Introduction to Linear Elasticity and Its Finite Element Approximation

INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION

LONG CHEN

CONTENTS 1. Introduction1 2. Physical Interpretation4 3. Tensors7 3.1. Forces7 3.2. Stress Tensor8 3.3. Strain Tensor8 3.4. Relation between Trace, Strain and Stress8 4. Variational Forms9 4.1. Displacement Formulation9 4.2. Mixed Formulation of Hellinger and Reissner 10 4.3. Mixed formulation of Hu and Washizu 12 5. Finite Element Methods 12 5.1. Locking and Nearly Incompressible Material 13 5.2. Symmetric stress and discontinuous displacement 13 References 13

The theory of elasticity is to study the deformation of elastic solid bodies under external load. A body is elastic if, when the external forces are removed, the bodies return to their original (undeformed) shape.

1. INTRODUCTION How to describe the deformation of a solid body? Let us denoted a solid body by a bounded domain Ω ⊂ R3. Then the deformed domain can be described as the image of a vector-valued function Φ:Ω → R3, i.e. Φ(Ω), which is called a conﬁguration or a placement of a body. For most problems of interest, we can require that the map be 1-1 and differentiable. Since we are interested in the deformation, i.e., the change of the domain, we let Φ(x) = x + u(x) or equivalently u(x) = Φ(x) − x and call u the displacement. Here we deal with continuum mechanics, i.e., systems have properties deﬁned at all points in space, ignoring details in atom and molecules level. The displacement is not the deformation. For example, a translation or a rotation, so- called rigid body motions, of Ω will lead to a non-trivial displacement u =6 0, but the shape and volume of Ω does not change at all. How to describe the shape mathematically? Vectors. For example, a cube can be described by 3 orthogonal vectors. The change of the

Date: Created Mar 19, 2011, Updated September 16, 2016. 1 2 LONG CHEN shape will then be described by a mapping of these three vectors, which can be represented by a 3 × 3 matrix. For example, a rigid motion has the form

(1) Φ(x) = x1 + Q(x − x0), T where x0, x1 are fixed material points and Q is a rotation (unitary) matrix (Q Q = I). Consider a point x ∈ Ω and a vector z pointing from x to x + z. Then the vector z will be deformed to Φ(x + z) − Φ(x) ≈ ∇Φ(x)z provided kzk is small. So the deformation gradient F (x) := ∇Φ(x) is a candidate of the mathematical quantity to describe the deformation. The deformation gradient can be decomposed into the summation of a symmetric part and an anti-symmetric part. The symmetric part will ... the change of the geometric metric and the antisymmetric part is for the rotation. Let us compute the change of the distance. For two points x and x+z in Ω, the squared distance in the deformed domain will be kΦ(x + z) − Φ(x)k2 ≈ k∇Φ(x)zk2 = zT F T F z. Comparing with the squared distance kx + z − xk2 = zT Iz in the original domain, we then define a symmetric matrix function 1 (2) E = (F T F − I) 2 and call it strain. Very crudely, strain may be thought of as a relative displacement per unit distance between two points in a body. When Φ is a rigid body motion, F = Q with QT Q = I, and thus E = 0. A nonzero strain E will describe change in shape. Strain is the geometrical measure of the deformation representing the relative displacement between points in the material body, i.e. a measure of how much a given displacement differs locally from a rigid-body displacement. This is the right quantity to describe the deformation. What is the relation between the strain E and the displacement u? Substituting ∇Φ = I + ∇u into (2), we get 1 (3) E = E(u) = (∇u + (∇u)T + (∇u)T ∇u). 2 We shall mostly studied the small deformation case, i.e. ∇u is small. Then we can drop the quadratic part and approximately use 1 (4) E(u) ≈ ε(u) = ∇su := (∇u + (∇u)T ), 2 where ∇s will be called the symmetric gradient. The strain tensor E defined in (3) is called the Green-St. Venant Strain Tensor or the Lagrangian Strain Tensor. Its approximation ε in (4) is called the infinitesimal strain tensor. Remark 1.1. The approximation E(u) ≈ ε(u) is valid only for small deformation. Con- sider the rigid motion (1). The deformation gradient is F = Q and thus E(u) = 0 but ε(u) = (Q + QT )/2 − I =6 0. Hence the infinitesimal strain tensor ε does not measure the correct strain when there are large rotations though the strain tensor E can. When considering the linear elasticity theory, we also call the kernel of ε rigid body motion. Consider the rotation in 2-D which can be written as Φ(x) = eiθx. Then u(x) = (eiθ − 1)x = iθx + O(θ2). Skipping the high order term when |θ| 1, we obtain the rigid motion for ε is a rotation of x = (x, y) by 90◦ counterclockwise, i.e. x⊥ := (−y, x). Exercise 1.2. Derive the rigid body motion for ε, i.e., the ker of ε in 3-D. INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION 3

