## Linear Elasticity Equations

### Summary of the Equations of Linear Elasticity

The stiffness matrix of linear elastic isotropic material contains two parameters:

• E, Young's modulus (elastic modulus)

• ν, Poisson’s ratio

Define the following quantities.

The equilibrium equation is

`$-\nabla ·\sigma =f$`

The linearized, small-displacement strain-displacement relationship is

`$\epsilon =\frac{1}{2}\left(\nabla u+\text{​}\nabla {u}^{T}\right)$`

The balance of angular momentum states that stress is symmetric:

`${\sigma }_{ij}={\sigma }_{ji}$`

The Voigt notation for the constitutive equation of the linear isotropic model is

`$\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{33}\\ {\sigma }_{23}\\ {\sigma }_{13}\\ {\sigma }_{12}\end{array}\right]=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\begin{array}{cccccc}1-\nu & \nu & \nu & 0& 0& 0\\ \nu & 1-\nu & \nu & 0& 0& 0\\ \nu & \nu & 1-\nu & 0& 0& 0\\ 0& 0& 0& 1-2\nu & 0& 0\\ 0& 0& 0& 0& 1-2\nu & 0\\ 0& 0& 0& 0& 0& 1-2\nu \end{array}\right]\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ {\epsilon }_{33}\\ {\epsilon }_{23}\\ {\epsilon }_{13}\\ {\epsilon }_{12}\end{array}\right]$`

The expanded form uses all the entries in σ and ε takes symmetry into account.

 $\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{12}\\ {\sigma }_{13}\\ {\sigma }_{21}\\ {\sigma }_{22}\\ {\sigma }_{23}\\ {\sigma }_{31}\\ {\sigma }_{32}\\ {\sigma }_{33}\end{array}\right]=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\begin{array}{ccccccccc}1-\nu & 0& 0& 0& \nu & 0& 0& 0& \nu \\ •& 1-2\nu & 0& 0& 0& 0& 0& 0& 0\\ •& •& 1-2\nu & 0& 0& 0& 0& 0& 0\\ •& •& •& 1-2\nu & 0& 0& 0& 0& 0\\ •& •& •& •& 1-\nu & 0& 0& 0& \nu \\ •& •& •& •& •& 1-2\nu & 0& 0& 0\\ •& •& •& •& •& •& 1-2\nu & 0& 0\\ •& •& •& •& •& •& •& 1-2\nu & 0\\ •& •& •& •& •& •& •& •& 1-\nu \end{array}\right]\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{12}\\ {\epsilon }_{13}\\ {\epsilon }_{21}\\ {\epsilon }_{22}\\ {\epsilon }_{23}\\ {\epsilon }_{31}\\ {\epsilon }_{32}\\ {\epsilon }_{33}\end{array}\right]$ (1)

In the preceding diagram, • means the entry is symmetric.

### 3D Linear Elasticity Problem

The toolbox form for the equation is

`$-\nabla ·\left(c\otimes \nabla u\right)=f$`

But the equations in the summary do not have ∇u alone, it appears together with its transpose:

`$\epsilon =\frac{1}{2}\left(\nabla u+\text{​}\nabla {u}^{T}\right)$`

It is a straightforward exercise to convert this equation for strain ε to ∇u. In column vector form,

`$\nabla u=\left[\begin{array}{c}\partial {u}_{x}/\partial x\\ \partial {u}_{x}/\partial y\\ \partial {u}_{x}/\partial z\\ \partial {u}_{y}/\partial x\\ \partial {u}_{y}/\partial y\\ \partial {u}_{y}/\partial z\\ \partial {u}_{z}/\partial x\\ \partial {u}_{z}/\partial y\\ \partial {u}_{z}/\partial z\end{array}\right]$`

Therefore, you can write the strain-displacement equation as

`$\epsilon =\left[\begin{array}{ccccccccc}1& 0& 0& 0& 0& 0& 0& 0& 0\\ 0& \frac{1}{2}& 0& \frac{1}{2}& 0& 0& 0& 0& 0\\ 0& 0& \frac{1}{2}& 0& 0& 0& \frac{1}{2}& 0& 0\\ 0& \frac{1}{2}& 0& \frac{1}{2}& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& \frac{1}{2}& 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{2}& 0& 0& 0& \frac{1}{2}& 0& 0\\ 0& 0& 0& 0& 0& \frac{1}{2}& 0& \frac{1}{2}& 0\\ 0& 0& 0& 0& 0& 0& 0& 0& 1\end{array}\right]\nabla u\equiv A\nabla u$`

where A stands for the displayed matrix. So rewriting Equation 1, and recalling that • means an entry is symmetric, you can write the stiffness tensor as

