airwalkbooks.comairwalkbooks.com/images/pdf/pdf_105_1.pdf · 2019-01-31 · Dr. S. Ramachandran, M.E., Ph.D., Professor - Mech Sathyabama Institute of Science and Technology Chennai

Dr. S. Ramachandran, M.E., Ph.D.,

Professor - Mech

Sathyabama Institute of Science and TechnologyChennai - 119

S. KumaranAirwalk Publications

(Near All India Radio)

80, Karneeshwarar Koil Street,

Mylapore, Chennai – 600 004.

Ph.: 2466 1909, 94440 81904Email: [email protected],

[email protected]

www.airwalkbooks.com, www.srbooks.org

As Per Lastest Syllabus of Anna University

FOR B.E IV SEMESTER MECHANICAL ENGINEERING STUDENTSFOR B.E IV SEMESTER MECHANICAL ENGINEERING STUDENTS

R - 2017

Strength of Materials for Mechanical Engineers

ISBN:978-93-88084-16-1

thFirst Edition : 16 , November 2018

450/-

978-93-88084-16-1

www.srbooks.orgwww.airwalkbooks.com

Cell: 9600003081, 9600003082

`

R 2017 ANNA UNIVERSITY SYLLABUS

CE 8395 STRENGTH OF MATERIALS FOR MECHANICALENGINEERS

UNIT I: STRESS, STRAIN AND DEFORMATION OF SOLIDS 9

Rigid bodies and deformable solids – Tension, Compression and ShearStresses – Deformation ofsimple and compound bars – Thermal stresses –Elastic constants – Volumetric strains – Stresseson inclined planes – principalstresses and principal planes – Mohr’s circle of stress.

UNIT II: TRANSVERSE LOADING ON BEAMS AND STRESSES INBEAM 9

Beams – types transverse loading on beams – Shear force and bendingmoment in beams – Cantilevers – Simply supported beams and over – hangingbeams. Theory of simple bending – bending stress distribution – Load carryingcapacity – Proportioning of sections – Flitched beams – Shear stressdistribution.

UNIT III: TORSION 9

Torsion formulation stresses and deformation in circular and hollowsshafts – Stepped shafts – Deflection in shafts fixed at the both ends – Stressesin helical springs – Deflection of helical springs, carriage springs.

UNIT IV: DEFLECTION OF BEAMS 9

Double Integration method – Macaulay’s method – Area momentmethod for computation of slopes and deflections in beams - Conjugate beamand strain energy – Maxwell’s reciprocal theorems.

UNIT V: THIN CYLINDERS, SPHERES AND THICK CYLINDERS9

Stresses in thin cylindrical shell due to internal pressure circumferentialand longitudinal stressesand deformation in thin and thick cylinders – sphericalshells subjected to internal pressure – Deformation in spherical shells –Lame’s theorem.

TOTAL: 45 PERIODS

Table of Contents

Unit I: Stress, Strain and Deformation of Solids

1.1 Introduction to Strength of Materials . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

1.2 Rigid and Deformable Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1

1.3 Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2

1.4 Unit of Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4

1.5 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

1.6 Types of Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

1.7 Normal Stress: Axially Loaded Bar . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5

1.8 Tensile Stress and Tensile Strain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6

1.9 Compressive Stress and Compressive Strain . . . . . . . . . . . . . . . . . . . 1.7

1.10 Shear Stress and Shear Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.8

1.11 Bearing Stress (Crushing Stress) in Connections. . . . . . . . . . . . . . 1.10

1.12 Stress-strain Behaviour of Materials . . . . . . . . . . . . . . . . . . . . . . . . 1.10

1.12.1 Stress Strain Curves (Tension) . . . . . . . . . . . . . . . . . . . . . . 1.11

1.12.2 Stress - Strain Curve for Ductile Materials . . . . . . . . . . . 1.11

1.12.3 Stress Strain Curves for Brittle Materials . . . . . . . . . . . . . 1.14

1.13 Stress Strain Curves (Compression). . . . . . . . . . . . . . . . . . . . . . . . . 1.15

1.14 Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.15

1.14.1 Factor of Safety . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.15

1.15 Deformation of a Body Under Axial Load at Free End . . . . . . . 1.16

1.16 Stiffness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.17

1.16.1 Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.18

1.17 Deformation of a Bar Under Axial Load at Different Sections . 1.29

1.17.1 Deformation in Simple Bar Subjected to Axial Load. . . . 1.31

1.18 Deformation for a Bar of Varying Section . . . . . . . . . . . . . . . . . . 1.32

1.19 Deformation of a Body Due to Self Weight . . . . . . . . . . . . . . . . . 1.41

1.20 Principle of Superposition. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.43

1.21 Stress in Bars of Uniformly Tapering Cross Section . . . . . . . . . . 1.52

1.22 Deformation of Uniformly Tapering Rectangular Bar. . . . . . . . . . 1.55

1.23 Deformation in Compound or Composite Bars . . . . . . . . . . . . . . . 1.57

Contents C.1

1.24 Bar of Uniform Strength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.74

1.25 Thermal Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.76

1.25.1 Thermal Stresses in Simple Bars . . . . . . . . . . . . . . . . . . . . 1.76

1.25.2 Thermal Stresses in Composite Bars . . . . . . . . . . . . . . . . . 1.79

1.25.3 Thermal Stress in Taper Bar - Circular Section . . . . . . . 1.85

1.25.4 Thermal Stress in Varying Section Bar . . . . . . . . . . . . . . . 1.87

1.26 Elastic Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.93

1.26.1 Modulus of Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.93

1.26.2 Rigidity Modulus (or) Shear Modulus . . . . . . . . . . . . . . . . 1.93

1.26.3 Bulk Modulus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.93

1.26.4 Linear Strain and Lateral Strain . . . . . . . . . . . . . . . . . . . . 1.93

1.27 Poisson’s Ratio. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.94

1.28 Volumetric Strain. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.96

1.28.1 Rectangular Body Subjected to Axial Loading . . . . . . . . . 1.97

1.29 Rectangular Bar Subjected to 3 Mutually Perpendicular Forces . 1.99

1.30 Cylindrical Rod Subjected to Axial Load . . . . . . . . . . . . . . . . . . 1.103

1.31 Shear Stress and Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.104

1.31.1 Shear Modulus or Modulus of Rigidity . . . . . . . . . . . . . . 1.105

1.32 Bulk Modulus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.106

1.33 Relationship Between Elastic Constants . . . . . . . . . . . . . . . . . . . . 1.106

1.33.1 Relation between Bulk Modulus and Young’s Modulus . 1.106

1.33.2 Relation between Modulus of Elasticity and Modulus ofRigidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.107

1.34 Principal Stresses and Principal Planes . . . . . . . . . . . . . . . . . . . . . 1.118

1.35 Analysis of Plane Stresses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.119

1.35.1 Stress on a inclined plane (A member subjected to anaxial load) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.119

1.35.2 Biaxial state of stress - Member subjected to biaxial stress 1.122

1.35.3 A member subjected to a simple shear stress. . . . . . . . . 1.129

1.35.4 Member subjected to a simple shear and a biaxial stress 1.131

C.2 Strength of Materials for Mechanical Eng., - www.airwalkbooks.com

1.35.5 A member subjected direct stress in one plane and asimple shear stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.150

1.36 Mohr’s Circle for Biaxial Stresses . . . . . . . . . . . . . . . . . . . . . . . . 1.154

1.36.1 A body subjected to a biaxial perpendicular unequaland like principal stresses. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.154

1.36.2 A body subjected to a biaxial perpendicular unequaland unlike principal stress. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.159

1.36.3 A body subjected to a Biaxial perpendicular unequal,like stresses with an simple shear stress. . . . . . . . . . . . . . . . . . . 1.162

UNIT II: Transverse Loading on Beams and Stresses in Beam

2.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

2.2 Types of Beams. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1

(i) Simply supported Beam:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2

(ii) Cantilever Beam: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2

(iii) Overhanging Beam: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2

(iv) Fixed Beam:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3

(v) Continuous Beam:. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3

(vi) Propped Cantilever Beam: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4

2.3 Supports and Support Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4

2.4 Types of Supports and Their Reactions . . . . . . . . . . . . . . . . . . . . . . . 2.5

(i) Simple support or Knife Edge support . . . . . . . . . . . . . . . . . . . . 2.5

(ii) Roller Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5

(iii) Hinge or Pin-Jointed support . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6

(iv) Fixed or Built - in support . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7

(v) Smooth surface support or Frictionless support . . . . . . . . . . . . 2.7

2.5 Static Equilibrium Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.8

2.6 Determinate And Indeterminate Beams . . . . . . . . . . . . . . . . . . . . . . . . 2.9

2.7 Types of Loading in Beams. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9

(a) Point or Concentrated load . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.9

(b) Uniformly Distributed Load (UDL) . . . . . . . . . . . . . . . . . . . . . 2.10

(c) Uniformly Varying Load (UVL) . . . . . . . . . . . . . . . . . . . . . . . . 2.10

Contents C.3

2.8 Shear Force in Beams (S.F). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.11

2.9 Sign Convention for Shear Force in Beam . . . . . . . . . . . . . . . . . . . 2.13

2.10 Couple . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.14

2.11 Bending Moment in Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.14

2.12 Sign Convention for Bending Moment in Beams . . . . . . . . . . . . . 2.15

2.13 Shear Force (S.F) and Bending Moment (B.M) Diagrams. . . . . . 2.15

2.12. Relation Between Shear Force and Bending Moment . . . . . . . . . 2.17

2.13 Method of Drawing Shear Force and Bending Moment Diagrams . . 2.18

2.14 Points to be Remembered For Drawing S.F.D and B.M.D . . . . . 2.19

Nature of SFD and BMD for Different Types of Loads . . . . . . . 2.20

2.15 Shear Force and Bending Moment Diagram for Cantilever Beams . 2.20

1. Cantilever beam with Point load at Free end . . . . . . . . . . . . . 2.20

2. Cantilever beam with UDL. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.21

3. Cantilever beam with UVL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.23

2.16 Shear Force and Bending Moment Diagram for Simply Supported Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.24

1. Simply Supported Beam with point load. . . . . . . . . . . . . . . . . . 2.24

2. Simply Supported Beam with UDL . . . . . . . . . . . . . . . . . . . . . . 2.26

3. Simply Supported Beam with UVL . . . . . . . . . . . . . . . . . . . . . . 2.28

2.17 Shear Force and Bending Moment Diagram for Overhanging Beams . 2.30

1. Overhanging beam with point load . . . . . . . . . . . . . . . . . . . . . . 2.30

2. Overhanging Beam with UDL . . . . . . . . . . . . . . . . . . . . . . . . . . 2.32

2.18 Stresses in Beams - Theory of Simple Bending . . . . . . . . . . . . . 2.111

2.19 Theory of Simple Bending or Pure Bending . . . . . . . . . . . . . . . . 2.111

2.19.1 Assumption in Theory of Simple Bending . . . . . . . . . . . . 2.111

2.19.2 Theory of simple bending - Bending stress equation . . . 2.111

2.20 Section Modulus or Modulus of Section . . . . . . . . . . . . . . . . . . . 2.114

2.20.1 Flexural Strength of a Section . . . . . . . . . . . . . . . . . . . . . 2.116

2.21 Load Carrying Capacity - Problems . . . . . . . . . . . . . . . . . . . . . . . 2.126

2.22 Proportioning of Sections - Problems . . . . . . . . . . . . . . . . . . . . . . 2.131

2.23 Beam of Uniform Strength . . . . . . . . . . . . . . . . . . . . . . . . . . 2.135


2.24 Composite Section Beams (or) Flitched Beams. . . . . . . . . . . . . . 2.137

2.25 Shear Stresses in Beams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.141

2.26 Shear Stress Distribution for a Rectangular Section . . . . . . . . . . 2.142

2.27 Shear Stress Distribution Over I - Section. . . . . . . . . . . . . . . . . . 2.145

2.28 Shear Stress Distribution Over T Section. . . . . . . . . . . . . . . . . . 2.152

2.29 Shear Stress Distr ibution Over a Circular Section . . . . . . . . . . . 2.154

2.30 Shear Stress Distr ibution Over a Triangular Section. . . . . . . . . . 2.157

2.31 Shear Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.165

Unit III: Torsion

3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.2 Pure Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1

3.2.1 Assumptions made in theory of Pure Torsion . . . . . . . . . . . . 3.1

3.3 Shear Stress - (Resistance Concept) . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2

3.4 Shear Strain - (Deformation Concept). . . . . . . . . . . . . . . . . . . . . . . . . 3.2

3.5 Analysis of Torsion of Circular Bars-derivation of Torsional Equations . 3.2

3.5.1 Theory of Torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2

3.6 Bar of Solid Section . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4

3.7 Polar Modulus Zp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6

3.8 Bars of Hollow Circular Section-strength Equation for Hollow Circular Shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.6

3.9 Torsional Rigidity and Stiffness of the Shaft . . . . . . . . . . . . . . . . . . 3.8

3.10 Power Transmitted by the Shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.8

3.11 Important Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.9

3.12 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11

3.13 Problems on Replacing a Solid Shaft by a Hollow Shaft . . . . . . 3.29

3.13 Stepped Shafts (or) Shafts in Series . . . . . . . . . . . . . . . . . . . . . . . . 3.41

3.13.1 Deflection in Shafts fixed at one end. . . . . . . . . . . . . . . . . 3.42

3.13.2 Deflection in Shafts fixed at both the Ends . . . . . . . . . . . 3.43

3.14 Compound Circular Shafts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.45

3.14.1 Shaft in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.45

3.14.2 Shafts in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.46

Contents C.5

3.15 Shaft Subjected to Number of Torques . . . . . . . . . . . . . . . . . . . . . 3.47

3.16 Problems Based on Special Conditions of Shaft . . . . . . . . . . . . . . 3.52

3.17 Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.68

3.18 Helical Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.69

3.18.1 Closely coiled helical springs . . . . . . . . . . . . . . . . . . . . . . . 3.69

3.18.2 Open coiled helical springs . . . . . . . . . . . . . . . . . . . . . . . . . 3.69

3.21 Comparison Between Closely Coiled and Open Coiled HelicalSpring. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.70

3.22 Closely Coiled Helical Spring Subjected to an Axial Load. . . . . 3.70

3.23 Shear Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.71

3.24 Deflection of Helical Springs . . . . . . . . . . . . . . . . . . . . . . . . . . 3.72

3.25 Stiffness of The Spring: k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.72

3.26 Strain Energy Stored U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.72

3.27 Energy Stored U . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.72

3.28 Close Coiled Helical Spring Subjected to Axial Twisting Couple . . 3.73

3.29 Maximum Bending Stress, b . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.74

3.30 Spring Under Impact Load. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.74

3.31 Springs in Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.75

3.32 Springs in Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.75

3.33 Open - Coiled Helical Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.76

3.34 Spring Index C . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.78

3.35 Angular Rotation in Open Coil Helical Spring . . . . . . . . . . . . . . . 3.78

3.36 Open - Coil Spring Subjected to Axial Twisting Couple Mo . . 3.78

3.37 Axial Deflection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.79

3.38 Maximum Shear Stress In Spring Section Including Wahl’s Factor . 3.80

3.39 Stresses in Spring Wire [Without Wahl’s Factor] . . . . . . . . . . . . . 3.82

3.40 Stresses in Helical Coil Spring Under Torsional Load. . . . . . . . . 3.83

3.41 Design of Helical Coil Springs . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.84

3.42 Leaf Springs (or) Carriage Springs (or) Laminated Springs . . . 3.117

3.42.1 Expression for Bending stress, deflection and StrainEnergy in a Semi - Elliptical Leaf Spring . . . . . . . . . . . . . . . . . 3.118


3.43 Quarter Elliptical Leaf Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.127

Unit IV: Deflection of Beams

4.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

4.2 Definition of Deflection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

4.2.1 Importance of deflection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1

4.3 Elastic Curve of Neutral Axis of the Beam Under Normal Loads. 4.2

4.4 Evaluation of Beam Deflection and Slope . . . . . . . . . . . . . . . . . . . . . 4.2

4.4.1 Flexural rigidity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4

4.4.2 Stiffness of beam. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4

4.5 Double Integration Method. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5

4.5.1 Simply supported beam carrying a point load at thecentre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.6

4.5.2 Simply supported beam carrying a UDL (UniformlyDistributed Load) over a whole span . . . . . . . . . . . . . . . . . . . . . . 4.13

4.6 Macaulay’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.18

4.6.1 Deflection of a simply supported beam with an eccentricpoint load . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.18

4.6.2 Problems on SSB - Uniformly Distributed Load (UDL) . . 4.43

4.6.3 Problems on SSB - Uniformly Varying Load (UVL) . . . . . 4.56

4.7 Deflection of Cantilevers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.58

4.7.1 Deflection of a cantilever with a point load at the freeend by double integration method . . . . . . . . . . . . . . . . . . . . . . . . . 4.58

4.7.2 Deflection of a cantilever with a point load at a distance‘a’ from the fixed end . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.60

4.7.3 Deflection of a cantilever with a uniformly distributed load. 4.61

4.7.4 Deflection of a cantilever with a uniformly distributedload for a distance ‘a’ from the fixed end . . . . . . . . . . . . . . . . . . 4.64

4.7.5 Deflection of a cantilever with a uniformly distributedload for a distance ‘a’ from the free end. . . . . . . . . . . . . . . . . . . 4.65

4.7.6 Deflection of a cantilever with a gradually varying load . 4.67

4.8 Deflection of Fixed Beam . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.91

Contents C.7

4.8.1 Slope and deflection for a fixed beam carrying a pointload at the centre . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.91

4.8.2 Slope and deflection for a fixed beam carrying a uniformlydistributed load over the entire length. . . . . . . . . . . . . . . . . . . . . . 4.96

4.9 Moment Area Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.103

4.9.1 First moment - area theorem (or) Mohr’s I theorem (or)Mohr’s theorem for slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.103

4.9.2 Second moment - area theorem (or) Mohr’s II theorem(or) Mohr’s theorem for deflection. . . . . . . . . . . . . . . . . . . . . . . . 4.105

4.9.3 MEI

diagram by parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.107

4.9.4 Deflection and slope of a simply supported beam carryinga point load at the centre by moment area method . . . . . . . . . 4.109

4.9.5 Deflection and slope of a simply supported beam carryinga uniformly distributed load by moment area method. . . . . . . . 4.110

4.9.6 Deflection and slope of a cantilever by moment area method. 4.111

Problems on Moment area method. . . . . . . . . . . . . . . . . . . . . . . . 4.115

4.10 Conjugate Beam Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.140

4.10.1 Simply supported beam with point load W acting at Centre. 4.149

4.10.2. Simply supported Beam with UDL . . . . . . . . . . . . . . . . . 4.151

4.11 Strain Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.158

4.11.1 Strain Energy Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.160

4.12 Deflection Due to Shear Using Strain Energy Method . . . . . . . 4.162

I. Cantilever carrying a concentrated point load W at thefree end . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.163

II. Cantilever carrying an UDL . . . . . . . . . . . . . . . . . . . . . . . . . . 4.163

III. Simply Supported Beam carrying a central point load W . 4.164

IV. SSB carrying an eccentric load . . . . . . . . . . . . . . . . . . . . . . . 4.165

V. SSB Carrying UDL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.165

4.13 Castigliano’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.166

4.14 Maxwell’s Reciprocal Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.185


Unit V: Thin Cylinders, Spheres and Thick Cylinders

5.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

5.1.1 Thin Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1

5.2 Stresses in Thin Cylindrical Shell Due to Internal Pressure . . . . . . 5.2

5.2.1 Hoop Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3

5.2.2 Longitudinal Stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4

5.3 Deformation in Thin Cylinder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6

5.4 Cylindrical Shell with Hemispherical Ends . . . . . . . . . . . . . . . . . . . 5.23

5.5 Thin Pipe Bounded by Wire . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.26

5.5.1 Joints in cylindrical shells . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.28

5.6 Spherical Pressure Vessels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.32

5.6.1 Spherical Shells subjected to internal pressure. . . . . . . . . . 5.32

5.7 Deformation in Spherical Shells . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.33

5.8 Thick Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.41

5.9 Lame’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.42

5.9.1 To Obtain Lame’s Equations. . . . . . . . . . . . . . . . . . . . . . . . . 5.42

5.9.2 Special Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.45

Condition 1: Internal pressure p1; External pressure p

2. . 5.45

Condition 2: Internal pressure p1; external pressure zero: 5.46

Condition 3: Internal pressure zero: External pressure p2

5.47

Condition 4: Solid circular shaft subjected to external pressure p2: 5.48

5.9.3 Longitudinal and Shear Stresses . . . . . . . . . . . . . . . . . . . . . . 5.48

5.9.4 Design of Thick Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.71

5.10 Compound Cylinders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.76

5.10.1 Difference of Radii for Shrinkage. . . . . . . . . . . . . . . . . . . . 5.79

Contents C.9

UNIT I

STRESS, STRAIN AND DEFORMATION

OF SOLIDS

Rigid bodies and deformable solids – Tension, Compression and ShearStresses – Deformation of simple and compound bars – Thermal stresses –Elastic constants – Volumetric strains – Stresses on inclined planes –principal stresses and principal planes – Mohr’s circle of stress

1.1 INTRODUCTION TO STRENGTH OF MATERIALSMaterials are very important for every application in all engineering

disciplines and before they can be used for any application, their behaviourunder the loads or forces under which the materials are to work must beknown. Strength of materials (Mechanics of materials) deals with thisbehaviour of solid materials by studying the distribution of internal forces,the stability and deformation of the materials under the applied loads or forces.In design of machine members and structures, in addition to strength, stiffnessand stability of materials, one has to consider factors like manufacturing, cost,life, utility, market demands etc. However, the most important role is playedby the factors like strength, stiffness and stability which are covered by thesubject of strength of materials.

Materials which we come across are generally classified as:

(A) Rigid bodies

(B) Deformable bodies

1.2 RIGID AND DEFORMABLE BODIESDeformation is the change in the shape and, or size of the body under theapplication of a force or a load. Deformable bodies are those which undergodeformation when subjected to external loading. Deformable bodies are furtherclassified into Plastic and Elastic bodies.

Define Elasticity (AU. Nov/Dec 2015, May/June 2012)An Elastic Material or body is one which undergoes deformation whensubjected to an external loading such that the deformation disappears on theremoval of the load.

Elasticity is the property by virtue of which the material return backto its original position after the removal of external force.

A Plastic material is one which undergoes a continuous deformation duringthe period of loading and the deformation is permanent and the material doesnot regain its original dimensions on the removal of the loading.

Plasticity is the property by virtue of which the material never returnback to its original position after the removal of external force.

Rigid body or material is one which does not undergo any deformation whensubjected to an external loading.

In practice, no material isabsolutely elastic nor plastic nor rigid.These properties are attributed whenthe deformations are within certainlimits.

Deformation can be understoodby a simple example, consider a barwhich is fixed at one end and isloaded by a force (F) as shown inFig.1.1 After the load is applied on thebar, there is change in the length of the bar as shown. The difference in theoriginal and final length is CC which is equal to called deflection. Thischange in length of bar is one form of deformation.

1.3 STRENGTH

Strength: Strength is the internal resistance offered by the body against thedeformation caused due to the application of an external load system.

A material when subjected to an external load system undergoes adeformation. Against this deformation, the material will offer a resistancewhich tends to prevent the deformation. This resistance is offered by thematerial as long as the member is forced to remain in the deformed condition.This resistance is offered by the virtue of its strength of material.

Consider a rod AB subjected to an external load F at the ends asshown in Fig.1.2(i). In order to keep the body in equilibrium, the body partC and D offers Reactions at section X X. In other words, we may say that

section X X offers resistance R against possible separation

The intensity of resistance offered is due to the strength of the bodyor material. The force of resistance per unit area offered by a body againstthe deformation is called stress. It is denoted by symbol ‘’

A B

C

C �

F

F

BAR

A B

Fig 1.1

1.2 Strength of Materials for Mechanical Eng., - www.airwalkbooks.com

RA

or

PA

Stress LoadArea

N

m2

In S.I unit,

Load is in N, area is in m2, unit of stress N/m2

The external force acting on the body is called load. The load isapplied on the body by which stress is induced in the material of the body.A loaded member remains in equilibrium when the resistance offered by themember against the deformation and the applied load are in equilibrium. Whenthe member is incapable of offering the necessary resistance against theexternal forces, the deformation will continue, leading to the failure of themember.

If the resistance offered by the section against the deformation isassumed to be uniform across the section, then the intensity of resistance perunit area of the section is called the intensity of stress or Unit stress.

C D

D

D

C

B

B

B

F

F

F

A

A

A

F

F

F R R

R

R (R esis tance)

X

X

C

Fig 1.2

(i)

(ii)

(iii)

www.srbooks.org Stress, Strain and Deformation of Solids 1.3

1.4 UNIT OF STRESS

The unit of stress is N/m2, which is known as pascal. 1 N/m2 1 Pa.For engineering materials, this is a small value. For larger values, we usekPa.

Kilo pascal (kPa) 103 Pa

Mega pascal (MPa) 106 Pa

Giga pascal (GPa) 109 Pa

For Engineering materials, the cross section, we use is in only

millimeters ie N/mm2

1 MPa 1 106 Pa 106 N/m2 or 1 MN/m2

106 N

103 mm 103 mm 1 N

mm2

1 MPa 1 N/mm2

1 GPa 109 Pa 109 N/m2 or 1 GN/m2

1000 106 N/m2

1000 MPa

1000 N/mm2

1 GPa 1000 N/mm2 1 kN/mm2

Intensity of stress RA

FA

in N/m2

where R: Reaction in Newton

F: Force in Newton

A: Area of cross section in m2


1.5 STRAINDue to the application of load, the length

of the member changes from l to l dl. The ratioof change in the length to the original length ofthe member is called strain.

Strain e change in lengthoriginal length

dll

1.6 TYPES OF STRESSBasically classified into 2 types

1.7 NORMAL STRESS: AXIALLY LOADED BAR

Stress which is normal to the cross section of the member (e.g stressdue to elongation of a bar) is called Normal stress.

Consider a bar of cross sectional area A, subjected to an axial loadP. To determine stress, a free body diagram is prepared either for left orright part of the bar, divided by the cutting plane as shown in Fig. 1.4(b).At any section the force vector P passes through the centroid of the bar. Thereaction on the left end is equilibrated at section a a by a uniformlydistributed normal stress . The sum of these stresses multiplied by theirrespective areas generate a stress resultant that is statically equivalent to theforce P Fig. 1.4(c). A thin slice of the bar with equal uniformly distributednormal stresses of opposite sense on the two parallel sections is shown inFig. 1.4(d). The uniaxial state of stress may be represented on an infinitesimalcube Fig. 1.4(e). However simplified diagram shown in Fig. 1.4(f) iscommonly used.

�

d�

Fig 1.3

Tangential Stress

STRESS

Shear Stress

Punchingshear Stress

Compressive Stress

Tensile Stress

Normal Stress

Fig. 1.3


Normal stress is defined as ratio of force applied to the cross section of areaof the bar.

Normal stress ForceArea

PA

in N/m2

In the integral form, the load applied is given by Load P A

dA

A material is capable of offering the following types of stresses.

1. Tensile stress2. Compressive stress3. Shear stress.

1.8 TENSILE STRESS AND TENSILE STRAINTensile stress is defined as theresistance offered by the section of amember/body against an increase inlength. For example consider the stressoffered by the section XX of a rod asshown in Fig. 1.5.

The intensity of tensile stress isgiven by

Tensile stress ForceArea

PA

RA

a

dx P

A(d )

(f)

Fig. 1.4 Successive steps in determ in ing the largest normal stress in an axia lly loaded bar.

a

a

P

C utting p lane

Bar axis(a )

P PP

a

aC entroid

(b )

a

aAdA = P

(c)

P dy

dx dz

(e )PA

P

�

P

d �

X

X

R RP P

X

XFig. 1.5


When the rod is subjected to tensile load, there is an increase in thelength of rod and the corresponding strain is called the tensile strain.

The ratio of change of dimensions of the body to the originaldimension is known as “strain”. It has no unit.

Tensile strain: Ratio of change in length to original length is known astensile strain Fig. 1.5 (b)

Tensile strain e Increase in lengthOriginal length

dLL

1.9 COMPRESSIVE STRESS AND COMPRESSIVE STRAIN

Compressive stress is the resistance offered by the section of member orbody against a decrease in length due to applied pushing load. For exampleconsider a bar subjected to pushing axial load as shown in Fig. 1.6. Intensityof compressive stress is given by

Compressive stress PA

RA

Due to the external loading, the length of the member decreases by dl.The ratio of the decrease in length to the original length is called compressivestrain

Compressive strain e Decrease in length

original length

dll

STRAIN

Tensile strain

com pressive strain

Lateralstrain

Longitudinalstrain

Volumetricstrain

Shear strain

Fig. 1.5 (a)

L dL

P

Fig 1 .5 (b)

P


1.10 SHEAR STRESS AND SHEAR STRAINConsider a bar AB subjected to transverse forces as shown in Fig.

1.7(a).

Two forces are passing at section C as shown in Fig. 1.7(b). Internalforce must exist in the plane of the section and their resultant is equal toP. These elementary forces are called shear forces with magnitude P.Dividing the shear force by Area of cross section we get shear stress

When a body subjected to two equal and opposite forces which areacting tangentially on any cross-sectional plane of a body, tending to slideone part of the body over the other part, then the body is said to be in astate of shear. It is denoted by ‘’.

Shear stress total tangential force

crosssectional area of resisting section

PA

in N/m2

Fig 1.6

Fig 1.7


Shearing stresses are commonly found in bolts, pins and rivets which areused to connect various structural members and machine components.Consider two plates A and B connected by rivets as shown in Fig. 1.8.

The shear stress shear forceshear Area

FA

PA

Consider a block of height l, length L and width unity Fig.1.9.

Shear Stress Shear force or Resistance

Shear Area

RL l

P

L l

Consider the block subjected to shear force P on its top and bottomfaces. When the block does not fail in shear, its shear deformation is shownin Fig. 1.9(a). When the block has deformed from the positionABCD to A B CD through an angle , its shear deformation is shown in Fig.1.9(b).

BCB ADA

Fig 1.8

A B

x

P

d x

A

dx

d lA’

PB B ’

x

d l

Fig 1.9(a ) (b )


Let the horizontal displacement of the upper face of the block be dl.Then the ratio of transverse displacement to the distance from the lower faceis called shear strain.

Shear strain Transverse displacement

Distance of the block from lower face

dll

At the section XX

The strain dxx

Since is very small, tan dll

dxx

shear strain

The Angular deformation in radians measures the shear strain.

1.11 BEARING STRESS (CRUSHING STRESS) IN CONNECTIONSBolts, pins and rivets create

stresses in the members they connect,along the bearing surface, or surface ofcontact.

For example consider again twoplates A and B connected by the rivetCD which we have discussed inprevious section. The rivet exerts onplate a force P equal and opposite tothe force F exerted by the plate on therivet.

The force P represents theresultant of elementary forcesdistributed on the inside surface of thehalf cylinder of diameter d and oflength t equal to the thickness of plate. The average nominal bearing stressis obtained by dividing load P by the area of rectangle representing projection.

