Principal stresses, plane BE, Mohr's circle

Reading of the figure: the normal stresses are $\sigma_x = 50$ N/mm² (right/left faces) and $\sigma_y = 40$ N/mm² (top face); the stress 30 N/mm² is the shear stress $\tau_{xy}$ on the faces. The plane BE is assumed to make $60^\circ$ with the horizontal (rising to the left from B), so its normal is at $\theta = 30^\circ$ from the $x$-axis. If the figure gives another angle, replace $\theta$ in the formulas below.

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

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 45.00 + (5.00)\cos 60^\circ + (30.00)\sin 60^\circ = 73.48\ \text{N/mm}^2 \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(5.00)\sin 60^\circ + (30.00)\cos 60^\circ = 10.67\ \text{N/mm}^2 \end{aligned}$$

Resultant stress on the plane $= \sqrt{73.48^2+10.67^2} = 74.25\ \text{N/mm}^2$, inclined at $8.26^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 45.00 \pm \sqrt{(5.00)^2+(30.00)^2} = 45.00 \pm 30.41$$

$\sigma_1 = 75.41\ \text{N/mm}^2$ and $\sigma_2 = 14.59\ \text{N/mm}^2$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(30.00)}{10.00} \Rightarrow 2\theta_p = 80.54^\circ$$

So $\theta_p = 40.27^\circ$ (normal of the plane carrying $\sigma_1$) and $130.27^\circ$ (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

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 30.41\ \text{N/mm}^2$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 85.27^\circ$ and $175.27^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 45.00\ \text{N/mm}^2$.

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 $Y(40, -30)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (45.00, 0)$.
• Draw the circle with centre C and radius $CX = 30.41$.
• It cuts the $\sigma$-axis at $\sigma_1 = 75.41$ and $\sigma_2 = 14.59$; $2\theta_p = \angle XCP_1 = 80.54^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm30.41$ with $\sigma = 45.00$.
• Turn CX through $2\theta = 60^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 73.48$ and $\tau \approx 10.67$, which match the analytical values.

Answer: $\sigma_n = 73.48$, $\tau = 10.67$ on the inclined plane; $\sigma_1 = 75.41$, $\sigma_2 = 14.59$ at $\theta_p = 40.27^\circ$ and $130.27^\circ$; $\tau_{max} = 30.41$ with normal stress $45.00$ (N/mm²).