Question 4 of 8: Plane Stress — Mohr's Circle

Question 4: Plane Stress — Mohr's Circle (20 marks)

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. From the printed stress element: $\sigma_x=-240$ MPa (compression, left/right faces), $\sigma_y=-40$ MPa (compression, top/bottom faces), $\tau_{xy}=-60$ MPa (top face shear points $-x$; right face shear points $-y$).

Find. $\sigma_1,\sigma_2$ and $\theta_p$; $\tau_{max}$, its associated normal stress, and $\theta_s$ — all via the Mohr's-circle construction.

Approach. Plot the two points $X(\sigma_x,\tau_{xy})$ and $Y(\sigma_y,-\tau_{xy})$ on the $\sigma$-$\tau$ plane; the line XY is a diameter of Mohr's circle. Read the centre, radius, and the physical rotation angles directly from the circle's geometry (remembering angles on the circle are $2\theta$, twice the physical angle).

1. Centre and radius of the circle. $$C=\frac{\sigma_x+\sigma_y}{2}=\frac{-240-40}{2}=-140.0\ \text{MPa}$$ $$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=\sqrt{(-100)^2+(-60)^2}=116.6\ \text{MPa}$$
2. (a) Principal stresses (the circle crosses the $\sigma$-axis at $C\pm R$): $$\sigma_1=C+R=-23.4\ \text{MPa}\qquad \sigma_2=C-R=-256.6\ \text{MPa}$$
3. (a) Orientation of the principal planes. The angle $2\theta_p$ from point X to the $\sigma_1$ point, measured on the circle, gives (by circle trigonometry, confirmed against the transformation equations as the WARNING permits): $$\theta_p = -74.5^\circ\ \text{(physical rotation from the x-axis, CCW positive)}$$
4. (b) Maximum in-plane shear. $\tau_{max}$ is the circle's radius, occurring 90° around the circle from the principal points (i.e. 45° physically) with associated normal stress equal to the centre $C$: $$\tau_{max}=R=116.6\ \text{MPa}\qquad \sigma_{avg}=C=-140.0\ \text{MPa}\qquad \theta_s=\theta_p-45^\circ=-119.5^\circ\ (\equiv 15.5^\circ)$$