Sizing a Square Bar Under Axial Load and Torque: 22-Mec-A7 Advanced Strength of Materials, May 2015, Q7

Question 7: Sizing a Square Bar Under 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. Square bar, side b, carrying axial P and torque T; design by Tresca with a factor of safety of 3.

Given data
Yield stress, σ_Y	310 MPa
Axial force, P	195 kN
Torque, T	17 kN·m
Factor of safety, N	3
Square-torsion constant	τ_max = T/(0.208 b³)

Find. The minimum side b for (a) axial (normal) P and (b) P applied parallel to the section.

Square-section bar built into a rigid support, loaded by centroidal axial force P and torque T.

Approach. Combine the normal stress from P with the torsional shear from T at the most-stressed point; the Tresca criterion limits the maximum in-plane shear to $\sigma_Y/(2N)$. Solve the resulting nonlinear equation for b.

Allowable shear. The maximum-shear-stress criterion with safety factor 3 gives $$\tau_{\text{allow}}=\frac{\sigma_Y}{2N}=\frac{310}{6}=51.7\ \text{MPa}.$$
(a) Axial + torsion. The axial stress is $\sigma=P/b^2$ and the torsional shear $\tau=T/(0.208b^3)$. The largest shear in the element combining a normal and a shear stress is $\tau_{\max}=\sqrt{(\sigma/2)^2+\tau^2}$, so $$\sqrt{\Big(\frac{P}{2b^2}\Big)^2+\Big(\frac{T}{0.208b^3}\Big)^2}=51.7\ \text{MPa}\ \Rightarrow\ \boxed{b=117\ \text{mm}.}$$ At this size $\sigma=14.3$ MPa and $\tau=51.2$ MPa—torsion dominates.
(b) P applied parallel to the section. Now P is a transverse shear force, not an axial load; its peak (centroidal) shear in a rectangular section is $\tau_P=1.5P/b^2$, acting on the same faces as the torsional shear, so they add: $$\frac{1.5P}{b^2}+\frac{T}{0.208b^3}=51.7\ \text{MPa}\ \Rightarrow\ \boxed{b=133\ \text{mm}.}$$ A larger section is needed because a transverse shear stress superposes directly on the torsional shear rather than combining under a square root.

Check: Part (b) superposes the maximum transverse-shear stress on the maximum torsional shear. Both peak at the mid-point of the side faces parallel to P (the neutral axis for transverse shear, the mid-side for torsion of a square), and on one of those faces they act in the same direction, so the arithmetic sum is the elementary-theory maximum, not an over-estimate. Bending from P is neglected because no member length or eccentricity is given; if a length were supplied, the bending stress at the support would have to be added to part (b).

Final results — Question 7
Case	Minimum side b
(a) P normal (axial) + torque	117 mm
(b) P parallel (transverse shear) + torque	133 mm

Topic: Combined loading and design by the maximum-shear-stress (Tresca) criterion.

Key relations: $\sigma=P/A$; square-shaft torsion $\tau=T/(0.208b^3)$; transverse shear $\tau=1.5V/A$; $\tau_{\max}=\sqrt{(\sigma/2)^2+\tau^2}\le\sigma_Y/(2N)$.

Why this works: Tresca limits the peak shear stress in the element. When P is axial it produces a normal stress that combines with the torsional shear through Mohr’s circle (the square-root form); when P is transverse it produces a shear stress that adds arithmetically to the torsional shear. The two load orientations therefore give different governing equations and different required sizes.

Common pitfall: using the circular-shaft $T r/J$ instead of the square-section constant 0.208, and treating a transverse shear like an axial stress (or vice-versa) when combining with torsion.

Source: Boresi, Advanced Mechanics of Materials, Ch. 6 (torsion of non-circular sections); Hibbeler, Ch. 8–10.

Question 7 of 8: Sizing a Square Bar Under Axial Load and Torque

Question 7: Sizing a Square Bar Under Axial Load and Torque (20 marks)