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Home , Elasticity (physics), Linear elasticity

ES240 Fall 2011

ES 240

Joost J. Vlassak

School of Engineering and Applied Sciences

Harvard University

These notes are largely based on course notes put together by Prof. Suo when he taught ES 240 in 2006 and on course notes developed by Prof. Vlassak for ES 246 and ES240 (2007, 2009)

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1. Elements of

Reference Text: 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 Any configuration of the material particles can be used as a reference configuration. The 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 , 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 u(x, y, z,t) in the x-direction, v(x, y, z,t) in the y-direction, and w(x, 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 u(x, y, z,t), and that of

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particle B is u(x + dx, y, z,t). Consequently, the distance between the two particles elongates by u(x + dx, y, z,t)− u(x, y, z,t). The axial strain in the x-direction is

u(x + dx, y, z,t)− u(x, y, z,t) ∂u ε = = . x dx ∂x

This is a strain of material particles in the vicinity of (x, y, z) at time t. The function ε x (x, y, z,t) is a component of the strain field of the body. u(x, y + dy, z,t) The shear strain is defined as follows. Consider two lines of material particles. In the reference configuration, the two lines are perpendicular to each other. The changes the included C angle by some amount. This change in the angle defines the shear dy strain, γ . We now translate this definition into a strain-displacement u(x, y, z,t) relation. Consider three material particles A, B, and C. In the reference configuration, their coordinates are A (x, y, z), A B dx B(x + dx, y, z), and C(x, y + dy, z). In the deformed configuration, in the x-direction, particle A moves by u(x, y, z,t) and particle C by u(x, 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:

∂u ∂v γ = + . xy ∂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

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∂u ∂v ∂w ε = , ε = , ε = x ∂x y ∂y z ∂z

∂u ∂v ∂v ∂w ∂w ∂u γ = + , γ = + , γ = + xy ∂y ∂x yz ∂z ∂y zx ∂x ∂z

Another definition of the shear strain relates the definition here by ε xy = γ xy / 2 . With this new definition, we can write the six strain-displacement relations neatly as

1 ⎛ ∂u ∂u j ⎞ ε = ⎜ i + ⎟ . ij 2 ⎜ x x ⎟ ⎝ ∂ j ∂ i ⎠

More generally, consider a body referred to an orthogonal set of axes as shown in the figure. The displacements at any point in this body during deformation are fully described by the displacements components (u1, u2, u3), which are assumed to be continuous functions of the coordinates (x1, x2, x3). Consider two arbitrary neighboring points P1 (x1, x2, x3) and P2 (x1+dx1, x2+dx2, x3+dx3) in the unstrained body and assume P1 and P2 undergo displacements (u1, u2, u3) and (u1+du1, u2+du2, u3+du3), respectively. Expanding the displacement at P2 in a Taylor series about P1, we find:

∂u1 ∂u1 ∂u1 du1 = dx1 + dx2 + dx3, ∂x1 ∂x2 ∂x3 ∂u2 ∂u2 ∂u2 du2 = dx1 + dx2 + dx3, ∂x1 ∂x2 ∂x3 ∂u3 ∂u3 ∂u3 du3 = dx1 + dx2 + dx3. ∂x1 ∂x2 ∂x3

The matrix [ui,j] is the relative displacement matrix and is in general not symmetric. Here the comma indicates differentiation with respect to the corresponding coordinate. These equations

x P' (x +dx +u +du ,x +dx +u +du ,x +dx +u +du ) 3 2 1 1 112 2 2 2 3333 P2 ds dso (x +u ,x +u ,x +u ) P'1 1 1 2 2 3 3 P1

x2

x1

Displacements in deformed body.

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can be rewritten as follows:

∂u1 1 ⎛ ∂u1 ∂u2 ⎞ 1 ⎛ ∂u1 ∂u3 ⎞ 1 ⎛ ∂u1 ∂u2 ⎞ 1 ⎛ ∂u1 ∂u3⎞ du1 = dx1 + ⎜ + ⎟ dx2 + ⎜ + ⎟ dx3 + ⎜ − ⎟ dx2 + ⎜ − ⎟ dx3, ∂x1 2 ⎝ ∂x2 ∂x1 ⎠ 2 ⎝ ∂x3 ∂x1 ⎠ 2 ⎝ ∂x2 ∂x1 ⎠ 2 ⎝ ∂x3 ∂x1 ⎠

1 ⎛ ∂u2 ∂u1 ⎞ ∂u2 1 ⎛ ∂u2 ∂u3 ⎞ 1 ⎛ ∂u2 ∂u1 ⎞ 1 ⎛ ∂u2 ∂u3⎞ du2 = ⎜ + ⎟ dx1 + dx2 + ⎜ + ⎟ dx3 + ⎜ − ⎟ dx1 + ⎜ − ⎟ dx3, 2 ⎝ ∂x1 ∂x2 ⎠ ∂x2 2 ⎝ ∂x3 ∂x2 ⎠ 2 ⎝ ∂x1 ∂x2 ⎠ 2 ⎝ ∂x3 ∂x2 ⎠

1 ⎛ ∂u3 ∂u1⎞ 1 ⎛ ∂u3 ∂u2 ⎞ ∂u3 1 ⎛ ∂u3 ∂u1 ⎞ 1 ⎛ ∂u3 ∂u2 ⎞ du3 = ⎜ + ⎟ dx1 + ⎜ + ⎟ dx2 + dx3 + ⎜ − ⎟ dx1 + ⎜ − ⎟ dx2. 2 ⎝ ∂x1 ∂x3⎠ 2 ⎝ ∂x2 ∂x3 ⎠ ∂x3 2 ⎝ ∂x1 ∂x3 ⎠ 2 ⎝ ∂x2 ∂x3 ⎠ Each of the terms in these expressions has a simple physical meaning and represents either a component of the strain matrix [εij] or a component of the rotation matrix [ωij], where: 1 εij = (ui, j + uj,i), 2 1 ω = u − u . ij 2 ( i,j j,i ) Note that the strain matrix is symmetric, while the rotation matrix is anti-symmetric. It follows therefore that if we decompose the relative displacement matrix into symmetric and anti- symmetric parts, the symmetric part represents pure deformation, while the anti-symmetric part represents a rigid body motion.

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The state of 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 ). Imagine a material particle inside the body. What state of stress does the material particle suffer from?

