Strength of Materials [4 ed.] 3030596664, 9783030596668

This fourth edition focuses on the basics and advanced topics in strength of materials. This is an essential guide to st

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Strength of Materials [4 ed.]
 3030596664, 9783030596668

Table of contents :
Preface to the Fourth Edition
Preface to the Third Edition
Preface to the Second Edition
Preface to the First Edition
Strength of Materials
Contents
About the Author
1
Simple Stresses and Strains
1.1 Introduction
1.2 Stress-strain curves in tension
1.3 True stress-strain curve
1.4 Poisson’s Ratio
1.6 Elongation produced in a test specimen
1.7 Shear Stress and Shear Strain
1.8 Volumetric strain
1.9 Bulk modulus of elasticity
1.10 Elastic constants and their relationships
1.10.1 Relationship between E and G
1.10.2 Relationship between E and K
1.10.3 Relationship between G and K
1.11 Factor of safety
1.12 Thermal stress and strain
1.12.1 Thermal Stress and Strain in a Simple Bar
1.12.2 Thermal Stress and Strain in a Compound Bar
2
Principal Stresses
2.1 Introduction
2.2 Stresses on an inclined plane (Principal planes and principal stresses)
2.3 Mohr’s circle of plane stress
3
Centroid and Moment of Inertia
3.1 Centre of gravity
3.2 First moment of area and centroid
3.3 Moment of inertia
3.3.1 Mass Moment of Inertia
3.3.2 Radius of Gyration w.r.t. Mass Moment of Inertia
3.3.3 Second Moment of Area
3.3.4 Radius of Gyration w.r.t. Second Moment of Area
3.4 Product of inertia
3.5 Principal axes and principal moments of inertia
3.6 Mohr’s circle for second moments of area
3.7 Parallel-axes theorem
3.8 Moment of inertia of a rectangular section
3.9 Moment of inertia of a solid circular section
3.10 Moment of inertia of a hollow circular section
3.11 Moment of inertia of a semi-circle
3.12 Moment of inertia of a quarter-circle
4
Shear Forces and Bending Moments in Beams
4.1 What is a beam?
4.2 Classification of beams
4.3 Types of loadings
4.4 Calculation of beam reactions
4.5 Shear forces in a beam
4.6 Bending moments in a beam
4.7 Sign conventions for shear force and bending moment
4.8 Shear force and bending moment diagrams (SFD and BMD)
4.9 Point of Contraflexure
4.10 SFD and BMD for cantilever beams
4.10.1 Cantilever Beam carrying a Point Load at its Free End
4.10.2 Cantilever Beam carrying Uniformly Distributed Load (udl) throughout the Span
4.10.3 Cantilever Beam carrying Uniformly Distributed Load over a certain Length from the Free End
4.10.4 Cantilever Beam carrying Uniformly Distributed Load over a certain Length from the Fixed End
4.10.5 Cantilever Beam carrying Uniformly Distributed Load over its Entire Span and a Point Load at its Free End
4.10.6 Cantilever Beam carrying several Point Loads
4.10.7 Cantilever Beam carrying Uniformly Varying Load
4.11 SFD and BMD for simply supported beams
4.11.1 Simply Supported Beam carrying a Central Point Load
4.11.2 Simply Supported Beam carrying an Eccentric Point Load
4.11.3 Simply Supported Beam carrying Uniformly Distributed Load (udl) over its Entire Span
4.11.4 Simply Supported Beam carrying Uniformly Varying Load which varies from Zero at Each End to w per unit length at the Midpoint
4.11.5 Simply Supported Beam carrying Uniformly Varying Load which varies from Zero at One End to w per unit length at Other End
4.11.6 Simply Supported Beam subjected to a Couple
4.12 Relations among load, shear force and bending moment
4.13 SFD and BMD for overhanging beams
4.13.1 Overhanging Beam with equal Overhangs on Each Side and loaded with Point Loads at the Ends
4.13.2 Overhanging Beam with equal Overhangs on Each Side and loaded with a Uniformly Distributed Load over its Entire Span
5
Stresses in Beams
5.1 Pure bending in beams
5.2 Simple bending theory
5.3 Position of the neutral axis
5.4 Section modulus
5.5 Composite beam
5.6 Beams of uniform strength
5.7 Shear stresses in beams
5.8 Shear stress distribution (general case)
5.9 Shear stress distribution in a rectangular cross-section
5.10 Shear stress distribution in a circular cross-section
5.11 Shear stress distribution in an I-section
6
Deflections of Beams
6.1 Introduction
6.2 Differential equation of flexure
6.3 Sign conventions
6.4 Double integration method
6.4.1 Cantilever Beam carrying a Point Load at its Free End
6.4.2 Cantilever Beam carrying udl over its Entire Span
6.4.3 Cantilever Beam subjected to a Pure Couple at its Free End
6.4.4 Cantilever Beam carrying a Point Load anywhere on its Span
6.4.5 Cantilever Beam carrying Gradually Varying Load
6.4.6 Simple Beam carrying a Central Point Load
6.4.7 Simple Beam carrying udl over its Entire Span
6.5 Macaulay’s method
6.6 Moment-area method
6.6.1 Cantilever Beam carrying a Point Load at its Free End
6.6.2 Cantilever Beam carrying a udl over its Entire Span
6.6.3 Simple Beam carrying a Central Point Load
6.6.4 Simple Beam carrying a udl over its Entire Span
6.7 Conjugate beam method
6.7.1 Simple Beam carrying a Point Load at its Centre
6.7. 2 Simple Beam carrying a Point Load not at the Centre
6.8 Method of superposition
7
Torsion of Circular Members
7.1 Introduction
7.2 Torsion equation
7.3 Torsional rigidity
7.4 Polar modulus
7.5 Power transmitted by a shaft
7.6 Effect of stress concentration
7.7 Torsion of a tapered shaft
7.8 Torsion of a thin circular tube
7.9 Strain energy due to torsion
8
Springs
8.1 Introduction
8.2 Spring terminology
8.3 Classification of springs
8.4 Load-deflection curve
8.5 Leaf spring
8.6 Quarter-elliptic leaf spring
8.7 Spiral spring
8.8 Helical spring
8.8.1 Close Coiled Helical Spring subjected to an Axial Load
8.8.2 Close Coiled Helical Spring subjected to an Axial Twist
8.8.3 Open Coiled Helical Spring subjected to an Axial Load
8.8.4 Open Coiled Helical Spring subjected to an Axial Twist
8.9 Combination of springs
8.9.1 Series Combination
8.9.2 Parallel Combination
9
Strain Energy
9.1 Introduction
9.2 Strain energy due to direct loads
9.2.1 Strain Energy due to Gradually Applied Load
9.2.2 Strain Energy due to Suddenly Applied Load
9.2.3 Strain Energy due to Impact or Shock Load
9.3 Strain energy due to shear
9.4 Strain energy due to pure bending
9.5 Strain energy due to principal stresses
9.6 Strain energy due to volumetric strain
9.7 Shear strain energy due to principal stresses
9.8 Castigliano’s theorem
10
Theory of Elastic Failure
10.1 Introduction
10.2 Maximum Normal stress theory
10.3 Maximum Normal strain theory
10.4 Maximum Total strain energy theory
10.5 Maximum shear stress theory
10.6 Maximum Distortion energy theory
11
Buckling of Columns
11.1 Introduction
11.2 Important terminology
11.3 Classification of columns
11.4 Euler’s theory
11.4.1 Euler’s Formula (when Both Ends of the Column are Hinged or Pinned)
11.4.2 Euler’s Formula (when Both Ends of the Column are Fixed)
11.4.3 Euler’s Formula (when One End of the Column is Fixed and Other End Hinged)
11.4.4 Euler’s Formula (when One End of the Column is Fixed and Other End Free)
11.4.5 Crippling Stress
11.5 Empirical formulae
11.5.1 Rankine-Gordon Formula
11.5.2 Johnston’s Parabolic Formula
11.5.3 Straight Line Formula
11.6 IS Code formula (IS: 800-1962)
11.7 Secant formula (for eccentric loading)
12
Pressure Vessels
12.1 Introduction
12.2 Stresses in A thin cylindrical shell
12.3 Volumetric strain for A thin cylindrical shell
12.4 Wire wound thin cylinders
12.5 Stresses in a thin spherical shell
12.6 Volumetric strain for A thin spherical shell
12.7 Cylindrical shell with hemispherical ends
12.8 Stresses in thick cylinders (Lame’s theory)
12.8.1 General Case (when Internal and External Pressures both are acting)
12.8.2 When only Internal Pressure is acting
12.8.3 When only External Pressure is acting
12.8.4 When a Solid Circular Shaft is subjected to External Pressure
12.9 Longitudinal stress
12.10 Strains in thick cylinders
12.11 Compound cylinders
12.11.1 Stress due to Shrinkage
12.11.2 Stresses due to Fluid Pressure
12.11.3 Resultant Stresses
12.11.4 Shrinkage Allowance
12.12 Stresses in A thick spherical shell
13
Plane Trusses
13.1 Introduction
13.2 Types of trusses
13.3 Forces in the truss
13.4 Analysis of Trusses
13.4.1 Analysis of Trusses by Method of Joints
13.4.2 Analysis of Trusses by Method of Sections
13.5 Zero-force members
14
Combined Loadings
14.1 Introduction
14.2 Combined Bending And Axial Loads
14.3 Combined bending and torsion of circular shafts
14.4 Combined Torsion and axial loads
14.5 Combined Bending, Torsion and Direct Thrust
14.6 Other Cases of combined loadings
14.6.1 Eccentric Loading on One Axis (Single Eccentricity)
14.6.2 Eccentric Loading on Two Axes (Double Eccentricity)
14.6.3 Biaxial Bending
14.6.4 Loading on a Chimney
14.6.5 Loading on a Dam
14.6.6 Loading on Retaining Walls
15
Unsymmetrical Bendingand Shear Centre
15.1 Symmetrical bending and simple bending theory
15.2 Unsymmetrical Bending
15.3 Doubly symmetric beaMs with skew or inclined loads
15.4 Pure Bending of Unsymmetrical Beams
15.5 Deflection in Unsymmetrical Bending
15.6 Shear Centre
15.6.1 Shear Centre for a Channel Section
15.6.2 Shear Centre for an Equal-leg Angle Section
16
Fixed Beams
16.1 Introduction
16.2 Shear Force and Bending Moment diagrams
16.3 Fixed Beam carrying a central point load
16.4 Fixed Beam carrying an eccentric point Load
16.5 Fixed Beam carrying Uniformly Distributed Load (UDL) over the Entire span
16.6 Fixed Beam carrying Uniformly Varying Load
16.7 Fixed Beam subjected to a couple
16.8 Sinking of a support
17
Rotating Rings, Discs and Cylinders
17.1 Introduction
17.2 Rotating Ring
17.3 Rotating Thin Disc
17.3.1 Hoop and Radial Stresses in a Rotating Solid Disc
17.3.2 Hoop and Radial Stresses in a Rotating Disc with a Central Hole
17.3.3 Hoop and Radial Stresses in a Rotating Disc with a Pin Hole at the Centre
17.4 Rotating Disc of uniform strength
17.5 Rotating long cylinder
17.5.1 Hoop and Radial Stresses in a Rotating Solid Cylinder or a Solid Shaft
17.5.2 Hoop and Radial Stresses in a Rotating Hollow Cylinder
18
Mechanical Testing of Materials
18.1 Introduction
18.2 Hardness test
18.2.1 Brinell Test
18.2.2 Rockwell Test
18.2.3 Vickers Test
18.3 Fatigue
18.4 Creep
18.5 Tension test
18.6 Compression test
18.7 Stiffness test
18.8 Torsion test
18.9 Bend test
18.10 Impact test
Model Multiple Choice Questions for Competitive Examinations
Appendix A
Appendix B
References
Subject Index

Citation preview

D. K. Singh

Strength of Materials 4th Edition

Strength of Materials

D. K. Singh

Strength of Materials Fourth Edition

123

D. K. Singh Department of Mechanical Engineering Netaji Subhas University of Technology New Delhi, Delhi, India

ISBN 978-3-030-59666-8 ISBN 978-3-030-59667-5 https://doi.org/10.1007/978-3-030-59667-5

(eBook)

Jointly published with Ane Books Pvt. Ltd. In addition to this printed edition, there is a local printed edition of this work available via Ane Books in South Asia (India, Pakistan, Sri Lanka, Bangladesh, Nepal and Bhutan) and Africa (all countries in the African subcontinent). ISBN of the Co-Publisher’s edition: 9789385462061 1st–4th editions: © The Author(s) 2007, 2009, 2013, 2021 This work is subject to copyright. All rights are reserved by the Publishers, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publishers, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publishers nor the authors or the editors give a warranty, express or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publishers remain neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

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Preface to the Fourth Edition

The fourth edition of the book is presented before you. The new edition includes four new chapters and has increased by about 300 pages. Now the book covers the subject in totality and takes care of both the basics and the advanced topics of the strength of materials. Besides including four new chapters, three existing chapters have been rewritten to make them more relevant and meaningful. This has further increased the scope of the book. I thankfully acknowledge the support and encouragement from two of my best ever friends, Jai Shankar and Shakil Ahmad. My parents and in-laws especially Babuji and Mummi are gratefully remembered for their blessings. My lovely wife Alka and daughters Shivangi and Shalvi deserve special mention for giving their affectionate support that keeps me always elevated and in high spirit. Finally, I appreciate the response from the readers of the book, which has helped tremendously to further improve the book. I hope readers will continue to give their valuable suggestions and remarks in future also. I promise to honour their suggestions.

New Delhi

D. K. Singh

vii

Preface to the Third Edition

In this revised and augmented third edition, many new solved problems have been added especially in the chapters 1, 2, 5, 6, 7, 10 and 13. Simultaneously, a large number of model objective questions have also been added to test the understanding of the students from the viewpoints of competitive examinations. The biggest advantage of this book is its simplicity in expressing the subject matter in the most simplified manner. The material covered is so designed that any beginner can follow it easily, and get a complete picture of the subject after having gone through the material covered in the text. I deeply appreciate the many comments and suggestions that I received from the users of the previous editions of this book. I may be contacted on my email for any further suggestion, comment or criticism. NSIT, New Delhi

D. K. Singh

ix

Preface to the Second Edition

The thoroughly revised edition of the book ‘Strength of Materials’ is in your hand. This new edition has one more chapter on Mechanical Testing of Materials, which further increases the scope of the book, while retaining the flavour of the first edition. A sincere attempt has been made to make the book error-free, but still can’t be claimed in totality. I hope, readers will enjoy this new edition with the same spirit. Any suggestion for the improvement of the book will be gratefully acknowledged.

D. K. Singh

xi

Preface to the First Edition

It gives me immense pleasure to present this book before you. There are many books on this subject DQGHDFKFODLPVWKHEHVWRQH,FDQ¶WPDNHVXFKFODLPEXWFDQDVVXUH\RXWKDW\RXZLOOUHDOO\¿QGWKLV ERRNLQWHUHVWLQJDQGXVHIXO7KHUHDUHWKLUWHHQFKDSWHUVLQDOODQGHDFKFKDSWHUKDVVXI¿FLHQWQXPEHU of solved examples for easy understanding of the subject. A number of multiple choice questions has been added at the end of each chapter, making the book useful for competitive examinations. I am thankful to all the people who helped in the compilation of this book including my publisher Mr. Sunil Saxena and Mr. Jai Raj Kapoor. Thanks are also due to my parents and in-laws for their continuous encouragement. Finally, I want to thank my wife, Alka, for her continuing encouragement, support, and affection and my daughters, Shalu and Sheelu, for making the atmosphere lively and workable. D.K. Singh New Delhi March 2007.

xiii

13 Strength of Materials Fourth Edition By D. K. Singh

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xv

Contents Preface to the Fourth Edition

vii

Preface to the Third Edition

ix

Preface to the Second Edition

xi

Preface to the First Edition 1.                 

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xvii

xviii

Strength of Materials

2. Principal Stresses

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4. Shear Forces and Bending Moments in Beams

143–206

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Contents xix

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xx

Strength of Materials

5. Centroid and Moment of Inertia

207–250



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6. 'HÀHFWLRQVRI%HDPV

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Contents xxi



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7. Torsion of Circular Members

319–364

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8. Springs

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409–432           

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Contents xxiii

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xxiv

Strength of Materials

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621-–692

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Contents xxv

15. Unsymmetrical Bending and Shear Centre

693–742



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743–798



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17. Rotating Rings, Discs and Cylinders

799–856



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857-–866



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±

Appendix A

±

Appendix B

±

References

±

Subject Index

±

About the Author

Dr. D. K. Singh is a Professor and Head of Mechanical Engineering at the Netaji Subhas University of Technology, New Delhi, India. He has an extremely rich experience of both teaching and research for nearly three decades. He has contributed over 45 papers to various journals of national and international repute. He has also participated and presented papers at several national and international conferences. He has also authored 12 books. He is a life member of the Indian Society for Technical Education (ISTE), New Delhi.

xxvii

1 Simple Stresses and Strains

Thomas Young (1773-1829)

Thomas Young, born on 13 July 1773, was an English physician, physicist and polymath. By the age of fourteen, he is said to be well acquainted with many languages like Latin, Greek, French, Italian, Hebrew, Arabic and Persian, and hence was nicknamed the English Leonardo. He made QRWDEOHFRQWULEXWLRQVWRWKH¿HOGVRIYLVLRQOLJKWVROLGPHFKDQLFVHQHUJ\ SKLORVRSK\ODQJXDJHPXVLFDOKDUPRQ\DQG(J\SWRORJ\+HGLVFRYHUHG WKHFDXVHRIDVWLJPDWLVPDYLVLRQGHIHFWLQ7KH Vy . Normals are dropped at the points M and N above and below x-axis such that MY = + Wxy and NX = – Wxy. Points X and Y are joined together to get a straight line XY. This line intersects x-axis at C. Taking C as centre and CX or CY as radius, draw a circle. The resulting circle is called Mohr’s circle. The circle cuts x-axis at two points K and L. OL = Vmax = V1 = Maximum (major) principal stress OK = Vmin = V2 = Minimum(minor) principal stress σx + σ y

.

 z

The centre C of the Mohr’s circle is located at

 z

²XCL (measured anticlockwise) = 2Tp, half of this angle i.e., TpLVGH¿QHGDVWKHDQJOHPDGHE\ the major principal plane with plane AB of the element under consideration in Fig. 2.12.

 z

The radius of the Mohr’s circle gives the maximum shear stress.

2

Fig. 2.13 The Mohr’s circle.  z

To know the position of the inclined plane AE, a point S is taken on the circle such that ²LCS = 2T (measured anticlockwise). The perpendicular distance from S on x-axis gives shear stress W„xy and the distance of this perpendicular from y-axis gives normal stress Vxc on the inclined plane. The complete sequence of construction of the Mohr’s circle is shown in Fig. 2.13.

Principal Stresses

75

Sign Conventions z z z

The tensile stresses are considered positive and hence they are plotted on right side of the origin. The compressive stresses are considered negative, hence they are plotted on left side of the origin. If the shear stress acting on a face has the tendency to rotate the element clockwise, then it is considered to be positive and is taken above x-axis (Fig. 2.14 (a)).

Fig. 2.14 z

If the shear stress acting on a face has the tendency to rotate the element counterclockwise, then it is considered to be negative and is taken below x-axis (Fig. 2.14 (b)).

Example 2.12 'UDZ0RKU¶VFLUFOHIRUDWZRGLPHQVLRQDOVWUHVV¿HOGVXEMHFWHGWR a) pure shear (b) pure biaxial tension (c) pure uniaxial tension and (d) pure uniaxial compression. Solution: Conditions

Mohr’s Circle

(a) Pure shear

(b) Pure biaxial tension

Contd...

76

Strength of Materials

(c) Pure uniaxial tension

(d) Pure uniaxial compression

Example 2.13 Using normal and shear stresses given in Example 2.1, (a) draw the Mohr’s circle (b  ¿QGWKHSULQFLSDOVWUHVVHV (c  ¿QGWKHPD[LPXPVKHDUVWUHVVDQG (d  ¿QGWKHSRVLWLRQRISULQFLSDOSODQHV Solution:

(a) Refer Fig. 2.15.

Selection of Scale Take Hence,

0.5 cm Vy Vx Wxy

= 5 MPa = 5 MPa = 0.5 cm = 25 MPa = 2.5 cm = 30 MPa = 3.0 cm

The Mohr’s circle is shown in Fig. 2.16.

Fig. 2.15

Principal Stresses

77

Fig. 2.16

(b) From Fig. 2.16, we have Vmax = V1 = 4.35 cm = 43.5 MPa

Ans.

Vmin = V2 = – 2.31 cm = – 23.1 MPa

Ans.

= 23.1 MPa (Compressive)

Ans.

(c) The maximum shear stress is Wmax = 3.3 cm = 33 MPa

Ans.

(d) The position of principal planes are given as: 2Tp = 63.5’ or

Tp = 31.75’

and

2Tp = 243.5’

or

Tp = 121.75’

1

2

Ans.

Example 2.14 At a point in the cross-section of a loaded member, the maximum principal stress is 15 N/mm2 tensile and maximum shear stress of 8 N/mm28VLQJ0RKU¶VFLUFOH¿QG a) the magnitude and nature of direct stress on the plane of maximum shear stress (b) the state of stress on a plane, making an angle 30º with the plane of maximum principal stress.

78

Strength of Materials

Solution:

Selection of Scale 1 cm = 5 N/mm2

Take

The Mohr's circle is shown in Fig. 2.17.

Fig. 2.17

OL = V1 = Maximum principal stress = 15 N/mm2 CL = Wmax= Maximum shear stress = Radius of Mohr's circle = 8 N/mm2 The direct stresses on the plane of maximum shear stress are: OC = Vx = 1.4 cm = 7 N/mm2 (Tensile) OK = Vy = – 0.2 cm = – 1 N/mm2 (Compressive)

Ans.

The stresses on a plane making an angle 30º with the plane of maximum principal stress are: ON = Vx = 2.2 cm = 11 N/mm2 (Tensile) OM = Vy = 0.6 cm = 3 N/mm2 (Tensile) YN = Wxy = 1.4 cm = 7 N/mm2

Ans.

Example 2.15 A circle of 100 mm diameter is drawn on a mild steel plate before it is subjected to direct tensile stresses of 80 N/mm2 and 20 N/mm2 in two mutually perpendicular directions and a shear stress of 40 N/mm2. Find the major and minor axes of the ellipse formed as a result of deformation of the circle. Assume, E = 200 GPa and Poisson’s ratio as 0.25. Solution:

Refer Fig. 2.18.

Given, Diameter of the circle on the plate, d = 100 mm Tensile stress in x-direction is

Vx = 80 N/mm2

Principal Stresses

Vy = 20 N/mm2 Wxy = 40 N/mm2 E = 200 GPa = 200 × 109 Pa v = 0.25

Tensile stress in y-direction is Shear stress, Modulus of elasticity, Poisson’s ratio

Fig. 2.18

The principal stresses are given as V1, 2 =

σx + σ y 2

±

(σ x − σ y )2 + 4τ 2xy 2

(80 − 20)2 + 4 × 402 80 + 20 ± = 2 2 = (50 ± 50) N/mm2 Hence,

V1 = 50 + 50 = 100 N/mm2

and

V2 = 50 – 50 = 0

Strain produced in the direction of V1 is

Strain produced in the direction of V2 is

σ σ 100 × 106 Pa 1 –ν 2 = E E 200 × 109 Pa –4 = 5 × 10 σ2 σ1 –ν 2 = E E 1 =

= – 1.25 × 10–4 Change in diameter of the circle along V1 is

'd1 = 1d = 5 × 10–4 × 100 = 0.05 mm

Change in diameter of the circle along V2 is

'd2 = 2d = – 1.25 × 10–4 × 100 = – 0.0125 mm

79

80

Strength of Materials

As a result of the stresses applied in x and y directions, the circle takes the form of an ellipse (Fig. 2.18). The diameter of the circle increases along V1 and decreases along V2. Hence, major diameter of the ellipse = (100 + 0.05) mm = 100.05 mm Ans. and minor diameter of the ellipse = (100 – 0.0125) mm = 99.9875 mm Ans. Example 2.16 A thin cylinder with closed ends has an internal diameter of 50 mm and a wall thickness of 2.5 mm. It is subjected to an axial pull of 10 kN and a torque of 500 N.m, while under an internal pressure of 6 MN/m2 . (a) Determine the principal stresses in the tube and the maximum shear stress. (b  5HSUHVHQW WKH VWUHVV FRQ¿JXUDWLRQ RQ D VTXDUH HOHPHQW WDNHQ LQ WKH ORDG GLUHFWLRQ ZLWK direction and magnitude indicated (schematic). (c) Sketch the Mohr’s stress circle. Solution:

Given,

Inside diameter of the cylinder,

d = 50 mm

Wall thickness of the cylinder,

t = 2.5 mm

Axial pull,

P = 10 kN

Torque applied,

T = 500 N.m

Internal pressure,

p = 6 MN/m2

The cross-sectional area of the cylinder is A = Sd × t = S × 50 × 10–3 × 2.5 × 10–3 = 3.927 × 10–4 m2 (a) The hoop stress is given as Vh =

–3 pd = 6 × 50 × 10 –3 = 60 MPa 2t 2 × 2.5 × 10

The longitudinal stress is Vl =

Vh 60 = = 30 MPa 2 2

The direct stress due to axial pull is V=

P = 10 × 103 × 1 MPa = 25.46 MPa A 3.927 × 10 –4 106

Principal Stresses

81

The longitudinal stress acts along the axis of the cylinder and the hoop stress acts perpendicular to it. The direct stress also acts along the axis of the cylinder. Hence, the total stress acting along the axis is Vl + V = 30 + 25.46 = 55.46 MPa = Vx (say) and Vh = Vy (say) The shear stress is Wxy =

2T 2 × 500 1 × 6 MPa = 50.93 MPa 2 = −3 2 −3 πd t π × (50 × 10 ) × ( 2.5 × 10 ) 10

The principal stresses are given as V1, 2 = = = V1 = = V2 = =

Hence, and

σx + σ y 2

±

(σ x − σ y )2 + 4τ 2xy 2

(55.46 − 60)2 + 4 × (50.93)2 55.46 + 60 ± 2 2 (57.73 ± 50.98) MPa Major principal stress 57.73 + 50.98 = 108.71 MPa Minor principal stress 57.73 – 50.98 = 6.75 MPa

Ans. Ans.

The maximum shear stress is obtained as Wmax =

 (b) and (c)

σ1 − σ 2 108.71 – 6.75 = = 50.98 MPa 2 2

Ans.

The stress configuration on a square element and the related Mohr’s circle is shown in Fig. 2.19. Vh

W Wxy Wmax

V

Vl

Vl

V

Wxy M

O

C

N

V2 Vx

Wxy

Vy

Vh

V1

(a  6WUHVVFRQ¿JXUDWLRQRQDVTXDUHHOHPHQW Fig. 2.19

b) Mohr’s circle

L

V

82

Strength of Materials

Example 2.17 A circular shaft is subjected to combined loads of bending M and torque T. With the help of Mohr’s circle diagram, represent the stresses on an element of the shaft surface. From this diagram or by FDOFXODWLRQ ¿QG WKH PD[LPXP VKHDU VWUHVV GXH WR WKH FRPELQHG HIIHFW RI WKHVH JUDGXDOO\ DSSOLHG loads of M and T. Solution:

The bending stress is 32 M πd 3 M = Bending moment

Vb =

(using bending equation)

d = Diameter of the shaft The shear stress is W=

16 M πd 3

(using torsion equation)

The principal stresses are given as σb2 + 4τ 2 σb ± 2 2 16 (M ± M 2 + T 2 ) = πd 3

V1, 2 =



 7KHPD[LPXPVKHDUVWUHVVLVREWDLQHGDV WPD[ =

σ1 − σ 2 2

σb2 + 4τ 2 2 16 M M2 −T2 = πd 3 =

The Mohr’s circle is shown in Fig. 2.20. ON = Vb OL = V1 OK = V2 OY = NX = W CC c = WPD[

Fig. 2.20

Principal Stresses

83

SHORT ANSWER QUESTIONS 1. 2. 3. 4. 5. 6.

What is principal stress? How many principal stresses are there for a plane stress condition? What is a principal plane? What is its significance? Who invented Mohr’s circle? What does the radius of Mohr’s circle indicate? How is the plane of maximum shear stress located? When is the shear stress positive or negative?

MULTIPLE CHOICE QUESTIONS 1. The stresses are said to be compound, when (a) (b) (c) (d)

normal and shear stresses are acting simultaneously torsion and bending stresses are acting simultaneously normal and bending stresses are acting simultaneously bending and stresses are acting simultaneously.

2. The principal planes are the planes of (a) maximum shear stress (c) maximum normal stress

(b) minimum shear stress (d) zero shear stress.

3. The principal stresses are given as (a)

(c)

σx +σy 2

σx −σy 2

±

±

2 (σ x + σ y )2 − 4τ xy

2 2 (σ x + σ y )2 + 4τ xy

2

(b)

σx +σy 2

(d)

σx −σy 2

±

±

2 (σ x − σ y ) 2 + 4τ xy

2 2 (σ x + σ y )2 − 4τ xy

.

2

4. The principal stresses are basically (a) shear stresses (c) normal stresses

(b) bending stresses (d) none of these.

5. The planes of maximum shear stress are located at which of the following angle to the principal SODQHV" (a) 90’

(b) 45’

(c) 60’

(d) 30’

84

Strength of Materials

6. The maximum shear stress is equal to (a) ±

2 (σ x − σ y )2 + 4τ xy

(b) ±

2

2 2 (c) ± (σ x − σ y ) − 4τ xy 2 7. The principal planes are separated by

(a) 180’

(d) ±

(b) 45’

2 (σ x + σ y )2 + 4τ xy

2 2 (σ x + σ y )2 − 4τ xy

.

2

(c) 90’

(d) 60’

8. The maximum shear stress is equal to (a) one-half of the algebraic difference of the principal stresses (b) the algebraic difference of the principal stresses (c) the sum of the principal stresses (d) the difference of the principal stresses. 9. For uniaxial loading condition, the maximum shear stress is equal to (a) uniaxial stress

(b) two times the uniaxial stress

(c) three times the uniaxial stress

(b) one-half of uniaxial stress.

10. For a complex stress system, the total number of principal planes is (a) two

(b) four

(c) three

(d) none of these.

11. The radius of the Mohr’s circle indicates the (a) maximum principal stress

(b) minimum principal stress

(c) maximum shear stress

(d) minimum shear stress.

12. In case one principal stress is zero, the other principal stress is equal to (a) maximum shear stress

(b) two times the maximum shear stress

(c) three times the maximum shear stress

(d) none of these.

ANSWERS 1. (a)

2. (d)

3. (b)

4. (c)

9. (d)

10. (c)

11. (c)

12. (b).

5. (b)

6. (a)

7. (c)

8. (a)

Principal Stresses

85

EXERCISES 1. A rectangular element is subjected to a tensile stress of 60 MPa, a compressive stress of 20 MPa and a shear stress of 30 MPa. Find (a) the principal stresses, (b) the maximum shear stress and (c) the location of the principal planes. (Ans. (a) 70 MPa, – 30 MPa; (b) 50 MPa; (c) 18.4o and 108.4o). 2. The stresses at a point on two perpendicular planes BC and AC are as shown in Fig. 2.21. Determine the position of the plane AB such that the shear stress on it is equal to zero. What ZLOOEHWKHQRUPDO SULQFLSDO VWUHVVHVRQVXFKSODQH"

Fig. 2.21

(Ans.

T = 18.43’, 90 MPa (Tensile), 10 MPa (Compressive)).

3. The principal stresses at a point within a strained material are 90 MPa and 70 MPa as shown in Fig. 2.22. Find the normal, the tangential and the resultant stress on a plane inclined at 30° to the axis of the major principal stress.

Fig. 2.22

(Ans.

75 MPa, 8.66 MPa, 75.5 MPa).

4. At a certain point in a strained material, two normal stresses, one a tensile stress of 100 MPa and another a compressive stress of 90 MPa are applied at two mutually perpendicular planes. If maximum diUHFWVWUHVVLVOLPLWHGWR03D WHQVLOH ¿QGWKHVKHDUVWUHVVRQWKHSODQHVRQ which two normal stresses are acting. (Ans. 109.54 MPa).

86

Strength of Materials

5. For a biaxial stress system, Vx = 65 MPa, Vy = 75 MPa and Wxy = 40 MPa. Find (a) the principal stresses and their planes and (b) the maximum shear stresses and the planes on which they occur. (Ans. (a) 110.31 MPa, 29.68 MPa, 48.56°, 138.56° (b) 40.315 MPa, 3.56°, 93.56°). 6. At a point in a strained material, the resultant intensity of stress across a plane is 100 MPa tensile inclined at 45° to its normal. The normal component of stress intensity across the plane at right angle is 30 MPa compression. Find the position of principal planes and stresses across them. $OVR¿QGWKHYDOXHRIPD[LPXPVKHDUDWWKDWSRLQW (Ans. 27.25°, 117.25°, 100.7 MPa, – 66.35 MPa, 86.6 MPa). 7. At a point in a loaded specimen, the principal stresses acting on two mutually perpendicular planes are 60 MPa and 40 MPa, both being compressive. Determine the resultant stress acting on a plane inclined at 60° measured clockwise to the plane on which the larger of the normal stresses is acting. (Ans. 45.82 MPa). 8. A steel shaft is subjected to a torque of 20 kN˜m and a bending moment of 10 kN˜m. The diameter of the shaft is 100 mm. Calculate the maximum and the minimum principal stresses and the maximum shear stress in the shaft at its surface. (Ans. 165 MPa, – 63 MPa, 114 MPa). 9. At a point in a strained material, the vertical shear stress is 15 MPa and the horizontal tensile VWUHVV LV  03D 8VLQJ 0RKU¶V FLUFOH PHWKRG ¿QG WKH SULQFLSDO VWUHVVHV DQG GLUHFWLRQ RI principal planes. (Ans. 32 MPa, – 7 MPa, 25°, 115°). 10. At a point in a material, there is a horizontal tensile stress of 270 MPa, a vertical tensile stress of 130 MPa and shearing stress of 40 MPa downward on left. With the aid of Mohr’s circle or RWKHUZLVH¿QGRXWWKHPD[LPXPDQGWKHPLQLPXPSULQFLSDOVWUHVVHVDQGWKHSODQHVRQZKLFK they act. Determine also the maximum shearing stress in magnitude and direction. (Ans. 280.62 MPa, 119.38 MPa, 14.87°, 104.87°, 80.62 MPa, 59.87°, 149.87°). 11. For a plane stress condition in which Vx = 140 MPa, Vy = 20 MPa and Wxy ±03D¿QGWKH principal stresses and the locations of the principal planes. (Ans. 165 MPa, –5 MPa, – 22.5°, – 112.5°). 12. For the plane stress condition shown in Fig.¿QGVx and the position of the principal planes, if the maximum principal stress is – 7MPa.

Fig. 2.23

(Ans.

105 MPa, – 26.6º, – 116.6°).

Principal Stresses

87

13. )RUWKHSODQHVWUHVVFRQGLWLRQVKRZQLQ)LJ¿QGWKHSULQFLSDOVWUHVVHVDQGWKHSRVLWLRQRI the principal planes.

Fig. 2.24

(Ans.

102.5 MPa, – 62.5 MPa, 36°, 126°).

14. A tensile stress V1 and a shear stress W act on a given plane of a material. Show that the principal stresses are always of opposite sign. If an additional tensile stress V2 acts on a plane perpendicular to that of V1¿QGWKHFRQGLWLRQWKDWERWKSULQFLSDOVWUHVVHVPD\EHRIWKHVDPHVLJQ (Ans.

W=

(σ1σ 2 ) ).

‰‰‰

3 Centroid and Moment of Inertia

Figure shows an I-section, which is one of the most widely used cross-sections for beams.

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ :KDWLVFHQWURLG"  ‡ +RZGRFHQWUHRIJUDYLW\DQGFHQWURLGGLIIHU"  ‡ :K\LVWKHDUHDPRPHQWRILQHUWLDRIWHQFDOOHGWKHVHFRQGPRPHQWRIDUHD"  ‡ :KDWLVFHQWURLGDOD[LV"  ‡ :KHUHLVSDUDOOHOD[HVWKHRUHPXVHG"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_3

89

90

Strength of Materials

3.1 CENTRE OF GRAVITY 7KH FHQWUH RI JUDYLW\ RI D ERG\ LV WKH SRLQW ZKHUH WKH HQWLUH ZHLJKW RI WKH ERG\ LV DVVXPHG WR EH concentrated. In other words, it is the point where the weight of the body acts. It may or may not be located within the body and its position depends upon the shape of the body. It is denoted by G. If the mass is uniformly scattered in the body, then the body can be assumed to be made of many elemental masses m1, m2, m3, ... . Suppose these elemental masses are located at distances x1, x2, x3, ... from yD[LVDQGy1, y2, y3, ... from xD[LVWKHQFHQWUHRIJUDYLW\RIWKHERG\LVGH¿QHGE\ x and y JLYHQE\

and

x =

m1 x1 + m2 x2 + m3 x3 + … = m1 + m2 + m3 + …

∑ mx ∑m

... (3.1)

y =

m1 y1 + m2 y2 + m3 y3 + … = m1 + m2 + m3 + …

∑ my ∑m

... (3.2)

3.2 FIRST MOMENT OF AREA AND CENTROID 7KH¿UVWPRPHQWRIDSODQHDUHDGH¿QHVWKHFHQWURLGRIWKHDUHD7KHFHQWURLGLVGH¿QHGDVWKHFHQWUHRI JUDYLW\RIDSODQHDUHDZKHUHWKHHQWLUHDUHDLVDVVXPHGWREHFRQFHQWUDWHG&RQVLGHUDQDUHDA located in the xy-plane as shown in Fig. 3.1.

Fig. 3.1 Plane area A with the centroid G.

Further we consider a small element of area dA in the plane area AZLWKDFHQWURLGKDYLQJFRRUGLQDWHV (x, y UHODWLYHWRWKHx and yD[HV 7KH¿UVWPRPHQWRIWKHDUHDA with respect to the xD[LVLVGH¿QHGDV Qx =

∫ y dA

... (3.3)

A

6LPLODUO\WKH¿UVWPRPHQWRIWKHDUHDA with respect to the yD[LVLVGH¿QHGDV Qy =

∫ x dA A

... (3.4)

Centroid and Moment of Inertia

91

The values of Qx and Qy may be positive, negative or zero, depending on the location of the origin O of the coordinate axes. In SI units, Qx and Qy are expressed in m3, and their dimension is [L3]. The centroid G of the area ALVGH¿QHGDVWKHSRLQWLQWKHxy-plane that has coordinates Qy Q and y = x x = A A 7KH¿UVWPRPHQWRIDQDUHDKDVWKHIROORZLQJSURSHUWLHV

... (3.5)

‡ ,I WKH FHQWURLG RI WKH DUHD FRLQFLGHV ZLWK WKH RULJLQ RI WKH xy FRRUGLQDWH V\VWHP IRU ZKLFK x = y = 0, then Qx = Qy = 0. ‡ )RUDQDUHDKDYLQJDQD[LVRIV\PPHWU\WKHFHQWURLGRIWKHDUHDOLHVRQWKDWD[LV

  

‡ ,IWKHDUHDFDQEHVXEGLYLGHGLQWRVLPSOHJHRPHWULFVKDSHVOLNHUHFWDQJOHVWULDQJOHVFLUFOHVHWF of areas A1, A2, A3WKHQHTXDWLRQV  DQG  FDQEHH[SUHVVHGLQWKHIROORZLQJIRUPV ⎛ ⎞ ... (3.6) Qx = ∑ ⎜ ∫ ydA⎟ = Σ (Qx )i = Σ Ai yi ⎝ Ai ⎠ Qy =





⎝ Ai



∑ ⎜ ∫ xdA⎟ = Σ(Q

) = Σ Ai xi

y i

... (3.7)

ZKHUH xi , yi) are the coordinates of the centroid of area Ai. )RUWKHSXUSRVHRILOOXVWUDWLRQFRQVLGHUDQDUHDA consisting of a number of elemental areas a1, a2, a37KHLUORFDWLRQVZUWxySODQHDUHVKRZQLQ)LJ

Fig. 3.2 Location of the centroid.

A = a1 + a2 + a3 + ... = Let

∑a

x = Distance of the centroid of the area A from the y-axis y = Distance of the centroid of the area A from the x-axis

8VLQJWKHPRPHQWSULQFLSOHRIDUHDZHKDYH a x + a2 x2 + a3 x3 + … x = 1 1 a1 + a2 + a3 + …

92

Strength of Materials

=

= and

y = =

=

∑ ax ∑a ∑ ax A

...(3.8)

a1 y1 + a2 y2 + a3 y3 + … a1 + a2 + a3 + …

∑ ay ∑a ∑ ay

A x and y WDNHQWRJHWKHUJLYHVWKHSRVLWLRQRIWKHFHQWURLGRIWKHDUHDA.

...(3.9)

Example 3.1 Find the centroid of a T-section shown in Fig. 3.3.

Fig. 3.3

Solution: Refer Fig. 3.3. The entire TVHFWLRQ LV GLYLGHG WZR SDUWV   DQG   7KH VHFWLRQ LV symmetrical about YY D[LV KHQFH WKH FHQWURLG ZLOO OLH VRPHZKHUH RQ WKLV OLQH /HW AB is chosen as the reference line. For part (1) Area

a1 = 200 – 30 = 6 – 103 mm2

Distance of its C.G. from AB y1 = ⎛ 250 + 30 ⎞ = 265 mm ⎜ ⎟ 2 ⎠ ⎝

Centroid and Moment of Inertia

93

For part (2) Area

a2 = 250 – 20 = 5 – 103 mm2

Distance of its C.G. from AB y2 =  %\GH¿QLWLRQ

y = a1 y1  a2 y2 a1  a2 =

and

250 = 125 mm 2

(6 × 103 × 265) + (5 × 103 × 125) (6 × 103 ) + (5 × 103 )

= 201.36 mm

x =0

 +HQFHWKHFHQWURLGRIWKHJLYHQVHFWLRQOLHVDWDGLVWDQFHRIPPIURPAB on line YY.

Ans.

Example 3.2 Find the centroid of an I-section shown in Fig. 3.4.

Fig. 3.4

Solution: 5HIHU)LJ7KHHQWLUHVHFWLRQLVGLYLGHGLQWRWKUHHSDUWV    DQG  7KHVHFWLRQLV symmetrical about YYD[LVKHQFHFHQWURLGRIWKHVHFWLRQZLOOOLHRQLW/HWAB be the reference line. For part (1) Area

a1 = 150 – 20 = 3 – 103 mm2

Distance of its C.G. from AB 20 ⎞ ⎛ y1 = ⎜ 30 + 200 + ⎟ = 240 mm 2 ⎠ ⎝

94

Strength of Materials

For part (2) Area

a2 = 200 – 25 = 5 – 103 mm2

Distance of its C.G. from AB

For part (3) Area

200 ⎞ ⎛ y2 = ⎜ 30 + ⎟ = 130 mm 2 ⎠ ⎝ a3 = 250 – 30 = 7.5 – 103 mm2

Distance of its C.G. from AB y3 =

30 = 15mm 2

y =

a1 y1  a2 y2  a3 y3 a1  a2  a3

 %\GH¿QLWLRQ

= and

(3 × 103 × 240) + (5 × 103 × 130) + (7.5 × 103 × 15)

x =0

(3 × 103 ) + (5 × 103 ) + (7.5 × 103 )

 +HQFHWKHFHQWURLGRIWKHJLYHQVHFWLRQOLHVDWDGLVWDQFHRIPPIURPAB.

= 95.64 mm Ans.

Example 3.3 Find the centroid of the angle section (L-section) shown in Fig. 3.5.

Fig. 3.5

Solution: 5HIHU)LJ7DNHAB and ACDVWKHUHIHUHQFHD[HV7KHJLYHQVHFWLRQLVGLYLGHGLQWR two parts (1) and (2). For part (1) Area

a1 = 250 – 15 = 3750 mm2

Centroid and Moment of Inertia

95

Distance of its C.G. from AC x1 =

15 = 7.5 mm 2

Distance of its C.G. from AB y1 =

250 = 125 mm 2

For part (2) Area

a2 = (120 – 15) – 15 = 1575 mm2

Distance of its C.G. from AC ⎡ ⎛ 120 − 15 ⎞ ⎤ x2 = ⎢15 + ⎜ ⎟⎥ 2 ⎝ ⎠⎦ ⎣ = 67.5 mm Distance of its C.G. from AB y2 =

15 = 7.5 mm 2

x =

a1 x1  a2 x2 a1  a2

 %\GH¿QLWLRQ

=

(3750 × 7.5) + (1575 × 67.5) 3750 + 1575

= 25.24 mm and

y =

=

a1 y1  a2 y2 a1  a2 (3750 × 125) + (1575 × 7.5) 3750 + 1575

= 90.24 mm  +HQFHWKHFHQWURLGRIWKHJLYHQVHFWLRQLVORFDWHGDWDGLVWDQFHRIPPIURPAC and at a distance of 90.24 mm from AB.

Ans.

96

Strength of Materials

Example 3.4 Find the centroid of the channel section shown in Fig. 3.6.

Fig. 3.6

Solution: 5HIHU)LJ7KHFKDQQHOVHFWLRQLVGLYLGHGLQWRWKUHHSDUWV    DQG  /HWAB and ACEHWKHD[HVRIUHIHUHQFH For part (1) Area

a1 = 50 – 15 = 750 mm2

Distance of its C.G. from AC x1 = Distance of its C.G. from AB

For part (2) Area

50 = 25 mm 2

15 ⎤ ⎡ y1 = ⎢(150 − 15) + ⎥ = 142.5 mm 2⎦ ⎣ a2 = [150 – (15 – 2)] – 10 = 1200 mm2

Distance of its C.G. from AC x2 = Distance of its C.G. from AB y2 = For part (3) Area

10 = 5 mm 2 150 = 75 mm 2

a3 = 50 – 15 = 750 mm2

Centroid and Moment of Inertia

97

Distance of its C.G. from AC x3 = Distance of its C.G. from AB y3 =

15 = 7.5 mm 2

x =

a1 x1  a2 x2  a3 x3 a1  a2  a3

 %\GH¿QLWLRQ

= and

50 = 25 mm 2

(750 × 25) + (1200 × 5) + (750 × 25) = 16.11 mm 750 + 1200 + 750

y = a1 y1  a2 y2  a3 y3 a1  a2  a3 =

(750 × 142.5) + (1200 × 75) + (750 × 7.5) = 75 mm 750 + 1200 + 750

 +HQFHWKHFHQWURLGRIWKHVHFWLRQOLHVDWDGLVWDQFHRIPPIURPAC and at a distance of 75 mm from AB. Ans. Example 3.5 A circular part of diameter 70 mm is cut out from a circular plate of diameter 200 mm as shown in Fig. 3.7. Find the centroid of the remaining part. Solution: Refer Fig. 3.7. Let AB and CDEHWKHD[HVRIUHIHUHQFH G &HQWURLGRIWKHFLUFXODUSODWH G„ &HQWURLGRIWKHFLUFXODUSDUWEHLQJFXW VKDGHGDUHD G„„ = &HQWURLGRIWKHUHPDLQLQJSDUW

Fig. 3.7

98

Strength of Materials

For circular plate Area of the plate,

a1 = π – 2002 = 3.141 – 104 mm2 4

Its C.G. lies on AB i.e., y1 = 0 and the distance of the C.G. from CD is 200 = 100 mm x1 = 2 For part being cut

π

– 702 = 3.848 – 103 mm2 Area of the shaded portion, a2 = 4 Its C.G. from CD

For remaining part

70 ⎤ ⎡ x2 = ⎢(200 − 70) + ⎥ = 165 mm 2⎦ ⎣

The distance of the C.GRIWKHUHPDLQLQJSDUWIURP&'LV a x  a2 x2 x = 1 1 a1  a2 =

(3.141 × 104 × 100) − (3.848 × 103 × 165) 3.141 × 104 − 3.848 × 103

= 90.92 mm and

y =0

 +HQFHWKHFHQWURLGRIWKHUHPDLQLQJSDUWOLHVRQXX line at a distance of 90.92 mm from CD.

Ans.

Example 3.6 Find the centroid of the section shown in Fig. 3.8.

Fig. 3.8

Solution: 5HIHU)LJ7KHHQWLUHVHFWLRQLVGLYLGHGLQWRWKUHHSDUWV    DQG  7DNHAB and BC as the reference lines. For part (1) 1 π – 502 = 982.14 mm2 Area a1 = – 2 4

Centroid and Moment of Inertia

99

Distance of its C.G. from AB x1 = 25 –

4 × 25 = 14.39 mm 3π

Distance of its C.G. from BC y1 = For part (2) Area

50 = 25 mm 2

a2 = 70 – 50 = 3500 mm2

Distance of its C.G. from AB 70 ⎞ ⎛ x2 = ⎜ 25 + ⎟ = 60 mm 2 ⎠ ⎝ Distance of its C.G. from BC y2 =

50 = 25 mm 2

a3 =

1 – 50 – 50 = 1250 mm2 2

For part (3) Area

Distance of its C.G. from AB 50 ⎞ ⎛ x3 = ⎜ 25 + 70 + ⎟ = 111.66 mm 3 ⎠ ⎝ Distance of its C.G. from BC y3 =  %\GH¿QLWLRQ

100 = 33.33 mm 3

x  'LVWDQFHRIWKHFHQWULRGRIWKHJLYHQVHFWLRQIURPAB

and

=

a1 x1  a2 x2  a3 x3 a1  a2  a3

=

(981.74 × 14.39) + (3500 × 60) + (1250 × 111.66) = 63.45 mm 981.74 + 3500 + 1250

y  'LVWDQFHRIWKHFHQWULRGRIWKHJLYHQVHFWLRQIURPBC =

a1 y1  a2 y2  a3 y3 a1  a2  a3

=

(981.74 × 25) + (3500 × 25) + (1250 × 33.33) = 26.82 mm 981.74 + 3500 + 1250

 +HQFHWKHFHQWURLGRIWKHJLYHQVHFWLRQOLHVDWDGLVWDQFHRIPPIURPAB and at a distance of 26.82 mm from BC. Ans.

100

Strength of Materials

Example 3.7 Find the centroid of the area bounded by the x-axis, the line x = a and the parabola y2 = k x as shown in Fig. 3.9.

Fig. 3.9

Solution: Consider a vertical elementary strip of thickness dx at a distance x from the y-axis as shown in Fig. 3.9. The area of the strip is dA = y dx

y from the x-axis. 2 The distance of the centroid of the whole area from the x-axis is given as

The centroid of the strip is located at distance of y„ =

y =

∑ dA . y′ ∑ dA

=



ydx . A

y 2 =



y 2 dx 2 A

a

∫ Kx dx =

0

(y2 = kx)

2A a

K ⎛ x2 ⎞ ⎜ ⎟ = 2 A ⎜⎝ 2 ⎟⎠0 =

Ka 2 4A

= K.

a2 1 . 4 A

b 2 . a 2 . 3b = a 4 8 =

3b 8

Centroid and Moment of Inertia

101

Now consider a horizontal strip of thickness dy at a distance x from the y-axis as shown in Fig. 3.10.

Fig. 3.10

The length of the strip

= (a – x)

The area of the strip is

dA = (a – x) dy

ax a− x⎞ ⎛ The centroid of the strip is located at a distance of x„ = ⎜ x + ⎟ = 2 from the y-axis. 2 ⎠ ⎝ The distance of the centroid of the whole area from the y-axis is given as x =

∑ dA . x ′ ∑ dA b

=

∫ (a − x) dy . 0

a+x 2

A b

=

1 (a 2 − x 2 ) dy 2 A ∫0 b

1 = 2 A ∫0 =

⎧ 2 2⎫ ⎪ 2 ⎛y ⎞ ⎪ ⎨a − ⎜⎜ ⎟⎟ ⎬ dy ⎝ k ⎠ ⎪⎭ ⎪⎩

b b ⎤ y4 1 ⎡ 2 ⎢ ∫ a dy − ∫ 2 dy ⎥ 2 A ⎢⎣ 0 0 K ⎦⎥

b 1 ⎡⎢ 2 b ⎛ y 5 ⎞ ⎤⎥ (a y )0 − ⎜⎜ = ⎟ 2A ⎢ 5 K 2 ⎟⎠0 ⎥ ⎝ ⎦ ⎣

(y2 = kx)

102

Strength of Materials

⎤ ⎡ ⎢ ⎥ 5 b ⎥ 1 ⎢ 2 a b− = 2A ⎢ ⎛ b4 ⎞ ⎥ ⎢ 5⎜ 2⎟ ⎥ ⎢⎣ ⎝ a ⎠ ⎥⎦  :KHQy = b, then x = a, hence b2 y2 K = x = a +HQFH

x =

=

a 2b ⎤ 1 ⎡ 2 1 .4 2 a b ⎢a b − ⎥ = 2A ⎣ 5 ⎦ 2A 5 1 . 4 a 2b = 3a ⎛ 2ab ⎞ 5 5 2⎜ ⎝ 3 ⎟⎠

3 3  +HQFHWKHFHQWURLGLVORFDWHGDWWKHSRLQW x , y ), that is, ⎛⎜ a, b ⎞⎟ . ⎝5 8 ⎠

Ans.

Example 3.8 Find the coordinates of the centroid of a quarter-ellipse shown in Fig. 3.11, using direct integration method.

Fig. 3.11

Solution: The equation of the ellipse is x2 y2 + =1 a 2 b2 where a and bDUHWKHVHPLPDMRUDQGVHPLPLQRUD[HVRIWKHHOOLSVH

Centroid and Moment of Inertia

103

From the equation of ellipse, we have x =

a 2 b − y2 b

...(1)

y =

b 2 a − x2 a

...(2)

Now consider a vertical strip of height y and width dx at a distance x from the y-axis as shown in Fig. 3.11. Area of the strip

dA = y ·dx

The distance of the centroid of the strip from the y-axis is x =

∫ xdA ∫ dA a

=

∫ x. ydx 0

a

∫ ydx 0

a

=

b

∫ x. a

a 2 − x 2 dx

0

a

b 2 2 ∫ a a − x dx

(on substituting y from equation (2))

0

a

or

x =

∫x

a 2 − x 2 dx



a − x dx

... (3)

0 a

2

2

0

Let

x = a sin T then dx = a cosT dT

when

x = 0, T= 0 π 2

x = a, T= Equation (3) can now be expressed as π 2

x =

∫ a sin θ · 0

∫ 0

π 2

( a 2 − a 2 sin 2 θ) · a cos θd θ

a 2 − a 2 sin 2 θ · a cos θd θ

... (4)

104

Strength of Materials π 2

=

∫ a sin θ · a cos θ · a cos θ d θ 0

π 2

∫ a cos θ · a cos θ d θ 0

π 2

=

∫a

3

0

sin θ cos2 θ d θ π 2

∫a

cos2 θ d θ

2

0

π 2

= a·

∫ sin θ (1 − sin 0



π 2

2

(cos 2T = 2 cos2 T – 1)

⎛ 1 + cos 2θ ⎞ ⎟⎠ d θ ⎜⎝ 2

0

π 2

= 2a ·

θ) d θ

∫ (sin θ − sin 0

3

θ) d θ

π 2

∫ (1 + cos 2θ) d θ 0

π 2

⎛ 3 sin θ − sin 3θ ⎞ ⎟⎠ d θ 4

∫ sin θ − ⎜⎝

= 2a · 0

π 2

∫ (1 + cos 2θ) d θ 0



= 2a · 0

π 2

3 sin θ sin 3θ ⎞ ⎛ + ⎟ dθ ⎜⎝ sin θ − 4 4 ⎠ π 2

∫ (1 + cos 2θ) d θ 0

π

= 2a ·

3 cos θ cos 3θ ⎤ 2 ⎡ ⎢⎣ − cos θ + 4 − 12 ⎥⎦ 0 π

sin 2θ ⎤ 2 ⎡ ⎢⎣θ + 2 ⎥⎦ 0

Centroid and Moment of Inertia

105

π 3 π 1 3π 3 1 ⎞ ⎛ ⎜⎝ − cos + cos − cos + cos 0 − cos 0 + cos 0⎟⎠ 2 4 2 12 2 4 12 = 2a · 1 ⎞ ⎛π 1 ⎜⎝ + sin π − sin 0 − sin 0⎟⎠ 2 2 2 3 1⎞ ⎛ ⎛ 4⎞ ⎜⎝ − 0 + 0 − 0 + 1 − + ⎟⎠ ⎜⎝ ⎟⎠ 4 12 12 = 2a · = 2a · ⎞ ⎛π ⎛ π⎞ ⎜⎝ + 0 − 0 − 0⎟⎠ ⎜⎝ ⎟⎠ 2 2

or

x =

4a 3π

Similarly, the distance of the centroid of the strip from the xD[LVLV ⎛ y⎞

y =

∫ ⎜⎝ 2 ⎟⎠ dA ∫ dA

As the centroid of the area dA lies a distance (y/2) from the xD[LV a

or

y =

⎛ y⎞

∫ ⎜⎝ 2 ⎟⎠ y · dx 0

a

∫ ydx

(dA = ydx)

0

a

∫ y dx 2

=

1 0 · 2 a

∫ ydx 0

a

b2 2 ( a − x 2 ) dx 2 1 ∫0 a = · a 2 b 2 2 ∫ a a − x dx 0

a

=

∫ (a

b 0 · 2a a

∫ 0

2

− x 2 ) dx

a 2 − x 2 dx

(on substituting y from equation (2))

106

Strength of Materials π 2

=

∫ (a

2

b 0 · π 2a 2



− a 2 sin 2 θ) a cos θ d θ (using equation (4))

a 2 − a 2 sin 2 θ a cos θ dx

0

π 2

b ∫0 · = 2a

a 2 cos2 θ · a cos θ d θ π 2

∫ a cos θ · a cos θ d θ 0

π 2

cos3 θ d θ b ∫0 = · π 2 2 2 ∫ cos θ d θ 0

π 2

=

⎛ cos 3θ + 3 cos θ ⎞ ⎟⎠ d θ 4

∫ ⎜⎝

b 0 · 2

π 2

⎛ 1 + cos 2θ ⎞ ⎟⎠ d θ 2

∫ ⎜⎝ 0

π 2

b ∫0 = · 4

(cos 3θ + 3 cos θ ) d θ π 2

∫ (1 + cos 2θ) d θ 0

π

⎡ sin 3θ ⎤2 + 3 sin θ ⎥ ⎢ b ⎣ 3 ⎦0 = · π 4 sin 2θ ⎤ 2 ⎡ ⎢⎣θ + 2 ⎥⎦ 0

π sin 0 3π ⎞ ⎛1 × sin + 3 sin − − 3 sin 0⎟ ⎠ b ⎜⎝ 3 2 2 3 = · sin 0 ⎞ 4 ⎛π 1 ⎜⎝ + × sin π − 0 − ⎟ 2 2 2 ⎠

(cos3T = 4cos3T – 3cosT)

Centroid and Moment of Inertia

107

⎞ ⎛1 × ( −1) + 3 × 1 − 0 − 0⎟ ⎠ b ⎜⎝ 3 = · 4 ⎞ ⎛π 1 ⎜⎝ + × 0 − 0 − 0⎟⎠ 2 2 ⎛ 1 ⎞ ⎛ 8⎞ − + 3⎟ ⎜ b ⎝ 3 ⎠ b ⎜⎝ 3⎟⎠ = · = · π 4 4 ⎛ π⎞ ⎜⎝ ⎟⎠ 2 2 y =

or

4b 3π

⎛ 4 a , 4b ⎞ . +HQFHWKHFHQWURLGRIWKHTXDUWHUHOOLSVHLV x , y ), that is, ⎜ ⎝ 3π 3π ⎟⎠ 7KHFHQWURLGVRIVRPHLPSRUWDQWJHRPHWULHVDUHJLYHQLQ7DEOH Table 3.1 Figure

Centroid

Area

7UDSH]RLG

x =

h (a  2b) 3 ( a  b)

2 2 y = a + ab + b 3 (a + b)

( a  b) h 2

Triangle

x =

( a  b) 3

y = h 3

1 bh 2

Rectangle

x = b 2

bd

y = d 2 &RQWG

108

Strength of Materials

Parallelogram

x =

(a cos θ + b) 2

y =

ab sin T

a sin T 2

Semi-circle

x =0

4r y = 3π

π r2 2

Quarter-circle

x =

4r 3π

y =

4r 3π

x =

3 a 8

π r2 4

Semi-parabola

y =

2 h 5

2ah 3

Parabola

x =0 2 y = h 5

4ah 3

&RQWG

Centroid and Moment of Inertia

109

Quarter-ellipse

x =

4a 3π

y =

4b 3π

x =

3 a 4

π ab 4

Parabolic segment (2nd degree)

y =

3 h 10

ah 3

General spander

⎛ n +1 ⎞ x =⎜ ⎟a ⎝n + 2⎠ ⎛ n +1 ⎞ h y =⎜ ⎟ ⎝ 2n + 1 ⎠ 2

3.3

ah (n  1)

MOMENT OF INERTIA

$ERG\URWDWLQJDERXWDQD[LVLQDUHIHUHQFHIUDPHUHVLVWVDQ\FKDQJHLQLWVURWDWLRQDOPRWLRQ DQJXODU YHORFLW\ 2QDFFRXQWRIWKLVSURSHUW\WKHERG\LVVDLGWRKDYHDPRPHQWRILQHUWLDRUURWDWLRQDOLQHUWLD DERXWWKDWD[LV The moment of inertia is of two types: z

Mass moment of inertia

z

Area moment of inertia, also called Second moment of area

3.3.1

Mass Moment of Inertia

If a particle of mass m is located at a distance rIURPDQD[LVRIURWDWLRQWKHQLWVPRPHQWRILQHUWLDI DERXWWKDWD[LVLVJLYHQDV I = mr2

...(3.10)

If m1, m2, m3DUHWKHPDVVHVRIWKHYDULRXVSDUWLFOHVFRQVWLWXWLQJWKHERG\DQGr1, r2, r3 ... are their GLVWDQFHVIURPWKHD[LVRIURWDWLRQ )LJ WKHQWKHPRPHQWRILQHUWLDRIWKHERG\DERXWWKDWD[LVLV JLYHQDV

110

Strength of Materials

Fig. 3.12

I = m1 r12 + m2 r22 + m3 r32 + ž = 6mr2 If the mass is continuously distributed in the body, then I =



r 2 dm

... (3.11)

... (3.12)

where dmLVWKHPDVVRIDQLQ¿QLWHVLPDOO\VPDOOHOHPHQWRIWKHERG\DWDGLVWDQFHrIURPWKHD[LVRI rotation.

3.3.2 Radius of Gyration w.r.t. Mass Moment of Inertia The radius of gyration, KRIDERG\LVGH¿QHGDVWKHUDGLDOGLVWDQFHIURPWKHJLYHQD[LVRIURWDWLRQ WKHVTXDUHRIZKLFKRQEHLQJPXOWLSOLHGE\WKHWRWDOPDVVRIWKHERG\JLYHVWKHPRPHQWRILQHUWLDRI WKHERG\DERXWWKDWD[LV+HQFH I = MK2 = 6mr2 where M is the total mass of the body. )URP(TXDWLRQ  ZHKDYH K =

I M

... (3.13)

... (3.14)

3.3.3 Second Moment of Area The moments of inertia of a plane area, with respect to the xyD[HVDVVKRZQLQ)LJDUHGH¿QHGDV

and

Ix =

∫y

dA

... (3.15)

Iy =

∫ x dA

... (3.16)

2

2

where x and y are the coordinates of the differential elements of area dA. Because dA is multiplied by the square of the distance, the moments of inertia are also called second moments of the area. Further, the integrals Ix and Iy are also referred to as rectangular moments of inertia, as they are computed from the rectangular coordinates of the element dA:KLOHHDFKLQWHJUDOLVDFWXDOO\DGRXEOHLQWHJUDOLWLV possible in many applications to select elements of area dALQWKHVKDSHRIWKLQKRUL]RQWDORUYHUWLFDO VWULSVDQGWKXVUHGXFHWKHFRPSXWDWLRQVWRLQWHJUDWLRQLQDVLQJOHYDULDEOH7KHPRPHQWVRILQHUWLDDUH DOZD\VSRVLWLYHDQGKDYH6,XQLWVRIP4, and their dimension is [L4].

Centroid and Moment of Inertia

111

3.3.4 Radius of Gyration w.r.t. Second Moment of Area 7KHUDGLXVRIJ\UDWLRQRIDQDUHDDERXWDQD[LVLVGH¿QHGDVLWVGLVWDQFHIURPWKHJLYHQD[LVWKHVTXDUH RIZKLFKRQEHLQJPXOWLSOLHGE\WKHWRWDODUHDRIWKHERG\JLYHVWKHVHFRQGPRPHQWRILQHUWLDDERXWWKDW D[LV,WLVJLYHQDV I = AK2 or

K =

I A

... (3.17)

Kx =

Ix A

... (3.18)

For the xD[LV

For the yD[LV Iy

Ky =

A

... (3.19)

 7KHFRQFHSWRIWKHUDGLXVRIJ\UDWLRQLVXVHIXOLQWKHDQDO\VLVRIEHKDYLRXURIVWUXFWXUDOPHPEHUVOLNH beams, columns etc.

3.4 PRODUCT OF INERTIA The product of inertia of a plane area is also called the product second moment of area or the product PRPHQWRILQHUWLDRUVLPSO\WKHSURGXFWRILQHUWLD,WLVGH¿QHGZLWKUHVSHFWWRDVHWRISHUSHQGLFXODU D[HVO\LQJLQWKHSODQHRIWKHDUHD:LWKUHVSHFWWRWKHxyD[HVDVVKRZQLQ)LJWKHSURGXFW PRPHQWRILQHUWLDLVGH¿QHGDV Ixy =

∫ xydA

... (3.20)

A

As each element of the area dA is multiplied by the product of its coordinates, hence the product of LQHUWLDFDQEHSRVLWLYHQHJDWLYHRU]HUR,WLVLPSRUWDQWWRQRWHWKDWSURGXFWRILQHUWLDRIDQDUHDLV]HUR ZLWKUHVSHFWWRDQ\SDLURID[HVLQZKLFKRQHD[LV HLWKHUx or yD[LV LVDQD[LVRIV\PPHWU\RIWKHDUHD As most of the structural members used in bending applications consist of cross-sections of a FRPELQDWLRQRIUHFWDQJOHVWKHYDOXHRIWKHSURGXFWRILQHUWLDIRUVXFKVHFWLRQVLVGHWHUPLQHGE\WKH addition of the IxyYDOXHIRUHDFKUHFWDQJOH,WKDV6,XQLWRIP4 and dimension [L4].

3.5 PRINCIPAL AXES AND PRINCIPAL MOMENTS OF INERTIA 3ULQFLSDOD[HVDUHWKHD[HVDERXWZKLFKWKHSURGXFWRILQHUWLDRIWKHFURVVVHFWLRQRIDEHDPLV]HUR$OO SODQHVHFWLRQVZKHWKHUWKH\KDYHDQD[LVRIV\PPHWU\RUQRWKDYHWZRPXWXDOO\SHUSHQGLFXODUD[HV DERXWZKLFKWKHSURGXFWRILQHUWLDLV]HUR6LPSOHEHQGLQJRIDEHDPWDNHVSODFHDERXWDSULQFLSDOD[LV DQGWKHPRPHQWVDUHDSSOLHGLQDSODQHSDUDOOHOWRRQHVXFKD[LV7KHPRPHQWVRILQHUWLDRIDQDUHDDERXW WKHSULQFLSDOD[HVDUHFDOOHGWKHSULQFLSDOPRPHQWVRILQHUWLDDQGWKHLUPD[LPXPDQGPLQLPXPYDOXHV DUHUHVSHFWLYHO\UHIHUUHGWRDVWKHPD[LPXPSULQFLSDOPRPHQWRILQHUWLDDQGWKHPLQLPXPSULQFLSDO moment of inertia.

112

Strength of Materials

The expression for the principal moments of inertia can be obtained using the expression for the principal stresses given below. V 1, V 2 =

σx + σ y 2

2

⎛ σx − σ y ⎞ ± ⎜ + τ 2xy ⎟ ⎝ 2 ⎠

... (3.21)

Substituting Vx = Ix , Vy = Iy and Wxy = Ixy in equation (3.21), we can write the expression for the principal moments of inertia as I 1, I 2 = where I1 I2 Ix Iy Ixy

= = = = =

Ix + I y 2

2

⎛ Ix − I y ⎞ ± ⎜ + I xy2 ⎝ 2 ⎟⎠

... (3.22)

Maximum principal moment of inertia Minimum principal moment of inertia Moment of inertia of the section about the x-axis Moment of inertia of the section about the y-axis. Product moment of inertia of the section with respect to x and y axes.

 7KHRULHQWDWLRQRIWKHSULQFLSDOD[HVLVGH¿QHGDV 2I xy tan 2Tp = – Ix − I y

... (3.23)

The two values of TpGLIIHUE\ƒDQGGH¿QHWKHWZRSHUSHQGLFXODUSULQFLSDOD[HVRIZKLFKRQH corresponds to the maximum moment of inertia and the other corresponds to the minimum moment of Inertia. There are no shear stresses on the principal planes, hence the product of inertia is zero with respect to the principal axes.

3.6 MOHR’S CIRCLE FOR SECOND MOMENTS OF AREA The transformation equations for the normal and shear stresses for a plane stress condition can be represented in a graphical form, known as Mohr’s circle, as discussed in the chapter of principal stresses. The coordinates of each point on the circle correspond to the normal and shear stresses acting RQVSHFL¿FVHWRISHUSHQGLFXODUSODQHV0RKU¶VFLUFOHFDQDOVREHDSSOLHGWRWKHVHFRQGPRPHQWVRIDUHD to represent the transformation equations for the second moments of area and the product of inertia of area, and the coordinates of each point on the circle represent the moment of inertia and the product of LQHUWLDZLWKUHVSHFWWRDVSHFL¿FVHWRISHUSHQGLFXODUD[HV)RUFRQVWUXFWLQJWKHFLUFOH )LJ b)), the second moments of area are plotted on the horizontal axis (the x-axis) and the product of inertia on the vertical axis (the y-axis). Two points, say A and B are considered. Point A with coordinates (Ix , Ixy) is plotted above the horizontal axis, if Ixy is positive. Similarly, point B with its coordinates (Iy , – Ixy) is plotted below the horizontal axis. The two points A and B are now joined by a straight line, and a circle is constructed with this line (AB) as the diameter. The resulting circle is called Mohr’s circle for second moments of area with centre C. OR (IP ) and OS (IP ) are the principal second moments of area with 2 1 WKHLUPLQLPXPDQGPD[LPXPYDOXHVUHVSHFWLYHO\7R¿QGWKHVHFRQGPRPHQWRIDUHDDERXWVRPHRWKHU axis, say the neutral axis (NA), which is inclined at an angle D to the x-axis, a point N is chosen on the circle such that < ACN = 2Din the counterclockwise direction. The horizontal coordinate of N gives the value of second moment of area about the neutral axis, that is, INA.

Centroid and Moment of Inertia

(a) Mohr’s circle for a plane stress condition

113

(b) Mohr’s circle for second moments of area

Fig. 3.13 Comparison of Mohr’s circles of plane stress and second moment of area.

2QFRPSDULQJ0RKU¶VFLUFOHVIRUSODQHVWUHVVDQGVHFRQGPRPHQWVRIDUHDDVVKRZQLQ)LJZH ¿QGWKDWWxy is associated with – Ixy. The point A with its coordinates (Ix , Ixy LQ)LJ b) is plotted above the horizontal axis, if Ixy is positive. In contrast, the corresponding point P with its coordinates (Vx , – Wxy LQ)LJ a) is plotted below the horizontal axis for positive Wxy. Otherwise, the procedure to construct Mohr’s circle for the second moments of area is identical to Mohr’s circle for plane stress.

