Question 1 of 8: Mohr's Circle for a Plane-Stress Element

Question 1: Mohr's Circle for a Plane-Stress Element (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 principal stresses and the orientation of the principal planes. (b) The maximum in-plane shear stress, its associated normal stress, and the orientation of the planes on which it acts.

Approach. Read σ_x, σ_y, τ_xy off the element using the standard sign convention (positive τ_xy on the +x face points in +y); plot the two face-points on the σ–τ plane, draw the circle through them by trigonometry (centre and radius from the plotted coordinates — no transformation equation), then read every requested quantity as a point on that circle.

1. Sign the stresses from the sketch. Both normal stresses point INTO the element (compression): $$\sigma_x=-85\ \text{MPa}, \qquad \sigma_y=-5\ \text{MPa}$$ On the right (+x) face the 30 MPa shear arrow points downward ($-y$); by the complementary-shear rule the top (+y) face's arrow then points leftward ($-x$) — both consistent with a single signed value $$\tau_{xy}=-30\ \text{MPa}$$
2. Construct the circle: centre and radius. Plot point $X(\sigma_x,\tau_{xy})= (-85,-30)$ and point $Y(\sigma_y,-\tau_{xy})=(-5,30)$; these are diametrically opposite, so the centre is their midpoint and the radius is half their separation — both found by simple trigonometry on the two plotted coordinates, not by the transformation formula: $$C=\frac{\sigma_x+\sigma_y}{2}=\frac{-85-5}{2}=\boxed{-45\ \text{MPa}}$$ $$R=\sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2}= \sqrt{(-40)^2+(-30)^2}=\sqrt{1600+900}=\boxed{50\ \text{MPa}}$$
3. (a) Principal stresses — circle intersects the σ-axis. $$\sigma_1 = C+R = -45+50=\boxed{5\ \text{MPa}}, \qquad \sigma_2 = C-R = -45-50=\boxed{-95\ \text{MPa}}$$ The angle from point X to the $\sigma_1$ point, measured around the circle, is $2\theta_{p}=\tan^{-1}\!\left(\dfrac{30}{40}\right)$ referenced from the circle geometry (a $40$–$30$–$50$ right triangle formed by X, C and the σ-axis): $$\beta=\tan^{-1}\!\left(\frac{30}{40}\right)=36.87^\circ$$ Working around the circle from X gives the physical principal-plane angle (measured CCW from the x-face) for $\sigma_2$: $\theta_{p2}=18.4^\circ$; the orthogonal plane, 90° further, carries $\sigma_1$: $\theta_{p1}=108.4^\circ$ (equivalently $-71.6^\circ$). Checked directly against the transformation equation at each angle (permitted only as a check): both reproduce $\sigma(18.4^\circ)=-95$ MPa and $\sigma(108.4^\circ)=+5$ MPa exactly.
4. (b) Maximum in-plane shear — top/bottom of the circle. $$\boxed{\tau_{max}=R=50\ \text{MPa}}, \qquad \sigma_{avg}=C=\boxed{-45\ \text{MPa (on both faces of that element)}}$$ occurring on planes 45° from the principal planes: $$\theta_{s}=\theta_{p2}-45^\circ= \boxed{-26.6^\circ}\ \text{(equivalently } 63.4^\circ\text{, the orthogonal plane)}$$

Final Results.