- #1
Dustinsfl
- 2,281
- 5
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?
$$
[t_{ij}] = \begin{bmatrix}
1 & 0 & 2\\
0 & 1 & 0\\
2 & 0 & -2
\end{bmatrix}.
$$
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?