3.7

PARALLEL-AXES THEOREM

7KHSDUDOOHOD[HVWKHRUHPLVXVHGWR¿QGWKHPRPHQWRILQHUWLDRIDSODQHDUHDDERXWDQD[LVSDUDOOHO to its centroidal axis. Consider an area A with its centroid G having coordinates ( x , y ) as shown in )LJ7KHKRUL]RQWDOFHQWURLGDOD[LVLVWKHxc-axis, which is parallel to the x-axis and the vertical centroidal axis is the yc-axis, which is parallel to the yDL[V)XUWKHUDQHOHPHQWRIDUHDdA is considered, where x and yFRRUGLQDWHVDUHGH¿QHGDV x = x + xc y = y + yc where x is the x-coordinate of the centroid G, which represents the perpendicular distance between two vertical parallel axes, namely the y-axis and the yc-axis. Similarly, y is the y-coordinate of the centroid G, which represents the perpendicular distance between two horizontal parallel axes, namely the x-axis and the xc-axis.

114

Strength of Materials

Fig. 3.14

7KHPDWKHPDWLFDOVWDWHPHQWVRIWKHSDUDOOHOD[HVWKHRUHPIRUWKHPRPHQWRILQHUWLDDUHQRZH[SUHVVHGDV 2 Ix = I x + Ay (for the xD[LVDQGWKHxcD[LV    Iy = I y + Ax 2 where

(for the yD[LVDQGWKHycD[LV 

 

I x = Ixc = Moment of inertia about the centroidal xD[LVWKDWLVWKH  xcD[LV I y = Iy c = Moment of inertia about the centroidal yD[LVWKDWLVWKH  y cD[LV

7KHSDUDOOHOD[HVWKHRUHPIRUWKHSURGXFWRILQHUWLDRIDQDUHDFDQEHREWDLQHGE\VXEVWLWXWLQJx and y in WKHH[SUHVVLRQIRUWKHSURGXFWRILQHUWLDWKDWLV Ixy = = = where

∫ x ′y ′dA ∫ y ′dA

and

∫ dA

∫ xydA ∫ ( x + x ′)( y + y ′)dA ∫ x ′y ′dA + x ∫ y ′dA + y ∫ x ′dA + x y ∫ dA

= I xy = Ixcyc = Product of inertia of the cross-sectional area with UHVSHFWWRWKHFHQWURLGDOD[HVx any y, that is, the D[HVxc and yc. =

∫ x ′dA 

 DVWKH\DUHWKH¿UVWPRPHQWRIDUHDDERXWWKH FHQWURLGDOD[HVx and y.)

=A

 +HQFHHTXDWLRQRIWKHSURGXFWRILQHUWLDRIWKHDUHDUHGXFHVWR Ixy = I xy + A x y

... (3.26)

Centroid and Moment of Inertia

115

 (TXDWLRQ  UHSUHVHQWVWKHPDWKHPDWLFDOVWDWHPHQWRIWKHSDUDOOHOD[HVWKHRUHPIRUWKHSURGXFWRI inertia of an area.  1RZWKHSRODUPRPHQWRILQHUWLDRIDQDUHDLVH[SUHVVHGDV Jo = Ix + Iy where

... (3.27)

Jo = Polar moment of inertia about the origin O.

 2QVXEVWLWXWLQJIx and Iy from equations (3.24) and (3.25), we get Jo = ( I x + Ay ) + ( I y + Ax 2) 2

2 2 = I x + I y + A( x + y )

 +HQFH

Jo = J G + Ar 2

... (3.28)

where J G = I x + I y = Polar moment of inertia of the area about the centroid G 2 2 r 2 = x + y , and its square root r = G as shown in Fig. 3.14.

x 2 + y 2 represents the distance between O and

(TXDWLRQ   UHSUHVHQWV WKH PDWKHPDWLFDO VWDWHPHQW RI WKH SDUDOOHOD[HV WKHRUHP IRU WKH SRODU moment of inertia of an area.

3.8 MOMENT OF INERTIA OF A RECTANGULAR SECTION &RQVLGHUDUHFWDQJXODUVHFWLRQOABC (Fig. 3.15). Let b :LGWKRIWKHVHFWLRQ

Fig. 3.15

d = Depth of the section XX +RUL]RQWDOFHQWURLGDOD[LV YY 9HUWLFDOFHQWURLGDOD[LV &RQVLGHUDVWULS SDUDOOHOWRWKHxD[LV RIWKLFNQHVVdy at a distance y from the XX-D[LV

116

Strength of Materials

Area of the strip = dA = bdy Moment of inertia of the strip about the XX-axis = dA – y2 = by2 dy The moment of inertia of the entire section about the XX-axis is found as d /2

IXX =



by 2 dy

−d / 2 d /2

⎛ y3 ⎞ = b ⎜⎜ ⎟⎟ ⎝ 3 ⎠− d / 2 b = 3

⎡ ⎛ d ⎞3 ⎛ d ⎞3 ⎤ ⎢⎜ ⎟ − ⎜ − ⎟ ⎥ ⎢⎣⎝ 2 ⎠ ⎝ 2 ⎠ ⎥⎦

=

b ⎛ d3 d3 ⎞ + ⎜ ⎟ 3 ⎜⎝ 8 8 ⎟⎠

=

bd 3 12

... (3.29)

The moment of inertia about the YY-axis is given as IYY =

db3 12

... (3.30)

The moment of inertia about the x-axis is given as Ix = IXX + Ah2 bd 3 ⎛d ⎞ = + (bd) – ⎜ ⎟ 12 ⎝2⎠

=

2

bd 3 3

d⎞ ⎛ ⎜⎝ h = ⎟⎠ 2 ... (3.31)

The moment of inertia about the y-axis is given as Iy = IYY + Ah2 =

db3 + (bd) – 12

=

db3 3

⎛b⎞ ⎜ ⎟ ⎝2⎠

2

b⎞ ⎛ ⎜h = ⎟ 2⎠ ⎝ ... (3.32)

Centroid and Moment of Inertia

117

3.9 MOMENT OF INERTIA OF A SOLID CIRCULAR SECTION First method Consider a circular section of radius r (Fig. 3.16). An elemental strip of thickness dy is considered at a distance y from the x-axis. Area of the strip, dA = 2r cos T . dy Now Hence,

y = r sinT dy = r cosT dT dA = 2r cosT . r cosT dT = 2r2 cos2 TdT

Fig. 3.16

The moment of inertia of the section about the centroidal axis XX is given as IXX =



y 2 dA

π/2

=



( r sin θ)2 . 2r 2 cos2 θ d θ

−π /2 π/2

= 4r

4



sin 2 θ cos2 θ d θ

0

π r4 π ⎛d⎞ πd4 = = ⎜ ⎟ = 4 ⎝ 2⎠ 4 64 where d is the diameter of the circular section. 4

Also

πd4 64 Ix = IXX and Iy = IYY

and

Iz =

Similarly,

IYY =

πd4 32

(d = 2r) ... (3.33)

... (3.34)

... (3.35)

118

Strength of Materials

Second method  &RQVLGHUDQHOHPHQWDOULQJRIWKLFNQHVVdx at a radius x ( Fig. 3.17). Area of the elemental ring, dA = 2S xdx

Fig. 3.17

 7KHPRPHQWRILQHUWLDRIWKHULQJDERXWDQD[LVSHUSHQGLFXODUWRWKHSODQHRIWKHFLUFXODUVHFWLRQ (zD[LV FDOOHGSRODUPRPHQWRILQHUWLDLVJLYHQE\ Iz = Area of the ring – Radius2 ring

= 2S xdx – x2 = 2Sx3dx

 7KHSRODUPRPHQWRILQHUWLDRIWKHHQWLUHVHFWLRQLVJLYHQDV r

Iz =



I zring

0 r

=

∫ 2πx dx 3

0

r

⎛ x4 ⎞ πr 4 = 2S ⎜ ⎟ = ⎜ 4 ⎟ 2 ⎝ ⎠0 = Now

πd4 32

d⎞ ⎛ ⎜r = ⎟ 2⎠ ⎝

Iz = IXX + IYY = 2IXX

( IXX = IYY)

Iz πd = = Ix 2 64 4

IXX = Similarly,

IYY

πd4 = = Iy 64

... (3.36) ... (3.37)

Centroid and Moment of Inertia

119

3.10 MOMENT OF INERTIA OF A HOLLOW CIRCULAR SECTION Refer Fig. 3.18.

Fig. 3.18

IXX = IYY

(also Ix = Iy , Ix = IXX and Iy = IYY)

π π 4 d D4 – 64 64 π = (D4 – d 4) 64 π Iz = (D4 – d4) 32 D = 2R d = 2r =

and where

3.11 MOMENT OF INERTIA OF A SEMI-CIRCLE Refer Fig. 3.19. π 4 d Ix = 128 Also

Iy =

π 4 d 128

... (3.38) ... (3.39)

... (3.40) ... (3.41)

Fig. 3.19

120

Strength of Materials

and where or

πd4 64 d = 2r Ix = IXX + Ah2

IZ =

... (3.42)

XVLQJSDUDOOHOD[HVWKHRUHP

IXX = Ix – Ah2 =

π 4 1 ⎛ π 2 ⎞ ⎛ 4 d ⎞2 d – ⎜ d ⎟ – ⎜ × ⎟ 128 2⎝4 ⎠ ⎝ 3π 2 ⎠

= 0.00686d 4 = 0.11r4

... (3.43)

3.12 MOMENT OF INERTIA OF A QUARTER-CIRCLE Refer Fig. 3.20.

Fig. 3.20

Ix = =

1 ⎛ π 4⎞ d ⎟ – ⎜ 2 ⎝ 128 ⎠ π 4 d = Iy 256

Ix = IXX + Ah2

... (3.44) XVLQJSDUDOOHOD[HVWKHRUHP

IXX = Ix – Ah2 =

2 π ⎛π 2 ⎞ 1 4 d⎞ d4 – – ⎜ d ⎟ – ⎛⎜ × ⎟ 256 ⎝4 ⎠ ⎝ 3π 2 ⎠ 4

= 0. 00343d4 = 0.0549r4 = IYY and

Iz =

πd4 128

 7KHVHFRQGPRPHQWVRIDUHDRIVRPHPRUHSODQH¿JXUHVDUHJLYHQLQ7DEOH

... (3.45) ...(3.46)

Centroid and Moment of Inertia Table 3.2 Figure

Second moment of area

Triangle

IXX =

bh3 36

hb3 48 bh3 Ix = 12

IYY =

Ellipse

π 3 ba 4 π 3 ba Iy = 4 Ix =

Iz =

πab 2 ( a + b2 ) 4

Ix =

2ah3 15

Parabola

16a 3h Iy = 15

Quarter-ellipse

πab3 Ix = 16 πba 3 Iy = 16

Parallelogram

a 2b sin 3 T 3 ab Iy = – 6 sin T Ix =

(2a2 cos2 T + 3ab cos T + 2b2)

121

122

Strength of Materials

Example 3.9 Find the second moment of area of an LVHFWLRQDERXWWKHFHQWURLGDOD[HVX X and YY as shown in Fig. 3.21. Solution: Refer Fig. 3.21. x and y DUHJLYHQDV and

x = 25.24 mm y = 90.24 mm

The moment of inertia of the section about the XXD[LVLVWKHVXPRIWKHPRPHQWVRILQHUWLDRIWKH WZRSDUWVDERXWWKHVDPHD[LVDQGLVJLYHQE\XVLQJWKHRUHPRISDUDOOHOD[HVDV IXX = IXX + IXX 1

2

2⎤ ⎡ = ⎢ 1 × 15 × 2503 + 250 × 15 × ⎛⎜ 250 − 90.24 ⎞⎟ ⎥ ⎝ 2 ⎠ ⎥⎦ ⎢⎣12 2 ⎡1 15 ⎞ ⎤ ⎛ 3 ( ) {( ) } . × − × + − × × − 120 15 15 120 15 15 90 24 ⎢ + ⎜ ⎟ ⎥ 2 ⎠ ⎥⎦ ⎝ ⎢⎣12

= 30.47 – 106 mm4

Ans.

Fig. 3.21

The moment of inertia of the section about the YYD[LVLVJLYHQDV IYY = IYY + IYY 1

2

2 ⎡1 15 ⎞ ⎤ ⎛ 3 . × × + × × − 250 15 250 15 25 24 = ⎢ ⎜ ⎟ ⎥ 2 ⎠ ⎥⎦ ⎝ ⎢⎣12 ⎡1 3 2⎤ + ⎢ × 15 × 105 + 105 × 15 × (67.5 − 25.24) ⎥ ⎣12 ⎦ = 55.1 – 105 mm4 Ans.

Centroid and Moment of Inertia

123

Example 3.10 Find the second moment of area of a TVHFWLRQVKRZQLQ)LJDERXWWKHFHQWURLGDOD[HVXX and YY. Solution: Refer Fig. 3.22. The section is symmetrical about the YYD[LVKHQFH and

x =0 y = 201.36 mm

Using thHRUHPRISDUDOOHOD[HVWKHPRPHQWRILQHUWLDRIWKHVHFWLRQDERXWWKHXX-D[LVLVJLYHQDV IXX = IXX + IXX 1

2

⎡1 3 2⎤ = ⎢ × 200 × 30 + 200 × 30 × (265 − 201.36) ⎥ ⎣12 ⎦ ⎡1 3 2⎤ + ⎢ × 20 × 250 + 250 × 20 × (201.36 − 125) ⎥ ⎣12 ⎦ = 7.99 – 107 mm4

Ans.

Fig. 3.22

The moment of inertia of the section about the YY-D[LVLVJLYHQDV IYY = IYY + IYY 1

2

1 ⎛1 ⎞ = ⎜ × 30 × 2003 + × 250 × 203 ⎟ = 2.016 – 107 mm4 12 ⎝ 12 ⎠

Ans.

Example 3.11 Find the second moment of area of an IVHFWLRQVKRZQLQ)LJDERXWWKHFHQWURLGDOD[HVXX and YY. Solution: Refer Fig. 3.23. The section is symmetrical about the YYD[LVKHQFH x =0 and

y = 95.64 mm

124

Strength of Materials

 8VLQJWKHRUHPRISDUDOOHOD[HVWKHPRPHQWRILQHUWLDRIWKHVHFWLRQDERXWWKHXXD[LVLVJLYHQDV IXX = IXX + IXX 1

2

⎡1 3 2⎤ = ⎢ × 150 × 20 + 150 × 20 × (240 − 95.64) ⎥ + 12 ⎣ ⎦

= 7.55 – 107 mm4

⎡1 3 2⎤ ⎢12 × 25 × 200 + 200 × 25 × (130 − 95.64) ⎥ ⎣ ⎦ 1 ⎡ 3 2⎤ + ⎢ × 250 × 30 + 250 × 30 × (95.64 − 15) ⎥ 12 ⎣ ⎦ Ans.

Fig. 3.23

The moment of inertia of the section about the YYD[LVLVJLYHQDV IYY = IYY + IYY 1

2

1 1 ⎡1 3 3 3⎤ = ⎢ × 20 × 150 + × 200 × 25 + × 30 × 250 ⎥ = 4.49 – 107 mm4 12 12 ⎣12 ⎦ Since IXX > IYY , hence the section is more stronger about the XXD[LV Example 3.12 Find the second moment of area of the shaded section shown in Fig. 3.24 about its centroidal D[LVSDUDOOHOWRWKHEDVH Solution: 7KH JLYHQ VHFWLRQ LV V\PPHWULFDO about the YYD[LVDQGLVHTXLYDOHQWWRDVHFWLRQ made of two triangles and one square minus one semi-circle as shown in Fig. 3.25. Let XX and YY EHWKHFHQWURLGDOD[HV Fig. 3.24

Ans.

Centroid and Moment of Inertia

125

Fig. 3.25

x =0

 +HUH Calculation of y Part (1)

Area, a1 =

Part (2)

1 × 25 × 50 2

= 625 mm

a2 = 625 mm2

Part (3)

a3 = 50 × 50 = 2500 mm2

2

Distance of C.G. from AB y2 =

50 y1 = 3

50 3

= 16.67 mm

= 16.67 mm

y3 =

50 2

= 25 mm

Part (4)

a4 =

1 × S u 252 2

= 981.74 mm2

y4 =

4 × 25 3π

= 10.61 mm

The distance of the centroid of the section from ABLVJLYHQDV y =

=

a1 y1 + a2 y2 + a3 y3 − a4 y4 a1 + a2 + a3 − a4 (625 × 16.67) + (625 × 16.67) + (2500 × 25) − (981.74 × 10.61) mm 625 + 625 + 2500 − 981.74

= 26.34 mm The second moment of area of the section about the XXD[LVLVIRXQGE\XVLQJSDUDOOHOD[HVWKHRUHP

126

Strength of Materials

IXX = IXX + IXX + IXX 1

2

3

2 ⎡ 25 × 503 1 ⎤ 50 ⎞ ⎤ ⎡ 50 × 503 ⎛ . + × 25 × 50 × 26 34 − + 50 × 50 × (26.34 − 25) 2 ⎥ =2– ⎢ ⎟⎠ ⎥ + ⎢ ⎜⎝ 2 3 ⎥⎦ ⎣ 12 ⎢⎣ 36 ⎦ 2⎤ 2 ⎡ × 25 4 × 25 π ⎛ ⎞ mm4 – ⎢0.11 × 254 + × ⎜ 26.34 − ⎟ ⎥ 2 3 π ⎝ ⎠ ⎥⎦ ⎢⎣ = 5.3 – 105 mm4 Ans.

Example 3.13 Find the second moment of area of the section shown in Fig. 3.26 about its centroidal axes.

Fig. 3.26

Solution: The section is symmetrical about both XX and YY axes. The second moment of area of the section about the XX-axis is given as IXX = IXX for I-section + IXX for two plates ⎛1 3 2 ⎞⎤ ⎡ 1 1 = ⎢⎛⎜ × 70 × 103 + 70 × 10 × 702 ⎞⎟ + ⎛⎜ × 10 × 1303 + 0 ⎞⎟ + ⎜ × 70 × 10 + 70 × 10 × 70 ⎟ ⎥ ⎠⎦ ⎠ ⎝ 12 ⎠ ⎝ 12 ⎣⎝ 12 2 ⎡1 5⎞ ⎤ ⎛ + 2 ⎢ × 90 × 53 + 90 × 5 × ⎜ 65 + 10 + ⎟ ⎥ 2 ⎠ ⎥⎦ ⎝ ⎢⎣12 7 4 = 1.41 × 10 mm Ans. The second moment of area of the section about the YY-axis is given as

IYY = IYY for I-section + IYY for two plates ⎡ ⎛1 1 ⎡1 3⎞ 3⎤ 3⎤ = ⎢ 2 ⎜ × 10 × 70 ⎟ + × 130 × 10 ⎥ + 2 ⎢ × 5 × 90 ⎥ = 1.19 – 106 mm4 ⎣12 ⎦ ⎠ 12 ⎣ ⎝ 12 ⎦

Ans.

Centroid and Moment of Inertia

127

Example 3.14 Find the expressions for the moments of inertia of a rectangle shown in Fig. 3.27 about the x-axis, the y-axis, the horizontal centroidal axis (XX) and the vertical centroidal axis (YY). Solution:

Fig. 3.27

An elementary area dA (horizontal) is considered as shown in Fig. 3.27. The moment of inertia of the rectangle about the x-axis is given as Ix =

∫y

2

dA

h

=

∫y

2.

(where dA = b.dy)

b dy

o

h

= b ∫ y 2 dy o

h

⎡ y3 ⎤ bh3 = b⎢ ⎥ = 3 ⎣ 3 ⎦o

Ans.

For moment of inertia of the rectangle about the y-axis, we consider a vertical elementary area dA„. Hence,

Iy =

∫x

2

dA′

b

=

∫ x . h dx 2

o

b

⎡ x3 ⎤ hb3 = h ∫ x dx = h ⎢ ⎥ = 3 ⎣ 3 ⎦o 0 b

2

Ans.

128

Strength of Materials

)RUWKHPRPHQWRILQHUWLDRIWKHUHFWDQJOHDERXWLWVKRUL]RQWDOFHQWURLGDOD[LV IXX), we use parallelD[HVWKHRUHPDV Ix = IXX + Ad 2 where +HQFH

d 3HUSHQGLFXODUGLVWDQFHEHWZHHQWZRD[HV IXX = Ix – Ad 2 bh 3 ⎛ h⎞ − bh . ⎜ ⎟ = ⎝ 2⎠ 3

2

bh3 bh3 bh3  = = 12 3 4

Ans.

6LPLODUO\WKHPRPHQWRILQHUWLDRIWKHUHFWDQJOHDERXWLWVYHUWLFDOFHQWURLGDOD[LV IYY LVJLYHQDV IYY = Iy – Ad 2 hb3 ⎛b⎞ −b⋅h⋅⎜ ⎟ = 3 ⎝2⎠

2

hb3 hb3 hb3  = = 3 4 12

Ans.

Example 3.15 )LQGWKHH[SUHVVLRQVIRUWKHPRPHQWVRILQHUWLDRIDFLUFOHDERXWLWVGLDPHWUDOD[HV Solution: Refer Fig. 3.28. &RQVLGHUDQHOHPHQWDU\DUHDdA. dA = rdT¹dr 7KHPRPHQWRILQHUWLDRIWKHFLUFOHDERXWLWVKRUL]RQWDOGLDPHWHU WKHKRUL]RQWDOFHQWURLGDOD[LV LVJLYHQDV IXX = =

∫y

2

dA

∫ ∫ (r sin θ) R 2π

=

∫ ∫ sin

2

2

⋅ rd θ ⋅ dr

θ ⋅ d θ ⋅ r 3dr

o o

R 2π

=

∫∫ o o

FIG. 3.28

1 − cos 2θ ⎞ ⎛ 2 ⎛ 1 − cos 2θ ⎞ 3 ⎟⎠ ⎟⎠ d θ ⋅ r dr ⎜⎝ sin θ = ⎜⎝ 2 2

R 2π 1 ⎡ = ∫ ⎢ ∫ dθ − 2 o ⎢⎣ o



∫ o

⎤ cos 2θ d θ ⎥ ⋅ r 3dr ⎥⎦

Centroid and Moment of Inertia 2π

R

1 sin 2θ ⎤ ⎡ 3 ⎢⎣θ − 2 ⎥⎦ ⋅ r dr = 2 o

1 = ∫ 2o

R

∫ 2π ⋅ r dr 3

o

R

⎛ r4 ⎞ πR 4 = π ∫ r dr = π ⎜⎜ ⎟⎟ = 4 ⎝ 4 ⎠o o R

129

3

Ans.

7KHPRPHQWRILQHUWLDRIWKHFLUFOHDERXWLWVYHUWLFDOGLDPHWHU WKHYHUWLFDOFHQWURLGDOD[LV LVJLYHQDV IYY = =

=

∫x



∫ o

+HQFH

2π R

∫ ∫ r dr ⋅ cos 3

R4 8

2

θ dθ

o o

2π R



=

dA

2 ∫ ∫ (r cosθ ) ⋅ rdθ ⋅ dr =

o

=

2

⎛ 1 + cos 2θ ⎞ 3 ∫ r dr ⋅ ⎜⎝ 2 ⎟⎠ d θ o

1 + cos 2θ ⎞ ⎛ 2 ⎜ cos θ = ⎟ 2 ⎝ ⎠

2π 2π ⎤ R4 ⎡ R4 1 θ + cos 2θ ⋅ dθ ⎥ d ⎢ ⋅ ⋅ (1 + cos 2θ ) d θ = ∫ ∫ 8 ⎢⎣ o 4 2 ⎥⎦ o 2π

sin 2θ ⎤ ⎡ ⎢θ + 2 ⎥ ⎣ ⎦o

IXX = IYY =

πD 4 πR 4 = 64 4

=

R4 πR 4 ⋅ 2π = 8 4

Ans.

(D = 2R)

Example 3.16 )LQG WKH H[SUHVVLRQV IRU WKH PRPHQWV RI LQHUWLD RI D WULDQJOH VKRZQ LQ )LJ  DERXW LWV EDVH WKH KRUL]RQWDOFHQWURLGDOD[LVDQGWKHKRUL]RQWDOD[LVSDVVLQJWKURXJKLWVYHUWH[ Solution: Refer Fig. 3.29.

Fig. 3.29

130

Strength of Materials

Consider a triangle of base b and height h. G represents its centroid. An elemental area dA, parallel to the base of the triangle, is considered. dA = x ˜dy

... (1)

From similar triangles OCD and ABC, we have h b x = (h  y ) b (h  y ) h On substituting x in equation (1), we get b (h − y ) ⋅ dy dA = h x =

Now the moment of inertia of the triangle about its base is given as Ix =

∫y h

2

dA

2 = ∫y ⋅ o

h h ⎤ b⎡ b (h − y ) ⋅ dy = ⎢ h ∫ y 2 dy − ∫ y 3 dy ⎥ h ⎢⎣ o h ⎥⎦ o

h h b ⎡⎢ ⎛ y 3 ⎞ ⎛ y 4 ⎞ ⎤⎥ b ⎛ h3 h 4 ⎞ − h − ⎟⎟ = ⎜ ⎟ ⎜ ⎟ = ⎜⎜ h ⋅ h ⎢ ⎜⎝ 3 ⎟⎠o ⎜⎝ 4 ⎟⎠o ⎥ 3 4 ⎠ h⎝ ⎦ ⎣

bh3 bh 3 bh 3 − = = 12 3 4 Using parallel-axes theorem, we have Ix = IXX + A.h„2 where

Ans.

h„ = Perpendicular distance between the x-axis and the XX-axis A = Area of the triangle

Now

IXX = Ix – A.h„2 bh3 1 ⎛h⎞ − ×b×h×⎜ ⎟ = 12 2 ⎝3⎠

2

bh3 bh3 3bh3  2bh3 bh3  Ans. = = 12 18 36 36 The moment of inertia of the triangle about the axis passing through its vertex IV , using parallel-axes theorem, is given by IV = IXX + Ah„2 =

where h„ = Perpendicular distance between the axis passing through vertex and the XX-axis =

2h 3

Centroid and Moment of Inertia

bh3 1 ⎛ 2h ⎞ + ×b×h×⎜ ⎟ IV = 36 2 ⎝ 3 ⎠ =

bh3 2 3  bh 36 9

=

bh3  8 bh3 36

=

131

2

bh3 4

Ans.

Example 3.17 Find the moment of inertia of the area shown in Fig. 3.30 about the x-axis. Solution: Consider an elemental area dA parallel to the xD[LVDVVKRZQLQWKH¿JXUH dA = (30 – x) dy The moment of inertia of the area about the x-axis is given as Ix =

∫y 30

=

2

∫y

dA

2

(30 − x) dy

o

Fig. 3.30 30

= 30

∫y

30 2

dy −

0

∫ 0

30

y2 y ⋅ ⋅ dy 30 2

2⎞ ⎛ ⎜⎜ x = y ⎟⎟ ⎟ ⎜⎜⎝ 30 ⎟⎠

30

⎛ y3 ⎞ 1 ⎛ y5 ⎞ = 30 ⎜ ⎟ − ⎜ ⎟ = 270000 – 162000 = 108000 cm4 ⎜ 3 ⎟ 30 ⎜⎝ 5 ⎟⎠0 ⎝ ⎠0

Ans.

132

Strength of Materials

Example 3.18 Find the moment of inertia of the shaded section shown in Fig. 3.31 about its centroidal axes.

Fig. 3.31

Solution: The section is symmetrical about both x and y axes, hence these two axes are also the centroidal axes with ODVRULJLQDVZHOODVWKHFHQWURLGRIWKHVHFWLRQ+HQFHZHDUHUHTXLUHGWR¿QGIx and Iy. Diameter of each semi-circle = 2 × 70 = 140 mm Calculation of Ix The moment of inertia of the shaded section about the horizontal centroidal axis is given as Ix

section

= IXX =

square ABCD

– IXX

semi-circle EFG

– IXX

π 1 × ( 200)4 − 2 × × (140)4 12 128

semi-circle HIJ

1 π ⎞ ⎛ × (140)4 ⎟ ⎜⎝ I XX semi-circle EFG = I XX semi-circle HIJ = × ⎠ 2 64

= 1.14 × 108 mm4 Calculation of Iy IYY

square

=

1 × ( 200) 4 12

= 1.34 × 108 mm4 The moment of inertia of the semi-circle EFG about its diameter EG is IEG =

1 π × (140) 4 × = 9.43 × 106 mm4 2 64

Ans.

Centroid and Moment of Inertia

133

The distance of the centroid of the semi-circle EFG from EG is h =

=

4r 3π

(r = Radius of the semi-circle)

4 × 70 = 29.7 mm 3π

The area of the semi-circle EFG is πr 2 A = 2 =

π × (70) 2 = 7696.9 mm2 2

The moment of inertia of the semi-circle EFG about its vertical centroidal axis Y1Y1, using parallel-axes theorem, is given as IEG = IY

1Y1

or

IY

1Y1

+ Ah2

(h = Distance between EG and Y1Y1)

= IEG – Ah2 = 9.43 × 106 – 7696.9 × (29.7) 2 = 2.64 × 106 mm4

Now the moment of inertia of the semi-circle EFG about the y-axis, that is, the vertical centroidal axis YY, is given as IYY

EFG

= IY

1Y1

+ Ah12

(h1 = Distance between YY and Y1Y1)

= 2.64 × 106 + 7696.9 × (100 – 29.7) 2

(h1 = 100 – h)

= 4.06 × 107 mm4 Also

IYY

HIJ

= IYY

EFG

= 4.06 × 107 mm4

Finally, the moment of inertia of the shaded section about the vertical centroidal axis is given as IYY

section

= IYY

square

– IYY

semi-circles

= (1.34 × 108 – 2 × 4.06 × 107 ) mm4 = 5.28 × 107 mm4

Ans.

134

Strength of Materials

Example 3.19 Calculate the product of inertia Ixy for the plane area (an angle section) with respect to the axes x and y as shown in Fig. 3.32

Fig. 3.32

Solution: The angle section consists of two rectangles (1) and (2) having dimensions of 20 mm × 150 mm and 170 mm × 20 mm respectively. The product of inertia Ixy for each rectangle is calculated separately and then they are added to get Ixy for the angle section, that is, (Ixy) section = (Ixy)1 + (Ixy)2 Ixy for rectangle (1) (Ixy)1 = ( I xy )1 + A1 x1 y1

(using parallel-axes theorem) where ( I xy )1 is the product of inertia of the rectangle about its own centroidal axes, and –x 1 and –y1 are the distances of its centroid from y and x axes respectively. Since the rectangle is symmetrical about its both centroidal axes, hence ( I xy )1 = 0 Hence,

(Ixy)1 = 0 + (150 × 20) × (10) × (75) = 2.25 × 106 mm4

Ixy for rectangle (2) It is also symmetrical about its both centroidal axes, hence its ( I xy )2 = 0 Now

(Ixy)2 = ( I xy )2 + A2 –x 2 –y2 = 0 + (170 × 20) × (105) × (10) = 3.57 × 106 mm4

Hence,

(Ixy) section = (Ixy)1 + (Ixy) 2 = (2.25 × 106 + 3.57 × 106) mm4 = 5.82 × 106 mm4

Ans.

Centroid and Moment of Inertia

135

Example 3.20 Determine the product of inertia Ixy of the Z-section with respect to the axes x and y as shown in Fig. 3.33.

Fig. 3.33

Solution: The Z-section is split into three rectangles (1), (2) and (3) having cross-sections (90 mm × 10 mm), (10 mm × 250 mm) and (90 mm × 10 mm) respectively, and their areas are found as A1 = 90 × 10

= 900 mm2

A2 = 10 × 250 = 2500 mm2 A3 = 90 × 10

= 900 mm2

The product of inertia Ixy of each rectangle is calculated separately and they are added to get Ixy for the Z-section, that is, (Ixy)section = (Ixy )1 + (Ixy )2 + (Ixy )3 Ixy for rectangle (1) (Ixy)1 = (Ixy)1 + A1 –x 1 –y1 where (Ixy)1 is the product of inertia of the rectangle about its own centroidal axes, and –x 1 and –y1 are the distances of its centroid from y and x axes respectively. Since the rectangle (1) is symmetrical about its own centroidal axes, hence (Ixy)1 = 0 Now

(Ixy)1 = 0 + (90 × 10) × (50) × (120) = 5.4 × 106 mm4

(using parallel-axes theorem)

136

Strength of Materials

Ixy for rectangle (2) 6LQFHWKHUHFWDQJOH  LVDOVRV\PPHWULFDODERXWLWVFHQWURLGDOD[HVVRLWV Ixy)2 = 0. Also, the centroid O of the Z-section coincides with the centroid of the rectangle, that is, –x 2 = 0 and –y2 = 0. XVLQJSDUDOOHOD[HVWKHRUHP  +HQFH I ) = (I ) + A –x –y  xy 2

xy 2

2

2

2

= 0 + (10 × 250) × (0) × (0) =0 Ixy for rectangle (3) 7KHUHFWDQJOH  LVDOVRV\PPHWULFDODERXWLWVRZQFHQWURLGDOD[HVVRLWV Ixy)3 = 0. XVLQJSDUDOOHOD[HVWKHRUHP  +HQFH I ) = (I ) + A –x –y  xy 3

xy 3

3

3

3

= 0 + (90 × 10) × (– 50) × (–120) = 5.4 × 106 mm4 Therefore, the product of inertia Ixy of the entire Z-section is calculated as (Ixy)section = (Ixy)1 + (Ixy)2 + (Ixy)3 = 5.4 × 106 + 0 + 5.4 × 106 = 10.8 × 106 mm4

SHORT ANSWER QUESTIONS 1. 2. 3. 4. 5. 6. 7.

What is the difference between centre of gravity and centroid? What is the significance of radius of gyration? What is centroidal axis? Do the centroidal axis same as the neutral axis? Why is the area moment of inertia also called the second moment of area? What is product of inertia? What is its value about a principal axis? What is the use of parallel-axes theorem? What is principal moment of inertia?

Ans.

Centroid and Moment of Inertia

137

MULTIPLE CHOICE QUESTIONS 1. The centroid of a hollow cone of height h and radius r placed on its base lies at the following distance. h h from the base (b) from the base 4 3 2h 2h from the base (d) IURPWKHDSH[ (c) 3 3 2. The centroid of a solid cone of height h and radius r placed on its base lies at the following distance. (a)

(a)

h from the base 3

(b)

h from the base 4

(c)

2h from the base 3

(d)

2h IURPWKHDSH[ 3

3. The centroid of a semi-circle of radius r lies at the following distance from its base. (a)

3z 4r

(b)

4r 3π

(c)

4π 3r

(d)

3r . 4π

4. The centroid of an equilateral triangle of side l lies at the following distance (perpendicular) from any side. (a)

2 3

l

(b)

2 3 l

(c)

l

(d)

2 3

3 . l 2

5. The centroid of a right-angled triangle with base b and height h is

b h⎞ ⎟ 3 3⎠

⎛ (c) ⎜ , ⎝

⎛ 2 , 1 ⎞ ⎛ b h⎞ (a) ⎜ b h⎟ (b) ⎜ , ⎟ ⎠ ⎝ 3 3 ⎝ 3 2⎠

⎛ 1 , 1 ⎞ (d) ⎜ b h⎟ . ⎝ 3 3 ⎠

6. The moment of inertia of a triangular section of base b and height hDERXWWKHFHQWURLGDOD[LVSDUDOOHO to its base is (a)

b3 h 36

(b)

b3 h 18

(c)

bh3 36

(d)

bh3 . 18

7. The moment of inertia of a square section of side aDERXWWKHFHQWURLGDOD[LVSDUDOOHOWRLWVVLGHLV a4 (a) 4

a4 (c) 8

a4 (b) 12

(d)

ANSWERS 1. (a)

2. (b)

3. (b)

4. (c)

5. (c)

6. (c)

7. (b).

a4 . 16

138

Strength of Materials

EXERCISES 1. Find the centroid of the Z-section shown in Fig. 3.34.

Fig. 3.34

(Ans.

x = 39.23 mm, y = 44.62 mm).

2. Find the centroid of the channel section shown in Fig. 3.35.

Fig. 3.35

(Ans.

x = 38.75 mm, y = 110 mm).

Centroid and Moment of Inertia

139

3. Find the centroid of the angle section shown in Fig. 3.36.

Fig. 3.36

(Ans.

x = 13.82 mm, y = 43.82 mm).

4. Find the centroid of the shaded portion of the section shown in Fig. 3.37 after a square is cut out from it.

Fig. 3.37

(Ans.

y = 45.74 mm).

140

5.

Strength of Materials

Find the centroid and the area moment of inertia of the I-section shown in Fig. 3.38 about its FHQWURLGDOD[HV

Fig. 3.38

(Ans. x = 0, y = 45.64 mm, IXX = 1.47 – 107 mm4, IYY = 2.31 – 106 mm4).

6. Find the coordinates of the centroid of a quarter-ellipse shown in Fig. 3.39, using direct integration method. y x2 y2 + =1 a2 b2

b

O

x a Fig. 3.39

(Ans.

x =

4a 4b . ) , y = 3π 3π

Centroid and Moment of Inertia

141

7. Find the centroid of the shaded area formed by a straight line y = mx and a curve y = kx2 as shown in Fig. 3.40, using direct integration method.

Fig. 3.40

(Ans.

x =

2 a , y = b ). 5 2

8. Determine the second moment of area of a triangle of base b and height h about its centroidal axes and base. hb3 bh3 bh3 , , I = (Ans. IXX = I = ). 48 base 12 36 YY 9. Determine the second moment of area of a semi-circle of radius r about its horizontal centroidal axis. (Ans. 0.11r4). 10. Find the moment of inertia of the shaded area shown in Fig. 3.41 about the x-axis and the horizontal centroidal axis.

Fig. 3.41

(Ans.

6.16 × 105 mm4, 5.76 × 105 mm4).

Centroid and Moment of Inertia

11. Find the product of inertia Ixy of the I-section shown in Fig. 3.42.

142

.

Fig. 3.42

(Ans. 12.

– 3.23 × 105 mm4).

Determine the product of inertia Ixy of the Z-section shown in Fig. 3.43.

Fig. 3.43

(Ans.

5.85 × 106 mm4). ‰‰‰

4 Shear Forces and Bending Moments in Beams

Figure shows a simple beam, one of the widely used statically determinate beams loaded with a central point load W.

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ :KDWLVWKHXVHRIVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPV"  ‡ :K\LVWKHSRVLWLYHEHQGLQJPRPHQWRIWHQFDOOHGVDJJLQJ"  ‡ :KDWLVSRLQWRIFRQWUDÀH[XUH"  ‡ +RZDUHWKHVKHDUIRUFHDQGWKHEHQGLQJPRPHQWUHODWHG"  ‡ +RZLVWKHQDWXUHRIDEHQGLQJPRPHQWFXUYHGHFLGHG"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_4

143

144

Strength of Materials

4.1 WHAT IS A BEAM? Beam iVDVWUXFWXUDOPHPEHUGHVLJQHGWRVXSSRUWYHUWLFDOORDGV)ORRUVRIDEXLOGLQJDUHVXSSRUWHGRQ beams. Many shafts of machinery act simultaneously as torsion members and as beams. Application RIORDGSURGXFHVVKHDUDQGEHQGLQJLQWKHEHDP:KHQORDGVDUHQRWDFWLQJDWULJKWDQJOHVWKH\ZLOO DOVRSURGXFHD[LDOIRUFHVLQWKHEHDP Beams are usually long, straight prismatic bars placed horizontally. There are two important points in the design of a beam; one is the determination of shear forces and bending moments as a result RIWKHDSSOLFDWLRQRIH[WHUQDOORDGVRQWKHEHDPDQGWKHRWKHULVWKHVHOHFWLRQRIWKHVXLWDEOHFURVV sections of the beam to resist shearing forces and bending moments. 4.2 CLASSIFICATION OF BEAMS %HDPVDUHFODVVL¿HGDFFRUGLQJWRWKHVXSSRUWVSURYLGHGDQGDUHGLVFXVVHGEHORZ z

A simple beam is supported at its two ends. One end is hinge supported and the other end is roller supported. A pin or hinge support is capable of resisting a force acting in any direction of the plane, that is, LWSUHYHQWVKRUL]RQWDORUYHUWLFDOPRYHPHQWEXWGRHVQRWSUHYHQWURWDWLRQKHQFHWKHEHDPFDQ rotate in the plane. Such support has two reaction force components, one in the horizontal and DQRWKHULQWKHYHUWLFDOGLUHFWLRQEXWKDVQRPRPHQWUHDFWLRQ



 $UROOHUDOORZVIUHHPRYHPHQWLQDKRUL]RQWDOSODQHDQGKHQFHKDVQRUHDFWLRQLQWKHKRUL]RQWDO GLUHFWLRQEXWSUHYHQWVYHUWLFDOPRYHPHQW$WWKHVDPHWLPHLWGRHVQRWSUHYHQWURWDWLRQKHQFH WKHEHDPLVIUHHWRURWDWH+HQFHLWKDVRQO\RQHUHDFWLRQIRUFHLQWKHYHUWLFDOGLUHFWLRQDQGQR moment reaction. Roller and pin supports taken together are termed as simple supports. z

An overhanging beamLVRQHZKRVHFHUWDLQSRUWLRQH[WHQGVEH\RQGWKHVXSSRUWRQRQHRUERWK sides.

A continuous beam has more than two supports. z A cantilever beamLV¿[HGDWRQHHQGDQGIUHHDWWKHRWKHUHQG z A ¿[HGEHDPLVFRQ¿QHGEHWZHHQWZRVXSSRUWV   )L[HGVXSSRUWFDQUHVLVWDIRUFHLQDQ\GLUHFWLRQDVZHOODVDPRPHQWRUDFRXSOH,WLVLQEXLOWDW WKHHQGDQGLVSUHYHQWHGIURPURWDWLQJ z A SURSSHGFDQWLOHYHUEHDPKDVDVXLWDEOHVXSSRUWLQRUGHUWRSUHYHQWLWVGHÀHFWLRQ z

6LPSO\ VXSSRUWHG RYHUKDQJLQJ DQG FDQWLOHYHU EHDPV DUH FDWHJRUL]HG DV VWDWLFDOO\ GHWHUPLQDWH EHDPVZKLOHFRQWLQXRXV¿[HGDQGSURSSHGFDQWLOHYHUDVVWDWLFDOO\LQGHWHUPLQDWHEHDPV,WVKRXOG EHQRWHGWKDWWKHUHDFWLRQVZLOOEHGHWHUPLQDWHLIWKHVXSSRUWVLQYROYHRQO\WKUHHXQNQRZQV,IPRUH XQNQRZQVDUHLQYROYHGWKHQWKHUHDFWLRQVZLOOEHVWDWLFDOO\LQGHWHUPLQDWHEHFDXVHWKHPHWKRGVRI VWDWLFVDUHQRWVXI¿FLHQWWRGHWHUPLQHWKHUHDFWLRQVDQGWKHSURSHUWLHV RIWKHEHDPZLWKUHJDUGWR its resistance to bending must be taken into consideration. The shear force and bending moment diagrams are drawn for statically determinate beams only. 'LIIHUHQW W\SHV RI EHDPV DUH VKRZQ LQ )LJ  ,Q ¿JXUHV a  b  DQG c), beams are hinge supported at A and roller supported at B, C, and BDQG&UHVSHFWLYHO\ZKHUHDVLQ¿JXUHV G  e), and f EHDPVDUH¿[HGDWA, A and B, and AUHVSHFWLYHO\

Shear Forces and Bending Moments in Beams

145

Fig. 4.1 Types of beams.

4.3 z

z

TYPES OF LOADINGS A FRQFHQWUDWHGORDG is acting at a poinWDQGKHQFHLWLVDOVRFDOOHGDSRLQWORDG )LJ a ,W LVH[SUHVVHGLQQHZWRQV 1 LQSI units. A GLVWULEXWHG ORDG DFWV RYHU D ¿QLWH OHQJWK RI WKH EHDP )LJ  b  ,W LV VSHFL¿HG E\ WKH intensity of loading per unit length, say w1PLQSIXQLWV:KHQLWKDVDFRQVWDQWYDOXHIRUD certain part of the beam, it is said to be uniformly distriEXWHGRYHUWKDWSDUWRIWKHEHDP XGO).

Fig. 4.2 Types of loadings.

8QLIRUPO\ YDU\LQJ ORDG implies the increase or decrease of loading intensity at a constant rate along the length of the beam. Distributed load may be represented by a parabolic, cubic or a higher RUGHUFXUYHIRUQRQXQLIRUPO\YDU\LQJORDG$XQLIRUPORDGLVVKRZQE\DUDFWDQJXODUGLVWULEXWLRQ D XQLIRUPO\YDU\LQJ ORDG E\ D WULDQJOH DQG WKH FRPELQDWLRQ E\ D WUDSH]LXP GUDZQ RQ WKH EHDP )LJ b) and Fig. 4.3).

146

Strength of Materials

Fig. 4.3 Types of uniformly-varying loads.

4.4 CALCULATION OF BEAM REACTIONS The beam reactions are calculated by using the three fundamental equations of static equilibrium. 6Fx = 0 6Fy = 0 and 6Mz = 0 The beam is usually placed horizontally, so it is the x-axis and the load on the beam is placed vertically, so it is the y-axis.

4.5

SHEAR FORCES IN A BEAM

Consider a section XX of the beam at its certain distance (Fig. 4.4). The part to the left of the section is in equilibrium under the action of three forces: the vertical reaction force at A, that is, RA, the external load W1, and force V induced at the section. For the equilibrium of the part to the right of the section, an opposite force V acts at the section, so that its equilibrium is decided by the reaction force at B i.e., RB, the external load W2, and the force V. The force V is called vertical shear force. Numerically, the shear force is equal to the algebraic sum of all the vertical components of the external forces to the left or to the right of the section. Whether the right-hand segment or the left-hand segment is used to determine the shear force at a section is immaterial and only arithmetical simplicity governs.

Fig. 4.4 Shear force in a beam.

Shear Forces and Bending Moments in Beams

4.6

147

BENDING MOMENTS IN A BEAM

:KHQDORDGLVDSSOLHGRQWKHEHDPEHQGLQJPRPHQWLVGHYHORSHGZKLFKWHQGVWREHQGWKHEHDPLQ WKHSODQHRIWKHORDGV7RFRXQWHUDFWWKHPRPHQWFDXVHGE\H[WHUQDOORDGLQWHUQDOUHVLVWLQJPRPHQW is produced to satisfy the static equilibrium condition for moment. Magnitude of the internal resisting PRPHQWLVHTXDOWRWKHH[WHUQDOPRPHQW7KHEHQGLQJPRPHQWDWDVHFWLRQLVHTXDOWRWKHDOJHEUDLF sum of moments to the left or to the right of the section.

4.7 SIGN CONVENTIONS FOR SHEAR FORCE AND BENDING MOMENT  a) For shear force: ,IWKHSRUWLRQRIWKHEHDPWRWKHOHIWRIWKHVHFWLRQPRYHVXSZDUGDQGWKHSRUWLRQ WRWKHULJKWRIWKHVHFWLRQFRPHVGRZQZDUGWKHQLWLVGXHWRSRVLWLYHVKHDU5HYHUVHRILWLVGXH WRQHJDWLYHVKHDU )LJ 

Fig. 4.5

 b) For bending moment: :KHQ EHQGLQJ PRPHQW FDXVHV XSZDUG FRQFDYLW\ LQ WKH EHDP LW LV WHUPHGDVSRVLWLYH%0 VDJJLQJ $VDUHVXOWRISRVLWLYH%0WKHXSSHUSDUWRIWKHEHDPLVLQ compression and the lower part in tension, thereby increasing the length of the bottom surface and decreasing the length of the top surface of the beam. On the other hand, if bending moment SURGXFHV GRZQZDUG FRQFDYLW\RUSURGXFHV FRPSUHVVLRQ LQWKH ORZHU SDUWDQGWHQVLRQLQWKH XSSHUSDUWRIWKHEHDP¶VFURVVVHFWLRQWKHQLWLVFDOOHGDQHJDWLYHEHQGLQJPRPHQW KRJJLQJ  )LJ 

Fig. 4.6

148

Strength of Materials

4.8 SHEAR FORCE AND BENDING MOMENT DIAGRAMS (SFD AND BMD) Shear force and bending moment diagrams are the pictorial representation of shear forces and EHQGLQJPRPHQWVUHVSHFWLYHO\7KHD[LVRIWKHEHDPLVVKRZQE\DKRUL]RQWDOOLQHDQGYDOXHVRIVKHDU IRUFHDQGEHQGLQJPRPHQWDUHVKRZQRQLWDVYHUWLFDOOLQHVDFFRUGLQJWRDVXLWDEOHVFDOH3RVLWLYH YDOXHVDUHVKRZQDERYHWKHD[LVDQGQHJDWLYHYDOXHVEHORZWKHD[LVRIWKHEHDP$OOWKHYDOXHVDUH WKHQMRLQHGE\DVWUDLJKWOLQHRUDFXUYHGHSHQGLQJXSRQWKHQDWXUHRIORDGLQJ %HWZHHQWZRSRLQWORDGVWKHEHQGLQJPRPHQWGLVWULEXWLRQLVVKRZQE\DVWUDLJKWOLQH,IWKHUHLV XQLIRUPO\GLVWULEXWHGORDGEHWZHHQWZRSRLQWVWKHQWKHFXUYHLVSDUDERODRIVHFRQGRUKLJKHURUGHU The bending moment MLVPD[LPXPRUPLQLPXPZKHQ dM = 0. Thus at sections where shear dx IRUFHLV]HURRUFKDQJHVLWVVLJQ IURPPD[LPXPWRPLQLPXP WKH%0LVHLWKHUPD[LPXPRUPLQLPXP There may be a point in the bending moment diagram, where the bending moment is zero and the sign RIWKHEHQGLQJPRPHQWVDUHFKDQJHG7KLVSRLQWLVFDOOHGLQÀHFWLRQSRLQWRUSRLQWRIFRQWUDÀH[XUH The shear force diagram consists of horizontal straight lines in case of point loads and inclined straight lines in case of uniformly distributed loads. The corresponding portions in the bending PRPHQW GLDJUDP DUH LQFOLQHG VWUDLJKW OLQHV DQG SDUDEROLF FXUYHV 7KH VKHDU IRUFH GLDJUDP KDV SDUDEROLF FXUYHV DQG EHQGLQJ PRPHQW GLDJUDP KDV FXELF SDUDEROLF FXUYHV IRU XQLIRUPO\ YDU\LQJ loads. 4.9

POINT OF CONTRAFLEXURE

,WLVDOVRFDOOHGSRLQWRILQÀHFWLRQ,WLVWKHSRLQWZKHUHWKHFXUYDWXUH RIWKHGHÀHFWLRQFXUYH DQG EHQGLQJPRPHQWFKDQJHVLJQV7KHEHQGLQJPRPHQWDQGWKHFXUYDWXUHRIWKHGHÀHFWLRQFXUYHERWK DUH]HURDWWKLVSRLQW+RZHYHUDSRLQWZKHUHERWKEHQGLQJPRPHQWDQGFXUYDWXUH GHÀHFWLRQ DUH ]HURLVQRWQHFHVVDULO\DSRLQWRIFRQWUDÀH[XUHDVWKH\PD\EH]HURZLWKRXWFKDQJLQJVLJQVDWWKDW point.

4.10 SFD AND BMD FOR CANTILEVER BEAMS 4.10.1 Cantilever Beam carrying a Point Load at its Free End 5HIHU)LJ a). AB is the length of the beam. Calculations for shear forces Consider a section XX of the beam at a distance [ from the free end B. Shear force at the section is V = + W and it is independent of the distance [,WUHPDLQVFRQVWDQW WKURXJKRXWWKHOHQJWKRIWKHFDQWLOHYHU )LJ b)). Calculations for bending moments Bending moment at the section is M[ = – :[ KRJJLQJ  Bending moment at the free end, MB  IRU[ = 0) %HQGLQJPRPHQWDWWKH¿[HGHQGMA = – Wl IRU[ = l) The equation of B.M. at the section represents a straight line, hence the bending moments at A and BDUHFRQQHFWHGE\DVWUDLJKWOLQH )LJ c)).

Shear Forces and Bending Moments in Beams

149

Fig. 4.7

Alternative method (for bending moment) A section of the beam may be selected at a distance [IURPWKH¿[HGHQG )LJ  Bending moment at the section is, M[ = – W l – [) %HQGLQJPRPHQWDWWKH¿[HGHQGA, MA = – W l – 0) = – Wl

IRU[ = 0)

Fig. 4.8

Bending moment at the free end B, MB = – W l – l   IRU[ = l)

4.10.2 Cantilever Beam carrying Uniformly Distributed Load (udl) throughout the Span 5HIHU)LJ a). Consider a section XX of the beam at C, a distance [ from the free end B )LJ a)).

150

Strength of Materials

Calculations for shear forces Shear force at the section is V = + Z[  IRU[ = 0) Shear force at the free end B, VB   6KHDUIRUFHDWWKH¿[HGHQGA, VA = + wl   IRU[ = l) The equation of the shear force at the section represents a VWUDLJKWOLQHKHQFHWKHYDULDWLRQRIWKHVKHDUIRUFHEHWZHHQA and BLVVKRZQE\DVWUDLJKWOLQH )LJ b)). Calculations for bending moments Bending moment at the section is wx  M[ = – Z[¹ [ = –   MB   IRU[ = 0)

Bending moment at B, Bending moment at A,

MA

=–

wa     IRU[ = l) 

Fig. 4.9

The equation of the bending moment at the section represents DSDUDERODKHQFHWKHEHQGLQJPRPHQWYDULDWLRQLVSDUDEROLFEHWZHHQA and B )LJ c)). Alternative method Calculations for shear forces Consider a section XX of the beam at a distance [IURPWKH¿[HGHQGA )LJ 

Fig. 4.10

Shear force at the section is V = + w l – [) Shear force at the free end, VB = + w l – l   IRU[ = l) 6KHDUIRUFHDWWKH¿[HGHQG VA = + wl  IRU[ = 0) Calculations for bending moments Bending moment at the section is M[ = – w l – [) ¹

w (l  x)2 (l  x) =– 2 2

Bending moment at the free end is  w (l  l ) 2   IRU[ = l) MB = 2

Shear Forces and Bending Moments in Beams

151

%HQGLQJPRPHQWDWWKH¿[HGHQGLV wl  w (l  0)2 MA =  IRU[ = 0) =–  2

4.10.3 Cantilever Beam carrying Uniformly Distributed Load over a certain Length from the Free End 5HIHU)LJ a). Consider a section XX of the beam at a distance [ from the free end B )LJ a)).

Fig. 4.11

Calculations for shear forces Shear force at the section is

V = + Z[

Shear force at the free end, VB  IRU[ = 0) Shear force at C is VC = + wa IRU[ = a) The shear force between A and C remains constant at wa7KHVKHDUIRUFHYDULDWLRQEHWZHHQ B and C is shown by an inclined straight line and between A and C by a horizontal straight line )LJ b)).

152

Strength of Materials

Calculations for bending moments Bending moment at the section is M[ = – Z[ ¹

[ wx  =–  

Bending moment at free end, MB  IRU[ = 0) Bending moment at C is wa  – MC =  IRU[ = a)  %HQGLQJPRPHQWDWWKH¿[HGHQGLV a⎞ ⎟ ⎠ 7KHYDULDWLRQRIWKHEHQGLQJPRPHQWEHWZHHQB and CLVVKRZQE\DSDUDEROLFFXUYHDQGEHWZHHQ A and CE\DQLQFOLQHGVWUDLJKWOLQH )LJ c)). MA = – wa ⎛⎜ l − ⎝

4.10.4 Cantilever Beam carrying Uniformly Distributed Load over a certain Length from the Fixed End Consider a section XX of the beam at a distance [ from C )LJ a)). Calculations for shear forces Shear force at the section is V = + Z[ Shear force at C, VC  IRU[ = 0)

Fig. 4.12

Shear Forces and Bending Moments in Beams

153

Shear force at A, VA = wa IRU[ = a) 6KHDUIRUFHYDULDWLRQEHWZHHQA and C is shown by an inclined straight line and there is no shear force for the portion BC )LJ b)). Calculations for bending moments Bending moment at the section is M[ = – Z[ ¹

wx  [ =–  

Bending moment at C, MC  IRU[ = 0) Bending moment at A, MA = –

wa   IRU[ = a) 

7KHYDULDWLRQRIWKHEHQGLQJPRPHQWEHWZHHQA and C is parabolic and there is no bending mo ment on BC )LJ c)).

4.10.5 Cantilever Beam carrying Uniformly Distributed Load over its Entire Span and a Point Load at its Free End 5HIHU)LJ a). Calculations for shear forces Consider a section XX of the beam at a distance [ from the free end B )LJ a)).

Fig. 4.13

154

Strength of Materials

Shear force at the section is V  Z[ + W) Shear force at B is VB  w . 0 + W) = + W IRU[ = 0) 6KHDUIRUFHDWWKH¿[HGHQGA is VA  wl + W  IRU[ = l) The SFDLVVKRZQLQ)LJ b). Calculations for bending moments Bending moment at the section is x⎞ ⎛ M[ = – ⎜ Wx + w ⋅ x ⋅ ⎟ ⎠ ⎝

Bending moment at B is

⎛ wx  ⎞ = – ⎜⎜ Wx + ⎟  ⎟⎠ ⎝

⎛ w ⋅ 02 ⎞ MB = – ⎜⎜ W ⋅ 0 + ⎟   IRU[ = 0) 2 ⎟⎠ ⎝ Bending moment at A is  ⎞ ⎛ MA = – ⎜ Wl + wl ⎟  IRU[ = l) ⎜  ⎟⎠ ⎝ Since the equation for bending moment at the section represents a parabola, hence the bending PRPHQWYDULDWLRQEHWZHHQA and BLVSDUDEROLF )LJ c)).

4.10.6 Cantilever Beam carrying several Point Loads Let us consider that three point loads W1, W and W3DUHDFWLQJDWUHVSHFWLYHGLVWDQFHVRIl1, l and l IURPWKH¿[HGHQGA )LJ a)). Part BD Consider a section XX of the beam between B and D, at a distance [1 from free end B. Shear force at the section is V = + W3 and it remains constant between B and D. Bending moment at the section is M[ = – W3 [1 Bending moment at B, MB  IRU[1 = 0) Bending moment at D, MD = – W3 l – l  IRU[1 = l – l) 7KHYDULDWLRQRIWKHVKHDUIRUFHEHWZHHQB and D is shown by a horizontal straight line and that of bending moment by an inclined straight line. Part CD Consider a section XX of the beam between C and D, at a distance [ from B.

Shear Forces and Bending Moments in Beams

Shear force at the section is V

 W + W3), and it remains constant between C and D.

Fig. 4.14

Bending moment at the section is M[ = – [W3 [ + W {[± l – l)}] Bending moment at C is MC = – [W3 l – l1) + W {l – l1 – l + l`@ IRU[ = l – l1) = – [W3 l – l1) + W l – l1)]

155

156

Strength of Materials

Part AC Consider a section XX of the beam between A and C, at a distance [3 from B. Shear force at the section is V  W1 + W + W3), and it remains constant between A and C. Bending moment at the section is M[ = – [W3 [3 + W {[3± l – l)} + W1 {[3± l – l1)}] Bending moment at A is MA = – [W3 l + W {l – l + l} + W1 {l – l + l1`@ IRU[3 = l) = – [W3 l + W l + W1 l1] The SFD and the BMDDUHVKRZQLQ)LJ b DQG c UHVSHFWLYHO\

4.10.7 Cantilever Beam carrying Uniformly Varying Load 7KHORDGRQWKHFDQWLOHYHUYDULHVXQLIRUPO\IURP]HURDWIUHHHQGWRwXQLWOHQJWKDWWKH¿[HGHQG )LJ a)). Calculations for shear forces Consider a section XX of the beam at a distance [ from the free end B. 's ABC and BDEDUHFRPSDUHGWR¿QGWKHUDWHRIORDGLQJDWWKHVHFWLRQ AC AB = DE BD or

DE =

AC.BD w.x AB = l

Shear force at the section is V = + Triangular load BDE = +

1 × base × height 2

= +

wx 1 wx × x× =+ l 2 l



Shear force at B, FB  IRU[ = 0) Shear force at A is VA = +

wx   IRU[ = l) l

Shear Forces and Bending Moments in Beams

157

Fig. 4.15

Calculations for bending moments Bending moment at the section is M[ = – Triangular load BDE – =–

[ 3

wx 2 x wx3 u =– 2l 3 6l

Bending moment at the free end, MB  IRU[ = 0) %HQGLQJPRPHQWDWWKH¿[HGHQGA is MA = –

wl 2  IRU[ = l) 6

7KHYDULDWLRQRIWKHVKHDUIRUFHEHWZHHQA and BLVVKRZQE\DSDUDEROLFFXUYHDQGRIWKHEHQGLQJ PRPHQWE\DFXELFFXUYH )LJ b DQG)LJ c UHVSHFWLYHO\ 

158

Strength of Materials

Example 4.1 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUDFDQWLOHYHUEHDPORDGHGDVVKRZQLQ )LJ a). Solution:

Calculations for shear forces

Shear force at G is VG N1DQGLWUHPDLQVFRQVWDQWXSWRE. Shear force just to the right of D 

  –  N1

Fig. 4.16

Shear Forces and Bending Moments in Beams

159

Shear force just to the left of D    –  N1 Shear force at C is   

VC  –   N1DQGLWUHPDLQVFRQVWDQWXSWRB.

Shear force just to the left of B   –   N1DQGLWUHPDLQVFRQVWDQWXSWRA.

7KHVKHDUIRUFHYDULDWLRQEHWZHHQC and D, and between D and E are shown by inclined straight OLQHV )LJ b)). Calculations for bending moments Bending moment at G is MG = 0 Bending moment just to the right of F = – – 1) = –N1¹m Bending moment just to the left of F = –   VLQFHDFRXSOHRIPDJQLWXGHN1¹m is acting at F) = –N1¹m Bending moment at E is ME = –>–  @ –N1¹m Bending moment at D is 1 MD = – ⎡ 2 × 3 + 20 + 2 × 1 × ⎤ = –N1¹m ⎢ 2 ⎥⎦ ⎣ Bending moment at C is MC = – ⎡ 2 × 4 + 20 + 2 × 2 × 2 + 3 × 1⎤ = –N1¹m ⎢ ⎥ 2 ⎣ ⎦ Bending moment at B is ⎡ 2⎞ ⎤ ⎛ MB = – ⎢ 2 × 5 + 20 + 2 × 2 × ⎜1 + ⎟ + 3 × 2 ⎥ = –N1¹m 2⎠ ⎝ ⎣ ⎦ Bending moment at A is MA = – ⎡⎢ 2 × 6 + 20 + 2 × 2 × ⎛⎜ 2 + 2 ⎞⎟ + 3 × 3 + 5 × 1⎤⎥ = –N1¹m 2⎠ ⎝ ⎣ ⎦ The BMDLVVKRZQLQ)LJ c).

160

Strength of Materials

Example 4.2 )LQGWKHUHDFWLRQDWWKH¿[HGHQGRIWKHFDQWLOHYHUORDGHGDVVKRZQLQ)LJ a). Draw also shear force and bending moment diagrams.

Fig. 4.17

Solution: Reaction at A Total downward load on the beam    N1 

+HQFHWKHYHUWLFDOXSZDUGUHDFWLRQDWWKH¿[HGHQGA  N1 Taking moments of forces about AWKHUHVXOWDQWPRPHQWRQWKHEHDPLVJLYHQDV M R = – 30 –± &RXSOHDWE) – 30 –±– &RXSOHDWB) = –N1¹P &ORFNZLVH

+HQFHDUHDFWLRQPRPHQWRIN1¹P DQWLFORFNZLVH ZLOODFWDWWKH¿[HGHQGA along with a YHUWLFDOUHDFWLRQRIN1

Shear Forces and Bending Moments in Beams

161

Calculations for shear forces Shear force between D and F N1 Shear force just to the left of D N1DQGLWUHPDLQVFRQVWDQWXSWRC. Shear force just to the left of C N1DQGLWUHPDLQVFRQVWDQWXSWRA. The SFDLVVKRZQLQ)LJ b). Calculations for bending moments Bending moment at F is

MF = 0

Bending moment just to the right of E = –N1¹m Bending moment just to the left of E = –   –N1¹m   6LQFHDFRXSOHRIPDJQLWXGHN1¹m is acting at E) Bending moment at D is MD = – –  –N1¹m Bending moment at C is MC = – – 3 + 10 + 30 – 1) = –N1¹m Bending moment just to the right of B = – – 4 + 10 + 30 ––  ±N1¹m Bending moment just to the left of B = – –N1¹P 6LQFHDQRWKHUFRXSOHRIPDJQLWXGH N1¹m is acting at B) Bending moment at A is M A = – –––±  –N1¹m The BMDLVVKRZQLQ)LJ c).

4.11

SFD AND BMD FOR SIMPLY SUPPORTED BEAMS

4.11.1 Simply Supported Beam carrying a Central Point Load A simply supported beam AB of length l with a point load W acting at its centre is shown in )LJ a). Reactions at A and B 7R¿QGWKHUHDFWLRQVDWWKHWZRVXSSRUWVA and B, take moments of the forces about A. RB – l = W – or Also

RB =

l 

W ) 

RA + RB = W = Load on the beam

162

or

Strength of Materials

RA = W – RB =

W () 2

Calculations for shear forces Consider a section XX of the beam in BC, at a distance x from B. Shear force at the section is V =–

W 2

Fig. 4.18

Shear force just to the right of C is – Shear force just to the left of C is –

W, and it remains constant for the portion BC. 2

W W, and it remains constant for the portion AC. +W=+ 2 2

Shear Forces and Bending Moments in Beams

163

Calculations for bending moments Bending moment at the section is

Bending moment at B,

W ¹[  MB  IRU[ = 0)

Bending moment at C,

MC =

M[ = +

W l l ˜  IRU[ = )   

=+ Bending moment at A is MA = +

Wl 4 W l ¹ l – W¹ = 0  

The bending moments between B and C, and between A and C are joined by inclined straight lines because B.M. at the section represents a straight line. The SFD and the BMDDUHVKRZQLQ)LJ b DQG c UHVSHFWLYHO\

4.11.2 Simply Supported Beam carrying an Eccentric Point Load A simply supported beam AB is carrying a point load W at a distance a from A and b from B )LJ a)). Reactions at A and B Taking moments of the forces about AZHKDYH RB – l = W – a RB =

or But or

Wa  ) l

RA + RB = W RA = W – RB =

Wb  ) l

Calculations for shear forces Shear force between B and C is –

Wa . l

Shear force just to the left of C is –

Wb Wa , and it remains constant for the portion AC. +W = l l

164

Strength of Materials

Calculations for bending moments The bending moments at A and B are zero. Bending moment at C is MC = RB – b =

Wab l

Fig. 4.19

The SFD and the BMDDUHVKRZQLQ)LJ b DQG)LJ c UHVSHFWLYHO\

Shear Forces and Bending Moments in Beams

165

4.11.3

Simply Supported Beam carrying Uniformly Distributed Load (udl) over its Entire Span The intensity of loading on the beam is wXQLWOHQJWK )LJ a)). Reactions at A and B Taking moments of the forces about AZHKDYH l RB – l = w – l –  l 7RWDOORDGHTXLYDOHQWWRXGO is w – lDFWLQJDWLWVFHQWUHRIJUDYLW\i.e. at a distance from A.)  wl  ) or RB =  1RZ RA + RB = wl = Total load on the beam or

RA =

wl  ) 

Fig. 4.20

166

Strength of Materials

Calculations for shear forces Consider a section XX of the beam at a distance [ from B. Shear force at the section is V = – RB + Z[ = –

wl  wx 

Shear force at B,

VB = –

wl  IRU[ = 0) 

Shear force at A,

VA = –

wl wl + wl IRU[ = l) = +  

Shear force at midpoint C of the beam =–

wl wl + =0  

The SFDLVVKRZQLQ)LJ b). Calculations for bending moments Bending moment at the section is M[ = RB [ – Z[ – = Bending moment at B,

[ 

wl wx  x  

MB  IRU[ = 0)

Bending moment at A is MA =

wl wl  l  IRU[ = l)  

=0 Bending moment at midpoint C of the beam is 

MC

wl l w ⎛ l ⎞ l ⋅ − ⎜ ⎟  IRU[ = ) =    ⎝⎠  2 wl = 8

The equation of the bending moment at the section represents a parabola, hence the bending mo PHQWFXUYHLVSDUDEROLF )LJ c)).

4.11.4 Simply Supported Beam carrying Uniformly Varying Load which varies from Zero at Each End to w per unit length at the Midpoint The beam is of length l and carries wXQLWOHQJWKRIORDGDWLWVPLGSRLQWC )LJ a)).

Shear Forces and Bending Moments in Beams

Reactions at A and B Taking moments of the forces about AZHKDYH RB – l = 6LQFHWRWDOORDGRQWKHEHDP  or But or

RB = RA + RB = RA =

wl l ˜   wl l , and it acts at a distance from A.)   wl ) 4 wl  wl wl wl  ) = 4 2 4

Fig. 4.21

167

168

Strength of Materials

Calculations for shear forces Consider a section XX of the beam in BC at a distance [ from B. Shear force at the section is V = – RB + Triangular load BEF Compare 's BCD and BFE. CD EF = BC FB EF = =

The triangular load BEF =

CD ˜ FB BC wx wx = ⎛l⎞ l ⎜ ⎟  ⎝ ⎠ 1 × base × height 2

1 2 wx wx 2 × x× = l l 2  +HQFHVKHDUIRUFHDWWKHVHFWLRQLV wl wx 2  V =– 4 l =

Shear force at B,

VB = –

wl  IRU[ = 0) 4 2

Shear force at C is

VC

wl w ⎛ l ⎞ l + ⎜ ⎟  IRU[ = ) =– 4 l ⎝2⎠  =–

Shear force at A is

VA = –

wl w l 2 + ⋅ = 0 4 l 4 wl wl wl  =+ 4 4 2

The shear force diagram is parabolic, since the equation of the shear force at the section represents DSDUDEROD )LJ b)). Calculations for bending moments Bending moment at the section is M[ = RB [ – Triangular load BEF – wl wx3 wl wx 2 x  x − ⋅ x = = 4 3l 4 l 3

[ 3

Shear Forces and Bending Moments in Beams

169

Bending moments at A and B are: MA = 0 MB = 0 )RUEHQGLQJPRPHQWWREHPD[LPXP dMx =0 dx wl wx 2  =0 4 l ,WJLYHVWKHYDOXHRI[ =

l QHJDWLYHYDOXHKDVQRVLJQL¿FDQFH 

+HQFHWKHEHQGLQJPRPHQWLVPD[LPXPDW[ =

l . 

l Substituting [ = LQ WKH HTXDWLRQ IRU EHQGLQJ PRPHQW DW WKH VHFWLRQ WR JHW PD[LPXP B.M.,  ZHKDYH MPD[ = MC =

wl l w ⎛ l ⎞ ⋅ − ⋅⎜ ⎟ 4 2 3l ⎝ 2 ⎠

3

=+

wl 2 12

The BMDLVVKRZQLQ)LJ c).

