Mechanics of Solids (APJ ABDUL KALAM TECHNOLOGICAL UNIVERSITY) And As per Revised Syllabus of Leading Universities in India AIR WALK PUBLICATIONS

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1 Advanced Mechanics of Solids (APJ ABDUL KALAM TECHNOLOGICAL UNIVERSITY) And As per Revised Syllabus of Leading Universities in India Dr. S. Ramachandran Prof. R. Devaraj Professors School of Mechanical Engineering Sathyabama University Chennai Mrs. Pradeepa Structural Engineer - Civil Mr. Kumaran.S Chennai AIR WALK PUBLICATIONS (Near All India Radio) 80, Karneeshwarar Koil Street Mylapore, Chennai Ph.: , aishram006@gmail.com, airwalk800@gmail.com

2 First Edition : ISBN: ISBN :

3 Syllabus S.1 Course Plan Module Contents Hours (I) (II) (III) Introduction to stress analysis in elastic solids stress at a point stress tensor stress components in rectangular and polar coordinate systems Cauchy s equations stress transformation principal stresses and planes hydrostatic and deviatoric stress components, octahedral shear stress equations of equilibrium Displacement field engineering strain strain tensor (basics only) analogy between stress and strain tensors strain displacement relations (small strain only) compatibility conditions. Constitutive equations generalized Hooke s law equations for linear elastic isotropic solids relation among elastic constants Boundary conditions St.Venant s principle for end effects uniqueness theorem -D problems in elasticity Plane stress and plane strain problems stress compatibility equation Airy s stress function and equation polynomial method of solution solution for bending of a cantilever with and end load FIRST INTERNAL EXAM Equations in polar coordinates (D) equilibrium equation, strain displacement relations, conversion of Airy s equation and definition of stress function and stress components Application of stress function to Lame s problem stress concentration problem of a small hole in a large plate. Sem. Exam Marks 6 15 % % 4 15 %

4 S. Advanced Mechanics of Solids - Module Contents Hours Axisymmetric problems governing equations application to thick cylinders, interference fit and rotating discs. Unsymmetrical bending of straight beams curved beams (rectangular c/s) shear (IV) center shear stresses in thin walled open sections Strain energy of deformation special cases of a body subjected to concentrated loads, moment or torque reciprocal relation strain energy of a bar subjected to axial force, shear force, bending moment and torque Second Internal Exam Maxwell reciprocal theorem Castigliano s first and second theorems virtual work (V) principle minimum potential energy theorem complementary energy theorem This topic is Torsion of non circular bars: Saint Venant s discussed in theory solutions for circular and elliptical Module 6. cross sections Prandtl s method solutions for circular and elliptical cross-section membrane analogy (VI) approximate solution methods for non-circular shafts. Torsion of thin walled tubes, thin rectangular sections, rolled sections and multiply connected sections. 4 Sem. Exam Marks 6 15 % 5 0 % % 5

5 Contents C.1 CONTENTS Module I ANALYSIS OF STRESS AND STRAIN 1.1 Introduction to Stress Analysis in Elastic Solids Solid Body (or) Solid Element Scalar, Vector and Tensor Stress at a Point Stress vector Normal and Shear stress components Stress Tensor Some Common Terms Components of a stress tensor Stress tensor conventions Symmetry in a stress matrix Plane stress Stress Components in Rectangular Co-ordinates Stress Component on an Arbitrary Plane Equality of cross shear Hydrostatic State of Stress Stress Transformation Equations Maximum Shear Stress Stress Invariants Standard method of solution for a cubic equation Principal Stresses and Planes Stress invariants in terms of principal stresses Principal stresses from cubic equation Principal directions State of Stress referred to principal axes Mohr s Circle for D State of Stress Procedure for constructing D Mohr s circle

6 C. Advanced Mechanics of Solids Special cases State of Pure Shear Hydrostatic and Deviatoric Stress Components Octahedral Stresses Lame s Ellipsoid Equations of Equilibrium in Cartesian Co-ordinates Equations of Equilibrium in Polar Co-ordinates Equations of Equilibrium in Cylindrical Coordinates Analysis of Strain Displacement Field Engineering Strain Strain Displacement Relations Infinitesimal (small) strain Strain Tensor Components of strain tensor Properties of strain tensor Spherical and Deviatorial strain tensor Pure shear part of the strain Strain invariants Analogy between Stress and Strain Tensors Some general analogies Strain - Directions Homogeneous equations Standard method of solution for a cubic equation Compatibility Conditions St.Venant s equations of compatibility Module - ELASTIC SOLIDS.1 Introduction Constitutive Equations

7 Contents C.. Generalized Hooke s Law Equations for Linear Elastic Isotropic Solids Relation Among Elastic Constants Displacement Equations of Equilibrium Saint - Venant s Principle for End Effects Cantilever beam with various end loads Uniqueness Theorem Two-dimensional Problems Plane Stress Plane Strain Stress Compatibility Equation Airy s Stress Function Expressions for the stress components Polynomial Method of Solution Cantilever Beam With End Load Bending of a Cantilever beam with an end load..41 Module - III AXISYMMETRY.1 Introduction Axisymmetry Stresses in Polar Coordinates Equilibrium Equation in D Polar Coordinates Stress Function Lame s Theorem Stress Concentration Stress concentration of small hole in a large plate..1.7 Axisymmetric Problems Governing equation for thick cylinders Shrink-fit Cylinder Characteristics of Shrink-Fit

