Question 5 of 8: Three-element (0°/60°/120°) strain rosette

Question 5: Three-element (0°/60°/120°) strain rosette

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. $\varepsilon_0=400\mu$, $\varepsilon_{60}=1100\mu$, $\varepsilon_{120}=500\mu$; $E=210$ GPa, $\nu=0.3$ ($\mu=\times10^{-6}$).

Find. (a) $\varepsilon_x,\varepsilon_y,\gamma_{xy}$; (b) principal strains and directions; (c) $\sigma_x,\sigma_y,\tau_{xy}$.

Approach. Invert the strain-transformation relations for the three gauge angles, then form Mohr’s-circle principal values and apply plane-stress Hooke’s law.

1. Cartesian strains (part a). Taking $x$ along the 0° gauge, $\varepsilon_x=\varepsilon_0=400\mu$. Adding and subtracting the 60°/120° equations, $$\varepsilon_y=\frac{2(\varepsilon_{60}+\varepsilon_{120})-\varepsilon_0}{3}\Big|_{\text{alg.}}=933\mu,\qquad \gamma_{xy}=\frac{\varepsilon_{60}-\varepsilon_{120}}{2\sin60^\circ\cos60^\circ}=693\mu$$ so $\boxed{\varepsilon_x=400\mu,\ \varepsilon_y=933\mu,\ \gamma_{xy}=693\mu}$ (the strain at $+45^\circ$ is $\varepsilon_{45}=1013\mu$).
2. Principal strains (part b). With centre $(\varepsilon_x+\varepsilon_y)/2=667\mu$ and radius $R=\sqrt{[(\varepsilon_x-\varepsilon_y)/2]^2+(\gamma_{xy}/2)^2}=437\mu$, $$\varepsilon_{1,2}=667\pm437\ \mu\;\Rightarrow\;\boxed{\varepsilon_1=1104\mu,\quad\varepsilon_2=230\mu}$$
3. Principal directions. $\tan2\theta_p=\gamma_{xy}/(\varepsilon_x-\varepsilon_y)=693/(-533)$, giving $\theta_p=63.8^\circ$ for $\varepsilon_1$ and $-26.2^\circ$ for $\varepsilon_2$ (measured from the x-axis).
4. Stresses by Hooke’s law (part c). For plane stress with $E/(1-\nu^2)=230.8$ GPa and $G=E/2(1+\nu)=80.8$ GPa, $$\sigma_x=\frac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)=\boxed{156.9\ \text{MPa}},\quad \sigma_y=\frac{E}{1-\nu^2}(\varepsilon_y+\nu\varepsilon_x)=\boxed{243.1\ \text{MPa}}$$ $$\tau_{xy}=G\,\gamma_{xy}=\boxed{55.9\ \text{MPa}}$$