Square cantilever bar — axial + torsion design (Tresca, SF = 3): 22-Mec-A7 Advanced Strength of Materials, December 2018, Q6

Question 6: Square cantilever bar — axial + torsion design (Tresca, SF = 3) (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 $a\times a$; $P=175\text{ kN}$, $T=20\text{ kN}\cdot\text{m}$; Tresca criterion, safety factor $N=3$, $\sigma_Y=350\text{ MPa}$.

Given data
Axial force $P$	175 kN
Torque $T$	20 kN·m
Yield strength $\sigma_Y$	350 MPa
Safety factor $N$	3
Square-shaft torsion factor	$\alpha=0.208$ ($\tau_\text{max}=T/\alpha a^3$)

Find. Minimum $a$ for (a) $P$ acting axially (normal to the section) and (b) $P$ acting parallel to the section (transverse shear).

Approach. Locate the critical surface point (mid-side of the square, where torsional shear peaks). For (a) that point sees axial normal stress plus torsional shear; for (b) it sees transverse shear plus torsional shear. Apply Tresca ($\tau_\text{max}=\sigma_Y/2N$) and solve for $a$.

The allowable shear is $\tau_\text{allow}=\dfrac{\sigma_Y}{2N}=\dfrac{350}{6}=58.33\text{ MPa}.$

(a) Axial + torsion. At the mid-side, $\sigma=P/a^2$ (normal) and $\tau=T/\alpha a^3$ (shear). The maximum shear stress for this $(\sigma,\tau)$ state is $$\tau_\text{max}=\sqrt{\left(\tfrac{\sigma}{2}\right)^2+\tau^2}=\sqrt{\left(\frac{P}{2a^2}\right)^2+\left(\frac{T}{\alpha a^3}\right)^2}=\tau_\text{allow}.$$ Solving numerically (torsion dominates: at the solution $\sigma=12.5$ MPa, $\tau=58.0$ MPa), $$\boxed{a_\text{(a)}=118.4\text{ mm}}.$$
(b) Transverse + torsion. If $P$ is applied parallel to the section it produces a transverse shear force. With no length given, bending is neglected (flagged below) and only the transverse shear is retained. For a rectangular section the peak transverse shear is $1.5V/A=1.5P/a^2$, which superposes on the torsional shear (both are pure shear at the critical point): $$\tau_\text{max}=\frac{T}{\alpha a^3}+\frac{1.5P}{a^2}=\tau_\text{allow}.$$ Solving (at the solution $\tau_\text{tor}=43.0$ MPa, $\tau_V=15.4$ MPa), $$\boxed{a_\text{(b)}=130.8\text{ mm}}.$$
Interpret. Case (b) requires a larger section: a transverse shear of $1.5P/a^2$ adds directly to torsion, whereas the axial normal stress of case (a) combines only through the $(\sigma/2)$ term in $\tau_\text{max}$ and so is less damaging here.

Check: Part (b) gives no member length, so the bending moment $PL$ from the offset/parallel load cannot be formed; the transverse-shear-only reading is the intended simplification. If a length were supplied, bending would dominate and control the design.

Final results — Question 6
Case	Minimum $a$
(a) $P$ axial + torsion	118.4 mm
(b) $P$ transverse + torsion	130.8 mm

Topic. Combined-stress design of a non-circular (square) shaft by the maximum-shear-stress theory.

Key relations. Square-shaft torsion $\tau_\text{max}=T/(\alpha a^3)$, $\alpha=0.208$; transverse shear $\tau=1.5V/A$; Tresca in-plane $\tau_\text{max}=\sqrt{(\sigma/2)^2+\tau^2}=\sigma_Y/2N$.

Why this works. The most stressed surface point of a torqued square bar is the mid-point of a side (torsional shear is maximal there and zero at the corners). Adding an axial or transverse load at that point gives a plane stress state whose Tresca measure is the in-plane maximum shear; setting it to the allowable $\sigma_Y/2N$ yields one equation in $a$. Because the two loads enter at different powers of $a$ ($a^2$ vs $a^3$), the equation is solved numerically.

Common pitfall. Using the circular-shaft $\tau=Tr/J$ for a square section, or adding the axial normal stress directly to shear. Normal and shear stresses combine through the Mohr radius, not by simple addition; only two shear contributions (torsion + transverse) add arithmetically.

Source. Boresi & Schmidt, Advanced Mechanics of Materials, Ch. 6 (Torsion of non-circular sections) & Ch. 4 (yield criteria); Ugural & Fenster §6.6.

Question 6 of 8: Square cantilever bar — axial + torsion design (Tresca, SF = 3)

Question 6: Square cantilever bar — axial + torsion design (Tresca, SF = 3) (20 marks)