advertisement

ES240 Solid Mechanics I Z. Suo Elements of Linear Elasticity Last modified on 17 September 2006 A classic textbook is Theory of Elasticity, by S.P. Timoshenko and J.N. Goodier, McGraw-Hill, New York. Material particle. A solid is made of atoms, each atom is made of electrons, protons and neutrons, and each proton or neutron is made of... This kind of description of matter is too detailed, and is not used in the theory of elasticity. Instead, we develop a continuum theory, in which a material particle contains many atoms, and represents their average behavior. Displacement field. Any configuration of the material particles can be used as a reference configuration. The displacement is the vector by which a material particle moves relative to its position in the reference configuration. If all the material particles in the body move by the same displacement vector, the body as a whole moves by a rigid-body translation. If the material particles in the body move relative to one another, the body deforms. For example, in a bending beam, material particles on one face of the beam move apart from one another (tension), and material particles on the other face of the beam move toward one another (compression). In a vibrating rod, the displacement of each material particle is a function of time. We label each material particle by its coordinates x, y, z in the reference configuration. At time t, the material particle x, y, z has the displacement ux, y, z, t in the x-direction, vx, y, z, t in the y-direction, and wx, y, z, t in the z-direction. A function of spatial coordinates is known as a field. The displacement field is a time-dependent vector field. It is sometimes convenient to write the coordinates of a material particle in the reference configuration as x1, x2 , x3 , and the displacement vector as u1 x1, x2 , x3 , t , u2 x1, x2 , x3 , t , u3 x1, x2 , x3 , t . If we place markers on a body, the motion of the markers visualizes the displacement field and its variation with time. Strain field. Given a displacement field, we can calculate the strain field. Consider two material particles in the reference configuration: particle A at x, y, z and particle B at x dx, y, z . In the reference configuration, the two particles are distance dx apart. At a given time t, the two particles move to new locations. The x-component of the displacement of particle A is ux, y, z, t , and that of particle B is ux dx, y, z, t . Consequently, the distance between the two particles elongates by ux dx, y, z, t ux, y, z, t . The axial strain in the x-direction is x u x dx, y, z , t u x, y, z , t u . dx x This is a strain of material particles in the vicinity of x, y, z at time t. The function x x, y, z, t is a component of the strain field of the body. ux, y dy, z, t C dy ux, y, z, t The shear strain is defined as follows. Consider two lines B A of material particles. In the reference configuration, the two lines dx are perpendicular to each other. The deformation changes the included angle by some amount. This change in the angle defines the shear strain, . We now translate this definition into a strain-displacement relation. Consider three material particles A, 5/31/16 Linear Elasticity-1 ES240 Solid Mechanics I Z. Suo B, and C. In the reference configuration, their coordinates are A x, y, z , B x dx, y, z , and C x, y dy, z . In the deformed configuration, in the x-direction, particle A moves by ux, y, z, t and particle C by ux, y dy, z, t . Consequently, the deformation rotates line AC about axis z by an angle u x, y dy, z, t u x, y, z, t u . dy y Similarly, the deformation rotates line AB about axis z by an angle v x dx, y, z , t v x, y, z , t v . dx x Consequently, the shear strain in the xy plane is the net change in the included angle: xy u v . y x For a body in the three-dimensional space, the strain state of a material particle is described by a total of six components. The strains relate to the displacements as u v w x , y , z x y z u v v w w u xy , yz , zx y x z y x z Another definition of the shear strain relates the definition here by xy xy / 2 . With this new definition, we can write the six strain-displacement relations neatly as 1 u u ij i j . 