A comprehensive textbook on the mechanics and strength of materials for students of engineering throughout their undergr
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English Pages 648 Year 1987
Table of contents :
TABLE OF CONTENTS:
1. Statically Determinate Force Systems.
2. Statically Determinate Stress Systems.
3. StressStrain Relations.
4. Statically Indeterminate Stress Systems.
5. Torsion.
6. Bending Stress.
7. Bending: Slope and Deflection.
8. Statically Indeterminate Beams.
9. Energy Methods.
10. Structural Analysis
11. Buckling Instability.
12. Stress and Strain Transformations.
13. Yield Criteria and Stress Concentration.
14. Variation of Stress and Strain.
15. Application of the Equilibrium and StrainDisplacement.
16. Elementary Plasticity.
17. Thin Plates and Shells.
18. Finite Element Method.
19. Tension, Compression, Torsion and Hardness.
20. Fracture Mechanics.
21. Fatigue.
22. Creep and Viscoelasticity.
MECHANICS OF ENGINEERING MATERIALS
Mechanics of • • eng1neer1ng materials P. P. BENHAM Professor of Aeronautical Engineering., The Queen's University of Belfast
R. J. CRAWFORD Reader in Mechanical Engineering, The Queen's University of Belfast
....... ,Longman I I::: ~ci~nti_fic ?c
Longman Scientific & Technical
Longman Group UK Limited Longman House, Burnt Mill, Harlow Essex CM20 2JE, England and Associated Companies throughout the world
Copublished in the United States with John Wiley & Sons, Inc., 605 Third Avenue, New York, NY 10158
© P.
P. Benham and R. J. Crawford 1987
All rights reserved; no part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechan~cal, photocopying, recording, or otherwise without either the prior written permission of the Publishers or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, 3334 Alfred Place, London, WC1E 7DP.
First published 1987 Fourth impression 1990 British Library Cataloguing in Publication Data
Benham, P. P. Mechanics of engineering materials. [Rev. ed.] 1. Materials 2. Mechanics, Applied I. Title II. Crawford, R. J. III. Benham, P. P. Mechanics of solids and structures 620.1'12 TA404.8
ISBN 0582286409 Library of Congress Cataloging in Publication Data
Benham, P. P. (Peter Philip), 1927Mechanics of engineering materials. Bibliography: p. Includes Index. 1. Strength of materials. 2. Structures, Theory of. I. Crawford, R. J. II. Title. 8620064 TA405.B484 1986 620.1'12 ISBN 0470207043 (USA only) Set in 10/12pt Linotron 202 Times Roman Produced by Longman Singapore Publishers Pte Ltd Printed in Singapore
Contents
Preface Notation Chapter 1
Chapter 2
Staticallydeterminate force systems
XV
xviii 1
Introduction Revision of statics Types of structural and solid body components Types of support and connection for structural components Statical determinacy Freebody diagrams Determinacy criteria for structures Determination of axial forces by equilibrium statements Bending of slender members Force and moment equilibrium during bending Sign convention for bending Shearforce and bendingmoment diagrams Cantilever carrying a concentrated load Cantilever carrying a uniformlydistributed load Simplysupported beam carrying a uniformlydistributed load Point of contraflexure Torsion of members Members subjected to axial force, bending moment and torque Torque diagram Principle of superposition Summary Problems
1 1 5 6 8 8 8 11 16 17 18 19 19 20 21 23 23 24 26 26 27 28
Staticallydeterminate stress systems
35
Introduction Stress: normal, shear and hydrostatic Staticallydeterminate stress system Assumptions and approximations Tie bar and strut (or column) Suspensionbridge cables
35 35 37 38 38 40
41 v
Contents
Chapter 3
Chapter 4
Chapter 5
Thin shells under pressure Shear in a coupling Torsion of a thin circular tube Joints Summary Problems
43 47 48 48 52 52
Stressstrain relations
56
Introduction Deformation Strain Elastic loaddeformation behaviour of materials Elastic stressstrain behaviour of materials Lateral strain and Poisson's ratio Thermal strain General stressstrain relationships Strains in a staticallydeterminate problem Elastic strain energy Strain energy from normal stress Strain energy from shear stress Plastic stressstrain behaviour of materials Viscoelastic stressstrain behaviour of materials Summary Problems
56 56 57 58 59 62 62 63 65 66 67 67 68 68 70 70
Staticallyindeterminate stress systems
72
Introduction Interaction of different materials Interaction of different stiffness components Restraint of thermal strain Volume changes Constrained material Maximum stress due to a suddenlyapplied load Maximum stress due to impact Summary Problems
72 72
Torsion
90
Introduction Torsion of a thinwalled cylinder Torsion of a solid circular shaft Relation between stress, strain and angle of twist Relation between torque and shear stress Torsion of a hollow circular shaft Torsion of nonuniform and composite shafts Torsion of a tapered shaft
vi
75 76 79 81 83 84 86 87
90 90 91 92 93 94 96 100 102
Contents
Chapter 6
Chapter 7
Chapter 8
Torsion of thin rectangular strip Effect of warping during torsion Torsion of solid rectangular and square crosssections Summary Problems
108
Bending: stress
112
Introduction Shear force and bendingmoment distributions Relationships between loading, shear force and bending moment Deformations in pure bending Stressstrain relationship Equilibrium of forces and moments The bending relationship The general case of bending Section modulus Beams made of dissimilar materials Combined bending and end loading Eccentric end loading Shear stresses in bending Bending and shear stresses in I section beams Asymmetrical or skew bending Shear stress in thinwalled open .sections and shear centre General case of bending of a thinwalled open section Bending of initially curved bars Beams with a small radius of curvature Summary Problems