8 C.4 Advanced Mechanics of Solids - Application of shrink fit cylinder Advantages of shrink fit cylinder General Equation for Radial Interference Rotating Disc Rotating disc with a small central hole Module 4 UNSYMMETRICAL BENDING 4.1 Introduction Bending Stress in Unsymmetrical Bending Method for finding bending stress in unsymmetrical bending Curved Beams Stresses in Curved Beam (Winkler - Batch Theory) Resultant Stesses in a Hook Shear Centre Properties of shear centre Determination of Shear Centre Determination of Shear Centre for Channel Section Determination of Shear Centre for I-Section Shear Flow Strain Energy Some Important Definitions Strain Energy Density Unit Strain Energy U Strain Energy of a Bar Subjected to Axial Force Strain Energy of a Bar Subjected to Shear Force Strain Energy in a Bar due to Bending General formula for the strain energy of a beam due to bending under gradually applied loads

9 Contents C Strain energy of a SSB loaded with a concentrated load placed at the mid-span Strain energy of a SSB loaded with uniformly distributed load throughout its length Strain energy of a cantilever loaded with a concentrated load placed at the free end Strain energy of a cantilever loaded with UDL throughout its length Strain Energy in a bar due to Torque Reciprocal Relation Module 5 ENERGY METHODS 5.1 Maxwell s Reciprocal Theorem Proof for Reciprocal Theorem Application of Maxwell Reciprocal Theorem Castigliano s Theorem Castigliano s First Theorem Castigliano s Second Theorem Application of Castigliano s Theorem Virtual Work Method For Deflection Principle of Virtual Work (Using Virtual Forces) Principle of Virtual Displacement Importance of Virtual Work Unit Load or Dummy Load Method Applications of virtual work Minimum Potential Energy Theorem - A Special Case of Virtual Work Principle Total, Elastic And Complementry Strain Energy Complementary Energy Principle

10 C.6 Advanced Mechanics of Solids - Module - VI TORSION 6.1 Torsion Saint Venant s Theory Torsion of Prismatic - Solid Section Prandtl s Method Solution For Circular and Elliptical Section Solution for Circular Section using Saint Venant s approach Solution for Elliptical Section using Saint Venant s approach Solution for circular section using Prandtl s method Solution for Elliptical Section using Prandtl s method Membrane Analogy Boundary Condition Approximate Solution Methods for Non-circular Shafts Thin Walled Tubes Torsion of thin wall section Torsion of thin rectangular section Torsion of Rolled Sections Torsion of Thin-wall Segmented Sections Torsion of Multiply Connected Sections

11 Module I ANALYSIS OF STRESS AND STRAIN Introduction to stress analysis in elastic solids - stress at a point - stress tensor - stress components in rectangular and polar coordinate systems - Cauchy s equations - stress transformation - principal stresses and planes - hydrostatic and deviatoric stress components, octahedral shear stress - equations of equilibrium Displacment field - engineering strain - strain tensor (basics only) - analogy between stress and strain tensors - strain - displacement relations (small-strain only) - compatibility conditions. 1.1 INTRODUCTION TO STRESS ANALYSIS IN ELASTIC SOLIDS The behaviour of an elastic solid body, when subjected to external forces, depends upon the magnitude of the forces, the direction and the inherent strength of the material. Structural and Mechanical construction units are usually subjected to various forces, with some having more detrimental effects than others. Therefore, it is often necessary, to consider and analyse the forces, that are transmitted through the material. In this chapter the concepts of stress and strain vectors acting on a surface, the state of stress at a point and the state of strain at a point will be introduced. It will be shown that the components of the stress vector can be obtained through a linear symmetric transformation with a matrix, whose elements are the components of a second rank tensor called the stress tensor. Further, it will be shown that the components of the stress tensor must satisfy three partial differential equations of the first order, called the differential equations of equilibrium. Therefore, in our analysis, we assume that the matter is continously distributed with no voids. Such a concept is called the continuum and the mechanics of such a body is called the continuum mechanics.

12 1. Advanced Mechanics of Solids SOLID BODY (OR) SOLID ELEMENT A solid body (or) solid element is defined as a region in space (or) an infinitesimal region of the solid continuum in isolation from its surroundings Forces acting on a solid body Generally, two types of forces can exist on a solid body. They are (i) Body force (ii) Surface force (ii) Body Force: An external force, acting throughout the mass of the body is called as body force. These forces act on each volume element of the body. In general, body force pulls the entire mass of the body. Ex: Gravitational force, Magnetic force, Inertia force etc. (ii) Surface force: A force applied to the surface of the body is called as surface force. i.e., generally acts on the surface or area elements of the body. Also, these forces arises when one body acts over another, by having a direct surface contact between them. Ex: Reaction forces, distributed forces (a force which acts over a region of length, surface or area), concentrated forces (a force which acts along a single line in space) etc. 1. SCALAR, VECTOR AND TENSOR Scalar A Scalar is a physical quantity which is represented by a particular point in space and time (magnitude alone). Ex: Temperature, Hydrostatic pressure 1... Vector A vector is a physical quantity that keeps track of magnitude and direction. Ex: Force, Velocity 1.. Tensor Many physical properties are direction dependent because of the arrangement of atoms in the crystal lattice. Some materials upon heating,

