Question 4 of 8: Design of a square bar under combined axial load and torque

Question 4: Design of a square bar under combined 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 solid square prismatic bar carrying a constant axial compression and a constant torque along its whole length, to be sized against yield.

Find. The minimum side $w$ (a) on the maximum-shear-stress criterion and (b) on the von Mises criterion.

Approach. Write the two stress components at the critical point as functions of $w$, express each yield criterion as a single equation in $w$, and solve each numerically for the allowable stress $\sigma_Y/n$.

1. Set the allowable stress. The safety factor is applied to the yield strength, so both criteria are checked against $$\sigma_{\text{allow}} = \frac{\sigma_Y}{n} = \frac{480}{1.5} = 320\ \text{MPa}.$$
2. Locate the critical point and write its stresses. For a solid square in torsion the shear stress is largest at the middle of each side (not at the corners, where it is zero), and the standard result for a square is $$\tau = \frac{T}{0.208\,w^3}.$$ The axial force is uniform over the section, $$\sigma = -\frac{P}{w^2},$$ and it acts along the bar axis, which is the same direction in which the torsional shear acts tangentially. The mid-side point is therefore in plane stress with one normal component and one shear component; the free surface carries no third stress.
3. Part (a) — reduce Tresca to one equation in $w$. For plane stress with $\sigma$ and $\tau$ the principal stresses are $\sigma/2 \pm \sqrt{(\sigma/2)^2 + \tau^2}$ together with zero. Because $\sigma$ is compressive here, the extreme pair straddles zero and Tresca becomes $$\sigma_1 - \sigma_3 = \sqrt{\sigma^2 + 4\tau^2} = \sigma_{\text{allow}},$$ that is $$\left(\frac{265\,000}{w^2}\right)^2 + 4\left(\frac{28 \times 10^6}{0.208\,w^3}\right)^2 = 320^2.$$
4. Solve for the Tresca size. The equation contains both $w^{-4}$ and $w^{-6}$, so it is solved numerically; bisection on $40 \le w \le 400$ mm converges to $$\boxed{w_{\text{Tresca}} = 94.5\ \text{mm}}.$$ At that size the two components are $\sigma = -29.6$ MPa and $\tau = 159.3$ MPa: the torque supplies almost all of the demand and the axial force is very nearly incidental.
5. Part (b) — repeat with the distortion-energy criterion. For the same plane-stress state the von Mises equivalent stress reduces to $$\sigma_e = \sqrt{\sigma^2 + 3\tau^2} = \sigma_{\text{allow}},$$ the only change being the factor 3 in place of 4 on the shear term. Solving the same way, $$\boxed{w_{\text{von Mises}} = 90.1\ \text{mm}}.$$
6. Compare the two answers and sanity-check the ratio. The von Mises bar is 4.6 % smaller on the side and 9.0 % smaller in cross-sectional area. That figure is not a coincidence: when the torque dominates, the axial term is negligible, both criteria reduce to $w^3 \propto \tau$ with the shear factors 4 and 3, and the size ratio tends to the fixed limit $$\frac{w_{\text{vM}}}{w_{\text{Tresca}}} \rightarrow \left(\frac{3}{4}\right)^{1/6} = 0.9532,$$ against the 0.9535 actually computed. A solve landing far from 4.6 % means the axial term has been mis-scaled.

Two features of this problem are worth recording. First, both criteria contain the axial force only as $P^2$, so a tensile force of the same magnitude gives an identical $w$ — the compressive sense stated in the question changes nothing about the yield check (it would matter for buckling, which is not asked). Second, the answer is very insensitive to $P$: dropping the axial force entirely would give $w_{\text{Tresca}} = 94.4$ mm, a change of about 0.1 %.