos kL L = 1 0 k + k 0 1 1 0 kcos kL kcos kL − ksin kL 1 =

k 1 1 k k 0 + sin kL [ + ksin kL] + cos kL +L kcos kL 1 1 kcos kL kcos kL − ksin kL = k − kcos kL + ksin 2 kL + kcos2 kL − k cos kL − Lk 2 sin kL 0 = k + k (sin 2 kL + cos2 kL) − 2kcos kL − k 2 L sin kL = 2k − 2k cos kL − k 2 L sin kL = 2k (1 − cos kL) = k 2 L sin kL

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558 Chapter 14

Let us take 1 − cos kL = 0, cos kL = 0 kL = 0, 2π , 4π − − − So, kL = 2p (minimum significant value) P L = 2π EI

or, Squaring both the sides,

P 2 L = 4π 2 EI P=

Buckling load,

4π 2 EI L2

Moreover, sin kL = sin 2π = 0 (also), which will also give P = 4π 2 EI /L2 . Example 14.6 Let us take another case of a column fixed at one end and hinged at the other end as shown in Fig. 14.6. End A is fixed as shown. End B is hinged.

P B

Solution Boundary conditions at

Hinged end

L

y

x = 0, end A, y = 0, dy /dx = 0

x = L, end B, y = 0, d2 y /dx2 = 0 (moment at at hinged end is zero). Equation of deflection, y = C1sin kx + C2 cos kx + C3 x + C4

(14. 13)

dy = C1kcos kx − C2 ksin kx + C3 dx d2 y = −C1k 2 sin kx − C2 k 2 cos kx dx2

x A

Fixed end P

Figure 14.6 (14. 14) (14. 15)

At x = 0, y = 0 0 = C1 × sin 0 + C2 cos 0 + C3 × 0 + C4 = 0 + C2 + 0 + C4 At

(14. 16)

x = 0, dy /dx = 0 0 = C1kcos 0 − C2 ksin 0 + C3

At

MTPL0259_Chapter 14.indd 558

0 = C1 k + C3 x = L, y = 0

(14. 17)

0 = C1sin kL + C2 cos kL + C3 L + C4

(14. 18)

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Struts and Columns

559

x = L, d2 y /dx2 = 0, that is, bending moment is zero at the hinged end

At

0 = −C2 k 2 sin kL − C2 k 2 cos kL

(14. 19)

Using Eqs. (14.16)–(14.19), the following matrix is made: 0 1 0 1 k 0 1 0 =0 sin kL cos kL L 1 − k 2 sin kL − k 2 cos kL 0 0 k 0 1 1 0 k = sin kL cos kL L+ L 1 sin kL 2 2 − k sin kL − k cos kL 0 0 0 − k 2 sin kL = k (+ k 2 L cos kL) + 1( − k 2 sin kL cos kL + k 2 sin kL cos kL) + 1( − k 2 sin kL) = k 3 L cos kL + 0 − k 2 sin kL or

0 = kL cos kL − sin kL

or

tan kL = kL

or

tan 0 = q kL = 2.04p 2 k 2 L2 = 2.04π 2 P × L2 = 2.04π 2 EI

Buckling load,

P=

2.04π 2 ∫ EI 2

L



2π 2 EI L2

Exercise 14.5 Using higher-order differential equation, determine the buckling load for a column hinged at both the ends. Exercise 14.6 Using higher-order differential equation, determine the buckling load for a column fixed at one end and free at the other end. [Hint: say, y = C1sin kx + C2 cos kx + C3 x + a, where a = deflection at free end]

Rankine Gordon Formula Euler has developed a theory for the buckling load of long columns, but short columns get crushed under the compressive load and crushing stress (sc) is much greater than the stress (se) given by Euler’s buckling load. Some columns are of medium length and cannot be classified as short columns. Rankine

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560 Chapter 14

has combined the two loads, that is, the crushing load for short column and the Euler’s buckling load for long columns to determine the buckling load for any long or short column. Rankine’s load, PR is 1 1 1 or = + PR PC + Pe Rankine’s load, PR =

Pc Pe = Pc + Pe

where sc = crushing stress,

σc A Pc = σc A Pc 1+ 1+ 2 × L2 Pe π I min

 L  σ c ALe2 σ c ALe2 σ = = 2c ×  e  2 2 2 π EI min π EAk min π E  kmin 

2

Rankine has taken σ c /π 2 E = a, a constant that is determined experimentally Rankine’s load, PR =

σc A

1 + a ( Le /kmin )

2

both sc and a are determined experimentally. Le = equivalent length of column kmin = minimum radius of gyration of the section of the column Le /kmin is known as slenderness ratio of the column, an important parameter, and the buckling load depends on the slenderness ratio; more the slenderness ratio, less is the buckling load. Table 14.1 gives the values of sc and a for various materials. Table 14.1

Rankine’s constants for different materials

Material

Cast iron

sc(N/mm2)

550

250

320

500

35

1/1,600

1/9,000

1/7,500

1/5,000

1/3,000

a

Wrought iron

Mild steel

Medium carbon steel

Timber

Example 14.7 A cast iron column of a hollow circular section with an external diameter of 250 mm and a wall thickness of 45 mm is subjected to an axial compressive load. The column is 7 m long with both ends hinged. Taking factor of safety (FOS) as 8, determine safe value of P. Rankine’s constants are, σ c = 560 N/mm2, a = 1/1, 600 Solution External diameter, Wall thickness, Internal diameter,

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D = 250 mm t = 45 mm d = 250 − 2 × 45 = 160 mm

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Struts and Columns

Area of cross-section,

561

A = π /4( D2 − d 2 ) =

π (2502 − 1602 ) = 289.8 × 102 mm2 4

L = 7 m = 7, 000 mm

Length,

End condition: both the ends are hinged Le = L k 2min =

D2 + d 2 2502 + 1602 = = 55.06 × 102 mm2 16 16

L2e 7, 0002 = = 8, 899.4 2 kmin 55.06 × 102 a×

L2e 1 = × 8, 899.4 = 55.62 2 1, 600 kmin

Rankine’s load, PR =

σc A 560 × 289.8 × 102 = 2 1 + 55.62 1 + a ( L2e /kmin )

= 286.6 kN Safe load,

PR′ = 286.6 8 = 35.83 kN

Exercise 14.7 A hollow cast iron column has 200 mm outside diameter, 150 mm inside diameter and is 6 m long with both the ends fixed. It is subjected to an axial compressive load P. Taking a FOS as 6, determine safe Rankine’s buckling load. Constants are σ c = 550 N/mm2, a = 1/1, 600 for both ends hinged.

Johnson’s Parabolic Formula We have learnt so far that buckling loads depend upon the slenderness ratio (Le /kmin) of the column. As the slenderness ratio increases, buckling required for the column decreases. Based on this principle, Johnson has suggested a formula: Working stress, σ w = σ c′ [1 − ϕ ( Le /k )] as function of Le /k , slenderness ratio Taking ϕ ( Le /k ) = b(L2e /k 2 ) , where b is a constant

σ w = σ c′ 1 − b(L2e /k 2 ) , an equation of a parabola. σ c′ = allowable stress in compression taking into account the FOS (factor of safety) = 110 N/mm2 for mild steel. Constant,

b = 0.00003 for hinged ends = 0.00002 for fixed ends.

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562 Chapter 14

Straight line formula σ w′ = σ c′[1 − C(Le /k )], applicable for column for which slenderness ratio is greater than 90. For structural steel, σ c′ = 140 N/mm2 in compression C = 0.0054 for hinged ends = 0.0038 for riveted ends. Example 14.8 A strut is built-up of two 100 mm × 45 mm channels placed back-to-back at a distance of 100 mm apart and riveted to two flange plates each 200 mm × 10 mm symmetrically, properties of one channel are: Area, A = 7.41 cm2 , I xx = 123.8 cm4 , I yy = 14.9 cm4 , and x = 1.4 cm (distance of CG from outer edge of web). If the effective length is 5 m, calculate the working load for the strut using Johnson’s parabolic formula: where σ w = σ c′[1 − b(Le /k )2 ] , where constant b = 0.00003 for hinged ends and σ c′ = 110 N/mm2 Solution Area of cross-section of built-up section, A = 2 × 200 × 10 + 2 × 741 = 4, 000 + 1, 482 = 5, 482 mm2 Ixx of built-up section = 2 × 123.8 × 104 + 2 × 200 × 10 × (55)2 = 247.6 × 104 + 1210 × 104 = 1457.6 × 104 mm4 Iyy of built-up section (Fig. 14.7) I yy = 2 × 14.9 × 104 + 2 × 741 × (50 + 14)2 + 2 × 10 × (2002/12) = 29.6 × 104 + 607 × 104 + 1, 333.33 × 104 = 1, 970 × 104 mm4 Y

45

10 mm

100 mm

X

G

G

G

x 14mm

100 10 mm

flange 200 mm Y

Figure 14.7 Example 14.8

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563

Ixx < Iyy

Now,

1, 457.6 × 104 = 2, 659 mm2 5, 482 L=5m

2 kmin =

L2 5, 0002 = = 9, 402 2 2, 659 kmin b = 0.00003 2

b

L = 0.00003 × 9402 = 0.282 2 kmin

(

)

σ w = σ c′ = 1 − b L2e /k 2  = 110(1 − 0.282) = 110 × 0.718 = 78.98 N/mm2 Pw = working load = σ w × A = 78.98 × 5482 = 433 kN Exercise 14.8 A stanchion is built up of 3 − 200 × 100 mm Rolled Steel Joist (RSJ) as shown in Fig. 14.8. If the length of the stanchion is 6 m, calculate the working load. The working stress is given by: 1  = 110 1 − ×  200

L 2  N/mm k

Properties of one RSJ, area = 2,527 mm2, I xx = 1, 696.6 × 104 mm4 , I yy = 115.4 × 104 mm4 and web thickness = 5.4 mm.

200

102.7 mm 100

Figure 14.8

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100

Exercise 14.8

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564 Chapter 14

Eccentric Loading of Columns

a

P B

A column AB of length L loaded eccentrically e (eccentricity with respect to axis of column) buckled under load P as shown in the Fig. 14.9. Say, deflection at end B is a, when the column has buckled. Consider a section at a distance x from end A, where deflection is y. Bending moment at the section, M = P(a + e − y ) or,

EI (d2 y /dx2 ) = P (a + e − y )

or,

d2 y Py P(a − e) + = . EI dx2 EI

e x L

y

x

x A

(14.20)

Figure 14.9

Solution of this differential equation is, y = Asin mx + Bcos mx + (a + e) where

m = P /EI , A and B are constants

At end A,

x = 0, y = 0, dy /dx = 0

Pulling

x = 0, y = 0 0 = Asin 0 + Bcos 0 + (a + e) 0 = 0 + B + ( a + e) B = − ( a + e)

or constant,

dy /dx = Am cos mx − Bm sin mx

Moreover,

dy /dx = 0, at x = 0 0 = Am cos 0 − Bm sin 0 0 = Am m ≠ 0, because m = P /EI , therefore, constant A = 0. Finally, At the end B,

y = −(a + e) cos mx + (a + e) x = L, y = a, putting this boundary condition a = −(a + e)cos mL + (a + e)

or,

−e = −(a + e) cos mL (a + e) = e sec mL

Maximum bending moment occurs at fixed end A. M max = P(a + e) = Pe sec mL

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565

sb = maximum stress due to bending = P e sec mL /Z , secant formula, where Z = section modulus = I /yc , I = moment of inertia yc = distance of extreme layer in compression from neutral layer sc = direct compressive stress. = P /A

Resultant stress at fixed end,

σ r = σb + σc =

P Pe + sec mL A Z

=

P P Pe + sec L A Z EI

In this case, we have considered that one end of the column is fixed and the other end is free. In this case, Le = 2L, that is, equivalent length, Le = 2L. Secant formula can be generalized for any column with any type of end conditions as follows: sr = resultant maximum stress =

L P Pe sec e + 2 A Z

P , where Le is the equivalent length. EI

Example 14.9 A steel tube of 80 mm outer diameter, 50 mm inner diameter and 3 m long is used as a strut with both ends hinged. The load is parallel to the axis of the strut but is eccentric. Find the maximum value of eccentricity so that the crippling load on strut is equal to 50 per cent of the Euler’s crippling load. Yield strength = 320 N/mm2, E = 210 kN/mm2. Solution End conditions: both ends hinged Length, Le = L = 3 m E = 210 kN/mm2 Steel tube D = 80 mm d = 50 mm I=

π π ( D4 − d 4 ) = (804 − 504 ) 64 64

= 170.38 × 104 mm4 Euler’s buckling load, Pe =

π 2 EI π 2 × 210 × 103 × 170.38 × 104 = 3, 000 × 3, 000 L2

= 3,92,369 N

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566 Chapter 14

Crippling load due to eccentricity, P = Pe / 2 = 1, 96,184.7 N 1, 96,184.7 2,10, 000 × 170.38 × 104

P = EI

= 54.83 × 10−8 = 7.404 × 10−4 Le = L = 3,000 mm Le P × = 1, 500 × 7.404 × 10−4 = 1.11 rad EI 2 = 63.64° Sec

Le 2

P = 2.252 EI A=

π (802 − 502 ) = 30.63 × 102 mm2 4

Z =

I 170.38 × 104 = = 4.2595 × 104 mm3 40 40

σ r = σ yp = 320 =

L P + Pe sec e A 2

P EI

1, 96,184.7 1, 96,184.7 + × e × 2.252 3, 063 4.2595 × 104

320 = 64.05 + e × 10.37 Eccentricity, e = (320 − 64.05) /10.37 = 24.68 mm Exercise 14.9 A steel tube of 80 mm outer diameter, 60 mm inner diameter and 2.8 m long is used as a strut with the ends hinged. The load is parallel to the axis of the strut but is eccentric. Find the maximum value of the eccentricity so that the crippling load on the strut is 60 per cent of the Euler’s buckling load. Given yield strength = 320 N/mm2. E = 210 kN/mm2

Professor Perry’s Approximate Formula Professor Perry has given an approximate formula for the buckling load of eccentrically loaded columns. In the secant formula for maximum stress due to buckling load,

σ max =

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L P + Pe sec e A Z

P EI

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567

Pe = π 2 EI /L2e

If Euler’s buckling load,

Pe L2e π2

or,

EI =

or,

P Pπ 2 = EI Pe L2e P π = EI Le

P Pe

putting the value in secant formula, Pe sec

Le 2

L π P P = Pe sec e × × 2 Le EI Pe = Pe sec

π 2

P Pe

Professor Perry found through experiments that the expression, π P 1.2 Pe sec = 2 Pe Pe − P where Pe = Euler’s buckling load (without eccentricity) P = buckling load of column with eccentricity. σ e = Pe /A, σ o = P /A Moreover, where A is the cross-sectional area of the column. sec Now,

π 2

1.2σ e P = Pe σ e − σ o L P Pe P + sec e 2 EI A Z P Pe 1.2σ e = + × A Z σe − σo

σ max =

Section modulus, Z = Ak 2 /yc , where yc is distance of extreme layer in compression from neutral layer. y 1.2σ e P σ max = + Pe × c 2 × A σc − σo Ak However,

P /A = σ o

σ max = σ o + σ o e

1.2σ e yc × k 2 σe − σo

 ey 1.2 σ e  = σ o 1 + 2c × σ e − σ o  k 

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568 Chapter 14

σ max = σ , allowable stress  ey 1.2 σ e  σ = σ o 1 + 2c ×  σ k  e − σo  σ  eyc 1.2σ e  σ − 1 = k 2 × σ − σ o e o or,

σ   σ o  1.2eyc  σ − 1 1 − σ  = k 2 . o e

This is Professor Perry’s approximate formula, knowing the value e, yc and k, one can determine the stress so due to load P on the eccentrically loaded column. Example 14.10 A column is of I-section, 125 mm × 250 mm. Find the safe load for this column of length of 5 m, hinged at both the ends using Professor Perry’s formula if maximum compressive stress is limited to 80 N/mm2 (Fig. 14.10). For RSJ, I xx = 37.18 × 106 mm4 , A = 3,553 mm2 , I yy = 1.93 × 106 mm4 E = 200 GPa, eccentricity from y-axis is 30 mm. Solution Eccentricity is about x-x axis, e = 30 mm, we will take Iyy as moment of inertia Length, L = 5,000 mm

Y

I yy = 1.93 × 106 mm4 k2 = =

I yy A 1.93 × 106 3, 553

250 mm

G e

= 543.2 mm2 yc = 125 / 2 = 62.5 mm e = 30 mm. eyc 30 × 62.5 = = 3.4517 543.2 k2

Y

Figure 14.10

σ = 80 N/mm2 , allowable stress σe =

π 2 Ek 2 π 2 × 2, 00, 000 × 543.2 = (both ends hinged) 5, 000 × 5, 000 L2

= 42.88 N/mm2

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569

 80   σo  ey  σ − 1 1 − 42.88  = 1.2 k 2 = 1.2 × 3.4517 o

 80   σo   σ − 1 1 − 42.88  = 4.142 o

(80 − σ o )(42.88 − σ o ) = 4.142 × 42.88 × σ o 80 × 42.88 − 48.88 − 80σ o + σ o2 = 177.61σ o 3, 430.4 − 122.88σ o + σ o2 = 177.61σ o

σ o2 − 300.49σ o + 3430.4 = 0 σo =

300.49 − (300.49)2 − 4 × 3, 430.4 2

=

300.49 − 90, 294.2 − 13, 721.6 2

=

300.49 − 276.7 76.21 = 2 2

σ o = 38.10 N/mm2 Safe load

= 38.10 × 3, 553 = 1, 35, 369 N = 135.37 kN

Exercise 14.10 A 400 mm × 140 mm RSJ is used as a sturt with hinged ends, having a length of 6 m. Using Prof. Perry’s formula, determine the safe load for: (a) eccentricity along x-x axis, e = 24 mm, (b) minimum allowable permissible compressive stress is 75 MPa and (c) for the joist A = 7,846 mm2, Ixx = 20,458.4 × 104 mm4, I yy = 622.1 × 104 mm4 , E = 210 kN/mm2

Long Columns with Eccentricity in Geometry A column AB of length L with both the ends hinged has initial eccentricity e′ in the centre. Assuming deflection curve to be sinusoidal (Fig. 14.11). Eccentricity at any section, y ′ = e′ sin

πx L

(14.21)

e′ = maximum initial eccentricity at centre.

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570 Chapter 14 P

Taking a section at a distance of x from end A, considering column buckled under load P. Final deflection = y Change in deflection = y-y′ (as shown) BM at the section at a distance x from A as shown

EI

x

y ′ = e′ sin

(14.22)

πx , as given in Eq. (14.21) L

A

x y

2

d y d y′ EI 2 − EI = −Py dx dx 2

c c

y

x

d2 (y − y ′) = − Py dx2

2

c e

L

M = −Py or

B

P

Figure 14.11 dy ′ π πx = e′ cos dx L L d2 y ′ π2 πx = − e sin ′ 2 2 L dx L

(14.23)

Putting this value in Eq. (14.22) d2 y π2 π x −P + e ′ 2 sin y = 2 L EI dx L d2 y P π2 πx + y e = − sin ′ 2 2 L dx EI L

or,

(14.24)

y = Ae′(sin π x /L)

Let us assume the solution as

dy π πx = A e′ cos dx L L

π2 πx d2 y = − Ae′ 2 sin 2 dx L L Putting this value in Eq. (14.24), − Ae′

π2 πx P π2 πx sin + y = − e ′ sin 2 2 L EI L L L

Putting the assumed value of y y = − Ae′ sin − Ae′ or,

MTPL0259_Chapter 14.indd 570

πx L

π2 πx P πx π2 πx sin + × Ae′ sin = −e′ 2 sin 2 L EI L L L L − Ae′

π2 P π2 + Ae ′ = − e ′ L2 EI L2

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571

 π2 P  π2 Ae′  2 −  = e′ 2 L  L EI 

or

 π2 P  π2 A 2 −  = 2  L EI  L

However,

 PL2  A 1 − 2  = 1  π EI  π 2 EI /L2 = Pe, Euler’s buckling load  P A 1 −  = 1  Pe  A = Pe ( Pe − P)

Constant, Finally, the deflection curve equation will be, y=

Pe πx e′ sin Pe − P L

Maximum deflection occurs at the centre, where x = L /2 ymax =

Pe × e′ Pe − P

M max = Pymax

Maximum bending moment,

=

P Pe e′ Pe − P

σ max = maximum compressive stress at central section of column M y P P Pe = σ o + max = + × e′ c 2 Z A Pe − P Ak where k = radius of gyration yc = distance of extreme layer  P e′ yc  σ max = σ o 1 + e  Pe − P k 2  However, Pe = σ e A, Euler ’s load. P = σ e A, putting these values; load applied.  e′ y  σe σ max = σ o 1 + × 2c  σ − σ k   e o or

 σ max  σe e′ yc  σ − 1 = σ − σ × k 2 e o o  σ max   σ o  e′ yc  σ − 1 1 − σ  = k 2 , the formula to obtain so. o e

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572 Chapter 14

Example 14.11 A 5-m-long hollow circular steel strut having an outside diameter of 120 mm and an inside diameter of 80 mm with both the ends hinged is initially bent. Assume that the centre line of the strut as sinusoidal with maximum deviation of 6 mm. Determine the maximum stress developed due to an axial load of 100 kN. E = 210 kN/mm2 Solution Length,

L=5m

Eccentricity,

e′ = 6 mm

Area of cross-section,

A = (π / 4)(1202 − 802 ) = 6283.2 mm2

Moment of inertia,

I = (p /64)(1204 − 804 ) = 816.816 × 104 mm4 k2 =

I 816.816 = × 104 = 1, 300 mm2 A 6, 283.2

Pe =

π 2 EI π 2 Ek 2 , σ = e L2 L2

σe =

π 2 × 210 × 1, 000 × 1, 300 = 107.8 N/mm2 5, 000 × 5, 000

P = 100 kN

σo = Distance,

100 × 1, 000 = 15.9 N/mm2 6, 283.2

yc = 120 2 = 60 mm e′ = 6 mm  σ max   σ o  e ′ yc  σ − 1 1 − σ  = k 2 o e 15.9  6 × 60  σ max    15.9 − 1 1 − 107.8  = 1, 300  σ max   15.9 − 1 × 0.9069 = 0.2769

σ max − 1 = 0.30535 15.9 σ max = (1 + 0.30535) × 15.9 = 20.7 N/mm2

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573

Exercise 14.11 A 4-m-long hollow circular aluminium strut having an outside diameter of 150 mm and an inside diameter of 100 mm with both ends hinged is initially bent. Assuming that the centre line of the strut as a sinusoidal curve with maximum deviation of 8 mm. Determine the maximum stress developed due to an axial load of 60 kN. E = 70 kN/mm2

Professor Perry Robertson Formula Professor Perry gave the relationship between σ e , σ o and σ max (discussed in the last section ‘Long Columns with Eccentricity in Geometry’) for a long column with eccentric loading as follows:  σ max   σ o  1.2eyc  60 − 1 1 − σ  = k2  e

(14.25)

where e = eccentricity in loading (load is not axial) Moreover,  σ max   σ o  e′ yc  σ − 1 1 − σ  = k 2 o e

(14.26)

where e′ = eccentricity in geometry of strut e1 = 1.2e + e′

say,

For an initially bent and eccentrically loaded column, formula has been modified as  σ max   σ o  e1 yc  σ − 1 1 − σ  = k 2 o e

(14.27)

σ   σo   σ − 1 1 − σ  = λ o e

(14.28)

where σ max = allowable stress, s say e1 yc /k 2 = λ , then or

(σ − σ o )(σ e − σ o ) = λ σ oσ e Rearranging the terms, we get,

σ o2 − σ o [σ + σ e (1 + λ )] + σ σ e = 0,

(14.29)

a quadratic in so. 2

σo =

σ + σ e (1 + λ )  σ + σ o (1 + λ )  −   − σ oσ e 2 2 

where λ = e1 yc / k 2 and σ e = π 2 E /( L2 /k 2 ) for both ends hinged.

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574 Chapter 14

Professor Robertson after having experimental observations came to the conclusion that λ = 0.003( L / k )is valid for a large number of observations. However, l depends on allowable stress in tons/in2 and σ e = π 2E /( L /k )2 in tons/in2. However, in the Eq. (14.28), there are ratios of σ / σ o and σ o / σ e, dimensionless and λ = e1 yc /k 2 also dimensionless, the expression for l remains the same for stresses in N/mm2. Example 14.12 Two 200 × 70 mm 6-m-long MS channels are welded together at their toes to form a box section of 200 × 140 mm. The box section is used as a strut with both the ends hinged. Estimate the safe load for this strut using Prof. Perry Robertson’s formula taking allowable stress, σ = 250 MPa, λ =0.003(L /K ) (Fig. 14.12). For each channel Area of cross-section, A = 1,777 mm2

Y′

Tve

I xx = 1,161.9 × 104 mm4 , I yy = 84.2 × 104 mm4 x = 19.7 mm, E = 210 kN/mm Solution Total,

G

x′

70 2

G′

x′ 200 mm

19.7 mm

Y′

Figure 14.12

70

Built-up section

A = 2 × 1, 777 = 3, 554 mm2 I x ′ x ′ = 2 × 1,161.9 × 104 mm4 = 2, 323.8 × 104 mm4 Iy ′ y ′ = 2 × 84.2 × 104 + 2 × 1, 777(70 − 19.7)2 = 168.4 × 104 + 899.2 × 104 =1, 067.6 × 104 mm4

Note that

Iy ′ y ′ < Ix ′ x ′ k2 =

Iy′ y′ A

=

1, 067.6 × 104 = 3, 004 mm2 1, 777 × 2

k = 54.8 mm Length of the strut,

L = 6,000 mm  6, 000  λ = 0.003  = 0.328  54.8 

Allowable stress,

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σ = 250 MPa

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π 2 Ek 2 π 2 × 210 × 1, 000 × 54.82 = = 172.84 N/mm2 L2 6, 0002 σ e (1 + λ ) = 172.84 (1 + 0.328) = 229.6 N/mm2

σe =

σ + σ e (1 + λ ) 250 + 229.6 = = 239.8 N/mm2 2 2 σ0 =

 σ + σ e (1 + λ )2  σ + σ e (1 + λ ) −   − σ σ e 2 2 

= 239.8 −

(239.8)2 − 250 × 172.84

= 239.8 − 119.5 = 120.3 N/mm2 Safe load = 120.3 × 3, 554 = 4, 27, 546 N = 427.546 kN Exercise 14.12 A stanchion is built up of a 500 mm × 180 mm RSJ with 200 mm × 20 mm plates riveted to each flange as shown in Fig. 14.7 of example 14.8. Estimate the safe load for this stanchion of a length of 5 m, with both the ends hinged using Prof. Perry Robertson’s formula, taking allowable stress, σ = 280 MPa, λ = 0.003( L /k ) for the joist. A = 9,550 mm2, I xx = 3.8529 × 106 mm4 I yy = 10.639 × 106 mm4 E = 208 GPa Hint: [Taking Iyy, calculate, k and se]

Lateral Loading of Strut with Point Load A strut or column has buckled under the combined action of axial thrust P and transverse load (at centre) as shown in Fig. 14.13. Lateral loading produces deflection in the strut and axial thrust produces a bending moment on account of deflection. Figure 14.13 shows a column AB of length L, hinged at both the ends and carrying a central transverse load W as shown. Due to W at C, there are horizontal support reactions, W /2 each. Consider a section at a distance x from end A Bending moment at the section, M = − Py − (Wx / 2) or,

MTPL0259_Chapter 14.indd 575

EI

d2 y Wx = − Py − 2 dx2

P W 2

13

B

C

L x

W

y

x x

W 2

A P

Figure 14.13 Lateral point load on a strut

5/23/2012 11:00:50 AM

576 Chapter 14

Wx d2 y P y=− + 2 2 EI dx EI

or

(14.30)

Solution of this differential equation is y = A cos mx + Bsin mx −

Wx 2P

where m = P /EI Boundary conditions are x = 0, y = 0, 0 = A cos 0 + Bsin 0 − 0 0 = A + 0−0 or constant, Moreover,

A=0 dy W = Bm cos mx − dx 2P x = L /2 , slope, dy /d x = 0 0 = Bm cos m B=

L W − 2 2P

L W sec m 2 Pm 2

Equation of deflection curve becomes y=

L Wx W sec m × sin mx − 2 Pm 2 2P

Maximum deflection occurs at centre at x = L /2 ymax =

W mL mL WL sec sin − 2 Pm 2 2 4P

ymax =

W mL WL tan − 2 Pm 2 4P

Maximum bending moment occurs at the centre M max = Pymax +

W L × 2 2

mL WL  WL  W = P tan − +  2 Pm 2 4P  4

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Struts and Columns

Mmax =

577

W mL , tan 2m 2

putting the value of m M max =

W 2

EI P tan L P EI

σ max = σ o + σ b = direct stress + bending stress W P = + 2 A

EI P tan L P EI Z

Z = section modulus = I /yc = Ak 2 /yc yc = distance of extreme layer in compression from neutral layer

σ max =

or,

P Wyc + A 2 Ak 2

EI tan P

P L . EI 2

Example 14.13 A 4-m-long horizontal pin-ended strut is formed from a standard T-section of 150 mm × 100 mm × 12.5 mm. The axial compressive load is 60 kN. A lateral concentrated load of 6 kN acts at the centre of the strut. Find the maximum stress developed if the xx axis is horizontal and the table of the T-section forms the compressive face. The centre of gravity is 24 mm away from the edge of the table (Fig. 14.14). Given Solution Axial load, Lateral load,

I xx = 250 × 104 mm4, A = 3,100 mm2, E = 200 GPa

x

P = 60 kN W = 6 kN I xx = 250 × 104 mm2

12.5 mm

Table Y 100

G

Y

x

A = 3,100 mm Length, T-section,

2

L=4m yc = 24 mm

24 150

Figure 14.14

Ak 2 = I xx = 250 × 104 mm2 P = EI

MTPL0259_Chapter 14.indd 577

60, 000 200 × 1, 000 × 250 × 104

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578 Chapter 14

= 3.464 × 10−4 EI = 2, 886.75 P Wyc EI 6, 000 × 24 × 2, 886.75 = 83.1384 × = 2 2 Ak P 2 × 250 × 104 P 60, 000 = = 19.35 A 3,100 tan

4, 000  P L  = tan  3.464 × 10−4 ×   EI 2 2  = tan(0.6928 rad ) = tan(39.695°) = 0.83

σ max =

P Wyc EI P L + × tan , putting the values A 2 Ak 2 P EI 2

= 19.35 + 83.1384 × 0.83 = 69.0 + 19.35 = 88.35 N/mm2 Exercise 14.13 A circular steel strut of a diameter of 25 mm and length of 1 m is subjected to an axial thrust of 12 kN. In addition a lateral load W acts at the centre of the strut. If the strut is to fail at σ max = 320 N/mm2 , determine the magnitude of W. Given E = 210 kN/mm2.

Strut with an Uniformly Distributed Lateral Load P

A strut subjected to axial thrust P and lateral load w per unit length is shown in Fig. 14.15. The length of the strut is L and it is hinged at both the ends A and B. Consider a section at a distance x from end A. The bending moment at the section, M = − Py − EI

or,

MTPL0259_Chapter 14.indd 578

wL 2

wL wx 2 x+ 2 2

d2 y wx = − Py − ( L − x) 2 dx2

d2 y P wx + y =− ( L − x) 2 EI dx2 EI

B w C

L

y x wL 2

A P

Figure 14.15

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Struts and Columns

579

Solution of this differential equation, y = complementary function + particular integral. Complementary function = A cos mx + Bsin mx, where P /EI = m. =

Particular integral,

wx2 wLx wEI − − 2 2P 2P P

or,

y = A cos mx + Bsin mx +

At

x = 0, y = 0 at end A

wx2 wLx wEI − − 2 2P 2P P

0 = A cos 0 + Bsin 0 + 0 − 0 −

wEI P2

or constant,

A = wEI /P2

Then,

dy wx wL = − Am sin mx + Bm cos mx + − dx P 2P

(14.31)

dy /dx = 0 at x = L / 2, at the centre of the beam due to symmetrical loading about centre C, x = L / 2, slope, dy /dx = 0, putting this value, 0 = − Am sin Am × sin m

mL L wL wL + Bm cos m + − 2 2 2 P 2P

L L = Bm cos m 2 2 B = A tan

mL 2

A = wEI /P2

However, So constant,

B=

wEI mL tan P2 2

Finally, the equation of y y=

wEI wEI L wx2 wLx wEI cos m x + tan m sin m x + − − 2 2 2P 2P P2 P2 P

At the centre, deflection is maximum, therefore, taking ymax =

MTPL0259_Chapter 14.indd 579

x=

L 2

wEI L wEI L L wL2 wL2 wEI cos m + 2 tan m sin m + − − 2 2 2 P 2 2 8P 4P P P

5/23/2012 11:00:59 AM

580 Chapter 14

L L  2 2 wEI  cos m 2 + sin m 2  wL2 wEI = 2   − 8P − P 2 L P cos m   2 =

wEI L wL2 wEI × sec m − − 2 2 8P P2 P

Bending moment at the centre, M max = − Pymax −

wL L wL2 × + 2 2 8

 wEI L wL2 wEI  wL2 = − P  2 sec m − − 2 − 2 8P 8 P   P =−

L wL2 wEI wL2 wEI sec m + + − 2 P 2 8P 8 P

=−

wEI mL wEI sec + P 2 P

=−

wEI  mL  − 1 , putting the value of m  sec  P  2

M max = − Maximum stress,

wEI P

 P L   sec EI × 2 − 1  

σ max = σ o + σ b =

P M max P M max yc + = + Z A A Ak 2

=

y P  wEI  + c2 ×  2  A Ak  P 

=

P wEyc  P L  + sec × − 1 as I = Ak 2  A P  EI 2 

 P L   sec EI × 2 − 1  

Example 14.14 A circular rod of diameter 50 mm is supported horizontally through pin joints at its ends and carries an uniformly distributed load of 1 kN/m run throughout its length and an axial thrust of 25 kN. If its length is 2.4 m, estimate the maximum stress induced in the rod, E = 200 GPa.

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581

Solution Axial thrust, P = 25,000 N Diameter, d = 50 mm Area, A = (π / 4)d 2 = 1963.5 mm2 P 25, 000 = = 12.73 N/mm2 A 1, 963.5 w = 1 kN/m = 1 N/mm E = 2,00,000 N/mm2 yc = 25 mm Length,

L = 2.4 m wEyc 1 × 2, 00, 000 × 25 = = 200 P 25, 000 I=

π d4 π × 504 = = 30.68 × 104 mm4 64 64

P 25, 000 = = 40.743 × 10−8 EI 2, 00, 000 × 30.68 × 104 P = 6.38 × 10−4 EI P L 2, 400 × = 6.38 × 10−4 × = 0.766 rad EI 2 2 = 43.89° sec 43.89° = 1.387

σ max =

P wEyc  P L  + sec × −1  A P  EI 2 

= 12.73 + 200(1.387 − 1) = 12.73 + 77.52 = 90.25 N/mm2

MTPL0259_Chapter 14.indd 581

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582 Chapter 14

Exercise 14.14 A rod of rectangular section 80 mm × 40 mm is supported horizontally through pin joints at its end and carries a vertical load of 3,000 N/m and an axial thrust of 100 kN. If its length is 2.0 m, estimate the maximum stress induced. E = 208 × 103 N/mm2

Energy Approach To determine the critical load at which a column ceases to be in a stable equilibrium, energy criterion can be used. Moreover, in a situation where the exact solution of a differential equation is not possible or difficult to obtain, energy approach can be used to get a good approximate solution. However, this approach converges to the exact solution if a large number of terms are taken to represent the deflection curve. Now, Energy stored in a system = work done by external loads

δ U + δ Ve = 0 δVe = external work done due to virtual displacement. δ ( U+Ve ) = 0 δ ( Kp ) = 0 U + Ve = K p = a constant referred to as total potential of the system. 2

L  ∂2 y  1 EI dx 2 ∫0  ∂x2 

Energy stored,

U=

External work done,

Ve = −

2

L

P  ∂y    dx 2 ∫O  ∂x 

L EI K p = U + Ve = − ∫  2 0 

2 2  ∂2 y  P  ∂y   −    dx  ∂x2  2  ∂x   

For small values of P, Kp is positive for any non-trivial admissible function y(x). The critical load can be obtained by assuming suitable admissible function. y ( x) = A sin

πx . L

Critical condition is reached when constant Kp = 0 2

2

L  ∂2 y  P L  dy  1 − ∫ EI  2  dx = cr ∫   dx 2 0  dx  2 0  dx 

MTPL0259_Chapter 14.indd 582

(14.32)

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Struts and Columns

Example 14.15 Let us consider a case of column that is fixed at one end and free at the other end, with boundary conditions (Fig. 14.16). Solution Deflection, y = 0 and x = 0 slope, dy /dx = 0, x = 0 Moment at end B is zero, that is, 2

EI

583

P B B′ Free end

y = f (x)

L

2

d y d y = 0 or EI 2 = 0 at x = L. dx 2 dx

x

Let us take a shape function, y ( x) = Ax2 dy = 2Ax dx

A

Fixed end

Figure 14.16

d2 y = 2A dx2 Putting these values in Eq. (14.32) L Pcr L 2 2 1 2 2 4A x dx EI ( A ) d x = 2 ∫0 2 ∫0 L

L

0

0

EIL = Pcr

L3 3

EI ∫ dx = Pcr ∫ x2 dx

Pcr = 3EI /L,2 but the exact value given by Euler’s theory is π 2 EI / 4 L2 = 2.467( EI /L2 ) Let us take more terms in the deflection shape function y ( x) = Ax2 (3L − x) = 3A x2 L − Ax3 dy = 6AxL − 3Ax2 dx d2 y = 6AL − 6Ax dx2 2

 dy  2 3 2 2 2 2 4   = 36A x L + 9A x − 36A x L dx

MTPL0259_Chapter 14.indd 583

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584 Chapter 14

2

 d2 y  2 2 2 2 2  dx2  = 36A L + 36A x − 72A Lx. Putting these values in the equation of total potential Kp, L

EI (36A2 L2 + 36A2 x2 − 72A2 Lx)dx 2 ∫O =

Pcr L (36A2 x2 L2 + 9A2 x4 − 36A2 x3 L)dx, (taking 9A2 common) 2 ∫0 L

L

0

0

2 2 2 2 4 3 or, EI ∫ (4L + 4x − 8 Lx) dx = Pcr ∫ (4 x L + x − 4 x L) dx

4 x2 8 L x2 EI 4 L x + − 3 2

L

2

0

L

4 x3 L2 x5 4 x4 L = Pcr + − 3 5 4 0

  4 L5 L5 4L3 − 4 L3  = Pcr EI  4L3 + + − L5 3 3 5    4  8 EIL3   = Pcr L5    3  15  Pcr =



EI 15 4 2.5 EI × × = L2 8 3 L2 2.467 EI (as given by Euler’s theory) L2

Exercise 14.15 Consider a column with both ends hinged as shown in Fig. 14.17. Take the shape function y = A( x4 − 2 L x3 + L 3 x), using energy approved, that is, total potential, derive expression for its buckling load. Problem 14.1 A column made of an aluminium alloy of rectangular section b × d and a length of 600 mm is fixed at end B as shown in Fig. 14.18. Two smooth and rounded fixed plates at the end A restraint the end A to move in one of the vertical planes of symmetry of the column. Determine the ratio of b /d of two sides of cross-section for the efficient design against buckling.

A

B

L

Figure 14.17

Exercise 14.15

E = 67 GPa, P = 15 kN, FOS = 2

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585

Solution For the efficient design, the buckling load for two possible modes of failure should be the same. There are two types of end conditions in the present case: (1) one end is fixed and the other end is free, I1 = bd 3 /12. (2) one end is fixed and the other end is hinged, I 2 = db3 /12. Taking Euler’s formula for buckling 2 EI 2 2 π 2 EI1 π = L2 4 L2

2

A b

or, 8 I 2 = I1 8 × db3 bd 3 = 12 12 8b2 = d 2

d B

Figure 14.18

b = 0.353 d FOS = 2 Load, P = 15 kN Design load = 2 × 15 = 30 kN Length = 600 mm

π 2 EI1 = 30, 000 4 L2 I1 =

=

4 × 30, 000 × 6002 = 65329.48 mm4 67, 000 × π 2 bd 3 0.353 d × d 3 = = 5, 329.48 12 12

d 4 = 22, 20, 832.168 d = 38.6 mm b = 0.353 × 38.6 = 13.63 mm Dimensions of the section are 13.63 mm × 8.6 mm.

MTPL0259_Chapter 14.indd 585

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586 Chapter 14

Problem 14.2 A long strut AB of length L is of uniform section throughout. A thrust P is applied at the ends eccentrically on the same side of the centre line with eccentricity at the end B twice than that at end A (Fig. 14.19). Show that the maximum bending moment occurs at a distance x from end A, where  2 − cos kL  tan kx =   sin kL  where k =

P F

B 2e

L y x

P EI

A

F

e P

Figure 14.19 Solution The strut is buckled under the thrust load P. P at two ends produce a couple P(2e − e) (cw) that is to be balanced by anticlockwise couple of horizontal reactions F each at A and B, as shown F × L (ccw). Force F is unknown. so 2 Pe − Pe − FL = 0 F=

or,

Pe L

(14.33)

Consider a section at a distance x from end A, Bending moment at the section, M = − P (2e + y ) + F ( L − x) EI

or

d2 y = − P (2e + y ) + F ( L − x) dx2

(14.34)

d2 y P 2 Pe Pe + y=− + ( L − x), putting the value of F EI LEI dx2 EI =−

2Pe Pex Pe − + EI LEI EI

(14.35)

Solution of this differential equation is, y = Acos kx + Bsin kx − e −

ex L

(14.36)

where A and B are constants, k = P /EI End conditions At end A, x = 0, y = 0, putting this in Eq. (14.36) 0 = A cos 0 + B sin 0 − e − 0 Constant, Moreover, at end B,

MTPL0259_Chapter 14.indd 586

A=e k = L, y = 0

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Struts and Columns

587

0 = A cos kL + B sin kL − 2e However,

A=e

so,

0 = e cos kL + B sin kL − 2e

Constant, B=

2e − e cos kL sin kL

Finally, the equation of deflection becomes, ex  2e − e cos kL  y = e cos kx +  sin kx − e −  sin kL  L Bending moment at the section, Pe x Pe L Pe x  2e − e cos kL  M x = −2 Pe − Pe cos kx − P  sin kx + Pe + + − , putting the value of y  sin kL  L L L  2 − cos kL  M x = − Pe cos kx − Pe  sin kx  sin kL  For maximum bending moment, dM x /dx = 0 dM x  2 − cos kL  = + Pe k sin kx − Pe  k cos kx = 0  sin kL  dx  2 − cos kL  sin kx =  cos kx  sin kL  tan kx =

2 − cos kL , where k = sin kL

P EI

Problem 14.3 A strut of length L is fixed at its lower end, its upper end is eccentrically supported against a lateral deflection (through a spring), so that the resisting force is k times the end deflection, k is spring constant. Show that the crippling load P is given by 1− ( P /kL) = tan mL/mL. where, m = P /EI Solution Figure 14.20 shows a strut AB of length L, fixed at end A and free at end B. At the end B, a horizontal reaction H = ka, given by a spring and P is the crippling load. Consider a section, at a distance x from end A. Bending moment at the section, M = P(a − y) − ka(L − x)

a

P B

ka

y L x

P

A

2

EI

MTPL0259_Chapter 14.indd 587

d y = P (a − y ) − ka ( L − x) dx2

Figure 14.20

5/23/2012 11:01:17 AM

588 Chapter 14

d2 y P Pa ka ( L − x) y= + − 2 EI EI EI dx

(14.37)

Solution of the differential equation y = A cos mx + B sin mx + a −

ka ( L − x) P

At end A, fixed end, x = 0, y = 0 0 = A cos 0 + B sin 0 + a −

kaL P

 kL  0 = A + a 1 −   P  kL  A = a  − 1  P  Then,

(14.38)

dy ka = − Am sin mx + Bm cos mx + dx P dy /dx = 0 at x = 0 0 = − Am sin 0 + Bm cos 0 +

ka P

or constant, B = −(ka/mP ) Finally, the equation of deflection curve is ka ( L − x ) ka  kL  y = a  − 1 cos mx − sin mx + a −  P  mP P At the end B,

y = a, x = L, putting the value ka  kL  a = a  − 1 cos mL − sin mL + a − 0  P  mP

or

ka  kL  a  − 1 cos mL = sin mL  P  mP tan mL = =

amP  kL   − 1 ka  P mP  kL   − 1 k  P

= mL −

MTPL0259_Chapter 14.indd 588

mP k

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589

P  = mL 1 −   kL  tan mL P = 1− mL kL Problem 14.4 A thin steel bar of a rectangular section 6 mm × 4 mm is axially compressed by 200-N load between two plates that are fixed at a constant distance of 160 mm apart as shown in Fig. 14.21. The assembly is made at 24°C, how high can the temperature of the bar rise so as to have an FOS of 2 with respect to buckling. E = 200 GPa, α =15 × 10−6 / °C ? Solution Section = 4 × 6 mm

6×4 rectangular section

I min =

6 × 43 = 32 mm4 12

160 mm

E = 2,00,000 N/mm2 End conditions: both ends hinged. L = 160 mn Pe = π 2 EI /L2

Euler’s buckling load,

Figure 14.21

π 2 × 2, 00, 000 × 32 = 2467.4 N 160 × 160 FOS = 2 =

Safe load =

2467.4 = 1233.7 N 2

Initial compressive load = 200 N Balance load to be applied through expansion of column (prevented by fixed ends) = 1,233.7 − 200 = 1,033.7 N Say temperature rise is DT Thus

α ∆T EA = 1033.7 N where A = area of cross-section of column 15 × 10−6 × ∆T × 2, 00, 000 × 24 = 1, 033.7 ∆T = Final temperature,

MTPL0259_Chapter 14.indd 589

1, 033.7 = 14.36°C 72

= 24 + 14.36 = 38.36°C.

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590 Chapter 14

Problem 14.5 A 6-m-long RSJ of a rectangular section of 325 mm × 165 mm is used as a strut, with one end is fixed and the other end is hinged. Calculate the crippling load by Rankine’s formula. Compare this with the load obtained by Euler’s formula. For what length of this strut will the two formulae give the same crippling load for the joist? Area of cross-section, A = 5,490 mm2 Ixx = 9,874.6 × 104 mm4 Iyy = 510.8 × 104 mm2 E = 210 kN/mm2 sc = 320 N/mm2 a=

1 for both hinged ends. 7, 500

Solution Iyy < Ixx L=6m Le =

6 2

m (one end is fixed and the other end is hinged)

Euler’s buckling load. Pe =

=

2π 2 EI min L2 2 ×π 2 × 2,10, 000 × 510.8 × 104 6, 0002

= 588.16 × 103 N Le =

6, 000 2

= 4, 243.3 mm

A = 5,490 mm2 2 = kmin

I yy A

=

510.8 × 104 = 930.42 mm2 5,490

L2e 4,243.32 = = 19, 352 2 930.42 kmin

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Struts and Columns

PR =

Rankine’s load,

=

σc A 330 × 5, 490 = 1 L2e × 19, 352 1+ a 2 1+ 7 500 , k 330 × 5, 490 = 506 kN 1 + 2.58

2π 2 EI yy Equating two loads,

591

L2

=

sc A L2 1 × × 2 1+ 7, 500 2 k

Putting the values, =

2 π 2 × 210 × 1, 000 × 510.8 × 104 = L2

21,17, 385.4 × 107 = 18,11,700

330 × 5, 490 L2 1+ 15, 000 × 930.42

L2 L2 1+ 13, 956, 300

  L2 1.1687 × 107 1 + = L2  13, 956, 300  1.1687 × 107 + L2 × 0.83742 = L2 0.16258 L2 = 1.1687 × 107 L2 = 71.88 × 106 L = 8.48 × 103 mm , for this length both Pe and PR are equal 15 mm

L = 8.48 m Problem 14.6 A round bar in a vertical position is clamped at the lower end and is free at the other end. The effective length is 2 m. If a horizontal force of 400 N at the top produces a horizontal deflection of 15 mm, what is the axial buckling load for the bar under the given conditions?

400 N

2m

Solution The bar is fixed as a cantilever beam (as shown Fig. 14.22). Deflection at the top

MTPL0259_Chapter 14.indd 591

=

400 × L3 3 EI

Figure 14.22

Problem 14.6

5/23/2012 11:01:26 AM

592 Chapter 14

15 =

400 × 2, 0003 3 EI

EI =

400 2, 0003 × = 0.711 × 1011 N mm2 15 3

Buckling load, Pe = π 2 × EI /4L2, as one end is fixed and the other end is free =

π 2 × 0.711 × 1011 4 × 2, 0003

= 0.438 × 105 N = 43.8 kN Problem 14.7 A 3-m-long straight steel bar of a rectangular section of 30 mm × 20 mm is used as a strut with both the ends hinged. Assuming that Euler’s formula is applicable and the material attains its yield strength at the time of buckling, determine the central deflection. E = 210 kN/mm2.Yield strength of steel is 320 N/mm2. Solution Section 30 × 20 mm2,

A = 600 mm2 Imin =

30 × 203 = 10, 000 mm4 12

E = 2,10,000 N/mm2 L = 3,000 mm (both the ends are hinged) Euler’s buckling load,

Pe =

π 2 EI π 2 × 2,10,000 × 104 = L2 3,0002 = 2,302 N

Pe 2, 302 = = 3.84 N/mm2 A 600 Say, central deflection, ymax.

σmax =

Z =

MTPL0259_Chapter 14.indd 592

P Pe ymax + A Z d × b2 30 × 202 = = 200 mm3 6 6

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Struts and Columns

320 = 3.84 +

593

23,002 × ymax 2,000

 320 − 3.84  ymax =  × 200 = 27.5 mm  2, 302  Problem 14.8 A steel column of a hollow circular section of 200 mm external diameter,160 mm internal diameter and 7-m-long has to take a 20-kN load at an eccentricity of 30 mm from the geometrical axis. If the ends are fixed, calculate the maximum and minimum stress intensities induced in the section, taking E = 210 kN/mm2. Moreover, calculate the maximum permissible eccentricity so that no tension is induced any where in the section. Solution Section D = 200 mm d = 160 mm A=

π (2002 − 1602 ) = 1.131 × 104 mm2 4

I=

π (2004 − 1604 ) = 0.4637 × 108 mm4 64

P = 200 kN E = 2,10,000 N/mm2 P = EI

2,00,000 = 1.433 × 10−4 2,10,000 × 0.4637 × 108

L=7m Ends are fixed,

Le = L/ 2 = 3.5 m = 3, 500 mm P Le 3, 500 × = 1.433 × 10−4 × = 0.2508 rad EI 2 2 = 14.37°

sec

P Le × = 1.0323 EI 2 yc = 100 mm

Eccentricity,

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e = 30 mm

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594 Chapter 14

P 2,00,000 = = 15.267 N/mm2 A 1.31 × 10 4

σ0 = Z =

I 0.4637 × 108 = = 0.4637 × 106 mm3 y 100 Pe P Le sec × Z EI 2

σb = ± =±

2,00,000 × 30 × 1.0323 = 13.357 N/mm2 0.4637 × 106

smax = 15.267 + 13.357 = 28.624 N/mm2 (compressive) smin = 15.267 − 13.357 = 1.91 N/mm2 (compressive) For no tension s0 = sb 15.267 = =

Pe′ P Le sec EI 2 Z 2,00,000 e ′ × 1.0323 0.4637 × 106

= 0.4637 × 106 15.267 = 0.4313 × 1.0323e′ 15.267 Eccentricity, e′ = 0.4313 × 1.0323 = 34.3 mm

Key Points to Remember  Euler’s buckling load, Pe = p 2 EI min /L2e where Imin = minimum moment of inertia Le = equivalent length of a column depending upon end conditions.  Euler’s formula is applicable for the slenderness ratio Le /K > π2 E/σ c ,where sc is ultimate compressive strength of the material.  Higher-order differential equation d4 y d2 y P + k 2 2 = 0, where k 2 = 4 dx dx EI

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Struts and Columns

595

 Solution, y = C1sin kx + C2 cos kx + C3x + C4 using different end conditions, the buckling load is determined.  Rankine’s load =

σcA  L  1+ a e   kmin 

2

where sc = crushing strength of column Le = equivalent length, kmin = minimum radius of gyration  Johnson’s parabolic formula for working stress  L2  σ w = σ c ′ 1 − b e2  k   where σ c ′ = allowable stress in compressive taking into account FOS b = constant  Eccentrically loaded column

σ max =

L P Pe + sec e 2 A Z

P EI

where P = axial applied load Z = section modulus e = eccentricity I = moment of inertia  Professor Perry’s formula σ   σ 0  1.2 eyc  σ − 1 1 − σ  = k 2 0 e where s = maximum stress allowed s0 = P/A, applied load/area se = Pe /A, Euler’s load/area e = eccentricity yc = distance of extreme layer in compression from neutral layer  If the column has initial eccentricity e′ along central section  σ min   σ 0  e′yc  σ − 1 1 − σ  = k 2 0 e  Energy approach, total potential, Kp = 0 L

L

2

 d2 y  P  dy  1 EI dx = cr ∫   dx 2 ∫  2 0  dx  2 0  dx  to determine Pcr, critical load.

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596 Chapter 14

Review Questions (1) (2) (3) (4) (5) (6) (7) (8) (9)

What are the drawbacks of Euler’s theory of buckling? What is limiting value of the slenderness ratio beyond which Euler’s formula is applicable? What are the merits of Rankine’s load over Euler’s load in buckling? What do you mean by equivalent length of a column? Discuss the effect of the slenderness ratio of a column over buckling load? Why the value of ‘a’ Rankine’s constant varies for different materials? What is straight line formula for working stress under buckling, where it is used? What approximation is taken by Professor Perry to modify the secant formula for eccentrically loaded column? What do you understand by total potential constant in energy approach?

Multiple Choice Questions 1. A column fixed at one end and free at the other end buckles at a load P. Now, both the ends of the column are fixed. What is the buckling load for these end conditions? (a) 16 P

(b) 8 P

(c) 4 P (d) 2 P 2. In Rankine’s formula ‘a’ is used, what is its value for cast iron? (a)

1 9, 000

(b)

1 7, 500

(c)

1 1, 600

(d) None of these

(c)

3 m 2

(b) 3 m (d) None of these

7. A column of a length of 2.4 m, an area of crosssection of 2,000 mm2 and moment of inertia of I xx = 720 × 104 mm4 and I yy = 80 × 104 mm4 is subjected to buckling load. Both the ends of the column are fixed. What is the slenderness ratio of column? (a) 120 (b) 80 (c) 60

3. What is the approximate value of sc for cast iron/sc for mild steel? (a) 0.6

(b) 1.0

(c) 1.7 (d) None of these 4. A hollow circular column, with D = 100 mm, d = 80 mm, what is radius of gyration? (a) 32

(b) 24

(c) 19.4 (d) None of these 5. In Johnson’s parabolic formula, allowable stress in compression for mild steel is 2

(a) 270 N/mm 2

2

(b) 110 N/mm

(c) 80 N/mm (d) None of these 6. A 3-m-long column is hinged at one end and fixed at the other end, what is its equivalent length?

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(a) 3 2 m

(d) 40 8. The ratio of equivalent length of a column with one end fixed and the other end free to its own length is (a) 2

(b) 1.0

(c) 0.5

(d) None of these

9. Euler’s buckling theory is applicable for (a) Short columns (b) Long columns (c) Medium long columns (d) All of them 10. In Johnson’s parabolic formula, what is constant b for hinged ends, for mild steel (a) 2 × 10−5

(b) 3 × 10−5

(c) 0.005

(d) None of these

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597

Practice Problems 1. What size of a steel pipe should be used for the horizontal member of a jib crane shown in Fig. 14.23 for supporting C the maximum force of 20 kN. Use an FOS of 2.5. The internal diameter of the pipe is 0.8 times the external diameter. E = 200 GPa. 308 2. Find the shortest length for a steel column with hinged B ends having a cross-sectional area of 30 mm × 60 mm for A which the elastic Euler’s formula is applicable. E = 200 2.5 m GPa and assume proportional limit to be 250 MPa. 3. A short-length of a tube of 30 mm internal diameter and 40 20 kN mm external diameter failed under a compressive load of 180 kN. When a 2-m length of the same tube is tested as a strut Figure 14.23 Practice problems 14.1 with both the ends hinged, the buckling load was found to be 40.8 kN. Assuming that sc, for the Rankine’s constant is given by first test, determine Rankines’ constant ‘a’. Moreover, estimate the cripP L1 L1 pling load for a piece of 3-m-long tube when used as strut with fixed ends. 2 2 4. A strut of a 2-m-long aluminium alloy has a rectangular section of 20 mm × 50 mm. A bolt through each end secures the strut so that it acts as a hinged column about an axis perpendicular to the 50-mm dimension and as fixed-ended column about an axis perpendicular to the 20-mm dimension. Determine the safe central EIB load using a FOS of 2. E = 70 GPa.   20 × 503 50 × 203 when both ends hinged, I = where both ends fixed   Hint: I = 12 12   5. A simple beam of flexural rigidity EIB is propped up at the middle by a slender rod of flexural rigidity EIC. Estimate the deflection of the beam at the centre if a force P double the Euler’s buckling load for the column is applied to the system as shown in Fig. 14.24. 6. A thin vertical strut of uniform section and length L is rigidly fixed at its bottom end and its top end is free. At the top end, there is a horizontal force H and a vertical load P acting through the centroid of the section as an axial load. Prove that horizontal deflection at the top is   H  tan µL P − L , where µ 2 =  Hint:   P µ EI    7. A horizontal bar CD is supported by a pin-ended column AB, 2.5 m long and 40 mm diameter as shown in Fig. 14.25. Calculate the allowable load P, if FOS with respect to buckling of column is 3.0. E = 200 GPa 8. A hollow cast iron column of external diameter of 200 mm and length of 4 m with both ends fixed supports an axial load of 800 kN. Find the thickness of the metal required, use Rankine’s constants, a = 1/1, 600 for both ends hinged, working stress, sw = 80 N/mm2. 9. A 1.2-m-long hollow circular steel pipe of 60 mm outside diameter and 50 mm inside diameter is fixed at both the ends so as to

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L2

EIC

Figure 14.24

0.5

1m

P D

C

A

40 mm j

2.5 m

B

Figure 14.25

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598 Chapter 14 prevent any expansion in its length. The pipe is unstressed at the normal temperature. Calculate the temperature stress in the pipe and the FOS against failure as a strut if the temperature rises by 40°C. Use Rankine’s formula. 1   −6  Hint: σ c = 330MPa, a = 7, 500 , α = 11.1 × 10 / °C     E = 208 kN/mm2  10. A 5-m-long steel strut, with I = 50 × 104 mm4, carries thrust load P = 20 kN, with eccentricities e = 10 mm on one side and 2e = 20 mm on the other side. E = 200 GPa. Determine the distance from one end, where the bending moment on strut is maximum.

Answers to Exercises π 2 EI L2 4p 2 EI Exercise 14.2: L2 Exercise 14.3: 400 kN

Exercise 14.8:

466 kN

Exercise 14.9:

12.07 mm

Exercise 14.4: L > 0.856 m

Exercise 14.12: 2,180 kN

Exercise 14.1:

Exercise 14.5:

Exercise 14.10: 128 kN Exercise 14.11: 8.05 N/mm2

Exercise 14.13: W = 1.340 kN

π 2 EI L2

Exercise 14.14: 161.70 N/mm2

π EI 4 L2 Exercise 14.7: 516.3 kN 2

Exercise 14.15: 9.88 EI/L2

Exercise 14.6:

Answers to Multiple Choice Questions 1. 2. 3. 4.

5. 6. 7. 8.

(a) (c) (c) (a)

9. (b) 10. (b)

(b) (c) (c) (a)

Answers to Practice Problems 1. D = 55.47 mm; d = 44.37 mm 2. L = 769 mm 3. a =

1 , 61.67 kN 7, 505

7. 8. 9. 10.

5.29 kN 32.3 mm 92.35 MPa, 3.17 x = 2.859 m

4. safe load = 11.50 kN 3 2 5. π I C L1 48 I B L23

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15 Theories of Failure CHAPTER OBJECTIVE The basic objectives of this chapter is to familiarize the students with various theories of failure for the efficient design of engineering components. Depending on the mechanical properties of raw materials and the applications of the engineering component or structure, selection of a particular theory for design is also discussed and students can use a particular theory of failure for designing their engineering components. Various theories are supplemented by practical examples to understand the concepts better.

Introduction There are various theories of failure depending on the types of stresses or types of strain energies alongwith the mechanical properties of the material used to design a component. Though the discussion on theories of failure should be based on three-dimensional response of any component due to external load applied on the component, but the analysis is based on the parameters of a simple tensile test on the material that is uniaxially loaded. It is preformed on a standard sample of the material, and mechanical properties, such as yield strength, ultimate tensile strength and percentage elongation, in case of ductile materials, and ultimate stress, in case of brittle materials, are established for use in the design procedure. Figure 15.1 shows load extension curves for a general ductile material and a brittle material. A ductile material is strong and tough and a majority of engineering components are made from them, such as

syp

• Ductile material

P

Texsile text

Tensile load

sut

Extension

•

sut

Brittle material

Extension

Figure 15.1

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600 Chapter 15

mild steel, medium carbon steel, high carbon steel, aluminium and copper. Brittle materials are weak in tension and poor in ductility or they are brittle in nature, resulting in poor toughness. So such materials are not used in tension. These materials are much stronger and possess higher ductility and toughness in compression. Therefore, these materials, such as grey cast iron, are used in machine components that sustain only compressive loads. There are five theories of failure, that is, maximum principal stress, maximum shear stress, maximum principal strain, total strain energy and distortional energy theory, all depending upon the magnitude of principal stresses at a critical section of the member to be designed. The maximum shear stress theory gives a conservative design, so it is safe to use this theory for ductile materials. The maximum principal stress theory is mainly used for brittle material components. For very brittle materials, such as rock and marble, Mohr’s theory of failure is used. It is difficult to understand physically a three-dimensional stress system. However, under plane stress conditions, two principal stresses are acting on the surface of a component. Therefore, graphical representation of all the theories in a two-dimensional stress system will be discussed in this chapter.