To talk about internal , 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 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.

Notation: σij . The first subscript signifies the direction of the vector normal to the face. The second subscript signifies the direction of the stress component. 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.

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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. 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:

⎡σ11 σ12 σ13 ⎤ ⎢ ⎥ . ⎢σ12 σ 22 σ 23 ⎥ ⎣⎢σ13 σ 23 σ 33 ⎦⎥

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 ⎢t ⎥ . = ⎢ 2 ⎥ ⎣⎢t3 ⎦⎥

Question: There are infinite many planes through a point. How do we determine the traction vectors on all these planes?

Answer: You can calculate the traction vector from

⎡t1 ⎤ ⎡σ11 σ12 σ13 ⎤⎡n1 ⎤ ⎢t ⎥ ⎢ ⎥⎢n ⎥ ⎢ 2 ⎥ = ⎢σ12 σ 22 σ 23 ⎥⎢ 2 ⎥ ⎣⎢t3 ⎦⎥ ⎣⎢σ13 σ 23 σ 33 ⎦⎥⎣⎢n3 ⎦⎥

We now see the merit of writing the components of a state of stress as a matrix. Also, the six components are indeed sufficient to characterize a state of stress, because the six components allow us to calculate the traction vector on any plane.

Proof of the stress-traction relation. 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:

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Ax = Anx , Ay = Any , Az = Anz

Balance of the forces in the x-direction requires that

tx A = σ xx Ax + σ xy Ay + σ xz Az .

This gives the desired relation

tx = σ xxnx + σ xyny + σ xznz .

This relation can be rewritten using the index notation:

t1 = σ11n1 + σ12n2 + σ13n3 .

It can be further rewritten using the summation convention:

t1 = σ1 jn j .

Similar relations can be obtained for the other components of the traction vector:

t2 = σ 21n1 + σ 22n2 + σ 23n3

t3 = σ 31n1 + σ 32n2 + σ 33n3

The three equations for the three components of the traction vector can be written collectively in the matrix form, as give in the beginning of this section. Alternatively, they can be written as

ti = σ ijn j .

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Here the summation is implied for the repeated index j. The above expression represents three equations.

Example: stress and traction. A material particle is in a state of stress with the following 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

nx (x − x0 )+ ny (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⎤ , , . Normalizing this vector, we obtain the unit vector normal to the plane: ⎣⎢1 2 3⎦⎥

⎡6/ 7⎤ n ⎢3/ 7⎥ . = ⎢ ⎥ ⎣⎢2/ 7⎦⎥

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(a) The traction vector on the plane is

⎡tx ⎤ ⎡1 2 5⎤⎡6 / 7⎤ ⎡22 / 7⎤ ⎢t ⎥ ⎢2 3 6⎥⎢3/ 7⎥ ⎢33/ 7⎥ ⎢ y ⎥ = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣⎢tz ⎦⎥ ⎣⎢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 σ = t ⋅n = × + × + × = 7 n 7 7 7 7 7 7

(c) and (d) The shear stress in the plane is a vector:

⎡− 20⎤ 1 τ = t −σ n = ⎢ 12 ⎥ . n 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.

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Stress field and balance Imagine the three-dimensional body again. At time t, the material particle (x, y, z) is under a

state of stress σ ij (x, y, z,t). Denote the distributed external force per unit volume by b(x, y, z,t).

An example is the gravitational force, bz = −ρg. The stress and the displacement are time- 2 2 dependent fields. Each material particle has the acceleration vector ∂ ui / ∂t . Cut a small differential element, of edges dx, dy and dz. Let ρ be the density. The mass of the differential element is ρdxdydz . Apply Newton’s second law in the x-direction, and we obtain that

dydz[σ xx(x + dx, y, z,t) − σ xx(x, y, z,t)] +dxdz[σ yx(x, y + dy, z,t) − σ yx(x, y, z,t)] ∂2u +dxdy σ x, y, z + dz,t − σ x, y, z,t + b dxdydz = ρdxdydz [ zx( ) zx( )] x 2 ∂t

Divide both sides of the above equation by dxdydz, and we obtain that 2 ∂σ ∂σ yx ∂σ ∂ u xx + + zx + b = ρ . x 2 ∂x ∂y ∂z ∂t

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This is the momentum balance equation in the x-direction. Similarly, the momentum balance equations in the y- and z-direction are

2 ∂σ xy ∂σ yy ∂σ zy ∂ v + + + b = ρ y 2 ∂x ∂y ∂z ∂t

2 ∂σ ∂σ yz ∂σ ∂ w xz + + zz + b = ρ z 2 ∂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

2 ∂σ ij ∂ ui + bj = ρ 2 . ∂x j ∂t

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Hooke's law for isotropic materials For an isotropic, homogeneous solid, only two independent constants are needed to describe its elastic property: Young’s modulus E and Poisson’s ratio ν. In addition, a coefficient α characterizes strains due to temperature change. When temperature changes by ΔT , thermal expansion causes a strain αΔT in all three directions. The combination of multi- axial stresses and a temperature change causes strains

1 ε = ⎡σ − ν σ + σ ⎤ + αΔT xx E ⎣ xx ( yy zz )⎦ 1 ε yy = ⎡σ yy − ν (σ zz + σ xx )⎤ + αΔT E ⎣ ⎦ 1 ε = ⎡σ − ν σ + σ ⎤ + αΔT zz E ⎣ zz ( xx yy )⎦

The relations for shear are

2(1+ν ) 2(1+ν ) 2(1+ν ) γ = σ ,γ = σ ,γ = σ . xy E xy yz E yz zx E zx

Recall the notation ε xy = γ xy / 2 , and we have

1+ν 1+ν 1+ν ε = σ ,ε = σ ,ε = σ xy E xy yz E yz zx E zx

Index notation and summation convention. The six stress-strain relation may be written as

1+ ν ν ε = σ − σ δ . ij E ij E kk ij

The symbol δij stands for 0 when i ≠ j and for 1 when i = j . We adopt the convention that a repeated index implies a summation over 1, 2 and 3. Thus, σ kk = σ11 + σ 22 + σ 33 .

Homogeneity. When talking about homogeneity, you should think about at least two length scales: a large (macro) length scale, and a small (micro) length scale. A material is said to be homogeneous if the macro-scale of interest is much larger than the scale of microstructures. A fiber-reinforced material is regarded as homogeneous when used as a component of an airplane, but should be thought of as heterogeneous when its fracture mechanism is of interest. Steel is

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usually thought of as a homogeneous material, but really contains numerous voids, particles and grains.

