Finding the normal stress of using rotated stress tensor

Question

I wish to find the normal stress at the point on an interface whose normal unit vector is

$$n= \begin{bmatrix} 0.6 & 0.8 & 0\\ \end{bmatrix}$$

given the stress tensor $$\tau_{ij}= \begin{bmatrix} 2000 & 500 & -1000\\ 500 & 3000 & 3000\\ -1000 & 3000 & -4000\\ \end{bmatrix}$$

my goal is to rotate the stress tensor so the rotated $x$ axis ($x'$) is coalesces with the normal vector $n$ such that the new rotated $\tau_{x'x'}$ should be the answer.

We have that

$$\alpha=\arctan\left(\frac{0.8}{0.6}\right)=53.13 $$ using the rotatin matrix about the $Z$ axis $$R_z=\begin{bmatrix} \cos(53.13) & -\sin(53.13)& 0\\ \sin(53.13) & \cos(53.13)& 0\\ 0& 0 & 1\\ \end{bmatrix}$$

the rotated tensor is: $$\tau_{i'j'}=R\tau_{ij}R^T=\\ \begin{bmatrix} \cos(53.13) & -\sin(53.13)& 0\\ \sin(53.13) & \cos(53.13)& 0\\ 0& 0 & 1\\ \end{bmatrix} \begin{bmatrix} 2000 & 500 & -1000\\ 500 & 3000 & 3000\\ -1000 & 3000 & -4000\\ \end{bmatrix} \begin{bmatrix} \cos(53.13) & \sin(53.13)& 0\\ -\sin(53.13) & \cos(53.13)& 0\\ 0& 0 & 1\\ \end{bmatrix}\\ =\begin{bmatrix} 2160 & -620 & -3000\\ -620 & 2840 & 1000\\ -3000 & 1000 & -4000\\ \end{bmatrix}$$

which yields a wrong answer: $\tau_{x'x'}=2160$.

But $\tau_{\textbf{i'j'}}=R^T\tau_{ij}R$ yields the correct answer:

$$\tau_{\textbf{i'j'}}=\\ \begin{bmatrix} \cos(53.13) & \sin(53.13)& 0\\ -\sin(53.13) & \cos(53.13)& 0\\ 0& 0 & 1\\ \end{bmatrix} \begin{bmatrix} 2000 & 500 & -1000\\ 500 & 3000 & 3000\\ -1000 & 3000 & -4000\\ \end{bmatrix} \begin{bmatrix} \cos(53.13) & -\sin(53.13)& 0\\ \sin(53.13) & \cos(53.13)& 0\\ 0& 0 & 1\\ \end{bmatrix}\\ =\begin{bmatrix} 3120 & 340 & 1800\\ 340 & 1880 & 2600\\ 1800 & 2600 & -4000\\ \end{bmatrix}$$

$\tau_{\textbf{i'j'}}=\tau_{x'x'}=3120$.

Why is that?

Answer 1

Rotation matrix $R_z$ represents the change from the new basis $\mathbb B$' to the old one $\mathbb B$, therefore

$$R_z\,T'=\tau \; R_z\,n' \implies T'=R_z^T\,\tau \, R_z\,n'$$

then in the new basis $\tau'=R_z^T\,\tau \, R_z$.

Refer also to

• Changing the basis of a linear transformation

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The key point is that matrix $R_z$ represents the change from the new basis $\mathbb B$' to the old one $\mathbb B$, indeed in the new basis the normal vector is $n'=[1\;\;0\;\;0]$ and then

$$R_z\,n' = n=[0.60\;\; 0.80\;\;0]$$

which is indeed the normal vector in the old basis.

Answer 2

$ \def\t{\theta} \def\o{{\tt1}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\trace#1{\op{Tr}\LR{#1}} \def\qiq{\quad\implies\quad} \def\qif{\quad\iff\quad} \def\p{\partial} \def\grad#1#2{\frac{\p #1}{\p #2}} \def\m#1{\left[\begin{array}{r}#1\end{array}\right]} \def\v#1{\begin{Bmatrix}#1\end{Bmatrix}} \def\mbrace#1{\left\lbrace\begin{array}{r}#1\end{array}\right\rbrace} \def\c#1{\color{red}{#1}} $Given $$\eqalign{ T = \m{ 2000 & 500 & -1000 \\ 500 & 3000 & 3000 \\ -1000 & 3000 & -4000 \\ }\qquad n = \m{0.6\\0.8\\0.0} \\ }$$ You are looking for the rotation which rotates the first basis vector $\{e_1\}$ into the $n$ direction.

So the first column of $R$ is given by $Re_1=n\:$ and the remaining columns can be determined using orthogonality and symmetry considerations, without any trig calculations $$\eqalign{ R = \m{ 0.6 & -0.8 & \;0.0 \\ 0.8 & 0.6 & \;0.0 \\ 0.0 & 0.0 & \;1.0 \\ } }$$ Now do the required stress calculation $$\eqalign{ \sigma &= n^TTn \;=\; \c{3120} \\ &= \LR{Re_1}^TT\LR{Re_1} \\ &= e_1^T\LR{R^TTR}e_1 \\ }$$ The rotated stress tensor (in parentheses) is $$\eqalign{ T' = R^TTR = \m{ \c{3120} & 340 & 1800 \\ 340 & 1880 & 2600 \\ 1800 & 2600 & -4000 \\} }$$