MECHANICS OF SOLIDS (For B.E. Mechanical Engineering Students) As per New Revised Syllabus of APJ Abdul Kalam Technological University Dr. S.Ramachandran, M.E., Ph.D., Mr. V.J. George, M.E., Mr. S. Kumaran, M.E., Mr. Ramkumar Raja, M.E., AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street Mylapore, Chennai - 600 004. Ph.: 2466 1909, 94440 81904 Email: aishram2006@gmail.com, airwalk800@gmail.com www.airwalkpublications.com
th First Edition: 9 August 2016 Price : Rs. 350/- ISBN:978-93-84893-51-4 ISBN : 978-93-84893-51-4 and
Syllabus S.1 ME201 MECHANICS OF SOLIDS Chapter 1: Stress, Strain and Deformation of Solids Introduction to analysis of deformable bodies internal forces method of sections assumptions and limitations. Stress stresses due to normal, shear and bearing loads strength design of simple members. Definition of linear and shear strains. Material behaviour uniaxial tension test stress-strain diagrams concepts of orthotropy, anisotropy and inelastic behaviour Hooke s law for linearly elastic isotropic material under axial and shear deformation Deformation in axially loaded bars thermal effects statically indeterminate problems principle of superposition - elastic strain energy for uniaxial stress. Chapter 2: Elastic Constants and Torsion Definition of stress and strain at a point (introduction to stress and strain tensors and its components only) Poisson s ratio biaxial and triaxial deformations Bulk modulus - Relations between elastic constants. Torsion: Shafts - torsion theory of elastic circular bars assumptions and limitations polar modulus - torsional rigidity economic cross-sections statically indeterminate problems shaft design for torsional load. FIRST INTERNAL EXAM Chapter 3: Beams Beams- classification - diagrammatic conventions for supports and loading - axial force, shear force and bending moment in a beam Shear force and bending moment diagrams by direct approach Differential equations between load, shear force and bending moment. Shear force and bending moment diagrams by summation approach elastic curve point of inflection.
S.2 Mechanics of Solids - www.airwalkpublications.com Chapter 4: Stresses in Beams Stresses in beams: Pure bending flexure formula for beams assumptions and limitations section modulus - flexural rigidity - economic sections beam of uniform strength. Shearing stress formula for beams assumptions and limitations design for flexure and shear. SECOND INTERNAL EXAM Chapter 5: Deflection of Beams Deflection of beams: Moment-curvature relation assumptions and limitations - double integration method Macaulay s method - superposition techniques moment area method and conjugate beam ideas for simple cases. Chapter 6: Transformation of Stress and Strains Transformation of stress and strains: Plane state of stress - equations of transformation - principal planes and stresses. Mohr s circles of stress plane state of strain analogy between stress and strain transformation strain rosettes Compound stresses: Combined axial, flexural and shear loads eccentric loading under tension/compression - combined bending and twisting loads. Chapter 7: Columns Theory of columns: Buckling theory Euler s formula for long columns assumptions and limitations effect of end conditions slenderness ratio Rankine s formula for intermediate columns. END SEMESTER EXAM
Contents C.1 Contents 1. Stress, Strain and Deformation of Solids 1.1 Introduction to Strength of Materials... 1.1 1.2 Introduction to Analysis of Deformable Bodies... 1.1 1.3 Internal Forces... 1.2 1.4 Method of Sections... 1.3 1.4.1 Limitations of the method of sections... 1.4 1.5 Stresses and Strains... 1.4 1.5.1 Stress... 1.4 1.5.2 Unit of Stress... 1.5 1.5.3 Strain... 1.5 1.6 Stress due to Normal, Shear and Bearing Loads... 1.6 1.6.1 Normal Stress: Axially Loaded Bar... 1.6 1.6.2 Tensile Stress and Tensile Strain... 1.7 1.6.3 Compressive Stress and Compressive Strain.. 1.8 1.6.4 Shear Stress and Shear Strain... 1.8 1.6.5 Bearing (Crushing) Stress in Connections... 1.10 1.7 Material Behaviour... 1.11 1.8 Stress-strain Diagram... 1.12 1.8.1 Stress-Strain Curve for Ductile Materials (Unaxial Tension Test)... 1.12 1.8.2 Stress-Strain Curves for Brittle Materials... 1.14 1.8.3 Stress Strain Curves (Compression)... 1.15 1.9 Concepts of Orthotropy, Anisotropy and Inelastic Behaviour... 1.15 1.10 Hooke s Law - Axial & Shear Deformation... 1.16 1.10.1 Factor of Safety... 1.16 1.10.2 Deformation of a body due to force... 1.17 1.11 Stiffness... 1.18 1.11.1 Stability... 1.19
C.2 Mechanics of Solids - www.airwalkpublications.com 1.12 Deformation in Axially Loaded Bars... 1.30 1.12.1 Deformation in Simple Bar Subjected to Axial Load... 1.31 1.13 Deformation for a Bar of Varying Section... 1.32 1.14 Deformation of a Body due to Self Weight... 1.45 1.15 Principle of Superposition... 1.47 1.16 Stress in Bars of Uniformly Tapering Cross Section 1.53 1.17 Deformation of Uniformly Tapering Rectangular Bar1.56 1.18 Deformation in Compound or Composite Bars... 1.57 1.19 Bar of Uniform Strength... 1.73 1.20 Thermal Stress and Thermal Effects... 1.74 1.20.1 Thermal Stresses in Simple Bars... 1.74 1.20.2 Thermal Stresses in Composite Bars... 1.78 1.20.3 Thermal Stress in Taper Bar... 1.83 1.20.4 Thermal Stress in Varying Section Bar... 1.84 1.21 Statically Indeterminate Problems in Tension and Compression... 1.90 1.22 Strain Energy... 1.95 1.22.1 Strain Energy Density... 1.97 1.22.2 Unit Strain Energy... 1.99 1.22.3 Strain Energy in Uniaxial Loading for uniaxial stress... 1.99 1.22.4 Expression for strain energy stored in a body when load is applied gradually... 1.99 1.22.5 Expression for strain energy stored in a body when the load is applied suddenly... 1.101 1.22.6 Expression for strain energy stored in body for impact loading... 1.102 1.22.7 Strain energy stored in varying section bar 1.103 1.22.8 Impact by shock... 1.103 1.23 Strain Energy in Pure Shearing... 1.124