How to find out the deformation? To cause a deformation, there should be forces applied to the solid body. In the static equilibrium, the deformation should be determined by the balance of forces. We now discuss possible external and internal forces. There are two types of external forces. Body forces are applied forces distributed throughout the body, e.g. the gravity or magnetic force. Surface forces are pressures applied to the surface or outer boundary of the body, such as pressures from applied loads. There are also internal forces. The deformation of a body will induces resistance forces like the forces produced when we stretch the spring or compress the rubber. How to describe such internal forces? Let us conceptually split the body along a plane through the body. Then we have a new internal surface across which further internal forces are exerted to keep the two parts of the body in equilibrium. These internal forces are called stresses. Mathematically we can define the stress at a point x as limA→{x} pA/A, where pA is the pressure applied to the cutting surface A, which contains x and shrinks to x. The pressure pA is a force (per unit area) which is not necessarily orthogonal to the surface. The normal part is called the normal stress and the tangential part is shearing stress. At a fixed point x of a body there are infinity many of different surface elements with centroid at x. Therefore the stress can be described as a function t :Ω × S2 → R3, where S2 is the unit 2-sphere used to representing a direction. At given a point x ∈ Ω, the stress t maps a unit vector n ∈ S2 to another vector (the force). A theorem by Cauchy simplifies the formulation of the stress function, by separating the variables, as t(x, n) = Σ(x) n, where the matrix function Σ(x) is called the stress tensor. Suppose f is the only applied external and body-type force. We can then write out the equilibrium equations Z Z (5) f dx + Σ n dS = 0 for all V ⊂ Ω, V ∂V Z Z (6) x × f dx + x × (Σ n) dS = 0 for all V ⊂ Ω. V ∂V Equation (5) is the balance of forces and (6) is the conservation of angular momentum. It can be derived from (6) that the stress tensor Σ is symmetric. Otherwise the body would be subjected to unconstrained rotation. Exercise 1.3. Show that equation (6) implies that Σ is symmetric. Using the Gauss theorem and letting V → {x}, the equation (5) for the balance of forces becomes (7) f + div Σ = 0. The 3 × 3 matrix function Σ has 9 components while (5)-(6) contains only 6 equations. Or the symmetric function Σ has 6 components while (7) contains only 3 equations. We need more equations to determine Σ uniquely. The internal force is a consequence of the deformation. Therefore it is reasonable to assume the stress is a function of the strain and in turn a function of the displacement, i.e. (8) Σ = Σ(E) = Σ(u), which is called constitutive equation. Then by substituting (8) into (7), it is possible to solve 3 equations to get 3 unknowns (3 components of the vector function u). It is then crucial to figure out the relation Σ(E) in (8). First we assume the stress tensor is intrinsic, i.e., independent of the choice of the coordinate. Second we consider 4 LONG CHEN

