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Shear stress vector

  1. Mar 2, 2013 #1
    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?
     
  2. jcsd
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