A TEXTBOOK OF
STRENGTH OF MATERIALS
(In S.I. Units)
[For Degree, U.P.S.C. (Engg. Services), Gate and
other Competitive Examinations]
By
Dr. R.K. Bansal
B.Sc. Engg. (Mech.), M. Tech., Hons. (I.I.T., Delhi)
Ph.D., M.I.E. (India)
Formerly Professor and Head
Department of Mechanical Engineering,
(University of Delhi)
Delhi College of Engineering, Delhi
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Contents
Pages
Chapter
j
Chapter 1. Simple Stresses and Strains
1.1. Introduction 4
1.2. Stress 2
2
1.3. Strain g
g.
1.4. Types of Stresses g
1.5. Elasticity and Elastic Limit
1.6. 
1.7. Hooke’s Law and Elastic Modulii
g
Modulus of Elasticity (pr Young’s Modulus) 6
1.8. Factor of Safety ^
1.9. Constitutive Relationship between Stress and Strain
1.10. ™ 24
1.11. Analysis of Bars of Varying Sections
1.12. Analysis of Uniformly Tapering Circular Rod 27
1.13. Analysis of Uniformly Tapering Rectangular Bar
™
Analysis of Bars of Composite Sections
3Q
1.14. Thermal Stresses
1.15. Thermal Stresses in Composite Bars ™
1.16.
1.17. Elongation of a Bar Due to its Own Weight 44
50
Analysis of Bar of Uniform Strength gj
g3
Highlights 54
Exercise 1
Chapter 2. . „Cons' tant,s 5984
Elastic 59
gg
2.1. Introduction 59
2.2. Longitudinal Strain 60
2.3. Lateral Strain 02
gg
2.4. Poisson’s Ratio 7g
70
2.5. Volumetric Strain
2.6. Volumetric Strain of a Cylindrical Rod ^
2.7. Bulk Modulus '
2.8.
Expression for Young’s Modulus in Terms of Bulk Modulus
29
2.10. Principle of Complementary Shear Stresses ?4
Stresses on Inclined Sections when the Element is Subjected to Simple
Shear Stresses _ .. 76
Diagonal Stresses Produced by Simple Shear on a Square Block
2 11  ^
2 12 Direct (Tensile and Compressive) Strains of the Diagonals 
2.13. Relationship between Modulus of Elasticity and Modulus of Rigidity ^...
Highlights _ 82
Exercise 2
Chapter Pages Chapter Pages
Chapter 3. Principal Stresses and Strains 85—142
85... Chapter 6. Shear Force and Bending Moment 235—291
... 85
3.1. Introduction ... 85 6.1. Introduction ... 235
3.2. ... 85 ... 235
3.3. Principal Planes and Principal Stresses ... 123 6.2. Shear Force and Bending Moment Diagrams ... 235
3.4. Methods of Determining Stresses on Oblique Section ... 128 6.3. Types of Beams 236
3.5. Analytical Method for Determining Stresses on Oblique Section ... 138 237
3.6. Graphical Method for Determining Stresses on Oblique Section ... 139 6.4. Types of Load ... 238
Mohr’s Circle
6.5. Sign Conventions for Shear Force and Bending Moment ... 239
Highlights
Exercise 3 6.6. Important Points for Drawing Shear Force and Bending Moment Diagrams ... 242
6.7. Shear Force and Bending Moment Diagrams for a Cantilever with a 250
Point Load at the Free End ... 252
Chapter 4. Strain Energy and Impact Loading 143169 ( 254
4.1. Introduction 143 6^8. Shear Force and Bending Moment Diagrams for a Cantilever with a 256
143
4.2. Some Definitions Uniformly Distributed Load ... 264
4.3. Expression for Strain Energy Stored in a Body when the Load is Applied 143
6.9. Shear Force and Bending Moment Diagrams for a Cantilever 266
Gradually 152 6.10.
165 6.11. Cartying a Gradually Varying Load ... .270
4.4. . Expression for Stain Energy Stored in a Body when the Load is Applied 166 6.12.
167 6.13. . 279
Suddenly
6.14. Shear Force and Bending Moment Diagrams for a Simply 284
4.5. Expression for Strain Energy Stored in a Body when the Load is Applied 287
6.15. Supported Beam with a Point Load at Midpoint ... 288
with Impact 6.16. 289
4.6. Expression for Strain Energy Stored in a Body due to Shear Stress Shear Force and Bending Moment Diagrams for a Simply
Highlights Supported Beam with an Eccentric Point Load ...
Exercise 4
Shear Force and Bending Moment Diagrams for a Simply
Supported Beam Carrying a Uniformly Distributed Load ...
Shear Force and Bending Moment Diagrams for a ...
Simply Supported Beam Carrying a Uniformly "
Varying Load from Zero at Each End to w Per Unit Length at the Centre
Shear Force and Bending Moment Diagrams for a Simply Supported Beam
Chapter 5. Centre of Gravity and Moment of Inertia 170—234 Carrying a Uniformly Varying Load from Zero at one End to w Per Unit
... 170 Length at the Other End ...
... 170
5.1. Centre of Gravity ... 170 Shear Force and Bending Moment Diagrams for Overhanging Beams ...
5.2. Centroid ... 170 Shear Force and Bending Moment Diagrams for
... 172
5.3. Centroid or Centre of Gravity of Simple Plane Figures ... 194 Beams Carrying Inclined Load ...
... 195 Shear Force and Bending Moment Diagrams for
5.4. Centroid (or centre of gravity) of Areas of Plane Figures ... 195 6.17. Beams Subjected to Couples ...
... 196 6.18. Relations between Load, Shear Force and Bending Moment ...
by the Method of Moments ... 197
... 211
5.5. Important Points ... 212
... 218
5.6. Area Moment of Inertia ... 219 Highlights ...
... 220
5.7. Radius of Gyration ... 228 Exercise 6 ...
5.8. Theorem of the Perpendicular Axis ... 229
.
5.9. Theorem of Parallel Axis Chapter 7. Bending' Stresses in Beams 292—341
5.10. Determination of area Moment of Inertia 7.1. Introduction ... 292
... 292
5.11. Mass Moment of Inertia 7.2. Pure Bending or Simple Bending ... 293
... 294
5.12. Determination of Mass Moment of Inertia 7.3. Theory of Simple Bending with Assumptions Made ... 295
... 297
5.13. Product of Inertia 7.4. Expression for Bending Stress ' ... 300
... 300
5.14. Principal Axes 7.5. Neutral Axis apd Moment of Resistance
... 3.12
5.15. Principal Moments of Inertia 7.6. Bending Stresses in Symmetrical/Sections
Highlights 7.7. Section Modulus
Exercise 5 7.8. Section Modulus for Various Shapes of Beam Sections
7.9. Bending Stress in Unsymmetrical Sections
Chapter Pages Chapter Pages
7.10. Strength of a Section Chapter 11. Analysis of Perfect Frames 465—510
7.11. Composite Beams (Flitched Beams) 320
Highlights  ... 327 11.1. rInt. rodJ uct.•ion  ... 465
Exercise 7
. 11.2. Types of Frames  4bb
11.3. Assumptions Made in Finding Out the Forces in a Frame
— 337 11.4. Reactions of Supports of a Frame 4o i
11.5. Analysis of a Frame 4gg
338 11.6. Method of Sections __ 4
11.7. Graphical Method g^4
Chapter 8. Shear Stresses in Beams 342—376
Highlights
8.1. Introduction — 342
8.2. Shear Stress at a Section Exercise 11
8.3. Shear Stress Distribution for Different Sections ... 342
Highlights
Exercise 8 — 348 1 511553
— 373
— 374 Chapter 12. Deflection of Beams
Chapter 9. Direct and Bending Stresses 377—408 Introd,uction ... 511
... 511
12.1. Deflection and Slope of a Beam Subjected to Uniform Bending Moment
9.1. Introduction — 377 12.2. ^
9.2. 377 12.3. Relation between Slope, Deflection and Radius of Curvature
9.3. Combined Bending and Direct Stresses ' 12.4. 519
Resultant Stress when a Column of Rectangular Section is Subjected to Deflection of a Simply Supported Beam Carrying a
9.4. an Eccentric Load ... ^
Resultant Stress when a Column of Rectangular Section is Subjected to a
9.5. Load which is Eccentric to both Axes ... 386 12.5. Point Load at the Centre ... g 4g
9.6. Resultant Stress for Unsymmetrical Columns with Eccentric Loading ... 393 12.6. Deflection of a Simply Supported Beam with an Eccentric Point Load ... g4g
9.7. Middle Third Rule for Rectangular Sections (t.e.. Kernel of section) ... 398
9.8. Middle Quarter Rule for Circular Sections (i.e. Kernel of section) ... 400 Deflection of a Simply Supported Beam with a Uniformly Distributed Load
9.9. , ... 401 12.7. Macaulay's Method
Kernel of Hollow Circular Section (or value of eccentricity ... 402 12.8. Moment Area Method
for hollow circular section)
Kernel of Hollow Rectangular Section (or value of — 405 12.9. Mohr’s Theorems of a Simply Supported Beam Carrying a _, Load at
eccentricity for hollow rectangular section) — 406 12. 10. Slope and Deflection Point
Highlights the Centre by Mohr’s Theorem .
12.11. Slope and Deflection of a Simply Supported Beam Carrying a
Exercise 9
Uniformly Distributed load by Mohr’s Theorem
Highlights
Exercise 12
'* Chapter 10. Dams and Retaining Walls 409—464 Chapter 13. Deflection of Cantilevers 554—577
10.1. Introduction  409 13.1. Introduction 554
10.2. 13.2. Deflection of a Cantilever with a Point Load at the Free end by Double
10.3. Types of Dams — 409 ^
10.4. Rectangular Dams — 409 Integration Method
10.5. Stresses Across the Section of a Rectangular Dam 13.3. Deflection of a Cantilever with a Point Load at a Distance ‘a’ from g5g
10.6. Trapezoidal Dam having Water Face Inclined ... 413
Stability of a Dam ... 424 the Fixed End ,.7
— 430 13.4. Deflection of a Cantilever with a Uniformly Distributed Load gg]L
— 443 13.5.
— 445 Deflection of a Cantilever with a Uniformly Distributed Load for a ggl
13.6.
10.7. Retaining Walls 455 Distance 7 from the Fixed End 
10.8. Rankine’s Theory of Earth Pressure 458 13.7.
10.9. Surcharged Retaining Wall 13.8. ‘a ^
10.10. Chimneys — 460
— 462 Deflection of a Cantilever with a Uniformly Distributed Load for a
Distance a‘ ’ from the Free End
Deflection of a Cantilever with a Gradually Varying Load
Highlights Deflection and Slope of a Cantilever by Moment Area Method
Exercise 10 Highlights
Exercise 13
Chapter Pages Chapter Pages
Chapter 14. Conjugate Beam Method, Propped 578—612 16.9. Flanged Coupling ... 695
Cantilevers and Beams 16. 10. Strength of a Shaft of Varying Sections ... 698
... 578 16.11. Composite Shaft ... 706
14.1. Introduction ... 578 16. 12. Combined Bending and Torsion ... 710
16.13. Expression for Strain Energy Stored in a Body Due to Torsion ... 713
14.2. Conjugate Beam Method ... 578 16.14. Springs ^ ... 721
14.3. Deflection and Slope of a Simply Supported Beam with a Point ... 734
Highlights
Load at the Centre. Exercise 16 — 736
14.4. Simply Supported Beam Carrying an Eccentric Point Load
... 580
14.5. Relation between Actual Beam and Conjugate Beam
14.6. Deflection and Slope of a Cantilever with a Point Load at the Free End ... 592 Chapter 17. Thin Cylinders and Spheres 740—780
14.7. Propped Cantilevers and Beams
... 592
14.8. S.F. and B.M. Diagrams for a Propped Cantilever Carrying a Point Load
at the Centre and Propped at the Free End 597... . 17.1. Introduction 740
740
14.9. S.F. and B.M. Diagram for a Propped Cantilever Carrying ... 741
a Uniformly Distributed Load and Propped at the Free End 741
17.2. Thin Cylindrical Vessel Subjected to Internal Pressure ... 742
14.10. S.F. and B.M. Diagrams for a Simply Supported Beam with 746
... 598 17.3. Stresses in a Thin Cylindrical Vessel Subjected to Internal Pressure 750
a Uniformly Distributed Load and Propped at the Centre 761
• 14.11. Yielding of a Prop ...
765.
Highlights 17.4. Expression for Circumferential Stress (or hoop stress) ...
Exercise 14 770
... 599 771
773
17.5. Expression for Longitudinal Stress ... 776
777
17.6. Efficiency of a Joint ...
17.7.
... 605 17.8. Effect of Internal Pressure on the Dimensions of a Thin Cylindrical Shell ...
... 609
... 610 A Thin Cylindrical Vessel Subjected to Internal Fluid Pressure and a Torque...
... 611
17.9. Wire Winding of Thin Cylinders ...
17.10. Thin Spherical Shells ...
Chapter 15. Fixed and Continuous Beams 613—671 17.11. Change is Dimensions of a Thin Spherical Shell Due to an Internal Pressure ...
17.12. Rotational Stresses in Thin Cylinders ...
15.1. Introduction 613 Highlights ...
614
... 618
622
15.2. Bending Moment Diagram for Fixed Beams ' Exercise 17 '
638 ...
... 648
651
15.3. Slope and Deflection for a Fixed Beam Carrying a Point Load at the Centre ... 652
652
15.4. Slope and Deflection for a Fixed Beam Carrying an Eccentric Point Load ... 669 Chapter 18. Thick Cylinders and Spheres 781—807
670
15.5. Slope and Deflection for a Fixed Beam Carrying a Uniformly Distributed 18.1. Introduction 781
18.2. Stresses in a Thick Cylindrical Shell ... 781
Load over the Entire Length ... ... 789
18.3. Stresses in Compound Thick Cylinders
15.6. Fixed End Moments of Fixed Beam Due to Sinking of a Support ... 18.4. Initial Difference in Radii at the Junction Df a Compound Cylinder for ... 794
800
15.7. Advantages of Fixed Beams ... Shrinkage
18.5. Thick Spherical Shells ... 805
15.8. Continuous Beams ... ... 806
15.9. Bending Moment Diagram for Continuous Beams ... Highlights
Exercise 18
Highlights ...
Exercise 15 ...
Chapter 16. Torsion of Shafts and Springs 672—739 808—870
Chapter 19. Columns and Struts
16.1. Introduction ... 672 19.1. Introduction " ... 808
16.2. ... 672 19.2. ... 808
16.3. Derivation of Shear Stress Produced in a Circular Shaft Subjected to Torsion ... 674 19.3. Failure, of a Column ... 809
16.4. ... 676 19.4. ... 809
16.5. Maximum Torque Transmitted by a Circular Solid Shaft ... 677 19.5. Assumptions Made in the Euler’s Column Theory ... 810
16.6. ... 687 19.6.
16.7. Torque Trasmitted by a Hollow Circular Shafts ... 688 End Conditions for Long Columns 811
16.8. Power Transmitted by Shafts ... 688
Expression for Crippling Load When Both the Ends of the Column are Hinged
Expression for Torque in Terms of Polar Moment of Inertia
Polar Modulus Expression for Crippling Load When One End of the
Strength of a Shaft and Torsional Rigidity Column is Fixed and the Other End is Free
.
Chapter Pages C/taPfer Pages
19.7. Expression for Crippling Load When Both the Ends of the Column are Fixed ... 813 22.4. Long Cylinders ... 940
Highlights " 953
19.8. Expression for Crippling Load When One End of the Column is Fixed and 816 Exercise 22 ... 955
818
the Other End is Hinged (or Pinned) ... 820
835
. 847
847
19.9. Effective Length (or equivalent length) of a Column ... 848
848
19.10. Limitation of Euler’s Formula ... 849 Chapter 23. Bending of Curved Bars 957—1004
853
19.11. Kankine’s Formula ... 858
866
19.12. Straight Line Formula ... 868 23.1. Introduction .. 957
23.2. ... 957
. 19.13. Johnson’s Parabolic Formula ... 23.3. Assumptions Made in the Derivation of Stresses in a Curved Bar ... gg7
23.4. ... 964
19.14. Factor of Safety ... 23.5. Expression for Stresses in a Curved Bar
Determination of Factor ‘h2’ for Various Sections ... 977
19.15. Formula by Indian Standard Code (I.S. Code) for Mild Steel ... 23.6. Resultant Stress in a Curved Bar Subjected to Direct Stresses and Bending ... 978
23.7. Stresses ... 987
19.16. Columns with Eccentric Load ... 23.8. Resultant Stress in a Hook ... 993
... 1000
19.17. Columns with Initial Curvature ... Stresses in Circular Ring ... 1002
Stresses in a Chain Link
19.18. Strut with Lateral Load (or beam columns) ...
Highlights
Highlights ...
Exercise 23
Exercise 19 . ...
,, Chapter 20. Riveted Joints 871—899 Chapter 24. Theories of Failure 1005—1037
20.1. Introduction ... 871 24.1. Introduction 1005
20.2. Types of Riveted Joints ... 871 1005
20.3. Chain Riveted Joint ... 872 24.2. Maximum Principal Stress Theory ... 1006
20.4. Zigzag Riveted Joint ... 872 24.3. Maximum Principal Strain Theory ... 1010
... 872 ... 1014
20.5. Diamond Riveted Joint ... 876 24.4. Maximum Shear Stress Theory ... 1018
... 879 24.5. Maximum Strain Energy Theory ... 1020
20.6. Failure of a Riveted Joint ... 880 \ 24.6. Maximum Shear Strain Energy Theory ... 1024
20.7. Strength of a Riveted Joint ... 892 ... 1033
20.8. Efficiency of a Riveted Joint ... 895 24.7. Graphical Representation of Theories for Two Dimensional Stress System ... 1036
20.9. Design of a Riveted Joint ... 897 24.8. ... 1036'
*. Highlights 24.9. Important Points from Theories of Failures used in Design
900—918 Energy of Distortion (or shear strain energy)
Exercise 20
... 900 Highlights
* ... 900
... 901 Exercise 24
Chapter 21. Welded Joints ... 905
... 907 Chapter 25. Objective Type Questions 1038—1088
21.1. Introduction ... 914
21.2. .Advantages and Disadvantages of Welded Connections ... 916 25.1. Objective Type Questions Generally Asked in Competitive Examinations ... 1038
 21.3. Types of Welded Joints 25.2. Answer of Objective Type Questions ... 1065
25.3. Objective Type Questions from Competitive Examinations ... 1066
21.4. Analysis of a Compound Weld 25.4. Answers with Explanations ... 1074
21.5. Analysis of Unsymmetrical Welded Sections which are Loaded Axially
1 Highlights
Exercise 21
Chapter 22. Rotating Discs and Cylinders 919—956 Subject Index 1089—1092
22 .1. Introduction 919
22 .2. Expression for Stresses in a Rotating Thin Disc 919
Disc of Uniform Strength 936
22:3:
PREFACE TO THE FIRST EDITION PREFACE TO THE FOURTH EDITION
I am glad to present the book entitled, ‘A Textbook of Strength of Materials’ to the The popularity of the third edition and reprints of the textbook of Strength of Materials
amongst the students and the teachers of the various Universities of the country, has prompted
engineering students of mechanical, civil, electrical, aeronautical and chemical and also to the the bringing out of the fourth edition of the book so soon. The fourth edition has been thoroughly
students of A.M.I.E. Examination of Institution of Engineers (India). The coursecontents have
been planned in such a way that the general requirements of all engineering students are revised and brought uptodate. A large number of problems from different B.E. degree
fulfilled. examinations upto 2005 of Indian Universities and other examining bodies, such as Institution
of Engineers U.P.S.C. (Engineering Services) and Gate have been selected and have been solved
During may long experience of teaching to the engineering students for the past 20 years, at proper places in this edition in S.I. Units.
