Question 6 of 8: Three-Element Strain Rosette

Question 6: Three-Element Strain 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 rectangular (0/45/90) rosette on a thin plate: $\varepsilon_0=800\,\mu$, $\varepsilon_{45}=600\,\mu$, $\varepsilon_{90}=400\,\mu$, with $E=75$ GPa and $\nu=0.32$.

Find. (a) strains rotated to $+60^\circ$; (b) principal strains and directions; (c) the in-plane stresses.

Approach. Convert the three gauge readings to $\varepsilon_x,\varepsilon_y,\gamma_{xy}$, apply the strain-transformation equations for part (a), read the principals directly (here $\gamma_{xy}=0$), and close with the plane-stress Hooke’s law for part (c).

1. Cartesian strains. $\varepsilon_x=\varepsilon_0=800\,\mu$, $\varepsilon_y=\varepsilon_{90}=400\,\mu$, and $\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=1200-1200=0$.
2. (a) Rotate to $+60^\circ$. With $\gamma_{xy}=0$: $\varepsilon_{x'}=\varepsilon_x\cos^2\theta+\varepsilon_y\sin^2\theta=800(0.25)+400(0.75)=500\,\mu$; $\varepsilon_{y'}=800(0.75)+400(0.25)=700\,\mu$; $\gamma_{x'y'}=-(\varepsilon_x-\varepsilon_y)\sin2\theta=-400\sin120^\circ=-346.4\,\mu$.
3. (b) Principal strains. Since $\gamma_{xy}=0$, the $x,y$ axes are already principal: $\boxed{\varepsilon_1=800\,\mu\ (0^\circ),\ \varepsilon_2=400\,\mu\ (90^\circ)}$.
4. (c) Plane-stress Hooke’s law. $\sigma_x=\dfrac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)=\dfrac{75000}{0.8976}(800+0.32\cdot400)\times10^{-6}=\boxed{77.5\ \text{MPa}}$; similarly $\sigma_y=\dfrac{E}{1-\nu^2}(\varepsilon_y+\nu\varepsilon_x)=\boxed{54.8\ \text{MPa}}$, and $\tau_{xy}=G\gamma_{xy}=0$.