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1 675(1*7+ 2) 0$7(5,$/6 (MECHANICS OF SOLIDS) (As per Leading Universities Latest syllabus including Anna University R2013 syllabus) Dr. S.Ramachandran, Professor and Research Head Faculty of Mechanical Engineering Sathyabama University Jeppiaar Nagar, Chennai Prof. George V.J. Structural Engineer Dr. R. Venkatasubramani Professor and Head Department of Civil Engineering Ms. A. Vennila Assistant Professor Sri Krishna College of Technology, Coimbatore AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street, Mylapore, Chennai Ph.: , aishram2006@gmail.com

2 First Edition: January 2006 Second Edition: January 2010 Third Edition: Novermber 2012 Fourth Edition: May 2014 Sixth Edition: 8th July 2017 All Rights Reserved by the Publisher This book or part thereof should not be reproduced in any form without the written permission of the publisher. ISBN: Price : ISBN: and Books will be door delivered after payment into AIRWALK PUBLICATIONS A/c No (IFSC: BKID ) Bank of India, Santhome branch, Mylapore, Chennai - 4 (or) S.Ramachandran, A/c.No (IFSC:IDIB000S201), Indian Bank, Sholinganallur Branch, Chennai Typeset by: aksharaa muthra aalayam, Chennai Ph.:

3 NME-302: MECHANICS OF SOLIDS UNIT-I Compound stress and strains: Introduction, normal stress and strain, shear stress and strain, stresses on inclines sections, strain energy, impact loads and stresses, state of plane stress, principal stress and strain, maximum shear stress, Mohr s stress circle, three dimensional state of stress & strain, equilibrium equations, generalized Hook s law, theories of failure UNIT II Stresses in Beams: Pure Bending, normal stresses in beams, shear stresses in beams due to transverse and axial loads, composite beams. Deflection of Beams: Equation of elastic curve, cantilever and simply supported beams, Macaulay s method, area moment method, fixed and continuous beams Torsion: Torsion, combined bending & torsion of solid & hollow shafts, torsion of thin walled tubes UNIT-III Helical and Leaf Springs: Deflection of springs by energy method, helical springs under axial load and under axial twist respectively for circular and square cross sections axial load and twisting moment acting simultaneously both for open and closed coiled springs, laminated springs. Columns and Struts: Buckling and stability, slenderness ratio, combined bending and direct stress, middle third and middle quarter rules, struts with different end conditions, Euler s theory for pin ended

4 columns, effect of end conditions on column buckling, Ranking Gordon formulae, examples of columns in mechanical equipments and machines. UNIT-IV Thin cylinders & spheres: Introduction, difference between thin walled and thick walled pressure vessels, Thin walled spheres and cylinders, hoop and axial stresses and strain, volumetric strain. Thick cylinders: Radial, axial and circumferential stresses in thick cylinders subjected to internal or external pressures, compound cylinders, stresses in rotating shaft and cylinders, stresses due to interference fits. UNIT-V Curved Beams: Bending of beams with large initial curvature, position of neutral axis for rectangular, trapezoidal and circular cross sections, stress in crane hooks, stress in circular rings subjected to tension or compression. Unsymmetrical Bending: Properties of beam cross-section, slope of neutral axis, stress and deflection in unsymmetrical bending, determination of shear center and flexural axis for symmetry about both axis and about one axis for I-section and channel section.

5 CONTENTS Chapter - 1 Stress, Strain and Deformation of Solids 1.1 Introduction to Strength of Materials Rigid and Deformable Bodies Strength Unit of Stress Strain Types of Stress Normal Stress: Axially Loaded Bar Tensile Stress and Tensile Strain Compressive Stress and Compressive Strain Shear Stress and Shear Strain Bearing Stress (Crushing Stress) in Connections Stress-strain Behaviour of Materials Stress Strain Curves (Tension) Stress - Strain Curve for Ductile Materials Stress Strain Curves for Brittle Materials Stress Strain Curves (Compression) Hooke s Law Factor of Safety Deformation of A Body Due to Force Acting on it Stiffness Stability Solved Problems Deformation of A Bar Under Axial Load Deformation in Simple Bar Subjected to Axial Load Deformation for A Bar of Varying Section