112 112 117 120 122 123 124 124 124 126 135 136 140 144 148 153 155 158 159 163 163
Bending: slope and deflection
172
Introduction The curvaturebendingmoment relationship Slope and deflection by the double integration method Discontinuous loading: Macaulay's method Superposition method Deflections due to asymmetrical bending Beams with varying load distribution Beams of uniform strength Deflection of a beam under impact loading Slope and deflection by the momentarea method Summary Problems
172 172
Staticallyindeterminate beams
209
Introduction Doubleintegration method
209
105
107 108 109
174
178 187 189
190 195 196 197 205 206
209 214 vii
Contents
Chapter 9
Chapter 10
Chapter 11
viii
Momentarea method Summary Problems
218 226 227
Energy theorems
229
Introduction Elastic strain energy: normal stress and shear stress Strain energy in torsion Strain energy in bending Helical springs Shear deflection' of beams Virtual work Strain and complementary energy solutions for deflections Bending deflection of beams Castigliano's second theorem The reciprocal theorem Summary Problems
229 229 229 230 232 236 239 240 244 249 250 252 252
Structural analysis
255
Introduction Forces by the method of tension coefficients Displacements by the change of geometry method Displacements by the Williot and WilliotMohr methods Displacements by the virtual work method Staticallyindeterminate frames Forces by the virtual work method Lack of fit Forces by the compatibility method Forces by the equilibrium method Summary Problems
255 255 258 258 262 265 266 269 269 271 274 274
Buckling instability
278
Introduction Stability of equilibrium Euler theory of buckling for slender columns Buckling characteristics for real struts Eccentric loading of slender columns Struts having initial curvature Empirical formulae for design Buckling under combined compression and transverse loading Pinended strut carrying a uniformlydistributed lateral load Other examples of instability Summary
278 278 279 283 286 288 291 293 294 298 299 299
Contents
Chapter 12
Chapter 13
Chapter 14
Stress and strain transformations
303
Introduction Symbols, signs and elements Stresses on a plane inclined to the direction of loading Element subjected to normal stresses Element subjected to shear stresses Element subjected to general twodimensional stress system Mohr's stress circle Principal stresses and planes Maximum shear stress in terms of principal stresses Maximum shear stress in three dimensions General twodimensional state of stress at a point States of strain Normal strain in terms of coordinate strains Shear strain in terms of coordinate strains Mohr's strain circle Principal strain and maximum shear strain Rosette strain computation and circle construction Relationships between the elastic constants Stress and strain transformations in composites Analysis of a lamina Analysis of a laminate Inplane behaviour of a symmetric laminate Flexural behaviour of a symmetric laminate Summary References Problems
303 304 305 306 307 308 310 313 314 317 318 319 320 321 322 325 326 329 331 332 338 339 343 343 344 344
Yield criteria and stress concentration
348
Introduction Yield criteria: ductile materials Fracture criteria: brittle materials Strength of laminates Concepts of stress concentration Concentrated loads and contact stresses Geometrical discontinuities Yield and plastic stress concentration factors Summary References Problems
348 348 356 358 360 362 365
Variation of stress and strain
376
Introduction Equilibrium equations: plane stress Cartesian coordinates '· ' · ' )rdinates
376 376 378
371 372
373 373
ix
Contents
Chapter 15
Chapter 16
Chapter 17
Strain in terms of displacement: Cartesian coordinates Strain in terms of displacement: cylindrical coordinates Compatibility equations: Cartesian coordinates Compatibility equations: cylindrical coordinates Stress functions Summary Problems
379 381 383 384 386 387 388
Application of the equilibrium and straindisplacement relationships
389
Introduction Shear stress in a beam Transverse normal stress in a beam Stress distribution in a pressurized thickwalled cylinder Methods of containing high pressure Stresses set up by a shrinkfit assembly Stress distribution in a pressurized compound cylinder Stress distribution in a thin rotating disc Rotational speed for initial yielding Stresses in a rotor of varying thickness with rim loading Summary Problems
389 389 391 392 399 400 402 405 409 411 415 415
Elementary plasticity
418
Introduction Plastic bending of beams: plastic moment Plastic collapse of beams Plastic torsion of shafts: plastic torque Plasticity in a pressurized thickwalled cylinder Plasticity in a rotating thin disc Residual stress distributions after plastic deformation Summary Problems
418 418 423 427 429 433 436 440 440
Thin plates and shells
443
Introduction Assumptions for small deflection of thin plates Relationships between moments and curvatures for pure bending Relationships between bending moment and bending stress Relationships between load, shear force and bending moment Relationships between deflection, slope and loading Plate subjected to uniform pressure Plate with central circular hole Solid plate with central concentrated force Other forms of loading and boundary condition Axisymmetrical thin shells Local bending stresses in thin shells
443 443 444 447 447 448 449 451 453 454 456 458 462
n~..J..:
X
__
~"~

,.
'.
