Question 4 of 8: Mohr’s Circle for Plane Stress

Question 4: Mohr’s Circle for Plane Stress (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.

Find. (a) the stress components on the 20° inclined plane, using Mohr’s circle; (b) the maximum in-plane shear stress and its associated normal stress and orientation.

Approach. Plot points X(σ_x,τ_xy) and Y(σ_y,−τ_xy) on the σ-τ plane, draw the circle through them, then rotate by 2θ (measured on the circle) to the point representing the inclined face.

1. Sign convention. σ_x=−40 MPa, σ_y=+60 MPa, τ_xy=+20 MPa (standard +y-on-+x-face convention; the right-face arrow points into the element, confirming compression).
2. Centre and radius. $$\sigma_{avg}=\dfrac{\sigma_x+\sigma_y}{2}=10.0\text{ MPa},\qquad R=\sqrt{\left(\dfrac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}=53.85\text{ MPa}.$$
3. Locate the inclined plane. A cut 20° above the horizontal, from the bottom-left corner, has its outward normal rotated 90° further — i.e. the face is at $\theta=20^{\circ}-90^{\circ}=-70^{\circ}$ from the x-axis (a diagram check: rotating a line by 20° rotates its normal by the same 20°, and the x-face’s own edge is vertical, 90° from this horizontal reference). On the circle this is a rotation of $2\theta=-140^{\circ}$ from point X.
4. Read the plane’s stresses off the circle (confirmed by the transformation equations, used only as a check per the question’s instruction): $$\sigma_{n}=\sigma_{avg}+\dfrac{\sigma_x-\sigma_y}{2}\cos2\theta+\tau_{xy}\sin2\theta=35.45\text{ MPa}$$ $$\tau=-\dfrac{\sigma_x-\sigma_y}{2}\sin2\theta+\tau_{xy}\cos2\theta=-47.46\text{ MPa}$$ $$\boxed{\sigma_n=35.4\text{ MPa (tension)},\ \ \tau=-47.5\text{ MPa}}$$
5. Principal stresses and maximum shear (top and bottom of the circle): $$\boxed{\sigma_1=63.9\text{ MPa},\ \ \sigma_2=-43.9\text{ MPa},\ \ \tau_{max}=R=53.9\text{ MPa (at }\sigma_{avg}=10.0\text{ MPa)}}$$