Strain-Rosette Reduction on a Thin Plate: 22-Mec-A7 Advanced Strength of Materials, May 2015, Q8

Question 8: Strain-Rosette Reduction on a Thin Plate (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 0–45–90° rosette: $\varepsilon_0=600\mu$, $\varepsilon_{45}=400\mu$, $\varepsilon_{90}=500\mu$; plate E = 60 GPa, v = 0.35.

Find. (a) strains on axes rotated +45°; (b) principal strains and their directions; (c) the in-plane stresses.

Approach. Reduce the rosette to Cartesian strains, transform to +45°, take the principal values from Mohr’s circle, and convert to stress with the plane-stress Hooke’s law.

Cartesian strains. For a rectangular rosette $\varepsilon_x=\varepsilon_0=600\mu$, $\varepsilon_y=\varepsilon_{90}=500\mu$, and $$\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=2(400)-600-500=-300\ \mu.$$
(a) Rotate to +45°. Applying the strain transformation at $\theta=45^\circ$, $$\varepsilon_{x'}=400\mu,\qquad \varepsilon_{y'}=700\mu,\qquad \gamma_{x'y'}=-100\ \mu.$$ ($\varepsilon_{x'}$ recovers the measured $\varepsilon_{45}$, a useful check.)
(b) Principal strains. With centre $\tfrac{\varepsilon_x+\varepsilon_y}{2}=550\mu$ and radius $R=\sqrt{50^2+150^2}=158\mu$, $$\boxed{\varepsilon_1=708\ \mu,\quad \varepsilon_2=392\ \mu,}\qquad \tan2\theta_p=\frac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y}=-3,$$ so $\theta_p=-35.8^\circ$ (to $\varepsilon_1$) and $+54.2^\circ$ (to $\varepsilon_2$).
(c) In-plane stresses. Plane-stress Hooke’s law with $\tfrac{E}{1-v^2}=68.4$ GPa and $G=22.2$ GPa gives $$\sigma_x=\frac{E}{1-v^2}(\varepsilon_x+v\varepsilon_y)=53.0\ \text{MPa},\ \ \sigma_y=48.6\ \text{MPa},\ \ \tau_{xy}=G\gamma_{xy}=-6.67\ \text{MPa}.$$

Final results — Question 8
Quantity	Value
ε_x′, ε_y′, γ_x′y′ (+45°)	400, 700, −100 μ
Principal strains ε₁, ε₂	708 μ, 392 μ
Principal direction (to ε₁)	−35.8°
σ_x, σ_y, τ_xy	53.0, 48.6, −6.67 MPa

Topic: Strain-rosette data reduction and plane-stress Hooke’s law.

Key relations: $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}$; $\varepsilon_{1,2}=\tfrac{\varepsilon_x+\varepsilon_y}{2}\pm\sqrt{(\tfrac{\varepsilon_x-\varepsilon_y}{2})^2+(\tfrac{\gamma_{xy}}{2})^2}$; $\sigma_x=\tfrac{E}{1-v^2}(\varepsilon_x+v\varepsilon_y)$; $\tau_{xy}=G\gamma_{xy}$.

Why this works: A gauge measures only the normal strain along its own axis, so three readings are exactly enough to recover the three plane-strain components. Mohr’s circle then delivers the principal strains and their orientation, and because the plate surface is free ($\sigma_z=0$) the plane-stress elastic law converts strains directly to the surface stresses.

Common pitfall: using the plane-strain (not plane-stress) elastic constants—a free surface is plane stress; and forgetting that shear stress uses $G$, not $E/(1-v^2)$.

Source: Hibbeler, Mechanics of Materials, Ch. 10 (strain transformation, rosettes); Ugural & Fenster §2.

Question 8 of 8: Strain-Rosette Reduction on a Thin Plate

Question 8: Strain-Rosette Reduction on a Thin Plate (20 marks)