Contents C.3 1.23.1 Expression for strain energy stored in a body due to shear stress... 1.124 2. Elastic Constants and Torsion 2.1 Tensor... 2.1 2.1.1 Introduction... 2.1 2.1.2 Definition... 2.1 2.1.3 Rank of a tensor... 2.1 2.2 Stress at a Point... 2.2 2.2.1 Stress vector... 2.2 2.2.2 Stress Tensor... 2.3 2.2.3. Components of a stress tensor... 2.3 2.2.4 Stress tensor conventions... 2.4 2.2.5 Symmetry in a stress matrix... 2.4 2.3 Strain at a point... 2.4 2.3.1 Strain tensor... 2.5 2.3.2 Components of strain tensor... 2.6 2.4 Elastic Constants... 2.7 2.4.1 Modulus of Elasticity... 2.7 2.4.2 Rigidity Modulus (or) Shear Modulus... 2.7 2.4.3 Bulk Modulus... 2.7 2.4.4 Linear Strain and Lateral Strain... 2.7 2.5 Poisson s Ratio... 2.8 2.6 Biaxial and Triaxial Deformations... 2.10 2.7 Volumetric Strain... 2.12 2.7.1 Rectangular Body Subjected to Axial Loading 2.12 2.8 Rectangular Bar Subjected to 3 Mutually Perpendicular Forces... 2.15 2.9 Cylindrical Rod Subjected to Axial Load... 2.16 2.10 Bulk Modulus... 2.17 2.11 Relationship Between Elastic Constants... 2.18
C.4 Mechanics of Solids - www.airwalkpublications.com 2.11.1 Relation between Bulk Modulus and Young s Modulus... 2.18 2.11.2 Shear Stress and Strain... 2.19 2.11.3 Shear Modulus or Modulus of Rigidity... 2.19 2.11.4 Relation between Modulus of Elasticity and Modulus of Rigidity... 2.20 2.12 Torsion... 2.26 2.12.1 Introduction... 2.26 2.13 Pure Torsion... 2.26 2.13.1 Assumptions made in theory of Pure Torsion 2.26 2.14 Shear Stress - (Resistance Concept)... 2.26 2.15 Shear Strain - (Deformation Concept)... 2.27 2.16 Analysis of Torsion of Circular Bars-derivation of Torsional Equations... 2.27 2.16.1 Theory of Torsion... 2.27 2.16.2 Limitations... 2.28 2.17 Economic Cross Sections... 2.29 2.17.1 Bar of Solid Circular section... 2.29 2.17.2 Bars of Hollow Circular Section... 2.30 2.18 Polar Modulus Z p... 2.32 2.19 Torsional Rigidity and Stiffness of the Shaft... 2.32 2.20 Power Transmitted by the Shaft... 2.33 2.21 Shaft Design for Torsional Load... 2.34 2.22 Problems on Replacing a Solid Shaft by a Hollow Shaft... 2.60 2.23 Stepped Shafts (Or) Shafts in Series... 2.75 2.23.1 Shafts fixed at one end... 2.75 2.23.2 Shafts fixed at both the Ends... 2.77 2.24 Compound Circular Shafts... 2.78 2.24.1 Shaft in Series... 2.78 2.24.2 Shafts in Parallel... 2.79
Contents C.5 2.25 Shaft Subjected to Number of Torques... 2.80 2.26 Statically Indeterminate Problems in Bars Subjected to Torsion... 2.99 3. Beams 3.1 Introduction... 3.1 3.2 Classification of Beams... 3.1 3.3 Diagrammatic Conventions For Supports and Loading 3.4 3.3.1 Supports and Support Reactions... 3.4 3.3.2 Types of Supports and their Reactions... 3.4 3.3.3 Static Equilibrium Equations... 3.8 3.3.4 Determinate and Indeterminate Beams... 3.8 3.3.5 Types of Loading in Beams... 3.8 3.4 Axial Force in Beams... 3.11 3.5 Shear Force in Beams (S.F)... 3.11 3.6 Sign Convention for Shear Force in Beam... 3.12 3.7 Couple Or Moment... 3.13 3.8 Bending Moment in Beams... 3.13 3.9 Sign Convention for Bending Moment in Beams... 3.14 3.10 Shear Force (S.F) and Bending Moment (B.M) Diagrams... 3.15 3.11. Differential Equation Between Load, Shear Force and Bending Moment... 3.16 3.12 Method of Drawing Shear Force and Bending Moment Diagrams by Summation Approach... 3.17 3.13 Points to be Remembered for Drawing S.F.D and B.M.D... 3.18 3.14 SFD and BMD for Cantilever Beam... 3.19 3.15 Shear Force and Bending Moment Diagram for Simply Supported Beams... 3.23 3.16 Shear Force and Bending Moment for Overhanging Beam... 3.29
C.6 Mechanics of Solids - www.airwalkpublications.com 3.16.1 Point of contraflexure or inflexion... 3.32 3.17 Elastic Curve... 3.33 3.18 Shear Force and Bending Moment Diagrams by Direct Approach.... 3.137 4. Stresses in Beams 4.1 Stresses in Beams - Theory of Simple Bending... 4.1 4.2 Simple Bending or Pure Bending... 4.1 4.2.1 Assumption in Theory of Simple Bending... 4.1 4.2.2 Theory of simple bending - Bending stress equation (or) Flexural Formula Derivation... 4.2 4.3 Section Modulus or Modulus of Section... 4.5 4.4 Flexural Strength of a Section... 4.6 4.5 Flexural Rigidity... 4.6 4.6 Beam of Uniform Strength... 4.23 4.7 Composite Section Beams (or) Flitched Beams... 4.24 4.8 Shear Stresses in Beams... 4.26 4.9 Economic Sections... 4.27 4.9.1 Shear Stress Distribution for a Rectangular Section... 4.27 4.9.2 Shear Stress Distribution over I - Section... 4.29 4.9.3 Shear Stress Distribution over T Section... 4.34 4.9.4 Shear Stress Distribution over a Circular Section... 4.36 4.9.5 Shear Stress Distribution over a Triangular Section... 4.39 4.10 Shear Flow... 4.46 4.10.1 Shear on a Horizontal plane... 4.46 4.11 Design for Flexure and Shear... 4.63 5. Deflection of Beams 5.1 Introduction... 5.1 5.1.1 Definition of Deflection... 5.1
