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

Question 3: Strain-Gauge Rosette on a Bar in Tension and Torsion (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-diameter bar loaded by axial P and torque T; a 0/45/90 rosette on the surface (0° along the axis).

Given data
Diameter / properties	d = 50 mm, E = 75 GPa, ν = 0.3
Rosette strains	ε₀ = 500μ, ε₄₅ = −100μ, ε₉₀ = −300μ (μ = 10⁻⁶)
Section properties	A = πd²/4 = 1963.5 mm², J = πd⁴/32 = 6.136×10⁵ mm⁴

Find. The axial load P and the torque T.

Surface element carries axial normal stress (from P) plus shear (from T); hoop stress is zero at the free surface.

Approach. At the free surface the state is uniaxial normal stress (axial) plus torsional shear, with zero hoop stress. The 0° strain gives the axial stress directly (hence P); the rosette combination 2ε₄₅−ε₀−ε₉₀ gives the engineering shear strain, hence τ and T.

Axial stress from the 0° gauge. With the hoop stress zero at a free surface, the axial gauge reads ε₀ = σ_axial/E, so $$\sigma_{\text{axial}}=E\varepsilon_0=75\,000(500\times10^{-6})=37.5\ \text{MPa}.$$
Axial load P. Multiply by the area: $$\boxed{P=\sigma_{\text{axial}}A=37.5\times1963.5=73.6\ \text{kN}.}$$
Shear strain from the rosette. For a rectangular (0/45/90) rosette the shear strain is $$\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(-100)-500-(-300)=-400\ \mu.$$
Shear stress. With G = E/[2(1+ν)] = 28.85 GPa, $$\tau=G\gamma_{xy}=28\,846(-400\times10^{-6})=-11.54\ \text{MPa}\ \Rightarrow\ |\tau|=11.54\ \text{MPa}.$$
Torque T. From the torsion formula τ = Tr/J with r = 25 mm, $$\boxed{T=\frac{|\tau|\,J}{r}=\frac{11.54\times6.136\times10^{5}}{25}=283\ \text{N}\cdot\text{m}.}$$

Check — gauge scatter. Pure axial + torsion predicts a hoop strain ε₉₀ = −νε₀ = −150μ, whereas the gauge reads −300μ. The difference is ordinary experimental scatter (misalignment / transverse sensitivity); it does not imply a real hoop stress. The axial stress is taken from the most direct reading (ε₀) and the shear from the rosette invariant, both of which are robust to this scatter.

Results — Question 3
Quantity	Value
Axial stress σ_axial	37.5 MPa
Axial load P	73.6 kN
Surface shear stress τ	11.54 MPa
Torque T	283 N·m

Topic: Combined axial–torsional loading and strain-gauge rosette reduction on a shaft surface.

Key relations: free surface ⇒ σ_hoop = 0, so σ_axial = Eε₀ and P = σA; rectangular rosette γ_xy = 2ε₄₅ − ε₀ − ε₉₀; torsion τ = Tr/J, G = E/[2(1+ν)].

Why this works: on the outer fibre of a bar the only stresses are the axial normal stress and the torsional shear; the radial and hoop stresses vanish. The rosette’s three readings over-determine the 2-D surface strain, and the γ_xy combination isolates the shear that the torque produces.

Common pitfall: inventing a hoop stress to force ε₉₀ = −νε₀, or using J = πd⁴/64 (that is the bending I, not the polar J). Keep J = πd⁴/32 for the shaft.

Source: Gere & Goodno, Mechanics of Materials, strain-rosette reduction (§7.7) and torsion of circular shafts (Ch. 3).

Question 3 of 7: Strain-Gauge Rosette on a Bar in Tension and Torsion

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