Question 6 of 8: Plane-Stress Plate — Inverse Problem

Question 6: Plane-Stress Plate — Inverse Problem (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 1 m square plate in biaxial plane stress.

Find. (a) $\sigma_y$ and $\nu$; (b) the thickness strain $\varepsilon_z$.

Approach. Write the two in-plane Hooke’s-law equations for plane stress and solve the pair simultaneously for the two unknowns $\sigma_y$ and $\nu$; then use the out-of-plane law for $\varepsilon_z$.

1. Hooke’s law, plane stress. $E\varepsilon_x=\sigma_x-\nu\sigma_y\Rightarrow \nu\sigma_y=160-106=54$; and $E\varepsilon_y=\sigma_y-\nu\sigma_x\Rightarrow \sigma_y-160\nu=132$.
2. Eliminate $\nu$. Substituting $\nu=54/\sigma_y$ gives $\sigma_y^2-132\,\sigma_y-8640=0$.
3. Solve (part a). The positive root is $$\boxed{\sigma_y=180\ \text{MPa},\qquad \nu=\frac{54}{180}=0.30}$$ which is admissible ($\nu\lt0.5$).
4. Thickness strain (part b). $$\varepsilon_z=-\frac{\nu}{E}(\sigma_x+\sigma_y)=-\frac{0.30}{200\,000}(160+180)=\boxed{-5.1\times10^{-4}}.$$