6 1.19 Deformation of A Body Due to Self Weight Principle of Superposition Stress in Bars of Uniformly Tapering Cross Section Deformation of Uniformly Tapering Rectangular Bar Deformation in Compound or Composite Bars Bar of Uniform Strength Thermal Stresses Thermal Stresses in Simple Bars Thermal Stresses in Composite Bars Thermal Stress in Taper Bar - Circular Section Thermal Stress in Varying Section Bar Elastic Constants Modulus of Elasticity Rigidity Modulus (or) Shear Modulus Bulk Modulus Linear Strain and Lateral Strain Poisson s Ratio Volumetric Strain Rectangular Body Subjected to Axial Loading Rectangular Bar Subjected to 3 Mutually Perpendicular Forces Cylindrical Rod Subjected to Axial Load Bulk Modulus Relationship Between Elastic Constants Relation Between Bulk Modulus and Young s Modulus Shear Stress and Strain Shear Modulus Or Modulus of Rigidity Relation Between Modulus of Elasticity and Modulus of Rigidity Solved Problems

7 Chapter 2 Beams - Loads and Stresses 2.1 Introduction Types of Beams Supports and Support Reactions Types of Supports and Their Reactions Static Equilibrium Equations Determinate and Indeterminate Beams Types of Loading in Beams Shear Force in Beams (S.F) Sign Convention for Shear Force in Beam Couple or Moment Bending Moment in Beams Sign Convention for Bending Moment in Beams Shear Force (S.F) and Bending Moment (B.M) Diagrams Relation between Shear Force and Bending moment Method of Drawing Shear Force and Bending Moment Diagrams Points to be Remembered for Drawing S.F.D and B.M.D SFD and BMD for Cantilever Beam SFD and BMD for Simply Supported Beams SFD and BMD for Overhanging Beam Solved Problems on Cantilever beams Solved Problems on SSB Solved Problems on Overhanging Beam Stresses in Beams - Theory of Simple Bending Simple Bending or Pure Bending Assumption in Theory of Simple Bending Section Modulus or Modulus of Section Flexural Strength of a Section

8 Solved Problems Beam of Uniform Strength Composite Section Beams (or) Flitched Beams Shear Stresses in Beams Shear Stress Distribution for a Rectangular Section Shear Stress Distribution over I - Section Shear Stress Distribution over T Section Shear Stress Distribution over Circular Section Shear Stress Distribution over Triangular Section Shear Flow Solved Problems Chapter 3 Torsion 3.1 Introduction Pure Torsion Assumptions Made in Theory of Pure Torsion Shear Stress - (Resistance Concept) Shear Strain - (Deformation Concept) Analysis of Torsion of Circular Bars-derivation of Torsional Equations Theory of Torsion Bar of Solid Section Polar Modulus Z p Bars of Hollow Circular Section-strength Equation for Hollow Circular Shaft Torsional Rigidity and Stiffness of the Shaft Power Transmitted by the Shaft Important Formula Solved Problems

9 3.13 Problems on Replacing a Solid Shaft by a Hollow Shaft Stepped Shafts (or) Shafts in Series Shafts Fixed at one end Shafts Fixed at both the ends Compound Circular Shafts Shafts in Series Shafts in Parallel Shaft Subjected to Number of Torques Problems based on special conditons of shaft SPRINGS Introduction Types of Springs Bending Spring Torsion Spring Different Forms of Springs Helical Spring Closely Coiled Helical Springs Open Coiled Helical Springs Compression Springs (or) Open Coiled Spring Tension Springs (or) Closely Coiled Helical Spring Comparison Between Closely Coiled and Open Coiled Helical Springs Closely Coiled Helical Spring Subjected to an Axial Load Shear Stress Deflection, Stiffness of the Spring: k Strain Energy Stored U Close Coiled Helical Spring Subjected to Axial Twisting Couple Maximum Bending Stress, b Spring Under Impact Load

