Question 4 of 8: Principal Stresses and Maximum Shear Stress by Mohr's Circle

Question 4: Principal Stresses and Maximum Shear Stress by Mohr's Circle

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. $\sigma_1,\sigma_2$ and their plane orientation; $\tau_{max}$ and its plane orientation — all via Mohr's circle geometry.

Approach. Plot points $X=(\sigma_x,-\tau_{xy})$ and $Y=(\sigma_y,\tau_{xy})$, the diameter of Mohr's circle; the center and radius give $\sigma_{1,2}$ directly, and the angle to $X$ gives the physical rotation to the principal plane.

1. Circle center and radius. $$\sigma_{avg}=\frac{\sigma_x+\sigma_y}{2}=\frac{150+30}{2}=90\text{ MPa}$$ $$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\sqrt{60^2+60^2}=84.85\text{ MPa}$$
2. Part (a) — Principal stresses. $$\boxed{\sigma_1=\sigma_{avg}+R=174.85\text{ MPa},\qquad \sigma_2=\sigma_{avg}-R=5.15\text{ MPa}}$$
3. Part (b) — Orientation of the principal plane. From the circle, the angle from point $X$ to the $\sigma_1$ axis is $$2\theta_p=\tan^{-1}\!\left(\frac{\tau_{xy}}{(\sigma_x-\sigma_y)/2}\right)=\tan^{-1}(1)=45^{\circ}\Rightarrow \theta_p=22.5^{\circ}$$ Rotating the stress element $22.5^{\circ}$ counterclockwise from the $x$-face aligns it with the $\sigma_1=174.85$ MPa plane (the $\sigma_2$ plane is $90^{\circ}$ away, perpendicular).
4. Part (c)&(d) — Maximum shear stress and its plane. The top of the circle gives $$\boxed{\tau_{max}=R=84.85\text{ MPa},\ \text{with an average normal stress }\sigma_{avg}=90\text{ MPa on that plane}}$$ located at $\theta_s=\theta_p-45^{\circ}=-22.5^{\circ}$ (i.e. $22.5^{\circ}$ clockwise from the $x$-face) — always $45^{\circ}$ from the principal planes.