Bearing stress b PA

Ptd

1.12 STRESS-STRAIN BEHAVIOUR OF MATERIALS

Young’s modulus StressStrain

Normal stressLinear strain

dP

t

F

F �

Fig AD

C

A

A d

Fig B

t

Fig 1.10

Fig.(a)

Fig.(b)


The value of Young’s modulus is determined from stress-straindiagram of the material. Some materials are equally strong incompression and tension (metals and alloys). Such materials areusually tested in tension. The test results usually pertain to a circularbar of uniform cross-section. The load on the test specimen isincreased gradually from zero, in suitable increments till thespecimen fails (breaks). The elongation of the specimen is measuredover a specific length known as “gauge length” (usually 50 to 200mm) at each load step. The stresses and the corresponding strainsare computed for the load and corresponding elongation readings.

Materials such as concrete, stones and bricks that are stronger incompression than in tension are tested in compression. Stress-strainvalues are plotted in the form of a graph and the value of Young’smodulus is determined from the slope of the curve for any stressvalue.

In the case of materials with linear stress-strain behaviour, young’smodulus is constant upto elastic limit. For materials with non-linearstress-strain relationship, the average value of slope is adopted foryoung’s modulus or the value is defined at a specified stress orstrain value.

Thus, the stress-strain diagram can be used to determine variousfactors like Young’s modulus, proportional limit, elastic limit, yieldstrength, ultimate strength etc.

1.12.1 Stress Strain Curves (Tension)When a bar or specimen is subjected to a gradually increasing axial

tensile load, the stresses and strains can be found out for number of loadingconditions and a curve is plotted upto the point at which the specimen fails.This curve is known as the stress strain curve.

1.12.2 Stress - Strain Curve for Ductile MaterialsA material is said to be ductile in nature, if it elongates appreciably

before fracture occurs. (Eg) Mild steel.

When a specimen of a mild steel is loaded gradually in tension, thestress is proportional to the strain in the initial stage and remains so upto apoint, known as limit of proportionality as shown in Fig. 1.11. The stressstrain diagram gives many important properties of materials.


Point P : Limit of proportionality

E : Elastic limit

Y : Upper yield point

L : Lower yield point

U : Ultimate point

B : Breaking point

Define Elastic limit (AU. Nov/Dec 2015)Next to the proportionality limit, we have a point called Elastic limit

(E) at which if the load is removed, the specimen will return to its originaldimensions. Beyond the elastic limit, the material enters into plastic rangeand removal of load does not return the specimen to its original dimensions,thus subjecting itself to a permanent deformation. On applying further loadthe specimen curve reaches upper yield point (Y) and corresponding stressis called upper yield stress. Beyond point Y, the load decreases withincrease in strain upto point (L) called lower yield point and correspondingstress is called lower yield stress. After lower yield point (L), the stress startsincreasing and reached maximum value at the point (U) called ultimate pointand the corresponding stress is called ultimate tensile stress. After theultimate point, the stress again starts decreasing, while the strain goes onincreasing until the material fractures at point B called Breaking Point andthe corresponding stress is called breaking stress. In the curve, all the stressesare calculated based on original cross section, hence, the curve is known asthe engineering stress strain curve.

Rupture Strength or Breaking Strength The co-ordinate of point B in the stress-strain diagram represents

the stress at failure and is known as “Rupture strength orBreaking strength”.

In a stress-strain diagram, note that this breaking strength is lowerthan the ultimate strength. This is because we find the breakingstrength by dividing the breaking load by original area of specimen.As noted earlier, while tensioning the specimen, its length increasesbut the diameter decreases.


For the ductile materials, as they yield, decrease in the diameter isalso more and more. The diameter of the specimen is considerablyreduced and we find the rupture strength with respect to the final(reduced) diameter, will be much more than the ultimate strength.We may call this as “Actual Rupture Strength” as indicated aspoint A in Fig 1.11.

Due to larger yielding in the material, a phenomenon called “Necking”occurs which is responsible for reducing the diameter of specimen. Theformation of necking shown in Fig 1.11(a) is more predominant in ductilematerials and at failure, a perfect cup and cone is formed. With loweringductility and increasing brittleness, the cup and cone failure slowly disappearand brittle failure with rough texture takes places.

**

* *

**

*

P-P roportiona lity Lim it

B- Breaking (Rupture streng th)

E - Elastic L im it

Y - Upper Yie ld point

L - Lower Yield point

U -U ltim ate strength

A - actual Rupture streng th

ST

RE

SS

STRAINO

Fig 1.11 Stress - Strain


Stresses and strain based on original dimensions are called as Engineeringor Nominal or Conventional stresses or strains. Stresses and strains based onactual dimensions are called True or Natural Stress and Strains.

Ductility of a material is measured by the percentage elongation of thespecimen (or) percentage reduction in cross sectional area of the specimenwhen failure occurs.

% increase in length l l

l 100

% Reduction in Area A A

A 100

1.12.3 Stress Strain Curves for Brittle Materials“Brittleness” is defined as

the property of material that willfail suddenly without undergoingnoticeable deformations.

For brittle materials and forthe materials with low ductility likehigher grades of steel, no definiteyield is observed.

Materials which show verysmall elongation before theyfracture are called brittle materials(Eg.) Cast Iron, concrete, high carbon steel etc.

Ultim atestreng th

S tress

L im it of P roportiona lity

Break ing oru ltima te Po in t

S trainFig 1.12


For Brittle materials, the stress strain curve is shown in Fig 1.12.

The ultimate tensile stress is defined as the ratio of ultimate load tothe original area of cross section and is taken as basis for determining thedesign stress for Brittle materials because there is no definite yield point.

1.13 STRESS STRAIN CURVES (COMPRESSION)

For ductile materials, stress strain curves in compression are identicalto those in tension at least upto the yield point for all practical purposes.

Brittle materials have compression stress strain curves of the same formas the tension test but the stresses at various points are generally considerablydifferent.

1.14 HOOKE’S LAW

Define Hooke’s Law.(AU. May/June 2013, Apr/May 2010)

Hooke’s Law states that within the Elastic limit the stress (compressive ortensile) is proportional to the strain

Mathematically, Hooke’s Law is

Stress Strain

StressStrain

Constant of proportionality.

Define Young’s Modulus(AU. Nov/Dec 2016)

Under normal (i.e., direct) stresses and strains, constant of proportionality iscalled Modulus of Elasticity or Young’s Modulus E

Young’s Modulus E Normal stress

Linear nominal strain

e

Under the shearing stresses and strain, the constant of proportionalityis called Modulus of rigidity and is denoted by G or C or N

Rigidity Modulus G Shearing stressShearing strain

1.14.1 Factor of SafetyFactor of safety is defined the ratio of ultimate stress to the permissiblestress (working stress)


Factor of safety Ultimate stress

Permissible stress

Factor of safety depends on so many factors like the type of material,its degree of reliability, workmanship, manufacturing method, nature ofloading, environmental conditions etc. and is “always greater than one”.The following values are commonly taken in practice.

Table 1

S.No Materials Factor of safety

1. Concrete 3

2. Steel 1.85

3. Timber 4 to 6

1.15 DEFORMATION OF A BODY UNDER AXIAL LOAD AT FREE END

Consider a body or rod BC of length L and uniform cross section ofArea A subjected to an axial load P

If the resultant axial stress induced is given by

Tensile stress PA

Within elastic limit, one may apply the Hooke’s law (stress) E e

where E Young’s modulus

e Strain

Strain e E

e P/A

E or e

PAE

Also we know that strain e L

. Substituting this in above equation.

[ change in length L]

e L

P

AE

B

C

P

L

l

Fig 1.13


Deflection (or) deformation or L PLAE

L Original Length

L Change in length

[ is also denoted as L (or) l.]

If the body is made up of different sections having Pi, Li, Ai and Ei as

internal force, length, Area of cross section and modulus of elasticityrespectively, then

Deflection PiLi

AiEi

If we consider a rod of variable cross section, then the strain ‘e’depends upon the position and is defined as e d/dx

d e dx PdxAE

By integrating over the entire length

Total deformation 0

L

P dx

AE

1.16 STIFFNESSConsider a bar BC of constant

cross-section area A and length L shownin the Fig. 1.14(a).

Let force ‘P’ is applied at thefree end. The deformed bar is shown inFig. 1.14 (b). Conceptually, it is oftenconvenient to think of such elasticsystem as spring as shown in Fig. 1.14(c)

The total deformation is PLAE

The deflection of rod is directlyproportional to the applied force andlength and is inversely proportional to A and E.

L

a

B a

B C

C ’

P

(a)

(b)

(c)

L

P

Fig 1.14


from the above equation we get P AEL

and also we get P

AEL

This equation is related to the familiar definition of the spring constantor stiffness k.

Stiffness, k P

AEL

in N/m

Stiffness k is defined as the ratio of force per unit deflection 1.

For an axially loaded ith bar or bar segment of length Li, the stiffness

is given by ki AiEi

Li

The reciprocal of stiffness k is defined as flexibility

i.e. f 1k

P

in m/N

For a particular case of ith bar of constant cross section

fi Li

AiEi

The concept of structural stiffness and flexibility are widely used instructural analysis.

1.16.1 StabilityAny structural or machine member loaded in compression is called a

column or strut or pillar. Generally columns are classified as shortcolumns,intermediate columns and long columns

The classification among the columns has been done on the basis oftheir behaviour in compression. The ability of a short column to take loadsdepends upon its cross sectional area and strength of material of column. Asthe length of column increases, the load carrying capacity depends upon crosssectional area, strength of material, length of column, geometry of section(radius of gyration), Young’s modulus E. A long or intermediate columnfails in compression by buckling sideways whereas a short column does notbuckle sideways as shown in Fig. 1.15. Therefore a short column can take


more load than long or intermediate column for same cross section and samematerial.

A column remains straight upto a certain load called the critical loadbeyond which a slight increase in load causes the column to buckle to a greatextent and fail.

A column under a load less than critical load is in stable equilibrium.At critical load, the column is in neutral condition. Beyond the critical load,the equilibrium is unstable.

Slenderness ratio is one of the important characteristics of the column onwhich the load carrying capacity of columns depends. It is defined as theratio of unsupported length of column to the least radius of gyration.

Slenderness ratio lk

where l Length of column

k Radius of gyration

Significance of percentage of Elongation & Reduction in Area

Let Lo Gauge length or initial length of the specimen

L Length at fracture

then Percentage Elongation L Lo

Lo 100

P P

45 o

M .S C.I

(i) Fa ilure by general y ie ld ing

(ii) Failu re due to shear

Inelasticbuckling

P P

Elasticbuckling

(a )

( )b ( )c

(a ) Short colum ns

(b ) In te rm ed ia te colum n (c) Long co lum nFig 1.15


Let Ao Original area of crosssection

A Area at neck when fracture occurs

Percentage reduction of area Ao A

Ao 100

SOLVED PROBLEMSProblem: 1.1: An elastic rod 25 mm in diameter, 200 mm long extends by0.25 mm under a tensile load of 40 kN. Find the intensity of stress, the strainand the elastic modulus for the material of the rod.

Given: diameter d 25 mm; Length L 200 mm

Load P 40 kN 40 103 N; Elongation L 0.25 mm

Area of cross section A d2

4 252

4 490.87 mm2

Solution:

Intensity of stress

LoadArea

PA

40 103

490.87 81.49 N/mm2

Strain e Elongation L

Length L LL

0.25200

0.00125

Elastic modulus E e

81.49

0.00125 65192 N/mm2

E 0.06519 N/m2

Problem 1.2: A rectangular wooden column of length 3 m and size 300mm 200 mm carries an axial load of 300 kN. The column is found to beshortened by 1.5 mm under the load. Find the stress and strain in the columnand state their nature.

Given

L 3m

Size of column Area of cross section, A 300 mm 200 mm 60000 mm2

Load P 300 kN 300 103 N

Shortening of column, L 1.5 mm


Solution:

Compressive stress PA

300 103

60000

5 N/mm2

Compressive strain in the column ec LL

1.5

3000

ec 0.0005

The nature of the stress is compression, since the column is shortened.

Problem 1.3: Find the maximum and minimum stress produced in the steppedbar shown in figure due to axially applied compressive load of 12 kN.

Given: d1 12 mm ; d2 25 mm ; Load, P 12 kN 12 103 N

Area of upper part

A1 d1

2

4 122

4 113.10 mm2

Area of Lower part

A2 d2

2

2 252

4 490.87 mm2

Solution:

Maximum stress,

max Load

Area A1

12 103

113.10

106.10 N/mm2

The upper bar takes the maximum stress because it is located in theimmediate vicinity (nearer) of the load.

Minimum stress min Load

Area A2

12 103

490.87 24.45 N/mm2

[. . . shortens]

12 m m

25 m m

1

2

12 kN


Problem 1.4: A steel wire of length 10 m and diameter 5 mm is used tohang a load at its bottom. The stress and strain in the wire are found to be

140 N/mm2 and 0.0007 respectively. Determine the load it carries and theelongation of wires.

Given: L 10 m 10000 mm ; d 5 mm ; 140 N/mm2 ; e 0.0007

Required Data: P ? L ?

Solution: Area of cross section of wire A 52

4 19.635 mm2

Stress LoadArea

PA

P A 19.635 140

P 2.749 kN

We know that, Strain, e LL

0.0007 LL

L 0.0007 10000

L 7 mm

Problem 1.5: A steel wire 6 mm diameter is used for lifting a load of 1.5kN at its lowest end, the length of the wire hanging vertically being 160

meters. Taking the unit weight of steel 78 kN/m3 and E 2 105 N/mm2,calculate the elongation of the wire. (AU.Nov/Dec 2011)

Given

Diameter of wire 6 mm

Hanging wire length 160 m

Unit weight of steel 78 kN/m2

Load P 1.5 kN E 2 105 N/mm2

Elongation of wire ?


Solution

(i) Stress PA

1.5 103

/4 62

Stress 53.05 N/mm2

(ii) Strain e E

53.05

2 105

Strain e 2.65 10 4

(iii) Elongation of wire dL

dL e L

2.65 10 4 160

0.0424 m

Elongation of wire 42.4 mm

Problem 1.6: A brass rod of 25 mm diameter and 1.3 m long is subjectedto an axial pull of 4 kN. Find the stress, strain and elongation of the bar.

If young’s modulus E 1 105 N/mm2.

Given:

P 4 kN 4000 N ; d 25 mm ; E 1 105 N/mm2 ; L 1.3 m 1300 mm

Required data: ?, e1 ?, L ?

Solution

Area A 4

d2 4

252 490.87 mm2

Stress Loadarea

PA

4000

490.87 8.15 N/mm2

Strain e E

8.15

1 105 82 10 6


Elongation L e L

82 10 6 1300

L 0.106 mm

Problem 1.7: A mild steel bar of 15 mm diameter and 400 mm lengthelongates 0.2 mm under an axial pull of 10 kN. Determine the young’smodulus of material.

Given: d 15 mm ; L 400 mm ; L 0.2 mm ; P 10 kN 10 103 N

Required Data: E ?

Solution

Young’s Modulus E StressStrain

stress PA

A 4

152 176.7 mm2

10 103

176.7

56.59 N/mm2

Strain e LL

0.2400

5 10 4

E 56.59

5 10 4

E 1.1318 105 N/mm2

Problem 1.8: A hollow cylinder 1.5 m long has an outside diameter of 45mm and inside diameter of 25 mm. If the cylinder is carrying a load of 20kN, find the stress in the cylinder. Also find the deformation of the cylinder.

Take E 100 GPa [1 Pa 1 N/m2].

Given: Length L 1.5 m; Outside diameter D 45 mm

Inside diameter d 25 mm ; Load P 20 kN 20 103 N


Modulus of Elasticity E 100 GPa 100 109 N/m2 100 103 N/mm2

Area of cross section, A 4

[D2 d2] 4

[452 252] 1099.5 mm2

Solution:

Stress LoadArea

PA

20 103

1099.5 18.2 N/mm2

Strain e Stress

Young’s modulus

E

18.2

100 103 1.82 10 4

Strain e Change in lengthOriginal length

LL

1.82 10 4 L

1500

Deformation L 1.82 10 4 1500 0.273 mm.

Problem 1.9: A specimen of a material having original diameter equal to13 mm and gauge length 50 mm is tested under tension, the final diameterbeing 9 mm at fracture and gauge length at fracture being 70 mm. Duringtesting, it is found that yielding occurs at a load of 35 kN (lower yield point)and the maximum load that the specimen can take is 60 kN (ultimate load).The specimen fractures or breaks under a load of 30 kN. Find yield strength,ultimate tensile strength, breaking strength, % elongation, % reduction inarea, young’s modulus if load corresponding to any point on the linearportion of the stress strain curve is 20 kN corresponding to an extension of0.0315 mm.

Solution:

Original cross section Area, A d2

4

4

132 132.73 mm

Yield strengthyield stress

Yield pointArea A

35 103

132.73 264 MPa

Ultimate strengthUltimate stress

Ultimate loadOriginal Area A

60 103

132.73 452 MPa


Breaking strengthBreaking stress

Breaking LoadOriginal Area

30 103

132.73 226 MPa

% elongation L L0

L0 100

70 5050

100 40%

% reduction in Area A0 A

A0 100

4

132 4

92

4

132 100 52 %

A0 Original area of cross-section

A Area at neck when fracture occurs

Young’s modulus E StressStrain

Stress LoadArea

20 103

0.1327 10 3 1.509 108 N/m2

Strain e Extension

Original length

0.031550

6.3 10 4

Young’s modulus E e

1.509 108

6.3 10 4 239 109 N/m2 239 GPa

Problem 1.10: In a tension test on mild steel specimen 10 mm diameterand 250 mm long gauge length, the following observations were madeElongation under 16 kN load 0.2 mm

Load at yield point 27 kN

Ultimate load 51 kN

Breaking load 36 kN

Length between gauge marks after fracture 290 mm

Diameter at Neck 7.5 mmCalculate (i) Nominal yield stress (ii) Nominal ultimate stress (iii) Nominalbreaking stress (iv) Young’s modulus (v) Percentage elongation(vi) Percentage reduction in area (vii) Safe stress with F.O.S = 2.

L length of fracture

L0 Gauge length


Solution:

(i) Nominal yield stress Load at yield point

Original cross sectional area

Nominal cross sectional area 4

102 78.54 mm2

Nominal yield stress 27 103

78.54 343.17 N/mm2

(ii) Nominal ultimate stress Ultimate load

Nominal cross sectional area

51 103

78.54 649.35 N/mm2

(iii) Nominal breaking stress Breaking load

area of c/s

36 103

78.54 458.37 N/mm2

(iv) Young’s modulus StressStrain

E e

PA

L0

L

Gauge length L0 250 mm

L 0.2 mm

P 16 kN

E 16 103 250

78.54 0.2 2.546 105 N/mm2

E 2.546 105 N/mm2

(v) Percentage elongation

% elongation L L0

L0 100

290 250250

100 16%

(vi) Percentage reduction in area A0 A1

A0 100


A0 original nominal area of cross section 4

d2 78.54 mm2

A1 area of cross section at neck 4

dn2

4

7.52 44.18

A1 44.18 mm2

Percentage reduction in area 78.54 44.18

78.54 100 43.75 %

(vii) Safe stress Yield stress

Factor of safety

343.172

171.89 N/mm2

Problem 1.11: A short hollow cast iron cylinder of external diameter 220 mmis to carry a compressive load of 600 kN. Determine the inner diameter of

the cylinder, if the ultimate crushing stress for the material is 540 MN/m2.Factor of safety of 6 is used.

Given: External Diameter D 220 mm ; Inner diameter d ?

Ultimate stress 540 MN/m2 540 N/mm2 ; Factor of safety 6

Solution:

Factor of safety Ultimate stressWorking stress

6 540

Working stress

Working stress work 5406

90 N/mm2

Also, Working stress work LoadArea

90 600 103

4

[D2 d2]

4

[2202 d2] 600 103

90 d 200 mm

Inner diameter d 200 mm


Problem 1.12: A load of 4 kN has to be raised at the end of a steel wire. If

the unit stress in the wire must not exceed 80 N/mm2, what is the minimumdiameter required? What will be extension of 3.50 m length of wire? Take

young’s modulus E 2 105 N/mm2

Given: Load 4000 N ; Stress 80 N/mm2 ; Length L 3.5 m ; L

Solution:

We know that stress LoadArea

80 4000

d2

4

d2

4

400080

or d2 4000 4

80

Diameter of wire d 7.979 mm

We know that strain e Extension

Original length LL

or L

Young’s modulus E StressStrain

80L/L

Extension L 80E

L 80

2 105 3500 1.4 mm

1.17 DEFORMATION OF A BAR UNDER AXIAL LOAD AT DIFFERENT SECTIONS

Consider an axially loaded bar as shown in the Fig. 1.16

The applied forces P1, P2, P3 are held in equilibrium by the force P4. The

cross sectional area A of the bar is permitted to change gradually. The change inlength that takes place in the bar between the point B and D due to applied forceis to be determined.

The normal strain ex or x in the x direction is given by

ex dudx


where, due to the applied forces, u is the absolute displacement of apoint on a bar from an initial fixed position in space and du is the axialdeformation of the infinitesimal element. This is the governing differentialequation for axially loaded bars. The initial length between the pointsB and D is L. On rearranging the above equation as du ex dx. Assuming

the origin of x at B and integrating

0

L

du u L u 0 0

L

ex dx

u L u 0 Change in length L 0

L

ex dx

For Elastic Materials according to Hooke’s law,

x or ex x

E

Also x Px

Ax substituting this in the above equation and simplifying

we get

Deformation L 0

L

Px dx

Ax Ex ...(1)

L+ L

L

uBuD

P 1 B DP2 P 3

P 4

xdx

(a )

P xP +dP xx

dx x+e dx(b )

Fig. 1.16


where Px Load in N

Ax Area in m2

Ex Young’s modulus

1.17.1 Deformation in Simple Bar Subjected to Axial LoadIf the bar is simple i.e., cross section of area (A) is uniform through

out the length L as shown in Fig. 1.17.

Modulus of Elasticity or Young’s modulus

E Axial stressAxial strain

E e

Direct stress in the bar, ForceArea

PA

Direct strain in the bar, e Stress

Young’s modulus

e P

AE

Elongation of bar Strain Length

L PLAE

(or)

Fig 1.17


From Hooke’s law,

E StressStrain

E P/AL/L

L PLAE

1.18 DEFORMATION FOR A BAR OF VARYING SECTIONConsider a bar of varying section with the internal force Pi, length

Li, cross sectional area Ai and modulus of elasticity Ei where i represents the

corresponding section.

Consider a section at 1. The Elongation due to load at section 1 isgiven by

L1 or 1 P1L1

A1E1

Similarly, we have Elongation at section 2 and 3

L2 or 2 P2L2

A2E2; L3 or 3

P3L3

A3E3.

Total Elongation due to the Load P

1 2 3 (or) L L1 L2 L3

i.e L or P1L1

A1E1

P2L2

A2E2

P3L3

A3E3

PiLi

AiEi

A 1 A 2 A3

12

3

L 1 L 2 L 3

P

Fig 1.18


Total Elongation L or PiLi

AiEi

If the load is uniformly distributed at all the sections, we have total

deformation L or P Li

AiEi

If the varying section is made up of same material with Young’smodulus E, then

Deformation L or PE

Li

Ai

Deformation L or PE

L1

A1

L2

A2

L3

A3

Problem 1.13: An axial pull of 35000 N is acting on a bar consisting of

three lengths as shown in Fig. If the Young’s modulus 2.1 105 N/mm2,determine(i) Stresses in each section and(ii) Total extension of the bar. (AU. Nov 2007)

Given:

l1 20 cm 200 mm ; d1 2 cm 20 mm ;

l2 25 cm 250 mm ; d2 3 cm 30 mm ;

l3 22 cm 220 mm ; d3 5 cm 50 mm ;

35000N2 cm D IA 3 cm D IA 5 cm D IA

20 cm 25 cm 22 cm

35000N

Section 1Section 2

Section 3


Solution:

Stress in Section (1), 1 PA1

350004

202 111.41 N/mm2


350004

302 49.51 N/mm2


350004

502 17.83 N/mm2

Total extension of the bar, L or 1E

1 l1 2 l2 3 l3

1

2.1 105 111.41 200 49.51 250 17.83 220

0.1837 mm

Problem 1.14: A bar ABCD of steel is 600 mm long and the two endsAB and CD are 30 mm and 40 mm in diameter and each is 150 mm inlength, the middle portion BC being 25 mm in diameter. Determine the finallength of the bar when subjected to an axial compressive load of

120 kN E 2.1 105 N/mm2. (AU. May 2007)

Given:

l1 150 mm ; d1 30 mm

l2 300 mm ; d2 25 mm

l3 150 mm ; d3 40 mm

A BC D

30m m

120 kN 120 kN

12

3

25m m

40m m

150m m

150m m

300m m


Area of cross section (1), A1 4

302 706.86 mm2


252 490.87 mm2


402 1256.64 mm2

Solution:

Compressive load, P 120 kN 120 103 N ; l 600 mm

Change in length (decrease in length because of compressive load).

l PE

l1A1

l2A2

l3A3

120 103

2.1 105

150706.86

300

490.87

1501256.64

l 0.54 mm

Final length of the bar l l 600 0.54

599.46 mm

Problem 1.15: An alloy circular bar ABCD (3 m long) is subjected to atensile force of 50 kN as shown in figure. If the stress in the middle portionBC is not to exceed 150 MPa, then what should be its diameter? Also findthe length of the middle portion, if the total extension of the bar should notexceed by 3 mm. Take E 100 GPa

(AU. Apr/May 2010)

Given: Length of circular bar, AD, L 3 m 3000 mm

Diameter of AB & CD 40 mm d1

Diameter of BC d


Length of BC l

Stress in middle portion BC, BC 150 MPa

Tensile Load, P 50 kN 50 103 N

Total extension l 3 mm

E 100 GPa 100 103 N/mm2

To find: d & l

Solution:

Stress in middle portion BC LoadArea

P

4

d2

150 50 103 4

d2

d 50 103 4

150 20.6 mm

Diameter of middle portion d 20.6 mm

A1 A3 4

402 1256.64 mm2

A 4

20.62 333.29 mm2

Total extension, L PE

l1A1

l2A2

l3A3

PE

l1 l3

A1

lA

[Because l2 l (given) and A2 A . . . d2 d and since

d1 d3 ; A1 A3 and l1 l3 3000 l]

l PE

3000 l

A1

lA


3 50 103

100 103 3000 l1256.64

l

333.29

3 12

3000 l 333.29 1256.54l

1256.64 333.29

6 999870 333.29l 1256.64l

418825.55

418825.55 6 923.35l 999870

l 1513083.3

923.35 1638.69 mm

Length of middle portion, l 1638.69 mm

Problem 1.16: Calculate the change in length of the rod ABCD carrying

axial loads as shown in Fig. E 2 107 N/cm2. The cross-sectional areasare given in figure.

Solution:

Portion AB

Length L1 50 cm ; Cross sectional area, A1 4 cm2

Portion is subjected to tensile force P1 30 kN

Elongation of AB, AB P1 L1

A1 E

30 103 50

4 2 107

AB 0.01875 cm

Portion BC

Length, L2 90 cm ; Sectional area, A2 6 cm2

BA C D

4cm 2

6cm 2

3cm 2

30kN80kN 30kN

50cm 90cm 40cm


Compressive force, P2 80 30 50 kN

Contraction of BC, BC P2 L2A2 E

50 103 90

6 2 107

BC 0.0375 cm

Portion CD

Length, L3 40 cm ; Sectional area, A3 3 cm2

Compressive force, P3 80 30 30 20 kN

Contraction of CD, CD P3 L3A3 E

20 103 40

3 2 107

CD 0.0133 cm

Total contraction in length of member ABCD,

L BC CD AB 0.0375 0.0133 0.01875

L 0.032 cm

Problem 1.17: A bar 1.5 m long is made up of two parts of aluminium andsteel and that cross sectional area of Aluminum bar is twice that of steel bar. Therod is subjected to an axial tensile load of 200 kN. If the elongation in aluminiumand steel bar are equal, find the length of the two parts of the bar. TakeEsteel 200GPa; EAluminium one third of Esteel.

Given: l 1.5 m 1500 mm ; AA 2As

P 200 kN 200 103 N ;

Es 200GPa 200 103 N/mm2

EA Es

3

200 103

3 N/mm2

Solution:

Elongation in aluminium

A P lA

AA EA

200 103 lA

2As 200 103

3


A 1.5 lA

As

Elongation in steel

s P ls

As Es

200 103 ls

As 200 103 lsAs

.

Elongation in Aluminium and steel are Equal.

A S

1.5lA

As

lsAs

or ls 1.5 lA

Total length of bar l lA ls

1.5 103 lA 1.5 lA 2.5 lA

lA 1.5 103

2.5 600 mm

ls l lA 1500 600 900 mm

Length of Aluminium bar 600 mm;

Length of steel bar 900 mm

Problem 1.18: A circular steel rod PQRS of different cross section is loadedas shown. Find the maximum stress induced in the rod and its deformation.E of material = 200 GPa

Alum in ium

1.5

m

�AA

B

200kN

Stee l

1m

2m

1m

1

2

50kN

100kN

3

25 kN

Q

R

S

70 m m

0 m m

O D 0 m m D 0 m m

P


Solution:

Area of part PQ, A1 D1

2

4 702

4 3848.5 mm2

Area of part QR, A2 D2

2

4 502

4 1963.5 mm2

Area of part RS, A3 4

[D2 d2] 4

[502 302] 1256.6 mm2

Stress in various parts:

PQ Load P1

A1

75 103

3848.5

19.49 N/mm2 Tensile

QR Load P2

A2

25 103

1963.5

12.73 N/mm2 Compressive

RS Load P3

A3

25 103

1256.6

19.89 N/mm2 Tensile

The maximum stress induced is 19.89 N/mm2 in the rod RS.

Total deformation

PiliAiEi

P1l1A1E

P2l2A2E

P3l3A3E

1E

P1l1A1

P2l2A2

P3l3A3

1E

[PQ l1 QR l2 RS l3] . . .

PA

1E

[PQ l1 QR l2 RS l3]

1

200 103 [19.49 1 103 12.73 2000 19.89 1000]

Deformation 0.069 mm

Loads in various pa rts

Compress ion Tensile25kN 25kN

R S

25kN 25kN

Q RP

75kN

Q

Tensile75kN


1.19 DEFORMATION OF A BODY DUE TO SELF WEIGHT

Deformations of prismatic bars due to selfweight

Apart from the external forces, the selfweight of structural elements also causeconsiderable amount of deformations inthe elements. The effect of self weightwith regards to deformation some timesseems to be even more than that of theapplied external forces in the case ofheavy structural elements.

Derive a relation for change in length of a bar hanging under its ownweight (AU. Nov/Dec 2014)

Consider a vertical bar of length ‘L’ and uniform cross sectionalarea A which is rigidly fixed at its upper end shown in Fig 1.19.

Let ‘w’ be the unit weight of the material and E be its Young’smodulus.

Consider a small strip PQRS of length dx at a distance ‘x’ from freeend.

Downward force acting at section

PQ weight of bar below the section PQ wAx

Tensile stress at section PQ w A x

A wx

Tensile strain in elemental strip wxE

Elongation of elemental strip wxE

dx

Total elongation, L o

L

wE

x dx

wE

o

L

x dx

B

A

RP

S

Q

Fig 1.19


wE

x2

2 o

L

L wL2

2E

L or WL2AE

[W wAL total weight]

From the above, we see that the deflection of the body due to selfweight is equal to half the deflection of a load attached to the body with aweight equal to the body.