Maximum Principal Stress Theory (Rankine’s Theory) There are three principal stresses at a point, that is, p1 > p2 > p3 on a component. As per this theory, if p1 ≥ σ ut (for brittle material) and p1  σ yp for ductile materials, then the component will fail. This theory is generally used for brittle materials. Let us consider any material, for which σ yp is the yield point stress, the point at which plastic deformation starts in the ductile material. Principal stress p1 is perpendicular to principal stress p2. So as per this theory, p1 p ≤ ±1 and 2 ≤ ±1 σ yp σ yp Figure 15.2 shows the graphical representation of the theory, that is, x = p1 /σ yp , y = p2 /σ yp. Boundary ABCD of the Fig. 15.2 defines the beginning of yielding in the material. In case of brittle materials, syp is to be replaced by sut. +y = B

+1.0

p2 syp A

+1.0 +x =

−x

p1 syp

−1.0

C

Figure 15.2

MTPL0259_Chapter 15.indd 600

−1.0

D −y

Graphical representation of the maximum principal stress theory

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Theories of Failure

601

Let us take the practical case of a thin cylindrical shell of diameter d, its wall thickness t, subjected to an internal fluid pressure, p. Circumferential stress,

σc =

pD (tensile) 2t

Axial stress,

σa =

pD (tensile) 4t

= −p (compressive)

Radial stress

Three principal stresses at a point in the cylindrical are pD pD , ,− p 2t 4t As per this theory, p1 ≤ σ yp (for no failure) pD ≤ σ yp 2t

or

Experimental results show that this theory is acceptable when principal stresses are of the same sign. It is unsafe to design a machine member on the basis of this theory, if the principal stresses p1 and p2 are of opposite nature. Example 15.1 A thick cylinder of internal diameter 250 mm is subjected to an internal pressure of 30 N/mm2; what should be the external diameter of the cylinder, if σ yp = 280 N /mm 2 ? Take factor of safety (FOS) as 4, inner radius of cylinder R1 = 125 mm. Solution

σ yp = 280 N /mm 2 FOS = 4

σ al =

Allowable stress,

280 = 70 N /mm 2 4

Maximum stress in the thick cylinder occurs at the inner radius of the cylinder due to the internal pressure. If R1 is the internal radius and R2 is the external radius and p is the pressure scmax = maximum hoop stress =p

R 22 + R 22 ≤ σ al R 22 − R12

Putting the value 30 ×

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R 22 + 1252 = 70 R 22 − 1252

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602 Chapter 15

R 22 + 125−2 70 = R 22 − 1252 30

or,

3R 22 + 3 × 1252 = 7 R 22 − 7 × 1252 4R22 = 10 × 1252 R22 = 2.5 × 1252 R2 = 197.64 mm D2 = 395.3 mm External diameter of the cylinder

= 395.3 mm

Exercise 15.1 A solid circular shaft of diameter d is subjected to a pure torque of 200 Nm. Determine the diameter of the shaft according to the maximum principal stress theory, taking a FOS of 2, if the yield strength of the material is 280 N/mm2. 16T    Hint: τ = π d 3 , p1 = +τ , p2 = −τ

Maximum Shear Stress Theory (Tresca Theory) As per this theory, yielding in the material is governed by the maximum shear stress developed in the material under the action of the principal stresses p1, p2 and p3. The maximum shear stress due to the principal stresses should not exceed the maximum shear stress developed in a simple tensile test, that is, σ yp /2 at yield point. Figure 15.3 shows a three-dimensional Mohr’s stress circle, where OA = p1, OB = p2 and OC = p3, principal stress, then tmax = maximum shear stress  p − p3  = 1 as a obvious from the diagram  2  As per this theory,  p1 − p3  σ yp  ≤ 2  2 or,

( p1 − p3 ) ≤ σ yp

However, in a two-dimensional case where p3 = 0 Then, as per the maximum shear stress theory,

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Theories of Failure Shear stress

603

p1 − p3

E

2

O

B

C

Normal stress D

A

OA = p1, OB = p2 OC = p3 DE =

Figure 15.3

p1 − p3 2

= tmax

Three-dimensional Mohr’s stress circle

p1 − p2 σ yp ≤ 2 2 ( p1 − p2 ) ≤ σ yp if p1 and p2 are of opposite sign p1 σ yp if p1 and p2 are of the same sign ≤ 2 2 or p1 ≤ σ yp (In this condition, it coincides with the maximum principal stress theory). Graphical representation of the maximum shear stress theory is shown in Fig. 15.4. In I quadrant, both p1 and p2 are positive, the theory coincides with the maximum principal stress theory. In second quadrant, p1 is negative and p2 is positive, line CD shows p1 – p2 = –σ yp. Line AF of IV quadrant shows p1 – p2 = +σ yp. In the third quadrant, both p1 and p2 are of the same sign and the theory coincides with the maximum principal stress theory. The designers very often use this theory because: 1. This theory provides more FOS, 2. A conservative design of the component is obtained and 3. Most engineering materials are ductile in nature, and this theory is most mutable for ductile materials. Example 15.2 A thin cylindrical shell of diameter 200 mm and wall thickness t is subjected to material pressure of 3 N/mm2. Yield strength of the material is 280 N/mm2. Taking an FOS of 3 and using the maximum shear stress theory, determine the wall thickness t of the cylinder. Solution

σ yp = 270 N /mm 2

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604 Chapter 15

y=

p2 s yp

C

−x

−x

+

y

=

B

1 I II

D

A

IV III

x E

−

y

=

x=

p1 s yp

1

F −y

Figure 15.4

Maximum shear stress theory (Graphical representation)

FOS = 3 Allowable normal stress, Allowable shear stress

σ al = =

270 = 90 N /mm 2 3 90 = 45 N /mm 2 2

Hoop stress in cylinder,

p1 =

pD 3 × 200 300 = = t 2t 2t

Axial stress in cylinder,

p2 =

pD 3 × 200 150 = = t 4t 4t

Radial stress, p3 = −p = −3 N/mm2 As per the maximum shear stress theory, 1  300  + 3 ≤ 45   2 t 300 ≤ 87 t

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Theories of Failure

605

300 87 ≥ 3.45 mm

t≥

or thickness,

Wall thickness of the cylinder should be minimum 3.45 mm. Exercise 15.2 A thick cylinder of internal diameter of 100 mm is subjected to an internal pressure of 50 N/mm2. Determine the thickness of the cylinder according to the maximum shear stress theory. If,

σ yp = 300 N /mm 2 , F OS = 2.5

Maximum Principal Strain Theory (St. Venant’s Theory) Strain is an actual physical quantity related to the change in dimensions, but stress is an abstract concept, that is, internal resistance per unit area. Even for the determination of stresses, strain gauges are used to measure the strains and the principal stresses are indirectly measured using the Young’s modulus and Poisson’s ratio of the material. In this theory, it is assumed that failure by yielding of a material takes place when the maximum principal strain in the material subjected to principal stresses is equal to the strain at the yield point in a simple tension or compression test. If p1 > p2 > p3 are the principal stresses, the maximum principal strain is

ε1 =

1 ( p1 − vp2 − vp3 ) E

where E is the Young’s modulus and n is the Poisson’s ratio of the material. σ yp As per the theory, ε1 ≤ E or,

( p1 − vp2 − vp3 ) ≤ σ yp

[p

1

− v( p2 + p3 )] = σ yp

However, this theory is not acceptable to designers. Taking a two-dimensional case p1 > p2 and p3 = 0, as per the maximum principal strain theory. ( p1 − vp2 ) ≤ σ yp or principal stress,

(15.1)

p1 ≤ σ yp + ν p2

Showing thereby that the principal stress p1 can be greater than σ yp. Equation (15.1) can be expressed as p1 π p2 − ≤, σ yp σ yp or,

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x −νy ≤ 1

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606 Chapter 15

p1 p = x, 2 = y σ yp σ yp

y=

C

or the equation of straight line is

nx y=1+

D

x = 1 + ny

(15.2)

y = 1 + nx

(15.3) −x = 1 + n

where n is the Poisson’s ratio of the material. These relationships are shown in Fig. 15.5. This theory assumes that the material obeys Hooke’s law. Experimental evidence shows that this theory is not acceptable. Example 15.3 A hollow shaft of 30 mm internal diameter and 50 mm external diameter is subjected to a twisting moment 800 Nm and an axial compressive force of 40 kN. Determine the FOS according to the maximum principal strain theory of failure, if the tensile and compressive yield strength of the material is 280 N/mm2 and the Poisson’s ratio = 0.3.

x=

p1 syp

y

or

p2 s yp

x = 1 + ny

where

A

B

−y = 1 + nx

Figure 15.5 Graphical representation of maximum principal strain theory

Solution Internal diameter of shaft, d = 30 mm External diameter of shaft, D = 50 mm Area of cross-section,

A=

π (502 − 302 ) = 12.5664 × 102 mm 2 4

Polar moment of inertia,

J=

π (504 − 304 ) = 53.4 × 104 mm 4 32

Maximum shear stress on surface of the shaft

τ= =

T D × where T = 800 Nm J 2 800 × 103 × 25 = 37.45 N /mm 2 53.4 × 10 4

40, 000 Axial compressive stress, σ = P = A 12.5064 × 10 4 = 31.83 N/mm2 Figure 15.6 shows the state of stress on the surface of the shaft.

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Theories of Failure t

Principal stress 2

p1 =

σ σ +   +τ2  2 2

s

2

=

607

s

s = 31.83 N/mm2 t = 37.45 N/mm2

t

31.83  31.83  +  + 37.452  2  2

Shaft axis

Figure 15.6

= 15.915 + 253.287 + 1402.50 = 15.915 + 40.691 = 56.60 N/mm2 (comp) p2 = 15.915 − 40.691 = −24.78 N/mm2 (tensile) Note that s is compressive stress syp = 280 N/mm2 n = 0.3 As per the maximum principal strain theory

σ yp p1 ν p2 − = E E E (F OS) 56.6 + 0.3 × 24.78 = F OS =

280 F OS

280 = 4.37 64.034

Exercise 15.3 A certain type of steel has a yield strength of 270 N/mm2. At a point in a strained region, the principal stresses are +120, +80 and −30 N/mm2. Determine the FOS as per the maximum principal strain theory. Given that the Poisson’s ratio is 0.285.

Strain Energy Theory (Beltrami, Haigh Theory) This theory is based on the strain energy absorbed during loading of component. As per this theory, the failure or yielding of a material occurs when the strain energy per unit volume stored in the machine component exceeds the strain energy stored in a unit volume at the yield point stress in a simple tensile test specimen. Figure 15.7 shows the principal stresses p1, p2 and p3 and the principal strains e1, e2 and e3 on an element.

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608 Chapter 15

Principal strains

2 p 2, e 2

p ν ( p2 + p3 ) ε1 = 1 − E E

ε2 =

p2 ν ( p3 + p1 ) − E E

ε3 =

p3 ν ( p1 + p2 ) − E E

p1, e1 0

1 u = ( p1ε1 + p2 ε 2 + p3ε 3 ) 2 1 [ p1 ( p1 − ν p2 − ν p3 ) + p2 = 2E × ( p2 − ν p3 − ν p1 ) + p3 ( p3 − ν p1 − ν p2 )]

1

•

where E and n are the Young’s modulus and the Poisson’s ratio of the material, respectively. Strain energy per unit volume,

=

p3, e3

p3, e3

p2, ε2

3

Figure 15.7

Principal stresses and strains on an element

1 [( p12 + p22 + p32 ) − 2ν ( p1 p2 + p2 p3 + p3 p1 )]. 2E

Strain energy at yield point,

2 σ yp 2E

u ≤ u′

σ yp 1 [( p12 + p22 + p32 ) − 2ν ( p1 p2 + p2 p3 + p3 p1 )] = 2E 2E 2

= or,

(

)

2  p12 + p22 + p32 − 2ν ( p1 p2 + p2 p3 + p3 p1 ) ≤ σ yp  

In a two-dimensional case, p3 = 0 ( p12 + p22 − 2 νp1 p2 ) ≤ σ 2yp or,

Taking

 p12 p22 2ν p1 p2   2 + 2 −  ≤ 1. 2 σ yp  σ yp σ yp  p1 p2 = x, =y σ yp σ yp ( x 2 + y 2 − 2ν xy ) ≤ 1 Let us take ν = 0.25, ( x 2 + y 2 − 0.5 xy ) ≤ 1

Graphical representation of the theory is shown in Fig. 15.8 through an ellipse. Experimental evidence available so far does not permit the use of this theory for design purposes.

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Theories of Failure

y=

609

p2 syp

+1 Ellipse

x=

−1

+1

p1 syp

−1

Figure 15.8 Graphical representation of the strain energy theory Example 15.4 The load on a bolt consists of an axial thrust of 8 kN together with a transverse shear load of 4 kN. Calculate the diameter of the bolt using the strain energy theory. Take 3 as FOS.

σ yp = 285 N /mm 2 , ν = 0.30 Solution Say, the diameter of the bolt is d

π 2 d 4

Area of cross-section,

A=

Axial compressive force,

P = 8 kN

Axial compressive shear,

σ=

Transverse load,

Q = 4 kN

Shear stress,

τ=

P 8, 000 = N /mm 2 A A

4, 000 N /mm 2 A t

State of shear on a bolt section is shown in Fig. 15.9. 8, 000 σ= A 4, 000 τ= A

MTPL0259_Chapter 15.indd 609

s

s

Bolt axis

t

Figure 15.9

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610 Chapter 15

Principal stresses are 2

4, 000  4, 000   4, 000  +  +  A   A  A 9, 656 = A

2

p1 =

2

p2 =

4, 000  4, 000   4, 000  −  +  A   A  A

=− Now,

2

1, 656 A

σ yp = 285 FOS = 3

σ al = allowable stress =

285 = 95 N/mm 2 3

Using the strain energy theory 2 2  9, 656  −1, 656  2   9, 656   −1, 656  2 = + − 2 ν +    = 95  A   A   A A  

A2 =

1 [93.24 + 2.74 + 9.594] × 106 2 95

A=

10.275 × 103 = 108.15 mm 2 95

=

π 2 d 4

Diameter of the bolt, d =

4 × 108.15 = 11.73 mm. π

Exercise 15.4 A thick cylinder of an internal diameter of 200 mm and an external diameter of 300 mm is subjected to an internal pressure p. Determine the maximum value of p as per the strain energy theory, if the yield point stress of the material is 180 N/mm2. Take an FOS of 2, Poisson’s ratio = 0.32.

Shear Strain Energy or Distortional Strain Energy Theory (Von Mises Theory) For the metal forming operations, it is the shear strain energy that is responsible for changing the shape of a metal. Metal forming dies are to be designed accordingly. In this theory, it is assumed that failure by yielding occurs when energy used in changing the shape of the unit volume of a component is equal to the distortion energy per unit volume at the yield stress of a specimen subjected to simple tension or

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compression test. Total strain energy consists of two parts, that is, (a) volumetric strain energy causing the change in volume and (b) shearing strain energy for changing the shape. So, 1 Total strain energy, u = [ p12 + p22 + p32 − 2ν ( p1 p2 + p2 p3 + p3 p1 )] 2E = uv + us = volumetric strain energy + shear strain energy Volumetric stress

=

p1 + p2 + p3 = pm, equal in all directions. 3

Principal stresses can be represented by p1 0 0

0 p2 0

0 pm 0 = 0 p3 0

Volumetric strain energy, uv = 3 ×

0 pm 0

0 ( p1 − pm ) 0 0 0 + 0 ( p2 − pm ) 0 pm 0 0 ( p3 − pm )

1 pm em 2

where strain,

εm =

pm (1 − 2ν ) E

so

uv =

p 3 p2 3 pm m (1 − 2ν ) = m (1 − 2ν ) 2 E 2E

=

3 ( p1 + p2 + p3 )2 (1 − 2ν ) 2E

=

1 (1 − 2ν )( p12 + p22 + p32 + 2 p1 p2 + 2 p2 p3 + 2 p3 p1 ) 6E

Strain energy given by ( p1 − pm ), ( p2 − pm ), ( p3 − pm ) is the Shear strain energy us = u − uv =

1 [( p12 + p22 + p32 ) − 2ν ( p1 p2 + p2 p3 + p3 p1 )] 2E 1 (1 − 2ν )[ p12 + p22 + p32 + 2 p1 p2 + 2 p2 p3 + 2 p3 p1 ] − 6E

= (1 + ν ) However, Therefore,

MTPL0259_Chapter 15.indd 611

(1 + ν ) =

1 [ p12 + p22 + p32 − ( p1 p2 + p2 p3 + p3 p1 )] 3E

E 2G

us = (1 + ν )

1 ( p12 + p22 + p32 − p1 p2 − p2 p3 − p3 p1 ) 6G

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612 Chapter 15

Multiplying the numerator and denominator by 2, we get us =

1 (2 p12 + 2 p22 + 2 p32 − 2 p1 p2 − 2 p2 p3 2 p3 p1 ) 12G

=

1 [( p1 − p2 ) 2 + ( p2 − p3 ) 2 + ( p3 − p1 ) 2 ] 12G

=

σ yp 1 [( p1 − p2 ) 2 + ( p2 − p3 ) 2 + ( p3 − p1 ) 2 ] ≤ 12G 6G

As per this theory 2

2 [( p1 − p2 ) 2 + ( p2 − p3 ) 2 + ( p3 − p1 ) 2 ] ≤ 2σ yp

or,

For a two-dimensional case, p3 = 0 2 ( p12 − 2 p1 p2 + p22 + p22 + p12 ) ≤ 2σ yp 2 ( p12 − p1 p2 + p22 ) ≤ 2σ yp

or, or,

p12 p1 p2 p22 − 2 + 2 ≤1 2 σ yp σ yp σ yp

or,

x 2 − xy + y2 ≤ 1 x=

where

p1 p ,y= 2 . σ yp σ yp

This is the equation of ellipse, as shown plotted in Fig. 15.10. For ductile materials, this theory is in good agreement with the experimental results and is used for the design of a shaft. Example 15.5 A thin cylindrical shell of a diameter of 200 mm and a wall thickness t is subjected to an internal pressure of 2.4 N/mm2. Neglecting the effect of p as principal stress, determine the thickness of the shell if syp = 270 N/mm2.

y=

p2 syp

+1 Ellipse

−1

+1

FOS = 2.5.

x=

p1 syp

Solution syp = 270 N/mm2. FOS = 2.5 Allowable stress, sal =

MTPL0259_Chapter 15.indd 612

270 = 108 N /mm 2 2.5

−1

Figure 15.10

Graphical representation of distortional energy theory

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613

D = 200 mm

Diameter,

p = 2.4 N/mm2 pD 2.4 × 200 240 p1 = = = t 2t 2 ×t

Hook stress,

p2 =

Axial stress,

pD 2.4 × 200 120 = = t 4t 4 ×t

(p

2 1

)

− p1 p2 + p22 ≤ σ al2

 240  2 240 240  120  2  2 × +   = 108  −  t   t t  t  57, 600 28, 800 14, 400 − + = 1082 t2 t2 t2 43, 200 = 11, 664 t2 t2 =

43, 200 = 3.7034 11, 664

Wall thickness, t = 1.924 mm. Exercise 15.5 A thin spherical shell has a diameter of 200 mm and a wall thickness t mm. It is subjected to an internal pressure of 5 N/mm2. The material has a yield strength of 265 N/mm2. Determine the thickness of the shell according to the distortional energy theory of failure. Take an FOS of 3. pD pD    Hint: the principle stresses are 4t , 4t , − p 

Mohr’s Theory of Failure Mohr has developed a failure theory for brittle materials, such as rock, marble, granite and cast iron, which are much stronger in compression than in tension. As an example, the ultimate compressive strength of grey cast iron is 550 MPa, while its tensile strength is only 150 MPa. This theory is based on stress circles drawn with suc and sut as diameters, where suc is the ultimate compressive strength and sut is the ultimate tensile strength of the material. Graphical representation of Mohr’s theory of failure is shown in Fig. 15.11. Circle with diameter OA = sut and circle with diameter OB = suc are drawn. Tangents to the circles, that is, CD and EF or envelopes to the circles provide combination of stresses s1 (tensile) and s2 (compressive) on the body as shown. Theory can be modified as

σ1 σ 2 + =1 σ ut σ uc

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614 Chapter 15 t (Shear Stress)

C s1

s2

OA = sut OB = suc

D s2

−s

B

O2

O

s1

O1

A

+s

F

State of stress E

Figure 15.11

Graphical representation of Mohr’s theory

where sut = ultimate tensile strength of the material suc = ultimate compressive strength of the material Example 15.6 For a certain brittle material sut = 160 MPa and suc = 500 MPa. Using Mohr’s theory of failure, determine the values of s1 and s2, if (σ 1 − σ 2 )/ 2 = 150 MPa Solution Draw circles with diameters OA = 160 MPa, = sut OB = −500 MPa = suc, so that O1A = +80 MPa O2B = −250 MPa to some suitable scale. Draw failure envelope CD, common tangent to the circles as shown in Fig. 15.12. Take shear stress, t = 150 MPa parallel to CD as shown. Draw a base EO3 parallel to CD, cutting the abscissa at point O3. From O3 as centre draw a circle of radius of 150 MPa. Thus, the circle cuts the abscissa at K and M such that OK = 96 MPa = s1 OM = −204 MPa = s2 Shear stress =

σ 1 − σ 2 96 + 204 = = 150 MPa (as given in the problem) 2 2

Therefore, the combinations of stress are p1 = +96 MPa p2 = −204 MPa

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615

sut C

15

t

0M

Pa

E

B

D

O2

•

M

• O3 O

O1

•

K

s A

OK = 96 MPa OM = 204 MPa

Figure 15.12

Exercise 15.6 For a certain material, sut = 160 MPa and suc = 550 MPa. A principal stress p1 = 55 MPa acts on the body. What is the permissible value of stress p2? Use Mohr’s theory of failure. Problem 15.1 In a steel drum that is subjected to axial compressive force, the internal pressure is 10 N/mm2. The maximum circumferential stress is 100 MPa and the longitudinal stress is 30 MPa. Find the equivalent tensile stress in a simple tensile test according to each of the five theories of failure (except Mohr’s theory). The Poisson’s ratio is 0.3. Solution Principal stresses at a critical point are +100, +30 and −10 MPa. p1 = 100, p2 = 30 and p3 = −10 MPa, say equivalent tensile stress = s, then (a) Maximum principal stress theory s = p1 = 100 MPa. (b) Maximum shear stress theory p1 − p3 σ = 2 2 100 + 10 = s s = 110 MPa (c) Maximum principal strain theory p1 ν p2 ν p3 σ − − = E E E E or,

100 − 0.3× 30 + 0.3×10 = s s = 100 − 9 + 3 = 94 MPa

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616 Chapter 15

(d) Strain energy theory

σ2 1  p12 + p22 + p32 − 2ν ( p1 p2 + p2 p3 + p3 p1 )  = 2E 2E  Putting the values,

σ 2 = 1002 + 302 + ( −10) 2 − 2 × 0.3[100 × 30 + 30( −10) + 100( −10)] s 2 = 11,000 − 0.6(3,000 − 300 − 1,000) = 11,000 − 0.6(1,700) = 11,000 − 1,020 = 9,980 s = 99.9 MPa (e) Distortion energy theory 2σ 2 = [( p1 − p2 ) 2 + ( p2 − p3 ) 2 ( p3 − p1 ) 2 ] = [(100 − 30) 2 + (30 + 10) 2 + ( −10 − 100) 2 ] = 4,900 + 1,600 + 12,100 = 18,600 s = 9,300 2

s = 96.43 MPa Problem 15.2 A metal bar is being compressed along the axes between two rigid walls such that strain e3 = 0 and stress p2 = 0. This process causes an axial stress p1 and no shear stress. Determine the apparent yield value of p1, if the material in a conventional compression test exhibits syp as yield stress and n is the Poisson’s ratio of the material. Assume that the (a) material is governed by Von Mises yield criterion and (b) if the Tresca criterion is applied, determine apparent yield value of p1. Solution Say, the principal stresses are p1, p2 and p3 p3 vp1 vp2 − − E E e

Strain

ε3 =

However,

p2 = 0,

Therefore,

0=

ε3 = 0

p3 vp1 − E E

p3 = vp1

or,

(15.4)

So, the principal stresses are p1, 0 and vp1 (a) Von Mises yield criterion 2 (p1 − p2)2 + (p2 − p3)2 + (p3 − p1)2 = 2 σ yp

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617

Putting the values (p1)2 + (+np1)2 + (np1 − p1)2 = 2 σ yp 2

2 p12 + n 2p12 + n 2p12 + p12 − 2np12 = 2 σ yp 2 or p12 + n 2p12 − np12 = σ yp 2 p12 (1−n + n 2) = σ yp

p1 =

σ yp 1− ν + ν2

(b) Tresca criterion Principal stresses are p1, 0 and np1

σ Maximum shear stress p1 − 0 = yp 2 2 p1 = syp Problem 15.3 A thin aluminium alloy tube bar has a mean diameter of 200 mm and a wall thickness of 2 mm. The tube is subjected to an internal pressure of 2 N/mm2 and a torque of 5 kN m. If the yield strength of the material is 240 N/mm2 and the Poisson’s ratio is 0.33, determine the FOS according to: (a) Maximum shear stress theory, (b) Maximum principal strain theory and (c) Strain energy theory. Solution Tube diameter, Wall thickness, Internal pressure,

D = 200 mm t = 2 mm p = 2 N/mm2 pD 2 × 200 Circumferential stress, σ c = = = 100 MPa 2t 2×2 pD 2 × 200 Axial stress, σa = = = 50 MPa 4t 4×2 Torque,

T = 12 kN m = 5 × 106 N mm

τ , where D /2 t is the maximum shear stress on the surface T=J

J = π Dt =

MTPL0259_Chapter 15.indd 617

D 2 π D 3t = mm4 4 4

π × 2003 × 2 = 12.56 × 10 −6 mm 6 4

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618 Chapter 15

sc = 100 MPa

D = 100 mm 2

τ=

19.9 6

TD 5 × 10 × 100 = 2 J 2 × 12.506 × 106

sa = 50 MPa

50

= 19.9 N/mm2 Neglecting the radial pressure p, stresses on an element are shown in Fig. 15.13. Principal stresses are

19.9 100

2

σ + σa  σ − σa  p1 , p2 = c ±  c +τ2  2  Z

Figure 15.13

2

=

100 + 50  100 − 50  2 ±   + (19.9)  2 2 

= 75 ± 252 + 19.92 p1 , p2 = 75 ± 625 + 396.01 = 75 ± 31.95 p1 = 106.95 MPa, p3 = 0 p2 = 43.05 MPa (a) Maximum shear stress theory

σ yp p1 240 = = 2 2 × F OS 2 × F OS 106.95 = FOS =

240 F OS 240 = 2.244 106.95

(b) Maximum principal strain theory n = 0.33

σyp p1 vp2 − = E E E × F OS or,

106.95 − 0.33 × 43.05 =

240 F OS

106.95 − 15.52 =

240 F OS

F OS =

MTPL0259_Chapter 15.indd 618

240 = 2.625 91.43

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(c) Strain energy theory σ   p12 + p22 − 2v( p1 p2 )  =  yp   FOS 

2

 σ yp  [106.952 + 43.052 − 2 × 0.33(106.95 × 43.05)] =   FOS 

2

 σ yp  11438.3 + 1853.3 − 0.66 × 4604.19 =   F OS 

F OS =

10252.8 =

σ yp F OS

101.256 =

σ yp F OS

2

240 = 2.370 101.256

Problem 15.4 A shaft is subjected to a bending moment and a twisting moment simultaneously and at a particular section, the bending moment is M and the twisting moment is T; show that the strain energy per unit volume is 4=

1 [σ 2 + 2τ 2 (1 + ν )] 2E

where s is the maximum bending stress, t is the maximum shear stress and v is the Poisson’s ratio. Solution

σ=

32 M 16T ,τ = 3 πd πd3

Stresses on an element of the shaft are shown in Fig. 15.14. Principal stresses are 2

p1 =

σ σ +   +τ2  2 2 t

σ 2  σ p2 =   −  +τ2  2 2  1 [ p12 + p 22 − 2ν ( p1 p2 )] 2E

t 2

p12 =

MTPL0259_Chapter 15.indd 619

s (Normal)

s

Strain energy per unit volume u=

(Stress)

σ2 σ2 σ σ 2 + +τ2 + 2   +τ 4 4 2  2

Figure 15.14

Stresses on an element of shaft

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620 Chapter 15

2

p 22 =

p12 + p 22 =

σ2 σ2 σ σ 2 + +τ2 − 2   +τ 4 4 2  2 σ2 σ2 × + 2τ 2 = σ 2 + 2τ 2 2 2

2 2 σ  σ  σ σ p1 p2 =  +   + τ 2  ×  −   + τ 2   2  2  2   2 

=

σ2 σ2 − − τ 2 = −τ 2 4 4

2np1 p2 = −2nt 2, putting value in the expression for strain energy per unit volume 4=

1 1 σ 2 + 2τ 2 (1 + ν )  . (σ 2 + 2τ 2 + 2ντ 2 ) = 2E 2E 

Problem 15.5 A mild steel shaft of 40 mm diameter when subjected to a pure torsion ceases to be elastic when torque reaches 2 kN m. A similar shaft is subjected to a torque of 1.2 kN m and a bending moment M kN m. If the maximum strain energy is the criterion for the elastic failure, find the value of M, Poisson’s ratio = 0.28. Solution Torque,

T = 2 × 106 Nm

Shaft diameter,

d = 40 mm

Maximum shear stress,

τ=

16T 16 × 2 × 106 = πd3 π × 403

= 159.15 N/mm2 Principal stresses on the surface of shaft are +τ , −τ , 0 or 159.15, −159.15, 0 N/mm2 Strain energy per unit volume, u =

MTPL0259_Chapter 15.indd 620

1 [ p12 + p 22 − 2ν ( p1 p2 )] 2E

=

1 [159.152 + ( −159.15) 2 − 2 × 0.28(159.15) ( −159.15)] 2E

=

(159.15) 2 2.56 × 159.152 (1 + 1 + 0.56) = 2E 2E

=

64845.5 2E

(15.5)

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Shaft under M and T We know that principal stresses are p1 =

16 [M + M 2 + T 2 ] πd3

p2 =

16 [M − M 2 + T 2 ] πd3

U′ =

1  16  [ p12 p22 − 2π ( p1 p2 )] ×  3   πd  2E

2

=

2 2 2 2 2 2 2 2   162  1  M + M + T + 2M M + T + M + M + T   2 E  −2 M M 2 + T 2 − 2π ( M + M 2 + T 2 )( M − M 2 + T 2 )   π d 3   

=

 162  1 [4 M 2 + 2T 2 − 2π ( M 2 − M 2 − T 2 )] ×  3  2E  πd 

=

1  16  [4 M 2 + 2T 2 − 2π ( −T 2 )] ×  3   πd  2E

2

2

1 1  16  [4 M 2 + 2T 2 (1 + π )] ×  3  = × 64845.5  πd  2E 2E  πd3  4 M 2 + 2.56T 2 = 64845.5 ×   16   π × 403  = 64845.5   16 

2

2

= 64845.3 × 153.9 × 106 = 10.239 × 1012

(15.6)

But, T = 1.2 × 10

6

T 2 = 1.44 × 1012 2.56 T 2 = 3.6864 × 1012, putting the value in Eq. (15.6) 4 M 2 = 10.239 × 1012 − 3.6864 × 1012 = 6.5526 × 1012 M 2 = 1.6385×1012 M = 1.28 × 106 M = 1.28 kN m

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622 Chapter 15

Problem 15.6 A circular steel cylinder of wall thickness of 10 mm and internal diameter of 200 mm is subjected to a constant internal pressure of 15 MPa. Determine how much axial compressive load can be applied to the cylinder before yielding commences according to the maximum shear stress theory. The yield stress of the material in simple tension is 240 MPa. Assume that the radial stress in the cylinder is negligible. Solution Cylinder Wall thickness,

t = 10 mm

Internal diameter,

D = 200 mm

Internal pressure,

p = 15 MPa

Hoop stress,

sc =

pD 15 × 200 = = 150 MPa 2t 2 × 10

Axial stress,

sa =

pD 15 × 200 = = +75 MPa 4t 4 × 10

Area of cross-section of the cylinder, A = π Dt = π × 200 × 10 = 2, 000π mm 2

σ c + σ a′ 240 σ yp = = as per the shear stress theory 2 2 2 where sa1 = net axial stress due to the internal pressure and axial compressive force. sc = 150 MPa 150 − σ a′ 240 = 2 2 sa′ = − 90 N/mm2 = sa + sac sa = +75 N/mm2 sac = axial compression stress due to axial face = −90 − sa = −90 − 75 = −165 MPa Axial compressive force,

Pc = p Dtsac = 2,000p × 165 = 1,036,725 N = 1.0376 MN

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Key Points to Remember  If p1, p2 and p3 are the principal stresses at a critical section of engineering component, if p1 > p2 > p3 and syp is the yield point stress of the material, when tested in simple tension or compression test, FOS is the factor of safety.  As per the maximum principal stress theory p1 ≤

σ yp F OS

 As per the maximum shear stress theory (i)

σ yp p1 − p3 ≤ 2 2 (F OS)

(ii) If p3 = 0, then

σ yp p1 − p2 ≤ , if p1 and p2 of opposite signs 2 2F OS

(iii) If p3 = 0, p1 and p2 of same signs

σ yp p1 ≤ 2 2 (F OS)  Maximum principal strain theory [ p1 − v ( p2 + p3 )] ≤ σ yp , where v is the Poisson’s ratio  Strain energy theory (two dimensional)

(p

2 1

 σ yp  + p 22 − 2vp1 p2 ≤   FOS 

2

 σ yp  + p 22 − p1 p2 ≤ 2   FOS 

2

)

 Shear strain energy theory

(p

2 1

)

Review Questions 1. What is the basic difference between ductile and brittle materials? Which theories of failure are used for brittle materials and which theories are used for ductile materials? 2. Why yield point stress is taken as failure stress for ductile materials? 3. Which theory of failure gives most conservative design? 4. Why is the maximum principal strain theory not used for general design purpose? 5. For brittle materials, such as rock and marble, which theory of failure is used and why? 6. Compare the maximum shear stress theory and the distortion energy theory for the design of a shaft made of ductile materials.

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624 Chapter 15

Multiple Choice Questions 1. Which of the following theories is used for brittle materials?

following theories of failure gives the smallest diameter of the shaft?

(a) Maximum shear stress theory

(a) Maximum principal stress theory

(b) Maximum principal strain theory

(b) Maximum principal strain theory

(c) Distortion energy theory

(c) Strain energy theory

(d) None of these 2. Graphical representation of which one of the following theories is by an ellipse? (a) Maximum principal strain theory (b) Distortion energy theory (c) Maximum shear stress theory (d) None of these 3. A shaft is subjected to a torque and an axial compressive force. Shear stress due to torque is 30 MPa and axial compressive stress due to force is 80 MPa. If syp = 270 MPa, what is FOS as per the maximum principal stress theory, Poisson’s ratio = 0.3? (a) 3.0

(b) 2.90

(c) 2.80 (d) None of these 4. Which one of the following theories gives conservative design of a component? (a) Maximum shear stress theory (b) Maximum principal strain theory (c) Distortion energy theory (d) None of these 5. A thin cylindrical shell with D/t ratio equal to 40 subjected to an internal pressure of 2 N/mm2. The yield point stress of the material is 210 N/mm2. Using the maximum shear stress theory for designing the thin shell, the FOS is: (a) 4.75

(b) 5.0

(c) 5.25 (d) 5.50 6. A shaft subjected to pure torsion is to be designed. The yield point stress of the material is 270 MPa and the Poisson’s ratio is 0.3, which of the

(d) None of these 7. A thick cylinder of an internal diameter of 100 mm and an external diameter of 200 mm is subjected to an internal pressure p. The yield strength of the material is 240 MPa. Taking an FOS of 2 and using the maximum principal stress theory of failure, the maximum value of the internal pressure p is (a) 120 N/mm2

(b) 90 N/mm2

(c) 72 N/mm2

(d) None of these

8. The principal stresses at a point are 70, 60 and −18 MPa; say syp is the yield point stress of the material. Using the maximum principal stress theory, FOS is 4. What is the FOS if the maximum principal strain theory is used, Poisson’s ratio of material is 1/3? (a) 4

(b) 4.5

(c) 5.0

(d) 5.25

9. The principal stresses developed at a point are +80, −80 MPa and 0, 0. Using the shear strain energy theory, the FOS is 3. The yield point stress of the material is (a) 80 ×

2 MPa

(c) 240 MPa

(b) 80 ×

3 MPa

(d) None of these

10. At a point in a strained material, the principal stresses are p1, p2 and 0.0. What combination of principal stresses will give same FOS by yielding according to the maximum shear stress theory and the distortion energy theory of failure? (a) p1 = −p2

(b) p1 = 0.5p2

(c) p1 = p2

(d) None of these

Practice Problems 1. Considering the principal stresses in a steam boiler as p, 0.5p, 0 and Poisson’s ratio = 0.28, equivalent stress in a simple tensile test as s, find p in each of the five theories of failure (except Mohr’s theory). 2. What combination of principal stresses will give the same FOS for failure by

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Theories of Failure

625

(i) the maximum shear stress theory and the distortion energy theory? Consider a two-dimensional stress system, that is, p1, p2, p3 = 0. 3. A hollow shaft of 30 mm internal diameter and 50 mm external diameter is subjected to a twisting moment of 800 Nm and an axial compressive force of 40 kN. Determine the FOS according to: (a) Maximum shear stress theory,

(b) Maximum principal strain theory and

(c) Strain energy theory. Given the yield point stress = 280 MPa. 4. A shaft is simultaneously subjected to a bending moment, M = 1.2 kN m and a twisting moment, T = 2.0 kN m at a particular section. Using the strain energy theory of failure determine FOS, if the diameter of the shaft is 50 mm n = 0.30, syp = 270 MPa. 5. A mild steel shaft of a diameter of 50 mm when subjected to pure torsion ceases to be elastic when the torque reaches 4.2 kN m. A similar shaft is subjected to a torque of 2.8 kN m and a bending moment of M kN m. If the maximum strain energy is the criterion for elastic failure, find the value of M, Poisson ratio is 0.3. 6. A ductile machine part has a yield strength of 270 MPa. It is loaded so that the following stresses result, sx = 100 MPa, sy = −100 MPa, txy = 0. Determine if the machine part will fail based on: (a) Distortion energy theory,

(b) Maximum shear stress theory.

Special Problems 1. A thin-walled cylindrical pressure vessel of ASTM 36 steel has an outer diameter of 100 mm and a wall thickness of 5 mm; determine the internal pressure that would fail the material by yielding based on the distortion energy theory. syp of A 36 steel is 250 MPa. [Hint: determine sc, sa] 2. A thick cylindrical shell of 150 mm internal radius and 200 mm external radius is subjected to an internal pressure of 30 MPa. Determine the FOS according to the distortion energy theory of failure, if the yield stress of the material is 280 MPa, take axial stress into account. [Hint: determine scmax, sa] 3. Determine the thickness of a thin cylinder of 600 mm diameter subjected to an internal pressure of 3 MPa according to (a) Maximum shear stress theory,

(b) Maximum principal stress theory and

(c) Maximum principal strain theory. Take FOS = 2, syp = 280 N/mm2, Poisson’s ratio = 0.28 4. A critical element develops principal stresses p1, p2 and p3 in the ratio of 5:2:-1, such that the stresses are 5p, 2p and -p, where p is a parameter. Determine the maximum magnitude of p, such that stresses may reach yielding according to (a) Tresca yield criterion (i.e., maximum shear stress theory) and (b) Von Mises criterion (distortion energy theory). syp = 400 MPa 5. The load on a bolt consists of an axial thrust of 8 kN together with a transverse shear force of 3 kN. Calculate the diameter of the bolt according to (a) Maximum principal stress theory and (b) Distortion energy theory. Take an FOS of 3. syp = 285 N/mm2

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626 Chapter 15 6. A shaft is subjected to a bending moment M and a twisting moment T. The value of M = 0.5T, show that strain energy per unit volume is equal to

512 M 2 (3 + v) when v is the Poisson’s ratio, d is the diameter of shaft and Eπ 2 d 6

E is the Young’s modulus.

Answers to Exercises Exercise 15.1: shaft diameter, d = 19.375 mm Exercise 15.2: 72.47 mm Exercise 15.3: 2.55

Exercise 15.4: 29.6 N/mm2 Exercise 15.5: t = 6 mm Exercise 15.6: P2 = −361 MPa

Answers to Multiple Choice Questions 1. 2. 3. 4.

(d) (b) (a) (a)

5. 6. 7. 8.

(b) (a) (c) (c)

9. (c) 10. (c)

Answers to Practice Problems 1. s, s, 1.162s, 1.015s, 1.154s 2. p1 > p2 both of same sign p1 = p2 p1 > 0, p2 < 0, p1 . p2 = 0 3. 3.44, 4.372, 4.10 Poisson’s ratio, v = 0.30

3. 2.33 4. 2.524 kN m 5. (a) will not fail (b) will not fail

Answers to Special Problems 1. 27.45 N/mm2 2. 2.36 3. (a) 6.293 mm, (b) 6.43 mm, (c) 5.50 mm

MTPL0259_Chapter 15.indd 626

4. (a) p = 66.67 MPa, (b) p = 76.98 MPa 5. (a) 11 mm, (b) 11.31 mm

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16 Strain Energy Methods CHAPTER OBJECTIVES The strain energy methods are most commonly used for solving the unknown factors in complicated structural problems. In this chapter, students will learn about:  Castigliano’s theorem for determination of displacement, angular rotation and angular twist by the partial derivatives of the strain energy absorbed by the body during deformation caused by external loads, moments etc. 

Strain energy in a body stored due to direct stress, shear stress, bending moment and twisting moment.



Deflection in beams due to shear strain energy.



Maxwell’s reciprocal theorem states that the total strain energy stored in a body, does not depend on the order in which forces are applied on it.



Principle of virtual force applied to trusses for determination of deflection at the joints.

Introduction Strain energy methods are used to determine the angular and linear displacements in complicated structural elements, such as a structure having members in different planes, straight or curved members. There is extensive use of these methods in determining: (a) displacements and forces in members of redundant structures, (b) displacement at any joint of a truss and (c) deflection and slopes in beams and cantilevers having variable section along length, etc. In this chapter, we will discuss Castigliano’s first theorem, Maxwell’s reciprocal theorem, strain energy expressions due to axial loads, bending moment, twisting moment and the principle of virtual force for determination of deflection in trusses.

Castigliano’s First Theorem According to Castigliano’s first theorem, ‘if a body is subjected to forces F1, F2, F3, …, Fn, and U is the strain energy stored in the body due to these forces, then partial derivative of the strain energy U, with respect to force Fi gives the displacement of the body in the direction of Fi, i.e., the displacement,

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628  Chapter 16

δ i = ∂U /∂Fi ’. This theorem is extremely useful in determining displacements in complicated structures subjected to external forces. w1 w2 w3 wn aa = y1 Proof Let us consider a beam AB of length L, initially straight and simply supported at the bb = y2 ends. Say, a number of transverse loads W1, W2, a b c n cc = y3 W3, …, Wn are gradually applied on the beam at a n b c nn = yn different points a, b, c, …, n and say deflections a  n under the loads are y1, y2, y3, …, yn, respectively, b c as shown in Fig. 16.1 by aa′, bb′, cc′, …, nn′, respectively. Figure 16.1 Beam subjected to various point Then, strain energy stored in the beam is: loads 1 U = (W1 y1 + W2 y2 + W3 y3 +  + Wn yn ) 2

(16.1)

That is, the area covered under the triangles shown shaded in Fig. 16.2. Now let us only increase load W1 by dW1, and due to this increase in the load, there is an increase in the deflection under the loads, W1, W2, W3, …, Wn, by dy1, dy2, dy3, …, dyn, which is indicated by the rectangular portion shown hatched in the diagram. Additional strain energy, 1 δU = δW1δ y1 + W1δ y1 + W2δ y2 + W3δ y3 + … + Wnδ yn 2 however, dW1dy1 is a very small quantity; therefore, it is neglected. Therefore, δU = W1δ y1 + W2δ y2 + W3δ y3 + … + Wnδ yn

(16.2)

Differentiating Eq. (16.1) partially with respect to load W1, (load, W2, W3, and W4 remain constant) ∂Wn ∂W3 ∂y ∂y ∂y ∂y ∂W2 2∂U + … + Wn n + yn = y1 + W1 1 + W2 2 + y2 + W3 3 + y3 ∂W1 ∂W1 ∂W1 ∂W1 ∂W1 ∂W1 ∂W1 ∂W1 ∂W2 ∂W3 ∂Wn = 0, because these loads remain unchanged = = ∂W1 ∂W1 ∂W1

but,

d W1

W1

a

W2

y1

a a d y1

Wn

W3 b

y2

b d y2

b

c

y3

c

c

d y3

n

yn

n n d yn

Figure 16.2 Graphs between loads and deflections

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Strain Energy Methods

∂y ∂y ∂y ∂y 2∂U = y1 + W1 1 + W2 2 + W3 3 + … + Wn n . ∂W1 ∂W1 ∂W1 ∂W1 ∂W1

629

(16.3)

Take Eq. (16.2) and divide throughout by ∂W1 ,

δy δy δy δy ∂U = W1 1 + W2 2 + W3 3 + … + Wn n ∂W1 ∂W1 ∂W1 ∂W1 ∂W1 in the limit ∂y ∂y ∂y ∂y ∂U = W1 1 + W2 2 + W3 3 + … + Wn n ∂W1 ∂W1 ∂W1 ∂W1 ∂W1

(16.4)

Subtracting Eq. (16.4) from Eq. (16.3) we get, ∂U = y1 . ∂W1

(16.5)

Similarly, it can be proved that ∂U ∂U ∂U = y2 , = y3 ,…, = yn. ∂W2 ∂W3 ∂Wn If a system of forces F1, F2, …, Fn, bending moments M1, M2, M3, …, Mn and twisting moments, T1, T2, T3, …, Tn act simultaneously on a body, then Castigliano’s first theorem can be extended to find angular rotation due to bending moment, angular twist due to twisting moment also, that is, ∂U = φi , regular rotation at location of Mi. ∂M i ∂U = θi , regular twist at the location of Ti. ∂Ti

Strain Energy Due to Axial Force Trusses are made from bars of different lengths joined at ends. Each bar carries some axial force, tensile or compressive depending on the external loads acting on the trusses. Figure 16.3 shows a bar of length L, area of cross-section A, subjected to a tensile force P. If E is the Young’s modulus of the bar, PL Change is length, dL = AE

A P

P L

Figure 16.3

1 P 2 L  P 2  AL P dL = = 2 2 AE  A2  2 E σ2 = × volume, where s is stress 2E developed in the bar

Strain energy, U =

∂U PL , axial extension, or displacement of bar along the force. = ∂P AE

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630  Chapter 16

Example 16.1 A truss ABCD is shown in Fig. 16.4. The cross section and material of each member is the same. Load P is applied at joint C. Length of the member CD is a. Using Castigliano’s first theorem, determine the deflection in the truss at the joint C. Solution Length of members DC = CB = BD = a a Length of AB = 2 Let us find forces in all the members of the truss by drawing Maxwell’s diagram. Give Bows’ notations E , F , G , H , I to spaces in truss as shown in the Fig. 16.5. Then, Maxwell’s diagram efigh is drawn. The following table lists the values of forces in members. a 2 A

B

H

e

E a

G

2P √3

a

h, g

I

608

2P √3

608

D

2P √3

C a

f

P

i P √3

F

Figure 16.4

Figure 16.5

Example 16.1

Maxwell’s diagram

Table 16.1 Forces in members of truss Force, F

Length, L

F 2L

EH

+2P (T ) 3

a 2

2P 2 a 3

BC

EI

+2P (T ) 3

a

4P 2 a 3

CD

FI

−

P (C) 3

a

P 2a 3

DB

HI

−

2P (C) 3

a

4P 2 a 3

Members

Bow’s notations

AB

∑F

MTPL0259_Chapter 16.indd 630

2

L=

11 2 P a 3

5/23/2012 11:41:00 AM

Strain Energy Methods

631

P

Note: C = Compressive T = Tensile Strain energy in members of the truss, 11P 2 a 11P 2 a U= = 3 × 2 AE 6 AE ∂U 11Pa Deflection under the load, = ∂P 3 AE

A 2m C

B

2m

Figure 16.6

2m

Exercise 16.1

Exercise 16.1 A structure ABC is shown in Fig. 16.6. Length of member CB is 4 m. Area of cross-section of each member is 5 cm2, determine the deflection under the load P. If P = 50 kN, E = 200 GPa.

Strain Energy Due to Shear Stress Consider a rectangular block of dimensions L × B × H as shown in Fig. 16.7. The block is fixed at the lower face and is subjected to a tangential force Q at the top face. The block is distorted due to this shear force Q and is subjected to shear strain f. ds

Q

f H

Q

B

L

Figure. 16.7 Rectangular block subjected to shear force Say, this displacement of the top face along force Q is ds as shown. 1 Work done on block = Qds 2 = Strain energy stored

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632  Chapter 16

1 Q × ds LBH (multiplying and dividing by LBH ) 2 LBH 1  Q   ds  =     LBH 2  LB   H  =

where Q = τ , shear stress developed LB ds = φ , shear strain, in reality angle f is very very small within the elastic hunt. H tan f = f 1 U = τφ LBH 2

therefore,

τ2 × LBH because shear strain, φ = shear stress , where G is the shear modulus 2G G τ2 Shear strain energy per unit volume, us = 2G In Chapter 9, we have learnt about distribution of transverse shear stresses in beams, that the distribution of shear stress across the section of the beam is non-uniform; therefore, the shear strain energy must be integrated over this whole section of a body. Yet shear strain energy due to shear deformation (particularly in beams) is very small and many a times ignored. =

Example 16.2 A beam of rectangular cross-section of breadth b, depth d and length l is simply supported at its ends. It carries a concentrated load W at its centre. Determine the shear strain energy in the beam and find the deflection due to shear. G = Modulus of rigidity for the beam Solution Figure 16.8(a) shows a beam AB of length l, simply supported at the ends and carrying a concentrated load W at its centre C. Shear force between A and C is +W/2 and between C and B, shear force is –W/2 .

dx A RA =

W 2

x

W

x x

C

B l 2

l 2

W 2

RB =

W 2

dy y

W 2

MTPL0259_Chapter 16.indd 632

N

d 2

A d 2 b (b)

(a)

Figure 16.8

y

Shear force diagram

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Strain Energy Methods

633

Consider a section X-X at a distance of x from the end A. W Shear force, Fx = + 2 Now, consider a small length dx. Let us determine the shear strain energy for the portion AC. Figure 16.8(b) shows the section of the beam. Consider a layer of thickness dy at a distance of y from the neutral axis. Fay Shear stress t at the layer = Ib where F = Shear force at the section = W/2 a y = First moment of the area above the layer under consideration about neural axis bd 3 I = Moment of inertia = 12 b = Breadth W 12 b  b d/ 2 − y   τ = × 3 ×  − y  y + Shear stress,     2 bd b 2 2  =

6W  b   d/ 2 + y  − y  3     2  bd 2

=

 3W  b 2 − y2  3  bd  4 

τ2 =

d 2 y2  9W 2  d 4 4 y + − 2  b2 d 6  16

Volume of the layer = bdxdy Shear strain energy in the layer τ2 = b dx dy 2G =

1 9W 2  d 4 d 2 y2  b × 2 6  + y4 − dydx 2G 2  b d  16

Total shear strain energy for the portion AC U s′ =

MTPL0259_Chapter 16.indd 633

d 2 y2  1 l / 2+ d /2 9W 2  d 4 4 + y − dydx 2G ∫0 −∫d /2 bd 6  16 2 

=

1 l / 2 9W 2 2G ∫0 bd 6

=

1 2G

∫

l/ 2

=

1 2G

∫

l/ 2

0

0

9W 2 bd 6

d /2

 d 4 y y5 d 2 y3  dx  16 + 5 − 6  − d /2  d 5 d 5   d 5 d5   d5 d5   +  +  +   −  +   dx  32 32   160 160   48 48  

9W 2  d 5 d 5 d 5  + − dx bd 6  16 80 24 

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634  Chapter 16

 9W 2 d 5  1  bd 6 × 30  dx = 2G

=

1 2G

=

1 9W 2 l 3W 2 l × × = 2G 30bd 2 40Gbd

∫

l/2

0

∫

l/2

0

9W 2 dx 30 bd

Since the beam is symmetrically loaded, shear force in the portion CB is the same, that is, −W/2. Shear strain energy for the beam, U s = 2U ′s =

3W 2 l 20Gbd

Deflection at the centre due to shear,

δs =

∂U s 3Wl = ∂W 10Gbd

Exercise 16.2 A beam of length 3 m is simply supported at ends. It carries a concentrated load of 6 kN at the midplane of the beam. The beam has a rectangular section of width 5 cm and depth 10 cm. If G = 84 kN/mm2, what is the deflection at the centre of the beam due to shear? Compare this deflection due to bending. E = 210 kN/mm2.

Strain Energy Due to Bending Consider a bar of length L, initially straight subjected to a gradually increasing bending moment, M. As the bending moment increases, curvature in the bar increases or the angular rotation f goes on increasing. Say, at a particular instant (Fig 16.9): Bending moment = M Angular rotation = f Radius of curvature = R 1 Work done on the bar = Mφ 2 L C however, f = since f is very small and the R stress in bar remains within the elastic limit. Work done = Energy stored in the bar 1 ML U = Mφ = 2 2R From the flexure formula, M E 1 M = or = I R R EI Therefore, strain energy, U=

MTPL0259_Chapter 16.indd 634

M 2L 2 EI

f

R M

M M

Length = L

Figure 16.9

U

f

Bar subjected to bending moment

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Strain Energy Methods

635

If we consider a beam subjected to transverse loads W1, W2, …, Wn, where the radius of curvature goes on changing from one section to the other, then the strain energy due to bending, U =∫

M x2 dx 2 EI

(16.6)

where Mx is the bending moment at any section X-X and dx is the small length along the axis of the beam. Say, di is the deflection under the load Wi. Then,

δi =

M ∂M x δU =∫ x × dx ∂Wi EI dWi

(16.7)

Example 16.3 A circular section cantilever of length L, free at one is fixed at the other end, with diameter d for half of its length and diameter 2d for the rest of its length carries a concentrated load W at the free end. If E is the Young’s modulus of the material, determine the deflection and the slope at the free end. Solution Figure 16.10 shows a cantilever ABC, fixed at end C and free at end A, with diameter d for half of its length AB and diameter 2d for next half of its length BC. Since we have to find out the slope at the free end A, let us apply a fictitious moment M = 0 at the free end. Portion AB l Bending moment, Mx = M + Wx, where x = 0 to taking 2 origin at A. Portion BC

W I2

C

B

I1

2d

d l 2

A M

l 2

M = Fictitious couple

Figure 16.10

l l  Mx = M + W  x +  , where x = 0 to , origin at B.  2 2 Strain energy,

U =∫

l /2

0

2 l /2  ( M + W ( x + l /2)  ( M + Wx )2 dx + ∫   dx 0  2 EI1 2 EI 2

where

I1 =

π d4 π ( 2 d )4 π d 4 = , I2 = 64 64 4

Deflection at A,

δ=

l/ 2 ( M + Wx )( x ) l/ 2  M + W ( x + l/ 2)   ∂U l =∫ dx + ∫   x +  dx  0 0 ∂W EI1 EI 2 2  

=

1 Mx 2 Wx3 + EI1 2 3

l /2

l /2

+ 0

1 Mx 2 Mlx Wlx 2 Wx3 Wl 2 x + + + + EI 2 2 2 2 3 4 0

however, M = 0

δ=

MTPL0259_Chapter 16.indd 635

1 1  Wl 3 Wl 3 Wl 3  Wl 3 × + + + 24 EI 2  8 24 8  EI1

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636  Chapter 16

=

1 Wl 3 1 7Wl 3 × + × 24 EI1 24 EI 2

Substituting the values of I1 and I2,

δ= =

φ=

Slope at end A,

92 Wl 3  64 7 × 4  Wl 3 = × 4  4 + 4  24 E pd pd  24 E pd 23Wl 3 6 Eπ d 4 l/2 M + Wx l/2 [ M + W ( x + l/2)](1)dx ∂U =∫ × (1) dx + ∫ 0 0 ∂M EI1 EI 2

1 Wx 2 φ= Mx + 2 EI1

l/2

0

l/2

1 Wx 2 Wl + + Mx + x 2 2 0 EI 2

however, M = 0

φ=

=

1 Wl 2 1  Wl 2 Wl 2  × + × + 8 4  EI1 EI 2  8 1 Wl 2 1 3 × + × Wl 2 8 EI1 EI 2 8

Substituting the values I1 and I2,

φ=

Wl 2  64 3 × 4  Wl 2 × 76 19Wl 2 + =  = 8 E  pd 4 pd 4  8Epd 4 2pEd 4

Example 16.4 A circular ring of mean radius R and second moment of area of its cross-section I, with a slit at one section is shown in Fig. 16.11. Points A and B are subjected for forces P as shown. Determine the relative displacement between points A and B. Only the strain energy due to bending is to be taken into account. Solution Consider an element of length ds = Rdq at angle q from the vertical axis. Bending moment of the force P on the element. Mx = P(R − Rcos q) = PR(1 − cos q) Strain energy, U = 2∫

π

0

=

MTPL0259_Chapter 16.indd 636

[ PR(1 − cos θ )]2 Rdθ 2 EI

P 2 R3 π (1 + cos2θ − 2 cos θ ) dθ EI ∫0

5/23/2012 11:41:10 AM

Strain Energy Methods

=

Y

 P 2 R3 π   1 + cos 2θ  1+   − 2 cos θ  dθ  ∫ 0  2 EI   π

P 2 R3 3 sin 2θ = θ+ − 2 sin θ EI 2 4 0

R dq

3π P 2 R3 = × 2 EI P

Relative displacement between points A and B,

δ=

637

∂U 3π 2 PR3 PR3 = × = 3π ∂P 2 EI EI

q

A

B Y

P

Figure 16.11

Exercise 16.3 A cantilever of a circular cross-section of length L, free at one end and fixed at the other end, with a diameter of 10 cm for half of its length, and 20 cm for the rest of the length carries a concentrated load of 3 kN at free end as shown in Fig. 16.12. If length of cantilever L = 2 m, E = 208 kN/mm2, determine the deflection and the slope at free end. Exercise 16.4 A circular ring of mean radius 30 cm and diameter of section 5 cm is subjected to loads P at ends A and B as shown in Fig. 16.13. Determine the value of P if the displacement between points A and B is 1 mm, if E = 208 kN/mm2.

O C

B

10 cm f

Figure 16.12

3 kN A

20 cm f 1m

30 cm

1m

Exercise 16.3

P

A

Figure 16.13

B

P

Exercise 16.4

Strain Energy Due to Twisting Moment Figure 16.14 shows a shaft of diameter d and length l subjected to a gradually increasing twisting moment. As the twisting moment increases, the angular twist also increases gradually. At a particular stage say, the torque applied is T and angular twist in the shaft is q.

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638  Chapter 16 T

Torque

d

T

Torque

q 0

U

d

T l

q

Figure 16.14 Angular twist

Work done on the shaft 1 = Tθ 2

= U, strain energy stored in the shaft From torsion formula (of Chapter 12), T Gθ Tl = or θ = J l GJ Strain energy due to twisting moment, U=

1 T 2l × 2 GJ

where J = Polar moment of inertia of the shaft

= U=

πd4 (for a solid shaft) 32 T 2l 2GJ

If the torque or the section vary along the length of the shaft, U=

1 2G

T 2 dl ∫0 J l

Angular twist due to the twisting moment,

θ=

∂U ∂T

Example 16.5 A circular bar of diameter d is bent at right angle. It is fixed at one end and a load W is applied at the other end as shown in Fig. 16.15. Determine the deflection under the load W, if E = Young’s modulus and G = Shear modulus of the material.

MTPL0259_Chapter 16.indd 638

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Strain Energy Methods

Solution Let us calculate the strain energy. Portion BC U1 = ∫

b

0

639

W C

Dia d

90

A

W 2 b3 (Wx )2 , due to bending moment dx = 2 EI 6 EI

B

b

a

Portion AB Figure 16.15

U 2 =Strain energy due to bending =∫

a

0

(Wx )2 W 2 a3 dx = 2 EI 6 EI

Straight portion AB is also subjected to a twisting moment, T = Wb U3 = Strain energy due to twisting moment U3 = where

I=

A

(Wb)2 a W 2 ab2 = 2GJ 2GJ

B

π d4 π d4 and J = 64 32

Total strain energy, U = U1 + U 2 + U 3 = Deflection under the load, ∂ =

W

a a D 2 3

2 3

2

W b W a W ab + + 6 EI 6 EI 2GJ

∂U Wb3 Wa3 Wab2 = + + ∂W 3EI 3EI GJ

2

a C

Figure 16.16 Exercise 16.5

=

W × 64 W × 32 ( a3 + b3 ) + ( ab2 ) 4 3E × π d G × πd4

=

32W  2( a 3 + b3 ) ab2  +  G  π d 4  3E

Exercise 16.5 A steel rod of diameter 5 cm and 30 cm long is bent at right angles at B and at C as shown in Fig 16.16. How much load W can be applied at end D so that the deflection under load is limited to 1 mm, G = 84 kN/mm2, E = 200 kN/mm2.

Maxwell’s Reciprocal Theorem This theorem states that ‘work done by the forces of the first state on the corresponding displacements of the second state is equal to the work done by the forces of the second state on the corresponding displacements of the first state’. In symbols P1 δ1II + P2 δ 2II = P3δ 3I + P4 δ 4I

MTPL0259_Chapter 16.indd 639

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640  Chapter 16

Figure 16.17(a) shows an elastic body supported and subjected to forces P1 and P2 at points 1 and 2, respectively, due to which displacements at four points are δ1I , δ 2I , δ 3I and δ 4I , respectively. Figure 16.17 (b) shows same elastic body subjected to forces P3 and P4 at points 3 and 4, respectively, due to which the displacements at four points are δ1II , δ 2II , δ 3II and δ 4II , respectively. P3 3 P2

3

P4

3

4

4

3

4

4

2

2 2

2 1

1

1

1

P1

11 = δ1I, 22 = δ2I,

33 = δ3I, 44 = δ4I,

11 = d1II, 33 = d3II,

(a)

22 = d2II, 44 = d4II, (b)

Figure 16.17 Let us say that forces P1 and P2 are applied first, then strain energy U′ =

1 1 P1 δ1I + P2 δ 2I 2 2

Now, in the second step, forces P3 and P4 are applied due to which displacements are δ1II , δ 2II , δ 3II and δ 4II , respectively, the strain energy U ′′ =

1 1 P3 δ 3II + P4 δ 4II + P1δ1II + P2 δ 2II 2 2

Total strain energy, U = U ′ + U ′′ =

(

)

1 P1 δ1I + P2 δ 2I + P3δ 3II + P4 δ 4II + P1 δ1II + P2 δ 2II 2

(16.8)

In the second manner of loading, we assume that P3 and P4 forces are applied first on the body and then the forces P1 and P2 are applied.

