Continuum Mechanics

Problem $5$

Consider the change in basis shown in Fig. $1.$ Here, the reference coordinate frame $\{e_i\}$ is rotated about

the $e_e-$axis about an angle $$ to yield a new coordinate frame $e_i^{*}.$ Note this is a simple two$-$dimensional

rotation of the coordinate frame such that the $e_3$ and $e_3^{*}$ directions remain aligned.

Figure $1$: Change in Basis

a$)$ For the given transformation, derive the rotation tensor $Q.$

b$)$ Show that the tensor $Q$ you derived satisfied $Q{Q}^{TT}=1$ and det$(Q)=1.$

c$)$ Let $A$ be a symmetric tensor. Assume in the reference $e_i$ frame the tensor $A$ is spherical. That is in

the reference frame it's components are given by

$[A]=\left[\begin{array}{ccc}{A}_{11}& 0& 0\\ 0& A_{22}& 0\\ 0& 0& A_{33}\end{array}\right]$

Using the transformation rule for tensors, find the components of $A^{*}$ as functions of the components

of A and the angle $.$

d$)$ Let $A_{11}=4$ and $A_{22}=2.$ Make a plot of $A_{11}^{*}$ in the horizontal axis versus $A_{12}^{*}$ in the vertical axis for

$in[0,2].$ Hint: the result should be a circle, commonly known as the Mohr's circle!