GATE 2023 CE (CE2) – Question 56
A 2D thin plate with modulus of elasticity, E = 1.0 N/m$^2$, and Poisson’s ratio, μ = 0.5, is in plane stress condition. The displacement field in the plate is given by $u=Cx^2y$ and $v=0$, where u and v are displacements (in m) along the X and Y directions, respectively, and C is a constant (in m$^{-2}$). The distances x and y along X and Y, respectively, are in m. The stress in the X direction is $\sigma_{XX}=40xy$ N/m$^2$, and the shear stress is $\tau_{XY}=\alpha x^2$ N/m$^2$. What is the value of α (in N/m$^4$, in integer)?
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: 5
Explanation
**Step 1: strains from the displacement field** ($u=Cx^2y$, $v=0$):
$$\varepsilon_x=\frac{\partial u}{\partial x}=2Cxy,\qquad\varepsilon_y=\frac{\partial v}{\partial y}=0,$$
$$\gamma_{xy}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=Cx^2 .$$
**Step 2: stress $\sigma_{xx}$ in plane stress.**
$$\sigma_{xx}=\frac{E}{1-\mu^2}\left(\varepsilon_x+\mu\varepsilon_y\right)=\frac{1}{1-0.25}\,(2Cxy)=\frac{8}{3}Cxy .$$
**Step 3: find C** from the given $\sigma_{xx}=40xy$:
$$\frac83C=40\;\Rightarrow\;C=15 .$$
**Step 4: shear stress.** With $G=\dfrac{E}{2(1+\mu)}=\dfrac{1}{3}$,
$$\tau_{xy}=G\gamma_{xy}=\frac13(15x^2)=5x^2 .$$
So $\alpha=\mathbf{5}$.