Torque and Axial Load from a Strain-Gauge Rosette: 22-Mec-A7 Advanced Strength of Materials, May 2014, Q3

Question 3: Torque and Axial Load from a Strain-Gauge 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. $d=50$ mm; rectangular (0°/45°/90°) rosette with $\varepsilon_0=250\,\mu\varepsilon$ (axial), $\varepsilon_{45}=-50\,\mu\varepsilon$, $\varepsilon_{90}=-150\,\mu\varepsilon$; $E=40$ GPa, $v=0.3$.

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

Approach. On the free (outer) surface the circumferential stress is zero, so the axial gauge alone fixes $\sigma_x$ and hence $P$. The rosette’s three readings give the surface shear strain $\gamma_{xy}$, which through $G$ gives the torsional shear stress and thus $T=\tau J/r$.

Axial stress and load. The bar surface is traction-free in the hoop direction ($\sigma_{\text{hoop}}=0$), so $\sigma_x=E\varepsilon_0$: $$\sigma_x=40\times10^{3}(250\times10^{-6})=10.0\ \text{MPa}\;\Rightarrow\;P=\sigma_x\!\cdot\!\tfrac{\pi}{4}d^{2}=10.0\,(1963.5)=\boxed{19.6\ \text{kN}}.$$
Surface shear strain. For a rectangular rosette the engineering shear strain is $$\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(-50)-250-(-150)=-200\ \mu\varepsilon.$$
Shear stress. With $G=\dfrac{E}{2(1+v)}=\dfrac{40}{2.6}=15.38$ GPa, $$\tau=G\,\gamma_{xy}=15.38\times10^{3}(200\times10^{-6})=3.08\ \text{MPa}.$$
Torque. Using $\tau=\dfrac{T\,r}{J}$ with $J=\dfrac{\pi d^{4}}{32}=6.136\times10^{5}\ \text{mm}^4$ and $r=25$ mm, $$T=\frac{\tau J}{r}=\frac{3.08\,(6.136\times10^{5})}{25}=\boxed{75.5\ \text{N}\cdot\text{m}}.$$

Check / note. A pure axial + torsion surface would give $\varepsilon_{90}=-v\varepsilon_0=-75\,\mu\varepsilon$, whereas the gauge reads $-150\,\mu\varepsilon$. This mismatch is gauge scatter / minor bending, not evidence of a hoop stress: the outer fibre is a free surface, so $\sigma_{\text{hoop}}=0$ is enforced and $\sigma_x=E\varepsilon_0$ is the correct axial reading. We do not invent a transverse stress to reconcile $\varepsilon_{90}$.

Quantity	Result
Axial stress $\sigma_x$	$10.0$ MPa
Axial load $P$	$19.6$ kN
Shear stress $\tau$	$3.08$ MPa
Torque $T$	$75.5$ N·m

Topic: Recovering the loads on a shaft from a surface strain-gauge rosette (combined axial + torsion).

Key relations: Rectangular rosette $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}$; free-surface axial $\sigma_x=E\varepsilon_0$; $\tau=G\gamma_{xy}$; $P=\sigma_x A$; $T=\tau J/r$.

Why this works: A rosette measures three normal strains at one point, enough to reconstruct the full in-plane strain tensor, hence $\gamma_{xy}$. On the bar’s outer surface the only stresses are the axial normal stress (from $P$) and the torsional shear (from $T$); the radial and hoop stresses vanish because the surface is unloaded. That decoupling lets the axial gauge give $P$ directly and the shear strain give $T$, with no simultaneous equations required.

Common pitfall: Using $\gamma_{xy}=\varepsilon_{45}$ (wrong—the 45° gauge is a normal strain, not the shear), or trying to force $\varepsilon_{90}$ into a nonexistent hoop stress. Also keep units consistent: $J$ in mm$^4$ with $\tau$ in MPa yields $T$ in N·mm.

Source: Hibbeler, Mechanics of Materials §10.4–10.5 (strain rosettes) and Ch. 5 (torsion); Boresi & Schmidt Ch. 2.

Question 3 of 8: Torque and Axial Load from a Strain-Gauge Rosette

Question 3: Torque and Axial Load from a Strain-Gauge Rosette (20 marks)