I have observed that the students face difficulty in understanding clearly the basic principles, Three advanced topics of Strength of Materials such as stresses due to rotation in thin
fundamental concepts and theory without adequate solved problems along with the text. To and thick cylinders, bending of curved bars and theories of failure of the material have been
meet this very basic requirement to the students, a large number of the questions taken from added. These chapters have been written in such a simple and easytofollow language that
the examinations of the various Universities of India and from other professional and competitive even an average student can understand easily by selfstudy.
examinations (such as Institution and Engineers and U.P.S.C. Engineering Service
Examinations) have been solved along with the text in S.I. units. In the chapter of ‘Columns and Struts’, the advanced articles such as columns with
eccentric load, with initial curvature and beam columns have been included.
The book is written in a simple and easytofollow language, so that even an average
students can grasp the subject by selfstudy. At the end of each chapter highlights, theoretical The notations in this edition have been used uptodate by the use of sigma and tau for
questions and many unsolved numerical problems with answer are given for the students to
solve them. stresses.
m/ I a thankful to my colleagues, friends and students who encouraged me to write this The objective type multiplechoice questions are often asked in the various competitive
examinations. Hence a large number of objective type questions with answers have been added
book. I am grateful to Institution of Engineers (India), various Universities of India and those in the end of the book.
authorities whose work have been consulted and gave me a great help in preparing the book.
Also a large number of objective type questions which have been asked in most of
I express my appreciation and gratefulness to my publisher, Shri R.K. Gupta (a Mechanical competitive examinations such as Engineering Services Examination and Gate with answers
and explanation have been incorporated in this edition.
Engineer) for his most cooperative, painstaking attitude and untiring efforts for bringing out
the book in a short period. With these editions, it is hoped that the book will be quite useful for the students of
different branches of Engineering at various Engineering Institutions.
Mrs. Nirmal Bansal deserves special credit as she not only provided an ideal atmosphere
at home for book writing but also gave inspiration and valuable suggestions. I express my sincere thanks to my collegues, friends, students and the teachers of different
Though every care has been taken in checking the manuscripts and proof reading, yet Indian Universities for their valuable suggestions and recommending the book to their students.
claiming perfection is very difficult. I shall be very grateful to the readers and users of this book Suggestions for the improvement of this book are most welcome and would be incorpo
for pointing any mistakes that might have crept in. Suggestions for improvement are most
welcome and would be incorporated in the next edition with a view to make the book more rated in the next edition with a view to make the book more useful.
useful. AUTHOR
AUTHOR
1
Simple Stresses and Strains
1.1. INTRODUCTION
When an external force acts on a body, the body tends to undergo some deformation.
Due to cohesion between the molecules, the body resists deformation. This resistance by which
material of the body opposes the deformation is known as strength of material. Within a
certain limit (i.e., in the elastic stage) the resistance offered by the material is proportional to
the deformation brought out on the material by the external force. Also within this limit the
resistance is equal to the external force (or applied load). But beyond the elastic stage, the
resistance offered by the material is less than the applied load. In such a case, the deformation
continues, until failure takes place.
Within elastic stage, the resisting force equals applied load. This resisting force per unit
area is called stress or intensity of stress.
1.2. STRESS
The force of resistance per unit area, offered by a body against deformation is known as
stress. The external force acting on the body is called the load or force. The load is applied on
the body while the stress is induced in the material of the body. A loaded member remains in
equilibrium when the resistance offered by the member against the deformation and the ap
plied load are equal.
—P
Mathematically stress is written as, a =
where a = Stress (also called intensity of stress),
P = External force or load, and
A = Crosssectional area.
1.2.1. Units of Stress. The unit of stress depends upon the unit of load (or force) and
munit of area. In M.K.S. units, the force is expressed in kgf and area in metre square (i.e.,
2
).
Hence unit of stress becomes as kgf/mz . If area is expressed in centimetre square cm(i.e., . 2
),
the stress is expressed as kgf/cm2 .
In the S.I. units, the force is expressed in newtons (written as N) and area is expressed
mas 2 Hence unit of stress becomes as N/m2 . The area is also expressed in millimetre square
.
then unit of force becomes as N/mm2 .
1 N/m2 = 1 N/(100 cm)2 = 1 N/104 cm2
= 10~4 N/cm2 or 10"fi N/mm2 mm2 2
10
SfMPLE STRESSES AND STRAINS 3
2 STRENGTH OF MATERIALS
1 N/mm2 = 10 8 N/m2 . Let P = Pull (or force) acting on the body,
A = Crosssectional area of the body,
Also 1 N/m2 = 1 Pascal = 1 Pa. L = Original length of the body,
dL = Increase in length due to pull P acting on the body,
The large quantities are represented by kilo, mega, giga and terra. They stand for :
a = Stress induced in the body, and
Kilo = 10s and represented by k
Mega = 106 and represented by e = Strain (i.e., tensile strain).
M
PFig. 1.1 (a) shows a bar subjected to a tensile force at its ends. Consider a section xx,
Giga = 109 and represented by G
which divides the bar into two parts. The part left to the section xx, will be in equilibrium if
Terra = 10 12 and represented by T.
P  Resisting force (R). This is shown in Fig. 1.1 (6). Similarly the part right to the section xx,
Thus mega newton means 10s newtons and is represented by MN. The symbol 1 MPa
stands for 1 mega pascal which is equal to 10® pascal (or 10 6 N/m2). will be in equilibrium ifP = Resisting force as showm in Fig. 1.1 (c). This resisting force per unit
area is known as stress or intensity of stress.
The small quantities are represented by milli, micro, nana and pica. They are equal to
Milli = 10"3 and represented by m
Micro = 1CT3 and represented by ...... p,
Nana = 109 and represented by r
Pica = 1012 and represented by p.
Notes. 1. Newton is a force acting on a mass of one kg and produces an acceleration of 1 m/s 2 he.,
N1 = 1 (kg) x 1 m/s2 .
2. The stress in S.I. units is expressed in N/m2 or N/mm2 .
3. The stress 1 N/mm2  10e N/m2 = MN/m2 Thus one N/mm2 is equal to one MN/m 2 .
.
4. One pascal is written as 1 Pa and is equal to 1 N/m2.
1.3. STRAIN
When a body is subjected to some external force, there is some change of dimension of
the body. The ratio of change of dimension of the body to the original dimension is known as
strain. Strain is dimensionless.
; Strain may be :
1. Tensile strain, 2. Compressive strain,
3. Volumetric strain, and 4. Shear strain.
If there is some increase in length of a body due to external force, then the ratio of Fig. 1.1
increase of length to the original length of the body is known as tensile strain. But if there is —Tensile stress = a = Resistin2g force (R) = Tensile load (P) (v P = R)
some decrease in length of the body, then the ratio of decrease of the length of the body to the r
original length is known as compressive strain. The ratio of change of volume of the body to the ACrosssectional area ...(1.1)
original volume is known as volumetric strain. The strain produced by shear stress is known
P f
as shear strain.
or A
And tensile strain is given by,
1.4. TYPES OF STRESSES Increase in length dL 2...( 1 . )
Original length L
The stress may be normal stress or a shear stress.
1.4.2. Compressive Stress. The stress induced in a body, when subjected to two equal
Normal stress is the stress which acts in a direction perpendicular to the area. It is and opposite pushes as shown in Fig. 1.2 (a) as a result of which there is a decrease in length
represented by a (sigma). The normal stress is further divided into tensile stress and compressive of the body, is known as compressive stress. And the ratio of decrease in length to the original
length is known as compressive strain. The compressive stress acts normal to the area and it
stress.
pushes on the area.
Tensile Stress. The stress induced in a body, when subjected to two equal and
opposite pulls as shown in Pig. 1.1 (a) as a result of which there is an increase in length, is Let an axial push P is acting on a body is crosssectional area A. Due to external push P,
known as tensile stress. The ratio of increase in length to the original length is known as let the original length L of the body decreases by dL.
tensile strain. The tensile stress acts normal to the area and it pulls on the area.
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
Consider a rectangular block of height h, length L and width unity. Let the bottom face
PAS of the block be fixed to the surface as shown in Fig. 1.4 (a). Let a force be applied
CDtangentially along the top face of the block. Such a force acting tangentially along a surface
ASis known as shear force. For the equilibrium of the block, the surface will offer a tangential
x («) reaction P equal and opposite to the applied tangential force P.
Resisting force <R)
XX XX Resistance
R
<R
Resistance
Resisting force (R)
<c) 77777777777777777777777
A *P B
J (M <0
(d) \ '' Fig. 1.4
Fig. 1.2 Consider a section xx (parallel to the applied force), which divides the block into two
parts. The upper part will be in equilibrium ifP = Resistance (Ft). This is shown in Fig. 1.4 (6).
Then compressive stress is given by,
PSimilarly the lower part will be in equilibrium if = Resistance (2?) as shown in Fig. 1.4 (c).
Resisting Force R( ) _ Push P( ) _ P This resistance is known as shear resistance. And the shear resistance per unit area is known
° Area (A) Area (A) A as shear stress which is represented by t.
And compressive strain is given by, 1 AR'Shear stress, x = Shear resistance _
Decrease in length _ d.L Shear area
Original length L
143 Shear Stress. The stress induced in a body, when subjected to two equal mad ixl RP A(v Land = x 1) ...(1.3)
Note that shear stress is tangential to the area over which it acts.
gential to the area. It is represented by x. As the bottom face of the block is fixed, the face
DABCD will be distorted to ABC through an angle <> as a
l1
result of force P as shown in Fig. 1.4 (d).
And shear strain (<j>) is given by.
Transversal displacement
Distance AD
DD (U
1
...(1.4)
AD h
Fig. 1.4 (d)
1.5. ELASTICITY AND ELASTIC LIMIT
When an external force acts on a body, the body tends to undergo some deformation. If
the external force is removed and the body comes back to its origin shape and size (which
means the deformation disappears completely), the body is known as elastic body. This property,
.
SIMPLE STRESSES AND STRAINS 7
6 STRENGTH OF MATERIALS
by virtue of which certain materials return back to their original position after the removal of stress to the corresponding strain is a constant within the elastic limit. This constant is repre
the external force, is called elasticity.
sented by E and is known as modulus of elasticity or Young’s modulus of elasticity.
The body will regain its previous shape and size only when the deformation caused by
the external force, is within a certain limit. Thus there is a limiting value of force up to and ! Normal stress _. = Constant or °£
within which, the deformation completely disappears on the removal of the force. The value of
stress corresponding to this limiting force is known as the elastic limit of the material. Corresponding strain e
If the external force is so large that the stress exceeds the elastic limit, the material I
loses to some extent its property of elasticity. If now the force is removed, the material will not where a = Normal stress, e = Strain and E = Young’s modulus
return to its origin shape and size and there will be a residual deformation in the material.
£or „ _ ...[17 CAM
E mThe
above \ gives the stress and strain relation for the normal stress one
equation
1.6. HOOKE’S LAW AND ELASTIC MODULII direction.
Hooke’s Law states that when a material is loaded within elastic limit, the stress is 1.9.2. For TwoDimensional Stress System. Before knowing the relationship be
proportional to the strain produced by the stress. This means the ratio of the stress to the tween stress and strain for twodimensional stress system, we shall have to define longitu i
corresponding strain is a constant within the elastic limit. This constant is known as Modulus nal strain, lateral strain, and Poisson’s ratio.
of Elasticity or Modulus of Rigidity or Elastic Modulii. 1. Longitudinal strain. When a body is subjected to an axial tensile load, there is an
1.7. MODULUS OF ELASTICITY (OR YOUNG’S MODULUS) increase in the length [of the body. But at the same time there is a decrease in other dimensions
of the body at right singles to the line of action of the applied load. Thus the body is laying
The ratio of tensile stress or compressive stress to the corresponding strain is a con
stant. This ratio is known as Young’s Modulus or Modulus of Elasticity and is denoted by E. axial deformation an( also deformation at right angles to the line of action of the applied load
— —E Tensile stress (i.e., lateral deformation).
Compressive stress
= or ; The ratio of axial deformation to the original length of the body is known as longitudinal
(or linear) strain. Thd longitudinal strain is also defined as the deformation of the body per
Tensile strain Compressive strain
unit length in the direction of the applied load.
1.8. 1.7.1. Modulus of Rigidity or Shear Modulus. The ratio of shear stress to the
corresponding shear strain within the elastic limit, is known as Modulus of Rigidity or Shear Let L = Lengih of the body,
P  Tensile force acting on the body,
Modulus. This is denoted by C or G or N.
.5—bL = Increase in the length of the body in the direction of P.
™ ——C(orGorIV)= Shear stress = —r ' „ L
Shear strain <j) ...(1.6) Then, longitudinal strain =
2. Lateral strain. The strain at right angles to the direction of applied load is known as
lateral strain. Let a rectangular bar of length L, breadth b and depth d is subjected to an axial
tensile load P as shoWn in Fig. 1.5. The length of the bar will increase while the breadth and
depth will decrease.
Let 1 SL = Increase in length,
bb = Decrease in breadth, and
1.9. Let us define factor of safety also. i bd = Decrease in depth.
FACTOR OF SAFETY \ ...[1.7 (B)]
mTihen ll ongitudinal1 s.ftpr,a;i,,n —=
It is defined as the ratio of ultimate tensile stress to the working (or permissible) stress. lateral strain = ...[1.7 (C)]
Mathematically it is written as
. Ultimate stress ..
Fact, or of safety =
Permissible stress
CONSTITUTIVE RELATIONSHIP BETWEEN STRESS AND STRAIN «(b6b)i*
1.9.1. For OneDimensional Stress System. The relationship between stress and I
strain for a unidirectional stress (i.e., for normal stress in one direction only) is given by Hooke’s
law, which states that when a material is loaded within its elastic limit, the normal stress
developed is proportional to the strain produced. This means that the ratio of the normal
Fig. 1.5
'. '
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
Note, (i) If longitudinal strain is tensile, the lateral strains will be compressive. The above two equations gives the stress and strain relationship for the twodimen
sional stress system. In the above equations, tensile stress is taken to be positive whereas the
(ii) If longitudinal strain is compressive then lateral strains will be tensile.
compressive stress negative,
(iii) Hence every longitudinal strain in the direction of load is accompanied by lateral strains of
the opposite kind in all directions perpendicular to the load. 1.9.3. For ThreeDimensional Stress System. Fig. 1.5 (b) shows a threedimensional
3. Poisson s ratio. The ratio of lateral strain to the longitudinal strain is a constant for body subjected to three orthogonal normal stresses av o o acting in the directions ofx, y and
a given material, when the material is stressed within the elastic limit. This ratio is called 2, 3
Poisson’s ratio and it is generally denoted by p. Hence mathematically,
z respectively.
Consider the strains produced by each stress sepa Y+
Pooi. sson,s ratio, u Lateral strain U mu—n 7 rately. —t
Longitudinal strain  / U'JJ ?>
The stress o will produce strain in the direction of x
or Lateral strain = ix Longitudinal strain x 1
and also in the directions ofy and z. The strain in the direc s'
As lateral strain is opposite in sign to longitudinal strain, hence algebraically, lateral —Etion ofx will be whereas the strains in the direction ofy ^/ j 1
J
strain is written as y)
X
Lateral strain =  p x Longitudinal strain ...[1.7 (£)] 01
4. Relationship between stress and strain. Consider a and z will be  p
twodimensional figure ABCD, subjected to two mutually perpen
dicular stresses Oj and o . 2 Similarly the stress o2 will produce strain —2 in the *z
2 f"
Refer to Pig. 1.5 (a).  Edirection ofy and strain of p —o Fig. 1.5 (i)
in the .. ofx andy
direction
Let Oj = Normal stress in xdirection ^
o = Normal stress y direction each.
2
in °i o,
Consider the strain produced by av Also the stress o will produce strain in the direction of z and strain of  p x A in
3
B
The stress o will produce
L strain in the direction of x and the direction ofx and y.
also in the direction ofy. The strain in the direction of x will be u2 _Total O On O3
. ] ^
ofx due stresses ov o o  .
strain in the direction to 2 and 3
longitudinal strain and will be equal to whereas the strain in Fig. 1.5 (a)
Similarly total strains in the direction ofy due to stresses o 1; a and o
2 3
the direction of y will be lateral strain and will be equal to  p x ^1. (v Lateral strain E'E= £2_ M„£El
E=  p x longitudinal strain)
Now consider the strain produced by ct2 . and total strains in the direction of a due to stresses a,, a and a
2 3
The stress a will produce strain in the direction ofy and also in the direction ofx. The
2
E—sti ain in the direction of y will be longitudinal strain and will be equal to — whereas the Let ev e2 and c are total strains in the direction of x, y and 2 respectively. Then
3
^Estrain in the direction of x will be lateral strain and will be equal to  p x E E** ...[1.7 CH)1
Let e  Total strain in xdirection
1
e = Total strain in ydirection —[1.7 (/)]
2
°
 *p
Now total strain in the direction of x due to stresses a, and o„ =  1 q i 02
E E1 2
^E^'E ...[1.7 («/)]
^Similarly The above three equations give the stress arid strain relationship for the three orthogonal
total strain in the direction ofy due to stresses a, and a„1 = rF
1
F.
normal stress system.
e= AProblem 1.1. rod 150 cm long and of diameter 2.0 cm is subjected to an axial pull of
l ETTPr N mm20 kN. If the modulus of elasticity of the material of the rod is 2 x 10s / 2 ; determine :
E ...[1.7 OP)]
...[1.7 (G)]
—e0  a„V2 Oi (i) the stress,
 pf F, (ii) the strain, and:
(iii) the elongation of the r.odv
:
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
Sol. Given : Length of the rod, L  150 cm Stress (a) is given by equation (1.1), as
Diameter of the rod, D = 2.0 cm = 20 mm _ P 50,000 = 101.86 N/mm2 .
Area, mm(20)2 = IOOji 2 °~a“ 490.87
Axial pull, P = 20 kN = 20,000 N Strain (e) is given by equation (1.2), as
Modulus of elasticity, E = 2.0 x 105 N/mm2 dL = O3 =00012
L 250
(i) The stress (a) is given by equation (1.1) as
EUsing equation (1.5), the Young’s Modulus ( ) is obtained, as
^a = = 200~ = 63.662 N/mm2. Ans. = Stress = 101.86 N/mm^ = 34333,33 N/mm2
A
100it Strain 0.0012 11r0t6xNT// m 2 \
)
(ii) Using equation (1.5), the strain is obtained as  84883.33 x 106 N/m 2 Ans. (v 1 N/mm 
.
B°. = 84.883 x 109 N/m2 = 84.883 GN/m2 . Ans. (v 109 = G)
a AProblem 1.4. tensile test was conducted on a mild steel bar. The following data was
Strain, g “_ E 63.662 0.000318. Ans. obtained from the test
~
{i ) Diameter of the steel bar = 3 cm
2 x 10 5 (:ii ) Gauge length of the bar
= 20 cm
(Hi) Elongation is obtained by using equation (1.2) as (:Hi ) Load at elastic limit
= 250 kN
dL (iv) Extension at a load of 150 kN
= 0.21 mm
Elongation, dL  e xL
(v ) Maximum load = 380 kN
 0.000318 x 150 = 0.0477 cm. Ans.
(vi) Total extension = 60 mm
Problem 1.2. Find the minimum diameter of a steel wire, which is used to raise a load (vii) Diameter of the rod at the failure = 2.25 cm.
Nof 4000
MN/mif the stress in the rod is not to exceed 95 2 (b ) the stress at ’astic limit.
.
NSol. Given : Load, P  4000 Determine : (a) the Young s modulus,
Stress, o = 95 MN/m2 = 95 x 106 N/m2 M(y = Mega = 10s) (c) the percentage elongation, and (d) the percentage decrease in area.
= 95 N/mm2 ( v 106 N/m2 = 1 N/mm 2) Sol. Area of the rod, A =  D2 =  2 cm2
(3)
mmDLet  Diameter of wire in 44
A = — D2 m: 7.0685 cm2 = 7.0685 x lO"4 2
4 .
