Question 5 of 8: Square Bar under Axial Force and Torque

Question 5: Square Bar under Axial Force 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 solid square shaft, $\sigma_Y=350$ MPa, $P=171$ kN, $T=19$ kN·m, Tresca design with safety factor $N=5$.

Find. The minimum side $b$ for (a) axial $P$ combined with torsion, and (b) $P$ acting transverse (parallel) to the section.

Approach. For a solid square, the peak torsional shear is $\tau=T/(0.208\,b^3)$ at the mid-side. Combine it with the direct stress under Tresca ($\sigma_1-\sigma_3=\sqrt{\sigma^2+4\tau^2}$) and solve for $b$; in case (b) the axial stress is replaced by a transverse-shear stress that adds directly to the torsional shear.

1. Allowable value. Tresca with $N=5$ gives $\sigma_Y/N=350/5=70$ MPa.
2. (a) Axial + torsion. $\sigma=P/b^2$, $\tau=T/(0.208b^3)$; the criterion $\sqrt{\sigma^2+4\tau^2}=70$ MPa is solved numerically for $\boxed{b=138.1\ \text{mm}}$ (torsion dominates: $\tau\approx65$ MPa vs $\sigma\approx9$ MPa at that size).
3. (b) $P$ parallel to the section. Now $P$ is a transverse shear force adding a peak $\tau_P=1.5P/b^2$ to the torsional shear (no normal stress). The most-stressed mid-side sees $\tau=T/(0.208b^3)+1.5P/b^2$, and Tresca $2\tau=70$ MPa gives $\boxed{b=155.3\ \text{mm}}$.
4. Compare. Case (b) needs a larger section because the two shear stresses add arithmetically at the critical point, whereas in case (a) the axial stress combines with shear under the square-root (root-sum) law.