Question 2 of 8: Plane Stress by Mohr's Circle

Question 2: Plane Stress by 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.

Find. (a) σ, τ on the 40° inclined plane; (b) the maximum in-plane shear stress, its associated normal stress, and the orientation of that plane — all via a numerically-constructed Mohr's circle.

Approach. Adopt the sign convention that shear is positive when it tends to rotate the element clockwise on the face under consideration; plot points A (x-face) and B (y-face), draw the circle through them, then read every requested quantity as a chord/point on that circle located by trigonometry (never scaled off).

1. Sign convention and circle parameters. On the +x face the given τ_xy acts downward (−y), i.e. it is a positive (clockwise-tending) shear by the adopted convention; the complementary shear on the +y face then acts leftward. Point A = (σ_x, τ_xy,CW) = (30, +30); point B = (σ_y, −τ_xy,CW) = (−110, −30). Centre and radius: $$C = \frac{\sigma_x+\sigma_y}{2} = \frac{30-110}{2} = -40\ \text{MPa}$$ $$R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = \sqrt{70^2+30^2} = \boxed{76.16\ \text{MPa}}$$
2. Part (a) — rotate 2θ = 80° (clockwise) from point A. The 40° physical rotation of the cut plane corresponds to an 80° rotation around the circle, in the same (clockwise) sense. Reading the new point by trigonometry on the circle (equivalently, evaluating the transformation equations at θ = −40°, which the question permits as a check): $$\sigma_n = \frac{\sigma_x+\sigma_y}{2} + \frac{\sigma_x-\sigma_y}{2}\cos(2\theta) + \tau_{xy}\sin(2\theta)\Big|_{\theta=-40^\circ} = \boxed{1.70\ \text{MPa (tension)}}$$ $$\tau_{nt} = -\frac{\sigma_x-\sigma_y}{2}\sin(2\theta) + \tau_{xy}\cos(2\theta) \Big|_{\theta=-40^\circ} = \boxed{63.7\ \text{MPa}}$$ with the companion face at $\sigma_{n'} = -81.70$ MPa (consistent check: $\sigma_n+\sigma_{n'} = -80.0 = \sigma_x+\sigma_y$, the stress invariant).
3. Part (b) — maximum in-plane shear. The maximum shear on the circle is simply the radius, occurring at the top/bottom of the circle: $$\tau_{max} = R = \boxed{76.16\ \text{MPa}}, \qquad \sigma_{avg} = C = -40\ \text{MPa (on both faces of that element)}$$ The principal-plane angle (needed to locate the max-shear planes, which sit 45° from the principal planes) is $$\theta_p = \tfrac12\arctan\!\left(\frac{\tau_{xy}}{(\sigma_x-\sigma_y)/2}\right) = -11.60^\circ \ \Rightarrow\ \theta_s = \theta_p - 45^\circ = -56.60^\circ$$ i.e. the max-shear face is 56.6° clockwise from the x-face. Principal stresses (bonus check): $$\sigma_{1,2} = C \pm R = 36.16\ \text{MPa},\ -116.16\ \text{MPa}$$

Final Results.