10 3.31 Springs in Series Springs in Parallel Open - Coiled Helical Spring Spring Index C Angular Rotation in Open Coil Helical Spring Open - Coil Spring Subjected to Axial Twisting Couple M o Axial Deflection Maximum Shear Stress in Spring Section Including Wahl s Factor Stresses in Spring Wire [Without Wahl s Factor] Stresses in Helical Coil Spring Under Torsional Load Design of Helical Coil Springs Solved Problems Leaf Springs Expression for bending stress, deflection, Strain energy in a Semi- Elliptical Leaf Spring Quarter Elliptical Leaf Spring Solved Problems Chapter 4 Beam Deflection 4.1 Introduction Definition of Deflection Importance of deflection Elastic Curve of Neutral Axis of the Beam Under Normal Loads Evaluation of Beam Deflection and Slope Flexural rigidity Stiffness of beam Double Integration Method

11 4.5.1 Simply supported beam with a concentrated load at the mid span Standard Formula for maximum slope and deflection Simply supported beam carrying a UDL (Uniformly Distributed Load) Macaulay s Method Problems on SSB - Point load Problems on SSB - Uniformly Distributed Load (UDL) Problems on SSB - Uniformly Varying Load (UVL) Problems on Cantilever Problems on overhanging beam Moment Area Method First moment - area theorem (or) Mohr s I theorem Second moment - area theorem (or) Mohr s II theorem Use of cantilever moment diagrams in moment area method Second moment - area theorem (or) Mohr s II theorem M EI diagram by parts Areas and centroids of various shapes Problems on Moment area method Conjugate Beam Method Conjugate beam theorem I (Mohr s theorem I) Conjugate beam theorem II (Mohr s theorem II) Simply supported beam with point load W acting at Centre Simply supported Beam with UDL Solved Problems

12 Chapter 5 Strain Energy and Energy Principles 5.1 Strain Energy Some Important Definitions Strain Energy Density Unit Strain Energy U Strain Energy in Uniaxial Loading Expression for strain energy stored in a body when load is applied gradually Expression for strain energy stored in a body when the load is applied suddenly Expression for strain energy stored in body for impact loading Strain energy stored in varying section bar Impact by shock Solved Problems Strain Energy in Pure Shearing Expression for strain energy stored in a body due to shear stress Torsional Strain Energy Torsional strain energy in a solid circular shaft Torsional strain energy in a hollow circular shaft Strain Energy due to Bending General formula for the strain energy of a beam due to bending under gradually applied loads Strain energy of a SSB with a concentrated load at the mid-span Strain energy of a SSB with uniformly distributed load Strain energy of a cantilever with a concentrated load at the free end Strain energy of a cantilever with UDL Solved Problems 5.46

13 5.4.6 Bending Under Impact Loads Deflection due to shear using strain energy method 5.55 I. Cantilever carrying a concentrated point load W at the free end II. Cantilever carrying an UDL III. SSB carrying a central point load W IV. SSB carrying with load - not at the centre V. SSB Carrying UDL Castigliano s Theorem Castigliano s theorem - Salient Points (i) Deflection due to axial load: (ii) Deflection due to bending: (iii) Deflection due to torsion (iv) Deflection due to shear (v) Deflection due to horizontal shear (vi) Rotation due to bending (vii) Rotation due to torsion Maxwell s Reciprocal Theorem Chapter - 6 Principal Stresses and Theories of Failures Introduction Analysis of Plane Stresses Stress on an inclined plane Normal stress Tangential stress Shear stress 6.3 Solved Problems Biaxial state of stress - Member subjected to biaxial stress Solved Problems A member subjected to a simple shear stress

14 6.1.4 Member subjected to a simple shear and a biaxial stress Solved Problems A member subjected to direct stress in one plane and a simple shear stress Solved Problems Mohr s Circle for Biaxial Stresses A body subjected to a biaxial perpendicular unequal and like principal stresses Guide lines for construction of the Mohr s circle Solved Problems on Mohr s Circle A body subjected to a biaxial perpendicular unequal and unlike principal stress Solved Problems A body subjected to a Biaxial perpendicular unequal, like stresses with an simple shear stress Solved Problems Combined Bending and Torsion Solved Problems on Combined Bending and Torsion: Principal Strains Strain Energy in Bending and Torsion Solved Problems Equivalent of Bending Moment M e and Equivalent Torque T e Solved Problems Theories of Failure (i) Maximum principal stress theory (or) Rankine theory 6.67 (ii) Maximum principal strain theory (or) St. Venant s theory (iii) Maximum shear stress theory (or) Guests theory (or) Tresca s theory (iv) Maximum strain energy theory (or) Haigh s theory 6.84

