Strain-Gauge Rosette on a Bar in Combined Torsion and Tension: 22-Mec-A7 Advanced Strength of Materials, May 2013, Q3

Question 3: Strain-Gauge Rosette on a Bar in Combined Torsion and Tension (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 50 mm solid shaft carrying axial force P and torque T; a 0°/45°/90° rosette (0° axial).

Given data
Diameter / geometry	d = 50 mm; A = 1963.5 mm², J = 6.136×10⁵ mm⁴
Gauge strains	ε₀ = 800×10⁻⁶, ε₄₅ = −200×10⁻⁶, ε₉₀ = −400×10⁻⁶
Elastic constants	E = 80 GPa, v = 0.28, G = E/[2(1+v)] = 31.25 GPa

Find. The axial load P and the torque T that produced these surface strains.

Approach. On the free surface of the bar the axial gauge reads the axial strain directly, giving the axial stress and hence P; the rosette’s shear strain follows from the three readings, giving the surface shear stress and hence T.

Axial stress and load. The bar surface is free (radial and hoop stresses zero), so the axial gauge measures $\varepsilon_{axial}=\sigma/E$: $$\sigma=E\,\varepsilon_0=(80\,000)(800\times10^{-6})=64\ \text{MPa},\qquad P=\sigma A=64(1963.5).$$ $$\boxed{P \approx 125.7\ \text{kN}}$$
Shear strain from the rosette. For a rectangular (0/45/90) rosette, $$\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(-200)-800-(-400)=-800\times10^{-6}.$$
Surface shear stress. Using the shear modulus, $$\tau=G\,\gamma_{xy}=(31\,250)(-800\times10^{-6})=-25\ \text{MPa}\quad(|\tau|=25\ \text{MPa}).$$
Torque. From the torsion formula $\tau=Tr/J$ with r = 25 mm, $$T=\frac{\tau J}{r}=\frac{25\,(6.136\times10^{5})}{25}=6.136\times10^{5}\ \text{N}\cdot\text{mm}.$$ $$\boxed{T \approx 0.614\ \text{kN}\cdot\text{m}\ (614\ \text{N}\cdot\text{m})}$$

Question 3 — results
Quantity	Value
Axial stress σ	64 MPa
Axial load P	125.7 kN
Shear strain γ_xy	−800×10⁻⁶
Shear stress τ	25 MPa
Torque T	0.614 kN·m

Check: for a free bar surface the transverse gauge should read −vε₀ = −224×10⁻⁶, whereas the measured ε₉₀ = −400×10⁻⁶. The excess is attributed to gauge/measurement scatter; because the hoop stress on the surface is genuinely zero, the axial stress is taken from ε₀ directly. (Treating the readings as a general plane-stress state would give σ_x = 59.7 MPa, P = 117 kN, plus a spurious hoop stress that cannot physically exist here.)

Topic: Reducing rosette strains to stresses; combined axial–torsional surface state.

Key relations: rectangular-rosette shear $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}$; uniaxial $\sigma=E\varepsilon$; torsion $\tau=Tr/J$, $J=\pi d^4/32$; $G=E/[2(1+v)]$.

Why this works: On the cylindrical surface of a bar loaded by axial force and torque, the only non-zero stresses are the axial normal stress and the torsional shear; the surface is traction-free in the radial and circumferential directions. Normal strains depend only on normal stresses, so the axial gauge isolates σ and hence P, while the 45° gauge (through the rosette identity) isolates the shear distortion and hence T. The two loads decouple cleanly in the strain readings.

Common pitfall: assuming a full biaxial stress state and manufacturing a non-existent hoop stress from ε₉₀; or dropping the factor of 2 between γ (engineering shear strain) and the tensorial shear.

Source: Ugural & Fenster, Advanced Strength and Applied Elasticity, Ch. 2 (strain transformation, rosettes); Boresi & Schmidt, Ch. 2–3.

Question 3 of 8: Strain-Gauge Rosette on a Bar in Combined Torsion and Tension

Question 3: Strain-Gauge Rosette on a Bar in Combined Torsion and Tension (20 marks)