Question 4 of 8: Biaxial Plate — Inverse Problem and Principal Stresses

Question 4: Biaxial Plate — Inverse Problem and Principal Stresses (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 biaxially stressed square plate of side 1.8 m; the two elongations and $\sigma_x$ are known, and $E$ is given.

Find. (a) $\sigma_y$ and $v$; (b) the principal stresses and their orientation once $\tau_{xy}=100\ \text{MPa}$ is added.

Approach. Write both plane-stress Hooke equations; eliminate $\sigma_y$ to get a quadratic in $v$, then back-substitute. For part (b) apply the in-plane principal-stress formula to $(\sigma_x,\sigma_y,\tau_{xy})$.

1. Hooke’s law, both directions. $E\varepsilon_x=\sigma_x-v\sigma_y$ gives $20=50-v\sigma_y$, so $v\sigma_y=30$. $E\varepsilon_y=\sigma_y-v\sigma_x$ gives $83.33=\sigma_y-50v$.
2. Eliminate $\sigma_y$. With $\sigma_y=30/v$, substitution yields $50v^2+83.33v-30=0$, whose admissible root is $$\boxed{v=0.304,\qquad \sigma_y=98.6\ \text{MPa}}$$ ($vt 0.5$, so the result is physically admissible).
3. Add the shear and locate the centre / radius. $\sigma_{\text{avg}}=\tfrac12(50+98.6)=74.3\ \text{MPa}$; $R=\sqrt{\left(\tfrac{50-98.6}{2}\right)^2+100^2}=102.9\ \text{MPa}$.
4. Principal stresses. $$\boxed{\sigma_1=177.2\ \text{MPa},\qquad \sigma_2=-28.6\ \text{MPa}}$$
5. Orientation. $\tan2\theta_p=\dfrac{2\tau_{xy}}{\sigma_x-\sigma_y}=\dfrac{200}{-48.6}$, so $\theta_{p1}=51.8^\circ$ from the x-axis to the $\sigma_1$ direction ($\sigma_2$ at $141.8^\circ$).