Question

A 2D thin plate (plane stress) has $E=1.0~\text{N/m^2$ and Poisson's ratio $\mu=0.5$. The displacement field is $u=Cx^2y$, $v=0$ (in m). Distances $x,y$ are in m. The stresses are $\sigma_{xx}=40xy~\text{N/m}^2$ and $\tau_{xy}=\alpha x^2~\text{N/m}^2$. Find $\alpha$ (in $\text{N/m}^4$, integer).}

Hint:

In plane stress: $\sigma_{xx}=\dfrac{E}{1-\mu^2}(\varepsilon_{xx}+\mu\varepsilon_{yy})$, $\tau_{xy}=G\gamma_{xy}$ with $G=\dfrac{E}{2(1+\mu)}$.
Use the given stress to back-calculate the displacement constant, then obtain the shear stress coefficient.

Answer 1

Step 1: Strain–displacement relations.
$\varepsilon_{xx}=\dfrac{\partial u}{\partial x}=2Cxy$, $\varepsilon_{yy}=\dfrac{\partial v}{\partial y}=0$, $\gamma_{xy}=\dfrac{\partial u}{\partial y}+\dfrac{\partial v}{\partial x}=Cx^2$.

Step 2: Constitutive law (plane stress).
$\displaystyle \sigma_{xx}=\frac{E}{1-\mu^2}\big(\varepsilon_{xx}+\mu\varepsilon_{yy}\big) =\frac{E}{1-\mu^2}\varepsilon_{xx}$.
With $E=1$ and $\mu=0.5$: $\displaystyle \frac{E}{1-\mu^2}=\frac{1}{1-0.25}=\frac{4}{3}$.
Hence $\sigma_{xx}=\dfrac{4}{3}\varepsilon_{xx} \Rightarrow \varepsilon_{xx}=\dfrac{3}{4}\sigma_{xx}=\dfrac{3}{4}(40xy)=30xy$.
But $\varepsilon_{xx}=2Cxy \Rightarrow 2C=30 \Rightarrow C=15$.

Step 3: Shear stress from shear strain.
$G=\dfrac{E}{2(1+\mu)}=\dfrac{1}{2(1.5)}=\dfrac{1}{3}$, and $\tau_{xy}=G\,\gamma_{xy}=G(Cx^2)=\dfrac{C}{3}x^2$.
With $C=15$: $\tau_{xy}=5x^2 \Rightarrow \alpha=5$.
\[ \boxed{\alpha=5~\text{N/m}^4} \]