13 expand by the same amount in each direction, but there are few other materials, where the expansion will differ in different directions. For this reason, the properties (such as the elasticity and thermal expansivity) are not expressed as scalars. Therefore, to deal with this complex situation, tensor tool is used. A tensor is a multidimensional array of numerical values that can be used to describe the physical state or properties of a material. Ex: stress Rank of a tensor Tensors are referred by their rank, which is a description of the tensor s dimension. (i) Zero rank tensor: A zero rank tensor is a scalar. (ii) First rank tensor: It is a one dimensional array of numbers. Ex: vector (iii) Second rank tensor: A second rank tensor looks like a typical square matrix. Ex: stress, strain, thermal conductivity, magnetic susceptibility and electrical permittivity. (iv) Third rank tensor: It looks like a three dimensional matrix. Ex: cube (v) Fourth rank tensor: A fourth rank tensor is a four-dimensional array of numbers. Ex. Elasticity of single crystal. 1.4 STRESS AT A POINT Analysis of Stress and Strain Stress vector Stress is a measure of the intensity of internal forces generated in a body. In general, stresses in a body vary from point to point. To understand the concept of stress at a point, the stress vector should be defined. For this, consider a body subjected to external forces as shown in Fig.1.1. Now, pass a cutting plane through the point P having a unit normal n. The cutting plane divides the body into two halves. On each half of the body, there are distributed internal forces acting on the cutting plane.

14 1.4 Advanced Mechanics of Solids - Cutting Plane F 1 F F P F 4 A (a) B F 1 F F A C s P n T (n ) n (b) F -n -T (n) P D B F 4 Fig:1.1.Stress Vector at a Point on a Plane. Let F be the resultant of the distributed internal forces and A be the area around point P of the cutting plane. Then, the stress vector at point P is defined as T n lim A 0 Similarly at point P, the action of D on C as A tends to zero, can be represented by a vector. T n lim A 0 F A F A Where T n and T n are called the stress vectors and they are represented as forces per unit area, acting at point P and P respectively. The stress vector T n represents the action of C on D at a point P, which is

15 Analysis of Stress and Strain 1.5 equal in magnitude and opposite in direction to the stress vector T n (represents the action of D on C at corresponding point P). This assumption is similar to Newton s third law of motion. Thus, we have T n T n...(1) Equation (1) is also called the cauchy reciprocal theorem Normal and Shear stress components Let T n be the resultant stress vector at point P, acting on a plane whose outward drawn normal is n. Now, the resultant stress vector T n can be resolved into two components, i.e., one along normal (n) and other perpendicular to normal (n). The component parallel to n is called the normal stress and the component perpendicular to n is known as the tangential stress (or) shear stress. Therefore T n Stress Tensor An infinite number of planes can be passed through point P to obtain infinite number of stress vectors. The totality of all stress vectors acting on every plane passing through a point describes the state of stress at that point, also called as stress tensor. 1.5 SOME COMMON TERMS Components of a stress tensor. The number of components of a stress tensor can be determined by No. of components n, Where n tensor rank No. of components 9 [n for stress tensor] Therefore, nine components are required to define the state of stress at a point. These stress components are represented in a matrix form as

16 1.6 Advanced Mechanics of Solids - x xy xz yx y yz zx zy z Out of nine, six components are independent. They are x, y, z, xy, yz, xz Note: From equilibrium principles, we have xy yx, yz zy, xz zx 1.5. Stress tensor conventions The nine stress components comprise of three normal and six shear components. Normal stress : The subscript on the normal stress component denotes the plane (face) at which the stress acts. Here, tension is positive and compression is negative. Shear stress : The standard convention for denoting the shear components is that, the first subscript refers to the plane on which the stress component acts, and the second subscript denotes the direction in which it acts. ex. xy Shear stress acts on a plane normal to x-axis and Shear stress acts in the direction of y-axis Symmetry in a stress matrix The stress matrix is symmetrical about the leading diagonal. i.e., x yx zx xy y zy xz yz z leading diag onal.

17 Analysis of Stress and Strain 1.7 In the matrix xy yz xz zx yz zy From this relation, it is clear that the state of stress at a point is defined completely by six independent components ( normal and shear) Note: A scalar quantity can be completely specified by 1 value and a vector by values but a tensor requires 6 values Plane stress A state of stress in which two faces of the cubic elements are free from stress. i.e., z 0, xz 0, yz 0 y y xy xy x x xy x y (a) Plane State of Strain Fig: 1. y (b) Conventional Representation Therefore, the stress components, acting on the element are limited to x, y, xy 1.6 STRESS COMPONENTS IN RECTANGULAR CO-ORDINATES Consider a body occupying a region in space, referred to a rectangular co-ordinate system oxyz. Let the body be cut by a plane parallel to the yz plane. Now, the resultant stress vector acting on the x plane will be T n.