.'. Area, stress = Load _ P
Area A
(a) To find Young’s modulus, first calculate the value of stress and strain within elast
4000 4000 x 4 The load at elastic limit is given but the extension corresponding to the load at elastic
limit and corresponding exten
or D2 = 4000 x 4 limit is not given. But a load of 150 kN (which is within elastic limit) and strain within elastic
sion are used for stress
0f o.21mm are given. Hence these values
limit
D = 7.32 mm. Ans. Stress = 150 x 1000 (v 1 kN = 1000 N)
‘ 7.0685 x 10‘ 4
mmProblem 1.3. Find the Young's Modulus of a brass rod of diameter 25 and of = 21220.9 x 104 N/m2
length 250 mm. which is subjected to a tensile load of 50 kN when the extension of the rod Increase in length (or Extension)
is equal to 0.3 mm. ^Stram = Original length (or Gauge length)
D mmSol. Given : Dia. of rod, = 25
Area of rod, mmA =  (25)2 = 490.87 2 mm_ _ 0.00105
20 x 10
Tensile load, 4
Extension of rod,
Length of rod, P = 50 kN = 50 x 1000 = 50,000 N .. Young’s Modulus,
dL = 0.3 mm 21220.9 x 10 =,20209523 x 104 N/m2
L = 250 mm 0.00105
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
(v 10 9 = Giga = G)
= 202.095 x 109 N/m2 2. lx 10 6
( ••• 10 6 = Mega = M)
= 202.095 GN/m2. Ans. 125 x 10® = or (,302  d2) = 4 x 2.1 x 10 6
(b) The stress at the elastic limit is given by, 2  2 Jt x 125 x 106
(,30
d )
Stress = Load at elastic limit 250 x 1000 0.09  d2 = 213.9 or 0.09  0.02139 = d2
Area ~~ 7.0685 x 10~4 m' d = ^0.09  0.02139 = 0.2619 = 26.19 cm. Ans.
= 35368 x 104 N/m2
= 353.68 x 10s N/m2 Problem 1.6. The ultimate stress, for a hollow steel column, which carries an axial load
MNof 1.9
= 353.68 MN/m2 is 480 N/mm2. If the external diameter of the column is 200 mm, determine the
. Ans.
in ternal diameter. Take the factor of safety as 4.
(c) The percentage elongation is obtained as,
Sol. Given :
Percentage elongation Ultimate stress, = 480 N/mm2
_ Total increase in length Axial load, / MNP = 1.9 N= 1.9 x 106 M(v = 10 6 )
X' = 1900000 N
Original length (or Gauge length)
—mmmm 60 x 100 = 30%. Ans. External dia., D = 200 mm
o20n x 1i0n
Factor of safety =4
(d) The percentage decrease in area is obtained as. Let
mmd  Internal diameter in
Percentage decrease in area Area of crosssection of the column,
_ (Original area  Area at the failure) mmA —=  =
Original area y4 (D2 d2 (200 2  d2 2
) )
4
Using equation (1.7), we get
—^ x 32  x 2.25 2 „ Ultimate stress
.4 4 1 , of safetA y
actor
Working stress or Permissible stress
—4 — 480 —* *
.
Working stress
x 100 = (9  5.0625) x 100 = 43.75%. Ans. —=
or „Wor rk, i. ng stress —480 = 120 N„//mm. mm2
:
"" 4
tT2.17 vx 10 a = 120 N/mm2
5s ess for a hollow steel column which carries an load of
’
1,2;5; MZNt/,m
kN is . If the external diameter of the column is 30 cm, determine the Now using equation (1.1), we get
internal diameter. —P 120 = 1900000 — —1900000 x4z..
.
Sol. Given 0 = A or
:
— (200 2 d2 "(40000 d 2 )
 )
Safe stress*, a = 125 MN/m2 = 125 x 10e N/m2
Axial load, P = 2.1 x 103 kN = 2.1 x 10« N or 40000  d2 = 900000X4 = 20159.6
D mExternal diameter, jc x 120
= 30 cm = 0.30
or d? = 40000  20159.6 = 19840.4
1^1 d  Internal diameter
d = 140.85 mm. Ans.
Area of crosssection of the column, AProblem 1.7. stepped bar shown in Fig. 1.6 is subjected to an axi
A=^(D )m2 d2) = » ally applied compressive load of 35 kN. Find the maximum and minimum
( . 302_ 2 2
stresses produced.
rf
44
Using equation (1.1), a= — Sol. Given : P  35 kN = 35 x 103 N
Axial load, D = 2 cm = 20 mm
A Dia. of upper part,
x
*Safe stress is a stress which is within elastic limit.
Fig. 1.6
I
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
— mmArea of upper part, Ai = (20 2) = 100 it 2 Similarly the strains of section 2 and of section 3 are,
a2 = P P
AE
E AE
2 3
.Sl.JLe
2
— mmArea of lower part, Oo
A„ = — DJ2 = (30 2 ) = 225 it 2
2 Eand, e‘~
24
4 But strain in section 1 = Change in length of section 1
Length of section 1
The stress is equal to load divided by area. Hence stress will be maximum where area is
minimum. Hence stress will be maximum in upper part and minimum in lower part.
/. Maximum stress Load _ 35 x 10 3  111.408 N/mm2. Ans.
Aj 100 x 3t where dL^ = change in length of section 1,
Load _ 35 x 103 = 49.5146 N/mm2. Ans. .. Change in length of section 1, dL = eL
t jl
Minimum stress An 225 x k
1.10. ANALYSIS OF BARS OF VARYING SECTIONS PLi I
_ \
A bar of different lengths and of different diameters (and hence of different crosssec
Ai E
tional areas) is shown in Fig. 1.6 (os). Let this bar is subjected to an axial load P.
Similarly changes in length of section 2 and of section 3 are obtained as :
Change in length of section 2, dL = e L
2 2
2
PL* (
[
_~ A%E
and change in length of section 3, dL = eL
3 xx
_ PL3 f
{
AE
3
Total change in the length of the bar,
— Ed,ht PL—3
Fig. 1.6 (a) = ,y + d,Lr„ dL+ , r PL PLo— + A3
3x A^E
dL. = r1 +
Though each section is subjected to the same axial load P, yet the stresses, strains and AE
change in lengths will be different. The total change in length will be obtained by adding the X
changes in length of individual section.
= P\h.
E [Aaix A A
2 3J
Let P = Axial load acting on the bar, Equation (1.8) is used when the Young’s modulus of different sections is same. If the
L  Length of section 1, Young’s modulus of different sections is different, then total change in length of the bar is
x
Aj = Crosssectional area of section 1, given by,
L2, A = Length and crosssectional area of section 2, dL=P + EJ2'A2 2 + L
2
. ?
AL _E\Ai ...(1.9)
3, E3A3J
3 = Length and crosssectional area of section 3, and
E = Young’s modulus for the bar. NProblem 1.8. Arc axial pull of 35000 is acting on a bar consisting of three lengths as
Then stress for the section 1, shown in Fig. 1.6 (b). If the Young’s modulus = 2.1 x 10s N/mm2 determine :
,
_ Load P (i) stresses in each section and
1 Area of section 1 Aj (ii) total extension of the bar.
Similarly stresses for the section 2 and section 3 are given as,
PP
a = anda ° =
2 ~"2r 3 ~ar
3
Using equation (1.5), the strains in different sections are obtained.
Strain of section 1, e, = fv o, =^—1
Fig. 1.6 (5)
. L
16 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
Sol. Given : Area of steel bar, mmA
Axial pull, 1
Length of section 1, = 5 x 5 = 25 cm2 = 250 2 pI
Dia. of section 1,
P = 35000 N Elastic modulus for steel bar, _i 5 cm x 5 cm
_
L = 20 cm = 200 mm T Steel bar
l Ej = 2,1 x 10° N/mm2
j
D = 2 cm = 20 mm Length of aluminium bar,
l 3D cm
L = 38 cm = 380 mm —2 J
2
mm^A, = (202 ) = 100 it 2
Area of section 1, Area of aluminium bar, 10 cm x 10 cm
4
Length of section . 2, mmA Aluminium bar
2 38 cm
Dia. of section 2, cm mm = 10 x 10 = 100 cm2 = 10000 2
L = 25 = 250
2
Elastic modulus for aluminium bar, 1
D cm = 30 mm
2 = 3 E 10 4 N/mm2
2
= 7 x
mm—A, = 2 = 2 mmTotal decrease in length, dL = 0.25 Fig. 1.,7
Area of section 2, (30 ) 225
it
Length of section 3,
4 Let P = Required force.
Dia. of section 3,
L = 22 cm = 220 mm As both the bars are made of different materials, hence total change in the lengths of
3
the bar is given by equation (1.9).
D = 5 cm = 50 mm
a
A — mm.. Area of section 3, s = 2 dL = P {j^ + J^)
(50 2 ) = 625 it
Young’s modulus, £ = 2.1x 10 s N/mm2. ^Por
(i) Stresses in each section 0.25 = +
„. o. = — Axial load ( 10 s x 2500 7 x 10 4 x j
Stress in section 1,
1 Area of section 1 1^2. lx 10000 J
=P (5.714 x 10 7 + 5.428 x 107 = P x 11.142 x 10"7
)
35000
100 it
N m= 0.25 xlO 7
£ = _ ln _4o8 /m 2 Ans. p _ 0.25
11.142
A . 11.142 x 10 7
x
N/mm~~Stress in section 2, 2 = 2.2437 x 10s = 224.37 kN. Ans.
a„ = =  = 49.5146 Ans.
.
1 A2 225 x it Problem 1.10. The bar shown in Fig. 1.8 is subjected to a tensile load of 160 kN. If
——A«„ = 7 = = 17.825 N/mm2. Ans. the stress in the middle portion is limited to 150 N/mm2 determine the diameter of the
,
Stress in section 3,
3 3 625 x it middle portion. Find also the length of the middle portion if the total elongation of the bar
(it) Total extension of the bar is to be 0.2 mm. Young’s modulus is given, as equal to 2.1 x 10s N/mm2
.
Using equation (1.8), we get
Sol. Given :
Tensile load, P = 160 kN = 160 x 103 N
— ~= —L42 Stress in middle portion, a = 150 N/mm2
P ( L, Lg 1 Total elongation,
~~~ 2
dL = 0.2 mm
Total extension E \A + A2 +
1
Ag j Total length of the bar. L = 40 cm = 400 mm
35000 ( 200 250 220 \ Young's modulus, E = 2.1 x 10s N/mm2
++ mmDiameter of both end portions, D = 6 cm = 60
t
2.1 x 10 s [ 100 it 225 x it 625 x it J
.". Area of crosssection of both end portions,
35000 (6.366 + 3.536 + 1.120) = 0.183 mm. Ans. mmA, = — x 602 = 900 it 2
= 2.1x10 s.
.
4
Problem 1.9. A member formed by connecting a steel bar to an aluminium bar is shown t 6 cm DIA
6 cm DIA
in Fig. 1.7. Assuming that the bars are prevented from buckling sideways, calculate the _L_
magnitude of force P that will cause the total length of the member to decrease 0.25 mm. The __L
values of elastic modulus for steel and aluminium are 2.1 x 105 N/mm2 and 7 x 104 N/mm2 Fig. 1.8
respectively
Sol. Given :
mmLength of steel bar, L  30 cm  300
l
j.
SIMPLE STRESSES AND STRAINS 19
STRENGTH OF MATERIALS A mmProblem 1.11. brass bar, having crosssectional area of 1000 3 is subjected to
where A, ,
D2 = Diameter of the middle portion axial forces as shown in Fig. 1.9.
Ll = Length of middle portion in mm.
Length of both end portions of the bar,
mmL
l
= (400  L2)
Using equation (1.1), we have
Stress = Load
.
Area Fig. 1.9
For the middle portion, we have Find the total elongation of the bar. Take E = 1.06 x 10s N/mm2 .
P Sol. Given : A = 1000 mm2
Area,
4°2=
160000 Value of E = 1.05 x 10s N/mm2
Let dL = Total elongation of the bar.
The force of 80 kN acting at B is split up into three forces of 50 kN, 20 kN and 10 kN.
kN BC
mm4 x 160000 2 Then the part AB of the bar will be subjected to a tensile load of 50 , part is subjected to
: 1358
a compressive load of 20 kN and part BD is subjected to a compressive load of 10 kN as shown
ji x 150
mmD = J 1358 = 36.85 = 3.685 cm. Ans. in Fig. 1.10.
Area of crosssection of middle portion,
mmA, = — x 36.85 = 1066 2
d4
Now using equation (1.8), we get
Total extension. AdL E _A
X
2J
160000 f (400  L%) Li 1
_,
2.1xlO s L 900it
1066
= (400  LA and A, = 1066]
0,2 x 2.1 x 10 s _ (400  L2) L2 Fig. 1.10
160000

900 it 1066 Part AB. This part is subjected to a tensile load of 50 kN. Hence there will be increase
1066(400  Lf) + 900 n L 2 in length of this part.
900 n x 1066 AB.. Increase in the length of
0.2625 x 900it x 1066 = 1066 x 400  1066 i + 900it x L A= xLl
2 a AE 1
791186 = 426400  1066 L + 2827 L2
2
791186  426400 = L (2827  1066)
2
50 x 1000
364786 = 1761 L = (v Pj = 50,000 N, Lj = 600 mm)
2
1000 x 1.05 x 105
mmT 364786 = 207.14 Ans.
= 20.714 cm. = 0.2857.
1.10.1. Principle of Superposition. When a number of loads are acting on a body, the Part BC. This part is subjected to a compressive load of 20 kN or 20,000 N. Hence there
resulting strain, according to principle of superposition, will be the algebraic sum of strains
will be decrease in length of this part.
caused by individual loads.
While using this principle for an elastic body which is subjected to a number of direct BC.. Decrease in the length of
forces (tensile or compressive) at different sections along the length of the body, first the free P 20,000 xlOOO (v L = 1 m = 1000 mm)
2* 2
body diagram of individual section is drawn. Then the deformation of the each section is obtained 2 ” 1000 x 1.05 x 10 5
The total deformation of the body will be then equal to the algebraic sum of deformations of the AE
individual sections. = 0rl904.
L. ,1
20 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS 2
Part BD. This part is subjected to a compressive load of 10 kN or 10,000 N. Hence there But Pj = 45 kN,
will be decrease in length of this part. P kN and P kN
a 4
Decrease in the length of BD = 450 = 130
—AE 45 + 450 = P + 130 or P = 495  130 = 365 kN
2 2
r 2200 The force of 365 kN acting at B is split into two forces of 45 kN and 320 kN {i.e., 36545
5
1000 X 1.05 X = 320 kN).
10
m( 3 = 1.2 + 1 = 2.2 or 2200 mm) The force of 450 kN acting at C is split into two forces of 320 kN and 130 kN {i.e. 450  320
= 0.2095. = 130 kN) as shown in Fig. 1.12.
Total elongation of bar = 0.2857  0.1904  0.2095 AB BCFrom Fig. 1.12, it is clear that part
is subjected to a tensile load of 45 kN, part is
(Taking +ve sign for increase in length and subjected to a compressive load of 320 kN and part CD is subjected to a tensile load 130 kN.
ve sign for decrease in length)
= 0.1142 mm.
Ans.
Negative sign shows, that there will be decrease in length of the bar.
Problem 1.12. A member ABCD is subjected to point loads P p P,„ P„ and P4 as shown in
Fig. 1.11.
BC j
Fig. 1.12
BCHence for part AB, there will be increase in length for part there will be decrease in
;
length and for part CD there will be increase in length.
AB.•. Increase in length of
—P 45000 P„ = 45 kN = 45000 N)
x . = 625 x 2.1 x 10 5r x 1200 (v
L,
Fig. 1.11 1
Calculate the force P2 necessary for equilibrium, if P, = 45 kN, P = 450 kN and mm= 0.4114
3
P
4 = 130 kN. Determine the total elongation of the member, assuming the modulus of elas Decrease in length of BC
N/mmticity to be 2.1 x 10s 2 P X L —32(1000 _ x 60Q _ 320
. A0 E 2500 x 2.1 x 10 s
Sol. Given : p. = 3200OO)
Part AB : Area, (.
Length, A mm= 625 2 and mm0.3657
1
Part BC : Area, = 120 cm = 1200 mm Increase in length of CD
Length, A = 2500 mm 2 and L= . 13Q 000 (v P = 130 kN = 130000)
2 x' ’ x 900
Part CD : Area, AE
L = 60 cm = 600 mm b 1250 X 2.1 xlO 5c
Length, 2
Value of = 0.4457 mm
mmA3 = 12.0 2 and
L = 90 cm = 900 mm Total change in the length of member
a
E = 2.1 x 10 s N/mm2 . = 0.41140.3657 + 0.4457
PValue of 2 necessary for equilibrium (Taking +ve sign for increase in length and
—ve sign for decrease in length)
Resolving the forces on the rod along its axis (i.e., equating the forces acting towards mm= 0.4914 (extension). Ans.
right to those acting towards left), we get mmProblem 1.13. A tensile load of 40 kN is acting on a rod of diameter 40 and of
P + P = P + P A mmlength 4 m. bore of diameter 20 is made centrally on the rod. To what length the rod
1 a 3 4
22 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS 23
should be bored so that the total extension will increase 30% under the same tensile had. Take
E = 2 x 10s N/mm2
.
Sol. Given :
Fig. 1.12 (a)
Tensile load, P = 40 kN = 40,000 N
Dia. of rod,
D = 40 mm
.. Area of rod,
mmA — (402) = 400re 2
IZ D Equating the equations (i) and (si),
2.6 _ 4x 4x
re 2re 6re
Fig. 1.12 (6) — —or
Length of rod, 2.6 = — — + or 2.6x6 = 3x(4x) + 4x
m mm 4 = 4 x 1000 = 4000
Dia. of bore, mm= 20 26
or 15.6 = 12  3x + 4x or 15.6  12 = x or 3.6 = *
.. Area of bore., mm—a = x 202 = 100 it 2 .. Hod should be bored upto a length of 3.6 m. Ans.
4 Problem 1.14. A rigid bar ACDB is hinged at A and supported in a horizontal position
Total extension after bore = 1.3 x Extension before bore A kNby two identical steel wires as shown in Fig. 1.12 (c). vertical load of 30 is applied at B.
Value of E =2x 105 N/mm2 TFind the tensile forces I and T2 inducedin these wires by the vertical load.
Let the rod be bored to a length of x meter or x x 1000 mm. Then length of unbored EF
mportion = (4  x) = (4  x) x 1000 mm. First calculate the extension before the bore is made.
The extension (5L) is given by,
p 40000 x 4000 = —2 nun
&L = AfE xL: r
400it x 2 x 10 6 it
Now extension after the bore is made
= 1.3 x Extension before bore
— mm—= 1. .q3 x 2 = 2.6 ... 30 kN
...(i)
re re Fig. 1.12 (c) Fig. 1.12 (rf)
The extension after the bore is made, is also obtained by finding the extensions of the
unbored length and bored length. Sol. Given :
For this, find the stresses in the bored and unbored portions. Rigid bar means a bar which will remain straight.
Stress in unbored portion Two identical steel wires mean the area of crosssections, lengths and value of E for
_ Load P 40000 _ 100 ^/rnm2 both wires is same.
Area A 400re re Aj = A Ej = E and L = L
2, z x 2
.•. Extension of unbored portion Load at B = 30 kN = 30,000 N
_ Stress x LLgejnjggtth o0f unbored portion Fig. 1.12 (c) shows the position of the rigid bar before load is applied at B. Fig. 1.12 id)
E shows the position of the rigid bar after load is applied.
SIMPLE STRESSES AND STRAINS 25
Fig. 1.13
Consider a small element of lengths* of the bar at a distance* from the left end. Let the
diameter of the bar be D. at a distance * from the left end.
Dr = D, L
= D kx where k = Di  D2
x
Area of crosssection of the bar at a distance * from the left end,
4=f 05,**)2
Now the stress at a distance * from the left end is given by,
Load
*(Dl k.xf x(Di~k.x)
4
The strain e in the small element of length dx is obtained by using equation (1.5).
x
Stress a.
4P 1 4P
E2 it E(D  2
1
7i{Dyk.x,) k.x)
Extension of the small elemental length dx
= Strain, dx = e, . dx
n E(D  k . xr
l
Total extension of the bar is obtained by integrating the above equation between the
limits O and L.