2 x j xi The state of stress at a material particle. Imagine a three-dimensional body. The body may not be in equilibrium (e.g., the body may be vibrating). The material property is unspecified (e.g., the material can be solid or fluid). Imagine a material particle inside the body. What state of stress does the material particle suffer from? To talk about internal forces, we must expose them by drawing a free-body diagram. Represent the material particle by a small cube, with its edges parallel to the coordinate axes. Cut the cube out from the body to expose all the forces on its 6 faces. Define stress as force per unit area. On each face of the cube, there are three stress components, one normal to the face (normal stress), and the other two tangential to the face (shear stresses). Now the cube has six faces, so there are a total of 18 stress components. A few points below get us organized. 5/31/16 Linear Elasticity-2 ES240 Solid Mechanics I Z. Suo zz yx yy zx zy xx xy yz x y xz xz zy z yz xx zx yy yx xy zz Notation: ij . The first subscript signifies the direction of the stress component. The second subscript signifies the direction of the vector normal to the face. We label the coordinates as x, y, z . It is sometimes convenient to write the coordinates as x1, x2 , x3 . Sign convention. On a face whose normal is in the positive direction of a coordinate axis, the stress component is positive when it points to the positive direction of the axis. On a face whose normal is in the negative direction of a coordinate axis, the stress component is positive when it points to the negative direction of the axis. Equilibrium of the cube. As the size of the cube shrinks, the forces that scale with the volume (gravity, inertia effect) are negligible. Consequently, the forces acting on the cube faces must be in static equilibrium. Normal stress components form pairs. Shear stress components form quadruples. Consequently, only 6 independent stress components are needed to describe the state of stress of a material particle. Write the six stress components in a 3 by 3 symmetric matrix: 11 12 13 12 22 23 . 13 23 33 Hooke's law. For an isotropic, homogeneous solid, only two independent constants are needed to describe its elastic property: Young’s modulus E and Poisson’s ratio . In addition, a thermal expansion coefficient characterizes strains due to temperature change. When temperature changes by T , thermal expansion causes a strain T in all three directions. The combination of multi-axial stresses and a temperature change causes strains 5/31/16 Linear Elasticity-3 ES240 Solid Mechanics I Z. Suo 1 x y z T E 1 y y z x T E 1 z z x y T E x The relations for shear are 21 21 21 xy , yz yz , zx zx . E E E Example: a rubber layer pressed between two steel plates. A very thin elastic layer, of Young's modulus E and Poisson's ratio , is well bonded between two perfectly rigid plates. A thin rubber layer between two thick steel plates is a good approximation of the situation. The thin layer is compressed between the plates by a known normal stress z . Calculate all the stress and strain components in the thin layer. Solution. The stress at the edges of the elastic layer is complicated. We will neglect this edge effect, and concentrate the field away from the edges, where the field is uniform. By symmetry, the field has only the normal components and has no shear components. Also by symmetry, we note that x y xy Because the elastic layer is bonded to the rigid plate, the two strain components vanish: x y 0 . Using Hooke’s law, we obtain that 1 0 x x y z , E or x z . 1 Using Hooke’s law again, we obtain that 1 1 2 . 1 z z y x z 1 E E Consequently, the elastic layer is in a state of uniaxial strain, but all three stress components are nonzero. When the elastic is incompressible, 0.5 , it cannot be strained in just one direction, and the stress state will be hydrostatic. Traction vector. Imagine a plane inside a material. The plane has the unit normal vector n, with three components n1 , n2 , n3 . The force per area on the plane is called the traction. The traction is a vector, with three components: t1 t t2 . t3 Question: There are infinite many planes through a point. How do we determine the traction vectors on all these planes? 5/31/16 Linear Elasticity-4 ES240 Solid Mechanics I Z. Suo Answer: You can calculate the traction vector from t1 11 12 13 n1 t 2 12 22 23 n2 t3 13 23 33 n3 z xx yx y tx zx x This traction-stress relation is the consequence of the equilibrium of a tetrahedron formed by the particular plane and the three coordinate planes. Denote the areas of the four triangles by A, Ax , Ay , Az . Recall a relation from geometry: Ax Anx , Ay Any , Az Anz Balance of the forces in the x-direction requires that t x A Ax xx Ay yx Az zx . This gives the desired relation t x nx xx n y yx nz zx . An example. The stress state at a material point is described by the stress components 1 2 5 2 3 6 . 