15 (v) Maximum shear strain energy theory (or) Distortion energy theory Solved Problems Chapter - 7 Cylinders and Spheres 7.1 Thin Cylindrical and Spherical Shells - Stresses Hoop Stress Longitudinal Stress Spherical Shells Deformation in Thin Cylinder Deformation in Spherical Shells Solved Problems on Thin Cylinder and Spherical Shells Thin Pipe Bounded by Wire Cylinderical Shell With Hemispherical Ends Thick Cylinders Lame s Theorem Lame s Equations - Derivation Special Conditions Longitudinal and Shear Stresses Solved Problems on thick cylinders with Lame s theorem Design of Thick Cylinders Compound Cylinders Difference of Radii for Shrinkage Solved Problems on compound cylinder Marks Question and Answers University Solved Problems

16 Chapter - 8 Columns 8.1 Introduction Slenderness ratio of a column Short columns Long columns Buckling load, Crippling or Critical load Equivalent length Buckling factor Safe load Euler Equation Assumptions made in Euler s equation Sign Conventions End Conditions Derivation of Euler s equation Slenderness Ratio Equivalent Length (or) Effective Length L e of a Column One end fixed and one end free column One end fixed and one end hinged column Two fixed ends column Limitations of Euler s Formula Rankine s Formula Eccentrically Loaded Columns Rankine-Gordon formula Middle Third Rule

17 Stress, Strain and Deformation of Solids 1.1 Chapter - 1 STRESS, STRAIN AND DEFORMATION OF SOLIDS 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 materials) 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 RIGID AND 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. 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 when subjected to an external loading.

18 1.2 Strength of Materials In practice, no material is absolutely elastic nor plastic nor rigid. These properties are attributed when the deformations are within certain BAR limits. A A B B C Deformation can be understood by a simple example, consider a bar F which is fixed at one end and is C loaded by a force (F) as shown in Fig.1.1 After the load is applied on Fig 1.1 F 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 STRENGTH 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. (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

19 Stress, Strain and Deformation of Solids 1.3 Consider a rod AB subjected to an external load F at the ends as shown in Fig.1.2(i). In order to keep the body in equilibrium, the body part C 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 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 A Stress Load 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. 1.4 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

20 1.4 Strength of Materials Giga pascal (GPa) 10 9 Pa For Engineering materials, the cross section, we use is in only millimeters ie N/mm 2 1 MPa Pa 10 6 N/m 2 or 1 MN/m 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 N/m MPa 1000 N/mm 2 1 GPa 1000 N/mm 2 1 kn/mm 2 Intensity of stress R A F A in N/m2 where R: Reaction in Newton F: Force in Newton A: Area of cross section in m 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. Strain e change in length original length dl l d Fig 1.3

21 Stress, Strain and Deformation of Solids TYPES OF STRESS Basically classified into 2 types STRESS Normal Stress Tangential Stress Tensile Stress Compressive Stress Shear Stress Punching shear Stress Fig 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. Consider a bar of cross sectional area A, subjected to an axial load 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 P a Cutting plane P P a P P a (a) Bar axis Centroid da = P A a a 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)

22 1.6 Strength of Materials respective areas generate a stress resultant that is statically equivalent to the force P Fig. 1.4(c). 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 Fig. 1.4(e). However simplified diagram shown in Fig. 1.4(f) is commonly used. Normal stress is defined as ratio of force applied to the cross section of 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. 1.8 TENSILE STRESS AND TENSILE STRAIN Tensile stress is defined as the resistance offered by the section of a d X member/body against an increase in P P length. For example consider the stress offered by the section XX of a rod as shown in Fig X X The intensity of tensile stress is P R R P given by Tensile stress Force Area P A R A Fig. 1.5 X 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.

23 Stress, Strain and Deformation of Solids 1.7 STRAIN Tensile strain compressive strain Lateral strain Longitudinal strain Volumetric strain Shear strain Fig. 1.5 (a) Tensile strain: Ratio of change in length to original length is known as tensile strain Fig. 1.5 (b) P P L dl Tensile strain e Fig 1.5 (b) Increase in length Original length dl L 1.9 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 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 dl l

24 1.8 Strength of Materials SHEAR STRESS AND SHEAR STRAIN Consider 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). 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 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 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 The shear stress Fig 1.7 shear force shear Area F A P A Fig 1.8

25 Stress, Strain and Deformation of Solids 1.9 Consider a block of height l, length L and width unity Fig.1.9. A P B A dl A P d l B B x x d x dx Shear Stress Fig 1.9 Shear force or Resistance Shear Area R 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 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 The strain dx x Since is very small, tan dl l dx shear strain x The Angular deformation in radians measures the shear strain.