Contents C.7 5.1.2 Importance of deflection... 5.1 5.2 Moment-curvature Relations... 5.2 5.3 Evaluation of Beam Deflection and Slope... 5.3 5.3.1 Flexural rigidity... 5.4 5.3.2 Stiffness of beam... 5.4 5.4 Double Integration Method... 5.5 5.4.1 Simply supported beam with a concentrated load at the mid span.... 5.6 5.4.2 Simply supported beam carrying a UDL over a whole span... 5.12 5.5 Superposition Methods for Beam Deflection... 5.16 5.6 Macaulay s Method... 5.23 5.6.1 Problems on SSB - Point load... 5.24 5.6.2 Problems on SSB - Uniformly Distributed Load (UDL)... 5.44 5.6.3 Problems on SSB - Uniformly Varying Load (UVL)... 5.56 5.6.4 Problems on Cantilever... 5.58 5.6.5 Problems on overhanging beam... 5.68 5.7 Moment Area Method... 5.77 5.7.1 First moment - area theorem (or) Mohr s I theorem... 5.78 5.7.2 Use of cantilever moment diagrams in moment area method... 5.78 5.7.3 Second moment - area theorem (or) Mohr s II theorem... 5.80 5.7.4 M EI diagram by parts... 5.81 5.7.5 Problems on Moment area method... 5.83 5.8 Conjugate Beam Method... 5.107 5.8.1 Simply supported beam with point load W acting at Centre... 5.115
C.8 Mechanics of Solids - www.airwalkpublications.com 5.8.2. Simply supported Beam with UDL... 5.117 6. Transformation of Stress and Strains 6.1 Introduction... 6.1 6.2 Analysis of plane state of stress... 6.1 6.2.1 Member subjected to a axial load (or) Stress on a inclined plane... 6.1 6.2.2 Biaxial state of stress - Member subjected to biaxial stress... 6.5 6.2.3 Member subjected to a simple shear stress... 6.13 6.2.4 Member subjected to a simple shear and a biaxial stress... 6.14 6.2.5 Member subjected direct stress in one plane and a simple shear stress (Equations of transformation)... 6.31 6.3 Mohr s Circle for Biaxial Stresses... 6.36 6.3.1 Body subjected to a biaxial perpendicular unequal and like principal stresses... 6.36 6.3.2 Body subjected to a biaxial perpendicular unequal and unlike principal stress... 6.41 6.3.3 Body subjected to a Biaxial perpendicular unequal, like stresses with an simple shear stress... 6.44 6.4 Plane State of a Strain... 6.48 6.4.1 Principal strains... 6.48 6.4.2 Maximum shear strain... 6.48 6.5 Strain Energy in Bending and Torsion... 6.49 6.6 Analogy Between Stress and Strain Transformation 6.54 6.7 Strain Gauge... 6.55 6.8 Strain Rosette... 6.56 6.8.1 Special cases of strain rosette layouts... 6.57 6.9 Combined Axial, Flexural and Shear Loads... 6.60
Contents C.9 6.10 Eccentric Load: (Tension and Compression)... 6.63 6.11 Combined Bending and Torsion... 6.66 6.11.1 Solved Problems on Combined Bending and Torsion... 6.67 6.11.2 Equivalent of bending moment M e and Equivalent torque T e... 6.70 7. Columns 7.1 Introduction... 7.1 7.1.1 Slenderness ratio of a column... 7.2 7.1.2. Short columns... 7.2 7.1.3 Long columns... 7.2 7.1.4 Buckling load, Crippling or Critical load... 7.2 7.1.5 Equivalent length... 7.3 7.1.6 Buckling factor... 7.3 7.1.7 Safe load... 7.3 7.2 Euler Equation... 7.3 7.2.1 Assumptions made in Euler s equation... 7.3 7.2.2 Sign Conventions... 7.4 7.3 End Conditions... 7.4 7.3.1 Derivation of Euler s equation... 7.5 7.4 Slenderness Ratio... 7.13 7.5 Equivalent Length (or) Effective Length L e of a Column... 7.14 7.5.1 One end fixed and one end free column... 7.15 7.5.2 One end fixed and one end hinged column... 7.15 7.5.3 Two fixed ends column... 7.15 7.6 Limitations of Euler s Formula... 7.17 7.7 Rankine s Formula... 7.33 7.8 Eccentrically Loaded Columns... 7.44 7.8.1 Rankine-Gordon formula... 7.44 7.9 Middle Third Rule... 7.55
Chapter - 1 STRESS, STRAIN AND DEFORMATION OF SOLIDS Introduction to analysis of deformable bodies - internal forces - method of sections - assumptions and limitations. Stress - stresses due to normal, shear and bearing loads - strength design of simple members. Definition of linear and shear strains. Material behavior - uniaxial tension test - stress-strain diagrams concepts of orthotropy, anisotropy and inelastic behavior - Hooke s law for linearly elastic isotropic material under axial and shear deformation Deformation in axially loaded bars - thermal effects - statically indeterminate problems - principle of superposition - elastic strain energy for uniaxial stress. 1.1 INTRODUCTION TO STRENGTH OF MATERIALS Materials are very important for every application in all engineering disciplines and before they can be used for any application, their behaviour (under the loads or forces) under which the materials are to work must be known. Strength of materials (Mechanics of solids) deals with this behaviour 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, stiffness and stability of materials, one has to consider factors like manufacturing, cost, life, utility, market demands etc., but most important role is played by the factors like strength, stiffness and stability which are covered by the subject of strength of materials. Materials which we come across are generally classified as: (A) Rigid bodies, (B) Deformable bodies 1.2 INTRODUCTION TO ANALYSIS OF DEFORMABLE BODIES Deformation is the change in the shape and, or size of the body under application of a force or a load. Deformable bodies are those which undergo deformation when subjected to external loading. Deformable bodies are further classified into Plastic and Elastic bodies.
