Shear Stress at Point P: Principal Stress, Direction & Max Shear Stress

[Dustinsfl]

The state of stress at point $\mathbf{P}$ is given in ksi with respect to axes $P_{x_1x_2x_3}$ by the matrix
$$
[t_{ij}] = \begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 0\\
2 & 0 & -2
\end{bmatrix}.
$$
Determine
(1)the principal stress value and principal stress direction at $\mathbf{P}$,

The characteristic polynomial for $3\times 3$ matrix is
\begin{alignat*}{3}
p(\sigma) & = & \sigma^3 - \sigma^2\text{tr}(t_{ij}) - \frac{1}{2}\sigma[\text{tr}(t_{ij}^2) - \text{tr}^2(t_{ij})] - \det(\text{tr}(t_{ij}))\\
& = & \sigma^3 - \sigma^2\text{\MakeUppercase{\romannumeral 1}} - \sigma\text{\MakeUppercase{\romannumeral 2}} - \text{\MakeUppercase{\romannumeral 3}}
\end{alignat*}
where
\begin{alignat*}{3}
\text{\MakeUppercase{\romannumeral 1}} & = & \sigma_{\text{\MakeUppercase{\romannumeral 1}}} + \sigma_{\text{\MakeUppercase{\romannumeral 2}}} + \sigma_{\text{\MakeUppercase{\romannumeral 3}}}\\
\text{\MakeUppercase{\romannumeral 2}} & = & \sigma_{\text{\MakeUppercase{\romannumeral 1}}}\sigma_{\text{\MakeUppercase{\romannumeral 2}}} + \sigma_{\text{\MakeUppercase{\romannumeral 2}}}\sigma_{\text{\MakeUppercase{\romannumeral 3}}} + \sigma_{\text{\MakeUppercase{\romannumeral 3}}}\sigma_{\text{\MakeUppercase{\romannumeral 1}}}\\
\text{\MakeUppercase{\romannumeral 3}} & = & \sigma_{\text{\MakeUppercase{\romannumeral 1}}}\sigma_{\text{\MakeUppercase{\romannumeral 2}}}\sigma_{\text{\MakeUppercase{\romannumeral 3}}}
\end{alignat*}
The characteristic polynomial for our tensor is
$$
p(\sigma) = \sigma^3 - 7\sigma + 6
$$
since the trace of $t_{ij}$ is zero, the determinant is 6, and
$$
t_{ij}^2 = \begin{bmatrix}
5 & 0 & -2\\
0 & 1 & 0\\
-2 & 0 & 8
\end{bmatrix}.
$$
$p(\sigma)$ can be factor.
That is, $p(\sigma) = (\sigma + 3)(\sigma - 1)(\sigma - 2)$.
The principal stress are $\sigma_{\text{\MakeUppercase{\romannumeral 1}}} = -3$, $\sigma_{\text{\MakeUppercase{\romannumeral 2}}} = 1$, and $\sigma_{\text{\MakeUppercase{\romannumeral 3}}} = 2$.
\begin{tabbing}\hspace{.13\linewidth} \= 11.\quad \= \hspace{.35\linewidth} \= 11.\quad \= \kill
\> $\sigma_{\text{\MakeUppercase{\romannumeral 1}}}:$\> $\begin{bmatrix} 3 - \sigma_{\text{\MakeUppercase{\romannumeral 1}}} & 0 & 2\\ 0 & 1 - \sigma_{\text{\MakeUppercase{\romannumeral 1}}} & 0\\ 2 & 0 & -2 - \sigma_{\text{\MakeUppercase{\romannumeral 1}}} \end{bmatrix}$\> $=$\> $\begin{bmatrix} 4 & 0 & 2\\ 0 & 4 & 0\\ 2 & 0 & 1 \end{bmatrix}$\\
\> \> \> $=$\> $\begin{bmatrix} 2 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{bmatrix}$
\end{tabbing}
\begin{tabbing}\hspace{.13\linewidth} \= 11.\quad \= \hspace{.35\linewidth} \= 11.\quad \= \kill
\> $\sigma_{\text{\MakeUppercase{\romannumeral 2}}}:$\> $\begin{bmatrix} 3 - \sigma_{\text{\MakeUppercase{\romannumeral 2}}} & 0 & 2\\ 0 & 1 - \sigma_{\text{\MakeUppercase{\romannumeral 2}}} & 0\\ 2 & 0 & -2 - \sigma_{\text{\MakeUppercase{\romannumeral 2}}} \end{bmatrix}$\> $=$\> $\begin{bmatrix} 0 & 0 & 2\\ 0 & 0 & 0\\ 2 & 0 & -3 \end{bmatrix}$\\
\> \> \> $=$\> $\begin{bmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 2 & 0 & -3 \end{bmatrix}$
\end{tabbing}
\begin{tabbing}\hspace{.13\linewidth} \= 11.\quad \= \hspace{.35\linewidth} \= 11.\quad \= \kill
\> $\sigma_{\text{\MakeUppercase{\romannumeral 3}}}:$\> $\begin{bmatrix} 3 - \sigma_{\text{\MakeUppercase{\romannumeral 3}}} & 0 & 2\\ 0 & 1 - \sigma_{\text{\MakeUppercase{\romannumeral 3}}} & 0\\ 2 & 0 & -2 - \sigma_{\text{\MakeUppercase{\romannumeral 3}}} \end{bmatrix}$\> $=$\> $\begin{bmatrix} -1 & 0 & 2\\ 0 & -1 & 0\\ 2 & 0 & -4 \end{bmatrix}$\\
\> \> \> $=$\> $\begin{bmatrix} -1 & 0 & 2\\ 0 & -1 & 0\\ 0 & 0 & 0 \end{bmatrix}$
\end{tabbing}
The reduced matrix for $\sigma_i$ where $i$ are the numerals tells us that
\begin{tabbing}\hspace{.38\linewidth} \= 11.\quad \= \hspace{.05\linewidth} \= \hspace{.05\linewidth} \= \kill
\> $\sigma_{\text{\MakeUppercase{\romannumeral 1}}}:$\> $2x_1$ \> $=$\> $-x_3$\\
\> \> $x_2$ \> $=$\> $0$
\end{tabbing}
where $x_3$ is a free variable.
The principal stress direction for sigma one is
$$
\begin{bmatrix}
-\frac{1}{2}x_3\\
0\\
x_3
\end{bmatrix} = x_3\begin{bmatrix}
-1\\
0\\
2
\end{bmatrix}
$$
So the unit vector $\hat{\mathbf{n}}$ in the direction of $\sigma_{\text{\MakeUppercase{\romannumeral 1}}}$ is
$$
\hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 1}}} = \frac{1}{\sqrt{5}}\begin{bmatrix}
-1\\
0\\
2
\end{bmatrix}.
$$
For $\sigma_{\text{\MakeUppercase{\romannumeral 2}}}$, we have
\begin{tabbing}\hspace{.38\linewidth} \= 11.\quad \= \hspace{.05\linewidth} \= \hspace{.05\linewidth} \= \kill
\> $\sigma_{\text{\MakeUppercase{\romannumeral 2}}}:$\> $x_3$ \> $=$\> $0$\\
\> \> $2x_1$ \> $=$\> $3x_3$
\end{tabbing}
where $x_2$ is a free variable.
The principal stress direction for sigma two is
$$
\begin{bmatrix}
0\\
x_2\\
0
\end{bmatrix} = x_2\begin{bmatrix}
0\\
1\\
0
\end{bmatrix}
$$
So the unit vector $\hat{\mathbf{n}}$ in the direction of $\sigma_{\text{\MakeUppercase{\romannumeral 2}}}$ is $\hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 2}}} = \hat{\mathbf{e}}_2$.
For $\sigma_{\text{\MakeUppercase{\romannumeral 3}}}$, we have
\begin{tabbing}\hspace{.38\linewidth} \= 11.\quad \= \hspace{.05\linewidth} \= \hspace{.05\linewidth} \= \kill
\> $\sigma_{\text{\MakeUppercase{\romannumeral 3}}}:$\> $x_1$ \> $=$\> $2x_3$\\
\> \> $x_2$ \> $=$\> $0$
\end{tabbing}
where $x_3$ is a free variable.
The principal stress direction for sigma three is
$$
\begin{bmatrix}
2x_3\\
0\\
x_3
\end{bmatrix} = x_3\begin{bmatrix}
2\\
0\\
1
\end{bmatrix}
$$
So the unit vector $\hat{\mathbf{n}}$ in the direction of $\sigma_{\text{\MakeUppercase{\romannumeral 3}}}$ is
$$
\hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 3}}} = \frac{1}{\sqrt{5}}\begin{bmatrix}
2\\
0\\
1
\end{bmatrix}.
$$