Isotropy. A material is isotropic when response in one direction is the same as in any other direction. and in polycrystalline form are isotropic at macro-scale, even though their constituents—grains of single crystals—are anisotropic. Wood, single crystals, uniaxial fiber reinforced composites are anisotropic materials.

Example: a rubber layer pressed between two steel plates. A very thin elastic layer, of Young's modulus E and Poisson's ratio ν, 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 σz. Calculate all the stress and strain components in the thin layer.

Solution. The stress state at the edges of the elastic layer is complicated. We will neglect this edge effect, and focus on the field away from the edges, where the field is uniform. This emphasis makes sense if we are interested in, for example, the displacement of one plate relative to the other. Of course, this emphasis is misplaced if we are concerned of debonding of the layer from the plates, as debonding may initiate from the edges, where stresses are high.

By , the field has only the normal components and has no shear components. Also by symmetry, we note that

σ x = σ y

Because the elastic layer is bonded to the rigid plate, the two strain components vanish:

ε x = ε y = 0 .

Using Hooke’s law, we obtain that

1 0 = ε x = (σ x −νσ y −νσ z ), E or

ν σ = σ . x 1−ν z

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Using Hooke’s law again, we obtain that

1 (1+ν )(1− 2ν ) ε z = (σ z −νσ y −νσ x )= σ z . E (1−ν )E

Consequently, the elastic layer is in a state of uniaxial strain, but all three stress components are nonzero. When the elastic layer is incompressible, ν = 0.5 , it cannot be strained in just one direction, and the stress state will be hydrostatic.

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Summary of Elasticity Concepts

Players: Fields Stress : stress state must be represented by 6 components (directional property). Strain tensor: strain state must be represented by 6 components (directional property). Stress field: stress state varies from particle to particle (positional property).

Stress field is represented by 6 functions, σ xx(x, y, z;t) , σ xy(x, y,z;t)… Displacement field is represented by 3 functions, u(x, y, z;t),v(x, y,z;t), w(x, y, z;t) .

Strain field is represented by 6 functions, ε xx (x, y,z;t),ε xy (x, y,z;t),.... Rules: 3 elements of solid mechanics Momentum balance Deformation geometry Material law

Complete equations of elasticity: Partial differential equations Boundary conditions - Prescribe displacement. - Prescribe traction. Initial conditions: For dynamic problems (e.g., and wave propagation), one also need prescribe initial displacement and velocity fields. Solving boundary value problems: ODE and PDE Idealization, analytical solutions: e.g., S.P. Timoshenko and J.N. Goodier, Theory of Elasticity, McGraw-Hill, New York Handbook solutions: R.E. Peterson, Stress Concentration Factors, John Wiley, New York, 1974. 2nd edition by W.D. Pilkey, 1997 Brute force, numerical methods: finite element methods, boundary element methods

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3D Elasticity: Collected Equations

Momentum balance

∂σ ∂σ ∂σ ∂ 2u x + xy + xz + b = ρ ∂x ∂y ∂z x ∂t2 ∂σ ∂σ ∂σ ∂ 2v yx + y + yz + b = ρ ∂x ∂y ∂z y ∂t2 2 ∂σ ∂σ zy ∂σ ∂ w zx + + z + b = ρ ∂x ∂y ∂z z ∂t2

Strain-displacement relation

∂u 1 ⎛ ∂v ∂w⎞ ε = , ε = ⎜ + x ∂x yz 2 ⎝ ∂z ∂y ⎠ ∂v 1 ⎛ ∂w ∂u ε = , ε = + ⎞ y ∂y zx 2 ⎝ ∂x ∂z ⎠ ∂w 1 ⎛ ∂u ∂v ⎞ ε = , ε = ⎜ + z ∂z xy 2 ⎝ ∂y ∂x ⎠

Hooke's Law

1 1 ε = σ − ν σ + σ + αΔT, ε = + ν σ x E [ x ( y z)] zx E zx 1 1 + ν ε = σ − ν σ + σ + αΔT, ε = σ y E [ y ( z x)] xy E xy 1 1 + ν ε = σ − ν σ + σ + αΔT, ε = σ z E [ z ( x y)] yz E yz

Stress-traction relation

⎡t1 ⎤ ⎡σ11 σ12 σ13 ⎤ ⎡n1 ⎤ ⎢t ⎥ = ⎢σ σ σ ⎥ ⎢n ⎥ ⎢ 2 ⎥ ⎢ 21 22 23 ⎥ ⎢ 2 ⎥ ⎣⎢t3 ⎦⎥ ⎣⎢σ 31 σ 32 σ 33 ⎦⎥ ⎣⎢n3 ⎦⎥

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3D Elasticity: Equations in other coordinates

1. Cylindrical Coordinates (r, θ, z)

Momentum balance u, v, w are the displacement components in the radial, circumferential and axial directions, respectively. Inertia and body force terms are neglected.

∂σ 1 ∂σ ∂σ σ − σ r + rθ + rz + r θ = 0 ∂r r ∂θ ∂z r ∂σ 1 ∂σ ∂σ 2σ rθ + θ + θz + rθ = 0 ∂r r ∂θ ∂z r ∂σ 1 ∂σ ∂σ σ rz + θz + z + rz = 0 ∂r r ∂θ ∂z r

Strain-displacement relation

∂u 1 ⎛ ∂v 1 ∂w⎞ ε = , ε = + r ∂r θz 2 ⎝ ∂z r ∂θ ⎠ 1 ⎛ ∂v 1 ⎛ ∂w ∂u ε = + u⎞ , ε = + ⎞ θ r ⎝ ∂θ ⎠ zr 2 ⎝ ∂r ∂z⎠ ∂w 1 ⎛ 1 ∂u ∂v v ε = , ε = + − ⎞ z ∂z rθ 2 ⎝ r ∂θ ∂r r ⎠

2. Spherical Coordinates (r, θ, φ)

θ is measured from the positive z-axis to a radius; φ is measured round the z-axis in a right- handed sense. u, v, w are the displacements components in the r, θ, φ directions, respectively. Inertia terms are neglected.

