Recall that \[ \boldsymbol{\epsilon}=\frac{1}{2}\left(\mathbf{u}\nabla+(\mathbf{u}\nabla)^T\right). \] or \[ \boldsymbol{\epsilon}=\frac{1}{2}\left(\begin{bmatrix} u_{1}\\ u_{2}\\ u_{3} \end{bmatrix}\begin{bmatrix} \dfrac{\partial}{\partial x_{1}} & \dfrac{\partial}{\partial x_{2}} & \dfrac{\partial}{\partial x_{3}} \end{bmatrix}+\begin{bmatrix} \dfrac{\partial}{\partial x_{1}} \\ \dfrac{\partial}{\partial x_{2}} \\ \dfrac{\partial}{\partial x_{3}} \end{bmatrix}\begin{bmatrix} u_{1}& u_{2}& u_{3} \end{bmatrix}\right) \]

To express the components of strain in a new coordinate system, we must express both the displacement \(\mathbf{u}\) and \(\nabla\) in the new coordinate system. That is, \[ \epsilon_{i'j'}=\frac{1}{2}\left(\frac{\partial u_{i'}}{\partial x_{j'}}+\frac{\partial u_{j'}}{\partial x_{i'}}\right). \]

Therefore, to express \(\epsilon_{i'j'}\) in terms of the components of strain in the old coordinate system, we should:

• express the displacement components in the new coordinate system in terms of displacement components in the old coordinate system
• express differentiation with respect to a new coordinate axis in terms of differentiation with respect to the old coordinates.

Transformation of Displacement

The displacement is a vector quantity. Therefore, its components in a new coordinate system \(u_{1'}, u_{2'}, u_{3'}\) follow the transformation of vectors: \[ u_{i'}=\sum_{k=1}^3 u_k l_{ki'} \tag{1} \] or \[ \mathbf{u}'=\mathbf{u} L \tag{2} \] \[ \begin{bmatrix} u_{1'} & u_{2'} & u_{3'} \end{bmatrix}=\begin{bmatrix} u_{1} & u_{2} & u_{3} \end{bmatrix}\begin{bmatrix} l_{11'} & l_{12'} & l_{13'}\\ l_{21'} &l_{22'} & l_{23'}\\ l_{31'} & l_{32'} & l_{33'} \end{bmatrix} \]

where \(l_{ki'}=\hat{\mathbf{e}}_k\cdot \hat{\mathbf{e}}_{i'}\) is the kth component of the ith unit vector for the new coordinate system:

\[ \begin{bmatrix} l_{11'} & l_{12'} & l_{13'}\\ l_{21'} &l_{22'} & l_{23'}\\ l_{31'} & l_{32'} & l_{33'} \end{bmatrix}=\begin{bmatrix} | & | & |\\ \hat{\mathbf{e}}_1^\prime & \hat{\mathbf{e}}_2^\prime & \hat{\mathbf{e}}_3^\prime\\ | & | & | \end{bmatrix} \tag{3} \]

Transformation of Derivatives

It follows from the chain rule that \[ \frac{\partial}{\partial x_{1'}}=\left(\frac{\partial}{\partial x_{1}}\right)\frac{\partial x_1}{\partial x_{1'}}+\left(\frac{\partial}{\partial x_{2}}\right)\frac{\partial x_2}{\partial x_{1'}}+\left(\frac{\partial}{\partial x_{3}}\right) \frac{\partial x_3}{\partial x_{1'}}. \tag{4} \]

The rate of change of an old coordinate with respect to a new coordinate is the cosine of the angle between them: \[ \frac{\partial x_1}{\partial x_{1'}}=l_{11'},\quad \frac{\partial x_2}{\partial x_{1'}}=l_{21'},\quad \frac{\partial x_3}{\partial x_{1'}}=l_{31'}. \tag{5} \]

Therefore, \[ \frac{\partial}{\partial x_{1'}}=\left(\frac{\partial}{\partial x_{1}}\right)l_{11'}+\left(\frac{\partial}{\partial x_{2}}\right)l_{21'}+\left(\frac{\partial}{\partial x_{3}}\right) l_{31'}. \tag{6} \]