1.2 Mechanics of Solids - www.airwalkpublications.com An Elastic Material or body is one which undergoes deformation when subjected to an external loading such that the deformation disappears on the removal of the load. A Plastic material is one which undergoes a continuous deformation during the period of loading and the deformation is permanent and the material does not regain its original dimensions on the removal of the loading. Rigid body or material is one which does not undergo any deformation A B when subjected to an external loading. In practice, no material is BAR absolutely elastic nor plastic nor rigid. A B These properties are attributed when C the deformations are within certain limits. C F Deformation can be understood by a simple example, consider a bar Fig 1.1 F which is fixed at one end and is loaded by a force (F) as shown in Fig.1.1 After the load is applied on the bar, there is change in the length of the bar as shown. The difference in the original and final length is CC which is equal to. is called deflection. This change in length of bar is one form of deformation. 1.3 INTERNAL FORCES (i) A X B F C D F (ii) A X F R C R(Resistance) D B F (iii) A F C R R D B F Fig 1.2
Stress, Strain and Deformation of Solids 1.3 Strength: Strength is the internal resistance offered by the body against the deformation caused due to the application of an external load system. A material when subjected to an external load system undergoes a deformation. Against this deformation, the material will offer a resistance which tends to prevent the deformation. This resistance is offered by the material 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. 1.4 METHOD OF SECTIONS Method of sections is used to determine the forces acting on a static body. If a body is in equilibrium, then any part of the body must also be in equilibrium. This method means that we can cut a body in two, in order to expose a cross section, on which the internal forces can be determined. Consider a homogeneous bar AB of a certain cross-sectional area (say A units) as shown in Fig.1.2. An external load F acts along the ends of the bar at A and B, intending to stretch the bar. To analyse the given region, consider a plane section X-X splitting the bar into two parts namely C and D. Under equilibrium conditions, pull at A is equal to pull at B. When the section is split, into parts C and D, to maintain the equilibrium, the cross section C exhibits internal resistances equal in magnitude and opposite in direction of the applied load F. The same phenomenon is observed with B and D. Thus, the process of segmenting a homogeneous body such as the rod AB, using an arbitary plane such as X-X, for the purpose of analysis is known as method of sections. A few assumptions were made in order to satisfy the method of section analysis. The body is homogeneous in nature The body is in equilibrium even after the application of the loads. Every infinitesimal part of the body is in equilibrium with itself. Sum of forces acting on the body is equal to the internal forces resisting them. The body exhibits the same physical, chemical and mechanical properties irrespective of the size and orientations of the cut sections. Frictional resistances between various layers of the material are neglected. Miscellaneous loads like external vibrations, magnetic and electric fields, and other forces are neglected.
1.4 Mechanics of Solids - www.airwalkpublications.com 1.4.1 Limitations of the method of sections Non-homogenous elements, when computed using same method, may result in wrong values. Complicated calculations are needed for irregular objects. In case of irregular objects, only an approximate value of the stress can be obtained. Does not account for miscellaneous external and internal forces. Can only be used when the body is in static equilibrium. 1.5 STRESSES AND STRAINS 1.5.1 Stress The intensity of resistance offered is due to the strength of the body or material. The force of resistance per unit area offered by a body against the deformation is called stress. It is denoted by symbol R A or P Load ; Stress A Area N mm 2 Load in N, area in mm 2, unit of stress N/mm 2 The external force acting on the body is called load. The load is applied 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 the member against the deformation and the applied load are in equilibrium. When the member is incapable of offering the necessary resistance against the external forces, the deformation will continue leading to the failure of the member. If the resistance offered by the section against the deformation is assumed to be uniform across the section, then the intensity of resistance per unit area of the section is called the intensity of stress or Unit stress.
Stress, Strain and Deformation of Solids 1.5 1.5.2 Unit of Stress The unit of stress is N/m 2, which is known as pascal. 1 N/m 2 1 Pa. For engineering materials, this is a small value. For larger values, we use kpa. Kilo pascal (kpa) 10 3 Pa Mega pascal (MPa) 10 6 Pa Giga pascal (GPa) 10 9 Pa For Engineering materials, the cross section, we use is in only millimeters ie N/mm 2 1 MPa 1 10 6 Pa 10 6 N/m 2 or 1 MN/m 2 10 6 N N 10 3 mm 10 3 mm mm 2 1 MPa 1 N/mm 2 1 GPa 10 9 Pa 10 9 N/m 2 or 1 GN/m 2 1000 10 6 N/m 2 1000 MPa 1000 N/mm 2 1 GPa 1000 N/mm 2 1 kn/mm 2 Intensity of stress R A P A in N/m2 where R: Reaction in Newton P: Force in Newton A: Area of cross section in m 2 1.5.3 Strain Due to the application of load, the length of the member changes from l to l dl. The ratio of change in the length to the original length of the member is called strain. change in length Strain e original length dl l d" " Fig 1.3
1.6 Mechanics of Solids - www.airwalkpublications.com 1.6 STRESS DUE TO NORMAL, SHEAR AND BEARING LOADS Basically stresses are classified into 2 types STRESS Normal Stress Tangential Stress Tensile Stress Compressive Stress Shear Stress Punching shear Stress Fig. 1.3 1.6.1 Normal Stress: Axially Loaded Bar Stress which is normal to the cross section of the member (e.g stress due to elongation of a bar) is called Normal stress. P a Cutting plane P P a P P A da = P a a (a) Bar axis a Centroid dy dz a dx dx P (c) P A (d) A (e) Fig. 1.4 Successive steps in determining the largest normal stress in an axially loaded bar. a (b) (f) Consider a bar of cross sectional area A, subjected to an axial load P. To determine stress, a free body diagram is prepared either for left or right 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. The reaction on the left end is equilibrated at section a a by a uniformly distributed normal stress. The sum of these stresses multiplied by their respective areas generate a stress resultant that is statically equivalent to the force P Fig. 1.4(c).