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Momentum balance

∂σ 1 ∂σ 1 ∂σ r 1 r + rθ + φ + 2σ − σ − σ − σ cotθ = 0 ∂r r ∂θ rsinθ ∂φ r ( r φ θ rθ )

∂σ 1 ∂σ 1 ∂σθφ 1 rθ + θ + + ⎡ σ − σ cotθ + 3σ ⎤ = 0 ∂r r ∂θ rsinθ ∂φ r ⎣( θ φ ) rθ ⎦ ∂σ 1 ∂σ 1 ∂σ 1 rφ + θφ + φ + 3σ + 2σ cotθ = 0 ∂r r ∂θ rsinθ ∂φ r ( rφ θφ )

Strain-displacement relation

∂u 1 ⎛ ∂w 1 ∂v ⎞ ε = ε = − wcotθ + r ∂r θϕ 2r ⎝⎜ ∂θ sinθ ∂ϕ ⎠⎟ 1 ⎛ ∂v ⎞ 1 ⎛ ∂w w 1 ∂u ⎞ εθ = ⎜ + u⎟ εϕr = − + r ⎝ ∂θ ⎠ 2 ⎝⎜ ∂θ r r sinθ ∂ϕ ⎠⎟ 1 ⎛ ∂w ⎞ 1 ⎛ 1 ∂u ∂v v⎞ εϕ = + usinθ + vcosθ εrθ = ⎜ + − ⎟ r sinθ ⎝⎜ ∂ϕ ⎠⎟ 2 ⎝ r ∂θ ∂r r ⎠

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Example: Lamé Problem.

As an example of a boundary value problem, consider a spherical cavity in a large body, subjected to remote hydrostatic tension. The symmetry of the problem makes the use of a spherical coordinate system convenient. a. List nonzero quantities.

• the radial displacement u ,

• the radial stress σr , two equal hoop stresses σθ = σφ ,

• the radial strain εr , two equal hoop strains εθ = εφ .

• They are all functions of r. b. List equations. Use the basic equation sheet and simplify taking into account the symmetry of the problem:

dσ σ −σ • Equilibrium equation: r + 2 r θ = 0 . dr r

du u • Deformation geometry: ε = ; ε = . r dr θ r

1 1 • Material law: εr = (σ r − 2νσθ ); εθ = ⎡(1− ν)σθ − νσ r ⎤ . E E ⎣ ⎦ c. 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. We'll take the approach below. We want to obtain a single equation in the radial stress, σr . From the equilibrium equation, we express σθ in terms of σr :

r dσ σ = σ + r . θ r 2 dr

Then we use the material law to express both strains in terms of σr :

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1 ⎡ dσ r ⎤ ε r = (1− 2ν )σ r −νr E ⎣⎢ dr ⎦⎥

1 ⎡ r dσ r ⎤ εθ = (1− 2ν )σ r + (1−ν ) E ⎣⎢ 2 dr ⎦⎥

We can eliminate u from the two equations for deformation geometry. This results in an equation in terms of the two strains:

ε r = d(rεθ )/ dr .

Express this equation in terms of the radial stress, and we get

d 2σ 4 dσ r + r = 0 . dr 2 r dr

m d. Solve the ODE. This is an equidimensional equation. The solution is of the form σr = r . m Substitute σr = r into the ODE, and we find two roots: m = 0 and m = -3. Consequently, the full solution is

B σ = A + , r r3 where A and B are constants to be determined by the boundary conditions. The hoop stress is given by

B σ = A − . θ 2r3

The boundary conditions are

• Prescribed remote stress: σr = S as r = ∞.

• Traction-free at the cavity surface: σr = 0 as r = a

Upon determining the two constants A and B, we obtain the stress distribution

3 3 ⎡ ⎛ a ⎞ ⎤ ⎡ 1 ⎛ a ⎞ ⎤ σ r = S⎢1− ⎜ ⎟ ⎥, σθ = S⎢1+ ⎜ ⎟ ⎥ . ⎣⎢ ⎝ r ⎠ ⎦⎥ ⎣⎢ 2 ⎝ r ⎠ ⎦⎥

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Stress concentration factor. Note that the tangential stress is nonzero near the cavity surface, where it stress reaches a 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.

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A note on compatibility

The six independent components of strain at each point are completely determined by the displacement field u (u, v, w). As a result it is not possible to choose the six strain components independently – they have to satisfy a number of equations that arise when the displacement components are eliminated from the equations defining the strain components. In other words, one can ask the following question: If I choose six strain components (or six functions corresponding to these components), does there exist a corresponding displacement field. The answer to this question is simple: if I choose the functions arbitrarily, there does not generally exist a corresponding displacement field.

The conditions that need to be satisfied by the strain components in order to get a corresponding displacement field can be found by eliminating the displacement components u, v, w from the definitions of the strains. From the definitions of the various strain components, we get:

2 3 2 3 2 3 3 ∂ ε ∂ u ∂ ε yy ∂ v ∂ γ xy ∂ u ∂ v xx = , = , = + , ∂y2 ∂x∂y2 ∂x2 ∂y∂x2 ∂x∂y ∂x∂y2 ∂y∂x2 from which we find that

2 2 2 ∂ ε ∂ ε yy ∂ γ xy xx + = . ∂y2 ∂x2 ∂x∂y

Two more relations of the same kind can be found by cyclical permutation of the letters x, y and z. From the derivatives

2 3 2 2 ∂ ε ∂ u ∂γ xy ∂ u ∂ v xx = , = + , ∂y∂z ∂x∂y∂z ∂z ∂y∂z ∂x∂z

2 3 2 2 ∂γ ∂ u ∂ w ∂γ yz ∂ v ∂ w xz = + , = + , ∂y ∂y∂z ∂x∂y ∂x ∂x∂z ∂x∂y it follows further that

2 ∂ ε ∂ ⎛ ∂γ yz ∂γ ∂γ xy ⎞ 2 xx = − + xz + . ∂y∂z ∂x ⎝⎜ ∂x ∂y ∂z ⎠⎟

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Two more equations of this kind can be derived through cyclical permutation of the coordinates. These equations are known as the compatibility equations or the conditions of compatibility because they tell you under what conditions a real displacement field exists. One can show that for simply connected regions (i.e., materials without holes), the compatibility equations are necessary and sufficient to assure that the displacements exist and are single-valued. If the region is multiply connected, additional conditions need to be imposed. It should be emphasized that these equations are completely superfluous when you treat the displacements as variables in the problem; you only need them if you want to use the strain components instead.