26 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
/. Total extension,
Using equation (1.10),
Jo Jt SO?!  A .*) 2 JtF Jo 1 4 PL 4 x 5000 x 400
tc E D\D2 jt x 2.1 x 10 5 x 40 x 20
^=_ 4FP fPt = 0.01515 nun. Ans.
nE Jo
(D k.x)*x(k) [Multiplying and dividing by ( ft )]
y
. eftc Problem 1.16. Find the modulus ofelasticity for a rod, which tapers uniformly from 30
mm mmto 15 diameter in a length of 350 mm. The rod is subjected to an axial load of 5.5 kN
4P (!>!  ft . x) and extension of the rod is 0.025 mm.
. Sol. Given : D = 30 mm
Larger diameter, 1
jar L (»*(*> Jo"*«k [0)ift.*)Jo Smaller diameter,
Length of rod,
_ _4P_ f 1 1 Axial load, D = 15 mm'
Extension, 2
~ kEU
_D  k.L UjftxO L = 350 mm
1
P = 5.5 kN = 5500 N
4P 1 1 dL = 0.025 mm
JiFft D k.L D Using equation (1.10), we get
1 x
^ in the above equation, we get ^,r _ iPL
X EDfD2
Substituting the value of ft =
Total extension. 4PL
4 x 5000 x 350
or E K ~
dL jt x 30 x 15 x 0.025
= 217865 N/mm2 or 2.17865 x 10s N/mm2. Ans.
4pl r i i_i 1.12. ANALYSIS OF UNIFORMLY TAPERING RECTANGULAR BAR
nE.(D D )[D 1 D + D2 dJ A bar of constant thickness and uniformly tapering in width from one end to the other
1 2 1
end is shown in Fig. 1.14.
4PL f 1 _1_'
D«n£E.(,!{?D!
1
Z»2 ) I»2 Uj
— — —— —D4PL 4PL
It E.(D
l
—x (D, D,) _
i
kED D2 ) Jt 1 2 ...(1.10)
("1111
DDIf the rod is of uniform diameter, then D, 
2 '
. Total extension, djtL = 4 PL
=
kE.D 2
mm mmProblem 1.15. A rod, which tapers uniformly from 40
mmin a length of 400 diameter to 20 diameter
is subjected to an axial load of 5000 N. IfE = 2.1 x 10s N/mm2 find the
,
extension of the rod.
Sol. Given :
Larger diameter. D = 40 mm
Smaller diameter, 1
Length of rod,
D = 20 mm
2
L = 400 mm
Axial load, P = 5000 N
Young's modulus, ~ E = 2.1 x 10s N/mm2
Let dL = Total extension of the rod Fig. 1.14
1
::
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
Let P = Axial load on the bar = e ...( 1 . 12)
Et(ab)
L Length of bar b
a = Width at bigger end m mmProblem 1.17. A rectangular bar made of steel is 2.8 long and 15 thick. The rod
b = Width at smaller end
mmis subjected to an axial tensile load of 40 kN. The width of the rod varies from 75 at one end
E = Young’s modulus
mmto 30 at the other. Find the extension of the tod ifE  2 x 10s N!mm2 .
t = Thickness of bar
Sol. Given L = 2.8 m = 2800 mm
Consider any section XX at a distance x from the bigger end. Length, t = 15 mm
Width of the bar at the section XX Thickness, P = 40 kN = 40,000 N
Axial load, a  75 mm
(a  b)x Width at bigger end, b = 30 mm
Width at smaller end,
where k =
Thickness of bar at section XX = t Value of E = 2 x 10s N/mm2
:. Area of the section XX Let dL = Extension of the rod.
= Width x thickness Using equation (1.12), we get
= (a  kx)t —dL =
Stress on the section XX PL rr ,log, a
Load P Et(a b) b
Area (a  kx)t 40000 x 2800
~ 2 x 10 B x 15(75  30) i°QSe V30 J
Extension of the small elemental length dx
= Strain x Length dx = 0.8296 x 0.9163 = 0.76 mm. Ans.
Stress mmProblem 1.18. The extension in a rectangular steel bar of length 400 and thickness
mm10 mm, is found to be 0.21 mm. The bar tapers uniformly in width from 100 to 50 mm. IfE
for the bar is 2 x 10s N/mm2 determine the axial load on the bar.
,
((qfer)f) Sol. Given
Extension,
(a  kx)t Length, dL = 0.21 mm
Thickness, LL 44<00 mm
E(a  kx)t Width at bigger end, mm( = 11<0
Total extension of the bar is obtained by integrating the above equation between the Width at smaller end, mma = 11(00
limits 0 and L. mmb  55<0
Total extension, Value of E = 2 x 10s N/mm2
P_ eL dx PLet = axial load.
Et Jo {a  kx)
E{a  kx)t Using equation (1.12), we get
^tp r t p PL
^.log ( ii = dL ~ Et(a  b )
e (akx) I°g (aH,) 1og o]
x^^J e e
P x 400
w Elog a  log (a  kL]] = log {^z) ' 5 t
k< * * 2 x 10 x 10(100  50)/
: 0.000004 Px 0.6931
0.21 75746 N
(ab\ 0.000004 x 0.6931

— —a L_
L JK J = 75.746 kN. Ans.
:'
30 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
But strain in bar 1 = Strain in bar 2
1.13. ANALYSIS OP BARS OF COMPOSITE SECTIONS
A bar, made up of two or more bars of equal lengths but of V71 E ~ (t°
2
different materials rigidly fixed with each other and behaving ' Ei
as one unit for extension or compression when subjected to an
axial tensile or compressive loads, is called a composite bar. For mFrom and
the composite bar the following two points are important equations (in) (u), the stresses a and a can be determined. By substituting
x 2
the values of 0 and o
X2 equations (it) and (iti), the load earned by different materials may be
computed.
1. The extension or compression in each bar is equal. Hence Jssj E
deformation per unit length i.e., strain in each bar is equal.
*2^2— ±_ Modular Ratio. The ratio of =L is called the modular ratio of the first material to the
2. The total external load on the composite bar is equal to Pi
the sum of the loads carried by each different material.
second.
Fig. 1.15 shows a composite bar made up of two different
P AProblem 1.19. steel rod of 3 cm diameter is enclosed centrally in a hollow copper tube
materials. i ia
of external diameter 5 cm and internal diameter of 4 cm. The composite bar is then subjected to
Let P = Total load on the composite bar,
an axial pull of 45000 N. If the length of each bar is equal to 15 cm, determine :
L = Length of composite bar and also length of bars of different materials, (i) The stresses in the rod and tube, and
A x = Area of crosssection of bar 1, (it) Load carried by each bar.
A 2 = Area of crosssection of bar 2, ETake for steel =2.1 x 10s N/mm2 and for copper = 1.1 x 10s N/mm2
.
Sol. Given :
Ei = Young’s Modulus of bar 1, mmDia. of steel rod = 3 cm = 30
E = Young’s Modulus of bar Area of steel rod, V7f7f7f7f7f7f7/7t7f7i7f7/7l7/7l7f^/7i7J7I7I7I7f7l7l7M7I7I7M7f7i*
2
2,
Pi = Load shared by bar 1, mmA =
s
P = Load shared by (30) 2 = 706.86 2
2
bar 2,
<Jj = Stress induced in bar 1, and External dia. of copper tube 15
J
a = Stress induced in bar 2. = 5 cm = 50 mm
2
Now the total load on the composite bar is equal to the sum of the load carried by the two Internal dia. of copper tube
= 4 cm = 40 mm
P=P +P Area of copper tube,
12
The stress in bar 1, Load carried by bar 1 mm mmA —=
Area of crosssection of bar 1 e
[50 2  40 2 2 = 706.86 2
]
NAxial pull on composite bar, P = 45000
Length of each bar, L = 15 cm Fig 1.16
.
EYoung’s modulus for steel, s = 2.1 x 106 N/mm2
Similarly stress in bar 2, A2 EYoung’s modulus for copper, c = 1.1 x 10s N/mm2
(i) The stress in the rod and tube
Substituting the values of P and P in equation (i), we get Let o = Stress in steel,
L 2 s
^ = ai^l + a2^2 ...(ill) Ps = Load carried by steel rod,
Since the ends of the two bars are rigidly connected, each bar will change in length by o = Stress in copper, and
the same amount. Also the length of each bar is same and hence the ratio of change in length c
to the original length (i.e., strain) will be same for each bar.
P = Load carried by copper tube.
c
Now strain in steel = Strain in copper
But strain in bar 1, Stress in bar 1 a
Young” s modulus of bar 1 x
E'
l
Similarly strain in bar 2, 2.1 x 10s
11 x 10 5 x a = 1.909 a
g;
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS 33
32
Now strain in steel = Strain in brass
Now stress = ~r , Load — Stress x Area Strain  
Area
Load on steel + Load on copper = Total load P('•' Total load = ) x—2 10 5
A Aax a +. „a „x a =_ fp  Ox. = 2„ a.
s c c
s 1x10s b b
or 1.909 a x 706.86 + o x 706.86 = 45000
c c
Now load on steel + Load on brass = Total load
o (1.909 x 706.86 + 706.86) = 45000
or Cc Ao,x x ho = 900000 (v Load = Stress x Area)
s
2056.25 a = 45000 A, + ct.o
c s
or
2 a x 4712.4 + a x 5340,7 = 900000 =(••
45000 21.88 N/nun2. or bb <r 2ct ,)
s
<= 2056.25 (
Ans. or 14765.5 a = 900000
b
Substituting the value of a in equation (i), we get a.6 = 900000 = 60A.a95_ NNr// mm2 . AAns.
c 14765.5
o = 1.909 x 21.88 N/mm 2
s
Substituting the value of pb in equation (i), we get
= 41.77 N/nun2 . Ans. o = 2 x 60.95 = 121.9 N/nun2. Ans.
(ti) Load carried by each bar. s
As load = Stress x Area
Load carried by brass tube
Load carried by steel rod, = Stress x Area
N=
P= * A, oj, xA = 60.95 x 5340.7
s t
= 41.77 x 706.86 = 29525.5 N. Ans. = N325515. = 325.515 kN. Ans.
Load carried by copper tube, Load carried by steel tube (
P = 45000  29525.5 = 900  325.515 = 574.485 kN. Ans.
Cc
Decrease in the length of the compound tube
= 15474.5 N. Ans.
mmProblem 1.20. A compound tube consists of a steel tube 140 internal diameter = Decrease in length of either of the tubes
= Decrease in length of brass tube
mm mmand 160 internal diameter and = Strain in brass tube x Original length
external diameter and an outer brass tube 160
mma1it8hs0haoxritaelneslx.otaeLdrenonafglt9dh0i0aomfkeNet.aecrFh.itnTudhbeethtieswos1t4tr0eusbsmeesms.aarneTdaokftehtehEelofsoaardmscetaerlerleinaegstdh2.byxTeh1ae(cFhcoNtmu/bpmeomuannddanttduhbefeoarcmaborrruainsetss 3x1ci . = 60.95 x . = 0.0853 nun. Ans.
140
\ 1 x 105 N/mm2. Problem 1.21. Two vertical rods one of steel and the other of copper are each rigidly
ak mfixed at the top and 50 cm apart. Diameters and lengths of each rod are 2 cm and 4 respec
Sol. Given : = 140 mm A Ntively. cross bar fixed to the rods at the lower ends carries a load of 5000 such that the cross
Internal dia, of steel tube = 160 mm
External dia. of steel tube bar remains horizontal even after loading. Find the stress in each rod and the position of the
mmA = — (1602  1402) = 4712.4 Eload on the bar. Take E for steel  2 x 10s N/mm2 and for copper = 1 x 10s N/mm 2.
2 __
.. Area of steel tube. s4 Sol. Given : ;
Internal dia. of brass tube = 160 mm Distance between the rods I I
Steel, Copper
= 50 cm = 500 mm
mmExternal dia. of brass tube = 180 2 cm dia 2 cm dia
mmAb =  Dia. of steel rod
 160 2 )
.. Area of brass tube, 2 5340.7
= Dia. of copper rod
^ (180
= 2 cm = 20 mm
Axial load carried by compound tube,
.. Area of steel rod
P = 900 kN = 900 x 1000 = 900000 N
Length of each tube, L = 140 mm = Area of copper rod
E for steel, E = 2 x 10 5 N/mm2
E for brass, s
mm= 4 x (20)2 = 100 n 2
Let E = 1 x 10s N/mm2
b 4
o = Stress in steel in N/mm2 and mmA A= = 100 it 2
s
a = Stress in brass in N/mm2 Fig. 1.17
b
38 STRENGTH OF MATERIALS simple stresses and strains 39
Problem 1.25. Two brass rods and one steel rod to 1.7 x 10 s N/mm2 which exceeds the safe stress of 1 x 105 N/mm2 for brass. Therefore let brass
gether support a load as shown in Fig. 1.20. If the stresses in be allowed to reach its safe stress of 1 x 10s N/mm 2 Then corresponding stress in steel will be
.
brass and steel are not to exceed 60 N/mrn2 and 120 N/mm2 1.176 x 105 N/mm2 which is less than 2 x 105 N/mm2.
,
Efind the safe load that can be supported. Take for steel Total load = P = Load on steel + Load on copper
~ 2 x 105 N/mm2 and for brass = 1 x 10s N/mm 2 The cross xAxA= <j + a
s b
.
mmsectional area ofsteel rod is 1500 2 and of each brass rod s b
mmis 1000 2 = 1.176 x 105 x 1500 + 1 x 10 5 x 2000
. N N= 3764 x 10s
Sol. Given : or 376.4 x 106
a N/mm2 = 376.4 MN. Ans. M(v = 106)
b
Stress in brass. = 60
Stress in steel, a = 120 N/mm2 Problem 1.26. Three bars made of copper, zinc and aluminium are of equal length and
s
E for steel, mmhave crosssection 500, 750 and 1000 square respectively. They are rigidly connected at
E for brass, E
s = 2 x 10 5 N/mm 2 their ends. If this compound member is subjected to a longitudinal pull of 250 kN, estimate the
Area of steel rod,
Area of two brass rods, E = lx 10 s N/mm2 Eproportional of the load carried on each rod and the induced stresses. Take the value of for
b
10s N/mm2.
A = 1500 mm2 copper = 1.3 x 10 s N/mm2 for zinc = 1.0 x10s N/mm2 and for aluminium = 0.8 x
s ,
A = 2 x 1000 Sol. Given :
b Total load,
P = 250 kN = 250 x 103 N
= 2000 mm2 Fig. 1.20
Length of steel rod, L = 170 mm For copper bar, E
s
Area, A =500 mm2 and E=l.Sx 105 N/mm2 £I
Length of brass rods, L = 100 mm c a.
b 0 2£
For zinc bar, & =
We know that decrease in the length of steel rod should be equal to the decrease in
mmArea, E 10s N/mm2
A = 750 2 and z  1.0 x
z
length of brass rods.
But decrease in length of steel rods For aluminium bar,
= Strain in steel rod x Length of steel rod mmAArea, a = 1000 2 and Ea = 0.8 x 10 s N/mm2 l/h/)//////////]h/)l////Ii//l/}ftm
, I
= e x L where e is strain in steel Let o = Stress induced in copper bar, = 250 kx
c FFijgg. 1j .221
ss s a = Stress induced in zinc bar,
z
Similarly decrease in length of brass rods
aa = Stress induced in aluminium bar,
= Strain in brass rods x Length of brass rods P = Load shared by copper rod,
c
= eb x L where e is strain in brass rod P  Load shared by zinc rod,
b z
b
Equating the decrease in length of steel rods to the decrease in length of brass rods, we get P = Load shared by aluminium rod, and
a
L = Length of each bar.
A ~e = e x L °r =1 Now, we know that the increase in length of each bar should be same, as length of each
ib
bar is equal hence strain in each bar will be same.
But Estress in steel = Strain in steel x (v Stress = Strain x E) Strain in copper = Strain in zinc = Strain in aluminium
S
...<0 Stress in copper Stress in zinc Stress in aluminium
E°s= e * =
s
s S'
%01 " Ea
Similarly stress in brass is given by,
Ea = eb y b
b
Dividing equation (i) by equation (it), we get
e.xE. 1.3 x 10 5
xEa — 100 2xl0 5 0.8 x 10 s : 1.625a,
s 1x10sT
TTTT
°b 170
 r. * —= 1.1I7U6
eb b
Suppose steel is permitted to reach its safe stress of 2 x 105 N/mm2 the corresponding —Ea = E, x aa = 1.0 x 10cs x o° = 1, .25 a ' —W
, 1 “
£ 0.8 xlOr5
stress in brass will be
a
2 x 10s total load = Load on copper + Load on zinc + Load on aluminium
1.176
xA A250 x 10 3 = Stress in copper
1.7 x 10s N/mm 1' c t Stress in zinc x z
+ Stress in aluminium x A
40 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
= x A + a x A + a„ x A external forces have been applied, the compressive load on the tube must be equal to the
c z 2 a
= 1.625aa x 500 + 1.25o x 750 + oa x 1000 tensile load on the rod.
a
= 2750aa <v a = 1.625oa and o. = 1.25oa ) Let o = Stress in the tube, and
c t
a = Stress in the rod
M°a = r
250 x 10 3 == 90.9 N/mm ,2 A„ns. ,
2750 Now, Tensile load on the rod = Compressive load on the tube
.
Substituting the value of o in equations (i) and (ii), we get a xA = a xA
a r r t t
a = 1.625 x 90.9 = 147.7 N/mm2. Ans. ^^xa=^xo,=or «
c o = 0.6875a
r ((
and ' = 1.25 x 90.9 = 113.625 N/mm 2 Ans.
.
Now load shared by copper =o xA (i) When the compressive load on the tube is 20 kN or 20,000 N.
cc
Then stress in the tube,
= 147.7 x 500 = 73850 N. Ans.
Load shared by zinc rod A= o x z = 113.625 x 750 Load 20000
2
<I(
= 85218 N. Ans.
Area of tube 68.75 it
Load shared by aluminium rod  92.599 N/mm2 (compressive). Ans.
= oa x A = 90.9 x 1000 (ii) Substituting this value in equation (i), we'get
a
x
= 90900 N. Ans. Stress in the rod, a = 0.6875 x o = 0.6875 92.599
r t
mmProblem 1.27. A steel rod 20 = 63.66 N/mm2 (tensile). Ans.
in diameter passes centrally through a steel tube of (iii) Stresses in the rod and tube, when one nut is tightened by one quarter of a turn.
mm mm mm25 internal diameter and 30 external diameter. The tube is 800 long and is closed
by rigid washers of negligible thickness which are fastened by nuts threaded on the rod. The Let o * = Stress in the rod and
r
nuts are tightened until the compressive load on the tube is 20 kN. Calculate the stresses in the o * = Stress in the tube due to tightening of the nut by onequarter of a turn.
f
tube and the rod.
As the stress in the tube is compressive and stress in the rod is tensile hence there will
Find the increase in these stresses when one nut is tightened by onequarter of a turn be decrease in the length of tube but there will be increase in the length of the rod.
relative to the other. There are 4 threads per 10 mm. Take E = 2 x 10s N/mm2. .. Decrease in the length of tube
Sol. Given : = Strain x L
Dia. of rod
= 20 mm Stress in tube Strain =
mm mmA —.'. Area of rod,
 (20) 2 2 = IOOjt 2
r 4 —h2 x x 800 = 0.004 x a*
‘
10 sr
mm mmArea of tube. A.  ~r (30 z  25 2 )
2 = 68.75:t 2
4 Increase in the length of the rod
mmLength of tube, L  800 Stress in rod
NCompressive load on tube, P = 20 kN = 20 x 10 3 E xL
t
Value of E = 2 x 105 N/mm2 o* , x 8o0n0A = (O.6875 x 0*4)x8OO
<v 0 = 0.68750.)
2 xx 10 5 2 x 10 5
= 0.00275 x o*
Axial advancement of the nut = Onequarter of a turn
Fig. 1.22 = ^ of a turn
When the nuts are tightened, the tube will be compressed and the rod will be elongated. But in one turn, the advancement of the nut is gth of 10 mm.