4.11.5 Simply Supported Beam carrying Uniformly Varying Load which varies from Zero at One End to w per unit length at Other End 5HIHU)LJ a). Load at A is zero and at B is w per unit length. Reactions at A and B Taking moments of the forces about AZHKDYH RB – l =

w u l 2l u 2 3

6LQFHWKHFHQWUHRIJUDYLW\RIDWULDQJOHOLHVDWDGLVWDQFH or and or

RB = RA + RB = RA =

wl ) 3 wl  wl wl wl –  ) =  3 

2l IURPWKHYHUWH[ 3

170

Strength of Materials

Fig. 4.22

Calculations for shear forces Consider a section XX of the beam at a distance [ from A. Shear force at the section is V = + RA – Triangular load ADE Compare 's ABC and ADE. DE BC = AE AB DE =

BC ˜ AE wx = AB l

Shear Forces and Bending Moments in Beams

171

+HQFHVKHDUIRUFHDWWKHVHFWLRQLV V =+

wl wx 2 wl 1 wx  − ×x × =+ 6 2l 6 2 l

Shear force at A is VA =

wl  IRU[ = 0) 

Shear force at B is VB = +

wl wl w .   l  IRU[ = l) = –  6 2l

7R¿QG[, equate F[ to zero. :HJHW[ =

l 3

LJQRULQJQHJDWLYHYDOXH

7KHYDULDWLRQRIWKHVKHDUIRUFHLVSDUDEROLFVLQFHWKHVKHDUIRUFHHTXDWLRQDWWKHVHFWLRQUHSUHVHQWV DSDUDEROD )LJ b)). Calculations for bending moments Bending moment at the section is M[ = RA – [ – Triangular load ADE –

[ 3

wl wx 2 x wl wx3 ⋅ = x− x 6 2l 3 6 6l Bending moment at A, where [ = 0, is =

MA = 0 and

MB  IRU[ = l)

7KHEHQGLQJPRPHQWLVPD[LPXPZKHUHVKHDUIRUFHLV]HURWKDWLVDW[ = 0D[LPXPEHQGLQJPRPHQWLV MPD[

wl l w⎛ l ⎞ ⋅ − ⎜ = ⎟ 6 3 6l ⎝ 3 ⎠ =

wl 2 6 3



wl 2 18 3

=

l . 3

3

wl 2 9 3

7KH EHQGLQJ PRPHQW GLDJUDP LV D FXELF FXUYH EHFDXVH WKH EHQGLQJ PRPHQW HTXDWLRQ DW WKH VHFWLRQLVDFXUYHRIWKLUGRUGHU )LJ c)).

172

Strength of Materials

4.11.6 Simply Supported Beam subjected to a Couple The couple is acting at point C )LJ a)). Reactions at A and B Take moments of the forces about B. RA – l = M or But

M l RA + RB = 0 RA =

or RB = – RA

Calculations for shear forces Consider a section of the beam at a distance [ from A. Shear force at the section is V = + RA The shear force between A and B remains constant at + The SFDLVVKRZQLQ)LJ b).

Fig. 4.23

M . l

Shear Forces and Bending Moments in Beams

173

Calculations for bending moments Bending moment at the section is M[ = + RA [ = + Bending moment at A is

M [ l

MA  IRU[ = 0)

Bending moment at C is MC = +

Ma  IRU[ = a) l

Bending moment just to the right of C = Bending moment at B is

M (a − l ) M Ma Mb –M= l – a) = – =– l l l l

M l−M =0 l The bending moments between A and C, and between B and C are shown by inclined straight lines )LJ c)). MB = RA l – M =

Example 4.3 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a). Solution:

Reactions at A and B

Take moments of the forces about A. RB – – 1 + 10 ––  or

RB =

and or

80  N1 ) 4

RA + RB  –  N1 RA N1 )

Calculations for shear forces Consider a section XX of the beam at a distance [ from A. The shear force at the section is V = + RA 

 N1 The shear force between A and CUHPDLQVFRQVWDQWDWN1 Shear force just to the right of C



 ± 

174

Strength of Materials

Fig. 4.24

The shear force remains zero for the portion CD. Shear force at B is VB ±± – = –N1 The shear force between A and C is connected by a horizontal straight line and between B and D by an inclined straight line due to XGOIRUWKLVSRUWLRQ )LJ b)). Calculations for bending moments Bending moment at the section is M [ = + RA ¹ [ 

 [ Bending moment at A is MA  IRU[ = 0) Bending moment at C is MC – IRU[ = 1 m)



 N1¹m

Shear Forces and Bending Moments in Beams

175

Bending moment at D is MD –  ±– 1 

 N1¹m Bending moment at B is

 MB –  ±–  ±–– ⎛⎜ ⎞⎟ = 0 ⎝⎠ The bending moment for AC is shown by an inclined straight line, for CD by a horizontal straight line and for BDE\DSDUDEROLFFXUYH )LJ c)). Example 4.4 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a). Solution:

Reactions at A and D

Take moments of the forces about A. 2 RD –   –– ⎛⎜ 0.5 + ⎞⎟ 2⎠ ⎝ or But or

⎛ 45 ⎞ RD = ⎜ ⎟  N1 ) ⎝ 4.5 ⎠ RA + RD – N1 RA N1 )

Calculations for shear forces Consider a section XX of the beam at a distance [ from A. Shear force at the section is V = + RA N1 The shear force for the portion ABUHPDLQVFRQVWDQWDWN1 Shear force at C is VC ±– ±N1 The shear force between C and DUHPDLQVFRQVWDQWDW ± N1 The shear force for AB and CD are shown by two horizontal lines and for BC, by an inclined VWUDLJKWOLQH )LJ b)). Calculations for bending moments Bending moment at the section is M [ = + RA ¹ [ 

 [

176

Strength of Materials

Fig. 4.25

Bending moment at A is MA  IRU[ = 0) Bending moment at B is MB = + RA – 

 –



 N1¹m Bending moment at C is MC = + RA –  ±––



 

 –±– N1¹m

Shear Forces and Bending Moments in Beams

Bending moment at D is ⎛ ⎞ MD = + RA –  ±–– ⎜ +  ⎟ = 0 ⎝ ⎠ Location of zero shear force Compare 's MNO and OPQ. PQ MN = OP MO 10 20 = (2  y ) y y = 1.33 m

or

+HQFHWKHVKHDUIRUFHLV]HURDWDGLVWDQFH  P=PIURPA. 7KHPD[LPXPEHQGLQJPRPHQWRFFXUVDWOJLYHQE\ MO = MPD[ = + RA –  ±îî î  N1¹m The BMDLVVKRZQLQ)LJ c). Example 4.5 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a). Solution:

Reactions at A and B

Take moments of the forces about A. RB – 3 = + 70 ––– 2QVROYLQJZHJHW But or

RB N1 ) RA + RB – N1 RA ± N1 )

Calculations for shear forces The shear force between A and C remains constant at RA N1 Shear force just to the right of C  ± –N1

 Shear force at B is

VB ±±– = –N1

177

178

Strength of Materials

Fig. 4.26

The SFDLVVKRZQLQ)LJ b). Calculations for bending moments The bending moments at A and B are zero, since the beam is simply supported. MA = MB = 0 Bending moment at C is MC = + RA – 1 

 – 1



 N1¹m The BMDLVVKRZQLQ)LJ c).

Shear Forces and Bending Moments in Beams

Example 4.6 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a).

Fig. 4.27

Solution:

Reactions at A and B

Take moments of the forces about A. RB –   –– 2QVROYLQJZHJHW But or

RB

 + 10 –– 

⎞ ⎛ ⎜ + ⎟ ⎠ ⎝

N1 )

RA + RB –– N1 RA = 30 – RB ± N1 )

179

180

Strength of Materials

Calculations for shear forces Shear force at A is VA = + RA N1 Sear force at C is VC = + RA±– 

 ± N1 Shear force at B is VB = + RA±–±– ±N1 The SFDLVVKRZQLQ)LJ b).

Calculations for bending moments The bending moments at A and B are zero. M A = MB = 0 Bending moment at C is MC = + RA –±–– 

 

 –±–– N1¹m

Location of zero shear force Compare 's MNO and OPQ. PQ MN = OP NO 2.5 17.5 = y 2 y 6ROYLQJIRUy, we get y P +HQFHWKHVKHDUIRUFHLV]HURDWDGLVWDQFH 

 P PIURPA. 7KHPD[LPXPEHQGLQJPRPHQWRFFXUVDWOJLYHQE\ 0.25 ⎞ ⎛2 MO = MPD[ = + RA –  ±–– ⎜ + 0.25⎟ – 10 –– ⎠ ⎝2 2

  –±–±–– N1¹m The BMDLVVKRZQLQ)LJ c).

Shear Forces and Bending Moments in Beams

Example 4.7 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a). Solution: 5HIHU)LJ

Fig. 4.28

Reactions at A and E Take moments of the forces about A. RE – – or 1RZ or

RE N1 ) RA + RE = 7 RA N1 )

181

182

Strength of Materials

Calculations for shear forces Shear force at A is VA = + RA N1 The shear force between A and BUHPDLQVFRQVWDQWDWN1 Shear force just to the right of B 

 ± ±N1 The shear force between B and EUHPDLQVFRQVWDQWDW±N1 The SFDLVVKRZQLQ)LJ b).

Calculations for bending moments Bending moments at A and E are zero, because the beam is simply supported. M A = ME = 0 Bending moment at B is

MB = + RA ¹ N1¹m

Bending moment at C is MC = + RA – 3 – 7 – ±N1¹m Bending moment just to the right of C ± N1P Bending moment at D is

MD = + RA –±– ±N1¹m

Bending moment just to the right of D ± N1¹m The BMDLVVKRZQLQ)LJ c). /RFDWLRQRISRLQWRIFRQWUDÀH[XUH There are two points where bending moments are zero, one point lies between B and C and other between C and D. Compare 's MNO and OPQ. PQ MN = OP NO PQ MN = NP  NO NO 4.25 1.25 = NO 2  NO or

NO P

+HQFHWKH¿UVWSRLQWRIFRQWUDÀH[XUHOLHVDWDGLVWDQFHRI   PIURPA.

Shear Forces and Bending Moments in Beams

183

1RZFRPSDUH's PRS and STU. UT PR = ST PS 3.75 1.75 = PS 2  PS or PS P +HQFHWKHVHFRQGSRLQWRIFRQWUDÀH[XUHOLHVDWDGLVWDQFHRI 

 P PIURPA.

Example 4.8 7KHORDGLQJRQDVLPSO\VXSSRUWHGEHDPRIOHQJWKPYDULHVJUDGXDOO\IURPN1PDWRQHHQGWR N1PDWWKHRWKHUHQG'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKLVEHDP Solution: 5HIHU )LJ  7KH JLYHQ EHDP LV FRQYHUWHG LQWR DQ HTXLYDOHQW EHDP VXFK WKDW LW FRQVLVWVRIXQLIRUPO\GLVWULEXWHGORDGRILQWHQVLW\N1PWKURXJKRXWLWVOHQJWKDQGDXQLIRUPO\YDU\ ing load with zero at BWRDPD[LPXPORDGRIN1PDWA. Reactions at A and B Take moments of the forces about A. RB – 10 = 3 – 10 – or Also or

1 10 10 + – 4 – 10 –   2 3 2

RB N1 ) RA + RB = 3 – 10 +

1 – 4 – N1 2

RA N1 )

Calculations for shear forces Consider a section XX of the beam at a distance [ from B. 7KHUDWHRIORDGLQJ YHUWLFDO IRUWKHWULDQJXODUORDGDWWKHVHFWLRQLV 4 [ . 10 Shear force at the section is V = – RB + 3[ + Triangular load at the section = – RB + 3[ +

1 4[ u [ ±[[ 2 10

Shear force at B is Shear force at A is

VB = – RB ±N1 IRU[ = 0)

VA ±–– IRU[ P  N1 The SFDLVVKRZQLQ)LJ c).

184

Strength of Materials

Calculations for bending moments Bending moment at the section is M[ = + RB [ – 3 – [ –

[ 1 4[ [ − × × [ ×  [±[±[3 2 2 10 3

Fig. 4.29

Shear Forces and Bending Moments in Beams

185

For shear force to be zero d Mx =0 dx ±[±[ = 0



2QVROYLQJZHJHW

[ RU±

:HDFFHSW [ P %HFDXVH[ = –PFDUULHVQRPHDQLQJ  %HQGLQJPRPHQWLVPD[LPXPDWWKHSRLQWZKHUHVKHDUIRUFHLV]HURi.e., at [ P M PD[ –±–±–3 N1¹m The bending moments at A and B are zero. MA = MB = 0 The BMDLVVKRZQLQ)LJ G). Example 4.9 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a). Solution: Reactions at A and B 7KHORDGRIN1LVDSSOLHGRQWKHEHDPWKURXJKDEUDFNHWRIOHQJWKP¿[HGWRLWDWDGLVWDQFHRIP from A'XHWRWKLVDORDGRIN1DQGDEHQGLQJPRPHQWRI –  N1¹m are acting at C. Take moments of the forces about A. RB – – 4 or Also or

RB N1 ) RA + RB N1 VA N1 )

Calculations for shear forces Shear force at A is VA = + RA N1 The shear force between A and CUHPDLQVFRQVWDQWN1 The shear force just to the right of C ± ±N1 The shear force between B and CUHPDLQVFRQVWDQWDW±N1 The SFDLVVKRZQLQ)LJ c).

186

Strength of Materials

Fig. 4.30

Calculations for bending moments The bending moments at A and B are zero. MA = MB = 0 Bending moment just to the left of C = + RA – – 4 



 N1¹m

Bending moment just to the right of C ± 



The BMDLVVKRZQLQ)LJ G).

 N1¹m

Shear Forces and Bending Moments in Beams

187

4.12 RELATIONS AMONG LOAD, SHEAR FORCE AND BENDING MOMENT AB is a simple beam supported at A and BDQGORDGHGZLWKYDU\LQJORDGZLWKLQWHQVLW\wXQLWOHQJWK )LJ 

Fig. 4.31

End A is chosen as the origin. A section CDPQ of length G[ of beam is considered. 7KHUHODWLRQVKLSEHWZHHQORDGDQGVKHDUIRUFHLVJLYHQDV dV W = dx i.e., the rate of change of shear force at any section is equal to the rate of loading at that section. 7KHUHODWLRQVKLSEHWZHHQVKHDUIRUFHDQGEHQGLQJPRPHQWLVJLYHQDV dM V = dx ,WPHDQVWKDWWKHUDWHRIFKDQJHRIEHQGLQJPRPHQWDWDQ\VHFWLRQLVHTXDOWRWKHVKHDUIRUFHDW that section.

4.13

SFD AND BMD FOR OVERHANGING BEAMS

4.13.1 Overhanging Beam with equal Overhangs on Each Side and loaded with Point Loads at the Ends The beam is supported at B and C7KHOHQJWKRIWKHEHDPLV la >)LJ a)]. Reactions at B and C Take moments of the forces about B. W – a + RC – l = W l + a) or But or

RC = W ) RB + RC = W + W W RB = W )

Calculations for shear forces Consider a section XX of the beam at a distance [ from A. Shear force at the section is V =–W

188

Strength of Materials

The shear force between A and B remains constant at – W. Shear force just to the right of B = – W + RB = – W + W = 0

Fig. 4.32

The shear force remains zero between B and C. = Shear force at C is

M. l−M =0 l

VC = 0 + RC = + W The shear force between C and D remains constant at + W. The shear forces between A and B, and between C and D are connected by horizontal straight lines )LJ b)).

Shear Forces and Bending Moments in Beams

189

Calculations for bending moments Bending moment at the section is M[ = – :[ Bending moment at A, where [ = 0, is MA = 0 Bending moment at B, where [ = a, is MB = – Wa Bending moment at C is MC = – W a + l) + RB l = – Wa – Wl + Wl = – Wa Bending moment at D is MD = – W a + l + a) + RB l + a) + RCa = – Wa – Wl – Wa + Wl + Wa + Wa = 0 The bending moment diagrams between A and B, and between C and D are shown by inclined VWUDLJKWOLQHV )LJ c)).

4.13.2

Overhanging Beam with equal Overhangs on Each Side and loaded with a Uniformly Distributed Load over its Entire Span 5HIHU)LJ a). Reactions at B and C Take moments of the forces about B. (l  a) w – a – a + RC – l = w l + a)  2 w  wa  + RC l = [l + ala]   or But or

RC =

w (l  2a) ) 2

RB + RC = w la) RB =

w (l  2a) ) 2

190

Strength of Materials

Fig. 4.33

Shear Forces and Bending Moments in Beams

Calculations for shear forces Consider a section XX of the beam at a distance [ from A. Shear force at the section is V = – Z[ Shear force at A, VA  IRU[ = 0) Shear force at B, VB = – wa IRU[ = a) Shear force just to the right of B = – wa + RB = – wa + =

w la) 

w l 

Shear force at C is VC =

w l – wl 

=–

wl 

Shear force just to the right of C

Shear force at D is

=–

wl + RC 

=–

w wl +  la)  

= wa

VD = wa – wa = 0 The SFDLVVKRZQLQ)LJ b). Calculations for bending moments Bending moment at the section is M[ = − w . x . Bending moment at A,

x 2

MA  IRU[ = 0)

191

192

Strength of Materials

Bending moment at B is MB = – w ¹ a ¹ a 2 =–

wa 2 2

Bending moment at C is MC = – w (a + l)

(a  l ) + RB – l 2

=–

w w (l + 2a) – l (a + l)2 + 2 2

=–

w 2 a 2

Bending moment at D is MD = 0 l⎞ ⎛ The shear force is zero at a distance ⎜ a + ⎟ from A. It is the position of the maximum bending 2⎠ ⎝ moment. The maximum bending moment is given as l⎞ ⎛ l⎞ 1 l ⎞ ⎛ ⎛ Mmax = − w ⎜ a + ⎟ ⋅ ⎜ a + ⎟ ⋅ + RB ⎜ a + − a ⎟ ⎠ ⎝ ⎝ 2⎠ ⎝ 2⎠ 2 2 2

= −

=

w⎛ l⎞ w l ⎜ a + ⎟ + (l + 2a ) ⋅ 2⎝ 2⎠ 2 2

w ⎛ l2 2⎞ ⎜⎜ − a ⎟⎟ 2⎝4 ⎠

Case I When a < l , then Mmax will be positive. 2 Case II When a = l , then Mmax = 0 2 w ⎛ l2 2⎞ ⎜⎜ − a ⎟⎟ = 0 2⎝4 ⎠

Shear Forces and Bending Moments in Beams

or

a =±

l 

:HDFFHSW

a = +

l 2

⎛w ⎞ ⎜ ≠ 0⎟ ⎝2 ⎠ ⎡ ⎛ ⎢ because ⎜⎝ − ⎣

l⎞ ⎤ ⎟⎠ carries no meaning.⎥ 2 ⎦

Case III :KHQa > l , then MPD[ZLOOEHQHJDWLYH  The BMDIRUDOOWKHFDVHVDUHVKRZQLQ)LJ c  G DQG e). Example 4.10 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ a). Solution: Reactions at B and E Take moments of the forces about B. 8⎞ ⎛  –±– 4 – 3 –– ⎜ 4 + 4 + ⎟ + RE –  2⎠ ⎝

 or Also

RE N1 ) RB + RE –

 or

 N1 RB N1  )

Calculations for shear forces Shear force at A is

VA = –N1

The shear force between A and BUHPDLQVFRQVWDQWDWN1 Shear force just to the right of B = –RB ± 

 N1 The shear force between B and CUHPDLQVFRQVWDQWDWN1 Shear force just to the right of C = + 74.13 – 70



 N1

193

194

Strength of Materials

Fig. 4.34

The shear force between C and DUHPDLQVFRQVWDQWDWN1 Shear force at E is VE = + 4.13 – 3 – ±N1 Shear force just to the right of E 

 ±RE



 ±  The SFDLVVKRZQLQ)LJ b).

Shear Forces and Bending Moments in Beams

Calculations for bending moments The bending moments at A and E are zero. MA = ME = 0 Bending moment at B is MB = –– –N1¹m Bending moment at C is MC = ––  RB – 4 

 N1¹m Bending moment at D is MD = ––  RB –  ±– 4 = –––±



 N1¹m The BMDLVVKRZQLQ)LJ c). Location of zero shear force Compare two triangles MNO and OPQ. MN PQ = NO OP PQ MN = NP  NO NO 4.13 19.87 = NO 8  NO or

NO = 1.37 m

+HQFHWKHVKHDUIRUFHLV]HURDWDGLVWDQFH ±  PIURPE. 7KHPD[LPXPEHQGLQJPRPHQWRFFXUVDWOJLYHQE\ MPD[ = MO = +RE –±–– 

 –±––



 N1¹m The BMDLVVKRZQLQ)LJ c).

6.63 2

6.63 2

195

196

Strength of Materials

Example 4.11 Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.35(a).

Fig. 4.35

Solution: Reactions at B and D Taking moments of the forces about A, we have RB – 2.5 + RD – 7.5 = 40 – 5.5 = 220 and

RB + RD = 20 + 40 = 60 kN

From equation (2) RB = 60 – RD

… (1) … (2)

Shear Forces and Bending Moments in Beams

Substituting RB in equation (1), we get (60 – RD) – 2.5 + RD – 7.5 = 220 or

RD = 14 kN ()

On using equation (2), we get RB = 60 – 14 = 46 kN () Calculations for shear forces Consider a section XX of the beam at a distance x from A. Shear force at the section is V = – 20 kN The shear force between A and B is constant at – 20 kN. Shear force just to the right of B = – 20 + RB = – 20 + 46 = + 26 kN The shear force between B and C is constant at 26 kN. Shear force just to the right of C = + 26 – 40 = – 14 kN The shear force between C and D is constant at – 14 kN. The shear force diagram is shown in Fig. 4.35 (b). Calculations for bending moments Bending moment at the section is Mx = – 20x Bending moment at A is MA = 0

(for x = 0)

Bending moment at B is MB = – 20 – 2.5

(for x = 2.5 m)

= – 50 kN¹m Bending moment at C is MC = – 20 – (2.5 + 3) + RB – 3 = + 28 kN¹m Bending moment D is MD = – 20 – (2.5 + 3 + 2) + 46 – (3 + 2) – 40 – 2 =0 The bending moment diagram is shown in Fig. 4.35 (c).

197

198

Strength of Materials

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a  WKHOHIWSDUWLVPRYLQJXSDQGWKHULJKWSDUWPRYLQJGRZQ



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a)

Wl  

b)

Wl  4

c) Wl

G) Wl . 

3. 7KHEHQGLQJPRPHQWGLDJUDPIRUDFDQWLOHYHUEHDPFDUU\LQJDSRLQWORDGDWLWVIUHHHQGLV JLYHQDV

Shear Forces and Bending Moments in Beams

199

4. 7KHPD[LPXPEHQGLQJPRPHQWIRUDFDQWLOHYHUEHDPFDUU\LQJDXGO of intensity wXQLWOHQJWK RYHULWVHQWLUHVSDQLVJLYHQDV 

a) wl  4

2 b) wl  8

 c) wl  

 G)



wl  . 

5. 7KHYDULDWLRQRIEHQGLQJPRPHQWIRUDFDQWLOHYHUEHDPFDUU\LQJDXGO of intensity wXQLWOHQJWK RYHULWVHQWLUHVSDQLVVKRZQE\ 

a  DVWUDLJKWOLQH





c  DWKLUGGHJUHHSDUDEROD

b) a second degree parabola G) an ellipse.

6. 7KHEHQGLQJPRPHQWGLDJUDPIRUDFDQWLOHYHUEHDPFDUU\LQJDXGORYHULWVFHUWDLQOHQJWKIURP the free end is shown as

7. 7KHVKHDUIRUFHDQGWKHEHQGLQJPRPHQWGLDJUDPVIRUDFDQWLOHYHUEHDPFDUU\LQJDXGORYHULWV FHUWDLQOHQJWKIURPWKH¿[HGHQGDUHJLYHQDV

200

Strength of Materials

8. The PD[LPXP EHQGLQJ PRPHQW IRU D VLPSO\ VXSSRUWHG EHDP VXEMHFWHG WR D SRLQW ORDG DW LWV FHQWUHLVJLYHQDV 

2 b) Wl  4

a) Wl  

G) Wl . 

c) Wl  4

9. The shear force diagram for a simply supported beam carrying a XGORYHULWVHQWLUHVSDQLVVKRZQ as

10. 7KH PD[LPXP EHQGLQJ PRPHQW IRU D VLPSO\ VXSSRUWHG EHDP VXEMHFWHG WR D XGO of intensity wXQLWOHQJWKRYHULWVHQWLUHVSDQLVJLYHQDV 

a)

wl 2  6

3 b) wl  24

c)

wl 2  8

G)

wl 2 . 12

11. The reactions at the two supports of a simply supported beam carrying a XGO of intensity wXQLW OHQJWKRYHULWVHQWLUHVSDQDUHJLYHQDV 

a)

w, w  2 2

b) wl , wl  4 4

wl , wl . 2 2

c) wl , wl  2

G)

c) G) = ZG[

G) F = ZG[

12. The load and shear force relationship is 

a) G F = ZG[

b) GZ = )G[

13. The shear force and bending moment relationship is 

a) F =

d M dx





b) M =

d F dx





c) M =

dF  dx

G) F =

dF . dx

14. 7KHPD[LPXPEHQGLQJPRPHQWIRUDVLPSO\VXSSRUWHGEHDPFDUU\LQJDXGOZKLFKYDULHVIURP zero at each end to wSHUXQLWOHQJWKDWLWVPLGSRLQWLVJLYHQDV 

a)

wl 2  8

b)

wl 3  12

2 c) wl  12

G)

wl 2 . 24

Shear Forces and Bending Moments in Beams

201

15. 7KHUHDFWLRQVDWWKHWZRVXSSRUWVIRUWKHEHDPLQ4XHVWLRQ1RDUHJLYHQDV 

a)

wl , wl  4 2

b)

wl , wl  4 4

c)

wl , wl  8 8

G)

wl , wl . 8 4

16. 7KHPD[LPXPEHQGLQJPRPHQWIRUDVLPSO\VXSSRUWHGEHDPFDUU\LQJDXGOZKLFKYDULHVIURP zero at one end to wSHUXQLWOHQJWKDWRWKHUHQGLVJLYHQDV 

a)

wl 2  24

2 b) wl  8

c)

wl 2 9 3



G)

wl 2 . 8 3

ANSWERS 1. a) 10. c)

2. c)

3. c)

11. G) 12. c)

4. G)

5. b)

13. G) 14. c)

6. c)

7. b)

15. b) 16. c).

8. c)

9. b)

202

Strength of Materials

EXERCISES 1. 'UDZ WKH VKHDU IRUFH DQG EHQGLQJ PRPHQW GLDJUDPV IRU WKH EHDP VKRZQ LQ )LJ $OVR LQGLFDWHWKHORFDWLRQDQGPDJQLWXGHRIPD[LPXPEHQGLQJPRPHQW,VWKHUHDQ\FRQWUDÀH[XUH"

Fig. 4.36

Ans.

MPD[ N1¹m at 3 m from E&RQWUDÀH[XUHH[LVWVDWPIURPA).





2.

Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.37.

Fig. 4.37





Ans.

MC = MPD[ N1¹m).

3. 'UDZ WKH VKHDU IRUFH DQG EHQGLQJ PRPHQW GLDJUDPV IRU WKH EHDP VKRZQ LQ )LJ $OVR GHWHUPLQHWKHORFDWLRQDQGPDJQLWXGHRIWKHPD[LPXPEHQGLQJPRPHQW

Fig. 4.38





Ans.

M ±N1PDWB; MPD[ N1¹m, 1.333 m left of C).

Shear Forces and Bending Moments in Beams

4.

203

Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.39.

Fig. 4.39

⎛ ⎜ Ans. F = wo ⎝ 5.

⎛l ⎜ ⎝π

⎞ ⎛π x ⎞ ⎟ cos ⎜ ⎟ ; M = wo ⎝ l ⎠ ⎠

 ⎛π x ⎞ ⎛l ⎞ ⎜ ⎟ sin ⎜⎝ l ⎟⎠ ; MPD[ = wo ⎝π ⎠

 l⎞ ⎛l ⎞ ⎜ ⎟ , at [ =  ⎟⎠ . ⎝π ⎠

Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.40.

Fig. 4.40

Ans.

B.M. just to the left of C N1¹m; B.M. just to the right of C N1¹m).





6.

Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.41.

Fig. 4.41





Ans. B.M. just to the right of D N1¹m; B.M. just to the left of D N1¹m).

204

Strength of Materials

7. 'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ

Fig. 4.42

Ans.

MB = MPD[ N1¹m).





8.

Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.43.

Fig. 4.43

Ans.

MPD[ N1¹m).





9.

Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.44.

Fig. 4.44





Ans.

MPD[ “N1¹m).

Shear Forces and Bending Moments in Beams

10.

205

'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ

Fig. 4.45





Ans.

MC N1¹m; MD N1¹m; ME N1¹m).

11.'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ

Fig. 4.46





Ans.

B.M. just to the left of C N1¹m B.M. just to the right of C N1¹m MD N1¹m, ME N1¹m).

12. Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.47.

Fig. 4.47

Ans. MC N1¹m; MD = N1¹m). 13.'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ

Fig. 4.48

Ans. MC N1¹m, B.M. just to the right of D N1¹m B.M. just to the left of D ±N1¹m).

206

Strength of Materials

14. Draw the shear force and bending moment diagrams for the beam shown in Fig. 4.49.

Fig. 4.49

Ans. FB ±N1FA N1 For ACVKHDUIRUFHLVFRQVWDQWDWN1 For BCVKHDUIRUFHLVFRQVWDQWDW±N1 M A = MB = 0 M OHIWRID  ±N1P M ULJKWRID  N1P MPD[ MC N1P  15.'UDZWKHVKHDUIRUFHDQGEHQGLQJPRPHQWGLDJUDPVIRUWKHEHDPVKRZQLQ)LJ

Fig. 4.50

Ans. FB = ±N1 For BDWKHVKHDUIRUFHLVFRQVWDQWDW±N1 F OHIWRID  N1 For CDWKHVKHDUIRUFHLVFRQVWDQWDWN1 F OHIWRIC  N1 For ACWKHVKHDUIRUFHLVFRQVWDQWDWN1 M A = MB = 0). ‰‰‰

5 Stresses in Beams

D. J. Jourawski, born in 1821, was a Russian bridge and railway engineer. In 1844, only two years after graduating from the Institute of Engineers of Ways of Communication in St. Petersburg, he was assigned the task of GHVLJQLQJ DQG FRQVWUXFWLQJ D PDMRU EULGJH RQ WKH ¿UVW UDLOZD\ OLQH IURP Moscow to St. Petersburg. He developed the now widely used approximate theory for shear stresses in beams, also called the shear formula and applied his theory to various shapes of beams.

D. J. Jourawski (1821-1891)

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ :KDWLVPHDQWE\SXUHEHQGLQJ"  ‡ :KDWLVDQHODVWLFFXUYH"  ‡ :KDWGRHVWKHVHFWLRQPRGXOXVRIDFURVVVHFWLRQLQGLFDWH"  ‡ :KDWLVDFRPSRVLWHEHDP"  ‡ :K\GRHVWKHQHXWUDOD[LVDOZD\VSDVVWKURXJKWKHFHQWURLG"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_5

207

208

Strength of Materials

5.1 PURE BENDING IN BEAMS When external loads are applied on a beam, it bends in a curve due to bending moments produced. Stresses produced in the beam may be shear and bending. If only bending stresses are considered and shear force is neglected, the beam is said to be under pure bending or simple bending. Bending stresses may be tensile and compressive. If a cantilever beam is loaded as shown in Fig. 5.1(a), its upper layer is under tension and the bottom layer is in compression as is shown in Fig. 5.1(b). In between the two layers, there is a plane where no stress is acting. Such a plane is known as neutral plane. The magnitude of bending stress at any depth of a beam is proportional to its distance from the neutral plane.

Fig. 5.1

5.2 SIMPLE BENDING THEORY The beam upon loading bends in the form of a curve, known as elastic curve. Following assumptions are considered for the simple bending equation: z The material of the beam is isotropic and homogeneous. z The transverse sections of the beam, which were plane before bending remain so even after bending. z The Young’s modulus of elasticity of the beam material remains same in tension as well as in compression. z The stresses produced are within the elastic limits. z The radius of curvature of the beam is very large as compared to its cross-sectional dimensions. z There is no resultant push or pull on the cross-section of the beam.  z The loads are applied in the plane of bending. Let us consider certain portion of a beam being subjected to pure bending as shown in Fig. 5.2, where GH represents the neutral plane. The beam bends in a curve of radius R, known as radius of curvature. The two transverse sections AC and BD meet at point O, known as centre of curvature, making an angle T between them. A layer EF of beam at a distance y from the neutral plane is considered. GH = RT and EF = (R + y) T The strain produced in EF is given as y EF − GH ( R + y ) θ − Rθ ° = = = GH R Rθ

... (5.1)

Stresses in Beams

209

Fig. 5.2

From Hooke’s law, we have

 RUҏ where

Stress =E Strain σb =E ∈ σb ° = E Vb = Bending stress

...(5.2)

E = Modulus of elasticity of the beam material From equations (5.1) and (5.2), we have σ y = b R E E σb or = R y

... (5.3)

E is constant. Hence, the bending stress varies directly proportional to R its distance from the neutral plane. Bending stress is maximum at outermost layer of beam, where y is maximum and zero at neutral plane, where y is zero. For a given loading condition,

The transverse section of beam is shown in Fig. 5.2 (b). An elemental area Ga is considered at a distance y from the neutral axis. Force acting on the elemental area = Vb – Ga Moment of this force about the neutral axis (NA) = Vb – Ga – y =

E 2 y Ga R

(using equation (5.3))

210

Strength of Materials

The moment of resistance is

E 2 y Ga R The moment of resistance is equal to bending moment M.



M = where





E E 2 y Ga = R R



y 2 δa

y 2 δa = Second moment of area about the neutral axis = I

Hence,

M =

E .I R

E M = R I Comparing equations (5.3) and (5.4), we have or

σb M E = = y I R Equation (5.5) is known as simple bending equation or ÀH[XUHIRUPXOD.

... (5.4)

... (5.5)

5.3 POSITION OF THE NEUTRAL AXIS Let us consider the cross-section of a beam (Fig. 5.2 (b)). For equilibrium of the section, net force acting on it must be zero. Force acting on the elemental area = E y Ga R Total force on the beam section = For equilibrium or

E

∫ R y Ga =

E y δa R∫

E y δa = 0 R∫ ∫ y δa = 0

E is a constant quantity. R In the above equation, Ga › 0 but y = 0, indicating that the distance from the neutral axis to the centroid of the cross-sectional area must be zero. Hence, the neutral axis always passes through the centre of the area i.e., its centroid.

since

Stresses in Beams

211

5.4 SECTION MODULUS From bending equation, we have σb M = y I The bending stress is maximum, when y is maximum. M I = =S σbmax ymax where

S = Section modulus =

I ymax

The unit of S is P3, and it is a measure of strength of beam section. The bending stress can be expressed as M Vb = ... (5.6) max S

5.5 COMPOSITE BEAM $FRPSRVLWH ÀLWFKHG EHDPLVPDGHRIWZRGLIIHUHQWPDWHULDOVIRUH[DPSOHVWHHODQGFRQFUHWHRUZRRG and iron. One of the materials of the beam is called reinforcing material and is used to increase the strength of the beam. The basic assumptions of simple bending theory are applicable to composite beams. A composite beam consisting of a wooden beam and two steel plates is shown in Fig. 5.3. The two materials are rigidly connected and hence they have the same radius of curvature and same strains are produced in both of them.

Fig. 5.3

Let

Vs Vw Es Ew

= Stress produced in the steel plate = Stress produced in wood = Modulus of elasticity of steel = Modulus of elasticity of wood

212

Strength of Materials

 7KHVWUHVVHVSURGXFHGDUHFRQVLGHUHGDWDFHUWDLQ¿[HGGLVWDQFHIURPWKHQHXWUDOD[LV Now Strain in steel = Strain in wood

σs σw = Es Ew or where 

σs σw

=

Es =P Ew

... (5.7)

P = Modular ratio

The moment of resistance of the steel plate is given as

Is y

Ms = Vs where

... (5.8)

Is = Moment of inertia of the steel plate about the neutral axis (NA) y = distance from neural axis

The moment of resistance of wood is given as Mw = Vw where

Iw y

... (5.9)

Iw = Moment of inertia of the wooden section

The moment of resistance of the composite beam is the sum of the two moments of resistance. ...(5.10) M = Ms + Mw = Vs = But

Iw =

σw [P,s + Iw] y

y =

...(5.11)

(using equation (5.7)) ... (5.12)

bd 3 12

Is = 2. and

Is Iw 1 + Vw = [V I + Vw Iw] y y y s s

td 3 td 3 = 12 6

d 2

Using Is, Iw and y in equation (5.11), we have M =

σwd 2 (b + 2PW) 6

... (5.13)

Stresses in Beams

or

M =

σ s d 2 ⎛ b + 2t ⎞ ⎜ ⎟ ⎠ 6 ⎝m

213

... (5.14)

Hence, the moment of resistance of the composite beam is similar to that of a wooden beam of width (b + 2PW) and depth d. The resulting wooden beam is called equivalent beam. Alternatively, it is also ⎛b ⎞ similar to that of an equivalent steel beam of width ⎜ + 2t ⎟ and depth d. m ⎝ ⎠

5.6 BEAMS OF UNIFORM STRENGTH We have seen that bending moment is maximum at the centre and zero at the two support ends of a simply supported beam. If the beam is designed in such a way that it has uniform bending stress over the entire span so that it has uniform strength throughout its length, then the resulting beam is called a beam of uniform strength. For that, beam is expected to have minimum cross-section at the two supports and gradually increasing towards the centre of the beam, thus preventing wastage of material. The change in cross-section is made effective by (a) Keeping width of beam constant, but reducing its depth (b) Keeping depth of beam constant, but reducing its width (a) When width is constant and depth is varying. Consider a simply supported beam of length l carrying a point load W at the centre (Fig. 5.4). The reactions at the two supports are found as RA = RB =

W 2

Fig. 5.4

Let D be a point at a distance [ from A, where depth is required to be found. The bending moment at D is given as MD =

W x 2

2 The moment of resistance is V – bd D . 6

(using bending equation)

214

Strength of Materials

where dD represents the diameter of the beam at D. Equating the two moments, we have V–

bd D2 6

=

dD =

or

W [ 2 3Wx σb

... (5.15)

Since W, V and b are constants for a given beam, hence depth at any section is proportional to thereby indicating that the variation of depth is parabolic between points A and B. The depth at C

(where [ =

l ) is 2

d =

3Wl 2 σb

[,

... (5.16)

(b) When depth is constant and width is varying. Consider the same loading condition. Depth and width are shown in Fig. 5.5. The bending moment at W [. D is again 2

Fig. 5.5

The moment of resistance is V –

bD d 2 . 6

On equating the two moments, we have  or

Vu

W bD d 2 × [ = 2 6 bD =

3Wx σd 2

... (5.17)

Stresses in Beams

215

W, V, and d are constants for the beam, hence width at any section is proportional to [. l⎞ ⎛ The width at C ⎜ where x = ⎟ is 2⎠ ⎝ b =

3Wl 2σ d 2

...(5.18)

Example 5.1 A 5 m cantilever beam of cross-section 150 mm – 300 mm weighing 0.05 kN/m carries an upward concentrated load of 30 kN at its free end (Fig. 5.6). Determine the maximum bending stress at a section 2m from the free end.

Fig. 5.6

Solution: The bending moment at the section (2 m from the free end) is given as 2⎞ ⎛ M = 30 – 2 – ⎜ 0.50 × 2 × ⎟ 2⎠ ⎝ = 59 kN.m = 59 – 106 N.mm The moment of inertia of the beam section about the neutral axis (NA) is given as I =

1 – 150 – 3003 = 3.375 – 108 mm4 12

Using bending equation, we have σb y or

=

Vb = =

M I M . y I 59 × 106 . ⎛ 300 ⎞ = 26.22 N/mm2 3.375 × 108 ⎜⎝ 2 ⎟⎠

Ans.

The top layer of the beam is under compression and the bottom layer is in tension. Both layers are subjected to an equal stress of 26.22 N/mm2.

216

Strength of Materials

Example 5.2 A 5m long steel beam having an I-section is simply supported at its ends (Fig. 5.7). The tensile stress in beam does not exceed 25 MPa. Determine the safe uniformly distributed load to be placed on the entire span of the beam.

Fig. 5.7

Solution: The moment of inertia of the beam section about the neutral axis is given as I = M.I. of rectangle 250 mm – 400 mm – 2 (M.I. of rectangle 105 mm – 320 mm) =

1 1 – 105 – 3203 mm4 – 250 – 4003 – 2 – 12 12

= 7.598 – 10–4 m4 If w kN/m be the uniformly distributed load to be placed on the entire beam, then the bending moment under this condition is M =

wl 2 w × 25 = kN¹m 8 8

Using bending equation, we have σ M = b I y w × 25 1 – = 8 7.598 × 10−4 Solving for w, we get

25 × 106 ⎛ 400 ⎞ 103 × ⎜ × 10−3 ⎟ ⎝ 2 ⎠

w = 30.392 kN/m

Ans.

Stresses in Beams

217

Example 5.3 A simple steel beam of 4 m span carries a uniform load of 6 kN/m over its entire span and a point load N1DWLWVFHQWUH,IWKHSHUPLVVLEOHVWUHVVGRHVQRWH[FHHG03D¿QGWKHFURVVVHFWLRQRIWKHEHDP assuming depth to be twice of breadth. Solution: Given, Length of the beam, Uniform load, Point load, Bending stress, Let Width of the beam Depth of the beam

l w W Vb

=4m = 6 kN/m = 6 – 103 N/m = 2 N = 2 kN = 2 – 103 N = 100 – 106 N/m2 =b =d d = 2b (Given) The maximum bending moment due to udl and point load is given as M =

wl 2 Wl + 8 4

6 × 103 × 42 2 × 103 × 4 + = = 14000 N¹m 8 4 The moment of inertia of the beam section about the neutral axis is calculated as I =

1 1 . . 2 b (2b)3 = b4 bd 3 = 12 12 3

y =

d 2b =b = 2 2

Using bending equation, we have σb M = y I 14000 100 × 106 = 2 4 b b 3 Solving for b, we get and

b = 0.0594 m = 59.4 mm d = 2b = 118.8 mm

Hence, the cross-section of the beam is 59.4 mm – 118.8 mm.

Ans.

Example 5.4 A simply supported beam of cross-section 50 mm by 50 mm having a length of 800 mm is capable of carrying a point load of 3 kN at its centre. This beam is required to be replaced by a cantilever beam of the same material having cross-section 50 mm by 100 mm and length 1500 mm. Determine the maximum point load that can be placed at the free end of the cantilever.

218

Strength of Materials

Solution: For simply supported beam Load on the beam,

W1 = 3 kN

Length of the beam,

l1 = 800 mm

Width of the beam,

b = 50 mm

Depth of the beam,

d = 50 mm

The distance of the neutral axis from the outermost layers is 50 = 25 mm 2 The moment of inertia of the beam section about the neutral axis is obtained as 1 – 50 – 503 = 520833.33 mm4 I1 = 12 The bending moment is given as W1 l1 3 × 800 M1 = = = 600 kN¹mm 4 4 Using bending equation, we have M1 σb = I1 y1 y1 =

or

Vb =

600 × 25 M1 . y1 = = 0.0288 kN/mm2 520833.33 I1

For cantilever beam Load on the beam,

W2 = ?

Length of the beam,

l2 = 1500 mm

Width of the beam,

b = 50 mm

Depth of the beam,

d = 100 mm

The distance of the neutral axis from the outermost layer is 100 = 50 mm 2 The moment of inertia of beam section about the neutral axis is

y2 =

1 – 50 – 1003 = 4166666.7 mm4 12 The bending moment is found as

I2 =

M2 = W2 l2 = 1500 W2 kN¹mm Since both beams are made of the same material, hence equal bending stresses will be developed in both of them.

Stresses in Beams

219

Again using bending equation, we have σb M2 = y2 I2 1500 W2 0.0288 = 50 4166666.7 or

W2 = 1.6 kN

Ans.

Example 5.5 A simple beam of length 5 m carries two types of loads : a udl of 6 kN/m is acting over the entire span and a point load of 2 kN at a distance 2 m from the left support. Cross-section of the beam is shown in Fig. 5.8. Calculate the maximum bending stress at a distance 3.5 m from the left support. Solution: Support reactions at A and C Taking moments of the forces about A, we get 5⎞ ⎛ RC – 5 = 2 – 2 + ⎜ 6 × 5 × ⎟ ⎝ 2⎠ or But

RC = 15.8 kN () RA + RC = 2 + (6 – 5) = 32 kN

Fig. 5.8

or

RA = 32 – RC = 32 – 15.8 = 16.2 kN ()

Bending moment at the desired section XX The bending moment at the section is given as M = RA – 3.5 – 2 – (3.5 – 2) – 6 – 3.5 –

3.5 = 16.95 kN¹m 2

220

Strength of Materials

Calculation of moment of inertia Component

Rectangle (1)

Circle (2) Total

Distance ‘y’ of the centroid of component from PQ (mm)

Area ‘a’ of the component (mm2)

400 = 200 2

400 × 250 = 100000 π × 1502 = 17671.46(–) 4

∑a

(400 – 100) = 300

= 82328.54



ay (mm3) 20000000

5301437.6 (–)

∑ ay =14698562

The distance of the centroid of the whole section or the neutral axis from PQ is given as y =

∑ ay ∑a

14698562 = 178.5 mm 82328.54 yt = 178.5 mm =

Hence,

yc = (400 – 178.5) mm = 221.5 mm Now

INA = Irectangle – Ihole ⎡⎧ 1 3 2⎫ = ⎢ ⎨ × 250 × 400 + 250 × 400 × (200 − 178.5) ⎬ ⎭ ⎣ ⎩12 π ⎧π 4 2 2 ⎫⎤ – ⎨ × 150 + × 150 × (221.5 − 100) ⎬⎥ – 10–12 m4 4 ⎩ 64 ⎭⎦ = 1.093 – 10–3 m4

Bending stress at XX Using bending equation, we have σb M = y I or

Vb = =

M .y I c

(since yc > yt )

16.95 × 103 1.093 × 10

= 3.43 MPa

−3

– 221.5 – 10–3 N/m2 = 3432343.1 N/m2 Ans.

Stresses in Beams

221

Example 5.6 Compare the weights of two equally strong beams of circular sections made of same material, one 2 of outside being of solid section and the other of hollow section with inside diameter being 3 diameter. Solution:

Let diameter of the solid cicular beam = D

Inside diameter of the hollow circular beam

= Di

Ouside diameter of the hollow circular beam

= D0

2 D 3 0 The section modulus of the solid beam is given as Given,

Di =

1 π D4 π = D3 = y 64 ⎛ D ⎞ 32 ⎜ ⎟ ⎝2⎠ The section modulus of the hollow beam is given as SS =

SH

4 D0 π π D03 ⎡ ⎛ Di ⎞ ⎤ 4 4 ⎢1 − ⎜ = = (D0 – Di )/ ⎟ ⎥ 2 64 32 ⎢ ⎝ Do ⎠ ⎥ ⎦ ⎣

π D03 = 32

⎡ ⎛ 2 ⎞4 ⎤ 65π D03 ⎢1 − ⎜ ⎟ ⎥ = 32 × 81 ⎢⎣ ⎝ 3 ⎠ ⎥⎦

Since the beams are of equal strength, hence their section modului are equal. SS = SH 65π D03 π D3 = 32 32 × 81 ⎛ D ⎞ ⎜ ⎟ ⎝ D0 ⎠ or

3

D D0

=

65 81

= 0.9292

Being same material for the two beams, their bending stresses are equal. In other words, ratio of their weights is equal to the ratio of their cross-sectional areas. If

WS = Weight of the solid beam WH = Weight of the hollow beam

Then

WS WH

=

π 2 D 4

π (D02 – Di2) 4

222

Strength of Materials

=

D2 ( Do2 − Di2 )

D2

= Do2

Using ratios D and Di , we get D0 D0 WS WH

⎡ ⎛ D ⎞2 ⎤ ⎢1 − ⎜ i ⎟ ⎥ ⎢ ⎝ Do ⎠ ⎥ ⎣ ⎦

(0.9292) 2 = 1.55 = ⎡ ⎛ 2 ⎞2 ⎤ ⎢1 − ⎜ ⎟ ⎥ ⎢⎣ ⎝ 3 ⎠ ⎥⎦

WS = 1.55 WH

Ans.

Example 5.7 A wire of diameter d is wound round a cylinder of diameter D. Determine the bending stress produced on WKHFURVVVHFWLRQRIWKHZLUH+HQFHRURWKHUZLVH¿QGWKHPLQLPXPUDGLXVWRZKLFKDPPGLDPHWHU circular and of high tensile steel can be bent without undergoing permanent deformation. Take yield stress = 1700 N/mm2 and E = 200 GPa. What is the magnitude of bending moment necessary for this? Solution: Using bending equation, we have σb E = y R

d D⎞ ⎛ ⎜⎝ y = and R = ⎟⎠ 2 2

σb E = ( D / 2) (d / 2) Vb = E × or

D =

d D

Ans.

E×d σb

200 × 10 × 10 × 10 9

=

1700 × 106

−3

m

⎛ σb = 1700 N/mm 2 ⎞ ⎜ ⎟ ⎝ d = 10 mm ⎠

= 1.17647 m = 1176.47 mm Now

R =

D 1176.47 = 588.23 mm = 2 2

Ans.

Again using bending equation , we have M = =

2σ b π 4 σb × d ×I = ( d / 2) d 64 πd 3 π × (10 × 10−3 )3 × 1700 × 106 × Vb = = 166.9 N.m 32 32

Ans.

Stresses in Beams

223

Example 5.8 The shear force diagram for a rectangular cross-section beam AD is shown in Fig. 5.9. Width of the beam is 100 mm and depth is 250 mm. Determine the maximum bending stress in the beam. Solution: Given, Width of the beam, b = 100 mm Depth of the beam, d = 250 mm Fig. 5.9 shows that shear force at D is zero and it gradually increases to 5 kN (15 kN – 10 kN) at C indicating that there is a udl of intensity 5 kN/m between C and D. At C, the shear force increases further by 5 kN. There is no load on BC, hence the SFD between B and C remains constant. At B, the shear force further increses by 5 kN so that between A and B, the shear force remains constant at 15 kN. The loaded beam is shown in Fig. 5.10.

Fig. 5.9 Shear Force Diagram (SFD).

Fig. 5.10 Loaded Beam.

The bending moments at different points are given as MD = 0 MC = −5 × 1 ×

1 2

= – 2.5 kN.m ⎡ ⎛ 1 ⎞⎤ MB = ⎢ −5 × 1 − 5 × 1 × ⎜1 + ⎟ ⎥ ⎝ 2 ⎠⎦ ⎣ = – 12.5 kN.m ⎡ 1 ⎞⎤ ⎛ MA = ⎢ −5 × 1 − 5 × (1 + 1) − 5 × 1⎜1 + 1 + ⎟ ⎥ 2 ⎠⎦ ⎝ ⎣ = – 27.5 kN.m = Mmax The moment of inertia of the cross-section of the beam is found as 1 3 1 bd = × (100 × 10−3 ) × (250 × 10−3 )3 m 4 I = 12 12 = 1.302 × 10–4 m4

224

Strength of Materials

The maximum bending stress is given by ıb =

 

=

M d u I 2 −27.5 × 103 1.302 × 10−4

(y = d/2) ×

250 × 10−3 1 × 6 MPa 2 10

= – 26.4 MPa = 26.4 MPa (Compressive)

Ans.

Example 5.9 A composite beam is made by placing two steel plates, 10 mm thick and 200 mm deep, one each on both sides of a wooden section 100 mm wide and 200 mm deep. Determine moment E of resistance of the section of beam. Given, s = 20. The stress in the wood should not exceed Ew 7.5 N/mm2. Solution: The composite beam is shown in Fig. 5.11.

Fig. 5.11

Given, Thickness of the steel plate, t = 10 mm Depth of the steel plate and the wooden section, d = 200 mm Width of the wooden section, b = 100 mm Modular ratio, P =

E σs = s = 20 Ew σw

Stress in wood,

Vw = 7.5 N/mm2

Stress in the steel plate,

Vs = mVw = 20 × 7.5 = 150 N/mm2

Stresses in Beams

225

The moment of resistance of the steel plate is given as M s = V s. I s y

(using equation (5.8))

⎛ td 3 ⎞ 1 = Vs ⎜ 2 . ⎟ . ⎝ 12 ⎠ (d / 2) =

σ s td 2 3

=

150 × 10 × (200)2 3

= 20000 kN¹mm The moment of resistance of the wooden section is given as Mw = Vw = Vw =

Iw y

(using equation (5.9))

bd 3 . 1 12 (d / 2)

σ w bd 2 7.5 × 100 × (200)2 = 6 6

= 5000 kN¹mm Hence, the moment of resistance of the composite beam is given as M = Ms + Mw = (20000 + 5000) kN.mm = 25000 kN¹mm

Ans.

Example 5.10 $ÀLWFKHGEHDPFRQVLVWVRIWZRWLPEHUMRLVWVHDFKPPZLGH– 250 mm deep, is reinforced by a steel SODWHPPZLGHDQGPPGHHSSODFHGV\PPHWULFDOO\LQWKHJURRYHPDGHLQWKHFHQWUHDQG¿[HG ¿UPO\ )LJ )LQGWKHPRPHQWRIUHVLVWDQFHRIEHDP7KHPD[LPXPEHQGLQJVWUHVVLQWLPEHULVQRW to exceed 8.5 N/mm2. Take Es = 2 – 105 N/mm2 and Ew = 104 N/mm2. Solution: For timber joist Width, Depth,

b = 80 mm d = 250 mm

226

Strength of Materials

Fig. 5.12

Maximum bending stress, Vw = 8.5 N/mm2 For steel plate Width,

b = 25 mm

Depth,

d = 180 mm

Es = 20 Ew The stress in the timber at a distance 90 mm from neutral axis (NA) is given as and

P =

8.5 – 90 = 6.12 N/mm2 125 Strain in the timber at 90 mm from NA is V„w =

σw′ 6.12 = 6.12 – 10–4 = Ew 104 At 90 mm from neutral axis, strains in timber as well as in steel are same. °w =

Maximum stress in steel is given as Vs = Es – °s = 2 – 105 – 6.12 – 10–4 = 122.4 N/mm2 The moment of resistance of timber is given as I Mw = Vw . w y = 8.5 –

(°s = °w) (using equation (5.9))

1 ⎛1 1 3 3⎞ ⎜ × 160 × 250 − × 25 × 180 ⎟ 125 ⎝ 12 12 ⎠

= 13340467 N¹mm

Stresses in Beams

227

The moment of resistance of steel is given as Ms = Vs .

Is y

= 122.4 –

(using equation (5.8)) 1 ⎛1 3⎞ ⎜ × 25 × 180 ⎟ = 16524000 N¹mm 90 ⎝ 12 ⎠

 7KHWRWDOPRPHQWRIUHVLVWDQFHRIWKHÀLWFKHGEHDPLVJLYHQDV M = Mw + Ms = (13340467 + 16524000) N¹mm = 29864467 N¹mm

Ans.

5.7 SHEAR STRESSES IN BEAMS In the derivation of bending equation, the beam is assumed to be subjected to pure bending and no shear force is considered. As a result, only longitudinal stresses, called bending stresses are produced in such cases. Now if beam is loaded in the transverse direction, as is the case most common in practice, shear stresses and bending moments both are produced. Shear stresses are due to shear force acting along depth of the transverse cross-section of beam. For equilibrium of the cross-section, complementary shear stresses of equal intensity are considered on longitudinal planes parallel to the axis of beam. Calculation of shear stresses are important although shear deformations are relatively small and bending stresses are dominating in majority of the cases. 5.8 SHEAR STRESS DISTRIBUTION (GENERAL CASE) Let us consider a small length AB = G[ of the beam (Fig. 5.13). M = Bending moment at left side, that is, at AAc M + GM = Bending moment at right side, that is, at BBc

Fig. 5.13

228

Strength of Materials

Consider an elementary area Ga at a height y from the neutral axis. From bending equation, we have VAA„ =

M y I

(bending stress at AA„)

( M + δM ) y I The force acting on face CC„ of the elemental area at A is VBB„ =

and

F1 =

(bending stress at BB„)

M y Ga I

The force acting on face DD„ of the elemental area at B is F2 = where

( M + δM ) y Ga I

Ga = b. Gy

Net unbalancing force (longitudinal) acting on the elementary area between A and B = F 2 – F1 ( M + δM ) M δM y Ga – y Ga = y Ga I I I Total unbalanced force acting on the area between A and B =

δM y Ga y1 I

y2

=



This force is balanced by shear force (longitudinal) acting on the area between A and B and is equal to W.b.G[ where

W = Shear stress at the section

Equating two forces, we have δM y Ga y1 I

y2

W . b . G[ . =

or



dM W = I . bdx 1 = V Ib

where

y2



y1

1 ⎛ dM ⎞ y δa = ⎟ ⎜ Ib ⎝ dx ⎠

y2



y1

y δa =

y2



y δa

y1

VQ V Ay = Ib Ib

V = Vertical shear force at the section

... (5.19)

Stresses in Beams y2



y1

229

y δa = First moment of area of the cross-section above C cDc about the neutral axis =Q =Ay y = Distance of C.G. of area above the plane C cDc where the shear stress is W

For a given section of beam, I and b are constants. Hence, the variation of shear stress depends upon A y . The shear stress is maximum for the maximum value of A y and is minimum for the minimum value of A y . Accordingly, the shear stress is maximum at the neutral axis and zero at extreme faces of beam.

5.9 SHEAR STRESS DISTRIBUTION IN A RECTANGULAR CROSS-SECTION Refer Fig. 5.14. Let b = Width of the beam d = Depth of the beam Wmax = Maximum shear stress W = Shear stress at CD y = Distance of plane at CD from the neutral axis Ay = Distance of C.G. of ABDC I = Moment of inertia of cross-section about the neutral axis =

1 bd 3 12

Fig. 5.14

230

Strength of Materials

⎛d ⎞ A = Area ABDC = ⎜ − y ⎟ b ⎝2 ⎠ Ay = y +

1 ⎛d 1 ⎛d ⎞ ⎞ ⎜ − y⎟ = ⎜ + y⎟ 2⎝2 2⎝2 ⎠ ⎠

Hence, the moment of area ABDC about the neutral axis is given as ⎞ b ⎛ d2 ⎞ d 1 ⎛d − y2 ⎟ Ay = ⎛⎜ − y ⎞⎟ b – ⎜ + y ⎟ = ⎜ 2⎝ 4 2 ⎝2 ⎠ ⎠ ⎝2 ⎠ Using equation (5.19), we have ⎞ b ⎛ d2 V V − y2 ⎟ A y W = = – ⎜ 2⎝ 4 Ib Ib ⎠ =

⎞ V ⎛ d2 − y2 ⎟ ⎜ 2I ⎝ 4 ⎠

... (5.20)

This is the equation of a parabola, hence the shear stresss variation is parabolic in nature (Fig. 5.14 (b)). For maximum shear stress, put y = 0. Equation (5.20) reduces to an equation giving maximum value of shear stress as Wmax = =

V . d2 2I 4 V 3V d2 = – 1 2bd 4 2 × bd 3 12

For minimum shear stress, put y =

... (5.21)

d 2

Equation (5.20) reduces to an equation giving minimum value of shear stress as Wmin = 0 Hence, the shear stress is zero at extreme faces of the cross-section for which y = The average shear stress is given as V Wav = bd

... (5.22) d . 2 ... (5.23)

 &RPSDULQJWKLVHTXDWLRQZLWKHTXDWLRQ  ZH¿QG Wmax =

3 W 2 av

... (5.24)

Stresses in Beams

231

5.10 SHEAR STRESS DISTRIBUTION IN A CIRCULAR CROSS-SECTION Refer Fig. 5.15. Let r = Radius of the circular section I = Moment of inertia of the beam cross-section about the neutral axis =

π 4 π 4 r = d , d being diameter 64 4

Fig. 5.15

Consider an elementary strip ABDC of thickness dy at a distance y from the neutral axis. b = Width of the strip = 2 r 2 − y2

...(5.25)

Moment of the strip about the neutral axis = Area of the strip – y = bdy – y = (2 r 2 − y 2 × dy ) – y = 2y

r 2 − y 2 dy

The moment of the area above CD about the neutral axis is Ay, given by y=r

Ay =



2 y r 2 − y 2 dy

y=y

From equation (5.25) b =

r 2 − y2

...(5.26)

232

Strength of Materials

b2 = 4 (r2 – y2)

or

= 4r2 – 4y2 Differentiating both sides w.r.t. y, we have 2b

db = 0 – 4 – 2y dy = – 8y

or

ydy = −

bdb 4

... (5.27)

Making necessary changes in the limits of integration, we have when

y = y, b = b y = r, b = 0

Using equations (5.25), (5.27) and changed limits of integration, equation (5.26) changes to 0

Ay =

∫ b

0

b 2 db 1 ⎛ b3 ⎞ 1 3 b =– ⎜ ⎟ = 4 4 ⎝ 3 ⎠ b 12

... (5.28)

Using equation (5.19), we have V V V b2 1 Ay = – b3 = 12 I Ib Ib 12

W = =

V V 4 (r 2 − y 2 ) = (r2 – y2) 12 I 3I

... (5.29)

This is an equation of parabola suggesting that shear stress variation is parabolic in nature. The shear stress is maximum, when y = 0 (at the neutral axis). Wmax =

V – r2 3I

... (5.30)

The shear stress is minimum, when y = r. Wmin = 0

... (5.31)

On substituting ILQHTXDWLRQ  ZH¿QG Wmax =

=

V 2 π 4 –r 3× r 4 4 V 3 πr2

... (5.32)

Stresses in Beams

The average shear stress is given as V Wav = πr2

233

...(5.33)

Comparing equations (5.32) and (5.33), we have Wmax =

4 W 3 av

... (5.34)

5.11 SHEAR STRESS DISTRIBUTION IN AN I-SECTION Refer Fig. 5.16.

Fig. 5.16

Let

D = Depth of I-section d = Depth of the web B :LGWKRIWKHÀDQJH b = Width of the web I = Moment of inertia of I-section about the neutral axis (NA) V = Shear force at the section

6KHDUVWUHVVGLVWULEXWLRQLQWKHÀDQJH For the hatched portion at a distance y from the neutral axis, we have ⎡ ⎛D 1⎛D ⎞⎤ ⎡ ⎞⎤ A y = ⎢ B ⎜ − y ⎟⎥ ⎢ y + ⎜ − y ⎟⎥ 2⎝ 2 ⎠⎦ ⎣ ⎠⎦ ⎣ ⎝2 where y is the distance of centroid of the hatched portion from the neutral axis. or

Ay =

⎞ B ⎛ D2 − y2 ⎟ ⎜ 2⎝ 4 ⎠

... (5.35)

234

Strength of Materials

Using equation (5.19) and making necessary changes, we have 2 V ⎛ D2 2⎞ V B ⎛ D − y2 ⎞ ⎟ ⎜ = 2I ⎜ 4 − y ⎟ W = – ⎠ ⎝ IB ⎠ 2 ⎝ 4

... (5.36)

 7KLV LV DQ HTXDWLRQ RI SDUDEROD KHQFH VKHDU VWUHVV YDULDWLRQ LQ ÀDQJH LV SDUDEROLF LQ QDWXUH D $WXSSHUSDUWRIWKHÀDQJHZKHUHy = 2 Wu = 0

... (5.37)

d $WORZHUSDUWRIWKHÀDQJHZKHUHy = 2 Wl =

V (D2 – d 2) 8I

... (5.38)

Shear stress distribution in the web Consider Fig. 5.17. For any section XX in the web, we have ⎡ ⎛ D − d ⎞⎤ ⎡ d 1 ⎛ D − d ⎞⎤ ⎡ ⎛d 1 ⎛d ⎞⎤ ⎡ ⎞⎤ A y = ⎢B ⎜ ⎟ ⎥ + ⎢b ⎜ − y1 ⎟ ⎥ ⎢ y1 + ⎜ − y1 ⎟ ⎥ ⎟⎥ ⎢ + ⎜ 2⎝2 ⎠⎦ ⎣ ⎠⎦ ⎣ ⎝ 2 ⎠⎦ ⎣ 2 2 ⎝ 2 ⎠⎦ ⎣ ⎝2

Fig. 5.17

⎞ b ⎛ d2 B 2 2 − y12 ⎟ ( − ) D d + ⎜ = 2⎝ 4 8 ⎠ Using equation (5.19), we have W =

V Ay Ib

...(5.39)

Stresses in Beams

=

V [B (D2 – d2) + b (d2 – 4y12)] 8 Ib

235

....(5.40)

The shear stress increases as y1 decreases and is maximum when y1 = 0. The variation of the shear stress is parabolic in the web section. V [B (D2 – d2) + bd2 ... (5. 41) Wmax = 8 Ib $WWKHMXQFWLRQRIZHEDQGÀDQJH ZKHUHy1 = Wj =

d ), the shear stress is found to be 2

VB (D2 – d2) 8 Ib

...(5.42)

The shear stress distributions for some other sections are given in Table 5.1. Table 5.1 Section 1. Triangular

2. Thin circular

6HFWLRQ¿JXUH

6KHDUVWUHVVGLVWULEXWLRQ

236

Strength of Materials

Example 5.11 A T-section beam shown in Fig. 5.18 is subjected to a shear force of 10 kN. Draw the shear stress distribution diagram. Solution: Distance of the neutral axis (NA IURPWKHWRSVXUIDFHRIWKHÀDQJHLV 500 × 20 × 10 + 600 × 20 × (300 + 20) = 179 mm 500 × 20 + 600 × 20 Distance of the neutral axis from PQ is 620 – 179 mm = 441 mm

` Fig. 5.18

The moment of inertia of the section about the neutral axis (NA) is given as I = I1 + I2 ⎡ 500 × 203 ⎤ ⎡ 20 × (600)3 ⎤ + 500 × 20 × (179 − 10)2 ⎥ + ⎢ + 20 × 600 (441 − 300)2 ⎥ = ⎢ 12 ⎣ 12 ⎦ ⎣ ⎦ (using parallel-axes theorem) 8

4

= 8.84 – 10 mm  'LVWDQFHRIWKHFHQWUHRIJUDYLW\RIWKHÀDQJHIURPWKHQHXWUDOD[LVLV y f = 179 – 10 = 169 mm  8VLQJHTXDWLRQ  VKHDUVWUHVVDWORZHUSDUWRIWKHÀDQJHLVJLYHQDV Wl =

V 10 × 103 × (500 × 20) × 169 AI yI = = 0.038 N/mm2 Ib 8.84 × 108 × 500

Maximum shear stress  )RUÀDQJHDQGZHEDERYHWKHQHXWUDOD[LVZHKDYH Ay = 500 – 20 – (179 – 10) + (600 – 441) – 20 –

(600 − 441) = 1.69 – 106 mm3 2

Stresses in Beams

237

The maximum shear stress occurs at the neutral axis (NA), given by VAy 10 × 103 × 1.69 × 106 Wmax = = 0.955 N/mm2 = Ib 8.84 × 108 × 20 7KHVKHDUVWUHVVDWWKHMXQFWLRQRIWKHÀDQJHDQGZHELV 500 = 0.95 N/mm2 0.038 – 20  6KHDUVWUHVVDWWRSRIWKHÀDQJHDQGERWWRPRIWKHZHELV]HUR7KHVKHDUVWUHVVGLVWULEXWLRQLVVKRZQLQ Fig. 5.19 (b).

Fig. 5.19

Example 5.12 An I-section beam shown in Fig. 5.20 is subjected to a bending moment of 50 kN.m at its certain section. Find the shear force at the section, if the maximum stress is limited to 100 N/mm2. Solution: Refer Fig. 5.20. y = Distance of the C.GRIWKHÀDQJHIURPWKHQHXWUDOD[LV PP A $UHDRIHDFKÀDQJH

Fig. 5.20

238

Strength of Materials

= 160 – 30 = 4800 mm2 I =

1 1 – 160 – 3603 – 2 – – 72.5 – 3003 = 2.9583 – 108 mm4 12 12

7KHVKHDUVWUHVVDWWKHORZHUSDUWRIWKHÀDQJHLVJLYHQDV W =

V × 4800 × 165 VAy = 1.7848 – 10–4 V = 2.9583 × 108 Ib

7KHEHQGLQJVWUHVVDWWKHORZHUSDUWRIWKHÀDQJHLVJLYHQDV M V = y I =

50 × 103 × 103

2.9583 × 108 The maximum stress (principal) is given as 100 =

σ + 2

... (1)

(using bending equation)

× 150 = 25.35 N/mm2

σ 2 + 4τ 2 2

100 = 12.675 + Solving for W, we get

642.62 + 4τ 2 2

W = 86.4 N/mm2 Also

W = 1.7848 – 10–4 V

which gives

V =

86.4 1.7848 × 10−4

= 4.84 – 105 N

(using equation (1)) Ans.

Example 5.13 A beam of square cross-section is placed on one of its diagonal horizontally (Fig. 5.21(a)). It is subjected to a shear force V. Draw the shear stress distribution diagram.

Fig. 5.21

Stresses in Beams

Solution: Let

239

Refer Fig. 5.21.

a = Side of the square PQRS d = Diagonal of the square = 2a Consider a section XX at a distance y from P. The Moment of inertia of the square section about the neutral axis (NA) is given as 3

⎛d ⎞ d ×⎜ ⎟ d4 ⎝2⎠ = I =2– 48 12 1 Cross-sectional area above XX, A = – 2y – y = y2 2 The shear stress at XX is given as W =

V . Ay Ib

(using equation (5.19))

⎛ d 2y⎞ V . y2 . ⎜ − ⎟ ⎝2 3⎠ = d4 × 2y 48 6V ⎛ a 2 ⎞ y − y⎟ = 4 ⎜ ⎝ 3 ⎠ 2 a

2

(in terms of a)

... (1)

This is the equation of a parabola, suggesting that the variation of the shear stress is parabolic in nature. For W to be maximum, we have dτ =0 dy 3a

3d 8 4 2 Position of the maximum shear stress from the neutral axis (NA) is a 3d d – = 4 2 8 8

which gives

y =

=

Using equation (1), we have Wmax =

6V ⎛ a 3a 2 9a 2 ⎞ 9 V . − . = – ⎟ 4 ⎜ a ⎝ 2 4 2 3 32 ⎠ 8 a2

The average shear stress occurs at the neutral axis, given by V Wav = 2 a τ max 9 Now = = 1.125 τ av 8 The shear stress distribution is shown in Fig. 5.21 (b).

... (2)

240

Strength of Materials

Example 5.14 The cross-section (Fig. 5.22) of a steel beam is subjected to a shear force of 20 kN. Draw the shear stress distribution diagram. Solution: The moment of inertia of cross-section about the neutral axis (NA) is given as π 1 I = – 120 – 1603 – (90)4 = 3.77 – 107 mm4 64 12

Fig. 5.22

The moment of interia is calculated by assuming that given cross-section is made of a rectangular cross-section 120 mm – 160 mm minus two semi circles equivalent to a full circle of diameter (45 + 45) mm = 90 mm. Shear stresses at L and Y (where y = 0) are zero. Shear stress at Q and S Area above Q, A = (120 – 35) mm2 = 4200 mm2 Distance of the centroid of this area from the neutral axis (NA) is y = 45 + Now

W = =

35 = 62 .5 mm 2

VAy Ib 20 × 103 × 4200 × 62.5 3.77 × 107 × 120

= 1.16 N/mm2

This is the shear stress at Q. By symmetry, the shear stress at S = 1.16 N/mm 2, that is, MN = XZ = 1.16 N/mm2. The variation of the shear stress between L and N, and between Y and Z is parabolic in nature.