18 1.8 Advanced Mechanics of Solids - This vector can be resolved into three components parallel to x, y and z axes. The component parallel to the x-axis is called normal stress component which is denoted as x and the components which are parallel to y and z axes are called shear stress components, and they and they are denoted as xy and xz respectively. Here xy is the stress component acting on the x plane in the y - direction and xz is the stress component acting on the x plane in the z-direction. The components acting on a small rectangular element are shown in Fig.1.4. In this element, the three mutually perpendicular planes, the x-plane, the y-plane and the z-plane can be drawn. The normal and shear stress components acting on these planes are z y o xz xy x Fig:1..Stress Components on x Plane x (n ) T X n

19 Analysis of Stress and Strain 1.9 x plane x, xy, xz y plane y, yx, yz z plane z, zx. zy 1.7 STRESS COMPONENT ON AN ARBITRARY PLANE Consider an arbitrary plane on a tetrahedron with outward drawn normal n, whose direction cosines are n x, n y, n z, and also consider three mutually perpendicular planes be x, y, and z planes. In this case, an arbitrary plane and perpendicular planes are considered, because, to prove that the stress components acting on the three mutually perpendicular planes while passing through a point will enable one, to determine the stress components acting on any plane passing through that point. Now, consider a small tedrahedron with the point P as the origin and the apices are labeled as A, B and C. Let h be the perpendicular distance from point P to the inclined face. If the tetrahedron is isolated from its parent body and a free body diagram is drawn, then it will be in equilibrium under the action of surface and body forces. The free body diagram of the tetrahedron is shown in Fig.1.5. Cauchy s formula Let T n be the resultant stress vector on face ABC. This can be n resolved into components T x, n Ty and n Tz which is parallel to x, y, and z axes. The other relavent quantities involved in the expression are defined as follows

20 1.10 Advanced Mechanics of Solids - F a h A Density Body force per unit mass Acceleration Height of the tetrahedron, measured perpendicular to ABC Area of the inclined face An x Area of BPC An y Area of CPA An z Area of APB The volume of the tetrahedron is 1 Ah. Also, for equilibrium of the tetrahedron, the sum of the forces in x, y and z. directions must be equal to zero. Thus, for equilibrium in x direction, apply Newton s second law to a free body in the shape of tetrahedron, F x ma x T x n A x An x yx An y zx An z 1 Ah a x 0 Divide throughout by A... m V 1 Ah T x n x n x yx n y zx n z 1 h x [... x ax ] Likewise, performing the same force balance in the other two co-ordinate directions leads to expressions for the three traction components on an arbitrary plane i.e., T y n xy n x y n y zy n z 1 h y T z n xz n x yz n y z n z 1 h z Since the size of the tetrahedron is very small, we let the height of the tetrahedron (h) shrink to zero. Therefore, the equation becomes

21 n T x n x x n y yx n z zx n T y n x xy n y y n z zy... (1) n T z n x xz n y yz n z z The set of these three equations is called Cauchy s stress formula or simply Cauchy s formula. The above equation can also be written as Analysis of Stress and Strain 1.11 T i n nx ix n y iy n z iz j n j ij z zz Where i and j stands for x or y or z and x xx, y yy and If T n is the resultant stress vector on plane ABC, we have T n n T n x T n y T z... () If n and n are the normal and shear stress components, then T n n n... () Since the normal stress is equal to the projection of T n along the normal n and also equals to the sum of the projections of its component along n, then from () and () we have n n x T x n ny T y n nz T z n Substituting the values of T x n, Ty n, Tz n we get n n x x n y y n z z n x n y xy n y n z yz n z n x zx... (4) Equality of cross shear. In Cauchy s equation there are nine rectangular stress components acting at P, which will enable one to determine, the stress components on any arbitrary plane passing through point P.

22 1.1 Advanced Mechanics of Solids - Among the nine rectangular stress components, only six are independent. i.e., xy yx, zy yz, zx xz This is known as the equality of cross shears. Problem 1.1: At a point P in a body, x N /cm, y 5000 N /cm, z 5000 N /cm, xy yz zx N /cm. Determine the normal and shearing stresses on a plane that is equally inclined to all the three axes. [Nov Calicut] Given: x N/cm, y 5000 N/cm, z 5000 N/cm xy yz zx N/cm To find: Normal and Shear stress Solution A plane that is equally inclined to all the three axes will have, n x n y n z 1 since nx n y nz 1 we know that, n n x x n y y n z z n x n y xy n y n z yz n z n x zx n n 0000 N/cm Also T x n x n x yx n y zx n z 1 h x - Cauchy s formula

23 Analysis of Stress and Strain 1.1 Since x 0 n T x [ ] N/cm Similarly T y n xy n x y n y zy n z ; n T y [ ] 5000 N/cm Simlarly, T z n xz n x yz n y z n z n T z [ ] 5000 N/cm Resultant stress vector [T n ] [T x n ] [Ty n ] [Tz n ] N/cm 4 We know that, [T n ] n n n [T n ] n N /cm 4 n N/cm 1.8 HYDROSTATIC STATE OF STRESS Consider a small sphere that is immersed in a liquid. The sphere inside the liquid would experience a state of stress at a point, where the resultant stress vector on any plane is normal to the plane and has the same magnitude. Hence this phenomena is known as hydrostatic or isotropic state of stress. Note: The word isotropy means independent of orientation or uniform in all directions We know, that an ideal fluid cannot sustain any shearing forces and the normal force on any surface that is compressive in nature. This can be represented by

24 1.14 Advanced Mechanics of Solids - T n Pn, P 0 Where P Pressure and (-ve) sign indicates that the pressure is compressive in nature. If n x, n y and n z are the direction cosines of n, then T x n Pnx ; T y n Pn y ; T z n Pn z... (1) Since all the shear stress components are zero, we have n T x n nx x ; T y n ny y ; T z nz z... () From (1) and () we get x y z P Thus, the plane n chosen arbitrarily shows that the resultant stress vector on any plane is normal and equal to P. Hence this type of stress is known as hydrostatic state of stress. 1.9 STRESS TRANSFORMATION EQUATIONS Consider a D plane with normal lying in the xy plane. If n x and n y are the direction cosines, then n x cos and n y sin (... n z 0) Therefore, from Cauchy s formula y n n xy x x x xy y y (a) (b) Fig: 1.6 Normal and Shear Stress Components on an Oblique Plane.