This means that the tube will be under compression and rod will be under tension. Since no
mmAxial advancement of the nut =  * j * 10 = 0.625
But axial advancement of the nut
= Decrease in length of tube + Increase in the length of rod
—* * :
simple stresses and strains 43
42 STRENGTH OF MATERIALS
0.625 = 0.004 x o* + 0.00275a* = 0.00675 x a 1.14.1. Stress and Strain when the Supports Yield. If the supports yield by an
0.625 amount equal to 8, then the actual expansion
= Expansion due to rise in temperature  8
^= 59NW=92
^0006
NW.ct * = 0.6875 x 92.59 = 63.65
 AnS ' = a.T.L  8.
Ans. Actual expansion T L(a . .  8)
1.14. THERMAL STRESSES ,. Actual strain = 0riginal length = L
And actual stress E= Actual strain x
Thermal stresses are the stresses induced in a body due to change in temperature. Ther (a.T.L 5) ..... .((11..116))
mal stresses are set up in a body, when the temperature of the body is raised or lowered and
the body is not allowed to expand or contract freely. But if the body is allowed to expand or _
L
mProblem 1.28. A rod is 2 long at a temperature of 10°C. Find the expansion of the rod
contact freely, no stresses will be set up in the body. when the temperature is raised to 80° C. If this expansion is prevented, find the stress induced
Consider a body which is heated to a certain temperature. E MN/min the material of the rod. Take = 1.0 x 10s 2 and ct = 0.000012 per degree centigrade.
Let L  Original length of the body, Sol. Given :
Length of rod,
T = Rise in temperature, Initial temperature, L = 2 m = 200 cm
E = Young’s Modulus Final temperature,
.. Rise in temperature, T = 10°C
a = Coefficient of linear expansion. Young’s Modulus, l
dL = Extension of rod due to rise of temperature. T = 80°C
2
If the rod is free to expand, then extension of the rod is given by
T=T T  80°  10°  70°C
dL = a. T.L. ...(1.13) 21 M(v = 106 )
This is shown in Fig. 1.23 (a) in which AB represents aA B B' E  1.0 x 105 MN/m2
1 = 1.0 x 105 x 106 N/m2
the original length and BB' represents the increase in length 5 ' = 10 11 N/m2
due to temperature rise. Now suppose that an external ®. t Coefficient of linear expansion, a = 0.000012
compressiye load,P is applied atB' so that the rod is decreased in '* B B' (i) The expansion of the rod due to temperature rise is given by equation (1.13).
its length from (L + aTL)toL as shown in Figs. 1.23 (6) and (c). A — p, = a.T.L
— —tirnuen v/3 —H .. Expansion of the rod
jL
compressive . = Decrease in len2gth L _*J = 0.000012 x 70 x 200
st, rain
Original length = 0.168 cm. Ans.
_ ct .T .L = aTL aa b (ii) The stress in the material of the rod if expansion is prevented is given by equation (1.15).
~ L
L + a.T.L ~ 3 .. Thermal stress, o a. T. E
Lf
= 0.000012 x 70 x 1.0 x 10 11 N/m2
Strain = 84 x 106 N/m2 = 84 N/mm 2 Ans. (v 106 N/m2 = 1 N/mm2)
.
Stress = Strain x E = a.T.E
Fig. 1.23 A mProblem 1.29. steel rod of 3 cm diameter and 5 long is connected to two grips and
AAnd load or thrust on the rod = Stress x Area = a.T.E x the rod is maintained at a temperature of 95°C. Determine the stress and pull exerted when the
If the ends of the body are fixed to rigid supports, so that its expansion is prevented, temperature falls to 30°C, if
then compressive stress and strain will be set up in the rod. These stresses and strains are
known as thermal stresses and thermal strain. (i) the ends do not yield, and
.'. Thermal strain, Extension prevented (ii) the ends yield by 0.12 cm.
Original length
Take E = 2 y. 10s MN/m2 and a = 12 x lO^TC.
dL a. T.L Sol. Given d = 3 cm = 30 mm
L L = °" T Dia. of the rod,
And thermal stress, a = Thermal strain x E
.. Area of the rod, mmx 302 = 225 it 2
= a.T.E.
Length of the rod, L = 5 m = 5000 mm
Thermal stress is also known as temperature stress. Initial temperature,
And thermal strain is also known as temperature strain. ...(1.15) Final temperature, T
1
= 95°C
n = 30°C
52
®
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS .. .. 45
:
.
Fall in temperature. T =  T2 = 95  30 = 65°C induced in the brass will be compressive whereas the stress in steel will be tensile as shown in
E = 2 x 105 MN/m2 Fig. 124 (c). Hence the load or force on the brass will be compressive whereas on the steel the
Modulus of elasticity,
= 2 x 105 x 10s N/m2 load will be tensile. Ab = Area of crosssection of brass bar
Let
= 2 x 10 11 N/m2
Coefficient of linear expansion, a = 12 x 10~®/°C. a = Stress in brass
(i) When the ends do not yield b
e = Strain in brass
b
The stress is given by equation (1.15). a = Coefficient of linear expansion for brass
h
Stress = a.T.E = 12 x lO"6 x 65 x 2 x 10 11 N/m2 E = Young’s modulus for copper
b
— 156 x 10® N/m2 or 156 N/mm2 (tensile). Ans.
A a e and a, = Corresponding values of area, stress, strain and coefficient of
,,
Pull in the rod = Stress x Area linear expansion for steel, and
= 156 x 225 jt = 110269.9 N. Ans. E — Young’s modulus for steel.
(«) When the ends yield by 0.12 cm
5 = Actual expansion of the composite bar
mm6 = 0.12 cm = 1.2
Now load on the brass = Stress in brass x Area of brass
The stress when the ends yield is given by equation (1,16).
~ ab* A
b
T L EStress = (ra    52 x
L And load on the steel = xA
s
——(12 x For the equilibrium of the system, compression in copper should be equal to tension in
10® x 65 x 5000  1.2)
„ 105 N/mm„2
x x the steel
2
5000
or Load on the brass = Load on the steel
= x 2 x 105 = 108 N/mm2. Ans. o xA = a xA .
b b s s
5000
Pull in the rod = Stress x Area Also we know that actual expansion of steel
= 108 x 225 — 76340.7 N. Ans. = Actual expansion of brass —
1.15. THERMAL STRESSES IN COMPOSITE BARS But actual expansion of steel
= Free expansion of steel + Expansion due to tensile stress
Fig. 1.24 (a) shows a composite bar consisting of two members, a bar of brass and
another of steel. Let the composite bar be heated through some temperature. If the members in steel
are free to expand then no stresses will be induced in the members. But the two members are
rigidly fixed and hence the composite bar as a whole will expand by the same amount. As the ^• T  L+ i' L
coefficient of linear expansion of brass is more than that of the steel, the brass will expand
more than the steel. Hence the free expansion of brass will be more than that of the steel. But And actual expansion of copper
both the members are not free to expand, and hence the expansion of the composite bar, as a
whole, will be less than that of the brass, but more than that of the steel. Hence the stress  Free expansion of copper  Contraction due to compressive
stress induced in brass
Substituting these values in equation (i), we get
—a x Tx L +  xL = a xTxL xL
s b
where T = Rise of temperature.
(a) (b) A mmProblem 1.30. steel rod of 20 diameter passes centrally through a copper tube of
Fig. 1.24 mm mm50 internal diameter. The tube is closed at each end by rigid,
nuts are tightened lightly home on the projecting parts of the
plates
external diameter and 40
of negligible thickness. The
: :
46 STRENGTH OF MATERIALS SIMPLE STRESSES AND. STRAINS 47
rod. If the temperature of the assembly is raised by S0°C, calculate the stresses developed in T
copper and steel. Take E for steel and copper as 200 GN/m2 and 100 GN/m2 and a for steel and and actual expansion of copper
copper as 12 x NT6 per °C and 18 x Kt6 per °C. = Free expansion of copper  Contraction due to compressive stress in copper
Sol. Given
Dia. of steel rod = 20 mm Substituting these values in equation (st), we get
A — mmArea of steel rod, — —T L +a. .
= x 20 2 = lOOic 2 .T .L  .L
«4 . E .L = a. E
s c
s
c
A mm mmArea of copper tube, —ft .TT °c
c
(502  402 ) 2 = 225ji a ^or
= 2 an .TT +4.
s
c
Rise of temperature, T = 50°C
E for steel, E = 200 GN/m2 12 x 106 x 50 + .. ?% = 18 x lO^5 x 50 2.25 aj
s
200 x 10 3 100 x 10
= 200 x 10 9 N/m2 G = 10(. . 9
)
= 200 x 103 x 106 N/m2 2 25 + £ = 18 x 106 x 50 12 x 106 x 50
= 200 x 103 N/mm2 ( v 10 6 N/m2 = 1 N/mm2 ) 200 x 10 3 100 x 10 s
E for copper, E GN/m2 1.125 x 10® o + 10~ 5 a = 6 x lO"6 x 50
c c c
= 100 = 100 x 10 9 N/m2
= 100 x 103 x 10 6 N/m2 = 100 x 103 N/mm2 2.125 x 10® a = 30 x IQ®
c
a for steel, a = 12 x 10~ 6 per °C 2.125c,, = 30
s
^a = =14.117 N/mma.
a for copper, a = 18 x 10"6 per °C. c 2.125 Ans.
c
As a for copper is more than that of steel, hence the free expansion of copper will be Substituting this value in equation (£), we get
more than that of steel when there is a rise in temperature. But the ends of the rod and the
tube is fixed to the rigid plates and the nuts are tightened on the projected parts of the rod. c = 14.117 x 2.25
Hence the two members are not free to expand. Hence the tube and the rod will expand by the s
same amount. The free expansion of the copper tube will be more than the common expansion, = 31.76 N/mm2. Ans.
whei eas the free expansion of the steel rod will be less than the common expansion. Hence the
copper tube will be subjected to compressive stress and the steel rod will be subjected to tensile A mm mmProblem 1.31. steel tube of 30 internal diameter
external diameter and20
mmencloses a copper rod of 15 diameter to which it is rigidly joined at each end. if, at a
stress. temperature of 10°C there is no longitudinal stress, calculate the stresses in the rod and tube
Let o = Tensile stress in steel Ewhen the temperature is raised to 200°C. Take for steel and copper as 2.1 x 10a Nlmm“ and 1
s
N/mmx 10s 2 respectively. The value of coefficient of linear expansion for steel and copper is
a, = Compressive stress in copper.
given as 11 x 10s per °C and 18 x Kt6 per °C respectively.
For the equilibrium of the system,
Tensile load on steel = Compressive load on copper Sol. Given  = 15 mm
Dia. of copper rod
or
.. Area of copper rod, mmA = — x 152 = 56.25a; 2
or c
Area of steel tube, mmA —= (30 2  20 2 ) = 125a; 2
s
Rise of temperature, T = (200  10) = lQO^C
E for steel, E = 2.1 x 10s N/mm2
E for copper, s
E = 1 x 105 N/mm2
c
Value of a for steel, a = 1 1 x 10fi per °C
$
Value of'a for copper, a = 18 x 10~6 per °C
c
As the value of a for copper is more than that of steel, hence the copper rod would
expand more than the steel tube if it were free. Since the two are joined together, the copper
M
2° STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS 49
will be prevented from expanding its full amount and will be put in compression, the steel N/mm1 x 10s 2 respectively. The linear coefficient of expansion for steel and gun metal is
being put in tension.
12 x NT6 per °C and 20 x ICt6 per °C.
Let o  Stress in steel
s
Sol. Given :
o = Stress in copper. Dia. of gun metal rod = 20 mm
c
For equilibrium of the system,
Compressive load on copper = Tensile, load on steel mmArea of gun metal rod, A —= x 20s = 100re 2
64
or ^A = o .A
e s s
A=~ mm‘
Aa<: = ° Area of steel tube, (30 2  25 2 ) = 68.75n 2
s
'4" =„ 125.it = 222 x 4
CTs
' 56.25 n
Fall in temperature, T = 140  30 = 110
We know that the copper rod and the steel tube will actually expand by the same amount. EValue of for steel, E  2.1 x 10s N/mm2
Now actual expansion of steel = Free expansion of steel + Expansion due to tensile stress Value of E for gun metal, s
= <x .r.L+^ .L Value of a for steel, E = lx 105 N/mm2
s g
E
a, = 12 x 10~6 per °C
s Value of a for gun metal, a = 20 x 10~6 per °C.
and actual expansion of copper = Free expansion of copper g
As ag is greater than a hence the free contraction of the gun metal rod will be more
s,
 Contraction due. to compressive stress
than that of steel when there is a fall in temperature. But, since the ends of the rods have been
= a .T.L^.L provided with nuts, the two members are not free to contract fully, each of the member will
c
contract by the same amount. The free contraction of the gun metal rod will be greater than
But actual expansion of steel = Actual expansion of copper
the common contraction, whereas the free contraction of the steel tube will be less than the
common contraction. Hence the steel tube will be subjected to compressive stress while the
a..T.L + %.L^.T.LS.. L gun metal rod will be subjected to tensile stress.
Let ct = Stress in steel tube and
s
or o = Stress in gun metal rod.
g
For the equilibrium of the system,
or 11 X 106 X 0S = 18 x 106 x 190  Total compressive force in steel = Total tensile force in gun metal
lx 10 s
or 190 + s, (v a. = 2.22aJ A
S
; 2.1 x 10 <V = VA?
or a, 2.22 a s A lOCto
or 2.1 xlO5 ' 1x10 s Va; Vor g=
or 0=
‘ 6&75^
as + 2.1 x 2.22 a. or o = 1.4545og —
s
s = 5 x 106 x 190
We also know that the steel tube and gun metal rod will actually contract by the same
2.1 x 10
a + 4.662t> amount.
s
s Actual contraction of steel = Actual contraction of gun metal rod
5.662a But actual contraction of steel = Free contraction of steel
o = —199.5 = 35.235 N/mm2. Ans. + contraction due to compressive stress in steel
5.662
—= a T L +. .
Substituting this value in equation (i), we get sE L.
N/mmct = 2.22 x 35.235 = 78.22 2 s
s
. Ans.
mm mmProblem 1.32. A steel tube of 30 Actual contraction of gun metal = Free contraction of gun metal
mmencloses a gun metal rod of 20 external diameter and 25 internal diameter  expansion due to tensile stress in gun metal
diameter to which it is rigidly joined at each end. The
temperature of the whole assembly is raised to 140°C and the nuts on the rod are then screwed V TL
lightly home on the ends of the tube. Find the intensity of stress in the rod when the common =
temperature has fallen to 30°C. The value of E for steel and gun metal is 2.1 x 10s N/mm2 and
}
:
SIMPLE STRESSES AND STRAINS °1
STRENGTH OF MATERIALS
Elongation of the element
Equating the two values, we get = Strain x Length of element
ct . T.L+ .L = a„ .T.L —= wx—x— x d,x
E
Total elongation of the bar is obtained by integrating, the above equation between limits
a* T zero and L.
ttra* T+
c 1* w xx _ w CL
1.4545 a„ , £>„ =
'
Jo
f12 x 10« x 110 * 2.1x10 s = 20 x 106 x 110  1x1f0 (v a = 1.4545 a ) w x2 Wl
E~E ~2jj _ ^X ...(1.17'
2
= 20 x 106 x 110  12 x 1CT6 x 110 a
a+ (y W=WXL
)
2.1 x 10°
—1.4545 a„ +2.= 1xo.1 = 8 x.lO'6 x 110 1.17. ANALYSIS OF BAR OF UNIFORM STRENGTH
5
2.1 x 10 s mIn the previous article we have seen that the stress due to self weight of the bar is
3.5545 a = 8 x 10“« x 110 x 2.1 x 10 5 = 184.8 constant but the stress increases with the increase of distance from the lower end. If the S'
weight is neglected and a bar of uniform section is subjected to an axial load, then the stress
g
the bar would be uniform.
a„ = = 51.99 N/mmz. Ans. Let us find the shape of the bar of which self weight of the bar is considered and
g 3.5545
having uniform stress on all sections when subjected to an axial P. Such bar is shown in Fig. 14
Substituting this value in equation (i), we get Ans. in which the area of the bar increases from the lower end to the upper end.
or = 1.4545 x 51.99 = 75.62 N/mm2.
1.16. ELONGATION OF A BAR DUE TO ITS OWN WEIGHT Let = Area of upper end,
AFig. 1.25 shows a bar AB fixed at end and hanging freely under //////////////// A = Area of lower end,
A 2
its own weight.
mm  w = Weight per unit volume of the bar,
Let L = Length of bar, *dx
 a  Uniform stress on the bar.
A = Area of crosssection, x
E = Young’s modulus for the bar material, Consider a strip of length dx at a distance x from \/
w = Weight per unit volume of the bar material. 
Athe lower end. Let be the area of the strip at section AB \ / ct(a + dA)
Consider a small strip of thickness dx at a distance x from the j
lower end. and (A + dA ) be the area at section DC. Consider the equi
librium of the strip ABCD. \/ D ft^tt^t'‘HL
j
The forces acting on the strip are :
i. OhrodwWC
B AT
Weight of the bar for a length of * is given by, (£) Weight of strip acting downward and equal to T AV I I;
P = Specific weight x Volume of bar upto length x ,, Axw wx volume, of strip i.e., x jj
= w xAxx B [_ dx. <rA + wAd>:
Fiigg. 11 .^255
AB(ii) Force on section due to uniform stress (a) Ii
I
i
This means that on the strip, a weight of w x A x x is acting in the downward direction. and is equm to a x A. This is acting downward. i 1
j
Due to this weight, there will be some increase in the length of element. But length of the (Hi) Force on section CD due to uniform stress (a) .
p
and is equal to c(A + dA). This is acting upwards. '
Fig. 1.26
element is dx. Now, Total force acting upwards
Now stress on the, element
Weight —acting on element wx. Axx v = Total force acting downwards
:
= = A =W XX or o(A + dA)  a x A + wAdx
Aor Acr x +ck£A'= o x + luAdx
Area of crosssection
or odA = wAdx
The above equation shows that stress due to self weight in a bar is not uniform. It dA w
depends on x. The stress increases with the increase of x.
or A ~ a ^x
Stress wxx
Strain in the element =
.
STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS
Integrating the above equation, we get
JT^O*dA w 0r log'Aw X + C Aj _ 0.0010667
400
ff
where C is the constant of integration.
= 0.0010667
At x = 0, A=A
2
Substituting these values in equation (i), we get ^2.3 log10 = 0.0010667
w C=, .
log
£ * x0+ Aog 0.0010667
io
C A• = log 1 _ 2.3 = 0.00046378
e2 “
400
Substituting the value of C in equation (t), we get
A=^log : Antilog of 0.00046378 = 1.00107.
e
x + log A2 mm= 400 x 1.00107 = 400.428 2
2 . Ans.
xAlog  =
e
log At or log
e e
WX HIGHLIGHTS
, 1. The resistance per unit area, offered by a body against deformation is known as stress. The
stress is given by
A=A e '
where P = External force or load A = Crosssectional area.
A, 2 ;
The above equation gives the area at a distance x from lower end.
At x = L, A = Aj
Substituting these values in equation («), we get
Ai=A2g a 2. Stress is expressed as kgf/m2 kgf/cm2 N/m2 and N/mm2 .
,,
...(1.19) 3. 1 N/m2 = 104 N/cm 2 or 10~6 N/mm2 .
Problem 1.33. A vertical bar fixed at the upper end and of uniform strength carries an 4. The ratio of change of dimension of the body to the original dimension is known as strain.
maxial tensile load of 600 kN. The bar is 20 long and having weight per unit volume as 0.00008
mmN/mm3. If the area of the bar at the lower end is 400 5. The stress induced in a body, which is subjected to two equal and opposite pulls, is known as
2 area
find the of the bar at the upper tensile stress.
,
end.
Sol. Given : 6. The stress induced in a body, which is subjected to two equal and opposite pushes, is known as
compressive stress.
Axial load, P = 600 kN = 600 x 10s N 7. Elasticity is the property by virtue of which certain materials return back to their original posi
Length, L = 20 m = 20 x 103 mm tion after the removal of the external force.
