Square Bar under Axial Load and Torque — Design by Max-Shear: 22-Mec-A7 Advanced Strength of Materials, May 2014, Q6

Question 6: Square Bar under Axial Load and Torque — Design by Max-Shear (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. Solid square section $a\times a$; $P=150$ kN, $T=25$ kN·m; max-shear criterion; $N=2$; $\sigma_Y=250$ MPa. Torsion of a square: $\tau_{\max}=T/(0.208\,a^3)$.

Find. (a) minimum $a$ with $P$ axial; (b) minimum $a$ with $P$ parallel to the section.

Approach. At the critical surface point combine the normal stress (axial $P$) with the torsional shear, form the maximum shear stress, and set it to $\sigma_Y/(2N)$. For part (b) the reoriented $P$ produces a transverse (direct) shear instead of a normal stress, which adds to the torsional shear.

Stresses (part a). Axial normal stress $\sigma=P/a^2$; torsional shear at the mid-side $\tau=T/(0.208\,a^3)$. The state is $\sigma_x=\sigma$, $\tau_{xy}=\tau$, $\sigma_y=\sigma_z=0$.
Max-shear criterion. The in-plane maximum shear is $\tau_{\max}=\sqrt{(\sigma/2)^2+\tau^2}$; Tresca requires $2\tau_{\max}=\sigma_Y/N$, i.e. $$\sqrt{\left(\tfrac{\sigma}{2}\right)^2+\tau^2}=\frac{\sigma_Y}{2N}=62.5\ \text{MPa}.$$
Solve for $a$. Substituting $\sigma=150\,000/a^2$ and $\tau=25\times10^{6}/(0.208\,a^3)$ (N, mm) and solving, $$\boxed{a_{(a)}=124.5\ \text{mm}}.$$ The torsion term dominates; the axial contribution is small.
Part (b): P parallel to the section. Now $P$ is a transverse force, so there is no axial normal stress; instead it gives a direct transverse shear whose maximum (rectangular section) is $\tau_V=1.5P/a^2$, acting at the mid-side in the same sense as the torsional shear. The point is now in pure shear: $$\frac{T}{0.208\,a^3}+\frac{1.5P}{a^2}=62.5\ \text{MPa}\;\Rightarrow\;\boxed{a_{(b)}=134.0\ \text{mm}}.$$ The section must be larger because the transverse load adds directly to the torsional shear rather than appearing as a (smaller-effect) normal stress.

Check / assumption. No bar length is given, so length-dependent bending ($M=PL$) cannot be evaluated. Part (b) is therefore taken at the loaded section, where the reoriented $P$ produces only direct transverse shear $V=P$; this shear and the torsional shear are superposed at the mid-side point where both act in the same direction. If a length were specified, the bending stress at the fixed end would govern instead.

Case	Minimum dimension $a$
(a) $P$ axial (normal to section)	$124.5$ mm
(b) $P$ parallel to section	$134.0$ mm

Topic: Design of a square shaft under combined loading using the maximum-shear-stress (Tresca) criterion.

Key relations: Square-section torsion $\tau=T/(0.208a^3)$; axial $\sigma=P/A$; transverse shear (rectangle) $\tau_V=1.5V/A$; Tresca design $\sqrt{(\sigma/2)^2+\tau^2}=\sigma_Y/(2N)$.

Why this works: A non-circular shaft in torsion has its peak shear at the middle of each side (not the corners, where shear is zero). The maximum-shear criterion compares the largest shear at that point to half the yield stress divided by the factor of safety. When the axial load is replaced by a transverse one, its effect changes character—from a normal stress that combines under the root, to a shear stress that adds arithmetically—so the required size grows.

Common pitfall: Using the circular-shaft formula $\tau=16T/(\pi d^3)$ for a square, placing the peak torsional shear at the corner, or forgetting the factor $2$ that converts “maximum shear” to the Tresca equivalent stress. In (b), do not re-introduce an axial normal stress—the load is now in-plane.

Source: Boresi & Schmidt, Advanced Mechanics of Materials 6th ed., Ch. 6 (torsion of non-circular sections) and Ch. 4 (failure criteria); Hibbeler §5.

Question 6 of 8: Square Bar under Axial Load and Torque — Design by Max-Shear

Question 6: Square Bar under Axial Load and Torque — Design by Max-Shear (20 marks)