`$\begin{array}{c}\sigma =\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\begin{array}{ccccccccc}1-\nu & 0& 0& 0& \nu & 0& 0& 0& \nu \\ •& 1-2\nu & 0& 0& 0& 0& 0& 0& 0\\ •& •& 1-2\nu & 0& 0& 0& 0& 0& 0\\ •& •& •& 1-2\nu & 0& 0& 0& 0& 0\\ •& •& •& •& 1-\nu & 0& 0& 0& \nu \\ •& •& •& •& •& 1-2\nu & 0& 0& 0\\ •& •& •& •& •& •& 1-2\nu & 0& 0\\ •& •& •& •& •& •& •& 1-2\nu & 0\\ •& •& •& •& •& •& •& •& 1-\nu \end{array}\right]A\nabla u\\ =\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left[\begin{array}{ccccccccc}1-\nu & 0& 0& 0& \nu & 0& 0& 0& \nu \\ 0& 1/2-\nu & 0& 1/2-\nu & 0& 0& 0& 0& 0\\ 0& 0& 1/2-\nu & 0& 0& 0& 1/2-\nu & 0& 0\\ 0& 1/2-\nu & 0& 1/2-\nu & 0& 0& 0& 0& 0\\ \nu & 0& 0& 0& 1-\nu & 0& 0& 0& \nu \\ 0& 0& 0& 0& 0& 1/2-\nu & 0& 1/2-\nu & 0\\ 0& 0& 1/2-\nu & 0& 0& 0& 1/2-\nu & 0& 0\\ 0& 0& 0& 0& 0& 1/2-\nu & 0& 1/2-\nu & 0\\ \nu & 0& 0& 0& \nu & 0& 0& 0& 1-\nu \end{array}\right]\nabla u\end{array}$`

Make the definitions

`$\begin{array}{l}\mu =\frac{E}{2\left(1+\nu \right)}\\ \lambda =\frac{E\nu }{\left(1+\nu \right)\left(1-2\nu \right)}\\ \frac{E\left(1-\nu \right)}{\left(1+\nu \right)\left(1-2\nu \right)}=2\mu +\lambda \\ \end{array}$`

and the equation becomes

`$\sigma =\left[\begin{array}{ccccccccc}2\mu +\lambda & 0& 0& 0& \lambda & 0& 0& 0& \lambda \\ 0& \mu & 0& \mu & 0& 0& 0& 0& 0\\ 0& 0& \mu & 0& 0& 0& \mu & 0& 0\\ 0& \mu & 0& \mu & 0& 0& 0& 0& 0\\ \lambda & 0& 0& 0& 2\mu +\lambda & 0& 0& 0& \lambda \\ 0& 0& 0& 0& 0& \mu & 0& \mu & 0\\ 0& 0& \mu & 0& 0& 0& \mu & 0& 0\\ 0& 0& 0& 0& 0& \mu & 0& \mu & 0\\ \lambda & 0& 0& 0& \lambda & 0& 0& 0& 2\mu +\lambda \end{array}\right]\nabla u\equiv c\nabla u$`

If you are solving a 3-D linear elasticity problem by using `PDEModel` instead of `StructuralModel`, use the `elasticityC3D(E,nu)` function (included in your software) to obtain the `c` coefficient. This function uses the linearized, small-displacement assumption for an isotropic material. For examples that use this function, see `StationaryResults`.

### Plane Stress

Plane stress is a condition that prevails in a flat plate in the x-y plane, loaded only in its own plane and without z-direction restraint. For plane stress, σ13 = σ23 = σ31 = σ32 = σ33 = 0. Assuming isotropic conditions, the Hooke's law for plane stress gives the following strain-stress relation:

`$\left[\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right]=\frac{1}{E}\left[\begin{array}{ccc}1& -\nu & 0\\ -\nu & 1& 0\\ 0& 0& 2+2\nu \end{array}\right]\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right]$`

Inverting this equation, obtain the stress-strain relation:

`$\left(\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right)=\frac{E}{1-{\nu }^{2}}\left(\begin{array}{ccc}1& \nu & 0\\ \nu & 1& 0\\ 0& 0& \frac{1-\nu }{2}\end{array}\right)\left(\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right)$`

Convert the equation for strain ε to ∇u.