5 6 4 (a) Compute the traction vector on a plane intersecting the axes x, y and z at 1, 2 and 3, respectively. (b) Compute the magnitude of the normal stress on the plane. (c) Compute the magnitude of the shear stress on the plane. (d) Compute the direction of the shear stress on the plane. Solution. We need to find the unit vector normal to the plane. This is a problem in analytical geometry. The equation of a plane intersecting the axes x, y and z at 1, 2 and 3 is x y z 1 1 2 3 Alternatively, a plane can be defined by a given point on the plane, x0, and a unit vector normal to the plane, n. For any point x on the plane, x – x0 is a vector lying in the plane, so that n x x0 0 , or 5/31/16 Linear Elasticity-5 ES240 Solid Mechanics I Z. Suo nx x x0 n y y y0 nz z z0 0 A comparison of the two equations of the plane shows that the normal vector is in the direction 1 1 1 1 , 2 , 3 . Normalizing this vector, we obtain the unit vector normal to the plane: 6 / 7 n 3 / 7 . 2 / 7 (a) The traction vector on the plane is t x 1 2 5 6 / 7 22 / 7 t 2 3 6 3 / 7 33 / 7 y t z 5 6 4 2 / 7 56 / 7 (b) Normal stress on the plane is the traction vector projected on to the normal direction of the plane 22 6 33 3 56 2 n t n 7 7 7 7 7 7 7 (c) and (d) The shear stress in the plane is a vector: 20 1 t nn 12 . 7 42 The direction of this vector is the direction of the shear stress on the plane. The magnitude of the shear stress is 6.86. zx x, y, z dz, t z dz xx x, y, z, t yx x, y, z , t x, y, z x yx x, y dy, z, t y zx x, y, z, t xx x dx, y, z, t dy dx Stress field. Imagine the three-dimensional body again. At time t, the material particle x, y, z is under a state of stress ij x, y, z, t . Denote the distributed external force per unit volume by bx, y, z, t . An example is the gravitational force, bz g . The stress and the 5/31/16 Linear Elasticity-6 ES240 Solid Mechanics I Z. Suo displacement are time-dependent fields. Each material particle has the acceleration vector 2ui / t 2 . Cut a small differential element, of edges dx, dy and dz. Let be the density. The mass of the differential element is dxdydz . Apply Newton’s second law in the x-direction, and we obtain that dydz xx x dx, y, z, t xx x, y, z, t dxdz yx x, y dy, z, t yx x, y, z, t dxdy zx x, y, z dz, t zx x, y, z, t bx dxdydz dxdydz 2u x t 2 Divide both sides of the above equation by dxdydz, and we obtain that xx yx zx 2u x . bx 2 x y z t This is the momentum balance equation in the x-direction. Similarly, the momentum balance equations in the y- and z-direction are yx x yy y zy z by 2u y t 2 xz yz zz 2u bz 2z x y z t When the body is in equilibrium, we drop the acceleration terms from the above equations. Using the summation convention, we write the three equations of momentum balance as ij 2 ui bj 2 . x j t Lamé Problem. As an example of boundary value problems, consider a spherical cavity in a large body, remote hydrostatic tension. The symmetry of the problem makes spherical coordinate system convenient. List nonzero quantities. the radial displacement u , the radial stress r , two equal hoop stresses , the radial strain r , two equal hoop strains . They are all functions of r. List equations. Use the basic equation sheet. Simplify to the special symmetry. d r 2 r 0. Equilibrium equation: dr r du u r , . Deformation geometry: dr r Material law (elasticity, Hooke’s law): 1 1 r r 2 1 r . E E 5/31/16 Linear Elasticity-7 ES240 Solid Mechanics I Z. Suo Reduce to a single ODE. The above are a set of 5 equations for 5 functions of r. You can follow a number of approaches to solve them. I’ll follow the approach below. I want to obtain a single equation for the radial stress, r . From the equilibrium equation, I express in terms of r : r d r r . 2 dr Then I use the material law to express both strains in terms of r : 1 d r 1 2 r r r E dr 1 r d r 1 2 r 1 E 2 dr I can eliminate u from the two equations of deformation geometry, and resulting equation is in terms of the two strains, r d r / dr . Express this equation in terms of the radial stress, and I have d 2 r 4 d r 0. dr 2 r dr Solve the ODE. This is an equidimensional equation. The solution is of form r rm . Substitute r rm into the ODE, and we find two roots: m = 0 and m = -3. Consequently, the full solution is B r A 3 , r where A and B are constants to be determined by the boundary conditions. The hoop stress is given by A B . 