Stresses in Beams

241

Shear stress at the neutral axis A y = 120 – 80 –

Now

WNA = Wmax =

2 4 × 45 ⎞ 80 ⎛ π × 45 × – ⎜⎜ ⎟⎟ = 323250 mm3 2 3 π 2 ⎠ ⎝

VAy Ib

=

20 × 103 × 323250 3.77 × 107 × 30

= 5.71 N/mm2

The shear stress distribution is shown in Fig. 5.22 (b). Example 5.15 A TVHFWLRQEHDPPPGHHSDQGPPZLGHKDVÀDQJHDQGZHEWKLFNQHVVRIPP,WFDUULHVD udlRIWKHLQWHQVLW\N1PRYHUWKHRYHUKDQJSRUWLRQRIWKHEHDPDQGDSRLQWORDGRIN1 )LJ  'UDZWKHEHQGLQJVWUHVVDQGVKHDUVWUHVVGLVWULEXWLRQGLDJUDPIRUWKHEHDP

Fig. 5.23

Solution: The beam ABKDVDWRWDOVSDQRIPZLWKRYHUKDQJSRUWLRQBCRIOHQJWKP Support reactions at A and C Using 6MA ZHKDYH 2⎞ ⎛ RC – 6 – 5 – 2 – ⎜ 3 + 3 + ⎟ ±– 3 = 0 2⎠ ⎝ or Now or

RC N1 ) RA + RC – N1 RA N1 )

 )RUDQ\VHFWLRQEHWZHHQC and D, at a distance xIURPA, the bending moment is Mx = RA x± x – 3) = 18.33x±x +120 = – 21.67x + 120 Bending moment at D, where x = 3 m, is MD = – 21.67 – N1¹m = Mmax Bending moment at C, where x = 6 m, is MC = – 21.67 – 6 + 120 = –N1¹m

242

Strength of Materials

Calculations for bending stresses Refer Fig. 5.24(a). If the neutral axis is located at a distance of y IURPWKHWRSVXUIDFHRIWKHÀDQJHWKHQ 30 ⎞ 270 ⎞ ⎛ ⎛ ⎜150 × 30 × ⎟ + 30 × 270 × ⎜ 30 + ⎟ 2 2 ⎠ ⎠ ⎝ y = ⎝ = 111.42 mm (150 × 30) + (30 × 270) The moment of inertia of whole section about the neutral axis (NA) is given as 30 ⎤ ⎡ 150 × 303 + 150 – 30 – ⎢111.42 − ⎥ I = 2⎦ 12 ⎣

2

270 ⎞ ⎛ 30 × 2703 ⎟ + + 30 – 270 – ⎜188.58 − 2 ⎠ ⎝ 12 = 1.146 – 108 mm4  7KHEHQGLQJVWUHVVDWWKHWRSVXUIDFHRIWKHÀDQJHLVFRPSUHVVLYHJLYHQE\ VC = =

M max ×y I 54.99 1.146 × 108

– 111.42 = 0.0534 MPa

Fig 5.24

The bending stress at the lower part of the beam is tensile, given by Vt =

54.99 1.146 × 108

– (300 – y )

2

Stresses in Beams

=

54.99 1.146 × 108

243

– (300 – 111.42) = 9.04 – 10–5 kN/mm2

= 0.0904 MPa Calculations for shear stresses Since RC > RA, hence maximum shear force is equal to RC = 31.67 kN, which occurs at C (= V). (Ay DWWKHORZHUSDUWRIÀDQJH  – 30 – (111.42 – 15) = 433890 mm3 ( y  'LVWDQFHRIWKHFHQWURLGRIWKHÀDQJHIURPWKHQHXWUDOD[LV b = 150 mm  +HQFHWKHVKHDUVWUHVVDWWKHORZHUSDUWRIWKHÀDQJHLVJLYHQDV Wl = =

VAy Ib 31.67 × 433890

kN/mm2

1.146 × 108 × 150

= 8 – 10–4 kN/mm2 = 0.8MPa  7KHVKHDUVWUHVVLQWKHZHE DWWKHMXQFWLRQRIWKHÀDQJHDQGZHE LVJLYHQDV Wj =

31.67 × 433890 1.146 × 108 × 30

kN/mm2

(b = 30 mm)

= 4 – 10–3 kN/mm2 = 4 MPa The shear stress at the neutral axis is maximum, given by WNA = Wmax =

31.67 × 533328.25 1.146 × 108 × 30

= 4.91 – 10–3 kN/mm2 = 4.91 MPa ⎛ 111.42 − 30 ⎞ [ b = 30 mm and Ay = 150 – 30 – (111.42 – 15) + 30 – (111.42 – 30) – ⎜ ⎟ 2 ⎝ ⎠ = 533328.25 mm2] The shear stress distribution diagram is shown in Fig. 5.24 (c) and the bending stress distribution diagram in Fig 5.24 (b).

244

Strength of Materials

SHORT ANSWER QUESTIONS        

       

:KDWLVPHDQWE\SXUHEHQGLQJ" :KDWLVÀH[XUDOULJLGLW\":KDWGRHVLWLQGLFDWH" :KDWLVWKHLPSRUWDQFHRIQHXWUDOD[LV" :KDWLVVHFWLRQPRGXOXV":KDWLVLWVLPSRUWDQFH" :KDWLVPHDQWE\DEHDPRIXQLIRUPVWUHQJWK" :KDWLVDFRPSRVLWHEHDP":KHUHLVLWXVHIXO" :KDWDUHWKHFRPPRQO\XVHGFURVVVHFWLRQVRIDEHDP" :KDWDUHWKHDGYDQWDJHVRIDQ,VHFWLRQEHDP"

MULTIPLE CHOICE QUESTIONS 1. The section modulus is the ratio of (a) Moment of inertia and bending stress (b) Moment of inertia and the distance from neutral axis (c) Moment of inertia and modulus of elasticity (d) Modulus of elasticity and moment of inertia. 2. The section modulus for a circular beam of diameter d is given as (a)

π 4 d 64

(b)

π 3 d 64

(c)

π 3 d 32

(d)

π 2 d . 64

3. The section modulus for a rectangular beam of width b and depth d is given as (a)

1 bd 3 12

(b)

1 3 bd 6

(c)

1 bd 2 12

(d)

1 2 bd . 6

4. The moment of inertia of a circular section of diameter d about the neutral axis is given as (a)

π 4 d 64

(b)

π 4 d 32

(c)

π 3 d 64

(d)

π 3 d . 32

5. The bending stress is proportional to (a) moment of inertia (c) its distance from neutral axis

(b) modulus of elasticity (d) radius of curvature.

6. The bending stress is maximum at (a) neutral axis (c) bottom layer of beam

(b) top layer of beam (d) top and bottom layer of beam.

Stresses in Beams

7. The bending stress is zero at (a) neutral axis (c) bottom layer of beam

(b) top layer of beam (d) top and bottom layer of beam.

8. The strength of a beam depends upon (a) modulus of elasticity (c) section modulus

(b) bending moment (d) radius of curvature.

245

9. The bending equation is valid for a beam subjected to (a) only bending moment and no shear force (b) combined bending and shear force (c) shear force only (d) normal force only. 10. A composite beam is made of (a) more than one material (c) more than one cross-section

(b) three materials (d) plastic material.

11. The shear stress at neutral axis of a rectangular section is (a) average shear stress (b) maximum shear stress (c) minimum shear stress (d) none of these. 12. For a triangular section, the shear stress is maximum at a height of H from the base 4 H (c) from the base 3

(a)

2H from the base 3 H (d) from the apex. 2 (b)

13. The shear stress variation across a rectangular section is (a) hyperbolic

(b) parabolic

(c) circular

(d) elliptical.

14. The relationship between maximum shear stress and average shear stress for a rectangular section is given as (a) Wav = 1.33Wmax

(b) Wmax = 1.33Wav

(c) Wav = 1.5Wmax

(d) Wmax = 1.5Wav.

15. The ratio of maximum shear stress and average shear stress for a triangular section is (a) 0.66

(b) 1.33

(c) 1.5

16. In the equation of shear stress, y LVGH¿QHGDVWKHGLVWDQFHEHWZHHQWKH (a) two extreme faces of section (b) neutral axis and the C.G. of the area above the neutral axis (c) neutral axis and the extreme face of the section (d) none of these.

(d) 0.75.

246

Strength of Materials

17. Compared to the bending deformation, the shear deformation is (a) large

(b) small

(c) very large

(d) zero.

18. The shear stress varies directly proportional to (a) moment of inertia about the neutral axis (b) width of the beam (c) distance between the neutral axis and the centroid of the area above the neutral axis (d) normal stress. 19. The shear stress is maximum, where the (a) bending stress is minimum (b) bending stress is maximum (c) bending stress is zero (d) bending stress is negative. 20. The shear stress variation for an I-section is represented as

21. For a beam of square cross-section and whose one of the diagonals is placed horizontally, the shear stress variation is shown as

Stresses in Beams

22. For a beam of cross-section shown in Fig. 5.25, the shear stress variation is shown as

Fig. 5.25

ANSWERS 1. (b)

2. (c)

3. (d)

4. (a)

5. (c)

6. (d)

7. (a)

8. (c)

9. (a)

10. (a)

11. (b)

12. (d)

13. (b)

14. (d)

15. (c)

16. (b)

17. (b)

18. (c)

19. (c)

20. (d)

21. (b)

22. (d).

247

248

Strength of Materials

EXERCISES 1. An I-section steel beam (Fig. 5.25) is 5 m long and simply supported at ends. Find the permissible uniform load to be placed on the beam. The maximum stress in tension does not exceed 25 N/mm2.

Fig. 5.25

(Ans. 32.93 N/mm). 2. A simply supported beam of length 4 m carries a point load of 40 kN at the centre and is supported at ends. Find the cross-section of beam assuming depth to be twice the width. The maximum bending stress in beam is not to exceed 200 N/mm2. (Ans. b = 67 mm, d = 134 mm). 3. Compare the weights of two equally strong beams of circular sections made of the same material, one is solid and the other hollow with inside diameter being 2/5 of outside diameter. (Ans.

Ws = 1.14 WH).

4. A rectangular section is to be cut from a circular log of wood of diameter D. Find the dimensions of the strongest section in bending. ⎛ ⎜⎜ Ans. ⎝

b=

D 3

,d =D

2 ⎞. ⎟ 3 ⎟⎠

5. A cast iron pipe having inside diameter 300 mm and outside diameter 350 mm is used to carry waWHUDQGLVVXSSRUWHGDWWZRHQGVPDSDUW7KHVSHFL¿FZHLJKWRIFDVWLURQDQGZDWHUDUH– 104 N/m3 and 1.0 – 104 N/m3 respectively. Determine the maximum bending stress induced in the pipe, if it is running full of water. (Ans. 1.447 N/mm2). 6. A timber beam of rectangular section section 60 mm wide – 150 mm deep is reinforced by two steel plates having 12 mm thickness on both sides. The composite beam is a simply supported beam of length 2.5 m and carries a load of 10 kN at its centre. Find the depth of steel plates. The maximum stress in the timber section is not to exceed 12 N/mm2. The modular ratio, P is 14. (Ans. 92.53 mm). 7. $ÀLWFKHGEHDPFRQVLVWVRIDZRRGHQVHFWLRQPPZLGH– 200 mm deep. It is to be reinforced by providing a steel plate at its bottom. The steel plate is 150 mm wide and 10 mm thick. The maximum stress in wood is not to exceed 8 – 106 Pa and the modular ratio, P is 15. Determine the moment of resistance of beam. (Ans. 1.33 – 108 N ¹mm).

Stresses in Beams

249

8. $ÀLWFKHGEHDPFRQVLVWVRIWZRWLPEHUMRLVWVHDFKPPZLGH– 250 mm deep, is reinforced E\DVWHHOSODWHPPZLGHDQGPPGHHSSODFHGV\PPHWULFDOO\LQWKHFHQWUHDQG¿[HGWR the timber joists. Find the maximum uniformly distributed load to be placed on beam of simply supported nature of length 6 m. The maximum stress in timber is not to exceed 10 N/mm2. Given, the modular ratio, P is 20. (Ans. 14 kN/m). 9. A TVHFWLRQEHDPV\PPHWULFDODERXWWKHYHUWLFDOD[LVKDVWKHÀDQJHPP– 10 mm and web 120 mm –PP:KDWLVWKHSHUFHQWDJHRIVKHDUIRUFHVKDUHGE\WKHZHE" Ans. 95.1%). 10. An I-section beam shown in Fig. 5.26 is subjected to a shear force of 50 kN. Find the magnitude and position of the maximum shear stress.

Fig. 5.26

(Ans. 12.2 N/mm2 at the neutral axis, which is located at 105 mm from the base). 11. A T-section beam shown in Fig. 5.27 is subjected to a shear force of 40 kN. Find the magnitude and position of maximum shear stress. Also draw the shear stress distribution diagram.

Fig. 5.27

(Ans. 12.81 N/mm2 at neutral axis, which is located at 76.3 mm IURPWKHWRSVXUIDFHRIWKHÀDQJH 

Stresses in Beams

250

12. Find the ratio of the maximum shear stress and mean shear stress for a hexagonal cross-section, whose one of the diagonals is placed horizontally. (Ans. 1.2). 13. An ornamental beam in the form of a cross-bar (Fig. 5.28 (a)) has a span of 4 m and carries a udl of 20 kN/m inclusive of its weight. Determine the maximum shear stress in the cross-section and draw the shear stress diagram.

Fig. 5.28

(Ans. 14.72 MPa, for second part of the problem refer Fig. 5.27 (b)). 14. A wooden beam of rectangular section 200 mm by 300 mm is used as a simply supported beam carrying a udl of w ton/m. What is the maximum value of w, if the maximum shear stress developed in the beam section is limited to 5 N/mm2DQGWKHVSDQOHQJWKLVP" Ans. 6.66 N/m). ‰‰‰

6 'HÀHFWLRQVRI%HDPV William Herrick Macaulay, born on 16 November 1853, was a British mathematician. He developed the Macaulay’s method, which is used LQVWUXFWXUDODQDO\VLVWRGHWHUPLQHWKHGHÀHFWLRQRIEHDPV7KHPHWKRG LV EDVHG RQ WKH XVH RI D VLQJXODULW\ IXQFWLRQ ZKLFK KHOSV WR IRUPXODWH D VLQJOH HTXDWLRQ RI PRPHQWV IRU DOO WKH ORDGV DFWLQJ RQ WKH EHDP IRU ZKLFKWKHERXQGDU\FRQGLWLRQVDSSO\XQLIRUPO\ZLWKRXWFRQVLGHULQJDQ\ GLVFRQWLQXLW\DQGLVSDUWLFXODUO\XVHIXOIRUFDVHVRIGLVFRQWLQXRXVDQGRU GLVFUHWHORDGLQJ

William Herrick Macaulay (1853-1936)

/($51,1*2%-(&7,9(6 $IWHUUHDGLQJWKLVFKDSWHU\RXZLOOEHDEOHWRDQVZHUVRPHRIWKHIROORZLQJTXHVWLRQV  ‡ +RZDUHVORSHDQGGHÀHFWLRQGH¿QHG"  ‡ :KDWLVWKHOLPLWDWLRQRIGRXEOHLQWHJUDWLRQPHWKRG"  ‡ :KHQLV0DFDXOD\¶VPHWKRGXVHIXO"  ‡ :KDWLV0RKU¶VWKHRUHP"  ‡ :KDWLVWKHSULQFLSOHRIVXSHUSRVLWLRQ"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_6

251

252

Strength of Materials

 ,1752'8&7,21 $ORDGHGEHDPLVLQLWLDOO\VWUDLJKWEXWUHPDLQVQRORQJHUVWUDLJKWLQGXHFRXUVHRIWLPHEHFDXVHRI GHÀHFWLRQRILWVD[LV'HÀHFWLRQLVWKHVKLIWLQJRIDSRLQWIURPLWVLQLWLDOSRVLWLRQLQWKHWUDQVYHUVH GLUHFWLRQZKLOHVORSHDWDSRLQWLVWKHDQJOHPDGHE\WKHWDQJHQWGUDZQRQWKHGHÀHFWHGEHDPDWWKDW SRLQWZLWKWKHRULJLQDOD[LVRIWKHEHDP 6ORSHDQGGHÀHFWLRQDUHUHTXLUHGWREHNQRZQIRUWKHSURSHUGHVLJQRIDEHDPLQRUGHUWRDYRLGLWV IDLOXUH)ROORZLQJPHWKRGVDUHXVHGWR¿QGWKHVORSHDQGGHÀHFWLRQRIDEHDP z

'RXEOHLQWHJUDWLRQPHWKRG

z

Macaulay’s method

z

Moment-area method

z

&RQMXJDWHEHDPPHWKRG

z

0HWKRGRIVXSHUSRVLWLRQ

z

6WUDLQHQHUJ\PHWKRG

 ',))(5(17,$/(48$7,212))/(;85( /HWXVFRQVLGHUDORDGHGVLPSO\VXSSRUWHGEHDPDVVKRZQLQ)LJ a ,WGHIOHFWVLQDPDQQHUDV VKRZQLQ)LJ b 7KHD[LVRIWKHEHDPEHQGVLQDFXUYHNQRZQDVHODVWLFFXUYH Consider two points A and B on the elastic curve.

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Let

T $QJOHPDGHE\WKHWDQJHQWGUDZQDWSRLQWAZLWKWKHD[LVRIWKHEHDP = Slope at point A dT $QJOHEHWZHHQWZRQRUPDOVGUDZQRQWKHHODVWLFFXUYHDWSRLQWVA and B R 5DGLXVRIFXUYDWXUHRIWKHEHQWEHDP

'HÀHFWLRQVRI%HDPV  253

dx = Small segment between A and B AB = RdT 1 dT dT =  = R AB dx

or

IRUVPDOOGHÀHFWLRQAB  dx)

... (6.1)

dT  LV 6  ORSH DW SRLQW B LV T – dT  ,W PHDQV WKDW VORSH GHFUHDVHV ZLWK LQFUHDVH RI dx DQG KHQFH dx QHJDWLYH (TXDWLRQ  PRGL¿HVWR 1 dT =– R dx

... (6.2)

dy = – tan T  – T dx

Now

IRUVPDOOYDOXHRIT)

'LIIHUHQWLDWLQJZUW xZHKDYH d2y dT 1 =– =  2 dx dx R

XVLQJHTXDWLRQ   

8  VLQJEHQGLQJHTXDWLRQZHKDYH M 1 = EI R

... (6.4)

6XEVWLWXWLQJHTXDWLRQ  LQHTXDWLRQ  ZHJHW M d2y = EI dx 2

or EI

d2y =M dx 2

... (6.5)

7KHSURGXFWEILVNQRZQDVWKHÀH[XUDOULJLGLW\DQGWKHHTXDWLRQ  LVFDOOHGGLIIHUHQWLDOHTXDWLRQ RIÀH[XUHorGLIIHUHQWLDOHTXDWLRQRIHODVWLFFXUYH 7KHVORSHLVJLYHQE\LQWHJUDWLQJHTXDWLRQ  DV 1 dy = ∫ Mdx EI dx 7KHGHÀHFWLRQLVREWDLQHGE\LQWHJUDWLQJHTXDWLRQ  DV

\ =



6.3

1 EI

∫ (∫ Mdx) dx

SIGN CONVENTIONS



zThe

x and \D[HVDUHSRVLWLYHWRWKHULJKWDQGXSZDUGUHVSHFWLYHO\



z7KHGHÀHFWLRQ\LVQHJDWLYHGRZQZDUG

... (6.6)

... (6.7)

254



Strength of Materials

dy is positive when measured counterclockwise with respect to the positive x D[LV dx   DQGLVQHJDWLYHZKHQPHDVXUHGFORFNZLVH z

7KHVORSH

z

7KHEHQGLQJPRPHQWM LVSRVLWLYHZKHQLWSURGXFHVFRPSUHVVLRQLQWKHXSSHUSDUWRIWKHEHDP DQGLVQHJDWLYHZKHQLWSURGXFHVWHQVLRQLQWKHXSSHUSDUW

,Q)LJM and TB are positive, whereas \and TADUHQHJDWLYH

)LJ

 '28%/(,17(*5$7,210(7+2' 7KLVPHWKRGLVEDVHGRQ¿QGLQJVORSHDQGGHÀHFWLRQXVLQJHTXDWLRQV  DQG    &DQWLOHYHU%HDPFDUU\LQJD3RLQW/RDGDWLWV)UHH(QG 5HIHU)LJ

)LJ

7KHEHQGLQJPRPHQWDWWKHVHOHFWHGVHFWLRQXX at a distance xIURPWKH¿[HGHQGALVJLYHQDV Mx = – W O – x 

 

8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH EI

dy =M dx  = – W O – x 

 

'HÀHFWLRQVRI%HDPV  255

2QLQWHJUDWLRQZHJHW EI

dy Wx 2 = – Wlx + + C1  dx 2

 

)XUWKHULQWHJUDWLRQJLYHV (,\ = – Wl

x 2 Wx3 + + C1 x + C2  2 6

  

where C1 and CDUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQVDUH At A, where x 

dy  DQG\  dx

ZKLFKJLYHVC1 DQGC  (TXDWLRQV  DQG  RQVXEVWLWXWLQJC1 and C are reduced to dy Wx 2 ⎤ 1 ⎡ − Wlx + = ⎢ ⎥ dx EI ⎣ 2 ⎦ and

\ =

1 EI

⎡ Wlx 2 Wx3 ⎤ + ⎢− ⎥ 2 6 ⎦ ⎣

 

 

 TXDWLRQ  LVDVORSHHTXDWLRQDQGHTXDWLRQ  DGHÀHFWLRQHTXDWLRQ ( 7KHVORSHDWSRLQWB ZKHUHx = O LVJLYHQE\ 1 ⎛ dy ⎞ TB = ⎜ ⎟ = EI ⎝ dx ⎠ x = l = −

⎛ Wl 2 ⎞ − Wl . l + ⎜⎝ 2 ⎟⎠

Wl    El

 

7KHGHÀHFWLRQDWB ZKHUHx = O LVIRXQGWREHPD[LPXPJLYHQE\ 

\B = \PD[ =

1 EI

⎛ Wl 3 l 2 Wl 3 ⎞ − − Wl ⋅ + =  ⎜ 3El 2 6 ⎟⎠ ⎝

7KHQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ 7KHGHÀHFWLRQDWA ZKHUHx  LV]HUR

 

256

Strength of Materials

 &DQWLOHYHU%HDPFDUU\LQJudlRYHULWV(QWLUH6SDQ 5HIHU)LJ

)LJ

7KHEHQGLQJPRPHQWDWWKHVHFWLRQLV Mx = − w (l − x)

(l − x) 2

w (l − x) 2  2 8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH =–

EI

dy dx 

 

=M =–

2QLQWHJUDWLRQZHKDYH

w (l  x)2 2

⎤ dy 1 ⎡ w (l − x)3 + C1 ⎥  =– ⎢− dx EI ⎣ 2 3 ⎦

 

 

)XUWKHULQWHJUDWLRQJLYHV \ =–



1 EI

⎡ w (l − x)4 ⎤ + C1 x + C2 ⎥  ⎢ 4 ⎣6 ⎦

where C1 and CDUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQVDUH At A, where x  and 

dy   dx \ 

(TXDWLRQ  RQVXEVWLWXWLQJERXQGDU\FRQGLWLRQJLYHV wl 3 C1 = 6

 

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(TXDWLRQ  RQVXEVWLWXWLQJERXQGDU\FRQGLWLRQJLYHV wl 4 24 +HQFHHTXDWLRQV  DQG  RQVXEVWLWXWLQJC1 and C are reduced to C = −

dy wl 3 ⎤ 1 ⎡ w 3 =– ⎥ ⎢ − (l − x) + dx EI ⎣ 6 6 ⎦ \ =–

and

1 EI

⎡ w (l − x)4 wl 3 wl 4 ⎤ + x − ⎢ ⎥ 24 6 24 ⎦ ⎣

 

 

(TXDWLRQV  DQG  DUHXVHGWR¿QGWKHVORSHDQGGHÀHFWLRQUHVSHFWLYHO\ )RUVORSHDWB, put x = OLQHTXDWLRQ   wl 3 ⎛ dy ⎞ – = T =  B ⎜ ⎟ 6 EI ⎝ dx ⎠ x = l

  

)RUGHÀHFWLRQDWB, put x = OLQHTXDWLRQ   wl 4 ⎤ 1 ⎡ wl 3 l− ⎥ ⎢ EI ⎣ 6 24 ⎦ wl 4 =–  8EI

\B = –



 

7KHQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ

 &DQWLOHYHU%HDPVXEMHFWHGWRD3XUH&RXSOHDWLWV)UHH(QG 5HIHU)LJ

)LJ

%HQGLQJPRPHQWDWWKHVHFWLRQLV Mx = – MR

 

8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH EI

dy dx 

=M = – MR

 

258

Strength of Materials

2QLQWHJUDWLRQZHKDYH EI )XUWKHULQWHJUDWLRQJLYHV

dy = – M R x + C 1 dx

x + C 1 x + C   where C1 and CDUHWKHFRQVWDQWVRILQWHJUDWLRQ (,\ = – MR

   

7KHERXQGDU\FRQGLWLRQVDUH dy  DQG\  dx

At A, where x   ZKLFKJLYHV





C1 DQGC 

+HQFHHTXDWLRQV  DQG  RQVXEVWLWXWLQJC1 and Care reduced to dy 1 =– [M o x]  dx EI \ =–

1 EI

⎡ x2 ⎤ M ⎢ o ⎥ 2⎦ ⎣

 

 

(TXDWLRQV  DQG  DUHXVHGWR¿QGWKHVORSHDQGGHÀHFWLRQUHVSHFWLYHO\ )RUVORSHDWB, put x = OLQHTXDWLRQ  

M l ⎛ dy ⎞ = − o . TB = ⎜ ⎟ ⎝ dx ⎠ x = l EI

 

)RUGHÀHFWLRQDWB, put x = OLQHTXDWLRQ   \B = −

Mo l   EI

 

7KHQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ

 &DQWLOHYHU%HDPFDUU\LQJD3RLQW/RDGDQ\ZKHUHRQLWV6SDQ 5HIHU)LJ

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'HÀHFWLRQVRI%HDPV  259

Consider a section XX at a distance xIURPA in the portion AC. %HQGLQJPRPHQWDWWKHVHFWLRQLV Mx = – W a – x 

 

8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH EI

dy =M dx  = – W a – x 

 

,QWHJUDWLRQRIWKHDERYHHTXDWLRQUHVXOWVLQ EI

W (a − x)2 dy + C1  = 2 dx

 

)XUWKHULQWHJUDWLRQJLYHV 

(,\ –

W ( a − x )3 + C1 x + C2 2 3

 

where C1 and CDUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQVDUH dy At A, where x   DQG\  dx ZKLFKJLYHV

C1 = –

Wa  

Wa3 6 (TXDWLRQV  DQG  RQVXEVWLWXWLQJC1 and C are reduced to C =

1 ⎡W (a − x)2 Wa 2 ⎤ dy − = ⎢ ⎥ 2 2 ⎦ EI ⎣ dx 

1 ⎡ W (a − x)3 Wa 2 Wa3 ⎤ − x+ ⎢− ⎥ EI ⎣ 6 2 6 ⎦  TXDWLRQ  LVDVORSHHTXDWLRQDQGHTXDWLRQ  DGHÀHFWLRQHTXDWLRQ ( \ =

 

 

)RUVORSHDWC, put x = aLQHTXDWLRQ   Wa  ⎛ dy ⎞ TC = ⎜ ⎟    =–  EI ⎝ dx ⎠ x = a Since there is no load on the portion BC, hence C cBcUHPDLQVVWUDLJKWLQGLFDWLQJWKDWVORSHVDWWKH points B and CDUHHTXDO TB = TC = –

Wa    EI

 

260

Strength of Materials

)RUGHÀHFWLRQDWC, put x = aLQHTXDWLRQ   

\C

1 = EI

In ' B„C„P tan TB =

⎡ Wa3 Wa3 ⎤ Wa3 − + –  ⎢ ⎥ = 6 ⎦ 3EI ⎣ 2

 

PB′ PB ′ = (l − a) PC ′

tan TB  TB IRUVPDOOYDOXHRITB) PBc  O – a) TB = – O – a)

Wa    EI

XVLQJHTXDWLRQ 

 HQFHWKHGHÀHFWLRQDWBLVJLYHQDV + \B = BP + PB„ = \C + PB„

 or

\B = \PD[ = −

Wa3 Wa 2 − (l − a ) 3EI 2 EI

⎡Wa 3 Wa 2 ⎤ + (l − a ) = −⎢ ⎥ 2 EI ⎦ ⎣ 3EI 7KHQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ  &DQWLOHYHU%HDPFDUU\LQJ*UDGXDOO\9DU\LQJ/RDG 5HIHU)LJ

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Consider a section XX at a distance xIURPWKH¿[HGHQGA. &RPSDULQJ'V ABC and EBD. AC AB = ED EB or

ED =

AC . w EB = (l − x) AB l

 

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%HQGLQJPRPHQWDWWKHVHFWLRQLV Mx = – $UHDRIWKHWULDQJOHEBD) –

EB 3

1 w (l − x) = – (l − x) × (l − x ) × 2 l 3 =–

w (l − x)3  6l

 

 VLQJHTXDWLRQ  ZHKDYH 8 EI

dy dx 

=M=–

w (l − x)3  6l

 

2QLQWHJUDWLRQZHJHW EI

w (l − x) 4 dy + C1  =+ . 4 6l dx

)XUWKHULQWHJUDWLRQJLYHV (,\ = –

w . ( l − x )5 + C1 x + C 24l 5

   

where C1 and CDUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQVDUH At A, where x  ZKLFKJLYHV

dy  DQG\  dx wl 3 C1 = – 24 wl 4 C = + 120

(TXDWLRQV  DQG  RQVXEVWLWXWLQJC1 and C are reduced to dy wl 3 ⎤ 1 ⎡ w (l − x)4 − = ⎢+ ⎥ EI ⎣ 24l 24 ⎦ dx and 

\ =

w wl 3 wl 4 ⎤ 1 ⎡ (l − x)5 − x+ ⎢− ⎥ EI ⎣ 120l 24 120 ⎦

 

 

 RUVORSHDWB, put x = OLQHTXDWLRQ   ) ⎛ dy ⎞ TB = ⎜ ⎟ ⎝ dx ⎠ x = l = −

wl 3  24 EI

 

262

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)  RUGHÀHFWLRQDWBSXWx = lLQHTXDWLRQ   wl 4 − yB =  30 EI

 

7KHQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ

6.4.6 Simple Beam carrying a Central Point Load 5HIHU)LJ

Fig. 6.8

8VLQJ

6 M A

ZHKDYH RB – l = W –

or 1RZ

l 2

W (n) 2 RA + RB = W RB =

or

RA = W – RB =

W (n) 2

&RQVLGHUDVHFWLRQXXDWDGLVWDQFHx from A %HQGLQJPRPHQWDWWKHVHFWLRQLV Mx = +

W x 2

8VLQJHTXDWLRQ  ZHKDYH EI

W d2y x = M= 2 2 dx

2QLQWHJUDWLRQZHJHW

   

EI

Wx dy = 4 dx

2

+ C1

Wx3 )XUWKHULQWHJUDWLRQJLYHV EIy = + C1 x + C2 12 ZKHUHC1DQGC2DUHWKHFRQVWDQWVRILQWHJUDWLRQ

   

'HÀHFWLRQVRI%HDPV  263

7KHERXQGDU\FRQGLWLRQVDUH O dy  DWC IRU x = )  dx 

\ DWA IRU x 

Wl 2 and C  16 (TXDWLRQV  DQG  RQVXEVWLWXWLQJC1 and C are reduced to

ZKLFKJLYHV

C1 = −

1 ⎡ W 2 Wl 2 ⎤ dy = ⎢+ x − ⎥ EI ⎣ 4 16 ⎦ dx \ =

1 ⎡ W 3 Wl 2 ⎤ x⎥  ⎢+ x − EI ⎣ 12 16 ⎦

 

 

 TXDWLRQ  FDQEHXVHGWR¿QGWKHVORSHDWDQ\SRLQWEHWZHHQA and C. ( )RUVORSHDWA, put x LQHTXDWLRQ   Wl 2 ⎛ dy ⎞ TA = ⎜ ⎟  =– 16 EI ⎝ dx ⎠ x =  and

TB = – TA =

Wl 2  16 EI

(  TXDWLRQ  FDQEHXVHGWR¿QGWKHGHÀHFWLRQDWDQ\SRLQWEHWZHHQA and & O )RUGHÀHFWLRQDWC, put x = LQHTXDWLRQ    1 ⎡ Wl 3 Wl 3 ⎤ Wl 3 −   \C = \PD[ = ⎥ =– ⎢+ 32 ⎦ EI ⎣ 96 48EI 7KUQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ

 6LPSOH%HDPFDUU\LQJudlRYHULWV(QWLUH6SDQ 5HIHU)LJ

)LJ

 

 

 

264

Strength of Materials

Support reactions at A and B Using 6MA = 0, we have RB – l = w – l – RB =

or Now

l 2

wl (n) 2

RA + RB = wl

wl (n) 2 Consider a section XX at a distance x from A within A and C. RA =

or

Bending moment at the section is Mx = RA x – wx ×

x 2

wlx wx 2 − 2 2 8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH

... (6.58)

=

EI

d2y dx 2

=M =

wlx wx 2 – 2 2

... (6.59)

On integration, we get dy 1 ⎡ wl 2 w 3 ⎤ x − x + C1 ⎥ = ⎢ dx EI ⎣ 4 6 ⎦ y =

Further integration gives

⎤ 1 ⎡ wl x3 wx 4 − + C1 x + C2 ⎥ ⎢ EI ⎣ 12 24 ⎦

where C1 and C2 are the constants of integration. The boundary conditions are:

which gives

l ) 2

dy = 0 at C dx

(for

x=

y = 0 at A

(for

x = 0)

wl 3 C1 = – 24

and

C2 = 0

... (6.60)

... ( 6.61)

'HÀHFWLRQVRI%HDPV  265

Equations (6.60) and (6.61) on substituting C1 and C2 become 1 ⎡ wl 2 w 3 wl 3 ⎤ dy = ⎢ x − x − ⎥ EI ⎣ 4 6 24 ⎦ dx

... (6.62)

dy ⎡ wlx3 wx 4 wl 3 − − y = ⎢ 24 24 dx ⎣ 12

... (6.63)

⎤ x⎥ ⎦

 TXDWLRQ  FDQEHXVHGWR¿QGWKHVORSHDWDQ\SRLQWEHWZHHQA and C. ( )  RUVORSHDWASXWx = 0 in equation (6.62). wl 3 ⎛ dy ⎞ =– TA = ⎜ ⎟ 24 EI ⎝ dx ⎠ x = 0 TB = – TA =



... (6.64)

wl 3 24 EI

... (6.65)

(TXDWLRQ  FDQEHXVHGWR¿QGWKHGHÀHFWLRQDWDQ\SRLQWEHWZHHQA and C. l in equation (6.63). )RUGHÀHFWLRQDWCSXWx = 2 1 ⎡ wl 4 wl 4 wl 4 ⎤ 5 . wl 4 − − yC = ymax = ⎢ ⎥ =– EI ⎣ 96 384 48 ⎦ 384 EI

... (6.66)

7KUQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ 6ORSHDQGGHÀHFWLRQXQGHUGLIIHUHQWORDGLQJFRQGLWLRQVDUHJLYHQLQ7DEOH Table 6.1 S.N. 1.

Loading condition

Slope (T) TA = 0 TB = TC =–

wa3 6 EI

'HÀHFWLRQ(y) yA = 0 4

yC = –

wa 8 EI

Remarks 7KH HODVWLF FXUYH LV VWUDLJKWEHWZHHQB and C EHFDXVH RI QR ORDG RQWKDWSRUWLRQ

⎡ wa 4

yB = − ⎢

⎣ 8 EI

⎤ wa3 (l − a)⎥ + 6 EI ⎦ 2.

yA = 0

TA = 0 TB = –

⎡ w ⎢⎣ 6 EI –

(l3 – a3)]

⎡ w

yB = − ⎢ – ⎣ 24 EI

7KHHODVWLFFXUYHLV FRQWLQXRXVEHWZHHQA and B.

(3l 4 – 4a3l + a4)]

Contd...

266 S.N.

Strength of Materials Loading condition

'HÀHFWLRQ(y)

Slope (T)

3. TA = –

\A \B = 

7 wl 3 360 El

4

\PD[ = –

wl 3 TB = 45EI

 TA = –

\C = –

Wb (l 2 − b 2 ) 6 EIl

TB = – TB

Wb (l 2 − b 2 ) = 6 EIl

13wl 2000 EI

147 wl 4 2340 EI

\A = \B  2 2

\C = –

Wa b

™ 7KHPD[LPXP GHÀHFWLRQRFFXUV at a distance x  OIURP A. ™ C LV FHQWUH RI WKH beam. ™ 7KHPD[LPXP GHÀHFWLRQRFFXUV at

3EIl x=

\PD[= 2



Remarks

l 2  b2 3

2 3/ 2

Wb ( l − b ) 9 3 EIl

5. TA

\A = \B 

5wl 3 =– 192 EI

\C = \PD[

TB = – TA =

5wl

=–

3

wl 4 120 EI

192 EI

™ 7RWDOORDGRQEHDP wl =  ™ RA = RB =

wl 

TC  6.

TA = –

\A = \B 

w u 6 EI

⎡ a 2l a 4 3⎤ − − (l − a ) ⎥ ⎢ 4l ⎣ 2 ⎦ TB = – TA = +

+

1 EI

⎡ wa 2 ⎢ ⎣ 4

wl (l − a ) 6

wa 2l wa 4 ⎤ − ⎥ 12 24l ⎦

\C = \PD[ =–

0.0056 wl 4 EI

7KHPD[LPXP GHÀHFWLRQRFFXUVDWC, where x  O – a).

'HÀHFWLRQVRI%HDPV  267

([DPSOH $PFDQWLOHYHUEHDPKDYLQJUHFWDQJXODUFURVVVHFWLRQLVORDGHGZLWKDSRLQWORDGRIN1DWLWVIUHH HQG)LQGWKHFURVVVHFWLRQRIWKHEHDP7KHPD[LPXPEHQGLQJVWUHVVLVQRWWRH[FHHG1PP  and WKHPD[LPXPGHÀHFWLRQLVUHVWULFWHGWRPP7DNHE –1PP. Solution:

Given,

  /HQJWKRIWKHEHDP Load on the beam,   %HQGLQJVWUHVV

O = 3 m W N1 –3 N Vb 1PP

  0D[LPXPGHÀHFWLRQ \PD[ = 8 mm /HWZLGWKRIWKHEHDP

 b

'HSWKRIWKHEHDP

 d

7KHPRPHQWRILQHUWLDRIWKHFURVVVHFWLRQRIEHDPDERXWWKHQHXWUDOD[LVLVJLYHQDV 1 bd 3 12 7KHPD[LPXPQHJDWLYHEHQGLQJPRPHQWLVJLYHQDV I =

M =W–O 

  – 3 –3 = 3 – N¹mm

7KHPD[LPXPEHQGLQJVWUHVVLVJLYHQDV VPD[ =

Muy  I 3 u 107 u

5 = or

XVLQJEHQGLQJHTXDWLRQ d 2

1 bd 3 12

bd  = 3.6 –

 

7KHPD[LPXPQHJDWLYHGHÀHFWLRQRFFXUVDWWKHIUHHHQGRIWKHFDQWLOHYHUJLYHQE\ 

\PD[ = 8 =

Wl 3  3EI 104 u (3 u 1000)3 3 u 2 u 104 u

or

XVLQJ(TXDWLRQ 

bd 3 –

bd 3 12  

6ROYLQJHTXDWLRQV  DQG  ZHJHW d PP and

b PP

Ans.

268

Strength of Materials

([DPSOH $PVLPSOHEHDPKDYLQJFURVVVHFWLRQPP–PPFDUULHVDSRLQWORDGRIN1DWDGLVWDQFH RIPIURPWKHOHIWHQG)LQGWKHVORSHDWWKHWZRHQGVGHÀHFWLRQXQGHUWKHORDGDQGWKHPD[LPXP GHÀHFWLRQ7DNHE –1PP. Solution: 5HIHU)LJ Given, /HQJWKRIWKHEHDP

O P PP

:LGWKRIWKHEHDP

 PP

'HSWKRIWKHEHDP

 PP

)LJ

Load on the beam,

W N1 – N

'LVWDQFHRIORDGIURPA,

a P

'LVWDQFHRIORDGIURPB,

b = 1.5 m

7KHPRPHQWRILQHUWLDRIWKHFURVVVHFWLRQRIEHDPDERXWWKHQHXWUDOD[LVLVJLYHQDV I = 

1 ––3 12

 – mm 7KHVORSHDWWKHHQGALVJLYHQDV TA = – =–

Wb (l 2  b 2 )  6 EIl

VHH7DEOH

2 × 104 × 1.5 × 103 × (20002 − 15002 ) 6 × 2 × 104 × 1.5625 × 109 × 2000

= – –± radian = –––3GHJUHH

Ans.

7KHVORSHDWWKHHQGBLVJLYHQDV TB =

Wb (l 2  b 2 )  6 EIl

VHH7DEOH

'HÀHFWLRQVRI%HDPV  269

=–

2 × 104 × 1.5 × 103 × (20002 − 15002 ) 6 × 2 × 104 × 1.5625 × 109 × 2000

= ––± radian = –––3GHJUHH

Ans.

7KHGHÀHFWLRQXQGHUWKHORDGLV \C = –



=–

Wa 2b2  3EIl

VHH7DEOH

2 u 104 u (0.5 u 1000)2 u (1.5 u 1000)2 3 u 2 u 104 u 1.5625 u 109 u 2000

= –PP

Ans.

7KHPD[LPXPGHÀHFWLRQLV \PD[ = –



=–

Wb (l 2  b 2 )3 / 2 9 3EIl



2 × 104 × 1.5 × 1000 (20002 − 15002 )3 / 2 9 3 × 2 × 104 × 1.5625 × 109 × 2000

VHH7DEOH = –PP

Ans.

7KHQHJDWLYHVLJQVKRZVWKHGRZQZDUGGHÀHFWLRQ ([DPSOH $FDQWLOHYHUEHDPRIXQLIRUPVHFWLRQDQGRIOHQJWKOFDUULHVWZRFRQFHQWUDWHGORDGVWDWWKHIUHHHQG DQGW at a distance ‘a¶IURPWKHIUHHHQG6WDUWLQJIURPWKH¿UVWSULQFLSOHVGHWHUPLQHWKHGHÀHFWLRQ XQGHUWKHORDGW. ,IWKHFDQWLOHYHULVPDGHIURPDVWHHOWXEHRIFLUFXODUVHFWLRQRIPPH[WHUQDOGLDPHWHUDQGPP thickness and O = 1.5 m, a  PGHWHUPLQHWKHYDOXHRIWVRWKDWWKHPD[LPXPEHQGLQJVWUHVVLV 03D&DOFXODWHWKHPD[LPXPGHÀHFWLRQIRUWKHORDGLQJ7DNHE *3D Solution: 5HIHU)LJ Consider a section XX in AC at a distance xIURPWKHIUHHHQG%

)LJ

7KHEHQGLQJPRPHQWDWWKHVHFWLRQLV Mx = – [WxW x – a)]

270

Strength of Materials

8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH dy EI = Mx dx  = – [Wx W x – a)] 2QLQWHJUDWLRQZHJHW ( x − a )2 Wx  dy – – + C1 W EI = 2  dx Wx  – W x – a )+ C1 =–  )XUWKHULQWHJUDWLRQJLYHV (,\ = –

 

( x  a )3 W . x3 + C 1x + C  –W. 3 2 3

3 W ( x − a )3 = – Wx – + C1x + C 6 3

 

where C1 and CDUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQVDUH At A, where [ O 

dy   dx

and

\ 

)URPHTXDWLRQ  ZHKDYH  –



Wl  – W O – a)+ C1 

Wl  – :O  – Wa:OD + C1  or C1 = 3 :O  + Wa –:OD 2 )URPHTXDWLRQ  ZHKDYH =–

 



– =–

W 3– W 3 3   3  O 3  O– a) + 2 :O + Wa O±:O a + C 6 W 3 W 3 O – O + 3 . W . O a – 3 . W OD + W a3 + 3 :O 3 + WaO±:O  a + C 6 3 3 3 2 3

⎛ 1 1 3⎞ = :O 3 ⎜ − − + ⎟ – :O  a + W a3 + C 3 ⎝ 6 3 2⎠ ⎛ −1 − 2 + 9 ⎞  W 3 = :O 3 ⎜ ⎟⎠ – :O a + 3 a + C ⎝ 6

'HÀHFWLRQVRI%HDPV  271

= :O 3 – :O  a + W a3 + C 3 or

C = – :O 3 – W a3 + :O  a 3

(TXDWLRQ  RQVXEVWLWXWLQJC1 and C becomes (,\ = –

⎛3 2 ⎞ W 3– W 2 x  x – a)3 + ⎜ Wl + Wa − 2Wla⎟ x – :O 3 – W a3 + :O D 6 3 3 ⎝2 ⎠

 

'HÀHFWLRQDWC IRUZKLFKx = a LVJLYHQE\ \ = =

W 3 1 ⎛ W 3 3 2 3 2 3 2 ⎞ ⎜⎝ − a + Wl a + Wa − 2Wla − Wl − a + Wl a⎟⎠ EI 6 2 3 ⎤ 1 ⎡ 3⎛ 1 1⎞ ⎛3 ⎞ Wa ⎜ − + 1 − ⎟ + Wl 2 a ⎜ + 1⎟ − 2Wla 2 − Wl 3 ⎥ ⎢ ⎝ ⎠ ⎝ ⎠ EI ⎣ 6 3 2 ⎦

⎞ W ⎛ a3 5 2 + l a − 2la 2 − l 3 ⎟ ⎜ EI ⎝ 2 2 ⎠ 7KHIROORZLQJSDUDPHWHUVRIWKHWXEHDUHJLYHQ =

 ([WHUQDOGLDPHWHU d0

PP

 7KLFNQHVV

W = 6 mm

 /HQJWK

O = 1.5 m

 /HQJWK

a P

 %HQGLQJVWUHVV

Ans.

Vb 03D 1PP

 ±–@ EI 1133.33 = –P –PP 105 7KHQHJDWLYHVLJQVDVVRFLDWHGZLWK\C and \BVKRZWKHGRZQZDUGGHÀHFWLRQVDWWKHVHSRLQWV =–

Ans.

([DPSOH )RUDVLPSOHEHDPORDGHGDVVKRZQLQ)LJ¿QGVORSHDQGGHÀHFWLRQDWLPSRUWDQWSRLQWV 7DNHÀH[XUDOULJLGLW\DVEI. Solution: 5HIHU)LJ Support reactions at A and B 8VLQJ6MA ZHKDYH RB – ––  –   or

RB N1 )

)LJ

Now or

RA + RB – N1 RA ±RB ± N1 )

8VLQJ 0DFDXOD\¶V PHWKRG WKH EHQGLQJ PRPHQW DW DQ\ VHFWLRQ DW D GLVWDQFH x IURP A LV IRXQG E\ H[WHQGLQJ XGO upto the point B DQG WKHQ DSSO\LQJ QHJDWLYH XGO RI WKH VDPH LQWHQVLW\ IRU WKH FRPSHQVDWLQJSDUWWRPDNHDEDODQFH )LJ 

)LJ

'HÀHFWLRQVRI%HDPV  279

The bending moment equation at the section is Mx = RA x – 25 (x – 2) – 3 (x – 4) = 20x | – 25 (x – 2) | −

(x − 8) (x − 4 ) + 3 (x – 8) + 82 (x – 10)0 2 2

3 3 2 2 (x − 4 ) + (x − 8) + 82(x – 10)0 2 2

... (1)

8VLQJGLIIHUHQWLDOHTXDWLRQRIÀH[XUHZHKDYH 3 3 2 2 d2y = M = 20x | – 25(x –2) | – ( x − 4) + ( x − 8) + 82(x –10)0 2 2 2 dx ,QWHJUDWLQJHTXDWLRQ  ZHJHW

EI

... (2)

dy ( x − 2) − 3 . ( x − 4) + 3 . ( x − 8) + 82 x − 10 x2 EI = 20 . + C1 − 25 . ( ) dx 2 2 2 3 2 3 2

3

3

⎤ 1 ⎡ 2 25 1 1 dy 10 x + C1 − ( x − 2) 2 − ( x − 4)3 + ( x − 8)3 + 82 ( x − 10)⎥ = ⎢ dx EI ⎣ 2 2 2 ⎦

or

...(3)

,QWHJUDWLQJHTXDWLRQ  ZHKDYH 3 4 4 2 ⎡10 x 3 82 ( x − 10) ⎤ 25 . ( x − 2 ) 1 . ( x − 4) 1 . ( x − 8) ⎢ ⎥ + C1 x + C2 − − + + 2 2 3 2 4 2 4 ⎢⎣ 3 ⎥⎦ ⎡10 3 25 1 1 4 2⎤ 3 4 = ⎢ x + C1 x + C2 − ( x − 2) − ( x − 4) + ( x − 8) + 41 ( x − 10) ⎥ 6 8 8 ⎣3 ⎦

1 y = EI

... (4)

The boundary conditions are: At AZKHUHx y = 0. (TXDWLRQ  RQXVLQJWKLVERXQGDU\FRQGLWLRQJLYHVC2 1HJDWLYHWHUPVZLWKLQVPDOOEUDFNHWV are omitted. At BZKHUHx Py = 0. 8VLQJWKLVERXQGDU\FRQGLWLRQZH¿QGC1ZLWKWKHKHOSRIHTXDWLRQ  DV 0 =

1 ⎡10 25 1 1 ⎤ × 123 + C1 × 12 + 0 − × 103 − × 84 + × 44 + 41 × 22 ⎥ ⎢ EI ⎣ 3 6 8 8 ⎦

6ROYLQJIRUC1ZHJHW C1 = – 106.44 Substituting C1LQHTXDWLRQ  ZHKDYH 1 dy = EI dx

⎡ 25 1 1 ⎤ 2 3 3 2 ⎢10 x − 106.44 − 2 ( x − 2) − 2 ( x − 4) + 2 ( x − 8) + 82 ( x − 10)⎥ ⎣ ⎦

7KLVLVWKHHTXDWLRQRIVORSHJLYLQJVORSHDWDQ\VHFWLRQRIWKHEHDP

... (5)

280

Strength of Materials

6XEVWLWXWLQJC1 and CLQHTXDWLRQ  ZHKDYH  

\ =

1 ⎡10 3 25 1 1 ⎤ x − 106.44 x − ( x − 2)3 − ( x − 4)4 + ( x − 8)4 + 41 ( x − 10)2 ⎥  ⎢ EI ⎣ 3 6 8 8 ⎦

 

7  KLVLVWKHHTXDWLRQRIGHÀHFWLRQJLYLQJGHÀHFWLRQDWDQ\VHFWLRQRIWKHEHDP Determination of slopes at various points using equation (5) Slope at A, where x LV TA = –

106.44  EI

QHJOHFWLQJQHJDWLYHWHUPV

Slope at C, where x PLV 1 ⎛ dy ⎞ >–±@ = TC = ⎜ ⎟ ⎝ dx ⎠ x =  EI

QHJOHFWLQJQHJDWLYHWHUPV

66.44 EI Slope at D, where x PLV =–

1 ⎛ dy ⎞ = TD = ⎜ ⎟ ⎝ dx ⎠ x =  EI

25 3.56 ⎡ 2 2⎤ ⎢⎣10 × 4 − 106.44 − 2 × 2 ⎥⎦ = EI 

QHJOHFWLQJQHJDWLYHWHUPV

Slope at E, where x = 6 m, is 1 ⎡ 25 1 ⎤ 49.56 ⎛ dy ⎞ × 42 − × 22 ⎥ = 10 × 62 − 106.44 − = TE = ⎜ ⎟ ⎢ ⎝ dx ⎠ x = 6 EI EI ⎣ 2 2 ⎦ Slope at F, where x = 8 m, is 1 ⎛ dy ⎞ = TF = ⎜ ⎟ EI ⎝ dx ⎠ x = 8

25 1 51.56 ⎡ 2 2 3⎤ ⎢⎣10 × 8 − 106.44 − 2 × 6 − 2 × 4 ⎥⎦ = EI

Slope at G, where x PLV 1 ⎛ dy ⎞ = TG = ⎜ ⎟ ⎝ dx ⎠ x = 10 EI

25 1 1 10.44 ⎡ 2 2 3 3⎤ ⎢⎣10 × 10 − 106.44 − 2 × 8 − 2 × 6 + 2 × 2 ⎥⎦ = – EI

Slope at B, where x PLV 1 ⎛ dy ⎞ = TB = ⎜ ⎟ ⎝ dx ⎠ x = 12 EI

25 1 1 ⎡ ⎤ 23.56 2 2 3 3 ⎢⎣10 × 12 − 106.44 − 2 × 10 − 2 × 8 + 2 × 4 + 82 × 2⎥⎦ = EI

'HÀHFWLRQVRI%HDPV  281

'HWHUPLQDWLRQRIGHÀHFWLRQVDWYDULRXVSRLQWVXVLQJHTXDWLRQ   'HÀHFWLRQDWA, where x LV yC =  IRUFKHFN 'HÀHFWLRQDWC, where x PLV yC =

79.77 1 ⎡10 ⎤ × 23 − 106.44 × 2⎥ = – ⎢ EI EI ⎣ 3 ⎦

QHJOHFWLQJQHJDWLYHWHUPV 

'HÀHFWLRQDWD, where x PLV yD =

1 ⎡10 25 ⎤ 245.76 × 43 − 106.44 × 4 − × 23 ⎥ = –  EI ⎢⎣ 3 6 EI ⎦

QHJOHFWLQJQHJDWLYHWHUPV

 'HÀHFWLRQDWE, where x PLV yE =

1 ⎡10 25 1 ⎤ 187.39 × 63 − 106.44 × 6 − × 43 − × 24 ⎥ = – ⎢ EI ⎣ 3 6 8 EI ⎦

 'HÀHFWLRQDWF, where x PLV yF =

1 EI

25 1 76.85 ⎡10 3 3 4⎤ ⎢⎣ 3 × 8 − 106.44 × 8 − 6 × 6 − 8 × 4 ⎥⎦ = – EI

 HÀHFWLRQDWG, where x PLV ' yG =

1 ⎡10 25 1 1 ⎤ 24.4 × 103 − 106.44 × 10 − × 83 − × 64 + × 24 ⎥ = – EI ⎢⎣ 3 6 8 8 EI ⎦

'HÀHFWLRQDWB, where x PLV yB  IRUFKHFN

6.6

MOMENT-AREA METHOD

7KH PRPHQWDUHD PHWKRG ZDV LQWURGXFHG E\ D IDPRXV *HUPDQ FLYLO HQJLQHHU 2WWR 0RKU ±  ,W LV D VHPLJUDSKLFDO PHWKRG LQ ZKLFK WKH LQWHJUDWLRQ RI WKH EHQGLQJ PRPHQW LV FDUULHGRXWLQGLUHFWO\XVLQJWKHJHRPHWULFSURSHUWLHVRIWKHDUHDXQGHUWKHEHQGLQJPRPHQWGLDJUDP ,WLVEDVHGRQWKHDSSOLFDWLRQRIWZRWKHRUHPVNQRZQDV0RKU¶VWKHRUHPVDQGLVHTXDOO\YDOLGIRU GHWHUPLQDWHRULQGHWHUPLQDWHEHDPV7KHPHWKRGWDNHVFDUHRIVORSHDQGGHÀHFWLRQE\SXUHEHQGLQJ DQGQRVKHDUFRQVLGHUDWLRQLVPDGH7KHWZRWKHRUHPVDUHVWDWHGEHORZ )LUVW 0RPHQW$UHD 7KHRUHP 0RKU¶V )LUVW 7KHRUHP  7KH FKDQJH LQ VORSH EHWZHHQ DQ\ WZR SRLQWVRQDEHDPLVHTXDOWRWKHQHWDUHDRIWKHEHQGLQJPRPHQWGLDJUDPEHWZHHQWKHVHWZRSRLQWV GLYLGHGE\WKHÀH[XUDOULJLGLW\ EI RIWKHEHDP

282

Strength of Materials

)LJ

Consider a beam ABRIFHUWDLQOHQJWKODQGORDGHGDUELWUDULO\ )LJ 7DNHDQHOHPHQW CDRI OHQJWKdx at a distance xIURPHQGBRIWKHEHDPZKHUHEHQGLQJPRPHQWLVM. Let

dT &KDQJHRIVORSHIRUWKHHOHPHQWCD between C and D T $QJOHEHWZHHQWZRWDQJHQWVGUDZQDWA and B

7KHGLIIHUHQWLDOHTXDWLRQRIÀH[XUHLV EI or

dy =M dx  dy M = dx  EI

,QWHUJUDWLQJWKHDERYHHTXDWLRQEHWZHHQWKHOLPLWVA and B, we have A

A M ⎛ dy ⎞ ⎜⎝ ⎟⎠ = − ∫ EI dx dx B B A

Mdx ⎛ dy ⎞ ⎛ dy ⎞ ⎜⎝ ⎟⎠ − ⎜⎝ ⎟⎠ = − ∫ dx A dx B EI B

A

or

Mdx EI B

TA – TB = − ∫

'HÀHFWLRQVRI%HDPV  283 A

or

TB – TA =

Mdx EI B



l

or

T =

∫ 

Mdx  EI

 

⎛ dy ⎞ TA = ⎜ ⎟ = Slope at A ⎝ dx ⎠ A

where

⎛ dy ⎞ TB = ⎜ ⎟ = Slope at B ⎝ dx ⎠ B (  TXDWLRQ  LVWKHPDWKHPDWLFDOH[SUHVVLRQRIWKH0RKU¶V¿UVWWKHRUHP ,IWKHÀH[XUDOULJLGLW\ EI RIWKHEHDPLVFRQVWDQWWKHQHTXDWLRQ  FDQEHZULWWHQDV 1 TB – TA = EI

A



M dx 

 

B

+HQFHWKHFKDQJHLQVORSHEHWZHHQA and B =

Net area of the bending moment diagram between A and B EI

Second Moment-Area Theorem (Mohr’s Second Theorem): 7KHYHUWLFDOGLVSODFHPHQWRIDSRLQW RQDVWUDLJKWEHDPPHDVXUHGIURPWKHWDQJHQWGUDZQDWDQRWKHUSRLQWRQWKHEHDPLVHTXDOWRWKH PRPHQWRIDUHDRIEHQGLQJPRPHQWGLDJUDPEHWZHHQWKHVHWZRSRLQWVDERXWWKHSRLQWZKHUHWKLV GHÀHFWLRQRFFXUVGLYLGHGE\WKHÀH[XUDOULJLGLW\RIEHDP 7KHWKHRUHPKHOSVWR¿QGGHÀHFWLRQRIDSRLQWZLWKUHVSHFWWRDQRWKHUSRLQWRI¿[HGSRVLWLRQIRU which x DQGVORSHTLVDOVR]HURDQGWKLVSRLQWLVFRQVLGHUHGDVUHIHUHQFHSRLQW7KDWLVZK\LW LVYHU\PXFKXVHIXOIRUFDQWLOHYHUEHDPV¿[HGEHDPVDQGV\PPHWULFDOO\ORDGHGVLPSO\VXSSRUWHG EHDPVIRUZKLFKSRVLWLRQRI]HURVORSHLVZHOOGH¿QHG )URPHTXDWLRQ  ZHKDYH dT $QJOHEHWZHHQWKHWDQJHQWVGUDZQDWC and D = )URP)LJZHJHW

Mdx  EI

 

LM = xdT = x.

Mdx  EI

)RUWRWDOOHQJWKRIWKHEHDPZHKDYH A

∫ LM = B

A



B

Mxdx EI

RQVXEVWLWXWLQJdT)

284

Strength of Materials l

or

BB„ =

∫ 

Hence,

\ =

+HQFHGHÀHFWLRQRIB w.r.t. A =

Mxdx  EI

1 EI

 

l



x . Mdx 

 

0

Moment of the area of the bending moment diagram between A and B EI

(TXDWLRQ  LVWKHPDWKHPDWLFDOH[SUHVVLRQRIWKH0RKU¶VVHFRQGWKHRUHP

 &DQWLOHYHU%HDPFDUU\LQJD3RLQW/RDGDWLWV)UHH(QG 5HIHU)LJ$SRLQWORDGWLVDFWLQJDWIUHHHQGBRIWKHFDQWLOHYHUEHDPRIOHQJWKO. 7  R¿QGWKHVORSH 7KHVORSHDW¿[HGHQGRIWKHFDQWLOHYHUEHDPLV]HUR TA 

)LJ

'HÀHFWLRQVRI%HDPV  285



6ORSHDWIUHHHQGB is TB &KDQJHLQVORSHRIWKHFDQWLOHYHUEHDPEHWZHHQA and B IURPB to A)

 8VLQJ0RKU¶V¿UVWWKHRUHPZHKDYH TB = =

Area of the bending moment diagram between A and B EI 1 ⎡1 Wl  ⎤ × Wl × l  0D[LPXPVORSH = ⎥⎦  EI EI ⎢⎣ 2

 

7R¿QGWKHGHÀHFWLRQ 8VLQJ0RKU¶VVHFRQGWKHRUHPWKHGHÀHFWLRQDWBLVJLYHQDV  

\B = =

Moment of the area of BMD between A and B about B EI Wl 3 1 ⎛ Wl 2 2l ⎞ × = 3EI EI ⎜⎝ 2 3 ⎟⎠

0D[LPXPGHÀHFWLRQ

 

 &DQWLOHYHU%HDPFDUU\LQJDudlRYHULWV(QWLUH6SDQ 5HIHU)LJ$FDQWLOHYHUEHDPRIOHQJWKOLVFDUU\LQJDXGORILQWHQVLW\wPOHQJWKRYHUWKHHQWLUH span.

)LJ

286

Strength of Materials

7R¿QGWKHVORSH Slope at A TA  and

TB = Slope at B =

Area of BMD between A and B  EI

=

1 ⎛ 1 wl 2 ⎞ wl 2 × × l = ⎟ 6 EI  0D[LPXPVORSH EI ⎜⎝ 3 2 ⎠

XVLQJ0RKU¶V¿UVWWKHRUHP  

7R¿QGWKHGHÀHFWLRQ 8VLQJ0RKU¶VVHFRQGWKHRUHPWKHGHÀHFWLRQDWBLVJLYHQDV 1 \B = \PD[ = 0RPHQWRIWKHDUHDRIBMD between A and B about B) EI 1 = EI

⎛ wl 3 3l ⎞ wl 4 ×  = ⎜ 6 8EI 4 ⎟⎠ ⎝

 

 6LPSOH%HDPFDUU\LQJD&HQWUDO3RLQW/RDG 5HIHU)LJ$SRLQWORDGWLVDFWLQJDWFHQWUHCRIWKHEHDPRIOHQJWKO.

)LJ

7R¿QGWKHVORSH 6LQFH WKH ORDG LV SODFHG DW WKH FHQWUH RI WKH EHDP KHQFH WKH GHÀHFWLRQ RQ ERWK VLGHV RI LW LV V\PPHWULFDODQGWKHVORSHDWWKDWSRLQWLV]HUR Area of BMD between A and C Slope at A is TA = EI 1 Wl l × × Wl 2 2 4 2    = = EI 16 EI

'HÀHFWLRQVRI%HDPV  287

Slope at B is Area of BMD between B and C EI

TB =

Wl 2 =  16 EI

 

7R¿QGWKHGHÀHFWLRQ 7KHGHÀHFWLRQVDWA and BDUH]HUR 'HÀHFWLRQDWC is 

\C =

Moment of the area of BMD between A and C about A EI

=

Wl 2 ⎛ 2 l ⎞ ×⎜ × ⎟ 16 EI ⎝ 3 2 ⎠

=

Wl 3  48EI

 

⎛2 O⎞ 6LQFHWKHFHQWURLGRIWKHDUHDRIBMD between A and COLHVDWDGLVWDQFHRI ⎜ × ⎟ IURPA„. ⎝ 3 2⎠  6LPSOH%HDPFDUU\LQJDudlRYHULWV(QWLUH6SDQ 5HIHU)LJ$XGORILQWHQVLW\w/mOHQJWKLVDFWLQJRYHUWKHHQWLUHVSDQO.

)LJ

7R¿QGWKHVORSH 6LQFHLQWHQVLW\RIXGODFWLQJRYHUWKHHQWLUHVSDQRIWKHEHDPLVXQLIRUPKHQFHWKHGHÀHFWLRQDWWKH PLGVSDQLVPD[LPXPDQGVORSHDWWKDWSRLQWLV]HUR Slope at A is TA =

$UHD RI BMD EHWZHHQ A DQG C EI

288

Strength of Materials

Slope at B is

=

1 1 ⎛ 4 l wl 2 ⎞ × × × EI 2 ⎜⎝ 3 2 8 ⎟⎠

=

wl 3  24 EI

 

wl 3 Area of BMD between B and C  = 24 EI EI

 

TB =

7R¿QGWKHGHÀHFWLRQ 7KHGHÀHFWLRQVDWA and BDUH]HUR 'HÀHFWLRQDWC is \C =



Moment of the area of BMD between A and C about A EI

wl 3 ⎛ l 3 l ⎞ 5 wl 4 ×⎜ − × ⎟ =  = 24 EI ⎝ 2 8 2 ⎠ 384 EI

 

 LQFHGLVWDQFHRIWKHFHQWURLGRISDUDEROLFVHFWLRQIURPC„ is 6 x =

3 O 3 ⎞ 3 ⎛O 5 × = O, and its distance IURPA„ is ⎜ − O ⎟ = O. ⎠ ⎝ 8 2 2 16 16 16

([DPSOH $  P VLPSOH EHDP LV FDUU\LQJ D XGO RI  N1P RYHU LWV HQWLUH VSDQ DQG WZR SRLQW ORDGV RI  N1 DW  P IURP ERWK HQGV )LQG WKH VORSH DQG GHÀHFWLRQ DW LPSRUWDQW ORFDWLRQV RI EHDP XVLQJ PRPHQWDUHDPHWKRG7DNHE *3DDQGI –± m. Solution:

5HIHU)LJ

)LJ

'HÀHFWLRQVRI%HDPV  289

Reactions at A and B 8VLQJ6 MA ZHKDYH RB × 6 = îî îî or and

6 2

RB N1 ) RA + RB – N1

or

RA ± N1 )

Bending moment diagram (BMD) 7KHEHQGLQJPRPHQWVDWA and BDUH]HUR 1.5  N1¹m 2 6  LPLODUO\WKHEHQGLQJPRPHQWDWD N1¹m

%HQGLQJPRPHQWDWE = RB –±– 1.5 –

7R¿QGWKHVORSH  KHVORSHDWWKHFHQWUHRIWKHEHDP SRLQWC LV]HUREHFDXVHWKHWDQJHQWWRWKHHODVWLFFXUYHDWWKDW 7 SRLQWLVKRUL]RQWDO Area of BMD between A and D Slope at A is TA =  XVLQJ0RKU¶V¿UVWWKHRUHP EI 1 × 1.5 × 106.875 1 ⎛1 ⎞ 2 × 1 . 5 × 106 . 875 = = ⎜ ⎟⎠ EI ⎝ 2 200 × 106 × 3.32 × 10−4  

 UDGLDQ

Ans.

Similarly, the slope at B is TB UDGLDQ 7R¿QGWKHGHÀHFWLRQ 7KHGHÀHFWLRQVDWA and BDUH]HUR 'HÀHFWLRQDWD is

\D =

Ans.

Moment of the area of BMD between A and D about A EI

⎞ 2 ⎛1 ⎜⎝ × 1.5 × 106.875⎟⎠ × × 1.5 2 3 = P ––3 m 6 200 × 10 × 3.32 × 10−4  PP

Ans.

6LPLODUO\WKHGHÀHFWLRQDWE is  \E PP

Ans.



290

Strength of Materials

'HÀHFWLRQDWC PLGVSDQ LV yC 

Moment of area of BMD between A and C about A EI 2 2 ⎡1 ⎤ ⎡ 1.5 + 1.5 × 3 + 3 ⎤ ( . ) . × + × 1 5 3 106 875 3 × − ⎥ ⎢⎣ 2 ⎥⎦ ⎢ 3 (1.5 + 3) ⎣ ⎦



 



 –±P PP

200 × 106 × 3.32 × 10−4

P Ans.

Example 6.8 $PSURSSHGFDQWLOHYHUEHDPFDUULHVDudlRIN1PRYHULWVHQWLUHVSDQDQGLVVXSSRUWHGDWLWVIUHH HQG)LQGWKHSURSUHDFWLRQZKHQWKHIUHHHQGGRHVQRW\LHOG7DNHEI N1¹P Solution:

5HIHU)LJ

Fig. 6.23

/HWRBEHWKHSURSUHDFWLRQDFWLQJDWWKHSRLQWB)LJ b JLYHVWKHBMDZKHQRQO\udlLV VXSSRVHGWREHDFWLQJRQWKHFDQWLOHYHUDQGWKHUHLVQRSURSUHDFWLRQ)LJ c JLYHVWKHBMD ZKHQRQO\SURSUHDFWLRQDFWVDQGWKHUHLVQRudl RQWKHFDQWLOHYHU

'HÀHFWLRQVRI%HDPV  291

7KHGHÀHFWLRQDWBZKHQQRSURSUHDFWLRQDFWVLVJLYHQDV \B =

 

Moment of the area of BMD between A and B about B  XVLQJ0RKU¶V¿UVWWKHRUHP EI

=

⎞ 3l 1 ⎛ 1 wl 2 × × l⎟ × ⎜ EI ⎝ 3 2 ⎠ 4

=

⎞ 3 1 ⎛ 1 2 × 32 × × × × 3  ––3P PP 3 ⎜ ⎟ 4 2 10 ⎝ 3 ⎠ 4

7KLVGHÀHFWLRQLVGLUHFWHGGRZQZDUG In case, when no XGOZHUHDFWLQJRQWKHFDQWLOHYHUEHDPWKHQWKHFDQWLOHYHUZRXOGKDYHGHÀHFWHG XSZDUGE\DQDPRXQWJLYHQE\ Moment of the area of BMD between A and B about B   \B„ = EI =

1 ⎛1 ⎞ 2 ⎜⎝ × RB × l × l ⎟⎠ × × l EI 2 3

%HFDXVHKHLJKWRIWKHWULDQJOHLVRB O and its C.GLVORFDWHGDWDGLVWDQFHRI \B„ =

or

1 ⎛1 ⎞ 2 × RB × 3 × 3⎟ × × 3 –± RB 4 ⎜ ⎠ 3 10 ⎝ 2

2 OIURPWKHDSH[ 3

Since end BGRHVQRW\LHOGLWPHDQVWKDWERWKGHÀHFWLRQV\B and \B„ are the same. ––3 –± RB or

RB =

2.025 × 10−3 9 × 10−4

 N1 )

Ans.

([DPSOH $FDQWLOHYHUEHDPRIOHQJWKOFDUULHVWZRSRLQWORDGVDVVKRZQLQ)LJ7KHÀH[XUDOULJLGLW\IRU the part ABLVEIDQGIRUWKHSDUW%&LV(,)LQGWKHVORSHDQGGHÀHFWLRQDWIUHHHQGRIWKHEHDP Solution:

5HIHU)LJ

 7KH M/EI GLDJUDPLVGLYLGHGLQWRWKUHHSDUWV WULDQJOHVA1, A and A3 7KHLUFHQWUHRIJUDYLWLHV 5O 2O 4O and , UHVSHFWLYHO\IURPWKHSRLQWC7KH M/EI) areas are G 1 , G  and G 3 are located at 3 3 3 IRXQGDV Wl 2 1 Wl ×l = A1 = × 2 EI 2 EI

and

A =

Wl 2 1 Wl × ×l= 2 2 EI 4 EI

A3 =

1 3Wl 3Wl 2 × ×l= 2 2 EI 4 EI

292

Strength of Materials

)LJ

7KHVORSHDQGGHÀHFWLRQDWWKH¿[HGHQGRIWKHEHDPDUH]HUR TA DQG\A  8VLQJ¿UVWPRPHQWDUHDWKHRUHPWKHVORSHDWCLVJLYHQDV TC $UHDRIWKH MEI GLDJUDPEHWZHHQA and C = A1 + A + A3 =

3Wl 2 Wl 2 Wl 2 3Wl 2 = + + 2 EI 2 EI 4 EI 4 EI

Ans.