Problem 1.19: A copper wire of 14 m length with a cross sectional area

5mm2 weighs 20 N. The wire is hanging freely. Determine the deflections at

the end and midway of the wire. Take E 2 105 N/mm2

Data Given:

Length l 14 m; CSA 5 mm2 ; Weight 20 N ; E 2 105 N/mm2

To find: Deflection at B and CSolution:

Deflection at CC Wl

2 AE

20 14 103

2 5 2 105 0.14 mm

Deflection at BB AB AB due to BC

The deflection at B is due to the self weight of AB and due to theweight of BC acting as an added load.

AB W/2 l/2

2AE

10 7 103

2 5 2 105 0.035 mm

AB due to BC W/2 l/2

AE

10 7 103

5 2 105

0.07 mm


Total deflection at BB 0.035 0.07

0.105 mm

1.20 PRINCIPLE OF SUPERPOSITIONWhat is principle of superposition? (AU. Nov/Dec 2015)

In all cases, it cannot be expected that the forces are always acting onthe ends of the body. Sometimes forces can also act at internal sections ofthe body as well as the outer edges.

In order to solve these problems, the forces acting on each of thesections should be taken individually. The deformation of the body will beequal to the algebraic sum of the individual deformations.

This principle of finding out the resultant deformation is called theprinciple of superposition. The resultant deformation can be calculated by

P1l1AE

P2l2AE

P3l3AE

1

AE [P1l1 P2l2 P3l3 ]

where P1, P2, P3, are forces acting on sections 1, 2, 3 ... and l1, l2, l3 are

the lengths of the respective sections.

Problem 1.20: A copper bar having cross sectional area of 1500 mm2 issubjected to an axial forces as shown in figure. Find the total elongation of a

bar. Take E 1 105 N/mm2

Given: Area A 1500 mm2 ; Young’s modulus E 1 105 N/mm2;

To find: Elongation for each section of the bar.


Solution:

Sign convention

Tensile Away from element ( ve)

Compressive towards element ( ve)

Force on AB

At A, 30 kN

Resultant At B, 70 kN

25 kN

15 kN

30 kN

Total force on AB 30 kN Tensile

Force on BC

At B, 70 kN 30 kN 40 kN

At C, 25 kN 15 kN 40 kN

Total force on BC 40 kN compressive

Forces on CD

At C, 30 kN 70 kN 25 kN 15 kN

At D, 15 kN

Total forces on CD 15 kN compressive

To calculate elongation

1

AE P1l1 P2l2 P3l3

1

1500 1 105 30 600 40 1000 15 1200

2.66 10 4 mm Reduction in length

A B

30kN 30kN

B C

40kN 40kN

C D

15kN 15kN


Problem 1.21: Find the value of ‘P’ and the change in length of eachcomponent and the total change in length of the bar shown in figure. Take

E 200 kN/mm2

(AU. Apr 2011)

Given: E 200 kN/mm2

To find: Force ‘P’, Elongation

Solution:

Since the rod is in equilibrium, forces acting towards left forcesacting towards right.

130 120 P 50

P 130 120 50 200 kN

To find individual elongation, use the principle of superposition

A 4

d2 4

252 490.87 mm2

At section AB

Force At A, 130 kN

At B, 200 50 120 130 kN Tensile

Change in length of AB

AB Pl1A1E

130 800

490.87 200 1.059 mm

25m m dia

130 kN 130 kN

800 m m

A B


At section BCForce at B 130 200 70 kN

C 50 120 70 kN

Force acting on BC 70 kN (compressive)Change in length of BC BC

A2 4

d2 4

302 706.858 mm2

BC Pl2A2E

70 1600

706.858 200 0.792 mm

At section CDForce on C 130 200 120 50 kN

Force on D 50 kN

Force acting on CD 50 kN Tensile

Change in length of CD CD

A3 4

d2 4

202 314.159 mm2

CD Pl3A3E

50 400

314.159 200 0.318 mm

Total change in length: AB BC CD 1.059 0.792 0.318

0.585 mm ElongationProblem 1.22: A member is subjected to point loads as shown in Fig.Calculate the force P, necessary for equilibrium if P1 45 kN, P3 450 kN

and P4 130 kN. Determine total elongation of the member, assuming the

modulus of elasticity to be E 2.1 105 N/mm2. (AU. Nov/Dec 2014)

400mm

50kN 50kN

20m m diaC D

30m m dia

B C

70kN70kN

1600m m


Given: P1 45 kN ; P3 450 kN ; P4 130 kN

Modulus of Elasticity E 2.1 105 N/mm2

To find: P2 ? Total elongation L ?

Solution:

P1 P2 P3 P4 For equilibrium

45 P2 450 130

P2 365 kN

Consider portion AB

elongation of AB LAB P1 LAB

E AAB

45 103 1200

2.1 105 625

0.4114 mm elongation

Consider portion BC

LBC PBC LBC

E ABC

320 103 600

2.1 105 2500 PBC 320 kN

PBC P1 P2

0.3657 mm contraction 45 365

320 kN

A

B C

D

P 1P 4

P 2 P 3

1 20 0m m

6 00m m

9 00m m

6 25 m m 2 1 25 0m m 2

250

0 m

m2

Fig.


Consider portion CD

LCD PCD LCD

E ACD

130 103 900

2.1 105 1250

0.4457 mm ElongationTotal Elongation L LAB LBC LCD

0.4114 0.3657 0.4457

0.49143 mm

Problem 1.23: A brass bar havinga cross sectional Area of

1000 mm2 is subjected to axial forceas shown in figure. Find the totalchange in length of the bar take

E 1.05 105 N/mm2

Solution:

Force at A 50 kN

Force at B 80 20 10

50 kN

Part AB is subjected to a tension of 50 kN

Extension of AB P1l1AE

50 1000 600

1000 1.05 105

0.2857 mm

Force at B 50 80 30 kN 30 kN compressive

Force at C 20 10 30 kN 30 kN compressive

Hence, Part BC is subjected to a compression of 30 kN

Contraction of

BC P2l2AE

30 1000 1000

1000 1.05 105 0.2857 mm

Force on D 10 kN 10 kN compressive

50kN 10kN80kN 20kN

600m m 1000 m m 1200 m m

A B C D


Force on C 50 80 20 10 kN 10 kN compressive

Hence, Part CD is subjected to a compression of 10 kN

Contraction of

CD P3l3AE

10 1000 1200

1000 1.05 105 0.1143 mm

Total change in length of bar 0.2857 0.2857 0.1143

l or 0.1143 mm

i.e. Total decrease in length of bar 0.1143 mm

Problem 1.24: A memberABCD is subjected to point loadsP1, P2, P3 and P4 as shown in

figure. Calculate the force P3

necessary for equilibrium ifP1 120 kN ; P2 220 kN and

P4 160 kN. Determine also the

net change in length of member.

Take E 2 105 N/mm2.

Solution:Resolving the forces on the axis of rod, we get

P1 P3 P2 P4

120 P3 220 160

P3 260 kN

Part AB: Tensile force on AB P1 120 kN

Elongation on ABL1

P1L1

A1E

120 1000 750

40 40 2 105

L1 or 1 0.28 mm

Part BC:

Force B 120 220 100 kN 100 kN compressive

40X40m m25X25m m 30X30m mA B C

P2 P3P1 P4

0 .75 m 1 m

120 220 160

1.2 m

D

[L ]


Force on C 160 260 100 kN 100 kN compressive

Compressive force on BC 100 kN

Contraction on BC PL2

A2E

100 1000 1000

25 25 2 105 0.80 mm

L2 or 2 0.80 mm contraction

Part CD: Tensile force on CD P4 160 kN

Elongaiton on CD P4 L3A3 E

L3 or 3 160 1000 1200

30 30 2 105 1.07 mm

Net change in length of bar, 1 2 3

0.28 0.80 1.07 0.55 mm extensionProblem 1.25 A steel member ABCD of uniform cross-section area

1000 mm2 is subjected to axial force as shown in fig. Calculate the force‘P’ required for equilibrium of the member and determine the change inlength of member.

SolutionFor equilibrium condition

Total force towards right Total force towards left

P 10 6 9

P 15 10 5 kN

A B C D

6kN 10kN 9kNP

1m 1.2m 8m


First consider the portion BC, on the left of A there is an axial pullof 6 kN. To the right of B also there is a net pull of 6 kN 10 9 5. Portion AB is under axial tension of 6 kN.

Elongation of portion AB P1 L1

AE

6000 1000

100 2 105 0.3 mm

Next consider portion BC On its left side (at B), there is a net axialpush of 4 kN 10 6. On its right side also at C there is a net push of 4

kN 9 5. Therefore portion AB is under axial compression of 4 kN.

Contraction of portion BC P2 L2

AE

4000 1200

100 2 105

0.24 mm

Similarly portion CD is under the axial tension 5 kN

Elongation of portion CD P3 L3

AE

5000 800

100 2 105

0.20 mm

Net elongation of member

L 0.30 0.24 0.20

L 0.26 mm

Force P at D 5 kN

Change in length 0.26 mm elongation


1.21 STRESS IN BARS OF UNIFORMLY TAPERING CROSS SECTIONDerive an expression for change in length of a circular bar with uniformlyvarying diameter and subjected to an axial load P. (AU. Nov/Dec 2014)Consider a circular bar of uniformly tapering cross section

P Axial Load

l length of bar

D1 Diameter of bigger end

D2 Diameter of smaller end

Consider a section of the bar dx at a distance of x from the largerend. The diameter of this section D can be calculated by

D D1 D1 D2

l x D1 kx

where k D1 D2

l

Cross sectional area of dx, Ax 4

D1 kx2

PD 1D

D 2

x

�

dx

P

Fig 1.20


Stress LoadArea

P

4

D1 kx2

4P

D1 kx2

Strain Stress

Youngs modulus

4P

D1 kx2

E

4P

E D1 kx2

The elongation of the element dx can be calculated by

dx strain dx 4P

E D1 kx2 dx

The total elongation of the bar can be calculated by integrating theabove equation between 0 and l.

Total elongation

0

l

4P dx

D1 kx2 E

4PE

0

l

dx

D1 kx2

4PE

D1 kx 1

1 k 0

l

4PEk

1D1 kx

0

l

4PEk

1D1 kl

1

D1

Substituting k D1 D2

l in the above equation

4P

E D1 D2

l

1

D1 D1 D2l

l

1

D1

4Pl

E D1 D2

1D2

1

D1


Deformation

4Pl

E D1 D2 D1 D2

D1D2

4PlED1D2

If the bar has uniform cross section i.e., D1 D2 D,

then 4Pl

ED2 Pl

4

D2E

PlAE

Problem: 1.26 A circular rod 3 m long tapers uniformly from 25 mmdiameter to 12 mm diameter. Determine the extension of the rod under a pull

of 30 kN. Assume the modulus of elasticity is 2 105 N/mm2

(Au. Nov 1996)

Given: Length l 3 m; Diameter D1 25 mm; D2 12 mm

Load P 30 kN; Young’s modulus E 2 105 N/mm2

To find: Extension of the rod for a tapered rod

Solution:

4Pl

ED1D2

4 30 103 3000

2 105 25 12 1.909 mm

Problem 1.27: Determine the modulus of elasticity of the rod which tapersfrom 40 mm to 15 mm in diameter. The rod is 400 mm long and is subjected

to an axial load of 5 103 N. The elongation of the rod is 0.035 mm.

Given: Large Diameter 40mm; Smaller Diameter 15mm;

Length l 400mm; {Load P 50 103 N ; Extension 0.035mm

To find: Young’s modulus

Solution:

We know 4Pl

ED1D2

E 4Pl

D1D2

4 5 103 400

40 15 0.035

Young Modulus E 1.21 105 N/mm2


1.22 DEFORMATION OF UNIFORMLY TAPERING RECTANGULAR BAR

A bar of uniform thickness ‘t’ tapers uniformly from a width of b1 at

one end to b2 at other end in a length ‘L’ as shown in Fig. 1.21 subjected

to an axial force ‘P’

Consider an elemental length dx at a distance ‘x’ from larger end. Rate

of change of breadth b1 b2

L

Hence, width of section x is b b1 b1 b2

L x

b b1 kx

where k b1 b2

L

Cross-sectional area of element A t b1 kx

Extension of element PdxAE

Pdx

b1 kx Et

Fig. 1.21

Pb1

b2

b1Pb

x dx

L

t


Total Extension of bar is L 0

L

Pdx

b1 kx tE

PtE

0

L

dx

b1 kx

PtE

1 k

[ln b1 kx]0

L

PtE

1 k

ln

b1

b1 b2

L L ln b1

PtE

1 k

ln b2 ln b1

P

tEk [ ln b2 ln b1]

P

tEk ln

b1

b2

or L PL

tE b1 b2 ln

b1b2

Problem: 1.25: A steel flat plate AB of 1 cm thickness tapers uniformlyfrom 10 cm to 5 cm width in a length of 40 cm. Determine the elongationof the plate, if an axial tensile force of 5000 N acts on it. Take

E 2 105 N/mm2. (AU. Nov 2013)

Given: Thickness t 1 cm; Bigger end b1 10 cm; Smaller end b2 5 cm;

Length L 40 cm; Force P 5000 N

To find: Elongation L (or)

Solution:

PL

Et b1 b2 ln

b1

b2

5000 400

2 105 10 100 50 ln

10050

0.0138

0.0138 mm


1.23 DEFORMATION IN COMPOUND OR COMPOSITE BARSA compound (or) composite bar is one

which is composed of two or more differentmaterials which are joined together in such a waythat they are elongated or compressed as a singleunit.

To solve problems in composite bars, thefollowing two points must be noted.

1. The extension or contraction of thematerials making up the bar are equal.Therefore, the strains i.e., deformation perunit length are also equal.

2. The total load on the bar is equal to the sum of the loads carriedby the different materials.

Consider a composite bar made up of two different materials.

P Total load on bar

l Length of bar

A1, E1, P1 Cross section area, Young’s Modulus and load on material 1.

A2, E2, P2 Cross sectional area, Young’s Modulus and load on material 2.

Total load on bar, P P1 P2

Stress in bar 1, 1 P1

A1

Strain in bar 1, e1 1

E1

P1

A1E1

Change in length of bar 1, 1 P1l1A1E1

Change in length of bar 2, 2 P2l2A2E2

Since change in length of both materials are equal

1 2

1 2 �

PFig 1.22


P1l

A1E1

P2l

A2E2

P2 P1 A2E2

A1E1

We know, P P1 P2 P1 P1 A2E2

A1E1

P1 1

A2E2

A1E1

P P1 A1E1 A2E2

A1E1

P1 P

A1E1

A1E1 A2E2

Similarly P2 P

A2E2

A1E1 A2E2

Using the above equations, we can find the load shared by the differentmaterials.

Also, since strain in the different materials are equal

e1 e2

1

E1

2

E2

1 E1

E2 2

2 E2

E1 1

From the above equations, we can find the stresses in the differentmaterials.


Problem 1.26: A reinforced concrete column 500 mm 500 mm in sectionis reinforced with 4 steel bars of 10 mm diameter, one in each corner. Thecolumn is carrying a load of 2 MN. Find the stresses in the concrete andsteel bars.

Take E for steel 2.1 105 N/mm2

and E for concrete 1.4 105 N/mm2 (AU. May/June - 2013)

Given: Size of concrete column 500mm 500mm;

Diameter of steel bar 10 mm

Load 2 106N

Esteel 2.1 105N/mm2

Econcrete 1.4 105 N/mm2

To find: Stresses in concrete and steelSolution:

We know that in compound bars, strain in steel strain in concrete

esteel econcrete

steel

Esteel

concrete

Econcrete

s Es

Ec c

s 2.1 105

1.4 105 c

s 1.5 c ...(1)

We know, P Ps Pc sAs cAc

As 4

102 4 314 mm2

Ac 500 500 314 249686 mm2

P sAs cAc

2 106 s 314 c 249686


2 106 314 1.5 c 249686 c [From (1)]

2 106 250157 c

c 7.9950 N/mm2

s 1.5 c 1.5 7.9950

and 11.9925 N/mm2

Problem 1.27: Two vertical rods one of steel and the other of copper are each

rigidly fixed to the top and 600 mm apart. Diameters and lengths of each rods are

25 mm and 5 m respectively. A cross bar is fixed to the rod at the lower end carries

a load of 7000 N such that the cross bar remains horizontal even after loading. Find

the stress in each rod and the position of the load on the bar. Take

‘E’steel 2 105 N/mm2 and Ecopper 1 105 N/mm2 (AU. Apr/May 2011)

Given Data:

Distance between bars 600 mm

Diameter of each bar 25mm Length of each bar 5m 5000mm

To find: Stress in each rod; Position of loadSolution:

Since the cross rod remains horizontal, the elongation of steel andcopper are equal. Thus the strain on the rods are also equal since the rodshave the same original length

es ec

s

Es c

Ec Strain

StressYoung’s modulus

s Es

Ec c

s 2 105

1 105 c

s 2 c ...(1)

We know, total load Load on steel Load on copper

600mm5mm S C

x

7000N


7000 sAs c Ac

7000 s 4

252 c 4

252

7000 s c156.25

2c c 156.25 [From (1)]

7000 3 c 156.25

c 7000

468.75 4.75 N/mm2

s 2 c 9.51 N/mm2

To find position of loadLet x be the distance of the load from the copper rod

Taking moments about copper rod

7000 x Ps 600 0

7000 x Ps 600

Ps sAs 9.51 156.25 4666.67 N

7000 x 4666.67 600

x 400 mm

i.e. 400 mm to left of copper rod

Problem 1.28: A steel rod of 25 mm diameter is placed inside a coppertube of 30 mm internal diameter and 5 mm thickness and the ends are rigidlyconnected. The assembly is subjected to a compressive load of 250 kN.Determine the stresses induced in the steel rod and copper tube and totalelongation in composite sections in length of 1m. Take the modulus ofelasticity of steel and copper as 200 GPa and 80 GPa respectively. (AU. Nov 2006)

Given:Diameter of steel rod, ds 25 mm

Length of both rod and tube 1 m 1000 mm

Inner diameter of Copper tube, dc 30 mm

Outer diameter of Copper tube, Dc 30 5 2 40 mm


Compressive load, P 250 kN 250 103N

Es 200 GPa 2 105 N/mm2 ; Ec 80 GPa 0.8 105 N/mm2

To find: Stresses in steel rod s and copper tube c

Solution:

We know that

Change in length of steel Change in length of copper

Strain in steel Strain in copper

s

Es

c

Ec [. . . Original length is same]

s c Es

Ec c

2 105

0.8 105

s 2.5 c

We know that, Total load P Ps Pc

250 103 s As c Ac

250 103 2.5 c 4

252 c 4

402 302

1227.19 c 549.8 c 1776.99 c

c 250 103

1776.99 140.68 N/mm2

s 2.5 c 2.5 140.68 351.72 N/mm2

Total elongation, l lc ls

l c l

Ec s l

Es

140.68 1000

0.8 105

1.7586 mm


Problem 1.29: A compound tube consists of a steel tube 140 mm internaldiameter and 160 mm external diameter and an outer brass tube 160 mminternal diameter and 180 mm external diameter. The two tubes are of samelength. The compound tube carries an axial compression load of 900 kN.Find the stresses and the load carried by each tube and the amount of itsshortens. Length of each tube is 140 mm. Take E for steel as

2 105 N/mm2 and for brass 1 105 N/mm2.

Data:ds Internal dia of steel tube = 140 mm

Ds External dia of steel tube = 160 mm

dc Internal dia of brass tube = 160 mm

Dc External dia of brass tube = 180 mm

Length of the tube = 140 mm

Compressive load; P 900 kN 9 105 N

Es 2 105 N/mm2 ; Eb 1 105 N/mm2

Find:

(i) Stress in steel tube s

(ii) Stress in brass tube b

(iii) Length of shortness L

Solution:Now we calculate the area of tube

Area of steel tube As 4

Ds2 ds

2 4

1602 1402

As 4712.4 mm2

Area of brass tube Ab 4

Dc2 dc

2 4

1802 1602

Ab 5340.7 mm2

W.K.T,

Change in Length of steel = change in length of brass


Strain in steel = strain in brass

s

Es b

Eb

Now,

s b Es

Eb

s b 2 105

1 105

s 2b

W.K.T

Total load P Ps Pb

P s As b Ab

9 105 s As b Ab

9 105 2b 4712.4 b 5340.7

9 105 9424.8 b 5340.7 b

9 105 14766 b

b 9 105

14766 60.95 N/mm2

b 60.95 N/mm2

s 2 b 2 60.95 121.9

s 121.9 N/mm2

Ps s As 121.9 4712.4 5.74 105 N

Pb b Ab 60.95 5340.7 3.26 105 N

Strain s

Es

121.9

2 105 6.095 10 4


Also, Strain Change in Length

Orginal Length

6.095 10 4 L140

L 0.085 mm

Problem 1.30: A steel rod of 3 cm diameter is enclosed centrally in a hollowcopper tube of external diameter 5 cm and internal diameter of 4 cm as shownin Fig. The composite bar is then subjected to axial pull of 45000 N. If thelength of each bar is equal to 15 cm, determine: (i) The stresses in therod and tube, and (ii) Load carried by each bar. Take E for steel

2.1 105 N/mm2 and for copper 1.1 105 N/mm2. (AU. Nov/Dec 2015)

3 cm

4 cm

5 cm

P = 450 00 N

15 cm

C opper Tu be

S tee l R od


Given:

Dia of steel rod 3 cm 30 mm

Area of steel rod As 4

302 706.86 mm3

External dia of copper tube 5 cm 50 mm

Internal dia of copper tube 4 cm 40 mm

Area of copper tube AC 4

502 402 mm2

706.86 mm2

Axial pull on composite Bar P 45 kN

Length of each Bar L 15 cm 150 mm

ec 1.1 105 N/mm2, es 2.1 105 N/mm2

(i) The stress in the rod & tube:

Strain in steel strain in copper

s

Es

cEc

s c

Ec Es

2.1 105

1.1 105 c

s 1.909 c

stress loadarea

; load stress area

Total load Load on steel load on copper

P s As c Ac

45000 c 1.909 706.86 c 706.86

2056.31c 45000

c 21.88 N/mm2

s 1.909 21.88 41.77 N/mm2


(ii) Load carried by each BarLoad Stress Area

Load carried by steel rod Ps s As

P 41.77 706.86 29531.25 N

Load carried by copper tube Pc 45000 29525.5

Pc 15468.75 N

Problem 1.31: A steel rod and two copper rods together support a load of400 kN as shown in figure. The cross sectional area of steel rod is

2000 mm2 and each copper rod is 1500 mm2. Find the stress in the rods.

Take Es 2 105 and Ec 1 105 N/mm2

Given Data:

Load 400kN 400 103 N

Area of steel rod As 2000 mm2

Area of copper rod Ac 2 1500 3000 mm2

Es 2 105 N/mm2; Ec 1 105 N/mm2

Length of steel Ls 15 10 25 cm 250 mm

Length of copper Lc 15cm 150mm


To find: Stresses in copper and steelSolution:

We know change in length of steel Change in length of copper.

Strain of steel original length Strain in copper original length.

s

Es Ls

c

Ec Lc

s

2 105 250

c

1 105 150

s 150

1 105 2 105

250 c

s 1.2 c

Also, total load P Ps Pc sAs cAc

400 103 s 2000 c 3000

400 103 1.2 c 2000 c 3000

400 103 5400 c

c 74.07 N/mm2

s 1.2 c 1.2 74.07 88.889 N/mm2

s 88.889 N/mm2

Problem 1.32: Three pillars, two of aluminium and one of steel support arigid platform of 20 kN as shown in Fig. If area of each aluminium pillar

is 1000 mm2 and that of steel pillar is 800 mm2, find the stresses developed

in each pillar. Take Ea 1 105 N/mm2 and Es 2 105 N/mm2. What

additional load ‘P’ can it take if working stresses are 65 N/mm2 in aluminium

and 150 N/mm2 in steel?


Solution:

Step 1Due to 200 kN load only.

Let Pa be force in each of aluminium pillars and Ps be the force in

steel pillar.

F force in vertical direction 0

Pa Ps Pa 200

2Pa Ps 200 ...(1)

Ls La

Ps Ls

As Es

Pa La

Aa Ea

Ps 250

800 2 105 Pa 200

1000 1 105

Ps 1.28 Pa ...(2)From equation (1) and (2), we get

250mm

200m m

Stee l

Alum in ium

Alu

min

ium

W = 200kN

P = ?


Ps 2 Pa 200

Ps 1.28 Pa 0

3.28 Pa 200

Pa 60.97 kN

Ps 200 2 60.97 78.048

Ps 78.048 kN

Stress developed

in Aluminium a 60.9751000

1000 60.976 N/mm2

in Steel s 78.048

800 1000 97.56 N/mm2

Step 2 Additional load carrying capacity

(a) If a governs the load carrying capacity

a 65 N/mm2

Pa 65 1000 65000 N ; Ps 1.28 65000 83200 N

Total load carrying capacity 2Pa Ps 213.2 kN

(b) If s governs the load carrying capacity

s 150 N/mm2

Ps 150 800 120000 N

Pa 120000

1.28Pa 93750 N

Total load capacity 2Pa Ps

307500 N 307.5 kN

Selecting a as the governing stress for load carrying capacity,

Pmin 213.2 N

Additional load P Pmin W

P 213.2 200

P 13.2 kN


Problem 1.33 A steel rod 18 mm in diameter passes centrally through asteel tube 25 mm in internal diameter and 30 mm in external diameter. Thetube is 750 mm long and is closed by rigid washers of negligible thicknesswhich are fastened by nuts threaded on the rod. The nuts are tightened untilthe compressive load on the tube is 20 N. Calculate the stresses in the tubeand rod. Find the increase in these stresses when one nut is tightened byone quarter of a turn relative to the other. There are 4 threads per 10 mm.

Take E 200 kN/mm2

Solution:

Note: When the nuts are tightened, the tube will be compressed and the rodwill be elongated. Since no external forces have been applied, the compressiveload on the tube must be equal to the tensile load on the rod

Area of tube At 4

D2 d2 4

302 252 216 mm2

Area of Rod AR 4

dr2

4

182 254.5 mm2

Tensile load on rod Compressive load on the tube

r Ar t At

r t At

Ar 0.8487 t

(i) Stress in the tube when Load is 20 kN

Stress in tube t 20216

92.6 10 3 N/mm2 (compressive)

Stress in rod

r t AtAr

0.8487 t 0.8487 92.6 10 3 78.6 10 3 N/mm2

(ii) When one nut is tightened by one quarter of turn. Let t and r be

stresses in tube and rod due to tightening of nut.

Obviously r 0.8487 t

Reduction in length of tube

Stress, Strain and Deformation of Solids 1.71

t tE

l t 750

200 103 0.00375 t mm

Extension in length of Rod

r rE

l 0.8487 t200 103 750 0.003183 t mm

Reduction in tube length + Extension in Rod Length Axial movement innut

0.00375 t 0.003183 t 14

14

10

0.006933 t 0.625

t 90.15 N/mm2 (compressive)

r 0.8487 90.15 76.52 N/mm2 tensile

Problem 1.34 Three bars made of copper, zinc and aluminium are of equallength and have cross-section 250, 375 and 500 square mm respectively. Theyare rigidly connected at their ends. If this compound member is subjected toa longitudinal pull of 125 kN estimate the proportion of the load carried oneach rod and induced stresses. Take the value of E for copper

1.3 105 N/mm2, for zinc 1 105 N/mm2 and for aluminium

0.8 105 N/mm2 (AU. Nov/Dec 2013)

P=125kN P=125kN

250m m 2

375m m 2

500m m 2

�

Copper

Zinc

A lum in ium


Given:

Total load, P 125 kN 125 103 N

For copper bar,

Area, Ac 250 mm2 ; Ec 1.3 105 N/mm2

For zinc bar,

Area Az 375 mm2 ; Ez 1 105 N/mm2

For aluminium bar,

Area, Aa 500 mm2 ; Es 0.8 105 N/mm2

Solution:

Strain in copper Strain in zinc Strain in aluminium

Stress in copperEc

Stress in zinc

Ez

Stress in aluminiumEa

c

Ec

z

Ez aEa

c Ec

Ea a

1.3 105

0.8 105 a 1.625 a

...(1)

z Ez

Ea a

1 105

0.8 105 a 1.25 a

...(2)

Total load Load oncopper

Load on zinc

Load onAluminium

125 103 stress in copper Ac stress in zinc Az

stress in Aluminium Aa

c Ac z Az a Aa

1.625 a 250 1.25 a 375 a 500

1375 a

a 125 103

1375 90.909 N/mm2

Substitute the value of a in equation (1) and (2), we get


c 1.625 90.9 147.73 N/mm2

z 1.25 90.9 113.64 N/mm2

Now load shared by copper Pc c Ac

147.73 250 36931.82 N

Load shared by zinc Pz z Az

113.625 375 42613.64N

Load shared by aluminium rod Pa a Aa

90.909 500 45454.54 N

1.24 BAR OF UNIFORM STRENGTHFig. 1.23 shows a bar subjected to an external tensile load P. If the

bar had been of uniform cross section, the tensile stress intensity at anysection would be constant, only if the self weight of the member is ignored.If the self weight is considered, the intensity of stress increases for sectionsat higher level.

It is possible to maintain a uniform stress for all the sections byincreasing the area from the lower end to the upper end.

A 1

A

A 2

A = Area

Fig. 1.23


If A1 and A2 are the area at upper and lower ends. If w is the weight

per unit volume of member and be the uniform stress intensity, then wehave

A1 A2 e wl/

i.e ln A1

A2

w

l

At any location x from bottom with Area A

ln AA2

w

x

Problem 1.35: A vertical tie of uniform strength is 18 m long. If the area

of the bar at lower end is 500 mm2; find the area at the upper end when

the tie is to carry a load of 700 kN. Take w 8 10 5 N/mm3

Given: A2 500 mm2 ; l 18 m ; Load P 700 kN ;

w 8 10 5 N/mm3

To find: Area at upper endSolution:

Intensity of stress 700 103

500 1400 N/mm2

We know that, ln A1

A2

w

l

ln A1

A2

8 10 5 180001400

A1500

e1.029 10 3

A1500

1.001029

A1 1.001029 500 500.514 mm2

Area at upper end A1 500.514 mm2


1.25 THERMAL STRESSES

When a body is subjected to an increase in temperature, the body tendsto expand. This is due to the tendency of the molecules to move further apartfrom each other. When the body is cooled it will contract. If this body isallowed to expand or contract freely with the increase and decrease intemperature, the body will not undergo any stresses. However, if thedeformation of the body is prevented by some means, stresses will be inducedin the body. This type of stress is called thermal stress or temperaturestress. The corresponding strain is called thermal strain or temperaturestrain.

1.25.1 Thermal Stresses in Simple BarsThe thermal stress in a body can be calculated by the following

method.

(a) Calculate the amount of deformation, the body will undergo due tothe temperature change

(b) Calculate the force used to bring the body back to its original length

(c) Calculate the stress induced by the above load

Consider a body subjected to increase in temperature

Let ‘l’ be the length of the rod; ‘t’ be the increase in temperature and‘’ be the coefficient of linear expansion

The increase in length l due to the change in temperature ‘t’

l l t

If the ends of the bar are fixed so that the bar cannot expand, thereis compressive stress and strain induced.