25 Analysis of Stress and Strain 1.15 T x n x cos xy sin T y n y sin xy cos T z n 0 Since the normal stress is equal to the projection of T n along the normal n and it also equals to the sum of the projection of its components T x n, Ty n and Tz n then we get n n x T x n ny T y n nz T z n cos x cos xy sin sin y sin xy cos x cos xy sin cos y sin xy cos sin x cos y sin xy sin cos Using the following trignometric identities cos 1 cos ; sin 1 cos ; sin cos We get the transformation equations for plane stress. x 1 cos y 1 cos x x cos y y cos xy sin sin xy sin cos x y x y cos xy sin... (1) Likewise T n x y sin xy cos Equations (1) and () are the transformation equations of stress.

26 1.16 Advanced Mechanics of Solids - Further, differentiating the equation (1) with respect to we get, d d x y sin xy cos 0 sin cos xy x y tan p xy x y...() The above equation gives two planes at right angle to each other. and p is the angle at which the maximum normal stress will occur. Equation () can also be written as tan p xy x y By formula we know that, tan opp adj opp,cos, sin adj hyp hyp Therefore, cos p x y R sin p xy R Thus, by applying Pythagoras theorem, we ger x y xy R x y xy xy R x y P Substituting the values of cos and sin in equation (1)

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28 1.18 Advanced Mechanics of Solids - max 1 For three dimensional plane, the principal stresses 1, and are arranged such that 1, then the maximum shear stress acting at the point will be max STRESS INVARIANTS In engineering, it is often necessary to calculate the normal and shear stress components, acting on any arbitrary plane, with unit normal vector n, that crosses a rigid body in equilibrium. But, we have seen earlier that these normal and shear stress components can be determined on any plane with normal n using Cauchy s equations. However, these equations, cannot determine the planes on which the normal and the shear stresses are maximum. Therefore, to deal with this situation, let us assume that there is a plane n with direction cosines n x, n y and n z on which the stress is wholly normal. The components of the stress vector T acting on any plane crossing an arbitrary point inside a rigid body may be calculated as, T i n ni The components along x, y and z axes are T x n nx, T y n ny, T z n nz... (1) Also we know that from Cauchy s formula, T x n x n y xy n y xz n z... () T y n xy n x y n y yz n z T z n xz n x yz n y z n z

29 Analysis of Stress and Strain 1.19 Subtracting (1) from () we get x n x xy n y xz n z 0 xy n x y n y yz n z 0... () xz n x yz n y z n z 0 Equation () contains a set of homogeneous equations If the solution for the above equation is trivial, then n x n y n z 0 If the solution for the above equation is non-trivial, then, the determinants of the coefficients of n x, n y and n z must be equal to zero. x xy xz xy y yz 0 xz yz z Expanding the above determinant, we get the following cubic equation x y z x y y z z x xy yz xz x y z xy yz xz x yz y xz z xy 0 Now one coefficients of, and the last term can be written as l 1 l l 0 where l 1 x y z l x y y z z x xy yz xz x xy xy y y yz yz z x xz xz z x xy xz l x y z xy yz xz x yz y xz z xy xy y yz xz yz z The quantities l 1, l and l are known as the first, second and third invariants of stress respectively.

30 1.0 Advanced Mechanics of Solids - Note: 1. An invariant is one, whose value does not change even when the frame of reference is changed.. The stress invariant is an important parameter, which is commonly used in plasiticity, failure criteria and Mohr s circle etc Standard method of solution for a cubic equation Consider a cubic equation y py qy r 0 where p, q and r are constant. Substitute y x 1 p, in the above equation. The equation becomes, x 1 p p x 1 p q x 1 p r 0 x x p xp 1 7 p p x 1 9 p xp q x 1 p r 0 x x p p q 1 7 p 1 9 p 1 pq r 0 x x 1 q p 1 7 p 9pq 7r 0 The above equation can be written as, x ax b 0 where a 1 q p, b 1 7 p 9pq 7r Also put cos b a 7 1/ and g a

31 Analysis of Stress and Strain 1.1 On knowing the values of and g, one can determine the standard solution for cubic equation as, 1 y 1 g cos p y g cos 10 p y g cos 40 p 1.11 PRINCIPAL STRESSES AND PLANES In order to avoid the trivial solution, the following condition is used n x ny nz 1 It is used along with any two equations from the set of three homogeneous equations i.e., x n x xy n y xz n z 0 xy n x y n y yz n z 0 xz n x yz n y z n z 0 Hence, with each value of, there will be an associated plane. These planes on each of which the stress vector is normal are called the principal planes and the corresponding stresses are called the principal stresses. Note: The principal plane is also known as shearless plane. Since the tangential (or) shear stress component on a principal plane is zero Stress invariants in terms of principal stresses In terms of principal stresses, the stress invariants are l 1 1 l 1 1 l 1