8VLQJVHFRQGPRPHQWDUHDWKHRUHPWKHGHÀHFWLRQDWCLVJLYHQDV 

\F 0RPHQWRIWKH MEI GLDJUDPEHWZHHQA and C about C = A1 ×

2l 4l 5l + A2 × + A3 × 3 3 3

23Wl 3 Wl 2 2 Wl 2 4l 3Wl 2 5l × l+ × + × = = 12 EI 2 EI 3 4 EI 3 4 EI 3

Ans.

'HÀHFWLRQVRI%HDPV  293

([DPSOH ,Q([DPSOHLIW = 5 kN, O PDQGEI = 1 – kN¹m¿QGWKHVORSHDQGGHÀHFWLRQDWIUHH HQGRIWKHEHDP Solution: 7KHVORSHDWWKHIUHHHQGLVJLYHQDV TC = =

3Wl 2 2 EI 3 × 5 × 22 2 × 104

UDGLDQ ’

Ans.

7KHGHÀHFWLRQDWWKHIUHHHQGLVJLYHQDV \C

3 23Wl 3 23 × 5 × 2  P PP = = 12 × 104 12 EI

Ans.

([DPSOH )RUWKHEHDPVKRZQLQ)LJ¿QGWKHGHÀHFWLRQDWSRLQWA. Solution: 5HIHU)LJ7KH MEI GLDJUDPLVVKRZQLQ)LJ F ,WFRQVLVWVRIDSDUDEROLF VHJPHQWDQGDWULDQJOH

)LJ

294

Strength of Materials

7KH MEI DUHDIRUWKHSDUDEROLFVHJPHQWLVIRXQGDV A1 =

1 ⎛ wa 2 ⎞ wa3 ×⎜ × a = 3 ⎝ 2 EI ⎟⎠ 6 EI

7KH M/EI DUHDIRUWKHWULDQJOHLVIRXQGDV 1 wa 2 ×l A = × 2 2 EI =

wa 2l 4 EI

7KHUHIHUHQFHWDQJHQWLVGUDZQDWB. 7KHWDQJHQWLDOGHÀHFWLRQRIC w.r.t. BXVLQJVHFRQGPRPHQWDUHDWKHRUHPLVJLYHQDV 

\CB 0RPHQWRIWKHDUHDRI M/EI GLDJUDPEHWZHHQB and C about C =

wa 2l 2 wa 2l 2l × = 6 EI 4 EI 3

&RPSDULQJ'V A„BA„„ and BCC„we have yC / B × a

wa 2l 2 a wa3l × = l l 6 EI 6 EI 7KHWDQJHQWLDOGHÀHFWLRQRIA w.r.t. BLVJLYHQDV A„A„„ =

=

\A/B 0RPHQWRIWKHDUHDRI M/EI GLDJUDPEHWZHHQA and B about A =

wa 4 wa3 3 × a = 8 EI 6 EI 4

+HQFHWKHGHÀHFWLRQDWALVJLYHQDV 

\A = \AB + A„A„„ =

wa 4 wa3 l wa3 ⎡ a l ⎤ + + = 8 EI 6 EI 2 EI ⎢⎣ 4 3 ⎥⎦

Ans.

 &21-8*$7(%($00(7+2' 7KHFRQMXJDWHEHDPPHWKRGLVUHJDUGHGDVWKHPRGL¿FDWLRQRIPRPHQWDUHDPHWKRG7KHPRPHQW DUHDPHWKRGLVFRQYHQLHQWIRUWKHEHDPVRIFRQVWDQWÀH[XUDOULJLGLW\EXWWKLVPHWKRGLVSDUWLFXODUO\ XVHIXOIRUWKHEHDPVRIYDULDEOHÀH[XUDOULJLGLW\7KHPHWKRGVWDUWVZLWKWKHIRUPDWLRQRIDFRQMXJDWH EHDPZKLFKLVDQLPDJLQDU\EHDPRIOHQJWKHTXDOWRWKHOHQJWKRIWKHDFWXDOEHDPEXWWKHORDGRQ M DFWLQJLWLVQRWWKHDFWXDOORDGUDWKHULWLVWKHHODVWLFZHLJKW FRUUHVSRQGLQJWRWKHDFWXDOORDG EI DFWLQJDWWKHSRLQWRIDFWXDOORDG 7KHFRQMXJDWHEHDPPHWKRGLVEDVHGRQWZRWKHRUHPVVWDWHGEHORZ

'HÀHFWLRQVRI%HDPV  295

Conjugate beam theorem I 7KHVORSHDWDQ\VHFWLRQRIDORDGHGEHDPLVHTXDOWRWKHVKHDUIRUFHDWWKHFRUUHVSRQGLQJVHFWLRQRI WKHFRQMXJDWHEHDP Conjugate beam theorem II 7KHGHÀHFWLRQDWDQ\VHFWLRQRIDORDGHGEHDPLVHTXDOWRWKHEHQGLQJPRPHQWDWWKHFRUUHVSRQGLQJ VHFWLRQRIWKHFRQMXJDWHEHDP  6LPSOH%HDPFDUU\LQJD3RLQW/RDGDWLWV&HQWUH 5HIHU)LJ )LJXUH F VKRZVDORDGHGFRQMXJDWHEHDPDQGWKHORDGLQJRYHULWLVWULDQJXODU7KH M/EI) GLDJUDPIRUWKHDFWXDOEHDPLVWKHHODVWLFZHLJKW M/EI IRUWKHFRQMXJDWHEHDP Let RA' = Reaction at A' RIWKHFRQMXJDWHEHDP RB' = Reaction at B' RIWKHFRQMXJDWHEHDP RA„ = RB„ =

Total load on the conjugate beam 2

1 Wl ×l× 2 4 EI = Wl = 2 16 EI 2 $FFRUGLQJWRWKH¿UVWWKHRUHPWKHVKHDUIRUFHDWDQ\VHFWLRQRIWKHFRQMXJDWHEHDPLVHTXDOWRWKH VORSHDWWKDWVHFWLRQIRUWKHDFWXDOEHDP TA = Slope at ARIWKHDFWXDOEHDP 

 6KHDUIRUFHDWARIWKHFRQMXJDWHEHDP 

Similarly 



Wl 2 16 EI

TB = Slope at BRIWKHDFWXDOEHDP  6KHDUIRUFHDWBRIWKHFRQMXJDWHEHDP 

Wl 2 16 EI

Wl 2       16 EI $FFRUGLQJWRWKHVHFRQGWKHRUHPWKHEHQGLQJPRPHQWDWDQ\VHFWLRQRIWKHFRQMXJDWHEHDPLVHTXDO WRWKHGHÀHFWLRQDWWKDWVHFWLRQIRUWKHDFWXDOEHDP   \C  'HÀHFWLRQDWCRIWKHDFWXDOEHDP TA = TB =

Hence,





 %HQGLQJPRPHQWDWCRIWKHFRQMXJDWHEHDP = RA′ × =

l 1 l Wl ⎛ 1 l ⎞ − × × ×⎜ × ⎟ 2 2 2 4 EI ⎝ 3 2 ⎠

Wl 3 Wl 2 l Wl 3 × − =  48EI 16 EI 2 96 EI



 

296

Strength of Materials

)LJ

'HÀHFWLRQVRI%HDPV  297

 6LPSOH%HDPFDUU\LQJD3RLQW/RDGQRWDWWKH&HQWUH 5HIHU)LJ

)LJ

Support reactions for actual beam AB Wa Wb RA = and RB = l l 7KHEHQGLQJPRPHQWGLDJUDPIRUDFWXDOEHDPLVVKRZQLQ)LJ b). Wab . 7KHPD[LPXPEHQGLQJPRPHQWDWC is l 7KHFRQMXJDWHEHDPZLWKHTXLYDOHQWORDGLVVKRZQLQ)LJ F). Support reactions for conjugate beam A„B„ 8VLQJ6MA„ ZHKDYH Wab ⎞ 2 Wab ⎞ ⎛ b⎞ ⎛1 ⎛1 RB„– O = ⎜ × a × ⎟ × a + ⎜⎝ × b × ⎟ × ⎜ a + ⎟⎠ ⎝2 EIl ⎠ 3 2 EIl ⎠ ⎝ 3 Wab Wa3b Wab 2 [2a 2 + 3ab + b 2 ] + (3a + b) = = 6 EIl 3EIl 6 EIl

298

Strength of Materials

or

RB„ = =

and

Wab 6 EIl

2

[2a 2 + 3ab + b 2 ]

Wab (l + a)  6 EIl

RA′ + RB ′ =

1 Wab ×l × 2 EIl

RA„ =

Wab (l + b) 6 EIl

a + b = O)

2QVROYLQJZHJHW

8VLQJ¿UVWWKHRUHPZHKDYH TA = Slope at ARIWKHDFWXDOEHDP 

 6KHDUIRUFHDWARIWKHFRQMXJDWHEHDP =

and

Wab (l + b)  6 EIl

 

TB = Slope at BRIWKHDFWXDOEHDP



 6KHDUIRUFHDWBRIWKHFRQMXJDWHEHDP =

Wab (l + a )  6 EIl

 

8VLQJVHFRQGWKHRUHPZHKDYH \C 'HÀHFWLRQDWCRIWKHDFWXDOEHDP 

 %HQGLQJPRPHQWDWCRIWKHFRQMXJDWHEHDP = RB′ × b −

1 Wab b ×b× × 2 EIl 3

Wab Wab3 Wa 2b2 ( + ) × − l a b =  = 6 EIl 6 EIl 3EIl

 

([DPSOH )RUWKHEHDPVKRZQLQ)LJ¿QGWKHVORSHDWWKHVXSSRUWSRLQWVDQGGHÀHFWLRQXQGHUWKHJLYHQ ORDG$OVR¿QGWKHGHÀHFWLRQDWWKHFHQWUHRIWKHEHDP7DNHE *3DDQG I ––5 m. Solution:

5HIHU)LJ

'HÀHFWLRQVRI%HDPV  299

)LJ

Support reactions for the actual beam 8VLQJ™ MA ZHKDYH or Now and

RB – – 5 RB = N1 ) RA + RB N1 RA ±N1 N1 )

BMD for the actual beam %HQGLQJPRPHQWDWC is MC = RB – – 3 

 N1¹m %HQGLQJPRPHQWDWD is MD = RA – – N1¹m 7KHBMDIRUDFWXDOEHDPLVVKRZQLQ)LJ b).

300

Strength of Materials

&RQMXJDWHEHDP )LJ c)) 6.25 1 = EI E (3I )

/RDGLQWHQVLW\MXVWWRWKHOHIWRIC – /RDGLQWHQVLW\MXVWWRWKHULJKWC = 18.75 ×

1 18.75 = EI EI

= RA × 4 ×

Load intensity at D

5 1 3.75 × 4 = = EI E (3I ) 3EI

Support reactions at A„ and B„ 8VLQJ6MB„ ZHKDYH ⎡⎛ 1 6.25 ⎞ ⎛ 5⎞ ⎛ 1 18.75 ⎞ ⎛ 2 ⎞⎤ RA„ – 8 = ⎢⎜ × 5 × ⎟⎠ × ⎜⎝ 3 + ⎟⎠ + ⎜⎝ × 3 × ⎟⎠ × ⎜⎝ × 3⎟⎠ ⎥ ⎝ 2 EI 3 2 EI 3 ⎣ ⎦ 72.91 56.25 129.16 + = EI EI EI 16.14 = EI

= or

RA„

RA′ + RB′ =

and

RB„ =

or

43.75 1 18.75 6.25 1 + –3– = –5– EI 2 EI EI 2 27.61 43.75 16.14 – = EI EI EI

(a) Slopes at A and B Slope at A is TA 6KHDUIRUFHDWA„ = RA„ = 

16.14 16.14 = EI 200 × 106 × 4 × 10−5

 UDGLDQ ƒ

Ans.

Slope at B is TB 6KHDUIRUFHDWB„ = RB„ = 

27.61 27.61 = EI 200 × 106 × 4 × 10−5

 UDGLDQ ƒ

Ans.

'HÀHFWLRQVRI%HDPV  301

(b) 'HÀHFWLRQXQGHUWKHJLYHQORDG 'HÀHFWLRQDWC is 

\C %HQGLQJPRPHQWDWCRIWKHFRQMXJDWHEHDP 6.25 5 1 – –5– EI 3 2

= RA„– 5 –

= =

16.14 25 × 6.25 ×5– EI 6 EI 16.14 × 5 200 × 10 × 4 × 10 6

−5



25 × 6.25 6 × 200 × 106 × 4 × 10−5

= 6.83 ––3 m = 6.83 mm

Ans.

(c) 'HÀHFWLRQDWWKHPLGSRLQWD 'HÀHFWLRQDWD is \D %HQGLQJPRPHQWDWDRIWKHFRQMXJDWHEHDP = RAc–±

=

40 4 16.14 1 5 î 4 − = – –– 3EI 3 EI 2 EI

16.14 × 4 200 × 106 × 4 × 10−5



 ––3P PP



40 3 × 200 × 106 × 4 × 10−5 Ans.

([DPSOH )RUWKHEHDPVKRZQLQ)LJILQGWKHGHIOHFWLRQDWWKHFHQWUHRIWKHEHDP7DNHE *3D and I – –5 m . Solution: 5HIHU)LJ Support reactions for the actual beam 8VLQJ6MA ZHKDYH RB – – 3 or Now and 

RB N1  RA +RB

N1

RA ±  N1 

302

Strength of Materials

)LJ

BMD for the actual beam %HQGLQJPRPHQWDWD is MD – N1¹m %HQGLQJPRPHQWVDWC and E are MC = ME – N1¹m 7KHBMDLVVKRZQLQ)LJ b). CRQMXJDWHEHDP )LJ c)) Load at D„

=

375 375 187.5 = = E (2 I ) 2EI EI

/RDGMXVWWRWKHOHIWRIC„

=

187.5 = PCc EI



 /RDGMXVWWRWKHULJKWRIEc = TE„

'HÀHFWLRQVRI%HDPV  303

=

/RDGMXVWWRWKHULJKWRIC„

187.5 93.75 = EI 2 EI

 /RDGMXVWWRWKHOHIWRIE„ = SE„

 Hence,

PQ = QC„ =

93.75 EI

and

TS = SE„ =

93.75 EI

Support reactions at A„ and B„ 6LQFHORDGRQWKHFRQMXJDWHEHDPLVV\PPHWULFDOKHQFHWKHVXSSRUWUHDFWLRQVDUHHTXDOHDFKEHLQJ HTXDOWRRQHKDOIRIWKHWRWDOORDGRQWKHFRQMXJDWHEHDP RA„ = RB„ $UHDRI'A„ PC„$UHDRIWUDSH]LXPC„QRD„ 187.5 ⎤ 1 ⎛ 93.75 187.5 ⎞ 351 ⋅ 5625 ⎡1 + ⎜ + × 1.5 = = ⎢ × 1.5 × ⎟ ⎥ EI ⎦ 2 ⎝ EI EI ⎠ EI ⎣2 'HÀHFWLRQDWWKHPLGSRLQWD of the actual beam 'HÀHFWLRQDWD is   \D %HQGLQJPRPHQWDWD„RIWKHFRQMXJDWHEHDP ⎛ QC ′ + RD′ ⎞ 1.5 ⎞ C ′ D′ ⎛ – area C„QRD„ – – ⎜ = RA„ – 3 – area A„ PC„ – ⎜1.5 + ⎟ ⎝ QC ′ + RD′ ⎟⎠ ⎝ 3⎠ 3 93.75 187.5 ⎞ ⎛ 2× + ⎜ 1 . 5 140.625 351.5625 EI EI ⎟ 210.9375 – ⎜ 93.75 187.5 ⎟ –3– –± = – 3 EI EI EI + ⎟ ⎜ ⎠ ⎝ EI EI =

=

632.8125 1054.6875 281.25 140.625 – – = EI EI EI EI 632.8125  ––3P PP ⎛ 200 × 109 ⎞ –5 ⎜ 103 ⎟ × 300 × 10 ⎝ ⎠

Ans.

([DPSOH )RUWKHEHDPVKRZQLQ)LJ¿QGWKHVORSHDWWKHVXSSRUWSRLQWVDQGWKHGHÀHFWLRQDWWKHFHQWUH7DNH E *3D I –± m

304

Strength of Materials

Solution:

5HIHU)LJ

)LJ

Support reactions for the actual beam 8VLQJ™ MA ZHKDYH RB –  or Now or 

100  N1 ) 4 RA + RB  RB =

RA = – RB = – N1  N1 ‘)

'HÀHFWLRQVRI%HDPV  305

BMD for the actual beam 7KHEHQGLQJPRPHQWVDWA and BDUH]HUR MA = MB  %HQGLQJPRPHQWMXVWWRWKHULJKWRIC = RB – – N1¹m %HQGLQJPRPHQWMXVWWRWKHOHIWRIC = RB –± 

 ± ±N1¹m

&RQMXJDWHEHDP )LJ c)) 50 EI 50 = EI

/RDGMXVWWRWKHULJKWRI C„

=

/RDGMXVWWRWKHOHIWRIC„

Support reactions at Ac and Bc  8VLQJ™ MAc ZHKDYH RBc –

50 50 1 1 2 ⎛ –  –– – – ⎜2 + –– EI EI 2 2 ⎝ 3

RB„ –

2⎞ ⎟ 3⎠

200 400 = 3EI 3EI 400 − 200 50 (↑) = 3EI × 4 3EI 50 = ‘ 3EI

or

RB„ =

and

RA„

Slope at the support points A and B of the actual beam Slope at A

6KHDUIRUFHDWA„RIWKHFRQMXJDWHEHDP = RAc = =

50 3EI 50 = 3.33 ––6 radian ⎛ 200 × 109 ⎞ −4 3×⎜ ⎟ × 250 × 10 ⎝ 103 ⎠

Ans.

306

Strength of Materials

Similarly

Slope at B = Shear force at B„ of the conjugate beam = 3.33 × 10–6 radian

Ans.

'HÀHFWLRQDWC 'HÀHFWLRQDWC is yC = Bending moment at C„ of the conjugate beam = RA„ × 2 – =

50 2 1 × ×2× 3 EI 2

100 50 ×2– =0 3EI 3EI

Ans.

Example 6.15 )LQGWKHVORSHDQGGHÀHFWLRQDWIUHHHQGRIDFDQWLOHYHUEHDPRIOHQJWKl carrying a point load W at its free end. Solution:

Refer Fig. 6.31.

Fig. 6.31

'HÀHFWLRQVRI%HDPV  307

7KHIUHHDQG¿[HGHQGVRIWKHFDQWLOHYHUEHDPDUHFRQYHUWHGWR¿[HGDQGIUHHHQGVUHVSHFWLYHO\IRU WKHFRQMXJDWHEHDP

Slope at B of the actual beam Slope at B is TB 6KHDUIRUFHDWB„RIWKHFRQMXJDWHEHDP =

1 Wl îOî 2 EI

Wl   EI 'HÀHFWLRQDW%RIWKHDFWXDOEHDP =

Ans.

'HÀHFWLRQDWB is 

\B %HQGLQJPRPHQWDWB„RIWKHFRQMXJDWHEHDP = =

2 1 Wl îOî î îO 3 2 EI Wl 3 3EI

Ans.

([DPSOH )LQGWKHVORSHDQGGHÀHFWLRQDWWKHIUHHHQGRIDFDQWLOHYHUEHDPRIOHQJWKOFDUU\LQJDSRLQWORDGDW a certain distance O1IURPWKH¿[HGHQG Solution: 5HIHU)LJ Slope at C is TC

6KHDUIRUFHDWC „RIWKHFRQMXJDWHEHDP =

1 Wl îO1î 1 2 EI

=

Wl12 2 EI

'HÀHFWLRQDWC is 

Ans.

\F %HQGLQJPRPHQWDWC„RIWKHFRQMXJDWHEHDP =

2 1 Wl îO1î 1 î O1 3 2 EI

Wl13 = 3EI

Ans.

308

Strength of Materials

)LJ

([DPSOH )RU D FDQWLOHYHU EHDP ORDGHG DV VKRZQ LQ )LJ  ILQG WKH GHIOHFWLRQ DW LWV IUHH HQG 7DNH E *3DDQGI = 1 – ± m . Solution:

5HIHU)LJ

%HQGLQJPRPHQWDWC is MC

– N1¹m

%HQGLQJPRPHQWDWA is MA –– N1¹m 7KHGHÀHFWLRQDWB is obtained as  

\B %HQGLQJPRPHQWDWB„RIWKHFRQMXJDWHEHDP 20 2 60 ⎛ 2 ⎡1 ⎤ ⎞⎤ ⎡ 20 2 ⎤ ⎡1 × × 2 ⎥ + ⎢ × 2 × ⎛⎜ 2 + ⎞⎟ ⎥ + ⎢ × 2 × = ⎢ × 2× × ⎜ 2 + × 2⎟ ⎥ ⎝ ⎠⎦ EI 3 2 ⎝ EI 3 ⎣ ⎦ 2⎠ ⎦ ⎣ 2 ⎣ EI =

 

1 1040 î EI 3

 P PP

Ans.

'HÀHFWLRQVRI%HDPV  309

)LJ

([DPSOH )LQGWKHVORSHDQGGHÀHFWLRQDWIUHHHQGRIDPFDQWLOHYHUEHDPFDUU\LQJDXGORILQWHQVLW\N1/m RYHULWVHQWLUHVSDQ7DNHE *3DDQGI ––5 m. Solution:

5HIHU)LJ

%HQGLQJPRPHQWDWA is

2 × 32  N1¹m 2 7KHYDULDWLRQRIEHQGLQJPRPHQWEHWZHHQA and B is parabolic. 7KHFHQWURLGRIWKHORDGHGFRQMXJDWHEHDPDFWVDWDGLVWDQFHRI 3 – PIURPBc 4 7KHVORSHDWBLVIRXQGDV TB 6KHDUIRUFHDWB„RIWKHFRQMXJDWHEHDP 1   – 3 – = = 3 EI EI MA =

=

9 9⎞

⎛ 200 × 10 –5 ⎜ 103 ⎟ × 2 × 10 ⎝ ⎠

 UDGLDQ ƒ

Ans.

310

Strength of Materials

)LJ

7KHGHÀHFWLRQDWB is obtained as 

\B %HQGLQJPRPHQWDWB„RIWKHFRQMXJDWHEHDP = =



20.25 1 9 ×3× × 2.25 = EI 3 EI 20.25 ⎛ 200 × 109 ⎞ −5 ⎜ 103 ⎟ × 2 × 10 ⎝ ⎠

 ––3P PP

Ans.

 0(7+2'2)683(5326,7,21 7KLV PHWKRG LV PDLQO\ VXLWDEOH IRU OLQHDUO\ HODVWLF GHIRUPDWLRQ SUREOHPV ZKHUH ORDGGHIRUPDWLRQ UHODWLRQVKLSLVOLQHDUDQGWKHGHIRUPDWLRQVSURGXFHGDUHYHU\VPDOODVFRPSDUHGWRWKHWUDQVYHUVH GLPHQVLRQVRIWKHEHDP7KHPHWKRGFDQEHXVHGWRVWXG\WKHGHIRUPDWLRQEHKDYLRXURIDVWUXFWXUH XQGHUDJLYHQVHWRIORDGV$FFRUGLQJWRWKLVPHWKRGWKHWRWDOGHÀHFWLRQRIDEHDPDWDQ\VHFWLRQ LVHTXDOWRWKHDOJHEUDLFVXPRIWKHGHÀHFWLRQVSURGXFHGVHSDUDWHO\DWWKDWVHFWLRQGXHWRHDFKORDG &RQVLGHUDFDQWLOHYHUEHDPFDUU\LQJWKUHHSRLQWORDGVDWFHUWDLQGLVWDQFHVIURPWKHIL[HGHQG )LJ 

'HÀHFWLRQVRI%HDPV  311

)LJ

Let

\1 'HÀHFWLRQSURGXFHGDWB due to load W1 only \ 'HÀHFWLRQSURGXFHGDWB due to load W only \3 'HÀHFWLRQSURGXFHGDWB due to load W3 only 7KHWRWDOGHÀHFWLRQSURGXFHGDWDDFFRUGLQJWRPHWKRGRIVXSHUSRVLWLRQLV \ = \1 + \ + \3

 

([DPSOH )LQGWKHVORSHDQGGHÀHFWLRQDWWKHIUHHHQGRIDFDQWLOHYHUEHDPRIOHQJWKPDVVKRZQLQ)LJ XVLQJPHWKRGRIVXSHUSRVLWLRQ Solution:

5HIHU)LJ

)LJ

Let 

TB and \B 6ORSHDQGGHÀHFWLRQDWIUHHHQGD due to 3WRQO\ DFWLQJDWB)



TC and \C 6ORSHDQGGHÀHFWLRQDWDGXHWRWRQO\ DFWLQJDWC)



TD and \D  6ORSHDQGGHÀHFWLRQDWD due to WRQO\ DFWLQJDWD)

7KHVORSHVDQGGHÀHFWLRQVDWYDULRXVSRLQWVRIWKHEHDPDUHJLYHQDV TB =

3W × 22 6W = 2 EI EI

\B =

8W 6W 80W 3W × 23 + ×4= + θ B × (2 + 2 ) = 3EI EI 3EI 3EI 2

TC

2W × (2 + 2) 16W = = 2 EI EI 3



\C =

2W × (2 + 2 ) 3EI

+ θc × 2 =

128W 16W 224W + ×2= 3EI EI 3EI

312

Strength of Materials 2

TD =

W × (2 + 2 + 2) 18W = EI 2 EI

W × (  +  + ) W + θD ×  = 3EI 3EI +HQFHWKHVORSHDWWKHIUHHHQGLV 6W 16W 18W 40W TB + TC + TD = + + = EI EI EI EI $QGWKHGHÀHFWLRQDWWKHIUHHHQGLV 520W 80W 224W 216W = \B + \C + \D = + + 3EI 3EI 3EI 3EI 3



\D =

6+257$16:(548(67,216       

      

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Ans. Ans.

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08/7,3/(&+2,&(48(67,216



1. 7KHGHÀHFWHGQHXWUDOVXUIDFHRIDEHDPDIWHUEHQGLQJLVFDOOHG 

a  GHÀHFWHGVXUIDFH b  EHQWVXUIDFH

F  HODVWLFFXUYH

d) plastic curve.

 7KHGLIIHUHQWLDOHTXDWLRQRIÀH[XUHLV dy

=M 

dy =M  dx



a) EI



a  HTXDOWRWKHGHÀHFWLRQSURGXFHGE\VKHDU



b  OHVVWKDQWKHGHÀHFWLRQSURGXFHGE\VKHDU



F  JUHDWHUWKDQWKHGHÀHFWLRQSURGXFHGE\VKHDU



d) unpredictable.

b) EI

dx   7KHGHÀHFWLRQSURGXFHGE\EHQGLQJLV

F) EI

d3y dx3

=M 

d) EM

dy = I. dx

4. 7KHÀH[XUDOULJLGLW\LVWKHSURGXFWRI 

a  PRGXOXVRIHODVWLFLW\DQGPDVVPRPHQWRILQHUWLD



b  PRGXOXVRIULJLGLW\DQGDUHDPRPHQWRILQHUWLD



F  PRGXOXVRIULJLGLW\DQGPDVVPRPHQWRILQHUWLD



d  PRGXOXVRIHODVWLFLW\DQGDUHDPRPHQWRILQHUWLD

5. 7KHVORSHDQGGHÀHFWLRQDWWKH¿[HGHQGRIDFDQWLOHYHUEHDPDUH 

a  ]HURPD[LPXP





F  PD[LPXPPLQLPXP  

b  ]HUR]HUR d  PD[LPXP]HUR

 7KHVORSHDQGGHÀHFWLRQDWWKHFHQWUHRIDVLPSOHEHDPFDUU\LQJDFHQWUDOSRLQWORDGDUH 

a  ]HUR]HUR



b  ]HURPD[LPXP



F  PD[LPXP]HUR



d  PLQLPXPPD[LPXP

7. :KLFKPHWKRGXVHV0RKU¶VWKHRUHPIRU¿QGLQJWKHVORSHDQGGHÀHFWLRQRIDEHDP" 

a  0DFDXOD\¶VPHWKRG  

b  ,QWHJUDWLRQPHWKRG



F  0RPHQWDUHDPHWKRG  

d  &RQMXJDWHEHDPPHWKRG

 7KHVORSHDWDQ\VHFWLRQRIDEHDPLVHTXDOWRZKLFKSDUDPHWHURIWKHFRQMXJDWHEHDP" 

a  EHQGLQJPRPHQW



b) slope



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a  VKHDUIRUFH

b  VORSH

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314

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⎛ ⎞ wl 4 at C ⎟ . Ans . ⎜ 120 EI ⎝ ⎠  8VLQJPRPHQWDUHDPHWKRG¿QGWKHVORSHDQGGHÀHFWLRQDWWKHIUHHHQGRIWKHFDQWLOHYHUEHDP VKRZQLQ)LJ

)LJ

⎛ 7 Wl 2 ⎞ Ans . θ = c ⎜ 24 EI ⎟ . ⎟ ⎜ 25Wl 3 ⎟ ⎜ yc = ⎜⎝ 72 EI ⎟⎠

316

Strength of Materials

4. 8VLQJPRPHQWDUHDPHWKRG¿QGWKHVORSHDQGGHÀHFWLRQDWWKHIUHHHQGRIWKHFDQWLOHYHUEHDP VKRZQLQ)LJ

)LJ

⎛ ⎜⎝ Ans.

wl ⎞ . ⎟ 6⎠

5. 8VLQJPRPHQWDUHDPHWKRGRURWKHUZLVH¿QGWKHGHÀHFWLRQDWSRLQWAIRUWKHEHDPVKRZQLQ )LJ7DNHEI = 1 –3 kN¹m.

)LJ





$QVPP 

 8VLQJPRPHQWDUHDPHWKRGRURWKHUZLVH¿QGWKHGHÀHFWLRQDWSRLQWDIRUWKHEHDPVKRZQLQ )LJ7DNHEI = 1 –3 kN¹m.

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$QV PP 

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'HÀHFWLRQVRI%HDPV  317

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⎛ ⎜⎝ Ans.

wl ⎞ . ⎟ 6⎠  8VLQJFRQMXJDWHEHDPPHWKRGILQGRXWVORSHDWADQGGHIOHFWLRQDWCIRUWKHEHDPVKRZQLQ )LJ7DNHI AC I BC

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⎛ ⎜⎝ Ans.

129 , 63 ⎞ . ⎟ 20 EI 5EI ⎠

 )LQGWKHPD[LPXPGHIOHFWLRQRIDVLPSOHEHDPKDYLQJIOH[XUDOULJLGLW\EI and loaded with DGLVWULEXWHGORDGZKLFKYDULHVIURP]HURDWERWKHQGVWRwXQLWOHQJWKDWWKHPLGVSDQ ⎛ ⎞ wl 4 at midspan⎟ . Ans . ⎜ 120 EI ⎝ ⎠  )LQGWKHGHÀHFWLRQDWWKHIUHHHQGRIDFDQWLOHYHUEHDPKDYLQJÀH[XUDOULJLGLW\EI and loaded with DWULDQJXODUORDGDVVKRZQLQ)LJ

)LJ

⎛ ⎜ Ans. ⎝

wl 4 ⎞ . 30 EI ⎟⎠

318

Strength of Materials

11. )LQGWKHFHQWUDOGHÀHFWLRQRIDVLPSOHEHDPKDYLQJÀH[XUDOULJLGLW\EIGXHWRXQLIRUPORDGw as VKRZQLQ)LJ

)LJ

⎛ ⎜ Ans. ⎝

5 wl 4 ⎞ . ⋅ 768 EI ⎟⎠

 )LQGWKHFHQWUDOGHÀHFWLRQRIDVLPSOHEHDPKDYLQJÀH[XUDOULJLGLW\EI due to two point loads, ERWKHTXDOWRWDVVKRZQLQ)LJ

)LJ

⎛ ⎜⎝ Ans.

Wa ⎞ (3l 2 − 4a 2 )⎟ . ⎠ 24 EI

 )LQGWKHGHÀHFWLRQRIDFDQWLOHYHUEHDPKDYLQJÀH[XUDOULJLGLW\EIGXHWRXQLIRUPORDGwDFWLQJ RYHUWKHPLGGOHKDOIRIWKHEHDPDVVKRZQLQ)LJ

)LJ

⎛ ⎜ Ans. ⎝

7 wl 4 ⎞ . ⋅ 64 EI ⎟⎠

14. )LQGWKHFHQWUDOGHÀHFWLRQRIDVLPSOHEHDPRIOHQJWKPZKLFKFDUULHVDXQLIRUPORDGWKDW YDULHVIURPN1PDWRQHHQGWR.1PDWWKHRWKHUHQG7KHFURVVVHFWLRQRIWKHEHDPLV PPGHHSDQGWKHPD[LPXPEHQGLQJVWUHVVLV03D7DNHE *3D $QVPP  ‰‰‰

7 Torsion of Circular Members Charles-Augustin de Coulomb, born on 14 June 1736, was a French physicist and engineer. He is best known for his Coulomb’s law used in HOHFWURVWDWLFVIRU¿QGLQJWKHHOHFWURVWDWLFIRUFHRIDWWUDFWLRQRUUHSXOVLRQ The SI unit of electric charge, the Coulomb (C), was named in his honour. He derived the torsion formula in about 1775. Coulomb leaves a legacy DVDSLRQHHULQWKH¿HOGRIJHRWHFKQLFDOHQJLQHHULQJIRUKLVFRQWULEXWLRQWR the design of retaining walls. His name is one of the 72 names inscribed on the Eiffel Tower.

Charles-Augustin de Coulomb (1736-1806)

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ :KDWLV7RUVLRQ"  ‡ +RZPXFKSRZHULVWUDQVPLWWHGE\DURWDWLQJVKDIW"  ‡ :KDWGRHVWRUVLRQDOULJLGLW\VLJQLI\"  ‡ :KDWLVSRODUPRGXOXV"  ‡ :K\LVSRODUPRPHQWRILQHUWLDXVHGLQWKHWRUVLRQIRUPXOD"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_7

319

320

Strength of Materials

7.1 INTRODUCTION The members in torsion are subjected to twisting action about their longitudinal axes. It has wide engineering applications, most common among them is the transmission shaft which is used to transmit power from one point to another e.g., from the prime mover (steam turbine, gas turbine etc.) to a machine or from a motor to a machine tool, or from the engine to rear axle of an automobile. Solid as well as hollow shafts can be used for the transmission of power. The torsion formula was derived by a French physicist Charles A. Coulomb in about 1775. His VWXG\ZDVFRQ¿QHGWRFLUFXODUPHPEHUV7RUVLRQVRIQRQFLUFXODUPHPEHUV¿UVWLQWURGXFHGE\/RXLV Navier were further improved by St.Venant and Prandtl. 7.2 TORSION EQUATION The following assumptions are made for circular members under torsion: z

The shaft is straight and has uniform cross-section throughout its length.

z

The plane sections remain plane and do not get distorted after being subjected to torsion.

z

Stresses induced in the shaft are within elastic limit.

z

The twist along the shaft is uniform.

z

The shaft material is homogeneous and isotropic.

Consider a solid circular shaft which is rigidly connected at one end. If a torque T is applied to the other end, the shaft will twist, with its free end rotating through an angle T, called the angle of twist (Fig. 7.1). The angle T varies in proportion to T for a certain range of values of T. Also, it is proportional to the length of the shaft.

Fig. 7.1 A solid circular shaft subjected to torsion.

/HW

l /HQJWKRIWKHVKDIW R = Radius of the shaft

Torsion of Circular Members

321

G = Modulus of rigidity or shear modulus of the shaft material I = Shear strain W = Shear stress :LWKWKHDSSOLFDWLRQRIWRUTXHSRLQW A at the radial distance shifts to A„ making angle T at the centre. ² ABAc = I ² AOA„ = T I and T both are expressed in radians.

In ' ABA„ or

Again

AAc AAc = AB l AAc I = l AA„ = Il

tan I =

T =

(tan I  I for small angle) ... (7.1)

AAc AAc = OA R

AAc= TR

...(7.2)

From equations (7.1) and (7.2), we have 

Il = TR

TR ... (7.3) = Imax l It shows that the shear strain I at a given point of a shaft under torsion is proportional to the angle of twist T and is maximum at the surface of the shaft. Similarly for any point at a radial distance r, we can write Tr  I = ... (7.4) l It simply means that the shear strain varies linearly with the distance from the axis of the shaft. )URPGH¿QLWLRQRIVKHDUPRGXOXV τ G = ... (7.5) φ W TR = (using equation (7.3)) ... (7.6) or  I = G l I =

or

W max W GT = = R r l

(using equation (7.4))

... (7.7)

Equation (7.7) shows that the shear stress varies linearly with the radial distance r, and is maximum at the surface of the shaft, but the relation holds good, if Hooke’s law is obeyed (Fig. 7.2).

322

Strength of Materials

Fig. 7.2

Now consider an elemental ring at a radial distance r from the axis of the shaft (Fig. 7.3). The YDULDWLRQRIVKHDUVWUHVVLVDOVRVKRZQLQWKHVDPH¿JXUH

Fig. 7.3

Shear force acting on the ring = (2S r dr)W = (2Sr dr – W) – r = 2Sr2drW

Torque acting on the ring

... (7.8)

From equation (7.7), we have W =

tmax . r R

Using W in equation (7.8), we get

W max 3 r dr R Hence, total torque acting on the entire cross-section is

Torque acting on the ring

= 2S

R

T =

∫ 0

R

2π τ max ⎛ r 4 ⎞ τ 2π max r 3 dr = ⎜ ⎟ R ⎝ 4 ⎠0 R

Torsion of Circular Members

τ 2π τ max R 4 = . = max R 4 R where

323

⎛ π R 4 ⎞ W max .⎜ .J ⎟ = R ⎝ 2 ⎠

π 4 R = Polar moment of inertia of the cross-section 2 = IXX + IYY

J =

W T = max R J From equations (7.7) and (7.9), we have

or

... (7.9)

W T GT = max = ... (7.10) R J l This is known as torsion equation and is also valid for a hollow circular shaft with suitable changes in diameter.

7.3 TORSIONAL RIGIDITY :HKDYHIURPWRUVLRQIRUPXODZHKDYH T GT = J l Tl or JG = Torsional rigidity = ... (7.11) T Hence, the product of polar moment of inertia and modulus of rigidity is known as torsional rigidLW\,WLVDPHDVXUHRIVWUHQJWKRIVKDIWDJDLQVWWRUVLRQDQGLVGH¿QHGDVWKHWRUTXHDFWLQJRQDVKDIW of unit length producing unit twist. 7.4 POLAR MODULUS From torsion formula, we have W T = max R J J T or = Zp = Polar modulus = W max R For a solid circular shaft of diameter d, the polar modulus is given as π 4 d π 3 Zp 32 d = = (Solid shaft) 16 ⎛d ⎞ ⎜ ⎟ ⎝2⎠

... (7.12)

... (7.13)

For a hollow circular shaft of external diameter d0 and internal diameter di, the polar modulus is given as π (d 04 − di4 ) Zp π (d 04 − di4 ) 32 . = ... (7.14) (Hollow shaft) = d0 16 ⎛ d0 ⎞ ⎜ 2⎟ ⎝ ⎠

324

Strength of Materials

7.5 POWER TRANSMITTED BY A SHAFT 7KHSRZHUJHQHUDWHGE\DSULPHPRYHUKDVWREH¿QDOO\WUDQVPLWWHGWRVRPHRWKHUGHYLFHVE\PHDQV of rotating shafts of solid or hollow circular sections. /HWWKHVKDIWURWDWLQJDWN rpm be subjected to a torque T. Angle turned by the shaft in one rotation = 2S radian Angle turned by the shaft in N rotations or angle turned per minute = 2S N radians Power transmitted by the shaft is given as P = Angle turned per minute – torque = 2SNT ... (7.15) z If we consider angle turned per second and T in N¹m, then equation (7.15) can be rewritten as P = = z

2πNT  :DWW 60 π NT 2π NT  N:  = 30000 60 × 1000

 (7.16)

If torque is expressed in kgf¹m and number of revolutions are considered on second basis, then equation (7.15) can be expressed as 2π NT 2πNT P = (hp) (1 hp = 75 kgf.m/s) ... (7.17) = 60 × 75 4500

Example 7.1 Fig. 7.4 shows the attachment of four pulleys to two different types of shafts. Pulleys B and C are connected by a hollow shaft with inside and outside diameters of 50 mm and 80 mm respectively. Other pulleys are connected by solid shafts of equal diameters. The torque acting on each pulley is VKRZQLQWKH¿JXUH)LQGWKHIROORZLQJSDUDPHWHUV (a) the maximum and minimum shear stress induced in the hollow shaft. (b) the diameter of the solid shaft, if the maximum shear stress is not to exceed 50 MPa.

Fig. 7.4

Solution:

Given,

Inside diameter of the hollow shaft, di = 50 mm Outside diameter of the hollow shaft, do = 80 mm

Torsion of Circular Members

325

(a) The net torque acting on the hollow shaft BC is given as TBC = (5 + 10) = 15 kN¹m (Clockwise) The polar moment of inertia of the section is found as J =

π (d o4 − di4 ) 32

π ⎡⎛ 80 ⎞ ⎛ 50 ⎞ = ⎢⎜ ⎟ ⎟ −⎜ 32 ⎢⎣⎝ 1000 ⎠ ⎝ 1000 ⎠ 4

4

⎤ 4 ⎥ m = 3.4 – 10–6 m4 ⎥⎦

Using torsion formula, we have TBC τ = max J ⎛ do ⎞ ⎜ 2 ⎟ ⎝ ⎠ Hence, the maximum shear stress induced in the shaft BC is given as TBC d o u J 2 15 1 ⎞ ⎛ 80 ×⎜ × = ⎟ −6 3.4 × 10 ⎝ 2 1000 ⎠

Wmax =

= 1.76 × 105 kN/m2

Ans.

The minimum shear stress induced in the shaft BC is given as Wmin =

TBC ⎛ di ⎞ × J ⎜⎝ 2 ⎟⎠ 15 1 ⎞ ⎛ 50 5 2 ×⎜ × = ⎟ = 1.1 – 10 kN/m −6 3.4 × 10 2 1000 ⎝ ⎠

Ans.

(b) Shafts AB and CD are solid shafts. 

/HW d = Diameter of the two shafts Again from torsion formula T W = (d / 2) J Both solid and hollow shafts are subjected to the same torque of 5 kN¹m. 5 π 4 d 32 Solving for d, we get

50 u 106 1 u = 103 (d / 2)

d = 0.08 m = 80 mm

Ans.

326

Strength of Materials

Example 7.2 Three pulleys are connected by two shafts as shown in Fig. 7.5. Assuming that both shafts are solid, ¿QGWKHPD[LPXPVKHDUVWUHVVLQGXFHGLQWKHVKDIWAB and BC. Solution:

Given,

Diameter of the shaft AB,

d1 = 25 mm

Diameter of the shaft BC,

d2 = 35 mm

/HQJWKRIWKHVKDIW AB /HQJWKRIWKHVKDIWBC

= 1.0 m = 1.5 m

Fig. 7.5

The torque exerted on the shaft AB is TAB = 0.5 kN¹m (clockwise) Using torsion formula, we have τ max TAB = ⎛ d1 ⎞ J ⎜ ⎟ ⎝2⎠ or

τ TAB = max π ⎛ d1 ⎞ × d14 ⎜ ⎟ 32 ⎝2⎠

Hence, the maximum shear stress in the shaft AB is given as Wmax = =

TAB × 32 π d13 × 2 0.5 × 32 3

⎛ 25 ⎞ π ×⎜ ⎟ ×2 ⎝ 1000 ⎠

= 162.97 MPa

The torque exerted on the shaft BC is TBC = 1.5 kN.m (clockwise).

Ans.

Torsion of Circular Members

327

The maximum shear stress in the shaft BC is given as T × 32 Wmax = BC 3 π d2 × 2 =

1.5 × 32 3

⎛ 35 ⎞ π ×⎜ ⎟ ×2 ⎝ 1000 ⎠

= 178.18 MPa

Ans.

Example 7.3 Four pulleys are connected by solid shafts as the shown in Fig. 7.6. Torque acting on each pulley is VKRZQLQWKH¿JXUH:KLFKVKDIWH[SHULHQFHVWKHPD[LPXPVKHDUVWUHVV"$OVR¿QGWKHPDJQLWXGH of that stress.

Fig. 7.6

Solution:

Refer Fig. 7.6.

Given, Shaft parameters Diameter  /HQJWK Torque



Shaft AB

Shaft BC

15 mm P 5 N¹m

20 mm P (20 – 5) = 15 N¹m



(clockwise)

(anticlockwise)

Shaft CD 25 mm  P (20 + 40 – 5) = 55N¹m (anticlockwise)

Using torsion formula, W is given as T d W = u J 2 Hence, the shear stress induced in the shaft AB is WAB =

⎛ 15 ⎞ ×⎜ ⎟ π ⎛ 15 ⎞ ⎝ 2 × 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ 5

= 7.54 MPa

4

Shaft DE 

30 mm P (20 + 40 – 5 – 50) = 5N¹m (anticlockwise)

328

Strength of Materials

Similarly, the shear stress induced in the shaft BC is WBC =

⎛ 20 ⎞ ×⎜ ⎟ = 9.55 MPa π ⎛ 20 ⎞ ⎝ 2 × 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ 15

4

Shear stress in the shaft CD is WCD =

⎛ 25 ⎞ ×⎜ ⎟ = 17.92 MPa π ⎛ 25 ⎞ ⎝ 2 × 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ 55

4

And the shear stress in the shaft DE is ⎛ 30 ⎞ ×⎜ ⎟ = 0.94 MPa π ⎛ 30 ⎞ ⎝ 2 × 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ Since WCD is the biggest among all the values, hence the maximum shear stress is induced in the shaft CD and its magnitude is 17.92 MPa. Ans. 5

WDE =

4

Example 7.4 An electric motor exerts a torque of 2.5 kN¹m at D as shown in Fig. 7.7. Find the maximum shear stress induced in the shafts AB, BC and CD, assuming that they are all solid. Solution:

The details of each shaft are given below.

Fig. 7.7 Shaft parameters

Shaft AB

Shaft BC

Shaft CD

Diameter /HQJWK Torque

30 mm P 0.6 kN¹m (clockwise)

35 mm P (0.6 + 0.7) = 1.3 kN¹m (clockwise)

40 mm P  (0.6 + 0.7 + 1.2) = 2.5 kN¹m (clockwise)







Torsion of Circular Members

329

For shaft AB The maximum shear stress induced in the shaft is T d W = u J 2 0.6 1 ⎞ ⎛ 30 ×⎜ × = ⎟ 4 π ⎛ 30 ⎞ ⎝ 2 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ = 113.17 MPa

π 4⎞ ⎛ ⎜ as J = d ⎟ 32 ⎝ ⎠ Ans.

For shaft BC The maximum shear stress induced in the shaft is W =

1 ⎞ ⎛ 35 ×⎜ × ⎟ π ⎛ 35 ⎞ ⎝ 2 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ 1.3

4

= 154.42 MPa

Ans.

For shaft CD The maximum shear stress induced in the shaft is W =

1 ⎞ ⎛ 40 ×⎜ × ⎟ = 198.94 MPa π ⎛ 40 ⎞ ⎝ 2 1000 ⎠ ×⎜ ⎟ 32 ⎝ 1000 ⎠ 2.5

4

Ans.

Example 7.5 An electric motor running at 600 rpm exerts a torque of 3 kN¹m at D as shown in Fig. 7.8. Find angle of twist between (a) B and C and (b) between B and D. Take G = 80 GPa.

Fig. 7.8

330

Strength of Materials

Solution:

Refer Fig. 7.8.

The details of each shaft are given below. Shaft parameters Diameter  /HQJWK Torque

Shaft BC 

Shaft CD

40 mm P 1.2 kN¹m (clockwise)

G = 80 GPa =



60 mm P 1.2 + 1.8 = 3 kN¹m (clockwise)

80 × 109 = 80 × 106 kPa 103

Using torsion formula, T is given as Tl T = JG Hence, the angle of twist produced in the shaft BC is given as 1.2 × 2 TBC = 4 π ⎛ 40 ⎞ 6 ×⎜ ⎟ × 80 × 10 32 ⎝ 1000 ⎠ = 0.1193 radian = 6.84o

(anticlockwise)

Ans.

(anticlockwise)

Ans.

(anticlockwise)

Ans.

The angle of twist produced in the shaft CD is given as 3 × 1.5 TCD = 4 π ⎛ 60 ⎞ 6 ×⎜ ⎟ × 80 × 10 32 ⎝ 1000 ⎠ = 0.0442 radian = 2.53o Hence, the angle of twist between B and D is TBD = TBC + TCD = (6.84 + 2.53)o = 9.37o Example 7.6 Find angle of twist between A and B and between A and C for the arrangement of pulleys and shafts shown in Fig. 7.9. Take G = 80 GPa.

Fig. 7.9

Torsion of Circular Members

Solution:

331

Refer Fig. 7.9.

The details of each shaft are given below. Shaft parameters

Shaft AB

Shaft BC

Diameter Length Torque

40 mm 2.5 m 2 kN¹m (clockwise)

50 mm 3.5 m 3 kN¹m (clockwise)

80 u 109 = 80 – 106 kPa 103 Using torsion formula, the angle of twist for the shaft AB is given as Tl TAB = JG 2 × 2.5 = 4 π ⎛ 40 ⎞ 6 ×⎜ ⎟ × 80 × 10 32 ⎝ 1000 ⎠ G = 80 GPa =

= 0.248 radian = 14.25o (clockwise)

Ans.

The angle of twist for the shaft BC is given as 3 × 3.5 TBC = 4 π ⎛ 50 ⎞ 6 ×⎜ ⎟ × 80 × 10 32 ⎝ 1000 ⎠ = 0.214 radian = 12.25o (clockwise) Hence, the angle of twist between A and C is given as TAC = TAB + TBC = 14.25o + 12.25o = 26.5o

(clockwise)

Ans.

Example 7.7 Determine angle of twist between A and C and between A and E for the problem given in Example 7.3. Take G = 220 GPa. Solution:

The angles of twist for different shafts are given in the following table.

Shaft AB

Angle of twist (T) TAB

=

5 × 1.8 π ⎛ 15 ⎞ 9 ×⎜ ⎟ × 220 × 10 32 ⎝ 1000 ⎠ 4

×

Tl ⎞ 180 ⎛ ⎜⎝ using θ = ⎟ π JG ⎠

= 0.471o (anticlockwise) BC

TBC

=

15 × 1.2 4

π ⎛ 20 ⎞ 9 ×⎜ ⎟ × 220 × 10 32 ⎝ 1000 ⎠

×

180 = 0.298o (clockwise) π Contd...

332

Strength of Materials

CD

TCD

=

DE

TDE

=

55 × 1.0 4

π ⎛ 25 ⎞ 9 ×⎜ ⎟ × 220 × 10 32 ⎝ 1000 ⎠ 5 × 1.0 4

π ⎛ 30 ⎞ 9 ×⎜ ⎟ × 220 × 10 32 ⎝ 1000 ⎠

×

180 = 0.373o (clockwise) π

×

180 = 0.016o (clockwise) π

The angle of twist between A and C is given as TAC = TAB – TBC = (0.471– 0.298)o = 0.173o (anticlockwise)

Ans.

The angle of twist between A and E is given as TAE = TAB – (TBC + TCD + TDE) = 0.471o – (0.298o + 0.373o + 0.016o) = – 0.216o = 0.216o (clockwise)

Ans.

Example 7.8 A stepped steel shaft is shown in Fig. 7.10. A torque of 100 N¹m is acting at C and another torque of 200 N¹m is acting at a distance 2 m from A. Determine angular displacement of the free end, if the maximum shear stress in the shaft is limited to 50 MPa. Take G = 100 GPa.

Fig. 7.10

Solution: For shaft BC  /HQJWK

l1 = 3 m

Diameter,

d1 = 40 mm

Torque,

T1 = 100 N¹m (clockwise)

Using torsion equation, we have T GT = J l T l or TBC = 1 1 J1 G

Torsion of Circular Members

=

100 × 3 4

π ⎛ 40 ⎞ 9 ×⎜ ⎟ × 100 × 10 32 ⎝ 1000 ⎠

For shaft AB /HQJWK

l2 = 4 m

Diameter,

d2 = 80 mm

Torque,

T2 = (200 – 100) = 100 N¹m

Now

TAB = =

T2 u 2 J2 G

×

333

180 degree = 0.684o (clockwise) π

(anticlockwise)

(Since torque is acting at a distance of 2 m from A ) 100 × 2 4

π ⎛ 80 ⎞ 9 ×⎜ ⎟ × 100 × 10 32 ⎝ 1000 ⎠

×

180 degree π

= 0.028o (anticlockwise) Hence, the angular displacement of the free end or angle of twist between A and C is given as TAC = (0.684 – 0.028)o = 0.656o (clockwise)

Ans.

Example 7.9 A stepped steel shaft shown in Fig. 7.11, is subjected to a torque of 100 N¹m (anticlockwise) at C and another torque of 200 N¹m (clockwise) at B. Determine angle of twist at the free end. Shear stress in the shaft is not to exceed 60 MPa and modulus of rigidity is 84 GPa.

Fig. 7.11

Solution: For shaft BC /HQJWK

l1 = 2 m

Diameter,

d1 = 40 mm

Torque,

T1 = 100 N¹m (Anticlockwise)

Using torsion equation, the angle of twist in the shaft BC is given as TBC =

T1 l1 J1 G

334

Strength of Materials

=

100 × 2 4

π ⎛ 40 ⎞ 9 ×⎜ ⎟ × 84 × 10 32 ⎝ 1000 ⎠

×

180 = 0.542o (anticlockwise) π

For shaft AB /HQJWK

l2 = 4 m

Diameter,

d2 = 80 mm

Torque,

T2 = 200 – 100 = 100 N¹m (clockwise)

The angle of twist in the shaft AB is given as TAB = =

T2 l2 J2 G 100 × 4 4

π ⎛ 80 ⎞ 9 ×⎜ ⎟ × 84 × 10 32 ⎝ 1000 ⎠

×

180 = 0.067o (clockwise) π

Hence, the net angle of twist at the free end of the shaft = TBC – TAB = (0.542 – 0.067)o = 0.475o (anticlockwise)

Ans.

Example 7.10 $FRPSRVLWHVKDIWPDGHRIVWHHODQGDOXPLQLXPLV¿[HGDWA and C. A torque of 100 N¹m is applied at point B (Fig. 7.12). Determine the following: (a) the resisting torques induced at the supports (b) the maximum shear stress induced in each shaft. Take Gs = 80 GPa and GAl = 30 GPa.

Fig. 7.12

Solution: For shaft AB /HQJWK

lS = 4 m

Diameter,

dS = 80 mm

For shaft BC /HQJWK

lAl = 2 m

Diameter,

dAl = 40 mm

Torsion of Circular Members

335

/HW TA and TC be the torques induced (anticlockwise) at ends A and C respectively. Given,

TA + TC = 100 N¹m

... (1)

6LQFHWKHHQGVRIWKHVKDIWDUH¿[HGKHQFHWKH\FDQ¶WURWDWH TC/A = Angle of twist of end C w.r.t. A =0 TC × l Al TC × lS 100 × lS + − =0 J Al GAl J S GS J S GS

... (2)

⎤ ⎡ ⎥ ⎢ 4 2 100 × 4 ⎥ = ⎢ + TC 4 4 4 ⎥ ⎢ π ⎛ 40 ⎞ π ⎛ 80 ⎞ π ⎛ 80 ⎞ 9 9 9 80 10 30 10 × × × × × × ⎥ ⎢ ×⎜ ⎟ ⎜ ⎟ ⎜ ⎟ × 80 × 10 32 ⎝ 1000 ⎠ 32 ⎝ 1000 ⎠ ⎥⎦ ⎢⎣ 32 ⎝ 1000 ⎠ On solving, we get

TC = 4.47 N¹m

Ans.

From equation (1), we have TA = 100 – TC = 100 – 4.47 = 95.53 N¹m

Ans.

The maximum shear stress induced in the aluminium part is given as

WAl

⎛d ⎞ TC × ⎜ Al ⎟ ⎝ 2 ⎠ = J Al ⎛ 40 ⎞ 4.47 × ⎜ ⎟ ⎝ 2000 ⎠ 5 = 4 = 3.55 – 10 Pa π ⎛ 40 ⎞ ×⎜ ⎟ 32 ⎝ 1000 ⎠

Ans.

The maximum shear stress induced in the steel part is given as ⎛d ⎞ TA × ⎜ s ⎟ ⎝ 2⎠ Ws = Js ⎛ 80 ⎞ 95.53 × ⎜ ⎟ ⎝ 2000 ⎠ = = 9.50 – 105 Pa 4 π ⎛ 80 ⎞ ×⎜ ⎟ 32 ⎝ 1000 ⎠

Ans.

336

Strength of Materials

Example 7.11 $  P ORQJ KROORZ VWHHO VKDIW RI RXWVLGH GLDPHWHU  PP LV WUDQVPLWWLQJ D SRZHU RI  N: DW 1000 rpm. Find thickness of the shaft, if the maximum shear stress in shaft is limited to 40 MPa. Take G = 80 GPa. Solution: Given, l =3m

/HQJWKRIWKHVKDIW Outside diameter of the shaft,

d0 = 75 mm

Power to be transmitted,

P N:

Revolutions per minute,

N = 1000

Maximum shear stress in the shaft,

Wmax = 40 MPa

Using power equation, T is given as P × 60 × 1000 T = 2π N =

τ T = max J ⎛ d0 ⎞ ⎜ 2⎟ ⎝ ⎠ τ T = max π ⎛ d0 ⎞ (d o4 − di4 ) ⎜ 2⎟ 32 ⎝ ⎠

Now

or

150 × 60 × 1000 = 1432.4 N¹m 2π × 1000

1432.4 4 ⎤ π ⎡⎛ 75 ⎞ 4 ⎢⎜ ⎟ − di ⎥ 32 ⎢⎣⎝ 1000 ⎠ ⎥⎦

Solving for di, we get

=

40 × 106 = 1.06 × 109 ⎛ 75 ⎞ ⎜ ⎟ ⎝ 2 × 1000 ⎠

di = 0.0651 m = 65.1 mm

Hence, the thickness of the hollow shaft is t=

75  65.1 d o  di = 4.95 mm = 2 2

Ans.

Example 7.12 $PORQJVROLGVKDIWWUDQVPLWVN:DWUSP)LQGWKHUHTXLUHGGLDPHWHURIWKHVKDIWDVVXPLQJ that maximum shear stress in the shaft is limited to 25 MPa and angle of twist is not to exceed 5 o. Take G = 80 GPa. Solution: Given, /HQJWKRIWKHVKDIW

l =3m

Torsion of Circular Members

337

Power being transmitted, P N: Revolutions per minute,

N = 1200

Maximum shear stress, Wmax = 25 MPa = 25 – 106 Pa

π × 5 = 0.087 radian 180

Angle of twist,

T = 5o =

Modulus of rigidity,

G = 80 GPa = 80 – 109 Pa

/HWd be diameter of the shaft. Using power equation, T is given as T =

15 × 60 × 1000 P × 60 × 1000 = = 119.36 N¹m 2π × 1200 2π N

Using torsion formula, we have T GT = J l

T GT = π 4 l d 32 119.36 80 u 109 u 0.087 = π 4 3 d 32 Solving for d, we get

d = 26.9 mm

τ max T = ⎛d ⎞ J ⎜ ⎟ ⎝2⎠ 2 × τ max T = π 4 d d 32

Again

T × 32 = 2 – Wmax πd3 119.36 × 32 = 2 – 25 – 106 πd3 Solving for d, we get d = 29 mm Of the two calculated values of dZHDFFHSWWKHODUJHVWRQH:K\"+HQFHGLDPHWHURIWKHVKDIW is 29 mm. Ans.

338

Strength of Materials

Example 7.13 A solid shaft of length 3.5 m and diameter 25 mm rotates at a frequency of 40 Hz. Find the maximum power to be transmitted by the shaft, assuming that maximum shear stress in the shaft is limited to 40 MPa and angle of twist does not exceed 6o. Take G = 80 GPa. Solution:

Given,

/HQJWKRIWKHVKDIW

l = 3.5 m

Diameter of the shaft,

d = 25 mm

Revolutions per minute,

N = 40 – 60 = 2400

Maximum shear stress, Wmax = 40 MPa = 40 – 106 Pa π × 6 = 0.104 radian 180

Angle of twist,

T = 6o =

Modulus of rigidity,

G = 80 GPa = 80 – 109 Pa

From torsion formula, we have T GT = J l π 4 d ×G×θ JGT 32 = T = l l 4

Again

π ⎛ 25 ⎞ 9 ×⎜ ⎟ × 80 × 10 × 0.104 32 ⎝ 1000 ⎠ = = 91.16 N¹m 3.5 Wmax T = (d / 2) J

or

J × 2 × τmax d π 4 d × 2 × τmax π 3 d × 2 × τmax = 32 = 32 d

T =

3

π ⎛ 25 ⎞ 6 ×⎜ = ⎟ × 2 × 40 × 10 = 122.71 N¹m 32 ⎝ 1000 ⎠ Of the two calculated values of T, we take the lowest one i.e., the accepted torque is 91.16 N¹m. Now

P = =

2π NT 60 × 1000 2π × 2400 × 91.16  N: 60 × 1000

Ans.

Torsion of Circular Members

339

Example 7.14 $PORQJVROLGVWHHOVKDIWWUDQVPLWVDSRZHURIN:DWUSP)LQGWKHVPDOOHVWSHUPLVVLEOH GLDPHWHURIWKHVKDIWLIWKHPD[LPXPVKHDUVWUHVVLVQRWWRH[FHHG03D$OVR¿QGWKHFRUUHVSRQGLQJ angle of twist. Take G = 80 GPa. Solution: Given, /HQJWKRIWKHVKDIW

l =3m

Power to be transmitted,

P N:

Revolutions per minute,

N = 2000

Maximum shear stress, Wmax = 50 MPa Modulus of rigidity,

G = 80 GPa = 80 – 109 Pa

/HWT be the torque applied on the shaft. Using power equation, T is given as P × 60 × 1000 2πN 20 × 60 × 1000 = = 95.5 N¹m 2π × 2000

T =

Using torsion formula, we have Wmax T (where, d is diameter of the shaft) = ( d / 2) J or

or

W T = max d π 4 d 2 32 50 u 106 95.5 = π 4 (d / 2) d 32

Solving for d, we get d = 21.34 mm Again

Ans.

T GT = J l

GT T = π 4 l d 32 95.5 80 × 109 × θ = 4 3 π ⎛ 21.34 ⎞ ⎜ ⎟ 32 ⎝ 1000 ⎠ Solving for T, we get T = 0.175 radian = 10o

Ans.

340

Strength of Materials

Example 7.15 $ VROLG FLUFXODU VKDIW WUDQVPLWV D SRZHU RI  N: DW  USP )RU D OLPLWLQJ VKHDU VWUHVV RI 03D¿QGGLDPHWHURIWKHVKDIW7KHVROLGVKDIWLVUHSODFHGE\DKROORZVKDIWZLWKLQVLGHGLDPHWHU EHLQJHTXDOWRWLPHVWKHRXWVLGHGLDPHWHU:KDWSHUFHQWRIVDYLQJLQZHLJKWFDQEHDFKLHYHG by the replacement, if both shafts are of equal length, made of the same material, being subjected to HTXDOPD[LPXPVKHDUVWUHVVDQGDUHWUDQVPLWWLQJHTXDOSRZHUVDWVDPHVSHHG" Solution: Given, Power to be transmitted, Revolutions per minute,

P N: N = 1500

Maximum shear stress, Wmax = 70 MPa = 70 – 106 Pa. /HWT be the torque applied on the shaft. Using power equation, T is given as T = =

P × 60 × 1000 2πN 100 × 60 × 1000 = 636.62 N¹m 2π × 1500

... (1)

From torsion formula, we have Wmax T = ( d / 2) J 2 × τmax T = π 4 d d 32 636.62 6 π 3 = 2 – 70 – 10 d 32 Solving for d, we get d = 35.91 mm  /HW

Ans.

do = Outside diameter of the hollow shaft di = Inside diameter of the hollow shaft di = 0.75do (Given)

Since two shafts are transmitting the same power at equal speed, hence they are being subjected to same torque. For hollow shaft

J × τmax × 2 do π 4 (d 0 − di4 ) × τmax × 2 32 = d0

TH =

Torsion of Circular Members

341

4 π 4 ⎡ ⎛ di ⎞ ⎤ d 0 ⎢1 − ⎜ ⎟ ⎥ × τmax × 2 32 ⎢ ⎝ d 0 ⎠ ⎥ ⎣ ⎦ = d0

=

π 3⎡ 4 d 0 1 − (0.75 ) ⎤ × 70 × 106 × 2 = 9395632.3d03 ⎣ ⎦ 32

9395632.3 d03 = 636.62

Now

(using equation (1))

Solving for do, we get d0 = 40.76 mm and

di = 0.75 d0 = 30.57 mm

If density of the material for the two shafts is U, then :HLJKWRIKROORZVKDIW WH = Density – Area – length – g

 :HLJKWRIVROLGVKDIW

π 2 2 = ρ × (d 0 − di ) × l × g 4 π 2 WS = ρ × d × l × g 4

The percentage saving in the weight is WS  WH – 100 WS d 2 − (d 02 − di2 ) × 100 = d2 =

(35.91) 2 − (40.762 − 30.57 2 ) × 100 = 43.63% (35.91)2

Ans.

Example 7.16 A hollow shaft of diameter ratio 0.6 running at 150 rpm is required to drive a screw propeller fitted to a vessel, whose speed is 10 m/s for an expenditure of 12000 shaft horse power. The efficiency of the propeller is 70%. Determine the shaft diameter, if the maximum shearing stress in the shaft is 80 N/mm 2. Solution:

Given

Revolution per minute,

N = 150 rpm

Speed of the vessel,

V = 10 m/s

Power consumption,

P = 12000 HP

(I¿FLHQF\RIWKHSURSHOOHU K = 70% = 0.70 Maximum shear stress,

Wmax = 80 N/mm2

342

Strength of Materials

The power equation is P = 12000 = or

2πNT 1 × 4500 η 2π × 150 × T 1 × 4500 0.70

12000 × 4500 × 0.70 kgf.m 2π × 150 = 40107 kgf.m

T =

= 40107 × 9.81 N.m = 393449.7 N.m 

/HW

do = Outside diameter of the shaft di = Inside diameter of the shaft

di = 0.6 (Given) do The polar moment of inertia of the shaft is found as J =

π 4 (d 0 − di4 ) 32

4 π 4 ⎧⎪ ⎛ di ⎞ ⎫⎪ d 0 ⎨1 − ⎜ ⎟ ⎬ = 32 ⎪ ⎝ d 0 ⎠ ⎪ ⎩ ⎭ π 4 4 d 0 {1 − (0.6) } = 0.0854 d 04 = 32 The maximum shear stress is given as

T d0 u J 2 393449.7 u 1000 d 0 u 80 = 0.0854d 04 2 d0 = 306.5 mm

Wmax = or Solving for d0, we get

Ans.

di = 0.6 d0 = 183.9 mm

Ans.

Example 7.17 Two shafts are made of same material and are of equal lengths. One of them is solid and another one is hollow. The ratio of inside and outside diameters for the hollow shaft is 0.65. They are subjected to the same torque and same maximum shear stress. Compare the weights of the two shafts. Solution: /HW

d = Diameter of the solid shaft do = Outside diameter of the hollow shaft di = Inside diameter of the hollow shaft

Torsion of Circular Members

di = 0.65 do

343

(Given)

From torsion formula, we have Wmax T = R J Since the torque and maximum shear stress for the two shafts are same, hence T J = Constant = Wmax R In terms of respective diameters, we have JS JH = ⎛d ⎞ ⎛ do ⎞ ⎜ ⎟ ⎜ 2 ⎟ 2 ⎝ ⎠ ⎝ ⎠ π 4 π 4 d (d o − di4 ) 32 32 = ⎛d ⎞ ⎛ do ⎞ ⎜ ⎟ ⎜ 2 ⎟ 2 ⎝ ⎠ ⎝ ⎠ ⎡ ⎛ d ⎞4 ⎤ i d3 = d ⎢1 − ⎜ ⎟ ⎥ d ⎢⎣ ⎝ o ⎠ ⎥⎦ 3 o

⎛d ⎞ ⎜ ⎟ ⎝ do ⎠ or

3

= 1 – (0.65)4 = 0.821

d = 0.936 do

... (1)

For the same material and same length, weight is proportional to the cross-sectional area of the respective shaft. π 2 d WS 4 = π 2 WH (d o − di2 ) 4 d2 = ⎡ ⎛ d ⎞2 ⎤ 2 d o ⎢1 − ⎜ i ⎟ ⎥ ⎢⎣ ⎝ d o ⎠ ⎥⎦ =

(d / d o )2 ⎛d ⎞ 1− ⎜ i ⎟ ⎝ do ⎠

= 1.517

2

=

(0.936) 2 1  (0.65) 2

(using equation (1))

Ans.

344

Strength of Materials

Example 7.18 A solid circular shaft 200 mm in diameter has the same cross-sectional area as a hollow circular shaft of the same material with inside diameter of 150 mm. For the same maximum shear stress, determine the ratio of torque transmitted by the hollow shaft to that by the solid shaft. Also, compare the angle of twist in the above shaft for equal length and same maximum shear stress. From the above results, ZKDWZLOOEH\RXUFRQFOXVLRQVUHJDUGLQJVWUHQJWKDQGVWLIIQHVVRIWKHWZRVKDIWV" Solution: Given, Diameter of the solid shaft,

d = 200 mm

Inside diameter the hollow shaft, Di = 150 mm /HW



D0 = Outside diameter of the hollow shaft l

/HQJWKRIWKHWZRVKDIWV

G = Modulus of rigidity for the two shafts TS = Torque transmitted by the solid shaft TH = Torque transmitted by the hollow shaft TS = Angle of twist produced in the solid shaft TH = Angle of twist produced in the hollow shaft Wmax = Maximum shear stress for the two shafts JS = Polar moment of inertia for the solid shaft JH = Polar moment of inertia for the hollow shaft The cross-sectional areas of the solid and hollow shafts are equal. π 2 π 2 2 = ( D0 − Di ) d 4 4 d 2 = D 02 – D i2 2002 = D02 – 1502 Solving for D0, we get

D0 = 250 mm

The torque transmitted by the solid shaft is given as Ts =

Wmax · Js ( d / 2)

=

πd 3 τmax 16

τ max ⎞ ⎛ T ⎜⎝ using J = (d / 2) ⎟⎠ ... (1)

The torque transmitted by the hollow shaft is given as TH =

τmax · JH ⎛ D0 ⎞ ⎜ 2 ⎟ ⎝ ⎠

π( D04 − Di4 ) τmax = 16 D0

... (2)

Torsion of Circular Members

345

Dividing equation (2) by equation (1), we have D04  Di4 TH = D0 · d 3 TS = Now

TS

and

TH TH TS

Hence,

2504 − 1504

250 × 2003 τ l = max × ( d / 2) G τmax l × = ( D0 / 2) G =

= 1.7

Ans.

d 200 = = 0.8 D0 250

Ans.

Conclusions: (a) More torque transmission means more stronger the shaft is. Hence, the hollow shaft is 1.7 times stronger than the solid shaft. (b /HVVHUDQJOHRIWZLVWSURGXFHGLQWKHVKDIWPHDQVPRUHVWLIIHUWKHVKDIWLV6LQFHTH < TS, hence the hollow shaft is stiffer than the solid shaft. Example 7.19 A solid steel shaft of diameter 40 mm is placed inside an aluminium tube and both of them DUH FRQQHFWHG WR D ¿[HG VXSSRUW DW RQH HQG 7KH RWKHU HQG RI WKH WZR PHPEHUV DUH FRQQHFWHG by a rigid plate as shown in Fig. 7.13. Find maximum torque to be applied to the plate, if maximum shear stresses in the steel shaft and the aluminium tube are limited to 110 MPa and 65 MPa respectively. Take GS = 80 GPa and GAl = 28 GPa.