MTPL0259_Chapter 16.indd 640

U′ =

1 1 P3δ 3II + P4δ 4II 2 2

U ′′ =

1 1 P1δ1I + P2δ 2I + P3δ 3I + P4 δ 4I 2 2

5/23/2012 11:41:16 AM

Strain Energy Methods

641

Total strain energy,

(

)

1 P1δ1I + P2δ 2I + P3δ 3II + P4δ 4I + P3δ 3I + P4δ 4I (16.9) 2 Total energy stored in the body does not depend on the order in which the forces are applied; therefore, from Eqs (16.8) and (16.9) U = U ′ + U ′′ =

P1δ1II + P2δ 2II = P3δ 3I + P4δ 4I Maxwell’s reciprocal theorem can be proved for any number of forces. Example 16.6 A beam of length l is simply supported at the ends. Loads W1 and then W2 are applied in two states on points 1 and 2 as shown in Fig. 16.18. (a) When load W1 is applied at point 1, in the first state equation for deflection, 3

W1 x3 W1  l 7 − x−  − W l 2 x (can be 8 6  4  128 1 derived) l deflection at position (2) δ 2I , at x = 2 EIy =

EIδ 2l =

W1 A

1 4

1 4 3W1 RA = 4

3

Deflection at position (1) δ1II at x =

x

A

W2

1 4

3

1 2

Figure 16.19

11W2 l W2 l Wl − omitted term − 2 = − 768 64 768

W2δ 2I = −

1 2 II-State

3

11W2 l 3 768 EI

B

2

1

l 4

δ1II = −

MTPL0259_Chapter 16.indd 641

W1 4

(16.10)

(can be derived)

Now,

RB =

I-State

11 W1l 3 × 768 EI

W2 x3 W2  W2 l 2 x l − x − − x   12 6  2 16

EIδ1II =

1 2

Figure 16.18

(b) When load W2 is applied at point 2, Fig. 16.19 in the second state, the equation for deflection, EIy =

B

2

1

11W1l 3 W1l 3 W1l 3 7W1l 3 − − =− 64 384 256 768

δ 2I = −

or,

x

3

(16.11)

11 W2W1l 3 × 768 EI

5/23/2012 11:41:18 AM

642  Chapter 16

W1δ1II = −

l3 11 WW × 1 2 768 EI

W B

2

A

1 C

From these expressions it is obvious Wδ =W δ , II 1 1

I 2 2

which goes to prove Maxwell’s reciprocal theorem.

L 2

L 2

L 2

Figure 16.20

Exercise 16.6 Figure 16.20 shows a beam M of length 3L , hinged at A and a roller supported at B. 2 In I-state, apply load W at point 1 and determine deflection at point 2. In the second state apply load W at point 2 and determine deflection at point 1, then prove Maxwell’s reciprocal theorem.

Principle of Virtual Forces Applied to Trusses The principle of virtual forces can be used to determine the displacement of any point in a truss subjected to external loads. To illustrate this principle, vertical displacement of the joint C caused by real load P at joint B will be determined, in Example 16.7. The load P at B can cause only the axial forces in members as shown for member AB; it is only necessary to consider the internal virtual work due to axial load. To obtain this virtual work, we will assume that each member has constant cross-sectional area A, and the virtual load n and real load N are constant throughout. To obtain the internal virtual work so as to get displacement of the joint C, we apply a unit load at the joint C. Due to this unit load at C, axial loads are determined in each member of the truss, as in member AB, axial force due to unit load, n = 0. Example 16.7 Due to the real load P at B, axial forces are determined with the help of a force polygon for the truss as shown in the Fig. 16.21(c). Solution Say, for member AB, axial force due to real load, N = −P (compressive load) Internal virtual work for a member L nN nNL ∫0 AE dx = AE Virtual work for the entire truss is, therefore, nNL 1∆ = ∑ AE Here, 1 = External virtual load acting on the truss joint, in the direction of displacement D D = Joint displacement caused by the real loads on the truss n = Internal virtual force in a truss member caused by external virtual unit load N = Internal axial force in a truss member due to real loads L = Length of a member A = Cross-sectional area of a member E = Young’s modulus of elasticity of a member

MTPL0259_Chapter 16.indd 642

5/23/2012 11:41:19 AM

Strain Energy Methods

643

C

D 45

l 45

45

A l

B A

A

l

P

N

(a)

N = −P

B N

(b)

2P

P

√2 P

√2 P Force polygon (due to real load, P at B) (c)

P 1 D

C

n

1

A

B

n

n=o 1

A

√2 (Force polygon for virtual unit load at C )

B (d)

Figure 16.21 Table 16.2

Forces in members due to real load and unit load

Member

n

N

L

nNl

AB

0

−P

2l

0

BC

0

− 2P

2l

0

CD

+1

+ 2P

CA

− 2

− 2P

l 2l

2Pl 2 2P l

∑ nNl = (2 + 2

MTPL0259_Chapter 16.indd 643

2) Pl

5/23/2012 11:41:21 AM

644  Chapter 16

Virtual work equation for the truss shown is =

2(1 + 2)Pl , if AE is constant for the truss for all members AE

For truss let us take

20 kN

30 kN

A

P = 10 kN

C

B

l = 2 m = 2,000 mm A = 500 mm2 E = 200 kN/mm2

D

4m E



=

(

)

2 1 + 2 × 10 × 2,000 500 × 200

= 0.966 mm

3m

3m

D = Vertical displacement at the joint C

Figure 16.22

Excercise 16.7

Exercise 16.7 A frame ABCDE is shown in Fig. 16.22. It is subjected to vertical loads of 20 kN at B and 30 kN at C for each member A = 4 cm2, E = 200 GPa. Determine the vertical displacement of point C using the principle of virtual force. Problem 16.1 A bar ABCD of rectangular section having uniform width b throughout but thickness varying as 3t for 3a length, 2t for length 2a and t for the length a is of the shape shown in Fig. 16.23. A load W is applied at the end D. Determine the deflection under the load. Given E = Young’s modulus of the material (consider only the strain energy due to bending). Solution Let us determine the strain energy due to bending Portion AB U1 = ∫

2a

0

B

2a

a B 2t

A I1

3t

I2

2a

W I3

C

2 a (Wx ) (Wx )2 dx + ∫ dx 0 2 EI 2 EI1 1

D a

Figure 16.23

Taking the origin for x at B′ as shown U1 =

8W 2 a3 W 2 a3 9 W 2 a3 3 W 2 a3 + = = 6 EI1 6 EI1 6 EI1 2 EI1

Portion BC Bending moment is constant Wa. U2 = ∫

2a

0

MTPL0259_Chapter 16.indd 644

(W 2 a2 )dx W 2 a3 = EI 2 2 EI 2

5/23/2012 11:41:22 AM

Strain Energy Methods

645

Portion CD Bending moment = Wx (taking origin for x at D) W 2 x 2 d x W 2 a3 = 0 2 EI3 6 EI3

U3 = ∫ Total strain energy,

a

U = U1 + U 2 + U 3 =

1 1  W 2 a3  3 + + E  2 I1 I 2 6 I3 

2Wa3  3 1 1  ∂U + + = vertical deflection at D =  ∂W E  2 I1 I 2 6 I3  I1 =

where

δD = =

b b bt 3 27 3 8 3 bt ; I 2 = ( 2t )3 = bt ; I3 = (3t )3 = 12 12 12 12 12 2Wa3  3 12 12 12  + +  ×  E  2 27bt 3 8bt 3 6bt 3  24Wa3  3 1 1  24Wa3 25 25Wa3 × =  + + = 72 3Ebt 3 Ebt 3  54 8 6  Ebt 3

Problem 16.2 A cantilever of length l, fixed at one end and propped at the other end, carries a concentrated load W at its centre and a uniformly distributed load w per unit length from the centre up to the fixed end. If EI is the flexural rigidity of the cantilever determine the reaction at the prop. Solution Figure 16.24 shows a cantilever ABC of length l, fixed at the end C and propped at end A, carrying loads as given in the problem. Let us first determine the strain energy due to bending. Say the reaction at the prop = R. Portion AB (Origin at A) U1 = ∫

l/2

0

2 3 l/2

(Rx )2 dx R x = 2 EI 6 EI

0

=

R2 l 3 48 EI

W w

B l 2

A l 2

R

Figure 16.24

Portion BC (Taking origin at B) Bending moment,

l wx 2  Mx = R  x +  − Wx −  2 2 U2 = ∫

l/2

0

Total strain energy = U1 + U2

(M x )2 dx 2 EI

∂U = 0, at the propped end ∂R therefore,

MTPL0259_Chapter 16.indd 645

l/2  M dM  ∂U 2 Rl 3 x = +∫  x dx 0  EI dR   ∂R 48 EI

5/23/2012 11:41:24 AM

646  Chapter 16

=

l/2   Rl 3 l wx 2   l 1 + ∫  R  x +  − Wx −   x +  dx × 0   EI 24 EI 2 2 2  

=

1 Rl 3 + 24 EI EI

∫

l/2

0

2   2 l2  l   2 xl  wx   R  x + + lx  − W  x +  −  x +   dx 4 2 2 2     l /2

Rl 3 Rlx 2 Wx3 Wx 2 l wx 4 wx3 l  1  Rx3 Rl 2 x 0= + + + − − − − 24 EI EI  3 4 2 3 4 8 12  0 or,

Rl 3 Rl 3 Rl 3 Rl 3 Wl 3 Wl 3 wl 4 wl 4 + + + − − − − =0 24 24 8 8 24 16 128 96 Rl 3 Rl 3 5Wl 3 7wl 4 + + − =0 12 4 48 384 R 5W 7wl − − =0 3 48 384

Reaction at the prop, R =

5W 7wl + 16 128

Problem 16.3 A bar is bent in the shape as shown in Fig. 16.25, with radius of the bend R and length of the straight portion l. Determine the horizontal deflection due to the force P applied at the end A, if EI is the flexural rigidity of the bar. Consider only the strain energy due to bending.

C

π

U1 = ∫

0

Rdq

dq q O

Solution Portion AC Consider an element of length = Rdθ = ds The bending moment on ds = PRsin q Strain energy

A

P

l

B

( PR sinθ ) P R Rd θ = 2EI 2 EI 2

R

2

3

∫

π

0

sin2θ d θ

Figure 16.25 π

=

P 2 R3 π  1 − cos 2θ  P 2 R3 θ sin 2θ − θ d =   2 EI ∫0  2 2 EI 2 4 0

=

P 2 R 3 π π P 2 R3 × = 2 EI 2 4 EI

(16.12)

Portion CB (origin at C), Bending moment, Mx = Px Strain energy,

P 2 x2 P 2l 3 dx = 0 2 EI 6 EI

U2 = ∫

l

2 3 2 3 Total strain energy, U = U1 + U 2 = π P R + P l 4 EI 6 EI

MTPL0259_Chapter 16.indd 646

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Strain Energy Methods

647

Horizontal deflection at A, P  π R3 l 3  ∂U π PR3 Pl 3 = + = +  2 EI 3EI EI  2 3 ∂P

δA =

Problem 16.4 Figure 16.26 shows a steel rod bent into the form of three quarters of a circle of radius r. End A of the rod is fixed, while end B is constrained to move vertically. If a load W is applied at the end B, determine the vertical deflection at the end B. Given EI is the flexural rigidity of the rod. Solution Figure 16.26 shows a rod bent into the form of three quarters of a circle. Since the end B is constrained to move only vertically, a horizontal reaction R will be offered by the constraint. Consider an element of length ds = rdq at an angle q to the vertical.

O A q rdq

R B W

Figure 16.26

Problem 16.4

Mθ = W × r sin θ − R × ( r − r cos θ ), where R is reaction at B

Bending moment,

U =∫

Strain energy,

3π /2

0

2 3π /2 [Wr sin θ − Rr (1 − cos θ )] Mθ2 r dθ = ∫ r dθ 0 2EI 2EI

Since there is no horizontal deflection, 0=−

r

1 2 EI

∫

3π /2

0

=−

r3 EI

∫

=−

r3 EI

∫

3 π /2

=−

r3 EI

∫

3π / 2

=−

r3 EI

∫

3π / 2

3π /2

0

0

0

0

∂U =0 ∂R

2[Wr sin θ − Rr (1 − cos θ )](1 − cosθ )r 2 dθ

[W sin θ − R + R cos θ )](1 − cos θ ) dθ (W sin θ − W sin θ cos θ − R + R cos θ + R cos θ − R cos2θ )dθ W sin 2θ R(1+cos 2θ )   − R + 2 R cos θ − W sin θ −  dθ 2 2   3π /R

Wsin 2θ 3R Rcos 2θ   − + 2 R cos θ − W sin θ − 2 2 2 0  3π /R

or

W cos 2θ 3R R sin 2θ   0 =  −W cos θ + − θ + 2 R sin θ − 2 4 4 0 

or

3p 3R 3p   W −W  cos − cos 0° + (cos 3p − cos 0) − ×   4 2 2 2  3p  R +2 R  sin − sin 0° − (sin 3p − sin 0) = 0   4 2

MTPL0259_Chapter 16.indd 647

5/23/2012 11:41:28 AM

648  Chapter 16

W+ or

W 9 Rπ ( −2) − + 2 R( −1) = 0 4 4 W 9π − R − 2R = 0 2 4 W= R=

Reaction,

9π R + 4R 2 W 9π / 2 + 4

(16.13)

Vertical deflection,

δB =

3π /2 [Wr sin θ − Rr (1 − cos θ )] r sin θ rd θ ∂U =∫ 0 ∂W EI

=

r3 EI

∫

=

r3 EI

∫

=

r3 W W R θ − sin 2θ + R cos θ − cos 2θ 4 4 EI 2 0

=

r3 EI

=

r 3  3πW R  r3  3πW R  −R+  = −    4 2  EI  4 2 EI

3π /2

0

(W sin2 θ − R sin θ + R sin θ cos θ ) dθ

3π /2

0

R sin 2θ  W dθ  2 (1 − cos 2θ ) − R sin θ + 2   3π /2

r3 = EI

 W 3π W 3π   R  2 × 2 − 4 (sin 3π − sin 0) + R  cos 2 − cos 0 − 4 (cos 3π − cos 0)   

W   3πW −   , putting the value of reaction, R 4 9π + 8  =

1  Wr3  3π −   EI  4 9π + 8 

Problem 16.5 A circular bar of diameter d is bent into the shape of a ‘U’ of radius R and straight portion of length l as shown in Fig. 16.27. Equal loads P each are applied at the ends A and E. Determine the relative shift between points A and E. E = Young’s modulus of the material

MTPL0259_Chapter 16.indd 648

x

x

Rdq dq C

P A

x

B X

q

R

D

E l

P

Figure 16.27

5/23/2012 11:41:30 AM

Strain Energy Methods

649

Solution Let us first determine the strain energy due to bending. Portion AB Bending moment, Mx = Px Strain energy, U1 = ∫

l

0

=

2 2 l P x dx M x2 dx πd4 , where I = =∫ ( moment of intertia) 0 2 EI 2 EI 64

P 2l 3 6 EI

Portion BC Taking origin at B, x = Rsin q where q varies from 0 to p/2 Length of the element consider, ds = Rdq Bending moment,

M x = P (l + R sin θ )

Stain energy,

U2 = ∫

π /2

0

[ P (l + R sin θ )]2 Rdθ 2 EI

=

P2 2 EI

∫

π/2

=

P2 2 EI

∫

π/2

=

P2 2 R3 P 2 R3 sin 2θ P 2 lR2 π/2 l Rθ + θ − + −2 cos θ 0 2 EI 2 0 4 EI 2 0 2 EI

=

P2 π  2 R3  2 P 2 lR2 × l R +  − 0 + 2 EI 2  2 2 EI

0

0

(l 2 R + R3 sin2θ + 2lR2 sin θ ) dθ  2  R3 (1 − cos 2θ ) + + 2lR2 sin θ  dθ l R  2  π/2

Total strain energy,

π/2

U = 2U1 + 2U 2 =

P 2 l 3 P 2π  2 R3  2 P 2 lR2 + l R+  +  EI 3EI 2 EI  2

=

4 P 2 l 3 + 6 P 2π l 2 R + 3P 2π R3 + 24 P 2 lR2 12 EI

Relative shift between points A and E

δ= =

MTPL0259_Chapter 16.indd 649

∂U 8 Pl 3 + 12 Pπ l 2 R + 6 Pπ R3 + 48 PlR2 = 12 EI ∂P 2 Pl 3 + 3Pπ l 2 R + 1.5 Pπ R3 + 12 PlR2 3 EI

5/23/2012 11:41:32 AM

650  Chapter 16

however,

I=

πd4 64

therefore,

δ=

64( 2 Pl 3 + 3Pπ l 2 R + 1.5 Pπ R3 + 12 PlR2 ) 3 Eπ d 4

=

32 P ( 4l 3 + 6π l 2 R + 3π R3 + 24 lR2 ) 4 3E π d

Problem 16.6 The frame, as shown in Fig. 16.28, is supported by a fixed end at G and is free at end A. A vertical load 12 kN acts at point C. Determine the horizontal deflection at A using Castigliano’s theorem, E = 200 kN/mm2 and I = 2,000 cm4. Consider only the bending energy. Solution Apply imaginary load, At the fixed end G, upward reaction, Horizontal reaction, Fixing moment, Member AB,

P @ 0 at A R1 = 12 kN R2 = P MG = 12 × 40 = 480 kN cm

Forces and moments as 3 portions are shown in Figure 16.29

80 P B

C

My = −Py (taking A as origin)

Moment,

∂M y ∂P

x

= − y , y varies from 0 to 80 cm 12 kN

Member BC

x C 40 cm

B

12 kN 40 cm

x

D

− 80 P D D

C

80

80 cm

y

MG G R2

P=0 A

y 480 P

R1

Figure 16.28

MTPL0259_Chapter 16.indd 650

12 kN

Figure 16.29

Problem 16.9

5/23/2012 11:41:34 AM

Strain Energy Methods

651

Mx = −80P ∂M x = −80 ∂P Member CD Mx = −80P − 12x, x varies from 0 to 40 cm ∂M x = −80 ∂P Member DG My = −Py − 480 (origin at G and y positive upwards) ∂M y

= −y ∂P Using Castigliano’s theorem,

δ H at end A =

40 1  80 ( − Py )( − y ) dy + ∫ ( −80 P )( −80) dx ∫ 0 EI  0

+ ∫ ( −80 P − 12 x )( −80) dx + ∫ ( − Py − 480)( − y ) dy   0 0 40

however,

80

P=0 80 1  40 δH = ( +960x ) dx + ∫ (1 + 480 y ) dy  ∫ 0  EI  0 =

 402  1  802  1 × 960  + × 480   EI  2  EI  2 

=

1 2, 304 × 103 (768 × 103 + 1536 × 103 ) = EI EI

E = 200 kN/mm 2 = 200 × 102 kN/cm 2 I = 2, 000 cm 4

δH =

2, 304 × 103 = 0.0576 cm = 0.576 mm 200 × 102 × 2, 000

= Horizontal deflection at point A

Note that for the calculation of strain energy for any member, that is, AB, BC, CD or DG, the origin for calculation of bending moment can be taken at any end. Problem 16.7 The semi-circular curved frame with radius R = 120 cm is subjected to a force P at its apex B. Determine the vertical deflection at B. The cross-section is a wide flange with moment of inertia I, cross-sectional area A and web area Aw. Consider the strain energy due to bending, normal force and shear force. Compare the deflection distribution from three points. E = 200 GPa, G = 84 GPa

MTPL0259_Chapter 16.indd 651

5/23/2012 11:41:35 AM

652  Chapter 16

Solution Free body diagram for a portion of the frame is shown in Fig. 16.30 on a section at an angle q, P Normal force, F = − cos θ 2 P Shear force, Q = − sin θ 2 P PR Moment, M = ( R − R cos θ ) = (1 − cos θ ) 2 2 ∂F cos θ Now, =− ∂P 2 ∂Q sin θ =− ∂P 2 ∂M R = + (1 − cos θ ). ∂P 2 Vertical deflection at B (considering one quadrant only and then multiply by 2)  B M ∂M δB = 2  ∫ ds +  A EI ∂P  π / 2 PR2 (1 − cos θ )2 Rdθ + δ B = 2 ∫ EI 4  0



∫

π/2

0

∫

B A

B Q F ∂F ∂Q  ds + ∫ × ds . A AE ∂P Aw G ∂P 

P cos2 θ × Rdθ + 4 AE

∫

π/2

0

 P × sin2 .Rdθ  4 Aw G 

where ds = Rdq. 1 + cos 2θ Now, (1 − cos q)2 = 1 − 2cos q + cos2 q = 1 − cos q + 2 = 1.5 − 2cos q + 0.5cos 2q cos2 θ =

1 + cos 2θ 2

P B F Q M

R A

q

C

R

q P 2

Figure 16.30 Problem 16.7

MTPL0259_Chapter 16.indd 652

5/23/2012 11:41:37 AM

Strain Energy Methods

sin2 θ = Therefore,

δB =

1 − cos 2θ . 2 PR3 2 EI +

∫

π /2

0

(1.5 − 2 cos θ + 0.5 cos 2θ )dθ +

PR 2 AE

∫

π /2

0

(0.5 + 0.5 cos 2θ )dθ

PR π /2 (0.5 − 0.5 cos 2θ )dθ 2 Aw G ∫0 π /2

=

653

π /2

1 0.5 PR3 PR 1.5 θ − 2 sin θ + sin 2θ + 0.5θ + sin 2θ 2 EI 4 2 AE 2 0 0 +

0.5 PR × 0.5‚ + sin 2‚ 2 Aw G 2

π /2

0

π  PR  π PR   π  1.5 2 − 2 + 2 AE 0.5 × 2  + 2 A G 0.5 × 2        w 0.178 3 0.3927 0.3927 PR = PR + PR + . EI AE Aw G =

PR3 2 EI

Problem 16.8 For the two cases given in Fig. 16.31, verify Maxwell’s reciprocal theorem. Solution In the first state of loading, a load P is applied at the middle and in the second state a couple M Pl 2 is applied at the end. The load P at position (1) produces slope at the ends The couple M 16 EI Ml 2 applied at the end produces the deflection at the middle. 16 EI P I

1

2

1 2

1 2

1

ll

2

M

ll

Figure 16.31 Problem 16.8

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654  Chapter 16

M2 Pl 2 , slope at position (2) by load P applied in first state δ1II = , deflection 16 EI 16 EI at position (1) by the couple M in the second state Therefore, i2I =

Pδ1II = Mi2I P×

Ml 2 Pl 2 =M× which verifies the reciprocal theorem. 16 EI 16 EI

Problem 16.9 Determine the horizontal displacement of point B of the truss shown in Fig. 16.32. Each member has a cross-sectional area of 400 mm2 and E = 200 GPa.

RCV 2

RC = 0.5

C

C

RCH 5m 3m

2

3m

3

1

1

B 1

4

5 3m

3m

60 kN

5m

RA = 40 kN

3

4

A

RA

B Unit load at B

A

RA = 0.5 4m

4m

(a)

(c)

3 50

4

2 3

4

50

= 60 kN 4

25

5

40

0.6

30

0.375

3

1 2

1

Force polygon for real load

2

3

Force polygon for unit load due to virtual load at B (d)

(b)

Figure 16.32

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Strain Energy Methods

655

Solution Force polygon shown in Fig. 16.32(b) shows the axial loads in various members of the truss. Figure 16.32(c) shows a unit load applied at point B, in the horizontal direction. Table 16.3

Virtual work equation

Member

N

N

L

nNL

AB

−0.625

−50

5m

+156.25

BC

−0.625

−50

5m

+156.25

CA

+0.375

+30

6m

+67.50



SnNL = 380 A = 400 mm2 = 4 cm2 E = 200 kN/mm2 AE = 4 × 10−4 × 200 × 106 = 80,000 kN. Horizontal deflection of point B =



380 80,000

= 4.75 × 10−3 m = 4.75 mm

Problem 16.10 A bent beam ABCDEF, as shown in Fig.16.33, carries a central load P. If EI is the A

a

a

B

a

E

F

b G C

D

p a 2

a 2

Figure 16.33 flexural rigidity of the beam and considering strain energy due to bending only, determine deflection under load P. Solution P . Considering portions AB, BC and CG. Free body 2 diagrams of AB, BC and CG of the beam are shown in Fig. 16.34. Strain energy due to bending

Due to symmetry reactions RA = RF =

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656  Chapter 16

P 2

P 2 A

B

B

P 2

P 2

Pa 2

G

C Pa 2

b

P 2

a a 2

Pa 2

C P 2

Figure 16.34

Forces and moments on various members

UAB =

1 2 EI

UBC =

1 2 EI

1 UCG = 2 EF

2

P 2 a3 P 2 a3 1 P  x d x = × × = ∫0  2  2 EI 4 3 24 EI a

∫

b

0

∫

P 2 a2 P 2 a2 b dx = 4 8 EI

a2

0

2

1  Pa P  + x  dx =  2 2  2 EI

1 P2 2 P 2 x3 P 2 ax 2 = + a x+ 2 EI 4 12 4 Total strain energy,

a 2

∫

a/ 2

0

=

0

 P 2 a2 P 2 x 2 P 2 ax   4 + 2 + 2  dx

19 P 2 a3 × 192 EI

U = 2 (U AB + U BC + U CG )  P 2 a3 P 2 a2 b 19 P 2 a3  = 2 + + × EI  8 EI 192  24 EI =

27 P 2 a3 P 2 a2 b × + 96 4 EI EI

∂U 27 Pa3 Pa2 b P  27 3 a2 b  a + = δG = + = EI  48 ∂P 48 EI 2 EI 2 

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Strain Energy Methods

657

Multiple Choice Questions 1. A beam of length L, simply supported at the ends, carries a concentrated load W at its centre. If EI is the flexural rigidity of the beam, strain energy due to bending in beam is:

(a)

W 2 L3 96 EI

(b)

WL4 48 + EI

(a)

2 3

(c) W L (d) None of these 24 EI 2. A beam of length L, simply supported at ends, carries a load W at a distance of a from one end and at a distance of b from other end such that a + b = L. If EI is the flexural rapidly of the beam, how much strain energy is absorbed in beam? 2 3 (a) W ab

3EIL

(b)

W 2 a2b3 6 EIL

Wa3b (d) None of these 6 EIL 3. A shaft of length L and a polar moment of inertia J is subjected to a twisting moment T. If G is the shear modulus, the strain energy stored in shaft is 2GJ

(c)

T 2L 4GJ

∂U ∂U ∂U + + ∂M ∂T ∂F

(b) ∂U + ∂U ∂M ∂F

∂U (d) None of these ∂F 5. A cantilever of length L is fixed at one end and a couple M is applied at the other end so as to bend the cantilever. If EI is the flexural rigidity of the cantilever, then slope at the free end of cantilever is ML (a) ML (b) EI 2 EI (c)

2ML (d) None of these EI 6. A cantilever is subjected to a load W at its free end, the deflection produced in centre of cantilever is d. Now the load W is applied only at the centre of cantilever, what is the deflection at free end? (c)

(c)

2 (a) TL

4. A body is subjected to a direct force F, twisting moment T and a bending moment M. The energy stored in the body is U. What is the displacement in the direction of force F ?

2 (b) T L 2GJ

(a) 2d (c) d

(b) 1.5d

(d) 0.5d

(d) None of these

Practice Problems M 1. A beam AB of length L, hinged at end A and roller supported at end B, is subjected to a couple M at point C, such that AC = L/3 as shown A B  in Fig. 16.35. If EI is the flexural rigidity of the beam, determine the C rotation in beam at point C. 2L L 2. A beam AB of length L, simply supported at ends, carries a load W at C, 3 3 such that AC = a, as shown in Fig. 16.36. If EI is the flexural rigidity of the beam, what is the deflection in the beam under the load? Figure 16.35 3. A steel ring of 20 mm in diameter (in section) is bent into a quadrant of 1.5 m radius. One end of ring is rigidly fixed in the ground, and at the other end, a vertical load P is applied. Determine the value of P so that the vertical deflection at the point of loading is 16 mm (E = 208 kN/mm2). 4. For a cantilever made of steel with length L, breadth b and depth d, show that if yb and ys are the deflections due to bending and shear at the free end due to a concentrated load W at the free end, then 2

ys  d = k   , where k is a constant  L yb

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658  Chapter 16 W Determine the value of k for steel and the least value of L/d, if the deflection due to shear is not to exceed 1.5 per cent of the total deflecA B tion. Take E/G = 2.6. C 5. A frame ABCD shown in Fig. 16.37 of lengths AB = BC = CD = L a b each of flexural rigidity EI throughout is subjected to horizontal and vertical loads W each at end D. Considering only the strain energy b due to bending, determine the horizontal and vertical deflections at end A. Figure 16.36 Hint (Take horizontal load, W = P, and analyse and finally put P = W) 6. A structure of horizontal length 2a and vertical height a carries a load P at its end as shown in Fig. 16.38. Area of cross-section of each member is A and E is the Young’s modulus of the material of the structure. Determine deflection of the structure under the load P. W C D

W

L

p A

C

B



A a

B E L

L

D



a

Figure 16.37

a

Figure 16.38

7. A beam AB is fixed at both the ends. It carries a concentrated load W at its centre. By using Castigliano’s theorem determine fixing moments at the ends (Fig. 16.39). 8. A frame ABC is subjected to a vertical force W at point C (Fig. 16.40). Area of cross-section of the frame throughout is A′, flexural rigidity is EI. Considering the effect of bending moment and axial forces, show that vertical deflection of point C is

Wa3 W 2 b Wb + + 3EI EI AE

9. A frame ABC has constant flexural rigidity EI throughout. Considering only the bending energy, determine the slope and the vertical deflection at point C, due to horizontal force P as shown in Fig. 16.41. 10. A semi-circular ring ABC fixed at end C, as shown in Fig. 16.42, is held in horizontal plane xy. Determine the vertical deflection under the vertical load P at point A. J = Polar monument of inertia I = Moment of Inertia E = Young’s modulus G = Shear modulus of ring

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Strain Energy Methods

659

a C

B MA

a

W

W

MB

B

b

A b

L 2

L 2

A

Figure 16.39

C

Figure 16.40

P

Figure 16.41

11. A 24 cm × 16 cm I-section beam with web 1 cm thick Z and flanges 2 cm is simply supported over a span of 5 m. A concentrated load of 20 kN acts at a distance of 2 m from one end of the beam. Assuming that the C shear force is carried by web only and shearing stress is p y uniformly distributed over the web, determine the total O deflection produced under the concentrated load [E = R 208 kN/mm2, G = 80 kN/mm2]. B A 12. Figure 16.43 shows a frame BC, DE and FA. At the ends A and B, two equal and opposite forces P are applied. Considering strain energy due to bending, Figure 16.42 determine the relative displacement between points A and B, if EI is the flexural rigidity of the frame. 13. Determine the horizontal displacement of point B of the truss shown in Fig. 16.44. There is a horizontal force of 4 kN at B and a vertical force of 5 kN at C. Each steel member has a cross-sectional area of 500 mm2, E = 200 GPa.

5 kN 4 kN P

b 2

F

B

C

b 2 C

A

1.5 m

B P

a A

E

D

Figure 16.43

MTPL0259_Chapter 16.indd 659

D 2m

Figure 16.44

5/23/2012 11:41:45 AM

660  Chapter 16

Answers to Exercises Exercise 16.1: 1.914 mm

Exercise 16.5: 20.82 kN

Exercise 16.2: 0.013 mm; 3.857 mm

3 Exercise 16.6: WL 32 EI Exercise 16.7: 19.3 mm

Exercise 16.3: 1.44 mm, 0.1° Exercise 16.4: 250 N

Answers to Multiple Choice Questions 1. (a) 2. (b)

3. (b) 4. (c)

5. (b) 6. (c)

Answers to Practice Problems WL 8

1. ML 9 EI

7. −

2 2 2. Wa b 3EIL

2 9. Pb ( 2a + b) , Pba 2 EI 2 EI

3. 9.96 N

L 4. k = 0.78, = 7.21 d 3 3 5. δ = 10 WL , δ = 17 WL H V 3 EI 3 EI

(

6. Pa 7 + 4 2 AE

)

MTPL0259_Chapter 16.indd 660

3 10. PR π  3 + 1  2  GJ EI 

11. 2.737 + 0.15 = 2.887 mm. 12.

Pb3 Pb2 a + 6 EI 2 EI

13. 0.294 mm

5/23/2012 11:41:47 AM

17 Bending of Curved Bars CHAPTER OBJECTIVES There are many engineering components which are not straight but curved and possessing large initial curvatures. Flexure formula of theory of simple bending cannot be used to determine stresses in such components. A theory has been developed for the determination of stresses and deflections in such components. In this chapter, students will learn about stresses and deflections in components as: 

Curved bars of different sections as rectangular, circular, trapezoidal and triangular.



Chain links.



Rings.



Crane hooks.



Frames of power presses, C-clamps.



Components with hollow section.

Since the derivations are lengthy and cumber some, students are advised to derive the relationships by themselves after studying the chapter.

Introduction Components with a large curvature are of engineering use in lifting machine and conveyor equipments. Components such as crane hook, chains and links used for lifting machines are designed with safety considerations. Therefore, determination of stresses in such components and location of critical sections are of utmost importance. The analysis of stresses in such components is quite complex. The theory of simple bending cannot be used for such components because in simple bending, the component is considered as initially straight. In this chapter, we will analyse stresses and deflection in components such as rings, links, crane hooks, frame of punch presses and eccentrically loaded members with large initial curvature, etc. In the chapter on ‘Theory of simple bending’, we assumed the beam to be initially straight before application of a bending moment and derived the relationship M/I = E/R = s/y, and studied about the stresses and deflections developed in beams. But, in this chapter, we will study about the effect of bending moment on bars of large initial curvature.

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662 Chapter 17

Stresses in a Curved Bar Figure 17.1 shows a portion of a curved bar of initial radius of curvature R, subtending an angle q at the centre of curvature O. This curved bar is subjected to a bending moment M tending to increase the curvature of the bar. To find out the stresses developed in the bar, a relationship between bending moment M, radius of curvature R and dimensions of the section of the bar are derived by taking the following assumptions: O

q R O1

M

+y

I E C A

G

E

I

q1 J F

C A

J D B

G

F D H

G

y

R −y

y1

dA

M

R1

B H

dy CD-Centroidal layer

Figure 17.1

1. The transverse sections of the bar, which are plane before application of a bending moment, remain plane after the application of bending moment. 2. The material obeys Hooke’s law and stress is directly proportional to strain. Consider a small portion IJHG of the curved bar in its initial unstrained position, where AB is a layer at a radial distance of y from the centroidal layer CD, that is, a layer passing through the centroidal axis of the sections. At layer AB, stresses due to the bending moment M are to be determined. After application of the bending moment, say, I ′J ′H ′G ′ is the final shape of the bar. The centroidal layer is now C ′D ′ and the layer AB takes the new position A′ B ′. Say, the final centre of curvature is O1 and final radius of curvature is R1 and q1 is the angle subtended by the length C ′D ′ at the centre. Say, s is the stress in the strained layer A′ B ′ under the bending moment M tending to increase the curvature (or tending to reduce the radius of curvature), and e is the strain in the same layer.

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Bending of Curved Bars

Strain,

ε=

663

A′ B ′ − AB ( R1 + y1 )θ1 − ( R + y )θ = AB ( R + y )θ

where y1 is the distance between centroidal layer C ′D ′ and layer A′ B ′, in the final position. or,

ε=

R1 + y1 θ1 × −1 R+ y θ

Moreover, e0 is the strain at the centroidal layer, that is, when y = 0 = or,

1+ ε =

and

1 + ε0 =

R1 θ1 × −1 R θ R1 + y1 θ1 R+ y θ

(17.1)

R1 θ1 × . R θ

(17.2)

Dividing Eq. (17.1) by Eq. (17.2), R +y R 1+ ε = 1 1× = R + y R1 1 + ε0

or,

y1 R1 y 1+ R

1+

 y1  1 + R  1 −1 ε = (1 + ε0 ) y  1 +   R

y1 y1 y + + ε0 − R1 R1 R ε= or, y 1+ R Now, y1 ≅ y considering that change in thickness is negligible.

ε0

ε0 Therefore, strain,

ε=

y y y y y + + ε0 − + ε0 − ε0 R1 R1 R R R . y 1+ R

Here, we have added and subtracted a term

or,

MTPL0259_Chapter 17.indd 663

ε0 y R

1 (1 + ε0 )  −  R1 ε = ε0 + y 1+ R

1 y R 

(17.3)

5/23/2012 11:35:57 AM

664 Chapter 17

The stress in the layer AB, which is tensile as is obvious from the diagram, that is, layers below the centroidal layer are in tension and layers above the centroidal layer are in compression for the bending moment shown.   1 1  (1 + ε0 )  −  y    R1 R   σ = Eε = E ε0 +   y 1+   R  

Stress,

where E is the Young’s modulus of the material. Total force on the section, F = ∫ σ dA Considering a small strip of elementary area dA, at a distance of y from the centroidal layer CD.  1 1 (1 + ε0 )  −  y dA  R1 R  F = E ∫ ε0 dA + E ∫ y 1+ R  1 1 y = E ∫ ε0 dA + E (1 + ε0 )  −  ∫ dA  R1 R  1 + y R  1 1 y = Eε0 A + E (1 + ε0 )  −  ∫ dA  R1 R  1 + y R where A is the area of cross-section of the bar. Now, the total resisting moment will be given by M = ∫ σ ydA = E ∫ ε0 ydA + E ∫

(17.4)

 1 1 (1 + ε0 )  −   R1 R  2 y dA y 1+ R

 1 1 y2 dA, = E (1 + ε0 )  −  ∫  R1 R  1 + y /R Because

∫ ydA = 0,

Therefore,

that is, first moment of any area about its centroidal layer is zero.  1 1 y2 dA M = E (1 + ε0 )  −  ∫  R1 R  1 + y /R

y2 2 ∫ 1+ y /R dA = Ah , a quantity which depends upon the disposition of section and the radius of curvature.  1 1 M = E (1 + ε0 )  −  Ah2 Therefore, (17.5)  R1 R  Let us assume that

MTPL0259_Chapter 17.indd 664

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Bending of Curved Bars

665

From Eq. (17.4),  1 1 y F = Eε0 A + E (1 + ε0 )  −  ∫ dA = 0  R1 R  1 + y /R because the bar is in equilibrium and the net force on the section is zero as no force is applied, only moment is applied. Now,

y

∫ 1 + y/R dA = ∫

Ry − y 2 + y 2 1 Ry 2 dA = ∫ ydA − ∫ dA R+ y R R+ y

=0−

1 y2 Ah2 = − A d R R ∫ 1 + y/R

Considering Eq. (17.4) again,  1 1  Ah2 Eε0 A = E (1 + ε0 )  −  × (since F = 0) R  R1 R  or,

 1 1 ε0 R = (1 + ε0 )  −  h2  R1 R 

(17.6)

 1 1 Substituting this value of (1 + ε0 )  −  in Eq. (17.5)  R1 R   ε R M = E  0 2  Ah2 = Rε0 EA  h  or,

ε0 =

M , strain at centroidallayer EAR

(17.7)

Substituting the value of e0 in the equation for stress,

σ=E σ=

εR M y + E 02 × EAR 1 + y /R h

M y R +E× × × ε0 . 1 + y /R h2 AR

Substituting the value of e0 again from Eq. (17.7),

σ=

M y R M +E× × 2× AR 1 + y /R h EAR

=

M M Ry 1 + × × 2 AR AR 1 + y /R h

=

M  R2 y  1+ 2 × ( tensile)  AR  h R + y 

(17.8)

On the other side of the centroidal layer y will be negative as for the layer EF shown in the figure.

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666 Chapter 17

σ ′ = stress where y is negative M  y R2  × − 1 (compressive) AR  R − y h2 

σ′ =

(17.9)

The expressions given in Eqs (17.8) and (17.9) are for the stresses due to the bending moment which tends to increase the curvature. If the bending moment tends to straighten the bar or tends to decrease the curvature, then q1 < q and R1 >R and the stresses will be reversed. Bending moment tending to decrease the curvature For y to be positive (away from centre of curvature)

σ=

M  R2 y  1+ 2 × (Compressive)  AR  h R + y 

(17.10)

On the other side of the centroidal layer, where y is negative (towards centre of curvature)

σ′ =

M  y R2  × − 1 (tensile). AR  R − y h2 

(17.11)

Ah 2 for a Rectangular Section Figure 17.2 shows the rectangular cross-section of breadth B and depth D of a curved bar with radius of curvature R, i.e., the radius from the centre of curvature C to the centroid G of the section. Consider a strip of thickness dy at a distance y from the centroidal layer. Area of the strip, dA = Bdy. Ah2 = ∫

Ry 2 Ry 2 Bdy y 2 + Ry − Ry dA = ∫ = BR∫ dy R+ y R+ y R+ y

= BR∫ ydy − BR2 ∫ =∫

+D/2 −D/2

ByRdy − ∫

y

G

B D 2

y R+ y−R = BR∫ ydy − BR2 ∫ dy R+ y R+ y

+D/2 −D/2

Ah2 = 0 − R2 BD + BR3 ln Now, area,

dy

BR2 dy + ∫

+D/2 −D/2

R D 2

R1 R2

Figure 17.2

BR3 dy R+ y

R + D/2 R − D/2

A = BD h2 = − R2 ×

BD BR3 2 R + D + ln BD BD 2 R − D

h2 R 2R + D = ln −1 R2 D 2 R − D =

MTPL0259_Chapter 17.indd 666

R  R R2 BR R2 R ln − 1 = ln − 1 =  B ln 2  − 1. D R1 A R1 A R1 

5/23/2012 11:36:02 AM

Bending of Curved Bars

667

where R is the radius of curvature up to the centroidal layer, R1 is the radius up to the inner surface of the curved bar and R2 is the radius up to the outer surface of the curved bar. Example 17.1 A circular ring of rectangular section with a slit is loaded as shown in Fig. 17.3. Determine the magnitude of the force P if the maximum resultant stress along the section ab does not exceed 100 N/mm2. Draw the stress distribution diagram along ab.

14

cm

P

8

cm

6 cm a 4 cm

G a

b c R = 11 cm

G

y1

b

R

y1

Section along ab

Figure 17.3 Solution Mean radius of curvature, Radius of curvature of inner surface, Radius of curvature of outer surface, Breadth, Depth,

R = 11 cm R1 = 8 cm R2 = 14 cm B = 4 cm D = 6 cm h2 R R = ln 2 − 1 2 D R1 R =

or,

11 14 11 ln − 1 = × 0.5596 − 1 = 0.02593 6 8 6

R2 = 38.56 h2

Maximum resultant stress will occur at the inner radius along section ab, that is, at the point b. Bending moment, M = P × R = 11P N-cm P P Direct stress, σd = = N/cm 2 (compressive) 4 × 6 24 Resultant stress at the point

b=

 P y2 M  R2 × − 1 + 2  AR  h R − y2  A

y2 = 3 cm

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668 Chapter 17

σ max =

Therefore,

y2  P P PR  R2 × − + 2  AR  h R − y2  A A

100 × 100 =

y2  P  P  R2 3  × =  38.56 ×  2   A h R − y2  24 11 − 3 

P=

24 × 10, 000 × 8 = 16597.5 N = 16.6 kN 3 × 38.56

Stress distribution (along Gb)

σ=

P R2 y × , where y varies from 0 to 3, compressive stressees A h2 R − y

= 0 at y = 0 cm =

1 16.6 × 38.56 × = 2.66 kN/cm 2 at y = 1 cm 11 − 1 24

16.6 2 × 38.56 × = 5.93 kN/cm 2 at y = 2 cm 24 11 − 2 16.6 3 = × 38.56 × = 10.13 kN/cm 2 at y = 3 cm 24 11 − 3 =

Stresses distribution Along Ga where y varies from 0 to 3 Resultant stress, σ =

M  R2 y  P 1+ 2 × − , tensile stresses AR  R + y  A h

Since, the direct stress is compressive =

P R2 y × × = 0 at y = 0 cm A h2 R + y

16.6 1 = × 38.56 × = 2.222 kN/cm 2 at y = 1 cm 24 11 + 1 =

16.6 2 × 38.56 × = 4.103 kN/cm 2 at y = 2 cm 24 11 + 2

=

16.6 3 = 5.715 kN/cm 2 at y = 3 cm × 38.56 × 24 11 + 3

Figure 17.4 shows the stress distribution along the radial thickness ab of the section which has maximum bending moment PR. In this case the resultant stress at centroid G is zero.

MTPL0259_Chapter 17.indd 668

5.715 kN/cm2

G

a

b

Resultant stress distribution along ab

10.13 kN/cm2 Stress distribution along depth

Figure 17.4

5/23/2012 11:36:06 AM

Bending of Curved Bars

669

Exercise 17.1 A curved bar of rectangular cross-section 40 mm × 60 mm is subjected to a bending moment, tending to increase the curvature of the bar. Radius of curvature of the bar is 200 mm. If the maximum stress developed in the bar is 100 N/mm2, what is the magnitude of bending moment? Due to this bending moment what is the stress at the CG of the section?

Value of h 2 for Sections Made Up of Rectangular Strips Sections such as T, I and channel section are made up of rectangular strips, the value of h2 for each section can be determined by considering each strip separately. For a single strip, R  h2 R  =  B ln 2  − 1 2 R A R1  where B is the breadth, R2 is the radius at outer fibres and R1 is the radius of the inner fibres of the section from the centre of curvature. Using this expression, let us determine h2/R2 for various sections. (1) T-section: Figure 17.5 shows a T section with following R3 dimensions: R2 Breadth of the flange = B Web R1 Breadth of the web = b Radius of curvature up to centroid G of the section = R B G b C Radius up to extreme outer edge of web = R1 R Radius up to inner edge of flange = R2 Radius up to outer edge of flange = R3 Flange Area of cross-section of T section = A A = B( R3 − R2 ) + b( R2 − R1 )

Figure 17.5

R R h2 R  B ln 3 + b ln 2  − 1. = R2 A  R2 R1  (2) I-section: Figure 17.6 shows an I section R3 with flange and web of breadths B and b, Outer respectively. flange R = Radius of curvature up to centroid G of the section G b R1 = Radius up to outer edge of inner flange B R2 = Radius up to inner edge of inner flange Web R3 = Radius up to inner edge of outer flange R4 = Radius up to outer edge of outer flange A = area of cross-section = B( R4 − R3 ) + b( R3 − R2 ) + B( R2 − R1 ) B( R4 − R3 ) + b( R3 − R2 ) + B( R2 − R1 ) R R R h2 R  = B ln 4 + b ln 3 + B ln 2  − 1. R2 A  R3 R2 R1 

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R4 Inner flange C

R1 R2

Centre line

R

I-Section

Figure 17.6

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670 Chapter 17

(3) Channel Section: Figure 17.7 shows a channel section with B = breadth of web b = breadth of flanges R = Radius of curvature up to centroid G of the section R1 = Radius up to inner surface R2 = Radius up to outer edge of web R3 = Radius up to the outer edge of flange A = Area of cross-section = B(R2 − R1) + 2b(R3 − R2)

Flange

Web

b

C

G B b

R R h R =  2b ln 3 + B ln 2  − 1. 2 R A R2 R1  2

R1 R2 R

Example 17.2 A curved beam whose centroidal line is a circular arc of 12 cm radius. The cross-section of the beam is of T shape with dimensions as shown in Fig. 17.8. Determine the maximum tensile and compressive stresses set up by a bending moment of 70,000 N cm; tending to decrease the curvature.

R3 Channel section

Figure 17.7

Solution Figure 17.8 shows the curved bar with T section subjected to a bending moment M tending to decrease the curvature. Therefore, there will be tensile stresses between A to G, and compressive stresses between G to B. Let us first calculate the distance of centroid from the outer edge of web. 5 × 1 × 2.5 + 6 × 1 × 5.5 12.5 + 33 = = 4.136 cm 5+6 11 y2 = 6 − 4.136 = 1.864 cm y1 =

M Flange y2

R2 R1

A

G

B

6 cm

y1 G

1 cm

Web

R = 12

1 5 cm cm

R3 M

Section along A−B

M = 60 kN cm

Figure 17.8

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671

R = 12 cm (given). R1 = 12 − 1.864 = 10.136 cm R2 = 11.136 cm. R3 = R1 + 6 = 10.136 + 6 = 16.136 cm

Radius of curvature, Radius up to inner surface, Radius up to outer edge of flange, Radius up to outer edge of web,

B = 6 cm, b = 1 cm, Area, A = 6 × 1 + 1 × 5 = 11 cm 2 R R h2 R  B ln 2 + b ln 3  − 1 = R2  R2 A  R1 h2 12  11.136 16.136  =  6 × ln + 1 × ln  −1 2  R 11 10.136 11.136  =

12 (6 × 0.0935 + 0.3712) − 1 = 1.01694 − 1 = 0.01694 11

R2 = 59.03 h2 Maximum compressive stress at point B =

y1  70, 000  59.03 × 4.136  M  R2 1+ 2 × = 1 +   AR  h R + y1  11 × 12  16.136 

= 8554 N/cm 2 = 85.54 MPa Maximum tensile stress at point A =

59.03  M  y2 R2  70, 000  1.864 × 2 − 1 = × − 1    11 × 12 12 − 1 . 884 1 AR  R − y2 h 

= 5, 227 N/cm 2 = 52.27 MPa Example 17.3 Figure 17.9 shows a press applying a 150-kN force on a job. Determine the stresses at the points a and b. The section is hollow as shown. 60 cm y2

R4 = 54 cm 150 kN

R1 a

b

150 kN

y1 4 cm

24 cm

G

a

b

8 cm 4 cm

R2

6 cm

R = 37 cm

20 cm

4 cm

R1 = 24 cm R2 = 30 cm R3 = 50 cm R4 = 54 cm Punch press

Figure 17.9

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672 Chapter 17

Solution Let us first determine the position of the centroid y1 =

16 × 4 × 2 + 2 × 20 × 14 × 4 + 24 × 6 × 27 16 × 4 + 2 × 20 × 4 + 24 × 6

128 + 2240 + 3, 888 = 17 cm 368 y2 = 30 − 17 = 13 cm =

R = 24 + 13 = 37 cm A = 24 × 6 + 2 × 4 × 20 + 4 × 16 = 398 cm2

Radius of curvature, Area of cross-section,

30 50 54  h2 R  =  24 ln + 2 × 4 × ln + 16 × ln  − 1 2  24 30 50  R A 37 ( 24 × 0.2232 + 2 × 4 × 0.5105 + 16 × 0.0769) − 1 368 37 = (5.3568 + 4.084 + 1.2304) − 1 368 =

=

37 × 10.6712 − 1 = 1.07292 − 1 = 0.07292 368

R2 = 13.714 h2 Bending moment,

M = Force × (60 + R) = 150 × 97 kN cm, where R = 37 cm

Direct tensile stress,

σd =

Bending stress due to M at

150 = 0.408 kN/cm 2 368 a= =

M  y2 R2  × −1 AR  R − y2 h2  150 × 97  13  × 13.714 − 1   368 × 37  37 − 13

= 6.869 kN/cm 2 ( tensile) Bending stress due to M at

b= =

y1 M  R2  + × 1 AR  R + y1 h2  150 × 97  17  × 13.714 1 +  368 × 37  37 + 17

= 5.682 kN/cm 2 (compressivee)

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673

a = 6.869 + 0.408 = 7.277 kN/cm2

Resultant stress at the point

= 72.77 N/mm2 (tensile) b = 5.682 − 0.408 = 5.274 kN/cm2

Resultant stress at the point

= 52.74 N/mm2 (compressive) Exercise 17.2 An open ring of channel section is subjected to a compressive force of 25 kN as shown in Fig. 17.10. Determine the maximum tensile and maximum compressive stress along the section ab. Exercise 17.3 A load P = 5 kN is applied on a C-clamp as shown is Fig. 17.11. Determine the stresses at the points a and b.

50 kN

1.5 cm

25 16 cm cm

3 cm b

a

b

a

Centre line

c

1.5 cm 3 6 cm cm 9 cm

Figure 17.10 Example 17.2

10 20 mm O

b

10 mm

10 a

30 mm Section at a−a

R2 b

30 mm

R1 a

c 60 mm

Load line

c 30 mm Load line

P P = 5 kN P 0

R1 = 30 mm R2 = 80 mm ac = 30 mm

Figure 17.11 Example 17.3

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674 Chapter 17

Ah 2 for a Trapezoidal Section Figure 17.12 shows a trapezoidal section of a curved bar with breadths B1 and B2, depth D and radius of curvature R. Say, C is the centre of curvature and G is the centroid of the section. Then, B + 2 B2 D × y1 = 1 3 B1 + B2 y2 =

D y1

y2 B2

b B1

C

G

B2 + 2 B1 D × 3 B1 + B2

y R

A = area of cross-section B + B2 Figure 17.12 = 1 × D. 2 Consider a strip of depth dy at a distance of y from the centroidal layer. If b is the breadth of the strip dy

B − B1   Area of the strip, dA = bdy =  B1 + 2 ( y1 − y ) dy D   Ah2 = ∫

Now

3 y1 R dA Ry 2 dA = − AR2 + ∫ − y2 R + y R+ y

= − AR2 + R3 ∫ = AR2 + R3 ∫

y1

− y2

y1 − y2

B2 − B1  1   B1 + D ( y1 − y )  R + y dy  

y1 B − B + y1  B − B  B1 y1 y 2 1 1 dy + R3 ∫ dy − R3 ∫  2 dy  − y2 − y2  R+ y D R+ y D  R+ y

= − AR2 + R3 B1 ln

y1  B − B  y1  B − B  R + y1 R + y1 R  B − B1  1 1 y ln − R3 ∫  2 + R3  2 dy + R3 ∫  2 dy − y2  − y2   D  1 R − y2 R − y2 D  D  R + y

= − AR2 + R3 B1 ln

R + y1 R + y1  B − B1   B − B1  4 R + y1  B − B1  y1 ln ( y1 + y2) +  2 R ln − R3  2 + R3  2     D     R − y2 D R − y2 D R − y2

R + y1 R  B2 − B1  R + y1 R R2  B2 − B1  R + y1 h2 R = − 1 + ) + ln B ln + ln − ( − y B B   1 1 2 1 A  D  R − y2 R2 A R − y2 A  D  R − y2 A =

 B2 − B1 R   R + y1 ( y1 + R)  ln − ( B2 − B1 ) − 1   B1 + A  D  R − y2 

Example 17.4 Determine the maximum compressive and tensile stresses in the critical section of a crane hook lifting a load of 40 kN. The dimensions of the hook are shown in Fig. 17.13. The line of application of the load is at a distance of 8 cm from the inner fibre (rounding off the corners of the crosssection are not taken into account).

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Bending of Curved Bars

675

P

4 cm

A

G

B

8 cm

C K

K y1 A

12 cm

B K C

y2

6 cm 8 cm

K

C = Centre of curvature K = Load - axis Section along A−B

Figure 17.13 Example 17.4 Solution Figure 17.13 shows a crane hook and the trapezoidal section. The load line KK′ is away from the centre of the curvature C. Position of CG of the section y1 = =

B1 + 2 B2 D × , where B1 = 4 cm; B2 = 8 cm; D = 12 cm 3 B1 + B2 4 + 16 12 20 × = cm 4+8 3 3

So,

y2 =

16 cm 3

Radius of curvature,

R = 6+

Area of cross-section,

A=

Now,

16 34 = cm 3 3

B1 + B2 4+8 ×D= × 12 = 72 cm 2 2 2

 B − B1 h2 R   R + y1 = −1 +   B1 + 2 ( y1 + R)  ln − ( B2 − B1 ) 2 R A  D  R − y2 

Substituting the values, 34 20   +  h2 34   8 − 4  34 20   3 3 = −1 +  4+  +   ln 34 16 − (88 − 4) 3 × 72   12  3 3  R2  − 3 3   34 34 (10 ln 3 − 4) = −1 + (10 × 1.09876 − 4) 216 216 = −1 + 1.0999 = +0.0999 = −1 +

R2 = 10.01 h2

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676 Chapter 17

KG = y2 + 8 =

Distance,

16 40 +8 = cm 3 3

40 1, 600 = kN-cm. 3 3 The bending moment tends to reduce the curvature, so the portion GA will be in compression and portion GB will be in tension. Bending moment,

M = 40 ×

Direct stress,

σd =

40 = 0.555 kN/cm 2 72

Maximum compressive stress at A,

σA = =

y1  M  R2 1+ 2 × − σd AR  R + y1  h 1, 600 × 3  20 1  ×  − 0.555 as R + y1 = 18 cm  1 + 10.01 × 3 × 72 × 34 3 18 

= 0.6536 [4.7074] − 0.555 = 2.52 kN/cm 2 = 25.2 MPa Maximum tensile stress at B

σB =

=

M  y2 R2  × − 1 + σ d AR  R − y2 h2 

1, 600 × 3  16 10.01  − 1 + 0.555  ×  3 × 72 × 34  3 6

= 0.6536 (7.898) + 0.555 = 5.717 kN/cm 2 = 57.17 MPa Exercise 17.4 The section of a crane hook is a trapezium. At the critical section, the inner and outer sides are 40 mm and 25 mm, respectively and depth is 75 mm. Centre of curvature of the section is at a distance of 60 mm from the inner fibres and the load line is 50 mm from the inner fibres. Determine the maximum load the hook can carry if the maximum stress does not exceed 120 N/mm2. d 2

Ah 2 for a Circular Section Figure 17.14 shows the circular section of diameter d of a curved bar of radius of curvature R, from the centre of curvature C up to the centroid G of the section. Area of cross section, A =

MTPL0259_Chapter 17.indd 676

C

G

b



π 2 d 4

Consider a strip of depth dy at a distance of y from the centroidal layer as shown. Breadth of the layer,

d 2

y

R

dy

Figure 17.14

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Bending of Curved Bars

677

2

 d b = 2   − y2  2 Area of the strip, 2

 d dA = bdy = 2   − y 2 dy  2 Ah2 = ∫

Now,

h2 =

or,

Ry 2 dA y+R

R y2 dA A∫ R+ y 2

=

4R +d /2 y2  d × ×   − y 2 dy 2 2 ∫− d / 2  2 R+ y πd

=

4 2  5 d  d2  1  d  1 +   +   +  16  2  2 R  16  2 R  

Example 17.5 A curved bar is formed of a tube of an outside diameter of 8 cm and a thickness of 0.5 cm. The centre line of this beam is a circular arc of radius 15 cm. Determine the greatest tensile and compressive stresses set up by a bending moment of 1.2 kN m tending to increase its curvature.

M

8 cm dia

Solution Figure 17.15 shows the cross-section of a curved bar of radius of curvature R = 15 cm. Area of cross-section, A=

b

G

C

a

7 cm dia R=

m

15 c

M

Figure 17.15

Example 17.5

π 2 2 (8 − 7 ) = 11.781 cm 2 4

Area of inner circle, A1 =

π 2 (7 ) = 38.485 cm 2 4

Area of outer circle, A2 = Bending moment,

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p 2 (8 ) = 50.266 cm 2 4

M = 1.2 kNm = 1.2 × 105 N cm

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678 Chapter 17

Now,

Ah2 = ∫

Ry 2 dA R+ y

Ah2 = A

2 4  5 d  d2  1  d  1 +   +   +      16  2 2 R 16 2 R 

For a circular section

For inner circle, h12 = =

2 4  d12  1  7  5 7 1 +   +   +  16  2  30  16  30  

49 (1 + 0.0272 + 0.009) = 3.14856 16

For outer circle, h22 =

2 4  d22  1  8  5 8 1 +   +   +  16  2  30  16  30  

82 (1 + 0.03555 + 0.00158) = 4.14852 16 Ah2 = A2 h22 − A1h12 = 50.266 × 4.14852 − 38.485 × 3.14856 =

= 208.53 − 121.17 = 87.36 87.36 = 7.415 h2 = 11.781 R2 15 × 15 = = 30.344 7.415 h2 Maximum tensile stress at b,

σb =

y1  M  R2 1+ 2 × , where y1 = 4 cm  AR  R + y1  h 1.2 × 10  30.344 × 4  1+ (15 + 4)  11.781 × 15  5

=

R P

G

P

= 679.06 (1 + 6.388) = 5016.9 N/cm 2 = 50.17 MPa Maximum compressive stress at a,

σa =

 y2 M  R2 × − 1 2  AR  h R − y2  1.2 × 10  30.344 × 4  − 1 = 6813.8 N/cm 2 = 68.14 MPa 11.781 × 15  (15 − 4)  5

=

MTPL0259_Chapter 17.indd 678

20 cm P = 15 kN R = 8 cm

Figure 17.16

Example 17.5

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Bending of Curved Bars

679

Exercise 17.5 A bar of circular cross-section is bent in the shape of a horse shoe. The diameter of the section is 8 cm and mean radius R is 8 cm as shown in Fig. 17.16. Two equal and opposite forces P = 15 kN each are applied so as to straighten the bar. Determine the maximum tensile and compressive stresses along the central section.

Ring Subjected to a Diametral Load Figure 17.17 shows a circular ring of mean radius R subjected to a diametral pull P. Consider a section CD at an angle q from the line of application of the load, that is, Y1Y2, and determine the bending moment and stresses in this section. Due to symmetry, the ring can be divided into four equal quadrants. Say, M1 is the bending moment on the section AB along the line of symmetry X1X2. Taking moments about CD, Bending moment at the section CD, M = M1 +

P K F

Y1

J q

R X2

A2 O

M D C

K R sin q

X1 A

B M1

Y2

P ( R − R sin θ ) 2

P 2

From Eq. (17.5) of the section ‘Stresses In A Curved Bar’,

P

P 2

Figure 17.17

 1 1 M = E (1 + ε0 )  −  Ah2  R1 R  where E is the Young’s modulus, e0 is the strain in centroidal layer, R is the initial radius of curvature and R1 is the radius of curvature after bending Ah2 = ∫

Ry 2 dA R+ y

Therefore,  1 1 PR PR E (1 + ε0 )  −  Ah2 = M1 + − sin θ 2 2  R1 R  Multiplying this equation throughout by Rdq and integrating for one quadrant

θ = 0 to

That is, =∫

π /2

0

MTPL0259_Chapter 17.indd 679

p 2

2 2 π /2 π / 2 PR π / 2 PR  1 1 E (1 + ε0 )  −  Ah2 Rdθ = ∫ M1 Rdθ + ∫ dθ − ∫ sin θ dθ 0 0 0 2 2  R1 R 

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680 Chapter 17

=∫

\

π /2

0

E (1 + ε0 ) 2 π π PR2π PR2 Ah Rdθ − E (1 + ε0 ) Ah2 = M1 R + − R1 2 2 4 2

(17.12)

R1θ1 [Eq. (17.2) of the section ‘Stresses in a Curved Bar’] Rθ Now for one quadrant, initial angle q = 90° = p/2 and final angle q1 = 90° = p/2 due to symmetry, that is, ∠Y1OX1 remains 90° even after the application of diametral load. However,

(1 + ε0 ) =

(1 + ε0 )

Therefore,

R =1 R1

Substituting this in Eq. (17.12), we get,

∫

x/2

0

EAh2 dθ − E (1 + ε0 ) Ah2

p p PR2  p  = M1 R +  − 1 2 2 2 2 

− Eε0 Ah2

or,

p p PR2  p  = M1 R +  − 1 2 2 2 2 

(17.13)

Again by the Eq. (17.4) of the section ‘Stresses in a Curved Bar’ Normal force on the section,  1 1 y F = Eε0 A + E (1 + ε0 )  −  ∫ dA  R1 R  1 + y R

∫

where

y 1+

y R

dA = ∫

Ry Ry + y 2 − y 2 y2 dA = ∫ dA = ∫ ydA − ∫ dA R+ y R+ y R+ y

= 0−

(17.14)

Ah2 1 Ry 2 d A = − R∫ R+ y R

 1 1  Ah2 Therefore, normal force, F = Eε0 A − E (1 + ε0 )  −   R1 R  R Normal force on the section CD, F= \

P sin θ 2

 1 1  h2 P sin θ = EAε0 − EA(1 + ε0 )  −  2  R1 R  R = EA ε0 −

MTPL0259_Chapter 17.indd 680

M (refer to Eq. (17.5) of the section ‘Stresses in a Curved Bar’ again) R

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M1 P ( R − R sin θ ) ( putting the value of M ) − R 2R M P P P sin θ = EA ε0 − 1 − + sin θ 2 R 2 2 = EA ε0 −

Therefore, ε0 =

M1 P + EAR 2 EA

(17.15)

Substituting the value of e0 in Eq. (17.13) − EAh2 ×

P  M1 Rp PR2 p  M1 = + + 2  EAR 2 EA  2 2

p   − 1 2

−

M1 M Rp PR2  p  p P × h2 × − × h2p = 1 +  − 1 2 4 2 2 2  R

−

P 2 PR2 PR2 p M p hp− p+ = M1 R + 1 h2 4 4 2 2 R 2 PR2 PR2 Ph2 M1 2 − − = [ R + h2 ] R p 2 2 M1 =

Now,

2  PR3 PR PR  R2 − = × − 1 2 2 2 2  2 R +h p  p (R + h ) 2

M = M1 +

PR (1 − sin θ ) 2

=

PR R2 2 PR PR PR sin θ × 2 × − + − 2 2 R +h p 2 2 2

=

 PR  R2 2 × − sin θ  2 2  2 R +h p 

M will be maximum when q = 0°. M max = M will be zero when

PR3 p ( R2 + h2 )

θ = sin −1

R2 2 × . 2 2 R +h p

So, there will be four sections, one in each quadrant where the bending moment M will be zero and consequently the stress due to bending moment will be zero. Now, substituting the value of M1 in the following equation.