Weight per unit volume, w = 0.00008 N/mm3 8. Hooke’s law states that within elastic limit, the stress is proportional to the strain.
mmArea of bar at lower end. A, = 400 3 9. The ratio of tensile stress (or compressive stress) to the corresponding strain is known as Young’s
modulus or modulus of elasticity and is denoted by E.
A = Area of bar at upper end.
x Tensile or compressive stress
Now' the uniform stress* on the bar, _
600 x IQ3 = 1500 N/mm2 Corresponding strain
400 10. The ratio of shear stress to the corresponding shear strain within the elastic limit, is known as
Using equation (1.19), we get Gmodulus of rigidity or shear modulus. It is denoted by C (or or N).
1 1. Total change in the length of a bar of different lengths and of different diameters when subjected
to an axial load P, is given by
0.00008 X 20 x 10 1 — — —P Li Lo L3 when E is same
7— + + + 
= 400 x e
[Aj A2 a
= 400 X g0. 0010667 AdL = E
*The stress on lower end = p = p Al_ + Ah + ESAS + Ewhen is different.
. We want that the stress in the bar should be uniform i.e., £A E2 A2
—equal to 1l .
Aq
^2
D D12 . The total extension of a uniformly tapering circular rod of diameters when the rod is
and
2,
1
subjected to an axial load P is given by
.
54 STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS 55
—dL = 7. Prove that the total extension of a uniformly tapering rod of diameters and D when the rod
2,
is subjected to an axial load P is given by
' '< where L = Total length of the rod.
k hninDrYDt<i' —4PL
13. A composite bar is made up of two or more bars of equal lengths but of different materials rigidly deddL =
ji
1 fixed with each other and behaving as one unit for extension or compression. 12
14. In case of a composite bar having equal length : (i) strain in each bar is equal and (ii) total load where L = Total length of the rod.
on the composite bar is equal to the sum of loads carried by each different materials. 8. Define a composite bar. How will you find the stresses and load carried by each member of a
15. The stresses induced in a body due to change in temperature are known as thermal stresses. composite bar ?
16. Thermal strain and thermal stress is given by 9. Define modular ratio, thermal stresses, thermal strains and Poisson’s ratio.
thermal strain, e = a .T and thermal stress, p = a.T.E 10 . A rod whose ends are fixed to rigid supports, is heated so that rise in temperature is T C. Prove
where a = Coefficient of linear expansion , that the thermal strain and thermal stresses set up in the rod are given by,
T =. Rise or fail of temperature, Thermal strain  a.T and
E = Young’s modulus. Thermal stress = a.T.E
17. Total elongation of a uniformly tapering rectangular bar when subjected to an axial load P is where a = Coefficient of linear expansion.
11 . What is the procedure of finding thermal stresses in a composite bar ?
given by
12. What do you mean by ‘a bar of uniform strength’ ?
———dL = log £b 13. Find an expression for the total elongation of a bar due to its own weight, when the bar is fixed
Et(ab)
' at its upper end and hanging freely at the lower end.
14. Find an expression for the total elongation of a uniformly tapering rectangular bar when it is
where L = Total length of bar t = Thickness of bar
; 6 = Width at smaller end subjected to an axial load P.
a = Width at bigger end
;
E = Young’s modulus.
18. In case of a composite bar having two or more bars of different lengths, the extension or compres (B) Numerical Problems
sion in each bar will be equal. And the total load will be equal to the sum of the loads carried by
each member. 1 . A rod 200 cm long and of diameter 3.0 cm is subjected to an axial pull of 30 kN. If the Young’s
2 determine : (i) stress, ( ii ) strain and (iii) the
N/mmmodulus of the material of the rod is 2 x 10“ ,
19. In case of nut and bolt used on a tube with washers, the tensile load on the bolt is equal to the elongation of the rod. N/mm[Axis, (i) 42.442 (ii) 0.000212 (iii) 0.0424 cm)
compressive load on the tube.
mm mm2 . Find the Young’s modulus of a rod of diameter 30 which is subjected
and of length 300
[Ans. 63.6 GN/m2 ]
20. Elongation of a bar due to its own weight is given by to a tensile load of 60 kN and the extension of the rod is equal to 0.4 mm.
w— — WLosrL = I? 3. The safe stress, for a hollow steel column which carries an axial load of 2.2 x 10 3 kN is 120 MN/m".
E 2
x or 2zrE= If the external diameter of the column is 25 cm, determine the internal diameter.
[Ans. 19.79 cm]
where w = Weight per unit volume of the bar material N4. An axial pull of 40000 is acting on a bar consisting of three sections of length 30 cm, 25 cm and
L = Length of bar. 20 cm and of diameters 2 cm, 4 cm and 5 cm respectively. If the Young’s modulus = 2 x 105 N/mm,
determine :
(ji) stress in each section and Jlp^C ii) total extension of the bar.
r N/mm[Ans. (i) 127.32, 31.8, 20.37 2 (ii) 0.025 cm]
,
MN5. The ultimate stress for a hollow steel Column which carries an axial load of 2 is 500 N/mm 2 .
(A) Theoretical Questions If the external diameter of the columns 250 mm, determine the internal diameter. Take the
[Ans.  205.25 mm]
1. Define stress and strain. Write down the S.I. and M.K.S. units of stress and strain. factor of safety as 4.0. ?&£.•
2. Explain clearly the different types of stresses and strains.
3. Define the terms : Elasticity, elastic limit. Young's modulus and modulus of rigidity. 6. A member formed by connecting a steehBwfto an aluminium bar
4. State Hooke’s law. is shown in Fig. 1.27. Assuming that the bars are prevented from 6 cm x 6 cm
5. Three sections of a bar are having different lengths and different diameters. The bar is subjected
buckling sideways, calculate the magnitude of force P, that will Steel bar
to an axial load P Determine the total change in length of the bar. Take Young’s modulus of
cause the total length of the member to decrease 0.30 mm. The 1 0 cm x 1 0 cm
different sections same.
values of elastic modulus for steel and aluminium are 2 x 10 Aluminium bar
6. Distinguish between the following, giving due explanation :
(i) Stress and strain, N/mm2 and 6. 5 x 104 N/mm2 respectively. [Ans. 406.22 kN]
(if) Force and stress, and
Fig. 1.27
(iff) Tensile stress and compressive stress.
56 7. STRENGTH OF MATERIALS SIMPLE STRESSES AND STRAINS 57
The bar shown in Fig. 1.28 is subjected to a tensile load of 150 kN. If the stress in the middle A14. steel rod of 2 cm diameter is enclosed centrally in a hollow copper tube of external diameter
portion is limited to 160 M/mm2 determine the diameter of the middle portion. Find also the 4 cm and internal diameter of 3.5 cm. The composite bar is then subjected to an axial pull of
,
50000 N. If the length of each bar is equal to 20 cm, determine :
length of the middle portion if the total elongation of the bar is to be 0.25 cm. Young’s modulus is
given as equal to 2.0 x 10 5 N/mm2 [Ans. 3.45 cm, 29.38 cm] (!) the stress in the rod and tube, and
.
.
(ii) load carried by each bar.
Take E for steel = 2 x 10 5 N/mm2 and for copper = 1 x 10 s N/mm2.
NN/mm[Ans. (!) 54.18 ; 108.36
2 (ii) 34043.4 and 15956.6 N]
mm mm15. A mild steel rod of 20
diameter and 300 long is enclosed centrally inside a hollow copper
mmtube of external diameter 30 and internal diameter of 25 mm. The ends of the tube and rods
Eare brazed together, and the composite bar is subjected to an axial pull of 40 kN. If for steel
and copper is 200 GN/m2 and 100 GN/m2 respectively, find the stresses developed in the rod and
Fig. 1.28 tube. Also find the extension of the rod. [Ans. 94.76 N/mm2 47.38 N/mm2 and 0.142 mm]
,
A mm98. A MN mm16. load of 1.9 x 200 mm. The column is rein
brass bar, having crosssection area of 900 2 is subjected to axial forces as shown in is applied on a short concrete column 300
, mmforced with four steel bars of 10
mFig. 1.29 in which AB = 0.6 m, BC = 0.8 and CD = 1,0 m. diameter, one in each corner. Find the stresses in the
Econcrete and steel bars. Take for steel as 2.1 x 105 N/mm2 and for concrete as 1.4 x 104 N/mm2 .
[Ans. 20.13, 301.9 N/mm2]
mm mm17. A reinforced short concrete column 250
x 250 in section is reinforced with 8 steel bars.
mmThe total area of steel bars is 1608.50 2 The column carries a load of 270 kN. If the modulus
.
of elasticity for steel is 18 times that of concrete, find the stresses in concrete and steel.
Fig. 1.29 If the stress in concrete shall not exceed 4 N/mm2 find the area of steel required so that the
,
Find the total elongation of the bar. Take E = 1 x 10s N/mm2 [Ans.  0.111 mm] mmcolumn may support a load of 400 kN. [Ans. o = 3 N/mm 2 o = 54 N/mm2 and = 2206 2
. c ,r
]
A member ABCD is subjected to point loads Fj, Pv P and P4 as shown in Fig. 1.30. Calculate 18. Two vertical rods one of steel and other of copper are each rigidly
3
P
the force 3 necessary for equilibrium if P = 120 kN, P2  220 kN and P = 160 kN. Determine fixed at the top and 60 cm apart. Diameters and length of each
l 4
Arod are 3 cm and 3.5 cm respectively. cross bar fixed to the rods
also the net change in the length of the member. Take E = 200 GN/m 2 . [Ans. 0.55 mm] Nat the lower ends carries a load of 6000 such that the cross bar
remains horizontal even after loading. Find the stress in each 200
Erod and the position of the load on the bar. Take for steel = 2 x
10 6 N/mm2 and for copper = 1 x 10“ N/mm2
.
[Ans. 2.829 and 5.658 N/mm2 39.99 cm]
;
A mm19. steel rod of crosssectional area 1600 2 and two brass rods
mmeach of crosssectional (area of 1000 2 together support a load
of 50 kN as shown in Fig. 1.31.
Fig. 1.30 Find the stresses in the rods. Take E for steel = 2 x 10^ N/mm2 1.31
and E for brass = 1 x 10 s N/nun2 .
10. A rod, which tapers uniformly from 5 cm diameter to 3 cm diameter in a length of 50 cm, is !
Esubjected to an axialload of 6000 N. If = 2 x 10s N/mm2 find the extension of the rod. [Ans. ct6 = 12.1 N/mm2 and a = 16.12 N/mm2 ]
s
.
, m20. A rod is 3 long at a temperature of 15“C. Find the expansion of the rod, when the temperature
[Ans. 0.00127 cm] is raised to 95°C. If this expansion is prevented, find the stress induced in the material of the
mm mm11. Find the modulus of elasticity for a rod, which tapers uniformly from 40 rod. Take E = lx 105 N/mm2 and a = 0.000012 per degree centigrade.
to 25
diameter in a length of 400 mm. The rod is subjected to a load of 6 kN and extension ofthe rod is [Ans. 0.288 cm, 96 N/mm2 ]
° 04 mm
 [Ans. 76.39 kN/mm2] mA21. steel rod 5 cm diameter and 6 long is connected to two grips and the rod is maintained at a
A m mm12. rectangular bar made of steel is 3 long and 10 thick. The rod is subjected to an axial temperature of 100°C. Determine the stress and pull exerted when the temperature falls to 20°C
mm mmtensile load of 50 kN. The width ofthe rod varies from 70
at one end to 28 at the other. if (!) the ends do not yield, and (!!) the ends yield by 0.15 cm.
Find the extension of the rod Eif = 2 x 10 5 N/mm2 [Ans. 1.636 mm] Take E = 2 x 10 5 N/mm2 and a = 12 x 10"6/', C.
.
mm13. The extension in a rectangular steel bar of length 800 and of thickness 20 mm, is found to be N/mm2 N N/mm2
,
mm0.21 mm. The bar tapers uniformly in width from 80 to 40 mm. If E for the bar is 2 x 105 [Ans. (!) 192 and 376990 (ii) 142 278816.3 Nj
N/mm2 A mm mm22. steel rod of 20
, determine the axial tensile load on the bar. [Ans. 60.6 kN] diameter passes centrally through a copper tube 40 external diameter
mmand 30 internal diameter. The tube is closed at each end by rigid plates of negligible thickness.
cboo STRENGTH OF MATERIALS

.
The nuts are tightened lightly home on the projected parts of the rod. If the temperature of the
assembly the stresses developed
is raised fay 60°C, calculate Ein copper and steel. Take for steel
and copper as 200 GN/m2 and 100 GN/m2 and a for steel and copper as 12 x 10s per °C and
[Ans. 16.23, 28.4 N/mm]
18 x 10'® per °C.
23. A vertical bar fixed at the upper end and of uniform strength carries an axial tensile load of 500
mkN. The bar is 18 long and having weight per unit volume as 0.00008 N/mm2 . If the area of the
mm mmbar at the lower end is 500
2 find the area of the bar at the upper end. [Ans. 500.72 ]
,
A24. straight circular rod tapering from diameter ‘D’ at one end to a diameter ‘d’ at the other end is 2.1.
subjected to an axial load ‘P\ Obtain an expression for the elongation of the rod.
INTRODUCTION
Ans. 51~jiE4P. Dj; 
When a body is subjected to an axial tensile load, there is an increase in the length of the
. d\] body. But at the same time there is a decrease in other dimensions of the body at right angles
to the line of action of the applied load. Thus the body is having axial deformation and also
deformation at right angles to the line of action of the applied load (i.e., lateral deformation).
2.2.
This chapter deals with these deformations, Poisson’s ratio, volumetric strains, bulk modulus,
relation between Young’s modulus and modulus of rigidity and relation between Young’s modu
lus and bulk modulus.
LONGITUDINAL STRAIN
When a body is subjected to an axial tensile or compressive load, there is an axial defor
mation in the length of the body. The ratio of axial deformation to the original length of the
body is known as longitudinal (or linear) strain. The longitudinal strain is also defined as the

2d.e3f.ormation of the body per unit length in the direction of the applied load.
Let L = Length of the body,
P = Tensile force acting on the body,
 6L = Increase in the length of the body in the direction of P.
Then, longitudinal strain = y5L .
LATERAL STRAIN
The strain at right angles to the direction of applied load is known as lateral strain. Let
a rectangular bar of length L, breadth b and depth d is subjected to an axial tensile load P as
shown in Fig. 2.1. The length of the bar will increase while the breadth and depth will
decrease.
Let SL = Increase in length,
6b = Decrease in breadth, and
6d = Decrease in depth.
Then longitudinal strain =y ...(2.1)
and lateral strain XV ...(2.2)
——= or
bd
59
.
60 STRENGTH OF MATERIALS
—But longitudinal strain =
L
6L
y = 0.00025.
eL (or change in length) = 0.00025 x L
UTsi. ng equation (2.3), = 0.00025 x 4000 = 1.0 mm. Ans.
Fig. 2.1 Poisson’s ratio _ = Lateral strain
Longitudinal strain
Note, (i) If longitudinal strain is tensile, the lateral strains will be compressive.
n ^ _ Lateral strain
(«) If longitudinal strain is compressive then lateral strains will be tensile.
0.00025
™ *ngi*udinal Strai " in the directi of ''°ad accompanied by lateral strains of
Lateral strain = 0.3 x 0.00025 = 0.000075.
owolSlTZTnlltchue opposite kind in all !d?irections perpendicular to the load.
We know that
2.4. POISSON’S RATIO —Lateral strain = or —for—)
ra idd0‘e'an°fotflaetfd!.rbrayelspsS.ctfdH’e'vnn,ctthe0lnmtahttehhe^eMmgaa^sttidicicalnlliaymli,t.stTrhaiisn is a d\t)b
ratio
when them f constant for a given material, bb = b x Lateral strain
is called
ittiSs ggeenneerraTlllSy Poisson’s ratio and = 30 x 0.000075 = 0.00225 mm. Ans.
Similarly, 5f = t x Lateral strain
Poisson’s ratio, g = —Lateral atrain __ ...U.o; = 20 x 0.000075 = 0.0015 mm. Ans.
Longitudinal strain ZTomnc^mblnf' f ZTlDbar
a3d0ocfm4,0°brkeNradTthhe4dhececmrUeCaadsnUede,i°ndfelpeYnto8hutnh4Z’icss gmmiovwdehunelanuss0tah.e0n7db5aPcromisisasonsndu’bsijnrecacrtteieaodsoetfoianambnerteaaaxldiltaihlc
01 Lateral strain = g x longitudinal strain of length
°
As lateral strain is opposite in sign to longitudinal strain, hence algebraically, the
lat
eral strain is written as
Lateral strain =  g x longitudinal strain Sol. Given :
...[2.3 (A)] Length, L = 30 cm Breadth,
™Th ValUe °f P° iSSOn,S rati° VaHes from °' 25 t0 0 3S  Fot bber, its value ranges from ;
b = 4 cm ; and Depth, d = 4 cm.
0.45 to 0 50 Area of crosssection. A=6xd=4x4
m m mm E mmN/mm, . ,Pr bI2 1’ * mm= 10 cm2 = 16 x 100 = 1600 2
.° Determine the changes in length, breadth and
thickness of a steel bar
, ic is lfong, 30 wide and 20 axial pull of 30 in the Axial compressive load, P = 400 kN = 400 x 1000 N
’ 0.3. Decrease in length.
Increase in breadth.
thick and is subjected to an
kNdirect, Longitudinal strain 5L = 0.075 cm
on of its length. Take  2 x 10s 2 and Poisson’s ratio 
Sol. Given : 6b = 0.003 cm
Length of the bar, L = 4 m = 4000 mm T ^bL 0.075
= =0.0025
Breadth of the bar, b = 30 mm
Thickness of the bar. t = 20 mm Lateral strain T66 = 0.003 == „„
Using equation (2.3), 0.00075.
~r
Area of crosssection, 64
Axial pull, A = b x t — 30 x 20 = 600 mm**
P = 30 kN = 30000 N Lateral strain 0.00075
Young’s modulus, E = 2 x 10s N/mm2 Poisson’s ratio = 0.3. Ans.
Longitudinal strain
Poisson’s ratio, g = 0.3. Longitudinal strain 0.0025
Stress = 
Now strain in the direction of load (or longitudinal strain), Stress = P
Stress Load E AxE
__ 400000
E Area x E Stress = _
’ 1600 x B
30000 m400000 . _ 1 Xx 1^05 N/m 2 Ans.
600 x 2 x 10 5 = 0.00025.
' 1600 x 0.0025' .
*
elastic constants 63
2.5. VOLUMETRIC STRAIN = Longitudinal strain (1  2p)
The ratio of change in volume to the original volume of a body (when the body is sub = j (1  2p) ...(2.5)
jected to a single force or a system of forces) is called volumetric strain. It is denoted by e„.
Lj
Mathematically, volumetric strain is given by
Problem 2.3. For the problem 2.1, determine the volumetric strain and final volume of
the given steel bar.
8V Sol. Given :
ve ~ The following data is given in problem 2.1. :
v
where 8V = Change in volume, and L = 4000 mm, 6 = 30 mm, t or d = 20 mm, p = 0.3.
V = Original volume. mmOriginal volume, V = L.b.d = 4000 x 30 x 20 = 2400000 3
2.5.1. Volumetric Strain of a Rectan d— The value of longitudinal strain fi.e.,—! in problem 2.1 is calculated
gular Bar which is Subjected to an Axial
^< I P = 0.00025
Load P in the Direction of its Length. Con
P —L — as, ^
sider a rectangular bar of length L, width b and
L~' —M*' Now using equation (2.5), we have
depth d which is subjected to an axial load P in
n_
the direction of its length as shown in Fig. 2.2.
Let bL = Change in length, Fiigg. 2.2 Volumetric strain, e = 5L „  „,
56 = Change in width,
j (1 dp.)
v
ancl bd = Change in depth. = 0.00025(1  2 x 0.3) = 0.0001. Ans.