Fig. 7.13

Solution: Given, For steel shaft Diameter, Shear modulus,

d = 40 mm GS = 80 GPa = 80 – 109 Pa

For aluminium tube Outside diameter, Thickness, Inside diameter,

d0 = 60 mm t = 5 mm di = do – 2t = 60 – 2 – 5 = 50 mm Contd...

346

Strength of Materials

Maximum shear stress, Wmax = 110 MPa s = 110 – 106 Pa



/HQJWK ls PP 

Shear modulus,



GAl = 28 GPa = 28 – 109 Pa Maximum shear stress, Wmax = 65 MPa Al = 65 – 106 Pa /HQJWK lAl = 450 mm

The torque applied is distributed to the shaft and tube.  /HW T1 = Torque exerted by the tube on the plate T2 = Torque exerted by the shaft on the plate T = Net torque to be applied on the plate T1 = Angle of twist produced in the tube T2 = Angle of twist produced in the shaft J1 = Polar moment of inertia of the tube J2 = Polar moment of inertia of the shaft π 4 (d o − di4 ) J1 = 32 π (604 − 504 ) × 10−12 m 4 = 6.58 – 10–7 m4 = 32 π 4 π d = × 404 × 10−12 m4 = 2.51 – 10–7 m4 J2 = 32 32 Now and

T = T1 + T2

... (1)

T1 = T2

... (2)

From torsion formula, we have Tl JG T2 lS = J 2 GS

T = T1 l Al J1 GAl

⎛ 450 ⎞ ⎛ 450 ⎞ T2 × ⎜ T1 × ⎜ ⎟ ⎟ ⎝ 1000 ⎠ ⎝ 1000 ⎠ = 2.51 × 10−7 × 80 × 109 6.58 × 10−7 × 28 × 109 or

T1 = 0.917 T2

... (3)

Check for shear stresses Using torsion formula, the torque T1 =

2 × τmax Al × J1 do

=

2 × 65 × 106 × 6.58 × 10−7 60 × 10−3

= 1427.3 N¹m

Torsion of Circular Members

347

The corresponding value of T2, by using equation (3), is given as T1 1427.3 T2 = = = 1556.48 N¹m 0.917 0.917 The maximum shear stress induced in the shaft, by using T2, is given as T ud Wmax = 2 s 2J2 1556.48 × 40 × 10−3 = N/m2 = 124 MPa −7 2 × 2.51 × 10 Since 124 MPa > 110 MPa i.e., maximum shear stress induced in the shaft is more than its allowable value, which is wrong. Hence, we accept Wmaxs = 110 MPa. :H¿QGT2 corresponding to Wmaxs = 110 MPa T2 = =

2 × τmax s × J 2 d 2 × 110 × 106 × 2.51 × 10−7 40 × 10−3

= 1380.5 N¹m

T1 = 0.917T2 = 0.917 – 1380.5 = 1265.9 N¹m

and

(using equation (3))

Hence, the net torque to be applied on the plate is T = T1 + T2 = (1265.9 + 1380.5) = 2646.4 N¹m

Ans.

Example 7.20 $KRUL]RQWDOVKDIWPLQOHQJWKLV¿[HGDWLWVHQGV )LJ :KHQYLHZHGIURPWKHOHIWHQGD[LDO couples of 50 kN.m clockwise and 75 kN.m counter clockwise act at 5 m and 9 m from the left end UHVSHFWLYHO\'HWHUPLQHWKHHQG¿[LQJFRXSOHVDQGWKHSRVLWLRQZKHUHWKHVKDIWVXIIHUVQRDQJXODUWZLVW

Fig. 7.14

Solution: Refer Fig. 7.14. *LYHQ

/HQJWKRIWKHVKDIW

l = 12 m

Couple at B,

TB = 50 kN.m (Clockwise)

Couple at C,

TC = 75 kN.m (Anticlockwise)

/HWT1 and T2 be the reactive torques at A and D respectively. Hence,

T1 + T2 = 75 – 50 = 25 kN.m

... (1)

348

Strength of Materials

Using torsion formula, we have T =

Tl JG

Angle of twist at C (from left end) is T1 =

5T1 (T1 + 50) × 4 + JG JG

Angle of twist at C (from right end) is T2 =

3T2 JG

T1 = T2

Now

5T1 (T1 + 50) × 4 3T2 + = JG JG JG 5T1 + 4T1 + 200 = 3T2 or and

9T1 – 3T2 + 200 = 0

... (2)

T1 + T2 – 25 = 0

(from equation (1))

Solving equations (1) and (2), we get T2 = 35.42 kN.m and

T1 = – 10.42 kN.m

The negative sign associated with T1 is indicative of its anticlockwise nature. /HWDWDGLVWDQFHx (in BC) from A, angular twist is zero, which means −

10.42 × 5 (50 − 10.42) × ( x − 5) = 0 + JG JG – 52.1 + 39.58x – 197.9 = 0

or

x = 6.32 m

Hence, the shaft suffers no angular twist at 6.32 m from A.

Ans.

Example 7.21 For the shaft loaded as shown in Fig. 7.15, calculate the maximum shear stress induced and the angle of twist for cross-section at A. The modulus of rigidity is G.

Fig. 7.15

Torsion of Circular Members

349

Solution: The algebraic sum of the torques acting on the shaft is zero. TA and TB are the reactive torques at A and B respectively. TA + M – 4M + TB = 0 or

TA + TB = 4M – M = 3M

... (1)

Using principle of superimposition, angular twists produced independently by M, 4M and TB are considered. Then these twists are added algebraically to get the total twist of ‘free end’ BZUW¿[HG end A. This should be zero as the end BLV¿[HG TB × l TB × l TB × 2l TB × 2l M × 2l 4M × 2l 4M × 2l =0 + + + + − − J1G J 2G J 3G J 4G J 4G J 3G J 4G l ⎡ TB TB 2TB 2TB 2 M 8M 8M ⎤ = 0 + + + − − ⎢ + ⎥ G ⎣ J1 J 2 J3 J4 J4 J3 J4 ⎦

or

TB TB 2TB 2TB 2M 8M 8M + + + + − − J1 J 2 J3 J4 J4 J3 J4

=0

⎛1 ⎛ 2 1 2 2 ⎞ 8 8 ⎞ TB ⎜ + + + ⎟+M ⎜ − − ⎟ =0 ⎝ J1 J 2 J 3 J 4 ⎠ ⎝ J 4 J3 J 4 ⎠ ⎡ ⎤ ⎡ ⎤ ⎢ ⎢ 2 8 8 ⎥ =0 1 1 2 2 ⎥ + + + TB ⎢ − − ⎥ +M ⎢ π π π π π π π 4 ⎥⎥ 4 4 4 4 4 4 ⎢ (3d ) ⎢ d ( 2d ) ( 2d ) ( 2d ) d d ⎥ ⎢⎣ 32 ⎢⎣ 32 32 ⎥⎦ 32 32 32 32 ⎦⎥ 2 8 ⎛ ⎞ ⎛1 1 ⎞ TB ⎜ + + + 2 ⎟ + M ⎜ 2 − − 8 ⎟ = 0 16 ⎝ ⎠ ⎝ 81 16 16 ⎠ 2.1998 TB – 6.5 M = 0 Hence,

TB = 2.954 M

and

TA = 3M – TB

(using equation (1))

= 0.046 M Since TB > TA , hence maximum shear stress is induced on account of TB . The maximum shear stress is

Wmax = =

=

TB d u J4 2 TB d × π 4 2 d 32 2.954 M × 32 2πd

3

=

15.04 M d3

Ans.

350

Strength of Materials

The angle of twist is

TA =

= =

TB u 2l = J 4G

TB × 2l π 4 d ×G 32

32 × 2TB × l πd 4G 64 × 2.954 M × l πd 4G

= 60.18 Example 7.22

Ml Gd 4

(on substituting TB) Ans.

The compound shaft shown in Fig. 7.16 is built-in at the two ends. It is subjected to a twisting moment TDWWKHPLGGOH:KDWLVWKHUDWLRRIWKHUHDFWLRQWRUTXHVT1 and T2DWWKHHQGV"

Fig. 7.16

Solution: Refer Fig. 7.16. T1 + T2 = T (Given)

... (1)

:HZLOOXVHSULQFLSOHRIVXSHULPSRVLWLRQIRUVROYLQJWKLVSUREOHP:HFRQVLGHUWKHDQJXODUWZLVWV produced by T and T2 independently, and then add them together algebraically to get the total twist of ‘free end’ BZUW¿[HGHQGA. This should be zero, as the end BLV¿[HGDQGKHQFHFDQ¶WURWDWH T2l Tl Tl + 2 − J 2G J1G J1G

=0

l ⎡ T2 T2 T ⎤ ⎢ + − ⎥ =0 G ⎣ J 2 J1 J1 ⎦ T2 T2 T1 + T2 + − =0 J 2 J1 J1 T2 T2 T1 T2 + − − J 2 J1 J1 J1

=0

(using equation (1))

Torsion of Circular Members

T2 T1  =0 J 2 J1 or

T1 T2

J = 1= J2

π 4 d 1 32 = π 16 ( 2d ) 4 32

351

Ans.

7.6 EFFECT OF STRESS CONCENTRATION In case of stepped shaft, where cross-section of the shaft abruptly changes at any point along its length, stress concentrations are found to occur near the discontinuity (Fig. 7.17).

Fig. 7.17 $VWHSSHGVKDIWXVLQJD¿OOHW

,QWKH¿JXUHA and B are the locations of high stress concentration. These are the weakest points in the shaft. A factor, known as stress-concentration factor (K), is introduced in the torsion analysis RIVXFKVKDIWV )LJ 6WUHVVFRQFHQWUDWLRQVDUHUHGXFHGE\WKHXVHRI¿OOHWVZKHUHFKDQJHLQ cross-section takes place gradually. The stress concentration factor depends upon two factors. (a) The ratio of the diameters of the shaft at two ends (D/d), and r (b  7KHUDWLRRIWKHUDGLXVRIWKH¿OOHWWRWKHVPDOOHVWGLDPHWHURIWKHVKDIW ⎛⎜ ⎞⎟ . ⎝d ⎠

Fig. 7.18 6WUHVVFRQFHQWUDWLRQIDFWRUDQGXVHRI¿OOHW

352

Strength of Materials

7KHPD[LPXPVKHDUVWUHVVLQGXFHGDWWKH¿OOHWORFDWLRQVPD\EHH[SUHVVHGDV K . T. d Wmax = 2J where K = Stress-concentration factor

... (7.18)

d = Diameter at smallest end of the shaft J = Polar moment of inertia based on smallest diameter of the shaft T = Torque Equation (7.18) is valid only if Hooke’s law is obeyed, that is, the maximum shear stress does not exceed the proportional limit of the shaft material. Example 7.23 A stepped shaft shown in Fig. 7.19 is subjected to a torque of 500 N¹m. Find the maximum shear stress LQWKHVKDIWZKHQWKH¿OOHWUDGLXVLVPP

Fig. 7.19

Solution:

Refer Fig. 7.16.

Given, Diameter at the bigger end, Diameter at the smallest end, 5DGLXVRIWKH¿OOHW Torque to be applied,

D d r T

= 50 mm = 25 mm = 2.5 mm = 500 N¹m

r 2.5 = 0.10 = d 25 50 D = = 2.0 25 d

The ratio The ratio

r D = 0.10 and = 2.0, the value of stress concentration factor, K from Fig. 7.15 d d is 1.45. Hence, the maximum shear stress induced in the shaft is given as Corresponding to

Wmax = =

K .T . d 2J

(using equation (7.18))

1.45 × 500 × (25 /1000) π ⎛ 25 ⎞ 2× ×⎜ ⎟ 32 ⎝ 1000 ⎠

4

= 236.31 MPa

Ans.

Torsion of Circular Members

353

Example 7.24 $ VWHSSHG VKDIW VKRZQ LQ )LJ  WUDQVPLWV  N: RI SRZHU DW  USP )LQG WKH VPDOOHVW permissible radius of the fillet, assuming that maximum shear stress in the shaft is limited to 60 MPa.

Fig. 7.20

Solution: Given, Diameter at the bigger end, D = 60 mm Diameter at the smaller end, d = 50 mm Revolutions per minute, N = 850 Power to be transmitted, P N: Maximum shear stress, Wmax = 60 MPa = 60 – 106 Pa /HWT be the torque applied on the shaft. The equation of power is P =

2π NT 60 × 1000

P × 60 × 1000 100 × 60 × 1000 = = 1123.44 N¹m 2πN 2π × 850 The maximum shear stress in the shaft is or

T =

Wmax =

K .T . d 2J

⎛ 50 ⎞ K × 1123.44 × ⎜ ⎟ ⎝ 1000 ⎠ 6 60 × 10 = 4 π ⎛ 50 ⎞ 2× ×⎜ ⎟ 32 ⎝ 1000 ⎠ Solving for K, we get K = 1.31 The ratio

D 60 = = 1.2 d 50

Corresponding to K = 1.31 and

or

D r = 1.2, ratio is found to be 0.11. d d

(using Fig. 7.15)

r = 0.11 d r = 0.11 – d = 0.11 – 50 = 5.5 mm

+HQFHWKHPLQLPXPUDGLXVRIWKH¿OOHWWREHXVHGLVPP

Ans.

354

Strength of Materials

7.7 TORSION OF A TAPERED SHAFT Consider a tapered shaft of diameter d 1 at the bigger end and diameter d 2 at smaller end (Fig. 7.21).  /HW W1 = Shear stress at the bigger end W2 = Shear stress at the smaller end G = Shear modulus of the shaft material T = Torque applied at the two ends l /HQJWKRIWKHVKDIW

Fig. 7.21

Consider a small length dx of diameter d at a distance x from bigger end. The entire section of shaft is subjected to same torque T.

where

π 3 π 3 π 3 d 2 τ2 = d τ d1 τ1 = 16 16 16 W = Shear stress at x

7KHDERYHHTXDWLRQRQVLPSOL¿FDWLRQEHFRPHV d13 W1 = d 23 W2 = d3W The angle of twist produced in the length dx is Tdx dT = JG Tdx 32T dx = π = 4 4 π d G d G 32 The diameter d can be expressed as d = d1 –

(d1 − d 2 ) x l

= d1 – Kx,

⎛ d − d2 ⎞ where K = ⎜ 1 ⎟ ⎝ l ⎠

... (7.19)

T Gθ ⎞ ⎛ ⎜⎝ using = ⎟ J l ⎠ ... (7.20)

Torsion of Circular Members

355

On substituting d, equation (7.20) becomes dT =

32T dx 4

π [d1 − Kx ] G

The total angle of twist for the entire shaft of length l is T =

32T πG

l

∫ (d 0

1

l

dx 4

− Kx )

=

32T −4 (d1 − Kx ) dx ∫ πG 0

l

32T = πG =

⎡ (d − Kx )−3 ⎤ l ⎥ = 32T ⎡(d1 − Kx )−3 ⎤ ⎢ 1 ⎦0 3K 3π GK ⎣ ⎥⎦ ⎢⎣ 0

32T ⎡ −3 −3 (d1 − Kl ) − d1 ⎤ ⎣ ⎦ 3πGK

−3 ⎤ ⎡⎛ d1 − d 2 ) ⎞ 32Tl −3 d − ⋅ l ⎢ = ⎜ 1 ⎟ − d1 ⎥ 3πG (d1 − d 2 ) ⎢⎣⎝ l ⎠ ⎥⎦

=

⎡1 32Tl 32Tl 1 ⎤ ⎡ d 2−3 − d1−3 ⎤ = ⎢ 3 − 3⎥ ⎣ ⎦ 3πG (d1 − d 2 ) 3πG (d1 − d 2 ) ⎢⎣ d 2 d1 ⎥⎦

=

⎡ d13 − d 23 ⎤ 32Tl ⎢ ⎥ 3πG (d1 − d 2 ) ⎢⎣ d13 d 23 ⎥⎦

⎡ (d1 − d 2 ) (d12 + d1 d 2 + d 22 ) ⎤ 32Tl ⎥ = 3πG (d1 − d 2 ) ⎢⎢ d13 d 23 ⎥⎦ ⎣ 32Tl ⎡ d12 + d1 d 2 + d 22 ⎤ = ⎢ ⎥ 3πG ⎢⎣ d13 d 23 ⎥⎦

... (7.21)

Maximum shear stress is induced at smaller end of the shaft because of smallest diameter at that point, and its value is given as Wmax =

16T πd 23

... (7.22)

7.8 TORSION OF A THIN CIRCULAR TUBE Consider a circular tube of outside diameter d 0 and inside diameter d i, being subjected to a torque T (Fig. 7.22). Thickness of the tube is d  di t = 0 2 The polar moment of inertia of the cross-section is found as π π 4 ⎡ d 04 − ( d 0 − 2t )4 ⎤ ( d 0 − d i4 ) = J = ⎦ 32 ⎣ 32

356

Strength of Materials

Fig. 7.22

⎛ π ⎡ 4 2t ⎞ ⎢ d 0 − d 04 ⎜1 − ⎟ = 32 ⎢ ⎝ d0 ⎠ ⎣ πd 04 = 32 =

The torque T is given as

4 πd 04 ⎡ ⎛ 2t ⎞ ⎤ ⎢1 − ⎜1 − ⎟ ⎥ ⎥ = 32 ⎢ ⎝ d 0 ⎠ ⎥ ⎥ ⎣ ⎦ ⎦

4⎤

⎡ ⎛ ⎞⎤ 8t + ... ⎟ ⎥ ⎢1 − ⎜1 − ⎢⎣ ⎝ d 0 ⎠ ⎥⎦

πd 04 8t × 32 d 0

(ignoring higher powers of t)

πd 03 t = 4 T =

2 × τmax × J d0

2 × τmax × πd 03 t 4 × d0 π 2 = d 0 t Wmax 2 The angle of twist produced in the tube is given as =

T =

=

Tl Tl = JG ⎛ πd 03 t ⎞ ⎜⎜ ⎟⎟ × G ⎝ 4 ⎠ 4Tl πd 03 t G

(using torsion formula) (on substituting J) ... (7.23)

(on substituting J)

... (7.24)

7.9 STRAIN ENERGY DUE TO TORSION The strain energy or the workdone due to torsion is given by 1 ... (7.25) U = ×T × θ 2 where torque is assumed to be applied gradually and is increasing from zero to its maximum value.

Torsion of Circular Members

357

From torsion formula, for a solid circular shaft, we have

Hence,

Wmax T GT = = (d / 2) J l J × τmax Ts = (d / 2) =

and

Ts =

π 4 τmax × 2 πd 3 d × τmax = 16 d 32 τmax × l 2 × τmax × l = G × (d / 2) Gd

Substituting Ts and Ts in equation (7.25), we have for a solid shaft 2 × τmax × l 1 πd 3 τmax × Us = × Gd 2 16 τ2max ⎛ π 2 ⎞ τ2max ⎜ d × l ⎟ = 4G × VS 4G ⎝ 4 ⎠ π 2 Vs = 4 d × l = Volume of the solid shaft =

where

... (7.26)

For a hollow shaft of inside diameter di and outside diameter do, the values of T and T are: 4 πd 03 ⎡ ⎛ di ⎞ ⎤ ⎢ 1 − TH = 16 ⎢ ⎜ d ⎟ ⎥⎥ τmax ⎣ ⎝ o⎠ ⎦

and

TH =

2 × τmax × l Gd 0

Substituting T and T in equation (7.25), we get 1 πd 03 UH = × 2 16

⎡ ⎛ d ⎞4 ⎤ 2 × τmax × l ⎢1 − ⎜ i ⎟ ⎥ τmax × Gd o ⎢ ⎝ do ⎠ ⎥ ⎣ ⎦

4 πd 02 ⎡ ⎛ di ⎞ ⎤ τ2max × l ⎢ 1 − ⎜ ⎟ ⎥× = G 16 ⎢ ⎝ d o ⎠ ⎥ ⎣ ⎦

=

where

πd 02 (d o2 − di2 ) (d o2 + di2 ) τ2max × l × × G 16 d o4

2 2 2 ⎡π 2 ⎤ τmax (d o + di ) 2 τ2 × × ( d − d ) × l o i = V × max × 1 = ⎢ ⎥ 2 2d o ⎣4 ⎦ 2G 2G π 2 2 VH = (d o − di ) × l = Volume of the hollow shaft 4

358

Strength of Materials

and

d o2  di2 2d o2

Hence,

= 1, when do ~ di

UH =

τ 2max × VH 2G

... (7.27)

&RPSDULQJHTXDWLRQ  ZLWKHTXDWLRQ  ZH¿QGWKDWDKROORZVKDIWFDQVWRUHDSSUR[LPDWHO\ twice strain energy as compared to a solid shaft, and it is a very important point in favour of a hollow shaft. Example 7.25 A shaft circular in section (Fig. 7.23) and of length l is subjected to a variable torque given by Kx2/l2, where x is the distance measured from one end of the shaft and K is a constant. Find the angle of twist for the shaft by use of Castigliano’s theorem. Torsional rigidity of the shaft is G. Solution: The torque is

T =

Kx 2 . l2

Consider a small length dx of the shaft at a distance x from the right end. The angle of twist, using the Castigliano’s theorem, is given by

Fig. 7.23 l

T =

∫ 0

Tdx = 1 JG JG

1 = JG

l

∫ 0

K = JGl 2

l

∫ Tdx 0

Kx 2 dx l2

(on substituting T)

l



x 2 dx

0

l

K ⎛ x3 ⎞ Kl = ⎟⎟ = 2 ⎜ ⎜ JGl ⎝ 3 ⎠0 3JG

Ans.

Torsion of Circular Members

359

SHORT ANSWER QUESTIONS 1. :KDWHIIHFWLVREVHUYHGZKHQDVKDIWLVWZLVWHG" 2. +RZGRHVWKHVKHDUVWUHVVYDU\ZLWKUDGLXVRIDVKDIW" 3. :KDWDUHWKHYDOXHVRIVKHDUVWUHVVDWWKHQHXWUDOD[LVDQGRQWKHVXUIDFHRIDVKDIW" 4. :KDWLVSRODUPRGXOXV":KDWLVLWV6,XQLW" 5. :KDWLVWRUVLRQDOULJLGLW\":KDWLVLWVVLJQL¿FDQFH" 6. :KDWLVSRODUPRPHQWRILQHUWLD"+RZGRHVLWGLIIHUIURPVHFRQGPRPHQWRIDUHD" 7. :KDWLVWKHUHODWLRQVKLSEHWZHHQSRODUPRPHQWRILQHUWLDDQGVHFRQGPRPHQWRIDUHDIRUDVROLG FLUFXODUVKDIW" 8. +RZGRHVWKHSRZHUWUDQVPLVVLRQYDU\ZLWKWKHUSPRIDVKDIW" 9. :KDWGRHVWKHKRUVHSRZHUPHDQ" 10. +RZDUHNLORZDWWDQGKRUVHSRZHUUHODWHG"

MULTIPLE CHOICE QUESTIONS 1. The shear stress produced in a circular shaft due to pure torsion is (a) proportional to radius of the shaft (b) inversely proportional to diameter of the shaft (c) inversely proportional to radius of the shaft (d) inversely proportional to the square of radius of the shaft. 2. 7KHWRUVLRQDOULJLGLW\LVGH¿QHGDVWKH (a) product of torque and length (b) product of polar moment of inertia and modulus of rigidity (c) sum of polar moment of inertia and modulus of rigidity (d) product of polar moment of inertia and angle of twist. 3. 7KHWRUVLRQDOVWLIIQHVVLVGH¿QHGDVWKH (a) ratio of torque to angle of twist (b) product of torque and angle of twist (c) sum of modulus of rigidity and angle of twist (d) ratio of torque to polar moment of inertia.

360

Strength of Materials

4. The torsion equation is (a)

T GT W = = I L r

T GT W = = J L r J Gθ τ (d) T = L = r . (b)

W T GT = = D J L where the symbols have their usual meanings. (c)

5. $ VROLG VWHHO VKDIW RI OHQJWK  P LV UXQQLQJ DW  USP WUDQVPLWWLQJ D SRZHU RI  N:7KH DQJOHRIWZLVWSURGXFHGLQWKHVKDIWLVƒDQGWKHSHUPLVVLEOHVKHDUVWUHVVLV03D:KDWLVWKH diameter of the shaft, if G = 8 – 104 N/mm2 " (a) 68.7 mm

(b) 82.55 mm

(c) 67.8 mm

(d) 65.7 mm.

6. The ratio of the torques transmitted by a hollow and a solid shaft, both made of same material, length and weight is n2 + 1 n2  1 n2  1 n2 − 1 (b) (a) (c) (d) n n 1 n n +1 n n2 − 1 n n2  1 where n =

do . di

7. The ratio of the weights of two shafts in Question No. 6 is 1 ⎞⎛ 1 ⎞ ⎛ (a) ⎜1 − 2 ⎟ / ⎜1 − 4 ⎟ n ⎠⎝ n ⎠ ⎝

2/3

1 ⎞⎛ 1 ⎞ ⎛ (b) ⎜1 − 2 ⎟ / ⎜1 + 4 ⎟ n ⎠⎝ n ⎠ ⎝

2/3

1 ⎞⎛ 1 ⎞ ⎛ (c) ⎜1 + 2 ⎟ / ⎜1 − 4 ⎟ n ⎠⎝ n ⎠ ⎝

2/3

1 ⎞⎛ 1 ⎞ ⎛ (d) ⎜1 + 2 ⎟ / ⎜1 + 4 ⎟ n ⎠⎝ n ⎠ ⎝

2/3

8. The power transmitted by a shaft is given as (a) πNT :  b) πNT (hp) (c) 2πNT (hp) 30 4500 75 where the symbols have their usual meanings.

.

(d) πNT  N:  30

9. 7KH IRUPXOD XVHG IRU ¿QGLQJ DQJOH RI WZLVW SURGXFHG LQ D WDSHUHG VKDIW RI OHQJWK L being subjected to a torque T is TL ⎡ r1 + r2 + 2r1 r2 ⎤ TL ⎡ r1 + r2 + 2r1 r2 ⎤ ⎢ ⎥ ⎢ ⎥ (b) (a) 3 3πG ⎣ (r1 r2 )3 3πG ⎣ (r1 r2 ) ⎦ ⎦ (c)

2TL ⎡ r12 + r22 + r1 r2 ⎤ ⎢ ⎥ 3πG ⎣ (r1 r2 )3 ⎦

3TL ⎡ r12 + r22 + r1 r2 ⎤ ⎢ ⎥. (d) 2πG ⎣ (r1 r2 )3 ⎦

where r1 and r2 are the radii at smaller and bigger end respectively.

Torsion of Circular Members

361

10. For a stepped shaft where two or more shafts are connected end-to-end, the angle of twist produced and the torques transmitted are given respectively as (a) T1 – T2, T1 + T2 (c)

(b) T1 + T2, T1 = T2

T2 θ1 + θ2 , T1 = 2 2

(d) θ1 − θ2 , T1 = T2 . 2

11. For a composite shaft made of two concentric shafts one solid and other hollow, the torque T is given as T  T2 T (c) T1 + T2 (a) T1 – T2 (b) 1 (d) 1  T2 . 2 2 12. For the composite shaft in Question No. 11, the angle of twist produced is equal to θ1 − θ2 (d) T = T1 = T2. 2 13. For a shaft being subjected to a torque T, the variation of the shear stress w.r.t. its radius is (a) T = T1 + T2

(b) T = T1 – T2

(b) T =

(a) linear

(b) parabolic

(c) hyperbolic

(d) none of these.

14. The shear stress for a shaft being subjected to a torque T is minimum at (a) half of radius from the axis (b) axis of the shaft (c) equal radial distances from the axis (d) its both ends. 15. The polar modulus of a shaft of diameter d is given as 3 4 (b) πd (a) πd 32 64 16. The shear strain in a circular shaft varies

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

3 (c) πd 16

4 (d) πd . 32

linearly with the distance from axis of the shaft inversely proportional to the distance from axis of the shaft inversely proportional to the square of the distance from axis of the shaft linearly with the square of the distance from the shaft.

17. The shear strain is maximum (a) (b) (c) (d)

at a distance equal to one-third from axis of the shaft at the centre of shaft on the surface of shaft none of these.

ANSWERS 1. (a)

2. (b)

10. (b)

11. (c)

3. (a)

4. (b)

5. (b)

6. (b)

12. (d) 13. (a) 14. (b) 15. (c)

7. (a)

8. (a)

16. (a) 17. (c)

9. (c)

362

Strength of Materials

EXERCISES 1. Find the loZHVWVSHHGDWZKLFKN:FRXOGEHWUDQVPLWWHGWKURXJKDVKDIWRIGLDPHWHUPP 7KHPD[LPXPVKHDUVWUHVVLVOLPLWHGWR03D,IOHQJWKRIWKHVKDIWLVP¿QGWKHDQJOHRI twist. Take G = 80 GPa. (Ans.

972.5 rpm, 6.82o).

2. $VROLGFLUFXODUVKDIWWUDQVPLWVDSRZHURIN:DWUSP,IWKHPD[LPXPVKHDUVWUHVVLV OLPLWHGWR3D¿QGGLDPHWHURIWKHVKDIW:KDWSHUFHQWRIVDYLQJLQZHLJKWFDQEHDFKLHYHG if solid shaft is replaced by a hollow shaft with inside diameter being equal to 0.6 times the outside diameter. Both shafts are of equal length, made of same material and are being subjected to the same maximum shear stress. (Ans.

122.15 mm, 29.72%).

3. A hollow shaft of inside diameter 80 mm and outside diameter 120 mm is twisted by 1.5 o. Find WKHUHTXLUHGWRUTXHLIOHQJWKRIWKHVKDIWLVP$OVR¿QGVKHDUVWUHVVDWWKHLQQHUDQGRXWHU surfaces of the shaft. Take G = 80 GPa. (Ans.

17 kN¹m, 41.6 MPa, 62.85 MPa).

4. (a) A hollow circular shaft of inner radius 30 mm and outer radius 50 mm and length 1 m is subjected to a twisting moment so that the angular twist in the shaft is 0.57 o. Find the maximum shear angle in the shaft. (b) A hollow shaft of outer radius 100 mm and inner radius 40 mm is subjected to a twisting moment. The maximum shear stress developed in the shaft is 50 MPa. Find the shear stress at the inner radius of the shaft. (Ans.

(a) 0.028o; (b) 20 MPa).

5. $KROORZVWHHOVKDIWRIRXWVLGHGLDPHWHUPPLVWRWUDQVPLWDSRZHURIN:DWUSP Find its inside diameter, if the maximum shear stress in the shaft is limited to 50 MPa. (Ans. 82.96 mm). 6. Two pulleys are attached to different shafts as shown in Fig. 7.24. The motor attached exerts a torque of 3.0 N¹m at C. Find the maximum shear stress induced in the shafts AB and BC, assuming them to be solid.

Torsion of Circular Members

363

Fig. 7.24

(Ans.

0.586 MPa, 0.356 MPa).

7. Part AB of the shaft shown in Fig. 7.25 has 25 mm diameter and is subjected to a maximum shear stress of 75 MPa. Part BC has 60 mm diameter and is subjected to a maximum shear stress of 03D1HJOHFWLQJWKHHIIHFWRIVWUHVVFRQFHQWUDWLRQ¿QGWKHWRUTXHT1 applied at A and the reaction torque exerted by the support.

Fig. 7.25

(Ans.

230 N¹m, 2120.5 N¹m (anticlockwise)).

8. Two shafts, one solid and another hollow, are made of same material and are of equal length. They are subjected to same torque. The ratio of inside and outside diameters for the hollow shaft is 0.75. If both shafts are subjected to the same maximum shear stress, compare their weights. ⎛ ⎞ WS = 1.77 ⎟ . ⎜ Ans. WH ⎝ ⎠ 9. Find the maximum torque that can be applied safely to a shaft of diameter 300 mm. The permissible angle of twist is 1.5o in a length of 7.5 m and the maximum shear is limited to 42 MPa. Take G = 84.4 GPa. (Ans.

222.6 kN¹m).

364

Strength of Materials

10. $KRUL]RQWDOVKDIWRIOHQJWKPLV¿[HGDWLWVHQGV:KHQYLHZHGIURPWKHOHIWHQGD[LDO couples of 50 kN¹m clockwise and 75 kN¹m anticlockwise act at 5 m and 9 m from the left HQGUHVSHFWLYHO\'HWHUPLQHWKHHQG¿[LQJFRXSOHVDQGWKHSRVLWLRQZKHUHWKHVKDIWVXIIHUVQR angular twist. (Ans. 10.42 kN¹m, Clockwise (left end); 35.42 kN¹m, Anticlockwise (right end); 6.32 m from the left end). 11. A solid circular shaft is subjected to an axial torque T and a bending moment M. If M = kT, ¿QGWKHUDWLRRIWKHPD[LPXPSULQFLSDOVWUHVVWRWKHPD[LPXPVKHDUVWUHVVLQWHUPVRIk. Find the power transmitted by a 50 mm diameter shaft at a speed of 300 rpm, when k = 0.4 and the maximum shear stress is 75 MPa. ⎛ ⎞ k , 57.6 kW⎟ . ⎜ Ans. 1 + ⎜⎝ ⎟⎠ (1 + k 2 )

‰‰‰

8 Springs Robert Hooke, born on 18 July 1635, was an English inventor, microscopist, physicist, surveyor, astronomer, biologist and polymath. He founded the law of elasticity, widely known as Hooke’s Law, in 1660, which forms the basis of design for many engineering structures and components. He originated the word ‘Cell’ in biology, which describes the tiniest component of a living system. He is the inventor of the iris diaphragm in cameras, the universal joint used in the motor vehicles and the balance wheel in a watch. He discovered the red spot of Jupiter and ZDV WKH ¿UVW WR UHSRUW DERXW WKH URWDWLRQ RI WKH SODQHW +H GHYLVHG WKH ZRUOG¶V¿UVWFRPSRXQGPLFURVFRSHWKDWXVHGGLDSKUDJP Robert Hooke (1635-1703)

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ +RZGRHVWKHVSULQJGHÀHFWLRQYDU\ZLWKWKHORDGDSSOLHG"  ‡ :K\LVOHDIVSULQJDOVRFDOOHGFDUULDJHVSULQJ"  ‡ :KDWLV:DKOFRUUHFWLRQIDFWRU"  ‡ :K\LVKHOLFDOVSULQJDOVRFDOOHGWRUVLRQVSULQJ"  ‡ +RZGRHVWKHVSULQJFRPELQDWLRQLQVHULHVDQGSDUDOOHOGLIIHU"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_8

365

366

Strength of Materials

8.1

INTRODUCTION

A spring is a device used to absorb or store energy and release it when required. The deformation produced in the spring is not of permanent nature and vanishes on removal of load due to its elastic QDWXUH ,W ¿QGV DSSOLFDWLRQ LQ PHFKDQLFDO FORFNV VKRFN DEVRUEHUV UHFLSURFDWLQJ PHFKDQLVPV automobile elements etc.

8.2 SPRING TERMINOLOGY z Proof Load : It is the maximum load a spring can be subjected without undergoing permanent deformation. z Proof stress : It is the stress corresponding to the proof load. z Resilience : It is the strain energy stored in the spring when loaded within elastic limit. Once the load is removed, the energy is given up or released. z Proof resilience : It is the maximum strain energy stored in the spring when loaded within elastic limit. z Modulus of Resilience : It is the maximum strain energy stored per unit volume. z Stiffness ,WLVWKHORDGUHTXLUHGWRSURGXFHXQLWGHÀHFWLRQLQWKHVSULQJ 8.3

CLASSIFICATION OF SPRINGS

Various types of springs are employed for different applications. Important types are discussed below. Helical springs are in the form of a helix made of a rod of circular, square or rectangular crosssection, but the circular section is most frequently used. Square or rectangular sections are used in heavy duty springs. The main disadvantages of non-circular wire section springs include non-uniform stress distribution and less energy absorbing capacity of the spring. Helical springs are also known as torsion springs, because they are subjected to torsion. Helical springs may be in the cylindrical or conical form. Cylindrical helical springs are subjected to tension or compression. A conical helical spring is used where there is a space problem or a single spring with a variable stiffness is desired. They are used for KHDY\ORDGVDQGGHÀHFWLRQV+HOLFDOVSULQJVDUHXVHGLQVKRFNDEVRUEHUVUHFLSURFDWLQJPHFKDQLVPVIURQW D[OHVXVSHQVLRQRIYHKLFOHVHWF7KH\DUHFODVVL¿HGLQWRWZRJURXSVGHSHQGLQJXSRQDQJOHRIKHOL[ In the close coiled helical spring, angle of helix is very small and can be approximated to zero so that plane of the coil is at right angle to the axis of the spring. In the open coiled helical spring, plane of the coil of the spring makes certain angle with the horizontal. Depending upon the nature of applications, helical springs can be of compression and tension type. Laminated springs or leaf springs are made of a number of thin curved plates of uniform thickness but different lengths which are placed over each other and clamped together at the centre. They may be of cantilever, semi-elliptical or elliptical type. The most common type of leaf spring is semielliptical leaf spring. For very heavy loads to be supported, it can be elliptical leaf spring. Laminated VSULQJVDUHDOVRFDOOHGFDUULDJHVSULQJVDQG¿QGDSSOLFDWLRQLQWKHVXVSHQVLRQV\VWHPRIIURQWRUUHDU axles of cars, trucks, trains etc. Sometimes they are also known as bending springs because being subjected to bending only.

Springs

367

Flat springs DUHPDGHRIDÀDWVWULSDQGPD\EHLQWKHIRUPRIDFDQWLOHYHURUDVLPSOHEHDP7KH HQGRIWKHÀDWFDQWLOHYHUVSULQJFDQEHJXLGHGDORQJDGH¿QLWHSDWKDVLWGHÀHFWV7KXVWKHVSULQJPD\ act as a structural member as well as an energy absorbing device. Spiral spring FRQVLVWVRIDÀDWVWULSZRXQGLQWKHIRUPRIDVSLUDODQGLVORDGHGLQWRUVLRQ7KH major stresses produced in the spring are tensile and compressive due to bending by the applied load. Various types of springs are shown in Fig. 8.1.

Fig. 8.1 Types of springs.

Sometimes springs are used to support a body that has vibrational problems. Sofa springs are the H[DPSOHVRIFRQLFDOKHOLFDOVSULQJV'HPHULWVRIWKHVSULQJLQFOXGHGLI¿FXOW\LQLWVPDQXIDFWXUHDQG buckling during the use.

8.4

LOAD-DEFLECTION CURVE

,IZHSORWORDGGHÀHFWLRQGLDJUDPIRUDVSULQJDVWUDLJKWOLQHLVREWDLQHGVXJJHVWLQJWKDWGHÀHFWLRQ LVGLUHFWO\SURSRUWLRQDOWRWKHORDGDSSOLHG)RUDQH[DPSOHLIWKHORDGLVGRXEOHGWKHQGHÀHFWLRQ will also be doubled (Fig. 8.2). The straight relationship is valid only if the spring is stressed within elastic limit. It is equally applicable in case when a torque or moment is acting in place of a load and OLQHDUGHÀHFWLRQLVUHSODFHGE\DQJXODUGHÀHFWLRQ

368

Strength of Materials

Fig. 8.2 /RDG9HUVXVGHÀHFWLRQFXUYHIRUDVSULQJ

8.5

LEAF SPRING

A leaf spring, also called laminated spring, is shown in Fig. 8.3. The long leaf or plate fastened to the supports is called the main leaf or master leaf. Its ends are bent to form an eye. If heavy loading is to be applied, one or more full length leaves are provided below the master leaf. The bundle of leaves having same thickness but different length are clamped together at the centre. All the leaves are bent to same curvature. The spring rests on the axle of the vehicle and is pin jointed to the chasis at the eyes.

Fig. 8.3 A leaf spring.

Consider a leaf spring hinged at both ends and carrying a load W at the centre. The load W is distributed equally at the two eyes. Let Span of the spring (biggest leaf) = l   :LGWKRIHDFKSODWH b Thickness of each plate = t Number of plates in the spring = n Bending stress in the plate = Vb   'HÀHFWLRQRIWKHPDVWHUSODWH G Radius of curvature of each plate = R The bending moment is maximum at the centre and goes on decreasing towards the ends (eyes); therefore maximum resistance of moment is required at the centre and less towards the ends. That

Springs

369

is why leafs of gradually reducing lengths are used in the spring. The law of variation of bending moment is linear. Using bending equation, the moment of resistance is found to be nVb bt 2 Mr = 6 The maximum bending moment due to load applied on the spring is given as

... (8.1)

Wl ... (8.2) 4 Equation (8.2) is obtained on comparing the spring with a simply supported beam loaded with a point load at the centre. M =

From the two equations (8.1) and (8.2), we have nVb ¸ bt 2 Wl = 6 4 3Wl or Vb = 2nbt 2 The radius of curvature can be obtained by considering Fig. 8.4.

... (8.3)

Fig. 8.4

2

⎛ l ⎞⎟ ⎜⎜ ⎟ + (R – G)2 = R2 ⎜⎝ 2 ⎟⎠ l2 or G = 8R From bending equation

... (8.4)

Vb E = (t / 2) R or

R =

Et 2Vb

... (8.5)

Substituting R in equation (8.4), we get 

G =

Vb l 2 4 Et

... (8.6)

370

Strength of Materials

where

E = Modulus of elasticity of the spring material

 $QRWKHUHTXDWLRQIRUGHÀHFWLRQFDQEHREWDLQHGE\VXEVWLWXWLQJVb in equation (8.6) as 

G =

3Wl 3

... (8.7) 8n Ebt 3 The strain energy stored in the spring is due to the work done by the load acting on it, given by 1 ˜ W ˜ G 2 Substituting G in equation (8.8), we get strain energy as U =

... (8.8)

3

U = or

U =

3Wl 1 ×W× 8n Ebt 3 2 3W 2l 3 16n Ebt 3

... (8.9)



The strain energy can also be expressed in terms of bending stress and volume of spring, given by Vb2 × Volume of the spring 6E n Volume of spring = lbt 2 U =

where

... (8.10)

Example 8.1 A steel carriage spring is 800 mm long and carries a central load of 6 kN. The plates are 70 mm wide and 5 mm thick. Determine the number of plates in the spring to sustain a maximum bending stress of 200 N/mm2:KDWZLOOEHWKHGHÀHFWLRQLQWKHVSULQJ"7RZKDWUDGLXVVKRXOGHDFKSODWHEHFXUYHG VRWKDWLWEHFRPHVVWUDLJKWXQGHUWKHJLYHQORDG"7DNHE = 200 kN/mm2. Solution: Given, Length of the biggest plate, Central load on the spring, :LGWKRIWKHSODWH Thickness of the plate, Bending stress induced in the spring, Modulus of elasticity of spring material, Let Number of plates in the spring 'HÀHFWLRQRIWKHVSULQJ Radius of curvature of each plate

l = 800 mm W = 6 kN = 6000 N b = 70 mm t = 5 mm Vb = 200 N/mm2 E = 200 kN/mm2 = 2 × 105 N/mm2 =n =G =R

Using equation (8.3) for bending stress, we have 3Wl  Vb = 2nbt 2

Springs

200 = or

371

3q 6000 q 800

2 q n q 70 q 52 n = 20.57

Hence, the number of plates can be taken to be 21. Ans.  1RZXVLQJHTXDWLRQ  IRUGHÀHFWLRQZHKDYH G =



3Wl 3 8n E bt 3 3

or

3× 6000 ×(800)

G =

8 × 21× 2 ×105 × 70 × 53 For the radius of curvature, use equation (8.4). G =



= 31.34 mm

Ans.

l2 8R 2

or

(800) l2 R = = = 2.552 m 8G 8 × 31.34

Ans.

Example 8.2 A laminated steel spring of length 900 mm carries a central proof load of 8 kN. The maximum GHÀHFWLRQDWWKHFHQWUHLVPPDQGWKHEHQGLQJVWUHVVVKRXOGQRWH[FHHGN1PP 2. Determine the thickness, width, number of plates, and the radius to which the plates should be bent. Assume the plate width to be ten times its thickness. Take E = 200 kN/mm2. Solution: Given, Length of the biggest plate, Load on the spring,  'HÀHFWLRQRIWKHVSULQJ Bending stress induced in the spring, Modulus of elasticity,  /HW :LGWKRIWKHSODWH Thickness of the plate

l W G Vb E

= 900 mm = 8 kN = 8000 N = 60 mm = 0.5 kN/mm2 = 500 N/mm2 = 200 kN/mm2 = 2 × 105 N/mm2  b =t

Number of plates in the spring = n Given,

b = 10t

... (1)

 8VLQJHTXDWLRQ  IRUGHÀHFWLRQZHKDYH G =



3Wl 2 8nEbt 3 3

60 =

or

3× 8000 ×(900)

8n × 2 ×105 ×10t × t 3 nt4 = 18225

...(2)

372

Strength of Materials

Using equation (8.3) for bending stress, we have Vb =



500 = or

3Wl 2nbt 2

3q 8000 q 900 2n q10t q t 2

nt3 = 2160

...(3)

Dividing equation (2) by equation (3), we get 18225 = 8.43 mm 2160 Hence, the thickness of the plate can be chosen to be 8 mm. t =

From equation (1)

Ans.

b = 10t = 10 × 8 = 80 mm

Ans.

From equation (3) n =

2160

83 Hence, the number of plates can be 5.

= 4.21 Ans.

 7KHDFWXDOGHÀHFWLRQXQGHUWKHSURRIORDGLVJLYHQDV 3

Gp =



3× 8000 ×(900)

3

8 × 5 × 2 ×105 × 80 ×(8)

= 53.4 mm

(using equation (8.7))

The radius of curvature is found by using equation (8.4). l2 R = 8δ p 2

(900) = = 1.896 m 8 × 53.4 Hence, the radius of curvature of each plate = 1.896 m.

Ans.

Example 8.3 A steel carriage spring of length 1.5 m having plate width 150 mm and thickness 10 mm is subjected to a bending stress of 200 N/mm2. The spring during its straightening absorbs 150 joule of energy. Find the number of plates and their radius of curvature. Take E = 200kN/mm2. Solution: Given, Length of the biggest plate,

l = 1.5 m = 1500 mm

 :LGWKRIWKHSODWH

b = 150 mm

Thickness of the plate,

t = 10 mm

Modulus of Elasticity of spring material, E = 200 kN/mm2 = 2 × 105 N/mm2

Springs

373

Bending stress induced in the spring, Vb = 200 N/mm2 Strain energy stored in the spring,

U = 150 joules

Let, Number of plates,

=n

Radius of curvature of each plate,

=R

The strain energy stored in the spring is given by the equation (8.10) as U =

Vb2 n q lbt 6E 2 2

3

150 × 10 = or

(200) × n ×1500 ×150 ×10 6 × 2 ×105 × 2

n =4

Ans.

 7R¿QGWKHUDGLXVRIFXUYDWXUHXVHHTXDWLRQ   2 ×105 ×10 Et = R = =5m 2Vb 2 × 200

Ans.

Example 8.4 A steel carriage spring of length 1 m has 15 leaves, each 150 mm wide and 10 mm thick. It is subjected to a maximum bending stress of 300 N/mm2. Find the curvature of each leaf. A load of 200 N is dropped on the spring without exceeding the given bending stress. Neglecting the ORVVRIHQHUJ\DWLPSDFW¿QGWKHPD[LPXPKHLJKWWKURXJKZKLFKWKHORDGIDOOVRQWKHVSULQJ7DNH E = 200 kN/mm2. Solution: Given, Length of the biggest leaf,

l = 1 m = 1000 mm

Number of leaves in the spring,

n = 15

Width of the leaf,

b = 150 mm

Thickness of the leaf,

t = 10 mm

Bending stress induced in the spring,  Vb = 300 N/mm2 Load to be dropped on the spring,

W = 200 N

Modulus of elasticity,

E = 200 kN/mm2 = 2 × 105 N/mm2

Let Radius of curvature of each plate

=R

Height of load

=h

The radius of curvature R can be found by using equation (8.5). R =

2 ×105 ×10 Et = = 3.34 m 2V b 2 × 300

Ans.

374

Strength of Materials

 7KHGHÀHFWLRQLQWKHVSULQJLVJLYHQE\HTXDWLRQ   2



(1000)

G =

8 × 3.34 ×103 The potential energy at a height h is given as

= 37.5 mm

P.E = W (G + h) = 200 (37.5 + h)

... (1)

The potential energy is stored in the spring as strain energy given by equation (8.10). Vb2 n q lbt U = 6E 2 2

=

(300)

6 × 2 ×10 Equating equations (1) and (2), we have

5

×

15 × 1000 × 150 × 10 = 843.75 joule 2

... (2)

200 (37.5 + h) = 843.75 × 103 or

8.6

h = 4.18 m

Ans.

QUARTER-ELLIPTIC LEAF SPRING

The analysis of a quarter-elliptic leaf spring (also called cantilever laminated spring) can be made by comparing the spring with a cantilever beam loaded with a point load at its free end (Fig. 8.5). In this case, maximum bending moment is given as M = Wl

... (8.11)

Fig. 8.5 $TXDUWHUHOOLSWLFOHDIVSULQJ

Using equations (8.1) and (8.11), we get the equation for maximum bending stress. 

Vbmax =

6Wl

nbt 2 where the symbols have their usual meanings.

... (8.12)

Springs

 'HÀHFWLRQLQWKHVSULQJLVJLYHQDV 6Wl 3  G = nE bt 3 Radius of curvature of each leaf is found to be R =

375

... (8.13)

l2 Et  2δ 2σbmax

... (8.14)

Example 8.5 A quarter-elliptic leaf spring has 10 leaves with cross-section of each leaf being 100 mm × 10 mm. The spring is 500 mm long. The bending stress is not to exceed 300 N/mm2. Determine the following parameters of the spring. (a) the maximum load which can be applied on the spring (b  WKHGHÀHFWLRQSURGXFHGDQG (c) the radius of curvature of leaves Take E = 200 kN/mm2. Solution: 

Given,

Number of leaves in the spring,

n = 10

:LGWKRIWKHSODWH

b = 100 mm

Thickness of the plate,

t = 10 mm

Lenth of the biggest plate,

l = 500 mm

Maximum bending stress induced in the spring, Vbmax = 300 N/mm2 Modulus of elasticity, E = 200kN/mm2 = 2 × 105 N/mm2 (a) Use equation (8.12) to get the load W acting on the spring. 6Wl  Vbmax = nbt 2 6W q 500 300 = 10 q100 q102 or

W = 10 kN

Ans.

(b  7KHGHÀHFWLRQFDQEHREWDLQHGE\XVLQJHTXDWLRQ   

6Wl 3

3

6 ×10 ×103 ×(500)

= = 37.5 mm nE bt 3 10 × 2 ×105 ×100 ×103 (c) Use equation (8.14) for the radius of curvature. G =

R =

5002 l2 = = 3.34 m 2δ 2 × 37.5

Ans.

Ans.

376

Strength of Materials

Example 8.6 $TXDUWHUHOOLSWLFOHDIVSULQJRIOHQJWKPPLVVXEMHFWHGWRDORDGRIN1SURGXFLQJDGHÀHFWLRQ of 100 mm. The cross-section of the spring is 70 mm × 10 mm. Determine the following parameters of the spring: (a) the number of leaves in the spring (b) the maximum bending stress induced in the spring and (c) the height through which if the given load is dropped on the spring, is to produce a maximum bending stress of 1 kN/mm2 Take E = 200 kN/mm2. Solution:

Given,

Length of the biggest plate,

l = 700 mm W = 3 kN = 3 × 103 N

Load on the spring,

G = 100 mm

 'HÀHFWLRQSURGXFHGLQWKHVSULQJ

E = 200 kN/mm2 = 2 × 105 N/mm2

Modulus of elasticity,

b = 70 mm

 :LGWKRIWKHSODWH Thickness of the plate,

t = 10 mm

(a) Use equation (8.13) for number of leaves in the spring. G =



6Wl 3 nE bt 3 3

100 = or

6 × 3×103 ×(700)

10 × 2 ×105 × 70 ×103 n = 4.41

Hence, the number of leaves can be chosen to be 5.

Ans.

(b) The maximum bending stress induced in the spring can be obtained by using equation (8.12). Vbmax =



6Wl nbt 2

=

6 × 3×103 × 700 5 × 70 ×102

= 360 N/mm2

Ans.

(c) If We is the equivalent gradually applied load to produce the given bending stress. Equation (8.12) LVDJDLQXVHGWR¿QGWe. 6We q 700 1000 = 5 q 70 q102 or

We = 8.34 kN

This is the equivalent impact load corresponding to the given load of 3 kN.  7KHGHÀHFWLRQSURGXFHGE\WKHLPSDFWORDGLVJLYHQDV 3



Ge =

6 × 8.34 ×103 ×(700) 5 × 2 ×105 × 70 ×103

= 245 mm.

Springs

377

Using principle’s of energy, we have Loss of potential energy = Gain of strain energy 1 qWe q Ge 2 where, h is the height through which the load falls. W (h + Ge) =

3 × 103 (h + 245) =

h = 95.2 mm

or

8.7

1 q 8.34 q103 q 245 2 Ans.

SPIRAL SPRING

$VSLUDOVSULQJFRQVLVWVRIDÀDWVWULSRIUHFWDQJXODUVHFWLRQZRXQGLQWRDVSLUDO )LJ ,WLVXVHG to produce torsion on the axle to which one of its end (inner one) is attached. If a force W, acting at a radius x, is applied to the axis of the spring, a twisting moment Wx is set up. Let Bending stress induced in the spring = Vb  :LGWKRIVSULQJVWULS

b



Length of spring strip

=l

Thickness of spring strip =t The maximum bending stress in the spring is found to be 

Vbmax =



= where

12Wx bt 2

(using bending equation) ...(8.15)

12M

...(8.16)

bt 2

M = Wx :LQGLQJWRUTXHRUWXUQLQJPRPHQWRQWKHVSLQGOH

 7KHDQJXODUGHÀHFWLRQRIWKHVSULQJLVJLYHQDV Wxl Ml  EI EI E = Modulus of elasticity of the spring material

...(8.17)

T (radian) =

 where

I = Moment of inertia of the spring cross-section The strain energy stored in the spring is given as U = =

1 q M qT 2 1 Wxl qWx q 2 EI

W 2 x 2l = 2 EI

...(on substituting T ) ...(8.18)

378

Strength of Materials

The number of turns given to the spindle is given as n =

θ 2π

...(8.19)

Example 8.7 $ÀDWVSLUDOVSULQJZLWKUHFWDQJXODUVHFWLRQPPîPPLVPHWUHORQJ2QHHQGRILWLVDWWDFKHG WRDVPDOOVSLQGOHDQGRWKHUWRD¿[HGSRLQW7KHEHQGLQJVWUHVVLVQRWWRH[FHHGN1PP 2. Find the following parameters of the spring: (a) the maximum turning moment applied to the spindle (b) the number of turns and (c) the strain energy stored in the spring The modulus of elasticity of the spring material can be taken to be 200 kN/mm 2. Solution:

Given, b = 10 mm

 :LGWKRIWKHVSULQJVHFWLRQ Thickness of the spring section,

t = 0.5 mm

Length of the spring,

l = 5 m = 5 × 103 mm

Bending stress induced in the spring,

Vbmax = 0.5 kN/mm2 = 500 N/mm2

Modulus of elasticity of spring material,

E = 200 kN/mm2 = 2 × 105 N/mm2

(a) The maximum turning moment applied to the spindle is given by equation (8.16). Vbmax = 500 = or

12M bt 2 12M 10 q(0.5) 2

M = 104.16 N.mm

Ans.

(b  7KHDQJXODUGHÀHFWLRQT of the spring is given by using equation (8.17). 

T = =

Ml EI 104.16 × 5 ×103 = 25 radian 5 ⎡ 1 3⎤ 2 ×10 × ⎢ ×10 × (0.5) ⎥ ⎢⎣12 ⎥⎦

For number of turns, use equation (8.19). θ 25  = 3.97 2π 2π Hence, the number of turns can be chosen to be 4. n =

Ans.

Springs

379

(c) Strain energy stored in the spring can be found by using equation (8.18). 1 U = q M qT 2 1 q104.16 q 25 2 = 1.3 joules =

Ans.

Example 8.8 $ÀDWVSLUDOVSULQJKDYLQJFURVVVHFWLRQPPîPPLVVXEMHFWHGWRDPD[LPXPEHQGLQJVWUHVV of 1 kN/mm27KH VSULQJ FDQ VWRUH DQ HQHUJ\ RI  -RXOH:LWK E = 200 kN/mm2, determine the following parameters of the spring: (a) the maximum torque required (b) the number of turns given to the spindle and (c) the length of the spring. Solution:

Given,

 :LGWKRIWKHVSULQJVHFWLRQ Thickness of the spring section,

b = 20 mm t = 0.5 mm Vbmax = 1 kN/mm2 = 1000 N/mm2

Maximum bending stress, Energy stored in the spring,

U = 10 Joule

(a) The maximum torque required can be obtained by using equation (8.16). 12M  Vbmax = bt 2 12M 1000 = 20 q(0.5) 2 or

M = 416.6 N˜mm

Hence, the required torque is 416.6 N˜mm.

Ans.

(b) The strain energy is given as 1 q M qT 2 1 10 × 103 = q 416.6 q T 2 T = 48 radian U =

or

The number of turns is given by equation (8.19). n =

θ 48  = 7.64 2π 2π

Ans.

380

Strength of Materials

(c) The length of the spring can be found by using equation (8.17). T =



Ml EI

416.6 × l ⎡1 ⎤ 2 ×105 × ⎢ × 20 × (0.5)3 ⎥ ⎢⎣12 ⎥⎦ l = 4.8m

48 = or

8.8

Ans.

HELICAL SPRING

8.8.1 Close Coiled Helical Spring subjected to an Axial Load Fig. 8.6 shows a close coiled helical spring subjected to an axial load W. Let R = Mean radius of the coil d = Diameter of the spring wire n = Number of turns or coils in the spring  T = Angle of twist produced in the spring wire  G 'HÀHFWLRQSURGXFHGE\WKHDSSOLHGORDGW  W = Shear stress induced in the wire of spring G = Modulus of rigidity of spring material Each section of the wire of spring is subjected to torsional shear stress and hence such springs are also called torsion springs. Bending effect, on account of its negligible value is neglected. Effect of direct stress is also neglected due to similar reasons. Total length of the spring wire, l = 2S Rn

Fig. 8.6 $FORVHFRLOHGKHOLFDOVSULQJVXEMHFWHGWRDQD[LDOORDG

Springs

381

Using torsion formula, we have GT T = l J where

... (8.20)

T = WR = Torque applied on the spring J = Polar moment of inertia of the spring wire

π 4 d 32 Substituting T, J and l in equation (8.20), we have =



T =

64WR 2 n

Gd 4 This is the required expression for the angle of twist expressed in radian.

...(8.21)

Shear stress induced in the spring is found by using torsion formula. T W = J r where r is radius of the spring wire. On substituting T and JWKHDERYHHTXDWLRQJLYHVVKHDUVWUHVVLQWKHRXWHUPRVW¿EUHRIWKHVSULQJ wire as 16WR  W = ...(8.22) πd 3  (TXDWLRQ   JLYHV PD[LPXP VKHDU VWUHVV LQGXFHG LQ WKH VSULQJ ZLUH :KLOH GHULYLQJ WKLV equation, the effect of the curvature of the wire is neglected. This is true only when the spring is subjected to static loads, because yielding of the spring material will relieve the stresses. Also, the effect of direct stress has been neglected. Curvature of the spring wire in the calculation of maximum shear stress was introduced by A.M. :DKOZKRLVIDPRXVIRUKLVERRNµ0HFKDQLFDO6SULQJV¶FRQVLGHUHGDVWKHELEOHRIVSULQJGHVLJQ $FFRUGLQJO\WKHHTXDWLRQIRUWKHVKHDUVWUHVVLVPRGL¿HGDV  where

Wmax =

8WD q KW πd 3

... (8.23)

KW :DKOFRUUHFWLRQIDFWRU =

8C − 1 0.615 + 4C − 4 C

C = Spring index = The strain energy stored in the spring is given as U =

1 TT 2

D for circular spring wires d

382

Strength of Materials

1 64WR 2 n 32W 2 R3 n WR × = = 2 Gd 4 Gd 4 The strain energy in terms of shear stress can be expressed as

... (8.24)

W2 U = × Volume of the spring 4G

... (8.25)

The workdone on the spring is equal to the strain energy stored in it. 1 1 Wδ  Tθ 2 2 2U G = W

... (8.26)

U = or

=

2 ⎛⎜ 32W 2 R3 n ⎞⎟⎟ ⎜ ⎟ W ⎜⎜⎝ Gd 4 ⎟⎠

(on substituting U)

64WR3 n

... (8.27) Gd 4  7KLVLVWKHUHTXLUHGH[SUHVVLRQIRUGHÀHFWLRQSURGXFHGLQWKHVSULQJXQGHUWKHD[LDOORDGW. =

Axial stiffness of the spring, KLVGH¿QHGDV K =

8.8.2

W Gd 4  G 64 R3 n

... (8.28)

Close Coiled Helical Spring subjected to an Axial Twist

Let the spring shown in Fig. 8.7 is subjected to an axial couple at its free end. The effect of this may be to open or close the spring, resulting in change in number of turns/coils of the spring. The number of turns in the spring increases if the torque resulting from the applied couple closes the spring and number of turns decreases if the torque opens the spring. During the analysis of such spring, effect of torsional shear stress is neglected and only bending stress is considered.

Fig. 8.7 $FORVHFRLOHGKHOLFDOVSULQJVXEMHFWHGWRDQD[LDOWZLVW

Springs

The length of the spring is obtained as l = 2S R1 n = 2SR2 nc where

383

...(8.29)

R1 and R2 ,QLWLDODQG¿QDOPHDQUDGLXVRIWKHFRLOUHVSHFWLYHO\ n and nc ,QLWLDODQG¿QDOQXPEHURIFRLOVLQWKHVSULQJUHVSHFWLYHO\

Length of the spring remains constant, but radius of coil and number of coils change when axial couple is applied. Using bending equation, the bending moment M is given as ⎡1 1⎤ M = EI ⎢⎢ − ⎥⎥ ⎣ R1 R2 ⎦ Using equation (8.29), equation (8.30) transforms to 2πEI [n – nc] l The angle of twist (in radian) at the free end of spring is given as I = 2S (n – nc) M =

 8VLQJHTXDWLRQ  ZH¿QGI to be Ml I = EI If the spring is made of circular wire of diameter d, then π 4 d I = 64 Equation (8.33) on using equations (8.29) and (8.31) becomes 128MRn  I = (R1 |R2 = R) Ed 4 The angular stiffness of the spring is given as K1 =

...(8.30)

... (8.31) ... (8.32) ... (8.33)

... (8.34)

M EI  φ l

Ed 4 128Rn The bending stress induced in the spring is calculated by using bending equation. Vb M = y I where  Vb = Maximum bending stress induced in the spring =

d 2 32 M Vb = πd3

... (8.35)

y =

Hence, 

... (8.36)

384

Strength of Materials

The strain energy stored in the spring is given as U = =

1 Mφ 2 64M 2 Rn

Ed 4 The strain energy can also be expressed in terms of bending stress, given as U =

8.8.3

Vb2 × Volume of the spring 8E

... (8.37)

... (8.38)

Open Coiled Helical Spring subjected to an Axial Load

7KH¿JXUHIRUDQRSHQFRLOHGKHOLFDOVSULQJXQGHUWKHDFWLRQRIDQD[LDOORDGW is similar to as shown in Fig. 8.6. The load produces two types of moments, one has twisting action and another has bending effect. The two moments are : ⎫ Twisting moment, T = WR cos D⎪ ⎪ ⎬ Bending moment, M = WR sin D ⎪ ⎪ ⎭

... (8.39)

D = Angle of helix

where 

The equivalent twisting moment is obtained as Te = =

T2 M2 W 2 R 2 cos 2 D W 2 R 2 sin 2 D = WR

... (8.40)

The equivalent bending moment is obtained as 1 2 2 Me = [ M T M ] 2 =

1 [WR sin D W 2 R 2 cos 2 D W 2 R 2 sin 2 D] 2

=

1 [WR sin α + WR ] 2

WR [1 sin D] 2 The maximum direct stress induced in the spring wire is given as =





32 M e (using bending equation ) πd3 The maximum shear stress induced in the spring wire is given as Vb =

16Te πd 3 The total length of wire in the spring is given as l = 2S nR sec D W =

(using torsion equation)

... (8.41)

... (8.42)

... (8.43)

... (8.44)

Springs

385

The strain energy stored in the spring is due to torsion and bending effect both. The strain energy due to torsion is given as Ut = where 

1 T 2l Tθ  2 2GJ

(using torsion equation)

... (8.45)

T = Angle of twist produced in the spring wire =

64WR 2 n sin D ⎡ 1 2 ⎤ ⎢ − ⎥ ⎢⎣ G E ⎥⎦ d4

... (8.46)

The strain energy due to bending moment is found as 1 Ub = M θ 2 M 2l = 2 EI The total strain energy stored in the spring is given as

⎛ ⎞ ⎜⎜using M = E and l = θ⎟⎟ ... (8.47) ⎟⎠ ⎜⎝ I R R

UT = Ut + Ub =

l ⎡⎢ T 2 M 2 ⎤⎥ + 2 ⎢⎢⎣ GJ EI ⎥⎦⎥

W 2 R 2l ⎡⎢ cos 2 D sin 2 D ⎤⎥ + = 2 ⎢⎢⎣ GJ EI ⎥⎥⎦

... (8.48)

... (8.49)

 :RUNGRQHRQWKHVSULQJLVDOVRIRXQGDV 1 W G = UT 2

... (8.50)

 &RPSDULQJHTXDWLRQV  DQG  ZH¿QGWKHYDOXHRIGHÀHFWLRQG as ⎡ 2 sin 2 D ⎤⎥ 2 ⎢ cos D WR l +  G = ⎢ GJ EI ⎥⎥⎦ ⎢⎣ For a spring made of circular wire of diameter d, we have πd 4 πd 4 I= and J = 64 32 Now equation (8.51) changes to 

G =

64WR 3n sec α ⎡⎢ cos2 α 2 sin 2 α ⎤⎥ ⎢ G + E ⎥ d4 ⎢⎣ ⎥⎦

... (8.51)

... (8.52)

On comparing this equation with close-coiled helical spring, where D is zero, equation (8.52) reduces to equation (8.27).

386

Strength of Materials

8.8.4

Open Coiled Helical Spring subjected to an Axial Twist

Let an open coiled helical spring be subjected to a torque M which produces both twisting and bending. ⎫ Twisting moment, T' = M sin D ⎪ ⎪ ⎬ Bending moment, M' = M cos D⎪ ... (8.53) ⎪ ⎭ The equivalent twisting moment is found as Te = =

T' 2 M' 2 M 2 sin 2 D M 2 cos 2 D = M

... (8.54)

The equivalent bending moment is found as 1 [ M' T' 2 M' 2 ] 2 1 2 2 2 2 = [ M cos D M sin D M cos D] 2

Me =

M [1 cos D] 2 The maximum direct stress induced in the spring is given as

... (8.55)

=





32 M e (using bending equation) πd 3 The maximum shear stress induced in the spring is given as Vb =

16Te πd 3 The total strain energy stored in the spring is given as W =

(using torsion equation)

... (8.56)

... (8.57)

1 1 T ′ θ′ + M ′ φ′ ... (8.58) 2 2 where T c and Ic are the angles of twist due to T c and M c respectively, and are expressed as UT =

T„c =



T al M al and I„ = GJ EI

... (8.59)

1 MI 2 where I is the net angle of twist because of the combined effects.

Also

... (8.60)

UT =

Comparing equations (8.58) and (8.60) and using equations (8.53) and (8.59), we have 

I =

64MRn sec D ⎡⎢ sin 2 D 2 cos 2 D ⎤⎥ ⎢ G + E ⎥⎥⎦ d4 ⎢⎣

... (8.61) (on using l = 2S Rn sec D)

Springs

387

On comparing this equation with close coiled helical spring, where D = 0 and subjected to an axial twist MZH¿QGWKDWHTXDWLRQ  LVVLPLODUWRHTXDWLRQ    7KHGHÀHFWLRQSURGXFHGLQWKHVSULQJZLWKFLUFXODUZLUHLVIRXQGWREH G =



64MR 2 n sin D ⎡ 1 2 ⎤ ⎢ − ⎥ ⎢⎣ G E ⎥⎦ d4

...(8.62)

Example 8.9 A close coiled helical spring of circular section having a mean coil diameter of 60 mm is subjected to an axial load of 80 N applied at the end of spring producing a shear stress of 100 N/mm 2 and a GHÀHFWLRQRIPP)LQGWKHGLDPHWHUWKHQXPEHURIFRLOVWKHOHQJWKRIWKHVSULQJZLUHDQGWKH strain energy stored in the spring. Take G = 80 kN/mm2. Solution:

Given,

Mean diameter of the coil, Axial load on the spring, Shear stress induced in the spring, 'HÀHFWLRQSURGXFHGLQWKHVSULQJ Modulus of rigidity of spring material,

D W W G G

= 60 mm = 80 N = 100 N/mm2 = 50 mm = 80 kN/mm2 = 8 × 104 N/mm2

The diameter of the spring wire can be found by using equation (8.22). 16WR  W = πd 3 or

⎛16WR ⎞⎟1 / 3 ⎟ d = ⎜⎜⎜ ⎝ πτ ⎟⎠

⎛ ⎞1/ 3 ⎜⎜16 ×80 × 60 ⎟⎟ ⎜ 2 ⎟⎟⎟ = ⎜⎜ = 4.96 mm ⎜⎜ π ×100 ⎟⎟⎟ ⎟ ⎜⎝ ⎠ Hence, the diameter of the spring wire is 4.96 mm.

Ans.

Using equation (8.27), number of coils in the spring can be found. G =

50 = or

64WR3 n Gd 4 ⎛ 60 ⎞⎟3 16 × 80 ×⎜⎜ ⎟⎟ × n ⎜⎝ 2 ⎠

8 ×104 × (4.96) 4 n = 17.5

Ans.

The length of the spring wire is given by l = 2S Rn = 2πq

60 × 17.5 = 3.29 m 2

Ans.

388

Strength of Materials

The strain energy stored in the spring can be found using equation (8.24). U =

=

32W 2 R3 n Gd 4 ⎛ 60 ⎞3 32 × 802 ×⎜⎜ ⎟⎟⎟ ×17.5 ⎜⎝ 2 ⎠ 8 ×104 × (4.96) 4

= 1.99 joule

Ans.

Example 8.10 A close coiled helical spring made of steel wire of diameter 6 mm has 15 coils. The spring has mean coil diameter of 100 mm and is subjected to an axial load of W producing a maximum shear stress of 100 N/mm2. Find the load WWKHGHÀHFWLRQSURGXFHGDVDUHVXOWRIWKHORDGDSSOLHGDQGWKHZRUN GRQHWRSURGXFHWKHGHÀHFWLRQ7DNHG = 80 kN/mm2. Solution:

Given,

Spring wire diameter,

d = 6 mm

Number of coils,

n = 15

Mean coil diameter,

D = 100 mm

Shear stress induced in the spring,

W = 100 N/mm2

Modulus of rigidity of the spring material, G = 80 kN/mm2 = 8 × 104 N/mm2 Load W can be calculated by using equation (8.22). W =



16WR πd 3

⎛100 ⎞⎟ 16 ×W ×⎜⎜⎜ ⎟ ⎝ 2 ⎟⎠ 100 = π × 63 or

W = 84.8 N

Ans.