Compressive strain is e ll

ltl

t

StressStrain

E

Stress Strain E

t E

The compressive stress eE t E

t E


In some cases where the supporting structures of the rod are not rigidlyfixed, they might yield to a certain distance due to the stress on the rod.Then the actual expansion of the rod would be

l l t

The strain here is e ll

lt

l t

l

and stress induced eE t

l E

Here, change in temperature ‘t’ or ‘T’ will not be affected with unitsin K (or) C.

Problem 1.36: A rod is 2 m long at a temperature of 10C. Find the

expansion of the rod, when the temperature is raised to 80C. If thisexpansion is prevented, find the stress induced in the material of the rod.

Take E 1.0 105 MN/m2 and 0.000012 per degree centrigrade.

Given

Length of rod 2m 2000 mm

Initial temperature t1 10 C

Final temperature t2 80 C

Rise in temperature t t2 t1 80 10 70 C

Young’s modulus, E 1.0 105 MN/m2 1 105 N/m2

Co-efficient of linear expansion 0.000012/ C

Solution:

Step 1: Expansion of rod due to temperature rise tL

0.000012 70 2000 1.68 mm

Step 2: Stress in the material of rod if expansion is prevented by equation

Thermal stress, tE

0.000012 70 1.0 105 N/mm2

P 84 N/mm2


Problem 1.37: A copper rod 2 m long is supported at its ends. If the thermal

stresses should not exceed 60 N/mm2, calculate the temperature through whichthe rod should be heated if: (a) The supports do not yield; (b) The supports

yield by 1 mm: Take E 1 105 N/mm2 and 18 10 6 / C.

Data Given: Length ‘l’ 2 m; Maximum stress 60 N/mm 2;

Coefficient of expansion 18 10 6 / C ; Young’s Modulus E 1 105 N/mm2

To find: Temperature change

Solution:

(a) When the supports do not yield

Thermal Stress t E

60 18 10 6 t 1 105

Rise in temperature t 60

18 10 6 1 105 33.33 C

(b) When the supports yield by 1 mm

Thermal Stress t

ll

E

60 18 10 6 t

12000

1 105

60

1 105 18 10 6 t 1

2000

60

1 105 1

2000 18 10 6 t

18 10 6 t 1.1 10 3

Rise in Temperature t 1.1 10 3

18 10 6 61.11 C

Problem 1.38: A steel rod of 3 cm diameter and 5 m long is connected totwo grips and the rod is maintained at a temperature of 95C. Determine

the stress and pull exerted when the temperature falls to 30C if(i) the ends do not yield (ii) the ends yield by 0.12 cm

Take E 2 105 MN/m2 and 12 10 6/ C


Given: Dia of rod, d 3cm 30 mm

Area of rod, A 4

302 225 mm2

Length of rod, l 5 m 5000 mm

Initial temperature t1 95 C

Final temperature t2 30 C

Change in temperature, t t1 t2 95 30 65C

Modulus of elasticity, E 2 105 MN/m2

2 105 106 N/mm2

2 1011 N/mm2

Co-efficients of linear expansion, 12 10 6 / CSolution:

Step 1 When the ends do not yield

Thermal Stress t E 12 10 6 65 2 105

156 N/mm2 Tensile

Pull in rod stress area 156 225

110269.9 N

Step 2 When the ends yield by 0.12 cm

0.12 cm 1.2 mm

Stress t l

l E

12 10 6 65 5000 1.2

5000 2 105

3.9 1.2

5000 2 105 108 N/mm2

Pull in the rod stress area 108 225

76340.7 N

1.25.2 Thermal Stresses in Composite BarsWhen two different materials are subjected to the same temperature

change, they will expand or contract to different lengths. This is because eachmaterial has its own coefficient of expansion. Suppose two materials of


different coefficients of expansion are rigidly fixed together and is subjectedto a rise in temperature. If the bars were free to expand, no stress would beinduced, but since the bars are rigidly fixed together, they would have toexpand by the same amount. This would cause a compressive stress in thematerial with a higher coefficient of expansion and a tensile stress in thematerial with a lower coefficient of expansion

Let 1 and 2 be the stress induced in the material 1 and 2 respectively

e1, e2 be the strains induced in materials 1 and 2 and 1, 2 be the

coefficients of expansion of materials 1 and 2.The compressive load on one material is equal to the tensile load on

the other1A1 2A2 ...(1)

Since the materials are rigidly fixed together the expansion is the same. Actual expansion in material 1 Actual expansion in material 2

Actual Expansion in material 1

Free expansionin Material 1

Expansion due to tensilestress in Material 1

1TL 1E1

L

Actual expansion in material 2


Contraction due to compressivestress in 2

2TL 2

E2 L

Actual Expansion in 1 Actual Expansion in 2

1TL 1

E1 L 2TL

2

E2 L

1T 1

E1 2T

2

E2

On rearranging the above equation we get

2T 1T 1

E

2

E


if E

strain e per unit length

1

E 2

E e1 e2 T 2 1

Total strain e1 e2 T 2 1

Problem 1.39: A steel rod of 20 mm passes centrally through a copper tubeof 50 mm external diameter and 40 mm internal diameter. The tube is closedat each end by rigid plates of negligible thickness. The nuts are tightenedlightly on the projecting parts of the rod. If the temperature of the assemblyis raised by 50C, calculate the stresses developed in copper and steel. Take

Es 200 kN/mm2, Ec 100 kN/mm2,

s 12 10 6 perC, c 18 10 6perC (AU. Nov/Dec 2016)

Given: Dia. of steel rod, ds 20 mm

Dia. of copper tube, Dc 50 mm & dc 40 mm

Change in Temperature, T 50C

Es 200 kN/mm2 200 103 N/mm2

Ec 100 kN/mm2 100 103 N/mm2

s 12 10 6/C ; c 18 10 6/C

Solution:

Since c s, copper will expand more than steel. But, here the copper

tube is closed at each end by rigid plates, the steel rod and copper tube arenot free to expand. The members expand by the same amount. The coppertube, supposed to expand more, being restricted, will be subjected tocompressive stress c while the steel rod will be subjected to tensile stress

s.

For the equilibrium of the system,

Tensile load on steel Compressive load on copper.

i.e s As c Ac


s 4

ds2 c

4

Dc2 dc

2

s c

4

Dc2 dc

2

4

ds2

c Dc

2 dc2

ds2

c 502 402

202

s 2.25 c

We know that

Actual expansion of steel Actual expansion of copper

Free expansion of steel Expansion due to tensile stress in

steel

Free expansion of copper Contraction due to compressive

stress in copper

i.e., s T L s

Es L c T L

c

Ec L

s T s

Es c T

c

Ec

Substituting the values,

12 10 6 50 2.25 c

200 103 18 10 6 50 c

100 103

0.0006 1.125 10 5 c 0.0009 0.1 10 4 c

1.125 10 5 c 0.1 10 4 c 0.0009 0.0006

2.125 10 5 c 0.0003

c 0.0003

2.125 10 5 14.117 N/mm2

s 2.25 c 2.25 14.117 31.76 N/mm2

c 14.117 N/mm2 & s 31.76 N/mm2


Problem 1.40: A steel tube of 30 mm external diameter and 20 mm internaldiameter encloses a copper rod of 15 mm diameter to which it is rigidlyjoined at each end. If, at a temperature of 10C there is no longitudinalstress, calculate the stresses in the rod and tube when the temperature is

raised to 200C. Take E for steel and copper as 2.1 105 N/mm2 and

1 105 N/mm2 respectively. The value of co-efficient of linear expansion for

steel and copper is given as 11 10 6perC and 18 10 6 per Crespectively. (AU. May/June 2012)

Given data,Diameter of rod 15 mm

Area of copper rod Ac 4

152 176.71 mm2

Diameter of steel tube Ds 30 mm, ds 20 mm

Area of steel tube, As 4

302 202 392.69 mm2

Rise of temperature 200 10 190C

E for steel Es 2.1 105 N/mm2

E for copper, Ec 1 105 N/mm2

Value of for steel s 11 10 6 perC

Value of for copper c 18 10 6 perCSolution:

Let s , Stress in steel; c , Stress in copper

For equilibrium of the system

Compressive load on copper = Tensile load on steel

c Ac s As

c s AsAc

c s 125

56.25

c 2.22 s ...(1)


Now actual expansion of steel Free expansion of steel Expansiondue to tensile stress.

sT.L sEs

L

Actual expansion of copper expansion of copper - contraction dueto compressive stress.

c T.L cEc

L

But actual expansion of steel Actual expansion of copper.

s T.L s

Es L c T.L

c

Ec L

s T s

Es c T

c

Ec

11 10 6 190 s

2.1 105 18 10 6 190

2.22 s

1 105

s

2.1 105

2.22 s

1 105 18 10 6 190 11 10 6 190

s 2.1 2.22 s

2.1 105 7 10 6 190

s 4.462 s 7 10 6 190 2.1 105

5.662 s 279.3

s 279.35.662

s 49.328 N/mm2

sub s in equn (1) we get

c 2.22 49.328

c 109.51 N/mm2


1.25.3 Thermal Stress in Taper Bar - Circular Section

l – Length of bar

d1 – diameter at Bigger end

d2 – diameter at Smaller end

T – Increase in temperature

– Coefficient of linear expansion

If the bar AB is subjected to rise in temperature, the bar AB will tendto expand. But since both ends are fixed, compressive stress is induced.

Elongation due to rise in temperature

l l T ...(1)

P Load required to bring to original length

then l 4Pl

Ed1d2

...(2)

Equating (1) and (2) we get

l T 4Pl

Ed1d2

Load P Ed1d2 T

4

Maximum stress

max Load

Area A2 Ed1d2T

4 4

d22

TEd1

d2

max TEd1

d2

If d1 d2 ; max TE As same for simple bar

Problem 1.41: A circular bar rigidly fixed at its both ends uniformly tapersfrom 75 mm at one end to 50 mm at the other end. If its temperature israised through 26 K, then what will be the maximum stress developed in the

bar? Take E as 200 GPa and as 12 10 6 / K for the bar material. (AU. Apr/May 2010)

�

A B

d 1 d2

Fig 1.24


Given: Larger dia of taper bar, d1 75 mm

Smaller dia of taper bar, d2 50 mm

Temperature raise, T 26 K

E 200 GPa 200 103 N/mm2 ; 12 10 6 / K

Solution:

Maximum stress induced in the taper bar, max TE d1

d2

12 10 6 26 200 103 75

50

93.6 N/mm2

Problem 1.42: A rigid fixed bar 1.75 m long uniformly tapers from 250mm diameter at one end to 200 mm diameter at the other. Maximum stressin bar is limited to 216 MPa. Find the temperature through which it can be

heated. Take E 100 GPa ; 18 10 6 / C

Solution:

Given L 1.75 m ; d1 250 mm ; d2 200 mm ;

max 216 MPa 216 N/mm2;

18 10 6 / C ; E 100 GPa 100 103 N/mm2

We know that the maximum stress induced in a taper bar is given by

max T E d1

d2

216 18 10 6 T 100 103 250

200

Rise in Temperature, T 216 200

18 10 6 100 103 250 96 C


1.25.4 Thermal Stress in Varying Section BarPQR is varying section barsubjected to increase in temperature

l1, 1, A1 length, stress, area in

section 1

l2, 2, A2 length, stress, area in

section 2

Coefficient of Expansion

T Increase in Temperature

Thermal load is shared equally inboth the sections

1A1 2A2

Total deformation of the bar (Free expansion)

l l1 l2 1l1E1

2l2E2

(For Different E )

l 1E

[1l1 2l2] [for same E]

Problem 1.43: A steel rod ABC is fixed between two supports A and C asshown in figure. Find the stress developed in the two portions of the rod,

when it is heated through 15C. Take 12 10 6 /C E 200 GPa

Solution:

If 1 and 2 are stresses in section (1) and

(2)

We know that 1A1 2A2

1 A2

A1 2

600400

2 1.5 2

Expansion (free) in part one due to rise in temperature

l1 l1 T 500 12 10 6 15 0.09 mm

�1 �2

P

Q

R

A 1 A 2

Fig 1.25


Similarly

l2 l2 T 800 12 10 6 15 0.144 mm

Total Expansion of rod, l l1 l2 0.09 0.144 0.234 mm

Also We know that

l 1E

1l1 2l2

0.234 1

200 103 1.52 500 2 800

0.234 7.75 10 3 2

2 0.234

7.75 10 3 30.2 N/mm2 30.2 MPa

Also 1 1.5 2 1.5 30.2 45.3 MPa

Problem 1.44: A composite bar made up of aluminium and steel, is heldbetween two supports as shown in figure. The bar is stress free at atemperature of 40C. What will be the stresses in the two bars when the

temperature is 23C, if (a) the supports are not yielding (b) the supportscome nearer to each other by 0.1 mm? Temp change is uniform along thelength of bar. Take Esteel 200 GPa, EAluminium 75 GPa,

S 11.7 10 6 / C ; Al 23.4 10 6 /C

Stee l A lum in ium

600m m 300m m

A =1000 m mst2

A =500m mAl2


Solution:

Given lst 600 mm ; lAl 300 mm ; Ast 1000 mm2 ;

AAl 500 mm2 ; Decrease in temp T 40 23 17C ;

If st, Al are stresses in Steel and Aluminium bar,

Free expansion in bars due to the temperature

lst lst st T 600 11.7 10 6 17 0.119 mm

lAl lAl st T 300 23.4 10 6 17 0.119 mm

Total Contraction in bar l lst lal 0.119 0.119 0.238 mm

We know that l st lst

Est Al lAl

EAl

(a) When supports does not yield

st Ast Al AAl

st Al 5001000

0.5 Al

st 0.5 Al

l 0.238 0.5 Al 600

200 103 Al 300

75 103

0.238 5.5 10 3 Al

Stress in aluminium bar

Al 0.238

5.5 10 3 43.3 N/mm2

Stress in steel bar st 0.5 43.3 21.65 N/mm2

(b) When the supports come near by 0.1 mm

l 0.238 0.1 0.138 mm

0.138 st lst

Est Al lAl

EAl


0.138 0.5 Al 600

200 103 Al 300

75 103

Al 0.138

5.5 10 3 25.1 N/mm2 or 25.1 MPa

st 0.5 25.1 12.55 N/mm2 or 12.55 MPa

Problem 1.45: A flat steel bar 200 mm 20 mm 8 mm is placed between

two Aluminium bars 200 mm 20 mm 6 mm to form a composite bar. Allthree bars are fastened at room temperature. Find the stress in each bar whenwhole assembly is raised by 80C. Take Est 200 GPa ; EAl 80 GPa ;

st 12 10 6/ C ; Al 24 10 6 / C

Given: Size of steel 200 20 8

Size of aluminium 200 20 6

Area of steel Ast 20 8 160 mm2

Area of aluminium AAl 20 6 2 240 mm2

Solution:

We know that Al AAl st Ast

st Al AAl

Ast

240160

A 1.5A

st 1.5 A


Strain in each bar est stEst

st

200 103

eAl Al

EAl

Al

80 103

We know that total strain

est eAl T A S

st

200 103

Al

80 103 80 [24 10 6 12 10 6]

1.5 Al

200 103 Al

80 103 960 10 6

20 10 6 Al 960 10 6

Stress in Aluminium bar, Al 960 10 6

20 10 6 48 N/mm2 or 48 MPa

Stress in steel bar st 1.5A 1.5 48 72 N/mm2 or 72 MPa

Problem 1.46: A rigid slab weighing 1600 kN is placed upon two bronze rods

and steel rod each of 6000 mm2 area at a temperature of 25 C as shown infigure. Find the Temperature at which the stress in steel rod will be zero. Take:

st 12 10 6 / C ; Br 18 10 6 / C; E Br 80 GPa E st 200 GPa ;

Given: P 1600 kN; ABr Ast 6000 mm2

T Rise in temperature when stress in steel rod is zero

Solution:

Since Br st, Bronze will expand more than steel. If steel bar stress

is zero, then the loads should be shared only by Bronze bars. In others words,decrease in length of two bronze bars should be equal to the difference ofthe expansion of bronze rods and steel rod.


Free expansion of bronze rod,

lBr lBr Br T 250 18 10 6 T 4.5 10 3 T

Free expansion of steel rod,

lst lst st T 300 12 10 6 T 3.6 10 3 T

Difference in expansion of two rods

4.5 10 3 T 3.6 10 3 T 0.9 10 3 T ...(1)

Contraction in bronze rod due to load 1600 kN.

PlBr

ABr EBr

1600 103 250

2 6000 80 103 0.4166 mm

...(2)

[. . . ABr 2 rods area 2 6000]

Equating (1) and (2) 0.9 10 3 T 0.4166

or Rise in Temperature T 462.2 C

Br B r


1.26 ELASTIC CONSTANTSThe modulus of Elasticity, Modulus of Rigidity, Bulk Modulus and

Poisson’s ratio are the four elastic constants of the material.

1.26.1 Modulus of ElasticityDefine young’s modulus (AU. Nov/Dec 2016)

The Ratio of the axial stress to the corresponding axial strain is foundto be constant within elastic limit. This constant is called “Modulus ofElasticity”. It is denoted by letter ‘E’.

E Axial stressAxial strain

Unit for modulus of elasticity is N/mm2 (or) Pascal or MPa.

1.26.2 Rigidity Modulus (or) Shear ModulusThe Ratio of shear stress to the corresponding shear strain is found to

be a constant upto elastic limit of material. This constant is called “Modulusof Rigidity” or “Shear Modulus”.

It is denoted by G, C, N, etc.

Shear Modulus, G Shear stressShear strain

1.26.3 Bulk ModulusDefine Bulk Modulus (AU. Apr/May 2010)

The Ratio of direct stress to thecorresponding volumetric strain is found to beconstant upto elastic limit. This elasticconstant is called “Bulk Modulus”. It isdenoted by letter ‘K’.

Bulk Modulus, K Direct stress

Volumetric strain

K ev

1.26.4 Linear Strain and Lateral StrainA body under the axial deformation is

subjected to the following strains.

1. Primary or Linear strain

2. Secondary or Lateral strain

Fig. 1.26 (a)


When a bar AB is subjected to an axial load P, then the bar is subjectedto a deformation both axially and radially. The strain in axial direction iscalled primary or Linear strain and the strain in every direction at rightangle to primary strain is called Lateral strain

The bar is expanded by length l and the diameter of the bar reducesfrom d1 to d2

Linear strain e ll

1.27 POISSON’S RATIO

If a body is stressed within the elastic limit, the lateral strain bears a

constant ratio to the linear strain known as Poisson’s Ratio 1m

or

Lateral strainLinear strain

A constant known as Poisson’s Ratio 1m

or

Lateral strain 1m

e e

Also Lateral strain Change in diameterOriginal diameter

d2 d1

d1 dd

In case of a rectangular bar

Lateral strain Change in width or thickness

Original width or thickness bb

or tt

b width of bar

t thickness of barFor most of the materials, lateral strain is opposite in sign to

longitudinal strain because,

�

A A A

d1 d 1

B B

B �

P

d 2�

Fig 1.26 (b)


When the material is subjected to tensile stress, it experiences anincrease in length in the direction of the force with correspondingdecrease in diameter and width.

When the material is subjected to compressive stress, it experiencesa decrease in length in the direction of the force with correspondingincrease in diameter and width.

Note: In some books, the poisson’s ratio is given as,

lateral strain

longitudinal strain

However, in our book, for simplicity, this negative sign is conceptuallyunderstood and it is not mentioned in the formulae.

Problem 1.47: A steel bar 3 m long, 50 mm wide and 30 mm thick issubjected to an axial pull of 200 kN in the axial direction. Find the change

in length, width and thickness of the bar. Take E 2 105 N/mm2 and

poisson’s ratio 0.3

Given: Length 3m 3000 mm ; Width 50 mm ; Thickness 30 mm ;

Load 200 kN ; E 2 105 N/mm2; Possion’s ratio 1m

0.3

To find: Change in length l; Change in breadth (b; Change in thickness

t

Solution:

Change in length l PlAE

200 103 3000

30 50 2 105 2 mm

Longitudinal Strain ll

2

3000 6.667 10 4

Lateral strainLongitudinal strain

1m

Lateral strain 1m

Longitudinal strain 0.3 6.667 10 4 2 10 4

Change in width b b Lateral Strain 50 2 10 4 0.01 mm

Change in thickness t t Lateral strain 30 2 10 4 0.006 mm


Problem 1.48: Determine the values of Young’s modulus and poisson’s ratioof a metallic bar of length of 25 cm, breadth 5 cm and depth 5 cm whenthe bar is subjected to an axial compressive load of 1.75 MN. The decreasein length is 0.175 cm and increase in breadth is 0.013 cm.

Given: Length 25 cm; Breadth 5 cm 50 mm ;Load 1.75 MN; l 0.175 cm; b 0.013 cm

To find: Poisson’s Ratio; Young’s Modulus

Solution:

Longitudinal strain: ll

0.175

25 7 10 3

Lateral strain bb

0.013

5 2.6 10 3

Poisson’s Ratio: Lateral Strain

Longitudinal Strain

2.6 10 3

7 10 3

1m

0.37

Young’s Modulus StressStrain

LoadArea

1

Strain

E 1.75 106

50 50 7 10 3 1 105

E 1 105 N/mm2

1.28 VOLUMETRIC STRAIN

Define Volumetric strain (AU. Nov/Dec 2013, Nov/Dec 2011)Whenever a body is subjected to a force it will undergo a change in

its dimensions. This deformation of the body will also cause some change inthe volume. The ratio of the change in volume to the original volume iscalled volumetric strain.

Volumetric strain is given as ev VV

where V Change in volume

V Original volume


1.28.1 Rectangular Body Subjected to Axial LoadingFor a rectangular body subjected to an axial load, the volumetric strain

can be calculated by

VV

P

btE 1

2m

PbtE

1 2

or VV

e 1

2m

e 1 2

. . .

PbtE

Linear strain

where P Load

b Width

t Thickness,

Problem 1.49: A steel bar 350 mm long, 55 mm wide and 40 mm thick issubjected to a pull of 300 kN axially. Determine the change in volume. Take

E 2 105 N/mm2 and 1m

0.25

Given: Length 350 mm; E 2 105 N/mm2; Width 55 mm; 1m

0.25;

Thickness 40 mm

To find: Change in volume

Solution:

Original volume L b t 350 55 40 770 103 mm3

Volumetric strain VV

P

btE 1

2m

300 103

55 40 2 105 1 2 0.25

3.409 10 4

Change in volume V Volumetric strain V 3.409 10 4 770 103

V 262.5 mm3

Problem 1.50: A steel bar 400 mm long, 60 mm wide and 15 mm thick issubjected to an axial tension of 100 kN. Calculate the final dimensions and

change in volume of bar. Take E 2 105 N/mm2 & 0.3.

1m

or Poisson’s ratio

E Young’s modulus


Given:

Breadth of steel bar, b 60 mm

Thickness of steel bar, t 15 mm

Length of steel bar, l 400 mm

Axial load, P 100 kN 100 103 N

Young’s modulus, E 2 105 N/mm2

Poisson’s ratio, 1m

0.3

To find1. Change in length

2. Change in breadth

3. Change in thickness

4. Change in volume

Solution:

Volume of bar b t l 60 15 400 360000 mm3

Area of bar along longitudinal direction, A b t 60 15

A 900 mm2

Longitudinal strain, e P

AE

100 103

900 2 105 5.556 10 4

Change in length Longitudinal strain length 5.556 10 4 400

0.2222 mm

Poisson’s Ratio lateral strain

longitudinal strain

0.3 lateral strain

5.556 10 4

Lateral strain 0.3 5.556 10 4 1.667 10 4

Lateral strain Change in breadth

Breadth

1.667 10 4 Change in breadth

60


Change in breadth 60 1.666 10 4 0.01 10 3 mm

Lateral strain Change in thickness

thickness

Change in thickness 1.666 10 4 15 2.5 10 3 mm

Volumetric strain ev Change in volume

Volume e

1

2m

Change in volume360000

5.55 10 4 1 2 0.3

Change in volume 80 mm3 increase

1.29 RECTANGULAR BAR SUBJECTED TO 3 MUTUALLY PERPENDICULAR FORCES

Consider a rectangular body subjected totensile loading in 3 mutually perpendiculardirection i.e., x, y and z

Let x be the stress in the x direction

y be stress in the y direction

and z be stress in the z direction

Let ex, ey, ez denote the strain in the

direction of three coordinate axes.

The stress component x causes a strain equal to x/E in the

x-direction, and strains equal to x/E in each of the y and z directions.

Similarly, the stress component y, if applied separately, will cause a strain

y/E in the y-direction and strains y/E in the other two directions.

Finally, the stress component z causes a strain z/E in the z-direction and

strains z/E in the x and y directions. Combining the results obtained,

we conclude that the components of strain corresponding to the givenmultiaxial loading are:

ex x

E

y

E z

E

1E

x y z

1E

[x y z]

ey x

E y

E z

E

1E

[y x z] 1E

[y x z]

z

x

y Fig. 1.27


ez x

E

yE

z

E

1E

[z x y] 1E

[z x y]

The volumetric strain is given by the formula

ev ex ey ez

1E

[x y z 2x 2y 2z]

1E

[x y z 2 [x y z]

ev x y z

E 1 2

(or)

ev 1E

x y z 1

2m

Problem 1.51: A metallic bar 300mm 100mm 40mm is subjected to aforce of 5 kN (tensile), 6 kN (tensile) and 4 kN (tensile) along thex, y and z direction respectively. Determine the change in the volume of block.

Take E 2 105 N/mm2 and poisson’s ratio 0.25. (AU. Nov/Dec 2015, Nov/Dec 2014)

Solution:

Given: Dimension of bar 300mm 100mm 40mm;

Px 5 kN; Py 6 kN; Pz 4 kN ; 1m

0.25

To find: Change in volumeSolution:

Volumetric strain ev

VV

1E

x y z 1

2m

Stress in x direction

x Px

y z

5 103

100 40 1.25 N/mm2

Stress in y direction

40

z

y

x

300

6kN

4kN

5kN

100


y Py

x z

6 103

300 40 0.5 N/mm2

Stress in z direction

z Pz

x y

4 103

300 100 0.133 N/mm2

VV

1E

x y z 1

2m

1

2 105 1.25 0.5 0.133 [1 2 0.25]

4.71 10 6

Change in Volume V Volumetric strain V

4.71 10 6 300 100 40

V 5.65 mm3

Problem 1.52: A cast iron flat 300 mm long and 30 mm (thickness) 60mm (width) uniform cross section, is acted upon by the following forces:30 kN tensile in the direction of the length360 kN compression in the direction of the width240 kN tensile in the direction of the thicknessCalculate the direct strain, net strain in each direction and change in volumeof the flat. Assume the modulus of elasticity and Poisson’s ratio for cast iron

as 140 kN/mm2 and 0.25 respectively. (AU. Apr/May 2011)

Solution:

Cast Iron Length = 300 mm; Width = 60 mm

Thickness = 30 mm

E 140 103 N/mm2, 0.25

The stresses in the direction of the axes x, y, z are

x 360 103

300 30

40 N/mm2 compressive


y 30 103

60 30

16.67 N/mm2 tensile

z 240 103

300 60

13.33 N/mm2 tensile

The strain along the principal directions are

ex 40E

16.67

E

13.33E

1E

[40 16.67 13.33 ]

ex 1E

[40 30 ]

ey 16.67

E

40E

13.33

E

1E

[16.67 26.67]

ez 13.33

E

16.67E

40E

ez 1E

[13.33 23.33]

Volumetric strain ev ex ey ez

ev 1E

[ 10 20]

Hence, 0.25; E 140 103 N/mm2

ev 1

140 103 [ 10 20 0.25]

ev 3.57 10 5

Change in volume (decrease in volume)v ev V

3.57 10 5 300 30 60

v 19.28 mm3


1.30 CYLINDRICAL ROD SUBJECTED TO AXIAL LOAD

Consider a cylindrical rod of length l and diameter d subjected to anaxial load P. The volumetric strain of the rod can be calculated by the formula

ev VV

ll

2dd

Problem 1.53: A steel rod 4.5 m long and 35 mm in diameter is subjectedto an axial pull of 150 kN. Determine the change in the length, diameter and

volume of the rod. Take E 2 105 N/mm2 and poisson’s ratio 0.25.

Data Given: Length l 45m 45 103 mm ;

Diameter d 35mm ; Load 150kN

To find: Change in diameter, length and volumeSolution:

Longitudinal strain ll

stress

E

PAE

150 103

4

352 2 105 7.795 10 4

Change in length l Longitudianal strain l 7.795 10 4 4.5 103

3.51mm

Poisson’s ratio Lateral strain

Longitudinal strain

Lateral strain Longitudinal strain Poisson’s ratio

7.795 10 4 0.25 1.95 10 4

Lateral strain dd

1.95 10 4

Change in diameter d 1.95 10 4 d

1.95 10 4 35 6.82 10 4 mm

Volumetric strain VV

ll

2 dd

7.795 10 4 2 1.95 10 4

VV

3.895 10 4


Change in Volume V 3.895 10 4 V 3.895 10 4 4

d2 l

3.895 10 4 4

352 4.5 103

V 1686.49 mm3

1.31 SHEAR STRESS AND STRAINDefine shear strain (AU. Nov/Dec 2013)

When a body is subjected to two equal andopposite forces which do not have the same lineof action, the result will be that the body tendsto shear off across a section leading to shearstress. The resulting strain is called shear strain.

Consider a cube of length l fixed at thebottom face AB. A force P is applied tangentiallyat CD. The cube gets distorted from ABCD to

ABCD through angle .

Shear strain Deformation

Original length

CC l

Shear stress P

Area

Problem 1.54: Two members are connected to carry a tensile force of 80kN by a lap joint with two number of 20 mm diameter bolt. Find the shearstress induced in the bolt. (AU. Nov/Dec 2016)

Data: Force P 80 kN 80 103 N; dia of bolt D 20 mm

Find: Shear stress

Solution:

Shear stress loadArea

80 103

2 4

202

80 103

628.35 127.3 N/mm2

Shear stress 127.3 N/mm2

D �DC �C

BA

P


Result:

Shear stress 127.3 N/mm2

1.31.1 Shear Modulus or Modulus of RigidityWithin the elastic limit, the shear stress is proportional to the shear

strain

G

G

G Modulus of rigidity (or) shear modulus

80kN 80kN

80kN

80kN


1.32 BULK MODULUSDefine Bulb Modulus (AU. Apr/May 2010)

When a body is subjected to three mutually perpendicular stresses ofequal intensity, the ratio of direct stress to the corresponding volumetric strainis known as bulk modulus. It is usually denoted by K.

K Direct stress

Volumetric strain

VV

1.33 RELATIONSHIP BETWEEN ELASTIC CONSTANTSIt was already seen that each elastic material has 4 elastic constants.

These constants are inter-related with one another. If the values of any 2constants are known for a material, the other 2 constants may be easilydetermined using the relationship equations.

1.33.1 Relation between Bulk Modulus and Young’s ModulusConsider a cube ABCD A1 B1 C1 D1 as shown in Fig.1.28. The cube

is subjected to an axial tensile stress along the face BB1 CC1 and AA1 DD1.

Consider the deformation of one side of the cube (AB). This side issubjected to the following strains.