Final length of the bar  L + bL
Final width of the bar =6 + 86 or ^=0.0001
Final depth of the bar = d + bd 6V= 0.0001 xV
mm= 0.0001 x 2400000 = 240 3
Now original volume of the bar, V = L.b.d
Final volume — {L + bL)(b + 86)(d + 8dy Final volume = Original volume + bV
= L.b.d. + bdbL + Lbbd + Ld.bb mm= 2400000 + 240 3
(Ignoring products of small quantities)
Change in volume, mm= 2400240 3 Ans.
6V = Final volume  Original volume .
= (Lbd + bdbL + Lbhd + Ldbb ) — Lbd
mm mm mmAProblem 2.4. steel bar 300 thick is subjected to a
= bdbL + Lbbd + Ldbb long, 50 wide and 40
pull of 300 kN in the direction of its length. Determine the change in volume. Take E = 2 x 10s
N/mm2 and p. = 0.25.
.. Volumetric strain, Sol. Given :
Length,
5V Width, L = 300 mm
ev~ V Thickness, 6 = 50 mm
f = 40 mm
bdbL + Lbbd + Ldbb ...(2.4) Pull,
P = 300 kN = 300 x 103 N
= •(() Value of £
= 2 x 105 N/mm2
Lbd
Value of p = 0.25
M ^~ L + d + b Original volume,
^ —But —ir = Longitudinal strain and V=Lx 6 xt
or are lateral strains.
mm mm= 300 x 50 x 40 3
3 = 600000
/y (t 0
Substituting these values in the above equation, we get The longitudinal strain (i.e., the strain in the direction of load) is given by
e = Longitudinal strain + 2 x Lateral strain dL Stress in the direction of load
v
=
From equation (2.3A), we have
~L E
Lateral strain = (ix Longitudinal strain.
But stress in the direction of load
Substituting the value of lateral strain in equation (i), we get
PP
e =• Longitudinal strain  2 x u longitudinal strain
_
Area but
STRENGTH OF MATERIALS
300 x 10 = 150 N/mm2 Substituting these values in equation (2.6), we get
50x40
dV =c
T ^= =000075 *
y' + e + e*
Now volumetric strain is given by equation (2.5) as r
Now, Let a = Tensile stress in xx direction,
x
a = Tensile stress in yy direction, and
y
a Tensile stress in zz direction.
x
ITe (1  2fl) E = Young’s modulus
"
= p = Poisson’s ratio.
= 0.00075 (1  2 x 0.25) = 0.000375
Let 6V = Change in volume. Then —dV represents volumetric strain. Now a —will produce a tensile strain equal to in the direction of x, and a compressive
xE
—dV strain equal to 11 x in the direction of y and z. Similarly, o will produce a tensile strain
= 0.000375
E
—equal to in the direction ofy and a compressive strain equal to — — jn the direction ofx
EE
or dV = 0.000375x17 ~and z.
mm= 0.000375 x 600000 = 225 3 Ans. Similarly a will produce a tensile strain equal to in the direction of z and a comp
z
.
2.5.2. Volumetric Strain of a Rec —ressive strain equal to
tangular BarSubjected to Three Forces t2 —X in the direction of x and y. Hence a and a will produce
which are Mutually Perpendicular. Con E
j' y
sider a rectangular block of dimensions x, y
and z subjected to three direct tensile stresses i ** and °** x z in the direction of x.
along three mutually perpendicular axis as _ compressive strains equal to
shown in Fig. 2.3.
I Net tensile strain along xdirection is given by
,
X
ax 11 * Oy (A
Then volume of block, V = xyz. EE E E
I
Taking logarithm to both sides, we have „. An.o,, Similarly, a (ax +oz
r Ig. y
Vlog = log x + log y + log z.
~E~^l E
Differentiating the above equation, we get
a +arx '
_ ,. I y
dV1 1 dx, 1dy, 1 EH E
,,, ,.
= dz
+ + Adding all the strains, we get
V xyz
— — —dV dy 1 . 2p .
y
"VVr dx dz £+e + , + + az>~
x z y
or = + + «* = °y
„out, — — ——dV = (a + a + a Xl  2 p).
Change of volume y 2
= = VTTol, umetric x
; :
strain
..
V Original volume
But e + e + e_ = Volumetric strain = pr .
dx _ Change of dimension x x y
x Original dimension x
— ~= (a + a + a )(1  2p) ...(2.7)
= Strain in the xdi recti on = e
VE y
Equation (2.7) gives the volumetric strain. In this equation the stresses ax , a and ax are
v
Similarly, = Strain in ydirection = e all tensile. If any of the stresses is compressive, it may be regarded as negative, and the above
— = Strain in zdirection = e ~equation will hold good. If the value of is positive, it represents increase in volume whereas
z*
~C represents a decrease in volume.
the negative value of
0—
66 STRENGTH OF MATERIALS ELASTIC CONSTANTS
mm mm mmProblem 2.5. A metallic bar 300 kNis subjected to a force of 5 mmProblem 2.6. A metallic bar 250
x 100 x 40
kN kN(tensile), 6
(tensile) and 4 (tensile) along x, y and z directions respectively. Determine the mm mmx 100 /
x 50 is loaded as shown in
E 2 r N/mm2 /
change in the volume of the block. Take v.lO ' and Poisson’s ratio = 0.25.
Fig. 2.5.
Sol. Given : = 300 mm x 100 mm x 40 mm Find the change in volume. Take /
Dimensions of bar x = 300 mm, y = 100 mm and z = 40 mm
E=2x.l s N/mm2 and Poisson’s ratio = 0.25. H "r p
Volume, V = x xy x z = 300 x 100 x 40
Also find the change that should be mrrli ,
= 1200000 mm® MNmade in the 4 load, in order that there /
Load in the direction of x =5 kN = 5000 N should be no change in the volume of the » 2 MN
=6 kN = 6000 N
Load in the direction of y =4 kN = 4000 N bar. Fig. 2.5
Load in the direction of z
= 2 x 105 N/mm2 Sol. Given :
Value of £
mmLength, x = 250 mm, y = 100 and z = 50 mm
mmVolume, V = xyz = 250 x 100 x 50 = 1250000 3
Poisson’s ratio, it = 0.25 * 4 kN Load in ^direction = 400 kN = 400000 N (tensile)
Load in ydirection
Stress in the ^direction, Load in zdirection MN N= 2 = 2 x 10 6 (tensile)
MN N= 4 = 4 x 10® (compressive)
Load in x direction
Modulus of elasticity, E = 2 x 105 N/mm2
= 1.25 N/mm2 Poisson’s ratio, (x = 0.25.
100 x 40 Now o = Stress in ^direction
Similarly the stress in ydirection is given by, i
Load in ^direction Fig. 2.4 Load inxdirection
Area of crosssection
400000 400000 = 80 N/mm2„ (tension).
y x z 100 x 50
= 0.5 N/mm2 Similarly, Load in ydirection
And stress in zdirection 300x40
Load in zdirection
4000 •Ists> 160NW
2 _ 300 x 100
X0””
= 0.133 N/mm2
and „ ,
Using equation (2.9), we get
* 250 x 100
4r4<°* + o, + °,Xl2i)
= 160 N/mm2 (compression).
Using equation (2.7) and taking tensile stresses positive and compressive stresses nega
tive, we get
— V EdV =1 (, a + 0 + a,Xl  2„It),
^y *
2 x 10 5 (1.25 + 0.5 + 0.113X1  2 x 0.25) x
1.883 —dV 1
“ 2xl05 x2 V = 2 x 10 5r (80 + 160  160X1  2 x 0.25)
1.883 T_ — %= X 0.5 = 0.0002.
5
4 x 10® 2 x
10
—= 41x.8—18053s x 1200000 .'. Change in volume,
dV = 0.0002 x V
= 5.649 mm®. Ans. = 0.0002 x 1250000
= 250 mm®. Ans.
,
68 STRENGTH OF MATERIALS ELASTIC CONSTANTS 69
MNChange in the 4 load when there is no change in volume of bar
VE—Using equation (2.7), = (a + a.y, + a,)(l  2p) id2 x L + 6d2 x L  Id x L x 5d + d2 x 6L
dV + 8d2 x bL  2d x 6d x 6L)
—If there is no change in volume, then  = 0 = — (d2 x L  2d x L x td + d2 x bL)
(o + o + o )(l  2p) = 0. 4
2 Neglecting the products and higher powers of two small quantities.
x y
Change in volume, 8V = Final volume  Original volume
But for most of materials, the value of u lies between 0.25 and 0.33 and hence the term
(1  2(x) is never zero. ' = — {d2 xL2dxLxbd + d?x bL)  — d2 x L
44
ax + a + a = 0.
z
y
The stresses a and o are not to be changed. Only the stress corresponding to the load = — (d2 x bL2dxLx bd)
MN x y 4
4 (i.e. stress in zdirection) is to be changed.
 N/mm2 —,V,ol: umet, n.c
cr2 = ax  a =  80  160 =  240 (compressive) ,. = Chang2e in volume = 6V
y stram, e
_ Load Load Load Original volume V
But, o, = = or 240 =
Area xx y 250 x 100 M4— (d 2 x bL  2d x L x bd) xr
MNLoad = 240 x 250 x 100 = 6 x 106 N = 6 = „s ,
MNBut already a compressive load of 4 is acting. Ld (28)
Additional load that must be added *d 2 xL
4
MN= 6 MN  4 MN = 2 (compressive). Ans. — —&L &cl
where is the strain of length and is the strain of diameter.
Ld
2.6. VOLUMETRIC STRAIN OF A CYLINDRICAL ROD
Volumetric strain = Strain in length  Twice the strain of diameter.
Consider a cylindrical rod which is subjected to an axial tensile load P. A m mmProblem 2.7. steel rod 5 long and 30 in diameter is subjected to an axial tensile
Let d  diameter of the rod
load of50 kN. Determine the change in length, diameter and volume of the rod. Take E = 2 x 10s
L = length of the rod
Due to tensile load P, there will be an increase in the length of the rod, but the diameter N/mm2 and Poisson’s ratio = 0.25.
of the rod will decrease as shown in Fig. 2.6.
Sol. Given : L = 5 m = 5 x 103 mm
Length, d 30 mm
Diameter,
.. Volume, V= — d2 x L = — (30)2 x 5 x 103 = 35.343 x 10 5
Tensile load, 44
Value oiE P = 50 kN = 50 x 103
= 2 x 105 N/mm2
Poisson’s ratio,
Let p = 0.25
bd = Change in diameter
Fig. 2.6
bL = Change in length
Final length L + bL
8V = Change in volume
:. Final diameter = d  bd Now strain of length =
Now original volume of the rod,
L = — d2 x L —Load 1
x
4
Area E
Final volume = 74 (d  bd)2(L + 5L) _ 1 50
X~
j2 E
Ji
~= (d2 + 8 d2  2d x 6d)(L + 6L)
. .,
70 STRENGTH OF MATERIALS ELASTIC CONSTANTS 71
0.4 X 50 X IQ 3 E = Young’s modulus of the material of the cube 0 F
= 0.0003536 E
a = Tensile stress acting on the faces
k x 30 2 x 2 x 10 s (i = Poisson’s ratio. ySV /jy
But strain of length = Then volume of cube, V = L3 B
Now let us consider the strain of one of the sides of
1
the cube (say AB) under the action of the three mutually
—6L a1
perpendicular stresses. This side will suffer the following
L = 0.0003536 three strains : //Hi G
6L = 0.0003536 x 5 x 103 !
AEHD1. Strain of AB due to stresses on the faces
= 1.768 mm. Ans. 4C
and BFGC. This strain is tensile and is equal to
~ Lateral strain Fig. 2.7
Poisson s ratio =
Longitudinal strain
Lateral strain = Poisson’s ratio x Longitudinal strain 2. Strain ofAB due to stresses on the faces AEFB andDHGC. This is compressive lateral
: 0.25 x 0.0003536 v Longitudinal strain =  —a
: 0.0000884 &strain and is equal to  p
3. Strain of AB due to stresses on the faces ABCD and EFGH. This is also compressive
Lateral strain =
—a
= 0.0000884
a lateral strain and is equal to  p
sl
ABHence the total strain of is given by
bd  0.0000884 x d dL a a pxja = or a 2p)
mm= 0.0000884 x 30 = 0.002652 LE
Now using equation (2.8), we get Now original volume of cube, V L'3
n 2M, If dL is the change in length, then dV is the change in volume.
—Volumetric
. 57 = ffi Differentiating equation (ii), with respect to L,
VL d
strain, dV = 3L2 x dL
= 0.0003536  2 x 0.0000884 = 0.0001768 Dividing equation (Hi) by equation (ii), we get
6V = V x 0.0001768
= 35.343 x 105 x 0.0001768 dV 3L2 x dL 3dL
mm= 624.86 3 Alls. V ~ I? L
.
2.7. BULK MODULUS —Substituting the value of  from equation ft), in the above equation, we get
When a body is subjected to the mutually perpendicular like and equal direct stresses, JU
the ratio of direct stress to the corresponding volumetric strain is found to be constant for a —dV =30an2po)^
given material when the deformation is within a certain limit. This ratio is known as bulk From equation (2.9), bulk modulus is given by , —fv dV 3ct
modulus and is usually denoted by K. Mathematically bulk modulus is given by
LVE
K_ Direct stress a ao
Volumetric strain ( dV \ —(2.9) K..= WT3o"
E (1  21 ) ^J
(
kVJ
2.8. EXPRESSION FOR YOUNG’S MODULUS IN TERMS OF BULK MODULUS E ...(2.10)
ABCDEFGHFig. 2.7 shows a cube ~
3(1  2p)
which is subjected to three mutually perpendicu E = 3K (1  2p) 2 11...( .
lar tensile stresses of equal intensity. )
Let L  Length of cube KEFrom equation (2.11), the expression for Poisson’s ratio (p) is obtained as p = 3
dL = Change in length of the cube
72 STRENGTH OF MATERIALS ELASTIC CONSTANTS 73
f 0.004') 0.000133 0.266. Ans.
0.0005
v 30 )
0.0005
(lit) Bulk modulus (K)
Using equation (2.10), we get
E _1.6975 x 10 s
= 3(1 — 2jx) “ 3(1  0.266 x 2)
= 1.209 x 10 s N/mm2. Ans.
2.9. PRINCIPLE OF COMPLEMENTARY SHEAR STRESSES
It states that a set of shear stresses across a plane is al
ways accompanied by a set of balancing shear stresses (i.e., of
the same intensity) across the plane and normal to it.
Proof. Fig. 2.8 shows a rectangular block ABCD, sub
ABjected to a set of shear stresses of intensity x on the faces
and CD. Let the thickness of the block normal to the plane of
the paper is unity. Fig 2 8
The force acting on face AB
= Stress x Area
= x x AB x 1 = t.AB
Similarly force acting on face CD
= x x CD x 1 = x.CD CD = AB)
xAB
The forces acting on the faces AB and CD are equal and opposite and hence these forces
will form a couple.
The moment of this couple = Force x Perpendicular distance — (*)
= x.AB x AD
If the block is in equilibrium, there must be a restoring couple whose moment must be
equal to the moment given by equation (i). Let the shear stress of intensity t is set up on the
faces AD and CB.
ADAD ADThe force acting on face
 1 x x 1 =1
x x
ADThe force acting on face BC = x' x BC x 1 = x'BC = x' (' BC  AD)
ADAs the force acting on faces and BC are equal and opposite, these forces also forms a
couple.
ADMoment of this couple = Force x Distance = x' x AB ••(«)
For the equilibrium of the block, the moments of couples given by equations (t) and (u)
should be equal
ADx.AB x AD = x' ABx or x = x'.
The above equation proves that a set of shear stresses is always accompanied by a trans
verse set of shear stresses of the same intensity.
The stress x' is known as complementary shear and the two stresses (t and x ) at right
angles together constitute a state of simple shear. The direction of the shear stresses on the
block are either both towards or both away from a comer.
,
STRENGTH OF MATERIALS ELASTIC CONSTANTS 75
BDIn Fig. 2.8, as a result of two couples, formed by the shear forces, the diagonal will be o = Normal stress on plane CE
ti
subjected to tension and the diagonal AB will be subjected to compression.
o = Tangential stress on plane CE
2.10, STRESSES ON INCLINED SECTIONS WHEN THE ELEMENT IS SUBJECTED (
Normal force on plane CE
TO SIMPLE SHEAR STRESSES
" Area of section CE
Pn _ x x BC x sin 0 + t x EB x cos 0
~ CEx l
Fig. 2.9 shows a rectangular block ABCD which is in a CE x 1
state of simple shear and hence subjected to a set of shear d *• C BC EB
/
stresses of intensity x on the faces AB, CD and the faces AjD = xx p— x . 0 + xx pz x cos 0
.
and CB. Let the thickness of the block normal to the plane of La sin La
= t x cos 0 x sin 9 + x x sin 0 x cos 0
the paper is unity. XT / r v. In triangle EBC,  : cos 0 and EE \
= 2x cos 0 x sin 0 = x sin 20 = sin 0
It is required to find the normal and tangential stresses I / ’ CE )
across an inclined plane CE, which is having inclination 0 with
the face CB. A / = ......((22..112))
Consider the equilibrium of the triangular piece CEB of — B an a _ Tangential force on plane CE
thickness unity. The forces acting on triangular piece CEB are ‘ Area of plane CE
T
shown in Fig. 2.10 and they are C P _ x x BC x cos 0  x x EB x sin 0
: t
/1
(i) Shear force on face CB, P„ 4.~ /$ ~ CEx 1~ CE
= Shear stress x area of face CB X\ // — —BC EB '
= x x BC x 1 /7 x x bc  Q, =xx CE x cos 0  x x  .
= x x BC acting along CB
/ , CE x sin 0
xx cos 0 x cos 0  x x sin 0 x sin 0
(ii) Shear force on face EB, / = x cos2 0  x sin2 0
Q2 = Shear stress x area of face EB eZ b = x[cos z 0  sin2 0] = x cos 20 ...(2.13)
stx EB x 1 = x x EB acting along EB
>. For the planes carrying the maximum normal stress, an should be maximum. But from
equation (2.12) it is clear that an will be maximum when sin 20 = ± 1
t x EB = Q2
(Hi) A force P normal to the plane EC Fig. 2.10
n
A EC(iv) force P tangential to the plane i.e., 20 =± —Jt
t
The force Q, is acting along the face CB as shown in Fig. 2.11. This force is resolved into
two components, i.e., cos 0 and Q 1 sin 0 along the plane CE and normal to the plane CE or 0 = ± — which means 0 = 45° or  45°
respectively. 4
The force Q2 is acting along the face EB. This force is also resolved into two components, when0 = 45°, then from equation (2.12), we have
Q ECQi.e.,
sin 0 and cos 0 along the plane EC and normal to the plane respectively. a = x sin 90° = x D
n
2 2
For equilibrium, the net force normal to the plane CE c when 0 =  45°, then a =  x
n
should be zero. cP^/N. (Positive sign shows the normal stress is tensile
\o^<2 <.
P Q.'. n  Qj sin 0  2 cos 0 = 0 whereas negative sign shows the normal stress is
*n.
Por n = Q x sin 0 + Q2 cos 0 /\/' X?® compressive).
= x x BC x sin 0 + r x EB x cos 0 y' When 0 = * 45°, then from equation (2.13), we find
(V Q, = x x BC and Q2 = x x EB) "Q, that
B
• £04V\U{9Q—0) Q> o = x cos 2 x 45° A
t
For equilibrium, the net force along the plane CE should X® = x cos 90° = 0 Fig. 2.12
be zero.