2r 3 The boundary conditions are Prescribed remote stress: r S as r . Traction-free at the cavity surface: r 0 as r a Upon determining the two constants A and B, we obtain the stress distribution a 3 1 a 3 r S 1 , S 1 . r 2 r Plot the stresses as a function of r. Stress concentration factor. Note that the hoop stress is nonzero near the cavity surface, where the hoop stress reaches the maximum. The stress concentration factor is the ratio of the maximum stress over the applied stress. In this case, the stress concentration factor is 3/2. 5/31/16 Linear Elasticity-8 ES240 Solid Mechanics I Z. Suo Principal Stress For a given stress state at a material particle, if you cut a cube in a suitable orientation, the stress components on all the cube faces are normal to the faces, and there are no shear stresses on the cube faces. These cube faces are called the principal planes, the normal vectors to these faces the principal directions, and the stresses on them the principal stresses. Examples Uniaxial tension Equal biaxial tension Hydrostatic pressure Pure shear is the same stress state as the combination of pulling and pressing in 45°. Given a stress state, how to find the principal stresses? When a plane is the principal plane, the traction on the plane is normal to the plane, namely, the traction vector t must be in the same direction as the unit normal vector n. Let the magnitude of the traction be . On the principal plane, the traction vector is in the direction of the normal vector, t n . Write in the matrix notion, and we have t1 n1 t n . 2 2 t3 n3 Here both t and n are vectors, but is a scalar representing the magnitude of the principal stress. Recall that the traction vector is the stress matrix times the normal vector, so that 11 12 13 n1 n1 21 22 23 n2 n2 . 31 32 33 n3 n3 This is an eigenvalue problem. When the stress state is known, i.e., the stress matrix is given, solve the above eigenvalue problem to determine the eigenvalue and the eigenvector n. The eigenvalue is the principal stress, and the eigenvector n is the principal direction. Linear algebra of eigenvalues. Because the stress tensor is a 3 by 3 symmetric matrix, you can always find three real eigenvalues, i.e., principal stresses, a , b , c . We distinguish three cases: (1) If the three principal stresses are unequal, the three principal directions are orthogonal (e.g., pure shear state). (2) If two principal stresses are equal, but the third is different, the two equal principal stresses can be in any directions in a plane, and the third principal direction is normal to the plane (e.g., pure tensile state). (3) If all the three principal stresses are equal, any direction is a principal direction. This stress state is called a hydrostatic state. Maximum normal stress. Why do we care about the principal stress? Chalks are made of a brittle material: they break by tensile stress, not by shear stress. When a chalk is under bending, the tensile stress is along the axial direction of the chalk, so that the chalk breaks on a plane normal to the axial direction. When a chalk is under torsion, the maximum tensile stress is 45° from the axial direction, so that the chalk breaks in a direction 45° from the axial direction. (The fracture surface of the chalk under torsion is not a plane, because of some 3D effects.) We care about the principal stress because brittle materials fail by tensile stress, and we want to find the maximum tensile stress. 5/31/16 Linear Elasticity-9 ES240 Solid Mechanics I Z. Suo Let’s order the three principal stresses as a b c . This ordering takes into consideration the signs: a compressive stress (negative) is smaller than a tensile stress (positive). On an arbitrary plane, the traction vector may be decomposed into two components: one component normal the plane (the normal stress), and the other component parallel to the plane (the shear stress). Obviously, when you look at a plane with a different normal vector, you find different normal and shear stresses. You will be delighted by the following theorem: Of all planes, the principal plane corresponding to c has the maximum normal stress. Maximum shear stress. Why do we care about the maximum shear stress? Most metals are ductile materials: they fail by plastic yielding. When a material is under a complex stress state, it is known empirically that yielding first occurs on a plane with maximum shear stress. To find the maximum shear stress and the particular plane, you are helped by the following theorem: The maximum shear stress is max c a / 2 . max