22-Mec-A7 Advanced Strength of Materials · May 2018
Question 5 of 8: Square Bar under Axial Load and Torque — Tresca vs Von Mises
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, May 2018 — 16-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours, eight problems of equal value; a complete paper is any five. All eight problems are solved here as a study resource.
Reference texts. Boresi & Schmidt, Advanced Mechanics of Materials (6th); Ugural & Fenster, Advanced Mechanics of Materials and Applied Elasticity (5th); Timoshenko & Goodier, Theory of Elasticity (3rd); Hibbeler, Mechanics of Materials (10th).
Question 5: Square Bar under Axial Load and Torque — Tresca vs Von Mises (20 marks)
Q5. Square bar carrying centroidal axial compression P and torque T; the critical point is the mid-side of the square, where torsional shear peaks and axial stress is uniform.
Given. A square bar carries a centroidal axial force and a torque only (no transverse load, so no bending); design to a factor of safety of 2.
Given data
Axial force
$P=95\ \text{kN}$ (compressive)
Torque
$T=10\ \text{kN}\cdot\text{m}$
Yield strength
$\sigma_Y=390\ \text{MPa}$
Safety factor
$N=2$
Find. The minimum side $b$ from (a) the maximum-shear criterion and (b) von Mises.
Approach. At the mid-side the state is uniform axial stress $\sigma=P/b^2$ plus torsional shear $\tau=T/(0.208\,b^3)$ for a square. Impose each criterion at the allowable stress $\sigma_Y/N$ and solve the resulting equation for $b$.
Stress components (symbolic). $\sigma=\dfrac{P}{b^2}$ (compression), $\tau=\dfrac{T}{\alpha b^3}$ with the square-torsion factor $\alpha=0.208$.
Maximum-shear (Tresca) criterion. For a normal stress plus shear, $\sigma_1-\sigma_2=2\sqrt{(\sigma/2)^2+\tau^2}=\sigma_Y/N$, so $$\sqrt{\left(\frac{P}{2b^2}\right)^2+\left(\frac{T}{0.208\,b^3}\right)^2}=\frac{\sigma_Y}{2N}=97.5\ \text{MPa}.$$ Solving numerically, $$\boxed{b_{\min}=79.1\ \text{mm}}$$ (at which $\sigma=15.2$, $\tau=97.2\ \text{MPa}$ — torsion dominates).
Von Mises criterion. $\sigma_{vm}=\sqrt{\sigma^2+3\tau^2}=\sigma_Y/N=195\ \text{MPa}$, i.e. $$\sqrt{\left(\frac{P}{b^2}\right)^2+3\left(\frac{T}{0.208\,b^3}\right)^2}=195\ \text{MPa}\ \Rightarrow\ \boxed{b_{\min}=75.4\ \text{mm}}$$
Comparison. Maximum-shear is the more conservative theory, so it demands the larger section ($79.1$ vs $75.4$ mm); the shear-dominated state makes the ratio close to the pure-torsion factor $2/\sqrt3$.