32 1. Advanced Mechanics of Solids Principal stresses from cubic equation We know that the cubic equation in form is l 1 l l 0 This cubic equation has three real roots. Therefore the solutions obtained on solving the above cubic equation are 1, x y x y where 1, and are principal stresses xy Principal directions The principal directions associated with 1, and are obtained by substituting i, {i 1,, } in the following equations and solving for n x, n y and n z. x i n x xy n y xz n z 0 xy n x y i n y yz n z 0 n x ny nz 1 (i) (ii) If 1, and are distinct (if 1, and have different values) then the three associated principal axes i.e., n 1, n and n are unique and mutually perpendicular. If 1 and is distinct, then the axis of n is unique and every direction perpendicular to n is a principal direction associated with 1 (iii) If 1, then every direction is a principal direction and this is same as the hydrostatic or isotropic state of stress.

33 State of Stress referred to principal axes The state of stress can be characterised by the six rectangular stress components referred to a coordinate frame of reference. On considering the principal axes as the coordinate axes, we have [ ij ] We know that, on any plane with normal n, the components of the stress vector are T x n 1 n x ; T y n n y ; T z n n z Resultant stress T n T x n T y n T z n 1 nx ny nz If and are the normal and the shearing stress acting on this plane, then we have 1 n x n y n z We know that T n n x ny 1 n y nz n z nx 1 Problem 1.: The state of stress at a point is given by x 70 MPa, y 10 MPa, z 0 MPa, xy 40 MPa, yz xz 0 MPa. Determine the principal stresses and maximum shear stress. [Nov University of Kerala] Given x 70 MPa, y 10 MPa, z 0 MPa, xy 40 MPa, yz xz 0 MPa To find: (i) Principal stresses, (ii) Maximum shear stress (iii) Direction of maximum principal stress. Solution we know that Analysis of Stress and Strain 1.

34 1.4 Advanced Mechanics of Solids - x xy xz 70 ij xy y yz 40 xz yz z 0 l MPa l MPa l MPa f From y py qy r 0 cubic equation, we have y ; p 60 ; q 00 ; r a 1 q p b 1 7 p 9pq 7r

35 Analysis of Stress and Strain 1.5 Now cos b a 7 1/ / cos cos Also 4500 g a y 1 g cos p cos MPa y g cos 10 p y g cos 40 p since 1, the principal stress are cos MPa cos MPa (i) MPa, MPa, 4.65 MPa (ii) Maximum shear stress max : If 1, then max max MPa

36 1.6 Advanced Mechanics of Solids Problem 1.: At a point P, the stress tensor is T ij. All 1 4 units are in kpa. Find principal stress and check for invariance. [Feb 01 - CUSAT] Given 1 1 ij 1 4 Solution l l l l f Compare the above cubic equation with, y py qy r 0 we have y, p, q 0, r 4 a 1 q p 1 [ 60 ] 69 b 1 7 p 9pq 7r 1 7 [ ] 1 [ ] 7 b

37 Analysis of Stress and Strain 1.7 Now, cos cos b a 7 1/ / cos Now, g a g 5.54 The solutions are 1 y 1 g cos P 5.54 cos y g cos 10 p 5.54 cos y g cos 40 p 5.54 cos since 1 we have kpa, 1.95 kpa, 4. kpa

38 1.8 Advanced Mechanics of Solids - Check on the invariance The stress invariants are determined by the following stress matrix T ij l l l Problem 1.4: The state of stress at a point is characterised by the following rectangular stress components x 0, y 10, z 5, xy 10, yz 15, zx 0. Find the values of principal stress. [Dec 01 - MG University] We know that, x xy zx ij xy y yz zx yz z l 1 Sum of leading diagonals x y z l l f l 1 l l 0 f Compare y py qy r 0 with the above cubic equation, The standard method of solution is

39 y, p 5, q 75, r a 1 [q p ] 1 [ 75 5 ] Analysis of Stress and Strain [ ] b 1 7 [p 9 pq 7r] 1 7 [ ] cos b a 7 1 [ ] / / cos cos Now, g a. The solutions can be written 1 y 1 g cos p. cos kpa 5 y g cos 10 p. cos kpa

40 1.0 Advanced Mechanics of Solids - y g cos 40 p. cos kpa Since 1 we have principal stresses as 1.87 kpa, 0.9 kpa, kpa Problem 1.5: For the state of stress at a point characterised by the components (in 1000 kpa) x 1, y 4, z 10, xy, yz zx 0 Determine the principal stresses Solution x we know that [ ij ] xy zx 1 0 xy zx y yz yz z l 1 Sum of the leading diagonals l x xy xy y y yz yz z x zx zx z [48 9] [40 0] [10 0] x l xy zx xy y yz zx yz z [0 0] 0 [0 0] l 90

41 Analysis of Stress and Strain 1.1 Now, f l 1 l l 0 f Compare y py qy r 0 with the above cubic equation, we have y, p 6, q 199, r 90 a 1 q p b 1 7 p 9pq 7r cos b a 7 cos [ ] / cos / 6. Also g a/ y 1 g cos p 5.95 cos y g cos 10 p 5.95 cos y g cos 40 p 5.95 cos