the isotropic material, i.e., the stress vectors do not change if we rotate the non-deformed body. With such hypothesis, we can assume Σ = Σ(F T F ) = Σ(I + 2E). By Rivlin-Ericksen theorem •1 , we can then expand it as •1 reference or details Σ = λ trace(E)I + 2µE + o(E), where λ, µ are two positive constants, known as Lame´ constants. For the theory of linear elasticity, we drop the high order term and use the approximation (9) Σ ≈ σ = λ trace(E)I + 2µE, which is appropriate for • small strain known as Hook’s law for linear material. • St. Venant-Kirchhoff material (not just for small strain). We now summarize the unknowns and equations for linear elasticity as the follows. ———————————————————————————————————— • (σ, ε, u): (stress, strain, displacement). σ, ε are symmetric 3 × 3 matrix functions and u is a 3 × 1 vector function. • Kinematic equation 1 ε = ∇su = (∇u + (∇u)T ). 2 • Constitutive equation σ = λ trace(ε)I + 2µ ε = λ div u + 2µ∇su. • Balance equation f + div σ = 0. ———————————————————————————————————— With appropriate boundary conditions, the kinematic equation, constitutive equation, and balance equation will uniquely determine the deformation described by (σ, ε, u). Remark 1.4. For different mechanics, the kinematic and balance equations remain the same. The difference lies in the constitutive equation.

2. PHYSICAL INTERPRETATION In this section, we explain the physical meaning of the stress and strain and the physical interpretation of kinematic and constitutive equations. Roughly speaking, strain may be thought of as a relative displacement per unit distance between two points in a body. Besides the change of length, the deformation contains also the change of angles. We thus classify two types of strain: • normal strain defines the amount of elongation along a direction; • shear strain defines the amount of distortion or simply the change of angles. Mathematically, given three points x, x + y and x + z in the undeformed body, the geometry (lengths and angles) of the two vectors y and z is characterized by the inner product (y, z) = yT z. After the deformation, Φ(x + y) − Φ(x) ≈ ∇Φ(x)y = F y and similarly Φ(x + z) − Φ(x) ≈ F z, the geometry is change to (F y, F z) = yT F T F z. Taking 2 y = z, we get the change of the distance, especially (δe1) = 2(Ee1, e1) = 2ε11. The change in the angle is also encoded in the strain E = (F T F − I)/2. More specifi- cally, let us choose y = e1 and z = e2 as the two axes vectors of the coordinate. Then INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION 5

T e1 Ee2 ≈ ε12. On the other hand, let us calculate the change of angles. The original angle between e1 and e2 is π/2. The angle of the deformed vectors is denoted by θ and the change of the angle by δθ, i.e., θ = π/2 − δθ. Then

2(Ee1, e2) = (F e1, F e2) = kF e1kkF e2k cos θ = (1 + δe1)(1 + δe2) sin δθ ∼ δθ.