ε0 =

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M1 P P R2 + = × EAR 2 EA EA p ( R2 + h2 )

(17.16)

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682 Chapter 17

 1 1 M E (1 + ε0 )  −  =  R1 R  Ah2

and

 1 1 y Stress, σ = Eε0 + E (1 + ε0 )  −  [Refer to Eq. (17.3) of the section ‘Stresses in a Curved Bar’]  R1 R  1 + y R P R2 M Ry × + 2× 2 2 A p ( R + h ) Ah R+ y

σ=

=

 1  PR 2 PR P R2 Ry R2 × + × 2 × 2 × − sin θ  ( puttting the value of M ) 2 2 2 2 A p ( R + h ) R + y Ah  2 ( R + h ) p 

=

 y P R2 R2  R2 2 × 2 2 × − sin θ   2 2 A  p ( R + h ) 2h  R + h2 p  R+

  y

P sin θ 2A Resultant stress at any point on the section Direct stress at any section, σ d =

σR =

 y P R2 R2  R2 2 + 2 2 × − sin θ   2 2 A  p ( R + h ) 2h  R + h2 p  R+

 P sinn θ + y  2A

Stress along Y1Y2 axis, where q = 0°

σR =

at the point K, stress,

at the point J, stress,

y=

σk =

 R2 P R2 R2  R2 PR2 2 y  y  1+ 2 × + 2 × 2 ×  =   2 2 2 2 2  A  p ( R + h ) 2h  R + h p  R + y  Ap ( R + h )  h R+ y d where d = diameter of the rod of the ring 2  R2 PR2 d  1+ 2 × (tensile) 2 2  2 R + d  Ap ( R + h )  h

y=−

σj =

d 2

 R2 PR2 d  1− 2 × (compressive) 2 2  2 R − d  pA( R + h )  h

It can be observed that the maximum stress occurs at the point J, where the diametral load is applied. Stresses along X1X2 axis, where q = 90°

σ R′ =

MTPL0259_Chapter 17.indd 682

P R2 R2  R2 2  y + × − 1  2 2 2  2 2 A  p ( R + h ) 2h  R + h p  R +

 P + y  2A

5/23/2012 11:36:28 AM

Bending of Curved Bars

683

At the point B, d 2 2  d  P P R2 R2  R2 (compressive) σ B′ =  + × − 1 + 2 2 2  2 2 A  p ( R + h ) 2h  R + h p  2 R + d  2 A y=+

At the point A, d 2 2  d  P P R2 R2  R2 ( tensile) σ A′ =  − × − 1 + 2 2 2  2 2 A  p ( R + h ) 2h  R + h p  2 R − d  2 A y=−

Example 17.6 A ring is made of round steel bar, 2 cm diameter and the mean diameter of the ring is 12 cm. Determine the greatest intensities of tensile and compressive stresses along a diameter XX if the ring is subjected to a pull of 5 kN along diameter YY. Solution Figure 17.18 shows a ring of a mean diameter of 12 cm, a bar diameter of 2 cm, subjected to a diametral pull P. Radius of curvature, R = 6 cm P = 5 kN Bar diameter, d = 2 cm Y Pull, P = 10 kN Area of cross-section, A =

p 2 ( 2) = 3.1416 cm 2 4

2 4  5 d  d2  1  d  h2 = 1 +   +   +      16  2 2 R 16 2 R 

R = 6 cm X

7 cm

5 cm O

A

B

X

2 4  22  1  2  5  2 = 1 +   +   +  16  2  12  16  12  

Y

1 1 5 1 1  + × × +  1 +  4  72 16 36 36 1 = (1 + 0.01388 + 0.00024) = 0.25353 4

=

or

P

Figure 17.18

R2 6×6 = = 142 2 0.25353 h

Stresses

σA =

MTPL0259_Chapter 17.indd 683

2  d  P P R2 R2  R2 (tennsile) − 2 × 2 × − 1  + 2 2 2 A  p ( R + h ) 2h  R + h p  2 R − d  2 A

5/23/2012 11:36:30 AM

684 Chapter 17

=

5, 000  36 142  36 2  2  5, 000 − × − 1 +   3.1416  p (36 + 0.25353) 2  (36 + 0.25353) p  12 − 2  2 × 3.1416

= 1591.5 [0.316 − 14.2(0.632 − 1)] + 795.77 = 8819.45 + 795.77 = 9615.2 N//cm 2 = 96.15 N/mm 2

σB =

P R2 R2  R2 2  d  P (comppressive) + 2 2 × − 1  + 2 2 2 A  p ( R + h ) 2h  R + h p  2 R + d  2 A

142 2   = 1591.5 0.316 + + 795.77 (0.632 − 1) × + 2  2 12  = 1591.5 [0.316 − 3.7326] + 795.77 = −4641.75 N/cm 2 = −46.42 (compressiive stress) Exercise 17.6 A ring is made of round steel rod of a diameter of 24 mm. The mean diameter of the ring is 240 mm. The ring is pulled by a force of 1 kN. Determine the greatest intensities of tensile and compressive stresses along the diameter of the loading.

Chain Link Subjected to a Tensile Load Figure 17.19 shows a chain link of mean radius R, length of the straight portion l, subjected to a pull P. Consider a section CD at an angle q from the line of application Y1Y2 of the pull P. Determine the bending moment and stresses in this section. Due to symmetry, the ring can be divided into four equal parts as shown. Say, M1 is the bending moment on the section AB along the line OX1. Taking moments at the section CD, M = M1 +

P ( R − R sin θ ) 2

P M

K F

Y1

D

J q

R

C k

l 2

O

X2

M1

P 2

A X B 1 P 2 d

l 2 X2

X1

R

From Eq. (17.5) of the section ‘Stresses In A Curved Bar’.  1 1 M = E (1 + ε0 )  −  Ah2  R1 R  where E is the Young’s modulus, e0 is the strain in the centroidal layer, R is the initial radius of curvature and R1 is the final radius of curvature

MTPL0259_Chapter 17.indd 684

Y2 P

Figure 17.19

5/23/2012 11:36:32 AM

Bending of Curved Bars

685

Ry 2 dA R+ y

Ah2 = ∫

 1 1 P Therefore, E (1 + ε0 )  −  Ah2 = M1 + ( R − R sin θ ) 2  R1 R  Multiplying throughout by Rdq and integrating from 0 to p/2 =∫

π /2

=∫

π /2

∫

p /2

0

0

\

0

 1 1 E (1 + ε0 )  −  Ah2 Rdθ  R1 R  M1 Rdθ + ∫

π /2

0

2 π / 2 PR PR2 dθ − ∫ sin θ dθ 0 2 2

E (1 + ε0 ) p × RAh2 dθ − E (1 + ε0 ) Ah2 R1 2

= M1 R

π PR2 PR2 + ×π − 2 4 R

(17.17)

R1 θ1 [Eq. (17.2) of the section ‘Stresses In A Curved Bar’] R θ In this case, initial angle θ = ∠X1OY1 = 90°. Now,

(1 + ε0 ) =

The final angle q1 will be a slight change from 90°. =

Slope at X1 where I is the moment of inertia of the section.

∫

Therefore,

p/ 2

0

M1l EI × 2

R(1 + ε0 ) p Ml dθ = − 1 R1 2 2 EI

Substituting in Eq. (17.17) p p PR2  p  p M l EAh2  − 1  − E (1 + ε0 ) Ah2 = M1 R +  − 1  2 2 EI  2 2 2 2  − EAh2 ×

M1l p p PR2  p  − Eε0 Ah2 = M1 R +  − 1 2 EI 2 2 2 2 

(17.18)

Again by Eq. (17.4) of the section ‘Stresses In A Curved Bar’. Normal force on the section CD,  1 1  Ry F = Eε0 A + E (1 + ε0 )  −  ∫ dA  R1 R  R + y Now,

Ry Ry + y 2 − y 2 y2 d A = d A = y d A − ∫ R+ y ∫ R+ y ∫ ∫ R + y dA = 0−

MTPL0259_Chapter 17.indd 685

1 Ry 2 Ah2 dA = − ∫ R R+ y R

(17.19)

5/23/2012 11:36:34 AM

686 Chapter 17

Therefore, normal force,  1 1  Ah2 P F = Eε0 A − E (1 + ε0 )  −  = sin θ 2  R1 R  R Therefore,

 1 1 1 P sin θ = EA ε0 − EAh2 (1 + ε0 )  −  × 2  R1 R  R

Therefore,

= EAε0 −

M (by Eq. (17.5) of the section Stresses In A Curvedd Bar’) R

= ε0 EA −

1 PR  M1 + (1 − sin θ )  R  2 

= ε0 EA −

M1 PR P − + sin θ R 2R 2

ε0 =

Therefore,

M1 P + EAR 2 EA

(17.20)

Substituting the values of e0 in Eq. (17.18) − EAh2 ×

M1l p M P  M1 Rp PR2  p  − EAh2  1 + +  − 1 = 2 EI 2  EAR 2 EA  2 2 2  − Ah2

M1l p Ph2p M1 Rp PR2p PR2 − M1h2 − = + − 2I 2R 4 2 4 2  pR Ah2 l h2p  PR2  p  Ph2p M1  + + =  1 −  − 2l 2 R  2  2 4  2

where I = Ak2; k = radius of gyration of the section  pR h2 l p h2  PR2  p  2 p M1  + + × = 1 −  − Ph 2  2 4  2 2k 2 2 R  Dividing throughout by p/2  h2 l h2  PR2 M1  R + 2 +  = 2 R πk 

M1 =

2  2  Ph  − 1 − π 2

 R2 R2 h2  − −  P 2 2  π R+

However,

MTPL0259_Chapter 17.indd 686

M = M1 +

(17.21)

h2 l h2 + πk2 R

P ( R − R sin θ ) 2

5/23/2012 11:36:36 AM

Bending of Curved Bars

=

 R2 R2 h2  P − −  2 2  π 2

R+

2

hl h + πk2 R

+

PR (1 − sin θ ) 2

687

(17.22)

Substituting the value of M1 in Eq. (17.20)   R2 R2 h2   − −  P 2 2 1   π P + ε0 = 2 2   EAR 2 EA hl h  R+ 2 +  R π k   Stress at any layer at a distance of y from the neutral layer is  1 1 y σ = E ε0 + E (1 + ε0 )  −   R1 R  1 + y R  1 1 M Moreover, E (1 + ε0 )  −  =  R1 R  Ah2 M Ry = E ε0 + 2 × y R+ y Ah 1+ R Substituting the values of M and e0, stress due to bending moment Therefore,

σ = E ε0 +

M × Ah2

y

  R2 R2 h2   − −  P 2 2 P 1 1   π Ry × 2 σb = + + 2 2   AR R + y ) Ah 2 A (R hl h  R+ 2 +  R π k     R2 R2 h2   − −  P 2 2  π 1 Ry PR  (1 − sin θ ) × + × 2× 2 2   2 R + y Ah hl h + + R   π k 2 R   Direct stress due to F, σ d = Resultant stress,

P sin θ 2A

σ R = σb + σd  R2 R2 h2  − −  P  π 2 2 + PR × y =   AR  h2 l h2  Ah2 R + R+ 2 +  R πk

MTPL0259_Chapter 17.indd 687

 R2 R2 h2   π − 2 − 2   y h2 l h2 R+ 2 +   R πk

5/23/2012 11:36:38 AM

688 Chapter 17

Ry 1 PR P P × (1 − sin θ ) + + sin θ R + y Ah2 2 2A 2A

+

(17.23)

 R2 R2 h2  − −  2 P  R y  π 2 2 + PR (1 − sin θ ) y + P (1 + sin θ ) = + × 1   R + y 2A AR  2 Ah2 h2 R + y   h2 l h2 R+ 2 +    R πk 2

This is an equation for resultant stress in any section along the curved portions X2Y1X1 and X2′Y2 X1′ of the chain link. The bending moment M1 on the straight portion X1X1′ and X2X2′ will remain constant and for the straight portion bending stress will be found with the help of general flexural formula. To obtain the resultant stress in this straight portion, direct tensile stress P/2A will be added to the bending stress. On the inner surface of the ring which is also called intrados stress can be obtained by substituting y = −d/2 in Eq. (17.21) where d is the diameter of the bar of the chain link. Similarly, for the outer surface which is also known as extrados the resultant stress is obtained by replacing y by +d/2 in Eq. (17.22). Maximum stress along Y1OY2 axis,

θ = 0°  R2 R2 h2  − −  2 P  R y  π 2 2 + PR × y + P σR = + × 1   AR  R + y   2h2 A R + y 2 A h2 h2 l h2 R+ 2 +   R πk 2

At the intrados, y = −

d 2

 R2 R2 h2  − −  2 P  R d  π 2 2 − PR × d + P = − × 1   AR  2 R − d   2 Ah2 2 R − d 2 A h2 h2 l h2 R+ 2 +   R πk 2

σ RI

At the extrados, y = +

d 2

 R2 R2 h2  − −  2 P  R d  π 2 2 + PR × d + P = + × 1   AR  2 R + d   2 Ah2 2 R + d 2 A h2 h2 l h2 R+ 2 +   R πk 2

σ RE

Maximum stresses along X1OX2 axis, q = 90°  R2 R2 h2  − −  P  R y  π 2 2 +P σR = 1+ 2 ×   2   A + AR  R y h h l h2  R+ 2 +   R πk 2

MTPL0259_Chapter 17.indd 688

5/23/2012 11:36:39 AM

Bending of Curved Bars

At the intrados y = −

689

d 2

 R2 R2 h2  2  π − 2 − 2 P   P R d σ RI′ = − × 1  + AR  2 R − d   h2 h2 l h2  A R+ 2 +  R πk At the extrados y = +

d 2

 R2 R2 h2  − −  P  R d  π 2 2 +P σ RE ′ = + × 1   2 2   A AR  2R + d   h h l h2 R+ 2 +   R πk 2

Maximum stress in straight portion, X1 X1 ′ or X 2 X 2 ′

Bending moment,

M1 =

 R2 R2 h2  P − −  2 2  π R+

h2 l h2 + πk2 R

Stress due to bending,

σb =

32 M1 π d3

Direct stress,

σd =

P 2P = 2A π d2

32 M1 2 P ± π d3 π d 2 Example 17.7 A chain link is made of round steel rod of 1 cm diameter. If R = 3 cm and l = 5 cm, determine the maximum stress along the section where tensile load is applied. If P = 0.5 kN. Resultant stress at intrados and extrados, respectively, sR =

Solution R = 3 cm, d = 1 cm, l = 5 cm, and P = 1 kN h2 =

MTPL0259_Chapter 17.indd 689

2 4  5 d  d2  1  d  1 + +      +  16  2 2 R 16 2 R 

=

2 4 1  1  1 5  1  1 + +       16  2  6  16  6  

=

1 (1 + 0.01390 + 0.00024) = 0.06338 16

5/23/2012 11:36:41 AM

690 Chapter 17

R2 3× 3 = = 142 2 0.06338 h Area,

A=

p 2 (1) = 0.7854 cm 2 , 4

Radius of gyration,

k=

d 1 1 = cm, k 2 = 4 4 16

R2 R2 h2 9 9 0.06338 − − = − − π 2 2 π 2 2 = 2.8648 − 4.500 − 0.03169 = −1.66689 R+

Now,

h2 l h2 0.06338 × 5 × 4 × 4 0.06338 + = 3+ + pk 2 R p ×1 3 = 3 + 1.6139 + 0.0211 = 4.635

 R2 R2 h2   p − 2 − 2  1.66689 =− = − 0.3596 2 2 4.635  hl h   R + pk 2 + R 

Then,

q = 0°, Therefore, Maximum stress at intrados,  R2 R2 h2  − −  2 P  R d  p 2 2 − PR × d + P = − × 1  2 2 2    AR  2R − d   2 Ah2 2 R − d 2 A h hl h R+ 2 +   R pk 2

σ RI

=

500 × 142 1 500 500  1 × + 1 − 142 ×  ( −0.3596) − 0.7854 × 3  5 2 × 0.7854 5 0.7854 × 2

= 2090.875 − 9039.98 + 318.31 = −6,630 kN/cm2 = −66.30 MPa (compressive) Maximum stress at extrados,  R2 R2 h2  − −  2 P P  R d  p 2 2 + P ×R × d = + × + 1   2 2 2 2   2 A h 2R + d 2 A AR  2R + d   h hl h R+ 2 +   R pk 2

σ RE

MTPL0259_Chapter 17.indd 690

5/23/2012 11:36:43 AM

Bending of Curved Bars

=

691

500  1 500 1 500 × 142 × + 1 + 142 ×  ( −0.3596) + 0.7854 × 3 7 2 × 0.7854 7 0.7854 × 2

= −1624.25 + 6457.13 + 318.31= +5151.2 kN/cm2 = +51.51 N/mm2 (tensile) Maximum stress occurs at the intrados, i.e., where the load is applied. Exercise 17.7 A chain link is made of round steel rod, 10 mm diameter. If R = 30 mm and L = 50 mm, determine the maximum stresses along the section at the end of the straight portion. A load of 1 kN (tensile) is applied on the chain link.

Deflection of Curved Bar In order to estimate the stiffness of a curved beam, subjected to a bending moment. It is necessary to determine the deflection of the curved beam and in such cases the influence of the initial curvature of the beam on its deflection is considerable. Figure 17.20 shows the centre line ABCD of a curved bar subjected to variable bending moment. Consider a small portion BC of length ds along the centre line. Say, the bending moment at B is M and at C is M + δM. Due to the bending moment say the centre line of the curved beam takes new position ABF and the element BC rotates by an angle df = DBF at the point B. The angular rotation is small and the displacement of the point D is also small. Displacement DF ≅ BD dφ ∠BDF = 90° for very small displacement DF Components of the displacement are DE perpendicular to the chord AD and EF parallel to the chord AD, that is, the line joining the ends of the centre line of the curved beam considered. FE shows negative displacement towards the point A. Deflection of the point D with respect to A is dDA and considering the small length ds only, say the deflection is ∆δ DA = − EF = − DF cos α where ∠DEF = α = −( BD dφ ) cos α ∠ADF = α

M B A

ds C h G

M+

dM

90 − a

df

D a

ABCD − Centre line of curved bar AD − Chord BG − Perpendicular to the chord from the point B

Figure 17.20

MTPL0259_Chapter 17.indd 691

GDF = DFE =a FE II AD

F E

Deflection of curved bar

5/23/2012 11:36:44 AM

692 Chapter 17

∠BDG = 90° − α or ∠DBG = α

Therefore, and

BD cos α = BG ∆δ DA = −( BD cos α )dφ = − BG dφ = − h dφ

Therefore,

where h is the perpendicular distance of the point B from the chord AD. Mds EI

Moreover,

dφ =

Therefore,

∆δ DA = −

M

P

Mhds EI

ds = Rdq Y1 C1

C h

Total deflection of D with respect to A

δ DA = − ∫

A

D

hMds EI

(1) Deflection of a closed ring Figure 17.21 shows the quadrant of a ring of mean radius of curvature R subjected to diametral pull P along OY1. We have to determine the deflection along the load line or along the chord Y2Y1. (Point Y2 not shown in Fig. 17.19). OY1 is half of the chord Y2Y1. Consider a small length ds at C at an angular displacement q. CC1 = perpendicular distance on chord from the point C

q

dq

R

C2

X1

Y1 O

M

Y2

Figure 17.21

Closed ringquadrant

= h = Rsin q. Bending moment at the section C, M = M1 + = =

PR (1 − sin θ ) 2

PR3 PR PR − + (1 − sin θ ) 2 2 p (R + h ) 2 2

 PR  R2 2 × − sin θ  [see Eq. (17.15) of the section ‘Ring Subjected To A Diametral Load’] 2  R2 + h2 p 

δ y = deflection along the load 10

= −∫

Y1

0

δY Y2 = − 1

 ( R sin θ )  PR  R2 2 × − sin θ   R dθ  2 2  EI  2  R + h p 

p /2  2  PR3 R2 ×2∫  × 2 × sin θ − sin2 θ  dθ 2 0 2 EI p R +h 

Note that we have considered only one quadrant and when we consider the complete ring, the deflection along the load will be δY Y2 1

MTPL0259_Chapter 17.indd 692

5/23/2012 11:36:46 AM

Bending of Curved Bars

 2  R2   PR3 θ sin 2θ p /2 p /2  cos θ − −   2 0 + 2 4 0  p  R + h   EI 2

=−

PR3 EI

=−

PR3  R2  2 PR3 p 1 × + × × EI  R2 + h2  p EI 2 2

=−

PR3 2  R2  PR3 p + × × EI p  R2 + h2  EI 4 =−

693

PR3  p 2  R2     − EI  4 p  R2 + h2  

(2) Deflection perpendicular to load line Refer to Fig. 17.19 again, now the chord is OX1 and perpendicular distance CC2 = h on the chord OX1 from C; h = Rcos q δ x = deflection perpendicular to load line 10

Total deflection,

x1 2  2  R cos θ   PR  R = −∫  × − sin θ   R dθ   2 2   EI   2  R + h p  0

δ x1x2 = −2 ∫

π /2

0

PR3 2 EI

 2  R2   cos θ − sin θ cos θ  dθ   2 2 p  R + h   p/2

PR3 2  R2  cos 2θ =− sin θ + 2 2  EI p  R + h  4 0 =− =

 PR3  2  R2  1 × 1 + ( −1 − 1)  ×  2 2 EI  p  R + h  4 

PR3  1 2  R2     − EI  2 p  R2 + h2  

Example 17.8 A ring with a mean diameter of 120 mm and a circular cross-section of 40 mm diameter is subjected to a diametral compressive load of 20 kN. Calculate the deflection of the ring along the load line. E = 200 GN/m2. Solution Since the diametral load is compressive, there will be reduction in diameter along the load line and increase in diameter perpendicular to the load line. R = 60 mm = 6 cm; d = 40 mm = 4 cm; P = 20 × 103 N E = 2,000 GN/m2 = 200 × 105 N/cm2 2 4  5 d  d2  1 d  h = 1 +   +   +  16  2  2 R  16  2 R   2

MTPL0259_Chapter 17.indd 693

5/23/2012 11:36:48 AM

694 Chapter 17

=

2 4  5  4 42  1  4  1 +   +   +  − 1 16  2  12  16  12  

= 1[1 + 0.0555 + 0.0038] = 1.0593 R2 36 = = 0.97 R + h2 37.0593 2

I=

pd 4 p = ( 4)4 = 4p = 12.5664 cm 4 64 64

Deflection along the load line = −

=−

PR3  p 2  R2     − EI  4 p  R2 + h2   20 × 103 63  p 2  ×  − × 0.97 200 × 105 12.5664  4 p

= −0.0796 × 10−3 (0.7854 − 0.6175) × 216 cm = −2.68 × 10−3 cm (reduction in dia) = −0.0268 mm Exercise 17.8 A Ring with a mean diameter of 150 mm and a circular cross-section of 30 mm diameter is subjected to a diametral tensile load of 8 kN. Calculate the deflection of the ring along the direction perpendicular to load line E = 200 kN/mm2.

Deflection of a Chain Link Figure 17.22 shows the quarter of a chain link subjected to axial tensile load P. The radius of curvature of the link is R and length of the straight portion is l. Consider a section at an angle q from the axis OY1. We have to determine the deflection along the load P or along the chord Y2Y1. CC1 = h = perpendicular from C to the chord = Rsinq Bending moment at the section R R h  P − −  2 2  PR p + (1 − sin θ ) M= 2 2 2 lh h R+ 2 + pk R 2

MTPL0259_Chapter 17.indd 694

2

P M Rdq C

Y1 C1

h q dq

O C 2

R X1 B

A

2

O C3

M1

l 2 Centre line

X

Figure 17.22

5/23/2012 11:36:50 AM

Bending of Curved Bars

695

[Refer to Eq. (17.6) of article 19.7.] R2 R2 h2 − − PR 2 2 = PK ′ + (1 − sin θ ), where K ′ = p 2 lh2 h2 R+ 2 + R pk Over the length l/2, the bending moment is constant and is equal to M1 = PK′ where the value of K′ is as above. Deflection along the line of loading p /2

δY1Y ′2 = −2∫

R sin θ EI

PR 2R 1    PK ′ + 2 (1 − sin θ )  Rdθ − EI × PK ′ × 2  

(with the help of theory of simple bending) =−

2 R2 EI

∫

p /2

0

PR PR 2  RPK ′ l  sin θ − sin θ  dθ −  PK ′ sin θ + 2 2 EI   p /2

=−

=

2 R2 2 R2  PR  2 R2  PRp  RPK ′ l × ( − PK ′ ) − × 1 + −   EI  8  EI EI  2 EI

=−

=

p /2

RPK ′ l 2 R2 2 R2 PR 2 R2 PR  θ sin 2θ  p /2 − PK ′ cos θ 0 − − cos θ + −  −  EI EI EI 2 2 2 EI 2 0 0

PR PRp  RPK ′ l 2R 2  −  PK ′ + − EI EI 2 8 

PR 2  pR  RPK ′ l − 2 K ′ − R −   EI EI  4

Deflection due to direct load P/2 is

δ′=

Pl 2 AE

δY1Y2 = δY1′Y2 + δ’ =

Total deflection,

PR2  pR RPK ′ l Pl  − 2 K’ − R  − +   4 2 AE EI EI

R2 R2 h2 − − 2 2 where K′ = p lh2 h2 R+ 2 + R pk (2) Deflection perpendicular to load line In this case chord is XX and CC2 = h = Rcos q + l/2

δ xx = −2∫

p /2

0

MTPL0259_Chapter 17.indd 695

R cos θ  PR PK ′lR  PK ′ + (1 − sin θ )  Rdθ − EI  2 EI 

5/23/2012 11:36:52 AM

696 Chapter 17

From theory of simple bending deflection in straight portion M  =  1 × l R  EI  where M1 is bending moment on straight portion and l is the length of the straight portion.

δ xx = −

where

p / 2 PR p / 2 PR 2 R2  p / 2  PK ′lR cos θ d θ − ∫ sin 2θ dθ  −  ∫0 PK ′ cos θ d θ + ∫0 0  EI 2 4 EI

=−

2 PR2 EI

PK ′lR R R  p /2 p /2 π/2   K ′(sin θ ) 0 + (sin θ )0 + (cos 2θ )0  − 2 8 EI

=−

2 PR2 EI

R R   PK ′lR  K ′ + + ( −2) − EI 2 8

=−

R  PK ′lR 2 PR2  −  K′ +  −  EI EI 4

R2 R2 h2 − − 2 2 , where k = radius of gyration. K′ = p 2 lh h2 R+ 2 + R pk

Example 17.9 A chain link is made of a steel rod of 12 mm diameter. The straight portion is 60 mm in length and the ends are 60 mm in radius. Determine the deflection of the link along the load line when subjected to a load of 1 kN. Given E = 200 × 103 N/mm2. Solution Rod diameter, Area of cross-section, Length of straight portion, Radius of curvature, Load, Radius of gyration,

d = 1.2 cm A = pd 2/4 = 1.31 cm2 l = 6 cm R = 6 cm P = 1.0 kN, E = 200 × 105 N/cm2 k = d/4 = 0.3 cm h2 =

2 4  5 d  d2  1 d  1 +   +   +      16  2 2 R 16 2 R 

2 4  1.2 2  1  1.2  5  1.2  = 1 +   +   +  16  2  12  16  12  

= 0.09(1 + 0.005 + 0.00003) = 0.09045 R2 R2 h2 36 36 0.09045 − − − − 2 2 = p 2 2 K′ = p 6 × 0.09045 0.09045 lh2 h2 6+ + R+ 2 + p × 0.009 6 R pk

MTPL0259_Chapter 17.indd 696

5/23/2012 11:36:53 AM

Bending of Curved Bars

= I=

697

11.459 − 18 − 0.0452 −6.5862 = = −0.83 6 + 1.1919 + 0.0150 7.934 pd 4 p = (1.2)4 = 0.1018 cm 4 64 64

Deflection along the load line

δY1Y2 = =

PR2  pR Pl  PRK ′ l − 2 K ′ − R − +   2 AE EI 4 EI 3 1 × 103 × 6 1 × 103 × 62 p ×6  1 × 10 × 6 × 0.83 × 6 2 0 . 83 6 + × − + +   200 × 105 × 0.1018 2 × 1.131 × 200 × 105 200 × 105 × 0.1018  4

= 17.68 × 10−3[0.3724] + 14.676 × 10−3 + 0.1326 × 10−3 cm = 10−3(6.584 + 14.576 + 0.1326) = 0.213 mm Exercise 17.9 A chain link is made of steel rod, 12 mm diameter. The straight portion is 60 mm in length and the ends are 60 mm in radius. Determine the deflection in the link along the direction perpendicular to the load line if the chain link is subjected to a load of 1 kN. Given E = 200 kN/mm2. Problem 17.1 For the frame of a punching machine shown in Fig. 17.23. Determine the circumferential stresses at A and B on a section inclined at an angle q = 45° to the vertical. Force, P = 200 kN. Solution Force, P = 200 kN P Perpendicular force on the section AB = Psin 45° = 2 P Tangential force on the section AB = Pcos 45° = 2 R1 = 20 cm

15 cm

B

y2

y1 G

5 cm 10 20 10 cm

q 45

R2 30 cm

D

R2 = 60 cm

A C

O

R1

100 cm

P K P

P sin q P cos q

K

Figure 17.23

MTPL0259_Chapter 17.indd 697

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698 Chapter 17

Area of cross-section, A = 30 × 10 + 5 × 20 + 15 × 10 = 550 cm2 Location of G y2 = =

30 × 10 × 5 + 20 × 5 × 20 + 15 × 10 × 35 550 1, 500 + 2, 000 + 5, 250 = 15.9 cm 550

y1 = 40 − 15.9 = 24.1 cm Radius of curvature, R = R1 + y2 = 20 + 15.9 = 35.9 cm Bending moment on the section, M= =

P 2 200 2

× (OK ′ + OA + y2 ) × (100 × 2 + 35.9)

= 25080.8 kN cm Direct force on the section, P′ =

P 2

=

200 2

= 141.4 kN

h2 R  30 50 60  =  30 ln + 5 ln + 15 ln  − 1 20 30 50  R2 A  h2 35.9 = (30 × 0.4056 + 5 × 0.5108 + 15 × 0.1824) − 1 R2 550 = 0.0652(12.168 + 2.554 + 2.736) − 1 = 1.1382 − 1 = 0.1382 R2 = 7.236 h2 Tensile stress at point A = =

M  y2 R2  P ′ × − 1 + AR  R − y2 h2  A 25080.8 × 1, 000  15.9  70.7 × 2, 000 × 7.236 − 1 +   550 550 × 35.9  35.9 − 15.9

= 1270.23 (3.752) + 257.09 = 6293.2 N/cm 2 = 62.93 N/nm 2 Compressive stress at point B =

MTPL0259_Chapter 17.indd 698

y1  P ′ M  R2 1 + × − 2  AR  R + y1  A h

5/23/2012 11:36:57 AM

Bending of Curved Bars

=

699

25080.8 × 1, 000  7.236 × 24.1 70.7 × 2, 000 1 + − 550 550 × 35.9  35.9 + 24.1 

= 1270.23 [3.906] − 257.8 = 4961.54 − 257.08 = 4704.5 N/cm 2 = 47.04 N/mm2 Problem 17.2 The radius of the inner fibres of a curved bar of trapezoidal section is equal to the depth of the cross-section. The base of the trapezium on the concave side is four times the base on the convex side. Determine the ratio of the stresses in the extreme fibres of the curved bar to the stresses in the same fibres of a straight bar subjected to the same bending moment. Solution See Fig. 17.24 R2 = D Now, Therefore, R1 = 2 R2 = 2 D

=

B1 + 2 B2 D × 3 B1 + B2 B1 + 8 B1 D × = 0.6 D B1 + 4 B1 3

Therefore,

y2 = 0.4 D

Area,

A = ( B1 + B2 ) =



Y1

B2 = 4 B1 y1 =

G

B1

B2

R



C (Centre of curvature)

Y2 R2

D R1

Figure 17.24

D 2

5 B1 D = 2.5 B1 D 2

R = R2 + y2 = 1.4 D  B − B1 h2 R     R + y1  ( y1 + R)  ln  − ( B2 − B1 ) −1 =   B1 + 2 2  A   D R   R − y2   =

 3 B1 R    1.4D + 0.6 D  × R1  ln   B1 +  − (3 B1 ) − 1  − A   D D D 1.4 0 . 4  

1.4 D (7 B ln 2 − 3B1 ) − 1 A 1.4 D (7 B1ln 2 − 3B1 ) − 1 = 2.5 B1 D =

= 0.56 (7 × 0.693 − 3) − 1 = 0.036 2

R = 27.35 h2

MTPL0259_Chapter 17.indd 699

5/23/2012 11:37:00 AM

700 Chapter 17

Let us consider that this curved bar is subjected to a bending moment, M tending to reduce the curvature. The maximum tensile stress, (when y = −y2)

σ2 =

M  y2 R2  × − 1 AR  R − y2 h2 

=

M 0.4 D   × 27.35 − 1   2.5 B1 D × 1.4 D 1.4 D − 0.4 D

=

M (10.94 − 1) 3.5 B1 D 2

=

2.84 M B1 D 2

(17.24)

The maximum compressive stress, (when y = +y1)

σ1 =

M  y1 R2  + 1 AR  R + y1 h2 

=

M  0.6 D  × 27.35 + 1   3.5 B1 D 2  1.4 D + 0.6 D

=

2.63 M B1 D 2

(17.25)

Stresses in the straight bar Let us divide the section in two triangles as shown, so as to calculate the moment of inertia IYY. BD Area of triangle I, A1 = 1 2 B2 D = 2 B1 D Area of triangle II, A2 = 2 B D3 IYY of triangle I, about its CG = 1 36 8 Distance, h1 = (0.6 − 0.3333) D = 0.2667 D = D 30 IYY of triangle II, about its CG = Distance, IYY of the whole section

MTPL0259_Chapter 17.indd 700

B2 D3 4 B1 D3 = 36 36

h2 = (0.4 − 0.3333) D = 0.0667 D = =

2 D 30

 B1 D3 B1 D  B D3 (0.2667 D)2 + 1 + 2 B1 D (0.0667 D) +  36 2  9 

2

5/23/2012 11:37:02 AM

Bending of Curved Bars

=

σ2 ′ =

The maximum tensile stress,

701

B D 64 2 2 B1 D × D 2 × 4 11 5 B1 D3 + 1 × D + = B1 D3 36 2 900 900 60 M × y2 , where ( y = − y2 ) IYY

=

60 M 2.18 M × 0.4 D = B1 D 2 11B1 D3

(y = +y1)

σ1 ′ =

60 M 3.27 M × 0.6 D = 3 11B1 D B1 D 2

Therefore, the ratios

σ 2 ′ 2.84 = = 1.303 σ 2 ′ 2.18

and

σ1 ′ 2.63 = = 0.804 σ1 ′ 3.27

The maximum compressive stress,

Problem 17.3 A chain link is made of round steel rod, diameter 12 mm. If R = 40 mm and l = 60 mm, determine the extreme stresses along the intrados if the link is subjected to a tensile load of 1 kN. Solution Bar diameter, Radius of curvature, Length of straight portion, Load,

d = 12 mm R = 40 mm l = 60 mm P = 1,500 N h2 =

2  5 d  d2  1  d  1 +    +   +  16  2 2 R 16 2 R 

 1  12  2 5  12  4  1 +   +   +   2  80  16  80  

=

(12)2 16

=

144 (1 + 0.01125 + 0.00016) = 9.10269 16

R2 402 = = 175.77 2 9.10269 h Area of cross-section, A =

p (12)2 = 113.098 mm 2 4

P 1, 500 = = 13.26 N/mm 2 A 113.098 Radius of gyration,

MTPL0259_Chapter 17.indd 701

k2 =

d 2 (12)2 = = 9 mm 2 16 16

5/23/2012 11:37:03 AM

702 Chapter 17

Moreover, R2 R2 h2 1, 600 1, 600 9.10269 − − = − − = 509.29 − 800 − 4.55 = −295.26 p 2 2 p 2 2 R+

h2 l h2 9.10269 × 60 9.10269 + = 40 + + = 40 + 19.316 + 0.227 = 59.543 2 pk R p ×9 40

R2 R2 h2 − − p 2 2 = − 295.26 = −4.959 2 59.543 h l h2 R+ 2 + R pk Along the intrados y = −d/2, the equation for the resultant stress is  R2 R2 h2  − −  2 P  R d  p 2 2 − PR (1 − sin θ ) d + P (1 + sin θ ) σ RI = + × 1   2 2 2 2R − d 2 A AR  2 R − d   2 Ah2 h hl h R+ 2 +   R pk Substituting the values 2

σ RI =

1, 000 12   1 − 175.77 ×  ( −4.959) 80 − 12  113.098 × 40 −

12 1, 000 1, 000 × 175.77(1 − sin θ ) + (1 + sin θ ) 2 × 113.098 80 − 12 2 × 113.098

= 32.90 − 137.13 (1 − sin θ ) + 4.420 (1 + sin θ ) = − 99.81 N/mm 2 at θ = 0° = + 41.74 N/mm 2 at θ = 90° Problem 17.4 A chain coupling is made of a 20-mm-diameter steel rod bent into an ‘S’ shape as shown in Fig. 17.25. If the coupling is attached to a tensile load of 1.0 kN as shown; calculate the maximum stress developed in chain coupling. Solution Maximum tensile stress will occur either at point A of smaller loop or at point B of larger loop. Rod diameter, d = 20 mm Radius of curvature of smaller loop, R1 = 40 + 10 = 50 mm Radius of curvature of longer loop, R2 = 60 + 10 = 70 mm Load, W = 1.0 kN = 1,000 N p Area of cross-section of rod, A = ( 20)2 = 314.16 mm 2 4 5 d  d2  1  d  1 +  +   h =  16  2  2 R1  16  2 R   2

2 1

MTPL0259_Chapter 17.indd 702

4

  

R1

A

40 mm

B 60 mm R2

1.0 kN

20 mm

Figure 17.25

5/23/2012 11:37:05 AM

Bending of Curved Bars

=

703

2 4 202  1  200  5  20   + 1 +      16  2  100  16  100  

= 25 [1 + 0.02 + 0.0005] = 25.5125 R12 502 = = 97.99 h12 25.5125 5  20  d 2  1  20  +  1 +    16  2  140  16  140  2

and

h 22 = =

4

  

202 [1 + 0.01 + .0001] = 25.2525 16

R22 702 = = 194 2 25.2525 h1 Stress at A,

Stress at B,

σA =

W × R1  10 R12  W × − 1 + AR1  50 − 10 h12  A

=

W 10 × × 97.99 A 40

=

1, 000 1 × × 97.99 = + 77.98 N/mm 2 314.16 4

σB =

W × R2  10 R22  W × − 1 + A × R2  70 − 10 h2  A

=

W 1 R22 1, 000 1 × × = × × 194 A 6 h2 314.16 6

= +102.92 N/mm 2

Multiple Choice Questions 1. A bar of square section 6 cm × 6 cm is curved to a mean radius of 12 cm. A bending moment M is applied on the bar. The moment M tries to straighten the bar. If the stress at the innermost fibres is 60 N/mm2 tensile, then the stress at the outermost fibres is (a) 60 N/mm2 (compressive) (b) 60 N/mm2 (tensile)

MTPL0259_Chapter 17.indd 703

(c) More than 60 N/mm 2 (compressive) (d) Less than 60 N/mm2 (compressive). 2. The most suitable section of a crane hook is (a) Square

(b) Round

(c) Hollow round

(d) Trapezoidal

3. A bar of square section 4 cm × 4 cm is curved to a mean radius of 80 m. A bending moment M, tending

5/23/2012 11:37:07 AM

704 Chapter 17 to increase the curvature is applied on the bar. If the stress at the outermost fibres is 80 MPa tensile, then the stress at the innermost fibres will be approximately equal to (a) 120 MPa (compressive) (b) 100 MPa (compressive) (c) 90 MPa (compressive) (d) 80 MPa (compressive) 4. A ring is subjected to a diametral tensile load. The variation of the stress at the intrados surface from the point of loading up to the section of symmetry is

(a) Maximum tensile stress to maximum compressive stress. (b) Throughout tensile stress. (c) Maximum compressive stress to maximum tensile stress. (d) Throughout compressive stress. 5. The distribution of stress along a section of a curved bar subjected to a bending moment tend to increase its curvature is (a) Linear (b) Uniform (c) Parabolic (d) Hyperbolic

Practice Problems 1. A sharply curved beam of rectangular section is 10 mm thick and 50 mm deep. If the radius of curvature, R = 60 mm, compute the stress in terms of the bending moment M at a point 20 mm from the outer surface. 2. Determine the diameter d of a round steel rod that is used as a hook to lift a 9-kN load acting through the R centre of curvature of the centroidal axis of the hook. Assume that = 2, and the maximum stress permitted 2 d is 125 N/mm . 3. The cross-section of a triangular hook has a base of 5 cm and altitude of 7.5 cm and a radius of curvature of 5 cm at the inner face of the shank. If the allowable stress in tension is 100 N/mm2 and in compression is 80 N/mm2, what load can be applied along a line 7.5 cm from the inner face of the shank? 4. Three plates are welded to form the curved beam 75 B of or, I-section shown in Fig. 17.26. If the moment 12.5 M = 1 kN m, determine the stresses at points A and 50 B and at the centre of the section (Fig. 17.26). 12.5 5. A steel link of rectangular section 24 mm × 8 mm A is shown in Fig. 17.27. If the angle b = 90° and the M M 75 allowable stress in link is 100 MPa, determine the 12.5 largest value of P which can be applied on the link. C 6. A curved bar of rectangular section with breadth B Dimensions in mm and depth D = 2B, is bent to a radius of curvature equal to 1.2D. It is subjected to a bending moment Figure 17.26 8 mm B

B 12 mm P

A 36 mm

b

24 mm

P

12 mm

A

24 mm

75

Figure 17.27

MTPL0259_Chapter 17.indd 704

5/23/2012 11:37:08 AM

Bending of Curved Bars of 1 kN m, tending to increase its curvature. Determine the size of the section if the maximum stress does not exceed 80 N/mm2. 7. Determine the maximum stress along, the section A-A in the crane hook. Section of the hook is circular of diameter 25 mm. Load 2 kN (Fig. 17.28). 8. Section of a punch press is in a T section of dimensions shown in Fig. 17.29. Centre of curvature of section is at a distance of 400 mm from inner side and load line passes through O, at a distance of 600 mm from inner side. Punch press frame is made of CI. Determine the safe load for a factor of safety of 3 based on ultimate strength.

705

Ri = 45 mm

P A

A

sut of CI = 150 MPa suc of CI = 600 MPa

P = 2 kN

25

9. A proving ring is 250 mm diameter, 40 mm wide and 6 mm thick. Figure 17.28 The maximum allowable stress in ring is 200 N/mm2. Find the load to cause this stress and the load should give 1 mm deflection of the ring in the direction of loading. (E = 200 GPa). 10. The curved beam with a circular centre line has a trapezoidal crosssection as shown in Fig. 17.30 and is subjected to pure bending in its plane of symmetry. The face b1 is on concave side of the beam. If h = 100 mm and a = 100 mm, find the ratio of b1/b2 of base widths so that the extreme fibre stresses in tension and compression are numerically equal. 11. A chain link is made of round steel rod of diameter 12 mm. If R = 40 mm and l = 60 mm draw the stress distribution diagram along the extrados for an angle of 90° starting from the outermost edge (along the direction of loading) of the link, if the link is subjected to a tensile load of 10 kN.

b2

h

T-section C

80

O b1

180 120

80

400

Figure 17.29

MTPL0259_Chapter 17.indd 705

200 mm

a

Load line

Figure 17.30

5/23/2012 11:37:09 AM

706 Chapter 17

Answers to Exercises Exercise 17.1: Exercise 17.2: Exercise 17.3: Exercise 17.4: Exercise 17.5:

2.16 kN m, +4.50 N/mm2 −93.8; +84.05 MPa 86.70, −37.873 MPa 30.60 kN −39.22, +99.83 MPa

Exercise 17.6: Exercise 17.7: Exercise 17.8: Exercise 17.9:

+26.1 N/mm2, −30.33 N/mm2 42.99, −30.26 0.055 mm 0.090 mm

Answers to Multiple Choice Questions 1. (d) 2. (d)

3. (d) 4. (c)

5. (d)

Answers to Practice Problems 1. 2. 3. 4. 5. 6.

72.92 M 43.52 mm 15.7 kN +22.755, −16.71, −3.55 MPa 4.37 kN 29.7, 59.4 mm

MTPL0259_Chapter 17.indd 706

7. 8. 9. 10. 11.

93.6 MPa 148.433 kN 375 N, 496 N 1.87 79.56 to −17.37 MPa

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18 Unsymmetrical Bending and Shear Centre CHAPTER OBJECTIVES When a section of a beam is not symmetrical about the plane of bending, an unsymmetrical bending takes place, i.e., in addition to bending, due to applied loads twisting is observed in the beam. Then there are principal axes of the section where the product of inertia is zero. In this chapter students will learn about:  Principal axes and their directions with respect to centroidal axes of the section.  



Moment of inertia of the section about the principal axes. Stresses developed at various points of the section due to unsymmetrical bending. Product of inertia of the section about any co-ordinate axes.



Position of neutral axis with respect to centroidal axes of the unsymmetrical sections.



Deflection in a beam due to unsymmetrical bending.



Location of shear center of sections which are symmetrical about one centroidal axes and sections which are not symmetrical about any centroidal axes.

Introduction Every section is not symmetrical about both the centroidal axes. Some sections are symmetrical only about one axis, whereas many sections as angle sections are not symmetrical about both the centroidal axes. In theory of simple bending, the section of the beam is symmetrical about the plane of bending. The simple flexural formula derived in theory of simple bending is not applicable when the section is not symmetrical about the plane of bending. In such sections, the principal axes and principal moments of inertia and the product of inertia are determined. Stresses developed in such sections of a beam are dependent on these parameters. If the load line on a beam does not coincide with one of the principal axes of the section, the bending takes place in a plane different from the plane of principal axes. This type of bending is known as unsymmetrical bending. The two reasons of unsymmetrical bending are as follows:

MTPL0259_Chapter 18.indd 707

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708 Chapter 18

1. The section is symmetrical about two axes like I-section, rectangular section, circular section but the load-line is inclined to both the principal axes. 2. The section itself is unsymmetrical like angle section or a channel section (with vertical-web) and load-line along vertical any centroidal axes. Figure 18.1(a) shows a beam with I-section with load-line coinciding with YY principal axis. I-section has two axes of symmetry and both these axes are the principal axes. Section is symmetrical about YY plane, i.e., the plane of bending. This type of bending is known as symmetrical bending. Figure 18.1(b) shows a cantilever with rectangular section, which has two axes of symmetry which are principal axes but the load-line is inclined at an angle a with the YY axis. This is the first type of unsymmetrical bending. Then, Fig. 18.1(c) shows a cantilever with angle-section which does not have any axis of symmetry but the load-line is coinciding with the YY axis. This is the second type of unsymmetrical bending. Figure 18.1(d) shows a channel section subjected to a vertical load passing through its centroid G. This member has been subjected to bending and twisting under the applied vertical load W. Now, the question arises; is it possible to apply the vertical load W in such a way that the channel member will bend without twisting and, if so, where the load W should be applied? W Y

Y

a

A

X G X

G

X X Y (a)

Y

(b)

W Y

W X

X

G G Y

(c)

Figure 18.1

MTPL0259_Chapter 18.indd 708

(d)

(a) Symmetrical bending, (b) unsymmetrical bending symmetrical section but oblique load, (c) unsymmetrical bending (unsymmetrical section) and (d) unsymmetrical bending (section not symmetrical about bending plane)

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Unsymmetrical Bending and Shear Centre

709

Y F1

G

X

h

X

W

F2

e

F1

F1

O

Y (e)

(f)

Figure 18.1 (e) Channel section (not symmetrical about yy axis, (f ) Bending without twisting (Contd.) Shear force in the flanges and web of the channel section is F1, F2 and F1, respectively, as shown in Fig. 18.1(e). Forces F1 constitute a couple F1 × h about centroid G. This couple is responsible for twisting of the member. Now, if the vertical load W or the shear force in the section is shifted from G, such that W × e = F1 × h, then the twisting couple is eliminated. So, it can be concluded that if the vertical load W, or vertical shear F is moved to the left in the channel section through a distance e, such that, F1 × h = We = Fe, the member will bend without twisting as shown in Fig. 18.1(f ). Before proceeding further, let us study about the principal axes of a section.

Principal Axes Figure 18.2 shows a beam section which is symmetrical about the plane of bending Y–Y, a requirement of the theory of simple bending or symmetrical bending. G is the centroid of the section. XX and YY are the two perpendicular axes passing through the centroid. Say, the bending moment on the section (in the plane YY of Y the beam) is M, about the axis XX. Consider a small element of area dA with (x, y) co-ordinates. dA M Stress on the element, σ= (18.1) y y I xx X x Force on the element,

dF =

G

MydA I xx

Bending moment about YY axis, dM =

MTPL0259_Chapter 18.indd 709

My xdA I xx

Y

Figure 18.2

x

Plane of bending yy

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710 Chapter 18

M1 = ∫

Total moment,

My xdA I xx

(18.2)

If no bending take place about YY axis, then M1 = 0 My xdA =0 I xx

or

∫

or

M xydA = 0 I xx ∫

or

∫ xydA = 0

(18.3)

The expression ∫ xydA is called the product of inertia of the area about XX and YY axes, represented by Ixy. If the product of inertia is zero about the two co-ordinate axes passing through the centroid, then the bending is symmetrical or pure bending. Such axes (about which product of inertia is zero) are called principal axes of the section and moment of inertia about the principal axes are called principal moments of inertia. The product of inertia may be positive, negative or zero depending upon the section and co-ordinate axes. The product of inertia of a section with respect to two perpendicular axes is zero if either one of the axis is an axis of symmetry. Example 18.1 Show that product of inertia of a T-section about a centroidal axis is zero. Solution Figure 18.3 shows a T-section with flange B × h1 and web b × h2. The section is symmetrical about YY axis. Say G is the centroid of the section on the axis YY, and XX and YY are the centroidal axes. Ixy = I′xy for flange + I″xy for web For flange, x varies from −( B / 2) to + ( B / 2) For web, x varies from −(b / 2) to + (b / 2) Now, I xx = ∫

y3 y2

∫

+ B/2

xy dx dy + ∫

−B/2

y2 − y1

∫

+b/2 −b/ 2

Y3

=∫

y3 y2

x2 2

+B/2

−B/2

y dy + ∫

− y1

y3

y2

y2

− y1

= 0 × ∫ y dy + 0 × ∫ (for flange)

MTPL0259_Chapter 18.indd 710

y2

x2 2

+b/ 2

y dy −b / 2

Y2

X

•

B 2

h1 X

G

Y1

y dy = 0

(for web)

y

B 2

xy dx dy

h2

b web y

Figure 18.3

Example 18.1

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Unsymmetrical Bending and Shear Centre

Example 18.2 Determine the product of inertia about axes X and Y for a triangular section shown in Fig. 18.4. Solution Consider a small element of area dA at co-ordinates x, y. Product of inertia about XY axis, 20

I xy = ∫

0

∫

2y

0

y

−20

0

=∫

y

y = 2

2y

x2 2

4 20

0

2y

0

dx

x

O x

xy dx dy

0

20

0

(∫

dA = dx, dy

dy 20 mm

40 mm

Figure 18.4

Note that limiting value of x = 40 mm = 2 × limiting value of y I xy = ∫

711

)

x dx y dy and also = ∫

40

0

20

20

0

0

(∫

x/2

0

Example 18.2

y

)

y dy x dx

y dy = ∫ 2 y 2 y dy = ∫ 2 y3 dy

h x

O

204 = = 80, 000 mm 4 . 2

b

Figure 18.5

Exercise 18.2

Exercise 18.1 Consider an I-section with flanges B × t1, and web H × t2 and show that the product of inertia about the centroidal axes is zero. Exercise 18.2 Figure 18.5 shows a rectangular section with breadth, b and altitude, h. Determine the product of inertia of the section about X–Y axes. y

Parallel Axes Theorem for Product of Inertia Figure 18.6 shows a section with its centroid at G, and GX′ and GY′ are the two rectangular co-ordinates passing through G. Say, the product of inertia about X′Y′ is I xy . Let us determine the product of inertia about the axis OX and OY, i.e., Ixy. Say, distance of G from OX axis = y, and distance of G from OY axis = x. Consider a small element of area dA = dxdy.

MTPL0259_Chapter 18.indd 711

Section

Small element dx and dy

y

x

x G

y x

y X

O

Figure 18.6

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712 Chapter 18

Say, co-ordinates of the element about the centroidal axis GX′, GY′ are x′, y′. Then, co-ordinates of the element about X–Y axis are, x = x + x′ and y = y + y′ Therefore, the product of inertia, I xy = ∫ xy d A = ∫ ( x + x ′ )( y + y ′ )dxdy = ∫ x ′ y ′dA + x y ∫ dA + y ∫ x ′dA + x ∫ y ′dA = I xy + x yA + 0 + 0

(because ∫ x ′ dA = ∫ y ′ dA = 0, about centroidal axes) I xy = I xy + A x y i.e., the product of inertia of any section with respect to any set of co-ordinate axes in its plane is equal to the product of inertia of the section with respect to the centroidal axes parallel to the co-ordinate axes plus the product of the area and the co-ordinates of the centroid of the section with respect to the given set of co-ordinate axes. Example 18.3 Figure 18.7 shows an unequal channel section, determine its product of inertia Ixy and I xy. Solution Let us break up the section into three rectangular strips I, II and III as shown in the figure and write the coordinates of their centroids with respect to the given set of axes YOX. Strip

Area (cm2)

6 cm

Y

III

x (cm)

y (cm)

A xy (cm4)

I II

18 8

4.5 0.5

1 6

81 24

III

12

3

11

396

II

2 cm

8 cm

G

5 cm

Remember that the product of inertia of these rectangular strips about their principal axes passing through the respective centroids is zero, because these rectangular strips have two axes of symmetry. ( I xy ) I = 0 + 81 cm 4 (using the parallel axis theorem for product of inertia)

MTPL0259_Chapter 18.indd 712

I O

2 cm X

3.5 cm 9 cm

Figure 18.7

Example 18.3

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Unsymmetrical Bending and Shear Centre

713

( I xy ) II = 0 + 24 cm 4 ( I xy ) III = 0 + 396 cm 4 I xy = 501 cm 4 To determine I xy , let us first determine the position of the centroid of the section.

10 y

18 × 4.5 + 8 × 0.5 + 12 × 3 x= = 3.184 cm 18 + 8 + 12 18 × 1 + 8 × 6 + 12 × 11 y= = 5.21 cm 18 + 8 + 12

60 cm

Area of cross-section,

10

A = 18 + 8 + 12 = 38 cm I xy = I xy − Ax y = 501 − 38 × 3.184 × 5.21 2

x

80 cm

Figure 18.8

= 501 − 630.37 = −129.37 cm4.

Exercise 18.3

Exercise 18.3 Figure 18.8 shows one unequal angle section, determine its product of inertia Ixy and I xy (through the centroidal axes).

Determination of Principal Axes In the section ‘Introduction’, we have learnt that principal axes pass through the centroid of a section and product of inertia of the section about principal axes is zero. Figure 18.9 shows a section with centroid G and XX and XX are two co-ordinate axes passing through G. Say, UU and VV is another set of axes V

Y

V

q

Y

q

GA = x, GB = y GA = u, GB = v

P

B

B B

y

F G

D A

A E C

U q x

x

B

D

MTPL0259_Chapter 18.indd 713

U A E

A C

G

U

q x

x

y

Figure 18.9

P •

V

Co-ordinates along principal axes

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714 Chapter 18

passing through the centroid G and inclined at an angle q to the X–Y co-ordinate. Consider an element of area dA at point P having co-ordinates (x, y). Say, u, v are the co-ordinates of the point P in U–V co-ordinate axes. So, where

u = GA′ = GD + DA′ = GD + AE GD = GA cos θ = x cos θ AE = DA′ = y sin θ

or,

u = x cos θ + y sin θ v = GB ′ = PA′ = PE − A′ E = PE − AD since A′ E = AD = PA cos θ − x sin θ = y cos θ − x sin θ

Similarly, x, y co-ordinates can be written in terms of u, v co-ordinates. x = GC − AC = GC − A′ F = u cos θ − v sin θ (as PA′ = v and GA′ = u) y = GB = PA = AF + FP = A′ C + FP = u sin θ + v cos θ Second moment of area about U–U axis, I uu = ∫ v 2 dA = ∫ ( y cos θ − x sin θ )2 dA = ∫ y 2 cos2 θ dA + ∫ x 2 sin2 θ dA − ∫ 2 xy sin θ cos θ dA = I xx cos2 θ + I yy sin2 θ − ∫ sin 2θ xy dA = I xx cos2 θ + I yy sin2 θ − I xy sin 2θ 1 1 = ( I xx + I yy ) + ( I xx − I yy ) cos 2θ − I xy sin 2θ 2 2

(18.4)

Second moment of area about V–V, 2

I vv = ∫ u2 dA = ∫ ( x cos θ + y sin θ ) dA = ∫ x 2 cos2 θ dA + ∫ y 2 sin2 θ dA + ∫ 2 xy sin θ cos θ dA = I yy cos2 θ + I xx sin2 θ + I xy sin 2θ 1 1 = ( I xx + I yy ) + ( I yy − I xx ) cos 2θ + I xy sin 2θ 2 2

MTPL0259_Chapter 18.indd 714

(18.5)

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Unsymmetrical Bending and Shear Centre

715

From Eqs. (18.4) and (18.5), I uu + I vv = I xx (sin 2 θ + cos2 θ ) + I yy (sin 2 θ + cos2 θ ) = I xx + I yy

(18.6)

Product of inertia about UV axes, I uv = ∫ uv dA = ∫ ( x cos θ + y sin θ ) ( y cos θ − x sin θ )dA = ∫ xy (cos2 θ − sin2 θ ) dA + ∫ y 2 sin θ cos θ dA − ∫ x 2 sin θ cos θ dA I uv = I xy cos 2θ + I xx

sin 2θ sin 2θ − I yy 2 2

However, as per the condition of pure bending or symmetrical bending Iuv = 0, then U and V will be the principal axes 2 I xy cos 2q + ( I xx − I yy ) sin 2q = 0

or,

tan 2q =

or,

2 I xy I yy − I xx

=

I xy ( I yy − I xx )

(18.7)

2 Say, q1 and q2 are two values of q given by Eq. (18.7) q2 = q1 + 90° sin 2q1 =

( I yy − I xx )

I xy 2

 I yy − I xx  + I xy2  2 

and cos 2θ1 =

2 2

 I yy − I xx  + I xy2  2 

Substituting these values of sin 2q1 and cos 2q1 in Eq. (18.4) 1 1 ( I xx − I yy ) ( I yy − I xx ) 1 2 − ( I uu )θ1 = ( I xx + I yy ) + 2 2 2  1 2  2 ( I yy − I xx )  + I xy

( I xy × I xy ) 1 ( I − I xx )2 + I xy2 2 yy

2

( I uu ) θ 1 =

1 1  ( I xx + I yy ) −  ( I yy − I xx )  + I xy2 2 2 

(18.8) 2

Similarly,

MTPL0259_Chapter 18.indd 715

1 1  ( I vv )θ1 = ( I xx + I yy ) +  ( I yy − I xx )  + I xy2 2 2  

(18.9)

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716 Chapter 18

θ2 = θ1 + (p / 2)

Now, for

sin 2q2 = sin(2q1 + p) = −sin 2q1 cos 2q2 = cos(2q1 + p) = −cos 2q1. Substituting these values in Eqs. (18.4) and (18.5) 2

( I uu )θ2 =

1 1  ( I xx + I yy ) +  ( I yy − I xx )  + I xy2 2 2 

(18.10)

2

1 1  ( I vv )θ2 = ( I xx + I yy ) −  ( I yy − I xx )  + I xy2 2 2 

(18.11)

From Eqs. (18.8) to (18.11), we learn that ( I uu )θ1 = ( I vv )θ2 , and ( I vv )θ1 = ( I uu )θ2 Maximum and minimum values of Iuu and Ivv I uu =

1 1 ( I xx + I yy ) + ( I yy − I xx ) cos 2θ + I xy sin 2θ 2 2

For maximum value of Iuu, dI uu =0 dθ i.e., or,

1 ( I yy − I xx ) ( −2 sin 2q ) + I xy × 2 cos 2q = 0 2 tan 2q =

I xy ( I yy − I xx ) 2

This shows that the values of ( I uu )θ1 and ( I vv )θ1 are the maximum and minimum values of Iuu and Ivv. These values are called the principal values of moment of inertia as Iuv = 0. The directions q1 and q2 are called the principal directions.