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Principal Stress

Imagine a material particle in a state of stress. The state of stress is fixed, but we can represent the material particle in many ways by cutting cubes in different orientations. For any given state of stress, it is always possible to cut a cube in a suitable orientation, such that 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

• Pure shear is the same state of stress 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 σ . On the principal plane, the traction vector is in the direction of the normal vector, t = σn . Write in the matrix notion, and we have

⎡t1 ⎤ ⎡n1 ⎤ ⎢t ⎥ ⎢n ⎥ . ⎢ 2 ⎥ = σ ⎢ 2 ⎥ ⎣⎢t3 ⎦⎥ ⎣⎢n3 ⎦⎥

Here both t and n are vectors, but σ 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

⎡σ11 σ12 σ13 ⎤⎡n1 ⎤ ⎡n1 ⎤ ⎢ ⎥⎢n ⎥ ⎢n ⎥ . ⎢σ 21 σ 22 σ 23 ⎥⎢ 2 ⎥ = σ ⎢ 2 ⎥ ⎣⎢σ 31 σ 32 σ 33 ⎦⎥⎣⎢n3 ⎦⎥ ⎣⎢n3 ⎦⎥

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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 σ and the eigenvector n. The eigenvalue σ 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, σ a ,σ b ,σ 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 maximum normal stress? Chalk is made of a brittle material: it break by tensile stress, not by shear stress. When 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 chalk is under torsion, the maximum tensile stress is 45° from the axial direction, so that it 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.

Let’s order the three principal stresses as σ a ≤ σ b ≤ σ 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 σc 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

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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 τ max = (σ c − σ a ) / 2 . The maximum shear stress τ max acts on a plane with the normal vector 45° from the principal directions na and nc.

A proof of the above theorems is outlined below:

Consider a system of coordinates that coincide with three orthogonal directions of the principal stresses, σ a ,σ b ,σ 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

⎡σ a 0 0 ⎤ ⎢ 0 0 ⎥ . ⎢ σ b ⎥ ⎣⎢ 0 0 σ c ⎦⎥

Thus, on the plane with unit vector ( n1,n2 ,n3 ), the traction vector is (σ an1,σ bn2 ,σ cn3 ). The normal stress on the plane is

2 2 2 σ n = σ an1 + σ bn2 + σ cn3 .

2 2 2 We need to maximize σ n under the constraint that n1 + n2 + n3 = 1.

The shear stress on the plane τ is given by

2 2 2 2 2 2 2 2 τ = (σ an1 ) + (σ bn2 ) + (σ cn3 ) − (σ an1 + σ bn2 + σ cn3 ) .

2 2 2 We need to maximize τ under the constraint that n1 + n2 + n3 = 1.

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Coordinate transformations and

1. The direction-cosine matrix relating two bases.

In 3D-space, let e1, e2 and e3 be an orthonormal basis, namely,

ei ⋅e j = δij .

The base vectors are ordered to follow the right-hand rule. Let e'1 ,e'2 ,e'3 be a new orthonormal basis, namely,

eα′ ⋅ e′β = δαβ .

Let the angle between the two vectors ei and e'α be θiα . Denote the direction cosine of the two vectors by

liα = ei ⋅e'α = cosθiα .

We follow the convention that the first index of liα 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 liα 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 liα is the component of the vector e'α projected on the vector ei . We can express each new base vector as a linear combination of the three old base vectors:

e'α = l1α e1 + l2α e2 + l3α e3 .

If you are tired of writing sums like this, you abbreviate it as

e'α = liα ei , with the convention that a repeated index implies summation over 1,2,3. 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 +li2e'2 +li3e'3 .

Using the summation convention, we write more concisely as

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ei = liα e'α .

2. 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 = fiei ,

where f1, f2 , f3 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 'α e'α

Recall the transformation between the two sets of basis, ei = liα e'α , we write that

f = fiei = filiα e'α

A comparison between the two expressions gives that

f 'α = filiα .

In matrix notation, the component column in the new basis is the of the direction- cosine matrix times the component column in the old basis.

Similarly, we can show that

fi = liα f 'α , or in matrix notation, the component column in the old basis is the direction-cosine matrix times the component column in the old basis.

3. Transformation of stress components due to change of basis.

The stress tensor, σ, 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 , and is independent of your choice of the basis (i.e., how

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you cut a cube to represent the particle). However, the components of the stress tensor do 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. In the old basis e1, e2 and e3, denote the components of the stress state by σ 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 . Using the summation convention, we write the traction-stress equations as

ti = σ ijn j .

Recall that we obtained this relation by the balance of forces on a tetrahedron. In the language of linear algebra, we call the stress as a linear operator that maps the unit normal vector of a plane to the traction vector acting on the plane.

Similarly, in the new basis e'1 ,e'2 ,e'3 , denote the components of the stress state by σ 'αβ , the components of the unit vector normal to the plane by n'β , and the components of the traction vector on the plane by t'α . Force balance requires that

t'α = σ 'αβ n'β . (a)

We now examine the relations between the components in the old basis and those in the new basis. The traction is a vector, so that its components transform as t'α = liα ti . Insert ti = σ ijn j into the above, and we obtain that t'α = liασ ijn j . The unit normal vector transforms as n j = l jβ n'β . Consequently, we obtain that

t'α = liασ ijl jβ n'β . (b)

Equations (a) and (b) are valid for any choice of the plane. Consequently, we must require that

σ 'αβ = liασ ijl jβ .

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.

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A special case: the new basis and the old basis differ by an angle θ around the axis e3

The sign convention for θ follows the right-hand rule. The direction cosines are

e1 ⋅e'1 = cosθ,e1 ⋅e'2 = −sinθ,e1 ⋅e'3 = 0,

e2 ⋅e'1 = sinθ,e2 ⋅e'2 = cosθ,e2 ⋅e'3 = 0,

e3 ⋅e'1 = 0,e3 ⋅e'2 = 0,e3 ⋅e'3 = 1.