In the ∼, we use the linear approximation sin δθ ∼ θ and skip the quadratic and higher order terms since the change in length and angle are small. In summary, ε12 = δθ/2, i.e., the shear strain εij represents the change of angles and εii represents the change of length in the i-th coordinate direction. Now we understand the stress. Consider the deformation of a bar-like body. On the conceptually cutting plane, the deformation will introduce two types for stress • normal stress which is normal to the plane; • shear stress which is tangential to the plane. The normal stress is considered as positive if it produces tension, and negative if it produces compression. Normal or shear is a relative concept. For a horizontally stretched bar, there is no shear stress if the cutting plane is vertical or no normal stress if the cutting plane is horizontal. For a symmetric and positive definite matrix, we can use eigen-vectors to form an orthonomal basis and in the new coordinate, the stress will be diagonal. The eigen-vectors of the stress tensor will be called principal directions, which are functions of locations. Then along the principal directions, we only see the normal stress. Similarly there are principle directions for strain tensor. Note that strain and stress tensor do not necessarily share the same eigenvectors. Both stress and strain can be represented by symmetric matrix functions. We emphasize that as intrinsic quantities of the material, they should not depend on the coordinate. In other words, these matrix representation should follow certain rules when we change the coordinate. We shall explore more in the Section3 on Tensor. How to interpret the constitutive equation relating stress and strain? We first derive a simple relation between the normal stress and the normal strain. Consider a spring elon- gated in the x-direction. By Hook’s law, the stress induced by the elongation will be proportional to the strain, i.e., 1 σ11 = E ε11 or equivalently ε11 = σ11. E The positive parameter E is called the modulus of elasticity in tension which is a property of the material. Usually E ≈ 1011 is a huge number, which means a small strain will lead to a large stress. In the elongation/compression case, the finite strain can be defined as the relative change of the length δL/L, where L is the length of the spring. Stretching a body in one direction will usually have the effect of changing its shape in others – typically decreasing it. The strain in y- and z- directions caused by the stress σ11 can be described as ν ε22 = ε33 = −νε11 = − σ11, E where ν is also a property of the material and called Poisson’s ratio. It can be mathematically shown ν ∈ (0, 0.5) and usually takes values in the range 0.25 − 0.3. The strain in any direction will be made up of a combination of the norm strain in that direction and the shear strain in other directions. By the superposition, which holds for 6 LONG CHEN

FIGURE 1. Modulus of elasticity and Poisson ratio linear elasticity, we have the relation 1 ε11 = ( σ11 − νσ22 − νσ33), E 1 ε22 = (−νσ11 + σ22 − νσ33), E 1 ε33 = (−νσ11 − νσ22 + σ33). E For the shear stress and shear strain, one can show 1 E (10) ε = σ for i =6 j, where G = ij G ij 1 + ν is called the modulus of elasticity in shear or the modulus of rigidity.

Exercise 2.1. To derive (10), we consider the effect of applying the normal stress σ33 = ◦ σ, σ22 = −σ and σ11 = −σ. Along planes at 45 the normal stress is zero while the shear stress is σ.

In summary, we can write the relation between the strain ε and the stress σ in the matrix form       ε11 1 −ν −ν 0 0 0 σ11  ε22   −ν 1 −ν 0 0 0   σ22         ε33  1  −ν −ν 1 0 0 0   σ33  (11)   =     .  ε12   0 0 0 1 + ν 0 0   σ12    E      ε13   0 0 0 0 1 + ν 0   σ13  ε23 0 0 0 0 0 1 + ν σ23 We can write in the simple form (12) Aσ = ε(u), where for isotropic elasticity, 1 λ (13) Aσ = σ − tr(σ)I . 2µ 2µ + dλ Inverting the matrix, we have the relation (14) σ = Cε INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION 7 where  1 − ν ν ν 0 0 0   ν 1 − ν ν 0 0 0    E  ν ν 1 − ν 0 0 0  (15) C =   . (1 + ν)(1 − 2ν)  0 0 0 1 − 2ν 0 0     0 0 0 0 1 − 2ν 0  0 0 0 0 0 1 − 2ν Comparing (15) with the constitutive equation σ = λ trace(ε)I + 2µ ε, we obtain the relation µ(3λ + 2µ) λ E = , ν = , λ + µ 2(λ + µ) and Eν E λ = , µ = , (1 + ν)(1 − 2ν) 2(1 + ν) which implies ν ∈ (0, 0.5) and limν→0.5 λ = ∞. When ν is close to 0.5, it will cause difﬁculties in the numerical approximation. We have derived the constitutive equation under the assumption: the material is isotropic and the deformation is linear. In general, the constitutive equation can be still described by (14). Mathematically the 9 × 9 matrix C contains 81 parameters. But C is a 4-th rank tensor and σij = Cijklεkl. The symmetry of ε and σ reduces the number of independent parameters to 36. For isotropic material, C = C0, where C0 is the matrix in another coordinate obtained by rotation. By choosing special rotations, one can show that C depends only on two parameters: the pair (E, ν) or the Lame´ constants (λ, µ). Taking a small volume V , the volume change is Z Z δV = u · n dS = div u dx = tr(ε). ∂V V Therefore the Lame´ constant λ describes the stress due to change in density.

