Let the state of stress at a point in a body be the difference of two plane states of stress shown in the figure. Consider all the possible planes perpendicular to the x-y plane and passing through that point. The magnitude of the maximum compressive stress on any such plane is $k \sigma_0$ , where $k$ is equal to _____ (round off to one decimal place).

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📖 Explanation

The total stress is the difference between the two stress states shown. Left figure (normal stress): $\sigma_x = 3\sigma_0$ $\sigma_y = 3\sigma_0$ $\tau_{xy} = 0$ Right figure is a rotated element by $45^\circ$ showing pure shear: It represents: $\sigma_x = 0$ $\sigma_y = 0$ $\tau_{xy} = 3\sigma_0$ Net stress state (subtracting second from first): $\sigma_x = 3\sigma_0$ $\sigma_y = 3\sigma_0$ $\tau_{xy} = -3\sigma_0$ Now, calculate principal stresses: $\sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{ \left( \frac{\sigma_x - \sigma_y}{2} \right)^2 + \tau_{xy}^2 }$ $= \frac{6\sigma_0}{2} \pm \sqrt{ 0 + 9\sigma_0^2 }$ $= 3\sigma_0 \pm 3\sigma_0$ $\Rightarrow \sigma_1 = 6\sigma_0,\quad \sigma_2 = 0$ Now, compute maximum compressive stress: This is the maximum normal stress on any inclined plane: $\sigma_{\text{max}} = \sqrt{ \sigma_{\text{avg}}^2 + \tau_{xy}^2 }$ Where: $\sigma_{\text{avg}} = 3\sigma_0,\quad \tau_{xy} = -3\sigma_0$ $\Rightarrow \sigma_{\text{max}} = \sqrt{ (3\sigma_0)^2 + (-3\sigma_0)^2 } = \sqrt{18}\sigma_0$ $\Rightarrow \sigma_{\text{max}} \approx 4.24\sigma_0$ So, $k = \frac{4.24\sigma_0}{2\sigma_0} = \boxed{2.1}$

Q5GATE 2023NAT

The infinitesimal element shown in the figure (not to scale) represents the state of stress at a point in a body. What is the magnitude of the maximum principal stress (in $\mathrm{N} / \mathrm{mm}^{2}$ , in integer) at the point?

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📖 Explanation

On plane $x^{\prime} \rightarrow \quad \sigma_{x^{\prime}}=5 \mathrm{~N} / \mathrm{mm}^{2}$ $\tau_{x^{\prime} y^{\prime}}=4 \mathrm{~N} / \mathrm{mm}^{2}$ Using transformation equations:

\begin{aligned} & \sigma_{x^{\prime}}=\left(\frac{\sigma_{x}+\sigma_{y}}{2}\right)+\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right) \cos 2 \theta+\tau_{x y} \sin 2 \theta \\ & 5=\left(\frac{\sigma_{x}+6}{2}\right)+\left(\frac{\sigma_{x}-6}{2}\right) \cos \left(2 \times 45^{\circ}\right)+3 \sin \left(2 \times 45^{\circ}\right) \\ & \Rightarrow \sigma_{x}=-2 \mathrm{~N} / \mathrm{mm}^{2} \\ & \sigma_{\text {major } / \text { minor }}=\frac{\sigma_{x}+\sigma_{y}}{2} \pm \sqrt{\left(\frac{\sigma_{x}-\sigma_{y}}{2}\right)^{2}+\tau_{x y}^{2}} \end{aligned}

$\sigma_{\text {major } / \text { minor }}=\frac{-2+6}{2} \pm \sqrt{\left(\frac{-2-6}{2}\right)^{2}+3^{2}}$ $\sigma_{\text {major } / \text { minor }}=2 \pm 5$ $\Rightarrow \quad \sigma_{\text {major }}=7 \mathrm{~N} / \mathrm{mm}^{2}, \sigma_{\text {minor }}=-3 \mathrm{MPa}$ Hence, magnitude of maximum principal stress is $7 \mathrm{~N} / \mathrm{mm}^{2}$ .

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