Stress, Strain and Deformation of Solids 1.7 A thin slice of the bar with equal uniformly distributed normal stresses of opposite sense on the two parallel sections is shown in Fig. 1.4(d). The uniaxial state of stress may be represented on an infinitesimal cube as Fig. 1.4(e). However the simplified diagram shown in Fig. 1.4(f), is commonly used. Normal stress is defined as the ratio of force applied to the cross sectional area of the bar. Normal stress Force Area P in N/m2 A In the integral form, the load applied is given by Load P da A A material is capable of offering the following types of stresses. 1. Tensile stress 2. Compressive stress 3. Shear stress. 1.6.2 Tensile Stress and Tensile Strain Tensile stress is defined as the resistance offered by the section of a member/body against an increase in length. For example consider the stress offered by the section XX of a rod as shown in Fig. 1.5. P P Fig. 1.5 R " X X X R X d" P P The intensity of tensile stress is given by Tensile stress Force Area P A R A STRAIN Tensile strain compressive strain Lateral strain Longitudinal strain Volumetric strain Shear strain Fig. 1.5 (a)
1.8 Mechanics of Solids - www.airwalkpublications.com When the rod is subjected to tensile load, there is an increase in the length of rod and the corresponding strain is called the tensile strain. The ratio of change of dimensions of the body to the original dimension is known as strain. It has no unit. Tensile strain: Ratio of change in length to original length is known as tensile strain Fig. 1.5 (b) Tensile strain e Increase in length Original length dl L P P L dl Fig 1.5 (b) 1.6.3 Compressive Stress and Compressive Strain Compressive stress is the resistance offered by the section of member or body against a decrease in length due to applied pushing load. For example consider a bar subjected to pushing axial load as shown in Fig. 1.6. Intensity of compressive stress is given by compressive stress P A R A Due to the external loading, Fig 1.6 the length of the member decreases by dl. The ratio of the decrease in length to the original length is called compressive strain Compressive strain e Decrease in length original length 1.6.4 Shear Stress and Shear Strain Consider a bar AB subjected to transverse forces as shown in Fig. 1.7(a). dl l
Stress, Strain and Deformation of Solids 1.9 Fig 1.7 Two forces are passing at section C as shown in Fig. 1.7(b). Internal force must exist in the plane of the section and their resultant is equal to P. 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 is subjected to two equal and opposite forces which are acting tangentially on any cross-sectional plane of a body, tending to slide one part of the body over the other part, then the body is said to be in a state of shear. It is denoted by. total tangential force crosssectional area of resisting section Shear stress P in N/m2 A Fig 1.8 Shearing stresses are commonly found in bolts, pins and rivets used to connect various structural members and machine components. Consider two plates A and B connected by rivets as shown in Fig. 1.8. shear force The shear stress shear area F A P A Consider a block of height l, length L and width unity as shown in Fig.1.9.
1.10 Mechanics of Solids - www.airwalkpublications.com A P B A dl A P d l B B x x d x dx Fig 1.9 Shear force or Resistance R Shear Stress Shear Area L l P L l Consider the block subjected to shear force P on its top and bottom faces. When the block does not fail in shear, the shear deformation is shown in Fig. 1.9. The block has deformed from the position ABCD to A B CD through an angle. BCB ADA 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 face is called shear strain. Transverse displacement Shear strain Distance from lower face dl l At the section XX, Shear strain dx x Since is very small, tan dl l dx shear strain x The Angular deformation in radians measures the shear strain. 1.6.5 Bearing Stress (Crushing Stress) in Connections Bolts, pins and rivets create stresses in the members, to which they connect. (i.e) along the bearing surface, or surface of contact. For example consider again two plates A and B connected by the rivet CD which we have discussed in previous section. The rivet exerts a force P
on plate which is equal and opposite to the force F exerted by the plate on the rivet. The force P represents the resultant of elementary forces distributed on the inside surface of the half cylinder of diameter d and of length t equal to the thickness of plate. The average nominal bearing stress is obtained by dividing load P by the area of rectangle representing projection. Bearing stress b P A P td 1.7 MATERIAL BEHAVIOUR Stress, Strain and Deformation of Solids 1.11 Young s modulus Stress Normal stress Strain Linear strain The value of Young s modulus is determined from stress-strain graph of material. Some materials are equally strong in compression and tension (metals and alloys). Such materials are usually tested in tension. The test results usually pertain to a circular bar of uniform cross-section. The load on the test specimen is increased gradually from zero, in suitable increments till the specimen fails (breaks). The elongation of the specimen is measured over a specific length known as gauge length (usually 50 to 200 mm) at each load step. The stresses and the corresponding strains are computed for the load and corresponding elongation readings. Materials such as concrete, stones and bricks that are stronger in compression than in tension are tested in compression. Stress-strain values are plotted in the form of a graph and the value of Young s modulus is determined from the slope of the curve for any stress value. In the case of materials with linear stress-strain behaviour, young s modulus is constant upto elastic limit. For materials with non-linear stress-strain relationship, the average value of slope is adopted for young s modulus or the value is defined at a specified stress or strain value. A Fig A A Fig 1.10 d Fig B P t d F t C D F
1.12 Mechanics of Solids - www.airwalkpublications.com Thus, the stress-strain diagram gives many important properties of material like Young s modulus. 1.8 STRESS-STRAIN DIAGRAM 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 loading conditions and a curve is plotted upto the point at which the specimen fails. Hence, this curve is known as stress-strain curve. 1.8.1 Stress-Strain Curve for Ductile Materials (Unaxial Tension Test) A 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, the stress is proportional to the strain in the initial stage and remains so upto a point, known as limit of proportionality as shown in Fig. 1.11. The stress strain diagram gives many important properties of materials. STRESS Y - Upper Yield point * * * * E - Elastic Limit P-Proportionality Limit * L - Lower Yield point U -Ultimate streng th A - actual Rupture streng th * * B- Breaking (Rupture strength) O STRAIN Fig 1.11 Stress - Strain
Stress, Strain and Deformation of Solids 1.13 Point P : Limit of proportionality E : Elastic limit Y : Upper yield point L : Lower yield point U : Ultimate point B : Breaking point Near the proportionality limit, we have a point called Elastic limit (E) at which if the load is removed, the specimen will return to its original dimensions. Beyond the elastic limit, the material enters into plastic range and removal of load does not return the specimen to its original dimensions, thus subjecting itself to a permanent deformation. On applying further load the specimen curve reaches upper yield point (Y) and corresponding stress is called upper yield stress. Beyond point Y, the load decreases with increase in strain upto point (L) called lower yield point and corresponding stress is called lower yield stress. After lower yield point (L), the stress starts increasing and reached maximum value at the point (U) called ultimate point and the corresponding stress is called ultimate tensile stress. After the ultimate point, the stress again starts decreasing, while the strain goes on increasing until the material fractures at point B called Breaking Point and the corresponding stress is called breaking stress. In the curve, all the stresses are calculated based on original cross section. 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 or Breaking strength. In a stress-strain diagram, note that the strength is lower than the ultimate strength. This is because we find the breaking strength by dividing the breaking load by original area of specimen. As noted earlier, while tensioning the specimen, its length increases but the diameter decreases. For the ductile materials, as they yield, (decrease in diameter of the specimen becomes more and more) the diameter of the specimen is considerably reduced and we find rupture strength with respect to the final (reduced) diameter, it will be much more than the ultimate strength. We may call this as Actual Rupture Strength as indicated as point A in Fig 1.11.