Contents
Chapter 18
Chapter 19
Chapter 20
Summary Bibliography Problems
463 463 463
Finite element method
466
Introduction Principle of finite element method Analysis of spring elements Analysis of frameworks Analysis of continua Stiffness matrix for a triangular element Summary References Bibliography Problems
466 466 468
Tension, compression, torsion and hardness
488
Introduction Stressstrain response in a tension test Stressstrain response in a compression test Stressstrain response in a torsion test Plastic overstrain and hysteresis Hardness measurement Hardness of viscoelastic materials Factors influencing strength, ductility and hardness Summary Bibliography Problems
488 488 492 494 496 497 502 502 505 505 506
Fracture mechanics
508
Introduction Fracture concepts Linear elastic fracture mechanics Strain energy release rate Stress intensity factor Modes of crack tip deformation Experimental determination of critical stress intensity factor Fracture mechanics for ductile materials Toughness measurement by impact testing Relationship between impact testing and fracture mechanics Summary References Bibliography
508 508 510 511 513 515 516 520 529 532 533 533 534 534
471
475. 478 485 486 486 486
xi
Contents
Chapter 21
Chapter 22
Appendix A:
xii
Fatigue
538
Introduction Forms of stress cycle Test methods Fatigue data Micromechanisms of fatigue: initiation and propagation Fracture mechanics for fatigue Influential factors Cumulative damage Failure under multiaxial cyclic stresses Fatigue of plastics and composites Summary Bibliography Problems
538 539 541 542 547 549 552 558 559 562 562 563 563
Creep and viscoelasticity
567
Introduction Stressstraintimetemperature relationships Empirical representations of creep behaviour Creeprupture testing Tension creep test equipment Creep during pure bending of a beam Creep under multiaxial stresses Stress relaxation Stress relaxation in a bolt Creep during variable load or temperature Creepfatigue interaction Viscoelasticity Creep behaviour of plastics Designing for creep in plastics Creep rupture of plastics Summary References Bibliography Problems
567 568 569 572 573 574 576 578 579 581 582 583 585 587 590 591 592 592 592
Properties of areas
596
Introduction First moment of area Position of centre of area, or centroid Second moment of area Parallel axes theorem Product moment of area Transformation of moments of area Mohr's circle for moments of area
596 596 597 598 600 601 602 603 604
Contents
Introduction to matrix algebra
606
Matrix definitions Matrix multiplication Matrix addition and subtraction Inversion of a matrix Transpose of a matrix Symmetric matrix
606 606 607 608 608 608
Appendix C:
Table of mechanical properties of engineering materials
610
Appendix 0:
Answers to problems
612
Index
620
Appendix 8:
xiii
Preface
The S.l. edition of Mechanics of Solids and Structures by P. P. Benham and F. V. Warnock was first published in 1973. It appears to have been very well received over the ensuing period. This preface is therefore written both for those who are familiar with the past text and also those who are approaching this subject for the first time. Although the subject matter is still basically the same today as it has been for decades, there are a few developing topics which have been introduced into undergraduate courses such as finite element analysis, fracture mechanics and fibre composite materials. In addition, style of presentation and illustrations in engineering texts have changed for the better and certain limitations of the previous edition, e.g. the number of problems and worked examples, needed to be rectified. Professor Warnock died in 1976 after a period of happy retirement and so it was left to the other author and the publisher to take the initiative to construct a new textbook. In order to provide fresh thinking and reduce the time of rewriting Dr Roy Crawford, Reader in Mechanical Engineering at the Queen's University of Belfast, was invited, and kindly agreed, to join the project as a coauthor. Although the present text might be regarded as a further edition of the original book the new authorship team preferred to make a completely fresh start. This is reflected in the change of title which is widely used as an alternative to Mechanics of Solids. The dropping of the reference to structures does not imply any reduction in that topic as will be seen in the contents. In order that the book should not become any larger with the proposed expansion of material in some chapters, it was decided that the previous three chapters on experimental stress analysis should be omitted as there are several excellent texts in this field. The retention of that part of the book dealing with mechanical properties of materials for design was regarded as important even though there are also specialized texts in this area. The main part (Ch. 118) of this new book of course still deals with the •
... ...
..... ,.11:
..
•
'•~~L~£'1aterials,whichever
XV
Preface
title one may prefer, being the study of equilibrium and displacement systems in engineering components and structures to enable designs to be effected in terms of stress and strain and the selection of materials. These eighteen chapters cover virtually all that is required in the threeyear syllabus of a university or polytechnic degree course in engineering, or the examinations of the engineering Council, C.N.A.A. etc. Although there is a fairly natural ordering of the material there is some scope for variation and lecturers wili have their own particular detailed preferences. As in the previous text, the first eleven chapters are concerned with forces and displacements in staticallydeterminate and indeterminate components and structures, and the analysis of uniaxial stress and strain due to various forms of loading such as bending, torsion, pressure and temperature change. The basic concepts of strain energy (Ch. 9) and elastic stability (Ch. 11) are also introduced. In Chapter 12 a study is made of twodimensional states of stress and strain with special emphasis on principal stresses and the analysis of strain measurements using strain gauges. Chapter 13 combines two chapters of the previous book and brings together the topics of yield prediction and stress concentration which are of such importance in design. Also included in these two chapters is an elementary introduction to the stress analysis and failure of fibre composite materials. These relatively new advanced structural materials are becoming increasingly used, particularly in the aerospace industry, and it is essential for engineers to receive a basic introduction to them. These thirteen chapters constitute the bulk of the syllabuses covered in first and secondyear courses. Four of the next five chapters appeared in the previous text and deal with more advanced or specialized topics such as thickwalled pressure vessels, rotors, thin plates and shells and postyield or plastic behaviour, which will probably occupy part of finalyear courses. One essential new addition is an introductory chapter on finite element analysis. It may seem presumptuous even to attempt an introduction to such a broad subject in one chapter, but it is an attempt to provide initial encouragement and confidence to proceed to the complete texts on finite elements. Chapters 19 to 22 cover much the same ground as in the previous text, but have been brought up to date particularly in relation to fracture mechanics. Since these chapters have such importance in relation to design, a number of worked examples have been introduced, together with problems at the end of each chapter. Bibliographies have still been included for further reading as required. The first Appendix covers the essential material on properties of areas. The second deals with the simple principles of matrix algebra. A useful table of mechanical properties is provided in the third Appendix. One of the recommendations of the Finniston Report to higher education was that theory should be backed up by more practical industrial applications. In this context the authors have attempted to ~~ · · ·• · · · · at the end of each xvi
.[
Preface
chapter realistic engineering situations apart from the conventional examinationtype applications of theory. There had been a number of enquiries for a solutions manual for the previous text and this can be very helpful to both lecturer and student. Consequently this text is accompanied by another volume which contains worked solutions to nearly 300 problems. The manual should be used alongside the main text, so that steps in each solution can be referred back to the appropriate development in the relevant chapter. It is most important not to approach solutions on the basis of plucking the "appropriate formula" out of the text, inserting the numbers, and manipulating a calculator! Every effort has been made by the authors to ensure accuracy of text and solutions, but lengthy experience demonstrates human fallibility in this respect. When errors subsequently come to light they will be corrected at the next reprinting and readers' patience and comments will be appreciated! Some use has been made of data and diagrams from other published literature and, in addition to the individual references, the authors wish to make grateful acknowledgement to all persons and organizations concerned. P. P. Benham R. J. Crawford
1987
xvii
Notation
a {3 y fJ E
'fJ ()
).