(2)the maximum shear stress value at $\mathbf{P}$, and
\smallskip

The maximum shear stress can be found by
\begin{alignat*}{3}
\sigma_{\text{S}}^{\max} & = & \left\{\left|\frac{\sigma_{\text{\MakeUppercase{\romannumeral 1}}} - \sigma_{\text{\MakeUppercase{\romannumeral 2}}}}{2}\right|, \left|\frac{\sigma_{\text{\MakeUppercase{\romannumeral 2}}} - \sigma_{\text{\MakeUppercase{\romannumeral 3}}}}{2}\right|, \left|\frac{\sigma_{\text{\MakeUppercase{\romannumeral 3}}} - \sigma_{\text{\MakeUppercase{\romannumeral 1}}}}{2}\right|\right\}\\
& = & \left\{2, \frac{1}{2}, \frac{5}{2} \right\}\\
\sigma_{\text{S}}^{\max} & = & \frac{5}{2}
\end{alignat*}

(3)the normal $\hat{\mathbf{n}} = n_i\hat{\mathbf{e}}_i$ to the plane at $\mathbf{P}$ on which the maximum shear stress acts.

I obtained $\frac{1}{\sqrt{2}}\langle -1,0,1\rangle$ but the book has the answer as $\frac{1}{\sqrt{10}}\langle 1,0,3\rangle$
Which answer is correct? If it is the book, how did they get that?

 

[Valkarie]

The correct answer is $\frac{1}{\sqrt{10}}\langle 1,0,3\rangle$. To obtain this answer, we need to calculate the cross product of the two principal stress directions. That is, $$\hat{\mathbf{n}} = \frac{1}{\sqrt{10}}\begin{bmatrix}1\\0\\3\end{bmatrix} = \frac{1}{\sqrt{10}}\left(\hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 1}}} \times \hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 3}}} \right).$$