Question 2 of 8: Mohr's Circle — Principal Stresses and Maximum In-Plane Shear

Question 2: Mohr's Circle — Principal Stresses and Maximum In-Plane Shear (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. Reading the arrow directions on the source element (both normal stresses point into the element → compressive; the shear pair is consistent on all four faces):

[Figure not reproduced: Given stress element (arrows as printed on the source exam). See the official exam paper.]

Find. σ₁, σ₂ and their plane orientation; then τ_max,in-plane, the associated normal stress, and its plane orientation.

Approach. Plot the centre and radius of Mohr's circle from σ_x,σ_y,τ_xy, read the principal points and the top/ bottom points off the circle by trigonometry, then rotate the element by half the corresponding central angle.

1. Centre and radius of the circle. $$C = \frac{\sigma_x+\sigma_y}{2} = \frac{-80-30}{2} = -55\ \text{MPa}$$ $$R = \sqrt{\left(\frac{\sigma_x-\sigma_y}{2}\right)^2+\tau_{xy}^2} = \sqrt{(-25)^2+(-20)^2} = 32.02\ \text{MPa}$$
2. (a) Principal stresses (circle intersects the σ-axis at $C\pm R$): $$\boxed{\sigma_1 = C+R = -55.00+32.02 = -22.98\ \text{MPa}\qquad \sigma_2 = C-R = -55.00-32.02 = -87.02\ \text{MPa}}$$
3. Principal-plane angle. From the circle geometry, the angle from point X $(\sigma_x,\tau_{xy})$ to the nearest σ-axis intercept is $$2\theta_0 = \tan^{-1}\!\left(\frac{\tau_{xy}}{(\sigma_x-\sigma_y)/2}\right) = \tan^{-1}\!\left(\frac{-20}{-25}\right) = -141.34^{\circ}$$ Direct substitution back into the transformation equation shows this angle locates σ₂, so the physical rotation to the σ₁ plane is $\theta_0/2 - 90^\circ = -70.67^\circ - 90^\circ$, i.e. $\theta_{p1}= 19.33^\circ$ measured the other way, or equivalently $\boxed{\theta_{p1} = -70.67^{\circ}}$ (CW) from the x-axis to the σ₁ plane, with the σ₂ plane 90° away at $+19.33^\circ$.
4. (b) Maximum in-plane shear. τ_max is simply the circle's radius, with the associated normal stress equal to the centre (both planes of maximum shear carry the same average normal stress): $$\boxed{\tau_{max} = R = 32.02\ \text{MPa}\ ,\qquad \sigma_{avg}=C=-55.00\ \text{MPa}}$$ occurring at $\theta_s = \theta_{p2}-45^{\circ} = 19.33^{\circ}-45^{\circ} = -25.67^{\circ}$ from the x-axis (45° from each principal plane, as required by the circle's geometry).