42 1. Advanced Mechanics of Solids Since 1, we have principal stress as 1 1, 10, (in 1000 kpa ) 1.1 MOHR S CIRCLE FOR D STATE OF STRESS In the real world, all the objects are three dimensional (D). Therefore, the use of Mohr s circle in a three dimensional geometry can make visualization of the stress condition clearer to the designer. However, a general approach to Mohr s circle in three dimensions is complicated. But in many cases of mechanical design, this circle has played a major role in determining the stresses and the factor of safety Procedure for constructing D Mohr s circle D E F C B A 0 1 Fig 1.7 Mohr s Stress Plane 1. Draw the normal stress (horizontal) and shear stress (vertical) axes. Arrange the principal stresses algebraically, such that 1

43 Analysis of Stress and Strain 1.. Mark off 1, and values along the normal stress axis 4. Construct three circles with diameters 1, and 1 5. The maximum shear stress is equal to max The three extremum values for normal stress is 1,, and the three extremum values for the shear stress is 1,, 1 7. The planes on which the normal stresses act are called principal normal planes and the planes on which the shear stresses act are called principal shear planes. 8. The principal normal planes are free from shear stresses but the principal shear planes are not free from normal stresses. Therefore, the normal stresses associated with these principal shears are, 1 and 1 The above stresses are indicated by points D, E and F 1.1. Special cases 1. For hydrostatic state of stress, 1, the three circles collapse to a single point on the axis and every plane is a shearless plane. Fig.1.8(a) 1

44 1.4 Advanced Mechanics of Solids - When 1, then the three circles reduce to a single circle and the shear stress on any plane will not exceed 1 1 1= Fig.1.8(b). Some sketches of Mohr s circle for common stress states are given below Plane Stress Pure Shear Uniaxial Tension 1 1 Fig 1.8 (c) Uniaxial Compression Fig 1.8 (d) Plane Strain Fig 1.8 (e) Plane Stress 1 =0 1 Fig 1.8 (f) Fig 1.8 (g)

45 1.1 STATE OF PURE SHEAR Analysis of Stress and Strain 1.5 A state of pure shear is said to exit at a point, only if x y z 0 for a particular choice of the co-ordinate system. Therefore, such a state with that particular choice of frame of reference, will have the following stress matrix, 0 ij xy xz we know that, l 1 x y z xy 0 yz xz yz 0 Therefore, the state of pure shear will have l 1 0. Hence, this is the necessary condition for the state of pure shear to exist at that point. The state of pure shear is also known as deviatoric stress or simply stress deviator HYDROSTATIC AND DEVIATORIC STRESS COMPONENTS Let a coordinate system be x [ ij ] xy xz xy y yz xz yz z... (1) i.e., p 0 0 Also, let p 1 x y z 1 l 1... () The given state (equation (1)) can be resolved into two different states, x xy xz xy y yz p 0 0 x p xy xz 0 p 0 xy y p yz xz yz z 0 0 p xz yz z p The first state on the right hand side of the above equation 0 0 p 0 is a hydrostatic state. 0 p

46 1.6 Advanced Mechanics of Solids - The second state on the right hand side of the above equation is a deviatoric state. Let the first invariant for the deviatoric state be l 1 x p y p z p l 1 x y z p From equation () we have l 1 x y z 1 x y z l 1 0 If the given state is referred to principal axes, then the decomposition of the hydrostatic and deviatoric stress will be p p p 1 p p p where p l 1 Problem 1.6: The state of stress characterised by ij is given below. Resolve the given state into a hydrostatic state and deviatoric state. Determine the normal and shearing stresses on an octahedral plane. ij Solution We know that, p 1 x y z 1 Resolving into hydrostatic and deviatoric states The octahedral normal stress oct 1 l 1 6 p

47 Analysis of Stress and Strain 1.7 Octahedral shearing stress oct l 1 l 1/ l ; l oct / / 961/ oct 9 441/ OCTAHEDRAL STRESSES An octahedron is a solid having eight identical faces. All these faces are equilateral triangles and equally inclined to x, y and z axes. A plane that is equally inclined to these three axes is called an octahedral plane and it will 1 1 have n x n y n z The normal and shearing stresses occurring on these planes are called octahedral normal stress and octahedral shearing stresses. Fig.1.9 Octahedral planes

48 1.8 Advanced Mechanics of Solids - Since n x ny nz 1, an octahedral plane will have n x n y n z 1 We know that normal stress, 1 n x n y n z Also we know that shearing stress, oct 1 [ 1 ]... (1) n x ny 1 n y nz n z nx oct 9 [ 1 1 ] 1 1 oct (or) 9 oct oct l 1 6 l () oct l 1 6 l 9 oct l 1 l oct l 1 l 1/

49 Analysis of Stress and Strain 1.9 Equations (1) and () can also be expressed in terms of the six independent components of the stress i.e., x, y, z. xy, yz and zx oct 1 x y z 9 oct x y y z z x 6 xy yz xz Note: It is important to remember that the octahedral planes are defined with respect to the principal axes and not with reference to an arbitrary frame of reference. Problem 1.7: The state of stress at a point is characterised by the component x 100 MPa, y 40 MPa, z 80 MPa ; xy yz zx 0 Determine the extremum values of the shear stresses, their associated normal stress, octahedral shear stress and its associated normal stress. Solution Arranging the term algebraically such that MPa, 80 MPa, 40 MPa Extremum values of the shear stress MPa 70 MPa 10 MPa The associated normal stresses are MPa 0 MPa 90 MPa