1.14 Mechanics of Solids - www.airwalkpublications.com Due to larger yielding in the material, a phenomenon called Necking occurs which is responsible for reducing the diameter of specimen. The formation of necking shown in Fig 1.11(a) is more predominant in ductile materials and at failure, a perfect cup and cone is formed. With lowering ductility and increasing brittleness, the cup and cone failure slowly disappear and brittle failure with rough texture takes places. Stresses and strains based on original dimensions are called as Engineering or Nominal or Conventional stresses or strains. Stresses and strains based on actual dimensions are called True or Natural Stress and Strains. Ductility of a material is measured by the percentage elongation of the specimen (or) percentage reduction in cross sectional area of the specimen when failure occurs. % Increase in length l l l 100 % Reduction in area A A 100 A 1.8.2 Stress-Strain Curves for Brittle Materials Brittleness is defined as the property of material that will fail suddenly without undergoing noticeable deformations. For brittle materials and for the materials with low ductility like higher grades of steel, no definite yield is observed. Materials which show very small elongation before they fracture are called brittle materials (Eg.) Cast Iron, concrete, high carbon steel etc.
Stress, Strain and Deformation of Solids 1.15 For Brittle materials, the stress strain curve is as shown in Fig 1.12. The ultimate tensile stress is defined as the ratio of ultimate load to the original area of cross section and is taken as basis for determining the design stress for Brittle materials because there is no definite yield point. Ultimate streng th Stress Fig 1.12 Breaking or ultimate Point Limit of Proportionality Strain 1.8.3 Stress Strain Curves (Compression) For ductile materials, stress strain curves in compression are identical to those in tension at least upto the yield point for all practical purposes. Brittle materials have compression stress strain curves of the same form as the tension test but the stresses at various points are generally considerably different. 1.9 CONCEPTS OF ORTHOTROPY, ANISOTROPY AND INELASTIC BEHAVIOUR Isotropic materials are the ones which exhibit same material properties in every single direction. They are homogeneous in nature and react the same way, irrespective of this size. Eg.Mild steel. These materials often have a linear stress-strain relationship upto a limit. Orthotropy is the phenomenon by which materials have different properties in different orthogonal axes. A classic example of an orthotropic material is wood. At any given point the properties differ in three mutually perpendicular directions. Anisotropic materials are directionally dependent. That is, their properties change along different directions. These properties include Young s modulus, tensile strength, absorptivity etc. For anistropic and orthotropic materials, the stress - strain curves as well the Young s modulus, change with respect to the direction of application and the nature of the applied load. Hence, for anisotropic materials, it is quite difficult to establish a definite stress-strain relationship. Hence, one cannot define the elastic limit of such materials. Almost every anisotropic material exhibits an inelastic behaviour and the physical properties also vary across the area of the material. As discussed before, inelasticity or plasticity is the phenomenon by which the material can not regain its original shape when the load is removed. The stress
1.16 Mechanics of Solids - www.airwalkpublications.com - strain curves for an inelastic material are often quite erratic and may not follow any definite rule (or) algorithms. They may fail sharply without any warning just like brittle materials, or they could exhibit enormous stretching capabilities in one direction and fail for the same load in another direction. 1.10 HOOKE S LAW - AXIAL & SHEAR DEFORMATION Hooke s Law states that within the Elastic limit the stress (compressive or tensile) is proportional to the strain Mathematically, Hooke s Law is, Stress Strain Stress Constant of proportionality. Strain Under normal (i.e., direct) stresses and strains, constant of proportionality is called Modulus of Elasticity or Young s Modulus E Normal stress Youngs Modulus E Linear nominal strain e Under the shearing stresses and strain, the constant of proportionality is called Modulus of rigidity and is denoted by G or C or N Rigidity Modulus G Shearing stress Shearing strain 1.10.1 Factor of Safety Factor of safety is defined the ratio of ultimate stress to the permissible stress (working stress) Ultimate stress Factor of safety Permissible stress Factor of safety depends on so many factors like the type of material, its degree of reliability, workmanship, manufacturing method, nature of loading, 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
1.10.2 Deformation of a body due to force acting on it Consider a body or rod BC of length L and uniform cross section of area A subjected to an axial load P If the resultant axial stress induced is given by Tensile stress P A Within elastic limit, one may apply the Hooke s law (stress) E e where E Young s modulus e Strain Stress, Strain and Deformation of Solids 1.17 B L Strain e E e P/A E or e P AE C Fig 1.13 l P Also we know that strain e. Substituting this in above equation. L [ change in length L] e L P AE Deflection or L PL AE L Original Length; L Change in length If the body is made up of different sections having P i, L i, A i and E i as internal force, length, area of cross section and modulus of elasticity respectively, then Deflection P i L i A i E i 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 Pdx AE L By integrating over the entire length Total deformation 0 [ is also denoted as L (or) l.] P dx AE
