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 (ε_zz) is ____________ (rounded off to 2 decimal places).

Use the Young’s Modulus E=200 GPa and Poisson’s ratio ν=0.27.

The correct answer is -0.27.

From the given state of stress shown in the image, we can identify the following values along the coordinate axes:

Normal stress in the x-direction (tensile, pointing outwards):
${\sigma }_{\mathrm{xx}}=350\text{GPa}$

Normal stress in the y-direction (compressive, pointing inwards):
${\sigma }_{\mathrm{yy}}=-150\text{GPa}$

Shear stress in the xy-plane:
${\tau }_{\mathrm{xy}}=50\text{GPa}$

Young's modulus:
$E=200\text{GPa}$

Poisson's ratio:
$\nu =0.27$

Under the plane stress assumption in the xy-plane, the normal stress along the thickness direction (z-direction) is zero:

$${\sigma }_{\mathrm{zz}}=0$$

Using the generalized Hooke's law, the normal strain along the thickness direction (${\epsilon }_{\mathrm{zz}}$) is expressed as:

$${\epsilon }_{\mathrm{zz}}=\frac{1}{E}\left[{\sigma }_{\mathrm{zz}}-\nu \left({\sigma }_{\mathrm{xx}}+{\sigma }_{\mathrm{yy}}\right)\right]$$

Substituting ${\sigma }_{\mathrm{zz}}=0$ into the equation simplifies the relation to:

$${\epsilon }_{\mathrm{zz}}=-\frac{\nu }{E}\left({\sigma }_{\mathrm{xx}}+{\sigma }_{\mathrm{yy}}\right)$$

Now, substituting the numerical values:

$${\epsilon }_{\mathrm{zz}}=-\frac{0.27}{200\text{GPa}}\left(350\text{GPa}+\left(-150\text{GPa}\right)\right)$$

Simplifying the sum of the in-plane normal stresses:

$$350\text{GPa}-150\text{GPa}=200\text{GPa}$$

Substituting this result back to calculate the strain:

$${\epsilon }_{\mathrm{zz}}=-\frac{0.27}{200}\times 200=-0.27$$

Therefore, the normal strain along the thickness direction is -0.27.