3. TENSORS We choose a Cartesian coordinate to describe the deformation. If we change the coordinate, the description will be different. The deformation itself is, however, intrinsic, i.e., independent of the choice of coordinates. Thus different description in different coordinates should be related by some rules. Mathematically speaking, the stress and strain tensors are equivalent class of matrix functions. The equivalent class is deﬁned by changing of coordinates.

3.1. Forces. Recall that Φ:ΩR → Φ(ΩR) is the the deformation mapping. Here we use ΩR to indicate it is the reference domain. The body force f is a vector composed by 3-forms. Therefore Z Z f(x) dx = f R(xR) dxR, V VR together with dx = det(∇Φ) dxR, its transformation satisﬁes −1 f(x) = det(∇Φ) f R(xR). 8 LONG CHEN

3.2. Stress Tensor. Two different Cartesian coordinates can be obtained from each other by a rigid motion. Since the origin of the Cartesian coordinate systems has no bearing on the definition of σ, we assume that the new coordinate system is obtained by rotating the old one about its origin, i.e., x0 = Qx, where Q is unitary. Recall that σ is a linear mapping of vectors. In the original coordinate, it is t = σn. Applying Q both sides, we get t0 = Qt = QσQ−1Qn = QσQT n0. Namely (16) σ0 = QσQT . We can thus think the stress is a linear mapping between linear spaces (of vectors). The stress matrix is a representation of this mapping in a specific coordinate. For a given linear operator T , the eigenvector v and corresponding eigenvalue λ is defined as T v = λv. By the definition of eigenvalue, it depends only on the linear structure not the representation. Therefore the eigenvalues of σ, and their combination, e.g. tr(σ) = λ1 + λ2 + λ3 and det(σ) = λ1λ2λ3, are invariant in the change of coordinate.

3.3. Strain Tensor. The geometry (length and angle) is also intrinsic, i.e., independent of the choice of the coordinate. Mathematically suppose x0 = Qx. Then x0T ε0y0 = xT QT ε0Qy = xT εy. Namely (17) ε0 = QεQT . Therefore tr(ε) = div u is invariant in the change of coordinate.

3.4. Relation between Trace, Strain and Stress. Let M be the linear space of second order tensor (d × d matrix) and S ⊂ M the subspace of symmetric tensor. We introduce a Frobenius inner product in M as T T (A, B)F = A : B := tr(B A) = tr(A B), which induces the Frobenius norm k · kF of a matrix. For a scalar p, we could embed it into S by pId and denote this mapping by

π : R → S, π(p) = pId, and the imagine is denoted by RId, where Id is the identity matrix. −1 Deﬁne TR : M → RId as the composition TR = d π ◦ tr which is indeed the projection in F -product:

(18) (TRσ, pId)F = (σ, pId)F , ∀p ∈ R. D The orthogonal complement (I − TR)σ is called the deviation and denoted by σ . We summarize the decomposition as

σ = (I − TR)σ ⊕ TRσ = σ − tr(σ)Id/d + tr(σ)Id/d. The differential operators ε, grad , and div are related by the identity: tr ε(u) = div u, div πp = grad p. which can be veriﬁed by direct computation and summarized as the triangle diagram INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION 9

1 d 2 H (K, R ) L (K, Rd) ε div div grad tr 2 2 π L (K, S) L (K) H(div,K, S) H1(K)

FIGURE 2. Relation of differential operators and trace operator

4. VARIATIONAL FORMS We deﬁne the energy Z 1 Z E(u, ε, σ) = ( ε : σ − f · u) dx + g · u dx. Ω 2 ΓN and consider the optimization problem inf E(u, ε, σ) (u,ε,σ) subject to the relation ε = ∇su, and σ = Cε = λ tr(ε)I + 2µε.