Consequently, the matrix of the direction cosines is

⎡cosθ − sinθ 0⎤ l = ⎢sinθ cosθ 0⎥ [iα ] ⎢ ⎥ ⎣⎢ 0 0 1⎦⎥

The components of a vector transform as

⎡ f '1 ⎤ ⎡ cosθ sinθ 0⎤⎡ f1 ⎤ ⎢ f ' ⎥ ⎢ sin cos 0⎥⎢ f ⎥ ⎢ 2 ⎥ = ⎢− θ θ ⎥⎢ 2 ⎥ ⎣⎢ f '3 ⎦⎥ ⎣⎢ 0 0 1⎦⎥⎣⎢ f3 ⎦⎥

Thus,

f '1 = f1 cosθ + f2 sinθ

f '2 = − f1 sinθ + f2 cosθ

f '3 = f3

The components of a stress state transform as

⎡σ '11 σ '12 σ '13 ⎤ ⎡ cosθ sinθ 0⎤⎡σ11 σ12 σ13 ⎤⎡cosθ − sinθ 0⎤ ⎢ ' ' ' ⎥ ⎢ sin cos 0⎥⎢ ⎥⎢sin cos 0⎥ ⎢σ 21 σ 22 σ 23 ⎥ = ⎢− θ θ ⎥⎢σ 21 σ 22 σ 23 ⎥⎢ θ θ ⎥ ⎣⎢σ '31 σ '32 σ '33 ⎦⎥ ⎣⎢ 0 0 1⎦⎥⎣⎢σ 31 σ 32 σ 33 ⎦⎥⎣⎢ 0 0 1⎦⎥

Thus,

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σ + σ σ −σ σ ' = 11 22 + 11 22 cos2θ + σ sin 2θ 11 2 2 12 σ + σ σ −σ σ ' = 11 22 − 11 22 cos2θ −σ sin 2θ 22 2 2 12 σ −σ σ ' = − 11 22 sin 2θ + σ cos2θ 12 2 12

σ '13 = σ13 cosθ + σ 23 sinθ

σ '23 = −σ13 sinθ + σ 23 cosθ

σ '33 = σ 33

State of plane stress.

An even more special case is that the stress components out of the plane are absent, namely,

σ13 = σ 23 = σ 33 = 0 .

σ + σ σ −σ σ ' = 11 22 + 11 22 cos2θ + σ sin 2θ 11 2 2 12 σ + σ σ −σ σ ' = 11 22 − 11 22 cos2θ −σ sin 2θ 22 2 2 12 σ −σ σ ' = − 11 22 sin 2θ + σ cos2θ 12 2 12

In Beer’s Section 7.4, these equations are represented by graphically (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 σ '12 = 0 , so that the two principal directions are at the angle

θ p from the e1 - and e2 -directions. This angle is given by

2σ12 tan2θ p = . σ11 − σ 22

2 σ11 + σ 22 ⎛σ11 −σ 22 ⎞ 2 The two principal stresses are given by ± ⎜ ⎟ + σ12 . 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.

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4. 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 'α = filiα , and the components of a tensor transform as

σ 'αβ = σ ijliα l jβ .

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.

Multilinear algebra. The rule of the above transformation is based on only one fact: the stress is a linear map from one vector to another vector. You can generate a new tensor from a linear map from one tensor to another tensor. You can also generate a tensor by a bilinear form, e.g., a bilinear map from two vectors to a scalar. Of course, a multi-linear map of several tensors to a tensor is yet another tensor. Consequently, all tensors follow a similar rule under a change of basis. We write this rule again for a third-rank tensor:

g'αβγ = gijkliα l jβ lkγ .

Invariants of a tensor. A scalar is invariant under any change of basis. When the basis changes, the components of a vector change, but the length of the vector is invariant. Let f be a vector, and fi be the components of the vector for a given basis. The length of the vector is the square root of

fi fi .

The index i is dummy. Thus, this combination of the components of a vector is a scalar, which is invariant under any change of basis. For a vector, there is only one independent invariant. Any other invariant of the vector is a function of the length of the vector.

This observation can be extended to high-order tensors. By definition, an invariant of a tensor is a scalar formed by a combination of the components of the tensor. For example, for a symmetric second-rank tensor σ ij , we can form three independent invariants:

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σ ii , σ ijσ ij , σ ijσ jkσ ki .

In each case, all indices are dummy, resulting in a scalar. Any other invariant of the tensor is a function of the above three invariants.

Exercise. For a nonsymmetric second-rank tensor, give all the independent invariants. Write each invariant using the summation convention, and then explicitly in all its terms.

Exercise. Give all the independent invariants of a third-rank tensor.

The state of strain of a material particle is 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'α (x'1 , x'2 , x'3 ,t).

Recall that u'α = liα ui and x j = l jβ x'β , so that

x ∂u'α ∂ui ∂ j ∂ui = liα = liα l jβ . ∂x'β ∂x j ∂x'β ∂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

ε 'αβ = ε ijliα l jβ .

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Helmholtz free energy of a rod under uniaxial tension.

Consider a rod, 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. We record the function f (L) experimentally, which need not be linear. When the length changes from L to L + dL , the loading machine does work fdL to the rod.

We model an elastic solid with a , F(L,T ). When the length increases by dL, and the temperature increases by dT, the free energy increases by

dF = fdL − SdT .

Here S is the , which can also be measured experimentally by measuring the heat absorbed by the rod in a slow loading process, dS = δQ/T.

Define the stress and strain as

f L − L σ = , ε = 0 . A0 L0

Define the free energy density, w, as the free energy per unit volume, namely

F w = . A0L0

We can similarly define the entropy density

S s = . A0L0

Here we have used the initial area and initial length to define the stress, the strain, and the free energy density. With these definitions, we can rewrite dF = fdL − SdT as

dw = σdε − sdT .

The free energy density is a function of the strain and the temperature:

w = w(ε,T ).

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For a given solid, the function w(ε,T ) is determined experimentally. Once we know this function, we can obtain the stress-strain relation by taking the differentiation:

∂w(ε,T ) σ = . ∂ε T

Similarly, we can also calculate the entropy density by

∂w(ε,T ) s = − . ∂T ε

We next restrict ourselves to small strains, so that we can expand the function w(ε,T ) into a Taylor series in the strain:

1 2 w(ε,T )= w0 + w1ε + w2ε . 2

The coefficients w0 ,w1,w2 are functions of the temperature. We will only go up to the quadratic term in strain.

The stress is obtained by taking partial differentiation:

σ = w1 + w2ε .

