Principal planes, stresses on plane AC

Reading of the figure: $\sigma_x = 180$ N/mm² (right face, tensile), $\sigma_y = 120$ N/mm² (top face), $\tau_{xy} = 40$ N/mm². The plane AC rises at $60^\circ$ to the horizontal, so its normal is at $\theta = 60^\circ - 90^\circ = -30^\circ$ from the $x$-axis (anticlockwise positive).

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

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 150.00 + (30.00)\cos(-60^\circ) + (40.00)\sin(-60^\circ) = 130.36\ \text{N/mm}^2 \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(30.00)\sin(-60^\circ) + (40.00)\cos(-60^\circ) = 45.98\ \text{N/mm}^2 \end{aligned}$$

Resultant stress on the plane $= \sqrt{130.36^2+45.98^2} = 138.23\ \text{N/mm}^2$, inclined at $19.43^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 150.00 \pm \sqrt{(30.00)^2+(40.00)^2} = 150.00 \pm 50.00$$

$\sigma_1 = 200.00\ \text{N/mm}^2$ and $\sigma_2 = 100.00\ \text{N/mm}^2$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(40.00)}{60.00} \Rightarrow 2\theta_p = 53.13^\circ$$

So $\theta_p = 26.57^\circ$ (normal of the plane carrying $\sigma_1$) and $116.57^\circ$ (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

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 50.00\ \text{N/mm}^2$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 71.57^\circ$ and $161.57^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 150.00\ \text{N/mm}^2$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(180, 40)$ and $Y(120, -40)$ (the two shear stresses go on opposite sides of the $\sigma$-axis); join XY. It cuts the $\sigma$-axis at the centre $C = (150.00, 0)$.
• Draw the circle with centre C and radius $CX = 50.00$.
• It cuts the $\sigma$-axis at $\sigma_1 = 200.00$ and $\sigma_2 = 100.00$; $2\theta_p = \angle XCP_1 = 53.13^\circ$ (measured with a protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm50.00$ with $\sigma = 150.00$.
• Turn CX through $2\theta = 60^\circ$ clockwise (the same sense as $\theta = -30^\circ$) to reach point Q; its coordinates read $\sigma_n \approx 130.36$ and $\tau \approx 45.98$, which match the analytical values.

Answer: $\sigma_n = 130.36$, $\tau = 45.98$ on plane AC; $\sigma_1 = 200.00$, $\sigma_2 = 100.00$ at $\theta_p = 26.57^\circ$ and $116.57^\circ$; $\tau_{max} = 50.00$ with normal stress $150.00$ (N/mm²).