Question

A thin steel plate is loaded in the $x$-$y$ plane as shown in the figure. Take Poisson’s ratio $\nu = 0.3$ and modulus of elasticity $E = 200 \, GPa$. The strain along $z$-direction is $\varepsilon_{zz} = -3 \times 10^{-4}$. The value of $\sigma_{yy}$ (in MPa) is .......... (rounded off to one decimal place).

Hint:

For plane stress, always use $\varepsilon_z = -\nu (\sigma_x + \sigma_y)/E$. A negative strain in $z$ indicates contraction due to Poisson’s effect.

Answer 1

Correct Answer: 79.5

Step 1: Strain-stress relation in 3D.
For isotropic elasticity: \[ \varepsilon_z = \frac{1}{E} \Big(\sigma_z - \nu (\sigma_x + \sigma_y)\Big) \]
Step 2: Given conditions.
- $\varepsilon_z = -3 \times 10^{-4}$ - $\sigma_x = 120 \, MPa$, $\sigma_z = 0$ (plane stress) So: \[ -3 \times 10^{-4} = \frac{1}{200 \times 10^3} \Big(0 - 0.3(120 + \sigma_y)\Big) \]
Step 3: Simplify.
\[ -3 \times 10^{-4} = \frac{-0.3(120 + \sigma_y)}{200 \times 10^3} \] Multiply: \[ -60 = -0.3(120 + \sigma_y) \] \[ 60 = 0.3(120 + \sigma_y) \] \[ 60 = 36 + 0.3\sigma_y \] \[ 0.3\sigma_y = 24 \Rightarrow \sigma_y = 80 \, MPa \] Correction: Factor error—recheck: \[ -3 \times 10^{-4} . 200 \times 10^3 = -60 \] \[ -60 = -0.3(120 + \sigma_y) \] \[ 60 = 0.3(120 + \sigma_y) \] \[ \sigma_y = 80 \, MPa \] Final Answer: \[ \boxed{80 \, MPa} \]