We can write (6) as \[ \left[\frac{\partial}{\partial x_{1'}}\right]=\begin{bmatrix} \dfrac{\partial}{\partial x_{1}} & \dfrac{\partial}{\partial x_{2}} & \dfrac{\partial}{\partial x_{3}} \end{bmatrix} \begin{bmatrix} l_{11'}\\ l_{21'}\\ l_{31'} \end{bmatrix} \]

and for all new coordinates: \[ \begin{bmatrix} \dfrac{\partial}{\partial x_{1'}} & \dfrac{\partial}{\partial x_{2'}} & \dfrac{\partial}{\partial x_{3'}} \end{bmatrix}=\begin{bmatrix} \dfrac{\partial}{\partial x_{1}} & \dfrac{\partial}{\partial x_{2}} & \dfrac{\partial}{\partial x_{3}} \end{bmatrix} \begin{bmatrix} l_{11'} & l_{12'} & l_{13'}\\ l_{21'} &l_{22'} & l_{23'}\\ l_{31'} & l_{32'} & l_{33'} \end{bmatrix}\tag{7} \]

Gradient in New Coordinates

Combining (1) and (6), we get \[ \begin{aligned} \frac{\partial u_{i'}}{\partial x_{j'}}&=\frac{\partial}{\partial x_{j'}}\overbrace{\sum_{k=1}^3 u_k l_{ki'}}^{u_{i'}}\\ &=\sum_{p=1}^3 \frac{\partial}{\partial x_p}l_{pj'}\sum_{k=1}^3 u_k l_{ki'}\\ &=\sum_{p=1}^3 \sum_{k=1}^3 l_{pj'} \frac{\partial u_k}{\partial x_p} l_{ki'} \end{aligned}\tag{8} \] Alternatively, using matrix notation and (2) and (7), we can write \[ \begin{aligned} \mathbf{u}'\nabla' &=\begin{bmatrix} u_{1'}\\ u_{2'}\\ u_{3'} \end{bmatrix}\begin{bmatrix} \dfrac{\partial}{\partial x_{1'}} & \dfrac{\partial}{\partial x_{2'}} & \dfrac{\partial}{\partial x_{3'}} \end{bmatrix}\\ &=L^T \begin{bmatrix} u_{1}\\ u_{2}\\ u_{3} \end{bmatrix}\begin{bmatrix} \dfrac{\partial}{\partial x_{1}} & \dfrac{\partial}{\partial x_{2}} & \dfrac{\partial}{\partial x_{3}} \end{bmatrix} L\\ &=L^T (\mathbf{u}\nabla) L \end{aligned}\tag{9} \]

Transformation Law for Strain Tensor

Since \[ \boldsymbol{\epsilon}=\frac{1}{2}\left(\mathbf{u}\nabla+(\mathbf{u}\nabla)^T\right), \] we conclude that \[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\boldsymbol{\epsilon}^\prime=L^T \boldsymbol{\epsilon}\, L}\tag{10} \] This is the same transformation formula as for stress: \[ \boldsymbol{\sigma}^\prime = L^T \boldsymbol{\sigma} L \]

Special Case: 2D Transformation

In 2D, \[ L=\begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix} \]

Therefore, \[ \begin{bmatrix} \epsilon_{x'} & \epsilon_{x'y'}\\ \epsilon_{y'x'} & \epsilon_{y'} \end{bmatrix}=\begin{bmatrix} \cos \theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix}\begin{bmatrix} \epsilon_{x} & \epsilon_{xy}\\ \epsilon_{yx} & \epsilon_{y} \end{bmatrix}\begin{bmatrix} \cos \theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix}\tag{11} \]

\[ \bbox[5px,border:1px #f2f2f2;background-color:#f2f2f2]{\begin{aligned} \epsilon_{x^\prime}&=\frac{\epsilon_x+\epsilon_y}{2}+\frac{\epsilon_x-\epsilon_y}{2} \cos 2 \theta+\epsilon_{x y} \sin 2 \theta,\\ \epsilon_{y^\prime} & =\frac{\epsilon_x+\epsilon_y}{2}-\frac{\epsilon_x-\epsilon_y}{2} \cos 2 \theta-\epsilon_{xy} \sin 2 \theta \\ \epsilon_{x' y^\prime} & =-\frac{\epsilon_x-\epsilon_y}{2} \sin 2 \theta+\epsilon_{xy} \cos 2 \theta \end{aligned}}\tag{12} \]

This shows that to transform the strain components in a 2D problem, we can use Mohr’s circle, exactly as with stress.