v p
a t'
cp w
A
c
D E F G H I J K
L M N p Q R xviii
angle, coefficient of thermal expansion angle shear strain deflection, displacement direct strain efficiency, viscosity angle, angle of twist, coordinate lack of fit Poisson's ratio radius of curvature, density direct stress shear stress angle, coordinate, stress function angular velocity area complementary energy diameter Young's modulus of elasticity force shear or rigidity modulus of elasticity, strain energy release rate force second moment of area, product moment of area polar second moment of area bulk modulus of elasticity, fatigue strength factor, stress concentration factor, stress intensity factor length bending moment number of stress cycles, speed of rotation force shear force
Notation
S T U V W X
Y Z
a b c d
e g h j
k 1 m
n p q r
s t u v w x y
z
cyclic stress temperature, torque strain energy volume weight, load body force body force body force, section modulus area, distance breadth, distance distance depth, diameter eccentricity, base of Napierian logarithms gravitational constant distance number of joints diameter ratio of cylinder length mass, modular ratio, number of members number pressure shear flow coordinate, radius, radius of gyration length thickness, time displacement in the x or rdirection deflection, displacement in they or 0direction, velocity displacement in the zdirection, load intensity coordinate, distance coordinate, distance coordinate, distance It should be noted that a number of these symbols have also been used to
denote constants in various equations.
xix
Chapter 1
Staticallydeterminate force systems
INTRODUCTION
Structural and solidbody mechanics are concerned with analysing the effects of applied loads. These are external to the material of the structure or body and result in internal reacting forces, together with deformations and displacements, conforming to the principles of Newtonian mechanics. 1Henc~a familiarity with the principles of statics, the cornerstone of whicliTsthe concept of equilibrium of forces, is essential. Forces result in four basic forms of deformation or displacement of structures or solid bodies and these are tension, compression, bending and twisting. The equilibrium conditions in these situations are discussed so that the forces may be determined for simple engineering examples. REVISION OF STATICS
A particle is in a state of equilibrium if the resultant force and moment acting on it are zero, and hence according to Newton's law of motion it will have no acceleration and will be at rest. This hypothesis can be extended to clusters of particles that interact with each other with equal and opposite forces but have no overall resultant. Thus it is evident that solid bodies, structures, or any subdivided part, will be in equilibrium if the resultant of all external forces and moments is zero. This may be expressed mathematically in the following six equations which relate to
en is the inclination of nn, the neutral surface, to the zaxis. The neutral surface is perpendicular to the plane of bending if either lz = ly or 8 = 0.
The angle section shown in Fig. 6.36 is subjected to a bending moment of 2 kNm about the zaxis. Determine the bending stress distribution. 10 mm
Fig. 6.36
~I
c
1,n /
100 mm
2 kNm
E)\,fftrt z 35 m
Considering the first moments of area about vertical and horizontal edges respectively, _
(90 X 10 X 5) + (60 X 10 X 30) z= (90 + 60)10
15mm
v = (90 X 10 X 55)+ (60 X 10 X 5) = 35 mm
151
Bending: stress
which gives the coordinates of the centroid. 10 X 903 60 X 103 lz = + (10 X 90 X 20Z) + + (60 X 10 X 302 ) u 12
= 151.2 x 104 mm4 ly =
90
103 X
12
+ (90 X 10 X 10Z) +
10
603 X
12
+ (10 X 60 X 152 )
= 41.3 x 104 mm4 lzy = (90 X 10 X 10 X ( 20)] + (60 X 10 X ( 15) X 30]
= 45 x 104 mm4 The position of the neutral axis may now be found using eqn. (6.45), from which, with My = 0, y(2000 X 41.3 X 10 8) + z( 2000 X 45 X 108) = 0
~ = 1.09 and z
 (R 1 + y)8 (R 1 +y)8
'''='"''
but for an element AB at the neutral surface there is no change in length, so that R 1 8 = R 2 8. Therefore y(l'j>8) CD (Rl +y)8
e

Making the substitution Ecv =
q, = R 1 8 IR 2 gives
y[(Rt!Rz) 1] y(Rt R 2 ) = :........:...=Rl +y R 2 (Rt +y)
(6.54)
For the slender beam, y can be neglected compared with R 1 and (6.55)
158
ot:arn:s with a small radius of curvature
For R 1 infinite, i.e. a straight beam, the expression reduces to that found previously. By using the same concept as for the straight beam it can be shown that for no applied end load the centroidal axis and the neutral axis coincide, and for equilibrium of the bending moment with the internal resisting moment (6.56)
BEAMS WITH A SMALL RADIUS OF CURVATURE
For the beam in which y is not negligible compared with R 11 the strain at distance y from the neutral axis is given by eqn. (6.54). This is no longer a linear distribution of strain across the section as for the slender beam, and hence the distribution of stress is nonlinear, as indicated in Fig. 6.43 and the centroidal and neutral axes no longer coincide, as will now be shown. Fig. 6.43 M
~
Centroidal axis
Assuming that there is no applied end load,
L
adA =0
and from eqn. (6.54) a= Ee
= Ey(R1 Rz) Rz(Rl + y)
(6.57)
Therefore
E(RlRz)J _Y_dA =O Rz ARl + Y ;t moment of area 159
Bending: stress
about the centroid, therefore the centroidal and neutral axes do not coincide. For equilibrium of internal and external moments,
M=Layd.A M
= E(R 1 R2
R 2)
f ____t__ +y
dA
AR1
(6.58)
The integral term may be reexpressed as
since the second integral is zero as shown above. Now let the distance between the centroidal and neutral axes be n, then, for the crosssection of a curved beam shown in Fig. 6.44, y = y' + n; hence
Lyd.A= Ly'd.A+ L nd.A=nA Fig. 6.44
since fAy' dA is zero, being the first moment of area about the centroid. Thus
y2 d.A=nA AR1 +y
f
Substituting into eqn. (6.58), ~=E(R1R 2)
160
~~u .. '"
with a small radius of curvature
and further substituting from eqn. (6.57), M a =(Rt +y)
nA
y
or
a M =y nA(R 1 + y)
(6.59)
which is in a form similar to the bendingstress relationship for slender beams, the second moment of area term, I, being replaced by
nA(R 1 + y). In order to determine the magnitude of bending stress it is first necessary to find the values of nand R 1 for the particular shape of crosssection. From the condition that
f
_Y_dA=O +y
AR1
it follows that
1 y +n dA=O ARI+yl
f or
f(
R 1+ y 1
R1 n ) 1 1 1 dA0 1 1 1 + A R +Y R +Y R +Y
f f A
J
1
dA+ 1R 1n 1dA=O AR +Y 1 AR +Y
dA
Hence A
n=R~
(6.60)
Rt=R~n
(6.61)
LR~~Y1
and
For rectangular and circular sections the values of nand R 1 are obtained from eqns. (6.60) and (6.61). Rectangular section
nR
d
1
1
R + d/2] loge [ Rl d/2
(6.62)
An accurate value of the denominator is necessary and may be expressed as lnn
JR + d /21 = .!!:._ r1 ~ ! (_.!!:.__ \2 ~ ! (_.!!:.__) 4 + ... ] I
161
Bending: stress
Circular section of radius r
= R' [! 2
(.!_)z +!8 (.!_)4 +... J R'
(6.63)
2 R'
A crane hook as illustrated in Fig. 6.45(a) is designed to carry a maximum force of 12 kN. Calculate the maximum tensile and compressive stresses set up on the crosssection AB shown at (b).
