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

• Dustinsfl
In summary, the state of stress at point ##\mathbf{P}## is given by the matrix [t_{ij}] with values of 1, 0, 2, 0, 1, 0, 2, 0, -2. The principal stress values at ##\mathbf{P}## are -3, 1, and 2, with corresponding principal stress directions of ##\hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 1}}} = \frac{1}{\sqrt{5}}\begin{bmatrix}-1\\0\\2\end{bmatrix}, \hat{\mathbf{n}}_{
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##.
\> ##\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}
\> ##\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}
\> ##\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?

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, $$\hat{\mathbf{n}} = \frac{1}{\sqrt{10}}\begin{bmatrix}1\\0\\3\end{bmatrix} = \frac{1}{\sqrt{10}}\left(\hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 1}}} \times \hat{\mathbf{n}}_{\text{\MakeUppercase{\romannumeral 3}}} \right).$$

## What is shear stress at point P?

Shear stress at point P is the stress that acts tangentially to the surface of a material at a specific point, causing the material to deform or shear. It is a result of forces acting parallel to the surface of the material.

## What is the difference between principal stress and maximum shear stress?

Principal stress is the maximum and minimum stress values that act on a material at a specific point, while maximum shear stress is the maximum difference between the principal stresses. In other words, maximum shear stress is the maximum amount of stress that the material experiences due to forces acting parallel to the surface.

## How is the direction of shear stress at point P determined?

The direction of shear stress at point P is perpendicular to the direction of the applied force. This means that if the force is applied horizontally, the shear stress will act vertically.

## What factors affect the shear stress at point P?

The shear stress at point P is affected by the magnitude and direction of the applied force, as well as the properties of the material, such as its elasticity and strength.

## What is the significance of shear stress at point P in material engineering?

Shear stress at point P is an important factor to consider in material engineering as it can cause a material to deform or fail. It is crucial to understand the shear stress at a specific point in order to design materials that can withstand different types of forces and loads.

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