1. Tensile strain E

due to stress on

faces BB1 CC1 and AA1 DD1

2. Compressive lateral strain 1m

E

due to stress on facesAA1 BB1 and DD1 CC1.

3. Compressive lateral strain equal to

1m

E

due to stress on faces

ABCD and A1B1C1D1

The net tensile strain that AB will suffer is

ll

E

1m

E

1m

E

E

1

2m

...(1)

A

CD

B

A 1 B 1

C 1

Fig 1.28


We know, the original volume of the cube

V l3

Differentiating with respect to l

V

l 3l2

V 3l2 l 3l3 ll

Substituting ll

from (1)

V 3l3 E

1

2m

or VV

3l3

l3 E

1

2m

3E

1

2m

VV

3E

1

2m

E3

1

1 2m

E3

1

m 2m

K mE

3 m 2 (or) K

E

3 1

2m

E

3 1 2

1.33.2 Relation between Modulus of Elasticity and Modulus of RigidityConsider a cube of length l subjected to

a shear stress of The cube will get distorted and the

diagonal AD will get elongated to AD and the

diagonal BC will get shortened to BC

Strain of AD AD AD

AD

D D2

AD

DD cos 45BD 2

DD 2BD

2

We see that the linear strain of the diagonal AD is half of the shearstrain and is tensile. It can also be seen that the linear strain of AC is equalto half the shear but is compressive

A B

C DC �

D 2

D �


Linear strain of AD 2

2G ...(2)

where Shear stress

G Modulus of rigidity

Consider the shear stress acting on AB, CD, CB and AD. The effectof this stress will cause a tensile stress on diagonal AD and compressivestress on diagonal BC

The tensile strain on AD due to stress on AD E

and tensile strain

on BC due to compressive stress on BC 1m

E

The combined effect of the above two stresses on diagonal

Strain on AD E

1m

E

E

1

1m

E

m 1

m ...(3)

Equating equations 2 and 3 we get

2G

E

m 1

m

Write the relationship between shear modulus and modulus of elasticity(AU. Nov/Dec 2014, May/June 2012)

G mE

2 m 1 E

2 1 1

m

E2 1

Problem 1.55: An alloy specimen has modulus of elasticity of 120 GPa andmodulus of rigidity of 45 GPa. Determine the Poisson’s ratio of the material. (AU. Apr/May 2010)

Given

Modulus of elasticity, E 120 GPa 120 103 N/mm2

Modulus of rigidity, G 45 GPa 45 103 N/mm2

Poisson’s ratio, 1/m ?

Solution

Modulus of rigidity, G E

2 1

1m


1 1m

E

2G

Poisson’s ratio, 1/m E

2G 1

120 103

2 45 103 1

0.333 .

Problem 1.56: In an experiment, a bar of 30 mm diameter is subjected toa pull of 60 kN. The measured extension on gauge length of 200 mm is 0.09mm and the change in diameter is 0.0039 mm. Calculate the Poisson’s ratioand the values of the three modulli. (AU. Apr/May 2010)

Given: Diameter of bar, d 30 mm ; Length of bar, l 200 mm

Extension, dl 0.09 mm ; Change in dia, d 0.0039 mm

Pulling load, P 60 kN 60 103 N

Solution:

Linear strain ll

0.09200

4.5 10 4

Lateral strain dd

0.0039

30 1.3 10 4


Lateral strainLinear strain

1.3 10 4

4.5 10 4

1m

0.289

Change in length, l PlAE

(or) E Pl

A l

A d2

4

302

4 706.86 mm2


Young’s Modulus E 60 103 200

706.86 0.09

E 188628.08 N/mm2

Bulk Modulus K E

3 1

2m

188628.08

3 1 2 0.289

K 148995.3 N/mm2

Rigidity Modulus, G E

2 1

1m

188628.08

2 1 0.289 73168.38 N/mm2

Problem 1.57: The following data relate to a bar subjected to a tensile test:Diameter of the bar = 30 mmTensile load P 54 kN

Gauge length L 300 mm

Extension of the bar 1 = 0.112 mm

Change in diameter d = 0.00366 mm

Calculate(i) Poisson’s ratio(ii) The values of three modulii (AU. Nov/Dec 2013)

Given

Diameter of the bar 30 mm; Gauge length L 300 mm

P 54 kN 54 103 N ; L 0.112 mm ; d 0.00366 mm

Solution:

(i) Poisson’s ratio: Lateral strainLinear strain

e

d

d

0.0036630

1.22 10 4

e LL

0.112300

3.73 10 4


e

1.22 10 4

3.73 10 4

0.327

(ii) Moduli:

Young’s Modulus E e

E P

A e

54 103

4

302 3.73 10 4

E 2.046 105 N/mm2

Now, E 3K 1 2

2.046 105 3K 1 2 0.327

3K 0.346

Bulk Modulus, K 1.971 105 N/mm2

Now, E 2G 1

2.04 105 2G 1 0.327

Modulus of rigidity, G 0.771 105 N/mm2

Problem 1.58: A rod of length 1 m and diameter 20 mm is subjected to atensile load 20 kN. The increase in length of the rod is 0.30 mm and decreasein diameter is 0.0018 mm. Calculate the Possion’s ratio and three moduli. (AU. Apr2013)

Given: d 20 mm, l 1 m 1000 mm;

P 20 kN 20 103 N; l 0.30 mm; d 0.0018 mm

Solution:

Lateral strain dd

0.0018

20 9 10 5

Linear or longitudinal strain


e ll

0.3

1000 3 10 4

(i) Poisson’s ratioWe know that Poisson’s ratio

1m

or

Lateral strainlongitudinal str ain

e

1m

or 9 10 5

3 10 4 0.3

(ii) Young’s modulus (E)Tensile stress

LoadArea

20 103

d2

4

20 103 4

202 63.65 N/mm2

Youngs modulus

E Tensile stress

Longitudinal strain

63.65

3 10 4 2.12 105 N/mm2

(iii) Rigidity modulus GWe know that E 2G 1 2.12 105 2 G 1 0.3

G 2.12 105

2 1.3 8.162 104 N/mm2

(iv) Bulk modulus KWe know that E 3K 1 2

2.12 105 3 K 1 2 0.3

Bulk modulus K 2.12 105

3 1 2 0.3 1.768 105 N/mm2


Probelm 1.59: A bar of cross section 8 mm 8 mm is subjected to an axialpull of 7000 N. The lateral dimension of the bar is found to be changed to7.9985 mm 7.9985 mm. If the modulus of rigidity of material is

0.8 105 N/mm2, determine the poisson’s ratio and modulus of elasticity. (AU. May/June 2012)

Given: Area of section 8 8 64 mm2

Axial pull, P 7000 N

Lateral dimensions 7.9985 mm 7.9985 mm

Modulus of rigidity, G 0.8 105 N/mm2

Solution

Let ‘1’/m Poisson’s ratio

E Modulus of elasticity

Lateral Strain Change in lateral dimensionOriginal lateral dimension

8 7.9985

8

0.0015

8 0.0001875

Axial Stress, P/A

7000

64 109.375 N/mm2

But Lateral Strain,

mE

0.0001875 109.375

mE

mE 109.375

0.0001875

mE 583333.33 ...(1)

E 2G 1

E 2G 1

1m


E 2G m 1

m

mE 2G m 1

2 0.8 105m 1

583333.33 2 0.8 105 m 1

m 1 583333.33

2 0.8 105

m 1 3.6458

m 3.6458 1

m 2.6458

Poisson’s Ratio 1/m 1

2.6458 0.378

0.378

Sub the value of m in equn (1) we get

2.6458 E 583333.33

E 583333.33

2.6458

E 2.2047 105 N/mm2

Problem 1.60: A brass bar 10 mm 10 mm size and 500 mm length issubjected to an axial compression of 15 kN (bar is prevented from bending).The decrease in length and increase in lateral dimensions are found to be0.5 mm and 0.004 mm respectively. Determine the elastic constants of brass.

Given

P 15 kN 15000 N

Area of cross section A 10 10 100 mm2

Axial stress in bar, 15000

100 150 N/mm2

Original length of bar, L 500 mm

Decrease in length, L 0.5 mm


To find: Elastic constants E, G, K,

Solution:

Longitudinal strain el 0.5500

1 10 3

Lateral dimension of bar, b 10 mm

Increase in lateral dimension b 0.004 mm

Lateral strain eb 0.004

10 0.4 10 3

Young’s modulus, E el

150

1 10 3 1.5 105 N/mm2

Poisson’s Ratio 1m

eb

el

0.4 10 3

1 10 3 0.4

Using the relation,

Rigidity modulus, G E

2 1

1m

1.5 105

2 [1 0.4] 0.53571 105

G 0.5357 105 N/mm2

Bulk modulus, K E

3 [1 2m

]

1.5 105

3 [1 2 0.4]

K 2.5 105 N/mm2

Elastic constants

Poisson’s Ratio 0.4

Young’s modulus 1.5 105 N/mm2

Rigidity modulus 0.5357 105 N/mm2

Bulk modulus 2.5 105 N/mm2


Problem 1.61: A solid circular bar of diameter 20 mm when subjected toan axial tensile load of 40 kN, the reduction in diameter of the rod was

observed as 6.4 10 3 mm. The bulk modulus of the material of the bar is67 GPa. Determine the following:(i) Young’s modulus,(ii) Poisson’s ratio,(iii) Modulus of rigidity,(iv) Change in length per metre and (v) Change in volume of the bar permetre length. (AU. May/June 2013)

Given

Dia of bar, d 20 mm;

Load, P 40 kN 40 103 N

Change in dia, d 6.4 10 3 mm

Solution

Cross sectional Area, A 4

202 314.16 mm2

Lateral strain dd

6.4 10 3

20 3.2 10 4

Axial stress, PA

40 103

314.16 127.32 N/mm2


Lateral strain

Longitudinal strain

Lateral strain 1m

Longitudinal strain

1m

E

3.2 10 4 1m

127.32

E

or E 127.32

3.2 10 4

1m

397875 1m

We know that, E 3K 1 2

1m

... (1)

Bulk modulus,

K 67 GPa 67 103 N/mm2

l 1 m 1000 mm [... per meter length]


or E 3K 6K 1m

397875 1m

3 67 103 6 67 103 1m

397875 1m

402 103 1m

201 103

i.e 799875 1m

201 103

i.e 799875 1m

201 103


201 103

799875

1m

0.25

From equation (1),

E 3K 1 2

1m

3 67 103 1 2 0.25

E 100500 N/mm2

Longitudinal strain E

127.32100500

1.267 10 3

We know that, K VV

VV

K

127.32

67 103 1.9 10 3

Using the relation,

E 2G 1

1m

(or) G E

2 1

1m

100500

2 1 0.25

Modulus of r igidity, G 40200 N/mm2

but, Longitudinal strain e ll

Change in length,

l e l

1.267 10 3 1000

l 1.267 mm

Change in volume,

V 1.9 10 3 V

1.9 10 3 A l

1.9 10 3 314.16 1000

V 596.9 mm3


1.34 PRINCIPAL STRESSES AND PRINCIPAL PLANESIn the previous sections, we have discussed about the direct tensile and

compressive stress as well as simple shear stress. Also we have referred thestress in a plane which is at right angles to the line of action of the force.But it has been observed that at any point in a strained material, there arethree planes, mutually perpendicular to each other which carry direct stressesonly and no shear stress. Out of these three direct stresses, one will bemaximum and other will be minimum. These perpendicular planes which haveno shear stress are known as Principal planes and the direct stresses alongthese planes are known as Principal stresses. The planes on which themaximum shear stress act are known as planes of maximum shear. Thinwalled pressure vessels provide an important application of the analysis ofplane stress.

It is obvious that a 3-Dimensional general stress system can berepresented by three normal stresses and three shearing stresses. The systemsin which direct stresses and shearing stresses act simultaneously are calledcombined stresses or compound stresses. A system consisting of two normalstresses and two shearing stresses are called Biaxial state of stress system.

Before discussing the analysis of plane stresses, it is essential to knowvarious planes and axes of a member.

Note: When the member is subjected to axial stresses alone i.e., in theabsence of shear stress, the normal plane is also called major principal plane.

Axis o f M ajo r s tress

Axi

s o

f Min

or

stre

ss

Nor

ma

l Pla

ne

Obl ique P

lane

Fig. 1.31


1.35 ANALYSIS OF PLANE STRESSESWhat do you mean by principal planes and principal stress

(AU. Nov/Dec 2016, Nov/Dec 2013, May/June 2013, May/June 2012,Apr/May 2011)

The Planes which have no shear stress are known as Principal planesand the corresponding normal stress acting on a principal plane is known asPrincipal stress.

1.35.1 Stress on a inclined plane (A member subjected to an axial load)Consider a rectangular bar of uniform cross sectional Area A. Let the

member be subjected to axial load P as shown in Fig. 1.32.

The load P produces a tensile stress

PA

LoadArea

on the sections normal to loading. On such sections only normal stressesare induced and no tangential stresses are induced.

Now pass an oblique plane BS forming an angle with a normalplane. On the plane BS, normal and tangential stresses are induced.

Area on plane BS A A/cos

Fig. 1.32. Stress on Inclined Plane


Stress normal to BS

n Force on BS

Prea along BS

P cos A

P cos A/cos

PA

cos2

Normal stress n cos2

Tangential stress on BS

t Tangential force on BS

Area along BS

P sin A

P sin A/cos

PA

sin cos

P

2A sin 2

Tangential stress t 2

sin 2

n cos2 : t 2

sin 2

When 0; n maximum value

; t 0[only normal force]

90; n 0 ; t 0

45; n 2

; t 2 [Maximum condition for t]

For the planes corresponding to 0 and 90 : for the sectionalplanes, we find that there are no shear or tangential stresses. Such planes arecalled Principal planes. Any stress in principal plane should be normal to theplane. The only normal stress occurring on these plane is called Principal stress.

Shear Stress

Shear Stress will be maximum when sin 2 is maximum.

From the equations 2 90, 45

At 45 Maximum Shear Stress 2

P

2A.

Hence a direct tensile stress induces a shear stress of half its intensityon a plane 45 to the plane carrying direct stress. Hence on any inclinedcross section of a specimen subjected to direct stress, shearing stress is alwayspresent and if a material is such that its shear length is less than half of itstensile length, then the material will fail by shear when subjected to directtensile or compressive load.


Problem 1.62: A short metallic column of 500 mm2 cross section areacarries an axial load of 100 kN. For a plane inclined at 60 with the directionof load, calculate normal stresses, tangential stress, resultant stress, maximumshear stress and obliquity of resultant stress. (AU. Nov/Dec 2011)

Given: Area A 500 mm2; Load P 100 kW 100 103 N ; 60with axis of load

Solution:

The column is shown in figureNormal area of cross section

a 500 mm2

Direct stress

LoadArea

100 103

500 200 N/mm2

AC is inclined at 60 with respectto load axis Now 90 60 30 withx-axis.

(i) Normal Stress

n cos2 200 cos230n 150 N/mm2

(ii) Tangential Stress 2

sin 2 2002

sin 2 30 86.6 N/mm2

(iii) Resultant stress (R)

R t2 n

2 1502 86.62 173.2 N/mm2

(iv) Maximum Shear Stress max)

max 2

2002

100 N/mm2

(v) Angle of Obliquity

tan 1 t

n tan 1

86.6150

30

60o

30o

x x

P=100kN

P=100kN

A

C


1.35.2 Biaxial state of stress - Member subjected to biaxial stressConsider a rectangular block WXYZ whose thickness perpendicular to

the plane of paper is unity as shown in Fig. 1.33.

The block is subjected to stresses x and y as shown.

Consider an oblique section YE at an angle with the principal planeXY.

Normal stress on EY

n Total force on EY

Section Area along EY

W E X

Z Y

x

x

y

y

n

t

Fig. 1.33. Biaxial Stress

A = x A cos A

Y

EA = y A sin

Y

E

x xA cos

y yA s in

n

Fig. 1.34.


n x Ax cos y Ay sin

A

. . . Force x Area

x A cos cos y A sin sin

A

n xcos2 y sin2

x

2 1 cos 2

y

2 1 cos 2

Normal stress

n x y

2 x y

2 cos 2

Now Tangential stress on the plane EY.

Tangential stress

t Total force tangential on EY

Area of section along EY

x Ax sin y Ay cos

A x.

A cos A

sin y A sin

A cos

t x cos sin y sin cos

x y sin cos

Tangential stress t x y

2 sin 2

Resultant stress on EY

Res t2 n

2

xcos2 ysin2 2 x y2 sin2 cos2

This gets simplified to x2cos2 y

2sin2

. . . Ax A cos Ay A sin

Y

E

x xA sin

y yA cos

t

Fig. 1.35


The angle that the line of action of the resultant stressmakes with the normal to plane is called the obliquity.

If the obliquity is , we have tan t

n as shown in

Fig. 1.36.

To find the Principal Planes

Equating t x y

2 sin 2 0

We have 2 0 or 180

or 0 or 90

We see that there are two Principal Planes at right angles to each other.

When 0

n x y

2

x y

2 cos 2

x y

2

x y

2 cos 0

n x

2

y

2

x

2

y

2 x

n1 x 1

When 90

n x y

2

x y

2 cos 2 90

x y

2 x y

2 y

n2 y 2

One Principal Plane carries maximum direct stress whereas the othercarries minimum.

n

t

Fig. 1.36. Obliquity


Maximum shear stress

We know t x y

2 sin 2

When sin 2 1, t x y

2

2 90 or 270

or 45 or 135

max x y

2

Problem 1.63: The principal stresses at a point across two perpendicular

planes are 180 N/mm2 and 140 N/mm2. Find the normal and tangentialstress, the resultant stress and its obliquity on a plane at 20 with majorprincipal plane. Find also the intensity of stress which acting alone canproduce the same maximum strain. Take 0.25

Given: Principal stresses: 1 180 N/mm2 ; 2 140 N/mm2; 20;

1m

0.25

Solution:

We know that for a biaxial stress on an oblique plane:

(i) Normal stress is given as

n 1 2

2

1 22

cos 2

n 180 140

2

180 1402

cos 2 20 175.32 N/mm2

Normal stress n 175.32 N/mm2 Ans.

(ii) Tangential stress is given by

t 1 2

2 sin 2

180 1402

sin 2 20 12.855 N/mm2

Tangential stress t 12.855 N/mm2 Ans.


(iii) Resultant stress

Res t2 n

2 12.8552 175.322

Res 175.792 N/mm2(Ans.)

(iv) Obliquity

tan 1 t

n tan 1

12.855175.32

4.194

(v) Maximum strain emax

emax 1E

2

mE

1E

[180 0.25 140] 145E

We know that stress Strain Young modulus E

Intensity of stress E 145E

145 N/mm2(Ans.)

Problem 1.64: Show that for a member subjected to two dimensionalstresses, the sum of normal components of stresses on any two mutuallyperpendicular planes is constant.

Solution:

Let 1 and 2 be two Principal stresses. On any plane at with major

principal plane we have

Normal stress n 1 2

2

1 2

2 cos 2

...(1)

On the second plane 90 with principal plane

Normal stress

n1

1 2

2

1 2

2 cos 2 90

1 2

2

1 22

cos 2...(2)

Adding (1) and (2) we get

n n1

1 2

2 1 2

2 1 2 constant


Problem 1.65: If the principal stresses at a point in a strained material are1 and 2. Find the resultant stress on the plane carrying maximum shear stress.

Solution:

The plane carrying maximum shear stress is at 45 with MajorPrincipal plane.

Normal stress n 1 2

2 1 2

2 cos 2

n 1 2

2 1 2

2 cos 2 45

1 22

Tangential stress

t 1 2

2 sin 2

1 2

2 sin 2 45

1 22

t 1 2

2

Resultant stress

Res t2 n

2

1 2

2

2

1 2

2

2

1

2

2

2 1 2

2 1

2

2

1

2

2

2 12

2 2

2

2

Res 1

2

2

2

2

2

12 2

2

2

Res 1

2 22

2

1/2


Problem 1.66: A point in a strained material is subjected to two mutuallyperpendicular tensile stresses 200 MPa and 100 MPa. Determine theintensities of normal, shear and resultant stresses on a plane inclined at30 with the axis of the minor tensile stress. (AU. Apr/May 2010)

Given: x 200 MPa 280 N/mm2, y 100 MPa 100 N/mm2, 30

Solution:

For Biaxial stress system

(i) Normal Stress n

n x y

2

x y

2 cos 2

200 100

2

200 1002

cos 2 30 175 N/mm2

n 175 N/mm2 (Ans.)

(ii) Tangential Stress (Shear Stress) t

t x y

2 sin 2

200 1002

sin 2 30 43.3 N/mm2

t 43.3 N/mm2

(iii) Resultant Stress

n2 t

2 1752 43.32 180.28 N/mm2

Resultant Stress 180.28 N/mm2(Ans.)

(iv) Obliquity

tan 1 t

n tan 1

43.3175

13.897

Problem 1.67: At a point in a strained material, the principal stresses are

100 N/mm2 (tensile) and 60 N/mm2 (compressive). Determine normal, shearstress, resultant stress on a plane inclined at 50 to the axis of major principalstress. Also determine the maximum shear stress at the point (AU. Dec 2010)


Given: Principal stresses: 1 100 N/mm2, 2 60 N/mm2,

90 50 40

Solution:

(i) Normal Stress (n

n 1 2

2 1 2

2 cos 2

100 602

100 60

2 cos 2 40

n 33.89 N/mm2 (Ans.)

(ii) Tangential Stress (t)

t 1 2

2 sin 2

100 602

sin 2 40 78.79 N/mm2

t 78.79 N/mm2

(iii) Resultant Stress (R)

R t2 n

2 33.892 78.792 85.76 N/mm2

Resultant Stress R 85.76 N/mm2 Ans

Obliquity tan 1 t

n tan 1

78.7933.89

66.73

(iv) Maximum Shear Stress ( max)

max 1 2

2

100 602

80 N/mm2 Ans

1.35.3 (iii) A member subjected to a simple shear stressConsider a section PQRS of uniform cross sectional area A and of unit

thickness and subjected to a simple shear stress . Consider a plane RE asshown in the Fig. 1.38.


The normal stress on ER

n Ax sin Ay cos

A

sin cos sin cos

2 sin cos

n sin 2

. . . Ax A cos ; and Ay A sin

Similarly Tangential stress

t .Ax cos Ay sin

A cos2 sin2

t cos2 sin2 cos 2

t cos 2

For Principal stresses

Equating t 0 cos 2

cos 2 0

2 90 or 270 or 45, 135

Hence the two principal planes are at 45 and 135 with the planeQR. Principal planes are also at right angles to each other

S R

P QE

n

t

Fig. 1.38. M em ber Subjected to Shear Stress.

R

E

A cos x

A s in y

t

Fig. 1.39

R

E

A x s in

Ay cos

n

A = x A cos A

R

EA = y A sin

x

y

Fig. 1.37


The two principal stresses are

1 sin 2

sin 90

2 sin 2

sin 270 Thus on one plane there will be a tensile stress and the other plane

there will be a compressive stress as shown in Fig. 1.40.

1.35.4 (iv) Member subjected to a simple shear and a biaxial stressConsider a rectangular block ABCD whose thickness is unity. Let it

be subjected to normal stresses x and y and tangential stress as shown in

the Fig. 1.41

P Q

= -

=

S R

45o

Fig. 1.40

A B

D C

x x

y

y

E

Fig. 1.41


Consider a sectional plane EC at an angle with plane BC

Normal stress on the plane EC

n x Ax .cos y Ay .sin Ax sin Ay cos

EC

n x cos2 y sin2 sin cos cos sin

n x

2 1 cos 2

y

2 1 cos 2 sin 2

n x y

2 x y

2 cos 2 sin 2

...(1)

Now Tangential stress on the plane EC

. . . Ax A cos Ay A sin

A = x A cos A

C

E A = y A sin

x

y

C

E

A x s in

Ay cos

n

x xA cos

y yA s in

Fig. 1.42

C

E

A cos x

A s in y

t

x xA s in

y yA cos

Fig. 1.43.


t x Ax sin y Ay cos Ax cos Ay sin

A

t x cos sin y sin cos cos2 sin2

t x y

2 sin 2 cos 2

...(2)

Principal plane

For principal plane t 0

x y

2 sin 2 cos 2 0

tan 2 2

x y

There are thus two values of 2 differing by 180 satisfying the aboverelation.

Let 21 and 22 be the solution of the above equation.

We have now sin 21 2

x y2 42

And sin 22 2

x y2 42

;

cos 21 x y

x y2 42

; cos 22 x y

x y2 42

For determining the Principal stresses n1 and n2

substitute the values

of 21 and 22 in the expression for normal stress n

Substituting above values in (1)

n1 1

x y

2 x y

2 cos 21 sin 21

x y

2 x y

2

x y

x y2 42

.2

x y2 42

x y-

2

(-

)+4

xy

2

2

Fig. 1.44.


1 n1

x y

2

1 y2 42

2 x y2 42

Therefore, the Major Principal stress is

1 n1 x y

2 1

2 x y

2 42

1 n1 x y

2

x y

2

2

2

Similarly substituting 22 in (1)

2 n2 x y

2 x y

2 cos 2 2 sin 2 2

x y

2

x y2

2 x y2 42

22

x y2 42

2 n2 x y

2

1

2 x y2 42

x y2 42

Therefore, the Minor Principal stress is,

2 n2

x y

2

12

x y2 42

(or)

2 n2

x y

2

x y

2

2

2


max n1

n2

2

max 12

x y2 42


(or)

max x y

2

2

2

The maximum shear stress will be at 1 45 and 1 135 with the

plane BC.

The principal stresses are shown in Fig. 1.45

Problem 1.68: For a biaxial stress system, the normal stress on twomutually perpendicular planes are x and y (both alike) and the shear stress

is . Find the Principal stresses if 2 xy.

Given: 2 xy; 1 and 2 normal stresses,

Solution:

We know that the Principal stresses are given as

n1, n2

x y

2

x y

2

2

2

y

2

y

1

n 1

n 2

n 2

n 1

P rinc ipal

p lanex

Fig. 1.45.

x

Princ ipa l s treses


Now 2 xy

n1, n2

x y

2 x y

2 4 xy

4

x y

2 x

2 2 xy x2 4 xy

4

n1, n2

x y

2 x y

2

4 x y

2 x y

2

n1 x y and ny

0

Major Principal stress, n1 or1 x y

Minor Principal stress, n2 or 2 0

Problem 1.69: A rectangular block of material is subjected to a tensile

stress of 220 N/mm2 on a plane and a tensile stress of 94 N/mm2 on a plane

at right angles, together with a shear stress of 126 N/mm2 on the sameplanes. Find (i) The direction of the principal planes (ii) The magnitudeof the principal stresses (iii) Magnitude of the greatest shear stress.

Given: x 220 N/mm3 : y 94 N/mm2; 126 N/mm2

Solution:

The inclination of principal planes is given by

tan 2 2

x y

2 126

220 94 2

2 tan 1 2 63.43 or 243.43

Inclination 31.71or 121.71(Ans.)


Major principal stress, 1 n1

x y

2

x y

2

2

2

220 94

2

220 94

2

2

1262

1 n1 297.87 N/mm2

Minor principal stress, 2 n2 x y

2

x y

2

2

2

220 94

2

220 94

2

2

1262

2 n2 16.13 N/mm2 Ans.


max n1

n2

2

297.87 16.132

140.87 N/mm2

max 140.87 N/mm2 Ans.

This will occur at an angle, 31.715 45 76.715

and 76.715 90 166.715The stresses are shown in the Figure.


Problem 1.70: The intensity of the resultant stress on a plane AB as shown

in Fig (a) at a point in a material under stress is 8 N/mm2 and is inclinedat 30 to the normal to that plane. The normal component of stress on another

plane BC, at right angles to the plane AB is 6 N/mm2. Determine thefollowing (a) The resultant stress on the plane BC. (b) The principal stressesand their directions. (c) The maximum shear stresses and their planes. (AU. Nov/Dec 2016)

Solution: The principal stresses are shown below

Normal stress on plane AB x 8 cos 30 6.93 N/mm2

Shear stress on plane, AB 8 sin 30 4 N/mm2

y=94N/m m2

=126N/m m2

x=24N/m m2

=126N

/mm

2

x=220N/m m2

=12

6N/m

m2

1

n2

n 2

n 1

n 1 2

=126N/m m2

y=94N/m m2


Normal stress on plane, CB y 6 N/mm2

The various stresses are shown in Fig.(b)

(i) Resultant stress on plane, BC 62 42 7.211 N/mm2

(ii) The principal stresses are given by

1 and 2 x y

2

x y

2

2

2

6.93 6

2

6.93 6

2

2

42

1 and 2 6.465 4.027

Major Principal stress 1 10.492 N/mm2;

Minor Principal stress 2 2.438 N/mm2

Position of Principal stresses

tan 2 2

x y

2 4

6.93 6 8.602

2 83.37 or 263.37

41.68 or 131.68

(iii) Maximum shear stress

max 1 2

2

10.492 2.4382

4.027 N/mm2

Angular position of maximum shear stress,

41.68 45 86.68 and 86.68 90 176.68

Problem 1.71: Two planes AB and BC which are at right angles carry shear

stress of intensity 22.5 N/mm2 while the plane AB carries a tensile stress of

160 N/mm2 as shown in figure. Find the normal and shear stresses on theplane AC inclined at 25 to the plane AB


Given: x 160 N/mm2 ; y 0 ; 22.5 N/mm2; 25

Solution:

Normal stress an plane CA is

n x y

2 x y

2 cos 2 sin 2

[Here y 0 ]

n 1602

1602

cos 50 22.5 sin 50

n 148.66 N/mm2

Tangential stress on plane CA

t x y

2 sin 2 cos 2

160

2 sin 50 22.5 cos 50

46.82 N/mm2

Tangential stress t 46.82 N/mm2

The stresses are shown in the Figure.

Problem 1.72: A rectangular block of material is subjected to a tensile

stress of 110 N/mm2 on one plane and a tensile stress of 47 N/mm2 on aplane at right angles to the former. Each of the above is accompanied by a

shear stress of 63 N/mm2. Determine the principal stresses, principal planeand the maximum shear stress. (AU. Nov 2012)

Given: x 110 N/mm2 ; y 47 N/mm2; 63 N/mm2

Solution:

We know that the Major and minor principal stresses are given by

n1 and n2

x y

2

x y

2

2

2

C B

A

n=148

.66

N/m

m2

25o

t = 46.82 N/m

m 2

x=160N /m m 2

=22.5N/mm 2

C B

A

25O

22.5N /m m2

160 N /m m 2


n1 and n2

110 47

2

110 47

2

2

632

n1 and n2

78.5 70.436

Major principal stress, 1 n1 148.936 N/mm2

Minor principal stress, 2 n2 8.064 N/mm2

Angular position of principal plane, tan 2 2

x y

tan 2 2 63

110 47 2

2 63.43 or 243.43

Angular position 31.715 or 121.715

Maximum shear stress,

max n1

n2

2 1 2

2

148.936 8.0642

70.436 N/mm2

Maximum shear stress max 70.436 N/mm2

Problem 1.73: A point in a strained material is subjected to the stress asshown in fig. Locate the principle plane and find the principle stresses.

