Question 6 of 8: Sizing a square bar under combined axial force and torque

Question 6: Sizing a square bar under combined 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.

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

Approach. Write both stress components as functions of $w$, form each criterion as a single equation in $w$, and solve — the equations are not polynomial in a convenient power, so solve them numerically.

1. Fix the allowable stress. Dividing yield by the safety factor,$$\sigma_{\text{allow}} = \frac{\sigma_Y}{N} = \frac{320}{3} = 106.67\ \text{MPa}$$Both criteria will be written as an equivalent stress set equal to this value.
2. Express the two stress components. The axial stress is uniform over the section, and for a solid square in torsion the peak shear stress occurs at the mid-point of each side face:$$|\sigma| = \frac{P}{w^2},\qquad\tau_{\max} = \frac{T}{0.208\,w^3}$$The coefficient 0.208 is the standard elasticity result for a square section (Saint-Venant); the corners, being free surfaces intersecting at a right angle, carry no shear at all.
3. Identify the critical element. At the mid-side point the element sees one normal stress and one shear stress and nothing else, so it is in plane stress with in-plane principal stresses$$\sigma_{1,2} = \frac{\sigma}{2} \pm\sqrt{\left(\frac{\sigma}{2}\right)^2+\tau^2},\qquad \sigma_3 = 0$$Because the radical always exceeds $|\sigma/2|$, one principal stress is positive and one negative, and the zero third principal stress lies between them.
4. Form the maximum-shear-stress criterion (part a). The governing difference is between the two in-plane principal stresses, and it reduces to a compact form:$$\sigma_1-\sigma_2 = 2\sqrt{\left(\frac{\sigma}{2}\right)^2+\tau^2} = \sqrt{\sigma^2+4\tau^2} = \sigma_{\text{allow}}$$Substituting the two expressions in $w$,$$\sqrt{\left(\frac{235\,000}{w^2}\right)^2+4\left(\frac{2.25\times10^{7}}{0.208\,w^3}\right)^2} = 106.67$$Solving numerically,$$\boxed{w_{\text{Tresca}} = 127.0\ \text{mm}}$$At that size $|\sigma| = 14.57$ MPa and $\tau = 52.83$ MPa.
5. Form the von Mises criterion (part b). For the same plane-stress element the distortion-energy criterion collapses to$$\sqrt{\sigma^2+3\tau^2} = \sigma_{\text{allow}}$$and solving the corresponding equation in $w$ gives$$\boxed{w_{\text{von Mises}} = 121.1\ \text{mm}}$$with $|\sigma| = 16.02$ MPa and $\tau = 60.89$ MPa. Each design sits exactly on its own yield surface, which is the check on both roots.
6. Compare the two answers against the theoretical limit. The ratio is $121.1/127.0 = 0.9538$. When the torque dominates, the axial term becomes negligible and the two criteria reduce to $2\tau = \sigma_{\text{allow}}$ and $\sqrt3\,\tau = \sigma_{\text{allow}}$, so the side ratio tends to the fixed value$$\left(\frac{3}{4}\right)^{1/6} = 0.95318$$The computed 0.9538 sits just above that limit, which confirms that the axial force here really is a minor player — it accounts for only 13.7 per cent of the Tresca equivalent stress. A result far from 0.953 would signal an error in scaling the axial term.

Two design observations follow. First, von Mises permits a bar 4.6 per cent smaller on the side, which is a 9.0 per cent saving in cross-sectional area and therefore in material cost; for a ductile steel that is the better-supported criterion, and the saving is real rather than a licence to be less careful. Second, the compressive sense of $P$ turns out not to matter. Both criteria contain $\sigma$ only as $\sigma^2$, so a tensile force of the same magnitude would give exactly the same $w$. What a compressive force does introduce is a buckling question, which this problem does not ask — but at 127 mm square the bar is extremely stocky and column instability would only govern for a very long member.