Question

The state of stress at a point is shown in the figure given below. Under plane stress assumption, the normal strain along the thickness direction (\(\epsilon_{zz}\)) is _________ (rounded off to 2 decimal places).

 

Hint:

To calculate normal strain under plane stress, use the formula: \(\epsilon_{zz} = -\frac{\nu}{E} (\sigma_{xx} + \sigma_{yy})\), where \( \nu \) is Poisson’s ratio and \( E \) is Young’s Modulus.

Answer 1

Step 1: Understanding the formula for normal strain in the thickness direction. Under the plane stress assumption, the normal strain along the thickness direction (\(\epsilon_{zz}\)) can be calculated using the following formula: \[ \epsilon_{zz} = -\frac{\nu}{E} (\sigma_{xx} + \sigma_{yy}) \] Where:
\( \nu = 0.27 \) is Poisson’s ratio,
\( E = 200 \, {GPa} = 200 \times 10^3 \, {MPa} \) is the Young's Modulus,
\( \sigma_{xx} = 150 \, {GPa} = 150 \times 10^3 \, {MPa} \) is the stress in the \(x\)-direction,
\( \sigma_{yy} = 50 \, {GPa} = 50 \times 10^3 \, {MPa} \) is the stress in the \(y\)-direction.
Step 2: Substituting the given values into the formula. \[ \epsilon_{zz} = -\frac{0.27}{200 \times 10^3} \left( 150 \times 10^3 + 50 \times 10^3 \right) \] \[ \epsilon_{zz} = -\frac{0.27}{200 \times 10^3} \times 200 \times 10^3 = -0.27 \]