Strain rosette on a bar under torsion + axial load: 22-Mec-A7 Advanced Strength of Materials, December 2018, Q2

Question 2: Strain rosette on a bar under torsion + axial load (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. Solid shaft, $d=55\text{ mm}$; rectangular (0/45/90) rosette on the free surface with the 0° gauge axial.

Given data
Diameter $d$	55 mm
$\varepsilon_0$ (axial)	$230\times10^{-6}$
$\varepsilon_{45}$	$-40\times10^{-6}$
$\varepsilon_{90}$ (hoop)	$-120\times10^{-6}$
$E,\ \nu$	40 GPa, 0.3

Find. The axial force $P$ and the torque $T$ that produce these surface strains.

Approach. On the traction-free surface the hoop stress is zero, so the axial stress follows directly from $\varepsilon_0$ (giving $P$); the engineering shear strain from the rosette gives the surface shear stress (giving $T$).

Check: A pure axial state would give $\varepsilon_{90}=-\nu\varepsilon_0=-69\times10^{-6}$, but the gauge reads $-120\times10^{-6}$. The mismatch is gauge scatter/misalignment; the free surface still has $\sigma_\text{hoop}=0$, so we take $\sigma_\text{axial}=E\varepsilon_0$ and do not invent a hoop stress.

Section properties. $$A=\tfrac{\pi}{4}d^2=\tfrac{\pi}{4}(55)^2=2376\text{ mm}^2,\quad J=\tfrac{\pi}{32}d^4=8.984\times10^{5}\text{ mm}^4,\quad r=\tfrac{d}{2}=27.5\text{ mm}.$$
Axial stress and force (free surface: $\sigma_\text{hoop}=0$). With one non-zero normal stress, $\varepsilon_0=\sigma_\text{axial}/E$, so $$\sigma_\text{axial}=E\varepsilon_0=(40\,000)(230\times10^{-6})=9.20\text{ MPa},$$ $$P=\sigma_\text{axial}\,A=(9.20)(2376)=\boxed{21.86\text{ kN}}.$$
Shear strain from the rosette. For a 0/45/90 rosette, $$\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(-40)-230-(-120)=-190\times10^{-6}.$$
Shear stress via the shear modulus. $$G=\frac{E}{2(1+\nu)}=\frac{40}{2(1.3)}=15.38\text{ GPa},\qquad \tau=G\,\gamma_{xy}=(15\,385)(190\times10^{-6})=2.923\text{ MPa}.$$
Torque from the torsion formula. $\tau=Tr/J$ gives $$T=\frac{\tau J}{r}=\frac{(2.923)(8.984\times10^{5})}{27.5}=\boxed{95.5\text{ N}\cdot\text{m}}.$$

Final results — Question 2
Quantity	Value
Axial stress	9.20 MPa
Axial force $P$	21.86 kN (tension)
Surface shear $\tau$	2.92 MPa
Torque $T$	95.5 N·m

Topic. Combined axial–torsion loading recovered from a surface strain rosette.

Key relations. Free-surface uniaxiality $\sigma_\text{axial}=E\varepsilon_0$; rosette shear $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}$; $\tau=G\gamma$; torsion $\tau=Tr/J$.

Why this works. The bar surface carries only axial normal stress and torsional shear — the radial and hoop stresses vanish at a free surface. Hence the axial strain maps one-to-one to axial stress, and the 45° gauge (rotated into the shear direction) captures the pure-shear part. The rosette identity for $\gamma_{xy}$ is exact for the rectangular layout and isolates shear from the two normal readings.

Common pitfall. Trying to force $\varepsilon_{90}=-\nu\varepsilon_0$ and inventing a hoop stress to reconcile the discrepancy — this contradicts the free-surface condition. Treat the mismatch as measurement scatter and keep $\sigma_\text{hoop}=0$.

Source. Ugural & Fenster, Advanced Strength and Applied Elasticity, §2.9 & §6.2; Hibbeler §10.5 (strain rosettes).

Question 2 of 8: Strain rosette on a bar under torsion + axial load

Question 2: Strain rosette on a bar under torsion + axial load (20 marks)