40 N/mm2

40 N/mm2

60 N/mm2

60 N/mm2

60o

60o

A B

D C


Data: x 60 N/mm2 ; y 40 N/mm 2 ; p 60Find: Principal stress

Solution:

x 60 sin 60 51.96 N/mm2

y 40 N/mm2

60 cos 60 30 N/mm2

Principle stresses;

1, 2 x y

2

x y

2

2

xy2

51.96 N /m m2

30N/m m2

51.96N /m m2

30N/m m2

40N/m m2

40N/m m2

60 cos 60

60 sin 60

6060


51.96 40

2

51.96 40

2

2

302

1 45.98 30.59 76.57 N/mm2

max 76.57 N/mm2

2 45.98 30.59 15.39 N/mm2

min 15.39 N/mm2

tan 2 2 xy

x y

2 30

51.96 40 5.017

2 78.73

39.36 and 129.36

Problem 1.74: Two planes AB and AC which are at right angles carry a

shear stress of intensity 17.5 N/mm2 while these planes also carry a tensile

stress of 70 N/mm2 and a compressive stress of 35 N/mm2 respectively.Determine (i) Direction of principal plane (ii) The magnitude of theprincipal stresses (iii) Maximum shear stress. (AU. Nov 2017)

Given: x 70 N/mm2 ; y 35 N/mm2; 17.5 N/mm2

Solution:

(i) Direction of principal stresses

tan 2 2

x y

2 17.5

70 35 0.333

2 tan 1 0.333

2 18.435 or 198.435

9.2175 or 99.2175


(ii) Major principal stressesMajor and Minor principal stresses

n1 or n2

x y

2

x y

2

2

2

70 35

2

70 35

2

2

17.52 17.5 55.339

Major principal stress n1 1 72.83 N/mm2

Minor principal stress n2 2 37.839 N/mm2

(iii) Maximum shear stress

max n1

n2

2 1 2

2

72.83 37.8392

55.33 N/mm2

max 55.33 N/mm2

This will occur at angle 9.2175 45 54.21

and 54.21 90 144.21

y=3 5 N /m m2

=1 7 .5 N /m m2

y=3 5 N /m m2

=1 7 .5 N /m m2

x=7 0 N /m m2 x=7 0 N /m m

2

=17

.5 N

/mm

2

=17

.5 N

/mm

2

A B

D C


Problem 1.75: A point is subjected to a tensile stress of 250 MPa in thehorizontal direction and another tensile stress of 100 MPa in the verticaldirection. The point is also subjected to simple shear stress of 26 MPa, suchthat when it is associated with the Major tensile stress, it tends to rotate theelement in the clock wise direction. What is the magnitude of the normal andshear stresses on a section inclined at an angle 20 with the Major tensilestress? (AU. Apr/May 2010)

Given: x 250 MPa 250 N/mm2, y 100 MPa 100 N/mm2,

25 MPa 25 N/mm2 , 90 20 70

Solution:

(i) Normal Stress ( n

n x y

2 x y

2 cos 2 sin 2

250 100

2

250 1002

cos 2 70 25 sin 2 70

n 133.62 N/mm2

(ii) Tangential Stress t

t x y

2 sin 2 cos 2

250 100

2 sin 2 70 25 cos 2 70

t 67.36 N/mm2

(iii) Maximum Shear Stress (max

max x y

2

250 1002

75 N/mm2

Problem 1.76: An element in a stressed material has tensile of 500 MPaand a compressive stress of 350 MPa acting on two mutually perpendicularplanes and equal shear stresses of 100 MPa on these planes. Find principalstresses and position of the principal planes. Find also maximum shearingstress. (AU. Nov/Dec 2013)


Given that:

x 500 MPa

y 350 MPa

100 MPa

Solution:

(i) Major principal stress:

n1 x y

2

12

x y2 42

500 350

2 1

2 500 3502 4 [100]2

75 12

873.2

n1 511.6 N/mm2

Maximum shear stress:

max 12

x y2 42

12

500 3502 4 1002

max 436.6 N/mm2

Position of principal plane:

tan 2 2

x y

2 100x y

2 100

500 350

Problem 1.77: The state of stress in N/mm2) acting at a certain point ofa strained material is shown in Fig. compute (i) Magnitude and nature ofprincipal stresses and orientation of principal planes. (AU. Apr/May 2011)

Given: x 45 N/mm2, y 75 N/mm2, 45 N/mm2

Minor principal stress

n2 x y

2 1

2 x y

2 42

500 350

2 1

2 500 3502 4 [100]2

75 12

873.2

n2 361.6 N/mm2

2 tan 1 0.2352 1313 636


Solution:

(i) Magnitude of Principal Stresses

1 or 2 x y

2

x y

2

2

2

45 75

2

45 75

2

2

452

1 or 2 15 602 452

15 75

1 90 N/mm2 Maximum Principal Stress Tensile

2 60 N/mm2 Minimum Principal Stress Compressive

(ii) Orientation of Principal Stresses

tan 2 2

x y

2 45

45 75 0.75

75

75

45

45

45

45

45

45

71.57o

161 .57o

90

90

60

60

45

75

45

45

45

45

45

75


2 tan 1 0.75

2 36.86

2 P1 323.13, 2 P2

143.13

P1 161.57

P2 71.57

The direction of the principal stresses are shown in Fig.

Problem 1.78: The normal stress at a point on two mutually perpendicularplanes are 140 MPa (Tensile) and 100 MPa (Compressive). Determine theshear stress on these planes if the maximum principal stress is limited to 150MPa (Tensile). Determine also the following:(i) Minimum principal stress,(ii) Maximum shear stress and its plane and(iii) Normal, shear and resultant stresses on a plane which is inclined at30 anticlockwise to X plane. (Au. May/June 2013)

Given: Major tensile stress, x 140 N/mm2 Tensile

Minor compressive stress y 100 N/mm2

Maximum principal stress 1 150 N/mm2

Solution:

(ii) Maximum shear stress and its plane

We know that,

Maximum principal stress, 1 x y

2

x y

2

2

2


1 avg max

1 x y

2 max

150 140 100

2 max

max 150 20 130 N/mm2

(i) Minimum principal stress

2 x y

2

x y

2

2

2

140 100

2

140 100

2

2

502

402

202 502

20 53.85 33.85 N/mm2

(ii) Normal, shear and Resultant stress, @ 30

n x y

2 x y

2 cos 2

140 1002

140 100

2 cos 2 30

n 80 N/mm2

Shear stress (tangential stress), t x y

2 sin 2

140 1002

sin 2 30

t 103.92 N/mm2

max 140 100

2

2

2

130 2402

2

2

130 1202 2

Squaring on both sides, 1302 1202 2

2500 2

50 N/mm2


Resultant stress, R n2 t

2 802 103.922

R 131.15 N/mm2

1.35.5 Case (v) A member subjected direct stress in one plane and a simple shear stress

From the previous derivation we have obtained Normal stress on planeEC as shown in Fig. 1.46.

n x y

2 x y

2 cos 2 sin 2

and Tangential stress on plane EC

t x y

2 sin 2 cos 2

Substituting y 0 in the above two equations, we get

n x

2

x

2 cos 2 sin 2

x2

1 cos 2 sin 2

and t x

2 sin 2 cos 2

A C

12

A BE

Fig. 1.46.


Principal plane

We know that tan 2 2

x y

Substituting y 0, we get tan 2 2x

We also have sin 2 2

x2 42

; cos 2 x

x2 42

Principal stresses

n1 x2

x

2

2

2 or x

2

12

x2 42

n2 x

2

x

2

2

2 or x

2

12

x2 42


max n1

n2

2

12

x2 42 or

x

2

2

2

Problem 1.79: A plane element in a body is subjected to a tensile stress of120 MPa accompanied by a shear stress of 35 MPa. Find (a) the normaland shear stress on a plane inclined at an angle of 25 with the tensile stressand (b) the maximum shear stress on the plane.

Given: Tensile stress along x-x axis x 120 MPa ; Shear stress

xy 35 MPa ; angle made by plane with tensile stress 90 25 65

2

2

x

Fig. 1 .47.

x

2 +42


Solution:

(a) Normal and Shear stressesWe know that the normal stress on the plane.

n x

2 x

2 cos 2 xy sin 2

120

2

1202

cos 2 65 35 sin 2 65 MPa

60 60cos 130 35 sin 130 MPa

60 38.57 26.81 5.38 MPa

and shear stress on the plane,

x

2 sin 2 xy cos 2

1202

sin 2 65 35 cos 2 65

60 sin 130 35 cos 130

45.96 22.49 68.45 MPa

(b) Maximum shear stress on the planeWe also know that maximum shear stress on the plane,

max x2

2

xy2

1202

2

352 69.46 MPa

Problem 1.80: An element in a strained body is subjected to a tensile stressof 140 MPa and a shear stress of 40 MPa tending to rotate the element inan anticlockwise direction. Find (a) the magnitude of the normal and shearstresses on a section inclined at 35 with the tensile stress and (b) themagnitude and direction of maximum shear stress that can exist on theelement.


Given: Tensile stress along horizontal x x axis x 140 MPa; Shear stress

xy 40 MPa (Minus sign due to anticlockwise) and angle made by section

with the tensile stress 90 35 55

Solution:

(a) Magnitude of the normal and shear stress on the sectionWe know that magnitude of the normal stress on the section.

n x

2 x

2 cos 2 xy sin 2

1402

1402

cos 2 55

40 sin 2 55 28.47 MPa

and shear stress on the section

x2

sin 2 xy cos 2 140

2 sin 2 55 40 cos 2 55

65.78 13.68 52.1 MPa

(b) Magnitude and direction of the maximum shear stress that can existon the element

We know that magnitude of the maximum shear stress.

max x2

2

xy2

1402

2

402 80.62 MPa

Let x Angle which plane of maximum shear stress makes with x x axis

We know that, tan 2 x

2 xy

1402 40

1.75 or 2 s 60.25

s 30.12 or 120.12


1.36 MOHR’S CIRCLE FOR BIAXIAL STRESSESWhat is Mohr’s circle? (AU. Nov/Dec 2011)

Mohr’s circle method is a graphical method, which is frequently usedto find out the normal, tangential and resultant stresses on an oblique plane

1.36.1 Case 1: A body subjected to a biaxial perpendicular unequal and like principal stresses.

Consider a rectangular body subjected to two unequal and like principalstresses x and y which are perpendicular to each other as shown in Fig.

1.48

Let 1 Major Principal stress

2 Minor Principal stress and

Angle between the major pr incipal stress and the oblique section.

Guide lines for construction of the Mohr’s circleRefer the Mohr’s circle diagram shown in Fig. 1.48

1. Draw the hor izontal line OA and OB to scale to represent1 and 2 respectively.

2. Bisect BA at C, with C as centre draw a circle whose radius is equalto BC or AC. This circle is known as Mohr’s circle.

3. Through C, draw a line CD making an angle 2 with CA,

4. From D, Draw DE perpendicular to AB.

5. Join OD which represents the Resultant stress on an oblique plane.

OB

C2

1

n

2Resultan t ( )

AE

D

t

1

2

2

1

Fig. 1.48. M ohr’s Circle - Biaxial Stress for like and U nequal S tress.


6. The angle DOA is called the angle of obliquity which is denoted by.

7. OE and DE represent Normal and Tangential stress on oblique planerespectively.

Points to be remembered:1. The normal stress is maximum when the point D is at A and

minimum when D is at B. It is obvious from above, that thetangential stress is zero when the normal stress is maximum (or)minimum.

2. The tangential stress is maximum when the line E is at C andminimum when E is at A or B, since the tangential stress isperpendicular to AB.

Maximum shear stress max 1 2

2

3. The angle of obliquity is maximum when the line OD is tangentto the Mohr’s circle.

SOLVED PROBLEMS ON MOHR’S CIRCLEProblem 1.81: A material is subjected to the following principal stresses

100 N/mm2 and 40 N/mm2 both tensile. Find the normal, tangential andresultant stresses across a plane through the point at 48 to the majorprincipal plane, using Mohr’s circle of stresses.

Given: 1 100 N/mm2 ; 2 40 N/mm2

Solution:

Procedure:

(i) Take a scale 10 N/mm2 1 cm

(ii) Draw a line OA 100 N/mm2

10 10 cm 1

(iii) Mark the point OB 40 N/mm2

10 4 cm 2

(iv) Locate centre point of AB 10 4

2 3 cm AC BC


(v) With C as centre and BC or AC as radius draw a circle. This circle iscalled Mohr’s circle.

(vi) Draw line BP such that PBA 48

(vii) Draw PQ perpendicular to OA

(viii) Measure PQ and OQ. On measuring we get PQ 2.98 cm, therefore

Tangential stress t 2.98 10 29.8 N/mm2

OQ 6.69 cm, therefore

Normal stress n 6.69 10 66.9 N/mm2

(ix) Join OP Measure OP 7.32 cm; therefore

Resultant stress 7.32 10 73.2 N/mm2

(x) Measure POQ obliquity 24

(xi) Measure OPB 48 24 24

Problem 1.82: At a certain point in a strained material, the intensities of

stresses on two planes at right angles to each other are 20 N/mm2 and

10 N/mm2 both tensile. They are accompanied by a shear stress of Magnitude

10 N/mm2. Find graphically or otherwise, the orientation of principal planesand evaluate the principal stresses. (AU. May/June 2012)

P

Q CO A

Resultant = 73.2 N /m m2

B

=24 o

2 = 40 N /m m2

n = 66 .9 N /m m2

1 = 100 N /m m2

=48 o


Given: Major tensile stress, x 20 N/mm2

Minor tensile stress, y 10 N/mm2

Shear stress, 10 N/mm2

This problem may be solved analytically or graphically.

Graphically

Take 1 cm 2 N/mm2

Then x 202

10 cm

y 102

5 cm

102

5 cm

By Measurements, we have, Length AM 13.1 cm ; Length AL 1.91 cm

FOB (or) 2 63.7

Major principal stress 1 Length AM Scale 13.1 2

26.2 N/mm2

2O

B

CA L

x= N /m m 22 0

M

F

G

y= 1 N /m m 20


Minor Principal stress 2 Length AL Scale 1.91 2

3.82 N/mm2

Location of Principal planes, 2 63.7

63.7

2 31.85

The second principal plane is given by 90 or 31.85 90 or121.85

Or

Alternate Solution

Major tensile stress, x 20 N/mm2

Minor tensile stress, y 10 N/mm2

Shear stress, 10 N/mm2

Location of principal planes, tan 2 2

x y

2 2 10

20 10

2 tan 1[2]

31 43 or 121 43

2 0 N /m m2

2 0 N /m m2

= 1 0 N /m m2

1 0 N /m m2

= 1 0 N /m m2

1 0 N /m m2

P r in c ip a lp lan es

M a jo r P rin c ip a l

stre ss

1 2 1 4 3o

’

3 1 43o

’

M in o r P rin c ip a l

stre ss


Magnitude of Principal stresses

Major Principal Stress 1 x y

2

x y

2

2

2

20 10

2

20 10

2

2

102

15 52 100 15 25 100

1 26.18 N/mm2

Minor Principal Stress 2 x y

2

x y

2

2

2

20 10

2

20 10

2

2

102 15 11.18

2 3.82 N/mm2

1.36.2 Case 2: A body subjected to a biaxial perpendicular unequal and unlike principal stress

Consider a rectangular body subjected to two unequal and unlikePrincipal stresses 1 and 2 which are perpendicular to each other as shown

in Fig. 1.49.

Let 1 Major Principal tensile stress

2 Minor Principal compressive stress

Angle between the major pr incipal stress and the oblique plane.


Construction of the Mohr’s circle:1. Draw the horizontal line OA and OB opposite to each other to

represent 1 and 2 respectively, since the stresses are unlike (i.e. one

is tensile where as other is compressive)

2. Bisect BA at C, with C as centre draw a circle whose radius is equalto BC or AC. This circle is known as Mohr’s circle.

3. Through C, draw a line CD making an angle 2 with CA,

4. From D, draw DE perpendicular to OA.

5. Join OD which represents the Resultant stress on an oblique plane.

6. The angle DOA is called the angle of obliquity .

7. OE and DE represent Normal and Tangential stresses on obliqueplane respectively.

8. The maximum shear stress is equal to the radius of the Mohr’s circle.

max 1 2

2

Problem 1.83: A member is subjected to two principal stresses of

80 N/mm2 tensile and 50 N/mm2 compressive. Find the resultant stresses onthe planes making 22 and 64 with the Major principal plane. Find thenormal and tangential stresses on these planes.

1

2

Q

R

E

2 (Com pressive)

1

P

S

(Ten

sile

)

B O

t

n

21

C EA

D

Fig. 1.49. Mohr’s Circle - Biaxial Stress for Unequal and Unlike Stresses.


Given: 1 80 N/mm2 (tensile); 2 50 N/mm2 (compressive)

22 and 64

Solution:

The Mohr’s circle is drawn as shown

(i) Take a scale 10 N/mm2 1 cm.

1 8 cm; 2 5 cm

(ii) Draw a line BOA such that BO 5 cm and OA 8 cm

(iii) Bisect the line AB at C. BC 5 8

2 6.5 cm

(iv) With C as centre and BC as radius draw a circle. This circle is calledMohr’s circle.

For 22 (v) Draw line P1B such that P1BA 22

(vi) Draw P1Q1 perpendicular to BA

COB A

P 1

P 2

Q 2 Q 1

-57 N/m

m2

76 .5 N/m

m2

t1

=

45.2

N/m

m2

2 2 O64 O

50 N /m m 2 8 0 N /m m 2

t2

=

51.2

N/m

m2

t2

=

51.2

N/m

m2

t1

=

45.2

N/m

m2

n1 = 61 .8 N /m m 2

n2

= 25 N /m m 2


(vii) Measure P1Q1 tangential stress

t1 4.52 10 45.2 N/mm2

OQ1 Normal stress

n1 6.18 10 61.8 N/mm2

OP1 Resultant stress

Res 7.65 10 76.5 N/mm2

BP1O 1 14 11

For 64(viii) Draw line P2B such that P2BA 64

(ix) Draw P2Q2 perpendicular to BA

(x) Measure P2Q2 tangential stress

t2 5.12 10 51.2 N/mm2

OQ2 Normal stress

n2 2.5 10 25 N/mm2 compressive

OP2 Resultant stress

Res 5.7 10 57 N/mm2 compressive

1.36.3 Case 3: A body subjected to a Biaxial perpendicular unequal, like stresses with an simple shear stress.

Consider a body subjected to two mutually perpendicular stresses1 and 2 with a simple shear stress as shown in Fig. 1.50.

Let 1, 2 Major and Minor principal stresses.

Simple shear stress.

Angle between the major principal stress and oblique plane


Construction of Mohr’s circle:1. Draw a horizontal line OA and OB equal to 1 and 2 on some scale

on same side since both are tensile.

2. Bisect BA at C

3. From A, draw a perpendicular line AS which is equal to shear stress to the same scale.

4. With C as centre and CS as radius draw a circle which meets theline at G and H. This circle is called as Mohr’s circle.

5. Draw line CD which makes an angle 2 with CS.

6. Join point D with O and also draw a line DE which is perpendicularto OA.

7. OE and DE represents Normal and Tangential stresses on the obliquesection respectively. OD represents Resultant stress .

8. OG and OH will give the maximum and minimum value of normalstresses to scale and CS will give the maximum shear stress value.

Problem 1.84: A member is subjected to stresses on two mutually

perpendicular planes which are 120 N/mm2 (tensile) and 60 N/mm2 (tensile).

The shear stress across these planes is 30 N/mm2. Find the principal stressesand maximum shear stress at point by using Mohr’s circle.

Given x 120 N/mm2; y 60 N/mm2; 30 N/mm2

B

C

E

2

1

2

A

D

1

1

BO CG

n

2

2 AE H

D

t

S

m a x

(Resultan t)

Fig. 1.50. Mohr’s Circle - Biaxial Stress with Unequal Like Stress with Simple Shear Stress.


Solution:


x 12 cm ; y 6 cm ; 3 cm

(ii) Draw a line OBA such that OA 120 N/mm2 = 12 cm and

OB 6 cm 60 N/mm2

(iii) Draw AG perpendicular to OA such that AG 30 N/mm2 3 cm

(iv) Bisect BA at C.

(v) With C as centre and CG as radius draw a circle. This circle is calledMohr’s circle.

(vi) The circle cuts the line OBA at H and R

(vii) Measure OH Minor principal stress

4.757 10 47.57 N/mm2 n2 Ans

OR Major principal stress

13.2 10 132 N/mm2 n1 Ans

CD CG Radius of circle max 4.5 10 45 N/mm2(Ans.)

B C A

D

G

H

2 n 2 = = 47.57 N /m m2

y = 60 N/m m2

x = 120 N/m m2

1 n 1 = = 132 N/m m2

S

m ax=45 N/m m2

OR

=30 N/m m2


Problem 1.85: The stresses at a point in a material is

x 200 N/mm2 and y 150 N/mm2 and 80 N/mm2. Find the principal

plane and principal stresses using Mohr’s circle method and verify usinganalytical method. (AU. Nov 2014)

Given: x 200 N/mm2, y 150 N/mm2, 80 N/mm2

Solution:

1. The Mohr’s circle is drawn below.


x 200 4 cm ; y 150 3 cm ; 80 1.6 cm

(ii) Draw a line BOA such that BO = 3 cm and OA 4 cm

(iii) Draw AS perpendicular to BA of length 1.6 cm

(iv) Bisect BA at C

(v) With C as centre and CS are radius draw a circle. This circle is Mohr’scircle.

(vi) Draw RB perpendicular to OB.

(vii) The circle cuts the line OBA produced at G and H.

(ix) Measure OG Minor principal stress

2 n2 3.34 50 167.14 N/mm2

OH Major principal stress

1 n1 4.348 50 217.41 N/mm2

2 SCA 24.5; 12.25

2. Analytical method:

Principal stresses, n1 and n2

x y

2

x y

2

2

2


200 150

2

200 150

2

2

802 25 192.41}

n1 1 217.41 N/mm2 ; n2

2 167.41 N/mm2

Locating principal plane, tan 2 2

x y

2 80

200 150 0.457

2 24.56 or 12.28 or 102.28

R

H

S

GB A

y=150 N/m m 2

2 24.5 o

1 =217.41 N /m m 22=167.14 N /m m 2

(-)(+ve)

O C

x=200 N/m m 2

=80


IMPORTANT FORMULAE AND POINTSUnit I: Stress, Strain and Deformation of Solids

1. Stress LoadArea

N

m2

PA

2. 1 MPa 1 N/mm2

3. 1 GPa 1000 N/mm2 1 kN/mm2

4. Strain e Change in lengthOriginal length

dll

Tensile strain e Increase in lengthOriginal length

dLL

Compressive strain e Decrease in length

original length

dll

5. Shear stress total tangential force

crosssectional area of resisting section

PA

in N/m2

6. Shear strain Transverse displacement

Distance of the block from lower face

dll

7. Bearing stress b PA

Ptd

8. Hooke’s Law states that within the Elastic limit the stress (compressiveor tensile) is proportional to the strain

Mathematically, Hooke’s Law is

Stress Strain

StressStrain

Constant of proportionality.

9. Young’s Modulus E Normal stress

Linear nominal strain e

10. Rigidity Modulus G Shearing stressShearing strain

11. Factor of safety Ultimate stress

Permissible stress

www.srbooks.org Stress, Strain and Deformation of Solids IF.1

12. Deflection (or) deformation or L PLAE

13. Stiffness, k P

AEL

in N/m

14. Slenderness ratio lk

15. Percentage Elongation L Lo

Lo 100

16. Percentage reduction of area Ao A

Ao 100

17. Deformation for a bar of varying section

Deformation L or PE

L1A1

L2A2

L3

A3

18. Deformation of a body due to self weight

L wL2

2E

L or WL2AE

19. Principle of superposition

The forces acting on each of the sections should be taken individually.The deformation of the body will be equal to the algebraic sum of theindividual deformations.

This principle of finding out the resultant deformation is called theprinciple of superposition. The resultant deformation can be calculated by

P1l1AE

P2l2AE

P3l3AE

1

AE [P1l1 P2l2 P3l3 ]

where P1, P2, P3, are forces acting on sections 1, 2, 3 ... and l1, l2, l3 are

the lengths of the respective sections.

where l Length of column;

k Radius of gyration

IF.2 Strength of Materials for Mechanical Eng., - www.airwalkbooks.com

20. Deformation of uniformly tapering circular bar

4Pl

E D1 D2 D1 D2

D1D2

4PlED1D2

21. Deformation of uniformly tapering rectangular bar

or L PL

tE b1 b2 ln

b1

b2

22. Deformation in compound (or) composite bars

Strain in the different materials are equal

e1 e2

1

E1

2

E2

1 E1

E2 2

2 E2

E1 1

Change in length of both materials are equal

1 2

P1l

A1E1

P2l

A2E2

P2 P1 A2E2

A1E1

The load shared by composite bar of different materials is,

P P1 P2 P1 P1 A2E2

A1E1

P1 1

A2E2

A1E1

P P1 A1E1 A2E2

A1E1


P1 P

A1E1

A1E1 A2E2

P2 P

A2E2

A1E1 A2E2

23. Bar of uniform strength

A1 A2 e wl/

ln A1

A2

w

l

At any location x from bottom with Area A

ln AA2

w

x

24. Thermal Stresses

(i) Bar with one end fixed and other end free

Thermal stress, 0

Thermal strain, e ll

ltl

t

(ii) Bar with both ends fixed rigidly.

e ll

ltl

t

Compressive stress eE t E

(iii) Bar with one end fixed rigidly and other end not fixed rigidly (Barmight yield to a certain distance)

Actual expansion of the rod would be

l l t

The strain here is e ll

lt

l t

l

and stress induced eE t

l E


25. Thermal stresses in composite bars

The compressive load on one material is equal to the tensile load onthe other

1A1 2A2 ...(1)

Since the materials are rigidly fixed together the expansion is the same. Actual expansion in material 1 Actual expansion in material 2

Actual Expansion in material 1


Expansion due to tensilestress in Material 1

1TL 1

E1 L

Actual expansion in material 2


Contraction due to compressivestress in 2

2TL 2

E2 L

Actual Expansion in 1 Actual Expansion in 2

1TL 1

E1 L 2TL

2

E2 L

1T 1

E1 2T

2

E2

On rearranging the above equation we get

2T 1T 1

E

2

E

if E

strain e per unit length

1

E 2

E e1 e2 T 2 1

Total strain e1 e2 T 2 1


26. Thermal stresses in circular taper bar

max TEd1

d2

27. Thermal stresses in varying section bar

Thermal load is shared equally in both the sections

1A1 2A2

Total deformation of the bar (Free expansion)

l l1 l2 1l1E1

2l2E2

(For Different E )

l 1E

[1l1 2l2] [for same E]

28. Elastic constants

(i) Modulus of elasticity, E Axial stressAxial strain

(ii) Shear Modulus, G Shear stressShear strain

(iii) Bulk Modulus, K Direct stress

Volumetric strain

(iv) Poisson’s ratio, 1m

(or) lateral strain

longitudinal strain

Linear strain e ll

Lateral strain Change in diameterOriginal diameter

d2 d1

d1 dd

In case of a rectangular bar

Lateral strain Change in width or thickness

Original width or thickness bb

or tt

29. Relationship between Elastic constants

(i) Relation between Bulk Modulus and Young’s Modulus

K mE

3 m 2 (or) K

E

3 1

2m

E

3 1 2


(ii) Relation between shear Modulus and Young’s Modulus

G mE

2 m 1

E

2 1

1m

E

2 1

30. Principal planes and principal stress. The Planes which have no shearstress are known as Principal planes and the corresponding normal stressacting on a principal plane is known as Principal stress.

31. Member subjected to uniaxial stress

(i) Normal stress n cos2

(ii) Tangential stress t 2

sin 2

(iii) At 45 Maximum Shear Stress 2

P2A

.

32. Member subjected to biaxial stress

(i) n x y

2 x y

2 cos 2

Fig. 1.32. Stress on Inclined Plane


(ii) Tangential stress

t x y

2 sin 2

(iii) When 0

(iv) When 90

n2 y 2

(v) max x y

2

33. Member subjected to simple shear stress (Pure shear)

(i) n sin 2

(ii) t cos 2

(iii) Maximum principal stress, 1

(iv) Minimum principal stress, 2

34. Member subjected to simple shear and biaxial stress

(i) n x y

2 x y

2 cos 2 sin 2

(ii) t x y

2 sin 2 cos 2

(iii) tan 2 2

x y

(iv) 1 n1 x y

2

12

x y2 42

1 n1 x y

2

x y

2

2

2

(v) 2 n2 x y

2

12

x y2 42

Y

E

x xA sin

y yA cos

t

Fig. 1.35


(or)

2 n2 x y

2

x y

2

2

2

(vi) max 12

x y2 42

(or)

max x y

2

2

2

35. Mohr’s circle method is a graphical method, which is frequently usedto find out the normal, tangential and resultant stresses on an obliqueplane

36. Radius of Mohr’s circle for biaxial stresses

R max 12

x y2 42

(or)

R max x y

2

2

2


E3136B.E./B.Tech. Degree Examination, April/May 2010

Fourth SemesterMechanical Engineering

MEE2254 – STRENGTH OF MATERIALS

(Regulations 2008)(Common to Automobile Engineering, Production Engineering)

Time: Three hours Maximum: 100 marks

Answer ALL questions

PART A – 10 2 20 marks)

1. State Hooke’s lawRefer Page No.1.15

2. Define bulk modulus.Refer Page No.1.106

3. What do you understand by the term ‘Point of contraflexure’?Refer Page No.2.34

4. What is the value of bending moment corresponding to a point havinga zero shear force?The value of bending moment is the maximum bending moment

where shear force changes sign or becomes zero on a beam.

5. Write the assumption for finding out the shear stress of a circular shaftsubjected to torsion.Refer Page No.3.1

6. Define the term stiffness of a spring.Refer Page No.3.72

7. What is the relation between slope, deflection and radius of curvatureof a beam?Refer Page No.4.4

9. List out the modes of failure in thin cylindrical shell due to an internalpressure.When the internal pressure in thin cylindrical shells increases, the

developed hoop stress and longitudinal stresses also increase. When thesestresses exceed the permissible limit, the shell may fail in two ways

University Question Answers UQA.1

(i) It may split into semi-circular halves along longitudinal axis.

(ii) It may split into two cylinders.

10. What do you mean by principal plane?Refer Page No.1.118

PART B – 5 16 80 marks)

11. (a) (i) An alloy circular bar ABCD 3 m long is subjected to a tensileforce of 50 kN as shown in figure. If the stress in the middle portionBC is not to exceed 150 MPa, then what should be its diameter? Alsofind the length of the middle portion, if the total extension of the bar

should not exceed by 3 mm. Take E 100 GPa.