26 1.10 Strength of Materials BEARING STRESS (CRUSHING STRESS) IN CONNECTIONS Bolts, pins and rivets create stresses in the members they connect, along the bearing surface, or surface of t C contact. P For example consider again two A d F plates A and B connected by the rivet F CD which we have discussed in Fig A previous section. The rivet exerts on D plate a force P equal and opposite to the force F exerted by the plate on the t rivet. The force P represents the A resultant of elementary forces Fig 1.10 Fig B 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. d Bearing stress b P A P td 1.12 STRESS-STRAIN BEHAVIOUR OF MATERIALS Young s modulus Stress Strain Normal stress 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.

27 Stress, Strain and Deformation of Solids 1.11 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. Thus, the stress-strain diagram gives many important properties of material like Young s modulus STRESS STRAIN CURVES (TENSION) When a bar or specimen is subjected to a gradually increasing axial tensile load, the stresses and strain can be found out for number of loading conditions and a curve is plotted upto the point at which the specimen fails, giving what is known as stress strain curve STRESS - STRAIN CURVE FOR DUCTILE MATERIALS 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 The stress strain 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 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,

28 1.12 Strength of Materials STRESS Y - Upper Yield point * * * * E - Elastic Limit P-Proportionality Limit * L - Lower Yield point U -U ltim ate strength A - actual Rupture strength * * B- Breaking (Rupture strength) O STRAIN Fig 1.11 Stress - Strain 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.

29 Stress, Strain and Deformation of Solids 1.13 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 this 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 the diameter is also 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 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 strain based on original dimensions are called as Engineering or Nominal or Conventional stresses or strains. Stresses and

30 1.14 Strength of Materials 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 100 l % Reduction in Area A A 100 A 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. For Brittle materials, the stress strain curve is as shown in Fig Ultimate streng th Stress Fig 1.12 Breaking or ultimate Point Limit of Proportionality Strain 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 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.

31 Stress, Strain and Deformation of Solids HOOKE S LAW 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 Youngs Modulus E Normal stress 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 FACTOR OF SAFETY Factor of safety is defined the ratio of ultimate stress to the permissible stress (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 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 Timber 4 to 6

32 1.16 Strength of Materials 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 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] Deflection L Original Length L Change in length e L P AE or L PL AE or L PL AE [ is also denoted as L (or) l.] 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

33 Stress, Strain and Deformation of Solids 1.17 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 By integrating over the entire length L Total deformation 0 P dx AE 1.16 STIFFNESS Consider a bar BC of constant cross-section area A and of length L B shown in the Fig. 1.14(a). (a) Let force P is applied at the free end. The deformed bar is shown in Fig (b). Conceptually, it is often (b) convenient to think of such elastic B system as spring as shown in Fig (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. L L a a Fig 1.14 C C P P

34 1.18 Strength of Materials 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 i.e. f 1 k in m/n P For a particular case of i th bar of constant cross section f i L i A i E i The concept of structural stiffness and flexibility are widely used in structural analysis 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. 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 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) (b) Intermediate column Fig 1.15 ( c) (c) Long column

35 Stress, Strain and Deformation of Solids 1.19 (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 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 on 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. where l Length of column k Radius of gyration Slenderness ratio l k 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 100 A o

36 1.20 Strength of Materials 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 N; Elongation L 0.25 mm Area of cross section A d mm 2 4 Intensity of stress Strain e Load Area P N/mm2 A Elongation L Length L L L Elastic modulus E e N/mm E N/m 2 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 Solution Size of column 300 mm 200 mm Load P 300 kn Shortening of column, L 1.5 mm P 300 kn N