acts on a plane with the normal vector 45° from the principal directions n a and n c . A proof of the above theorems is outlined below. Consider a system of coordinates that coincide with three orthogonal directions of the principal stresses, a , b , c . Then consider an arbitrary plane whose unit normal vector has components n1 , n2 , n3 in this coordinate system. The components of the stress tensor in this coordinate system is a 0 0 0 0 . b 0 0 c Thus, on the plane with unit vector ( n1 , n2 , n3 ), the traction vector is ( a n1, b n2 , c n3 ). The normal stress on the plane is n a n12 b n22 c n32 . We need to maximize n under the constraint that n12 n22 n32 1 . The shear stress on the plane is given by 2 a n1 2 b n2 2 c n3 2 a n12 b n22 c n32 . 2 We need to maximize under the constraint that n12 n22 n32 1 . 5/31/16 Linear Elasticity-10 ES240 Solid Mechanics I Z. Suo Change of Coordinates The direction-cosine matrix relating two bases. In the 3D space, let e1, e2 and e3 be an orthonormal basis. The base vectors are ordered to follow the right-hand rule. Let e'1 , e'2 , e'3 be a new orthonormal basis. Let the angle between the two vectors e i and e' be i , and denote the direction cosine of the two vectors by li cosi . We follow the convention that the first index of li refers to a coordinate in the old basis, and the second to a coordinate in the new basis. For the two bases, there are a total of 9 direction cosines. We can list li as a 3 by 3 matrix. By our convention, the rows refer to the old basis, and the columns to the new basis. Note that li is the component of the vector e' projected on e i . We can express each new base vector as a linear combination of the three old base vectors: e' l1 e1 l2 e2 l3 e3 . If you are tired of writing sums like this, you abbreviate it as e' li ei , with the convention that a repeated index implies summation over 1,2,3. (This is called the Einstein summation convention.) Because the sum is the same whatever the repeated index is named, such an index is called a dummy index. Similarly, we can express the old basis as a linear combination of the new basis: ei li1e'1 li 2e'2 li 3e'3 . Using the summation convention, we write more concisely as ei li e' . Transformation of components of a vector due to change of basis. Let f be a vector. It is a linear combination of the base vectors: f f i ei , where f1 , f 2 , f 3 are the components of the vector, and are commonly written as a column. Consider the vector pointing from Cambridge to Boston. When the basis is changed, the vector between Cambridge and Boston remains unchanged, but the components of the vector do change. Let f '1 , f '2 , f '3 be the components of the vector f in the new basis, namely, f f ' e' Recall the transformation between the two sets of basis, ei li e' , we write that f fiei fili e' A comparison between the two expressions gives that f ' fili . Thus, the component column in the new basis is the transpose of the direction-cosine matrix times the component column in the old basis. Similarly, we can show that fi li f ' The component column in the old basis is the direction-cosine matrix times the component column in the old basis. Transformation of stress components due to change of basis. The stress tensor, , 5/31/16 Linear Elasticity-11 ES240 Solid Mechanics I Z. Suo describes the state of stress suffered by a material particle. Represent the material particle by a cube. The stress components are the force per unit area on 6 faces of the cube. The stress tensor is represented by a 3 by 3 symmetric matrix. The state of stress of a material particle is a physical object, and is independent of your choice of the basis (i.e., how you cut a cube to represent the particle). However, the stress components depend on your choice of the basis. How do we transform the stress components when the basis is changed? Consider the stress state of a material particle, and the traction vector on a given plane, first in the old basis e1, e2 and e3 and then in the new basis e'1 , e'2 , e'3 . In the old basis, denote the components of the stress state by ij , the components of the unit vector normal to the plane by n j , and the components of the traction vector on the plane by ti . convention, we write the traction-stress equations as ti ij n j . Using the summation Similarly, in the new basis, denote the components of the stress state by ' , the components of the unit vector normal to the plane by n ' , and the components of the traction vector on the plane by t ' . Force balance requires that t ' ' n' . (a) The traction is a vector, so that its components transform as t ' li ti . Insert ti ij n j into the above, and we obtain that t ' li ij n j . The unit normal vector transforms as n j l j n' . Consequently, we obtain that t ' li ijl j n' . (b) Equations (a) and (b) are valid for any choice of the plane. Consequently, we must require that ' li ijl j . Thus, the stress-component matrix in the new basis is the product of three matrices: the transpose of the direction-cosine matrix, the stress-component matrix in the old basis, and the direction-cosine matrix. Scalars, vectors, and tensors. When the basis is changed, a scalar (e.g., temperature, energy, and mass) does not change, the components of a vector transform as f ' fili , and the components of a tensor transform as ' ijli l j . This transformation defines the second-rank tensor. By analogy, a vector is a first-rank tensor, and a scalar is a zeroth-rank tensor. We can also similarly define tensors of higher ranks. A special case: the new basis and the old basis differ by an angle around the axis e3. (See Beer’s Chapter 7 for an alternative presentation.) The sign convention for follows the right-hand rule. The direction cosines are e2 e2 ' e1 e'1 cos , e1 e'2 sin , e1 e'3 0, e 2 e'1 sin , e 2 e'2 cos , e 2 e'3 0, e3 e'1 0, e3 e'2 0, e3 e'3 1. Consequently, the matrix of the direction cosines is 5/31/16 Linear Elasticity-12 e1 ' e1 ES240 Solid Mechanics I Z. Suo cos sin 0 li sin cos 0 0 0 1 The components of a vector transform as f '1 cos sin 0 f1 f ' sin cos 0 f 2 2 f '3 0 0 1 f3 Thus, f '1 f1 cos f 2 sin f '2 f1 sin f 2 cos f '3 f 3 The components of a stress state transform as '11 '12 '13 cos sin 0 11 12 13 cos ' ' '23 sin cos 0 21 22 23 sin 22 21 '31 '32 '33 0 0 1 31 32 33 0 Thus, '11 11 22 sin cos 0 0 0 1 11 22 cos 2 12 sin 2 2 2 22 11 22 '22 11 cos 2 12 sin 2 2 2 22 '12 11 sin 2 12 cos 2 2 '13 13 cos 23 sin '23 13 sin 23 cos '33 33 State of plane stress. A even more special case is that the stress components out of the plane are absent, namely, 13 23 33 0 . 22 11 22 '11 11 cos 2 12 sin 2 2 2 22 11 22 '22 11 cos 2 12 sin 2 2 2 22 '12 11 sin 2 12 cos 2 2 In Beer’s Section 7.4, these equations are represented by a graph (i.e., Mohr’s circle). In this case, e3 is one principal direction, and the principal stress in this direction is zero. To find the other two principal directions, we set '12 0 , so that the two principal directions are at the angle p from the e1 and e2 directions. This angle is given by 5/31/16 Linear Elasticity-13 ES240 Solid Mechanics I tan 2 p Z. Suo 2 12 . 11 22 The two principal stresses are given by 11 22 22 2 11 12 . 2 2 We need to order these two principal stresses and the zero principal stress in the e3 direction. The maximum shear stress is the half of the difference between the maximum principal stress and the minimum principal stress. 2 The state of strain of a material particle is characterized by a second-rank tensor. In the old basis, the coordinates of a material particle are x1, x2 , x3 , and the components of the displacement field are ui x1 , x2 , x3 , t . In the new basis, the coordinates of the same material particle are x'1 , x'2 , x'3 , and the components of the displacement field are u ' x'1 , x'2 , x'3 , t . Recall that u ' li ui and x j l j x' , so that u ' u x j u li i li l j i . x' x j x' x j Thus, the gradients of the displacement field, ui / x j , form the components of a second-rank tensor. Consequently, the state of strain, being the symmetric part of the displacement gradient, is a second-rank tensor. When the basis is changed, the components of the strain state transform as ' ijli l j . Elastic energy density function for a rod. Note that the stiffness matrix is symmetric and positive-definite. To appreciate deeper meaning of this statement, we need to consider energy stored in the elastic body. First consider a rod, of initial length L0 and cross-sectional area A0 . When a loading machine applies a force F to the rod, the length of the rod becomes L. The function FL can be determined experimentally, and need not be linear in general. When the length changes from L to L dL , the loading machine does work FdL to the rod. For an elastic solid, this work is stored as the elastic energy in the rod. That is, when the force is removed gradually, the rod will recover its initial length, and does work to the loading machine. (For a plastic solid, the work done by the machine is mainly dissipated as heat. When the