Question 5 of 8: Sizing a square bar under combined compression and torque

Question 5: Sizing a square bar under combined compression 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 bar carrying a uniform axial compression together with a uniform torque, to be sized against yield with a factor of safety of 4.

Find. The minimum side dimension $w$ under (a) the maximum-shear-stress (Tresca) criterion and (b) the maximum-distortion-energy (von Mises) criterion.

Approach. Both loads produce their maxima at the mid-point of a face, where the state of stress is one normal component plus one shear component; write the two effective stresses for that state, set each equal to the allowable stress and solve the resulting algebraic equation for $w$.

1. Allowable stress. The factor of safety is applied to the yield strength, so the design limit on the chosen effective stress is $$\sigma_{\text{all}} = \frac{\sigma_Y}{n} = \frac{325}{4} = 81.25\ \text{MPa}$$
2. The critical stress state. On the mid-point of any face the axial load gives $\sigma = P/w^{2}$ (compressive) and the torque gives the peak shear of a square shaft, $$\tau_{\max} = \frac{T}{0.208\,w^{3}}$$The 0.208 coefficient is the standard Saint-Venant result for a square cross-section; a square is not a circle and the polar second moment cannot be used. There is no stress on the face normal, so the state is plane stress with principal values $\sigma_{1,2} = \sigma/2 \pm \sqrt{(\sigma/2)^{2}+\tau^{2}}$ and $\sigma_3 = 0$.
3. Part (a) — the Tresca condition. For this state the two in-plane principals straddle zero, so the largest principal difference is $\sigma_1-\sigma_2$ and the criterion reduces to $$\sqrt{\sigma^{2}+4\tau^{2}} = \sigma_{\text{all}}$$Substituting the two expressions gives one equation in $w$: $\sqrt{(178\,000/w^{2})^{2} + 4(17\times10^{6}/0.208w^{3})^{2}} = 81.25$. Solving numerically, $$\boxed{w_{\text{Tresca}} = 126.6\ \text{mm}}$$At that size $\sigma = 11.10$ MPa and $\tau = 40.24$ MPa, so the torque is overwhelmingly the governing load.
4. Part (b) — the von Mises condition. For the same plane state the distortion-energy criterion is $$\sqrt{\sigma^{2}+3\tau^{2}} = \sigma_{\text{all}}$$differing from Tresca only in the coefficient on the shear term. Solving the same way, $$\boxed{w_{\text{von Mises}} = 120.8\ \text{mm}}$$with $\sigma = 12.20$ MPa and $\tau = 46.38$ MPa at that size.
5. Comparing the two answers. The von Mises bar is 5.8 mm smaller, a saving of 4.6 per cent on the side and about 9 per cent on the cross-sectional area. That is the expected relationship: in pure shear Tresca and von Mises differ by the maximum $2/\sqrt{3} = 1.155$, and this problem is very nearly pure shear, so the gap sits near its theoretical ceiling of $(1.155)^{1/3} = 1.049$ on the dimension.
6. A useful observation about the sign of $P$. Both criteria contain $\sigma$ only through $\sigma^{2}$, so the answer is unchanged if the axial load is tensile rather than compressive. What the compressive sense does demand is a separate buckling check on the cantilever, which the question does not supply data for.