Fig. 6.45 n
54 Section AB
(dimensions in mm)
The position of the centroid is given by d _ [48 + (2 X 24)] 54_ 124mm 72 3 Hence
d2 =54 24 = 30mm The value of n is given as follows:
n=R'
!(b 1 +b 2 )d [ b2 + b
'7Q_
162
1
~ bz (R' + dz) Jloge(~:~~:) (bt bz)
36 X 54

':l
Problems
and R 1 =78 3 =75mm A= 36 x 54= 1944mm2 M= 12000 X 0.064= 768Nm
(negative sinc;e curvature is reduced) }\ = 24 + 3 = 21 mm
Yz =
+ 30 + 3 = + 33 mm
n·uect stress on section . AB =
_12000 = +6.17 MN I m z 0 001944
768 X 0.021 . Bendmg stress at B =            0.003 X 0.001944(0.075 0.021) = +51.1 MN/m2 Total stress at B = +57.27 MN/m 2 . 768 X 0.033 Bendmg stress at A= 0.003 x 0.001 944(0.075 + 0.033) = 40.1 MN/m2 Total stress at A= 33.93 MN/m2
SUMMARY
Many important practical problems of members subjected to bending have been analysed in this chapter and if this summary is to be succinct it must pick out the most important principles. Firstly, little progress can be made unless there is a clear understanding of the calculation of shearforce and bendingmoment distributions. Then the properties of areas are always present in order to find the position of the neutral axis and the second moment of area of the crosssection. From this point in order to analyse a wide range of situations involving bending the same basic procedure is required for every case: (i) to write the equilibrium statement relating applied forces and internal reactions (as a function of stress), (ii) with appropriate assumptions to decide on the geometry of deformation of a suitable free body of the beam, and (iii) to link (i) and (ii) through the elastic stressstrain relationship. The final solution will also entail the use of the specific boundary conditions of the problem.
PROBLEMS
6.1
Draw the bendingmoment and shearforco:: diagrams for the beam loaded as · · · ' " ·· ... · · '• •  :h diagram. 163
Bending: stress
Fig. 6.46 5 kN
c 2m
6.2
2m
A gear wheel is mounted on a 25 mm diameter shaft which is driven through a 600 mm diameter pulley as in Fig. 6.47. During operation the gear wheel is subjected to a horizontal force of 3 kN and a vertical force of 5 kN. Calculate the value of the maximum bending moment on the shaft and sketch the bendingmoment diagram. Pulley weight =
Fig. 6.47
6.3
5 kN
1
400 mm
200 mm
200 mm
The portal frame shown in Fig. 6.48 is pinned at A, C and E. Sketch the bendingmoment diagrams for the sides and top of the frame and insert the principal values. 10
Fig. 6.48
10
10
10
4 D
10 kN 10 m
E
A
~ 6.4
A, E, C pinned B, D fixed
A dam 12m in length retains water at a depth of 3m as shown in Fig. 6.49. Sketch the bendingmoment and shearforce diagrams for the dam. The density of. water may be taken as 1000 kg/m3 •
Fig. 6.49
Dam
3m Water
164
Problems
6.5
6.6
A floor joist 6 m in length is simplysupported at each end and carries a varying distributed load of grain over the whole span. The loading distribution is represented by the equation w = ax 2 + bx + c, where w is the load intensity in kN/m at a distance x along the beam, and a, band care constants. The load intensity is zero at each end and has its maximum value of 4 kN/m at midspan. Apply the differential equations of equilibrium relating load, shear force and bending moment to obtain the maximum values and distributions of shear force and bending moment. A horizontal beam, the weight of which may be ignored, is loaded as shown in Fig. 6.50. The distributed load varies linearly from zero at the lefthand end to 6 kN/m at a distance of 3m from the lefthand end. Sketch the shearingforce and bendingmoment diagrams for the loading shown and insert the principal values of each on these diagrams.
Fig. 6.50
6 kN/m
1 kN
&
1~'=4~r m
5m
6.7
6.8
6.9
6.10
•
A cantilevered balcony which projects 2m out from a wall is constructed of timber joists at 0.33 m spacing supporting a boarded floor 12 mm thick. The design loading on the floor is 4.5 kN/m< and the selfweight of a joist and its associated boarding may be neglected. If each joist is to be 120 mm deep, determine the required width so that the maximum tensile bending stress does not exceed 10 MN/m 2 • For the purpose of calculating the second moment of area, it may be assumed that the neutral axis of the combined joist crosssection and its associated strip of boarding is at the middepth of the whole section. Also find where the neutral axis is correctly located when the joist width is determined. A channel which carries water is made of sheet metal 3 mm thick and has a crosssection of 400 mm width and 200 mm depth. If the maximum allowable depth of water is 150 mm determine the maximum simplysupported span. The maximum bending stress (tension or compression) is not to exceed 35 MN I m2 , and selfweight may be neglected. Loading due to water, 9.81 kN/m3 • The roof of a petrol station is made up of two main beams and eleven crossbeams (purlins) as shown in Fig. 6.51. The main beams have an !section 200 mm wide, 600 mm deep with a web thickness of 10 mm and a flange thickness of 15 mm. The purlins also have an !section 125 mm wide, 250 mm deep with a web thickness of 6 mm and a flange thickness of 10 mm. Calculate the maximum stresses in each type of beam if there is a snow loading of 120 kg/m2 on the roof. The density of the beam steel is 7850 kg/m3 and you should allow for the weight of the beams. A tapered shaft of length 1is built in at the larger end of diameter d2 , and is free at the smaller end of diameter d 1 • A force W is applied at the free end perpendicular to the axis of the shaft. Show that the maximum bending stress at any section distant x from the free end of the shaft is given by
Wx Hence determine the distance x at which point the greatest value of the maximum
165
Bending: stress
Fig. 6.51
~~lJ
,,nt:tt::::::tt~
1 I     I
~
==!=i=====:::::r:::l== __ LL ____ _j_J __ rrj,~~ll
,,An
15
m
==l::::::t:::::====:::J::j== I I I I ==~=~======i=l==
==P=F====~== 12 m
6.11
Fig. 6.52
166
A small trailer has a suspension sy:stem as shown in Fig. 6.52. If the weight of the trailer is 4 kN and its centre of gravity is 0.5 m forward of the. wheels, calculate the bending moments and torques in sections AB and BC. Calculate also the maximum bending and shear stresses in these sections. Ignore any effects at comers or changes in section.