force is removed, the rod will not recover to its initial length.) We will consider elastic solid here. Denote U as the elastic energy stored in the rod. The work done by the loading machine equals the increment of the elastic energy in the rod, namely, dU FdL. Define the stress and strain as F L L0 , . A0 L0 Define the elastic energy density, i.e., elastic energy per unit volume, as U w . A0 L0 5/31/16 Linear Elasticity-14 ES240 Solid Mechanics I Z. Suo Here we have used the initial area and initial length to define the stress, strain and energy density. We will be concerned with small strains, so that the difference between the initial and the final lengths is small. With these definitions, we can write dw d . The energy density is a function of the strain, w . When the strain changes from to d , the energy density changes by dw w d w . The stress is just the differential coefficient of the energy density function, namely, w . Once we know the energy density function w , we can obtain the stress-strain relation by taking the differentiation. When the rod is made of a linear elastic solid, under uniaxial load, the stress-strain relation is E . Consequently, the energy density function is 1 w E 2 2 This result holds only for linear elastic solid in uniaxial stress state. Elastic energy density function of a block under a simple shear. Consider a block of height H0 and cross-sectional area A0 . Subject to a shear force F, the block shear by an angle . When the shear angle changes from to d , the loading machine does the work FH0d to the block. For an elastic solid the work is stored as the elastic energy in the block. Define the shear stress and the (engineering) shear strain as F , A0 The energy per unit volume is a function of the shear strain, w . From the above, we have dw d . The shear stress is the differential coefficient of the energy density function. When the block is made of a linear elastic solid, under shear load, the stress-strain relation is G . Consequently, the energy density function is 1 w G 2 2 This result holds only for linear elastic solid in pure shear condition. Energy density function under multiaxial stress state. The advantage of the energy density function becomes clear under the multiaxial stress state. The state of strain is specified by the six strain components: x , y , z , yz , zx , xy In this order we will label them as 1 , 2 , 3 , 4 , 5 , 6 . The six strain components can vary independently. The elastic energy per unit volume is a function of all the six strain components, w 1 , 2, 3 , 4 , 5 , 6 . This is the energy density function. When each strain component changes by a small amount, d i , the energy density changes by dw 1d1 2d 2 3d 3 4d 4 5d 5 6d 6 Each stress component is the differential coefficient of the energy density function: w i . i 5/31/16 Linear Elasticity-15 ES240 Solid Mechanics I Z. Suo If the function w 1 , 2, 3 , 4 , 5 , 6 is known, we can determine the six stress-strain relations by the differentiations. Consequently, by introducing the energy density function, we only need to specify one function, rather than six functions, to determine the stress-strain relations. Linear elastic solids. The above considerations apply to solids with linear or nonlinear stress-strain relations. We now examine linear elastic solids. For the stress components to be linear in the strain components, the energy density function must be a quadratic form of the strain components: 1 1 w cij i j c111 1 c12 1 2 c 21 2 1 .... 2 i, j 2 Here cij are 36 constants. The cross terms come in pairs, e.g., c12 c21 1 2 . Only the combination c12 c21 will enter into the stress-strain relation, not c12 and c21 individually. We can call c12 c21 by a different name. A convenient way to say that there is only one independent constant is to just let c12 c21 . We can do the same for other pairs, namely, cij c ji . The matrix cij is symmetric, with 21 independent elements. Consequently, 21 constants are needed to specify the elasticity of a linear anisotropic elastic solid. Because the elastic energy is positive for any nonzero strain state, the matrix cij is positive-definite. Recall that each stress component is the differential coefficient of the energy density function, i w / i . The stress relation becomes i cij j . j In the matrix notation, we write 1 c11 c12 c13 c14 c15 c16 1 c 2 21 c22 c23 c24 c25 c26 2 3 c31 c32 c33 c34 c35 c36 3 4 c41 c42 c43 c44 c45 c46 4 5 c51 c52 c53 c54 c55 c56 5 6 c61 c62 c63 c64 c65 c66 6 The physical significance of the constants cij is now evident. For example, when the solid is under a uniaxial strain state, 1 0, 2 3 4 5 6 0 , the six stress components on the solid are 1 c111, 2 c211,... The matrix cij is known as the stiffness matrix. Inverting the matrix, we express the strain components in terms of the stress components: 1 s11 s12 s13 s14 s15 s16 1 s 2 21 s22 s23 s24 s25 s26 2 3 s31 s32 s33 s34 s35 s36 3 4 s41 s42 s43 s44 s45 s46 4 5 s51 s52 s53 s54 s55 s56 5 6 s61 s62 s63 s64 s65 s66 6 The matrix s ij is known as the compliance matrix. The compliance matrix is also symmetric and positive definite. 