Refer Page No.1.35, Problem 1.15

(ii) A circular bar rigidly fixed at its both ends uniformly tapers from 75mm at one end to 50 mm at the other end. If its temperature is raisedthrough 26 K, then what will be the maximum stress developed in the

bar. Take E as 200 GPa and as 12 10 6/K for the bar material.Refer Page No.1.85, Problem 1.41

Or

(b) (i) In an experiment, a bar of 30 mm diameter is subjected to a pullof 60 kN. The measured extension on gauge length of 200 mm is 0.09mm and the change in diameter is 0.0039 mm. Calculate the Poisson’sratio and the values of the three modulliRefer Page No.1.109, Problem 1.56

(ii) An alloy specimen has modulus of elasticity of 120 GPa and modulusof rigidity of 45 GPa. Determine the Poisson’s ratio of the material.Refer Page No.1.108, Problem 1.55

12. A simply supported beam of 4 m span is carrying loads as shown infigure. Draw the shear force and bending moment diagrams for thebeam.Refer Page No.2.59, Problem 2.14

UQA.2 Strength of Materials for Mechanical Eng., - www.airwalkbooks.com

(b) A horizontal beam 10 m long is carrying a uniformly distributed loadof 1 kN/m. The beam is supported on two supports 6 m apart. Findthe position of the supports, so that bending moment on the beam isas small as possible. Also draw the shear force and bending momentdiagrams.Refer Page No.2.106, Problem 2.36

13. (a) (i) Obtain a relation for the torque and power, a solid shaft cantransmit.Refer Page No.3.8, Section 3.10

Consider a shaft subjected to torque T as shown in figure. Consider aelemental area da at a distance x from the axis of the shaft.

Let be the maximum shear stress induced in the outer most layer ofthe shaft and r be the radius of the shaft.

Shear stress at this section x xr

Area of the element da 2x dx

Shear force on the element area stress area

xr 2x dx

2r

x2 dx

Turning moment of this force about this element

dT shear force distance of the element from the axis

2r

x2 dx x 2r

x3 dx

Total moment of resistance offered by this element

T 0

r

2r

x3 dx

T 2r

x4

4 0

r

Td a

xr

d x


r 2r4

4 r r2

2 r d4

32 . . . r

d2

But r4

2 d4

32 is the polar moment of inertia of the cross section

and it is denoted by J

T r J

T r D4

32

D/2

D4

32

Torque T 16

D3

... r

D2

Torque for solid shaft T 16

D3

Let a shaft running at N rpm, transmits power “P” watts.

Let T be the mean Torque, in Nm

Power transmitted,

P Mean Torque Angle turned/s

T N60

2

P 2 NT

60 watts .

(ii) A solid steel shaft has to transmit 100 kW at 160 r.p.m. Taking allowableshear stress as 70 MPa, find the suitable diameter of the shaft. Themaximum torque transmitted in each revolution exceeds the mean by20%.Refer Page No.3.23, Problem 3.15

Or

(b) (i) Derive an equation for deflection of an open coiled helical spring.Refer Page No.3.76, Section 3.33


(ii) A closely coiled helical spring is made up of 10 mm diameter steel wirehaving 10 coils with 30 mm mean diameter. If the spring is subjectedto an axial twist of 10 kN-mm, determine the bending stress andincrease in the number of turns. Take E as 200 GPa.Refer Page No.3.106, Problem 3.60

14. (a) A cantilever AB, 2 m long, is carrying a load of 20 kN at free endand 30 kN at a distance 1 m from the free end. Find the slope and

deflection at the free end. Take E 200 GPa and I 150 106 mm4.Refer Page No.4.75, Problem 4.23

Or

(b) A simply supported beam AB of span 4 m, carrying a load of 100 kNat its mid span C has cross sectional moment of inertia

24 106 mm4 over the left half of the span and 48 106 mm4 over theright half. Find the slope at two supports and the deflection under the

load. Take E 200 GPa.Refer Page No.4.153, Problem 4.51

15. (a) (i) A cylindrical vessel 2 m long and 500 mm in diameter with10 mm thick plates is subjected to an internal pressure of 3 MPa.

Calculate the change in volume of the vessel. Take E 200 GPa and

Poisson’s ratio 0.3 for the vessel material.Refer Page No.5.16, Problem 5.11

(ii) A spherical shell of 2 m diameter is made up of 10 mm thick plates.Calculate the change in diameter and volume of the shell, when it is

subjected to an internal pressure of 1.6 MPa. Take E 200 GPa and

1/m 0.3.Refer Page No.5.40, Problem 5.29

Or

(b) (i) A point in a strained material is subjected to two mutuallyperpendicular tensile stress of 200 MPa and 100 MPa. Determine theintensities of normal, shear and resultant stresses on a plane inclined

at 30 with the axis of the minor tensile stress.Refer Page No.1.128, Problem 1.66


(ii) A point is subjected to a tensile stress of 250 MPa in the horizontaldirection and another tensile stress of 100 MPa in the verticaldirection. The point is also subjected to a simple shear stress of 25MPa, such that when it is associated with the major tensile stress, ittends to rotate the element in the clockwise direction. What is themagnitude of the normal and shear stresses on a section inclined at

an angle of 20 with the major tensile stress?



Question Paper Code: 11410B.E/B.Tech Degree Examination, April/May 2011

Fourth Semester – Mechanical EngineeringME 2254 — STRENGTH OF MATERIALS

(Common to Autonomous Engineering and Production Engineering)(Regulation 2008)

(Common to PTME 2254 Strength of Materials for B.E (Part-Time)Third Semester Mechanical Engineering Regulation 2009)


Answer All questionsPART — (10 2 20 marks)

1. A rod of diameter 30 mm and length 400 mm was found to elongate0.35 mm when it was subjected to a load of 65 kN. Compute themodulus of elasticity of the material of this rod.

CitronDiameter of rod 30 mm; L 0.35 mm

Length 400 mmLoad P 65 kN

Stress PA

65 103

4

302

91.96 N/mm2

Strain Ll

0.35400

8.75 10 4

E stressstrain

91.96

8.75 10 4

Modulus of elasticity of material

E 1.051 105 N/mm2

2. What is strain energy and write its unit in S.I system?

Strain energyThe energy, which is absorbed in the body due to staining effect is

known as strain energy.


The straining effect may be due to gradually applied load or suddenlyapplied load or impact load

Unit N m

3. Mention and sketch any two types of supports for the beams.Refer Page No.2.5

4. Sketch the bending stress as well as shear stress distribution for a beamof rectangular cross section.

5. Write down the simple torsion formula with the meaning of each symbolfor circular cross section.Refer Page No.3.5

6. Define stiffness of spring and mention its unit in SI system.Refer Page No.3.72

7. List any four methods of determining slope and deflection of loadedbeam.Refer Page No.4.5

9. What are assumptions involved in the analysis of thin cylindrical shells.Refer Page No.5.3

10. Define principle planes.Refer Page No.1.118

Bending Stress

q m ax

Shear - Stress distribution

Rectangular Section


PART - B

11. (a) Two vertical rods one of steel and other of copper are each rigidityfixed at the top and 600 mm apart. Diameters and lengths of the rodsare 25 mm and 5 m respectively. A cross bar fixed to the rods at thelower end carries a load of 7 kN such that the cross bar remainshorizontal even after loading. Find the steps in each rod and theposition of the load on the cross bar. Assume the modulus of elasticity

for steel and copper as 200 kN/mm2 and 100 kN/mm2 respectively.Refer Page No.1.60, Problem 1.27

11. (b) A cast iron flat 300 mm long and 30 mm (thickness) 60 mm(width) uniform cross section, is acted upon by the following forces:30 kN tensile in the direction of the length360 kN compression in the direction of the width240 kN tensile in the direction of the thicknessCalculate the direct strain, net strain in each direction and change involume of the flat. Assume the modulus of elasticity and Poisson’s ratio

for cast iron as 140 kN/mm2 and 0.25 respectively.Refer Page No.1.101, Problem 1.52

12. (a) Draw the shear force and bending moment diagrams for the beamshown in Fig. Q.12 (a). Also determine the maximum bending momentand location of point of contra flexure.

Solution

Taking moment of all forces about A, we get

RB 10 100 12 6

RB 720 N

100N /m

10m 2m

Fig.Q.12(a)


RA Total load RB

1200 720RA 480 N

Refer Similar Problem No. 2.25in Page No. 2.82

(a )

(b)

(c )


Step 2. Shear force diagram (S.F.D)S.FA 0 (without considering RA)

S.FA RA 480 kN (Considering RA)

S.FB 100 2 200 kN (without RB)

S.FB 200 720 520 N [with considering RB]

S.FC 0 [from Right]

S.F diagram is shown in the Fig (b).

Step 3. Bending moment diagram (B.M.D)Taking individual moments at various sections.

B.MA 0

B.MB 100 2

22

200 kNm

B.MC 0

Step 4. Maximum Bending Moment B.Mmax

S.FX RA w x 0 B.Mmax occurs whenS.F changes sign at X

480 100 x 0 x 480100

4.8 m

B.Mmax 480 4.8 100 4.82

2 1152 Nm

Step 5. Point of contraflexureThe point where bending moment changes its sign (or) where bending

moment is zero, is called the point of contraflexure (or) point of inflexion.

Taking moment about P RA y w y y2

0

RA w y

2 y

2RA

w

2 480100

9.6 m

The point of contraflexure occurs at 9.6 m from left end.

Or


12. (b) A cast iron pipe 300 mm internal diameter, metal thickness 15 mm,is supported at two points 6 m apart. Find the maximum bending stressin the metal of the pipe when it is running full of water. Assume the

specific weight of cast iron and water as 72 kN/m3 and 10 kN/m3

respectively.Refer Page No.2.120, Problem 2.42

13. (a) A steel shaft is required to transmit 75 kW power at 100 r.p.m andthe maximum twisting moment is 30% greater than the mean. Find the

diameter of the steel shaft if the maximum stress is 70 N/mm2. Alsodetermine the angle of twist in a length of 3 m of the shaft. Assume

the modulus of rigidity for steel as 90 kN/mm2.Refer Page No.3.27, Problem 3.20

Or

13. (b) A helical spring, in which the man diameter of the coils is 12 timesthe wire diameter, is to be designed to absorb 300 J energy with anextension of 150 mm. The maximum shear stress is not to exceed

140 N/mm2. Determine the mean diameter of the spring, diameter ofthe wire which forms the spring and the number of turns. Assume the

modulus of rigidity of the material of the spring as 80 kN/mm2.Refer Page No.3.104, Problem 3.59

14. (a) A beam of length 6 m is simply supported at the ends and carriestwo point load so 48 kN and 40 kN at a distance of 1 m and 3 mrespectively from the left support. Compute the slope and deflection

under each load. Assume EI 17000 kN m2.Refer Page No.4.29, Problem 4.7


15. (a) A cylindrical shell 800 mm in diameter, 3 m long is having 10 mmmetal thickness. If the shell is subjected to an internal pressure of

2.5 N/mm2

(i) the change in diameter (ii) the change in length and (iii) the change in volumeAssume the modulus of elasticity and Poisson’s ratio of the material

of the steel as 200 kN/mm2 and 0.25 respectively.Refer Page No.5.20, Problem 5.16

15. (b) The state of stress (in N/mm2) acting at a certain point of thestrained material is shown in Fig.Q.15 (b). Compute(i) The magnitude and nature of principal stresses and(ii) The orientation of principal planesRefer Page No.1.147, Problem 1.77

45 45

45

75

75

45

Fig. Q. 15(b)


Question Paper Code: 55461B.E/B.Tech. Degree Examination, Nov/Dec 2011


ME 2254 — STRENGTH OF MATERIALS(Common to Production Engineering and Automobile Engineering)

(Regulation 2008)Common to PTME 2254 — Strength of Materials for B.E (Part-Time)

Third Semester Mechanical Engineering, Regulation 2009)


Answer ALL questionsPART A — (10 2 20 marks)

1. Define ‘Volumetric strain’.Refer Page No.1.96

2. What is proof resilence?

Proof resilence The strain energy stored by the body ‘within’ elastic limits, when

loaded externally is called ‘Resilience’ and the maximum energy which abody stores ‘up to’ elastic limit is called ‘Proof Resilience’.

Proof Resilience is the mechanical property of materials and it indicatestheir capacity to bear shocks.

3. Mention the various types of beams with respect to support conditions.Refer Page No.2.5

4. What is meant by shear centre?Refer Page No.SQA 21, Q.No. 2.35

5. List the loads normally acting on a shaft.

Loads normally acting on a shaft 1. Torsional load

2. Bending load

3. Axial load

4. Combination of above three loads


6. Write the equation of torsion acting in a circular shaft.Refer Page No.3.5

7. Write the expression for the deflection at the free end of a cantileverof length ‘L’ carries an UDL of w kN/m. Assume uniform cross sectionthroughout.Refer Page No.4.156

9. What are the stresses developed in thin cylinders when they are subjectedto internal fluid pressures?Refer Page No.SQA 40, Q.No. 5.1

10. What are the uses of a Mohr’s circle?Refer Page No.1.154

PART B

11. (a) A steel wire 6 mm diameter is used for lifting a load of 1.5 kN atits lowest end, the length of the wire hanging vertically being 160

meters. Taking the unit weight of steel 78 kN/m3 and

E 2 105 N/mm2, calculate the elongation of the wire.Refer Page No.1.22, Problem 1.5

Or

11. (b) A steel bar 4 cm 4 cm in section, 3 meters long is subjected to

an axial pull of 128 kN. Taking E 20 1010 N/m2 calculate thealteration in the length of the bar. Calculate also the amount of energystored in the bar during extension.

Given

Steel bar cross section 4 cm 4 cm

L 3 m

P 128 kN

E 20 1010N/m2 2 105 N/mm2

l ? and Energy stored ?


Solution

(i) Stress P/A

128 103

40 40

80 N/mm2

(ii) Strain e E

80

2 105

e 4 10 4

ll

4 10 4 l 4 10 4 3000

l 1.2 mmEnergy stored in the bar during elongation

U 2

2E 4l

802

2 2 105 40 40 3000

76,800 N mmAns. Alteration in length of bar 1.22 mm

Energy stored 76.8 N m or J

12. (a) A simply supported beam of span 6 meters carries a point load of1 kN at 1 metre from the left end and another point load of 4 kN at1 metre from the right end. It carries a uniformly distributed load of2 kN/m run over a distance of 2 metres at midspan of the beam. Findthe SF and BM values. Draw SFD and BMD.Refer Nov/Dec 2013 Q.No. 12(a)

Or

12. (b) A timber beam 150 mm 250 mm in cross section is simplysupported at its ends and has a span of 3.5 m. The maximum safe

allowable stress is bending is 7500 kN/m2. Find the maximum safeUDL which the beam can carry. What is the maximum shear stress inthe beam for the UDL calculated?Refer Page No.2.177, Problem 2.77


13. (a) A hollow steel shaft of 100 mm internal diameter and 150 mmexternal diameter is to be replaced by a solid alloy shaft. If the polarmodulus has the same value for both, calculate the dimeter of the latterand ratio of their torsional rigidities.Refer Page No.3.32, Problem 3.24

Or

13. (b) A close coiled helical spring of 100 mm mean diameter is madeof 10 mm diameter rod and has 20 turns. The spring carries an axialload of 200 N. Determine the shearing stress. Taking the value of

modulus of rigidity as 84 109 N/m. Determine the deflection whencarrying this load. Also calculate the stiffness of the spring andfrequency of free vibrations for a mass hanging from it.Refer Page No.3.100, Problem 3.56

14. (a) A steel girder of uniform section, 14 meters long is simplysupported at its ends. It carries concentrated loads of 90 kN and 60kN at two points 3 metre and 4.5 metre from the two ends respectively.Calculate the deflection of the girder at the points under the two loads

and find the maximum deflection. Take I 64 10 4 m4 and

E 210 106 kN/m2.Refer Page No.4.27, Problem 4.6

15. (a) A cylindrical shell 3 metre long which is closed at the ends hasan internal diameter of 1 m and a wall thickness of 15 mm. Calculatethe circumference and longitudinal stress induced and also change inthe dimensions of the shell if it is subjected to an internal pressure of

1.5 106 N/m2. Take E 20 1010 N/m2 and Poisson’s ratio = 0.3.


Or

15. (b) A short metallic column of 500 mm2 cross sectional area carries

an axial load 100 kN. For a plane inclined at 60 with the direction

of load, calculate normal stress, tangential stress, resultant stress,maximum shear stress and obliquity of the resultant stress.Refer Page No.1.121, Problem 1.62


Question Paper Code: 10414Strength of Materials May/June 2012

Answer All QuestionsPart - A

1. Define ElasticityRefer Page No.1.1

2. Give the relationship between Modulus of elasticity and modulus ofRigidity.Refer Page No.1.108

3. List out the types of Beams.Refer Page No.2.1

4. What is point of contraflexure?Refer Page No.2.34

5. Write the expression for torsional rigidity of solid circular shaft.Refer Page No.3.8

TJ

Gl

GJ Tl

GJ Torsional rigidity

where T 16

d3

d diameter of shaft

6. State the difference between closed coil and open coil helical springs.Refer Page No.3.70

7. Give the expression for deflection of a simply supported beam carryinga point load at the centre.Refer Page No.4.8

ymax Wl3

48EI

9. List out the stresses induced in thin cylindrical shell due to internalpressure.Refer Page No.5.2

10. What are principal planes and stresses?Refer Page No.1.119


Part - B 16 Marks

11. (a) A steel tube of 30 mm external diameter and 20 mm internaldiameter encloses a copper rod of 15 mm diameter to which it is rigidly

joined at each end. If, at a temperature of 10C there is no longitudinalstress, calculate the stresses in the rod and tube when the temperature

is raised to 200C. Take E for steel and copper as 2.1 105 N/mm2

and 1 105 N/mm2 respectively. The value of co-efficient of linear

expansion for steel and copper is given as 11 10 6perC and

18 10 6 per Crespectively.Similarly Problem Refer Page No. 1.83, Problem No. 1.40

11. (b) A bar of cross section 8 mm 8 mm is subjected to an axial pullof 7000 N. The lateral dimension of the bar is found to be changed

to 7.9985 mm 7.9985 mm. If the modulus of rigidity of material is

0.8 105 N/mm2, determine the poisson’s ratio and modulus ofelasticity.Refer Similarly Problem Page No. 1.113, Problem No. 1.59

12. (a) A cantilever 1.5 m long is loaded with a uniformly distributed loadof 2 kN/m run over a length of 1.25 m from the free end. It alsocarries a point load of 3 kN at a distance of 0.25 m from free end.Draw the shear force and Bending moment diagrams of cantilever.Similar Problem Page No. 2.39, Problem No. 2.4

12.(b) A timber beam of rectangular section is to support a load of 20 kNuniformly distributed over a span of 3.6 m when beam is simplysupported. If the depth of the section is to be twice the breadth, and

the stress in the timber is not to exceed 7 N/mm2, Find the dimensionsof the cross-section.Refer Page No. 2.176, Problem No. 2.76

13. (a) A solid circular shafts transmits 75 kW power at 200 rpm.Calculate the shaft diameter, if the twist in the shaft is not to exceed

1 in 2 m length of shaft, and shear stress is limited to 50 N/mm2.

Take Modulus of rigidity C 1 105 N/mm2

Refer Page No. 3.13, Problem No. 3.4


13.(b) The stiffness of a close coiled helical spring is 1.5 N/mm ofcompression under a maximum load of 60 N. The maximum shearing

stress produced in the wire of the string is 125 N/mm2. The solid lengthof spring (when the coils are touching) is given as 5 cm. Find(i) diameter of wire(ii) Mean diameter of the coils and(iii) Number of coils required. Take Modulus of rigidity

4.5 104 N/mm2


14. (a) A beam of length 6 m is simply supported at its ends and carriestwo point loads of 48 kN and 40 kN at a distance of 1 m and 3 mrespectively from the left support. Find(i) deflection under each load,(ii) Maximum deflection, and(iii) the point at which maximum deflection occurs. Refer Page No. 4.29, Problem No. 4.7

15. (a) A cylindrical thin drum 80 cm in diameter and 3 m long has ashell thickness of 1 cm. If the drum is subjected to an internal pressure

of 2.5 N/mm2, determine(i) Change in diameter(ii) Change in length

(iii) Change in volume Take E 2 105 N/mm2 and poisson’s ratio

0.25Refer Page No. 5.12, Problem 5.8

(b) At a certain point in a strained material, the intensities of stresses on

two planes at right angles to each other are 20 N/mm2 and

10 N/mm2 both tensile. They are accompanied by a shear stress of

Magnitude 10 N/mm2. Find graphically or otherwise, the orientationof principal planes and evaluate the principal stresses.Refer Page No. 1.156, Problem 1.82


B.E./B.Tech. Degree Examination, May/June 2013Fourth Semester Mechanical Engineering

ME 2254/ME 45/CE 1259/10122 ME 405/080120018 – STRENGTH OFMATERIALS

(Common to Production Engineering and Automible Engineering)(Regulation 2008/2010) (Common to PTME 2254 – Strength of

Materials for B.E. (Part-Time) Third Semester MechanicalEngineering, Regulation 2009)


Answer ALL questions.PART A – 10 2 20 marks)

1. Define Hooke’s law.Refer Page No. 1.15

2. Define the term modulus of resilience.It is defined as the proof resilience of a material per unit volume.

Modulus of Resilience Proof Resilience

Volume of the body

3. Define point of contra flexure.Refer Page No. 2.34

4. What is meant by shear flow?Refer Page No. 2.165

5. Compute the torsional rigidity of a 100 mm diameter, 4 m length shaft

C 80 kN/mm2.Refer Page No. 3.8

GivenDiameter, d 100 mm

Length, l 4 m 4000 mm

Torsion Equation

TJ

C

l

TlCJ


Torsional Rigidity CJ

Torque J 32

1004 9.82 106 mm4

Torsional rigidity 80 103 9.83 106 7.86 1011

K 7.86 1011 N mm2

6. Write any four differences between open and closed coiled springs.Refer Page No. 3.70

7. Describe the double integration method.Refer Page No. 4.5

9. Define principal stress and principal plane.Refer Page No. 1.119

10. Define circumferential and Hoop stress.Refer Page No. 5.2

PART B – 5 16 80 marks)

11. (a) A reinforced concrete column 500 mm 500 mm in section is

reinforced with 4 steel bars of 25 mm diameter; one in each corner,the column is carrying a load of 1000 kN. Find the stresses in the

concrete and steel bars. Take E for steel 210 103 N/mm2 and E for

concrete 14 103 N/mm2.Refer Similar Problem Page No. 1.59, Problem No. 1.26

Or

11. (b) A solid circular bar of diameter 20 mm when subjected to an axialtensile load of 40 kN, the reduction in diameter of the rod was observed

as 6.4 10 3 mm. The bulk modulus of the material of the bar is 67GPa. Determine the following:(i) Young’s modulus,(ii) Poisson’s ratio,(iii) Modulus of rigidity,(iv) Change in length per metre and (v) Change in volume of thebar per metre length.Refer Problem 1.61 Page No. 1.116


12. (a) A simply supported beam of span 6 m is carrying a uniformlydistributed load of 2 kN/m over the entire span. Calculate themagnitude of shear force and bending moment at every section, 2 mfrom the left support. Also draw shear force and bending momentdiagrams.Refer Page No. 2.51, Problem 2.10

Or

12. (b) State the assumptions made in the theory of simple bending andderive the simple bending equation.Refer Page No. 2.11

13. (a) A hollow shaft with diameter ratio 3/5 is required to transmit 450kW at 120 rpm. The shearing stress in the shaft must not exceed

60 N/mm2 and the twist in a length of 2.5 m is not to exceed 1.Calculate the minimum external diameter of the shaft.

C 80 kN/mm2.Refer Page No. 3.36, Problem No. 3.27

13. (b) Derive a relation for deflection of a closely coiled helical springsubjected to an axial downward load W.Refer Page No. 3.70

14. (a) Determine the maximum bending moment and deflection of a beamof length l and flexural rigidity “EI”. The beam is fixed at both endsand carries a point load “W” at the mid span.Refer Page No. 4.91

15. (a) A thin cylinder is 3.5 long, 90 cm in diameter, and the thicknessof metal is 12 mm. it is subjected to an internal pressure of

2.8 N/mm2. Calculate the change in dimensions of the cylinder and the

maximum intensity of shear stress induced. Given E 200 GPa and

Poisson’s ratio 0.3.Refer Page No. 5.14, Problem No. 5.9

Or


15. (b) The normal stress at a point on two mutually perpendicular planesare 140 MPa (Tensile) and 100 MPa (Compressive). Determine theshear stress on the stress on these planes if the maximum principalstress is limited to 150 MPa (Tensile). Determine also the following:(i) Minimum principal stress,(ii) Maximum shear stress and its plane and(iii) Normal, shear and resultant stresses on a plane which is inclined

at 30 anticlockwise to X plane.Refer Page No. 1.148, Problem No. 1.78


Question Paper Code: 31565B.E/B.Tech. Degree Examination, November/December 2013


ME 2254/ME 45/CE 1259/10122 ME 405/080120018 — STRENGTH OF MATERIALS

(Common to Production Engineering and Automobile Engineering)(Regulation 2008/2010) (Common to PTME 2254 – Strength of

Materials for B.E (Part-Time) Third Semester MechanicalEngineering, Regulation 2009)


Answer ALL questions.PART A — (10 2 20 marks)

1. Define shear strain and volumetric strain.Refer Page No. 1.104 and 1.96

2. What is meant by strain energy and proof resilience?

Proof Resilience:The maximum strain energy that can be stored in a material within its

elastic limit is known as proof resilience.

3. What are the types of beam and draw a neat sketch for each type?Refer Page No. 2.2

4. Write down an equation for shear stress distribution across the crosssection of a beam and draw a typical shear stress distribution diagramfor an I-section.Refer Page No. 2.142For a beam with rectangular section of depth d, the equation is

F2I

d2

4 y2

Shear stress distribution of beam (Refer Page No. 2.144)

Shear stress distribution diagram for I - section (Refer Page No. 2.144)

5. Write down the equation of torsion showing the various terms involvedin it.Refer Page No. 3.5


Equation of torsion:

TJ

Gl

r

T - Torque N.mmJ - Polar moment of inertia

G - Modulus of Rigidity - N/mm2

6. A closely coiled helical spring is to carry a load of 500 N. Its meancoil diameter is to be 10 times that of the wire diameter. Calculatethese diameter if the maximum shear stress in the material of the spring

is to be 80 MN/m2.

Given that:P 500 NLet D be the coil diameter and d, be the wire diameter.Then,D 10 d

80 MN/m2

80 N/mm2

Solution:

Shear stress 8WD

d3 . . . R

D2

80 8 500 10d

d3d2 159.155d 12.6 mm

7. Write down the relationship between slope, deflection and radius ofcurvature.Refer Page No. 4.4

9. Calculate the thickness of metal required for a cast iron main 800 mmin diameter for water at pressure head of 100 m if the maximum

permissible tensile stress is 20 MN/m2 and weight density of water is

10 kN/m3.Refer Page No. 5.9, Problem 5.2

- Angle of twist - radl - Length - mm

- Shear stress - N/mm2

r = Radius - mm


10. What is meant by principal planes and principal stresses?The planes which have no shear stress are known as principal planes,

the planes carry only normal stresses. The magnitude of normal stresses,acting on the principal planes are known as principal stresses.

PART - B

11. (a) Three bars made of copper, zinc and aluminium of equal lengthare rigidly connected at their ends as shown in Fig. (i).

They have cross-sectional areas of 250 mm2, 375 mm2 and 500 mm2

respectively. If the compound member is subjected to a longitudinalpull of 125 kN, estimate the proportion of load carried on each rod

and the induced stresses. Take ECu 130 GN/m2, EZn 100 GN/m2,

EAl 80 GN/m2.

Refer Page No. 1.72, Problem No. 1.34Or

11. (b) The following data relate to a bar subjected to a tensile test:Diameter of the bar = 30 mmTensile load P = 54 kNGauge length l = 300 mm

Extension of the bar 1 = 0.112 mm

Change in diameter d = 0.00366 mm

Calculate(i) Poisson’s ratio(ii) The values of three moduliiRefer Page No. 1.110, Problem 1.57

P=125kN P=125kN

250m m 2

375m m 2

500m m 2

�

Copper

Zinc

A lum in ium


12. (a) A simply supported beam is subjected to loads as shown. Find themaximum bending moment and determine its value.

Same Problem was asked in Nov/Dec 2011 Q.No. 12(a) V 0

RA RB 1 2 2 4 9 kN

MA 0

RB 6 4 5 2 2 3 1 1 0

RB 5.5 kN

RA 3.5 kN

SF: @B = – 5.5 kN

@F = 5.5 4 1.5 kN

@E = –1.5 kN

@D = 1.5 2 2 2.5 kN

@C = 2.5 1 3.5 kN

@A = 3.5 kN

BM:@B = 0

@F = RB 1 5.5 kN

@E = RB 2 4 1 7 kN.m

@D = RB 4 4 3 2 2 1 6 kN.m

@C = RB 5 4 4 2 2 2 3.5 kN.m

@A = 0

Max BM:. . . SF 0

5.5 4 2 x 2 0

x 2.75 mMax BM:Mm RB 2.75 4 1.75

2 0.75

0.752

7.56 kN.m

1kN 4kN2kN /m

A C D E F B

1m 1m 2m 1m 1m

Fig.(ii)


Or

12. (b) A steel bar 120 mm in diameter is completely encased in analuminium tube of 180 mm outer diameter and 120 mm inner diameterso as to make it a composite beam. The composite beam is subjectedto a bending moment of 15 kN-m. Determine the maximum stress in

steel and aluminium due to bending. Take Es 3Ea.


B

2kN/m1kN 4kN

A

R =3.5kN

A R =5.5kN

B

-5 .5kN -5 .5kN

-1 .5kN -1 .5kN

2.5kN2.5kN

3.5kN 3.5kN

x=2.75m

7.56kN .m

7kN.m

5.5kN .m

6kN.m

3.5kN .m

A C D E F B

C D E F

1m 1m 2m 1m 1m

SFD

BMD


13. (a) A solid circular shaft transmits 75 kW power at 200 r.p.m.Calculate the shaft diameter, if the twist in the shaft is not to exceed

1 in 2 metres length of shaft and shear stress is limited to

50 MN/m2. Take C 100 GN/m2.Refer Page No. 3.13, Problem No. 3.4

Or

13. (b) A closed coiled helical spring has stiffness of 10 N/mm; Its lengthwhen fully compressed with adjacent coils touching each other is 400mm; The modulus of rigidity of the material of the spring = 80 GPa.(i) Determine the wire diameter and mean coil diameter if their ratiois 1/10.(ii) If the gap between any two adjacent coil is 2 mm, what maximumload can be applied before the spring becomes solid (ie) adjacent coilstouch.(iii) What is the corresponding maximum shear stress in the spring?Refer Page No. 3.107, Problem No. 3.61

14. (a) A 2 meters long cantilever made up of steel tube of section 150mm external diameter and 10 mm thick is loaded as in Fig. (iii)

Take E 200 GPa. Calculate

(i) The value of W so that the max bending stress is 150 MPa

(ii) The maximum deflection for the loading.Refer Page No. 4.77, Problem No. 4.24

AC

B

2m

2W

( -a) = 1.5m� a = 0.5m

W

t =10m m

FreeEnd

130mm

150mm

Cross Section of Cantilever beam.

Fig. (iii)


15. (a) An element in a stressed material has tensile of 500 MPa and acompressive stress of 350 MPa acting on two mutually perpendicularplanes and equal shear stresses of 100 MPa on these planes. Findprincipal stresses and position of the principal planes. Find alsomaximum shearing stress.Refer Page No. 1.145, Problem No. 1.76

Or

15. (b) A boiler shell is to be made of 15 mm thick plate having tensilestress of 120 MPa. If the efficiencies of the longitudinal andcircumferential joints are 70% and 30% respectively, determine:(i) Maximum permissible diameter of the shell for an internal pressureof 2 MPa.(ii) Permissible intensity of internal pressure when the shell diameteris 1.5 m.

