Question 6 of 8: Strain Rosette — Transformation, Principal Strains and Yield

Question 6: Strain Rosette — Transformation, Principal Strains and Yield (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 rectangular rosette with the 0° gauge along $x$; plane-stress plate with $E=90\ \text{GPa}$, $v=0.28$.

Find. (a) $\varepsilon_{x'},\varepsilon_{y'},\gamma_{x'y'}$ at $+60^\circ$; (b) principal strains and directions; (c) the yield strength implied by the max-shear criterion.

Approach. Reduce the rosette to $(\varepsilon_x,\varepsilon_y,\gamma_{xy})$, rotate them to $60^\circ$, take the principal strains from the strain Mohr’s circle, convert to principal stresses by plane-stress Hooke’s law, and apply Tresca.

1. Rotate to $+60^\circ$. With $\tfrac{\varepsilon_x+\varepsilon_y}{2}=450\,\mu$, $\tfrac{\varepsilon_x-\varepsilon_y}{2}=150\,\mu$, $\tfrac{\gamma_{xy}}{2}=-50\,\mu$ and $2\theta=120^\circ$: $$\boxed{\varepsilon_{x'}=331.7\,\mu,\quad \varepsilon_{y'}=568.3\,\mu,\quad \gamma_{x'y'}=-209.8\,\mu}$$ (the sum $\varepsilon_{x'}+\varepsilon_{y'}=900\,\mu$ equals $\varepsilon_x+\varepsilon_y$, a useful check).
2. Principal strains. $\varepsilon_{1,2}=450\pm\sqrt{150^2+50^2}=450\pm158.1\,\mu$, so $$\boxed{\varepsilon_1=608.1\,\mu,\qquad \varepsilon_2=291.9\,\mu}$$
3. Principal directions. $\tan2\theta_p=\dfrac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y}=\dfrac{-100}{300}$, giving $\theta_p=-9.2^\circ$ (to $\varepsilon_1$); $\varepsilon_2$ lies at $80.8^\circ$.
4. Principal stresses (plane stress). $\sigma_{1,2}=\dfrac{E}{1-v^2}(\varepsilon_{1,2}+v\,\varepsilon_{2,1})$ with $\dfrac{E}{1-v^2}=97.66\ \text{GPa}$: $\sigma_1=67.4\ \text{MPa}$, $\sigma_2=45.1\ \text{MPa}$ (both tensile, $\sigma_3=0$).
5. Max-shear yield. With both in-plane principals positive, the extreme pair is $\sigma_1$ and $\sigma_3=0$, so $$\boxed{\sigma_Y=\sigma_1-\sigma_3=67.4\ \text{MPa}}$$