Moment of Inertia About Any Axis If the principal moments of inertia Iuu and Ivv are known, then moment of inertia about any axis inclined at an angle q to the principal axes can be determined. Say u, v are the co-ordinates of an element of area dA in the U–V principal axes system. X and Y are the co-ordinate axes inclined at an angle q to the U–V axes. x co-ordinate of element = u cos q − v sin q y co-ordinate of element = u sin q + v cos q Moment of inertia, 2 I yy = ∫ x 2 dA = ∫ (u cos θ − v sin θ ) dA

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Unsymmetrical Bending and Shear Centre

717

= ∫ u2 cos2 θ dA + ∫ v 2 sin2 θ dA − ∫ 2uv sin cos θ dA = I vv cos2 θ + I uu sin2 θ − 0, siince ∫ uv dA = 0

Similarly,

= I vv cos2 θ + I uu sin2 θ

(18.12)

I xx = I uu cos2 θ + I vv sin2 θ

(18.13)

From Eqs. (18.12) and (18.13), I xx + I yy = I uu + I vv = J, where J is polar moment of inertia about an axis passing through G and normal to the section. 1 cm

Example 18.4 Determine the principal moments of inertia for the equal angle shown in Fig. 18.10. Solution Let us consider the angle section in two portions, I and II, as shown and determine the position of the centroid x=y=

II V 10 cm

=

Ixx = Iyy

45 G

2.87

x I 1 cm

2.87 U

(due to symmetry x = y ) Moment of inertia,

G2

x

10 × 0.5 + 9 × 5.5 = 2.87 cm 19

U Y

Figure 18.10

G1 10 cm Y

V

Example 18.4

10 × 13 1 × 92 + 10( 2.87 − 0.5)2 + + 9 (5.5 − 2.87)2 12 12

= 0.833 + 56.169 + 60.75 + 62.252 = 180.004 cm4 Co-ordinates of centroid of portion I = [(5 − 2.87) − (2.87 − 0.5)] = (2.13, − 2.37) Co-ordinates of centroid of portion II = [ −( 2.87 − 0.5), (5.5 − 2.87)] = ( −2.37, 2.63) Product of inertia, I xy = 10( 2.13)( −2.37) + 9( 2.63)( −2.37) (as the product of inertia about their own centroidal axis is zero, since portions I and II are rectangles).

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718 Chapter 18

I xy = −50.481 − 56.098 = −106.579 cm 4

Therefore,

If q = angle of principal axis UU with respect to XX-axis tan 2θ =

I xy 106.579 =− =∞ ( I yy − I xx ) 0.0 2

or,

2q = 90°, or q = 45°

Principal angles are

q1 = 45°, q2 = 90° + 45° = 135°

Principal moments of inertia 1 1 I uu = ( I xx + I yy ) + ( I xx − I yy ) cos 2θ1 − I xy sin 2θ1 2 2 1 1 = (180.004 + 180.004) + × 0 × cos 90° + 106.579 × sin 90° 2 2 = 180.004 + 106.579 = 286.583 I vv = I xx + I yy − I uu = 2 × 180.004 − 186.583 = 73.425 Example 18.5 Figure 18.11 shows an I-section 15 cm × 20 cm. Axis X′X′ and Y′Y′ are inclined at an angle of 30° to the axis of symmetry. Determine the moment of inertia about these axes. Calculate also the product of inertia Ix′y′. Solution The I-section shown has two axes of symmetry, i.e., UU and VV passing through the centroid G. Therefore, these are the principal axes and Iuu and Ivv are the principal moments of inertia. The angle of inclinations of UU and VV axes with respect to X ′X ′ and Y ′Y ′ axes is q = 30°. sin q = 0.25, cos q = 0.75 2

2

Iy′y′ = Ivv cos q + Iuu sin q 2

since

15 cm V

Y

2 cm 30 U

x

x 30

G

16 cm

1 cm 2 cm V

Y

2

Iuv = 0

U

Figure 18.11

Example 18.5

Ix′x′ = Iuu cos2 q + Ivv sin2 q Now,

I uu =

15 × 203 14 × 163 − = 10, 000 − 4, 778.667 = 5, 221.333 cm 4 12 12

I vv =

2 × 153 16 × 13 2 × 153 + + = 562.5 + 1.333 + 562.5 = 1,126.333 cm 2 12 12 12

Therefore, Iy′y′ = 1,126.333 × 0.75 + 5,221.333 × 0.25 = 844.749 + 1,305.333 = 2,150.082 cm4

MTPL0259_Chapter 18.indd 718

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Unsymmetrical Bending and Shear Centre

Ix′x′ = 5,221.333 × 0.75 + 1,126.33 × 0.25

6 cm 1 cm

= 3,915.999 + 281.583 = 4,197.582 cm4 Now, I uu =

5, 221.333 =

719

1 ( I x ′x ′ + I y ′y ′ ) 2 1 + ( I x ′x ′ − I y ′y ′ ) cos 2θ − I x ′y ′ sin 2θ 2

1 cm 9 cm

1 ( 4,197.582 + 2,150.082) 2 1 + ( 4,197.582 − 2,150.082) cos 60° − I x ′y ′ × 0.866 2

6 cm

Figure 18.12

Exercise 18.5

5, 221.333 = 3,173.832 + 511.875 − I x ′y ′ × 0.866 I x ′y ′ = −

1, 535.626 = −1, 733.24 cm 4 0.866

Exercise 18.4 Consider a rectangular section of 6 cm width and 12 cm depth. Determine Ixx, Iyy and Ixy about XX and YY axes inclined at an angle of 45° to the principal axes. Exercise 18.5 Determine the principal angles and principal moments of inertia of a z-section shown in Fig. 18.12.

Stresses Due to Unsymmetrical Bending When the load-line on a beam does not coincide with one of the principal axes of the section, unsymmetrical bending takes place. Figure 18.13(a) shows a rectangular section, symmetrical about P

v

y, v

y f

x, u

y, v

G C

f

f

u q

x

u, x

P

G

x

x

C

G

x, u

u v, y

y

(a)

(b)

Rectangular section

Angle section

v

y, v

(c) Channel section

Figure 18.13

MTPL0259_Chapter 18.indd 719

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720 Chapter 18

XX and YY axes or with UU and VV principal axes. Load-line is inclined at an angle f to the principal axis VV, and passing through G (centroid) or C (shear centre) of the section. Figure 18.13(b) shows an angle section which does not have any axis of symmetry. Principal axes UU and VV are inclined to axes XX and YY at an angle q. Load-line is inclined at an angle f to the vertical or at an angle (90 − f − q) to the axis UU. Load-line is passing through G (centroid of the section). Figure 18.13(c) shows a channel section which has one axis of symmetry, i.e., XX. Therefore, UU and VV are the principal axes. G is the centroid of the section while C is the shear centre. Load-line is inclined at an angle f to the vertical (or the axis VV ) and passing through the shear centre of the section. Shear centre for any transverse section of a beam is the point of intersection of the bending axis and the plane of transverse section. If a load-line passes through the shear centre there will be only bending of the beam and no twisting will occur. If a section has two axes of symmetry, then shear centre coincides with the centre of gravity or centroid of the section as in the case of a rectangular, circular or I-section. For sections having one axis of symmetry only, shear centre does not coincide with centroid but lies on the axis of symmetry, as shown in the case of a channel section. For a beam subjected to symmetrical bending only, following assumptions are made: 1. The beam is initially straight and of uniform section throughout. 2. Load or loads are assumed to act through the axis of bending. 3. Load or loads act in a direction perpendicular to the bending axes and load-line passes through the shear centre of transverse section. Figure 18.14, shows the cross-section of a beam subjected to bending moment M, in the plane YY. G is the centroid of the section and XX and YY are the two co-ordinate axes passing through G. Moreover, UU and VV are the principal axes inclined at an angle q to the XX and YY axes, respectively. Let us determine the stresses due to bending at the point P having the co-ordinates u, v corresponding to principal axes. Moment applied in the plane YY can be resolved into two components M1 and M2.

M2

y M

v q v x

u

P v A q

B G C

A ∠uGA = a, ∠uGx = q

y

v

M2, moment in the plane VV = M cos q The components M1 and M2 have their axis along VV and UU, respectively. Resultant bending stress at the point P,

σb =

M1= M sin q M2= M cos q GA= u GB= v

Figure 18.14

M1u M 2 v M sin θ u M cos θ v + = + I vv I uu I vv I uu

=M

MTPL0259_Chapter 18.indd 720

x

a

u

M1, moment in the plane UU = M sin q

M1

u

v cos θ u sin θ +  I uu I vv

(18.14)

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Unsymmetrical Bending and Shear Centre

721

The exact nature of stress (whether tensile or compressive) depends upon the quadrant in which the point P lies. In other words sign of co-ordinates u and v is to be taken into account while determining the resultant bending stress. The equation of the neutral axis can be determined by considering the resultant bending stress. At the neutral axis bending stress is zero, i.e.,  v cos θ u sin θ  =0 M + I vv   I uu v=−

or,

sin θ I uu × u cos θ I vv

= −u tan a tan α =

where

(18.15)

I  sin θ I uu = tan θ  uu  cos θ I vv  I vv 

This is the equation of a straight line passing through the centroid G of the section. All the points of the section on one side of the neutral axis have stresses of the same nature and all the points of the section on the other side of the neutral axis have stresses of opposite nature. Example 18.6 A 40 mm × 40 mm × 5 mm angle section shown in Fig. 18.15 is used as a simply supported beam over a span of 2.4 m. It carries a 0.100-kN load along the line YG, where G is the centroid of the section. Determine the resultant bending stresses on points A, B, C, i.e., outer corners of the section, along the middle section of the beam. Solution Let us first determine the position of the centroid

5

V

Y

A

U

W

II G2 40 mm x

45

x

G

y B

y x

U

Figure 18.15

5 mm I G1

C V

40 mm

Example 18.6

40 × 5 × 2.5 + 35 × 5 × 22.5 200 + 175 500 + 3937.5 = = 11.83 mm 375

x=y=

5 × 353 40 × 52 + 5 × 35 ( 22.5 − 11.83)2 + + 40 × 5 (11.83 − 2.5)2 12 12 = 17,864.583 + 19,923.557 + 416.667 + 17,409.780

Moment of inertia, I xx =

= 55,614.587 mm4 = 5.561 × 104 mm4 = Iyy (because it is equal angle section) Co-ordinates of G1 (centroid of portion I) = +(20 − 11.83), −(11.83 − 2.5) = (8.17, −9.33) mm

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722 Chapter 18

Co-ordinates of centroid G2 = −(11.83 − 2.5), +(22.5 − 11.83) = −9.33, +10.67 mm Product of inertia, Ixy = 40 × 5(8.17) × (−9.33) + 35 ×5(−9.33) × (10.67) (Product of inertia about their own centroidal axes is zero because portions I and II are rectangular strips.) Ixy = −15,245.22 − 17,421.44 = −32,666.6 mm4 = −3.266 ×104 mm4. Principal angle, q tan 2θ =

I xy

=

−3.266 × 104 = ∞ (infinity ) 0

1 ( I − I xx ) 2 yy = tan 90°, θ = 45°

Principal moments of inertia 1 1 I uu = ( I xx + I yy ) + ( I xx − I yy ) cos 90° − I xy sin 90° 2 2 1 1 (5.561 + 5.561) × 104 + × 0 × cos 90° + 3.266 × 104 2 2 4 = (5.561 + 3.266) × 10 = 8.827 × 104 mm 4 =

Ivv = Ixx + Iyy − Iuu = (5.561 + 5.561 − 8.827) × 104 = 2.295 × 104 mm4 Bending moment, M =

Wl 0.100 × 103 × 2.4 × 103 = = 0.060 × 106 N mm 4 4

Components of bending moment, M1 = M sin 45° = 0.060 × 0.707 ×106 = 42.42 × 103 N mm M2 = M cos 45° = 0.060 × 0.707 ×106 = 42.42 × 103 N mm u–v co-ordinates of the points Point A, x = −11.83, y = 40 − 11.83 = 28.17 mm u = x cos q + y sin q = −11.83 × 0.707 + 28.17 × 0.707 = 11.55 mm v = y cos q − x sin q = 28.17 × 0.707 + 11.83 × 0.707 = 28.28 mm Point B, x = −11.83, y = −11.83 u = −11.0083 × 0.707 − 11.83 × 0.707 = −16.727 mm, v = 0

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Unsymmetrical Bending and Shear Centre

Point C,

723

x = 40 − 11.83 = 28.17, y = −11.83 u = 28.17 × cos 45° − 11.83 sin 45° = 28.17 × 0.707 − 11.83 × 0.707 = 11.55 mm Y

v = −11.83 × 0.707 − 28.17 × 0.707 = −28.28 mm.

W

Resultant bending stresses at points A, B and C sA =

B

15 YV

M1u M 2 u + = 42.42 × 103 I vv I vv

28.28   11.55 2 +   = 34.94 N/mm 2.295 × 104 8.827 × 104  0  −11.627  σ B = 42.42 × 103  + = −21.49 N/mm 2 4  2.295 × 10 8.827 × 104  28.28   11.55 σ C = 42.42 × 103  − = +7.76 N/mm 2 4  2.295 × 10 8.827 × 104 

XU

x

G

A

4.5 cm xu

x

5 cm 2 mm Y. V 3 cm

Figure 18.16

D Y

Exercise 18.6

Exercise 18.6 Figure 18.16 shows I-section of a cantilever1.2 m long subjected to a load W = 40 N at free end along the direction Y′G inclined at 15° to the vertical. Determine the resultant bending stress at corners A and B, on the fixed section of the cantilever.

Deflection of Beams Due to Unsymmetrical Bending Figure 18.17 shows the transverse section of a beam with centroid G. XX and YY are two rectangular co-ordinate axes and UU and VV are the principal axes inclined at an angle q to the XY set of co-ordinate axes. Say the beam is subjected to a load W along the line YG. Load can be resolved into two components, i.e., Wu = W sinq (along UG direction) Wv = W cosq (along VG direction) Say, deflection due to Wu is GA in the direction GU i.e.,

GA = δ u =

KWu l 3 EI vv

where K is a constant depending upon the end conditions of the beam and position of the load along the beam. Deflection due to Wv is GB in direction GV

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724 Chapter 18 Y

V q

Wv N

u W = load

Wu q

G

X

du A

X

g d

dv B

C

u

A

GB = dv GA = du GC = d NA = Neutral axis

V

Y

Figure 18.17

i.e.,

Total deflection,

GB = δ v =

KWv l 3 EI uu

δ = δ u2 + δ v2 =

=

KWl 3 E

Kl 3  Wu2 Wv2  + 2  E  I vv2 I uu 

sin2 θ cos2 θ + 2 I vv2 I uu

Total deflection d is along the direction GC, at angle g to VV axis. tan g = =

CB GA Wu I uu = = × GB AC I vv Wv I W sin q I uu × = tan q uu W cos q I vv I vv

Comparing this with Eq. (18.15) of the section ‘Stresses Due to Unsymmetrical Bending’ tan α = tan θ

I uu , I vv

where a is the angle of inclination of the neutral axis with respect to UU axis

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Unsymmetrical Bending and Shear Centre

tan γ = tan θ

and

725

I uu I vv

where g is the angle of inclination of direction of d with respect to VV axis. g = a, showing thereby that resultant deflection d takes place in a direction perpendicular to the neutral axis. Example 18.7 A simply supported beam of a length of 2 m carries a central load of 4 kN inclined at 30° to the vertical and passing through the centroid of the section. Determine (a) maximum tensile stress, (b) maximum compressive stress, (c) deflection due to the load and (d) direction of neutral axis. Give, E = 200 ×105 N/cm2. Solution Let us first determine the position of the centroid of the T-section shown in Fig. 18.18 y=

15 × 1 × 7.5 + 10 × 2 × (15 + 1) = 12.36 cm 15 + 20

The section is symmetrical about vertical axis; therefore, the principal axes pass through the centroid G and are along U–U and V–V axes shown. Therefore,

I xx = I uu =

10 × 23 1 × 153 + 20( 4.64 − 1.0)2 + + 15(12.36 − 7.5)2 12 12

Y, V 30

W

10 cm N

V

A

B 2 cm 4.64 cm

G

U, X

U, X

NA Netrual axis

72 15 cm

12.36 cm A C

D V, Y

Wv 90

1 cm

Wu

Wu = horizontal load Wv = vertical load

T 1m

1m

Beam with central load

Figure 18.18 Example 18.7

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726 Chapter 18

= 6.667 + 264.992 + 281.250 + 354.294 = 907.203 cm4 I yy = I vv =

2 × 103 15 × 13 + = 166.667 + 1.250 = 167.917 cm 4. 12 12

W = 4,000 N

Load

Components of W, Wv = 4,000 × cos 30° = 4,000 × 0.866 = 3,464 N Wu = 4,000 × sin 30° = 4,000 × 0.500 = 2,000 N Bending moment,

Mv =

Wv × l 3, 464 × 200 = = 1, 73, 200 N cm 4 4

Bending moment,

Mu =

Wu × l 2, 000 × 200 = = 1,00,000 N cm 4 4

Due to Mv, there will be maximum compressive stresses at A and B and maximum tensile stress at C and D. Due to Mu, there will be maximum compressive stresses at B and D and maximum tensile stresses at A and D. Therefore, maximum compressive stress at B,

σB = =

M v × 4.64 M u × 5 + I uu I vv 1, 73, 200 × 4.64 1, 00, 000 × 5 + = 885.852 + 2, 977.661 907.203 167.917

= 3,863.5 N/cm2 = 38.63 N/mm2 Maximum tensile stress at C,

σC = =

M v × 12.36 M u × 0.5 + I uu I vv 1, 73, 200 × 12.36 1, 00, 000 × 0.5 + 907.203 167.917

= 2,359.727 + 297.766 = 2,657.493 N/cm2 = 26.57 N/mm2 Deflection, δ =

KWl 3 E

sin2 θ cos2 θ + 2 I vv2 I uu

where K = 1/48, as the beam is simply supported and carries a concentrated load at its centre. 2

I  KWl 3 sin2 θ ×  uu  + cos2 θ Therefore, δ = EI uu  I vv 

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Unsymmetrical Bending and Shear Centre

727

Now, sin q = 0.5, sin2 q = 0.25, cos q = 0.866, cos2 q = 0.75 2

δ=

1 4, 000 × ( 200)3  907.203  × 0.25 ×  + 0.75  169.917  48 200 × 105 × 907.203

= 0.0367 0.25 × 28.50 + 0.75 = 0.0367 × 2.8065 = 0.103 cm = 1.03 mm Position of the neutral axis tan α = tan θ

I uu 907.203 = tan 30° × = 0.5774 × 5.339 = 3.0828 I vv 169.917

α ≅ 72° Exercise 18.7 A cantilever 2.8 m long having T-section with flange 12 cm × 2 cm and web 13 cm × 2 cm carries a concentrated load W at its free end but inclined at an angle of 45° to the vertical. Determine the maximum value of W if the deflection at the free end in not to exceed 2 mm. Given that E = 200 kN/mm2, what is the direction of neutral axis with respect to the vertical axis.

Shear Centre We have studied about the distribution of shear stresses in the transverse section of a beam subjected to a bending moment M and a shear force F. Summation of the shear stresses over the section of the beam gives a set of forces which must be in equilibrium with the applied shear force F. In case of symmetrical sections such as rectangular and I-sections, the applied shear force is balanced by the set of shear forces summed over the rectangular section or over the flanges and the web of I-section and the shear centre coincides with the centroid of the section. If the applied load is not placed at the shear centre, the section twists about this point and this point is also known as centre of twist. Therefore, the shear centre of a section can be defined as a point about which the applied shear force is balanced by the set of shear forces obtained by summing the shear stresses over the section. For unsymmetrical sections such as angle section and channel section, summation of shear stresses in each leg gives a set of forces which should be in equilibrium with the applied shear force. Figure 18.19(a) shows an equal angle section with principal axis UU. We have learnt in previous examples that a principal axis of equal angle section passes through the centroid of the section and the corner of the equal angle as shown in the figure. Say this angle section is subjected to bending about a principal axis UU with a shearing force F at right angles to this axis. The sum of the shear stresses along the legs, gives a shear force in the direction of each leg as shown. It is obvious that the resultant of these shear forces in legs passes through the corner of the angle and unless the applied force F is applied through this point, there will be twisting of the angle section in addition to bending. This point of the equal angle section is called its shear centre or centre of twist. For a beam of channel section subjected to loads parallel to the web, as shown in Fig. 18.19(b), the total shearing force carried by the web must be equal to applied shear force F, then in flanges there are two equal and opposite forces say F1 each. Then, for equilibrium, F × e is equal to F1 × h and we can determine the position of the shear centre along the axis of symmetry, that is, e = ( F1 × h /F ).

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728 Chapter 18 F

F1

F

F 45°

U

U

h

F

e Shear centre

Shear centre F1 F1

F1 Equal angle section (a)

Shear centre T-section (c)

Channel section (b)

Figure 18.19 Similarly, Fig. 18.19(c) shows a T-section and its shear centre. Vertical force in the web F is equal to the applied shear force F and horizontal forces F1 in two portions of the flange balance each other at shear centre. Example 18.8 Figure 18.20(a) shows a channel section, determine its shear centre. Solution Figure 18.20(a) shows a channel section with flanges b × t1 and web h × t2. XX is the horizontal symmetric axis of the section. Say, F is the applied shear force, vertically downwards. Then, shear force in the web will be F upwards. Say, the shear force in the top flange = F1. Shear stress in the flange at a distance of x from right hand edge Fay τ= I xx t F

Top flange dA A

t1

t

F1 dx x

F

•SC

Web

h 2

x x

h

t t

t2 F1 e

b (a)

Bottom flange

t (b)

Figure 18.20 Shear stresses in channel sections (a) and (b)

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Unsymmetrical Bending and Shear Centre

729

where F = applied shear force h ay = (t1 x ) , 2 first moment of area about axis X–X t = t1 (thickness of the flange)

τ=

Shear stress

F t1 x h F xh = I xx t1 2 2 I xx

Figure 18.20(b) shows, the variation of shear stress in flanges and web. Shear force in elementary area (t1dx = dA) = τ dA = τ t1 dx b

Total shear force in top flange = ∫ τ t1dx, where b = breadth of the flange o

b

F1 = ∫

(say)

o

Fxt1 h 2 I xx

dx =

Ft1 h b2 I xx 4

There will be equal and opposite shear force in the bottom flange. Say shear centre is at a distance of e from web along the symmetric axis XX. Then, for equilibrium Fe = F1h = Moment of inertia,

I xx =

Ft1 h2 b2 4 I xx

or e =

t2 h3 2 × b × t13  h + + 2 × b × t1    2 12 12

(

t1 b2 h2 4 I xx

2

)

in which, the expression 2bt13 /12 is negligible when comparing to other terms \

I xx =

t2 h3 bt1 × h2 h2 + = (t h + 6bt1 ) 12 2 12 2

Substituting this in the expression for e e=

3t1b2 t1b2 h2 12 × 2 = 4 h (t2 h + 6bt1 ) (t2 h + 6bt1 )

if we take, bt1 = area of flange = Af ht2 = area of web = Aw Then,

MTPL0259_Chapter 18.indd 729

e=

3bAf Aw + 6 Af

=

3b A 6+ w Af

5/23/2012 11:34:40 AM

730 Chapter 18

Exercise 18.8 A channel section has flanges 6 mm × 1 mm and web 8 cm × 0.5 cm. Determine the position of its shear centre. Problem 18.1 Find the product of inertia of a quadrant of a circle about axes X and Y as shown in Fig. 18.21.

Y

Solution Figure 18.21 shows the quadrant of a circle of radius R. Consider a small element at radius r, radial thickness dr and subtending an angle dq at the centre. Area of the element,

R r dq

dr dq R

y

dA = rdqdr Co-ordinates of the element x = r cos θ and y = r sin θ

q x

O

x

Figure 18.21

Problem 18.1

Product of inertia, Irθ = ∫

R

=∫

R

0

0 R

∫

p /2

0

{r ∫ 3

= ∫ r3 − 0

( r cos θ r sin θ ) rdθ dr p /2 0

}

sin θ cos θ dθ dr π /2

R R4 cos 2θ  2 dr = ∫ r3   dr = 0  4 4 0 8

Problem 18.2 A beam of angle section shown in Fig. 18.22 is simply supported over a span of 1.6 m with 15 cm leg vertical. A uniformly distributed vertical load of 10 k N/m is applied throughout the span. Determine (a) maximum bending stress, (b) direction of neutral axis and (c) deflection at the centre. Given E = 210 k N/mm2. Solution Let us first determine the position of the centroid 10 × 1 × 0.5 + 14 × 1 × 8 y= 14 + 10 117 = = 4.875 cm 24 14 × 0.5 + 10 × 5 57 x= = = 2.375 cm 14 + 10 24

1 cm

V

A

N

II Y

15 cm x

G

I xx =

10 × 13 1 × 143 + 10 ( 4.875 − 0.5)2 + + 14(7 + 1 − 4.875)2 12 12

= 0.833 + 191.406 + 228.667 + 136.718 = 557.624 cm 4

MTPL0259_Chapter 18.indd 730

O a = 687 I

4.875 U

Moment of inertia

U

G2

G1 Y 10 cm

x 1 cm

2.375

Figure 18.22

Problem 18.2

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Unsymmetrical Bending and Shear Centre

731

14 × 13 1 × 103 + 10(5 − 2.375)2 + + 14( 2.375 − 0.5)2 12 12 = 83.333 + 68.9906 + 1.167 + 49.218 = 202.624 cm 4

I yy =

Co-ordinates of G2 and G1: [ −1.875 (8 − 4.875) and (5 − 2.375) − 4.375) ] I xy = 14(8 − 4.875)( −1.875) + 10(5 − 2.375) ( −4.375) = −82.031 − 114.843 = −196.874 cm 4 (Note that parallel axes theorem for product of inertia is used here and product of inertia of each rectangular strip about its own centroidal axis is zero.) Directions of principal axes tan 2θ =

I xy −196.874 196.874 = = = 1.1091 ( I yy − I xx ) ( 202.624 − 557.624) 177.5 2 2

2θ = 47°58′ or θ = 23°59′ cos 2θ = 0.6695 sin 2θ = 0.7428 Principal moments of inertia 1 1 I uu = ( I xx + I yy ) + ( I xx − I yy ) cos 2θ − I xy sin 2θ 2 2 1 1 = (557.624 + 202.624) + (557.624 − 202.624) × 0.6695 + 196.874 × 0.77428 2 2 = 380.124 + 177.5 × 0.6695 + 146.238 = 645.2 cm4 1 1 ( I xx + I yy ) + ( I yy − I xx ) cos 2q + I xy sin 2q 2 2 = 380.124 − 177.5 × 0.6695 − 196.874 × 0.7428

I vv =

= 380.124 − 118.836 − 146.238 = 115.05 cm4 (a) Maximum bending stress w = rate of loading = 10 k N/m = 10,000 N/m = 100 N/cm Components, wu = w sin q = 100 × 0.4065 = 40.65 N/cm wv = w cos q = 100 × 0.9138 = 91.38 N/cm. The beam is simply supported and carries uniformly distributed load, the maximum bending moment occurs at the centre of the beam. Bending moment M u = ( wu l 2 /8) = ( 40.65 × 160 × 160 /8) = 130, 080 N cm (as span length I = 160 cm)

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732 Chapter 18

Bending moment M v = ( wv l 2 /8) = (91.38 × 160 × 160 /8) = 292, 416 N cm As it is obvious, maximum bending stress occurs at the point A with co-ordinates, x = −2.375, y = 15 − 4.875 = 10.125. Co-ordinates u = x cos q + y sin q = −2.375 × 0.9138 + 10.125 × 0.4065 = −2.170 + 4.116 = +1.946 cm v = y cos q − x sin q = 10.125 × 0.9138 + 2.375 × 0.4065 = 9.252 + 0.965 = 10.217 cm. Maximum bending stress, Mu u Mv v + I vv I uu 130, 080 × 1.946 292, 416 × 10.217 = + = 2, 200.2 + 4, 630.5 115.05 645.2 = 6, 830.7 N/cm 2 = 68.30 N/mm2.

σb =

Direction of neutral axis, tan α = tan θ

Deflection at the centre

=

I uu 645.20 = 0.4448 × = 2.4944 or α = 68°9. I vv 115.05

Kl 4 w sin2 θ cos2 θ + 2 E I vv2 I uu

where

w = rate of loading

Constant,

k=

Span length,

l = 160 cm. E = 210 × 105 N/cm2

Deflection,

δ=

5 384

5 1604 1 × × 100 × 5 384 210 × 10 I uu

∫

sin2

I uu2 + cos2 θ I vv2

2

=

40.635  645.2  + (0.9138)2 (0.4065)2 ×   115.05  I uu

=

40.635 × 5.180 + 0.835 = 0.155 cm = 1.55 mm 645.20

(in the direction perpendicular to the neutral axis) Problem 18.3 A 3-m-long cantilever of I-section carries a load of 2 kN at the free end and 3 kN at its middle. Line of load 2 kN is passing through the centroid of the section and inclined at an angle of 30° to the vertical and the line of application of a load of 3 kN is also passing through the centroid but

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Unsymmetrical Bending and Shear Centre

733

inclined at 45° to the vertical on the other side of a load of 2 kN as shown in Fig. 18.23. I-section has two flanges 12 cm × 2 cm and web 16 cm × 1 cm. Determine the resultant bending stress at the corners A, B, C and D. 2 kN

3 kN Y, V

B

A 2 cm

45

30

U, X 3 kN

U, X G

2 kN

16 cm

1 cm Fixed end

C 1.5 m

1.5 m

Figure 18.23

Y,V 12 cm

D

2 cm

Problem 18.3

Solution Moment of inertia, I xx =

12 × 203 11 × 163 − = 8, 000 − 3, 754.667 = 4, 245.333 cm 4 12 12

I yy =

2 × 2 × 123 16 × 13 + = 576 + 1.333 = 577.333 cm 4 12 12

I-section shown is symmetrical about XX and YY axes, so principal axes UU and VV passing through the centroid of the section are along XX and YY axes. Loads applied can be resolved into components along U and V directions. Components of 2 kN load, Wu1 = 2, 000 × sin 30° = 1, 000 N Wv1 = 2, 000 × cos 30° = 1, 732 N Components of 3 kN load, Wu2 = 3, 000 × sin 45° = 3, 000 × 0.707 = 2,121 N Wv2 = 3, 000 × cos 45° = 3, 000 × 0.707 = 2,121 N Bending moments at the fixed end, M u = Wu1 × 3 + Wu2 × 1.5 = −1, 000 × 3 + 2,121 × 1.5 Nm = 181.5 Nm = 0.18 × 105 N cm M v = Wv1 × 300 + Wv2 × 150 = 1, 732 × 300 + 2,121 × 150 = 8.38 × 105 N cm

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734 Chapter 18

Resultant bending stresses

σA = −

M u × 6 M v × 10 0.18 × 105 × 6 8.38 × 105 × 10 + =− + I vv I uu 577.333 4, 245.333

= ( −1.870 + 19.739) × 102 N/cm 2 = 17.869 N/mm2 M × 6 M v × 10 sB = + u + = ( +1.870 + 19.739) × 102 N/mm 2 = 21.609 N/mm2 I vv I uu Due to Mu, there will be tensile stresses at points B and C and compressive stresses at points D and A. Due to Mv, there will tensile stress at points A and B, and compressive stress at points C and D. Stress, σ C = +

6 M u M v × 10 − = ( +1.870 − 19.739) × 102 N/cm 2 = −17.869 N/cm 2 I vv I uu

Stress, σ D = −

6 M u M v × 10 − = ( −1.87 − 19.739) × 102 N/cm 2 = −21.609 N/mm2 I vv I uu

Problem 18.4 For an extruded beam, the cross-section is shown in Fig. 18.24. Determine (a) location of the shear centre O and (b) the distribution of the shearing stresses caused by vertical shearing force, F = 35 kN applied at O. Given Iz = 0.933 × 106 mm4.

30 mm F1

A

B 4 mm

dx

30 mm

x

F2 6 mm D z

o 30 mm

dx 6 mm C

x

E

F2 6 mm F

35 kN

G

30 mm

I2 = 0.93 × 106 mm4 F1 4 mm

H e

I 30 mm

Figure 18.24 Problem 18.4

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Unsymmetrical Bending and Shear Centre

Solution Shear force, Moment of inertia,

735

F = 35,000 N Iz = 0.933 × 106 mm4

Shear stress at A,

τA =

F × (30 × 4)45 35, 000 × 120 × 45 = = 50.64 N/mm2 Iz × 4 4 × 0.933 × 106

Shear stress at D,

τD =

F [30 × 4 × 45 + 30 × 6 × 30 + 30 × 6 × 15] Iz × 6

= Shear stress at C,

τC =

(at the centre of web)

35, 000 × [5, 400 + 5, 400 + 2, 700] = 84.4 N/mm 2 6 × 0.93 × 106 35, 000 [5, 400 + 2, 700 + 45 × 6 × 22.5] = 88.6 N/mm 2 6 × 0.993 × 106

Shear force in the segment AB, F1 = ∫

30

0

F F [4 x × 45] 4 dx = Iz × 4 Iz

∫

30

0

180 x dx =

900 81, 000 F F × 180 × = . 2 Iz Iz

Shear force in the segment DE, F2 = ∫

30

0

90 F F [6 x × 15] 6 dx = Iz × 6 Iz

∫

30

0

x dx =

90 F 40, 500 F × 450 = . Iz Iz

Taking moments about the centre of the web Fe = 2 × F1 × 45 + 2 F2 × 15 = 90 × e=

81, 000 F 30 × 40, 500 F + Iz Iz

729 × 104 + 121.5 × 104 850.5 × 104 850.5 × 104 = = = 9.11 mm m. Iz Iz 0.933 × 106

Problem 18.5 Determine the location of the shear centre O of a beam of a uniform thickness of 3 mm having the cross-section shown in Fig. 18.25. Solution Shear force in segment AB and EF will be equal and let us say it is equal to F1. Shear force in segment BC will pass through the point C. Therefore, we need not calculate F2. Shear force

F1 = ∫

100

60

40 mm

5 mm

D

B

C

e 80 mm

Figure 18.25

60 mm

O E

τ t dy

Let us take dimensions in centimetre. Say, F is the vertical shear force on the section applied at O.

MTPL0259_Chapter 18.indd 735

A

40 mm 60 mm

F

Problem 18.5

5/23/2012 11:34:50 AM

736 Chapter 18 F1

A dy

40 mm

B

F

y dx

60 mm

t = 5 mm

x

F2 q

C

O

Z e

F Shear force on shear centre

Figure 18.26 Problem 18.5

Shear force,

10

10

6

6

F1 = ∫ τ t dy = ∫

Ft (10 + y ) dy t (10 − y ) Izt 2

=

Ft 2I z

=

Ft y3 100 y − 2I z 3

∫

10

6

(100 − y 2 ) 10

= 6

69.335 Ft Iz

We need not calculate the shear force F2 in segment BD as it is passing through the point D. Moment of inertia, Iz 6 3 3 length, BD = 62 + 82 = 10 cm, tanθ = = , sin θ = 8 4 5 10 10 I z = 2  ∫ t dx( x sin θ )2 + ∫ ty 2 dy    0 6 10  10 9 2  = 2 t ∫ x dx + t ∫ y 2 dy  0 25 6  

  9 103 t = 2 t × × + (103 − 63 )  = 2t × 381.33 25 3 3   Shear force,

MTPL0259_Chapter 18.indd 736

F1 =

Ft F 69.33 Ft = = 2t × 381.33 t × 11 11

5/23/2012 11:34:51 AM

Unsymmetrical Bending and Shear Centre

737

Taking moments of the forces about the point D 2 × F1 × 8 = F × e 16 F = F ×e 11 e = 1.454 cm = 14.54 mm

or

Problem 18.6 Determine the location e of the shear centre, point O for the thin-walled member having the cross-section shown in Fig. 18.27. The member segments have the same thickness t. Solution Moment of inertia of the section

B

A

d

I oz = t × 2 × b × ( d sin 45°)2 + 2∫ (tdx )( x sin 45°)2 d

dx

F

0

C 45

e d

Let us calculate the shear force in members AB or DE due to the vertical shear force F at shear centre O, then b

b

0

0

=

45

x

2 d x d2 + 2t ∫ dx o 2 2 td 2 td 2 = btd 2 + = [3b + d ] 3 3

= 2bt ×

F1 = ∫ τ t dx = ∫

F1

F1 D

Ftx d t dx × I oz t 2

E b

Ft d b Ft b2 d × , x dx = ∫ I oz 2 0 I oz 2 .2

Figure 18.27

Problem 18.6

substituting the value of Ioz =

Ft b2 d 2 2

×

3 1.5 Fb2 = td 2 (3b + d ) d 2(3b + d )

Taking moments of the forces about the point C Fe = = or

MTPL0259_Chapter 18.indd 737

e=

1.5 Fb2 1 d × ×2 (3b + d ) 2 d 2 1.5 Fb2 (3b + d ) 1.5b2 d + 3b

5/23/2012 11:34:53 AM

738 Chapter 18

Problem 18.7 A force P is applied to the web of the beam as shown in Fig. 18.28. If e = 250 mm, determine the height h of the right side flange so that the beam will deflect without any warping. The member segments have the same thickness t. P

Solution Moment of inertia about xx axis I xx =

t × 103 t × h3 t + = (1, 000 + h3 ) cm 4, 12 12 12

e

neglecting web.

100 mm

h   −y h / 2 Pt  h   2 P1 = 2∫ × t dy  − y   y + 0 I xx t  2 2     

=

h/2 Pt Pt 2 4 y3 × ∫ ( h2 − 4 y 2 ) dy = h y− 4 I xx 0 4 I xx 3

=

h/ 2

h/2

= 0

Pt 4 I xx

3

C

Problem 18.7

 h3 h3  −  2 6 

D

3

(

P1

)

Now, e = 250 mm = 25 cm Taking moment of the forces about the point E

h

e

Pth Pt × 12h Ph = = 3 12 I xx 12 × t 1, 000 + h 1, 000 + h3 3

D

x

Figure 18.28

Pt I xx

0

E x D

300 mm

( h − 2 y )( h + 2 y ) dy 4

=

∫

P

A

Shear force in the vertical segment CD (as in Fig.18.29),

dy

h 2

y

25 × P = 30 × P1

h 2

Ph3 P = 1.2 × 1, 000 + h3 or, or,

1, 000 + h3 = 1.2h3 0.2h3 = 1, 000 h = 17.099 cm =170.99 mm.

C t

Flange

Figure 18.29

Problem 18.8 Determine the location of the shear centre O of a thin-walled beam of uniform thickness having the cross-section as shown in Fig. 18.30. Solution Take the dimensions in centimetre. Let us calculate the shear forces in vertical segments DE and FG.

MTPL0259_Chapter 18.indd 738

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Unsymmetrical Bending and Shear Centre

t = thickness of each segment

e

A

Say, the shear force on beam = F at O. SF in vertical number DE,

D

Z

12.5 cm

10 cm O

6.25

= 0

G

E

Ft 6.25 = (39.0625 − y 2 ) dy I z ∫0 Ft y3 39.0625 y − Iz 3

F

H

 6.25 + y  F (6.25 − y )  t 6.25  2  F1 = 2∫ tdy 0 Izt

=

739

50 mm

Ft 162.76 Ft ( 244.140 − 81.380) = Iz Iz

Figure 18.30

Problem 18.8

where Iz = moment of inertia of the section about OZ axis Similarly, shear force in vertical member FG, F2 =

Moment of inertia,

Ft 5 83.33 Ft ( 25 − y 2 ) dy = ∫ 0 Iz Iz

 12.53 12.53 103  Iz = t ×  + = + t [ 408.854] cm 4 12 12   12

 12.53 12.53 103  Iz = t ×  + = + t [ 408.854] cm 4 12 12   12 The moments of the shear forces about the point H is taken as shown in Fig. 18.31.

F

H

Fe = F1 × 7.5 + (5 + 7.5) F2

F2

F1

O e

162.76 Ft × 7.5 83.33 Ft × 12.5 + t × 408.854 t × 408.854 e = 2.985 + 2.547 = 5.33 cm = 55.3 mm =

7.5 cm

5.0 cm

Figure 18.31

Problem 18.9 A beam with thin-walled semi-circular section is shown in Fig. 18.32. It is loaded in a principal plane xy, so as to produce simple bending in this plane. Find the distance e for the shear centre O as shown in figure. Solution The shear stress t at each section s = r·f along the centre line of radius r will be in the direction of the tangent to the line. t is tangential to the centre line at s′ or perpendicular to the line GS′ as shown. The magnitude of shear stress at s′,

τ=

F F φ y dA = r cos θ br dθ Izb ∫ I z b ∫0

Fr 2 Fr 2 φ = sin θ 0 = × sin φ Iz Iz

MTPL0259_Chapter 18.indd 739

(18.16)

5/23/2012 11:34:57 AM

740 Chapter 18

We note that shear stress is maximum at (φ = π /2) and is zero at f = 0 and p. dF, elementary shear force = t b ds = t br df, dT, Moment of dF about centre G = t br2 dt. Total moment π

p

0

0

T = ∫ τ br 2 dφ = ∫

Fr 2 sin φ br 2 dφ Iz

S

(18.17)

S

2 Fr 4 b Fr 4 b π = sin φ dφ = . ∫ Iz 0 Iz

t

f

y

The horizontal component of the elemental shear force, τ br dφ above the neutral axis OZ, cancel the horizontal components of those below the neutral axis, hence the shear stress resultant is a vertical force equal to the shear force F at the section. To produce twisting moment T, calculated above, this force must act through a point O such that, F ( r + e) = T =

q

dq

Fr 4 b Iz

G

O

e

e=

where

I z = 2∫

p /2

0

r

b

Figure 18.32

r3 b cos2 θ dθ =

Z

F

2r 4 b −r Iz

or,

y = r cos q

Problem 18.9

(18.18) pr3 b 2

Substituting the value of Iz, e=

2r 4 b 4r ×2−r = − r = 0.273r. 3 p pr b

Problem 18.10 Determine the location e of the shear centre point O for the thin-walled open pipe with the cross-section shown in Fig. 18.33. Solution The shear stress t at any section, s = rϕ along the centre line of radius r will be in a direction tangential to centre line as shown. Magnitude of shear stress

τ=

MTPL0259_Chapter 18.indd 740

F F ϕ y dA = tr dθ y ∫ Izt I z t ∫0

5/23/2012 11:34:59 AM

Unsymmetrical Bending and Shear Centre t

s

741

t

dq

q

f

r

a

F

db

(p - a)

b

0

Z a

Figure 18.33

e

Problem 18.10

y = r sin(π − α + θ ) = r sin(α − θ )

where,

τ=

Ft ϕ rr sin(α − θ ) dθ I z t ∫0

=

Fr 2 Iz

=

Fr 2 (sin α sin θ + cos α cos θ )ϕ0 Iz

=

Fr 2 (sin α sin ϕ + cos α cos ϕ − sin α .sin 0 − cos α .1) Iz

=

Fr 2 (sin α sin ϕ + cos α cos ϕ − cos α ) Iz

∫

ϕ

0

(sin α cos θ − cos α sin θ ) dθ

Torque about the centre, α

α

0

0

T = 2∫ τ tr dϕ r = 2tr 2 ∫ τ dϕ

MTPL0259_Chapter 18.indd 741

5/23/2012 11:35:00 AM

742 Chapter 18

= 2tr 2 ∫

α

0

Fr 2 (sin α sin ϕ + cos α cos ϕ − cos α ) dϕ Iz

=

2tFr 4 ( − sin α cos θ + cos α sin ϕ − ϕ cos α )α0 Iz

=

2 Ftr 2 [( − sin α cos α + sin α cos α − α cos α ) − ( − sin α + 0 × sin α − 0 × cos α )] Iz

=

2 Ftr 2 2 Ftr 4 ( −α cos α + sin α ) = (sin α − α cos α ) Iz Iz

For the moment of inertia, consider an element at an angular displacement of q, α

α

0

0

I z = 2∫ ( rdβ )(t )( r sin β )2 = 2 tr3 ∫ sin2 β dβ α  1 − cos 2β  = 2tr3 ∫   dβ 0  2

α

= tr3 ∫ (1 − cos 2β ) dβ = tr3 β − 0

=

α

sin 2β 2 0

tr3 tr3 ( 2β − sin 2β )α0 = [2α − sin 2α ] 2 2

Therefore, twisting moment T, T=

2 Ftr 4 2(sin α − α cos α ) × tr3 ( 2α − sin 2α )

= 4 Fr e=

(sin α − α cos α ) = Fe ( 2α − sin 2α )

4 r (sin α − α cos α ) ( 2α − sin 2α )

Problem 18.11 Determine the position of the shear centre of the section of a beam shown in Fig. 18.34 Solution Figure 18.34 shows the section for which the shear centre is to be determined. In the diagram, direction of shear flow is given. Due to symmetry shear forces, F1 = F5, shear forces, F2 = F4. The section is symmetrical about the axis XX; therefore, shear centre will lie on this axis. Let us determine shear force F1 or F5.

MTPL0259_Chapter 18.indd 742

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Unsymmetrical Bending and Shear Centre F5

Shear stress in the vertical portion AB,

τ=

=

A

b − y FAy F (b1 − y )t1  h = × + y+ 1 2 I xx t1 I xx t1 2 

F

x

D

0

=

Ft1 2 I xx

∫

b1

0

x

Ft hy 2 y3 = 1 hb1 y − + b12 y − 2 I xx 2 3

B

C

F (b1 − y ) (h + b1 + y )t1dy 2 I xx

( hb1 − hy + b12 − b1 y + b1 y − y 2 ) dy

h

F3 t3

F2 e

b1

F4

0 SC

dA = t1dy

Shear force F1 = ∫

y

F4

where F is the applied shear force on the section.

b1

dy

F5

F (b1 − y )  h + b1 + y    I xx 2

Now,

743

F1

b1

A t1 b2

Figure 18.34

Problem 18.11

b1

0

 2 b13  h 2 3 − + − b h b b  1  1 2 1 3 

=

Ft1 2 I xx

=

2 2 2 Ft1  b1 h 2b1  Fb1 t1 (3h + 4b1 ) + =   2 I xx  2 3  12 I xx

Shear force in the horizontal portion BC, F2 = ∫

b2

0

F = I xx

 b F h  × b1t1 ×  h + 1  + t1 x  t2 dx   I xx t2  2 2

∫

b2

0

 b1t1h b12 t1 t2 h  x dx + +  2 2   2

b2 t1 t h x2 F b1t1h x+ 1 x+ 2 = I xx 2 2 2 2

MTPL0259_Chapter 18.indd 743

b2

0

5/23/2012 11:35:04 AM

744 Chapter 18

=

F I xx

 b1b2 t1h b2 b12 t1 t2 b22 h  + +   2 4   2

= F4 (due to symmetry). Taking moments of the shear forces about the centre O of the vertical web, Fe + 2 F1b2 = F2 h Fe =

2 2 2 Fh  b1b2 t1h b1 b2 t1 t2 b2 h  Fb2 b1 t1 (3h + 4b1) + + −   6 I xx I xx  2 2 4 

e=

2 2 2 3 h2  b1b2 t1 t2 b2  hb1 b2 t1 b2 b1 t1h 2 b2 b1 t1 + + − − ×   4  2 I xx 2 I xx 3 I xx  2 I xx

=

2 2 3 h2 b1b2 t1 h t2 b2 2 b2 b1 t1 × + − I xx 2 4 I xx 3 I xx

=

2 2 b1b2 t1  h2 2 2  t2 h b2 , where − b + 4 I xx I xx  2 3 1  2

Moment of inertia Ixx =

2

t h3 2 × b13 × t1 2b t 3 h b   h + 2b1t1  + 1  + 2 2 + 2b2 t2   + 3  2 2 2 12 12 12

Problem 18.12 For a section shown in Fig. 18.35, determine the position of the shear centre. The thickness of the section is t throughout.

B a2

N

a1 dz

F2

F1

O

SC

z A

F3

F

F4

D

a2 e

y

a1

A NA − Neutral axis SC − Shear centre

C

Figure 18.35 Problem 18.12

MTPL0259_Chapter 18.indd 744

5/23/2012 11:35:05 AM

Unsymmetrical Bending and Shear Centre

745

Solution Due to symmetry Shear force in portion AB, F1 = F4, shear force in portion CD Shear force in portion BO, F2 = F3, shear force in portion OC Shear force F1 Fay Shear stress, τ= , where F = applied shear force I NA t a1

Shear force,

F1 = ∫ τ dA

where

a = z t (as shown) z is along AB

Therefore,

Distance,

Shear force,

0

dA = t dz y = ( a sin 45° − a sin 45°) + z sin 45° 2 1 2 2a − 2a1 + z z  =  a2 − a1 +  sin 45° = 2  2 2 2 F1 = ∫

a1

0

=

=

Fzt  2a2 − 2a1 + z   t dz I NA t  2 2

a1 Ft ( 2a2 z − 2a1 z + z 2 ) dz ∫ 2 2 I NA 0

Ft 2 2 I NA

 2 a13  Fta12 (3a2 − 2a1 ) 3 = a a a − + 2 1 1  3  6 2 I NA

shear forces F2 and F3 are passing through the point O and will not produce any moment about O. Moment of inertia Ixx Moment of inertia of AB, about their principal axes t a13 a1t 3 I uu = , I vv = , θ = 45° (as shown in Fig. 18.35) 12 12 I xx = I uu cos2 θ + I vv sin2 θ =

I uu + I vv a1t 2 2 = ( a + t ), substituting cos q = sin q = 0.707 2 24 1 2

  a t a ( 2 a − a )2 I NA′ = I xx + ta1  a2 − 1  sin 45° = I xx + 1 2 1 2 8  

MTPL0259_Chapter 18.indd 745

5/23/2012 11:35:07 AM

746 Chapter 18

U

= = =

a a aa  a1t 2 2 ( a + t ) + a1t  + − 1 2  24 1 2   2 8 2 2

2 1

B a2

a1t 2 2 ( a + t + 12a22 + 3a12 − 12 a1a2 ) 24 1

(

a1t 2 t + 4 a12 + 12a22 − 12a1a2 24

X V

N

)

0

Figure 18.36

a1 V q X q = 45 A U A

Problem 18.12

Similarly, the moment of inertia of BO

Now,

ta23 a2 t 3 I uu′ = , I vv ′ = 12 12 θ = 45° I xx ′ = I uu′ cos2 θ + I vv ′ sin2 θ = =

a2 t 2 2 ( a + t ) as 24 2

I uu ′ + I vv ′ 2

sin θ =

1 1 ,sin2 θ = cos2 θ = 2 2

2

as

a  I NA′′ = I xx ′ + a2 t  2  sin2 45°  2 =

 a2  a t a2 t 2 2 a2 + t + a2 t  2  = 2 4 a22 + t 2 24  8  24

(

)

(

)

Total moment of inertia of section, I NA = 2 I NA′ + 2 I NA′′ I NA =

(

)

(

a1t 2 at t + 4a12 + 12a22 − 12a1a2 + 2 4a22 + t 2 12 12

)

Taking moments of the shear forces about the point O F × e = F1 × a2 + F1 × a2 =

or,

MTPL0259_Chapter 18.indd 746

e=

t a12 a2 3 2 I NA

F t a12 a2 3 2 I NA

(3a2 − 2a1 )

(3a2 − 2a1 )

5/23/2012 11:35:09 AM

Unsymmetrical Bending and Shear Centre

747

Key Points to Remember  Unsymmetrical bending occurs in a beam: (i) if the section is symmetrical but load-line is inclined to the principal axes or (ii) if section itself is unsymmetrical.  Product of inertia, Ixy = ∫ xydA Product of inertia of a section about its principal axes is zero.  For a symmetrical section, principal axes are along the axes of symmetry.  Parallel axes theorem for product of inertia, Ixy = I xy + Axy where Ixy = Product of inertia about any co-ordinates axes X–Y I xy = Product of inertia about centroidal axes X − Y x , y = Coordinate of the centroid of the section about XY co-ordinates.  If Ixy, Iyy, Ixy are moments of inertia about any co-ordinates axes X–Y passing through the centroid of the section. Inclination of principal axis with respect to X–X axes.

θ=

2 I xy 1 tan −1 . I yy − I xx 2

Principal moments of inertia Iuu, Ivv =

2

1 1  ( I xx + I yy ) ±  ( I yy − I xx ) + I xy2 . 2 2  

 If principal moments of inertia of a section are Iuu, Ivv, then moment of inertia about an axis X–X inclined at angle q to U–V axis is Ixx = Iuu cos2 q + Ivv sin2 q.  Stresses due to unsymmetrical bending, if u, v are the co-ordinates of a point and M is the bending moment applied on the section and q is the angle of inclination of axis of M, with respect to the principal axes UU. Resultant bending stress at the point  v cos θ u sin θ  + . sb = M  I vv   I uu

MTPL0259_Chapter 18.indd 747

5/23/2012 11:35:10 AM

748 Chapter 18  Angle of inclination of neutral axis with respect to principal axis UU  I  α = tan −1  tan θ uu  . I vv    Deflection of a beam under load W causing unsymmetrical bending

δ=

KWl 3 sin 2 θ cos 2 θ + 2 E I vv2 I uu

where K = Constant depending upon end conditions of the beam and position of the load q = Angle of inclination of load W with respect to VV principal axis.  If the direction of the applied load on a beam passes through the shear centre of the section, no twisting takes place in the beam.  For a section symmetrical about two axes, shear centre lies at the centroid of the section.  For a section symmetrical about one axis only, shear centre lies along the axis of symmetry.  About the shear centre, the moment due to the applied shear force is balanced by the moment of the shear forces obtained by summing the shear stresses over the various portions of the section.

Multiple Choice Questions 1. The product of inertia of a rectangular section of a breadth of 4 cm and a depth of 6 cm about its centroidal axis is (a) 144 cm4 (b) 72 cm4 4 (c) 36 cm (d) None of the above 2. The product of inertia of a rectangular section of a breadth of 3 cm and a depth of 6 cm about the coordinate axes passing at one corner of the section and parallel to the sides is (a) 81 cm4 (b) 72 cm4 4 (c) 54 cm (d) None of these 3. For an equal angle section, co-ordinate axes XX and YY passing through centroid are parallel to its length. The principal axes are inclined to XY axes at an angle (a) 22.5° (b) 45.0° (c) 67.5°

(d) None of the above

4. For an equal angle section, moment of inertia Ixx and Iyy are both equal to 120 cm4. If one principal

MTPL0259_Chapter 18.indd 748

moment of inertia is 180 cm4, the magnitude of other principal moment of inertia is (a) 180 cm4 (b) 120 cm4 (c) 60 cm4

(d) 30 cm4

5. For a section, principal moments of inertia are Iuu = 360 cm4 and Ivv = 160 cm4. Moment of inertia of the section about an axis inclined at 30° to the U–U axis, is (a) 310 cm4 (b) 260 cm4 (c) 210 cm4

(d) 120 cm4

6. For an equal angle section, Ixx = Iyy = 32 cm4 and Ixy = −20 cm4. The magnitude of one principal moment of inertia is (a) 52 cm4 (b) 42 cm4 4 (c) 32 cm (d) 16 cm4 7. For a T-section, shear centre is located at (a) Centre of the vertical web (b) Centre of the horizontal flange

5/23/2012 11:35:11 AM

Unsymmetrical Bending and Shear Centre (c) At the centroid of the section (d) None of the above. 8. For an I-section (symmetrical about X–X and Y–Y axis) shear centre lies at (a) Centroid of top flange (b) Centroid of bottom flange (c) Centroid of the web (d) None of the above. 9. For a channel section symmetrical about X–X axis, shear centre lies at

(a) (b) (c) (d)

749

The centroid of the section The centre of the vertical web The centre of the top flange None of the above.

10. If the applied load passes through the shear centre of the section of the beam, then there will be (a) No bending in the beam (b) No twisting in the beam (c) No deflection in the beam (d) None of these

Practice Problems 1. Figure 18.37 shows Z-section of a beam simply appalled over a span of 2 m. A vertical load of 2 kN acts at the centre of the beam and passes through the centroid of the section. Determine the resultant bending stress at points A and B. 2. Figure 18.38 shows a section of a beam subjected to shear force F. Locate the position of the shear centre as defined by e.

t 6 cm A

F

1 cm

b1 h O

1 cm

e

9 cm

t

1 cm

b1

b2

B 6 cm

Figure 18.37

Practice Problem 1

Figure 18.38

Practice Problem 2

3. Determine the location e of the shear centre point C for the thin-walled member having the cross-section shows in Fig. 18.39, where b2 > b1, the member segments have the same thickness t. 4. For an extruded beam having the cross-section shown in Fig. 18.40, determine (a) location of shear centre and (b) distribution of shear stresses caused by vertical shear force F = 12 kN.

MTPL0259_Chapter 18.indd 749

5/23/2012 11:35:12 AM

750 Chapter 18

A 50 B

e

t

C

O

h

150 mm

F = 12 kN

e t

50 t

b2

b1

100

Figure 18.40

Figure 18.39 Practice problem 3

Practice problem 4

5. Determine the location e of the shear centre for the thin-walled member having the cross-section shown in Fig. 18.41. The member segments have the same thickness. 6. Determine the location of shear centre O of a thin-walled beam of uniform thickness having the equilateral triangular section as shown in Fig. 18.42.

t h1 B b1

t h O

h1

e

F

60 e b

MTPL0259_Chapter 18.indd 750

A

D

b1

Figure 18.41

Practice problem 5

a 2

60

t

a 2

60°

E

Figure 18.42

Practice problem 6

5/23/2012 11:35:13 AM

Unsymmetrical Bending and Shear Centre

751

7. Determine the location of shear centre of a thin-walled beam of uniform thickness having the cross-section shown in Fig. 18.43.

t

a O

e

Figure 18.43

Practice Problem 7

Answers to Exercises Exercise 18.2:

Exercise 18.4: 540 cm4, 540 cm4, −324 cm4

b2 h2 4

Exercise 18.5: 28° 31′, 118° 31′, 360 cm4, 38.33 cm4

2 2 4 4 4 4 Exercise 18.3: 147.25 × 10 mm , + 32.30 × 10 mm Exercise 18.6: sA = −491.67 N/cm , sB = +2,812.5 N/cm

147.25 × 104 mm4 , + 32.30 × 104 mm4 .

Exercise 18.7: 221.2 N, 15° 13′ Exercise 18.8: 2.7 cm

Answers to Multiple Choice Questions 1. 2. 3. 4.

(d) (a) (b) (c)

MTPL0259_Chapter 18.indd 751

5. 6. 7. 8.

(a) (a) (b) (c)

9. (d) 10. (b)

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752 Chapter 18

Answers to Practice Problems 1. 147.44 MPa, 17.42 MPa 2. e =

1 (8b13b2 + 3h 2b22 + 6b1b2 h 2 ) 12 I xx

3. e =

3(b22 − b12 ) h + 6b1 + 6b2

MTPL0259_Chapter 18.indd 752

4. 23.2 mm, 12.36 MPa at B, 25.2 MPa at neutral axis 5. e =

3  h 2b 2 − (h − 2h1 ) 2 b12 

h3 + 6bh 2 + 6b1 (h − 2h1 ) 2

6. e = 0.1443a 7. e = 2a

5/23/2012 11:35:14 AM

19 Three-Dimensional Stresses CHAPTER OBJECTIVES There is no structural component which is only two-dimensional. Only to simplify the analysis of stresses, a component is considered under plane stress condition, i.e., negligible thickness. In actual practice, there is effect of thickness on the stresses of an element. Even in a bar subjected to uniaxial load, presence of a notch creates triaxial stresses in the vicinity of notch. Therefore, three-dimensional stresses in a body will be studied in this chapter. Students will learn about: 

Components of stresses on three orthogonal planes of a body.



Generalized Hooke’s law for a threedimensional stress system.



Normal and Shear stresses on a plane inclined to three orthogonal planes.



Equilibrium equations.



Strain compatibility conditions.



Strain components on three orthogonal planes.





Plane stress and plane strain conditions.

Airy’s stress functions for determination of stresses in a body subjected to external forces and moments.



Normal and shear strains.

Introduction In most of the engineering applications studied so far, we have considered two-dimensional stress system, under plane stress conditions. But, actually a material is non-homogeneous by nature and it may contain irregularities, blow holes, cuts and notches, in its internal structure. To completely ascertain the behaviour of a material, it is necessary to analyse a three-dimensional stress system acting on a body. Components such as crankshaft of an engine, turbine blades of a jet engine, blades of a hydraulic turbine, wings of an aircraft and many more need to be analysed three-dimensionally by using special experimental techniques, such as three-dimensional photoelasticity. A common example of a notch in a mild steel sample of Izod/Charpy impact tests which causes three-dimensional stresses in the sample and the ductile mild stress behaves in a brittle manner due to notch.

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754 Chapter 19

Stress Tensor All stresses acting on a cube of infinitesimally small dimensions ∆x = ∆y = ∆z → 0 are identified by the diagram of stresses on a cube. First, subscript on normal stress s or shear stress t associates, the stress with the plane of the stress, that is, subscript defines the direction of the normal to the plane of and second subscript identifies the direction of stress as: sxx = normal stress on yz plane, in x direction txy = shear stress on yz plane, in y direction txz = shear stress on yz plane, in z direction Similarly, the stresses on planes xz and xy are identified as shown in Fig. 19.1 y

syy

xz plane (ADEF)

E

D tyx

F

∆y

A

tzy

C

tzx

x

∆z

szz z

sxx

txz

tzy O xy plane (GBAF)

yz plane (ABCD)

txy

tyz

G

∆x

B

Figure 19.1 Three-dimensional stress system Stress tensor tij can be written as

τ xx

τ xy

τ xz

τ ij = τ yx

τ yy

τ yz stress tensor

τ zx

τ zy

τ zz

Generally, the normal stress is designated by s. Therefore, τ xx = σ xx , τ yy = σ yy , τ zz = σ zz Stress tensor can now be written as

MTPL0259_Chapter 19.indd 754

σ xx

τ xy

τ xz

τ ij = τ yx

σ yy

τ yz

τ zx

τ zy

σ zz

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Three-Dimensional Stresses

755

Stress tensor is symmetric, that is, tij = tji. Complementary shear stresses are txy and tyx; tyz and tzy; tzx and txz. Complementary shear stresses meet at diametrically opposite corners of an element to satisfy the equilibrium conditions. Stress invariants, I1 = σ xx + σ yy + σ zz I 2 = σ xxσ yy + σ xxσ zz + σ yyσ zz − τ xy2 − τ 2yz − τ zx2 I 3 = σ xxσ yyσ zz − σ xxτ 2yz − σ zz τ xy2 − σ yyτ zx2 + 2τ xyτ yz τ zx .