50 1.40 Advanced Mechanics of Solids - Octahedral shear stress oct MPa The associated normal stress is oct MPa Problem 1.8: A solid shaft of diameter.16 m is subjected to a tensile force P N and torque T 5000 Nm. At point A on the surface, determine the principal stress, the octahedral shearing stress and the maximum shearing stress. [Nov Calicut] A P T Given Diameter, d.16 m ; Force, P N ; Torque, T 5000 Nm To find: (i) Principal stresses 1 and, (ii) Maximum shearing stress max (iii) Octahedral shearing stress oct

51

52 1.4 Advanced Mechanics of Solids - l x y y z z x xy yz zx N /m 4 oct [ ] 1/ oct N/m 1.16 LAME S ELLIPSOID Let Pxyz be a coordinate frame of reference at point P. On any plane with normal n, the components of the stress vector are T x n 1 n x ; T y n n y ; T z n n z... (1) Let PQ be the resultant stress vector and its length be equal to its magnitude i.e., PQ T n The coordinates x, y, z of the point Q are x T x n ; y Ty n ; z Tz n y... () n Q z P Fig: 1.10 Lame s Ellipsoid P x n Q n T

53 Analysis of Stress and Strain 1.4 Now equation (1) becomes n x T x n ; n y T n y ; n 1 z T n z n x x 1 ; n y y ; n z z... () Since n x ny nz 1, [we get (From equation ())] x 1 y z 1 This is the equation of ellipsoid referred to the principal axes. Hence, this equation of ellipsoid is called as the stress ellipsoid (or) Lame s ellipsoid. Note 1. If 1, then Lame s ellipsoid becomes an ellipsoid of revolution and the state of stress at a given point is symmetrical with respect to the third principal axis Pz.. If 1, then Lame s ellipsoid becomes a sphere EQUATIONS OF EQUILIBRIUM IN CARTESIAN CO-ORDINATES The equations of equilibrium for a continuous medium (body) are obtained by considering a very small volume element of the body, writing its equilibrium equations in terms of its dimensions and taking the limit as these dimensions shrink to zero. In cartesian coordinates, we consider a small rectangular elements with sides x, y, and z as shown in Fig.1.11(b). Since the volume element is a cubical element isolated from its parent body, we have the stress components as shown in Fig.1.11(c) Fig (a) Parent body

54 1.44 Advanced Mechanics of Solids - y + y y 4 z y x z (x,y,z) Fig (b) Cubical element isolated from parent body x z + z 1 z 6 y 5 + x Fig (c) x x The faces of the cuboid are marked as 1,, etc. On the left hand face i.e., face 1, the average stress components are x, xy and xz. On the right hand face i.e., face, the average stress components are hand face. x x x x ; xy xy x x ; xz xz x x This is because, the right hand face is x distance away from the left Following a similar procedure, the stress components on the six faces of the cubical element are as follows. Faces Face 1 (left face) Face (right face) Face (bottom face) Face 4 (Top face) x, xy, xz average stress components x x x x, xy xy x x, xz xz x x y, yx, yz y y y y, yx yx y y, yz yz y y

55 Analysis of Stress and Strain 1.45 Faces Face 5 (Back face) Face 6 (Front face) z, zx, zy average stress components z z z z, zx zx z z, zy zy z z Let the body force (with dimensions of force per unit volume) components in the x, y and z directions be x, y and z. Now, summing all the forces acting in the x-direction and setting the sum equal to zero for equilibrium, we obtain x x x x y z yx yx y y xz zx zx z z xy x x y z x y z yz x z zx x y 0 On dividing throughout by x y z, we get x x x x yx y yx y zx z zx z x x x yx y zx z 0 x x yx y zx z x 0 Performing the similar operation with the forces in the y and z direction leads to two more equations on the basis of the equality of cross shears, we have xy yx, yz zy, zx xz Therefore, the three equations of equilibrium are x x xy y xz z x 0 y y xy x yz y y 0 z z xz x yz y z 0

56 1.46 Advanced Mechanics of Solids Equations of equilibrium for a plane stress state For a plane stress state, the condition is z 0, xz 0, yz 0 and z 0 Therefore the differential equations of equilibrium become x x xy y x 0 y y xy x y Boundary conditions: Consider a two dimensional body in which the outward normal n is drawn at a boundary point P. Let F x and F y be the components of the surface forces per unit area. Then by using Cauchy s formula T x n x n x xy n y F x [For D] T y n y n y xy n x F y The above equations shows that the stress components x, y and xy are consistent with the externally applied forces at a boundary point 1.18 EQUATIONS OF EQUILIBRIUM IN POLAR CO-ORDINATES In the last section we have seen the rectangular or cartesian frame of reference for analyses. Such a frame of reference is useful, only if the body under analysis happens to possess rectangular or straight boundaries. But in many practical cases numerous problems exist where the bodies under analysis posses radial symmetry. Ex: Thick cylinder subjected to external or internal pressure. For the analysis of such problems, it is more convenient to use polar or cylindrical frame of reference. In polar coordinates, for a two dimensional plane stress, we consider the stresses acting on a small element of length dr and width rd. The thickness perpendicular to the plane is assumed to be unity. On the near