`$\epsilon =\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}& 0\\ 0& \frac{1}{2}& \frac{1}{2}& 0\\ 0& 0& 0& 1\end{array}\right]\nabla u\equiv A\nabla u$`

Now you can rewrite the stiffness matrix as

`$\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{12}\\ {\sigma }_{21}\\ {\sigma }_{22}\end{array}\right]=\left[\begin{array}{cccc}\frac{E}{1-{\nu }^{2}}& 0& 0& \frac{E\nu }{1-{\nu }^{2}}\\ 0& \frac{E}{2\left(1+\nu \right)}& \frac{E}{2\left(1+\nu \right)}& 0\\ 0& \frac{E}{2\left(1+\nu \right)}& \frac{E}{2\left(1+\nu \right)}& 0\\ \frac{E\nu }{1-{\nu }^{2}}& 0& 0& \frac{E}{1-{\nu }^{2}}\end{array}\right]\nabla u=\left[\begin{array}{cccc}\frac{2\mu \left(\mu +\lambda \right)}{2\mu +\lambda }& 0& 0& \frac{2\lambda \mu }{2\mu +\lambda }\\ 0& \mu & \mu & 0\\ 0& \mu & \mu & 0\\ \frac{2\lambda \mu }{2\mu +\lambda }& 0& 0& \frac{2\mu \left(\mu +\lambda \right)}{2\mu +\lambda }\end{array}\right]\nabla u$`

### Plane Strain

Plane strain is a deformation state where there are no displacements in the z-direction, and the displacements in the x- and y-directions are functions of x and y but not z. The stress-strain relation is only slightly different from the plane stress case, and the same set of material parameters is used.

For plane strain, ε13 = ε23 = ε31 = ε32 = ε33 = 0. Assuming isotropic conditions, the stress-strain relation can be written as follows:

`$\left(\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{22}\\ {\sigma }_{12}\end{array}\right)=\frac{E}{\left(1+\nu \right)\left(1-2\nu \right)}\left(\begin{array}{ccc}1-\nu & \nu & 0\\ \nu & 1-\nu & 0\\ 0& 0& \frac{1-2\nu }{2}\end{array}\right)\left(\begin{array}{c}{\epsilon }_{11}\\ {\epsilon }_{22}\\ 2{\epsilon }_{12}\end{array}\right)$`

Convert the equation for strain ε to ∇u.

`$\epsilon =\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& \frac{1}{2}& \frac{1}{2}& 0\\ 0& \frac{1}{2}& \frac{1}{2}& 0\\ 0& 0& 0& 1\end{array}\right]\nabla u\equiv A\nabla u$`

Now you can rewrite the stiffness matrix as

`$\left[\begin{array}{c}{\sigma }_{11}\\ {\sigma }_{12}\\ {\sigma }_{21}\\ {\sigma }_{22}\end{array}\right]=\left[\begin{array}{cccc}\frac{E\left(1-\nu \right)}{\left(1+\nu \right)\left(1-2\nu \right)}& 0& 0& \frac{E\nu }{\left(1+\nu \right)\left(1-2\nu \right)}\\ 0& \frac{E}{2\left(1+\nu \right)}& \frac{E}{2\left(1+\nu \right)}& 0\\ 0& \frac{E}{2\left(1+\nu \right)}& \frac{E}{2\left(1+\nu \right)}& 0\\ \frac{E\nu }{\left(1+\nu \right)\left(1-2\nu \right)}& 0& 0& \frac{E\left(1-\nu \right)}{\left(1+\nu \right)\left(1-2\nu \right)}\end{array}\right]\nabla u=\left[\begin{array}{cccc}2\mu +\lambda & 0& 0& \lambda \\ 0& \mu & \mu & 0\\ 0& \mu & \mu & 0\\ \lambda & 0& 0& 2\mu +\lambda \end{array}\right]\nabla u$`