 7KHGHÀHFWLRQSURGXFHGLQWKHVSULQJFDQEHIRXQGE\XVLQJHTXDWLRQ   

G =

64WR3 n Gd 4 3

⎛100 ⎞⎟ 64 × 84.8 ×⎜⎜ 15 ⎜⎝ 2 ⎟⎟⎠ × = = 98.1 mm 8 ×104 × 64 The workdone on the spring is given as 1 1 W G = q 84.8 q 98.1 = 4.16 Joule 2 2

Ans.

Ans.

Springs

389

Example 8.11 A close coiled helical spring made of steel wire of diameter 7 mm has an axial stiffness of 6 N per mm and an angular stiffness of 100 N˜mm per degree angle of twist. Find the mean radius of the coil, the number of turns in the spring and its length. Take E = 200 kN/mm2 and G = 80 kN/mm2. Solution: Given, Spring wire diameter,

d = 7 mm

Axial stiffness of the spring,

K = 6 N/mm

Angular stiffness of the spring,

K1 = 100 N˜mm/degree

The axial stiffness of the spring is given by using equation (8.28). K =

6 = or

W Gd 4  G 64 R3 n 8 q104 q 7 4 64 q R3 n

R3n = 500208.33

...(1)

The angular stiffness is given by equation (8.35).

or

K =

M Ed 4 π = × φ 128Rn 180

100 =

2 q105 q 74 π q 128q Rn 180

Rn = 654.77

...(2)

Using equations (1) and (2), we get or

R = 27.64 mm

Ans.

From equation (2) n =

654.77 = 23.7 R

Ans.

The length of the spring wire is given as l = 2S Rn = 2S × 27.64 × 23.7 = 4.11 m

Ans.

Example 8.12 A close coiled helical spring made of 15 mm steel wire is subjected to an axial load of 200 N. The VSULQJKDVFRLOVDQGLWVPHDQFRLOUDGLXVLVPP)LQGWKHGHÀHFWLRQWKHPD[LPXPVKHDUVWUHVV and the strain energy stored in the spring per unit voulme. Now the axial load is replaced by an axial torque of 11 N˜m. Find the axial twist, the maximum bending stress and the strain energy stored per unit volume of the spring. Take G = 80 kN/mm2 and E = 200 kN/mm2.

390

Strength of Materials

Solution:

Given,

Diameter of the spring wire, Axial load on the spring, Number of coils in the spring, Mean radius of the coil, Axial torque on the spring, Modulus of rigidity, Modulus of elasticity,

d W n R M G E

= 15 mm = 200 N = 15 = 100 mm = 11 N˜m = 80 kN/mm2 = 8 × 104 N/mm2 = 200 kN/mm2 = 2 × 105 N/mm2

 7KHGHÀHFWLRQSURGXFHGLQWKHVSULQJLVJLYHQE\HTXDWLRQ   G =



64WR3 n Gd 4 64 q 200 q1003 q15

= 47.4 mm 8 q104 q154  8VHHTXDWLRQ  WR¿QGWKHPD[LPXPVKHDUVWUHVVLQGXFHGLQWKHVSULQJ =

W =



16 q 200 q100 16WR 3 = = 30.18 N/mm2 3 π q153 πd

Ans.

Ans.

Use equation (8.25) for the strain energy stored in the spring. W2 × Volume of the spring (V) 4G 2 (30.18) U W2 = or = 2.84 × 10–3 N˜mm/mm3 = 4G 4 × 8 ×104 V Use equation (8.34) for the axial twist in the spring. U =

I =



=

128MRn Ed 4 128 q11q103 q100 q15

2 q105 q154 For the bending stress in the spring, use equation (8.36).

= 0.208 radian = 11.95o

32 q11q103 32 M = 33.19 N/mm2 = Vb = 3 πd 3 π q15



Ans.

Ans.

Ans.

Use equation (8.38) for the strain energy stored in the spring. U = or

U V

=

Vb2 × Volume of the spring (V) 8E (33.19)2 = 6.88 × 10–4 N˜mm/mm3 8 × 2 × 105

Ans.

Springs

391

Example 8.13 A close coiled helical spring made of steel wire is subjected to an axial load of 50 N. The maximum shear stress induced in the spring is limited to 100 N/mm2. The stiffness of the spring is 0.8 N/mm of compression and its solid length is 50 mm. Determine the following parameters: (a) the diameter of the spring wire (b) the mean radius of the coil and (c) the number of coils in the spring Take G = 80 kN/mm2. Solution: Given, Axial load on the spring,

W = 50 N

Maximum shear stress in the spring, 

W = 100 N/mm2

Stiffness of the spring,

K = 0.8 N/mm

Modulus of rigidity of spring material,

G = 80 kN/mm2 = 8 × 104 N/mm2

Let

d = Diameter of the spring wire n = Number of coils in the spring

Given, solid length of the spring, nd = 50 mm (a) Use equation (8.28) for the stiffness of the spring. K = or

0.8 =

Gd 4 64 R3 n 8 q104 qd 4

64 R3 n Use equation (8.22) for the shear stress in the spring.





W = 100 =

or Since

... (1)

16WR nd 3 16 q 50 q R

nd 3 R = 0.392d3

... (2)

nd = 50

50 d Using n and R in equation (1), we have or

n =

0.8 = Solving for d, we get

8 ×104 × d 4

⎛ 50 ⎞ 64 × (0.392d 3 )3 ×⎜⎜ ⎟⎟⎟ ⎜⎝ d ⎠ d = 4.76 mm

Ans.

392

Strength of Materials

(b) Using equation (2), we have R = 0.392d3 = 0.392 × (4.76)3 = 42.4 mm (c) The number of coils in the spring is given as n =

50 50  = 10.5 d 4.76

Ans.

Ans.

Example 8.14 A close coiled helical spring made of steel wire of 4 mm diameter has mean coil radius of 30 mm. The number of turns in the spring is 5 and its pitch, when no load is acting on it, is 20 mm. Find the axial load to be applied gradually on the spring so that the pitch reduces to minimum i.e, coils touch each other. An impact load of 10 N is allowed to fall on the spring through a certain height so that again the pitch is minimum. Find the height. Take G = 80 kN/mm2. Solution: Given, Diameter of the spring wire,

d = 4 mm

Mean coil radius of the spring,

R = 30 mm

Number of turns in the spring,

n =5

Pitch while unloaded

= 20 mm

Impact load on the spring,

= 10 N

Modulus of rigidity of spring material,

G = 80 kN/mm2 = 8 × 104 N/mm2

 7KHGHÀHFWLRQLQWKHVSULQJ GXULQJQRORDGFRQGLWLRQ IRURQHWXUQ )LJ LVJLYHQDV  G = Pitch – Spring wire diameter = 20 – 4 = 16 mm

Fig. 8.8

 8VLQJHTXDWLRQ  WKHGHÀHFWLRQLQWKHVSULQJLVJLYHQDV G =

64WR3 n Gd 4

Springs

16 =

393

64W q 303 q 5 8 q104 q 44

W = 37.9 N

Ans.

 7KHWRWDOGHÀHFWLRQLQWKHVSULQJLVJLYHQDV 

Gt 'HÀHFWLRQLQRQHWXUQîQXPEHURIWXUQV = 16 × 5 = 80 mm

The strain energy stored in the spring is given as 1 1 W Gt = × 37.9 × 80 2 2 = 1516 N˜mm = 1.51 joule

U =

Loss in potential energy, when load of 10 N is released from a height h, is = 10 (h + 80)

...(1) ...(2)

Equating equations (1) and (2), we have 10 (h + 80) = 1516 or

h = 71.6 mm

Hence, the height of the load is 71.6 mm.

Ans.

Example 8.15 A close coiled helical spring of free length 200 mm and mean coil radius 40 mm is subjected to a maximum stress of 100 N/mm2. Find the diameter of the spring wire and the number of turns in the spring, assuming that the spring can store a maximum strain energy of 30 joule, when the pitch is reduced to a minimum. Take G = 80 kN/mm2. Solution: Given, Mean coil radius of the spring, Maximum shear stress induced in the spring,  Free length of the spring, Strain energy stored in the spring, Modulus of rigidity of spring material,

R = 40 mm W = 100 N/mm2 = 200 mm U = 30 joule G = 80 kN/mm2 = 8 × 104 N/mm2

Free length of the spring is its total length in uncompressed state. The solid length is its total length in compressed state. They are related to each other as  6ROLGOHQJWK'HÀHFWLRQ )UHHOHQJWK nd + G = Free length or nd = 200 – G ... (1) The shear stress is given by equation (8.22). 

W =

16WR πd 3

394

Strength of Materials

100 =

16W q 40 πd3

W = 0.49d 3

or

... (2)

The strain energy stored in the spring is given by equation (8.25). U = 30 × 103 =

W2 × Volume of the spring (V) 4G (100) 2

×V

4 q 8 q104

V = 96 × 104 mm3

or

... (3)

But volume of the spring is given as V = Length of the spring × Cross-sectional area of the spring wire π 2 = 2πRn q d 4

96 × 104 = 2S × 40 × n ×

π 2 d 4

nd 2 = 4863.4 or

n =

4863.4

... (4)

d2

The strain energy is also given as 1 WG 2 1 × 0.49d 3 × G 30 × 103 = 2 U =

G =

(on substituting W from equation (2))

122448.98

d3 On substituting n and G in equation (1), we have 4863.4 d Solving for d, we get

2

q d = 200 –

122448.98 d3

d = 25.3 mm

Ans.

From equation (4) n =

4863.4 2

(25.3)

= 7.6

Ans.

Springs

395

Example 8.16 A body of mass 25 kg moving with a velocity of 3 m/sec is to be stopped by using a close coiled helical spring of 6 mm wire diameter and having a mean coil radius of 30 mm. The maximum shear stress induced in the spring is limited to 500 N/mm2. Find the number of turns in the spring and the GHÀHFWLRQSURGXFHG7DNHG = 80 kN/mm2. Solution:

Given,

Diameter of the spring wire, d = 6 mm Mean radius of the coil,

R = 30 mm

Maximum shear stress induced in the spring,  



W = 500 N/mm2

The strain energy stored in the spring is given as W2 × Volume of the spring (V) 4G The kinetic energy of the mass is given as U =

1 1 mV2 = × 25 × 32 = 112.5 jolues 2 2 Equating the two energies, we have K.E. =

112.5 × 103 =

(500) 2 4 q 8 q104

qV

or

V = 1.44 × 105 mm3

Also

π V = 2πRn q d 2 4

1.44 × 105 = 2S × 30 × n × or

n = 27

π × 62 4 Ans.

 7RGHWHUPLQHWKHGHÀHFWLRQSURGXFHGLQWKHVSULQJDQHTXLYDOHQWORDGLVFDOFXODWHGZKLFKZKHQ applied gradually on the spring may cause the given shear stress. Using equation (8.22), we have W =



500 = or

16WR πd 3 16W q 30 π q 63

W = 706.8 N

396

Strength of Materials

The strain energy stored in the spring is equal to kinetic energy of the moving mass, given as U =

1 qW q G × = 112.5 × 103 2

2 q112.5 q103 = 318.3 mm G = 706.8

or

Ans.

Example 8.17 An open coiled helical spring of 8 mm wire diameter has 10 turns and mean coil radius of 60 mm. It LVVXEMHFWHGWRDQD[LDOORDGRI1,IWKHSLWFKRIWKHVSULQJLVPP¿QGWKHIROORZLQJ (a) the maximum direct and shear stress induced in the spring wire (b) the angle of twist and (c  WKHGHÀHFWLRQRIWKHIUHHHQG Take E = 200 kN/mm2 and G = 80 kN/mm2. Solution:

Given,

Diameter of the spring wire,

d = 8 mm

Number of turns in the spring,

n = 10

Mean coil radius of the spring,

R = 60 mm

Axial load on the spring,

W = 150 N

Pitch of the spring,

p = 70 mm

The angle of helix D is found using Fig. 8.9.

Fig. 8.9

tan D =

p 2πR

70 = 0.185 2πq 60 D = 10.51o =

or Twisting moment,

T = WR cos D

Bending moment, M = WR sin D

Springs

397

The equivalent twisting moment is given by equation (8.40). Te = WR = 150 × 60 = 9000 N˜mm The equivalent bending moment is given by equation (8.41). Me =

WR [1 sin D] 2

150 × 60 [1 + sin 10.51º ] = 5320.8 N˜mm 2 (a) The maximum direct stress in the spring wire is given by equation (8.42). 32 M e  Vb = πd 3 =

32 q 5320.8 π q83 = 105.8 N/mm2 =

Ans.

The maximum shear stress in the spring wire is given by equation (8.43). 

16Te 16 q 9000 = = 89.5 N/mm2 3 πd3 πq 8 (b) The angle of twist of spring wire is given by equation (8.46). W =

I =



=

Ans.

64WR 2 n sin D ⎡ 1 2 ⎤ ⎢ − ⎥ ⎢⎣ G E ⎥⎦ d4 64 ×150 × 602 ×10 × sin 10.51º ⎡ 1 2 ⎤ ⎥ ⎢ − 4 4 ⎢⎣ 8 ×10 8 2 ×105 ⎥⎦

= 0.038 radian = 2.20o

Ans.

(c  7KHGHÀHFWLRQRIWKHVSULQJLVJLYHQE\HTXDWLRQ   64WR3 n sec D ⎡⎢ cos 2 D 2 sin 2 D ⎤⎥ G = ⎢ G + E ⎥ d4 ⎥⎦ ⎢⎣ 64 ×150 × 603 ×10 × sec10.51° ⎡⎢ cos 2 10.51º 2 sin 2 10.51º ⎤⎥ = ⎢ 8 ×104 + 2 ×105 ⎥ 84 ⎥⎦ ⎢⎣ = 63.9 mm



8.9

COMBINATION OF SPRINGS

Springs may be combined together in two ways. z

Series combination

z

Parallel combination

Ans.

398

Strength of Materials

8.9.1

Series Combination

The combination of springs in series is shown in Fig. 8.10 (a). Each spring carries the same load.

Fig. 8.10

Let

W K1 and K2 K G1 and G2

= Load applied on the spring = Stiffnesses of springs (1) and (2) respectively = Stiffness of the combined spring = Extensions produced in two springs respectively

 7KHWRWDOGHÀHFWLRQLQWKHFRPELQHGVSULQJLVWKHVXPRIWKHGHÀHFWLRQVSURGXFHGLQGLYLGXDOO\LQ the two springs.  G = G1 + G2 ... (8.63)  7KHGHÀHFWLRQSURGXFHGLQVSULQJ  LV 

G1 =

W K1

 7KHGHÀHFWLRQSURGXFHGLQVSULQJ  LV W  G2 = K2 Adding the two equations (8.64) and (8.65), we have 



... (8.64)

... (8.65)

⎡ 1 1 ⎤⎥ G1 + G2 = W ⎢⎢ + ⎥ ⎣ K1 K 2 ⎦ ⎡ 1 1 ⎤⎥ or G = W ⎢⎢ + ... (8.66) ⎥ ⎣ K1 K 2 ⎦ If the two springs are replaced by a single spring of stiffness KSURGXFLQJDGHÀHFWLRQG, then G =

W K

... (8.67)

Springs

399

Comparing equations (8.66) and (8.67), we get ⎡ 1 1 ⎤⎥ W = W ⎢⎢ + ⎥ K ⎣ K1 K 2 ⎦ 1 1 1 = K1 K 2 K

or

8.9.2

... (8.68)

Parallel Combination

Two springs connected in parallel are shown in Fig. 8.10 (b). Here, each spring carries different load and the load applied on the combined spring is the sum of the loads shared by the two springs. 'HÀHFWLRQSURGXFHGLQHDFKVSULQJLVWKHVDPHDQGVRLVWKHGHÀHFWLRQSURGXFHGLQWKHFRPELQHG spring. Let W1 = Load shared by the spring (1) W2 = Load shared by the spring (2) ... (8.69) Then W = W1 + W2 GK = GK1 + GK2 or K = K1 + K2 ... (8.70) Example 8.18 Two close coiled helical springs are connected in series. Both springs have 15 turns and the same mean coil diameter of 50 mm. Find the diameter of spring wire of one of the two springs if the GLDPHWHURIRWKHUVSULQJZLUHLVPP$OVR¿QGWKHORDGWREHDSSOLHGRQWKHHTXLYDOHQWVSULQJDQG WKHLWVFRUUHVSRQGLQJGHÀHFWLRQ7KHVWLIIQHVVRIWKHHTXLYDOHQWVSULQJLV1PPDQGWKHPD[LPXP shear stress is limited to 150 N/mm2. Given G = 80 kN/mm2. Solution:

Given,

Number of turns in both springs,

n1 = n2 = 15

Mean coil radius of both springs,

R1 = R2 =

'LDPHWHURIZLUHLQWKH¿UVWVSULQJ Stiffness of equivalent spring, Maximum shear stress induced in the equivalent spring, Let Diameter of wire in the second spring Load on equivalent spring

50 = 25 mm 2 d1 = 5 mm K = 0.5 N/mm W = 150 N/mm2 = d2 =W

 :KHQFRQQHFWHGLQVHULHVERWKVSULQJVZLOOFDUU\WKHVDPHORDGW, which will be acting on the HTXLYDOHQWVSULQJEXWZLOOKDYHGLIIHUHQWGHÀHFWLRQV  7KHGHÀHFWLRQLQWKH¿UVWVSULQJG1 is given as 

G1 =

64WR13 n1 Gd14

(using equation (8.27))

400

Strength of Materials

=

64 qW q 253 q15 8 q104 q 54

= 0.3 W

 7KHGHÀHFWLRQLQWKHVHFRQGVSULQJG2 is given as G2 =



=

64WR23 n2 Gd 24 64 ×W × (253 ) ×15 8 ×10

4

× d 24

=

187.5W d 24

 %RWKGHÀHFWLRQVDUHDGGHGWRJHWWKHWRWDOGHÀHFWLRQLQWKHHTXLYDOHQWVSULQJ G = G1 + G2



= 0.3W +

But 

G =

187.5W d 24

W W  K 0.5

... (1)

... (2)

Comparing equations (1) and (2), we have ⎡ 187.5 ⎤ W = W ⎢⎢0.3 + 4 ⎥⎥ 0.5 d 2 ⎥⎦ ⎢⎣ 187.5 ⎛ 1 ⎞ ⎜⎜ − 0.3⎟⎟⎟ = ⎜⎝ 0.5 d 24 ⎠ or

d2 = 3.24 mm

Ans.

Using equation (8.22), we have W =



150 = or

16WR π d 23 16W q 25 π q (3.24)3

W = 40 N

Ans.

 :KLOHXVLQJHTXDWLRQ  VSULQJZLWKVPDOOHUZLUHGLDPHWHULVXVHGWRDYRLGLWVIDLOXUH7KH mean coil diameter of the equivalent spring is also the same as for the two springs.  8VLQJHTXDWLRQ  ZHJHWGHÀHFWLRQLQWKHHTXLYDOHQWVSULQJDV 

G =

W 40  = 80 mm 0.5 0.5

Ans.

Springs

401

Example 8.19 Two close coiled helical springs are connected in parallel to take a load of 1500 N. Both springs have 16 coils and are made of equal wire diameter of 15 mm. Their mean coil diameters are 60 mm and 80 mm respectively. Find the loads shared by the two springs and the maximum shear stresses induced LQWKHP$OVR¿QGWKHGHÀHFWLRQLQWKHHTXLYDOHQWVSULQJ7DNHG = 80 kN/mm2 for both springs. Solution: Given, Load on the equivalent spring, Number of coils in two springs,  :LUHGLDPHWHURIWKHWZRVSULQJV

W = 1500 N n1 = n2 = 16 d1 = d2 = 15 mm

60 = 30 mm 2 80 Mean coil radius of the second spring, R2 = = 40 mm 2  :KHQFRQQHFWHGLQSDUDOOHOERWKVSULQJVSURGXFHWKHVDPHGHÀHFWLRQEXWVKDUHGLIIHUHQWORDGV The total load acting on the equivalent spring is the sum total of the loads shared by both springs. Let W1 and W2 are the respective loads on the two springs. Then, W = W1 + W2  0HDQFRLOUDGLXVRIWKH¿UVWVSULQJ

or

R1 =

1500 = W1 + W2

... (1)

 'HÀHFWLRQVLQWKHWZRVSULQJVDUHJLYHQDV G = G1 = G2



= or

64W1 R13 n1 Gd14



W1 × 303 = W2 × 403

64W2 R23 n2

... (2)

Gd 24

( n1 = n2 = 16 and d1 = d2 = 15 mm)

64 W2 27 Substituting W1 in equation (1), we have W1 =

64 W2 W2 27 W2 = 445 N

1500 = or

Ans.

 )URPHTXDWLRQ  ZH¿QGW1. 1500 = W1 + 445 or

W1 = 1055 N

Ans.

 7KHVKHDUVWUHVVLQWKH¿UVWVSULQJLVJLYHQE\XVLQJHTXDWLRQ   

W1 =

16W1R1 16 ×1055× 40 = = 47.76 N/mm2 πd13 π ×153

Ans.

402

Strength of Materials

The shear stress in the second spring is given as W2 =



16W2 R2 16 × 455× 40 = = 26.86 N/mm2 3 3 πd 2 π ×15

Ans.

From equation (2), we have G =



64W1 R13 n1 Gd14

=

64 q1055 q 303 q16 8 q104 q154

= 7.2 mm

 +HQFHWKHGHÀHFWLRQLQWKHHTXLYDOHQWVSULQJLVPP

Ans.

Example 8.20 A composite spring made of two close coiled helical springs is subjected to an axial load of 100 N. One of the two springs is placed inside the other and both springs are concentric. The inner spring has 10 coils and is 15 mm shorter than the outer spring of 4 mm wire diameter with number of coils 12 and mean coil diameter 50 mm. The radial clearance between the two springs is 1 mm. Find the stiffness and the wire diameter of the inner spring, if the axial load applied on the composite spring SURGXFHVDGHÀHFWLRQRIPPLQWKHRXWHUVSULQJ7DNHG = 80 kN/mm2 for both springs. Solution: Given, Axial load on the composite spring, W Number of coils in the inner spring, n1 Number of coils in the outer spring, n2 Mean coil diameter of the outer spring, D2  :LUHGLDPHWHURIWKHRXWHUVSULQJ d2  'HÀHFWLRQSURGXFHGLQWKHRXWHUVSULQJ G2

= 100 N = 10 = 12 = 50 mm = 4 mm = 30 mm

The axial load applied on the composite spring is shared by the two springs. Let W 1 and W2 be the load shared by the inner and outer spring respectively, then W = W1 + W2 or

W1 + W2 = 100 N (Given)

... (1)

Use equation (8.27) to get the load (W 2  VKDUHG E\ WKH RXWHU VSULQJ LQ RUGHU WR GHÀHFW LW E\ 30 mm. G2 =



64W2 R23 n2 Gd 24 G2 q G q d 23

= 51.2 N

or

W2 =

and

W1 = (100 – 51.2) N = 48.8 N

64 R23 n2

 7KHGHÀHFWLRQSURGXFHGE\WKHORDGW1 in the inner spring is given as G1 = (30 – 15) = 15 mm

(using equation (1))

Springs

403

The stiffness of the inner spring is given as W1 48.8  = 3.25 N/mm G1 15 The mean coil diameter of the inner spring is given as K1 =

Ans.

D1 = D2 – d2 – 2 × 1 – d1 = 50 – 4 – 2 × 1 – d1 = 44 – d1 Again using equation (8.27) for the inner spring, we have G1 =



64W1 R13 n1 Gd14

W1 Gd14 = K1 = = G1 64 R13 n1 =

or

3.25 =

8 ×104 d14 ⎛ 44 − d1 ⎞⎟3 64 ×⎜⎜ ×10 ⎜⎝ 2 ⎟⎟⎠

8 ×104 × 8 × d14 640(44 − d1 )3 103 d14 (44  d1 )3

(using K1)

Solving for d1, we get d1 = 3.8 mm

SHORT ANSWER QUESTIONS   :KDWLVWKHSXUSRVHRIXVLQJDVSULQJ"   :KDWLVD[LDOVWLIIQHVVRIDVSULQJ"   :K\LVOHDIVSULQJDOVRFDOOHGFDUULDJHVSULQJ"   :KHUHLVDVSLUDOVSULQJXVHG"   :K\LVKHOLFDOVSULQJDOVRFDOOHGWRUVLRQVSULQJ"   +RZLVFORVHFRLOHGKHOLFDOVSULQJGLIIHUHQWIURPRSHQFRLOHGKHOLFDOVSULQJ"   :KDWLVWKHSXUSRVHRIFRPELQLQJVSULQJV"   +RZLVD[LDOWZLVWLQJGLIIHUHQWIURPD[LDOORDGLQJ"   :KDWDUHWKHFRQVWUXFWLRQIHDWXUHVRIDOHDIVSULQJ"   +RZLVVHULHVFRPELQDWLRQRIVSULQJVGLIIHUHQWIURPSDUDOOHOFRPELQDWLRQ"

Ans.

404

Strength of Materials

MULTIPLE CHOICE QUESTIONS 1. The deformation produced in the spring is said to be (a) semi-elastic

(b) plastic

(c) elastic

(d) visco-elastic.

2. 7KHVWLIIQHVVRIWKHVSULQJLVGH¿QHGDVDUDWLRRI (a) load and angle of twist (b  ORDGDQGGHÀHFWLRQ (c) load and strain energy (d) load and strain. 3. In a close-coiled helical spring, the (a) plane of the coil and axis of the spring are closely attached (b) angle of helix is large (c) plane of the coil is normal to the axis of the spring (d  GHÀHFWLRQLVVPDOO 4. A conical helical spring is used, where (a) space is a problem (b) more stiffness is required (c) more load is to be taken (d  OHVVGHÀHFWLRQLVUHTXLUHG 5. 7KHORDGGHÀHFWLRQFXUYHRIDVSULQJLVDVWUDLJKWOLQHZKHQVSULQJLVVWUHVVHG (a) upto yield point (b) upto ultimate point (c) upto failure point (d) within the elastic limit. 6. The strain energy stored in a leaf spring is given as (a)

(c)

3W 2l 3

(b)

16nEbt 3 3W 2l

(d)

16nEbt 3

5Wl 3 16nEbt 3 3W 2l 3 . 16nEbt 2

where the symbols have their usual meanings. 7. The maximum bending stress developed in the wire of spiral spring is (a)

12M bt

(b)

6M bt

2

(c)

where the symbols have their usual meanings.

12M bt

2

(d)

12M . b2t

Springs

405

8. 7KHGHÀHFWLRQSURGXFHGLQDFORVHFRLOHGKHOLFDOVSULQJZKHQVXEMHFWHGWRDQD[LDOORDGLV (a) (c)

64W 3 Rn Gd 4 64WR3 n G4d

(b) (d)

64WR3 n Gd 4 64WR 2 n 2 . G4d

where the symbols have their usual meanings. 9. :DKO¶VFRUUHFWLRQIDFWRULVLQWURGXFHGWR (a) increase the number of coils in the spring (b) take care of extra load on the spring (c) take care of curvature of spring wire (d) take care of extra stiffness in the spring. 10. 7KHVSULQJLQGH[LVGH¿QHGDVDUDWLRRI (a  ORDGDQGGHÀHFWLRQ (b) mean coil diameter and spring wire diameter (c) load and angle of twist (d) mean coil diameter and length of spring wire. 11. )RUWZRVSULQJVEHLQJFRQQHFWHGLQVHULHVZKLFKRIWKHIROORZLQJVWDWHPHQWVLVFRUUHFW" (a  7KHGHÀHFWLRQSURGXFHGLQWKHHTXLYDOHQWVSULQJLVWKHVXPRIWKHGHÀHFWLRQVSURGXFHGLQWKH individual spring. (b) The total weight is the sum of the weights acting separately on the two springs. (c) The equivalent stiffness is the sum of the individual stiffness. (d) The equivalent stiffness is the product of the individual stiffness. 12. The equivalent stiffness, when two springs are connected in series, is given by the expression (a)

K1 K 2 1 = K K1 + K 2

(c) K =

K1 K 2 K1 + K 2

2 (b) K =

K1 + K 2 K1 K 2

(d) K = K1 + K2 .

13. )RUWZRVSULQJVEHLQJFRQQHFWHGLQSDUDOOHOZKLFKRIWKHIROORZLQJVWDWHPHQWVLVFRUUHFW" (a) The equivalent stiffness is the sum of the individual stiffness. (b) The equivalent load is the sum of the individual load. (c  7KHHTXLYDOHQWGHÀHFWLRQLVWKHVXPRIWKHLQGLYLGXDOGHÀHFWLRQ (d  7KHHTXLYDOHQWGHÀHFWLRQLVWKHSURGXFWRIWKHLQGLYLGXDOGHÀHFWLRQ

406

Strength of Materials

14. For parallel combination of two springs, the expression for the equivalent stiffness is (a)

K1 + K 2 1 (b) K = K K 1 2

K1 K 2 1 = K K1 + K 2

(c) K = K1 + K2

(d) K =

K1 K 2 .

ANSWERS 1. (c) 10. (b)

2. (b)

3. (c)

11. (a) 12. (c)

4. (a)

5. (d)

13. (a) and (b)

6. (a) 14. (c)

7. (c)

8. (b)

9. (c)

Springs

407

EXERCISES 1. A quarter-elliptic laminated spring of length 700 mm is loaded with a load of 3 kN producing a GHÀHFWLRQRIPP7KHFURVVVHFWLRQRIWKHVSULQJZLUHLVPPZLGHE\PPWKLFN)LQG the following: (a) the number of leaves in the spring (b) the maximum stress induced in the spring and (c) the height through which the load is released, giving a maximum stress of 1 kN/mm 2 Take E = 200 kN/mm2. (Ans.

(a) 13

(b) 282.57 N/mm2

(c) 269.3 mm).

2. A close coiled helical spring of mean coil radius equal to 6 times the wire diameter is subjected WRDQD[LDOORDGRI1,WFDQDEVRUEMRXOHVRIHQHUJ\SURGXFLQJDGHÀHFWLRQRIPP7KH maximum stress in the spring is not to exceed 80 N/mm2. Find the mean coil diameter, the wire diameter, the number of turns in the spring and the length of the spring. Take G = 80 kN/mm2. (Ans. 55 mm, 4.58 mm, 18.42, 3.18 m). 3. A close coiled helical spring of mean coil diameter 100 mm is subjected to a torque of 6.5 N˜m producing angular twist of 60o7KHGHÀHFWLRQRIWKHVSULQJXQGHUWKHDFWLRQRIDQD[LDOORDGRI 250 N is 120 mm. Find Poisson’s ratio of the spring wire. (Ans. 0.191). 4. A close coiled helical spring of 10 mm wire diameter has 15 turns with mean coil diameter of 120 mm. It is subjected to an axial load of 500 N. Find the shear stress in the spring wire (a) neglecting correction factor and (b) considering correction factor. (Ans.

(a) 152.78 N/mm2

(b) 171.03 N/mm2).

5. An open coiled helical spring of 8 mm wire diameter has mean coil radius of 120 mm and 10 turns. The angle of helix is 20o. It is subjected to an axial torque of 3 N˜m. Find the following: (a) the maximum direct and shear stress induced in the spring wire and (b) the angle of twist. Take G = 80 kN/mm2 and E = 200 kN/mm2. (Ans.

(a) 57.88 N/mm2, 29.84 N/mm2 (b) 17.65o).

6. :KDWLVWKHHUURULQWKHFDOFXODWLRQRIVWLIIQHVVLIRSHQFRLOHGKHOLFDOVSULQJLVPLVWDNHQWREHFORVH coiled helical spring. Given, angle of helix = 30 o and E/G = 4.0. Assume all the parameters of the two springs to be same. (Ans. 1% (negative)). 7. A close coiled helical spring is of 80 mm mean coil diameter. The spring extends by 37.75 mm when loaded axially by a weight of 500 N. There is an angular rotation of 45° when the spring is subjected to an axial couple of magnitude 20 N˜m. Determine Possion’s ratio of the material of the spring. (Ans. 0.2016).

408

Strength of Materials

8. A truck weighing 25 kN and moving at 2.5 m/s has to be brought to rest by a buffer. Find how many springs each of 25 coils will be required to store energy of motion during compression of 0.2 m. The spring is made of 25 mm diameter steel rod coiled to a mean diameter of 0.2 m. Take G = 100 GPa. (Ans. 16.3). 9. A close coiled helical spring is made of 10 mm diameter steel wire, the coil consisting of 10 complete turns with a mean diameter of 120 mm. The spring carries an axial pull of 200 N. Determine the shear stress induced in the spring neglecting the effect of stress concentration. 'HWHUPLQH DOVR WKH GHÀHFWLRQ RI WKH VSULQJ LWV VWLIIQHVV DQG WKH VWUDLQ HQHUJ\ VWRUHG E\ LW LI modulus of rigidity of spring material is 80 GPa. (Ans. 61.11 MPa, 34.56 mm, 5787 N/m, 3.456 joules). 10.)LQGWKHPD[LPXPVKHDUVWUHVVDQGWKHGHÀHFWLRQSURGXFHGLQDKHOLFDOVSULQJRIWKHIROORZLQJ VSHFL¿FDWLRQVLILWKDVWRDEVRUEWKHHQHUJ\RIN1P Mean diameter of the spring = 100 mm Diameter of the spring steel wire = 20 mm Number of coils = 30 Modulus of rigidity of steel = 85 GPa. (Ans.

338.87 MPa, 187.87 mm).

11. A close coiled helical spring made of 6 mm wire diameter and mean coil diameter 100 mm H[WHQGVE\PPXQGHUDQD[LDOORDGRI17KHVDPHVSULQJZKHQ¿UPO\¿[HGDWRQHHQG rotates through 90º under a torque of 5.7 N.m. Calculate the value of Poisson’s ratio for the material. (Ans. 0.3). 12. Find the load required to produce an extension of 8 mm in an open coiled helical spring made of 6 mm wire diameter, having 10 coils of mean diameter 76 mm with a helix angle of 20º. Also, calculate the bending and shear stresses produced in the surface of the spring wire. What would be the angular twist at the free end of the spring, when subjected to an axial torque of 1.5 N.m? Take E = 210 GPa and G = 70 GPa. (Ans. 20 N, 12.3 MPa, 16.8 MPa, 17.3°). 13. A compound spring is made of two close coiled helical springs connected in series, where each spring has 12 coils at a mean diameter of 25 mm. Find the diameter of the wire in one of the springs, if the diameter of wire in the other spring is 2.5 mm and the stiffness of the compound spring is 700 N/m. Estimate the greatest load that can be carried by the composite spring and the corresponding extension for a maximum shearing stress of 180 MPa. (Ans. 63.2 mm, 44.2 N). ‰‰‰

9 Strain Energy

Carlo Alberto Castigliano, born on 9 November 1847, was an Italian mathematician and physicist. He is well known for his Castigliano’s theorem, which is used for determining the displacements in a linear-elastic system using the partial derivatives of strain energy.

Carlo Alberto Castigliano (1847-1884)

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ :K\LVVWUDLQHQHUJ\VRQDPHG"  ‡ +RZGRHVVXGGHQO\DSSOLHGORDGGLIIHUIURPJUDGXDOO\DSSOLHGORDG"  ‡ :KDWLVVKHDUVWUDLQHQHUJ\"  ‡ :KDWLVYROXPHWULFVWUDLQHQHUJ\"  ‡ :KDWLV&DVWLJOLDQR¶VWKHRUHP"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_9

409

410

Strength of Materials

9.1

INTRODUCTION

:KHQDQH[WHUQDOORDGLVDSSOLHGRQDERG\LWWHQGVWRGHIRUPWKHERG\'XULQJWKHSURFHVVLQWHUQDO IRUFHVDUHLQGXFHGLQWKHERG\WRFRXQWHUWKHHIIHFWRIH[WHUQDOORDG,IWKHERG\UHJDLQVLWVVKDSH DQGVL]HRQUHPRYHGRIORDGLWLVVDLGWREHZLWKLQHODVWLFOLPLW7KLVOLPLWLVGH¿QHGE\+RRNH¶V law which states that within this limit, stress is directly proportional to strain. The internally induced force acting on unit area of the body is called stress. Energy is stored in the body during deformation process and this energy is called strain energy. The workdone to produce the deformation is equal to the strain energy stored in the body.

9.2

STRAIN ENERGY DUE TO DIRECT LOADS

The load acting on a body may be of the following types: (a) Gradually applied load (b) Suddenly applied load

9.2.1

(c) Impact load

Strain Energy due to Gradually Applied Load

Consider a bar of length l placed vertically and one end of it is attached at the ceiling (Fig. 9.1). This is the most general case.

Fig. 9.1

Let

P = Gradually applied load l = Length of the bar A = Cross-sectional area of the bar Gl 'HÀHFWLRQSURGXFHGLQWKHEDU V  $[LDO RU GLUHFW VWUHVV LQGXFHG LQ WKH EDU ,W PD\ EH WHQVLOH RU compressive, depending upon if the bar under consideration is under tension or compression load. E = Modulus of elasticity of the bar material

Strain Energy

The work done by the load is given as 1 W = ˜ P ˜ Gl 2 P %\GH¿QLWLRQ V = A or P =VA From Hooke’s law

Gl =

411

... (9.1)

Pl σl  AE E

On substituting Gl, equation (9.1) becomes

1 P 2l 2 AE

(in terms of P)

... (9.2)

=

1 σ 2 Al 2 E

(in terms of V)

... (9.3)

=

1 V2V 2 E

(in terms of V and V)

...(9.4)

W =

where

V = Volume of the bar = Al

The strain energy stored in the bar is equal to the workdone by the load. V2 U = ×V 2E 1 V 1 × V× × Stress × Strain × Volume = ×V= 2 E 2 The strain energy stored per unit volume is known as strain energy density, given as

...(9.5) ...(9.6)

V2 U = Strain energy density = ...(9.7) 2E V 7KHORDGYHUVXVGHÀHFWLRQFXUYHLVVKRZQLQ)LJGeVWDQGVIRUWKHHODVWLFGHÀHFWLRQDQGPe for the HODVWLFORDG7KHDUHDRIWKHFXUYHJLYHVWKHZRUNGRQHE\WKHORDGWRSURGXFHWKHGHÀHFWLRQ

Fig. 9.2

412

Strength of Materials

If the bar is not of uniform cross-sectional area as shown in Fig. 9.3, then strain energy stored in the bar is not represented by equation (9.6). The strain energy under this condition is given by the Equation (9.8) as 1 P2 U = 2 E

l

¨ 0

dx A

...(9.8)

Fig. 9.3

9.2.2

Strain Energy due to Suddenly Applied Load

:KHQWKHORDGLVDSSOLHGVXGGHQO\ZRUNGRQHE\WKHORDGLVJLYHQDV W = P ˜ Gl

... (9.9)

The applied load has the same value during the deformation of the bar, hence average load is P and not P . The workdone by the load is shown in Fig. 9.4. 2

Fig. 9.4

Strain Energy

413

The strain energy stored in the bar is given by equation (9.5). V2 U = ×V 2E Equating equations (9.5) and (9.9), we have P˜Gl =

V2 ×V 2E

V2 ⎛ σ ⋅ l ⎞⎟ × Al = ⎜⎝ E ⎟⎟⎠ 2E

P ⎜⎜

P =

⎛ ⎞ ⎜⎜δl = σl ⎟⎟ ⎜⎝ E ⎟⎠

V ¸A 2

2P ... (9.10) A Hence, the stress induced by the suddenly applied load is two times the stress produced by the same load but applied gradually.

or

V =

9.2.3

Strain Energy due to Impact or Shock Load

Consider the situation, when a load P is allowed to fall freely on the collar attached to the lower end of a bar. Let, h = Height through which the load falls on the collar The workdone by the falling load is given as W = P (h + Gl)

... (9.11)

It is important to note here that the load does not change and the same load acts throughout the process (Fig. 9.5).

Fig. 9.5

414

Strength of Materials

Equating workdone to the strain energy stored in the bar, we have P (h + Gl) =

V2 qV 2E

... (9.12)

⎛ σl ⎞ σ2 P ⎜⎜h + ⎟⎟⎟ = q Al ⎜⎝ 2E E⎠ ⎛ Al ⎞⎟ 2 ⎛ Pl ⎞⎟ ⎜⎜ ⎟ σ − ⎜⎜ ⎟ σ − Ph = 0 ⎜⎝ 2 E ⎟⎠ ⎜⎝ E ⎟⎠

or

⎛ ⎞ ⎜⎜δl = σl and V = Al ⎟⎟ ⎟⎠ ⎜⎝ E ...(9.13)

This is a quadratic equation in terms of V Its two values are given as 

V =

P ⎡⎢ 1± A ⎢⎢⎣

⎛ ⎞⎤ ⎜⎜1 + 2 AEh ⎟⎟ ⎥ ...(9.14) ⎜⎝ Pl ⎟⎠ ⎥⎥⎦ The effect of the load is to elongate the bar, hence it produces tensile stress in the bar. Therefore, possibility of negative stress does not arise. Equation (9.14) accordingly reduces to



V =

P ⎡⎢ 1+ A ⎢⎢⎣

⎛ ⎞⎤ ⎜⎜1 + 2 AEh ⎟⎟ ⎥ ⎜⎝ Pl ⎟⎠ ⎥⎦⎥

...(9.15)

when h = 0, then equation (9.15) becomes 2P V = ...(9.16) A The equation is same as equation (9.10). Hence, in case of zero height, stress induced in the bar is equal to stress produced by the same load but applied suddenly.  1RZVXSSRVHLIWKHGHÀHFWLRQGl is very small as compared to height h, equation (9.12) reduces to

σ2 q Al Ph = 2E or

V =

9.3

2EPh Al

...(9.17)

STRAIN ENERGY DUE TO SHEAR

Consider a cube ABCD of side l being subjected to a shear force P which is applied tangentially to the side CD. As a result, side CD changes to Cc Dc (Fig. 9.6). Let 

I = Angle turned by the side AD as a result of P being applied W = Shear stress induced on the side CD A = Area of each side = l2 G = Modulus of rigidity of the cube material

Strain Energy

415

Fig. 9.6

The shear force P is given as P = W˜ A = Wl 2

...(9.18)

In ' ADDc tan I = :KHQI is very small

DD ' AD

δl = Shear strain l Gl = Il

tan I a I = or

The workdone by the shear force is given as 1 1 W = Pδl = × τ l 2 × φ l 2 2

τ 1 1 = × τ×φ×l 3 = × τ× ×l 3 G 2 2

τ2l 3 = 2G Hence, the strain energy stored in the cube is given as W2 qV U = 2G where

9.4

...(9.19)

(on substituting P and Gl ) ⎛ ⎞ ⎜⎜φ = τ ⎟⎟ ⎜⎝ G ⎟⎠ ...(9.20)

...(9.21)

V = l 3 = Volume of the cube.

STRAIN ENERGY DUE TO PURE BENDING

Consider a small length dx of the beam at a distance x from A. Let the bending moment acting on the length be M. (Fig. 9.7).

416

Strength of Materials

Fig. 9.7

Bending stress acting on the elementary area (dA) at a distance yIURPWKHQHXWUDOD[LV NA) is given by using bending equation as M ¸y ... (9.22) I I 0RPHQWRILQHUWLDRIWKHFURVVVHFWLRQDERXWWKHQHXWUDOD[LV NA)

V = where

The strain energy stored in the elementary area is =

V2 × Volume of the elementary area 2E

=

V2 dx ˜ dA 2E

=

M 2 y2 dx dA I 2 2E

(on substituting V from equation (9.22))

The strain energy stored in the beam between 1 and 2 is given as dU =

M2 2 EI 2

M2 dx 2 EI 2 EI 2 The strain energy stored in the entire beam is obtained as =

M2

dx ¨ y 2 dA dx ⋅ I =

( I = ∫ y 2 dA )

l

U =

¨ dU 0 l

=

¨ 0

M2 dx 2 EI

...(9.23)

Strain energy stored in different beams due to bending under different loading conditions are given in Table 9.1.

Strain Energy

417

Table 9.1 Strain energy due to Bending Loaded beam 1.

A simple beam with a point load not at its centre

2.

A simple beam with a point load at its centre

3.

A simple beam with udl over the entire span

/RDGLQJ¿JXUH

6WUDLQHQHUJ\VWRUHG

P 2l12l22 (l1 > l2) 6 EIL

P 2l 3 96 EI

P 2l 3 240 EI where, P = wl = Total load on the beam

4.

A cantilever beam with a point load at free end

5.

A cantilever beam with udl over the entire span

P 2l 3 6 EI

P 2l 3 40 EI where, P = wl = Total load on the beam

418

Strength of Materials

9.5

STRAIN ENERGY DUE TO PRINCIPAL STRESSES

Consider a cube of side l being subjected to three mutually perpendicular principal stresses V 1, V2 and V3 along x, y and z axes respectively. (Fig. 9.8).

Fig. 9.8

If 1, 2 and 3 are the strains produced in x, y and z directions respectively due to V1, V2 and V3, then ⎫ V ⎪ V1 V −v 2 −v 3⎪ ⎪ E E E⎪ ⎪ ⎪ V3 V2 V1 ⎪⎪ ∈2 = −v −v ⎬ E E E⎪ ⎪ ⎪ V3 V1 V2 ⎪ ∈3 = −v −v ⎪ ⎪ E E E⎪ ⎪ ⎭ v = Poisson’s ratio ∈1 =

where

...(9.24)

E = Modulus of elasticity The strain energy stored in the cube is given as 1 × Stress × Strain × Volume 2 1 1 1 V  V = V1 1 V + V2 2 V + 2 2 2 3 3 V (V1 1 + V2 2 + V3 3) = 2 V = Volume of the cube = l3

U =

where

...(9.25)

Substituting 1, 2 and 3 in equation (9.25), we have U =

V ⎡ 2 V1 + V22 + V32 − 2v (V1V2 + V2 V3 + V3V1 )⎤⎥ ⎦ 2 E ⎢⎣

...(9.26)

Strain Energy

419

Strain energy stored per unit volume of the cube is given as 1 ⎡ 2 U V1 + V22 + V32 − 2v (V1V2 + V2 V3 + V3V1 )⎤⎥ ...(9.27) = ⎢ ⎣ ⎦ E 2 V Equation (9.27) does not contain any shear stress term because principal planes do not have shear stresses. For a uniaxial stress system 

V2 = 0



V3 = 0 and

V1 = V

Equation (9.26) reduces to For a biaxial stress system

U =

V2 ×V 2E

V3 = 0

 Equation (9.26) reduces to

U =

9.6

V ⎡ 2 V1 + V22 − 2v V1V2 ⎤⎥ ⎢ ⎣ ⎦ 2E

...(9.28)

STRAIN ENERGY DUE TO VOLUMETRIC STRAIN

Refer Fig. 9.9. Let

V = Volumetric strain produced in the cube K = Bulk modulus of elasticity V1 = V2 = V3=V= Stress of equal intensity applied on all sides of the cube

Fig. 9.9

The volumetric strain is produced as a result of changes in all sides of the cube due to V 1, V2 and V3.

420

Strength of Materials

%\GH¿QLWLRQ K = But

V ‰v

V = 1 + 2 + 3 = 3 =

3V (1 – 2v) E

(using equation (9.24))...(9.29)

Strain produced in all the sides of the cube is same, since equal stress intensity is acting on them. The workdone in any of the three directions is given as 1 W = î/RDGî'HÀHFWLRQ 2 1 1 = × V ˜ l 2 × l = V ×  × l3 2 2

⎛ ⎞ ⎜⎜∈= δl ⎟⎟ ⎜⎝ l ⎟⎠

The total workdone is given as Wt = 3W =

But or

3 × V ×  × l3 2

=

V × 3∈×V 2

=

V 3V q (1 – 2v) × V 2 E

3 V2 = (1 – 2v) × V 2 E E = 3K(1 – 2v) E =K 3(1 − 2v)

(using equation (9.29)) ...(9.30)

...(9.31)

Substituting K in equation (9.30), we get V2 1 1 × V2 × ×V = ×V K 2 2K Hence, the strain energy stored in the cuble is given as Wt =

U = Wt =

V2 × Volume 2K

...(9.32)

9.7 SHEAR STRAIN ENERGY DUE TO PRINCIPAL STRESSES Let a rectangular body be subjected to three mutually perpendicular principal stresses (Fig. 9.10). Strain energy stored in the body is given as U V

=

1 ⎡ 2 V1 + V22 + V32 − 2 v (V1V2 + V2 V3 + V3V1 )⎤⎥ ⎦ 2 E ⎢⎣ (using equation (9.27))

Strain Energy

421

Fig. 9.10

The average stress is given as 

V1 V2 V3 3 If V1 =V2 = V3 = V, then the strain energy equation reduces to Vav =

...(9.33)

2

3Vav U = (1 – 2v) ... (9.34) 2E V Equation (9.34) is same as equation (9.32) and it gives the strain energy due to change in volume of the body. The volumetric strain energy due to Vav is given as U V

3V2av (1 – 2v) = 2E 2 3 ⎛⎜ V1 + V2 + V3 ⎞⎟ ⎟ = ⎜ ⎟⎠ (1 – 2v) 2 E ⎜⎝ 3

=

(using equation (9.33))

1 (V1 + V2 + V3 )2 (1 – 2v) 6E

⎛1 − 2v ⎞⎟ ⎡ 2 2 2 ⎟ V + V2 + V3 + 2 (V1V2 + V2 V3 + V3V1 )⎤⎦⎥ = ⎜⎜⎜ ⎝ 6 E ⎟⎠ ⎣⎢ 1

... (9.35)

On per unit volume basis, we have Shear strain energy

= Total strain energy – Volumetric strain energy

... (9.36)

Using equations (9.27) and (9.35), the shear strain energy per unit volume is ⎛1 + v ⎞⎟ ⎡ 2 2 2 ⎟ (V − V2 ) + (V2 − V3 ) + (V3 − V1 ) ⎤⎦⎥ Us = ⎜⎜⎜ ⎝ 6 E ⎟⎠ ⎣⎢ 1 But

E = 2G (1 + v)

... (9.37)

422

Strength of Materials

On substituting E in equation (9.37), equation (9.37) becomes 1 ⎡ (V1 − V2 ) 2 + (V2 − V3 )2 + (V3 − V1 )2 ⎤⎥ ⎦ 12G ⎢⎣ This is the required expression for the shear strain energy on per unit volume basis. Us =

If 

V2 = V3 = 0

and 

V1 = V

... (9.38)

Hence, the shear strain energy per unit volume is V2 Us = 6G

9.8

... (9.39)

CASTIGLIANO’S THEOREM

The Castigliano’s theorem is named after the Italian engineer Alberto Castigliano (1847-1884), ZKR¿UVWSURSRVHGLW,WLVEDVHGRQWKHZRUNHQHUJ\SULQFLSOHDQGLVPDLQO\XVHGWRGHWHUPLQHWKH GHÀHFWLRQRIDVWUXFWXUHIURPWKHVWUDLQHQHUJ\RIWKHVWUXFWXUHXQGHUGLIIHUHQWORDGLQJFRQGLWLRQV There are two statements of this theorem as discussed below.

First theorem The partial derivative of the strain energy with respect to any displacement produced within elastic limit, as a result of the application of external forces on a given member, gives forces in the direction of displacements. Mathematically, Pk = where

sU sGk

... (9.40)

U = Strain energy Gk = Displacement in the direction of k Pk = Force in the direction of k

Second theorem The partial derivative of the strain energy with respect to a force produced, within elastic limit, gives the displacement in the direction of force. Mathematically, 

Gk =

sU sPk

... (9.41)

Example 9.1 $VROLGVWHHOURGRIPOHQJWKDQGPPGLDPHWHULVVXEMHFWHGWRDQD[LDOORDGRI N1)LQGWKH stresses induced in the rod if the load is applied (a) gradually (b) suddenly and (c) with impact after IDOOLQJWKURXJKDKHLJKWRIPP$OVR¿QGWKHVWUDLQHQHUJ\VWRUHGLQWKHURGXQGHUWKHJLYHQ FRQGLWLRQV7DNHE N1PP2.

Strain Energy

Solution:

423

Given,

Length of the steel rod,

l = 5 m = 5 × 1000 mm

'LDPHWHURIWKHURG

d = 10 mm

$[LDOORDG

P = 5 kN = 5 × 103 N

Height of the impact load,

h = 150 mm

(a  :KHQORDGLVJUDGXDOO\DSSOLHGWKHVWUHVVLQGXFHGLQWKHURGLVJLYHQDV V =

 where

P A

A = Cross-sectional area of the rod =

π 2 π d = ×102 = 78.54 mm2 4 4

Hence, the stress in the rod is 

5 q103 = 63.66 N/mm2 78.54 The strain energy stored in the rod is given by Equation (9.3). V =

U = =

Ans.

1 σ 2 Al 2 E 1 (63.66) 2 q 78.54 q 5 q1000 q N˜mm 2 2 q105

= 3978.88 N˜mm = 3.978 joule

Ans.

(b  :KHQORDGLVVXGGHQO\DSSOLHGWKHVWUHVVLQGXFHGLQWKHURGLVJLYHQDV 

V = 2 =

P A

2 q 5 q103 = 127.32 N/mm2 78.54

Ans.

The strain energy stored in the rod is U = =

V2 qV 2E (127.32) 2 q 78.54 q 5 q1000 2 q 2 q105 q103

= 15.91 joules

(V = Al) Ans.

424

Strength of Materials

(c  'XULQJWKHLPSDFWORDGLQJWKHVWUHVVLQGXFHGLVJLYHQE\XVLQJHTXDWLRQ   V =

=

P ⎡⎢ 2 AEh ⎤⎥ 1 + 1 + A ⎢⎢⎣ Pl ⎥⎥⎦ 5 ×103 78.54

⎤ ⎡ 5 ⎢1 + 1 + 2 × 78.54 × 2 ×10 ×150 ⎥ ⎥ ⎢ 5 ×103 × 5 ×103 ⎥⎦ ⎢⎣

= 940 N/mm2

Ans.

The strain energy stored in the rod is given as (940) 2 × 78.54 × 5 ×103 σ2 × V = U = 2E 2 × 2 ×105 ×103 = 867.47 joules

Ans.

Example 9.2 $FXEHRIVLGHPP¿[HGDWWKHERWWRPLVVXEMHFWHGWRDVKHDUIRUFHRIN1RQLWVWRSIDFH)LQG strain energy stored in the cube and modulus of resilience. Take G = 80 kN/mm2. Solution:

Given,

Side of the cube,

l = 100 mm

Shear force applied,

P = 50 kN = 5 × 104 N

Volume of the cube,

V = l3 = (100 mm)3 = 106 mm3

P 5 ×104 = The shear stress produced, W = = 5 N/mm2 A 100 ×100 The strain energy stored in the cube is given by equation (9.21). U =

W2 qV 2G

52 q106 N˜mm = 156.25 N˜mm = 4 2 q8q10 = 0.156 joule

Ans.

The modulus of resilience is the strain energy stored per unit volume of the cube, given by U V

=

156.25 106

N˜mm/mm3

= 1.5625 × 10–4 Nmm/mm3

Ans.

Strain Energy

425

Example 9.3 A rod of diameter 10 mm and length 1.5 m hangs vertically from the ceiling of a roof. A collar is attached at its lower end on which a load of 250 N falls from a height of 200 mm. Find the strain HQHUJ\DEVRUEHGDQGWKHLQVWDQWDQHRXVGHÀHFWLRQRIWKHURG7DNHE = 200 kN/mm2. Solution:

Given,

 'LDPHWHURIWKHURG Length of the rod,

d = 10 mm l = 1.5 m = 1.5 × 1000 = 1500 mm

Impact load to be applied, P = 250 N Height through which the load falls, h = 200 mm The cross-sectional area of the rod A given by A =

π 2 d 4

π q102 = 78.54 mm2 4 The stress induced in the rod is given by equation (9.15). =



⎡ ⎤ P⎢ 2 AEh ⎥ V = ⎢1 + 1 + ⎥ A⎢ Pl ⎥ ⎣ ⎦ ⎤ ⎡ 250 ⎢ 2 × 78.54 × 2 ×105 × 200 ⎥ × + + 1 1 = ⎥ ⎢ 78.54 ⎢ 250 ×1500 ⎥⎦ ⎣ = 415.22 N/mm2

Ans.

The strain energy stored in the rod is given by equation (9.5). U =

σ2 σ2 ×V = × Al 2E 2E 2

=

(415.22) × 78.54 ×1500 2 × 2 ×105 ×103

= 50.77 joules

Ans.

 7KHLQVWDQWDQHRXVGHÀHFWLRQRIWKHURGLVJLYHQDV 

GL = =

σl E 415.22 q1500 2 q105

= 3.11 mm

Ans.

426

Strength of Materials

Example 9.4 A rectangular block is subjected to three mutually perpendicular tensile stresses of magnitude 60, 70 and 80 N/mm 2. Calculate strain energy and shear strain energy. The Poisson’s ratio is 0.3. Take E = 200 kN/mm2. Solution: Given, Tensile stresses,      Poisson’s ratio,

V1 V2 V3 v

= 60 N/mm2 = 70 N/mm2 = 80 N/mm2 = 0.3

The strain energy stored per unit volume of the block is given by equation (9.27). U V

= =

1 2 [V1 + V22 + V32 − 2v (V1V2 + V2 V3 + V3V1 )] 2E 1 2 × 2 ×105

[602 + 702 + 802 − 2 × 0.3(60 × 70 + 70 × 80 + 80 × 60)]

= 0.0153 N˜mm/mm3

Ans.

The relationship between E and G is given as E = 2G (1 + Q) E 2 ×105 = or G = = 7.7 × 104 N/mm2 2(1 + v) 2(1 + 0.3) The shear strain energy per unit volume is given by equation (9.38). U V

1 [(V1 − V2 )2 + (V2 − V3 ) 2 + (V3 − V1 )2 ] 12G 1 [(60 − 70)2 + (70 − 80)2 + (80 − 60)2 ] = 12 × 7.7 ×104

=

= 6.5 × 10

–4

N˜mm/mm3

Ans.

Example 9.5 Find the weight which falls through a height of 5 m on a collar attached to the lower end of a vertical URGRIGLDPHWHUPPDQGOHQJWKP7KHGHÀHFWLRQSURGXFHGLQWKHURGLVPP Take E = 200 GPa. Solution:

Given,

Height of the weight, Diameter of the vertical rod, Length of the rod,  'HÀHFWLRQSURGXFHGLQWKHURG

h d l Gl

=5m = 40 mm =3m = 5 mm

Strain Energy

427

The cross-sectional area of the rod is A, given by 2 π 2 π ⎛ 40 ⎞⎟ A = d = ×⎜⎜⎜ ⎟⎟ 4 4 ⎝1000 ⎠ = 1.256 × 10–3 m2 The instantaneous stress produced in the rod is given as V =



= Using equation (9.15), we have V =



8

3.34 × 10 =

E δl l 200 ×109 × 5 ×10−3 = 3.34 × 108 N/m2 3 P ⎡⎢ 1+ A ⎢⎢⎣

⎛ ⎞⎤ ⎜⎜1 + 2 AEh ⎟⎟ ⎥ ⎜⎝ Pl ⎟⎠ ⎥⎦⎥

P 1.256 ×10−3

⎡ ⎢ ⎢1 + ⎢⎢ ⎣

−3 9 ⎛ ⎞⎤ ⎜⎜1 + 2 ×1.256 ×10 × 200 ×10 × 5 ⎟⎟ ⎥ ⎟⎟ ⎥ ⎜⎜ P ×3 ⎝ ⎠ ⎥⎥ ⎦

P = 210 N

Solving for P, we get

Ans.

Example 9.6 A cantilever beam of length l carries a udl of intensity w per unit length over its entire span and a point load PDWLWVIUHHHQG)LQGWKHPD[LPXPGHÀHFWLRQRIWKHEHDPXVLQJ Castigliano’s theorem. Solution:

Refer Fig. 9.11.

The bending moment due to load P at a distance x from the free end is given as M1 = – Px The bending moment due to udl at x is given as M2 = –

wx 2 2

Fig. 9.11

The combined bending moment at x is given as ⎛ wx 2 ⎞⎟⎟ ⎜ M = – ⎜⎜ Px + ⎟ ⎜⎝ 2 ⎟⎠

428

Strength of Materials

The strain energy stored in the beam is given as l

U =

¨ 0

M 2 dx 2 EI

 7KHGHÀHFWLRQLQWKHGLUHFWLRQRIORDGDSSOLHGXVLQJVHFRQG&DVWLJOLDQR¶VWKHRUHPLVREWDLQHGDV sU sP ⎡ l 2 ⎤ M dx ⎥ ∂ ⎢⎢ ∫ ⎥ l ⎢⎣ 0 2 EI ⎥⎦ 2 M ∂M = ⋅ ⋅ dx = ∫ ∂P 2 EI ∂P

G =



But

l

Hence, 

G =

∫ 0

1 = EI

M (−x )dx EI l

⎧ ⎛ ⎪ ⎪

∫ ⎨⎪−⎜⎜⎜⎜⎝ Px + 0

⎪ ⎩

l⎛

⎫ wx 2 ⎞⎟⎪ ⎟⎟⎪ ⎬ (−x )dx 2 ⎟⎠⎪ ⎪ ⎭ ⎟⎟ dx ⎟⎟ ⎠

wx 2 ∫ ⎜⎜⎜⎜⎝ Px + 2 0

1 = EI

4⎤ ⎡ Px 3 wx 4 ⎤ l ⎡ 3 ⎢ ⎥ = 1 ⎢ Pl + wl ⎥ + ⎢ 3 8 ⎥⎥⎦ 0 EI ⎢⎢⎣ 3 8 ⎥⎥⎦ ⎢⎣

G =

(on substituting M)

3⎞

1 = EI

when P = 0, then 

0

sM =–x sP

Ans.

wl 4 8EI

SHORT ANSWER QUESTIONS 1. :K\LVVWUDLQHQHUJ\VRFDOOHG" 2. :KDWLVVWUDLQHQHUJ\GHQVLW\" 3. :KDWLVWKHVWUDLQHQHUJ\VWRUHGLQDVKDIWZKHQWKHVKDIWLVVXEMHFWHGWRDWRUTXHT and produces an angle of twist T" 4. :KDWLV&DVWLJOLDQR¶VWKHRUHP":KHUHLVLWXVHG" 5. :KDWLVWKHEDVLFSULQFLSOHRI&DVWLJOLDQR¶VWKHRUHP"

Strain Energy

429

MULTIPLE CHOICE QUESTIONS 1. Energy stored in a material during its deformation is known as (a) elastic energy

(b) plastic energy

(c) strain energy

(d) potential energy.

2. Strain energy stored per unit volume in a bar subjected to a gradually applied load P on its cross-section A is given as P2 2P . P2 P (b) (c) (d) (a) 2 2A E 2 AE A2 E 2 A2 E 3. :RUNGRQHE\DJUDGXDOO\DSSOLHGORDGPRQDERG\SURGXFLQJDGHÀHFWLRQGl is given as 2 2 P ¸δ l (d) Pδ l (c) P δ l 2 2 2 4. 7KHPD[LPXPVWUDLQHQHUJ\VWRUHGLQDERG\DWWKHHODVWLFOLPLWLVFDOOHG (a) resilience (b) modulus of resilience (c) proof resilience (d) potential energy.

(a) P . Gl

(b)

5. The workdone by a suddenly applied load PSURGXFLQJDGHÀHFWLRQGl is given as P2 ¸ δ l (a) P ¸ δ l (d) P ˜ Gl 2. (b) P ˜ Gl (c) 2 2 6. The strain energy stored in a shaft of length l subjected to a torque T with its polar moment of inertia J is (a)

T 2l 2 JG

(b)

T 2l JG

(c)

2T 2l JG

(d)

Tl . 2 JG

7. The strain energy stored in a simply supported beam of span l loaded with a central point load P and moment of inertia I is Pl 3 (a) 48EI

P 2l 2 (b) 96 EI

P 2l 3 (c) 96 EI

P 2l 3 . (d) 24 EI

8. The strain energy stored in a cantilever beam of span l carrying a point load P at its free end with moment of inertia I is P 2l 3 (a) 48EI

P 2l 3 (b) 6 EI

P 2l 3 (c) 24 EI

P 2l 3 . (d) 96 EI

9. The strain energy stored per unit volume in a body subjected to two mutually perpendicular principal stresses V1 and V2 with Poisson’s ratio v is (a)

1 (V12 + V22 − 2v V1V2 ) 2E

(b)

1 (σ1 σ 2 σ1σ 2 ) E

(c)

1 2 (V1 + V22 − 2v V1V2 ) E

(d)

1 (V1 + V2 − V1V2 ) . E

430

Strength of Materials

10. The strain energy stored per unit volume in a cube subjected to a stress intensity V on its all sides with bulk modulus K is V V V2 . σ2 (c) (d) (b) 2K K2 2K 2 2K 11. The strain energy stored per unit volume in a cube subjected to three mutually perpendicular principal stresses V1, V2 and V3 with 1, 2 and 3 being the strains produced in the respective directions of the stresses is 1 (b) (a) V11 + V2 2 + V33 (V11 + V22 + V33) 2 1 2 1 (d) (V112 + V2 22 + V332). (c) (V1 1 + V22 2 + V333) 2 2 12. The stress produced by a suddenly applied load is how many times the stress produced by the JUDGXDOO\DSSOLHGORDG" (a)

(a) four times

(b) three times

(c) two times

(d) eight times.

13. The stress produced in a bar of length l and cross-section A when a load P is dropped on it from a height hSURGXFLQJQHJOLJLEOHGHÀHFWLRQLVJLYHQDV (a)

2EPh Al

(b)

EPh Al

(c)

EPh 2 Al

(d)

EPh . 3 Al

14. The strain energy stored per unit volume in a body subjected to a shear stress W is given as 2

W W 2W2 W2 . (d) (b) (a) (c) 2 2G 2G G G 15. Two bodies of equal cross-sections and equal lengths absorb equal strain energy when loaded GLIIHUHQWO\7KH¿UVWERG\LVFDUU\LQJDQD[LDOORDGP and the other one is acting as a cantilever beam with a point load P at its free end. The stresses produced in the two bodies are related as V V V2 (c) V1 = 2 (a) V1 = 2 (b) V1 = V2 (d) V1 = 2 . 3 2 2 16. For a straight bar of length l being subjected to a tensile load P, the strain energy is given as Pl 2 2 AE

Pl . Pl 2 (d) 2 A2 E AE 17. For a bar of length l being subjected to a bending moment M, the strain energy is given as Ml 2 Ml 2 . Ml 2 M 2l (b) (a) (d) (c) EI EI 2 EI 2 EI (a)

(b)

P 2l 2 AE

(c)

ANSWERS 1. (c) 10. (c)

2. (a)

3. (b)

11. (b) 12. (c)

4. (c)

5. (b)

6. (a)

7. (c)

8. (b)

13. (a) 14. (b) 15. (a) 16. (b) 17. (a).

9. (a)

Strain Energy

431

EXERCISES 1.  $  YHUWLFDOEDURIOHQJWKPLV¿[HGDWLWVORZHUHQG,WLVVWULNHGE\DKRUL]RQWDOO\PRYLQJERG\ of mass 5 kg at its upper end with a certain velocity VSURGXFLQJDPD[LPXPVWUHVVRI03D in the bar. Find V. Take E = 200 GPa. (Ans. 3.61 m/s). 2. $EDURIOHQJWKPLVVXEMHFWHGWRDQD[LDOORDGRIN17KHGLDPHWHURIWKHEDUIRURQHKDOI of its length is 30 mm and for the other half 60 mm. Calculate the strain energy stored in the bar. Take E = 200 GPa. (Ans. 2.76 joules). 3. Show that the strain energy stored in a beam of length l having rectangular cross-section P 2l 3 , where the symbols supported at the ends and loaded with a central point load P is given as 96 EI have their usual meanings. 4. A load of 20 kN falls through a height of 1 m on a rectangular beam of length 1 m with crosssection 20 mm wide × 30 mm deep supported at the ends. Find the instantaneous stress developed and the strain energy stored in the beam. Take E = 200 GPa. (Ans. 12747.18 N/mm2, 27 kJ). 5. A load of 2 kN is allowed to fall freely from a height of 1 m at the centre of a circular VLPSO\ VXSSRUWHG EHDP RI OHQJWK  P DQG GLDPHWHU  PP )LQG WKH GHÀHFWLRQ RI WKH EHDP Take E = 200 GPa. (Ans. 77.65 mm). 6. 8VLQJ &DVWLJOLDQR¶V WKHRUHP ¿QG WKH GHÀHFWLRQ DW WKH FHQWUH RI D VLPSO\ VXSSRUWHG EHDP RI length l carrying a triangular load which varies from zero at one end to w/unit length at another end. ⎛ 1 wl 4 ⎞⎟⎟ . ⎜⎜ Ans. ⎟ ⎜⎜ 156.25 EI ⎟⎠ ⎝ 7. $KROORZVKDIWKDYLQJWKHRXWVLGHDQGWKHLQVLGHGLDPHWHUUDWLRRILVVXEMHFWHGWRDPD[LPXPVKHDU 5W 2 , stress of W. Show that the strain energy stored in the shaft on per unit volume basis is given by 16G where G is the shear modulus of the shaft material. 8. $PORQJEDULVVXEMHFWHGWRDQD[LDOSXOOZKLFKLQGXFHVDPD[LPXPVWUHVVRI03D7KH area of cross-section of the bar is 2 × 10–4 m2 over a length of 0.95 m and for the central 0.05 m length the sectional area is equal to 1 × 10–4 m2. Assuming that E for the bar material is 200 GPa, calculate the strain energy stored in the bar. (Ans. 2.953 joules). 9. A 500 mm × 180 mm rolled steel beam is simply supported over a span of 6m. A load of 20 kN is GURSSHGRQWRWKHPLGGOHRIWKHEHDPIURPDKHLJKWRIPP)LQGWKHPD[LPXPLQVWDQWDQHRQV GHÀHFWLRQDQGWKHPD[LPXPVWUHVVLQGXFHG7KHVHFRQGPRPHQWRIDUHDRIWKHVHFWLRQHTXDOVWR 45218 × 10–8 m4 and the Young’s modulus is 200 GPa. (Ans. 6.026 mm, 36.715 MPa). 10. 8VLQJ WKH &DVWLJOLDQR¶V  WKHRUHP FDOFXODWH WKH YHUWLFDO GHÀHFWLRQ DW WKH PLGGOH RI D VLPSO\ supported beam which carries a uniformly distributed load of intensity w over the full span l. The ÀH[XUDOULJLGLW\EI of the beam is constant and only strain energy of bending is to be considered. 4 ⎞ ⎛ ⎜⎜ Ans. 5wl ⎟⎟ . ⎟ ⎜⎜ 384 EI ⎟⎠ ⎝

432

Strength of Materials

11. A steel tube having outside and inside diameter of 100 mm and 60 mm respectively is bent into the form of a quadrant of 2 m radius as shown in Fig. 9.12. One end is rigidly attached to a horizontal base plate to which a tangent to that end is perpendicular, and the free end supports DORDGRIN1'HWHUPLQHWKHYHUWLFDODQGKRUL]RQWDOGHÀHFWLRQVRIWKHIUHHHQGXQGHUWKLVORDG using the Castigliano’s theorem. Take E = 200 GPa.

Fig. 9.12

(Ans.

7.353 mm, – 4.681 mm).

12. A shaft circular in section (Fig. 9.13) and of length l is subjected to a variable torque given by kx 2 , where x is the distance measured from one end of the shaft and k is a constant. Find the i2 angle of twist for the shaft using Castigliano’s theorem. Torsional rigidity of the shaft is JG.

