Deformation Gradient - Strain

Computation of deformation gradient and strain tensor

This example shows a single linear finite element (triangle) which is deformed by a transformation matrix. This matrix is computed as $$\mathbf{T} = \mathbf{R} \mathbf{S} \mathbf{U},$$ with the rotation matrix $$\mathbf{R} = \begin{pmatrix} \cos{\alpha} & -\sin{\alpha} \\ \sin{\alpha} & \cos{\alpha} \end{pmatrix},$$ the scaling matrix $$\mathbf{S} = \begin{pmatrix} s_x & 0 \\ 0 & s_y \end{pmatrix},$$ the shearing matrix $$\mathbf{U} = \begin{pmatrix} 1 & u_x \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ u_y & 1 \end{pmatrix},$$ the scaling factors $s_x$, $s_y$, and the shear values $u_x$, $u_y$.

The deformation gradient for such an element can be determined as $$\mathbf{F} = \mathbf{D}_s \mathbf{D}_m^{-1},$$ where $$\mathbf{D}_s = \begin{pmatrix} \mathbf{x}_1 - \mathbf{x}_0 & \mathbf{x}_2 - \mathbf{x}_0 \end{pmatrix},$$ $$\mathbf{D}_m = \begin{pmatrix} \mathbf{X}_1 - \mathbf{X}_0 & \mathbf{X}_2 - \mathbf{X}_0 \end{pmatrix}.$$ Here $\mathbf{X}_i$ denote the reference positions of the triangle and $\mathbf{x}_i$ the deformed configuration.

This example shows that the deformation gradient extracts exactly the original transformation from the deformed configuration of the triangle vertices. Moreover, we can see that the linear Cauchy strain tensor is not zero for a pure rotation. This means that a rotation, which is a rigid body transformation, leads to a non-zero strain which can cause artifacts in the simulation.