Question 7 of 8: Combined Axial Load and Torque from a Strain Rosette

Question 7: Combined Axial Load and Torque from a Strain Rosette (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. A solid shaft under axial tension plus torsion, read by a $0^\circ/45^\circ/90^\circ$ rosette (gauge 0 axial).

Find. The torque $T$ and axial load $P$.

Approach. On the free surface the only stresses are axial $\sigma$ (from $P$) and shear $\tau$ (from $T$); the circumferential direction is unloaded, so $\sigma_y=0$. The axial gauge gives $\sigma$ directly; the rosette combination gives $\gamma_{xy}$ and hence $\tau$.

1. Axial stress and load. With $\sigma_y=0$, $\varepsilon_0=\sigma/E$, so $\sigma=E\varepsilon_0=40\,000(250\times10^{-6})=10\ \text{MPa}$ and $$P=\sigma A=10\cdot\frac{\pi}{4}(50)^2=\boxed{19.6\ \text{kN}}.$$
2. Shear strain from the rosette. For a rectangular rosette $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(-50)-250-(-150)=-200\ \mu$.
3. Shear stress. $G=\dfrac{E}{2(1+\nu)}=15.4\ \text{GPa}$, so $\tau=G\gamma_{xy}=15\,385(200\times10^{-6})=3.08\ \text{MPa}$ (magnitude).
4. Torque. With $J=\dfrac{\pi d^4}{32}=6.14\times10^{5}\ \text{mm}^4$ and $r=25\ \text{mm}$, $$T=\frac{\tau J}{r}=\frac{3.08(6.14\times10^{5})}{25}=\boxed{75.5\ \text{N}\cdot\text{m}}.$$