1
$\begingroup$

Given the following stress tensor matrix, determine the value of $\sigma_{22}$ such that there is a traction free plane and determine the unit normal to this plane. $$ \sigma_{ij} = \begin{bmatrix} 0 & 2 & 1\\ 2 & \sigma_{22} & 2\\ 1 & 2 & 0 \end{bmatrix} $$


The Cauchy stress tensor is $$ \mathbf{t}^{(n)} = \begin{bmatrix} 0 & 2 & 1\\ 2 & \sigma_{22} & 2\\ 1 & 2 & 0 \end{bmatrix} \hat{\mathbf{n}} $$ If we want a traction free plane, (this is just a guess) we want the traction tensor to be zero. I cant find anything online about this or in my book so I am not to sure about my method. Then $$ \mathbf{A}\mathbf{x} = \mathbf{0} $$ iff the determinant of $\mathbf{A}$ is zero. Therefore, $\sigma_{22} = 0$. Now we can solve for $\hat{\mathbf{n}}$. \begin{align} 2n_2 + n_3 &= 0\\ n_2 &= -n_3/2\\ 2n_1 + 2n_3 &= 0\\ n_1 &= -n_3\\ n_1 + 2n_2 &= 0\\ n_3 &= 0 \end{align} The magnitude of $\hat{\mathbf{n}}$ is $\lVert\hat{\mathbf{n}}\rVert = \frac{\sqrt{5}}{2}$ so $$ \hat{\mathbf{n}} = \frac{1}{\sqrt{5}}\langle 2, 1, 0\rangle $$

$\endgroup$
1
$\begingroup$

There is some value of $\sigma_{22}$ for which t=(0,0,0) on some (unknown) plane. Since $[t]=[\sigma][n]$, this is equivalent to the condition that the homogeneous linear equation $[\sigma][n]=[0]$ has a non-trivial solution.

This occurs when $\det[\sigma]=0$.

From the component form, evaluating the determinant gives you $0 + 4-\sigma_{22}+4=0$, so the only solution is $\sigma_{22}=8$.

Plug it back in to $[\sigma][n]=[0]$, writing $\hat{n}=(n_1,n_2,n_3)$ and you have the linear equations $$ 2 n_2 + n_3 = 0 \\ 2 n_1 + 8 n_2 + 2 n_3 = 0 $$ From (1) and (2) you have $n_3=-2n_2$, $n_1=-2n_2$.

Since $\hat{n}$ is a normal vector, you must have $|\hat{n}|=1$. This will finally give you the two unit vectors $\pm (\dfrac{2}{3},\dfrac{-1}{3},\dfrac{2}{3})$.

You can plug the value $\sigma_{22}=8$ and these values of $\hat{n}$ to check that you indeed get zero traction on this plane.

Edited to add: You calculated the determinant incorrectly, so your answer is off.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.