Stress at a Point At a point of interest within a body, the magnitude and the direction of the resultant stress depend on the orientation of a plane passing through the point. An infinite number of planes can pass through the point of interest so that there are infinite numbers of resultant stress vectors. Yet the magnitude and direction of each of these resultant stress vectors can be specified in terms of the nine stress components as shown on the three faces of elemental tetrahedron, Fig. 19.2. Let us take the altitude of plane abc, h → o and neglecting body forces, the three components of σ r (resultant stress) in x, y and z directions can be written as σ rx = σ xx cos (n, x) + τ yx cos( n, y ) + τ zx cos (n, z )

z c n sxx

sn

txy syy

tyz

p

sr

txz

tyx h tzy

tn tzX

b

y

tzy x

σ ry = τ xy cos (n, x) + σ yy cos (n, y ) + τ zy cos (n, z )

a

Figure 19.2

szz

Elemental tetrahedron abc altitude of tetrahedron h

σ rz = τ xz cos (n, x) + τ yz cos (n, y ) + σ zz cos (n, z ) From these three Cartesian components, resultant stress, sr can be determined as follows:

σ r = σ rx2 + σ ry2 + σ rz2 The three-direction cosines that define the line of action of resultant stress σ r are, σ ry σ σ cos(σ r, x ) = rx ; cos(σ r, y ) = ; cos(σ r, z ) = rz σr σr σr Normal stress σ n and shear stress τ n acting on the plane under consideration will be

σ n = |σ r | cos (σ r , n) τ n = |σ r | sin (σ r , n)

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756 Chapter 19

Angle between the resultant stress vector σ r and normal to the plane σ n can be determined by using cos (σ r , n) = cos (σ r, x ) cos (n, x ) + cos (σ r , y ) cos (n, y ) + cos (σ r , z ) cos (n, z ) The normal stress sn can also be determined by considering projections of srx, sry and srz onto the normal to the plane under consideration; therefore,

σ n = σ rx × cos ( n, x ) + σ ry cos ( n, y ) + σ rz cos ( n, z ) After the determination of normal stress sn, shear stress tn is obtained by

τ n = σ r2 − σ n2 Example 19.1 At a point in a stressed body, the Cartesian components of stresses are sxx = 60 MPa, syy = -30 MPa, szz = +30 MPa, txy = 40 MPa, tyz = tzx = 0. Determine the normal and shear stresses on a plane whose outer normal has the direction cosines of cos (n, x ) =

6 6 7 , cos (n, y ) = , cos (n, z ) = 11 11 11

Solution Let us say, normal and shear stresses on a plane are sn and tn, the resultant stress on the

σ r = σ n2 + τ n2

plane is

For the problem, the three components of σ r are as follows:

σ rx = σ xx cos ( n, x ) + τ yx cos ( n, y ) σ ry = τ xy cos ( n, x ) + σ yy cos ( n, y ) σ rz = σ zz cos ( n, z ), because τ yz = τ yx = 0. Putting the values of σ xx, σ yy, σ zz , τ ny, cos (n, x ),cos (n, y ) and cos (n, z ),

σ rx = 60 ×

6  6  600 + 40   = MPa = 54.54 MPa  11 11 11

σ ry = 40 ×

6  6  60 − 30   = MPa = 5.454 MPa  11 11 11

σ rz = 30 ×

7 210 = MPa = 19.09 MPa 11 11

σ r = σ rx2 + σ ry2 + σ rz2 = =

1 6002 + 602 + 2102 11

1 1 3, 60, 000 + 3, 600 + 44,100 = × 638.5 11 11

= 58.0467 MPa, resultant stress

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Three-Dimensional Stresses

757

Normal stress, σ n, on the plane is σ n = σ rx cos ( n, x ) + σ ry cos ( n, y ) + σ rz cos ( n, z ) (Putting the values) = 54.54 ×

6 6 7 + 5.454 × + 19.09 × 11 11 11

= 29.749 + 2.975 + 12.148 = 44.872 MPa.

τ n = σ r2 − σ n2 = 58.04672 − 44.8722

Shear stress,

= 3369.42 − 2013.5 = 36.82 MPa Example 19.2 At a point in a stressed body, the Cartesian components of stress are sxx = 70 MPa, syy = 60 MPa, szz = 50 MPa, txy = 20 MPa, tyz = -20 MPa and tzx = 0. Determine the normal and shear stresses on a plane whose outer normal has directions cosines, cos (n, x ) =

12 15 16 ; cos (n, y ) = ; cos (n, z ) = 25 25 25

Solution

σ rx = σ xx cos ( n, x ) + τ yx cos ( n, y ) σ ry = τ xy cos ( n, x ) + σ yy cos ( n, y ) + τ yz cos ( n, z ) σ rz = τ yz cos ( n, y ) + σ zz cos ( n, z ) σ rx = 70 ×

12 15 + 20 × = 45.6 MPa 25 25

σ ry = 20 ×

12 15 16 + 60 × − 20 × = 32.8 MPa 25 25 25

σ rz = −20 ×

15 16 + 50 × = −12 + 32 = +20 25 25

σ r = 45.62 + 32.82 + 202 = 2079.36 + 1075.84 + 400 = 59.625 MPa

σ n = σ rx cos ( n, x ) + σ ry cos ( n, y ) + σ rz cos ( n, z ) 12 15 16 + 32.8 × + 20 × 25 25 25 = 21.888 + 19.68 + 12.8 = 54.368 MPa = 45.6 ×

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758 Chapter 19

Shear stress,

τ n = 59.6252 − 54.3682 = 3, 555.14 − 2, 955.88 = 24.48 MPa

Exercise 19.1 Determine the normal and shear stresses on a plane whose outer normal makes equal angles with the x, y and z axes, if the Cartesian components of stress at the point are

σ xx = σ yy = σ zz = 0, τ xy = 80 MPa, τ yz = 0 and τ zx = 100 MPa Exercise 19.2 At a point in a stressed body, the Cartesian components of stresses are sxx = 40 MPa, syy = 60 MPa, szz = 20 MPa, txy = 60 MPa, tyz = 50 MPa and tzx = 40 MPa. Determine (a) normal and shear stresses on a plane whose outer normal has the direction cosines, 4 4 7 cos (n, x ) = , cos (n, y ) = , cos (n, z ) = , 9 9 9 (b) the angle between sr and outer normal n.

Plane Stress Condition Under plane stress condition, stresses sxx, syy and txy are represented in one plane, that is, the central plane of a thin element in yx plane, as shown shaded in Fig. 19.3. In this case stresses szz = txz = tyz = 0. syy tyx

y

t txy sxx

sxx

txy

tyx

syy x

o

Figure 19.3

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Plane stress condition

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Three-Dimensional Stresses

759

If E and v are Young’s modulus and Poisson’s ratio of the material, then strains are as follows:

σ yy σ σ yy ε xx = xx −σ vxx ε xx E= −Ev E E σ yy σ xx σ vyy σ ε yy = − ε yy E= −Ev xx E E v ε zz = − (σvxx + σ yy ) ε zz E = − (σ xx + σ yy ) E Shear strain, γ xy = τ xy /G But, shear strain γ xy is equally divided about both x and y axes. Strain tensor for plane stress condition will be as follows:

γ xy

ε xx Strain tensor =

0

2

γ xy 2 0

ε yy

0 .

0

ε zz

For a particular set of three orthogonal planes, where shear stresses are zero and normal stresses on these planes are termed as principal stresses s1, s2 and s3.

σ1 Stress tensor = 0 0

Strain Tensor Figure 19.4 shows a body of dimensions ∆x and ∆y subjected to shear stresses txy and tyx shear strain γ xy /2 develops from y to x and from x to y. Total shear strain about xy axes is γ xy = τ xy /G, where G is shear modulus. This shear strain γ xy is equally divided about both axes x and y. Strain tensor in a biaxial case

ε xx will be

γ yx 2

γ xy

MTPL0259_Chapter 19.indd 759

0

σ2 0

0 .

σ3

y C

txy

gxy 2 B

C

B

tyx ∆y

2 A

ε yy

gxy 2

Similarly strains, γ zx = τ zx /G and

0

γ yz = τ yz /G

o

∆x

Figure 19.4

A

x

Shear strains

5/23/2012 11:32:05 AM

760 Chapter 19

ε xx

γ xy 2

Three-dimensional strain tensor will be γ yx

γ xz 2 γ yz 2

ε yy

2 γ zx 2

γ zy 2

ε zz

In the plane stress case (sxx, syy, txy) the strains are

σ yy σ xx −v E E σ yy σ − v xx ε yy = E E τ xy γ xy = G v ε zz = − (σ xx + σ yy ) E ε xx =

ε xx Strain tensor for a plane stress state is

γ yx 2 0

γ xy

0

2

ε yy

0

0

ε zz

For strain condition shown in Fig. 19.5, the strain

ε xx tensor is

γ yx 2

MTPL0259_Chapter 19.indd 760

gyx 2 gxy

ε yy

2

(σ xx + σ yy ) σ zz −v E E =0

exx

exx

ε zz =

Therefore, σ zz = v(σ xx + σ yy ) to satisfy the condition of plane strain. Moreover,

eyy

2

ε zz = 0 In this, strain

y

γ xy

γ yz = γ zx = 0

gxy 2

o

gyx

x eyy

2

Figure 19.5

Plane strain condition

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Three-Dimensional Stresses

761

Example 19.3 A steel cube of 100 mm is subjected to a uniform pressure of 100 MPa acting on all faces. Determine the change in dimensions between the parallel faces of the cube, if E = 200 GPa, v = 0.3. Solution s = s1 = s2 = s3 = 100 MPa, hydrostatic stress. Principal strains,

ε = ε1 = ε 2 = ε3 = ε=

σ (1 − 2v ) E

100 (1 − 2 × 0.3) = 0.2 × 10−3 3 200 × 10

Change in dimensions, δ = δ x = δ y = δ z = ε a where

a = side of the cube

Therefore,

δ = 100 × 0.4 × 10-3 (change in dimension) = -0.04 mm (because s is the pressure, reducing the sides)

Exercise 19.3 A steel rectangular parallelopiped is subjected to stresses 100, -60 and +80 MPa as shown in Fig. 19.6. Dimensions of the body are 150, 100 and 75 mm in x, y and z directions. Determine the change in dimensions, if E = 200 × 103 MPa, v = 0.3.

y

syy = −60 MPa

100 MPa 100 mm

80 MPa

sxx = 100 MPa x

o 75 mm

80 MPa z

150 mm syy = −60 MPa

Figure 19.6

Exercise 19.3

Deformations There are two types of strains resulting from the deformation of a body, that is, (a) linear or extensional strains and (b) shear strains, resulting in change of shape. Let us consider an element of dimensions ∆x and ∆y with changes in directions of x and y on ∆x and ∆y as shown in Fig. 19.7. A body of dimensions ∆x and ∆y in the x–y plane, represented by ABCD, deforms to AB′C′D as shown in Fig. 19.7(a).

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762 Chapter 19

Strain in x direction, ε x = ∆u /∆x. Again it is deformed to A′B′CD as shown in Fig. 19.7(b), strain in y direction, ε y = ∆v /∆y. Now the body in the x–y plane is deformed to A′B′C′D as shown in Fig. 19.7(c). y

B B

A

A A

B

A'

B B

∆v

∆v A C

q2 ∆y

∆y

B

∆y C

C

D

C ∆x

x

D

D

C

∆u

∆x

(a)

q1 ∆x

(b)

Figure 19.7

C ∆u

(c)

Normal and shear strains

gxy = q1 + q2

Shear strain,

=

∆u ∆v γ + that is xy on both sides of co-ordinate axes: ∆y ∆x 2

Moreover direct strains are

εx =

∆u ∆v , εy = . ∆x ∆y

In order to study the deformation, or change in shape, one has to consider a displacement field(s) for a point in a body subjected to deformation. A point P is located at position vector op = r and displaced to position upper op′ = r′. The displacement vector s = r′ - r. The displacement vector with functions of x, y, z has components u, v, w in x, y, z directions, such that y S = ui + vj + wk, where S is a function of initial co-ordinates of point P, displacement components are u = u(x, y, z)

p

v = v(x, y, z)

r

w = w(x, y, z) function of x, y and z.

r

These strains at point P are defined by

MTPL0259_Chapter 19.indd 762

εx =

∂u ∂x

εy =

∂v ∂y

S

O

p x

z

Figure 19.8

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Three-Dimensional Stresses

763

∂w ∂z ∂u ∂v = + ∂y ∂x ∂v ∂w = + ∂z ∂y

εz = γ xy

and shear strains are

γ yz

γ zx =

∂w ∂ u + ∂x ∂z

In x–y plane, ox in a two-dimensional case, strains are ε x =

∂u ∂v ∂u ∂v , εy = and γ xy = + . ∂x ∂y ∂y ∂x

Example 19.4 The displacement field for a given point of a body is given by. S = [( x 2 + 2 y + 3)i + (3 x + 4 y 2 ) j + (2 x3 + 6 z )k ] × 10−4 at point P(1, -2, 3). Determine displacement components in x, y and z directions. What is the deformed position? Solution Displacement components are: u = ( x 2 + 2 y + 3) × 10−4 v = (3 x + 4 y 2 ) × 10−4 w = ( 2 x3 + 6 z ) × 10−4 Putting the values of initial co-ordinates (1, -2, 3) of point P, the components are u = 1 + 2( −2) + 3 = 0.0 v = 3 × 1 + 4( −2)2 = 19 × 10−4 w = 2 × 13 + 6 × 3 = 20 × 10−4 Deformed position is given as follows: x′ = x + u = 1 − 0 = 1 y ′ = y + v = −2 + 19 × 10−4 = −1.9981 z ′ = z + w = 3 + 20 × 10−4 = 3.002 Example 19.5 The displacement field for a body is given by S = [( x 2 +2 y )i + (4 + 2 z ) j + ( x2 + 3 y )k ] × 10−4 . Write down the strain components and strain matrix at point (2, 1, 2). Solution Displacement components are: u = ( x 2 + 2 y ) × 10−4 v = ( 4 + 2 z ) × 10−4 w = ( x 2 + 3 y ) × 10−4

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764 Chapter 19

∂u = ( 2 x ) × 10−4 ∂x ∂v =0 ∂y ∂w =0 ∂z ∂v ∂u + = (0 + 2) × 10−4 ∂x ∂y ∂v ∂w + = ( 2 + 3) × 10−4 ∂z ∂y ∂w ∂u + = (2 x) × 10−4. ∂x ∂z Putting the values of initial co-ordinates (2, 1, 2), we get

ε x = 4 × 10−4 , ε y = 01 , ε 2 = 0 γ xy = 2 × 10−4 , γ yz = 5 × 10−4 , γ zx = 4 × 10−4 γ xy 2

Strain matrix

= 1 × 10−4 ,

γ yz 2

= 2.5 × 10−4 ,

γ zx = 2 × 10−4 2

4 1 2 = 1 0 2.5 × 10−4 2 2.5 0

Example 19.6 Displacement field is S = ( x 2 + y 2 + 2)i + (3 x + 4 y 2 ) j  × 10−4 . What is strain field at point (1, 2)? Solution Displacement components are u = ( x 2 + y 2 + 2) × 10−4 v = (3 x + 4 y 2 ) × 10−4 Strain components

MTPL0259_Chapter 19.indd 764

εx =

∂u = ( 2 x ) × 10−4 = 2 × 1 × 10−4 = 2 × 10−4 ∂x

εy =

∂v = 8 y × 10−4 = 8 × 2 × 10−4 = 16 × 10−4 ∂y

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Three-Dimensional Stresses

yxy 2

Strain tensor =

=

1  ∂u ∂v  1 + = ( 2 y + 3) × 10−4 2  ∂y ∂x  2

=

1 ( 2 × 2 + 3) × 10−4 = 3.5 × 10−4 2

765

2 3.5 × 10−4 3.5 16

Exercise 19.4 The displacement field for a body is given by S = [( x2 + 2 y )i + (3 y + z ) j + ( x2 + z )k ] × 10−3 . What is the deformed position of point originally at (3, 1, -2)? Write down the strain matrix. Exercise 19.5 Displacement field is u = kx2 and v = k (4x + 2y2) where k is a very small quantity. What are the strains at (1, -2) point?

Generalized Hooke’s Law For a simple prismatic bar subjected to axial stress sxx and axial strain exx, Fig. 19.9. Hooke’s law states that stress m strain y sxx ’ exx

x z

Figure 19.9 Stress strain on simple bar sxx ∝ exx sxx = Eexx

(19.1)

where E is proportionality constant. E is Young’s modulus of elasticity of the material. But for an elastic body subjected to six stress components sxx, syy, szz, tyx, tyz and tzx and six strain components, that is, exx, eyy, ezz, gxy, gyz and gzx the generalized Hooke’s law can be expressed as

σ xx = A11 ε xx + A12 ε yy + A13 ε zz + A14 γ xy + A15 γ yz + A16 γ zx σ yy = A21 ε xx + A222 ε yy + A23 ε zz + A24 γ xy + A25 γ yz + A26 γ zx σ zz = A31 ε xx + A32 ε yy + A33 ε zz + A34 γ xy + A35 γ yz + A36 γ zx τ xy = A41 ε xx + A42 ε yy + A43 ε zz + A44 γ xy + A45 γ yz + A46 γ zx

(19.2)

τ yz = A51 ε xx + A52 ε yy + A53 ε zz + A54 γ xy + A55 γ yz + A56 γ zx τ xz = A61 ε xx + A62 ε yy + A63 ε zz + A64 γ xy + A65 γ yz + A66 γ zx

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766 Chapter 19

where A11, A12,..., A65, A66 are 36 elastic constants for a given material. For homogeneous linearly elastic material, above noted six equations are known as generalized Hooke’s law. Similarly, strains can be expressed in terms of stresses as follows:

ε xx = B11 σ xx + B12 σ yy + B13 σ zz + B14 τ xy + B15 τ yz + B16 τ zx • • • γ xz =

• • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • B61 σ xx + B62 σ yy + B63 σ zz + B64 τ xy + B65 τ yz + B66 τ zx

(19.3)

For a fully anisotropic material there are 21 constants [ 6 +(36 - 6)/ 2 = 21 ] in the generalized Hooke’s law and for orthotropic materials there are nine constants. Say, for an isotropic material having the same elastic properties in all directions and material as such has no directional property, say there are three principal stresses s1, s2 and s3 and three principal strains e1, e2 and e3. Then Hooke’s law can be written as

σ1 = Aε1 + Bε 2 + C ε3

(19.4)

where A, B and C are elastic constants. In the above equation, e1 is longitudinal strain along s1 but e2 and e3 are lateral strains so constants B = C, therefore,

σ1 = Aε1 + Bε 2 + Cε3 = Aε1 + Bε1 − Bε1 + Bε 2 + Bε3 = ( A − B)ε1 + B[ε1 + ε 2 + ε3 ] where

e1 + e2 + e3 = ∆, a cubical dilatation = first invariant of strain

σ1 = ( A − B )ε1 + ∆B Let us denote (A - B) by 2m and B by l, then

Similarly

σ1 = λ∆ + 2 µ ε1

(19.5)

σ 2 = λ∆ + 2 µ ε 2

(19.6)

σ 3 = λ∆ + 2 µ ε3

(19.7)

l and m are called Lame’s coefficients From Eqs (19.5)–(19.7) σ1 + σ 2 + σ3 = 3λ∆ + 2 µ(ε1 + ε2 + ε3 ) = 3λ∆ + 2 µ ∆ = (3λ + 2 µ) ∆

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Three-Dimensional Stresses

∆=

σ1 + σ 2 + σ 3 . (3λ + 2 µ )

σ1 = λ∆ + 2 µ ε1

Principal stress,

σ1 =

λ (σ1 + σ 2 + σ 3 ) + 2 µ ε1. 3λ + 2 µ

ε1 =

 λ+ µ  λ σ − (σ + σ 3 )  . µ (3λ + 2 µ )  1 2(λ + µ ) 2 

ε1 =

1 [σ1 − v (σ 2 + σ 3 )]. E

Therefore, Young’s modulus,

E=

µ (3λ + 2 µ ) . λ+µ

Poisson’s ratio,

v=

λ . 2(λ + µ )

767

(19.8) (19.9) (19.10)

Solving Eq. (19.9) for e1, we will get

From elementary strength of materials

Example 19.7 For steel Young’s modulus E = 208,000 MPa and Poisson’s ratio, v = 0.3. Find Lame’s coefficients l and m. Solution

E=

µ (3λ + 2 µ ) = 208, 000 λ+µ

(19.11)

v=

λ = 0.3 2( λ + µ )

(19.12)

Taking Eq. (19.11),

µ (3λ + 3µ − µ ) = 208, 000 λ+µ 3µ −

µ2 = 208, 000 λ+µ

(19.13)

From Eq. (19.12),

λ = 0.3 2( λ + µ ) or,

l = 0.6l + 0.6m l = 1.5m

MTPL0259_Chapter 19.indd 767

(19.14)

5/23/2012 11:32:19 AM

768 Chapter 19

Putting this value in Eq. (19.13) 3µ −

µ2 = 208, 000 1.5µ + µ

3m - 0.4m = 208,000. Lame’s coefficient,

m = 80,000 N/mm2

Coefficient,

l = 1.5m = 120,000 N/mm2

Exercise 19.6 For a concrete block, E = 27.5 × 103 MPa and Poisson’s ratio is 0.2. Determine Lame’s coefficients l and m.

Elastic Constants K and G From the previous article, we know that E=

µ (3λ + 2 µ ) λ+µ

v=

λ 2( λ + µ )

E = 2µ +

or

or

2v =

λµ λ+µ

λ λ+µ

(19.15)

(19.16)

Putting the value of λ /(λ + µ ) from Eq. (19.16) in Eq. (19.15) we get E = 2m + 2m v = 2m(1 + v) From elementary strength of material, we know that E = 2G (1 + v ) Therefore, Lame’s coefficient m = G, shear modulus (from Eq. (8) of the previous chapter). If s1 = s2 = s3 = p, hydrostatic stress or volumetric stress is ∆= or,

3p 3λ + 2 µ

3λ + 2 µ p = = K, bulk modulus 3 ∆ 3λ + 2 µ 3

Therefore, bulk modulus,

K=

Shear modulus,

G = m, Lame’s coefficient.

MTPL0259_Chapter 19.indd 768

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Three-Dimensional Stresses

769

Example 19.8 For aluminium E = 67,000 MPa, Poisson’s ratio, v = 0.33. Determine bulk modulus and shear modulus of aluminium. E=

Solution

µ (3λ + µ ) λ+µ

= 2µ + or,

3µ −

µλ λ+µ

(19.17)

µ2 = 208, 000 λ+µ

(19.18)

λ 2( λ + µ )

(19.19)

From Eq. (19.19), v=

Poisson’s ratio,

2v =

or,

λ = 2 × 0.33 λ+µ

λ = 0.66 λ + 0.66 µ 0.34 λ = 0.66 µ

λ = 1.94 µ  λ  E = 2µ + µ   λ + µ 

(19.20)

Putting the value of l in terms of m in Eq. (19.20), E = 2 µ + µ × 0.66 = 2.66 µ

µ=

E 67, 000 = = 25,188 MPa 2.66 2.66

= Lame’s coefficient = Shear modulus G = 25,188 MPa

(19.21)

λ = 1.94 µ = 1.94 × 25,188 = 48, 865 MPa Bulk modulus,

K=

3λ + 2 µ 3

=

3 × 48, 865 + 2 × 25,188 3

=

146, 595 + 50, 376 3

= 65,657 MPa

MTPL0259_Chapter 19.indd 769

5/23/2012 11:32:23 AM

770 Chapter 19

Exercise 19.7 For steel E = 200,000 MPa and Poisson’s ratio is 0.295. Determine Lame’s coefficients and bulk modulus K.

Equilibrium Equations Consider a small infinitesimal element of a body of dimensions ∆x, ∆y and thickness t = 1 subjected to stresses varying over distances ∆x and ∆y as shown in Fig. 19.10. X and Y are body forces per unit volume in x and y directions.

syy +

syy ∂y

y

× ∆y tyx +

tyx ∂y

× ∆y

t=1 D

C

txy +

Y sxx

∆y

X sxx +

txy ∂x

sxx ∂x

txy A

tyx

syy

× ∆x

∆y

B x

o ∆x

Figure 19.10

A small infinitesimal element in equilibrium

Volume of the body = ∆ x × ∆y × 1 Taking the summation of forces in x direction,

σ xx ∆y1 +

∂τ yx ∂σ xx ∆x∆y − σ xx ∆y1 + τ yx ∆x1 + ∆y∆ x − τ yx ∆ x1 + X∆ x∆ y = 0 ∂x ∂y

or,

∂τ yx ∂σ xx ∆x∆y + ∆y∆x + X∆x∆y = 0 ∂x ∂y

Simplifying further, we get ∂σ xx ∂τ yx + +X =0 ∂x ∂y

MTPL0259_Chapter 19.indd 770

(19.22)

5/23/2012 11:32:24 AM

Three-Dimensional Stresses

771

Taking the summation of forces in y direction, we get ∂ σ yy ∂ τ xy σ yy ∆x1 + ∆y∆ x1 − σ yy ∆x1 + τ xy ∆y1 + ∆y∆ x − τ xy ∆ y1 + Y ∆ x ∆ y = 0 ∂x ∂x ∂σ yy

or,

∂y

∆y ∆x +

∂τ xy ∂x

∆x ∆y + Y∆x ∆y = 0

Simplifying further, we get ∂σ yy ∂y

+

∂τ xy ∂x

+Y = 0

(19.23)

In a three-dimensional case, equilibrium equations can be written as ∂σ xx ∂τ yx ∂τ zx + + +X =0 ∂x ∂y ∂z ∂τ xy ∂σ yy ∂τ zy + + +Y = 0 ∂x ∂y ∂z ∂τ xz ∂τ yz ∂τ zz + + +Z =0 ∂x ∂y ∂z where X, Y and Z are body forces per unit volume. In these equations, mechanical properties have not been used. So these equations are applicable whether a material is elastic, plastic or viscoelastic. In a two-dimensional case, equillibrium eequations are: ∂σ xx ∂τ yx + = 0, where the body forces are zero. ∂x ∂y ∂τ xy ∂σ yy + = 0. ∂x ∂y We may permanently satisfy these equations by expressing stresses in terms of a function f, called Airy’s stress function as follows:

σ xx =

∂ 2ϕ ∂ 2ϕ ∂ 2ϕ ; σ yy = 2 ; τ xy = − 2 ∂ x∂y ∂y ∂x

In a plane stress case a body subjected to stresses σ xx, σ yy and τ xy, the strains are

MTPL0259_Chapter 19.indd 771

ε xx =

σ xx vσ yy 1  ∂2φ ∂ 2φ  − =  2 −v 2 E E E  ∂y ∂x 

(19.24)

ε yy =

σ yy σ ∂ 2φ  1  ∂ 2φ − v xx =  2 − v 2  E E E  ∂x ∂y 

(19.25)

5/23/2012 11:32:25 AM

772 Chapter 19

ε zz = −

Shear strain,

γ xy = −

v v  ∂ 2φ ∂ 2φ  (σ xx + σ yy ) = −  2 + 2  E E  ∂x ∂y 

τ xy G

=−

(19.26)

1 ∂ 2ϕ G ∂x∂y

(19.27)

Strain compatibility equations are 2 ∂2 ε xx ∂ ε yy ∂2γ xx + = ∂x∂y ∂y 2 ∂x 2

(19.28)

Substituting the values of exx, eyy, and gxy from Eqs (19.24) and (19.25) and Eq. (19.27) in Eq. (19.28) we will get 1  ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ  1 ∂ 4ϕ v v − + − = − E  ∂y 4 G ∂x 2 ∂y 2 ∂ x 2 ∂y 2 ∂ x 4 ∂x2 ∂y 2 

(19.29)

Putting the value of E /G = 2(1 + v ) in Eq. (19.29) we get ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ ∂ 4ϕ 2 2 1 + − v = − ( + v ) ∂x 4 ∂y 4 ∂x 2 ∂y 2 ∂x 2 ∂y 2 Simplifying further the equation becomes ∂4 φ ∂4 φ 2∂4φ + + = 0, a biharmonic equation ∂x 4 ∂y 4 ∂x 2 ∂y 2 ∆ 4φ = 0.

or,

(19.30)

Airy’s stress function f chosen for any problem must satisfy the above biharmonic equation. Example 19.9 Following strains are given

ε xx = 6 + x 2 + y 2 + x 4 + y 4 ε yy = 6 + 3x 2 + 3 y 2 + x 4 + y 4 γ xy = 5 + 4 xy ( x 2 + y 2 + 2) = 5 + 4 x3 y + 4 xy3 + 8 xy Determine whether the above strain field is possible. If it is possible, determine displacement components u and v, assuming u = v = 0 at origin. Solution Strain compatibility condition is ∂2 ε xy ∂y 2

MTPL0259_Chapter 19.indd 772

+

∂2 ε yy ∂y 2

=

∂2γ xy ∂x∂y

(19.31)

5/23/2012 11:32:27 AM

Three-Dimensional Stresses

773

∂ε xx = 2 y + 4 y3 ∂y ∂2 ε xx = 2 + 12 y 2 ∂y 2 ∂ε yy ∂x ∂2 ε yy ∂x 2 ∂γ xy ∂x ∂2γ xy ∂x∂y

(19.32)

= 6 x + 4 x3

= 6 + 12 x 2

(19.33)

= 0 + 12 x 2 y + 4 y3 + 8 y

= 12 x 2 + 12 y 2 + 8

(19.34)

2 ∂2 ε xx ∂2 ε yy 2 2 2 2 2 + ∂ ε = 2 + 12 y + 6 + 12 x = 8 + 12 y + 12 y ∂ ε xx2 ∂y + ∂xyy2 = 2 + 12 y 2 + 6 + 12 x 2 = 8 + 12 y 2 + 12 y 2 ∂y 2 ∂x 2 ∂2γ xy = ∂2γ xy = ∂x∂y ∂x∂y

Strain field is possible. Now,

But

∂u = 6 + x2 + y 2 + x4 + y 4 ∂x x3 x5 u = 6 x + + xy 2 + + xy 4 + C1 3 5 u = 0 at x = 0, y = 0, So, C1 = 0.

ε xx =

Displacement component is u = 6x +

ε yy =

x3 x5 + xy 2 + + xy 4 3 5

(19.35)

∂v = 6 + 3 x2 + 3 y 2 + x4 + y 4 . ∂y

Integrating, we get V = 6 y + 3 x 2 y + y3 + x 4 y +

MTPL0259_Chapter 19.indd 773

y5 + C2 5

5/23/2012 11:32:28 AM

774 Chapter 19

however, v = 0 at x = 0, y = 0, constant C2 = 0 v = 6 y + 3 x 2 y + y3 + x 4 y +

therefore,

y5 . 5

Exercise 19.8 Given the following system of strains:

ε xx = 8 + x 2 + 2 y 2 ε yy = 6 + 3 x 2 + y 2 γ xy = 10 xy Determine whether the above strain field is possible. If possible, determine displacement components u and v, if u = 0, v = 0 at origin.

Second-degree Polynomial Let us consider an Airy’s stress function:

φ=

A 2 C x + Bxy + y 2 2 2

∆4f = 0, satisfies the compatibility condition. Stress,

σ xx =

∂2φ =C ∂y 2

Stress,

σ yy =

∂2φ =A ∂y 2

∂2φ = − B (negative, tends to rotate the body in anticlockwise direction) ∂x∂y y syy = A This represents a plane stress condition as shown txy = −B in Fig. 19.11. Shear stress, τ xy = −

Example 19.10 Airy’s stress function f = 40x2-30xy + 60y2 satisfies the compatibility condition ∆4f = 0. Determine stresses sxx, syy and txy and show graphically the stress distribution. Stresses are in MPa.

sxx = C

sxx

Solution f = 40x2 - 30xy + 60y2 ∂ϕ = 80 x − 30 y ∂x

MTPL0259_Chapter 19.indd 774

o

x txy

Figure 19.11

syy

Plane stress condition

5/23/2012 11:32:30 AM

Three-Dimensional Stresses

∂ 2ϕ = 80 MPa ∂x 2

Stress,

σ yy =

Shear stress,

τ xy = −

Moreover,

∂ϕ = −30 x + 120 y ∂y

Stress,

σ xx =

775

(19.36)

∂ 2ϕ = +30 MPa ∂x∂y

(19.37)

∂ 2ϕ = 120 MPa ∂y 2

(19.38)

Note that shear stress, txy is +ve, that is, tending to rotate the body in clockwise direction. Figure 19.12 shows the stress distribution which is a plane stress state. tyx are shear stresses complementary to txy. y syy = 80 MPa tyx txy sxx = 120 MPa

sxx

x

o

txy = + 30 MPa tyx syy

Figure 19.12

Example 19.10

Exercise 19.9 An Airy’s stress function f = 50x2 - 40xy + 80y2 satisfies the compatibility condition ∇4 f = 0. Determine normal and shear stresses and state the type of state of stress.

A Beam Subjected to Pure Bending For a beam of depth d subjected to a pure bending moment M and no shear force, the Airy’s stress function can be a third-degree polynomial.

ϕ=

MTPL0259_Chapter 19.indd 775

A 3 B 2 C D y + y x + yx 2 + x3 6 2 2 6

5/23/2012 11:32:31 AM

776 Chapter 19

∇4 f = 0, for this function

σ xx =

∂2φ = Ay + Bx + 0 + 0. ∂y 2

Taking B = 0 as stress σ xx is independent of x,

σ xx = Ay y varies from −d/2 to +d/2 as shown in Fig. 19.11. Maximum value of σ xx in tension, d M σ t = A , where A = 2 I =

M d × . I 2

Maximum value of σ xx in compression,

σc = − A

d M d =− × I 2 2

where M = bending moment I = second moment of area (of cross-section of beam) about neutral plane. ∂2φ = Cy + Dx ∂x 2

Now,

σ yy =

however,

σ yy = 0, so constant C = D = 0

Shear stress

τ xy = −

∂ 2ϕ ∂x∂y

= By + Cx

however,

τ xy = 0, also because it is a case of pure bending. sc d 2

M

M

d 2

d st

Figure 19.13

MTPL0259_Chapter 19.indd 776

Beam under pure bending

Figure 19.14

Stress distribution

5/23/2012 11:32:34 AM

Three-Dimensional Stresses

Therefore, constants

777

B=C=0

Finally, the Airy’s stress function is φ = ( A / 6) y3. Example 19.11 A bar of circular section of diameter 30 mm is subjected to a pure bending moment of 3 × 105 N mm. What is the maximum bending stress developed in beam? What is Airy’s stress function for this case? Solution Moment of inertia of beam, I =

d4 = 64

× 304 64

= 3.976 × 104 mm4 M = 3 × 105 N mm d = 15 mm 2 Bending stress,

σ xxmax =

30 × 105 × 15 = 113.18 MPa. 3.976 × 104

Airy’s stress function for this case

where

φ=

A 3 y 6

A=

M Bending moment = I Moment of inertia

=

30 × 105 = 7.545 3.976 × 104

A 7.545 = = 1.257. 6 6 Airy’s stress function

f = 1.257y3

where y varies from -15 mm to +15 mm. Exercise 19.10 For a beam of rectangular section B = 20 mm, D = 30 mm, Airy’s stress function f = 1.6y3. Determine the magnitude of constant BM acting on beam section. Problem 19.1 A beam of rectangular section is subjected to shear force F = 1 kN and bending moment = -1 × 106 N mm. Section of beam is B = 20 mm, D = 60 mm. Write down the stress tensor for an element located at 15 mm below the top surface (see Fig. 19.15). Solution

M = -1 × 106 N mm (producing convexity) I = second moment of area about neutral plane

MTPL0259_Chapter 19.indd 777

5/23/2012 11:32:35 AM

778 Chapter 19

20 G

20

Area a 15 30

F 60 mm

22.5

y

Tension N

L

D 30 Compression F side M

M

Figure 19.15 Problem 19.1

=

Figure 19.16

Problem 19.1

BD3 20 × 603 = 12 12

= 36 × 104 mm 4

Bending stress,

y = 15 mm from neutral layer M y σ xx = I =

1 × 106 × 15 36 × 104

= 41.66 N/mm2 . To determine shear stress at y = 15 mm from neutral layer

τ xy =

Fay (Fig. 19.13) Ib

where area, a = 15 × 20 = 300 mm2 = area of section above the layer under consideration. y = 22.5 mm, distance of CG of area a from neutral layer b = Breadth = 20 mm I = 36 × 104 mm4 F = 1 kN = 1,000 N

τ xy =

1, 000 × 300 × 22.5 36 × 104 × 20

= 0.9375 N/mm 2

MTPL0259_Chapter 19.indd 778

5/23/2012 11:32:37 AM

Three-Dimensional Stresses

779

Stress tensor for this state of stress at a layer 15 mm below top surface: 41.66 0.9375 MPa 0.9375 0 Problem 19.2 At a point in a stressed body the Cartesian components of stress are sxx = 80 MPa, syy = 50 MPa, szz = 30 MPa, txy = 30 MPa, tyz = 20 MPa and tzx = 40 MPa. Determine (a) normal and shear stresses on a plane whose normal has the direction cosines 1 2 2 cos (n, x ) = , cos (n, y ) = , cos (n, z ) = , 3 3 3 (b) angle between resultant stress and outer normal n. Solution Components of resultant stress sr are

σ rx = σ xx cos ( n, x ) + τ yx cos ( n, y ) + τ zx cos ( n, z ) σ ry = τ xy cos ( n, x ) + σ yy cos ( n, y ) + τ zy cos ( n, z ) σ rz = τ xz cos ( n, x ) + τ yz cos ( n, z ) + σ zz cos ( n, z ) Substituting the values as above 2 2 220 1 σ rx = 80 × + 30 × + 40 × = = 73.33 MPa 3 3 3 3 1 2 2 σ ry = 30 × + 50 × + 20 × = 56.66 MPa 3 3 3 1 2 2 σ rz = 40 × + 20 × + 30 × = 46.66 MPa 3 3 3 Resultant stress, σ r = σ rx2 + σ ry2 + σ rz2 = (73.33)2 + (56.66)2 + (46.66)2 = 5377.77 + 3211.0035 + 2177.77 = 10766.576 = 103.76 MPa. Normal stress,

σ n = σ rx × cos (n, x ) + σ ry cos (n, y ) + σ rz cos (n, z ) 2 1 2 = 73.33 × + 56.66 × + 46.66 × 3 3 3 = 24.444 + 37.777 + 31.111 = 93.332 MPa.

Shear stress,

τ n = σ r2 − σ n2 = 103.762 − 93.3322 = 10766.576 − 8710.862 = 45.34 MPa.

MTPL0259_Chapter 19.indd 779

5/23/2012 11:32:38 AM

780 Chapter 19

Angle between resultant stress vector sr and normal to the plane n is given by cos(σ r , n) = cos(σ r , x ) cos( n, x ) + cos(σ r , y ) cos( n, y ) + cos(σ r , z ) cos ( n, z ) where

σ rx 73.33 = = 0.7067 |σ r | 103.76 σ ry 50.666 cos (σ r , y ) = = = 0.546 |σ r | 103.76 σ 46.66 cos (σ r , z ) = rz = = 0.4497. |σ r | 103.76 cos (σ r , x ) =

1 2 2 cos (σ r , n) = 0.7067 × + 0.546 × + 0.4497 × 3 3 3 = 0.2355 + 0.364 + 0.2998 = 0.8993

Therefore,

cos−1 0.8993 = 25°56′ . Problem 19.3 A cantilever beam of a rectangular section B × D and of length L carries a concentrated load W at free end as shown in Fig 19.17. Consider a section at a distance x from fixed end and a layer at a distance y from neutral layer zz. Derive an expression for Airy’s stress functions if B = 40 mm, D = 60 mm, W = 1 kN, x = 2 m, and L = 5 m. Write the stress tensor for the layer bc, if y = 20 mm. y

W y

X z

z

x z

x

D

B

x

L a

d G

b

c

y z

y

z

Figure 19.17

MTPL0259_Chapter 19.indd 780

5/23/2012 11:32:39 AM

Three-Dimensional Stresses

781

Solution Stresses,

σ xx =

W ( L − x) y I xx

σ yy = 0  D  −y  W  2  D  × B  − y  y +  τ xy = 2  I zz × B 2        D  +y W D   2 = ×  − y   I zz  2 2    =

 W  D2 − y2  .  2 I zz  4 

Let us assume the Airy’s stress function f = C1y3 + C2xy3 + C3xy. Boundary conditions are as follows: D (1) For y = ± , txy = 0, syy = 0 2 (2) For x = 0, sxx = 0 + D/ 2

(3) Everywhere

∫

τ xy Bdy = W , shear force is constant.

− D/2

Applying these boundary conditions, ∂ 2ϕ σ yy = 2 = 0 , everywhere f satisfies this condition ∂x (τ xy )

So,

or,

3C2

D y=+ 2

= 0, or

∂ 2ϕ = −(3C2 y 2 + C3 ) D y= + ∂ x∂ y 2

D2 + C3 = 0 4 3 C3 = − C2 D 2 4

(19.39)

 ∂2φ  (σ xx )x = L  2  = (6C1 y + 6C2 xy )x = L = 0  ∂y  x = L C1 = −C2 L

MTPL0259_Chapter 19.indd 781

(19.40)

5/23/2012 11:32:41 AM

782 Chapter 19

D/2

∫

 ∂ 2ϕ  − ∫  ∂x∂y  Bdy = W − D/2  D/2

τ xy Bdy =

−D/2

D/2

=

∫

−(3C2 y 2 + C3 )Bdy = W

−D/2 D/2

W = − B 3C2

Moreover,

y3 + C3 y 3 −D/2

  D3 D3   D D  W = − B C2  + + C3  +     2 2  8    8 3  D  = − B  C2 + C3 D 4   −C2 B

or,

D4 − C3 BD = W . 4

(19.41)

Putting the value of C3 = −(3/4) C2D2 in Eq. (19.41) −C2 B

D3 3 + C2 D 2 BD = W 4 4

(19.42)

1 3  C2  BD3 − BD3  = W 4  4 C2 =

then

W ×2 BD3

3 2W 3 W =− C3 = − × D2 × 4 4 BD BD3 C1 = −C2 L =−

2WL . BD3

Finally the Airy’s stress function is

φ=−

2WL 3 2W 3 3 W y + xy − − xy 4 BD BD3 BD3

At the layer where x = 2 m = 2,000 mm, L = 5,000 mm.

MTPL0259_Chapter 19.indd 782

5/23/2012 11:32:42 AM

Three-Dimensional Stresses

783

y = 20 mm BD3 40 × 603 = = 72 × 104 mm 4 12 12 W ( L − x) y σ xx = I xx I zz =

=

τ xy =

1, 000 × 3, 000 × 20 = 83.33 MPa 72 × 104  W  D2 − y2  21zz  4 

=

 1, 000  602 − 202  2 × 72 × 104  4 

=

1, 000 × 500 = 0.347 MPa 144 × 104

Stress tensor =

83.33 0.347 MPa 0.347 0

Problem 19.4 A solid circular shaft of steel is transmitting 100 HP at 200 rpm. Determine the diameter of the shaft if the maximum shear stress in the shaft does not exceed 80 MPa. Write down the stress tensor for the surface of the shaft. Show that surface of the shaft is under both plane stress and plane strain conditions. E = 200 × 103 N/mm2, v = 0.3. Solution

HP transmitted = 100 RPM, N = 200

Angular speed, Torque transmitted,

Torque transmitted,

Maximum shear stress,

MTPL0259_Chapter 19.indd 783

ω=

2π N 2π × 200 = 60 60

= 20.94 rad/s 100 × 746 = 3562.6 Nm 20.94 = 3562.6 × 103 Nmm

T=

τ = 80 MPa π T = d3 × τ 16

5/23/2012 11:32:43 AM

784 Chapter 19 tyx

−80

80

+80 MPa

txy = 80 MPa

80

Figure 19.18 Shear stresses ≡ principal stresses d3 =

Shaft diameter,

16T 16 × 3562.6 × 103 = = 226.8 × 103 mm3 πτ π × 80

d = 60.87 mm On the surface of the shaft,

t = ±80 MPa

Principal stresses on the surface of the shaft are:

σ1 = +80 MPa (a plane stress condition) σ 2 = −80 MPa 80 v80 80 + = (1 + v ) E E E 80 v80 80 ε2 = − − = − (1 + v ) E E E v80 v80 ε3 = − + =0 E E

ε1 =

Principal strains,

Since e3 = 0, this is a plane strain condition, Strains are

MTPL0259_Chapter 19.indd 784

80 (1 + 0.3) = 0.52 × 10−3 200 × 103 80 ε2 = − (1 + 0.3) = −0.52 × 10−3 200 × 103

ε1 =

Stress tensor =

80 0 MPa, 0 −80

Strain tensor =

0.52 × 10−3 0

or

0 +80 +80 0

0 −0.52 × 10−3

5/23/2012 11:32:44 AM

Three-Dimensional Stresses

785

Multiple Choice Questions 1. Principal stresses at a point are 120, -40 and -20 MPa. What is the maximum shear stress at the point? (a) 50 MPa (b) 70 MPa (c) 80 MPa (d) None of these 2. A bar is subjected to an axial load such that its length l is increased by 0.001l. If Poisson’s ratio is 0.3, what is the change in its diameter d? (a) -3 × 10-4d (b) +3 × 10-3d -4 (c) +3 × 10 d (d) None of these 3. Lame’s coefficient for a material are l and m. What is Poisson’s ratio? (a)

λ λ+µ

(b)

λ 2(λ + µ )

(c)

λ 3( λ + µ )

(d) None of these

4. Ratio of volumetric stress/volumetric strain is known as (a) Shear modulus (b) Bulk modulus (c) Young’s modulus (d) None of these 5. Airy’s stress function is φ = 50 x 2 − 40 xy + 80 y 2 φ = 50 x 2 − 40 xy + 80 y 2, what is the normal stress syy?

(a) 100 MPa (c) +160 MPa

(b) 40 MPa (d) 80 MPa

6. A rectangular-section beam of breadth b, depth d is subjected to shear force F. At what depth y from top surface transverse shear stress is the maximum? d (a) y = 0 (b) y = 4 (c) y =

d 3

(d) y =

d 2

7. A shaft is subjected to pure twisting moment M. Surface of the shaft represents (a) Plane strain condition (b) Plane stress condition (c) Both plane stress and plane strain conditions (d) Neither plane stress nor plane strain condition 8. A thin metallic sheet is subjected to a plane shear stress, what is the state of stress of thin sheet? (a) a plane strain state (b) a plane stress state (c) a hydrostatic state of stress (d) None of these

Practice Problems 1. Consider a beam of a rectangular section B = 25 mm, D = 60 mm subjected to a bending moment +1.5 × 106 N mm. Write down (a) the stress tensor for an element located at top surface and (b) the stress tensor for an element located in a plane 15 mm below the top surface. 2. For a material Lame’s coefficients are l = 1.2 × 105 MPa and m = 0.8 × 105 MPa. Determine E, v and G for the material. 3. A cantilever of a rectangular section B × D is of length L as shown in Fig. 19.19. Write down stress tensor to determine state of stress at section XX at a distance of x from free end and at a layer at a distance of y from neutral layer. 4. A cylindrical bar of length L, area of cross-section A is fixed at top end. Write down Airy’s stress function for stress due to self weight in bar, if w is the weight density of the bar (Fig. 19.20). 5. Consider the displacement field S = (y2i + 3yxj) × 10-2. Find whether, strain field is compatible. If yes, find strain components ε x , ε y and γ xy /2 at point (1, -1).

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786 Chapter 19

W y

L

X

x

A

x

y L

Figure 19.19

Figure 19.20

Answers to Exercises Exercise 19.1: σ n = 120 MPa, τ n = 43.20 MPa Exercise 19.2: sn = 117.77 MPa, tn = 58.547 MPa, angle 26°35 Exercise 19.3: +0.070 mm, -0.057 mm, +0.028 mm

3 6 1   Exercise 19.4: 3.011, 1.001, -1.993, 1 3 0.5 × 10−3    3 0.5 1  Exercise 19.5: 4k, -8k, +2k

Exercise 19.7: l = 111,120 MPa, m = 77,220 MPa, K = 162,600 MPa x3 + 2 xy 2 3 Exercise 19.8: strain field is possible y3 v = 6 y + 3 yx 2 + 3 u = 8x +

Exercise 19.9: σ xx = 160 MPa, σ yy = 100 MPa, τ xy = +40 MPa,

σ xx = 160 MPa, σ yy = 100 MPa, τ xy = +40 MPa, a plane stress condition

Exercise 19.6: 7639 N/mm2, 11458 N/mm2

Exercise 19.10: 432 Nm

Answers to Multiple Choice Questions 1. (c) 2. (a) 3. (b)

MTPL0259_Chapter 19.indd 786

4. (b) 5. (a) 6. (d)

7. (c) 8. (b)

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Three-Dimensional Stresses

787

Answers to Practice Problems 1 wy x 2 2

 −100 0   −50 0  1. (a )  ( b)   MPa 0   0  0 0

4. φ =

2. E = 208,000 MPa, v = 0.3, G = 80,000 MPa

5. Compatible

 Axy  3.  2  A  D − y2    2  4 

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 A D − y2  2  4  2

0

  12W  , where A =  BD3   

ε x = −2 × 10−2 , ε y = +3 × 10−2 ,

γ xy 2

= −1.5 × 10−2

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20 Mechanical Properties CHAPTER OBJECTIVES The utility of any material for the performance of an engineering component depends on its mechanical properties such as strength, hardness, ductility, resistance to cyclic loads, and resistance to time-dependent strain, etc. 



The strength of materials under various types of loading, such as tension, compression, twisting and bending has been discussed in this chapter. Behaviour of the materials under impact loading is also an important parameter.



The behaviour of the materials under fatigue and creep loading conditions has been explained in detail.

Introduction For successful designing of any machine components, a designer must know the mechanical properties of the material which are being used. A simple tensile test on a material establishes the basic properties of yield strength (i.e., at which yielding starts or plastic deformation begins), ultimate strength and percentage elongation (for ductility). There are many materials which are much stronger in compression than in tension. Such materials are used in components which are basically used for compressive loading as in the beds of machines. A compression test indicates the ultimate compressive strength of materials such as cast iron and concrete. Wear and tear of any material depends upon the hardness of the material. There are various hardness tests and various hardness numbers, such as Brinell’s Hardness Number (BHN), Vicker’s Pyramid Number (VPN) and Rockwell Hardness Number, which has been discussed in the text. There are various engineering parts which are subjected to cyclic loading over a long period of time, and the behaviour of the materials under such conditions must be ascertained for safety and optimum design as of turbines blades in an aero engine. Yet there are components which are subjected to time-dependent strain over longer periods (for months and years), and the creep behaviour of such components needs careful study and test. As shock loading of any material is the worst type of loading, so the impact testing of a standardized specimen will also be discussed.

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Mechanical Properties

789

Materials Behaviour Under Static Tension Mechanical behaviour of a material, that is, its response or deformations to an applied external load in terms of strength, toughness and ductility is important for a structural or design engineer. A simple tensile test is used to determine yield strength, ultimate strength, ductility and Young’s modulus of a material. Figure 20.1 shows a specimen of standard shape with L0 = gauge length, A0 = area of cross-section subjected to a tensile load, F. As the tensile load on the specimen increases, there is an increase in length, but at the same time, there is a decrease in the diameter of the specimen. If the axial strain dL/L is positive then lateral strain (in a direction perpendicular to axis) dd/d is negative. If a graph is plotted between tensile force F and change in length dL, a straight line portion OA and a curved portion ABC are observed for a ductile material as shown in Fig. 20.2. Collar

A0 F

F (Tensile load)

d

L0

Tensile test specimen

Figure 20.1 B (Ultimate load) •

Engineering stress,

σ=

F A0

δ L L − L0 = L0 L0 Final length − Initial length = Initial length

Engineering strain,

ε=

In the portion OA, the graph between force and change in length is linear and the slope of the curve gives information on Young’s modulus of the material.

F A•

• C (Fracture)

Tensile force

O

dL Change in length

Figure 20.2 F L 0 σ Engineering stress × = = A 0 δ L ε Engineering strain both engineering stress and engineering strain are defined on the basis of initial area of cross-section and initial length respectively. Moreover, the ratio of lateral strain (dd/d) to axial strain (dL/L0) is known as Poisson’s ratio Young’s modulus,

E =

Poisson’s ratio,

v =

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Lateral strain Longitudinal strain or axial strain

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790 Chapter 20

At point B, the maximum load, Fmax gives the value of ultimate tensile strength,

σ ut =

Fmax A0

At point A, the graph becomes, non-linear or yielding has started in the material. Yield strength, σ yp =

Load at yield point A0

Area under the curve OABC provides information on strain energy stored in the specimen up to fracture which defines toughness of the material. Moreover up to A, the graph is linear due to limiting stress at A. This is known as elastic limit stress, which is more pronounced in mild steel, and there is distinction between elastic limit stress and yield stress as shown in Fig. 20.3 for mild steel, in load extension graph, A is the limit of proportionality, B is the elastic limit, C is the upper yield point, D is the lower yield point, E is the ultimate load and F is the fracture load point. If the load applied on a sample is removed at the elastic limit point B, then there will not be any residual deformation in the sample. Whatever the strain energy is absorbed by the sample, the same amount of strain energy is released or the material possesses the property of resilience up to the elastic limit. The graph OABCDEF gives relationship between engineering stress and engineering strain. If a graph is plotted between true stress (taking into account the actual area at that load) and true strain (taking into account the length up to that stage), then the graph will be ABCD′E′F′ as shown in Fig. 20.3 The yielding marked by upper yield point and lower yield point is known as discontinuous yielding.

Non-linear Behaviour For the non-linear behaviour of a material such as copper and rubber, either tangent modulus (slope of TT) or secant modulus (slope of OA) can be used to determine Young’s modulus of elasticity as shown in Fig. 20.4. At the point A, Tangent modulus, Etan = Slope of tangent TT at A. Secant modulus, Esec = Slope of line OA. F E T

E •

F

B• A•

•F (Fracture)

C D

(Tangent)

F

• A Non linear curve

D T

Secant

O

Figure 20.3

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dL

Load-extension curve for mild steel

O

Figure 20.4

dL

Non-linear behaviour

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Mechanical Properties

791

On atomic scale, Young’s modulus is defined as  dF  E= ,  dx  x o

where x is the distance between two atoms, xo is the interatomic distance at equilibrium position of atoms and E is the slope of the curve AA at equilibrium distance xo at point P as shown in Fig. 20.5. On atomic scale, we get theoretical value of Young’s modulus E, which is about 100 times greater than the practical value of E for the same material in bulk form, as the bulk material contains lot of impurities, dislocations and imperfections in crystal structure.

A

F net = Force of attraction – Force of repulsion (between atoms)

F net

x

P

x

x0

A

Atoms

Figure 20.5

Shear Modulus If a rectangular block fixed at one face is subjected to shear stress, t on the opposite face, then the shear strain f is developed in the block as shown in Fig. 20.6. Relation between t and f is linear. Shear stress t t f

t

O

f Shear strain

Figure 20.6

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792 Chapter 20

Table 20.1

Elastic constants of different materials

Material

E (GPa)

G (GPa)

Aluminium

70

26

0.33

Brass

100

39

0.34

Cast iron

105

44

0.25

Copper

110

45

0.35

Nickel

45

17

0.29

Steel

207

77

0.31

Titanium

107

46

0.34

Tungsten

408

162

0.28

Modulus of rigidity,

G=

v, Poisson’s ratio

τ Shear stress = φ Shear strain

E, G and v for common engineering materials are given in Table 20.1.

Yield Strength

A

Yielding is characterized by initial departure from linearity of stress–strain curve. Deformation in any member after yielding has begun is permanent and generally structures are designed only for the elastic deformation. Therefore, it is very important to know the yield point, where there is onset of yielding as shown in Fig. 20.7. Yield point is not clearly defined for some materials, where there is non-linear elastic region. The usual practice is to define some specified amount of strain, that is, 0.002 or 0.2 per cent as shown in Fig. 20.7. A tangent is drawn at origin O on the nonlinear s–e curve. Taking OB = 0.002 strain, a line BA is drawn parallel to the tangent at O. Then stress at Y, YB intersecting the stress–strain curve at Y (i.e., syp) is defined as 0.2 per cent Proof Stress.

syp

Non-linear Y

Stress s

O

B e, Strain 0.002 or 0.2% Strain

Figure 20.7

Stress-strain graph

Ductility The ductility of a material is defined by percentage elongation in the sample up to fracture, % Elongation =

L f − L0 L0

× 100 =

Final length − Initial length ×100 Initial length

For common engineering materials, yielding strength, ultimate tensile strength and percentage elongation are given in Table 20.2

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Mechanical Properties

Table 20.2

793

Mechanical properties of various materials syp(MPa)

Metal

sut (MPa)

% Elongation

Aluminium

35

90

40

Copper

70

200

44

Brass

75

300

65

Iron (Wrought)

132

260

45

Nickel

140

480

40

Steel (mild)

280

400

25

Behaviour of Materials Under Static Compression For a material, stress–strain diagram for tension and compression generally differs. Similarly, the ductility and mode of failure exhibited by a material under tensile and compressive loading also differ. It is in the plastic range after yielding has started that differences between the behaviour under tension and compression are the greatest. Behaviour in the elastic range is important for brittle materials which do not exhibit yielding. In crystalline materials, the elastic action in compression is exactly the same as the elastic action in tension but in the reverse direction. So, the elastic stress–strain curve in compression is a linear extension of that in tension for many materials as shown in Fig. 20.8(a). If the ductile materials are subjected to compression, slipping of atoms on crystallographic planes leads to yielding at a stress approximately the same as the yield stress in tension. This also applies to discontinuous yielding as in the case of mild steel, which has also upper and lower yield points in compression that are usually the same as those for tension. In brittle materials, slip leads to fracture along a single shear plane or a multitude of small failures or shear (or slip) planes in all directions leading to fragmentation: +s

(s yp )T + s

+s

Tension

Tension

Tension Fracture −e

O

+e

−e

+e

O

−e

+e

Mild steel

Cast iron

Compression

−s

(s yp )c

Compression

−s

Compression −s

(a)

(b)

(c)

Figure 20.8

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794 Chapter 20

The axial compressive stress required to cause fracture in a brittle material is much greater than the required tensile stress. In tension, fracture is initiated by stress raisers in the form of cracks, holes and other imperfections even though the stress is well below the stress necessary to cause slip on the 45° shear plane. But, in compression, these imperfections cease to act as stress raisers. Instead, if cracks or holes are present in the material, these tend to close up under compressive force and their effect is reduced. The stress can then reach the larger values needed to initiate slip. Imperfections oriented along the shear planes act as shear stress raisers, but these are far less effective than the stress raisers in tension. So, the strength of a material in compression is often increased in cast iron. Concrete rocks are examples of this effect. The net result is that brittle materials are stronger in compression than in tension. Figure 20.8(c) shows the complete stress–strain diagram for a grey cast iron in tension and in compression. The tensile strength of a typical grey cast iron is 150–160 N/mm2 but its compressive strength is 500–550 N/mm2. The plastic range in compression extends from the end of the +s elastic stage to final fracture. Both the area of cross-section and the strength of material increase with compressive plastic strain, B the former due to Poisson’s effect and the latter due to strain hardA• ening. Therefore, the load which is the product of the area and the stress always increases throughout the plastic range. The plastic range is potentially much larger in compression than in tension. Say, a ductile material has equal strength in tension and in O C compression, that is, (syp)T = (syp)C as shown by the points A and −e +e A′ in Fig. 20.9. A specimen made of such a ductile material is loaded in tension up to the point B and then unloaded (as shown by the loading curve OAB and unloading curve BC). Now the •D specimen is loaded again but in compression, it is observed that A • Bauschinger’s the compressive yield strength has been decreased, that is, from effect the stress at the point A′, now it is reduced to the stress at the point D. This is the well-known Bauschinger’s effect. −s One of the reasons for Bauschinger’s effect is that yielding in a polycrystalline metal is non-uniform. The crystals are oriented at random and when the specimen is loaded in tension, these crys- Figure 20.9 Bauschinger’s effect tals yield by different amounts so that the stress varies slightly from crystal to crystal. When the specimen is unloaded, it contracts until the average stress becomes zero. The crystals that yielded the least do not quite return to zero stress and remain in tension while the crystals that yielded the most go beyond zero stress and are under compression. Therefore, there are microscopic residual stresses throughout the material, some in tension and some in compression. When the material is subjected to compressive load, the crystals that already have residual compressive stresses will yield at a lower than normal stress and so the overall yield stress is σ D < σ A ′.

Behaviour of Materials Under Bending In pure bending, no shear stress is present and only the normal stresses are present across the section. Figure 20.10 shows a beam ABCD carrying loads W each, at a distance a from each support. If the portion BC of the beam is subjected to pure bending, as is obvious from the SF diagram, the shear force is zero along the portion BC and the bending moment is constant and equal to Wa throughout its length of (l − 2a).

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Mechanical Properties

795

W W For convenience, the beam may be thought as it is composed of longitudinal elements of infinitesimal crosssection or fibres, each of which is in a state of simple tenA D B C sion or compression. In Chapter 8, we have studied about the relationship between bending moment, stress, section modulus and radius of curvature, and we have shown that ( − 2a) a a variation of strain along the depth of the section is always Wa Wa linear even when the extreme fibres of the beam go in to B C the plastic stage. +W If the materials are perfectly brittle, the flexure formula can be used all the way up to rupture, σ r = M r y lxx , is S. F. diagram −W called the modulus of rupture, where Mr is the bending moment causing rupture in the beam. Since no material is perfectly brittle, stress sr is never quite equal to the maximum stress in the beam at rupture. Wa However, it is a commonly used property for materials B. M. diagram such as ceramics, cast iron, concrete, wood and brittle plastics even though some of these materials have considFigure 20.10 Bending Moment erable plastic deformation before rupture. diagram Accompanying the change in length of the longitudinal fibres is a lateral strain, just as in simple tension and compression (due to Poisson’s effect). The fibres on the tension side of the beam contract laterally and those on the compression side expand laterally. Consequently, the beam becomes wider on the compression side and narrower on the tension side and hence a transverse curvature is produced in the opposite direction from the longitudinal curvature [as shown in Fig. 20.11(b)]. C

Compression face a

Compression M

b

a

b

Lateral strain

M

c Tension

c

d

d

Tension face

(a)

(b)

Figure 20.11

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796 Chapter 20

Yielding in Pure Bending The atomic mechanism of yielding in pure bending is the same as in simple tension; slip along planes is in the general direction of the maximum shearing stress at 45° with the axis of the beam. When the extreme fibres (those farthest from the neutral axis) reach the strain at which yielding begins in the simple tension, local yielding takes place. As bending continues, yielding progresses gradually inward towards the neutral surface. The stress in each fibre follows the stress–strain relationship for simple tension. Figure 20.12(a) shows the stress–strain diagram for the material of the beam, in simple tension and compression. Figure 20.12(b) shows the stress distribution across the section of the beam just before yielding and Fig. 20.12(c) shows the stress-distribution diagram across the section, after yielding has started in extreme fibres of the beam. −s

+sP

+sy

−e

+sy

+sP

+e

−sy

−sP

−sP

−sy

−s (a)

Figure 20.12

(b)

(c)

(a) Stress-strain diagram in simple tension and compression, (b) Elastic-stress distribution just before yielding and (c) Elastic-plastic-stress distribution after yielding.

Because of the concentration of the maximum stress in the extreme fibres and the support given by the inner fibres, the beam usually does not begin to yield until some higher stresses are reached than are ordinarily observed in tension. When yielding begins at some point, owing to imperfections it forms a small slip band starting at the extreme surface and progressing inward towards the neutral surface in the form of a wedge. This wedge acts like a notch having stress concentration at its tip and the inner fibres, therefore, yield at stresses much lower than the stress at extreme fibres. Final failure in beams made of ductile materials usually involves either excessive deformation or lateral buckling of some kind.