37 Stress, Strain and Deformation of Solids 1.21 Area of cross section A A mm 2 [... shortens] Compressive stress P A N/mm 2 Compressive strain in the column e c L L e c L 3 m 3000 mm Results Stress 5 N/mm 2 or 5 MPa comp Strain or comp 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 N Area of upper part A 1 d 1 2 Area of Lower part A 2 d 2 2 Maximum stress max mm mm 2 Load N/mm2 Area A

38 1.22 Strength of Materials 12 kn 1 12 mm 2 25 mm Minimum stress min Load N/mm2 Area A 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 respectively. Determine the load it carries and the elongation of wires. Given L 10 m mm Solution d 5 mm 140 N/mm 2 e Required Data P? L? Area of cross section of wire A 52 4 A mm 2

39 Stress, Strain and Deformation of Solids 1.23 Stress Load Area P A A P P P kn W.K.T Strain e L L L L L L 7 mm Results P kn 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 N/mm 2. Given P 4 kn 4000 N d 25 mm E N/mm 2 L 1.3 m 1300 mm Required data:? e 1? L?

40 1.24 Strength of Materials Solution Area A 4 d mm 2 Stress Load area P A N/mm Strain e E Elongation L e L L mm Results Stress 8.15 N/mm 2 Strain e Elongation L 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 N Required Data: E? Solution Young s Modulus E Stress Strain stress P A

41 Stress, Strain and Deformation of Solids 1.25 A mm N/mm 2 Strain e L L E e E 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 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 N Modulus of Elasticity E 100 GPa N/m 2 Area of cross section E N/mm 2 A 4 [D2 d 2 ] 4 [ ] mm 2 Stress Load Area P N/mm2 A

42 1.26 Strength of Materials Stress Strain e Young s modulus E Strain e Change in length Original length L L L 1500 Deformation L 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 mm. Solution: Original cross section Area A d m 2 Yield strength Yield point yield stress Area A N/m MPa Ultimate strength Ultimate stress Ultimate load Original Area A N/m MPa Breaking strength Breaking Load Breaking stress Original Area N/m MPa

43 Stress, Strain and Deformation of Solids 1.27 % elongation L L L length of fracture L 0 L 0 Gauge length % 50 % reduction in Area A 0 A A A 0 Original area of cross-section A Area at neck when fracture occurs Young s modulus E Stress Strain % Stress Load Area N/m 2 Extension Strain e Original length Young s modulus E e N/m GPa 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 Ultimate load Breaking load 27 kn 51 kn 36 kn Length between gauge marks after fracture 290 mm

44 1.28 Strength of Materials 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 Nominal cross sectional area Nominal yield stress (ii) Nominal ultimate stress (iii) Nominal breaking stress (iv) Young s modulus Stress Strain Load at yield point Original cross sectional area mm N/mm2 Ultimate load Nominal cross sectional area N/mm 2 E e P A L 0 L Gauge length L mm L 0.2 mm P 16 kn Breaking load area of c/s N/mm 2

45 Stress, Strain and Deformation of Solids 1.29 (v) Percentage elongation E N/mm E N/mm 2 L L 0 L % elongation % 250 (vi) Percentage reduction in area A 0 A 1 A A 0 original nominal area of cross section A 0 4 d mm 2 A 1 area of cross section at neck 4 d n 2 Percentage reduction in area A A mm % 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 mm Load at yield point kn Maximum load kn

46 1.30 Strength of Materials Length of specimen after failure 375 mm Neck diameter 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 d mm 2 A 4 0 At load of 15 kn Stress Load N/mm2 Area A Strain at this load e L L Youngs modulus E Stress Strain e N/mm 2 Yield strength Yield load N/mm 2 Yield stress Area A Ultimate stress Ultimate load Area A N/mm 2 Percentage Elongation L L % L Percentage reduction in Area A 0 A 1 A Safe stress % Yield stress Factor of safety N/mm 2 2

47 Stress, Strain and Deformation of Solids 1.31 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; Inner diameter d? Ultimate stress 540 MN/m N/mm 2 ; Factor of safety 6 [1MN/m 2 1 N/mm 2 ] Factor of safety 6 Ultimate stress Working stress 540 Working stress Working stress work Working stress work Load Area N/mm [D2 d 2 ] 4 [D2 d ] 90 4 [2202 d ] 90 Inner diameter d 200 mm 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/mm 2, what is the minimum diameter required? What will be extension of 3.50 m length of wire? Take young s modulus E N/mm 2