Question 4 of 8: Square bar under axial load and torque

Question 4: Square bar under axial load and torque (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. A cantilevered aluminium-alloy bar of solid square section a×a under a centroidal compressive axial force and a torque, designed to the maximum-shear-stress criterion.

Find. the minimum allowable side dimension a.

Approach. Combine the axial normal stress with the torsional shear of a square shaft, form the Tresca equivalent, set it to the factored yield, and solve for a.

1. Stress components. Axial $\sigma=P/a^2$; for a square section the peak torsional shear (mid-side) is $$\tau=\frac{T}{0.208\,a^{3}}.$$
2. Tresca condition. With only $\sigma$ and $\tau$ acting, the maximum shear stress is $\sqrt{(\sigma/2)^2+\tau^2}$, so design requires $$\sqrt{\left(\tfrac{\sigma}{2}\right)^2+\tau^2}=\frac{\sigma_Y}{2N}=\frac{350}{4}=87.5\ \text{MPa}.$$
3. Solve. Substituting $\sigma=177\times10^3/a^2$ and $\tau=23\times10^6/(0.208a^3)$ (a in mm) and solving numerically, $$a_{\min}=108.3\ \text{mm}\ \Rightarrow\ \boxed{a=109\ \text{mm}}.$$

Torsion controls the design: at the solution the shear term is roughly ten times the axial half-stress, so the required size is set almost entirely by T. Rounding up to 109 mm keeps the working stress just inside the allowable and honours the factor of safety.