Mild Steel Beams of mild steel are of particular interest because of their wide use as structural members and because of the discontinuous behaviour in yielding of mild steel. After the yielding has progressed some distance from the outer surfaces, the stress distribution has the appearance as shown in Fig. 20.13(a) the

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Mechanical Properties syl

797

syl

Depth of yielding h 2

h

syl

syl

(a)

(b)

Figure 20.13 maximum stress syl (stress at the lower yield point) is approximately constant over the depth of yielding. The limit, as the yielded region approaches the centre of the beam, the stress distribution can be represented by two rectangles as shown in Fig. 20.13(b). This distribution is referred to as the fully plastic condition and the corresponding moment can easily be calculated. So long as the strain hardening does not occur, the bending moment cannot increase beyond this value, which is therefore called the ultimate moment, Mu bh  h bh2  M u =  σ yl  = σ yl  22 4 where σ yl is the lower yield point stress, b is the breadth of the cross-section and h is the depth of the cross-section. The bending moment at which yielding begins, M y = σ yl

lxx h /2

M y = σ yl

bh2 6

Therefore, M u/ M y = 1.5 or the ultimate moment is 50 per cent more than the yield moment. The ratio Mu /My depends upon the shape of the cross-section of the beam, and therefore it is called as shape factor. For circular sections, the value of the shape factor is approximately 1.8. stR M

A

Residual Stresses After a beam has been bent into the plastic range, the removal of the load leaves the beam with internal residual stresses, because the stress–strain diagram for unloading is different than for loading. Figure 20.14 shows the distribution of residual stresses in the beam after unloading, AOB is the stress distribution after the beam has been loaded, producing stresses in the plastic range, that is, beyond the yield point. When the beam is unloaded, the stress distribution for unloading, that is,

MTPL0259_Chapter 20.indd 797

Compression side

sC

A

N

O Compressive stress (residual) Tension side

M

Tensile stress (residual)

A B st

B sCR

Figure 20.14 Residual stresses after complete unloading

5/23/2012 11:31:10 AM

798 Chapter 20

A′OB′ is linear as the strain distribution is always linear across the depth of the section during loading of the beam, so as to satisfy the assumption that plane sections remain plane in pure bending. When the load is completely removed, the moment of stress distribution must be zero, so as to maintain equilibrium. Consequently, the stress across the cross-section is reduced further (i.e., beyond zero) such that the stress in the outer fibres changes sign and produces an opposite moment to balance that of the remaining stress in the inner fibres. Line A′OB′ represents the necessary superimposed linear stress distribution. The result is that in most parts of the unloaded beam, the residual stress is not zero. On the tension side there are residual compressive stresses (s cr) in outer fibres and tensile residual stresses in the interior. On the compression side, there are residual tensile stresses (s tr) in the outer fibres and compressive residual stress in the interior, the net moment of the stress distribution is zero.

Behaviour of Materials Under Torsion Torsion tests are performed on materials to determine properties such as modulus of rigidity, yield strength and modulus of rupture. Parts such as shafts, axles and drills are subjected to torsional loading in service and the torsion test is performed on such full-sized members, otherwise a test specimen is made on which the test is performed. The specimen generally has a circular cross-section and in the elastic range, shear stress varies linearly from zero at the centre to the maximum at the surface. In the case of a thin walled tube, shear stress is nearly uniform over the cross-section of the specimen and it is preferable to use thin-walled tube specimens for the determination of yield strength and modulus of rupture. The torsion-test specimen shown in Fig. 20.15 (a) is gripped in the chucks of a torsion-testing machine. Twisting moment is gradually applied on the twisting head, gripping on one end of the specimen and torque T is measured on the weighing head connected to the other end of the specimen. Angular twist q is measured with the help of a troptometer near one end of the test section with respect to the test section of the specimen at the other end. A torque, T vs q (angular twist), diagram usually obtained for a ductile material is shown in Fig. 20.15(b). The elastic properties in torsion may be obtained by using the torque at the proportional limit or the torque at some offset angle of twist, generally 0.04 rad/m of gauge length, and calculating the shear stress at the twisting moment Typ, using the torsion formula, where T is the torque at the yield point.

keyway

(d)

Twisting moment

T yp

•

T Gauge length = l (a)

Offset Angular twist Q (b)

Figure 20.15 (a) Torsion test specimen and (b) twisting moment versus angular twist graph

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Mechanical Properties

799

Because of the stress gradient across the radius of the solid shaft, the surface fibres are restrained from yielding by the lesser stressed inner fibres, therefore, the first onset of yielding is not readily apparent. The use of a thin-walled tubular specimen minimizes this effect because the shear stress is nearly uniform in the section of tube. However, an ultimate torsional shear strength or modulus of rupture is frequently determined by using Tmax in the torsion formula. Tl Jθ where q = angular twist within the elastic limit corresponding to torque T. Modulus of rigidity,

G=

Polar moment of inertia,

J=

πd4 32

Tmax d × J 2 where d = diameter of solid circular section and l = gauge length of the specimen. Modulus of rupture,

τr =

Torsion Failure Figure 20.16 shows the state of stress at a point on the surface of a circular specimen tested under torsion. The maximum shear stress occurs on two mutually perpendicular planes, parallel and perpendicular to the longitudinal axis XX of the specimen. The principal stresses, p1 and p2, make an angle of 45° with the longitudinal axis and are equal in magnitude to the shear stresses, where p1 is a tensile stress and p2 is an equal compressive stress. Torsion failures are different from tensile failures. Ductile materials fail in tension after considerable elongation and reduction in area, and showing cup and cone type fracture while in torsion a ductile material fails by shear along one of the planes of maximum shear stress. Generally, the plane of fracture is normal to the longitudinal axis as shown in Fig. 20.16(b). A brittle material fails in torsion along a plane perpendicular to the direction of the maximum tensile stress. This plane bisects the angle between the two planes of maximum shear stress and makes an angle of 45° with the longitudinal axis, resulting in a helical fracture [as shown in Fig. 20.16(c)] due tensile principal stress, p1 = +t.

P1

P2

(a)

Figure 20.16

• ••••• ••••• • • •• • •• • •••• •• •

••••••••• •••••••••••• • • ••• ••••••••••• •••••••• ••

X

(b)

••

P1

••

X

P2

(c)

(a) Principal stresses, (b) shear (ductile) failure and (c) tensile (brittle) failure

Behaviour of Materials Under Impact The impact tests are used to study the toughness of a material, that is, the ability of the material to absorb strain energy during plastic deformation. In the static tensile test, the area under the load extension curve gives the strain energy absorbed by the specimen up to breaking. In order to have high toughness, the

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800 Chapter 20

material should possess high strength and large ductility. Brittle materials have low toughness since, they exhibit very small deformation before fracture. The use of such materials in structures or machines is dangerous as the fracture may occur suddenly without any noticeable deformation. In the case of polycrystalline materials, there are two types of fractures: (1) brittle fracture as in the case of cast iron and (2) shear fracture (or the ductile fracture) as in the case of mild steel and aluminium. The strength of the material can be described by two characteristics: (1) resistance of the material to separation and (2) resistance of the material to sliding. If the resistance to sliding is greater than the resistance to separation, the material is brittle and if the resistance to separation is greater than the resistance to sliding, the material is ductile. Three basic factors contribute to a brittle type fracture, that is, (1) a triaxial state of stress, (2) a low temperature and (3) a high strain rate or rapid rate of loading. All the three factors need not to be present at the same time to produce a brittle fracture. Triaxiality of stresses: Figure 20.17 shows a round bar with a groove (or notch) subjected to axial tensile force P. Due to the presence of the groove or notch, the stress at the root of the notch is very high due to the effect of stress concentration. So, maximum stress at the root of the notch depends upon the root-radius. The material in the centre of the bar carrying the tensile load tries to contract laterally (i.e., along the radius) because of Poisson’s effect, but it is hindered by the resistance of the unstrained material. The result is that there are tensile stresses acting radially outward on the inner portion of the material, which produce a state of triaxial tension. The triaxial stresses, sa, sr, and sr, lead to the brittle failure of the material along the notch. Therefore, the impact test on ductile materials is generally performed on bars with a notch, so as to have the effect of triaxiality of stresses. Effect of temperature: Steels are used for building purposes and the notch impact strength of the steel depends on temperature. The energy required for a given notched bar impact test falls rapidly and irregularly once the temperature drops below a critical level and then usually a ductile steel breaks in a brittle manner.

P

Unstrained material

sa

sa

r

sr

r = Notch root radius

Pr P A specimen with a notch of radius r

Figure 20.17

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Triaxiality of stresses

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801

0.01% C

400

I

Impact energy

Transition zone for II

300

0.22% C II

Nm 200

Ductile zone for II

Brittle zone for II

0.43% C III IV

100

O

−200

0

200

Temperature

0.63% C steel

400 F

Figure 20.18 In general, at high temperature, fractures in steel occur with a large deformation and high values of impact energy are obtained. The fracture is fibrous in character. As the temperature drops, the impact energy values fall more or less rapidly within a critical temperature range and brittle fracture occurs (i.e., fracture with a very small deformation). The fracture is granular having crystalline appearance. It can be easily seen that the transition curves flatten out as the carbon content is increased in steel and also the maximum impact energy at which only ductile fracture occurs falls as the carbon content is increased. In Fig. 20.18, three zones for 0.22 per cent carbon steel are shown, that is, brittle zone, transition zone and ductile zone. If the temperature of 0.22 per cent C steel is less than −140°F, brittle fracture occurs and if the temperature is more than 40°F, ductile fracture occurs. Effect of straining rate: The plastic stress–strain curve of a ductile metal is raised by increasing the strain rate. In other words, if a tensile load is applied on a metallic specimen with a very high strain rate, its yield point is increased in comparison to the yield point obtained in static tensile test. This effect is also temperature dependent and is more pronounced near the melting point of the metal. This effect is fairly small at room temperature. For example, increasing the strain rate by a factor of 100 increases the yield stress of copper by only 10 to 15 per cent at room temperature. But at the temperature near the melting point if the strain rate is increased from 10-6 to 10+3 per second, the yield stress is almost doubled. Especially in mild steel, the yield point is subjected to striking variations with strain rate, which is closely associated with the causes of discontinuous yielding. With high strain rates, the stress can reach much higher values before general yielding begins in mild steel. The importance of increased yield strength at higher strain rates lies in its effect on ductility. The result is a decreased ductility and a greater tendency to brittle fracture, so increasing the rate of loading has the same general effect on ductility as increasing the triaxiality of stress. Notch effect: The stress concentration at the root of the notch provides large stress necessary to raise the yield stress and a high local strain rate at the root. Say, Kt = theoretical elastic stress concentration factor and sav = average stress at the section containing notch.

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σ max = Ktσ av σ max σ = Kt av E E

or

ε max = Kt ε av . Differentiating both sides with respect to time, we get

εmax = Kt εav This shows that the local strain rate at the root of the notch is multiplied by the same concentration factor as the stress. If the combined effect of triaxiality, high strain rate, low temperature and stress concentration raises the yield stress above the fracture stress, a crack will form near the root of the notch, which has locally yielded. The immediate effect of crack formation is a sharp local increase in strain rate which further increases the yield stress and brittle fracture continues and the crack rapidly runs through the material. Notch sensitivity: The tendency of a ductile material to behave in a brittle manner in the presence of a notch is called notch sensitivity. This property also depends on strain rate, triaxiality and temperature. The effect of notch sensitivity is obtained by plotting a curve between impact energy and temperature for a notched bar impact test, keeping triaxiality constant by using a standard notch for all specimens and keeping strain rate constant at some high value by standard impact loading. The high overall strain rate multiplied by the stress concentration factor of the notch produces local strain rates as high as 103 mm/mm/s. Notch sensitivity is measured partly by the sharpness of the transition in the fracture energy or impact energy versus temperature curve as shown in Fig. 20.19 for a low carbon steel. The sharper the transition, the more notch sensitive is the material. In the case of low carbon steel, the transition is so abrupt that a single temperature T defines it. Notched bar impact test: This is a standard test on Impact notch sensitivity combining all the three factors, that is, energy triaxiality (notch), high strain rate (pendulum) and temU perature. High temperatures up to 2,000°F are obtained in ovens/furnaces. Low temperatures are obtained by (1) forced air circulation over dry ice (−109°F), (2) liquid Low carbon nitrogen (−319°F) and (3) liquid hydrogen (−423°F). steel The pendulum of the impact testing machine must be carefully constructed with the striking edge at its centre of percussion to minimize vibrations. In the case of Charpy impact test, standard specimen with a notch in the centre is supported like a beam loaded at the centre as shown in Fig. 20.20(a). The notch Temperature, T is on the tension side. While in the case of Izod impact test, the specimen is fixed as a cantilever loaded at the Figure 20.19 end [Fig 20.20(b)].

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Striking edge Specimen 10 × 10 × 55 mm

28 mm

22 mm

Notch 45

Direction of blow

45 40 mm

8 mm

2 mm Specimen 10 × 10 × 75 mm

10 mm

Figure 20.20

Charpy impact test

Cross-section at notch

Izod impact test

(a)

(b)

(c)

(a) Charpy impact test, (b) cross-section at notch and (c) izod impact test

The Charpy test has two advantages over the Izod test, that is, (a) it is easier to place the specimen in the machine, an important consideration in low temperature tests when the test must be performed within a few seconds after removing the specimen from a low temperature bath and (b) it is also free from compressive stresses around the notch, which are produced in the Izod specimen by the vice, when we consider the complexity of the stress distribution introduced by the notch itself. When the notched bar impact test is used to compare the notch sensitivities of materials, the significant information is obtained by simple tabulation of comparative impact energy values.

Hardness Hardness of a material is defined as resistance of the material to (a) plastic deformation, (b) wear and tear, (c) scratch, (d) abrasion, etc. It is the most important property of a material because wear and tear or life of a material depends on its hardness. Generally, it is measured by localized plastic deformation caused by an indenter (of hardened steel or diamond) pressed onto the surface of the material under a specified load. A German Scientist employed the criterion of scratch and qualitatively decided hardness on the Mohs scale of hardness ranging from 1 (for soft material) to 10 (for diamond). But Brinell and Vicker have developed quantitative measurement of hardness by using a hardened steel ball indenter and a diamond indenter (of rhombus pyramid shape), respectively. Hardness is measured by taking compressive load in kilograms-force (kgf) and area of impression in square millimetre (mm2). Measurement and calculation of area of indented portion takes same time, Rockwell has devised hardness measurement procedure by considering the depth of resulting indentation, and related the hardness to the depth of indentation, taking minimum time in hardness measurement.

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Brinell’s Number Indentors are made in various shapes such as spheres, cones and pyramids. The area over which the force acts increases with the depth of penetration. Figure 20.21(a)–(d) show the indentations produced on surfaces of sample by conical, square pyramid, knoop and ball indentors. Around the indentation produced by a ball, the stress distribution is highly complex, because the material is forced outward from the region of indentation; triaxial stresses are produced in specimen which vary from the centre to the edge of the indentation. Friction between ball and specimen produces hydrostatic stresses. In the case of pyramid indentors, the sharp corners produce even more complex stress conditions. P P

P

Cone Square pyramid

Knoop Indentor

Indent Indent Indent

(c)

(b)

(a) P

P

P

Ball

e d

Knoop Indentor

D

Indent Indent

Indent

(c)

(b) d (d)

Figure 20.21

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Pyramid Hardness Diamond pieces are ground in the shape of square or rhombus pyramids. =

Hardness Number

P A

Load, P = ld2, where l is a constant, d = diagonal of square or rhombus indent A = bd2, where b is another constant

λd2 λ = βd 2 β Hardness number is l/b, independent of both the load and size of identation. It is easier to measure the diagonal of a pyramid indent due to sharp edges than to measure the diameter of a circular impression. In the case of VPN, the angle between the opposite faces of the pyramid is 136°. Surface of contact area between indentor and impression, H=

Hardness number,

A=

d2 2 sin

VPN =

α 2

, where α = 136°

P P α × 2 sin = 1.8544 2 d2 2 d

P is load in kg and diagonal in mm. Knoop indentor is developed especially to study the microhardness, that is, the hardness of a microscopic area at selected site as in the individual metallic grain. Knoop hardness number is computed from the projected area of the impression. P = Knoop Hardness 0.0708d 2 where P = Applied load in kilogram, d = Long diagonal of impression in millimetre. Brinell’s hardness number J.A. Brinell used hardened steel ball to determine the hardness of the metals BHN = Area =

P Load in kg = Area Area of indentation 2P

π D ( D − D2 − d 2 )

where D = diameter of the ball in millimetre and d = diameter of the indentation in millimetre. BHN is dependent on the load used. For this reason, it is necessary to use the same load for all measurements with a given ball if a comparison of hardness of different metals is to be made.

Rockwell Hardness Test This method is used to determine the hardness of a wide range of materials. Rockwell Hardness is measured by the use of a hardened steel ball (1.59 mm diameter) or a cone shaped diamond indentor (120° cone angle). In this test, depth of impression is measured in place of diameter or diagonal of the impression. For the metallic specimen, three tests are used as:

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Rockwell A—For case hardened materials and thin metals such as razor blades. Rockwell B—For soft or medium hard metals such as mild steel, brass and copper. Rockwell C—For hard metals such as high speed steel, high carbon steel and tool steels. For Rockwell A and C, diamond indenter and for Rockwell B, steel ball indentor are used. A minor load of 10 kg is applied initially on the specimen through the indentor to overcome the thin oxide film on the metal. Then additional load of 50, 90 and 140 kg is applied on the indentor in the case of Rockwell A, B or C, respectively. t Rockwell Hardness number = H − 0.002 where H = a constant depending upon the scale = 130 for Rockwell B = 100 for Rockwell A and C t = depth of indent in mm and 0.002 mm corresponds to one unit of hardness number.

Mechanism of Indent Formation When the indentor is pressed onto the surface of a metal under a static compressive load, a large amount of plastic deformation takes place under the indentor. The deformed material flows out in all directions. The region affected extends to a distance of about three times the radius of indentation, surface surrounding the Ridging Sinking (a) (b) impression bulges out slightly to account for the volume of the metal displaced under the indentor. In some Figure 20.22 (a) Ridging and cases ridging occurs while in some other cases sinking (b) sinking occurs as shown in Fig. 20.22. In the case of ridging type impression, the diameter of the indentation is greater than the true value, where as with sinking type impression, the diameter of the impression is slightly less than the true value. Plastic deformation during indentation is accompanied by large amount of transient creep which is time dependent. For the harder materials such as iron and steel, the time required to reach the maximum deformation is nearly 15 s. Soft materials such as magnesium may require longer time, sometimes 2 min.

Rebound Hardness Hardness is sometimes measured by dropping a hard object on the surface of a specimen and observing the height of its rebound. Usually a diamond point at the end of a small cylindrical piece is used to strike the surface. As it falls, its potential energy is converted into kinetic energy and a part of this kinetic energy is dissipated in producing plastic deformation of surface and rest is stored in recoverable elastic energy. The amount of strain energy absorbed in specimen depends upon its yield point, stiffness and damping capacity. All the elastic strain energy is not recovered in the form of rebound of indentor due to the internal friction of the material. So, the rebound hardness measures a combination of hardness, stiffness and damping capacity of the material. In Shore Seleroscope test, a pointed hammer is allowed to fall from a height of 254 mm within a glass tube, which has a graduated scale inscribed on it. The standard hammer is approximately 6.35 mm in

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diameter, 19 mm long and 2.4 gm weight with a diamond tip of radius 0.25 mm. The scale is graduated in 140 divisions. A rebound of 100 is approximately equivalent to the hardness of a martensitic high carbon steel. Diamond is the hardest material and in terms of kg/mm2, its hardness is extrapolated to 8,000 BHN as shown in Table 20.3. Table 20.3

BHN of various materials

Material

BHN

Material

BHN

Diamond

8,000

Silica

820

Cubic boron nitride

5,000

Hardened tool steel

740

Silicon carbide

2,800

Annealed low carbon steel

80

Aluminium oxide

2,100

Annealed copper

40

Tungsten carbide

2,000

Superficial Hardness Test In this test, the minor load is reduced from 10 to 3 kg and indentors are of diamond,1 16 ″ diameter ball and 1 8 ″ diameter ball with major loads equal to 15, 30, and 45 kg applied as follows: Example 60 HR 30 W, indicates a superficial Rockwell Hardness of 60 on 30 W scale using 1 8″ ball ball indenter. An empirical relation exists between ultimate strength and BHN for most steels as follows: sut (MPa) = 3.45 × BHN of steel. Table 20.4

Minor and major loads for superficial hardness

Scale symbol

Indentor

Minor load (kg)

Major load (kg)

15 N

Diamond

3

15

30 N

30

45 N

45

15 T 30 T

1 ″ 16

45 T

ball

15 W 30 W

1 ″ 8

45 W

ball

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3

15 30 45

3

15 30 45

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808 Chapter 20

Fatigue Failure due to fatigue is characterized by cyclic stresses or fluctuating loads on structures such as bridges, aircraft, turbine blades, springs and machine components such as shafts, bearings, etc. Failure due to cyclic stresses occurs at a stress level much lower than the tensile strength or yield strength under static load. About 90 per cent of failures in metallic components causes due to fatigue fracture. The fracture due to fatigue is sudden and catastrophic and even a ductile material under fatigue loading behaves in a brittle manner. Fatigue fracture occurs in three stages as: 1. Nucleation of a fine crack on atomic scale. 2. Propagation and growth of crack under continued cycles of stresses. 3. Sudden fracture, generally in a direction perpendicular to applied stress.

Cyclic Stresses The applied stress may be axial (tension and compression), bending (completely reversed) or shear (torsional, twisting). In general, there are three different fluctuating stress versus time cycles possible (Fig. 20.23) in which mean stress,

Stress

Stress smax Tension (+)

sm

O

s min

O

Time, t Compression

(−) Random stress cycle

Stress

(+) σm

O (−)

Completely reversed cycle

Figure 20.23 Different types of stress cycles

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Mechanical Properties

σm =

809

σ max + σ min maximum stress + minimum stress = 2 2

σ r = σ max σ min

Stress range,

R=

Stress ratio,

σ max σ min

σ max − σ min 2 Random stress cycles are observed in aircraft wings during landing on the ground? Completely reversed cycles produce maximum fracture effect. Therefore, in laboratory, standard tests are performed on samples subjected to rotating and bending as shown in Figs. 20.24 (a) and (b). In Fig. 20.24(a), sample is loaded as a cantilever produces bending moment WL at critical section. If d is diameter of the sample, then the maximum bending stress, Stress amplitude,

σa =

32WL π d3 Cycle of stress as a particular point on the surface of the sample is also shown. Figure 20.24 (b) shows a sample mounted as simply supported beam and load W is applied through bearings on Sections A and B, the bending moment on beam is constant between Sections A and B and is equal to Wa. Bending Wa × 32 stress, smax on surface = π d3

σ max =

Specimen



Ball bearing

smax w

d 

Stress cycle

L W

Weight BB

BB 

 d



 A

B

a

a L

Weight

Figure 20.24

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(a) Cantilever specimen and (b) pure bending sample

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810 Chapter 20

At a particular load, the specimens are allowed to rotate at fre- Stress quency f and number of the revolutions up to fracture of sample is S noted on a revolution counter provided on the machine. For different stress levels, fatigue tests are performed and numbers of cycles required for fracture of sample are also noted down. Then a 32Wa 32Wa graph between stress amplitude, S = and a number or π d3 π d3 of cycles required for fracture at each stress level are plotted as shown Se in Fig. 20.25. Higher the magnitude of the stress, smaller the number of cycles and the material is capable of sustaining before failure. For N cycles some ferrous (iron alloys) and titanium alloys S–N curve becomes Number of cycles to failure S -N curve horizontal at higher values of N, then limiting stress is called fatigue limit or endurance limit, Se. Figure 20.25 S–N cycle The fatigue limit represents the largest value of fluctuating stress that will not cause fracture for essentially an infinite number of cycles. For many steels fatigue limit ranges between 35 and 60 per cent of the tensile strength. Most non-ferrous alloys (e.g., copper, aluminium, magnesium) do not have any fatigue limit. But the S–N curve continues its downward trend at increasingly greater N-values. For such materials, fatigue strength at a particular value of N, is defined as stress level at which failure will occur at some specified number of cycles. There always exists a considerable scatter in fatigue data, that is, variation in measured N-values for a number of specimens tested at the same stress level, due to number of factors as fabrication and surface preparation, metallurgical variables, specimen alignment etc. Several statistical techniques have been developed to specify fatigue life and fatigue limit in terms of probabilities. Generally these values are taken for 90 per cent survivals. (At a particular stress, a number of specimens are tested; N is that value of number of cycles at which 90 per cent of the total specimens tested have survived.) Final rapid failure may be either ductile or brittle. Generally, ductile materials show plastic deformation at final fracture.

Factors Affecting Fatigue Life Fatigue behaviour of materials is highly sensitive to following variables: (a) Mean stress, (b) Surface effects and (c) Design details. Dependence of fatigue life on stress amplitude is shown in Fig. 20.26. Increasing the mean stress Sa level leads to a decrease in fatigue life (Fig. 20.26). sm > sm > sm For many common loading conditions, the maximum stress within a component or a structure occurs at its surface. So, most cracks leading to fatigue sm failure originate at surface positions, specially at sm stress raiser sites. Fatigue life is sensitive to surface sm finish on component surface. Various surface treatStress ments lead to improvement in fatigue life. amplitude Fatigue life Sa Design details of a component have significant (Log N) influence on fatigue behaviour. Any notch (as keyway) or geometrical discontinuity (splines) acts Figure 20.26 Stress amplitude versus as stress raiser. Sharper the discontinuity, more fatigue life curves 3

2

1

1

2

3

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severe is the stress concentration. Fillet radius in a shaft is provided to reduce stress concentration at stepped section of shaft. Fatigue behaviour is classified into two types, that is 1. High loads—low cycle fatigue, producing not only elastic strain, but plastic strain also, number of cycles are less, 103–105 cycles. 2. Low loads—high cycle fatigue, wherein deformations are totally elastic, resulting in much longer lives.

Crack Initiation and Propagation The process of fatigue failure is characterized by three distinct stages: (1) crack initiation—a very fine crack forms at some point of high-stress concentration, (2) crock propagation—during which this crack advances incrementally at each stress level and (3) final fracture—which occurs very rapidly when the advancing crack has reached a critical size. Fatigue life, Nf = Ni + Np = Number of cycles to crack initiation +number of cycles to crack propagation. At low stress levels (high cycle fatigue) a large fraction of fatigue life is utilized in crack initiation. With increasing stress level, Ni decreases and cracks form more rapidly. Crack nucleation sites include surface scratches, sharp fillets, keyways, threads, dents, etc. In addition, cyclic loading can produce microscopic discontinuities resulting from dislocation slip steps which may also act as stress raisers at crack initiation sites. Once a stable crack is initiated, it then propagates very slowly and in polycrystalline metals, along crystallographic planes of high shear stress, this is first stage crack propagation. In the second propagation stage, crack propagation rate increases dramatically and direction of crack propagation roughly becomes perpendicular to applied tensile stress. During this stage of crack propagation, crack growth proceeds by a repetitive plastic blunting and sharpening process of the crack tip.

Surface Treatments During machining operations of components, due to tool action, scratches are invariably introduced on surface of components. These surface scratches reduce the fatigue life. Treatments like polishing of surface enhances the fatigue performance life of the components significantly. Another effective method of increasing fatigue performance is by imposing residual compressive stresses within a thin outer surface layer. Residual compressive stresses are Sa introduced into ductile metals by shot peening, which causes localized plastic deformation in outer layer of surface. Small Shot peening hard particles (shots) of diameters 0.1 to 1.0 mm are projected at high velocities on the surface to be treated. The resulting deformations induce compressive stresses to a depth of about one quarter and one half of shot diameter. Figure 20.27 shows Normal improvement in fatigue performance due to shot peening.

Case Hardening Case hardening enhances both surface hardness and fatigue life. In the case of carburising or nitriding process, a component is

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Stress amplitude

Figure 20.27

Log (N ) Fatigue life

Fatigue life

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812 Chapter 20

exposed to carbonaceous or nitrogenous atmosphere at an elevated temperature. A carbon or nitrogen rich outer surface layer (or case) is produced by atomic diffusion of carbon or nitrogen from the gaseous phase. Carbonaceous layer can be as thick as 1 mm and nitrogenous layer can be 0.2 mm thick.

Environmental Effects If fatigue loading takes place in corrosive atmosphere, chemical attack during fatigue process is termed as corrosion fatigue and fatigue life is drastically reduced.

Creep ASTM defines creep as ‘Time dependent strain under D constant stress’. Many materials exposed to ele- Creep vated temperatures are subjected to static mechani- strain e III cal stress, for example, turbine rotors in jet engines II C and steam generators which experience centrifugal stresses and high pressure steam or gases. DeformaI B tion under such circumstances is termed as creep. A Secondary Generally the creep phenomenon is predominant in ei a material, when operating temperature is more than 0.4 Tm, where Tm is the melting temperature of the O Time, t material. Amorphous polymers, plastics and rubbers Constant load creep behaviour are highly sensitive to creep deformation. I - Primary creep II - Secondary creep A typical creep test consists of subjecting a speciIII - Tertary creep men to a constant load or a constant stress while maintaining the temperature constant. Deformation (or Figure 20.28 Constant load creep strain) versus time is plotted as shown in Fig. 20.28. behaviour Most tests are of constant load type, which yield information of an engineering nature. But constant stress tests are employed to provide a better understanding of mechanism of creep. (In constant stress as extension continues, area of cross-section of sample continuously decreases and a mechanism provides reduction in applied load, producing constant stress in sample.) Figure 20.28 shows constant load creep behaviour: 1. An instantaneous deformation or strain, ei mostly elastic (OA) 2. A continuously decreasing creep rate—primary creep (AB) deformation becomes more and more difficult as the material is stressed. 3. Steady-state creep—rate is constant (BC), secondary creep—plot becomes linear. This creep is of the longest duration. During secondary creep, creep rate remains constant due to balance between the process of thermal softening due to temperature and strain hardening (due to plastic strain). In the tertiary stage, strain rate accelerates and final fracture takes place resulting from microstructural changes and grain boundary separation, formation of internal cracks, cavities and voids. Also under tensile load a neck may form at some point in the sample. All these factors lead to a decrease in the effective cross-sectional area and increase in strain rate.

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Mechanical Properties

For metallic materials, most creep tests are performed in uniaxial tension. For brittle materials, uniaxial compression tests are performed, on specimens having l/d (aspect ratio) ratio equal to 2 to 4. Important property from a creep test is creep rate, ε = ∆ε /∆t, slope of secondary portion of creep curve. It is a useful parameter in engineering design, for long life operations of components in nuclear reactor power plant operating for several decades. Many short-life creep tests are performed as on turbine blades in military aircraft, etc.

813

s 3 or T3 s 2 or T 2 Creep Strain

s1 or T 1

e s T < 1.4 Tm

Time, t

Figure 20.29

Stress and Temperature Effects

Both the temperature and stress level influence the creep behaviour. At temperatures considerably below 0.4 Tm, after initial instantaneous deformation, strain is virtually independent of time. With either increasing stress or increasing temperature, following will be noticed: 1. Instantaneous strain (at the time of stress application) increases, 2. Steady-state creep rate is increased and 3. Rupture life time is diminished. For some alloys and for over relatively large stress ranges, non-linear behaviour is observed. Steady-state creep rates,

εs = K1σ n where K1 is a material constant and n another material constant, neglecting the change in temperature  Q  εs = K2σ n exp  − a  , if temperature influence is included, K2 is constant, Qa, activation energy of creep.  RT 

Alloys for High-temperature Use Many factors as melting temperature, elastic modulus and grain size affect the creep behaviour of metals. In general, higher the melting temperature, greater the elastic modulus and larger the grain size, better is the resistance of metals to creep. Relative to grain size, smaller grains persist more grain boundary sliding, resulting in higher creep rate. Stainless steels, refractory metals, superalloys are especially resistant to creep. The creep resistance of the Cobalt and Nickel superalloys is enhanced by solid solution alloying and also by the addition of a dispersed phase which is virtually insoluble in matrix.

Stress Relaxation Bolts and fasteners required to hold two or more rigid plates in tight contact are frequently found to have relaxed considerably after long periods of time as a result of creep. This is called stress relaxation and defined as the time dependent decrease in stress in a member which is constrained to a certain fixed deformation.

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814 Chapter 20

Plates Stress s

Nut Bolt Stress relaxation

Time, t

Figure 20.30 Let us consider that two plates are joined by a bolt and nut and, ei = initial strain in bolt. If this initial strain is maintained constant, the strain caused by creep is simply subtracted from it, thereby reducing the elastic part of the total strain, ei: Elastic strain at any time, eel = ei−ecr (creep strain) The stress due to elastic strain eel, that is, eel × E goes on decreasing with time as shown in Fig. 20.30.

Review Questions 1. Differentiate between: (a) Elastic strain and plastic strain. (b) Tangent modulus and secant modulus. (c) Yield strength and 0.2 per cent proof stress. 2. Explain Bauschinger’s effect. 3. Draw stress–strain curve for cast iron in tension and in compression. 4. Mild steel and cast iron are tested up to destruction in tension. Compare their fractured surfaces. 5. Explain ductile to brittle transition of a material under impact loading. 6. Explain the purpose of a notch in a sample of impact testing. 7. Describe various types of indentors used in Hardness testing. 8. What is the mechanism of indentation when Brinell Ball is used for hardness measurement? 9. What are the three stages of fatigue failure? 10. Explain how a submicroscopic crack is initiated during fatigue loading of a component? 11. What is the difference between fatigue strength and fatigue limit? 12. Explain the methods used for improving fatigue strength of a material. 13. Explain the temperature dependence of creep strain–time behaviour.

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Multiple Choice Questions 1. The most important reason for Bauschinger’s effect in ductile materials is (a) Ductile materials weakness in shear (b) Compressive residual stress (c) Tensile residual stress (d) None of these 2. The notch angle in the Izod impact test specimen is (a) 25° (b) 30° (c) 45° (d) None of these 3. Length of specimen fixed in vice in Izod Impact test is (a) 40 mm (b) 45 mm (c) 47 mm (d) 50 mm 4. The angle between the opposite faces of the diamond pyramid in the case of Vicker’s Pyramid. Hardness Test is (a) 118° (b) 120° (c) 136° (d) 140° 5. Which indentor is used for Microhardness test? (a) Hardened Brinell ball (b) Vicker’s Diamond Pyramid (c) Conical shaped diamond (d) Knoop Indentor 6. The process which does not improve the fatigue strength of a material is (a) Shot peening of the surface (b) Cold rolling of surface (c) Electroplating of the surfaceg (d) None of these

7. The depth of penetration of the hardened steel ball in specimen is 0.140 mm. Rockwell Hardness of material is (a) 60 (b) 65 (c) 70 (d) 75 8. What is Poisson’s ratio of Nickel? (a) 0.33 (b) 0.31 (c) 0.30 (d) 0.28 9. Which of the following materials is remarkably brittle? (a) Polymers (b) Ceramics (c) Bronzes (d) None of these 10. Which of the following materials fail in tension by making a necking to a point? (a) Mild steel (b) Wrought iron (c) Lead (d) Aluminium 11. Which of the following statements is incorrect? (a) Fatigue limit exists for most ferrous alloys (b) For many steels, fatigue limit ranges from 35 to 60 per cent of tensile strength (c) Increasing the mean stress level leads to increase in fatigue limit (d) None of these 12. Which of the following creep is of longest duration? (a) Primary Creep (b) Secondary Creep (c) Tertiary creep (d) All the above have equal duration

Answers to Multiple Choice Questions 1. 2. 3. 4.

(b) (c) (c) (c)

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5. 6. 7. 8.

(d) (c) (a) (b)

9. 10. 11. 12.

(b) (c) (c) (b)

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21 Material Testing CHAPTER OBJECTIVES The objectives of this chapter to make students learn about conducting various laboratory tests so as to determine important mechanical properties of materials such as tensile strength, compressive strength, shear strength, ductility, modulus of elasticity, hardness, impact strength, torsional shear strength and stiffness. Different technical institutions have different types of testing machines; therefore, discussion on apparatus and features of the testing machines are beyond the scope this book. However, attachments used to perform a particular test on a machine has been discussed. Many reputed Indian institutions use devices, such as tensometer (a machine used to perform about a dozen of mechanical tests) and universal testing machines (UTMs), so brief description on both has been made. Most institutions use devices, such as simple UTM, torsion testing machine, hardness tester, Charpy/Izod impact-testing machine and spring testing apparatus. Therefore, in all 16 simple tests, which are performed at UG level has been explained in this chapter.

Tensile Test on a Specimen Using a Tensometer A tensometer by the trade name Hounsfield-Tensometer marketed by a UK company, is a versatile apparatus used as small UTM on which smaller test pieces are employed to derive data on mechanical properties of materials. Mild steel is the most common engineering material used in 90 per cent of engineering applications. Figure 21.1 shows a sketch of the tensile test specimen used on a tensometer and Fig. 21.2 shows a sketch of grips used for holding the sample.

Specimen

Collars

Hole for pin

d l

Figure 21.1

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Tensile test specimen

Grips

Figure 21.2

Grips

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Material Testing

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Depending on the size of samples and maximum force required for testing a sample, machine is equipped 90 80 70 60 50 40 30 Adjusting with deflection plates of various capacities between 20 10 0 screw 300–20,000 N. Different deflection plates and load Zero scales are adjusted on the tensometer machine. Set Before holding the specimen in grips, it is adjusted to zero reading in percentage reduction in area gauge Figure 21.3 Percentage reduction in and percentage elongation gauge as shown in Figs 21.3 area gauge and 21.4, respectively. Initially, the original sample is placed at zero reading and the screw is adjusted in the case of percentage reduction in area gauge. In the case of percentage elongation gauge, the slider is Main adjusted at the gauge length reading on the main scale. KeepScrew Scale ing the pointer at zero, the screw is tightened. Slider Now, the test piece is gripped in two circular halves (with cavity for the collar end of the specimen) and the collars are Slider fitted in the grips. The grips are further fitted in the loading Test piece sockets provided in between the spring plate and the lead Gage Pointer length screw with the help of pins as shown in Fig. 21.5. Before starting to extend the specimen, a graph paper is wrapped on the drum with a rubber lining. Proper magnifica100 50 0 tion of extension is selected (The extension of the sample is magnified). On the tensometer, a magnification of 4, 8 or 16 Figure 21.4 Percentage can be achieved. However, for best results, a magnification of elongation gauge 16 should be selected. Now, the handle is moved slowly (with the Test piece help of worm and wheel arrangement, a heavy Pin Spring plate Lead screw reduction of handle movement is obtained). S = Dispiacement Motion is transmitted to the lead screw and the Pull. P specimen is pulled as shown in Fig. 21.5. As the specimen is stretched, the spring dl = S − d = Change in length d Collar plate also gets deflected, and this deflection is of test piece added to the extension in the test piece when Figure 21.5 Fitting of test piece in between the graph between the load on the specimen the spring plate and the lead and its extension is plotted. However, the screw deflection d in the spring plate is linearly proportional to the pull P in the lead screw. Figure 21.6 shows the mechanism of converting the deflection in spring plate, or in other words, the pull P in the lead screw into the length of mercury thread in capillary. When the spring plate is deflected, it will push a lever hinged at one end as shown in Fig. 21.6. In turn, the lever pushes a piston inside the mercury chamber, and mercury is pushed out of it. The length of the mercury thread in a capillary tube is proportional to deflection of the spring plate or the pull P on the sample. As the lead screw moves, its movement is magnified and the drum with the graph rotates. Now, the marker is moved along the mercury thread and is coincided with the end of the mercury thread. Then, the end of the marker is pressed on the graph paper and a pin prick is obtained on the graph paper. The handle is moved slowly and with each movement, the drum rotates, the marker is adjusted with the load scale and a pin prick is obtained on the graph. In the same way, the test piece is tested up to its

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818 Chapter 21 Spring plate defection

P

Mercury thread in glass

Load scale Piston Mercury

Lever

Mercury chamber

Hinge Mechanism of measurement pull ´P

Figure 21.6

Mechanism of measurement pull P

breaking point. The mild steel specimen breaks with a S cup-and-cone fracture. Broken pieces of the test piece D are taken out of grips and joined together and the neck • • • • • • • B • is inserted in the percentage reduction in area gauge • • •E Pp • to the value of percentage reduction learn at the neck A P • • • C area. Then, the two pieces are adjusted in the groove Load of percentage elongation gauge and reading of percentage elongation is taken on the scale. Now, the graph is taken out from the drum and can be analysed for various properties as shown in Fig. 21.7. In the recently developed tensometers, the mercury 0.2% strain elongation chamber has been replaced by a load cell and an autoO matic load-extension graph is plotted. From this load•K Extension extension graph on mild steel the following loads can dl d l p be noted: PA = load at elastic limit Figure 21.7 Load extension curve for PB = load at upper yield point mild steel PC = load at lower yield point PD = maximum load PE = breaking load Say, d = diameter of the specimen, noted down initially. π A = d 2 , area of cross-section of specimen 4 a = area of cross-section at the neck Note that the area at the neck is calculated, if the percentage reduction in this area is noted down with the help of gauge. P Now elastic limit stress, σe = A A

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Material Testing

Upper yield point stress,

σ uyp =

Stress at lower yield point,

σ lyp =

Ultimate tensile stress,

σ ult =

Nominal breaking stress,

=

Actual breaking stress,

=

819

PB a PC A PD A PE A PE Breaking load = Area at neck a

0.2 × l = OK 100 on a graph and draw a line parallel to straight line portion OA, where l is the gauge length. Line from K cuts the load extension curve at P. Note down load at P, P 0.2 per cent, proof stress = P . A Precautions 1. Move the handle only in one direction during loading of the sample. 2. Apply the load slowly and gradually. 3. Move the handle to remove the initial gap of the specimen in the grips, and when it is just tight and the mercury has just started moving in the capillary, adjust the zero reading of the thread on the load scale. 4. At the maximum load, a neck is formed in the sample, after this stage, move the handle slowly and in smaller steps and keep marking the pinpricks on the graph. 5. Choose a proper spring plate and proper load scale. Many a times, it is desired to have 0.2 per cent proof stress for ductile materials. Take δlp =

Shear Punching Test on Sheet Sample Sheets of various metals, such as aluminium, brass, mild steel and copper, are available in the market. The ultimate shear strength of the material of the sheet can be obtained by punching a small hole in sheet and noting down the force required to punch the hole. Again this test can be easily performed on a tensometerlly by using an attachment, a punch holder and a die as shown in Fig. 21.8. The diameter of the punch is noted with the help of a micrometer. The punch and the die are fitted in the axial holes provided in the compression attachment. Now, the metallic sheet is taken and its thickness is measured with the help of micrometer. The sheet is placed between the punch and the die, and the handle of the tensometer is rotated till the punch just touches the sheet and a slight load is recorded on the scale. The zero reading of the mercury thread is adjusted and the load is slowly and gradually applied by moving the handle and is continuously recorded so as to note down the maximum load. At the maximum load, the sheet is pierced with a punch and the resistance of the sheet falls and the load is reduced on scale. At this stage, reverse the movement of the handle and take out the sheet with the punch and the die and the punch set.

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820 Chapter 21 Punch Hole

Die Compression attachment (tensometer)

Punch holder

Figure 21.8 Say,

d = diameter of the punch t = thickness of the sheet A = area under shear = pdt P = maximum load required to punch the hole τ ult = ultimate shearing strength of the material = P/pdt

For any sheet, make four or five punches, record the maximum load and calculate the mean value of the ultimate shearing strength. Precautions 1. 2. 3. 4.

When the punch just touches the sheet, adjust the zero of the load scale. Continuously note down the loads so as to record the maximum load. Once the load has started reducing, reverse the direction of movement of the handle. Apply the load slowly and gradually.

Tensile Test on a Sample Using UTM On a UTM, a bigger sample can be used as there are devices with capacity of 100–1,000 kN load. Sample for tensile test in a UTM are generally provided with collars at ends, in between there is a gauge length of smaller diameter. These collars are threaded or are plane circular, depending upon the type of grips used. In ordinary UTMs, a specimen taken from a simple rolled bar is fitted using wedge grips. These grips reduce the strength of the specimen in grips and many a times, the specimen breaks in these wedge grips. Fitting of the specimen in wedge grips is shown in Fig. 21.9. Sometimes for accurate measurement of extension for a few millimetres, an extensometer is fitted on the specimen of standard size (specimen made as per testing standards). Required capacity of the machine is selected and the minimum strain rate is adjusted before starting the experiment. Now, the specimen is slowly and gradually extended by using hydraulic pressure.

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Machine grip Wedge grips

d

L

Specimen

P Wedge Grips

Figure 21.9

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Material Testing

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Once the specimen is firmly gripped, the machine is put on and the readings of the load and extension are simultaneously recorded (Nowadays, an automatic graph is obtained between the load and extension.) While performing the tension test on a mild steel specimen, the load increment should be carefully observed, because at the upper yield point, there is slight reduction in the load and then the load again continuously increases, and at a particular instant the test piece breaks at the neck (in the case of a ductile material). From the automatic graph or graph plotted between the load and extension, various mechanical properties can be determined. To note down the percentage reduction in area, the two broken pieces of sample are joined together and the diameter at the neck is measured. Say,

D = diameter of the sample d = diameter at the neck (after fracture)

D2 − d 2 × 100 D2 To determine the value of percentage elongation, join the two broken pieces and measure the changed gauge length. % reduction in area =

l = initial gauge length

Say,

l′ = final gauge length l′ − l × 100 l It is necessary to note down the change in gauge length from the broken piece, because once the fracture occurs, the load is removed from the specimen and the elastic part of the deformation is recovered immediately as shown in Fig. 21.10 by the portion D′D. Line CD is parallel to the straight line portion OA of the graph. After the yield point, once the plastic deformation has started, the cross-sectional area of the specimen starts decreasing at a faster rate and the actual stress at any instant (in the plastic stage) is much more than the nominal stress (calculated on the basis of original area of cross-section). % increase in length =

True stress =

Load at a particular instant Actual area at that instant

Nominal stress =

Precautions 1. Fit the specimen firmly in grips. 2. Adjust the pointer on the zero of the load scale before starting the experiment. 3. Adjust suitable capacity of the machine depending on size of specimen and strain rate.

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B

Load at a particular instant Original area of specimen

Graph OAB′C′ is the true stress—strain curve for the material.

C

B

Load

O

A

Fracture C Ductile material

D

D

Extension Elastic recovery

Figure 21.10

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822 Chapter 21

4. Record the readings of the load and extension simultaneously. 5. Watch the movements of the pointer on the load scale continuously so as to get the upper and lower yield points (in the case of mild steel). 6. Once the neck is formed in the sample, decrease the strain rate of experiment.

Double Shear Test on a Specimen Using UTM To perform this test, a shear attachment is used in which three rollers (of hardened steel) are fitted in the attachment. The specimen passes through a central hole through all the three rollers as shown in Fig. 21.11. Diameter of the specimen is noted with the help of micrometer. The rollers are fitted in the shear attachment, which is placed in the top of moveable cross head of the UTM. Figure 21.11 shows the specimen and deformation in the roller and the final fractured surface of the specimen. The machine is put on and the dowel provided at the top of the shear attachment fits into a hole provided in the top fixed cross head. Once the gap between the dowel and bottom portion of the fixed cross head becomes zero, the load starts acting on the specimen. Initially, the hard rollers get penetrated into the metallic sample slowly and gradually, and the area of cross-section under the shear force goes on reducing (as shown by the dull portion of the section in Fig. 21.11). At a particular stage, straight fracture occurs, breaking the test piece into three pieces. Maximum load recorded by the pointer is noted down. maximum load = P

Say,

diameter of sample = d

1

3

2

Hardened steel rollers Specimen

d

Movable roller

Movable roller Shining portion

Stationary roller b

• • • ••• • • •• • • • • •• • • • •• • • • • • • • • • •• • •• • ••• •• •• •• • •• • • • •• • • • • • • • • • • • • •• • • • • •• • • ••• • • • • • • • • • • •• •• • • ••• •• •• • • • • • •

a

•

••

•

•

•

•

•

•

• • •

•

• •

• • • • •

•

•

•

•

•

a − a and b − b are planes under shear

• • •

b

•

a

Dull fracture

Fractured surface

Figure 21.11

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ultimate shearing strength of the material, 2P P = π 2 π d2 2× d 4 Take out the broken pieces from the shear attachment and analyse the fractured surface. The shining portion shows slow and gradual penetration of hard rollers in the sample, the dull portion shows the straight fracture under shear.

τ ult =

Precautions 1. Load should be applied slowly and gradually. 2. Select a proper load scale on the UTM, depending upon the type of metal and diameter of the specimen. 3. Specimen should be symmetrically placed in the rollers, that is, projection of the specimen should be equal on both the sides 4. There should not be any clearance between the rollers. 5. The clearance between the roller’s bore and the specimen should be minimum. 6. Continuously note down the load on the load scale; as after the fracture, the pointer returns to zero. 7. Measure the diameter of the sample at several sections and then take the mean. For samples of smaller diameters, say up to 2 mm in the form of wire, there are hand operated machines, wherein a wire is fitted in a die and a punch shears the wire at two sections. The load on the sample is noted down from a reading on a calibrated proving ring.

Compression Test on a Cast Iron Specimen Using UTM The aim of the experiment is to determine the ultimate compressive strength of cast iron sample. A cast iron sample of L/d ratio approximately equal to 1.5 is taken. The diameter of the sample is not more than 15 mm if it is to be tested on a UTM with a maximum capacity of 100 kN. Adjust the sample on the top surface of the movable cross head of the machine. On the top of the sample compression attachment as shown in Fig. 21.12 (two blocks with hemispherical cavities and a ball in between them) is concentrically placed. It is very difficult to apply compressive load on a sample such that the load line is passing exactly through the axis of specimen. Ball and hemispherical cavities is an attempt to achieve axial load on the specimen. The diameter of the sample is noted with the help of a micrometer. Keeping a low strain rate, the machine is put on. When the gap between the two blocks and a ball in between becomes zero, the load starts acting on the sample; continuously watch the

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Dowel

Compression attachment

•• • • • • •• • •• •• • •• • • •• • • • ••• ••• • • • •• • • • • •• • •• • • ••• • • •• • •• • • • •• •

Fractured surface

L

d

Specimen

Shear plane

Figure 21.12

Movable cross-head

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824 Chapter 21

movement of the pointer on the load scale, so that maximum load on the specimen can be recorded. Cast iron is a brittle material, strong in compression, but weak in shear; therefore, the sample is going to break along a shear plane as shown in the figure. Say,

d = diameter of cast iron sample

ultimate compressive load = P (say) Ultimate compressive strength of cast iron =

4P . π d2

Precautions 1. Select the proper load scale depending upon the size of the sample. 2. Measure diameter of the sample at several sections and then take the mean. 3. Place the specimen in the centre of the compression block. 4. Faces of the sample must be perpendicular to its axis. 5. Apply the load slowly and gradually. 6. Provide a protective cover around the sample so that after breaking, if the sample slips, it may not hurt the machine operator.

Deflection Test on a Bar Using UTM A bending attachment is required to perform this test, on which the length between the supports can be varied as shown in Fig. 21.13. The bending attachment is fitted with its dowel in a square hole provided on the bed of movable cross head. The diameter of the sample and the length between the supports are accurately measured. Now, a bending knee is fitted in a hole provided at the bottom of the fixed cross head of the UTM. The test piece is symmetrically placed over the supports. A proper load scale depending

Bending knee

Sample

d

Support

Support (Adjustable)

Dowel in square hole L

Figure 21.13 Bending attachment

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Material Testing

825

upon the material, diameter of the sample and Plastic stage length between the supports is selected on the machine. The machine is put on and the strain rate is suitably adjusted. As soon as the gap between the sample and bending knee becomes zero, the load starts acting on the sample. Continuously record Load, W dw the readings of deflection and load on beam (on some UTM, the graph between load and deflection is automatically recorded). After recording some dy readings of deflection, say up to 0.2 D (where D is the diameter of the sample), the machine is stopped and the load is released. Due to elastic recovery, Elastic some deflection is recovered and if the test has been recovery performed into the plastic stage of the material, O O there will be some residual deflection in the beam. Deflection, y From the readings obtained, plot a graph between Figure 21.14 the load and deflection as shown in Fig. 21.14. The load–deflection curve may show a slight curvature in the beginning instead of a straight line. This is due to the weak mill scale on the sample (sample taken from a rolled bar) and setting of the bending knee and the supports into the same. Say, the load is applied at the centre of the sample or the beam, then central deflection, y = where

WL3 48 EI

W = central load L = length between the supports E = Young’s modulus of elasticity 1 = moment of inertia =

π d4 64

Precautions 1. Put the sample symmetrically on the supports with equal overhang on both the sides. 2. Measure the diameter of the sample at several sections and then take the mean. 3. Apply the load slowly and gradually. 4. Select the proper load scale depending upon the type of the material, diameter of the sample and length between the supports. 5. Do not excessively bend the sample.

Compression Test on Brick The size of the brick is large; therefore, compression test will be performed by a spring testing machine. This machine is basically designed to provide compressive force on a specimen and experiments, such as determination of (a) stiffness of a helical spring, (b) stiffness of a leaf spring and (c) compressive strength of wood can be performed on this machine. The brick is placed at the centre of the bed of the machine and below the compression plate of lead screw of the machine as shown in Fig. 21.15. The steelyard of the

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826 Chapter 21

Lead screw

Compression plate Gap

Brick

Bed of machine

Figure 21.15

Brick under compression

machine is balanced horizontally with the help of a balancing weight provided for the purpose. The lead screw of the machine is slowly and gradually lowered and when the gap between the brick and the compression plate becomes zero, the load starts acting on the brick and the steelyard starts moving up. Apply the load further by moving a hand wheel. Now, balance the steelyard and take the reading of the load applied. In the same manner, continuously record the load applied after each rotation of the hand wheel. Once the brick starts crushing, its resistance is decreased and the load applied on the brick is decreased. Now, balancing weights are adjusted, so that the steelyard is again balanced and becomes horizontal. Reading on the scale tells us about the load acting on the sample. b = breadth of brick

Say,

t = thickness of brick h = height of the brick Pmax = maximum load applied on the brick at which the brick has started crushing. P Ultimate crushing strength of brick = max b×t Precautions 1. Adjust the brick at the centre of the bed. 2. Adjust the steelyard initially (with the help of a tare weight) with zero reading of balancing weight. 3. Never apply any external pressure on the bed.

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4. Operator should not lean against the bed. 5. Apply the load slowly and gradually. 6. Record the load continuously so as to record maximum value of crushing load.

Compression Test on a Wooden Sample The aim of this test is to determine the ultimate compressive strength of a wooden sample. The test can be performed by a UTM or a spring testing machine, depending upon the size of the sample. Say the test is being performed by a spring testing machine. Put the sample on a cylindrical block kept on the bed of the machine and adjust the sample so that the axis of the sample coincides with the axis of the lead screw of the machine. The sample is of the square section of side a and height h, the ratio h /a as is also shown. The sample’s aspect ratio is about 1.5. Balance the steelyard of the machine and rotate the hand wheel. When the gap between the sample and bottom surface of compression plate becomes zero, the load starts acting on the sample. Slowly and gradually rotate the hand wheel and note down the load continuously on scale as has been explained earlier. At the maximum load, the sample has failed and the resistance of the sample is decreased; therefore, the load recorded by steelyard is also less. Figure 21.16 shows the type of fracture obtained in a wooden sample. Pmax = maximum compressive load on the sample

Say

A=a×a

Area of cross-section,

Pmax a2 Samples of different types of wood, such as teak, deodar, sisal, kail, chir and shisham can be tested. Ultimate compressive strength =

a a

Lead screw

Compression plate Gap h

Wooden Sample block

Fractured wooden Sample Bed

Figure 21.16 Wooden sample on bed

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828 Chapter 21

Precautions 1. As far as possible adjust the axis of the sample along the axis of the lead screw. 2. Apply the load slowly and gradually. 3. If the load is eccentric load, then sample will deform due to eccentric load, causing distortion. 4. Place the wooden sample on steel block with flow of grains in the direction of the load (the wooden sample has maximum strengths along grain flow and minimum strength in a direction perpendicular to grain flow).

Hardness Test—Brinell Hardness Number The aim of this test is to determine Brinell hardness number (BHN) for Brinell ball 5-mm-diameter a given metallic sample. For this test, a sample of the material is taken in the form of a circular disc, say a diameter of 80 mm and a thickness of about 50 mm. A compression attachment is fitted in the tensometer cross heads, then a Brinell ball bolster is fitted in an axial circular hole Brinell ball bolster provided in one plate of the compression attachment. The bolster conFigure 21.17 tains a 5-mm-diameter hardened steel ball as shown in Fig. 21.17. The metallic specimen is placed against the Brinell ball, resting on another compression plate as shown in Fig. 21.18. Compression Now, the handle of the tensometer is rotated and attachment the load starts acting on the sample when the gap between the Brinell ball and the specimen becomes Plate zero. A specified load is applied and the load is maintained for 15 seconds on the sample so that edge of Brinell ball the impression on the sample is clearly formed. For Specimen bolster different types of metals, different loads are taken as follows: Mild steel 750 kg Brass 250 kg Figure 21.18 Aluminium 125 kg After maintaining the load for 15 seconds, it is released and the sample is taken out for noting down the diameter of the impression (a partly spherical cavity). The diameter of the impression is measured with the help of a travelling microscope. Say, diameter of the ball

= D mm

Diameter of the impression

= d mm

(

)

πD D − D2 − d 2 2 Load in kgf Brinell hardness number, BHN = Area of spherical indentation Area of the spherical cavity =

=

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P in kgf πD D − D2 − d 2 2

(

)

in mm2

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Material Testing

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Figure 21.19 shows the partly spherical cavity due to indentation in the specimen. Precautions 1. Apply the load slowly and gradually on the sample. 2. Distance between the previous impression and the location for new impression should be at least 3D, that is, three times the ball diameter. 3. After applying the specified load, wait for 15 seconds and then release the load. 4. During unloading there is elastic recovery of some deformation and diameter of plastically deformed spherical portion is to be measured. 5. Do not apply more than the specified load. Due to excessive load, Brinell ball will be deformed beyond elastic recovery.

Brinell ball diameter D

D

d

Cavity Specimen

d-Diameter of spherical indentation

Figure 21.19

d-Diameter of spherical indentation

Hardness Test—Rockwell Hardness Number The aim of this test is to determine Rockwell hardness number (RHN) of various metals. For metallic samples, the Rockwell hardness scales can be classified as: Rockwell A—for case hardened materials. Rockwell B—for soft materials, such as mild steel, brass and aluminium. Rockwell C—for hard materials, such as high carbon steel, high speed steel and tool steels. For Rockwell A and C, diamond indentor is used, and for Rockwell B, hardened steel ball indentor is used. For Rockwell A hardness test, full load on indentor is 60 kg; for Rockwell B, full load on indentor is 100 kg and for Rockwell C, full load on indentor is 150 kg. In Brinell hardness test, area of spherical indentation is calculated and then hardness is determined; therefore, determination of BHN takes time, but in Rockwell hardness test, hardness is related inversely to the depth of indentation; therefore, this test is quick, as soon as the indentor is removed from the specimen, reading of RHN is taken on a dial gauge. The load on indentor is applied through a lever as shown in Fig. 21.20. Using this machine, first of all samples with a flat surface is placed on the anvil of the lead screw. Now, the hand wheel is moved and the specimen is raised till it touches the indentor I. L F

•

S I

H Sample

Anvil Lead screw

H — Hanger L — Lever I — Indentor S — Support F — Fulcrum of lever

Figure 21.20

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830 Chapter 21

As the specimen touches the indentor, the pointer on dial gauge starts moving. The pointer on scale is adjusted at zero reading with the help of a setting screw. At this stage, load of the lever acts on the sample through the indentor. The equivalent load of the lever at the indentor is 10 kg. This initial load on indentor breaks any oxide layer present on the surface of the sample and sets the indentor on sample. This initial load of 10 kg remains the same for all tests, that is, Rockwell A, B and C. There are three loads provided at the end of the lever; if load one is applied on the hanger, it is equivalent to additional 50 kg load on sample. Loads one and two are equivalent to 90-kg load on indentor, and all the three loads are equivalent to 140 kg additional load on sample. Initially these loads are resting on a piston of a dash pot. Depending upon the test, loads are selected. Now, the hand wheel is rotated and the sample is raised by which the lever is raised and lever no longer is supported at S, but now it is supported at indentor, I. For Rockwell A and C, pointer on dial gauge is rotated by three revolutions, but for Rockwell B pointer on dial gauge is rotated by three revolutions and 30 divisions, while moving the hand wheel. Now, the main lever of the machine is operated, releasing the piston in the dash pot, the load along with the piston comes down slowly and at a certain stage the load acts on the hanger and the piston is separated from the load. When the load remains applied on the hanger, the lever L moves down with indentor penetrating into the sample and the pointer on the dial gauge rotates in the opposite direction, that is, opposite to the direction of rotation when the lever was raised. When the pointer comes to rest, wait for 15 seconds, so that edge of the indentation is clearly defined. Reverse the main lever of the machine and the load is lifted off the hanger or load is removed from the indentor. Take the reading of the hardness number on dial gauge. In the same manner, repeat the experiment for different types of sample and for different tests. Precautions 1. After the sample just touches the indentor, adjust the zero of the pointer with the help of a setting screw. 2. Surface of the sample must be ground and perpendicular to the axis of the indentor. 3. Main lever of the machine should be operated slowly and gradually. 4. Fresh indentation on the sample should be at least 3 mm away from the previous indentation. 5. Record the reading of the hardness number only after removal of the load from the indentor. 6. Indentation should not be made near the edge of the sample.

Hardness Test—Vickers Pyramid Number The aim of this test is to determine Vickers pyramid number (VPN) of any material. The indentor used in here is a diamond in the shape of a square pyramid, with nose angle equal to 136°, that is, the angle included between the opposite faces of the square pyramid. The test is performed on a Vickers hardness tester, with loads varying from 1 to 120 kgf on the indentor. In this case very fine and very small size of the indentation can be obtained; therefore, the specimen needs a glossy surface finish for testing. After the indentation, the diagonal of the indent is accurately measured with the help of a travelling microscope fitted on the tester. The diagonal is measured by focusing a cross-wire device in the optical equipment. Say,

Vickers pyramid number, VPN =

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Load applied in kgf Surface area of the pyramidal indentation in mm 2

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Material Testing

Figure 21.21 shows the square indentation obtained on the specimen during Vickers hardness test. Say, S = slant height of each of the four triangular faces

P

S Area, A = 4 a 2 D 2

Now, a =

831

136

Therefore, the area of indentation A=

2D × S = 2 DS 2

Moreover, S =

a

a 2 sin 68°

a

a 1 A = 2D × 2 sin 68° Area, A = however,

D

Da 2 sin 68° D a= 2

Therefore, A = VPN =

Figure 21.21

D2 D2 = 2 sin 68° 1.854 P in kgf P P = × 1.854 = 1.854 2 A in mm 2 D 2 D

Standard charts are also supplied along with the Vicker’s testing machine which gives VPN corresponding to diagonal under specific load. Precautions 1. After applying the load through the pedal, wait for 15 sec so that the edges of indentation are welldefined. 2. Take moderate load say 50–60 kgf for a steel sample. 3. Choose the load for sample depending upon the nature of the material. 4. Metal surface must be highly polished.

