Stresses on 30 degree plane; Mohr's circle

Reading of the figure: $\sigma_x = 50$ MPa on the left/right faces, $\sigma_y = 0$ and a shear stress $\tau_{xy} = 30$ MPa on all four faces. The plane is at $30^\circ$ to the vertical, so its normal is at $30^\circ$ to the $x$-axis.

Stresses on the inclined plane ($\theta = 30^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 25.00 + (25.00)\cos 60^\circ + (30.00)\sin 60^\circ = 63.48\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(25.00)\sin 60^\circ + (30.00)\cos 60^\circ = -6.65\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{63.48^2+-6.65^2} = 63.83\ \text{MPa}$, inclined at $5.98^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 25.00 \pm \sqrt{(25.00)^2+(30.00)^2} = 25.00 \pm 39.05$$

$\sigma_1 = 64.05\ \text{MPa}$ and $\sigma_2 = -14.05\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{50.00} \Rightarrow 2\theta_p = 50.19^\circ$$

So $\theta_p = 25.10^\circ$ (normal of the plane carrying $\sigma_1$) and $115.10^\circ$ (normal of the plane carrying $\sigma_2$), both measured anticlockwise from the $x$-axis. There is no shear stress on these planes.

Maximum shear stress

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 39.05\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 70.10^\circ$ and $160.10^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 25.00\ \text{MPa}$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(50, 30)$ and $Y(0, -30)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (25.00, 0)$.
• Draw the circle with centre C and radius $CX = 39.05$.
• It cuts the $\sigma$-axis at $\sigma_1 = 64.05$ and $\sigma_2 = -14.05$; $2\theta_p = \angle XCP_1 = 50.19^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm39.05$ with $\sigma = 25.00$.
• Turn CX through $2\theta = 60^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 63.48$ and $\tau \approx -6.65$, which match the analytical values.

Answer: $\sigma_n = 63.48$, $\tau = -6.65$ on the inclined plane; $\sigma_1 = 64.05$, $\sigma_2 = -14.05$ at $\theta_p = 25.10^\circ$ and $115.10^\circ$; $\tau_{max} = 39.05$ with normal stress $25.00$ (MPa).