Sizing a square cantilever under combined compression and torsion: 22-Mec-B9 Advanced Engineering Structures, Undated paper, Q3

Question 3: Sizing a square cantilever under combined compression and torsion (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 cantilever of solid square section $w \times w$ loaded at its free end by an axial compressive force and a torque.

Quantity	Symbol	Value
Axial compressive force	P	258 × 10³ N
Torque	T	27 × 10³ N·m = 27 × 10⁶ N·mm
Yield strength	σ_Y	480 MPa
Safety factor	n	2
Allowable equivalent stress	σ_allow	σ_Y/n = 240 MPa

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

Figure 3.1 — the cantilever with the end couple T and the axial compression P, and the square w × w section. The governing point is the mid-point of any side, where the torsional shear is greatest and the axial stress is uniform.

Approach. Write the two stress components at the critical point as functions of $w$, form each yield criterion in closed form, and solve the resulting single equation numerically after a torsion-only first estimate.

Locate the critical point and write the stresses. The axial force gives a uniform normal stress over the whole section, $$\sigma = -\frac{P}{w^2},$$ negative because $P$ is compressive. For a solid square in Saint-Venant torsion the shear stress is greatest at the mid-point of each side (not at the corners, where it is zero), and the standard result is $$\tau_{\max} = \frac{T}{0.208\,w^3}.$$ The mid-side point therefore carries the full axial stress and the full torsional shear together and is the point to design against.
Assemble the two-dimensional stress state there. With the bar axis as $x$, the element at the mid-side carries $\sigma_x = \sigma$, $\sigma_y = 0$ and $\tau_{xy} = \tau_{\max}$, so the in-plane principal stresses are $$\sigma_{1,2} = \frac{\sigma}{2} \pm \sqrt{\left(\frac{\sigma}{2}\right)^2 + \tau^2},$$ with the third principal stress zero. Because the square-root term always exceeds $\left|\sigma/2\right|$ when a torque is present, one root is positive and the other negative, so the zero principal stress is the intermediate one.
Part (a) — form the Tresca condition. With $\sigma_1 > 0 > \sigma_2$, the Tresca stress is $\sigma_1 - \sigma_2 = 2\sqrt{(\sigma/2)^2 + \tau^2}$, which simplifies to the familiar combined-loading form $$\sqrt{\sigma^2 + 4\tau^2} = \sigma_{\text{allow}} = \frac{\sigma_Y}{n} = 240 \text{ MPa}.$$
Get a first estimate, then solve. Torsion dominates here, so setting $\sigma = 0$ gives $2\tau = 240$, i.e. $\tau = 120$ MPa and $w \approx (T/0.208\tau)^{1/3} = 102.65$ mm. Restoring the axial term and solving $\sqrt{(P/w^2)^2 + 4(T/0.208w^3)^2} = 240$ by bisection gives $$\boxed{\;w_{\text{Tresca}} = 102.83 \text{ mm}\;}$$ At that size $\sigma = -24.40$ MPa and $\tau = 119.38$ MPa, giving principal stresses $+107.8$ and $-132.2$ MPa whose difference is 240.0 MPa — the allowable, as required.
Part (b) — form and solve the von Mises condition. For a state of one normal stress plus one shear the von Mises stress reduces to $$\sigma_{\text{vM}} = \sqrt{\sigma^2 + 3\tau^2} = \sigma_{\text{allow}}.$$ The torsion-only estimate is now $\tau = 240/\sqrt{3} = 138.56$ MPa, giving $w \approx 97.85$ mm, and solving the full equation gives $$\boxed{\;w_{\text{von Mises}} = 98.05 \text{ mm}\;}$$ with $\sigma = -26.83$ MPa and $\tau = 137.70$ MPa at that size.
Cross-check the ratio of the two answers. When the torque dominates, the two criteria differ only through the coefficient on $\tau^2$, so $w_{\text{vM}}/w_{\text{Tresca}} \to (3/4)^{1/6} = 0.9532$. The computed ratio is $98.0529/102.8312 = 0.9535$, within 0.04 % of that limit — a free check that the axial term has been scaled correctly in both equations. Von Mises therefore permits a side about 4.6 % smaller, which is roughly 9 % less cross-sectional area and hence 9 % less mass.

Criterion	Governing equation	Minimum side w	σ at that size	τ at that size
Maximum shear stress (Tresca)	$\sqrt{\sigma^2+4\tau^2} = 240$ MPa	102.83 mm	-24.40 MPa	119.38 MPa
von Mises	$\sqrt{\sigma^2+3\tau^2} = 240$ MPa	98.05 mm	-26.83 MPa	137.70 MPa

Topic. Combined axial load and torsion of a non-circular prismatic bar, sized against a yield criterion with a factor of safety.

Key relations. $\sigma = P/A$; $\tau_{\max} = T/(k\,w^3)$ with $k = 0.208$ for a square (Saint-Venant); Tresca $\sqrt{\sigma^2+4\tau^2} \le \sigma_Y/n$; von Mises $\sqrt{\sigma^2+3\tau^2} \le \sigma_Y/n$.

Why this works. A non-circular section warps out of plane when twisted, so the elementary $\tau = Tr/J$ result does not apply and the peak shear does not occur at the point farthest from the centroid. Saint-Venant's membrane analogy shows the shear stress is proportional to the slope of a soap film stretched over the section: the slope is steepest at the middle of each side and zero at the corners, giving the $0.208w^3$ torsion modulus for a square. Both yield criteria then reduce to a single equation in $w$ because $\sigma$ and $\tau$ scale as $w^{-2}$ and $w^{-3}$ respectively; no iteration on the stress state is needed, only on the size.

Common pitfall. Using $\tau = Tr/J$ with $J = w^4/6$ (the polar second moment) instead of the torsion constant $J_t = 0.1406w^4$ overestimates the capacity badly, and placing the critical point at a corner gets the answer exactly backwards. Also note that both criteria contain $P$ only through $\sigma^2$, so a tensile force of the same magnitude would give an identical $w$.

Source. Boresi & Schmidt, Advanced Mechanics of Materials, 6th ed., §6.2–6.4 (torsion of non-circular sections) and §4.4–4.6 (yield criteria).

Question 3 of 8: Sizing a square cantilever under combined compression and torsion

Question 3: Sizing a square cantilever under combined compression and torsion (20 marks)