Elasticity/Stress example 2

Example 2

Given: A homogeneous stress field with components in the basis $({𝐞}_1,{𝐞}_2,{𝐞}_3)$ given by

$\left[\mathit{\sigma }\right]=\left[\begin{array}{ccc}3& 1& 1\\ 1& 0& 2\\ 1& 2& 0\end{array}\right]\text{(MPa)}$

Find:

1. The traction ($𝐭$) acting on a surface with unit normal $\widehat{𝐧}=({\widehat{𝐞}}_2+{\widehat{𝐞}}_3)/\sqrt{2}$.
2. The normal traction (${𝐭}_n$) acting on a surface with unit normal $\widehat{𝐧}=({\widehat{𝐞}}_2+{\widehat{𝐞}}_3)/\sqrt{2}$.
3. The projected shear traction (${𝐭}_s$) acting on a surface with unit normal $\widehat{𝐧}=({\widehat{𝐞}}_2+{\widehat{𝐞}}_3)/\sqrt{2}$.
4. The principal stresses.
5. The principal directions of stress.

Solution

Here's how you can solve this problem using Maple.
with(linalg):

sigma := linalg[matrix](3,3,[3,1,1,1,0,2,1,2,0]);

$\sigma :=\left[\begin{array}{ccc}3& 1& 1\\ 1& 0& 2\\ 1& 2& 0\end{array}\right]$

e2 := linalg[matrix](3,1,[0,1,0]);

$𝑒2:=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right]$

e3 := linalg[matrix](3,1,[0,0,1]);

$𝑒3:=\left[\begin{array}{c}0\\ 0\\ 1\end{array}\right]$

n := evalm((e2+e3)/sqrt(2));

$n:=\left[\begin{array}{c}0\\ \frac{\sqrt{2}}{2}\\ [2ex]\frac{\sqrt{2}}{2}\end{array}\right]$

sigmaT := transpose(sigma);

$𝑠𝑖𝑔𝑚𝑎𝑇:=\left[\begin{array}{ccc}3& 1& 1\\ 1& 0& 2\\ 1& 2& 0\end{array}\right]$

t := evalm(sigmaT&*n);

$𝐭:=\left[\begin{array}{c}\sqrt{2}\\ \sqrt{2}\\ \sqrt{2}\end{array}\right]\text{Solution for Part 1}$

tT := transpose(t);

$𝑡𝑇:=\left[\begin{array}{ccc}\sqrt{2}& \sqrt{2}& \sqrt{2}\end{array}\right]$

N := evalm(tT&*n);

$N:=\left[\begin{array}{c}2\end{array}\right]\text{Solution for Part 2}$

tdott := evalm(tT&*t);

$𝑡𝑑𝑜𝑡𝑡:=\left[\begin{array}{c}6\end{array}\right]$

S := sqrt(tdott[1,1] - N[1,1]^2);

$S:=\sqrt{2}\text{Solution for Part 3}$

sigPrin := eigenvals(sigma);

$𝑠𝑖𝑔𝑃𝑟𝑖𝑛:=1,-2,4\text{Solution for Part 4}$

dirPrin := eigenvects(sigma);

$𝑑𝑖𝑟𝑃𝑟𝑖𝑛:=[1,1,\{[-1,1,1]\}],[-2,1,\{[0,-1,1]\}],[4,1,\{[2,1,1]\}]$

dirPrin[1];

$[1,1,\{[-1,1,1]\}]\text{Solution for Part 5}$

dirPrin[2];

$[-2,1,\{[0,-1,1]\}]\text{Solution for Part 5}$

dirPrin[3];

$[4,1,\{[2,1,1]\}]\text{Solution for Part 5}$