22-Mec-A7 Advanced Strength of Materials · December 2019
Question 6 of 7: Minimum shaft diameter under axial load and torque
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2019 — 16-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours. Seven problems of equal value; any five constitute a complete paper. All seven problems are solved as a study resource.
Given. A solid circular shaft under simultaneous axial tension and torsion, sized to the onset of yield by von Mises.
Given data
Quantity
Symbol
Value
Axial load
$P$
250 kN
Torque
$T$
50 kN·m $=5\times10^{7}$ N·mm
Yield strength
$\sigma_Y$
250 MPa
Criterion
Maximum distortion energy (von Mises)
Find. The minimum shaft diameter $d$ for which the von Mises stress just reaches yield.
Approach. Express the axial normal stress and the torsional shear in terms of $d$, combine them through the plane-stress von Mises form $\sqrt{\sigma^2+3\tau^2}=\sigma_Y$, and solve for $d$.
Stresses at the surface (worst point — torsion is maximal there; axial is uniform):
$$\sigma=\frac{4P}{\pi d^2},\qquad \tau=\frac{16T}{\pi d^3}$$
Von Mises with a normal + shear pair. For $\sigma_x=\sigma$, $\tau_{xy}=\tau$ (others zero),
$$\sigma_{vM}=\sqrt{\sigma^2+3\tau^2}=\sigma_Y$$
Substitute and solve. Writing everything in terms of $d$ (mm):
$$\sqrt{\left(\frac{4(250\,000)}{\pi d^2}\right)^2+3\left(\frac{16(5\times10^{7})}{\pi d^3}\right)^2}=250$$
Solving this single equation numerically (bisection) gives
$$\boxed{d_{\min}\approx121\ \text{mm}}$$
At $d=121$ mm, $\sigma=21.7$ MPa and $\tau=143.7$ MPa, so $\sqrt{\sigma^2+3\tau^2}=250$ MPa — torsion dominates the design.