1.18 Mechanics of Solids - www.airwalkpublications.com 1.11 STIFFNESS Consider a bar BC of constant cross-section area A and of length L shown in the Fig. 1.14(a). Let force P is applied at the free end. The deformed bar is shown in Fig. (b) 1.14 (b). Conceptually, it is often B convenient to think of such elastic system as spring as shown in Fig. 1.14 (c) The total deformation is PL (c) AE The deflection of rod is directly proportional to the applied force and length and is inversely proportional to A and E. from the above equation we get P AE L P and also we get AE L This equation is related to the familiar definition of the spring constant or stiffness k. k P AE in N/m L Stiffness k is defined as the ratio of force per unit deflection 1. For an axially loaded i th bar or bar segment of length L i, the stiffness is given by k i A i E i L i The reciprocal of stiffness k is defined as flexibility (a) i.e. f 1 k in m/n P B L L a a Fig 1.14 C C P P For a particular case of i th bar of constant cross section f i L i A i E i
Stress, Strain and Deformation of Solids 1.19 The concept of structural stiffness and flexibility are widely used in structural analysis. 1.11.1 Stability Any structural or machine member loaded in compression is called a column or strut or pillar. Generally columns are classified as short columns, long columns and intermediate columns. P (a) P M.S 45 o C.I P (i) Failure by general yielding (a) Short columns (ii) Failure due to shear P Inelastic buckling Elastic buckling ( b ) ( c) (b) Intermediate column Fig 1.15 (c) Long column The classification among the columns has been done on the basis of their behaviour in compression. The ability of a short column to take loads depends upon its cross sectional area and strength of material of column. As the length of column increases, the load carrying capacity depends upon cross sectional area, strength of material, length of column, geometry of section (radius of gyration), Young s modulus E. A long or intermediate column fails in compression by buckling sideways whereas a short column does not buckle 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 same material. A column remains straight upto a certain load called the critical load beyond which a slight increase in load causes the column to buckle to a great extent 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 in which the load carrying capacity of columns depends. It is defined as the ratio of unsupported length of column to the least radius of gyration.
1.20 Mechanics of Solids - www.airwalkpublications.com Slenderness ratio l k where l Length of column k Radius of gyration Significance of percentage of Elongation & Reduction in Area Let L o Gauge length or initial length of the specimen L Length at fracture then Percentage Elongation L L o L o 100 Let A o Original area of crosssection A Area at neck when fracture occurs Percentage reduction of area A o A A o 100 SOLVED PROBLEMS Problem: 1.1: An elastic rod 25 mm is diameter, 200 mm long extends by 0.25 mm under a tensile load of 40 kn. Find the intensity of stress, the strain and the elastic modulus for the material of the rod. Solution: Given: diameter d 25 mm; Length L 200 mm Load P 40 kn 40 10 3 N; Elongation L 0.25 mm Area of cross section A d2 4 252 490.87 mm 2 4 Intensity of stress Load Area P 40 103 81.49 N/mm2 A 490.87 Strain e Elongation L Length L L L 0.25 200 0.00125 Elastic modulus E e 81.49 65192 N/mm2 0.00125 E 0.06519 N/m 2
Stress, Strain and Deformation of Solids 1.21 Problem 1.2 A rectangular wooden column of length 3 m and size 300 mm 200 mm carries an axial load of 300 kn. The column is found to be shortened by 1.5 mm under the load. Find the stress and strain in the column and state their nature. Given L 3m; Size of column 300 mm 200 mm Load P 300 kn; Shortening of column, L 1.5 mm Solution P 300 kn 300 10 3 N Area of cross section A 300 200 A 60000 mm 2 [... shortens] Compressive stress P 300 103 A 60000 5 N/mm 2 Compressive strain in the column e c L L 1.5 3000 e c 0.0005 L 3 m 3000 mm Problem 1.3: Find the maximum and minimum stress produced in the stepped bar shown in figure due to axially applied compressive load of 12 kn. Solution: Given: d 1 12 mm ; d 2 25 mm; Load 12 kn 12 10 3 N Area of upper part A 1 d 1 2 4 122 113.10 mm 2 4 Area of Lower part A 2 d 2 2 2 252 490.87 mm 2 4 Maximum stress max Load 12 103 106.10 N/mm2 Area A 1 113.10 12 kn 1 2 12 mm 25 mm
1.22 Mechanics of Solids - www.airwalkpublications.com Minimum stress min Load 12 103 24.45 N/mm2 Area A 2 490.87 Problem 1.4 A steel wire of length 10 m and diameter 5 mm is used to hang a load at its bottom. The stress and strain in the wire are found to be 140 N/mm 2 and 0.0007 respectively. Determine the load it carries and the elongation of wires. Given L 10 m 10 1000 mm; d 5 mm 140 N/mm 2 ; e 0.0007 Solution Area of cross section of wire A 52 4 Stress Load Area A 19.635 mm 2 P A A P P 19.635 140 P 2.749 kn We know that, Strain e L L 0.0007 L L L 0.0007 10000 L 7 mm Problem 1.5 A brass rod of 25 mm diameter and 1.3 m long is subjected to an axial pull of 4 kn. Find the stress, strain and elongation of the bar. If young s modulus E 1 10 5 N/mm 2.
