Question 7 of 8: Strain rosette on a plate

Question 7: Strain rosette on a 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° strain rosette on a thin elastic plate in plane stress.

Find. (a) principal strains and directions; (b) $\sigma_x,\sigma_y,\tau_{xy}$ with x along the 0° gauge.

Approach. Convert the rosette readings to Cartesian strains, use Mohr's circle for the principal strains and angle, then apply the plane-stress Hooke's law.

1. Cartesian strains. For a 0/45/90 rosette $\varepsilon_x=\varepsilon_0=600\mu$, $\varepsilon_y=\varepsilon_{90}=500\mu$ and $$\gamma_{xy}=2\varepsilon_{45}-\varepsilon_0-\varepsilon_{90}=-300\ \mu.$$
2. Principal strains. With mean $550\mu$ and radius $R=\sqrt{50^2+150^2}=158\ \mu$, $$\varepsilon_{1,2}=\boxed{708\ \mu,\ 392\ \mu},\qquad \tan2\theta_p=\frac{\gamma_{xy}}{\varepsilon_x-\varepsilon_y}=-3\ \Rightarrow\ \theta_p=-35.8^\circ.$$
3. Stresses (plane stress). With $E/(1-\nu^2)=77.4$ GPa and $G=E/2(1+\nu)=26.7$ GPa, $$\sigma_x=\frac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)=\boxed{58.5\ \text{MPa}},\quad \sigma_y=\boxed{53.1\ \text{MPa}},\quad \tau_{xy}=G\gamma_{xy}=\boxed{-8.02\ \text{MPa}}.$$

Because the shear strain is only a tenth of the normal strains, the principal axes lie close to the gauge axes and the two normal stresses are nearly equal — the plate is in a mild, almost biaxial-tension state. The negative shear simply reflects the sense of the 45° reading relative to the 0° and 90° gauges.