5/31/16 Linear Elasticity-16 ES240 Solid Mechanics I Z. Suo Anisotropic elasticity. An isotropic, linear elastic solid is characterized by two constants (e.g., Young’s modulus and Poisson’s ratio) to fully specify the stress-strain relation. Some solids are anisotropic, e.g., fiber reinforced composites, single crystals. Each stress component is a function of all six strain components. Consequently, 21 constants are needed to specify the elasticity of a linear anisotropic elastic solid. For a crystal of cubic symmetry, such as silicon and germanium, when the coordinates are along the cube edges, the stress-strain relations are 0 0 1 1 c11 c12 c12 0 c 0 0 2 2 12 c11 c12 0 3 c12 c12 c11 0 0 0 3 0 4 4 0 0 0 c44 0 5 0 0 0 0 c44 0 5 0 0 c44 6 6 0 0 0 The three constants c11 , c12 and c44 are independent for a cubic crystal. Isotropic solid is a special case, in which the three constants are related, c44 c11 c12 / 2 . For a fiber reinforced composite, with fibers in the x3 direction, the material is isotropic in the x2 and x3 directions. The material is said to be transversely isotropic. Five independent elastic constants are needed. The stress-strain relations are 1 c11 c12 c13 0 c 2 12 c11 c13 0 3 c13 c13 c33 0 4 0 0 0 c44 5 0 0 0 0 0 6 0 0 0 0 0 0 0 c44 0 1 0 2 3 0 0 4 5 0 c11 c12 / 2 6 0 Example: stress in an epitaxial film. Both silicon (Si) and germanium (Ge) are crystals of cubic unit cell. The edge length of the unit cell of Si is aSi 5.428Å , and that of Ge is aGe 5.658Å . A Ge film 10 nm thick is grown epitaxially (i.e. with matching atomic positions) on the [100] surface of a 100 m thick Si substrate. Calculate the stress and strain components in the Ge film. The respective elastic constants are (in GPa) Si: c11 = 165.8, c12 = 63.9, c44 = 79.6. Ge: c11 = 128.5, c12 = 48.2, c44 = 66.7. Solution. Because the Si substrate is much thicker than the Ge film, the strains in the substrate are much smaller than those in the film. We will neglect these small strains, and assume that the substrate is undeformed. Let axis 3 be normal to the film surface, and axes 1 and 2 be in the plane of the film, parallel to the cube edges of the crystal cell. To register one atom on another, Ge must be compressed in directions 1 and 2 to conform to the undeformed atomic unit cell size of Si. The two in-plane strains in the Ge film are 5/31/16 Linear Elasticity-17 ES240 Solid Mechanics I 11 22 Z. Suo aSi aGe 4% . aGe There will be an elongation normal to the film, 33 0 . All shear strains vanish. According to the generalized Hooke’s law, the stress normal to the film surface relates to the strains as 33 c11 33 c1211 c12 22 . Physically it is evident that there is no stress normal to the surface of the film, 33 0 . Inserting into the above expression, we obtain that 33 2c12 11 3% . c11 The two in-plane stress components are equal, given by 11 22 c1111 c12 22 c12 33 , or 2c122 11 22 c11 c12 11 . c11 Inserting the numerical values, we obtain that 11 22 5.6GPa . This is a huge stress, it may generate defects in the film. 5/31/16 Linear Elasticity-18 ES240 Solid Mechanics I Z. Suo 3D Elasticity: Equations in other coordinates Cylindrical Coordinates (r, , z) u, v, w are the displacement components in the radial, circumferential and axial directions, respectively. Inertia terms are neglected. r 1 r rz r 0 r r z r r 1 z 2 r 0 r r z r rz 1 z z rz 0 r r z r r u , r 1 v u, r z w , z z zr 1 v 1 w 2 z r 1 w u 2 r z r 1 1 u v v 2 r r r Spherical Coordinates (r, , ) is measured from the positive z-axis to a radius; is measured round the z-axis in a righthanded sense. u, v, w are the displacements components in the r, , directions, respectively. Inertia terms are neglected. r 1 r 1 r 1 2 r r cot 0 r r rsin r r 1 1 1 cot 3 r 0 r r rsin r r 1 1 1 3 r 2 cot 0 r r rsin r r u , 1 w w cot 1 v r 2r sin 1 u r1 v u , r 1 w wr 2 r r sin 1 w u sin v cos , r 1 r1 u v vr r sin 2 r 5/31/16 Linear Elasticity-19