Here ΓN is an open subset of the boundary ∂Ω and the complement is denoted by ΓD, i.e., ΓD ∪ ΓN = ∂Ω.

4.1. Displacement Formulation. We eliminate ε and σ to get the optimization problem (19) inf E(u), 1 u∈HD

1 1 3 where HD = {v ∈ (H (Ω)) : v(x) = 0 for x ∈ ΓD}, and Z 1 Z E(u) = ∇su : C∇su − f · u dx + g · u dx Ω 2 ΓN Z λ Z = (div u)2 + µ∇su : ∇su − f · v dx + g · u dx. Ω 2 ΓN The classic formulation is − 2µ div ∇su − λgrad div u = f in Ω,

u = 0 on ΓD, σ(u) · n = g on ΓN .

1 The weak formulation is: ﬁnd u ∈ HD(Ω) such that s s 1 (20) (C∇ u, ∇ v) = (f, v) − hg, viΓN for all v ∈ HD(Ω). To establish the well posedness of (20), it is crucial to have the following Korn’s inequality Theorem 4.1. There exists a constant C(Ω, λ, µ) such that for all v ∈ H1(Ω) s s 2 2 (21) (C∇ v, ∇ v) + kvk ≥ C(Ω, λ, µ)kvk1. 10 LONG CHEN

Proofs of Korn’s inequality are non-trivial for general H1 functions. For pure Dirichlet 1 boundary condition, i.e. v ∈ H0(Ω), using the identity 2∇sv : ∇sv − ∇v : ∇v = div[(v∇)v − (div v)v] + (div v)2,

one can prove that √ s |v|1 ≤ 2k∇ vk. In general, we can use the norm equivalence 2 2 2 2 (22) kvk h k∇vk−1 + kvk−1 for all v ∈ L (Ω), and the identity ∂2u ∂ε ∂ε ∂ε (23) = ik + ij − jk , ∂xk∂xj ∂xj ∂xk ∂xi to prove the Korn equality. Exercise 4.2. Prove the identity (23) and use (22) to prove the Korn inequality. Exercise 4.3. Prove the well-posedness of (20) using Korn’s inequality and ﬁgure out the dependence of the stability constant on the Lame constants. 4.2. Mixed Formulation of Hellinger and Reissner. We keep the displacement and stress (u, σ) and eliminate the strain ε to get the saddle-point problem Z 1 Z inf sup E(u, σ) = inf Aσ : σ − f · u dx + g · u dS, σ u σ u ( , ) Ω 2 ΓN subject to •2 •2 check this energy s σ = C∇ u in Ω σ · n = g on ΓN . Or equivalently inf (Aσ, σ), σ∈H(div,Ω,S) subject to the constraint

− div σ = f, in Ω, σ · n = g, on ΓN . The displacement can be mathematically treat as the Lagrange multiplier to impose the d×d constraint. Note that σ is a symmetric matrix function taking values in S = Rsym . Differ- ential operators are applied row-wise to a matrix. The strong formulation is − div σ = f, σ = C∇su in Ω,

u = 0 on ΓD, σ · n = g on ΓN .

Deﬁne Hg(div, Ω, S) = {τ ∈ H(div, Ω, S), τ · n = g on ΓN }. The weak formulation is: ﬁnd σ ∈ Hg(div, Ω, S) such that

(Aσ, τ ) + (u, div τ ) = 0 for all τ ∈ H0(div, Ω, S) −(div σ, v) = (f, v) for all v ∈ L2(Ω).