Thus, we identify w2 as Young’s modulus E. We also identify w1 as the residual stress σo. Recalling the experimental observation of thermal expansion, we may wish to write this residual stress as

w1 = −Eα(T − Tref ),

where α is the coefficient of thermal expansion, and Tref is a reference temperature at which the stress vanishes. In most circumstances, E and α only depend on the temperature weakly, and are regarded as material constants. The energy density then becomes

1 2 w(ε,T ) = w0 + σ oε + Eε . 2

The entropy density is then given by

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∂w ⎛ dw dσ 1 dE ⎞ s = − o = − o + o ε + ε 2 . ∂T ⎝ dt dT 2 dT ⎠

Experimentally, the entropy density may be determined by measuring heat capacity c, namely, the heat absorbed by a unit volume of the solid per unit change in temperature, when the strain is held constant. Thus,

cdT ds = . T

Using the experimentally measured value of c, we can determine w0 ,w1,w2 . The above analysis can be generalized in many directions to account for diverse experimental observations. For the time being, we will focus on the stress strain relation of a linearly elastic solid, and make one important generalization: we will consider an anisotropic solid under a multiaxial stress state. We will drop the temperature dependence, and throw in thermal expansion strain at the end of the analysis. When we are not particularly interested in the temperature dependence of the free energy, following a large body of literature, we will call the Helmholtz free energy the .

Example: Elastic energy density 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 θ . When the shear angle changes from θ to θ + dθ , the loading machine does

the work FH0dθ 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 τ = , γ = θ A0

The energy per unit volume is a function of the shear strain, w(γ ). From the above, we have

dw = τdγ .

The shear stress is the differential coefficient of the energy density function.

When the block is made of a linearly elastic solid, under shear load, the stress-strain relation is τ = Gγ . Consequently, the energy density function is

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1 w(γ )= Gγ 2 2

This result holds only for linear elastic solid in pure shear condition.

Multi-axial stress states

The advantage of the energy density function becomes clear under the multi-axial stress state. The energy density is scalar, and the strain is a tensor. The energy density is a function of all components of the stain tensor:

w = w(ε11,ε12 ,...).

(For people familiar with thermodynamics, we deal energy at a constant temperature, so that the energy is the Helmholz free energy.)

The components of stress tensor are differential coefficients:

dw = σ pqdε pq .

That is

∂w(ε11,ε12 ,...) σ pq = . ∂ε pq

In linear elasticity, we assume that stress is linear in strain. Thus, the energy density is a quadratic form of the stain tensor, written as

1 w = C ε ε . 2 ijkl ij kl

Here the Cijkl coefficients are the components of a fourth-rank tensor called the tensor. Without losing any generality, we can assume the following :

Cijkl = C jikl = Cijlk = Cklij .

If we count carefully, we should have 21 independent components for a generally anisotropic elastic solid.

The components of the stress tensor are linear in the components of the strain tensor:

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σ pq = C pqijεij .

We can also invert this relation to express the strain in terms of the stress:

ε pq = S pqijσ ij

Here S pqij are the components of a fourth-rank tensor called the compliance tensor. They have the same symmetry properties.

Stress-strain relation in a matrix form

We can also write the above equations in another form. The state of strain is specified by the six components:

ε x ,ε y ,ε z ,γ yz ,γ zx ,γ xy

In this order we will label them as ε1,ε 2 ,ε3,ε 4 ,ε5 ,ε 6 . The six strain components can vary independently. The elastic energy per unit volume is a function of all the six strain components, w(ε1,ε2,ε3,ε4 ,ε5 ,ε6) . This is the energy density function. When each strain component changes by a small amount, dεi , the energy density changes by

dw = σ1dε1 + σ2dε2 + σ3dε3 +σ 4dε4 +σ5dε5 + σ6dε6 .

Here we use the engineering strains for the shear, rather than the tensor components. We do so to avoid the factor 2 in the above expression. Each stress component is the differential coefficient of the energy density function:

∂w σ i = . ∂εi

If the function w(ε1,ε2,ε3,ε4 ,ε5 ,ε6) 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 : 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:

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1 1 w = ∑cijεiε j = (c11ε1ε1 + c12ε1ε2 + c21ε2ε1...). 2 i, j 2

Here cij are 36 constants. The cross terms come in pairs, e.g.,(c12 + c21)ε1ε2 . 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 = cji .

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,

σ i = ∂w/ ∂εi . The stress relation becomes

σi = ∑cijε j . j

In the matrix notation, we write

⎡σ1 ⎤ ⎡c11 c12 c13 c14 c15 c16 ⎤⎡ε1 ⎤ ⎢ ⎥ ⎢c c c c c c ⎥⎢ ⎥ ⎢σ 2 ⎥ ⎢ 21 22 23 24 25 26 ⎥⎢ε 2 ⎥ ⎢σ 3 ⎥ ⎢c31 c32 c33 c34 c35 c36 ⎥⎢ε 3 ⎥ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ c c c c c c ⎢σ 4 ⎥ ⎢ 41 42 43 44 45 46 ⎥⎢ε 4 ⎥ ⎢σ ⎥ ⎢c c c c c c ⎥⎢ε ⎥ ⎢ 5 ⎥ ⎢ 51 52 53 54 55 56 ⎥⎢ 5 ⎥ ⎣⎢σ 6 ⎦⎥ ⎣⎢c61 c62 c63 c64 c65 c66 ⎦⎥⎣⎢ε 6 ⎦⎥

The physical significance of the constants cij is now evident. For example, when the solid is under a uniaxial strain state, ε1 ≠ 0, ε2 =ε3 =ε4 = ε5 = ε6 = 0 , the six stress components on the solid are σ1 = c11ε1, σ 2 =c21ε1,... The matrix cij is known as the stiffness matrix.

Inverting the matrix, we express the strain components in terms of the stress components:

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⎡ε1 ⎤ ⎡s11 s12 s13 s14 s15 s16 ⎤⎡σ1 ⎤ ⎢ ⎥ ⎢s s s s s s ⎥⎢ ⎥ ⎢ε 2 ⎥ ⎢ 21 22 23 24 25 26 ⎥⎢σ 2 ⎥ ⎢ε3 ⎥ ⎢s31 s32 s33 s34 s35 s36 ⎥⎢σ 3 ⎥ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ s s s s s s ⎢ε 4 ⎥ ⎢ 41 42 43 44 45 46 ⎥⎢σ 4 ⎥ ⎢ε ⎥ ⎢s s s s s s ⎥⎢σ ⎥ ⎢ 5 ⎥ ⎢ 51 52 53 54 55 56 ⎥⎢ 5 ⎥ ⎣⎢ε 6 ⎦⎥ ⎣⎢s61 s62 s63 s64 s65 s66 ⎦⎥⎣⎢σ 6 ⎦⎥

The matrix sij is known as the compliance matrix. The compliance matrix is also symmetric and positive definite.