Fig. 9.13

⎛ ⎞ ⎜⎜ Ans. kl ⎟⎟ . ⎜⎝ 3JG ⎟⎠ ‰‰‰

10 Theory of Elastic Failure

Henri Edouard Tresca, born on 12 October 1814, was a French mechanical engineer. He discovered the Tresca yield criterion, also called maximum shear stress theory, which is one of the two main failure criteria used today for ductile materials along with von Mises yield criterion. He became an honorary member of the American Society of Mechanical Engineers (ASME) in 1882. He is called the ‘father of plasticity’ for his contribution LQWKH¿HOGRIQRQHODVWLFGHIRUPDWLRQ

Henri Edouard Tresca (1814-1885)

LEARNING OBJECTIVES After reading this chapter, you will be able to answer some of the following questions:  ‡ :K\LVWKH\LHOGVWUHQJWKPRVWLPSRUWDQWFULWHULDRIDPDWHULDO¶VIDLOXUH"  ‡ :KLFKLVWKHVLPSOHVWWKHRU\RIDPDWHULDO¶VIDLOXUH"  ‡ :K\LVWKH7UHVFD¶V\LHOGFULWHULRQWKHPRVWSRSXODUWKHRU\RIDPDWHULDO¶VIDLOXUH"  ‡ :KDWLVDSULQFLSDOSODLQ"  ‡ :KDWLVWKHGLIIHUHQFHEHWZHHQVWUDLQHQHUJ\DQGVKHDUVWUDLQHQHUJ\"

© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_10

433

434

Strength of Materials

10.1 INTRODUCTION When a material is stressed, there is a possibility for it to fail. Sometimes it fails without giving any indication and sometimes the failure is delayed. The analysis of failure of materials helps to know WKHDFWXDOUHDVRQVRIWKHLUIDLOXUH7KHUHDUH¿YHWKHRULHVUHODWLQJWRDPDWHULDO¶VIDLOXUHEXWQRQHRI them give the exact reason of failure because of the complexity of compound stress system under multiaxial loading. Yield point is the stage in the material, beyond which permanent deformation occurs, and it is the most important failure criteria. The different theories of failure, also called failure criteria, are discussed below. 10.2 MAXIMUM NORMAL STRESS THEORY 7KLVWKHRU\LVDOVRNQRZQDVPD[LPXPSULQFLSDOVWUHVVWKHRU\RU5DQNLQH¶VWKHRU\QDPHGDIWHUD Scottish civil engineer William John Macquorn Rankine (1820–1872). It is the oldest, as well as the simplest, of all the theories of failure, and applies well to brittle materials in all ranges of stresses, provided a tensile principal stress exists. According to this theory, a material fails, when the maximum principal stress developed due to external load reaches the ultimate strength of the material under uniaxial tension condition. Consider a rectangular body being subjected to two normal stresses Vx and Vy along x and y directions respectively (Fig. 10.1). Let Vu be the ultimate stress in simple tension.

Fig. 10.1

The principal stresses are given as V1, 2 =

σx + σ y 2

±

(σ x − σ y )2 + 4τ 2 2

...(10.1)

The maximum principal stress is given as V1 =

σx + σ y 2

+

(σ x − σ y )2 + 4τ 2 2

...(10.2)

The minimum principal stress is given as V2 =

σx + σ y 2



(σ x − σ y )2 + 4τ 2 2

...(10.3)

Theory of Elastic Failure

435

According to this theory, V1 =Vu σx + σ y

2 1 2 (σ x − σ y ) + 4τ = Vu 2 2 The normalised form of the Rankine’s theory for a plane stress condition is

+

...(10.4)

σ σ1 = ± 1 or 2 = ± 1 ...(10.5) σu σu and is graphically represented as shown in Fig. 10.2, where V1 and V2 are the principal stresses and Vu LVWKHXOWLPDWHVWUHQJWKLQWKHXQLD[LDOWHQVLRQWHVW7KHERXQGDU\RIWKHVTXDUHGH¿QHVWKHIDLOXUH criterion. The material is safer for the principal stresses lying within the square, but fails when the stresses lie on or outside the square.

Fig. 10.2 Plotting of the maximum normal stress theory.

10.3 MAXIMUM NORMAL STRAIN THEORY This theory is also called maximum principal strain theory or Saint Venant’s criterion in honour of a French mathematician Saint Venant. The theory holds reasonably well for cast iron (a brittle material), but is not generally used these days as other theories give better result. According to this theory, a material fails, when the maximum principal strain reaches the strain at the yield point in the simple tension test. Consider a complex stress system in which a body is subjected to three mutually perpendicular stresses Vx, Vy and Vz (Fig. 10.3). Let the major principal strain occurs in the direction of V x and Vyp is the yield point stress.

Fig. 10.3

436

Strength of Materials

Principal strain in the direction of Vx is given as σy σx σ ±v ±v z  E E E where

 

E = Modulus of elasticity v = Poisson’s ratio

According to this theory,

σy σ yp σx σ ±v ±v z = E E E E Vx±v (Vy +Vz) = Vyp

or

 

10.4 MAXIMUM TOTAL STRAIN ENERGY THEORY This theory is also called Haigh’s criterion. According to this theory, for a body being subjected to a complex stress system, failure occurs, when the total strain energy per unit volume reaches the total strain energy at the yield point in the simple tension test. The theory gives fairly good results for ductile materials, but is not used these days as other improved theories are available. &RQVLGHUDERG\ZKLFKLVVXEMHFWHGWRWKUHHPXWXDOO\SHUSHQGLFXODUVWUHVVHV )LJ  The strain energy per unit volume under the loading condition is given as U 1    = [ 2 + σ 2y + σ 2z ±v (Vx Vy + Vy Vz + Vz Vx @ V 2E σ x The strain energy per unit volume at the yield point, when Vyp is the yield stress, is given as U yp V According to this theory,

=

σ 2yp 2E



 

σ 2yp 1 2 2 2 [ σ x + σ y + σ z ±v (Vx Vy + Vy Vz + Vz Vx)] = 2E 2E or

2 σ 2x + σ y + σ 2z ±v (Vx Vy + Vy Vz + Vz Vx) = V2yp

 

10.5 MAXIMUM SHEAR STRESS THEORY This theory is also known as Tresca’s yield criterion, named after a French engineer Henri Edouard 7UHVFD ± 7KLVWKHRU\JLYHVEHWWHUUHVXOWZKHQDSSOLHGWRGXFWLOHPDWHULDOVDQGYDOXHV obtained are very near to experimental values, hence oftenly used for ductile materials in machine design and is one of the widely used laws of plasticity. According to this theory, a material fails, when the maximum shear stress developed in the material equals to the maximum shear stress at the yield point in the uniaxial tension test. Consider a body being subjected to two mutually perpendicular normal stresses V x and Vy and a shear stress W )LJ 

Theory of Elastic Failure

437

Fig. 10.4

The maximum shear stress is given as Wmax = ± =± where

σ1 − σ 2  2 1 2

  2

2 (σ x − σ y ) + 4τ

V1 = Major principal stress =

σx + σ y

+

2

(σ x − σ y ) 2 + 4τ 2 2

V2 = Minor principal stress =

σx + σ y 2

±

(σ x − σ y ) 2 + 4τ 2 2

The maximum shear stress, when the body is subjected to a stress Vyp, is given as σ yp  Wmaxyp = 2

 

According to this theory, Wmax =Wmaxyp

or

±

σ yp σ1 − σ 2 = 2 2

±

1 2

±

2

2 (σ x − σ y ) + 4τ

σ yp 2

2

2 (σ x − σ y ) + 4τ = Vyp

 

The normalised form of the Tresca's yield criterion for a plane stress condition is σ1 σ σ σ = ± 1 or 2 = ± 1 or 1 ± 2  “ σ yp σ yp σ yp σ yp

 

438

Strength of Materials

DQG LWV JUDSKLFDO UHSUHVHQWDWLRQ IRUPV D KH[DJRQ FDOOHG 7UHVFD V KH[DJRQ DV VKRZQ LQ )LJ  where V1 and V2 are the principal stresses and Vyp is the yield point stress in the uniaxial tension test.

Fig. 10.5 Plotting of the maximum shear stress theory.

7KHERXQGDU\RIWKHKH[DJRQGH¿QHVWKHIDLOXQHFULWHULRQ7KHPDWHULDOLVVDIHUIRUWKHVKHDU stresses within the hexagon, but fails when the stresses lie on or outside the hexagon.

10.6 MAXIMUM DISTORTION ENERGY THEORY 7KLVWKHRU\LVDOVRNQRZQDVYRQ0LVHV\LHOGFULWHULRQQDPHGDIWHUD*HUPDQ$PHULFDQPDWKHPD WLFLDQ5LFKDUGYRQ0LVHV ± RURFWDKHGUDOVKHDUWKHRU\RUPD[LPXPVKHDUVWUDLQHQHUJ\ theory, and is the most popular theory for predicting yielding in ductile materials. According to this theory, a material begins yielding, when the maximum shear strain energy (also called distortion energy) per unit volume equals to the shear strain energy per unit volume at the yield point in the uniaxial tension test. )RUDFRPSOH[VWUHVVV\VWHPVKRZQLQ)LJWKHHQHUJ\RIGLVWRUWLRQLVJLYHQDV Us 1 = [(V1  V2)2 + (V2±V)2 + (V±V1)[email protected] 12G V

 

where V1, V2 and V are three mutually perpendicular principal stresses. The shear strain energy at the yield point, when the body is subjected to a stress Vyp, is given as U yp V

=

σ 2yp 6G



According to this theory, U yp Us = V V σ 2yp 1 2 2 2 [( V1±V2) + (V2±V) + (V±V1) ] = 6G 12G

 

Theory of Elastic Failure

(V1±V2)2 + (V2±V)2 + (V±V1)2 = V2yp

or

439

 

The normalised form of the maximum distortion energy theory for a plane stress condition is 2

2

⎛ σ1 ⎞ ⎛ σ2 ⎞ ⎛ σ1 ⎞ ⎛ σ 2 ⎞ ⎜ ⎟ +⎜ ⎟ −⎜ ⎟⎜ ⎟ =1 ⎝ σ yp ⎠ ⎝ σ yp ⎠ ⎝ σ yp ⎠ ⎝ σ yp ⎠

 

DQGLWVJUDSKLFDOUHSUHVHQWDWLRQIRUPVDQHOOLSVHDVVKRZQLQ)LJ7KHERXQGLQJRIWKHHOOLSVH GH¿QHVWKHIDLOXUHFULWHULD7KHPDWHULDOLVVDIHUIRUWKHSULQFLSDOVWUHVVHVO\LQJZLWKLQWKHHOOLSVH but fails if the stresses lie on or outside the ellipse. The corresponding Tresca's hexagon is shown by dashed lines.

Fig. 10.6 Plotting of the maximum distortion energy theory.

Example 10.1 Two direct stresses are acting at two mutually perpendicular planes in a material. Both of WKHP DUH WHQVLOH DQG DUH  1PP 2 DQG  1PP2 respectively. Find the shear stress acting on the planes to consider the material’s failure according to maximum principal stress theory, maximum shear stress theory and shear strain energy theory. Take yield stress to be equal to 1PP2. Solution:

Given,

Direct stress in xGLUHFWLRQ Vx 1PP2 (Tensile) Direct stress in yGLUHFWLRQVy 1PP2 (Tensile) Yield stress,

Vyp 1PP2

The major principal stress V1 is given as V1 =

σx + σ y 2

+

(σ x − σ y ) 2 + 4 τ 2 2

440

Strength of Materials

2

2 (150 − 80 ) + 4τ

150 + 80 + = 2

2

4900 + 4τ2    2 The minor principal stress V2 is given as V2 = 

σx + σ y 2

±

...(1)

(σ x − σ y ) 2 + 4 τ 2 2

4900 + 4τ2 2 According to the maximum principal stress theory, we have  ±

...(2)

V1 = Vyp 1PP2 4900 + 4τ2    2



Solving for W,we get W 1PP2

Ans.

The maximum shear stress Wmax is given as Wmax =

σ1 − σ 2 2

Using equations (1) and (2), we get Wmax =

4900 + 4τ2

According to the maximum shear stress theory, we have Wmax = Vyp 1PP2 or

4900 + 4τ2  

Solving for W, we get W 1PP2

Ans.

The shear strain energy per unit volume, for a biaxial stress system, is given as 1 Us = [(V1±V2)2 + V22 + V12] 12G V =

1 [V2 + V22±V1 V2] 6G 1

(V 

Theory of Elastic Failure

441

According to the shear strain energy theory, we have σ 2yp 1 2 2 [V + V2±V1 V2] = 6G 1 6G V12 + V22±V1 V2 = V2yp   2 or

⎡ ⎢115 + ⎢ ⎣

2

⎡ 4900 + 4τ2 ⎤ 4900 + 4τ2 ⎥ + ⎢115 − 2 2 ⎥ ⎢ ⎦ ⎣ ⎛ ± ⎜115 + ⎜ ⎝



⎤ ⎥ ⎥ ⎦

2

2 4900 + 4τ2 ⎞ ⎛ ⎟ ⎜115 − 4900 + 4τ ⎟⎜ 2 2 ⎠⎝

⎞ ⎟  –4 ⎟ ⎠

Solving for W, we get W 1PP2

Ans.

Example 10.2 $FLUFXODUVKDIWRIGLDPHWHUPPLVVXEMHFWHGWRDPD[LPXPEHQGLQJPRPHQWRI–1¹mm DQGDWZLVWLQJPRPHQWRI–1¹mm. Find the factor of safety of the shaft using maximum shear VWUHVVWKHRU\7KH\LHOGVWUHVVRIWKHVKDIWLVQRWWRH[FHHG1PP 2. Solution:

Given,

Diameter of the shaft,

d PP –1¹mm

Bending moment,

M

Twisting moment,

T –1¹mm 1PP2

Yield stress of the shaft, Vyp

The major principal stress is given as V1 = =

[M +

16 πd3

]

M2 T2

]

[

16  2 5 2 5 2 v 3 –  (3 × 10 ) + (5 × 10 )  1PP π × (50 )

The minor principal stress is given as V2 = = 

16 πd 3

[ M±

16

M2 T2

[ – ± 

3

π × (50 )

 ±1PP2

] (3 × 105 )2 + (5 × 105 ) 2

]

442

Strength of Materials

7KHQHJDWLYHVLJQVLJQL¿HVWKHFRPSUHVVLYHQDWXUHRIV2. The maximum shear stress is given as Wmax =

σ1 − σ2 36  11.53 =  1PP2 2 2

The working simple stress corresponding to Wmax is given as Vw = 2 –Wmax  1PP2

 Hence, the factor of safety is

n =

σ yp σw

=

90   47.52

Ans.

Example 10.3 For a complex stress system, three principal stresses are 2V„V„DQG±V„. The stress in simple WHQVLRQDWWKHHODVWLFOLPLWLV1PP2. Find the value of V„ according to (a) the maximum principal stress theory (b) the maximum principal strain theory (c) the total strain energy theory (d) the maximum shear stress theory and (e WKHGLVWRUWLRQHQHUJ\WKHRU\7DNH3RLVVRQ¶VUDWLRWREH Solution:

Given,

Major principal stress,

V1 = 2V„

Minor principal stress,

V2 ±V„

Third principal stress,

V V„

Stress at the elastic limit,

Ve 1PP2

Poisson’s ratio,

v 

The minor principal stress is compressive because of negative sign associated with it. (a) Using the maximum principal stress theory, we have Major principal stress V1 = Stress at the elastic limit Ve V1   2V„  or

V„ 1PP2

(b) Using the maximum principal strain theory, we have σ1 σ σ σ −v 2 −v 3 = e E E E E (On replacing Vx, Vy and Vz by V1, V2 and VUHVSHFWLYHO\LQHTXDWLRQ   or

V1±vV2±vV = V

Ans.

Theory of Elastic Failure

or

443

2V „±– ±V„ ± V „  

On solving, we get V„ 1PP 2

Ans.

(c) Using the total strain energy theory, we have 2 σ12 + σ22 + σ32 − 2υ (σ1 σ2 + σ2 σ3 + σ3σ1 ) = Ve

(On replacing Vx, Vy and Vz by V1, V2 and VUHVSHFWLYHO\LQHTXDWLRQ   ( 2σ ′ )2 + ( − 1σ ′ )2 + (1.5σ ′ )2 − 2 × 0.25 [( 2σ ′ × ( − 1σ ′ ) + ( − 1σ ′ ) × (1.5σ ′) + (1.5σ ′) × 2σ ′)]    2 On solving, we get V„ 1PP2

Ans.

(d) The maximum shear stress is given as Wmax =

σ1 − σ2 2

2σ′ − (− 1σ′)  V„ 2 According to the maximum shear stress theory, we have =

Wmax = 



σe 200 =   2 2

V„  V„ 1PP2

or

Ans.

(e) Using the distortion energy theory, we have 1 [(σ1 − σ2 )2 + (σ2 − σ3 ) 2 + (σ3 − σ1 )2 ] = Ve2 2 or

1 [(2σ′ + 1σ′)2 + (− 1σ′ − 1.5σ′)2 + (1.5σ′ − 2σ′) 2 ]    2 2

On solving, we get V„ 1PP2

Ans.

Example 10.4 $ FLUFXODU VKDIW RI  PP GLDPHWHU LV VXEMHFWHG WR FRPELQHG EHQGLQJ DQG WZLVWLQJ PRPHQWV the bending moment being four times the twisting moment. Find the allowable twisting moment according to (a) the maximum principal stress theory (b) the maximum shear stress theory and (c WKHGLVWRUWLRQHQHUJ\WKHRU\*LYHQWKHVWUHVVDWWKHHODVWLFOLPLWLV1PP 2 and the factor RIVDIHW\LV

444

Strength of Materials

Solution:

Given,

Diameter of the shaft, Elastic limit stress,

d PP Ve 1PP2

Factor of safety,

n 

/HW  

T = Twisting moment M = Bending moment M = 4T (Given)

The major principal stress is given as V1 = = =

16 πd 3

(M +

16 3

π × (80 ) 16 T

M2 T2 ) (4T  (4)2  T 2 )

± 17 ) = 8 – T

3

(4 +

3

(4 − 17 ) = ± 1.22 –±T

π × (80 )

The minor principal stress is given as V2 =

16T π × (80)

The working stress at the elastic limit is given as σe 4 =  1PP2 n 3 Using the maximum principal stress theory, we have Vw =

(a)

V1 = Vw 8 –± T  On solving, we get (b)

T –41¹mm The maximum shear stress is given as Wmax =

σ1 − σ2 8 × 10−5 T + 1.22 × 10−6 T =  –± T 2 2

Using the maximum shear stress theory, we have Wmax = 

–± T =

σw 2 1.33 2

Ans.

Theory of Elastic Failure

445

On solving, we get T –41¹mm For a biaxial stress system, the distortion energy theory reduces to

(c)

Ans.

2 σ12 + σ22 − σ1 σ2 = Vw

(8 × 10−5 T ) 2 + (− 1.22 × 10−6 T ) 2 − 8 × 10−5 T × (− 1.22 × 10−6 T )    2 Solving for T, we get T –41˜mm

Ans.

Example 10.5  FLUFXODUVKDIWRIGLDPHWHUPPLVVXEMHFWHGWRDEHQGLQJPRPHQWDQGDWZLVWLQJPRPHQW $ Find the relationship between the two moments assuming that they are causing failure alone according to (a) the maximum principal stress theory (b) the maximum principal strain theory (c) the total strain energy theory (d) the maximum shear stress theory and (e) the distortion HQHUJ\WKHRU\7DNH3RLVVRQ¶VUDWLRWREH Solution: Given, Diameter of the shaft, Poisson’s ratio, /HW

d PP v  M = Bending moment T = Twisting moment Ve = Stress at the elastic limit V1 = Major principal stress V2 = Minor principal stress

The bending stress, due to bending moment M acting alone, is given as 32 M Vb = πd 3 The maximum shear stress, due to twisting moment, that is, torque T acting alone, is given as W =

16T πd 3

Principal stresses are V1, 2 = ±

16T πd 3

(a) Using the principal stress theory, we have V1 = Ve 16T πd or

3

=

32 M πd 3

T =2M

(here Ve = Vb = bending stress) Ans.

446

Strength of Materials

(b) For a biaxial stress system, the maximum principal strain theory is expressed as V1±vV2 = Ve (On replacing Vx and Vy by V1 and V2UHVSHFWLYHO\LQHTXDWLRQ   32 M ⎛ 16T ⎞ − 0.25 ⎜ − 3 ⎟ = πd 3 πd ⎝ πd ⎠

16T

(Here Ve = Vb = Bending stress)

3



 2QVLPSOL¿FDWLRQZHJHW T M

Ans.

(c) The total strain energy theory, for a biaxial stress system, is expressed as 2 σ12 + σ 22 − 2v σ1 σ 2 = V e



  8VLQJHTXDWLRQ  DQGUHSODFLQJVx and Vy by V1 and V2 respectively.) 2

2

16T ⎛ 16T ⎞ ⎛ 32M ⎞ ⎛ 16T ⎞ ⎛ 16T ⎞ ⎜ 3 ⎟ + ⎜ − 3 ⎟ − 2 × 0.25 × 3 × ⎜ − 3 ⎟ = ⎜ ⎟ πd ⎝ πd ⎠ ⎝ πd ⎠ ⎝ πd ⎠ ⎝ πd 3 ⎠ 

2

 2QVLPSOL¿FDWLRQZHJHW T M

Ans.

(d) Using the maximum shear stress theory, we have W = 16T πd

3

=

V σe = b 2 2 1 32M × 2 πd 3

T =M

or

Ans.

(e) For a biaxial stress system, the distortion energy theory is expressed as σ12 + σ22 − σ1 σ2 = Ve2 2

2

⎛ 16T ⎞ ⎛ 16T ⎞ ⎛ 16T ⎞ ⎛ 32M ⎞ ⎛ 16T ⎞ ⎜ 3 ⎟ + ⎜− 3 ⎟ − ⎜ 3 ⎟ ⎜− 3 ⎟ = ⎜ ⎟ ⎝ πd ⎠ ⎝ πd ⎠ ⎝ πd ⎠ ⎝ πd ⎠ ⎝ πd 3 ⎠

2

 2QVLPSOL¿FDWLRQZHJHW T M

Ans.

Theory of Elastic Failure

447

Example 10.6 Find the maximum principal stress developed in a cylindrical shaft 8 cm in diameter and subjected WRDEHQGLQJPRPHQWRIN1PDQGDWZLVWLQJPRPHQWRIN1P,IWKH\LHOGVWUHVVRIWKHVKDIW PDWHULDO LV  01P2, determine the factor of safety according to the maximum shearing stress theory of failure. Solution:

Given,

Diameter of the shaft,

d FP P

Bending moment,

M N1P

Twisting moment,

T N1P

Yield stress,

Vyp 01P2

The bending stress is given as Vb =

32 M πd 3

=

32 × 25 × 103 π × (0.08)3 × 106

01P2 01P2

The shear stress is given as W = =

16T πd 3 16 × 4.2 × 103 π × (0.08)3 × 106

01P2 01P2

The maximum principal stress is given as V1 = = 

σb 1 + σb2 + 4τ 2 2 2

(' Vx = Vb and Vy 

49.73 1 + (49.73) 2 + 4 × (41.78) 2 2 2

 01P2

Ans.

The maximum shear stress is given as Wmax =

1 σb2 + 4τ 2 2

1 ( 49.73) 2 + 4 × (41.78) 2  01P2 2 According to the maximum shear stress theory, failure occurs when W maxHTXDOVWRRQHKDOIRIWKH yield stress. Hence, the factor of safety is =

n =

σ yp 2 × τ max

=

300   2 u 48.62

Ans.

448

Strength of Materials

Example 10.7 A thin walled circular tube of wall thickness t and mean radius r is subjected to an axial load P and a torque TLQDFRPELQHGWHQVLRQWRUVLRQH[SHULPHQW (a) Determine the state of stress exising in the tube in terms of P and T. (b) Using the von Mises failure criterion, show that the failure takes place, when σ2 + 3τ2 = Vyp where Vyp is the yield stress in uniaxial tension, V and W are respectively the axial and torsional shearing stresses in the tube. Solution: 7KHYRQ0LVHVWKHRU\IRUDWZRGLPHQVLRQDOVWUHVVV\VWHPLVJLYHQDV 2 σ12 + σ22 − σ1σ2 = Vyp

... (1)

where V1 and V2 are the two principal stresses and Vyp is the yield stress in uniaxial tension. 5HIHU)LJIRUWKHVWDWHRIVWUHVVLQWKHWXEH$VPDOOHOHPHQWRQWKHVXUIDFHRIWKHWXEHZLWK VWUHVVHVDFWLQJRQLWLVVKRZQLQWKH¿JXUH The direct stress due to axial load P is given as V =

P P = 2 A πr

... (2)

The shear stress due to torque T is W = Substituting J =

T ur J

π 4 r in the shear stress equation, we get 2 2T W =  πr 3

The principal stresses w.r.t. two axes system are given as

V1 =

σx + σ y 2 σx + σ y

+ −

(σ x − σ y )2 + 4τ2xy 2 (σ x − σ y )2 + 4τ2xy

and

V2 =

But

Vy Vx = V and Wxy = W

Hence,

V1 =

2

σ σ 2 + 4τ 2 + 2 2

2

 

Theory of Elastic Failure

449

Fig. 10.7

σ σ 2 + 4τ 2 − V2 = 2 2

and

Substituting V1 and V2 in equation (1), we have σ 2 σ 2 + 4τ2 σ σ 2 + 4τ 2 σ 2 σ 2 + 4τ 2 σ σ 2 + 4τ 2 ⎛ σ 2 σ 2 + 4τ 2 ⎞ 2 + +2· · + + −2· · −⎜ − ⎟⎟ = Vyp ⎜ 4 4 4 2 4 4 4 2 2 4 ⎝ ⎠ σ2 σ 2 + 4τ 2 2 + 3· = Vyp 4 4 4 τ2 σ2 3 2 2 + σ + 3· = Vyp 4 4 4 2 V2W2 = Vyp



σ2 + 3τ2 = Vyp

Hence,

Proved.

Example 10.8 7KUHHH[DFWO\VLPLODUVSHFLPHQVRIPLOGVWHHOWXEHDUHPPLQH[WHUQDOGLDPHWHUDQGPP in internal diameter. One of these tubes was tested in tension and the limit of proportionality was DFKLHYHGDWDORDGRIN17KHVHFRQGZDVWHVWHGLQWRUVLRQZKHUHDVWKHWKLUGZDVWHVWHGLQWRUVLRQ ZLWKDVXSHULPSRVHGEHQGLQJPRPHQWRI1P,IWKHIDLOXUHFULWHULDLVPD[LPXPVKHDUVWUHVV estimate the torque at which the two specimens would fail. Solution:

Given,

,QWHUQDOGLDPHWHURIWKHWXEH

di PP

External diameter of the tube,

d PP

Bending moment,

M 1P

450

Strength of Materials

The area of the tube is A = =

π 2 (d 0 − d12 ) 4 π (37.52 − 31.252 )  PP2 4

7KHSRODUPRPHQWRILQHUWLDRIWKHFURVVVHFWLRQRIWKHWXEHLVREWDLQHGDV J =

=

π 4 (d o − di4 ) 32 π (37.54 − 31.254 )  PP4 32

The stress at the limit of proportionality is given as Ve =

70 u 103 1PP2 337.47

 1PP2



According to the maximum shear stress theory, the maximum shear stress is given by Wmax = =

σe 2 207.42  1PP2 2

The equivalent torque Te is given by Te = 1RZ

M2 T2

=

= or

M2 T2 τmax × J ⎛ d0 ⎞ ⎜ 2 ⎟ ⎝ ⎠

(using torsion equation)

103.71 u 100517.7 u 2  1PP 37.5

(350 × 103 )2 + T 2  

Solving for T, we get T 1P

Ans.

Theory of Elastic Failure

451

Example 10.9 $PLOGVWHHOVKDIWRIPPGLDPHWHULVVXEMHFWHGWRDEHQGLQJPRPHQWRIN1PDQGDWRUTXHT. ,IWKH\LHOGSRLQWRIVWHHOLQWHQVLRQLV03D¿QGWKHPD[LPXPYDOXHRIWKHWKHRU\WRUTXHZLWKRXW causing yielding of the shaft material according to (a) the maximum principal stress theory, (b) the maximum shear stress theory and (c) the maximum distortion strain energy theory. Solution:

Given,

Diameter of the shaft, Bending moment, Yield point stress,

d PP M N1P î1PP Vyp 03D 1PP2

The bending stress is given by Vb =

M uy I

(using bending equation)

1 50 6 × = 2 × 10 × π 1PP2 1PP2 2 4 × 50 64 The shear stress is given by W =

T d u J 2

= T×

(using torsion equation) 1

π × 504 32

×

50 1PP2 î±T 1PP2 2

where TLVWKHWRUTXHLQ1PP The maximum principal stress is given by V1 =

σ1 − σ2 2

(162.975) 2 + 4 × (4.07 × 10−5 ) 2 T 2 162.975 + = 2 2 = 81.488 +

26560.85 + 6.626 × 10−9 T 2 2

The minimum principal stress is given by V2 = 81.488 −

26560.85 + 6.626 × 10−9 T 2 2

452

Strength of Materials

The maximum shear stress is Wmax =

=

σ1 − σ2 2 26560.85 + 6.626 × 10−9 T 2 2

(a) According to the maximum principal stress theory, we have V1 = Vyp 26560.85 + 6.626 × 10−9 T 2   2

81.488 +

or Solving for T, we get

T 1PP 1P

Ans.

(b) According to the maximum shear stress theory, we have Wmax =

or

σ yp 2

26560.85 × 6.626 × 10−9 T 2 200 = 2 2

Solving for T, we get T 1PP 1P

Ans.

(c) According to the distortion energy theory, we have 2 (σ1 − σ2 ) 2 + σ12 + σ22 = 2Vyp

or

(' V 

⎛ 26560.85 + 6.626 × 10−9 T 2 26560.85 + 6.626 × 10 T + ⎜ 81.488 + ⎜ 2 ⎝ −9

2

⎞ ⎟ ⎟ ⎠

⎛ 26560.85 + 6.626 × 10−9 T 2 + ⎜ 81.488 − ⎜ 2 ⎝

2

2

⎞ ⎟  î  2 ⎟ ⎠

Solving for T, we get T 1PP 1P

Ans.

Theory of Elastic Failure

453

Example 10.10 $FXEHRIPPVLGHLVORDGHGDVVKRZQLQ)LJ a) Determine the principal stresses V 1, V 2 and V . (b :LOOWKHFXEH\LHOGLIWKH\LHOGVWUHQJWKRIWKHPDWHULDOLV03D"8VHYRQ0LVHVWKHRU\ Solution: 5HIHU)LJ

Fig. 10.8

Side of the cube, a PP Yield strength, Vyp 03D 1PP2 Area of each face of the cube = a2 PP2 (a  1RUPDOVWUHVVLQxGLUHFWLRQLV Vx = 

 1RUPDOVWUHVVLQyGLUHFWLRQLV Vy =



2000  1PP2 25

1000 25

1PP2

 1RUPDOVWUHVVLQzGLUHFWLRQLV Vz =

500  1PP2 25

Shear stress in xy plane is Wxy = 

800  1PP2 25

 6LQFHQRVKHDUVWUHVVLVDFWLQJRQWKHSODQHRI1IRUFHKHQFHLWLVRQHRIWKHSULQFLSDO planes. Hence,

V = Vz 1PP2

Ans.

454

Strength of Materials

The other two principal stresses are given as V1, 2 =

σx + σ y 2

±

(σ x − σ y )2 + 4τ xy2 2

(80 − 40)2 + 4 × 322 80 + 40 ±  “1PP2 2 2 V1 1PP2 =

Hence, 

 1PP2

 and



Ans.

V2 ±1PP

2

 1PP2

 2

2

(b  1RZ  V1±V2) + (V2±V) + (V±V1)

Ans. 2

  ± 2 ± 2 ± 2 

 and

 1PP2 2 2Vyp  î2 1PP2

2 Since (V1±V2)2 + (V2±V)2 + (V±V1)2 > 2Vyp ,, hence according to the von Mises theory, the yielding will occur. Ans.

SHORT ANSWER QUESTIONS        

+RZLVDIDLOXUHFULWHULRQXVHIXOLQWKHDQDO\VLVRIDPDWHULDO¶VIDLOXUH"  :KLFKWKHRU\SUHGLFWVDEULWWOHIDLOXUHPRUHDFFXUDWHO\"  :KLFKWKHRU\SUHGLFWVDGXFWLOHIDLOXUHPRUHDFFXUDWHO\"  :KDWLV\LHOGSRLQW":KDWGRHVLWVLJQLI\"  :K\LVDIDFWRURIVDIHW\XVHGLQHQJLQHHULQJGHVLJQ"  +RZDUHWKHSULQFLSDOVWUHVVDQGWKHSULQFLSDOVWUDLQXVHIXOLQSUHGLFWLQJDPDWHULDO¶VIDLOXUH"  +RZGRHVGLUHFWVWUDLQHQHUJ\GLIIHUIURPVKHDUVWUDLQHQHUJ\"  :KLFKIDLOXUHFULWHULRQLVDOVRFDOOHG5DQNLQH¶VWKHRU\")RUZKLFKW\SHRIPDWHULDOVWKLVWKHRU\ LVDSSOLHG"   :KLFKIDLOXUHWKHRU\LVDOVRFDOOHG7UHVFD¶V\LHOGFULWHULRQ")RUZKLFKW\SHRIPDWHULDOVWKLV WKHRU\LVDSSOLHG"   :K\LVWKHPD[LPXPGLVWRUWLRQHQHUJ\WKHRU\DOVRFDOOHGWKHRFWDKHGUDOVKHDUWKHRU\"

Theory of Elastic Failure

455

MULTIPLE CHOICE QUESTIONS 1. The most important point in the consideration of a material’s failure is (a) ultimate point (b) yield point (c) failure point (d) elastic limit. 2. The maximum principal stress theory is also known as (a) Haigh’s theory (b) St.Venant’s theory (c) Rankine’s theory (d) von Mises theory. 3. )  DFWRURIVDIHW\LVGH¿QHGDVWKHUDWLRRI (a) ultimate stress to working stress (c) yield stress to ultimate stress

(b) working stress to ultimate stress (d) lateral strain to longitudinal strain.

4. The principal stresses are given as (a)

(c)

σx + σ y 2 σx + σ y 2

±

±

(σ x − σ y ) 2 + 4 τ 2 2 (σ x − σ y ) 2 − 4 τ 2 2

5. Hooke’s law is valid for (a) brittle materials (c) isotropic materials

(b)

σx − σ y 2

±

(d) σ x + σ y ±

(σ x + σ y ) 2 − 4 τ 2 2 (σ x − σ y ) 2 + 4 τ 2

.

2

(b) ductile materials (d) isotropic and homogenous materials.

6. For a body being subjected to three mutually perpendicular stresses V x, Vy and Vz, principal strain in the xGLUHFWLRQLV σy σy σ σ σ σ +υ z +υ z (a) x − υ (b) x + υ E E E E E E σy σx σ σ υ −υ −υ z (d) x − (σ y + σ z ) . E E E E 2 7. The maximum shear stress theory gives better results for (a) brittle materials (b) ductile materials (c) brittle and ductile materials both (d) nonmetallic materials. (c)

8. For a biaxial stress system, the strain energy per unit volume is (a)

1 (σ2x − σ2y + 2υ σ x σ y ) 2E

(b)

1 (σ2x + σ2y − 2υ σ x σ y ) 2E

(c)

1 (σ2x + σ2y − 2υ σ x σ y ) 2G

(d)

1 (σ2x − σ2y + 2υ σ x σ y ) . 2G

456

Strength of Materials

9. According to the Tresca’s theory, the failure occurs, when the (a) major principal stress exceeds the elastic limit stress (b) maximum principal strain exceeds the elastic limit strain (c) maximum shear stress exceeds the maximum shear stress at the elastic limit (d) distortion energy per unit volume exceeds the distortion energy per unit volume at the elastic limit. 10. The shear strain energy per unit volume, for a biaxial stress system, in which V 1 and V2 are the major and minor principal stresses respectively, is given as (a)

1 [σ12 + σ22 − σ1 σ2 ] 12G

(b)

1 2 [σ1 + σ22 − σ1 σ2 ] 6G

1 2 1 2 [σ1 − σ22 + σ1 σ2 ] [σ1 − σ22 + σ1 σ2 ] . (d) 6G 6G 11. 7KHYRQ0LVHVHTXLYDOHQWVWUHVVIRUDELD[LDOVWUHVVV\VWHPLVGH¿QHGDV (c)

(a)

σ12 − σ22 + σ1 σ2

(b)

σ12 + σ22 − σ1 σ2

(c)

σ1 + σ2 − σ1 σ2

(d)

σ1 − σ2 + σ1 σ2 .

where V1 and V2 are the major and minor principal stresses respectively. 12. The maximum shear stress is given as (a)

σ1 + σ2 2

(b)

σ1 − σ2 2

σ12 + σ22 (σ1 + σ2 ) 2 . (d) 2 2 where V1 and V2 are the major and minor principal stresses respectively. (c)

13. :KLFKRQHRIWKHJURXSVFRQVLVWLQJRIWZRWKHRULHVLVVDLGWREHPRGHUQEHFDXVHRIWKHLUFORVH QHVVWRH[SHULPHQWDOYDOXHV" (a) Rankine’s and St. Venant’s theory (b) Tresca’s and Rankine’s theory (c) Tresca’s and von Mises theory (d) Rankine’s and von Mises theory. 14. The shear strain energy per unit volume at the elastic limit for a body being subjected to a stress V is (a)

V2 6E

1 ⎞ ⎛ 1 + (c) σ2 ⎜ ⎟ ⎝ 6 E 6G ⎠

(b)

V2 6G

(d)

V . G

Theory of Elastic Failure

15. The maximum shear stress at the elastic limit for a body being subjected to a stress V is (a) 2V

(b)

V 

V 2

(d)

V. 

(c)

ANSWERS 1. (b) 10. (b)

2. (c)

3. (a)

4. (a)

11. (b) 12. (b) 13. (c)

5. (d)

6. (c)

14. (b) 15. (c)

7. (b)

8. (b)

9. (c)

457

458

Strength of Materials

EXERCISES 1. $ F\OLQGULFDO VKHOO PDGH RI PLOGVWHHO SODWH \LHOGV DW  1PP 2 in uniaxial tension. The GLDPHWHURIWKHVKHOOLVPDQGLWVWKLFNQHVVLVPP)LQGWKHSUHVVXUHDWZKLFKWKHIDLOXUH occurs according to (a) the maximum shear stress theory and (b) the distortion energy theory. (Ans.

(a 1PP2 (b 1PP2).

2. For a biaxial stress system, Vx 1PP2 and Vy 1PP2. Find the equivalent stress at the elastic limit assuming that the failure occurs due to the maximum principal strain theory. Take 3RLVVRQ¶VUDWLRWREH Ans. –1PP2). 3. $ SUHVVXUH YHVVHO RI LQVLGH UDGLXV  PP LV VXEMHFWHG WR D SUHVVXUH RI  1PP2. Find the WKLFNQHVVRIWKHYHVVHODFFRUGLQJWRWRWDOVWUDLQHQHUJ\WKHRU\7KHIDFWRURIVDIHW\LVDQG WKH\LHOGLQJRFFXUVLQVLPSOHWHQVLRQDW1PP27DNH3RLVVRQ¶VUDWLRWREH (Ans. PP  4. $F\OLQGULFDOWXEHRIRXWVLGHGLDPHWHUPPDQGWKLFNQHVVPPLVVXEMHFWHGWRDWRUTXHRI 2 –41¹P7KHVWUHVVDWWKHHODVWLFOLPLWLQVLPSOHWHQVLRQLV1PP 2. Calculate the factor of safety according to (a) the maximum shear stress theory and (b) the distortion energy theory. (Ans. (a  b) 2.18). 5. $VKDIWRIGLDPHWHUPPLVVXEMHFWHGWRDWRUTXHRI1¹m and an axial thrust. For a factor RIVDIHW\RI¿QGWKHPD[LPXPYDOXHRIWKHWKUXVWDFFRUGLQJWR a) the maximum shear stress theory and (b WKHGLVWRUWLRQHQHUJ\WKHRU\7KHIDLOXUHRFFXUVDWDVWUHVVRI1PP 2 at the elastic limit. (Ans. (a –41 b –41  6. $WKLFNVSKHULFDOSUHVVXUHYHVVHORILQQHUUDGLXVPPLVVXEMHFWHGWRDQLQWHUQDOSUHVVXUH RI03D&DOFXODWHLWVZDOOWKLFNQHVVEDVHGXSRQ a) the maximum principal stress theory and (b WKHWRWDOVWUDLQHQHUJ\WKHRU\3RLVVRQ¶VUDWLR DQG\LHOGVWUHQJWK 03D (Ans. PPPP  7. $FRPSRQHQWLQDQDLUFUDIWÀDSDFWXDWRUFDQEHDGHTXDWHO\PRGHOOHGDVDF\OLQGULFDOEDUVXEMHFWHG WRDQD[LDOIRUFHRIN1DEHQGLQJPRPHQWRI1˜PDQGWRUVLRQDOPRPHQWRI1˜P$PP GLDPHWHUVROLGEDURI7DOXPLQLXPKDYLQJVu 03DVyt 03DDQGVys  MPa is recommended for its use. Determine the factor of safety according to (a) the maximum principal stress theory and (b) the maximum shear stress theory. (Ans.   8. $WKLQF\OLQGULFDOSUHVVXUHYHVVHOZLWKFORVHGHQGVKDYLQJLQVLGHGLDPHWHUPPLVUHTXLUHGWR withstand an internal pressure of 4 MPa. Find the thickness of the vessel taking a factor of safety RIDQGDVVXPLQJWKH\LHOGVWUHVVRI03DDFFRUGLQJWRWKHIROORZLQJIDLOXUHFULWHULD (a) the maximum shear stress theory. (b) the shear strain energy theory.

(Ans. PPPP 

‰‰‰

11 Buckling of Columns

Leonhard Euler (1707-1783)

Leonhard Euler, born on 15 April 1707, was a great Swiss mathematician DQGSK\VLFLVW+HLVDVHPLQDO¿JXUHLQWKHKLVWRU\RIPDWKHPDWLFVDQG RQHRIWKHPRVWSUROL¿FPDWKHPDWLFLDQVZKRZRUNHGLQDOPRVWDOODUHDV RI PDWKHPDWLFV VXFK DV JHRPHWU\ LQ¿QLWHVLPDO FDOFXOXV WULJRQRPHWU\ algebra and number theory. He introduced several notations used in mathematics such as the letter eIRUWKHEDVHRIWKHQDWXUDOORJDULWKPZKLFK DSSUR[LPDWHO\HTXDOVWRDQGLVDOVRNQRZQDV(XOHUQXPEHUWKH *UHHNOHWWHU™IRUVXPPDWLRQDQGWKHOHWWHUi to denote imaginary unit. +HLQWURGXFHGWKHFRQFHSWRIDIXQFWLRQDQGXVHGf(x WRGHQRWHIXQFWLRQ f applied to a quantity x7KH(XOHUFRQVWDQWȖ JDPPD LVQDPHGDIWHU KLP+HLVDOVRNQRZQIRUKLVZRUNLQPHFKDQLFVÀXLGPHFKDQLFVRSWLFV DQGDVWURQRP\+HZDVHOHFWHGDIRUHLJQPHPEHURIWKH5R\DO6ZHGLVK $FDGHP\RI6FLHQFHVLQ

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© The Author(s) 2021 D. K. Singh, Strength of Materials, https://doi.org/10.1007/978-3-030-59667-5_11

459

460

Strength of Materials

11.1

INTRODUCTION

Columns are long, slender structural members designed to support axial compressive loads. Vertical SLOODUVXVHGLQWKHFRQVWUXFWLRQRIDEXLOGLQJDUHVXLWDEOHH[DPSOHVRIFROXPQV7KHORDGHGVOHQGHU FROXPQPD\GHÀHFWODWHUDOO\DQGIDLOE\EHQGLQJ EXFNOLQJ UDWKHUWKDQIDLOLQJE\GLUHFWFRPSUHVVLRQ %XFNOLQJLVRQHRIWKHPDMRUFDXVHVRIIDLOXUHVLQVWUXFWXUHV6RGHVLJQRIDFROXPQLVHVVHQWLDOWR HQVXUHWKDWORDGFDUULHGE\LWLVZLWKLQWKHVDIHSHUPLVVLEOHOLPLWDQGFDQVXVWDLQLWZLWKRXWEXFNOLQJ XQGHUWKHVSHFL¿HGORDG

11.2 z

z

z



11.3 z

IMPORTANT TERMINOLOGY Buckling or crippling or critical load is the maximum limiting load at which the column HLWKHUEXFNOHVRUWHQGVWREXFNOH%XFNOLQJRIFROXPQPHDQVLWVODWHUDOGLVSODFHPHQWGXULQJ the maximum loading condition and it depends upon the length and cross-sectional area RIWKHFROXPQ Slenderness ratio LVWKHUDWLRRIHIIHFWLYHOHQJWKRIWKHFROXPQWRLWVOHDVWUDGLXVRIJ\UDWLRQ7KH FROXPQVDUHXVXDOO\FODVVL¿HGDFFRUGLQJWRWKHLUVOHQGHUQHVVUDWLR7KHFULSSOLQJVWUHVVGHSHQGV on this ratio. The smaller the slenderness ratio, the higher is the crippling stress. Safe load LVWKHDFWXDOORDGWREHSODFHGRQWKHFROXPQZLWKRXWSURGXFLQJDQ\EXFNOLQJDQGLV given as Buckling load 6DIHORDG = Factor of safety (n )

CLASSIFICATION OF COLUMNS Short columns have length less than 8 times their least lateral dimension (diameter) or their VOHQGHUQHVVUDWLRLVOHVVWKDQ6KRUWFROXPQVGRQRWEXFNOHEXWIDLOE\\LHOGLQJDQGKHQFH FDQEHVXEMHFWHGWRPD[LPXPSHUPLVVLEOHFRPSUHVVLYHVWUHVV

zLong

columns have very large lengths as compared to their lateral dimensions and their VOHQGHUQHVVUDWLRLVJUHDWHUWKDQ6XFKFROXPQVIDLOPDLQO\E\EXFNOLQJDQGWKHHIIHFWRI direct compressive stress is almost negligible.

zMedium

columns KDYHVOHQGHUQHVVUDWLRLQEHWZHHQDQGDQGIDLOGXHWRWKHFRPELQHG HIIHFWRIEXFNOLQJDQGGLUHFWFRPSUHVVLYHVWUHVV

11.4

EULER’S THEORY

7KH(XOHU¶VWKHRU\LVDSSOLFDEOHIRUORQJFROXPQVZKHUHEXFNOLQJLVWKHPDMRUFDXVHIRUWKHLUIDLOXUH DQG WKH HIIHFW RI GLUHFW FRPSUHVVLYH VWUHVV LV LJQRUHG RQ DFFRXQW RI LWV QHJOLJLEOH YDOXH +HUH WKH EXFNOLQJORDGGHSHQGVRQWKHPRGXOXVRIHODVWLFLW\RIWKHFROXPQPDWHULDOEXWQRWRQLWV\LHOGVWUHQJWK

Buckling of Columns

461

7KHIROORZLQJassumptions DUHPDGHZKLOHXVLQJWKH(XOHU¶VWKHRU\ z  7KHFROXPQLVLQLWLDOO\SHUIHFWO\VWUDLJKWDQGKDVXQLIRUPFURVVVHFWLRQWKURXJKRXWLWVOHQJWK z  7KHPDWHULDORIWKHFROXPQLVKRPRJHQHRXVDQGLVRWURSLF z  7KH FROXPQ IDLOV GXH WR EXFNOLQJ EHQGLQJ  RQO\ DQG WKH HIIHFW RI GLUHFW FRPSUHVVLRQ LV neglected. z The column has very large length as compared to its lateral dimensions. z  +RRNH¶VODZLVREH\HGE\WKHPDWHULDORIWKHFROXPQi.e., the load acting on the column are within elastic limit. z  7KHFROXPQLVVXEMHFWHGWRSHUIHFWO\D[LDOORDG z  7KHZHLJKWRIWKHFROXPQLVQRWFRQVLGHUHG 7KHIROORZLQJend conditions DUHXVHGIRUWKHORQJFROXPQVLQWKH(XOHU¶VIRUPXOD z  %RWKHQGVRIWKHFROXPQDUHKLQJHGRUSLQQHG z  %RWKHQGVDUH¿[HG z  2QHHQGLV¿[HGDQGRWKHUHQGKLQJHG z  2QHHQGLV¿[HGDQGRWKHUHQGIUHH The equivalent length or the effective length RI WKH FROXPQ LV GH¿QHG DV WKH GLVWDQFH EHWZHHQ WKHDGMDFHQWSRLQWVRILQÀH[LRQRQWKHHODVWLFFXUYHV7KHSRLQWRILQÀH[LRQLVSRVLWLRQHGQHDUWKH HQGRIWKHFROXPQ7KHHIIHFWLYHFROXPQOHQJWKDFFRUGLQJWRGLIIHUHQWHQGFRQGLWLRQVDUHJLYHQLQ Table 11.1. Table 11.1

Effective length of Columns

End conditions

Effective length (le )

z

Both ends hinged/pinned

le = l (l LVWKHDFWXDOOHQJWKRIWKHFROXPQ

z

%RWKHQGV¿[HG

le = l/2

z

2QHHQG¿[HGDQGRWKHUHQGKLQJHG

le =

z

2QHHQG¿[HGDQGRWKHUHQGIUHH

le = 2l

11.4.1

l 2

Euler’s Formula (when Both Ends of the Column are Hinged or Pinned)

Consider a long column ABRIOHQJWKl loaded with an axial compressive load PDQGKLQJHG ¿[HGLQ position but not in direction) at A and B (Fig. 11.1). Let P EH WKH FULWLFDO ORDG ZKLFK LV UHVSRQVLEOH IRU EXFNOLQJ RI WKH FROXPQ DQG y be the lateral GHÀHFWLRQDWDGLVWDQFHxIURPWKHHQGB. Moment at the distance x is M = – Py

462

Strength of Materials

7KHGLIIHUHQWLDOHTXDWLRQRIÀH[XUHLV EI

d2y =M dx 2

EI

d2y = – Py dx 2

Fig. 11.1

EI or

d2y  Py = 0 dx 2

d2y ⎛ P ⎞ +⎜ ⎟ y =0 dx 2 ⎝ EI ⎠

...(11.1)

Solving the above equation, we get ⎛ P ⎞ ⎛ P ⎞ y = C1 cos ⎜ x + C sin x ⎟⎟ ⎟ ⎜⎜ 2 ⎜ EI ⎟ ⎝ ⎠ ⎝ EI ⎠ where C1 and C2DUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQLV At B, where x = 0, y = 0 On substituting boundary condition in equation (11.12), we have ⎛ P ⎞ ⎛ P ⎞ 0 = C1 cos ⎜⎜ × 0 ⎟⎟ + C2 sin ⎜⎜ × 0 ⎟⎟ ⎝ EI ⎠ ⎝ EI ⎠ = C1 – 1 + C2 – 0 = C1 which gives

C1 = 0

Also at A, where x = l, y = 0 On substituting boundary condition in equation (11.12), we get ⎛ P ⎞ l ⎟⎟ 0 = C2 sin ⎜⎜ ⎝ EI ⎠

...(11.2)

Buckling of Columns

463

⎛ P ⎞ l ⎟⎟ DUHUHTXLUHGWREH]HUR,IC2LV]HURQRGHÀHFWLRQ For LHS to be zero, either C2 or sin ⎜⎜ ⎝ EI ⎠ occurs in the column, and hence C2 can not be zero. ⎛ P ⎞ l ⎟⎟ = 0 sin ⎜⎜ ⎝ EI ⎠ P l = 0, S, 2S, 3S, ... EI

which means

Zero value is not admissible and values other than SFDUU\QRSUDFWLFDOVLJQL¿FDQFH P l =S EI Squaring both sides, we have P 2 l = S2 EI π 2 EI ... (11.3) = Pcr l2 7KLVLVWKHUHTXLUHGH[SUHVVLRQIRUWKHORDG FULWLFDO ZKLFKEXFNOHVWKHJLYHQFROXPQZLWKERWK RILWVHQGVKLQJHG l 7KHODWHUDOGHÀHFWLRQLVPD[LPXPDWx = . 2 or

11.4.2

P =

Euler’s Formula (when Both Ends of the Column are Fixed)

Consider a column ABRIOHQJWKl¿[HG ERWKLQSRVLWLRQDQGGLUHFWLRQ DWERWKHQGVDQGLVVXEMHFWHGWR a crippling load P )LJ 6LQFHWKHHQGVDUH¿[HGKHQFHWKH\KDYH]HURVORSHDQG]HURGHÀHFWLRQ But there are restraint moments say M at each end. 7RGHULYHWKHUHTXLUHGIRUPXODDVHFWLRQXX is considered at a distance xIURPB, where the lateral GHÀHFWLRQGXHWRDSSOLHGORDGLVy. The net bending moment at x is M – Py. Hence,

EI

d2y = M – Py dx 2 d2y M P  y = 2 dx EI EI

or

d2y ⎛ P ⎞ +⎜ ⎟ y = M 2 dx ⎝ EI ⎠ EI

...(11.4)

7KHJHQHUDOVROXWLRQRIWKHDERYHHTXDWLRQLV ⎛ P ⎞ ⎛ P ⎞ M y = C1 cos ⎜ x + C sin x ⎟⎟ + ⎟ ⎜⎜ 2 ⎜ EI ⎟ ⎝ ⎠ ⎝ EI ⎠ P

...(11.5)

464

Strength of Materials

Fig. 11.2

where C1 and C2DUHWKHFRQVWDQWVRILQWHJUDWLRQ 'LIIHUHQWLDWLQJHTXDWLRQ  ZUWx, we get dy = − C1 dx

⎛ P ⎞ P sin ⎜⎜ x ⎟⎟ + C2 EI ⎝ EI ⎠

⎛ P ⎞ P cos ⎜⎜ x ⎟⎟ EI ⎝ EI ⎠

...(11.6)

7KHERXQGDU\FRQGLWLRQLV dy =0 dx From equation (11.6), we have At B, where x = 0,

0 = 0  C2 or But or

C2

P EI

P =0 EI P › 0, as P › 0 EI C2 = 0

Also at B, where x = 0, y = 0. From equation (11.5), we have 0 = C1  or

C1 = 

M P

M P

Equation (11.5) on substituting C1 and C2 becomes y = −

⎛ P ⎞ M M cos ⎜⎜ x ⎟⎟ + P ⎝ EI ⎠ P

...(11.7)

Buckling of Columns

465

And at A, where x = l, y = 0 0 = − ⎛ P ⎞ cos ⎜⎜ l ⎟⎟ = 1 ⎝ EI ⎠

⎛ P ⎞ M M cos ⎜⎜ l ⎟⎟ + P ⎝ EI ⎠ P

P l = 0, 2S, 4S, 6S, ... EI 7DNLQJWKHOHDVWVLJQL¿FDQWYDOXHZHKDYH or

P l = 2S EI Squaring both sides, we have P 2 l = 4S2 EI or

P =

4π 2 EI = Pcr l2

...(11.8)

,WLVWKHUHTXLUHGH[SUHVVLRQIRUWKHFULWLFDOORDGZKLFKEXFNOHVWKHFROXPQ7KLVHTXDWLRQLVVDPH l DVIRUDFROXPQRIOHQJWK with both ends hinged. At the same time, on comparing equation (11.8) 2 ZLWKHTXDWLRQ  ZH¿QGWKDWWKHEXFNOLQJORDGIRUDFROXPQZLWKERWKHQGV¿[HGLVIRXUWLPHV as compared to a column with both ends hinged.

11.4.3 Euler’s Formula (when One End of the Column is Fixed and Other End Hinged) Consider a column ABRIOHQJWKl which is hinged at AEXW¿[HGDWB (Fig. 11.3). Let M be restraint moment at B 7R FRXQWHU WKLV DQRWKHU PRPHQW LV FRQVLGHUHG E\ DSSO\LQJ D KRUL]RQWDO IRUFH F at point A. The bending moment at a distance xIURPBZKHUHWKHODWHUDOGHÀHFWLRQLVy, is F (l – x) – Py. Hence, or

EI

d2y = F (l – x) – Py dx 2

F (l − x) d 2 y Py  = 2 EI dx EI

...(11.9)

7KHJHQHUDOVROXWLRQRIWKHDERYHHTXDWLRQLV ⎛ P ⎞ ⎛ P ⎞ F x ⎟⎟ + C2 sin ⎜⎜ x ⎟⎟ + (l − x) y = C1 cos ⎜⎜ ⎝ EI ⎠ ⎝ EI ⎠ P where C1 and C2DUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQLV

...(11.10)

466

Strength of Materials

At B, where x = 0, y = 0. From equation (11.10), we have 0 = C1  or

C1 = 

Fl P

Fl P

Fig. 11.3

Differentiating equation (11.10) w.r.t. x, we get dy = − C1 dx Also at B, where x = 0,

⎛ P ⎞ P sin ⎜⎜ x ⎟⎟ + C2 EI ⎝ EI ⎠

⎛ P ⎞ F P cos ⎜⎜ x ⎟⎟ − EI ⎝ EI ⎠ P

...(11.11)

dy = 0. dx

From equation (11.11), we have P F  EI P

0 = C2 or

C2 =

F P

EI P

And at A, where x = l, y = 0 Equation (11.10) on substituting the above boundary condition, and C1 and C2 reduces to 0 = − ⎛ P ⎞ Fl cos ⎜⎜ l ⎟⎟ = F P P ⎝ EI ⎠

⎛ P ⎞ F Fl cos ⎜⎜ l ⎟⎟ + P EI ⎝ ⎠ P ⎛ P ⎞ EI sin ⎜⎜ l ⎟⎟ P ⎝ EI ⎠

⎛ P ⎞ EI sin ⎜⎜ l ⎟⎟ P EI ⎝ ⎠

Buckling of Columns

⎛ P ⎞ l cos ⎜⎜ l ⎟⎟ = ⎝ EI ⎠ or

⎛ P tan ⎜⎜ ⎝ EI

⎞ l ⎟⎟ = ⎠

467

⎛ P ⎞ EI sin ⎜⎜ l ⎟⎟ P ⎝ EI ⎠ P l EI

...(11.12)

⎛ P ⎞ l ⎟⎟ to hold the above relationship good. These values are 7KHUHDUHWZRSRVVLEOHYDOXHVRI ⎜⎜ ⎝ EI ⎠ 0 and 4.493 radian. But

P l › 0, as P › 0 EI

or

P l = 4.493 EI

Squaring both sides, we have P 2 2 l = 20.187 = 2S (approx.) EI or

P =

2π 2 EI = P cr l2

...(11.13)

2QFRPSDULQJWKLVHTXDWLRQZLWKHTXDWLRQ  ZH¿QGWKDWEXFNOLQJORDGLVWZRWLPHVWKHORDG UHTXLUHGZKHQERWKHQGVRIWKHFROXPQDUHKLQJHGRUSLQQHG,QRWKHUZRUGVDFROXPQRIHIIHFWLYH l length with both ends hinged will have the same crippling load. 2

11.4.4 Euler’s Formula (when One End of the Column is Fixed and Other End Free) Consider a column ABRIOHQJWKlZKLFKLV¿[HGDWBEXWIUHH ERWKLQSRVLWLRQDQGLQGLUHFWLRQ DWA (Fig. 11.4). The bending moment at a distance xIURPBZKHUHWKHODWHUDOGHÀHFWLRQLVy, is P (d – y), d being WKHGHÀHFWLRQDWWKHIUHHHQG Hence,

EI

d2y = P (d – y) = Pd – Py dx 2

d 2 y Py Pd  = 2 dx EI EI 7KHJHQHUDOVROXWLRQRIWKHDERYHHTXDWLRQLV or

⎛ P ⎞ ⎛ P ⎞ x ⎟⎟ + C2 sin ⎜⎜ x ⎟⎟ + d y = C1 cos ⎜⎜ ⎝ EI ⎠ ⎝ EI ⎠ where C1 and C2DUHWKHFRQVWDQWVRILQWHJUDWLRQ 7KHERXQGDU\FRQGLWLRQLV At B, where x = 0, y = 0

...(11.14)

...(11.15)

468

Strength of Materials

From equation (11.15), we have 0 = C1 + d or C1 = – d 'LIIHUHQWLDWLQJHTXDWLRQ  ZUWx, we get ⎛ P ⎞ P sin ⎜⎜ x ⎟⎟ + C2 EI ⎝ EI ⎠ dy Also at B, where x = 0, =0 dx From equation (11.16), we have dy dx

= − C1

P × 0 + C2 EI

0 = d C2 But

P EI

=0

P EI

› 0, as P › 0

or

⎛ P ⎞ P cos ⎜⎜ x ⎟⎟ ...(11.16) EI ⎝ EI ⎠

P EI

Fig. 11.4

C2 = 0

Equation (11.15) on substituting C1 and C2 reduces to

And at A, where x = l, y = d

⎛ P ⎞ x ⎟⎟ + d y = − d cos ⎜⎜ ⎝ EI ⎠

Equation (11.17) on substituting the above boundary condition becomes ⎛ P ⎞ l ⎟⎟ + d d = − d cos ⎜⎜ ⎝ EI ⎠ ⎛ P ⎞ d cos ⎜⎜ l ⎟⎟ = 0 ⎝ EI ⎠ But

d ›0 ⎛ P cos ⎜⎜ ⎝ EI

or

⎞ l ⎟⎟ = 0 ⎠

P π 3π 5π l = , , ,… EI 2 2 2

7DNLQJWKHOHDVWVLJQL¿FDQWYDOXHZHKDYH P π l = EI 2

...(11.17)

Buckling of Columns

469

Squaring both sides, we get P 2 π2 l = 4 EI

or

P=

π 2 EI = Pcr 4l 2

...(11.18)

&RPSDULQJWKLVHTXDWLRQZLWKHTXDWLRQ  ZH¿QGWKDWFULSSOLQJORDGLQWKLVFDVHLVRQHIRXUWK RIWKHORDGZKHQDFROXPQZLWKERWKHQGVKLQJHGLVXVHG,QRWKHUZRUGVDFROXPQRIHIIHFWLYH OHQJWK l ZLWKERWKHQGVKLQJHGZLOOKDYHWKHVDPHFULSSOLQJORDG

11.4.5

Crippling Stress

7KH YDOXH RI WKH VWUHVV FRUUHVSRQGLQJ WR WKH FULWLFDO RU FULSSOLQJ ORDG LV FDOOHG FULSSOLQJ VWUHVV RU FULWLFDOVWUHVVDQGLVGHQRWHGE\Vcr. π 2 EI P Vcr = cr = 2 (using equation (11.3)) l A A =

π 2 EAr 2 l2 A

(I = Ar2, where r 5DGLXVRIJ\UDWLRQ 

π2 E ...(11.19) (l / r ) 2 Where (l/r LVFDOOHGWKHslenderness ratioRIWKHFROXPQ,IWKHVOHQGHUQHVVUDWLRLVEDVHGRQWKH HIIHFWLYHOHQJWKRIWKHFROXPQLWLVNQRZQDVHIIHFWLYHVOHQGHUQHVVUDWLR le/r). The equation (11.19) VKRZVWKDWFULSSOLQJVWUHVVYDULHVLQYHUVHO\SURSRUWLRQDOWRWKHVTXDUHRIWKHVOHQGHUQHVVUDWLRPHDQ LQJWKHUHE\KLJKHUWKHVOHQGHUQHVVUDWLRVPDOOHULVWKHFULSSOLQJVWUHVV+HQFHPLQLPXPYDOXHRIWKH UDGLXVRIJ\UDWLRQrVKRXOGEHXVHGLQWKHFDOFXODWLRQRIVOHQGHUQHVVUDWLRDQGWKHFULSSOLQJVWUHVVLQ DFROXPQ$FULSSOLQJVWUHVVYHUVXVVOHQGHUQHVVUDWLRSORWIRUVWUXFWXUDOVWHHOLVVKRZQLQ)LJ =

Fig. 11.5 Vcr versus (l/r) plot for structural steel.

470

Strength of Materials

,IWKHFULSSOLQJVWUHVVVcr is higher than the yield strength VYWKHQLWLVRIQRLQWHUHVWEHFDXVHWKH FROXPQZLOO\LHOGLQFRPSUHVVLRQDQGFHDVHWREHHODVWLFEHIRUHLWKDVDFKDQFHWREXFNOH

11.4.6

Limitations of Euler’s Formula

)URPWKH)LJLWLVFOHDUWKDWIRUKLJKHUYDOXHRIFULSSOLQJVWUHVVVOHQGHUQHVVUDWLRLVORZ$WWKH VDPHWLPHWKHFULSSOLQJVWUHVVRIWKHFROXPQFDQQRWEHPRUHWKDQLWV\LHOGVWUHQJWK+HQFHLWSXWVD OLPLWDWLRQRQWKHXVHRI(XOHU¶VIRUPXODDQGWKHIRUPXODLVQRWYHU\XVHIXOLIWKHVOHQGHUQHVVUDWLRLV less than a particular value. 8VLQJWKHFRQGLWLRQWKDWWKHFULSSOLQJVWUHVVVKRXOGEHOHVVWKDQWKH\LHOGVWUHQJWKIRUVWUXFWXUDO steel, we have Vcr … VY π2 E …250 – 106 (l / r ) 2

(using equation (11.19))

π2 × 210 × 109 …250 – 106 2 (l / r )

(using E = 210 GPa)

π2 × 210 × 109 … (l/r)2 250 × 106 8290.46 … (l/r)2 (l/r) • 91

...(11.20)

+HQFHWKH(XOHU¶VIRUPXODIRUDVWUXFWXUDOVWHHOFROXPQZLWKLWVERWKHQGVKLQJHGLVYDOLGRQO\ when the slenderness ratio (l/r) is greater than or equal to 91. Example 11.1 $SLQHQGHGVTXDUHFURVVVHFWLRQFROXPQRIOHQJWKPLVVXEMHFWHGWRDFRPSUHVVLYHVWUHVVRI 03D8VLQJDIDFWRURIVDIHW\RI¿QGWKHFURVVVHFWLRQLIWKHFROXPQLVWRVDIHO\VXSSRUW a) DN1ORDGDQG b DN1ORDG7DNHE = 15 GPa. Solution:

Given,

/HQJWKRIWKHFROXPQl = 3 m Compressive stress on the column,V = 10 MPa )DFWRURIVDIHW\  (a) The crippling load is given as P )DFWRURIVDIHW\–6DIHORDG = 2.5 – 100 

 N1 Using equation (11.3), we have P =

π 2 EI l2

Buckling of Columns

or

471

Pl 2 I = 2 π E =

250 × 103 × 32 4 m π2 × 15 × 109

= 1.52 – 10–5 m4 IIRUDVTXDUHFURVVVHFWLRQRIVLGHa is given as I =

a4 12

= 1.52 – 10–5 or

a = 0.1162 m = 116.2 mm

Hence, the acceptable cross-section is 117 mm – 117 mm.

Check The normal stress produced in the column is given as Vd =

100 1 – 3 MPa (0.117) 2 10

= 7.3 MPa Since V„03DKHQFHWKHFDOFXODWHGFURVVVHFWLRQLVZLWKLQVDIHOLPLWDQGLVDFFHSWDEOH (b) Crippling load is P = 2.5 – 200 

 N1 7KHPRPHQWRILQHUWLDI is given as 500 × 103 × 32 4 I = 2 m π × 15 × 109 = 3.04 – 10–5 m4

or

a4 = 12 a = 0.1382 m = 138.2 mm

Check The normal stress produced in the column is given as Vd =

200 1 – 3 MPa = 10.47 MPa (0.1382) 2 10

Ans.

472

Strength of Materials

Since V „ > 10 MPa, the permissible compressive stress in the column, hence the above crosssection is not acceptable. In that case, cross-section is found on the basis of given permissible compressive stress in the column. Area of the cross-section is A =

200 1 – 3 m2 10 10

= 0.02 m2 or

a = A = 0.1414 m = 141.4 mm

Hence, the acceptable cross-section is 142 mm – 142 mm.

Ans.

Example 11.2 A structural steel column is in the form of a tube of thickness 15 mm and external diameter 250 mm DQGLVPORQJ,WVRQHHQGLV¿[HGZKLOHWKHRWKHUHQGLVIUHH)LQGWKHPD[LPXPD[LDOORDGWREH DSSOLHGRQWKHFROXPQ+RZWKLVORDGYDULHVZLWKRWKHUHQGFRQGLWLRQV"7DNHE = 200 GPa. Solution:

Given,

7KLFNQHVVRIWKHFROXPQ t = 15 mm External diameter, Length of the column,

d0 = 250 mm l =3m

Let diEHWKHLQWHUQDOGLDPHWHURIWKHFROXPQ )LJ  1RZ

di + 2t = d0

Fig. 11.6

or

di = d0 – 2t = 250 – 2 – 15 = 220 mm

7KHPRPHQWRILQHUWLDRIWKHFURVVVHFWLRQRIWKHFROXPQLVJLYHQDV π ⎡⎛ 250 ⎞ ⎛ 220 ⎞ ⎢⎜ I = ⎟ ⎟ −⎜ 64 ⎢⎣⎝ 1000 ⎠ ⎝ 1000 ⎠ 4

4

⎤ ⎥ m4 – 10–5 m4 ⎥⎦

Buckling of Columns

473

7KHFULSSOLQJORDGZKHQRQHHQGRIWKHFROXPQLV¿[HGZKLOHWKHRWKHUHQGIUHHLVJLYHQDV P = =

π 2 EI 4l 2

(using equation (11.18))

π2 × 200 × 109 × 7.67 × 10−5 1 × 3 N1 2 4×3 10 Ans.

 N1



7KHFULSSOLQJORDGVZLWKRWKHUHQGFRQGLWLRQVDUHJLYHQDV End conditions

Crippling load

Both ends hinged

P =

=  %RWKHQGV¿[HG

π 2 EI l2 π2 × 200 × 109 × 7.67 × 10−5 1 × 3 N1 2 3 10

 N1Ans. P =

4π 2 EI l2 ⎛ π2 EI ⎞ ⎟ 2 ⎝ l ⎠

= 4×⎜

= 4 – 16822.2N1 N1Ans. 2QHHQG¿[HGDQGRWKHUHQGKLQJHG

P =

2π 2 EI l2 ⎛ π2 EI ⎞ ⎟ 2 ⎝ l ⎠

=2– ⎜

= 2 – 16822.2 N1Ans.

Example 11.3 $VWHHOFROXPQRIGLDPHWHUPPDQGKLQJHGDWERWKHQGVLVVXEMHFWHGWRDQD[LDOORDG&DOFXODWH WKHPLQLPXPOHQJWKRIWKHFROXPQIRUZKLFK(XOHU¶VIRUPXODLVYDOLG7DNHE *3DDQGOLPLWRI proportionality = 220 MPa. Solution: Given, 'LDPHWHURIWKHFROXPQ

d = 60 mm =

60 m = 0.06 m 1000

7KHPRPHQWRILQHUWLDRIWKHFURVVVHFWLRQRIWKHFROXPQLVJLYHQDV π 4 π d = × (0.06) 4 m4 I = 64 64 = 6.36 – 10–7 m4

474

Strength of Materials

7KH(XOHU¶VIRUPXODXQGHUWKHJLYHQHQGFRQGLWLRQLVJLYHQDV P = σ×

π 2 EI l2

(using equation (11.3))

π 2 EI π 2 d = l2 4

or

l =

=

4π EI σ d2 4π × 200 × 109 × 6.36 × 10−7 m = 1.42 m. 220 × 106 × (0.06) 2

Ans.

Example 11.4 A TVHFWLRQFROXPQRIOHQJWKPZLWKIODQJHGLPHQVLRQPP– 20 mm and web dimension 120 mm –PPLVVXEMHFWHGWRDQD[LDOORDG)LQGWKHFULSSOLQJORDGXVLQJ(XOHU¶VIRUPXOD DVVXPLQJWKDWWKHFROXPQLVKLQJHGDWERWKHQGV7DNHE = 200 GPa. Solution: 5HIHU)LJ

Fig. 11.7

*LYHQ/HQJWKRIWKHFROXPQ l = 2 m