P cos 0 + Q2 sin 0 = 0 C>\ This shows that the planes, which carry the maxi
t
mum normal stresses, are having zero shear stresses.
or P  Q cos Q Q2 sin 0
t 1
( ve sign is taken due to opposite direction) s Now from equation 2.13, it is clear that shear stress
will be maximum when cos 26 = ± 1,
= x x BC x cos 0  x x EB x sin 8 Fig. 2.11
ie., 20 = 0° or 180° or 0 = 0° or 90°
ELASTIC CONSTANTS
When 0 = 0° or 90°, the value of on from equation (2.12), is zero. Similarly it can be proved that on the plane BD, a direct compressive stress of magni
This shows that the planes, which carry the maximum shear stresses, are having zero BDtude x is acting. This compressive stress is perpendicular to the plane
normal stresses. These planes are known as planes of simple shear. or this compressive
Important points. When an element is subjected to a set of shear stresses, then : ACstress is along the diagonal AC. Hence the diagonal
(t) The planes of maximum normal stresses are perpendicular to each other. is subjected to compressive stress of
(U) The planes of maximum normal stresses are inclined at an angle of 45° to the planes magnitude x. The pure direct tensile and compressive stresses active on the diagonal planes
of pure shear. AC BDand are called diagonal tensile and diagonal compressive stresses. The stress on the
(Hi) One of the maximum normal stress is tensile while the other maximum normal stress AC BDdiagonal plane
(i.e., along diagonal BD) is tensile whereas on the diagonal plane ie’’
is compressive. ACalong the diagonal
is compressive.
(iv) The maximum normal stresses are of the same magnitude and are equal to the in
Hence the set of shear stresses x on the faces AB, CD and the faces AD and CB are
tensity of shear stress on the plane of pure shear. ACequivalent to a compressive stress x along the diagonal
and a tensile stress x along the
diagonal BD.
2.12. DIRECT (TENSILE AND COMPRESSIVE) STRAINS OF THE DIAGONALS
2.1 1. DIAGONAL STRESSES PRODUCED BY SIMPLE SHEAR ON A SQUARE BLOCK In Art. 2.11, we have proved that when a square block ABCD of unit thickness is sub
CD ADjected to a set of shear stresses of intensity q on the faces AB,
ABCDFig. 2.13 shows a square block of each side equal to ‘a’ and subjected to a set of BDdiagonal and the faces and CB, the
shear stresses of intensity x on the faces AB, CD and faces AD and CB. Let the thickness of the will experience a tensile stress of magnitude q D t CC
ACwhereas the diagonal
will experience a compressive stress of r’   ° j
,
block normal to the plane of the paper is unity. \~magnitude q. Due to these stresses the diagonalBD will be elon \
’,'\j
gated whereas the diagonal AC will be shorted. Let us consider \E 'v x. \
the joint effect of these two stresses on the diagonal BD. \\ \ \
Due to the tensile stress q along diagonal BD, there will \ 'n\. \
be a tensile strain in diagonalBD. Due to the compressive stress \6 \
q along the diagonal AC, there will be a tensile strain in the f 'v
BDdiagonal
due to lateral strain.* \ —
Let x = Poisson’s ratio A Tg
E = Young"s modulus for the material of the block
BDNow tensile strain in diagonal due to tensile stress x Fi8 214
along BD
_ Tensile stress along BD _ x
E ~~E
BD ACTensile strain in diagonal
Fig. 2.13 due to compressive stress x along
The normal stress (on) on plane AC is given by equation (2.12) as
a = x sin 20 ••(»)
n
But as shown in Fig. 2.13 (6) the angle made by plane AC with face BC is given by, BDTotal tensile strain along diagonal
tan 0 = = — ABCD[v is a square of side ‘a’] = T + JX X T = X <1 + ^ ...(2.14)
BC ci B ~B~ £
=i ACSimilarly it can be proved that the total strain in the diagonal will be compressive
e = 45° and will be given by
Substituting this value of 0 in equation (i), we get Total compressive strain in diagonal AC
and a = x x sin 2 x 45° = tx sin 90° = x ^E (1 + ‘
ft
ao = x x cos 20 = x x cos 2 x 45° BDThe total tensile strain in the diagonal is equal to half the shear strain. This is
t
= t x cos 90° = 0
Hence on the plane AC, a direct tensile stress of magnitude x is acting. This tensile proved as given below :
BDstress is parallel to the diagonal BD. Hence the diagonal is subjected to tensile stress of
magnitude x. Please refer to Art. 2.4, in which it is proved that every strain in the direction of load is accom
panied by lateral strain of the opposite kind perpendicular to the direction of load.
78 STRENGTH OF MATERIALS ELASTIC CONSTANTS
Due to the shear stresses acting on the faces, the square block ABCD will be deformed to diagonal strain due to shear stress x is given by equation (2.14) as
position ABCjDj as shown in Fig. 2.14. T
= (1 + q)
Now increase in the length of diagonal BD = BD  BD BD —Total tensile strain along diagonal
l
hi
BD.. Tensile strain in the diagonal From equation (2.15) also we have total tensile strain in diagonal BD
Increase in length BD BD ^ — shear strain = 1 x Shear stress ( Shear stress modulus of rigidity = C
1
Original length “ BD AA Shear strain
From D, draw a perpendicular DE on BDV 1 x
X
We know that the distortion DD is very small and hence angle DBD will be very small. ~ (v Shear stress = t)
1 y
2C
Hence we can take
Equating the two tensile strain along diagonal BD, we get
BD = BE
It
and ACDB = ACJOjB = 45° t d +
g)=*£
Now in triangle DD E, lDD E = 45° •
Xx
Length D E = DD cos (DD^E) (Cancelling t to both sides)
XX
...(2.16)
DD
DD 45° X E = 2C (1 + n)
= X cos ~
^2
In triangle ABD, BD = JAB 2 + AD 2 for C = ....(2.17)
2 (1 + g)
= VAD 2 + AD2 = yj2 x AD AB(*•* = AD) Problem 2.10. Determine the Poisson’s ratio and bulk modulus of a material, for which
[v BD = BE]
N/mm N/mmYoung’s modulus is 1.2 x 10s 2
2 and modulus of rigidity is 4.8 x 104
.
Now from equation (i), we have Sol. Given :
Tensile strain in diagonal Young’s modulus, E = 1.2 x 10 5 N/mm2
BD  BD Modulus of rigidity, C = 4.8 x 104 N/mm2
bd X
Let the Poisson’s ratio = p
^bd~ Using equation (2.16), we get
BD  BE E = 2C (1 + p)
l
BD or 1.2 x 10s = 2 x 4.8 x 104 (1 + p)
[v BD  BE = D~E] 1.2 x 10 5
1
or , = 2 x 4.8 x 10 t4 = 1.25 or u = 1.25  1.0 = 0.25. Ans.
(1 + li)
Iti DYE = ^and BD = 72 x AD Bulk modulus is given by equation (2.10) as
yj2 x AD J2 ^^ ~ E 1,2 x10 s
1 DD 1 DD 3(l2p) 3(1 0.25 x 2) .
(( v' gtx ~=°0.25)
X 1~ l
V2 x j2 AD 2 AD = 8 x 104 N/mm2. Ans.
= — Shear strain* •: Shear strain =  mm mmAProblem 2.11. bar of crosssection 8 x 8 is subjected to an axial pull of
mm7000 N. The lateral dimension of the bar is found to be changed to 7.9985
At ...(2.15) x 7.9985 mm. If
N/mmthe modulus of rigidity of the material is 0.8 x 10s 2 determine the Poisson’s ratio and
,
2.13. RELATIONSHIP BETWEEN MODULUS OF ELASTICITY AND MODULUS OF modulus of elasticity.
RIGIDITY Sol. Given :
We have seen in the last article that when a square block of unit thickness is subjected mmArea of section = 8 x 8 = 64 2
to a set of shear stresses of magnitude t on the faces AB, CD and the faces AD and CB, then the
Axial pull, P  7000 N
mm mmLateral dimensions = 7.9985
x 7.9985
*Please refer to Art. 1.4.3, for shear strain. Volume of C = 0,8 x 10s N/mm2
80 STRENGTH OF MATERIALS ELASTIC CONSTANTS 81
Let (i = Poisson’s ratio and Longitudinal strain = 4 x Lateral strain (v (i = 0.25)
E = Modulus of elasticity. Hydrostatic pressure, p = 100 N/mm2
— — —Now Lateral strain — =1 ,,
C—hange i—n lateral dimension
, strain = . U.Zi)
> Longitudinal strain 4
lateral .,
Original lateral dimension
— —= 8  7.9985 = 0.0015 == 0.0001875. Poisson’s ratio, 11 = 0.25
oo Let
C = Modulus of rigidity
To find the value of Poisson’s ratio, we must know the value of longitudinal strain. But
K = Bulk modulus
in this problem, the length of bar and the axial extension is not given. Hence longitudinal
E = Young’s modulus = 1 x 105 N/mm2
strain cannot be calculated. But axial stress can be calculated. Then longitudinal, strain will Using equation (2.16), we get
be equal to axial stress divided by E. E = 2Ca + p)
~:. Axial stress, a = —
= 77 = 109.375 N/mm2 and longitudinal strain = 1 x 10 s = 2C(1 + 0.25)
EArea 64
But lateral strain = p x longitudinal strain : *X = 4 x 104 N/mm2. Ans.
0.0001875 = p X 109.375 C=
2 x L25
E
E 109.375 ('.' Lateral strain = 0.0001875) For bulk modulus, using equation (2.11), we get
7 = 0^001875 = 583333 33 E = 3K(1 2p)
E = 583333.33p 1 x 105 = 3K(1  2 x 0.25)
Using equation (2.17), we get K= 1 = 0.667 x 10 5 N/mm2. Ans.
3 x 0.5
Now using equation (2.9), we get
77c = **or Jf= 1 + ,) K_ P =
2( i '
Volumetric strain ( dV \
= 2 x 0.8 X 105 (1 + p) C(v = 0.8 x 10s )
(v E = 583333.33p)
583333.33(1 = 2 x 0.8 x 10 5 (1 + p) where p = 100 N/mm2
1 + p= 583333.33p = 3.6458(1 0.667 x 105 =
qr
2 x 0.8 x 10s
l = 3.6458u i = 2.6458ii
Poisson’s ratio = p = z„.b45o = 0.378. Ans. dV = 100 =1 5xl0‘3
Modulus of elasticity (E) is obtained by substituting the value of (i in equation (i). 
IT 76^71^
E = 583333.33(1
VdV = X 1.5 x 10® = 1060287.52 x 1.5 x lO'3
—583333.33 mm= 1590.43 3 Ans.
E = = 2.2047 x 105 N/mm2. Ans. .
2.6458
Problem 2.12. Calculate the modulus of rigidity and bulk modulus of a cylindrical bar HIGHLIGHTS
mm mofdiameter 30
and of length 1.5 if the longitudinal strain in a bar during a tensile stress
is four times the lateral strain. Find the change in volume, when the bar is subjected to a 1. Poisson’s ratio is the ratio of lateral strain to longitudinal strain. It is generally denoted by (t.
Ehydrostatic pressure of 100 N/mm2. Take  1 x 105 N/mm2. 2. The tensile longitudinal stress produces compressive lateral strains.
Sol. Given : 8l .
Dia. of bar, d = 30 mm and
Length of bar, m mmL = 1.5 = 1.5 x 1000 = 1500 3. If a load acts in the direction of length of a rectangular bar, then longitudinal strain =
^
Lateral strain = — —6b 5d
or
ba
.'. Volume of bar, V= 2 xZ,= x 30x 1500
7 <i 7 where 8 1 = Change in length,
44
mm= 1060287.52 3 6b  Change in width,
5d = Change in depth.
.
82 STRENGTH OF MATERIALS
ELASTIC CONSTANTS 83
4. The ratio of change in volume to original volume is known as volumetric strain. 5. (a) Derive an expression for volumetric strain for a rectangular bar which is subjected to three
5. Volumetric strain (e^) for a rectangular bar subjected to an axial load P, is given by mutually perpendicular tensile stresses.
e„=(l2p). (6) A test element is subjected to three mutually perpendicular unequal stresses. Find the change
in volume of the element, if the aljebraic sum of these stresses is equal to zero.
6. Volumetric strain for a rectangular bar subjected to three mutually perpendicular stresses is
6. Explain briefly the term ‘shear stress’ and ‘complimentary stress’ with proper illustrations.
given by. g (ax + a + o )(l  2r)
s 7. State the principle of shear stress.
y
108.. What do you understand by ‘An element in a state of simple shear’ ?
where a and a, are stresses in x, y and z direction respectively. 9. When an element is in a state of simple shear then prove that the planes of maximum normal
x,
stresses are perpendicular to each other and these planes are inclined at an angle of 45° to the
7. Principle of complementary shear stresses states that a set of shear stresses across a plane is planes of pure shear.
always accompanied by a set of balancing shear stresses (i.e. of the same intensity) across the Derive an expression between modulus of elasticity and modulus of rigidity.
,
plane and normal to it.
8. Volumetric strain of a cylindrical rod, subjected to an axial tensile load is given by, (B) Numerical Problems
= Longitudinal strain  2 x strain of diameter
m mm1. Determine the changes in length, breadth and thickness of a steel bar which is 5 long, 40
_U mmwide and 30
g 5d kNthick and is subjected to an axial pull of 35 in the direction of its length.
~T~ d'
ETake = 2 x 10 5 N/mm2 and Poisson’s ratio = 0.32.
K9. Bulk modulus is given by, [Ans. 0.0729 cm, 0.000186 cm, 0.000139 cm]
2. For the above problem, determine the volumetric strain and the final volume of the given steel
bar. mm[Ans. 0.0000525, 6000317 3
]
10. The relation between Young’s modulus and bulk modulus is given by, 3. Determine the value of Young’s modulus and Poisson’s ratio of a metallic bar of length 25 cm,
breadth 3 cm and depth 2 cm when the bar is subjected to an axial compressive load of 240 kN.
E = 3K (1  2p). The decrease in length is given as 0.05 cm and increase in breadth is 0.002.
1 1. When an element is subjected to simple shear stresses then : [Ans. 2 x 10s N/mm 2 and 0.33]
(i) The planes of maximum normal stresses are perpendicular to each other.
(it) The planes of maximum normal stresses are inclined at an angle of 45° to the plane of pure mm mm mm4. A steel bar 320
long, 40 wide and 30 thick is subjected to a pull of 250 kN in the
shear.
mdirection of its length. Determine the change in volume. Take E = 2 x 10s N/mm2 and = 4.
(Hi) One of the maximum normal stress is tensile while the other maximum normal stress is
mm[Ans. 200 3
compressive. ]
(iti) The maximum normal stresses are of the same magnitude and are equal to the shear stress mm mm mm5. A metallic bar 250 kNis subjected to a force of 20
x 80 x 30 kN(tensile), 30 (tensile)
on the plane of pure shear.
and 15 kN (tensile) along x, y and z directions respectively. Determine the change in the volume
12. The relation between modulus of elasticity and modulus of rigidity is given by
Eof the block. Take = 2 x 105 N/mm2 and Poisson’s ratio = 0.25. mm[Ans. 19.62 3
C=—E = 2C(l + p) or
]
mm mm mmA6. metallic bar 300
x 120 x 50 is loaded as shown in Fig. 2.15.
Find the change in volume. Take E = 2 x 105 N/mm2 and Poisson’s ratio = 0.30.
EXERCISE 2
(A) Theoretical Questions
1. Define and explain the terms : Longitudinal strain, lateral strain and Poisson’s ratio. —/ 300 mm
2. Prove that the volumetric strain of a cylindrical rod which is subjected to an axial tensile load is 2.5 kN
equal to strain in the length minus twice the strain of diameter.
Fig. 2.15
3. What is a bulk modulus ? Derive an expression for Young’s modulus in terms of bulk modulus
and Poisson’s ratio.
4. Define volumetric strain. Prove that the volumetric strain for a rectangular bar subjected to an MNAlso find the change that should be made in 4.5 load, in order that there should be no change
axialload P in the direction of its length is given by
61 in the volume of the bar. [Ans. 450 min2 , 4.5 MN]
Tev = (1  2n) A m mmsteel rod 4 long and 20 diameter is subjected to an axial tensile load of 40 kN. Deter
~where p = Poisson's ratio and = Longitudinal strain. mine the change in length, diameter and volume of the rod. TakeE = 2 x 10 s N/mm2 and Poisson’s
ratio = 0.25. mm[Ans. 2.5464, 0.05092, 5598 3
]
STRENGTH OR MATERIALS
84
9. 10 s N/mm2 and Poisson’s ratio 0.28. Calculate
8 For a material, Young’s modulus is given as 1.4 x [Ana. 1.06 x 10* N/mm2]
the bulk modulus.
A mmbar of 20 diameter subjected to a pull of 50 kN. The measured extension on gauge length of
mm mm250 is 0.12 and change in diameter is 0.00375 mm. Calculate :
(i) Young’s modulus (»i> Poisson’s ratio and (Hi) Bulk modulus.
N/mm N/mm[Ans. (i ) 1.989 x 10 6 2
2 (it) 0.234, (Hi) 1.2465 x 10 s
]
,
10. Determine the Poisson’s ratio and bulk modulus of a material, for which Young’s modulus is Principal Stresses and Strains
[Ans. 0.33, 1.2 x 105 N/mm ]
1.2 x 105 N/mm2 and modulus of rigidity is 4.5 x 104 N/mm2
.
mm mma11. bar of crosssection 10 is subjected to an axial pull of 8000 N. The lateral dimen
x 10
mmsion of the bar is found to be changed to 9.9985 x 9.9985 mm. If the modulus of rigidity of the
material is 0.8 x 105 N/mm2 determine the Poisson’s ratio and modulus of elasticity. 3.1. INTRODUCTION
,
In chapter 2, the concept and definition of stress, strain, types of stresses (i.e., tensile,
[Ans. 0.45, 2.4 x 105 N/mm 2 compressive and simple shear) and types of strain (i.e., tensile, compressive, shear and volu
metric strains etc.) are discussed. These stresses were acting in a plane, which was at right
] angles to the line of action of the force. In many engineering problems both direct (tensile or
compressive stress) and shear stresses are acting at the same time. In such situation the re
mm12. Calculate the modulus of rigidity and bulk modulus of a cylindrical bar of diameter of 25 sultant stress across any section will be neither normal nor tangential to the plane. In this
chapter the stresses, acting on an inclined plane (or oblique section) will be analysed.
and of length 1.6 m, if the longitudinal strain in a bar during a tensile test is four times the
lateral strain. Find the change in volume,' when the bar is subjected to a hydrostatic pressure
of 100 N/mm2. Take E = 1 x 105 N/mm2 .
mm[Ans.
4 x 104 N/mm2 0.667 x 10 s N/mm2 1178 3
, , ]
mm13. A bar 30 in diameter was subjected to tensile load of 54 kN and the measured extension on
mm mm300 gauge length was 0.112 and change in diameter was 0.00366 mm. Calculate Poisson’s 3.2. PRINCIPAL PLANES AND PRINCIPAL STRESSES
ratio and values of three modulii. The planes, which have no shear stress, are known as principal planes. Hence principal
K[Ans. = 196 kN/mm 3
= 0.326, E ~ 204.6 kN/mm2 C = 77.2 kN/mm2 planes are the planes of zero shear stress. These planes carry only normal stresses.
[x , , The normal stresses, acting on a principal plane, are known as principal stresses.
14. Derive the relation between E and C. Using the derived relationship, estimate the Young’s modulus
(E) when the modulus of rigidity (C) is 0.80 x 1G5 N/mm2 and the Poisson’s ratio is 0.3.
E[Hint. = 2C (1 + p> = 2 x 0.80 x 10s (1 + 0.3) = 2.08 x 105 N/mm2 .]
3.3. METHODS FOR DETERMINING STRESSES ON OBLIQUE SECTION
The stresses on oblique section are determined by the following methods :
1. Analytical method, and 2. Graphical method.
3.4. ANALYTICAL METHOD FOR DETERMINING STRESSES ON OBLIQUE SECTION
The following two cases will be considered :
A1. member subjected to a direct stress in one plane.
2. The member is subjected to like direct stresses in two mutually perpendicular direc
tions.
A3.4.1. Member Subjected to a Direct Stress in one Plane. Fig. 3.1 (a) shows a
rectangular member of uniform crosssectional area A and of unit thickness.
Let P = Axial force acting on the member.
A = Area of crosssection, which is perpendicular to the line of action of the force P.
The stress al, ong . a= —P
xaxis,
Hence, the member is subjected to a stress along xaxis.
Consider a crosssection EF which is perpendicular to the line of action of the force P.
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