Stress, Strain and Deformation of Solids 1.23 Given P 4 kn 4000 N; d 25 mm E 1 10 5 N/mm 2 ;L 1.3 m 1300 mm Solution: Area A 4 d2 4 252 490.87 mm 2 Stress Load area P A 4000 8.15 N/mm2 490.87 Strain e E 8.15 6 82 10 5 1 10 Elongation L e L 82 10 6 1300 L 0.106 mm Problem 1.6 A mild steel bar of 15 mm diameter and 400 mm length elongates 0.2 mm under an axial pull of 10 kn. Determine the young s modulus of material. Given d 15 mm; L 400 mm L 0.2 mm, P 10 kn 10 10 3 N Solution: Young s Modulus E Stress Strain stress P A A 4 152 176.7 mm 2 10 103 176.7 56.59 N/mm2 Strain e L L 0.2 400 e 5 10 4 56.59 E 5 10 4 E 1.1318 105 N/mm 2 Problem 1.7: A hollow cylinder 1.5 m long has an outside diameter of 45 mm and inside diameter of 25 mm. If the cylinder is carrying a load of 20
1.24 Mechanics of Solids - www.airwalkpublications.com kn, find the stress in the cylinder. Also find the deformation of the cylinder. Take E 100 G Pa [1 Pa 1 N/m 2 ]. Solution: Given: Length L 1.5 m; Outside diameter D 45 mm Inside diameter {d 25 mm; Load P 20 kn 20 10 3 N Modulus of Elasticity E 100 GPa 100 10 9 N/m 2 E 100 10 3 N/mm 2 Area of cross section A 4 [D2 d 2 ] 4 [452 25 2 ] 1099.5 mm 2 Stress Load Area P 20 103 18.1 N/mm2 A 1099.5 Strain e Strain e Stress Young s modulus E 18.1 4 1.81 10 3 100 10 Change in length Original length L L 1.81 10 4 L 1500 Deformation L 1.81 10 4 1500 0.271 mm. Problem 1.8: A specimen of a material having original diameter equal to 13 mm and gauge length 50 mm is tested under tension, the final diameter being 9 mm at fracture and gauge length at fracture being 70 mm. During testing, 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 in area, young s modulus if load corresponding to any point on the linear portion of the stress strain curve is 20 kn corresponding to an extension of 0.0315 mm. Solution: Original cross section Area A d2 4 4 13 1000 2 0.1325 10 3 m 2
Yield strength Yield point yield stress Area A 35 10 3 0.1325 10 3 Stress, Strain and Deformation of Solids 1.25 264 10 6 N/m 2 264 MPa Ultimate strength Ultimate stress Ultimate load Original Area A 60 10 3 0.1325 10 3 452 10 6 N/m 2 452 MPa Breaking strength Breaking Load Breaking stress Original Area 30 10 3 0.1325 10 3 226 10 6 N/m 2 226 MPa % elongation L L 0 L 0 100 70 50 50 100 40% % reduction in Area A 0 A A 0 100 L length of fracture L 0 Gauge length A 0 Original area of cross-section A Area at neck when fracture occurs Young s modulus E Stress Strain 4 132 4 92 100 52% 4 132 Stress Load Area 20 10 3 0.1325 10 3 1.509 108 N/m 2 Extension Strain e Original length 0.0315 6.3 10 4 50 Young s modulus E e 1.509 108 6.3 10 4 239 10 9 N/m 2 239 GPa
1.26 Mechanics of Solids - www.airwalkpublications.com Problem 1.9 In a tension test on mild steel specimen 10 mm diameter and 250 mm long gauge length, the following observations were made Elongation 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 mm Calculate (i) Nominal yield stress (ii) Nominal ultimate stress (iii) Nominal breaking stress (iv) Young s modulus (v) Percentage elongation (vi) Percentage reduction in area Solution (i) Nominal yield stress Load at yield point Original cross sectional area Nominal cross sectional area 4 102 78.54 mm 2 27 103 Nominal yield stress 343.17 N/mm2 78.54 Ultimate load (ii) Nominal ultimate stress Nominal cross sectional area (iii) Nominal breaking stress (iv) Young s modulus Stress Strain 51 103 78.54 E e P A L 0 L Breaking load area of c/s 649.35 N/mm 2 36 103 78.54 Gauge length L 0 250 mm; L 0.2 mm; P 16 kn E 16 103 250 78.54 0.2 E 2.546 10 5 N/mm 2 2.546 10 5 N/mm 2 458.37 N/mm2
Stress, Strain and Deformation of Solids 1.27 (v) Percentage elongation L L 0 L 0 100 % elongation 290 250 250 100 16% (vi) Percentage reduction in area A 0 A 1 A 0 100 A 0 original nominal area of cross section A 0 4 d2 78.54 mm 2 A 1 area of cross section at neck 4 d n 2 A 1 4 7.52 44.18 A 1 44.18 mm 2 Percentage reduction in area 78.54 44.18 78.54 100 43.75% Problem 1.10: The following data refer to a mild steel specimen tested in laboratory. Diameter of specimen 25 mm; Length of specimen 300 mm Extension under load 15 kn 0.045 mm; Load at yield point 127.65 kn Maximum load 208.60 kn Length of specimen after failure 375 mm Neck diameter 17.75 mm Determine: Young s modulus, yield strength, ultimate stress, percentage elongation, percentage reduction in area, safe stress with a factor of safety 2. Solution: Area of specimen A d2 4 252 490.87 mm 2 A 4 0 At load of 15 kn Stress Load 15 103 30.56 N/mm2 Area A 490.87 Strain at this load e L L 0.045 300 1.5 10 4
1.28 Mechanics of Solids - www.airwalkpublications.com Youngs modulus E Stress Strain e 30.56 1.5 10 4 2.036 105 N/mm 2 Yield strength Yield load 127.65 103 260.05 N/mm 2 Yield stress Area A 490.87 Ultimate stress Ultimate load Area A 208.60 103 490.87 424.96 N/mm 2 Percentage Elongation L L 0 375 300 100 100 25% L 0 300 Percentage reduction in Area A 0 A 1 A 0 100 252 17.75 2 25 2 100 49.6% Yield stress Safe stress Factor of safety 260.05 130.025 N/mm 2 2 Problem 1.11: A short hollow cast iron cylinder of external diameter 220 mm is 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/m 2. Factor of safety of 6 is used. Solution: Given: external diameter D 220 mm; Ultimate stress 540 MN/m 2 540 N/mm 2 ; Factor of safety 6 [1MN/m 2 1 N/mm 2 ] Ultimate stress Factor of safety Working stress 540 6 Working stress Working stress work 540 90 N/mm2 6 Working stress work Load Area