Problems
6.12
6.13
A timber beam 80 mm wide by 160 mm deep is to be reinforced with two steel plates 5 mm thick. Compare the resisting moments for the same value of the maximum bending stress in the timber when the plates are: (i) 80 mm wide and fixed to the top and bottom surfaces of the beam, and (ii) 160 mm deep and fixed to the vertical sides of the beam. E for steel= 20 x E for timber. A composite beam is to be made up of a Ushaped steel sheet with a wooden board glued on top as in Fig. 6.53. What depth must the wood be to cause the neutral axis in pure bending to be at the horizontal diameter of the semicircle? Calculate the maximum bending moment which may be applied to this beam if the maximum stresses in the steel and wood are not to exceed 280 MN/m2 and 7 MN/m2 , respectively. The moduli for steel and wood are 210 GN/m2 and 7 GN/m2 respectively.
Fig. 6.53 60 mm
6.14
6.15
6.16
Fig. 6.54
A reinforcedconcrete beam has a rectangular crosssection 500 mm deep and 250 mm wide. The area of steel reinforcement is 1100 mm2 , and it is placed at 50 mm above the tension face. Calculate the resisting moment of the section and the stress in the steel if the compressive stress in the concrete is not to exceed 4.2 MN/m2 and the modular ratio is 15. A horizontal beam of rectangular crosssection 100 mm deep and 50 mm wide is simply supported at each end of a 1.5 m span. Vertical loads of 5 kN are applied at 0.5 m and 1m from one end, and a horizontal tension of 40 kN is applied at the ends 25 mm below the upper surface. Determine and plot the disttibution of longitudinal stress across the section at midspan. What eccentricity of end load is required so that there is just no resultant compressive stress at the outer surface? The crosssection through a concrete dam is illustrated in Fig. 6.54. Calculate the required width of the base AB so that there is just no tensile stress at B. What is the resultant compressive stress at A? Loading due to water= 9.81 kN/m3 • Weight of concrete= 22.7 kN/m3 • 10m
r  ~n
Ji=
60 m
Dam
A
6.17
B 
A concrete cooling tower may be assumed to consist of two truncated cones as in Fig. 6.55. If the estimated maximum horizontal wind pressure is 1.5 kN /m2 , calculate the wall thickness of the tower in order to avoid tensile stresses in the concrete. The den~ity ~f th! ~oncrete is 2400 kg/m3 • The volume of a truncated
167
Bending: stress
Fig. 6.55
6.18 6.19
6.20 6.21
Derive an expression for the shearstress distribution across a solid circular section rod of radius R, subjected to bending. Calculate the ratio of the maximum shear stress to the average shear stress on this section. A channelsection beam is 50 mm wide and 50 mm deep with a 5 mm wall thickness. It is simplysupported over a length of 1m and carries a uniformlydistributed load of 50 kN/m over its whole length. It also has a line load of 50 kN at midspan. Sketch the shear stress distribution across the beam section 0.25 m from one of the supports and indicate the important values. A circular tube of mean radius r and wall thickness t( «r) is subjected to a transverse shear force Q during bending. Show, by derivation, that the maximum shear stress, r, occurs at the neutral axis and is equal to QI :rcrt. A wooden beam section can be made up by one of the two methods shown in Fig. 6.56. If the shear force in the beam is constant at 3 kN, calculate which design is preferable to keep the shearing forces in the nails to a minimum. If the nails can withstand a shear force of 400 N, determine the maximum permissible spacing of the nails along the beam for the design selected. 60
Fig. 6.56
~~,
I
j,
r=
60
'
1
l
wo[
100 I

11 15
(a)
6.22
168
'
~
~
11
11
15
_I
15
(b)
A beam is made up of four 50 mm X 100 mm pieces of timber glued to a 25 mm x 500 mm web of the same wood as shown in Fig. 6.57. Calculate the maximum allowable shear force and bending moment that this section can carry. The maximum shearing stresses in the wood and glued joints must not exceed 500 and 250 ~N/m2 respectively, and the maximum oermissible direct stress is
Problems
Fig. 6.57
50
10{
50
MM 
'

~
500

r
(dimensions in mm)
6.23
A Zsection beam is 2m long and is supported as a cantilever with a 1 kN load at the free end. The direction of the 1 kN relative to the beam section is as shown in Fig. 6.58. Calculate the magnitude and position of the maximum tensile and compressive stresses on the section. 60
Fig. 6.58
L lO;l
_!
G
z~
100
1
___!Q
Z
ll
kN~ 30°
1
I]10 I
(dimenswns m mm)
60
100
Fig. 6.59
.JM,~~~~l,~N
.r:2=0'Fo/_ _ _ _ _'jjlO v3kN
z
zr100
_I  1   
Ll.