When ΓN = ∅, i.e., u = 0 on ∂Ω, the mixed formulation is in the pure Neumman type and solutions are not uniquely. We need to consider the quotient space Z Hˆ (div, Ω) = {τ ∈ H(div, Ω) : tr(τ ) dx = 0}. Ω INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION 11

The constraint comes from, for τ = ε(v), Z Z Z tr(τ ) dx = div v dx = v · n dS = 0. Ω Ω ∂Ω

Denote the space for stress by Σ equipped with norm k · kΣ and the space for displace- ∗ ment by V with norm k · kV . Let us introduce the linear operator L : Σ × V → (Σ × V ) hL(σ, u), (τ , v)i := (Aσ, τ ) − hε(u), τ i + hdiv σ, vi. Deﬁne bilinear forms a(σ, τ ) := (Aσ, τ ) and b(τ , v) = −(div τ , v). It is well known that L is an isomorphism from Σ×V onto (Σ×V )∗ if and only if the following so-called Brezzi conditions hold:

(1) Continuity of bilinear forms a(·, ·) and b(·, ·): there exist constants ca, cb > 0 such that

a(σ, τ ) ≤ cakσkΣkτkΣ, b(τ , v) ≤ cbkτ kΣkvkV , for all σ, τ ∈ Σ, v ∈ V . (2) Coercivity of a(·, ·) in the kernel space. There exists a constant α > 0 such that 2 a(σ, σ) ≥ αkσkΣ for all σ ∈ ker(div), where ker(div) = {τ ∈ Σ : b(τ , v) = 0 for all v ∈ V }. (3) Inf-sup condition of b(·, ·). There exists a constant β > 0 such that b(τ , v) inf sup ≥ β. v∈V ,v=06 τ ∈Σ,τ =06 kτ kΣkvkV The continuity condition usually can be easily satisfied by choosing appropriate norm. Inf-sup condition is also relatively easy. The difficulty is the coercivity of a(·, ·). 2 1 d To illustrate this difficulty, we first use the pair Σ × V = L (Ω; S) × H0(Ω; R ). Now −1 div σ ∈ H (Ω; Rd) is understood as the dual operator, i.e., b(τ , v) = −(ε(v), τ ). The inf-sup condition holds trivially by choosing τ = −ε(v) and Korn inequality. The coercivity of a(·, ·) in L2-norm is in trouble. The L2-inner product for two tensors indeed involves two inner product strcutures: one is (·, ·)F among tensors and another is 2 R the L -inner product, i.e., Ω f · g dx of scalar functions. The orthogonality in (·, ·)F will be inherit in (·, ·).

−1 Lemma 4.4. Let TRσ = d tr(σ)Id. Then 1 1 a(σ, τ ) = ((I − T )σ, (I − T )τ ) + (T (σ),T (τ )), 2µ R R dλ + 2µ R R Proof. Using the formulae of A, c.f. (13), with parameter ρ = dλ/(dλ + 2µ) ∈ (0, 1), we have 2µAσ = σ − ρTRσ = (I − TR)σ + (1 − ρ)TRσ.

Recall that TR is the projector in (·, ·)F inner product. Using the property of orthogonal projectors, we have

((I − TR)σ, τ ) = ((I − TR)σ, (I − TR)τ ), (TRσ, τ ) = (TRσ,TRτ ) and the identity then follows. This implies the coercivity on the whole space: for all σ ∈ Σ 1 1 (24) a(σ, σ) ≥ min , kσk2. 2µ dλ + 2µ 12 LONG CHEN

The constant α, unfortunately, is in the order of O(1/λ) as λ → +∞. Namely it is not robust to λ. 2 We then consider the pair Σ × V = Hˆ (div, Ω; S) × L (Ω; V). The inf-sup condition is also trivial. Now the coercivity is only needed in the kernel space. It is helpful to use the following triangle diagram connecting the linear elasticity with the inf-sup stability of Stokes equations

1 d H0(Ω, R ) ε div tr 2 Hˆ (div, Ω, S) L (Ω)