Given:Refer Page No. 5.30, Problem No. 5.23


Question Paper Code : 97029B.E./B.Tech. Degree Examination, Nov/ Dec 2014

Third Semester - Mechanical Engineering

CE 6306 — STRENGTH OF MATERIALS(Common to Mechatronics Engineering, Industrial Engineering and

Management, Industrial Engineering, Manufacturing Engineering, MechanicalEngineering (sandwich) and Material Science and Engineering)

(Regulation 2013)

Time : Three hours Maximum : 100 marks


1. Derive a relation for change in length of a bar hanging freely underits own weights.Refer Pg.No.1.41

2. Write the relationship between shear modulus and young’s modulus ofelasticity.Refer Pg.No.1.108

3. Draw SFD for a 6 m cantilever beam carrying a clockwise moment of6 kN-m at its free end.

4. What are flitched beams?Refer Pg.No.2.137

5. What is meant by torsional rigidity?Refer Pg.No.3.8

6. Differentiate open coiled and close coiled helical springs.Refer Pg.No.3.70

7. What are the limitations of double integration method?Refer Pg.No.SQA 32 / Q.No.4.6

A Bo o

S.F = 06m

6KN-mA

B


8. Define strain energy.Refer Pg.No.4.158

9. What is meant by circumferential stress?Refer Pg.No.5.2

10. Write down Lame’s equations.Refer Pg.No.5.45

Part B

11. (a) (i) Derive an expression for change in length of a circular barwith uniformly varying diameter and subjected to an axial tensile load‘P’.Refer Pg.No.1.52

11. (a) (ii) A member is subjected to point loads as shown in Fig. Q. 11(a). Calculate the force P, necessary for equilibrium if

P1 45 kN, P3 450 kN and P4 130 kN. Determine total elongation

of the member, assuming the modulus of elasticity to be

E 2.1 105 N/mm2.Refer Problem No. 1.22, Page 1.46

11. (b) A metallic bar 300 mm x 100 mm y 40 mm z is subjectedto a force of 5 kN (tensile), 6 kN (tensile) and 4 kN (tensile) alongx, y and z directions respectively. Determine the change in the volume

of the block. Take E 2 105 N/mm2 and Poisson’s ratio = 0.25.

Refer Pg.No.1.110, Problem No. 1.51

A

B C

D

P 1P 4

P 2 P 3

1 20 0m m

6 00m m

9 00m m

6 25 m m 2 1 25 0m m 2

250

0 m

m2

Fig.


12. (a) Draw SFD and BMD and find the maximum bending moment forthe beam given in Fig. Q.12 (a).Similar Problem in Page No. 2.67, Problem No. 2.18

12. (b) Prove that the ratio of depth to width of the strongest beam thatcan be cut from a circular log of diameter ‘d’ is 1.414. Hence calculatethe depth and width of the strongest beam that can be cut out of acylindrical log of wood whose diameter is 300 mm.Refer Page No. 2.131, Problem No. 2.50

13. (a) Derive torsion equation.Refer Pg.No.3.2 - 3.5

13. (b) The stiffness of a close-coiled helical spring is 1.5 N/mm ofcompression under a maximum load of 60 N. The maximum shearing

stress produced in the wire is 125 N/mm2. The solid length of thespring (when the coils are touching) is given as 50 mm. Find(i) The diameter of wire (ii) The mean diameter of the coils (iii) Number of coils required.

Take C 4.5 104 N/mm2

Refer Pg.No.3.101, Problem No. 3.57

14. (a) Determine the deflection of the beam at its mid span and also theposition of maximum deflection and maximum deflection. Take

E 2 105 N/mm2 and I 4.3 108 mm4. Use Macaulay’s method.The beam is given in Fig. Q.14 (a).Refer Pg.No.4.51, Problem No. 4.16

A B C D E

20kN6kN /m

2m 2m 2m 2m


15. (a) Derive relations for change in dimensions and change in volumeof a thin cylinder subjected to internal pressure P.Refer Pg.No.5.6, Section 5.3

15. (b) Find the thickness of metal necessary for a thick cylindrical shellof internal diameter 160 mm to withstand an internal pressure of

8 N/mm2. The maximum hoop stress in the section is not to exceed

35 N/mm2.Refer Pg.No.5.56, Problem No. 5.35

A B

1m 4m 3m

40kN /m

Fig.

DC


Question Paper Code: 27099

B.E./B.Tech. Degree Examination, Nov/Dec 2015Third Semester - Mechanical EngineeringCE 6306 - STRENGTH OF MATERIALS

(Common to Mechatronics Engineering, Industrial Engineering andManagement, Industrial Engineering, Manufacturing Engineering,

Mechanical Engineering (Sandwich) Material Science and Engineeringand also Common to Fourth Semester Automobile Engineering,

Mechanical and Automation Engineering and Production Engineering)

(Regulations 2013)


Answer ALL questions.PART A - 10 2 20 marks

1. Differentiate Elasticity and Elastic Limit.Refer Page No. 1.1 and 1.12

2. What is principle of super position?Refer Page No. 1.43

3. Write the assumption in the theory of simple bending?Refer Page No. 2.111, Section No. 2.19.1

4. What are the types of beams?Refer Page No. 2.1, Section No. 2.2

5. The shearing stress in a solid shaft is not to exceed 40 N/mm2 when

the torque transmitted is 20000 Nm. Determine the minimum diameterof the shaft.

Max shear stress 40 N/mm2

Torque transmitted T 20000 Nm

20 106 Nmm

Let D min Dia of shaft

T 16

D3


D 16T

13 16 20000000

40

1/3

136.5 mm

6. What are the various types of springs?Refer Page No. 3.68

7. What are the methods of determining slope and deflection at a sectionin a loaded beam?Refer Page No. 4.5

8. What is the equation used in the case of double integration method?Refer Page No. 4.5

9. State the expression for maximum shear stress in a cylindrical shell.Refer SQA 40, Q.No. 5.5

10. Define-hoop stress and longitudinal stress.

Refer SQA 40, Q.No. 5.2 and 5.3

Part - B

11. (a) A metallic bar 300 mm 100 mm 40 mm is subjected to a forceof 5 kN (tensile), 6 kN (tensile) and 4 kN (tensile) along x, y and zdirections respectively. Determine the change in the volume of the

block. Take E 2 105 N/mm2 and Poisson’s ratio 0.25.Refer Page No. 1.100, Problem No. 1.51

11. (b) A steel rod of 3 cm diameter is enclosed centrally in a hollowcopper tube of external diameter 5 cm and internal diameter of 4 cmas shown in Fig.1. The composite bar is then subjected to axial pullof 45000 N. If the length of each bar is equal to 15 cm, determine:(i) The stresses in the rod and tube, and (ii) Load carried by each

bar. Take E for steel 2.1 105 N/mm2 and for copper

1.1 105 N/mm2.Refer Page No. 1.65, Problem No. 1.30


12. (a) Draw the shear force and B.M diagrams for a simply supportedbeam of length 8 m and carrying a uniformly distributed load of10 kN/m for a distance of 4 m as shown in fig.2.Refer Page No. 2.57, Problem No. 2.13

12. (b) A steel plate of width 120 mm and of thickness 20 mm is bent intoa circular arc of radius 10 m. Determine the maximum stress inducedand the bending moment which will produce the maximum stress. Take

E 2 105 N/mm2.Refer Page No. 2.116, Problem No. 2.37

13. (a) A hollow shaft of external diameter 120 mm transmits 300 kWpower at 200 r.p.m. Determine the maximum internal diameter if the

maximum stress in the shaft is not to exceed 60 N/mm2.Refer Page No. 3.18, Problem No. 3.10

13. (b) A closely coiled helical spring of mean diameter 20 cm is made of3 cm diameter rod and has 16 turns. A weight of 3 kN is dropped onthis spring. Fine the height by which the weight should be droppedbefore striking the spring so that the spring may be compressed by

18 cm. Take C 8 104 N/mm2.


14. (a) A beam 6 m long, simply supported at its ends, is carrying a pointload of 50 kN at its centre. The moment of inertia of the beam is given

as equal to 78 106 mm4. If E for the material of the beam

2.1 105 N/mm2, calculate: (i) deflection at the centre of thebeam and (ii) slope at the supports.Refer Page No. 4.32, Problem No. 4.8

14 (b) A beam of length 6 m is simply supported at its ends and carriestwo point loads of 48 kN and 40 kN at a distance of 1 m and 3respectively from the left support as shown Fig-3.Using Macaulay’s method find:(i) deflection under each load, (ii) maximum deflection, and (iii) the point at which maximum deflection occurs,Refer Page No. 4.29, Problem No. 4.7


15. (a) A boiler is subjected to an internal steam pressure of 2 N/mm2.The thickness of boiler plate is 2.6 cm and permissible tensile stress

is 120 N/mm2. Find the maximum diameter, when efficiency oflongitudinal joint is 90% and that of circumference point is 40%.Refer Page No. 5.29, Problem No. 5.22

15.(b) Calculate: (i) the change in diameter, (ii) change in length and (iii)change in volume of a thin cylindrical shell 100 cm diameter, 1 cm thick

and 5 m long when subjected to internal pressure of 3 N/mm2. Take the

value of E 2 105 N/mm2 and Poisson’s ratio, 0.3.


��

��

��

�

�

��

��

�

�

��


Question Paper Code: 80197B.E./B.Tech. Degree Examination, Nov/Dec 2016

Third Semester – Mechanical EngineeringCE 6306 — STRENGTH OF MATERIALS

(Common to Mechatronics Engineering, Industrial Engineering andManagement, Agriculture Engineering, Industrial Engineering, Manufacturing

Engineering, Mechanical Engineering (Sandwich), Materials Science andEngineering and also Common to Fourth Semester Automobile Engineering,

Mechanical and Automation Engineering and Production Engineering)(Regulations 2013)

Time: Three hours Maximum : 100 marks


1. Define Young’s Modulus.Refer Page No. 1.15, 1.93

2. What do you mean by principal planes and principal stresses?Refer Page No. 1.119

3. Draw the shear force diagram and bending moment diagram for thecantilever beam carries uniformly varying load of zero intensity at thefree end and w kN/m at the fixed end.Refer Page No. 2.23 / Fig. 2.27

4. List out the assumptions used to derive the simple bending equation.Refer Page No. 2.111

5. Define torsional rigidity.Refer Page No. 3.8

6. What is a spring? Name the two important types of springs.Refer Page No. 3.68

7. List out the methods available to find the deflection of a beam.Refer Page No. 4.5

8. State Maxwell’s reciprocal theorem.Refer Page No. SQA 39, Q.No. 4.36


9. Name the stresses develop in the cylinder.Refer SQA Page No. 40, Q.No. 5.1

10. Define radial pressure in thin cylinder.Refer SQA Page No. 40, Q.No. 5.4

PART B — (5 13 65 marks)

11. (a) (i) A compound tube consists of a steel tube 140 mm internaldiameter and 160 mm external diameter and an outer brass tube 160mm internal diameter and 180 mm external diameter. The two tubesare of same length. The compound tube carries an axial compressionload of 900 kN. Find the stresses and the load carried by each tubeand the amount of its shortens. Length of each tube is 140 mm. Take

E for steel as 2 105 N/mm2 and for brass 1 105 N/mm2. (10)Refer Page No. 1.63, Problem No. 1.29

11. (a) (ii) Two members are connected to carry a tensile force of 80 kNby a lap joint with two number of 20 mm diameter bolt. Find the shearstress induced in the bolt. (3)Refer Page No. 1.104, Problem 1.54

11. (b) (i) A point in a strained material is subjected to the stress asshown in fig. Q.11.(b)(i). Locate the principle plane and find theprinciple stresses. (7)Refer Page No. 1.141, Problem 1.73

40 N/mm2

40 N/mm2

60 N/mm2

60 N/mm2

60o

60o

A B

D C


11. (b) (ii) A steel rod of 20 mm diameter passes centrally through acopper tube of 50 mm external diameter and 40 mm internal diameter.The tube is closed at the end by rigid plates of negligible thickness.The nuts are tightened lightly on the projecting parts of the rod. If the

temperature of the assembly is raised by 50C, calculate the stresses

developed in copper and steel. Take E for steel as 2 105 N/mm2 and

copper as 1 105 N/mm2 and for steel and copper as 12 10 6

per C and 18 10 6 per C. (6)Refer Page No. 1.81, Problem No. 1.39

12. (a) (i) A simply supported beam AB of length 5 m carries point loadsof 8 kN, 10 kN and 15 kN at 1.50 m, 2.50 and 4.0 m respectivelyfrom left hand support. Draw shear force diagram and bending momentdiagram. (8)Refer Similar Problem No. 2.8 - Page No. 2.46

12. (a) (ii) A cantilever beam AB of length 2 m carries a uniformlydistributed load of 12 kN/m over entire length. Find the shear stress

and bending stress, if the size of the beam is 230 mm 300 mm.

(5)Refer Page No. 2.118, Problem No. 2.40

12. (b) Construct the SFD and BMD for the beam shown in fig. Q.12.(b)(i).(6)Refer Page No. 2.48, Problem 2.9

12. (b) (ii) Two timber joist are connected by a steel plate, are used asbeam as shown in fig.Q.12.(b)(ii). Find the load W if, the permissible

stresses in steel and timber are 165 N/mm2 and 8.5 N/mm2

respectively. (7)Refer Page No. 2.138, Problem No. 2.55

A BC 2.00m2.00m

DE 0.75m0.50m

25kN Fig.


13. (a) (i) A solid shaft has to transmit the Power 105 kW at 2000 r.p.m.The maximum torque transmitted in each revolution exceeds the meanby 36%. Find the suitable diameter of the shaft, if the shear stress is

not to exceed 75 N/mm2 and maximum angle of twist is 1.5 in a

length of 3.30 m and G 0.80 105 N/mm2. (8)Refer Page No. 3.15, Problem No. 3.6

13. (a) (ii) A laminated spring carries a central load of 5200 N and itis made of ‘n’ number of plates, 80 mm wide. 7 mm thick and length500 mm. Find the numbers of plates, if the maximum deflection is 10

mm. Let E 2.0 105 N/mm2. (5)Refer Page No. 3.122, Problem No. 3.69

13. (b) (i) A stepped solid circular shaft is built in at its ends and subjectto an externally applied torque T at the shoulder as shown in fig.

Q.13.(b)(i). Determine the angle of rotation of the shoulder sectionwhen T is applied. (7)

Refer Page No. 3.43, Section 3.13.2 in Strength of material

T

TA

T B

ab

d ,J1 A d ,J2 B

A BC1m 1m

w w

150mm

Timber

S tee l

40m m 40m m

10m m

Cross Section


13. (b) (ii) A closed coiled helical spring is to be made out of 5 mmdiameter wire 2 m long so that it deflects by 20 mm under an axialload of 50 N. Determine the mean diameter of the coil. Take

C 8.1 104 N/mm2. (6)Refer Page No. 3.89, Problem No. 3.46

14. (b) (i) Derive the formula to find the deflection of a simply supportedbeam with point load W at the centre by moment area method. (8)Refer Page No. 4.117, Problem No. 4.34

14. (b) (ii) A simply supported beam of span 5.80 m carries a centralpoint load of 37.50 kN, find the maximum slope and deflection, let

EI 40000 kN m2. Use conjugate beam method. (5)Refer Page No. 4.149, Problem No. 4.49

15. (a) Calculate Change in diameter, Change in length and Change involume of a thin cylindrical shell 100 cm diameter, 1 cm thick and 5

m long when subjected to internal pressure of 3 N/mm2. Take the value

of E 2 105 N/mm2 and Poisson’s ratio = 0.30. (13)Refer from Page No. 5.21, Problem No. 5.17

15. (b) Calculate the thickness of metal necessary for a cylindrical shellof internal diameter 160 mm to with stand an internal pressure of

25 MN/m2, if maximum permissible shear stress is 125 MN/m2. (13)Refer Page No. 5.56, Problem No. 5.35

PART C — (1 15 15 marks)

16. (a) The intensity of resultant stress on a plane AB (Fig. Q.16(a)) at

appoint in a materials under stress is 8 N/mm2 and it is inclined at

30 to the normal to that plane. The normal component of stress on

another plane BC at right angles to plane AB is 6 N/mm2. Determinethe following:(i) The resultant stress on the plane BC(ii) The principal stresses and their directions(iii) The maximum shear stresses. (15)Refer Page No. 1.138, Problem No. 1.70


16. (b) A water tank vertical wall is stiffened by vertical beam, and theheight of the tank is 8 m. The beams are spaced at 1.5 m centre tocentre. If the water reaches the top of the tank, calculate the maximumbending moment. Sketch the shear force and bending moment diagrams.

Unit weight of water 9.8 kN/m3. (15)Refer Page No. 2.75, Problem No. 2.22

6 N /m m2

8 N /m m2

30o

A

BC


Question Paper Code: 40778B.E./B.Tech. DEGREE EXAMINATION, APRIL/MAY 2018

Third/Fourth SemesterMechanical Engineering

CE 6306 – STRENGTH OF MATERIALS(Common to Mechanical Engineering (Sandwhich)/Agriculture

Engineering/Automobile Engineering/Industrial Engineering/IndustrialEngineering and Management/Manufacturing Engineering/Materials

Science and Engineering/Mechanical and AutomationEngineering/Mechatronics Engineering/Production Engineering)

(Regulations 2013)

Time: Three Hours Maximum: 100 Marks

Answer ALL questionsPART - A 10 2 20 Marks)

1. Give the relation between modulus of elasticity and bulk modulus.Refer Page No. 1.107

2. What is principal stress?Refer Page No. 1.119

3. What do you mean by the point of contraflexure?Refer Page No. 2.83

4. What is meant by shear stresses in beam?Refer Page No. 2.142

5. What is polar moment of inertia for solid shaft?Refer Page No. 3.6

J 32

D4

6. Differentiate open coiled helical spring from the close coiled helicalspring and state the type of stress induced in each spring due to anaxial load.Refer Page No. 3.69 and 3.70

7. Write the equation giving maximum deflection in case of a simplysupported beam subjected to a point load at mid span.Refer Page No. 4.8


8. State the two theorems of conjugate beam method.Refer Page No. 4.140

9. Write the expression for circumferential stress and longitudinal stresswhen a thin cylinder is subjected to an internal fluid pressure of ‘P’Refer Page No. 5.4 and 5.5

10. Write down the Lame’s equations.Refer Page No. 5.45

PART - B 5 13 65 Marks

11. (a) A solid steel bar, 40 mm diameter, 2m long passes centrally througha copper tube of internal diameter 40 mm, thickness of metal 5 mmand length 2m. The ends of the bar and tube are brazed together anda tensile load of 150 kN is applied axially to the compound bar.

Assume Ec 100 GN/m2 and Es 200 GN/m2. Find the stresses and

load sheared by the steel and copper section.Refer Page No. 1.61, Problem No. 1.28

Given:

Diameter of steel rod, ds 40 mm

Length of both rod and tube 2 m 2000 mm

Inner diameter of Copper tube, dc 40 mm

Outer diameter of Copper tube, Dc 40 5 2 50 mm

Compressive load, P 150 kN 150 103N

Es 200 GPa 2 105 N/mm2 ; Ec 100 GPa 1 105 N/mm2

To find: Stresses in steel rod s and copper tube c

Solution:

We know that

Change in length of steel Change in length of copper

Strain in steel Strain in copper

s

Es

c

Ec [. . . Original length is same]


s c Es

Ec c

2 105

1 105

s 2 c

We know that, Total load P Ps Pc

150 103 s As c Ac

150 103 2 c 4

402 c 4

502 402

150 103 2513.3 c 706.9 c

c 150 103

3220.2 46.6 N/mm2

s 2 c 2 46.6 93.2 N/mm2

Total elongation, l lc ls

l c l

Ec s l

Es

46.6 2000

1 105

0.932 mm

Load shared by steel Ps

Ps s As

93.2 4

402

117.1 kN

Load shared by copper PL

Pc 32.9 kN

(Or)


(b) At a point within a body subjected to two mutually perpendicular

directions, the tensile stresses are 80 N/mm2 and 40 N/mm2

respectively. Each stress is accompanied by a shear stress of

60 N/mm2. Determine the normal stress, shear, stress and resultant

stress on an oblique plane inclined at an angle of 45 with the axisof minor tensile stress.Refer Page No. 1.128, Problem No. 1.66

Given: x 80 MPa 80 N/mm2, y 40 MPa 40 N/mm2, 45

Solution:

For Biaxial stress system

(i) Normal Stress n

n x y

2

x y

2 cos 2

80 40

2

80 402

cos 2 45 60 0 60 N/mm2

n 60 N/mm2 (Ans.)

(ii) Tangential Stress (Shear Stress) t

t x y

2 sin 2

80 402

sin 2 45 20 N/mm2

t 20 N/mm2

(iii) Resultant Stress

n2 t

2 602 202 63.25 N/mm2

Resultant Stress 63.25 N/mm2(Ans.)

(iv) Obliquity

tan 1 t

n tan 1

2060

18.43

[. . . cos 90 0]


12. (a) A simply supported beam of an effective span of 16 m carries pointloads of 4 kN, 5 kN and 3 kN at 3 m, 7 m and 11 m respectivelyfrom left hand support. Determine the reactions at the support anddraw the S.F.D and B.M.D.Refer Page No. 2.46, Problem No. 2.8

Solution:

Step 1. To find reactions at supports

Taking moment about A

RE 16 WB 3 WC 7 WD 11 12 35 33 80

RE 5 kN

Also RA RE WB WC WD

RA 5 4 5 3 12

RA 7 kN

Considering individual section at various points A to E. Move thesection from left to right.

S.FA 0 without considering RA

S.FA RA 7 kN (considering RA)

S.FB 7 kN without considering point load 4 kN

S.FB RA WB 7 4

3 kN considering 4 kN

S.FC 3 kN without considering 5 kN

S.FC 3 5 2 kN (considering 5 kN)

S.FD 2 kN without considering 3 kN

S.FD 2 3 5 kN (considering 3 kN)

S.FE 5 kN without considering RE

S.FE 5 5 0 (considering RE)

SFD is shown in Fig (b)


(a)

(b)

(c)

R =7A R =5E

(- e)vx

x

(+ e )vx

x S . F. D

B y S ign C onven tion

Fig.


Step 2. Bending moment diagram (B.M.D)By sign convention left upward moment is positiveTaking moment about the respective cross section from left sideB.MA RA 0 0

B.MB RA 3 7 3 21 kN m

B.MC RA 7 WB 7 3 49 16 33 kN m

B.MD RE 5 5 5 25 kN m

B.ME 0

The bending moment diagram is shown in Fig.(c).

(Or)

(b) A timber beam of rectangular section is support a load of 50 kNuniformly distributed over a span of 4.8 m when beam is simplysupported. If the depth of section is to be twice the breadth, and the

stress in the timber is not to exceed 7 N/mm2, find the dimensions ofthe cross section.Refer Page No. 2.133

13. (a) A solid cylindrical shaft is to transmit 300 kW power at 100 r.p.m.

(a) if the shear stress is not exceed 80 N/mm2, find its diameter. (b)What percentage saving in weight would be obtained if this shaft isreplaced by a hollow one whose internal diameter equals to 0.6 timesthe external diameter, the length, material and maximum shear stressbeing the same.Refer Page No. 3.31, Problem No. 3.23

(Or)

(b) A closed coil helical spring is to have a stiffness of 1.5 N/mm ofcompression under a maximum load of 60 N. The maximum shearing

stress produced in the wire of the spring is 125 N/mm2. The solidlength of the spring is 50 mm. Find the diameter of coil, diameter of

wire and number of coils C 4.5 104 N/mm2



14. (a) A cantilever 2m long is of rectangular section 120 mm wide and240 mm deep. It carries a uniformly distributed load of 2.5 kN permetre length for a length of 1.25 metres from the fixed end and apoint load of 1 kN at free end. Find the deflection at the free end.

Take E 10 GN/m2 (AU. Apr/May 2018)Refer Page No. 4.72, Problem No. 4.21

(Or)

(b) A beam AB of 8 m span is simply supported at the ends. It carries apoint load of 10 kN at a distance of 1 m from the end A and a unifomlydistributed load of 5 kN/m for a length of 2 m from the end B. If

I 10 10 6 m4, determine: (i) Deflection at the mid-span; (ii)Maximum deflection; (iii) Slope at the end A.Refer Page No. 4.47, Problem 4.15

15. (a) A cylinder shell 3m long which is closed at the ends has a internaldiameter of 1.5m and a wall thickness of 20 mm. Calculate thecircumferential and longitudinal stressess induced and also change inthe dimensions of the steel. If it is subjected to an internal pressure

of 1.5 N/mm2. Take Young’s modulus 200 kN/mm2 and Poisson’s

ratio 0.3.Refer Page No. 5.19, Problem No. 5.15

(Or)

(b) A compound cylinder, formed by shrinking one tube to another is

subjected to an internal pressure of 90 kN/m2. Before the fluid isadmitted, the internal and external diameters of the compound cylinderare 180 mm and 300 mm respectively and the diameter at the junctionis 240 mm. If after shrinking on the radial pressure at the common

surface is 12 MN/m2, determine the final stresses developed in thecompound cylinder.Refer Page No. 5.81, Problem No. 5.43


PART - C 1 15 15 Marks

16. (a) A beam ABCD, 10 m long, is simply supported at B and C whichare 4 m apart, and overhangs the support B by 3m. The overhangingpart AB carries UDL of 1 kN/m and the part CD carries UDL of 0.5kN/m. Calculate the position and magnitude of the least value of thebending moment between the supports. Draw the S.F. and B.M.diagrams.Refer Page No. 2.106, Problem No. 2.36

(Or)

(b) A bar 250 mm long, cross-sectional area 100 mm 50 mm, carries a

tensile load of 500 kN along lengthwise, a compressive load of 5000

kN on its 100 mm 250 mm faces and a tensile load of 2500 kN on

its 50 mm 250 mm faces. Calculate (i) the change in volume, (ii)

what change must be made in the 5000 kN load so that no change in

the volume of bar occurs. Take E 1.8 105 N/mm2; Poission’s ratio= 0.25.Refer Page No. 1.100, Problem No. 1.51


IndexA

Angle of Twist 3.50

Axial Deflection 3.79

B

Bar of Uniform Strength 1.74

Bar of Solid Section 3.4

Bearing Stress (Crushing Stress) in

Connections 1.10

Bending Moment in Beams 2.14

Bulk Modulus 1.93, 1.106

C

Cantilever Beam: 2.2

Castigliano’s Theorem 4.166

Closely coiled helical springs 3.69

Composite Section Beams (or) Flitched

Beams 2.137

Compound Circular Shafts 3.45

Compound Cylinders 5.76

Compressive Stress and Compressive

Strain 1.7

Conjugate Beam Method 4.140

Continuous Beam: 2.3

Couple 2.14

Cylindrical Shell with Hemispherical

Ends 5.23

D

Deflection of Cantilevers 4.58

Deflection of Fixed Beam 4.91

Deflection of Helical Springs 3.72

Deformation in Thin Cylinder 5.6

Design of Helical Coil Springs 3.84

Design of Thick Cylinders 5.71

Double Integration Method 4.5

E

Elastic Constants 1.93

Energy Stored U 3.72

F

Factor of Safety 1.15

Fixed or Built - in support 2.7

Fixed Beam: 2.3

Flexural rigidity 4.4

Flexural Strength of a Section 2.116

Free length of the spring 3.85

H

Helical Spring 3.69

Hinge or Pin-Jointed support 2.6

Hooke’s Law 1.15

Hoop Stress 5.3

L

Lame’s Theorem 5.42

Leaf Springs (or) Carriage Springs (or)

Laminated springs 3.117

Linear Strain and Lateral Strain 1.93

Load Carrying Capacity - Problems 2.126

Longitudinal and Shear Stresses 5.48

Longitudinal Stress 5.4

Index I.1

M

Macaulay’s Method 4.18

Maximum Bending Stress, b 3.74

Maxwell’s Reciprocal Theorem 4.185

Mean Diameter of coil D 3.84

Modulus of Elasticity 1.93

Mohr’s Circle for Biaxial Stresses 1.154

Moment Area Method 4.103

MEI

diagram by parts 4.107

N

Normal Stress: Axially Loaded Bar 1.5

Number of turns of coils n 3.85

O

Open - Coiled Helical Spring 3.69, 3.76

Overhanging Beam: 2.2

P

Pitch of coil 3.86

Point or Concentrated load 2.9

Poisson’s ratio 1.94

Polar Modulus Zp 3.6

Principal Stresses and Principal

Planes 1.118

Principle of Superposition 1.43

Proportioning of Sections - Problems 2.131

Propped Cantilever Beam: 2.4

Pure Torsion 3.1

R

Relationship Between Elastic

Constants 1.106

Rigid and Deformable Bodies 1.1

Rigidity Modulus (or) Shear Modulus 1.93

Roller Support 2.5

S

Section Modulus or Modulus of

Section 2.114

Shear Stress 3.71

Shear Stress and Shear Strain 1.8

Shear Stress 1.120

Shear Stresses in Beams 2.141

Shear Force in Beams (S.F) 2.11

Shear Stress and Strain 1.104

Shear Force (S.F) and Bending Moment

(B.M) Diagrams 2.15

Shear Modulus or Modulus of

Rigidity 1.105

Shear Stress - (Resistance Concept) 3.2

Shear Flow 2.165

Shear Strain - (Deformation Concept) 3.2

Sign Convention for Shear Force in

Beam 2.13

Simple support or Knife Edge support 2.5

Simply supported Beam: 2.2

Smooth surface support or Frictionless

support 2.7

Spherical Pressure Vessels 5.32

Spring Index C 3.78

I.2 Strength of Materials for Mechanical Eng., - www.airwalkbooks.com

Springs 3.68

Springs in Series 3.75

Springs in Parallel 3.75

Stability 1.18

Static Equilibrium Equations 2.8

Stiffness of The Spring: k 3.72

Stiffness of beam 4.4

Stiffness 1.17

Strain Energy Density 4.160

Strain Energy 4.158

Strain 1.5

Strain Energy Stored U 3.72

Strength 1.2

Stress - Strain Curve for Ductile

Materials 1.11

Stress Strain Curves (Compression) 1.15

Stress-strain Behaviour of Materials 1.10

Stress Strain Curves for Brittle

Materials 1.14

Stress Strain Curves (Tension) 1.11

Stresses in Beams - Theory of Simple

Bending 2.111

T

Tensile Stress and Tensile Strain 1.6

Theory of Torsion 3.2

Theory of Simple Bending or Pure

Bending 2.111

Thermal Stresses 1.76

Thick Cylinders 5.41

Thin Pipe Bounded by Wire 5.26

Thin Cylinders 5.1

Types of Beams 2.1

Types of Springs 3.68

U

Uniformly Varying Load (UVL) 2.10

Uniformly Distributed Load (UDL) 2.10

Unit of Stress 1.4

V

Volumetric Strain 1.96

Index I.3

Documents

airwalkbooks.comairwalkbooks.com/images/pdf/pdf_105_1.pdf · 2019-01-31 · Dr. S. Ramachandran, M.E., Ph.D., Professor - Mech Sathyabama Institute of Science and Technology Chennai