28.7
I
I
·ly
169
Bending: stress
6.24
6.25
A 100 mm X 100 mm angle section as shown in Fig. 6.59 is built in at one end of its 1 m length and subjected to a point load of 3 kN at the free end. The point load is applied at an angle of 20" to the vertical axis as indicated. Calculate the stresses at L, M and Nand the position of the neutral axis. A channelsection beam 1 m long is built in at one end and subjected to a point load of 1 kN at the free end. If the direction of the load is as in Fig. 6.60, calculate the direction of the neutral axis and the maximum values of the tensile and compressive stresses on the section.
Fig. 6.60
10
10
M 
M I
~~~
z
60
z 20
I I
100 (dimensions in mm)
6.26
Determine the location of the shear centre for the beam crosssection shown in Fig. 6.61. Also calculate maximum values of the horizontal and vertical shear stresses in a flange and the web respectively for a vertical load of 500 kN applied at the shear centre.
Fig. 6.61 ,L_ _
200
_lg I
z
z
I
10
n
J
50
(dimensions in mm)
6.27
170
!50
.I
A proving ring, used to calibrate a testing machine, has a mean diameter of 500 mm and a rectangular section 76 mm wide and 12.7 mm thick. If the maximum permitted stress under diametral comnr,..,.,;,...n i~ 'i5 MN/m2 , determine
Problems
6.28
A chain coupling is made up of a 20 mm diameter steel rod bent into an 'S' shape as shown in Fig. 6.62. If the coupling is subjected to a tensile load of 1 kN as indicated, calculate the maximum stress in the steel.
Fig. 6.62
171
Chapter 7
Bending: slope and deflection
INTRODUCTION
In Chapter 6 the stresses during bending were investigated. In this chapter the problem will be approached from an equally important direction, namely with regard to stiffness. The total deflection of a beam is due to a very large extent to the deflection caused by bending, and to a very much smaller extent to the deflection caused by shear. In practice it is usual to put a limit on the allowable deflection, in addition to the stresses. It is important that we should be able to calculate the deflection of a beam of given section, since for given conditions of span and load it would be possible to adopt a section which would meet a strength criterion but would give an unacceptable deflection. Various methods are available for determining the slope and deflection of a beam due to elastic bending, and examples of the use of each method will be found in this chapter. That part of the total deflection caused by shear will be discussed in Chapter 9.
, THE CURVATUREBENDINGMOMENT RELATIONSHIP
In Fig. 7.1, 0 is the angle which the tangent to the curve at C makes with the xaxis, and ( 0 dO) that which the tangent at D makes with the same axis. The normals to the curve at C and D meet at 0. The point 0 is the centre of curvature and R is the radius of curvature of the small portion CD of the deflection curve of the neutral axis. Numerically ds = R dO and 1/R = dO/ds. Using the sign convention on page 18 it will be seen that positive increments of ds from left to right are associated with a negative change in dO. Thus, when signs are taken into account, the above equation becomes
1
172
dO
(7.1)
rebendingmoment relationship
Fig. 7.1
l~·x v 0dO
0
'    Deflected shape
Deflections of the neutral axis are denoted by the symbol v, measured positive downwards, and are relatively small, giving a flat form of deflection curve; therefore no error is introduced in assuming that ds = dx, that()= tan()= dv/dx, and hence that d(J/ds = d2v/dx 2. Therefore
d2v
1
(7.2)
R= dx 2
On page 124 it was shown that, when elastic bending occurs,
1 R
M
=
EI
Therefore
I
d2v M =dx2 EI
(7.3)
This is the differential equation of the deflection curve. The exaCt relationship between the radius of curvature and geometry of the deformed beam is also obtained from 1/R = d() Ids by simple calculus, and is given by 1 d2v/dx 2 ±[1 + (dv/dx) 2 12
R=
P
For small deflections, this again may be reduced to the form 1/R = ±d2v/dx 2, since (dv/dx) 2 is small compared with unity. Therefore d2v M= ±El dx 2 The correct sign is chosen by observing that in the bea!ll shown M is positive and the rate of change of slope, d2v /dx 2, is negative; hence d 2v M= El dx 2 or (7.4)
173
Bending: slope and deflection
SLOPE AND DEFLECTION BY THE DOUBLE INTEGRATION METHOD
The first integration of eqn. (7 .3) gives the slope of the beam at a distance x along its length when the origin is taken at A. Therefore (7.5) The second integration gives the deflection of the beam at the above point, or (7.6) C and C11 the constants of integration, can be evaluated from the known conditions of slope and deflection at certain points, usually at the supports. Equations (7.5) and (7.6) are widely used for determining the slope and deflection of a beam at a given point. Examples of their use will be found in the following paragraphs. In each case the sign convention used to obtain eqn. (7 .3) will be adopted. BEAM SIMPLY SUPPORTED WITH DISTRIBUTED LOADING
The bending moment at Dis, from Fig. 7.2,
M = wLx wx D 2 2
2
and (7.7) and
dv
wL
wx 3
E l  =   x 2 ++C dx 4 6
Fig. 7.2
A
r
w/unit le11gth
B
,.,~~ (c)
174
~~
:ted shape
~oV ... ~
From symmetry dv I dx
[Lx
= 0,
2
Y"V
at x
3
3
' x dv _ w L  dx 2EI 2 3 12
V~"VV.OV"
(the double integration method
= !L; therefore C = wL 3 /24.
Hence
J
(7.8)
The slope at the ends of the beam is given by wL 3 /24EI at A, where x = 0, and wL3 /24EI at B, where x = L.