Izod Impact Test The aim of Izod impact test is to determine the toughness of the material under the conditions of impact. Many a times a simple tensile test does not reveal the brittle nature of the metals, and it is necessary to test the material under shock or sudden loading conditions. An ordinary ductile material behaves like a brittle

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832 Chapter 21

material if a discontinuity such as a notch is present in component. In an impact test specimen, a notch is provided and impact Hammer energy is supplied to break the sample at the notch section. 5 There are two types of most commonly employed impact tests, that is, (a) Izod and (b) Charpy impact tests. The machines are standardized in all respects, including the dimensions of 10 the specimens to be used. The principle employed in impact testing is that a material 28 mm absorbs a certain amount of energy before it breaks. The quans 22 mm tity of energy thus absorbed is a characteristic of the physical nature of the material. If the material is brittle it breaks more easily requiring less energy, and if it is tough it needs more 45 energy to break. Vice A swinging hammer strikes the specimen firmly gripped in a vice. The potential energy of the hammer is converted into 2 mm kinetic energy when it is released and the hammer breaks the specimen. The height of the hammer on the other side deter- 47 mm mines the residual potential energy in the hammer. The energy actually absorbed by the specimen is given by the difference between the initial and residual energies in the hammer. In the case of Izod impact test, the specimen is fixed as a cantilever. The maximum energy of the pendulum (or hammer) is 160 Nm. The standard specimen for Izod impact test is a square rod of 10 mm side, 75 mm long, 2 mm deep, 45° 10 mm notch made at a distance of 28 mm from one end of the specimen as shown in Fig. 21.22. Due to the notch, there is stress concentration at the root of the notch as shown. When the 10 mm hammer strikes the specimen, a high tensile stress is developed at the root of the notch. This high tensile stress is more Figure 21.22 than the ultimate tensile strength of the material and the specimen cracks at the root of the notch, and the crack propagates through the thickness depending upon the brittleness of the material (for a brittle material such as cast iron and notch is not needed in the sample). Sometimes, the pendulum not only breaks the specimen but also throws off the fragment. The fractured surface of the sample can be studied to know about the toughness of the material. In the case of mild steel, the crystalline nature of the fractured area is the region below the crack. Figure 21.23 reveals that the material possesses a high resistance to the initial propagation of the crack. Notch Mild steels and low carbon steels have impact values ranging from 50 to 120 Nm. Precautions 1. Use positioning gauge to correctly position the specimen in the vice. 2. Adjust the pointer at zero reading initially, or 160 Nm, depending upon the type of the machine.

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Fibrous structure

••••••••••••••• Crystalline structure

Figure 21.23

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Material Testing

833

3. Firmly grip the specimen in the vice. 4. Loss of energy due to air resistance should be accounted for. 5. After the test, pendulum should not be allowed to swing; it should be brought to rest in the vertical position.

Charpy Impact Test The aim of this test is to determine the impact strength or toughness of a material through the Charpy impact testing machine. This test is similar to the Izod impact test, with the only difference that the specimen is placed as a simply supported beam, while in Izod test, the specimen is held as a cantilever. Figure 21.24 shows the arrangement of Charpy impact test, sample is 60 mm long, supported over 40 mm span, with a square sides 10 mm × 10 mm and a notch of 2 mm depth, 45° angle. When the hammer strikes the specimen with a blow, crack starts at the root of the notch and the specimen breaks along the notch. Energy absorbed by specimen is noted down on the scale. Precautions 1. Adjust the notch in the centre along the line of strike of hammer. 2. The hammer should be adjusted to a sufficient height, so that enough energy is present in the hammer to break the sample.

Hammer Striking edge, radius

8

Specimen

80

45

40 mm

Figure 21.24

Torsion Test The aim of this test is to study the curve between torque and angular twist for a shaft of a given material and to determine the modulus of rigidity of the material. There are various types of torsion

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834 Chapter 21

testing machines marketed by several companies Tapered ends in UK, USA and India. Each machine has differMovable Head ent gripping arrangement and different shape and d size of samples. However, for a particular machine, Keyway a specimen of the shape, as shown in Fig. 21.25, Fixed is made on a lathe machine, and the keyways are head cut on a milling machine on the tapered ends of the i sample. The specimen is inserted in the two hubs Torsion test specimen of the machine, that is, movable hub and fixed hub. The steelyard of machine is balanced by keeping Figure 21.25 the balancing weight at zero reading. Initial backlash between the keyway and key is removed by moving the hand wheel of the machine and when the steelyard starts moving up, the steelyard is again balanced. Now, the hand wheel is rotated slowly and the angle of twist q and twisting moment T are measured on respective scales. Initial readings may be taken at an interval of 0.05° degree of angular twist but once the elastic limit is crossed the readings can be taken at an interval of 1° and then 2° or 5°. Balancing weight H Fulcrum

Steel yard

B

• •

E A

C

Hub F

D

Arm

Figure 21.26 Mechanism of machine Figure 21.26 shows the mechanism of the machine transmitting the applied torque T to the steelyard for its measurement. When torque is applied on the sample through movable hub, a resisting torque T is developed at the fixed end of the specimen or at fixed hub as shown and arm of the hub tries to rotate in one direction. Say the resisting torque tries to move the hub in the clockwise direction. Link CD will try to move down pushing the end A downwards. When the end A moves down the steelyard is lifted up as shown. The steelyard is balanced with the help of a sliding balancing weight and reading of twisting moment is taken. In the same manner, a number of readings between T and q are recorded and plotted as shown in Fig. 21.27.

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•

T

•

•

•

•

• A Typ

Twisting moment

•

∆T

• ∆q • O Angular Twist

q

Figure 21.27

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Material Testing

From A, onwards, the graph has become non-linear. Torque at A, that is, Typ gives the torque at yield point. The value of shear modulus G can be calculated by the expression,

835

Fractured surface Fractured surface (Failed in tension) (Failed in shear)

∆T J G = ∆θ l

π d4 where J = , polar moment of inertia of 32 the shaft G = shear modulus l = length of the shaft. Ductile material Brittle material The values of DT and Dq are to be taken Figure 21.28 from the straight line portion OA of the graph as shown. The specimen can be tested up to its destruction, that is, by moving the hand wheel continuously till the specimen breaks. A ductile material will take large values of q before it breaks in torsion and a brittle material will take much less value of q before it breaks in torsion. Figure 21.28 shows the fractured surfaces of samples tested in torsion. Precautions 1. Apply the twisting moment on the samples slowly and gradually. 2. Rotate the hand wheel only in one direction. 3. Adjust the zero reading of the pointer after removing backlash between the sample shaft and key. 4. Initial reading should be recorded at small intervals of q, that is, 0.05°.

Stiffness of a Helical Spring To determine the experimental value of stiffness, take a helical spring, count its number of active coils, measure the wire diameter and the outer diameter of the coils. Place it on the bed of the spring testing machine as shown in Fig. 21.29. Adjust the axis of the spring along the axis of the lead screw and balance the steelyard of the testing machine. Now, rotate the hand wheel till the compression plate just touches the spring and the steelyard of machine has started moving up. At this stage, again balance the steelyard with balancing pointers at zero reading on the load scale. Rotate the hand wheel till the compression plate moves down by, say, 2 mm or deflection in the spring is 2 mm. The reading of the deflection can be taken on a vertical scale provided for this purpose and by balancing the steelyards with the help of balancing pointers take the reading of the load. In the same manner, at regular intervals, record the readings of deflection in the spring and the load on the spring. Plot the readings of y and W as shown in Fig. 21.30. Pass a straight line through the mean positions of the points plotted. From the graph, stiffness of spring, k=

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∆W in N/mm ∆y

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836 Chapter 21

Lead screw

Compression plate

Helical spring

Bed

Figure 21.29 Spring machine testing

•

n = number of active coils of the spring

Say,

d = wire diameter D0 = outside diameter of the coils

• Load W

• •

∆W

D = D0 − d = mean coil diameter D0 − d = mean coil radius 2 G = shear modulus of the material of spring = 84 kN/mm2 (for steel) Gd 4 Theoretical value of stiffness, k ′ = 64 nR3 R=

Percentage error

=

k′ − k × 100 k′

• • •

∆y

Deflection, y

Figure 21.30

Precautions 1. Count the number of active coils in the spring, even if it is not in whole number (the part of the spring that is flattened and is in contact with the bottom of the compression plate and the top of the bed is not counted). 2. As far as possible adjust the axis of the spring along the axis of the lead screw of the machine.

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Material Testing

837

3. Apply the load slowly and gradually. 4. Do not apply excessive load on the spring (the coil should not touch each other). 5. Take diameter of the wire at several sections and then take the mean.

Stiffness of a Leaf Spring The aim of this test is to determine experimentally the value of stiffness of a leaf spring used in automobiles. Take the leaf spring and place it on the trolleys provided on the bed of the spring testing machine as shown in Fig. 21.31. Adjust the centre of the spring along the axis of the lead screw of the machine. Now, place a flat-faced roller at the centre of the leaf spring. Balance the steelyard of this machine with the help of a tare weight. Now, rotate the hand wheel and when the compression plate just touches the roller at the centre of the leaf spring, the steelyard has just started moving up. At this stage, balance the steelyard again with the help of the tare weight, keeping the balancing weights at zero reading. Now, rotate the hand wheel so as to produce deflection in the leaf spring. Take the readings of deflection on a vertical scale provided and note down the reading of the load by balancing the steelyard with the help of balancing weights. In the same manner, increase the deflection of the leaf spring in steps of say, 1 mm or 2 mm or 3 mm, depending upon the length of the spring and simultaneously record the load. Plot a graph between load W and deflection y as shown in Fig. 21.32. Stiffness of the leaf spring, k=

∆W (as shown in the graph) ∆y Lead screw

Compression plate Flat surface roller Leaf spring

Trolley

Bed L

Figure 21.31

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838 Chapter 21

Now, remove the load and take out the spring. Count the number of leaves and measure. (a) thickness of each leaf, (b) breadth of leaves, b, and (c) length of the longest leaf L (between the eye centres). If E = Young’s modulus of the material of the spring. Theoretical value of stiffness, 4 Enbt k= 3 L3 % Error =

• • W

• • •

Load

•

3

k′ − k × 100 k′

•

∆W

∆y

• • • y, deflection

Precautions Figure 21.32 Load versus deflection 1. Adjust the ends of the leaf spring at the centres of the graph trolleys. 2. Adjust the centre of the leaf spring along the axis of the lead screw of the machine. 3. Put a flat-faced roller at the centre of the spring. 4. Apply the load slowly and gradually. 5. Do not apply excessive load on the spring (otherwise, a leaf may break). 6. Measure the length of the longest leaf from centre to centre of eyes at the ends.

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22 Engineering Materials CHAPTER OBJECTIVES This chapter on engineering materials is intentionally added to the text so as to expose the students to various materials employed in engineering applications. The properties, composition and uses of such materials will be discussed. Students will be able to learn about  Plain carbon steels, for 90 per cent engineering applications plain carbon steels are used

of its negligible coefficient of thermal expansion 

Alloys of aluminum, copper, magnesium, nickel and titanium

 Stainless steels



Alloys used in bearings

 Effects of alloying elements on the properties of alloy steels



Plastics as thermosetting and thermoplastics

 Special alloys such as Invar steel used in measuring instruments because



Ceramics such as glasses



Natural and synthetic rubbers

 Alloy steels

Introduction A mechanical/structural engineer should be fully conversant with the mechanical and physical properties of various materials used for common engineering purposes. At the same time, the engineer must also have the knowledge of what materials are available in the market and their properties and applications, so that they can choose a particular material from market for their application. Mild steel is the most commonly used engineering material, and 90 per cent of metallic components are made from mild steel because of its properties of moderate strength but good ductility. Aluminium is light in weight and highly ductile, but poor in strength. To improve its strength, aluminium alloys have been developed, which are used in aerospace industry. Similarly Invar steel is used in measuring instruments and gauges for its property of very low coefficient of the thermal expansion. Phosphor bronze is used in such applications where its anticorrosive properties are utilized. Cast iron is weak in tension but strong in compression and is used in such applications where only compressive stress acts on a member as beds of machines. To improve the ductility of cast irons, spherical graphite (SG) iron is developed, which is used in many applications such as connecting rod of marine engines.

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840 Chapter 22

In this chapter we will discuss various materials such as plastics, elastomers and ceramics. Special alloys for aerospace applications and Babbits for bearing application will also be discussed.

Carbon Steels Plain carbon steels are classified as low-, medium- and high-carbon steels. These steels contain only residual concentration of impurities other than carbon, that is, a small amount of manganese, phosphorous and sulphur. Low-carbon steel contains less than 0.25 per cent by weight of carbon. They are unresponsive to heat treatment. Their applications are in automobile body components and rolled sections such as I, T, L sections and sheets. Mechanical properties are syp = 270–280 MPa, sut = 415–550 MPa, Percentage elongation 20–25 per cent. Medium-carbon steel contains carbon 0.25–0.6 per cent by weight. These can be heat treated. Their properties are improved by alloying elements such as Cr, Ni, Mo. Applications of medium-carbon steels include railway wheels, rail tracks, gears, crank shafts and other machine parts. They possess higher strength, wear resistance and toughness. High-carbon steels contain carbon 0.6–1.4 per cent by weight. They are hard, strong but least ductile and are always used in hardened and tempered condition. They are wear resistant and possess the property of holding sharp cutting edges. Tools and die steels are high-carbon steel alloys containing Cr, V, W, Mo as alloying elements. These elements along with carbon form very hard and wear resistant carbides. A particular tool steel contains C 0.85%, Cr 3.8%, Ni 1.3%, Mo 8.7%, W 1.20%, V 1.25% used for drills, saws, lathe and planer tools and punches.

Alloy Steels Plain carbon steels are used for general purpose engineering applications, and 90 per cent of total steel production amounts to plain carbon steel. The properties of plain carbon steel are mainly due to the amount of carbon element. In addition to carbon, the plain carbon steels contain manganses less than 1.65 per cent and silicon less than 0.6 per cent and copper less than 0.4 per cent. These elements are not added but are present as impurity atoms. Similarly a small amount of sulphur and phosphorous are present in plain carbon steels. However, manganese and silicon do not impair mechanical properties, but sulphur and phosphorous reduce ductility and toughness of steel. Alloying elements are purposely added in steel to achieve desired mechanical properties. The properties of alloy steel depend on the percentage of both carbon and alloying elements. Table 22.1 gives rough estimate about the general effects of alloying element.

Stainless Steel Stainless steels are highly corrosion resistant in a variety of environments. Addition of chromium of minimum 11 per cent by weight provides corrosion resistance. There are three types of stainless steel as given in Table 22.2:

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Engineering Materials

Table 22.1

Effects of alloying elements

Alloying element

General effects

Aluminium

Strong deoxidiser, refines grain size

Nitralloy

Chromium

Forms hard and stable carbides

Ball bearing steels Stainless steels

Cobalt

Retains hardness at elevated temperature

Magnet steel High speed steel

Copper

Raises yield strength and ultimate strength

Munz metal, gilding metal

Manganese

Stabilizes carbides, increases strength and hardenability

Hadfield steel

Molybdenum

Imparts high temperature strength, enhances resistance to creep

Case hardening steels High speed steels Creep resistant steels

Nickel

Stabilizes austenite; impact strength at low temperature

Case hardening steel Low temperature steel Non-magnetic steel

Silicon

Potent deoxidiser; resistance to scaling and corrosion

Spring steel Transformer steel

Titanium

Forms stable and hard carbides, raises creep strength

Creep resistant steel Permanent magnet steel

Tungsten

Forms hard carbide, increases hot strength and wear resistance at high temperature

High speed steel Wear resistant steel

Vanadium

Forms carbides and nitrides, increases hardness

High speed steel Wear resistant steel

Table 22.2

841

Alloy steel

Composition, properties and uses of stainless steels Composition (%)

Ferritic stainless steel

C 0.08–0.2 Cr 12–27 Mn 1–1.5, Si 1.0

Ferritic structure, high corrosion and oxidation resistance

Austenitic stainless steel

C < 0.2 Cr 16–24 Ni 8–22

Austenite is stabilized by Used in chemical plants, nickel roller and ball races

Martensitic stainless steel

C 1–1.2 Cr 12–18 Mn 1–2 Si 0.5–1.0 Ni 1–2

Hard martensitic structure

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Used in furnaces, storage for acids and cutlery

Turbine blades, tools, springs and knife edges

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842 Chapter 22

Table 22.3

Special alloy steels developed for specific purposes Composition (%)

Rail steel

C 0.4–0.6, Mn and Cr up to 1

High strength, ductility, fatigue strength

Spring steel

C 0.6, Mn 0.9, Si 2.0

High elastic limit for high resilience

High speed steel

C 0.8, W 18, Cr 4, V 1

Cutting tools, high hardness at high temperature

Creep resisting steel

Low-carbon steel Mo 0.4–0.6 V 0.25–1.0 Cr up to 6

Creep resistant, application in pipe lines at 400–550°C

Ball bearing steel

C 0.9–1.1, Cr 0.6–1.6 Mn 0.2–0.4

High hardness 61 HRC, Balls, inner and outer races of ball bearings

Hadfield steel

C 1–1.4, Si .03–1.0 Mn 1.0–1.4

Resistance to abrasion and shock, used in excavating and crushing machines

Maraging steel

Ni 18, Co 7, Mo 4.5, C 0.02, Mn 0.1, Si 0.1, Ti 0.2 Be 0.003

Aircraft under carriage parts, portable bridges, booster motor in missiles

Invar steel

Ni 36, Fe 64

Very low coefficient of thermal expansion, and for measuring instruments, tapes

Cast Iron Most cast irons contain 2.5–4 per cent carbon, sometimes containing other alloying elements to improve their mechanical properties. They are easily melted above 1,150°C, a temperature considerably lower than the melting point of steels. Some cast irons are extremely brittle, and casting is the most convenient fabrication technique. Graphite formation in cast iron is promoted by the addition of silicon of more than 1 per cent. There are three broad classifications of cast iron, that is, (1) grey cast iron, (2) white cast iron, (3) Malleable cast iron. Grey cast iron is weak in tension and brittle in nature, but its strength and ductility are much higher under compressive loads. Grey cast irons are very effective in damping vibrations energy and are most commonly used in beds of machines. In molten state, grey cast irons have high fluidity, permitting casting of intricate shapes and least expensive as cast metals. In grey cast iron, carbon is in the form of graphite flakes (free form). But in white cast iron, carbon is in combined form, that is, cementite (Fe3C). White cast iron is very hard and wear resistant and is used in rollers in rolling machines. Malleable cast iron is obtained by heating white cast iron at temperature between 860° to 900°C for a prolonged period in a neutral atmosphere, causing decomposition of cementite into pearlite + graphite or ferrite + graphite, depending upon the cooling rate. Graphite exists in the form of clusters. Malleable

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cast iron possesses high strength and fair ductility. It is commonly used in connecting rods, transmission gears and marine parts of heavy duty services.

Ductile a Nodular Cast Iron Adding a small amount of magnesium and/or cerium to the grey cast iron melt during casting produces SG iron, in which graphite is converted into nodules or sphere-like particles. SG iron has 10–20 per cent ductility and the ultimate tensile strength is 380–480 MPa. SG iron is used in the manufacture of pump bodies, crankshafts, gears, connecting rods etc.

Aluminium, Magnesium, Titanium and Their Alloys Aluminium is a low-density material possessing high ductility, resistance to corrosion, high electrical and thermal conductivity. Most commonly used aluminium alloys are as follows: Recently, aluminium–lithium alloys have been developed for aerospace industry. This material possesses excellent fatigue – resistance and low temperature toughness properties. Aluminium–lithium alloy T–651, with Cu 1.3%, Mg 0.95%, Li 2.0%, Zr 0.1% + rest Al is used for aircraft structure that must be highly damage tolerant. Magnesium is a very low-density material of 1.79 gm/cm3, in which aluminium, zinc, manganese and some rare earth elements are added as alloying elements. These alloys are fabricated by casting, as it is difficult to deform magnesium alloys. The composition of a few cast alloys is as follows: Al 9%, Mn 0.5%, Zn 0.7%, Mg 90.35%—used in die casting parts of automobiles and electronic devices. Al 4.3%, Si 1.0%, Mn 0.35%, Mg 94.35%—used in die casting parts of automobiles and electronic devices. Titanium is a new engineering material and possesses an extraordinary combination of properties. Titanium alloys are extremely strong having 1,400 MPa as strength. These alloys are highly ductile, easily forged and machined but are expensive. These are immune to air, seawater and many industrial environments. Composition, strength and applications of a few titanium alloys are given below: Al 8, Mo 1.0, V 1.0, Ti 90, sut = 950 MPa, used in jet engine components. Al 6%, Sn 2%, V 6%, Cu 0.75%, Ti 85.25%, sut = 1,050 MPa, used in rocket engine case, air frame structures. Table 22.4

Aluminium alloys Composition (%)

Duralumin

Cu 4, Mg 0.5, Si 0.7, rest Al

Electrical cables, aircraft components.

Y–alloy

Cu = 4, Mg = 1.5, Fe = 0.6, rest Al

Casting of engine parts.

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844 Chapter 22

Copper and its Alloys Unalloyed copper is so soft and ductile that it is difficult to machine it. It has unlimited capacity to be cold worked. It is highly resistant to corrosion under diverse conditions, including sea water. Copper is alloyed with zinc to produce brass. Some common brasses are as follows: Table 22.5

Composition and uses of various bassses

Brass

Composition (%)

Application

Admiralty brass

Cu 71, Zn 28, Su 1

Condenser, evaporator, heat exchanger tubes

Cartridge brass

Cu 70, Zn 30

Bullet cartridges, springs, automobile radiators

Munz metal

Cu 60, Zn40

Pumps, fittings, marine

Gilding metal

Cu 95, Zn 5

Coins, medals

Red brass

Cu 90, Zn 10

Rivets, plumbing, hardwares

Delta metals

Cu 55, Zn 40, Pb3, Fe 2

Motor brushes, parts of mining industry

Yellow brass

Cu 65, Zn 35

Screws, rivets, reflectors

Copper is alloyed with Ni or Mn to produce materials used in resistors. Table 22.6

Copper-Nickel alloys

Alloy

Composition (%)

Application

Nickel silver

Cu 65, Zn 23, Ni 12

Name plates, optical parts (camera), plumbing fixtures

Constantan

Cu 58.5, Ni 40, Mn 1.5

Thermocouples, electrical resistance strain gauges

Bronzes Bronze is an alloy of copper having constituents other than zinc (though small amount of zinc may be present). Tin, aluminium, silicon and beryllium bronzes are commercially available. Sn bronze is also called phosphor bronze as it contains 0.2% phosphorous. Bronzes are used as bearing materials and possess good corrosion resistance and high strength. Table 22.7

Composition and uses of various bronzes

Bronze

Composition (%)

Application

Phosphor bronze

Cu 89.8, Sn 10, P 0.2

Marine application, bushes, gears, springs

Gun metal

Cu 88, Sn 10, Zn 2

Gears, bearings

Aluminium bronze Cu 86, Al 10.5, Fe 3.5

High ultimate strength, condenser tubes, gears, nuts and bolts

Beryllium bronze

Very high mechanical strength, fatigue, wear and corrosion resistant, springs

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Cu 98, Be 1.7, Cu 0.3

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Nickel, Cobalt and Their Alloys Pure nickel is an excellent corrosion resistant against alkalis and many acids, and hence used in chemical and food industries. Nickel has a very good solid solubility in many metals; therefore, it is used extensively as an alloying element, Austenitic stainless steel, Ni-resist irons and cupronickels are such examples. Commonly added elements to nickel are copper, cobalt, chromium, iron, manganese, tungsten, zinc etc. Nickel alloys are popularly known by their trade names as shown in Table 22.8: Table 22.8

Nickel alloys Composition (%)

Application

Monels

Nickel and copper alloys. Various compositions are available; one such composition is Cu 30, Mn 1.0, Si 1.0, Fe 1.5, C 0.15, rest nickel

Corrosion resistance and strong used in vessels, pipe lines, filters, valves, Heat-exchanging tubes, storage of food products, marine propellers shafts

Nichrome

Ni 80, Cr 20

Inconel

Chromium 14–17, Fe 0–10, rest Nickel

Nimonics

Various compositions are available; one such composition is C 0.15, Cr 20, Co 20, Ti 2, Al 0.5, Mo 6, rest nickel

Heating elements for electric furnaces, hair dryers, toasters etc. Superior resistance to corrosion at elevated temperature, pyrometer sheaths, springs for high temperature core, heat exchangers Turbine blades, exhaust valves

Haste alloys

Various compositions (A, B, C, D), Haste alloy A Mo 20, Fe 20, Mn 2, Si 1.0, C 0.15, rest Nickel

They can be used at a high temperature of 700–1,000°C, high strength and a good resistance to corrosion abrasion, used in magnets

Cobalt Cobalt is mainly used as an alloying element rather than a base metal. Very few commercial alloys can be considered as cobalt-base alloys. Cobalt alloys are known by their trade names as Vitallium, Elgiloy, Stellites, Haynes alloy, Permalloy, Vicalloy. These are wear-resistant, corrosion-resistant and heat-resistant alloys; the composition of some cobalt alloys is as follows: Stellites: A group of cobalt-base alloys having chromium, tungsten, nickel, molybdenum, niobium, titanium and iron as alloying elements. Salient features are excellent resistance to abrasion at elevated temperature, up to 1,000°C, used for gas turbine blades in jet engine, super chargers. Vitallium: A cobalt-base alloy having chromium and molybdenum as the main alloying elements and is used in surgical implants. Vicalloy (Co–V–Fe alloy): Co 14%, V 34% + Fe 52% high-energy magnetic alloys—used in magnetic tapes, compass needles and in many electric motor and devices.

Babbits The alloys of tin, copper, lead and antimony are called Babbits. Tin provides compressive strength, copper provides toughness, antimony provides property of leakage proof while lead contributes to

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846 Chapter 22

ductility in a Babbit. Liners of bearings are made from Babbit for its antifriction behaviour. When Babbit is backed up by a solid metal of high-compressive strength, it gives good service under high speeds. Composition of two commonly used Babbits are given below: Tin Babbit Sn 80%, Sb 12%, Cu 8% High-speed bearing bushes. Lead Babbit Pb 85%, Sb 10%, Sn 5% Railway wagon bearings.

Plastics Materials derived from carbon are termed as organic materials. These organic materials can be natural, such as wood, cotton and rubber. And then there are synthetic organic materials such as plastics, adhesives and synthetic rubber. Plastics are known as polymers. A monomer is a small molecule that converts into a large molecule—called polymer—on polymerization. ‘Poly’ stands for many and ‘mer’ stands for unit, so polymers are composed of large number of monomers. Example below shows organic structure of an ethylene monomer and a polyethylene polymer. H H | | C=C | | H H

Ethylene monomer (C2H4)

H H H H H H H H | | | | | | | | C= C−C= C−C= C−C= C | | | | | | | | H H H H H H H H Polyethylene polymer (C2H4)n

Natural polymers are cellulose, shellac, lignin, silk, starch, while synthetic polymers are polyethylene, nylon, polyvinyl chloride (PVC), polymethyl methacrylate (PMMA) etc. Plastics can be broadly classified as (i) thermosetting and (ii) thermoplastics. Thermosetting plastics have a three-dimensional network of primary bonds in all the directions. On the application of heat, they become soft first and then harden; after setting they cannot be softened again. Permanent hardening is called curing. Commonly used thermosets are: 1. 2. 3. 4. 5.

Polyester—used as matrix with synthetic fibres. Epoxy—used as matrix in composites for structural components. Phenolics—used in telephone receivers and foams. Melamines—used in crockeries. Silicone—used in high-temperature-resisting components.

Thermoplastics Their long chain molecules are secondary bonded that break more easily due to increase in thermal energy. They become soft when heated. Since no chemical hardening takes place, the shaped articles re-soften on heating. They have excellent plasticity but low melting point. Common examples are:

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Engineering Materials

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

847

Polyethylene—used in bags, tubes and containers. Polypropylene—used in ropes and vacuum flasks. Polystyrene—used for sound proofing of refrigerators and building. PVC—used in electrical insulation, piping and gramophone records. Polytetrafluroethylene (PTFE–Teflon)—used in biomedical implants and frying pans. Polyhexamethylene–adipamide (Nylon 66)—used in ropes. Polyhexamethylene Sebacamide (Nylon 610)—used in flexible tubes. Polycaprolactum (Nylon 6)—used in synthetic fibres. Cellulose acetate—used in fibre. PMMA (Perspex)—used in sheets for tables, a photoelastic material.

Rubber Rubber is an organic polymer found in latex (sap) of certain plants. Latex is treated to get raw or natural rubber. Additives are blended in raw rubber to get desired properties. Carbon black and zinc oxide are added as reinforcing agents while sulphur, stearic acid and vegetable oils are used as plasticizers. Synthetic rubbers are superior to natural rubbers. These are manufactured from raw materials such as coal tar, petroleum, coke, limestone, natural gas, alcohol, ammonia, salt and sulphur. Some commonly used synthetic rubbers are: Neoprenes—used in oil seals, gaskets, adhesives and tank lining. Butyl—used in tyres, tube and insulation for cables and wires. Butadiene—used in belts and soles of shoes. Nitrile—used in conveyor belts and tanks lining. Poly urethane—used in tyres and expanded foams. Silicone—used in coatings, packaging, tubs, gaskets and cable insulation.

Elastomers Elastomers exhibit rubbery behaviour and undergo elastomeric (recoverable) deformation from 300 to 1,000 per cent. It is basically a natural rubber of monomer–isoprene. The production of elastomer requires heating of raw rubber in the presence of sulphur under a specified pressure—called vulcanization–cross-linking of chains. H H CH3 H | | | | C= C− C −C | | H H Isoprene monomer

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848 Chapter 22

More the number of cross-links, more will be the degree of vulcanization (Figure 22.1). Tensile modulus of vulcanized rubber is 100 times the tensile modulus of raw rubber (E = 0.01 GPa). Similarly the sut of vulcanized rubber is two times the sut of raw rubber (sut = 3 MPa).

Ceramic Materials

H

H CH3 H

C=C− C −C− H

S

H

S

H H

−C

C− C −C−

H

H

CH3 H

Ceramics are non-metallic, inorganic amorphous solids and Figure 22.1 Sulphur crossare mostly metallic oxides and carbides. They have poor linking chain during tensile strength and are brittle in nature. They are either vulcanization crystalline or non-crystalline. Most ceramics can sustain high temperature, while many are workable in cryogenic temperature. Special uses of ceramics are making artificial limbs, teeth, superconducting materials, memory cores of computers etc. Ceramics range from refractories, glasses, cements to rock, stones, abrasives etc., but in this chapter we will deal with only some special ceramics used as glasses, abrasives, pigments and lubricant, wear coatings. Si Silica is an important gradient of glasses. The structure of silica consists of repeating units of silicate tetrahedron. Each unit has a silicon cation at the centre Oxygen − of tetrahedron accompanied by four oxygen atoms at Silicon − four corners. Oxygen atoms are shared with adjoining units. Therefore, chemical symbol of silica is SiO2 Figure 22.2 Silica structure (Fig. 22.2). In each tetrahedral unit, effective number of Si atoms is 1 Effective number of oxygen atoms is 2 Silica exists in crystalline form as Quartz (having piezo electric effect) and in non-crystalline form as silica glass.

Glasses Glass is an inorganic fusion product, cooled to a condition in which crystallization does not occur. Silica is a perfect glass-forming material, having high melting point. To lower its fusion point and viscosity, some basic metal oxides are added to form: Sodium disilicate—by adding sodium oxide Soda lime glass—by adding calcium oxide Besides silica, oxides of boron, vanadium, germanium, phosphorous are other constituents of glass. Elements and compounds such as tellurium, selenium and Be F2 also form glasses.

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Engineering Materials

Table 22.9

849

Various types of glasses

Components

Soda lime glass

Lead glass (flint glass)

Borosilicate glass (Pyrex)

High silica glass vycor

%

%

%

%

S1 O2

70–75

35–58

73–82

96–97

Na2 O

12–18

5–10

3–10

—

K2 O

0–1

9–10

0.4–1

—

Ca O

5–14

0–6

0–1

—

Pb O

—

15–40

0–1

—

B2 O3

—

—

5–20

3–4

Al2 O3

0.5–2.5

0.2

2–3

—

Mg O

0–4

—

—

—

Compositions of typical glasses are given in Table 22.9. Special features and applications of these glasses are: Soda lime glass—used in window panes, electric bulbs, bottles, table wares. Lead glass (flint glass)—used in optical devices, neon sign tubing, extra dense optical glasses. Borosilicate glass (Pyrex)—low thermal, expansion, used in scientific apparatus. High silica glass—low thermal expansion, resistance to thermal shock; this is, however, much more expensive than other glasses. Photochromatic and Zena glass—silver chloride is mixed with ordinary glass to make Zena glass— used in lenses for goggles.

Polymorph of Carbon Diamond is a metastable carbon polymorph at room temperature and atmospheric pressure. A diamond cubic crystal structure contains 18 atoms as follows: eight atoms, one at each corner; six atoms, one each at centre of faces; and four inside the cube. Diamond is a polycrystalline and may consist of very small to very large grains. Graphite is a polymorph of carbon, more stable, and its structure is composed of layers of hexagonally arranged carbon atoms by strong covalent bonds. Fullerenes is another polymorphic form of carbon (discovered in 1985) existing in discrete molecular form, consisting of hollow spherical cluster of 60 carbon atoms C60. Table 22.10 gives a list of typical ceramics in engineering use, with class, example and application.

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850 Chapter 22

Table 22.10

Typical ceramics in engineering use

Class

Example

Applications

Single oxides

Alumina A12O3 Chromium oxide Cr2O3 Zirconium ZrO2 Magnesium oxide, MgO Silica, S1O2

Electrical insulation Wear coatings Thermal insulation, pigment, wear parts, abrasive, glass.

Mixed oxide

Kaolinite (Al2 O3–2S1 O2. 2H2O)

Clay products

Carbide

Vanadium carbide VC Tantalum carbide TaC Tungsten carbide WC Titanium carbide TiC Silicon carbide SiC Chromium carbide Cr3C2 Boron carbide B4C

Wear resistant Wear resistant Cutting tools Wear resistant Abrasive Wear coatings Abrasives

Sulphides

Molybdenum disulphide Mo S2 Tungsten disulphide WS2

Lubricant Lubricant

Nitrides

Boron nitride BN Silicon nitrides Si3 N4

Insulator Wear products

Metalloid elements

Germanium Ge Silicon Si

Electronic devices Electronic devices

Intermetallic

Nickel–Aluminide (Ni Al)

Wear coatings

Review Questions 1. What are plain carbon steels? Differentiate between low-, medium- and high-carbon steels. 2. Give at least two applications of each of low-, medium- and high-carbon steels. 3. What are the general effects of the following alloying elements in steels? (a) Chromium (b) Copper (c) Manganese (d) Tungsten (e) Nickel 4. Give composition of following steels: (a) Rail steel (b) Spring steel (c) High speed steel 5. What are the special features of the following steels? (a) Hadfield steel (b) Maraging steel (c) Invar steel 6. Differentiate clearly between ferritic, austenitic and martenstic stainless steels.

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Engineering Materials

851

7. Briefly describe the following: (a) Grey cast iron (b) Malleable cast iron (c) Nodular cast iron 8. What are the important applications of magnesium alloy? 9. What are the characteristics of titanium alloys? What are their applications? 10. Give the composition and uses of following brasses: (a) admiralty brass (b) cartridge brass (c) gilding metal (d) yellow brass 11. What are bronzes? What are the important applications of phosphor bronze? 12. What are the special properties of Inconels and Nimonics? Where are they used? 13. Name a few alloys used in heating elements in furnaces, hair dryers, toasters etc. What is the composition of these alloys? 14. What are Stellites? What are their salient properties? What are their applications? 15. What are tin Babbits and lead Babbits? Where these are used? 16. What are thermoplastic and thermosetting plastics? What are basic differences between them? 17. Name four thermosetting plastics? What are their applications? 18. Name at least eight thermoplastics along with their applications? 19. What is rubber? What is vulcanization of rubber? How the strength of rubber is increased by vulcanization? 20. What do you understand by ceramics? What is the difference between a refractory, a glass and an abrasive? 21. What do you mean by glass? Give composition and uses of (a) Sodalime glass (b) Lead glass or flint glass (c) Pyrex glass (d) Vycor glass 22. What do you understand by a polymorph of carbon, that is,? (a) diamond (b) graphite (c) fullerene

Multiple Choice Questions 1. What is the percentage of carbon in tool steel? (a) 0.35 (b) 0.65 (c) 0.9 (d) None of these 2. Which one of following statements is incorrect for high-carbon steel? (a) these are wear resistant (b) always used under hardened and tempered conditions (c) they are tough and ductile (d) used for rail tracks, gears

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3. What is the purpose of adding Nickel in steel? (a) Nickel stabilizes austenite (b) Nickel forms hard carbides (c) Nickel raises creep strength (d) None of above 4. For stone crushers which of the following steel is used: (a) Invar steel (b) High speed steel (c) Hadfield steel (d) None of these

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852 Chapter 22 5. Make proper combination A Ferritic stainless steel I Ball and roller races B Austenitic stainless II Turbine blades C Martensitic stainless steel III Storage for acids A B C (a) I II III (b) III I II (c) III II I (d) II III I 6. What is the percentage of nickel in invar steel? (a) 30 (b) 36 (c) 40 (d) None of these 7. Strength and ductility of SG iron are more than for grey cast iron, because of (a) Graphite is in nodular form (b) Graphite is in cluster form (c) Graphite is in flakes form (d) None of these 8. Titanium alloys are used in (a) Pump bodies (b) Crank shaft (c) Jet engine components (d) All the above 9. Brass with a composition of Cu 95%, zinc 5% is known as (a) Cartridge brass (b) Munz metal (c) Gilding metal (d) None of these 10. Which of following alloy is used in electrical resistance strain gauges? (a) Delta metal (b) Constantan (c) Nickel silver (d) None of these 11. Make proper combination of alloys and its applications: A Munz metal I Pyrometer sheaths B Inconel II Heating elements of furnaces C Nichrome III Turbine blades D Nimonics IV Pump fittings

A B C (a) IV II I (b) IV I II (c) I II III (d) II I IV 12. Identify the thermosetting following: I Nylon II Melamine III PMMA IV Epoxy V Teflon (a) I, III, IV (b) II, IV (c) II, IV, V

D III III IV III plastics from the

(d) I, IV, V 13. Make proper combination of rubber and its applications: A Neoprene I Shoe soles B Butadiene II Expanded foams C Polyurethanes III Tyres and tubes D Butyl IV Gaskets A B C D (a) I II III IV (b) II I IV III (c) IV I II III (d) I IV III II 14. In Vycor glass, what is the percentage of silica? (a) 35–58 (b) 70–75 (c) 96–97 (d) None of these

Answers to Multiple Choice Questions 1. 2. 3. 4. 5.

(c) (c) (a) (c) (b)

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6. 7. 8. 9. 10.

(b) (a). (c) (c) (b)

11. 12. 13. 14.

(b) (b) (c) (c)

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Index

A Airy’s stress function, 771–72 equilibrium equations, 771–72 Alloy steels, 840 aluminium, 843 elements, effects of, 841 for high temperature use, 813 magnesium, 843 titanium, 843 Angular twist, 476 shaft due to twisting moment, development, 469–72 Area moment method, slope and deflection BM diagram, 433 diagram, 434 Area of cross-section bolt and tube assembly, 69 composite bar, 63 Axial bursting force, 2 strain energy, 629 thin cylindrical shell, 167 Axial deflection conical spring, 527 Axial tensile force, 2

B Babbits, 845–46 Barba’s Law, 32 Bars, 18 axial stresses, 8–9 of different lengths, 67–68 in equilibrium, 16–17

MTPL0259_Index.indd 853

stepped bar, 16 of uniform strength, 20–21 of varying cross-sections, 10 Young’s modulus, 9, 16–17 Bauschinger’s effect, 794 Beam bending, curvature, 403 bending moment (BM), 403 loading, impact on, 425–27 of mild steel, 796–97 with point load, 243 relation between rate of loading in BM and SF, 273–74 SF diagrams, 249 subjected to pure bending, 775–77 types of, 242 UDL with simply supported ends, 407–409 uniform strength, 314–17 variable loading, 270–71 Bending moment (BM), 241, 402 area moment method, slope and deflection, 433 in beam, 403 beam/cantilever with load through crank, diagrams, 266–69 cantilevers/beams with point loads, diagrams, 258–59 curved bars, 666 diametral load, ring, 681 differential equation, 404 positive and negative, 246–48 relation between rate of loading, 273–75 types, 246–47

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854 Index

with UDL, cantilevers/beams diagrams, 262–66 variable loading on beam, 270–73 BHN. see Brinell’s Hardness Number (BHN) Bolt and tube assembly, 68, 70–71 area of cross-section, 69 compressive force in, 69 contraction in length of, 69 extension in length of, 69 strain in, 69 tensile force in, 69 Young’s modulus, 69 Brick, compression test material testing, 825–27 Brinell’s Hardness Number (BHN), 788, 804, 828–29 of various materials, 807 Brittle type fracture Charpy test, 802 notched bar impact test, 802 notch effect, 801–802 notch sensitivity, 802 straining rate, effect, 801 temperature, effect, 800 triaxiality of stresses, 800 Bronze composition and uses of, 844 Built up section, 313 rolled steel, 312 Bulk modulus volumetric stress and strain, 22 Young’s modulus, relationship between, 154–56 Bursting force thick spherical shell, 217

C Cantilever carrying point loads, 249 point load at free end, 410–11 UDL, 412–13 with uniform strength, beams, 315–16 Carbon, polymorph, 849 ceramics in engineering use, 850 Carbon steels, 840 Carriage springs. see Leaf springs

MTPL0259_Index.indd 854

Case hardening of materials, 811–12 Castigliano’s first theorem, 627, 629 beam, strain energy stored in, 628 loads and deflections, graphs between, 628 Cast irons grey, 842 malleable, 842–43 white, 842 Central point load bending moment and curvature, 405 deflection, 406 flexure curve of, 405 Centroidal layer curved bars, 662–65 Ceramic materials, 848 Chain link deflection, 694–96 tensile load, 684–89 Channel section moment of inertia, 308 section modulus, 308 Charpy impact test material testing, 833 Chimney shafts, wind pressure, 384–85 Circular section core of, 379–80 cross section, area, 676 strip, area, 677 Circular tapered bar, 13 Close-coiled helical spring, 512 axial load, 513–15 Closed ring-quadrant curved bars, 692 Cobalt, 845 Coil of helical springs, 512 Complementary shear stress distribution, 7 moment of couple, 5 rectangular block, 6 stress force, 5 Composite bar area of cross-section, 63 change in length of, 65 and coefficients of expansion, 72 compressive strain, 73 force shared, 64 free expansion, 72

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Index

grips, 63 Hooke’s law, 63 temperature stresses, 74 tensile strain, 73 Young’s modulus of elasticity, 72 Composite beam Hooke’s law, 318–19 neutral layer (NL), 319 Composite systems, 67 corresponding strains, 65 length, changes in, 65 ratio of stresses, 66 stress in bar, 65 Compound cylinder outer cylinder, 208, 210 resultant stresses inner, 209 shrinkage fitting, 208 stresses due to junction pressure, inner cylinder, 208 variation of hoop stresses, 209–10 Compound shaft torque, 478 Compressive force in bolt and tube assembly, 69 composite bar, 73 Concrete, 322 RCC structure, 323 Conical spring, 526 axial deflection, 527 total strain energy, 527 Conical water tank hoop stress, 180 meridional stress, 180–81 radius of curvature and pressure, 179 weight density of water, 179 Conjugate beam method, 435–36 Contraction in length of bolt and tube assembly, 69 Copper composition and uses, 844 nickel alloys, 844 Crack initiation and propagation fatigue life, 811 Creep, 813 define, 812 Cross-sectional area tapered bar of elementary strip, 15

MTPL0259_Index.indd 855

855

Curvature, 402 bending in beam, 403 differential equation, 404 Curved bars assumptions, 662 bending moment, 666 centroidal layer, 662–65 closed ring-quadrant, 692 deflection, 691–93 of principal stresses in beam, sets, 352 strain, 662–63 total resisting moment, 664 Young’s modulus, 664 Cyclic stresses cantilever specimen and pure bending sample, 809 S-N curve, 810 types of, 808

D Deflection, 436 of curved bars, 691–93 Deformations, 761 normal and shear strains, 762 two-dimensional case, 763 Diametral bursting force thin spherical shell, 171 Diametral load, ring bending moment, 681 normal force, 680 resultant and direct stress, 682 Young’s modulus, 679 Differential equation of curvature, 404 Dimensions and applications of unequal I-section, 309 Direct force, 1 Directional distribution of shear stresses I-section, 353 Distribution of complementary shear stress, 7 Double curved wall, pressure vessel circumferential stress, 179 element and stresses, 177–78 forces, 178–79 thin cylindrical shell, 179 thin spherical shell, 179

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856 Index

Ductile material load extension curves for, 599 Ductility mechanical properties of various materials, 792

E Eccentric axial thrust on column, 371 direct compressive load, 372 maximum bending stresses, 372 resultant stress, 373 section modulus and direct stress, 372 Eccentricity in geometry, long columns, 569–70 deflection curve equation, 571 Eccentric load on beam, 421–23 of columns, 564–65 Elastic constants, 148 Lame’s coefficient, 768 modulus of rigidity, determination, 150–51 relation between, 152–54 Young’s modulus and Poisson’s ratio, 149–50 Elastomers, 847–48 Ellipse of stresses, 113–15 Energy approach critical load and condition, 582 Engineering component application of FOS, 34 Environmental effects of materials, 812 Equilibrium equations Airy’s stress function, 771–72 bars, 16, 21 small infinitesimal element in, 770 three-dimensional case, 771 two-dimensional case, 771 Equivalent length of column Euler’s buckling load, 554 limitations of, 555 Euler’s theory of buckling, 551, 554 Extension in bar due to self-weight, 18–20 Extension in length of bolt and tube assembly, 69

F Factor of safety (FOS), 35 engineering component, application, 34

MTPL0259_Index.indd 856

Fatigue factors affecting, 810–11 failure of, 808 Flexure formula, 298, 302, 311 curve, 405 modulus of rupture, 311 Free expansion of composite bar, 72

G Glasses, 848 types, 849 Graphical representation, 111–13 Mohr’s theory, 106, 613–14 plane, strain components of, 116 principal directions, 110 procedure, 107–109 proof of principal stresses, 109 Rankine’s theory, 600 sign conventions, 106–107 strain energy, 608–609 stress on inclined plane, 109–110 St. Venant’s theory, 606 Tresca theory, 603–604 Von Mises theory, 612 Grey cast irons, 842

H Hardness of material defined, 803 Helical springs close-coiled, 512–15 coil of, 512 material testing, stiffness of, 835–37 Hemispherical ends for cylindrical shell, 172 hoop strain, 173 Higher-order differential equation, 556 Hooke’s law, 3, 301 channel section, 63 composite bar, 63 composite beam, 318–19 Lame’s coefficients, 766–67 stress strain on simple bar, 765–66 Hoop stress compound cylinder, variation of, 209–10 conical water tank, 180

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Index

Lame’s coefficients, 200 thick cylindrical shell, 206–207 wire winding of thin cylindrical shell, 174–75 Horse-power transmitted by shaft, 474 torque, 475 Hub and shaft assembly circumferential and radial stress, 214 keyways, 214 Lame’s constants, 214 radial strain in, 214 shrinkage allowance, 215

I Impact loads, 30 bar, 28 quadratic equation, 29 Inclined plane cross-sectional area, 8 forces on, 7 normal and tangential component, 7 Inclined plane (I), stresses, 100 internal resistances, 99 normal and shear stresses, 99 triangular element with, 98 Inclined plane (II), stresses, 103 expressions, 102 forces on planes, 100–101 Indent formation, mechanism, 806 Izod impact test material testing, 831–33

J Johnson’s parabolic formula, 561–62

K Key fitting shaft and hub, 487 stresses developed in, 486–87

L Lame’s coefficients, 768 axial stress, 200–201 boundary conditions, 202 circumferential stress, 202–203 Hooke’s law, 766–67

MTPL0259_Index.indd 857

857

hoop stresses, 200 radial stress, 200, 202–203 small element, stresses on, 200–201 Lateral strain axial stress versus axial strain graph, 11 bar of length and diameter, 10 elastic constants, 11 Leaf springs, 528 material testing, stiffness of, 837–38 plates, 530–31 theory for deflection, 529 Load eccentric, 375 bending stresses, 376 direct compressive stress, 376 section modulus, 376 Load extension curves for ductile material, 599 Load through crank SF and BM diagrams of beam/cantilever, 266–67 Longitudinal strain axial stress versus axial strain graph, 11 bar of length and diameter, 10 elastic constants, 11

M Macaulay’s method beam with several points loads, 414 steps, 414 Malleable cast iron, 842–43 Materials, behaviour under bending, 794 under impact, 799–803 modulus of rupture, 795 under torsion, 798–99 Material testing bending attachment, 824 BHN, 828–29 brick, compression test, 825–27 Charpy impact test, 833 compression test cast iron specimen using UTM, 823–24 deflection test on bar using UTM, 824–25 double shear test on specimen using UTM, 822–23

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858 Index

helical spring, stiffness of, 835–37 Izod impact test, 831–33 leaf spring, stiffness of, 837–38 machine, mechanism, 834 measurement pull, mechanism, 818 movable cross-head, 823 percentage elongation gauge, case, 817 RHN, 829–30 sample using UTM, tensile test, 820–22 shear punching test, 819–20 spring machine testing, 836 tensile test using tensometer, 816–19 torsion test, 833–35 VPN, 830–31 wooden sample, compression test, 827–28 Maximum principal strain theory. see St. Venant’s theory Maximum principal stress theory. see Rankine’s theory Maximum shear stress theory. see Tresca theory Maximum stress uniform strength, beams, 314 Maxwell’s reciprocal theorem, 639 total strain energy, 640–41 Meridional stress conical water tank, 180–81 Mild steel, 33, 816 Barba’s Law, 32 beams of, 796–97 load extension curve for, 818 mechanical properties, 30–31 tensile load and extension, 30 total extension, 32 Modulus of rigidity, 151, 792 extension in length of diagonal, 152 longitudinal strain, 152 Mohr’s stress circle, 152–53 shear strain, 152–53 shear stress, 5 torsion test, 150 Modulus of rupture flexure formula, 311 shear stress-shear strain curves, 474 torsion formula, 474 Mohr’s theory, 106 graphical representation of, 613–14

MTPL0259_Index.indd 858

normal strains, 120 principal angles, 121 Moment of couple complementary shear stress, 5 Moment of inertia, 716–17 channel section, 308 symmetrical I-section, 306 unequal I-section, 310 Moment of resistance bending stresses in, 303 flexure formula, 302 neutral layer (NL) and neutral axis, 302 section modulus, 303

N Neutral axis, 301 force, 302 Neutral layer (NL) composite beam, 319 and neutral axis, moment of resistance, 302 simple bending, theory, 300 Nickel alloys, 844-45 Nodular cast iron, 843 Non-linear behaviour atomic scale, 791 mild steel, load-extension curve for, 790 Young’s modulus, 791 Normal and shear strains deformations, 762

O Open-coiled helical spring, 511, 516–18 axial moment, 520–21 stresses developed in spring wire, 522–23 Outer cylinder compound, 208

P Parallel axes theorems area with centroidal co-ordinates, 305 moment of inertia, 305 Pascal’s law of fluid mechanics, 22 Perpendicular axes theorems area with centroidal co-ordinates, 305 moment of inertia, 305

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Index

Plane spiral spring, 511, 524 bending moment, 525 energy stored, 525 Plane stress condition, 758, 760 second-degree polynomial, 774 Young’s modulus and Poisson’s ratio, 759 Plastics, 846 Poisson’s ratio axial stress versus axial strain, 11, 149 bar of length and diameter, 10 elastic constants, 11 lateral strain versus axial strain, 149 plane stress condition, 758, 760 static tension, mechanical behaviour, 789 Principal axes, 709–10 determination of, 713–16 Principal planes, 106 tapered bar under compression, 107 Young’s modulus, 127–28 Principal stresses, 97, 128 angles of planes, 104 ellipse of, 113–15 graphical solution, 106–12 on inclined plane (I), 98–99 on inclined plane (II), 100–103 maximum and minimum, determination, 104–105 Mohr’s strain circle, 120–23 practical cases of, 106 rosettes, 129–33 strain components, 115–19 strain gauge rosettes, 129 stress invariant, 105 three-dimensional stresses, 123–27 values of, 105 Product of inertia parallel axes theorem, 711–12 Professor Perry’s approximate formula, 566–68 Robertson formula, 573–74 Proof resilience, 26 Propped cantilever, 427–28 Pyramid hardness Knoop hardness and BHN, 805

MTPL0259_Index.indd 859

859

Q Quarter elliptic spring, 531 equivalent cantilever spring, 532

R Radius of curvature and pressure conical water tank, 179 Rankine Gordon formula, 559 constants for different materials, 560 Euler’s buckling load, 560 Rankine’s theory brittle materials, 600 graphical representation, 600 thin cylindrical shell, case, 601 RCC beams economic section, 324 Hooke’s law, 323 stresses and strains in, 323 Young’s modulus, 324 Rebound hardness, 806–807 Rectangular block of complementary shear stress, 6 Rectangular cross-section, 666–67 Rectangular section core of, 378–79 Rectangular strips, value channel section, 670 T-section and I-section, 669 Residual stresses, 797–98 RHN. see Rockwell hardness number (RHN) Robertson formula, 573–74 Rockwell hardness number (RHN), 829–30 Rockwell hardness test, 805–806 Rolled steel sections symmetrical I-section, 306 Rosettes strain gauge, used, 129–30

S SCF. see Stress concentration factor (SCF) Second-degree polynomial plane stress condition, 774 Section modulus channel section, 308

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860 Index

Section with variable breadth, 344 forces, 345 shear stress, 345–46 Shafts, 483 inertia, polar moment, 476 maximum principal stress, 482 small element, stress, 482 stepped, 477 tapered circular, 476 torsional resilience, 484–85 twisting and bending moments, 482 uniformly tapered circular, 475–76 Shear centre, 727–28 Shear force (SF), 241 applied moment, diagrams of beam/ cantilevers, 256–57 diagrams of beams, 249–51 with load through crank, beam/cantilever diagrams, 266–69 positive and negative, 242–46 relation between rate of loading, 273–75 types of beams, 242 with UDL, cantilevers and beams diagrams, 252–56 variable loading on beam, 270–73 Shear modulus elastic constants of different materials, 792 modulus of rigidity, 792 Shear strain energy/distortional strain energy theory. see Von Mises theory Shear stress, 471 in circular section of beam, 349–51 directional, 353 due to variable bending moment, 344 hollow circular shaft, distribution in, 473 modulus of rigidity, 5 positive and negative shear force, 4–5 in rectangular section of beam, 347–48 shaft due to twisting moment, development, 469–72 solid circular shaft, distribution in, 473 strain energy, 631–32 Shock load. see Impact loads Shrinkage allowance inner cylinder, 212 outer cylinder, 212–13

MTPL0259_Index.indd 860

Shrinkage fitting of compound cylinder, 208 Sign conventions, 404 graphical method, 106–107 Simple bending, theory assumptions, 298–99 element of beam, 300 Hooke’s law, 301 linear stress distribution, 301 neutral layer (NL), 300 radius of curvature, 300 Simple stresses and strains, 1 bar of uniform strength, 20–21 bars subjected to various forces, 16–18 complementary shear stress, 6–7 cross-sections, bars, 8–10 extension in bar due to self-weight, 18–20 FOS, 34–35 impact loads, 28–30 on inclined plane, 7–8 lateral strain and Poisson’s ratio, 10–12 longitudinal strain, 10–12 mild steel, tensile test, 30–33 shear stress and strain, 4–5 statically indeterminate problems, 23–25 strain energy and resilience, 25–26 stress concentration, 33–34 sudden load, 26–28 tapered bar, 12–14 tapered flat, 15–16 tensile and compressive stresses, 2–4 volumetric, 22–23 Single bar, temperature stresses, 72 area of cross-section, 71 compressive strain, 71 Small element of thin spherical shell, 171 S-N curve cyclic stresses, 810 Spherical shells, 170–71 conical water tank, 179–80 derivation of axial and hoop stresses, 166 development of axial and hoop stresses, 165 with hemispherical ends, 172–73 pressure vessel with double curved wall, 177–79 subjected to internal pressure, 166–69 wire winding, 174–76

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Index

Spring plate, 817 Spring wire section resultant shear stress in, 513 Stainless steels, 840 composition, properties and uses, 841 Statically indeterminate problems, 25 bar field between two rigid supports, 23 deformation compatibility equation, 24 portions under axial loads, 23 Young’s modulus, 23–24 Static compression, behaviour of materials, 793 axial compressive stress, 794 Bauschinger’s effect, 794 Static tension, mechanical behaviour yield strength, 790 Young’s modulus and Poisson’s ratio, 789 Stepped beams, 431–32 slope and deflection of, 438–39 Stepped shaft, 477 Stiffness of helical springs, 835–37 Strain, 607 in bolt and tube assembly, 69 displacement, 118 due to bending, 634–35 due to twisting moment, 637–38 graphical representation, 608–609 on inclined plane, 122 method, 518–19 normal strain, 115–16 plane, graphical representation, 116 principal stresses, 608 and resilience, 25–26 and shear strain, 115–17, 119 on small element and enlarged view, 117 total shear angle, 119 two-dimensional case, 608 Strain gauge rosettes electrical resistance, 129 Stress and temperature effects of materials, 813 Stress concentration elliptical hole, plate with, 34 SFC, 33–34 stepped bar under axial load, 33 Stress concentration factor (SCF), 33–34 Stress concentration in bending, 326–27 Stress distribution

MTPL0259_Index.indd 861

861

symmetrical I-section, 306 Stress force complementary shear stress, 5 Stress invariant, 105 Stress relaxation, 813–14 Stress tensor, 756 elemental tetrahedron, faces, 755 plane stress case, 759-760 shear strains, 759 three-dimensional, 754, 760 Strut lateral loading with point load, 575–77 with uniformly distributed lateral load, 578–80 St. Venant’s theory graphical representation, 606 and Poisson’s ratio, 605–606 principal stresses, 605 Young’s modulus, 605 Sudden load, 26, 28 and deformation, graph, 27 Superficial hardness test minor and major loads, 807 Surface treatments of materials, 811 Symmetrical bending, 708 Symmetrical I-section, 305 moment of inertia, 306 rolled steel sections, 306 stress distribution in, 306

T Tapered bar axial stress and strain, 13 circular shaft, 476 cross-sectional area, 13, 15 elementary disc, diameter, 12 stress at elementary strip, 15 total change in the length of flat, 15–16 Temperature stresses in composite bar, 74 Tensile and compressive stresses axial stress in, 2–3 Hooke’s law, 3 lateral strain in, 3 Young’s modulus of elasticity, 3 Tensometer magnification, 817

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862 Index

tensile test specimen and grip, 816 Thermoplastics, 846–47 Thick cylindrical shell, 199 compound cylinder, 208–10 hoop stress, 206–207 hub and shaft assembly, 214–15 Lame’s equations, 200–203, 205 shrinkage allowance, 212–13 spherical shell, 216–20 subjected to external pressure, 205–207 Thick spherical shell, 216 bursting force, 217 circumferential and radial strain, 217–18 constant of integration, 219 hoop and radial stress, 219 principal stresses, 218 resisting force, 218 Thin cylindrical shell axial bursting force, 167 change in volume, 169 circumferential stress, 167–68 conical water tank, 179–80 derivation of axial and hoop stresses, 166 development of axial and hoop stresses, 165 diametral bursting force, 168 double curved wall, pressure vessel, 179 with hemispherical ends, 172–73 internal pressure, small element, 167 pressure vessel with double curved wall, 177–79 small element, subjected to internal pressure, 166 spherical shell, 170–72 strains, 168–69 subjected to internal pressure, 166–69 volumetric strain, 169 wire winding of, 174–76 Thin spherical shell, 170 diametral bursting force, 171 double curved wall, pressure vessel, 179 small element of, 171 strains, 171–72 Three-dimensional stresses equilibrium equations, 771 normal stress and shear stress, 123 octahedral plane, 125, 127

MTPL0259_Index.indd 862

stress invariants, 124–25 stress tensor defines, 123–24 Torque of compound shaft, 478 Torsion failure, 799 formula, 474 Torsional loading stress concentration in, 488 Total strain energy conical spring, 527 Total twisting moment, 472 Trapezoidal section area of strip, 674 Tresca theory, 605 graphical representation, 603–604 Mohr’s stress circle, three-dimensional, 602–603 T-section modulus, 307 Twisting moment stresses in shaft subjected, 481–82 Two-dimensional case deformations, 763 Types of cyclic stresses, 808

U Unequal I-section dimensions and applications of, 309 moment of inertia, 310 section modulus, 310 Uniform strength, beams cantilever with, 315–16 maximum stress, 314 rectangular section, 314 variable width of cantilever, 316 variation of depth, 315–17 with varying width, 314 Universal testing machines (UTM) deflection test on bar using, material testing, 824–25 double shear test on specimen using, material testing, 822–23 material testing, 823–24 tensile test, material testing, 820–22

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Index

Unsymmetrical bending and shear centre, 707 beams, deflection, 723–25 channel section, 709 plane of bending, 709 principal axes, 709–10 principal moments of inertia, 710 product of inertia of area, 710 stresses, 719–21 UTM. see Universal testing machines (UTM)

863

chimney shafts, 384–85 on walls, 380–84 Wire winding of thin cylindrical shell, 176 axial bursting force in, 175 circumferential strain in, 175 dimensional bursting force, 175 hoop stress, 174–75 tensile stress in, 174–75

Y V Vicker’s Pyramid Number (VPN), 788, 830–31 Virtual forces applied to trusses, principle, 642 real load and unit load, 643 truss for, 644 Volumetric stress and strain, 23 bulk modulus, 22 Pascal’s law of fluid mechanics, 22 Von Mises theory graphical representation, 612 metal forming dies, 610–11 principal stresses, 611 total strain energy, 611 two-dimensional case, 612 volumetric stress, 611 VPN. see Vicker’s Pyramid Number (VPN)

W Wahl’s factor, 512–13 Weight density of water conical water tank, 179 White cast iron, 842 Wind pressure

MTPL0259_Index.indd 863

Yield strength in pure bending, 796 stress-strain graph, 792 Young’s modulus, 791 axial stress versus axial strain, 149 bars, 9, 16–17 bolt and tube assembly, 69 bulk modulus, relationship between, 154–56 composite bar of elasticity, 72 curved bars, 664 diametral load, ring, 679 of elasticity, 3, 72 extension in length of diagonal, 152 lateral strain versus axial strain, 149 longitudinal strain, 152 Mohr’s stress circle, 152–53 non-linear behaviour, 791 shear strain, 152–53 statically indeterminate problems, 23–24 static tension, mechanical behaviour, 789 strain along diagonal, 153 St. Venant’s theory, 605 tensile and compressive stresses of elasticity, 3 Young’s modulus of elasticity, composite bar, 72

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