Normal/shear stress on 25 degree plane, principal stresses

Reading of the figure: $\sigma_x = 60$ MPa (tensile) on the vertical faces, $\sigma_y = 30$ MPa (tensile) on the horizontal faces, $\tau_{xy} = 50$ MPa. Tensile stress is taken as positive and compressive as negative. The plane AB is at $25^\circ$ to the vertical, so its normal is at $\theta = 25^\circ$ from the $x$-axis, measured anticlockwise.

Stresses on the inclined plane ($\theta = 25^\circ$)

$$\begin{aligned} \sigma_n &= \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \\ &= 45.00 + (15.00)\cos 50^\circ + (50.00)\sin 50^\circ = 92.94\ \text{MPa} \\ \tau_\theta &= -\frac{\sigma_x-\sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \\ &= -(15.00)\sin 50^\circ + (50.00)\cos 50^\circ = 20.65\ \text{MPa} \end{aligned}$$

Resultant stress on the plane $= \sqrt{92.94^2+20.65^2} = 95.21\ \text{MPa}$, inclined at $12.53^\circ$ to the normal of the plane.

Principal stresses

$$\sigma_{1,2} = \frac{\sigma_x+\sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 45.00 \pm \sqrt{(15.00)^2+(50.00)^2} = 45.00 \pm 52.20$$

$\sigma_1 = 97.20\ \text{MPa}$ and $\sigma_2 = -7.20\ \text{MPa}$.

Principal planes

$$\tan 2\theta_p = \frac{2\tau_{xy}}{\sigma_x-\sigma_y} = \frac{2(50.00)}{30.00} \Rightarrow 2\theta_p = 73.30^\circ$$

So $\theta_p = 36.65^\circ$ (normal of the plane carrying $\sigma_1$) and $126.65^\circ$ (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

$$\tau_{max} = \frac{\sigma_1-\sigma_2}{2} = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = 52.20\ \text{MPa}$$

It acts on planes at $45^\circ$ to the principal planes, i.e. at $\theta_s = 81.65^\circ$ and $171.65^\circ$. The normal stress on these planes is $\frac{\sigma_x+\sigma_y}{2} = 45.00\ \text{MPa}$.

Verification by Mohr's circle

• Draw the $\sigma$-axis (tension to the right) and the $\tau$-axis to a convenient scale.
• Plot $X(60, 50)$ and $Y(30, -50)$ (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 $CX = 52.20$.
• It cuts the $\sigma$-axis at $\sigma_1 = 97.20$ and $\sigma_2 = -7.20$; $2\theta_p = \angle XCP_1 = 73.30^\circ$ (measured with a protractor).
• The highest and lowest points of the circle give $\tau_{max} = \pm52.20$ with $\sigma = 45.00$.
• Turn CX through $2\theta = 50^\circ$ in the same sense as $\theta$ to reach point Q; its coordinates read $\sigma_n \approx 92.94$ and $\tau \approx 20.65$, which match the analytical values.

Answer: $\sigma_n = 92.94$ MPa, $\tau = 20.65$ MPa on AB; $\sigma_1 = 97.20$ MPa, $\sigma_2 = -7.20$ MPa; $\tau_{max} = 52.20$ MPa.