Torque and axial load from a strain-gauge rosette: 22-Mec-A7 Advanced Strength of Materials, December 2016, Q7

Question 7: 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. A circular bar under combined axial load and torque; a 0°/45°/90° rosette on the free surface reads three strains.

Given data
Diameter	d = 60 mm (r = 0.03 m)
Rosette strains	ε₀ = 240×10−⁶, ε₄₅ = −50×10−⁶, ε₉₀ = −140×10−⁶
Elastic constants	E = 60 GPa, v = 0.3 → G = 23.08 GPa

Find. The magnitudes of the axial load P and the torque T.

Approach. On the free surface the hoop stress is zero, so the 0° reading gives the axial stress (hence P) directly; the rosette shear-strain combination gives the shear stress (hence T).

Axial stress and P. With $\sigma_{hoop}=0$ on the surface, $\varepsilon_0=\sigma_{ax}/E$, so $\sigma_{ax}=E\varepsilon_0=60\times10^9(240\times10^{-6})=14.4$ MPa. Then $P=\sigma_{ax}A=14.4\times10^6\cdot\tfrac{\pi}{4}(0.06)^2=$ $$\boxed{P=40.7\ \text{kN.}}$$
Shear strain from the rosette. For a 0/45/90 rosette $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(-50)-240-(-140)=-200\ \mu\varepsilon$; the magnitude $200\times10^{-6}$ sets the shear.
Shear stress and torque. $\tau=G|\gamma_{xy}|=23.08\times10^9(200\times10^{-6})=4.62$ MPa. With $J=\tfrac{\pi}{32}d^4=1.272\times10^{-6}$ m$^4$ and $\tau=Tr/J$: $$\boxed{T=\tau J/r=196\ \text{N}\cdot\text{m.}}$$

Check: on an axial-only surface $\varepsilon_{90}$ should equal $-v\varepsilon_0=-72\ \mu\varepsilon$, but the gauge reads $-140\ \mu\varepsilon$. This is gauge scatter/misalignment, not a real hoop stress (a free surface carries none); the axial stress is taken from $\varepsilon_0$ as above and no hoop stress is invented.

Final results — Q7
Quantity	Value
Axial stress	14.4 MPa
Axial load P	40.7 kN
Surface shear stress	4.62 MPa
Torque T	196 N·m

Topic: Combined axial–torsion loading resolved from a surface rosette.

Key relations: Free surface $\sigma_{ax}=E\varepsilon_0$; rosette shear $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}$; $\tau=G\gamma$; $\tau=Tr/J$, $\sigma=P/A$.

Why this works: The bar surface is in plane stress with an axial normal stress and a torsional shear only. The axial gauge reads the axial strain, which on a stress-free surface maps to the axial stress through the uniaxial modulus. The 45° gauge, combined with the axial and transverse gauges, isolates the engineering shear strain, and elasticity plus the torsion formula convert it to the applied torque.

Common pitfall: Trying to force $\varepsilon_{90}$ to represent a hoop stress — a free lateral surface has zero hoop stress, so a mismatched 90° reading is gauge error, not data to be modelled.

Source: Hibbeler, Mechanics of Materials §10.5 (strain rosettes) and Ch. 5 (torsion); Boresi & Schmidt Ch. 6.

Question 7 of 8: Torque and axial load from a strain-gauge rosette

Question 7: Torque and axial load from a strain-gauge rosette (20 marks)