Principal stresses and Mohr's circle (30, 18, 75)

Reading of the figure: $\sigma_x = 75$ N/mm² tensile (left and right faces), $\sigma_y = -30$ N/mm² (compressive, top and bottom faces), $\tau_{xy} = 18$ N/mm².

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 22.50 \pm \sqrt{(52.50)^2+(18.00)^2} = 22.50 \pm 55.50$$

$\sigma_1 = 78.00\ \text{N/mm}^2$ and $\sigma_2 = -33.00\ \text{N/mm}^2$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(18.00)}{105.00} \Rightarrow 2\theta_p = 18.92^\circ$$

So $\theta_p = 9.46^\circ$ (normal of the plane carrying $\sigma_1$) and $99.46^\circ$ (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

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 55.50\ \text{N/mm}^2$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 54.46^\circ$ and $144.46^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 22.50\ \text{N/mm}^2$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(75, 18)$ and $Y(-30, -18)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (22.50, 0)$.
• Draw the circle with centre C and radius $CX = 55.50$.
• It cuts the $\sigma$-axis at $\sigma_1 = 78.00$ and $\sigma_2 = -33.00$; $2\theta_p = \angle XCP_1 = 18.92^\circ$ (scaled by protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm55.50$ with $\sigma = 22.50$.

Answer: $\sigma_1 = 78.00$, $\sigma_2 = -33.00$ at $\theta_p = 9.46^\circ$ and $99.46^\circ$; $\tau_{max} = 55.50$ with normal stress $22.50$ (N/mm²).