v
2~1 [
= 
=
+ c1
= 0; therefore cl = 0 and
At X= 0, v
v
Lt :~ ~;]
1;/ [Lx3 ~4 L;x]
(7.9)
The maximum deflection occurs at midspan M, where x
w [L
vmax
SIMPLYSUPPORTED BEAM WITH UNIFORM BENDING MOMENT
4
L
4
L
= !L. Therefore
4
4
5 wL El
]
(7· 10)
= 12El S 324 = 384
The beam shown in Fig. 7.3 is acted on by the terminal couples M. The bending moment at any point along the beam is M, and with the previous sign convention,
d2v EI dx 2 = M
(7.11)
Therefore dv dx

El= Mx+C and
Fig. 7.3 (a)
fF
Fl L
(b)
M~F"(
1,ID
B
+o
B
A (c)
~
•
)M
=
Fd
~ 
Deflected shape
175
Bending: slope and deflection
At X = 0, v = 0; therefore cl = 0; and at X = L, v C = !ML, and the slope dv/dx is given by
= 0; therefore
:=~I [Mx + ~L J
(7.12)
from which the slope at the ends is +ML/2EI at A, ML/2EI at Band at midspan is zero. The deflection at any point is given by 2
Mx ML v=EI z+Tx 1 [
J
(7.13)
The maximum deflection occurs at midspan, and hence
v CANTILEVER WITH UNIFORMI,.YDISTRIBUTED LOADING
ML 2
max
(7.14)
=
8£/
The bending moment at D in Fig. 7.4 is
wx 2
Mv=RAxMA 2 where RA = wL and MA. = !wL2 , the fixing moment at the support. Therefore
d2 v wL 2 wx 2 EI= M= wLx +  + 2
dx2
(7.15)
2
dv wL wL 2 wx 3 El=x 2 +x++C dx
2
2
6
At x = 0, dv/dx = 0; therefore C = 0. Hence 2
dv =2_ [ wL x 2 + wL x + wx dx EI 2 2 6 Fig. 7.4
3 ]
(7.16)
/w/unit
(a)
~
I
t== (~
B
(b)
M A
RA
(c)
A
~D
f1 )
176
1y the double integration method
At the free end, x = L. Therefore
wL 3 Slope at free end = EI 6
(7.17)
2
v
=
and at X
4
1 [ wL wL wx EI 6x3+4xz+ 24
= 0, v = 0; therefore cl = 0. 2
]
+ Ct
Hence 4
1 [  wL wx ] v= x 3+wL  x2 + EI 6 • 4 24
(7.18)
The deflection at the free end B, where w = L, is given by
vB
4
EI
CANTILEVER CARRYING A CONCENTRATED LOAD
4
=_.!._ [ wL + wL + wL 6
4
24
4 ]
=! wL
4
(7.19)
8 EI
The bending moment at D in Fig. 7.5 is given by
Mv=MA+RAx where MA = WI, the fixing moment at the support, and RA = W. Therefore
M= Wl+Wx d2v EI= dx2 M=WlWx
(7.20)
and
dv Wx 2 EI=Wlx+C dx 2 Fig. 7.5
(a)
(b)
~A M,( ~,ID
E
L
\ E
RA
(c)
A I
177
Bending: slope and deflection
At x = 0, dv I dx = 0; therefore C = 0. Hence
dv Wlx Wx 2 =dx EI 2EI At
X
dv dx
(7.21)
Wl 2 2El
(7.22)
=
=I,
and at X= 0,
V
= 0; therefore C1 = 0, and (7.23)
which is the equation for the deflection curve for the beam. For the deflection under the load we substitute x = I in eqn. (7 .23) and
1 Wl 3 VB= 3 EI
(7.24)
Deflection at free end E =deflection at B +slope at B x (L I). Therefore
1 Wl 3 Wl 2 Ve=+(L1) 3 EI 2EI = Wl2
2EI CANTILEVER WITH CONCENTRATED LOAD AT FREE END
(L _!_)
(7.25)
3
The slope and deflection under the load are obtained by substituting L for I in eqns. (7.22) and (7.24), and thus
v
1 WL 3 =    and m~ 3 EI
WL 2 2EI
8=
(7.26)
DISCONTINUOUS LOADING: MACAULAY'S METHOD
It was seen on pages 22 and 113, when considering the bending moment distribution for a beam with discontinuous loading, that a bending moment expression has to be written for each part of the beam. This means that in deriving slope and deflection a double integration would have to be performed on each bending moment expression and two constants would result for each section of the beam. A further example of discontinuous loading is shown in Fig. 7.6(a); in this case there would be three bending moment equations and thus six constants of integration. There are apparently only two boundary conditions, those of zero deflection at each end. However, at the points of discontinuity, B rom one section to 178
uous loading: Macaulay's method
Fig. 7.6
B
C
.
A
.
'YYYXYYY'
D
B
A
(a)
C
D
{b)
the next, so that At B
(:tB (:tc
At C (:) BC
=
and
VAB:: V 8 c
=(:)CD
and
V8c
=
Vcv
The above four plus the two end conditions enable the six constants of integration to be determined. The derivation of the deflection curve by the above approach is rather tedious; it is therefore an advantage to use the mathematical technique termed a step function, commonly known as Macaulay's method when applied to beam solutions. This approach requires one bending moment expression to be written down for a point close to the righthand end to cover the bendingmoment conditions for the whole length of beam, and hence., on integration, only two unknown constants have to be determined. The step function is a function of X of the form fn(x) = [x a such that for x a, fn(x) = (x at. Note the change in the form of brackets used: the square brackets are particularly chosen to indicate the use of a step function, the curved brackets representing normal mathematical procedure. The important features when using the step function in analysis are that, if on substitution of a value for x the quantity inside the square brackets becomes negative, it is omitted from further analysis. Square bracket terms must be integrated in such a way as to preserve the identity of the bracket, i.e.
r
J[x a]
2
d(x a)= Ux aP
Also, for mathematical continuity, distributed loading which does not extend to the righthand end, as in Fig. 7.6(a), must be arranged to Fig. 7.7
(a)
x =a
w (b)
M
~
=
~
x = a
x =a
M =
~~
T [x a)
2
_/
w (c)
W[x a] 1
x =a
179
Bending: slope and deflection
continue to x = I, whether starting from x = 0 or x =a. This may be effected by the superposition of loadings which cancel each other in the required portions of the beam as shown in Fig. 7 .6(b). An applied couple M 0 must be expressed as a step function in the form M 0 [x  a]0 so that the bracket can be integrated correctly (see example on page 182). The three common step functions for bending moment are shown in Fig. 7. 7 and several illustrative examples now follow.
BEAM SIMPLY SUPPORTED WITH CONCENTRATED LOAD
Taking moments about one end, the reactions at the supports in Fig. 7.8 are R1= W(La) L
and
R = Wa 2 L
When x >a, the bending moment at D isM= R 1x  W[x a]. Hence
d2 v EI dx 2 = M = R 1x + W[x a]
(7.27)
When x