FIGURE 3. Linear elasticity and the inf-sup stability of Stokes equations

Lemma 4.5. 2 2 2 (25) kTRσk ≤ C k(I − TR)σk + k div σk−1 , for all σ ∈ Σ. Proof. Let p = tr σ. By the inf-sup stability of Stokes equation, we have (div u, p) kpk ≤ C sup . 1 |u| u∈H0 1 Using div = tr ε and p = tr σ, we compute −1 −1 d (div u, p) = d (tr ε(u), tr σ) = (TRε(u),TRσ) = (ε(u),TRσ)

= (ε(u), (TR − I)σ) + (ε(u), σ) = (ε(u), (TR − I)σ) − (u, div σ). Then using Cauchy-Schwarz inequality, we get (25). We then obtain a robust coercivity of a(·, ·) restricted to the null space ker(div). Theorem 4.6. There exists a constant independent of λ such that (26) a(σ, σ) ≥ Ckσk2 for all σ ∈ Σ ∩ ker(div). Exercise 4.7. Prove Theorem 4.6. In applications, when we are mostly interested in computing the stresses, we shall use this mixed formulation.

4.3. Mixed formulation of Hu and Washizu. We keep all variables. The weak formulation is to impose the constitutive equation weakly.

5. FINITE ELEMENT METHODS • Rigid body motions has no effect on the stress or strain. This invariance must be preserved in the discretization. • Locking effect of displacement formulation. • Symmetric stress element in the mixed formulation. INTRODUCTION TO LINEAR ELASTICITY AND ITS FINITE ELEMENT APPROXIMATION 13

5.1. Locking and Nearly Incompressible Material. What is the locking effect: finite element approximation produces significantly smaller displacements than it should. Why: from the large parameter λ and null space of div operator. Recall that the linear elasticity equation using displacement u is 2µ(∇su, ∇sv) + λ(div u, div v) = (f, v). The Lame´ constant λ → ∞ as the Poisson ratio ν → 0.5. The error of naive finite element approximation is amplified by the constant λ/µ, which is not robust as λ → ∞. Locking is unavoidable if the finite element space Vh is not big enough. For example, one can show the locking exists if Vh ∩ ker(div) = {0} and k div vhk ≥ C(h)kvhk1. When λ 1, the material is called nearly incompressible. The quantity div u measures the incompressibility of the material. When λ 1, div u should be small. Indeed when the domain of interest is smooth or convex, we have the regularity result

kuk2 + λk div uk1 ≤ Ckfk. We can introduce an artificial pressure p = λ div u and rewrite the equation into a per- turbed Stokes equations s s 1 2µ(∇ u, ∇ v) + (p, div v) = (f, v), for all v ∈ HD(Ω), 1 (div u, q) − (p, q) = 0, for all q ∈ L (Ω). λ 2 Then we can use stable finite element methods developed for Stokes equations. For example, the finite element space for the displacement (now is the velocity in Stokes problem) should be fine enough and spaces (Vh,Ph) should satisfy the inf-sup condition. s s 1 2µ(∇ uh, ∇ vh) + (ph, div vh) = (f, vh), for all vh ∈ Vh ⊂ HD(Ω), 1 (div u , q ) − (p , q ) = 0, for all q ∈ P ⊂ L (Ω). h h λ h h h h 2 2 In general, div Vh 6⊆ Ph. We define Qh as the L projection to Ph. Taking qh = Qh div vh and addition these two equations, we obtain a modified formulation s s 2µ(∇ uh, ∇ vh) + λ(Qh div uh,Qh div vh) = (f, vh). The inner product of L2 projection can be thought of as a selective reduced integration. 5.2. Symmetric stress and discontinuous displacement. The trick to recover the coercivity (25) can be applied to the discretization provided that ker(Bh) ⊂ ker(B) which is equivalent to the requirement div(Σh) ⊂ Uh. With this consideration the natural finite ele- −1 ment spaces would be Pk+1(Th; S)−Pk (Th; V) for k ≥ 0. The discrete inf-sup condition is now more difficult to verify. Hu and Zhang’s work

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