The components of the stiffness tensor relate to the corresponding components of the stiffness matrix as

c11 = C1111, c14 = C1123, c44 = C2323 .

However, the corresponding relations for compliance are

s11 = S1111, s14 = 2S1123, s44 = 4S2323 .

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

⎡σ1 ⎤ ⎡c11 c12 c12 0 0 0 ⎤⎡ε1 ⎤ ⎢ ⎥ ⎢c c c 0 0 0 ⎥⎢ ⎥ ⎢σ 2 ⎥ ⎢ 12 11 12 ⎥⎢ε 2 ⎥ ⎢σ 3 ⎥ ⎢c12 c12 c11 0 0 0 ⎥⎢ε3 ⎥ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ 0 0 0 c 0 0 ⎢σ 4 ⎥ ⎢ 44 ⎥⎢ε 4 ⎥ ⎢σ ⎥ ⎢ 0 0 0 0 c 0 ⎥⎢ε ⎥ ⎢ 5 ⎥ ⎢ 44 ⎥⎢ 5 ⎥ ⎣⎢σ 6 ⎦⎥ ⎣⎢ 0 0 0 0 0 c44 ⎦⎥⎣⎢ε 6 ⎦⎥

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 .

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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

⎡σ1 ⎤ ⎡c11 c12 c13 0 0 0 ⎤⎡ε1 ⎤ ⎢ ⎥ ⎢c c c 0 0 0 ⎥⎢ ⎥ ⎢σ 2 ⎥ ⎢ 12 11 13 ⎥⎢ε 2 ⎥ ⎢σ 3 ⎥ ⎢c13 c13 c33 0 0 0 ⎥⎢ε3 ⎥ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ 0 0 0 c 0 0 ⎢σ 4 ⎥ ⎢ 44 ⎥⎢ε 4 ⎥ ⎢σ ⎥ ⎢ 0 0 0 0 c 0 ⎥⎢ε ⎥ ⎢ 5 ⎥ ⎢ 44 ⎥⎢ 5 ⎥ ⎣⎢σ 6 ⎦⎥ ⎣⎢ 0 0 0 0 0 (c11 − c12 )/ 2⎦⎥⎣⎢ε 6 ⎦⎥

Invariants and isotropic materials. If a material is isotropic, the elastic energy density w should be a function of the invariants of the strain tensor. For a linear material, w is quadratic in strain. The strain tensor can form only two invariants quadratic in its components: εijεij and 2 (ε kk ) . Consequently, for an isotropic, linearly elastic solid, the elastic energy density takes the form

1 2 w = µεijεij + λ(ε kk ) , 2 where µ and λ are known as the Lamé constants. Taking differentiation, we obtain the stress- strain relations:

σ ij = 2µεij + λε kkδij .

A comparison with the stress-strain relations in the uniaxial stress state shows that

E νE µ = , λ = . 2(1+ν ) (1+ν )(1− 2ν )

Exercise. It is reasonable to require that the elastic energy density be a positive-definite quadratic form of the strain tensor. That is, w > 0 for any state of strain, except that w = 0 when the strain tensor vanishes. For an isotropic, linearly elastic solid, confirm that this positive- definite requirement is equivalent to require that E > 0 and −1 <ν < 1/ 2.

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

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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 µm 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

aSi − aGe ε11 = ε 22 = = −4% . aGe

There will be an elongation normal to the film, ε33 > 0. All shear strains vanish. According to the generalized Hooke’s law, the stress normal to the film surface relates to the strains as

σ 33 = c11ε33 + c12ε11 + c12ε 22 .

Physically it is evident that there is no stress normal to the surface of the film, σ 33 = 0 . Inserting into the above expression, we obtain that

2c12 ε33 = − ε11 = +3% . c11

The two in-plane stress components are equal, given by

σ11 = σ 22 = c11ε11 + c12ε 22 + c12ε33 , or

2 ⎡ 2c12 ⎤ σ11 = σ 22 = ⎢c11 + c12 − ⎥ε11 . ⎣ c11 ⎦

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Inserting the numerical values, we obtain that σ11 = σ 22 = −5.6GPa . This is a huge stress that may generate dislocations in the film.

Ferroelastic phase transition. This part goes beyond linear elasticity. Suppose we have the following experimental observations. A crystal has a rectangular symmetry at a high temperature. When the temperature drops below a critical value, Tc , the crystal undergoes a phase transition. The crystal at a low temperature acquires a spontaneous strain in shear. Because of the symmetry, the shear strain can go both directions. (Sketch the geometry.)

We model this crystal with a free energy density

1 2 1 4 w(γ ,T )= A(T − Tc )γ + Bγ , 2 4 where A and B are positive constants. Due to symmetry, the crystal is equally likely to shear in two directions, so that we keep the even powers in the strain γ . (Sketch w as a function of γ at two temperatures, one above Tc and the other below Tc .)

2 When T > Tc , the coefficient of the γ term is positive, so that the crystal behaves like usual 4 elastic solid, with the A(T − Tc ). The γ term is unnecessary to describe the behavior of the crystal.

2 When T < Tc , the coefficient of the γ term is negative, and the energy is no longer minimal at γ = 0. Instead, the energy is minimal at two nonezero strains, known as the spontaneous strains, 4 ± γ s . In this case, the γ term will ensure that energy goes up again when the strain is large enough.

The stress-strain relation is

∂w(γ ,T ) 3 τ = = A(T − Tc )γ + Bγ . ∂γ

(Sketch this function. Mark the spontaneous strains and the hysteresis loop.) Setting τ = 0 , we find the spontaneous strains:

γ s = ± A(Tc − T )/ B .

Because the material is nonlinear, the shear modulus is no longer a constant, and is given by

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∂τ (γ ,T ) 2 µ = = A(T − Tc )+ 3Bγ . ∂γ

At the spontaneous strain, the shear modulus is given by

µ = 2A(Tc − T ).

Sketch the shear modulus as a function of the temperature, both below and above the critical temperature.

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