# Tensors: traction free planes

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 $$\hat{\mathbf{n}} = \frac{1}{\sqrt{5}}\langle 2, 1, 0\rangle$$

## 1 Answer

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 $[\sigma][n]=$ 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 $[\sigma][n]=$, 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.