Compute D matrix for plane problems

(a) Plane stress

$$\frac{E}{1 - \nu^2} = \frac{200\times10^3}{1 - 0.09} = 219\,780\ \text{MPa}$$ $$[D] = 219\,780\begin{bmatrix} 1 & 0.3 & 0 \\ 0.3 & 1 & 0 \\ 0 & 0 & 0.35 \end{bmatrix} = \begin{bmatrix} 219\,780 & 65\,934 & 0 \\ 65\,934 & 219\,780 & 0 \\ 0 & 0 & 76\,923 \end{bmatrix}\ \text{MPa}$$

Note that $\dfrac{1 - \nu}{2} = 0.35$ and $219\,780(0.35) = 76\,923 = G = \dfrac{E}{2(1 + \nu)}$.

(b) Plane strain

$$\frac{E}{(1 + \nu)(1 - 2\nu)} = \frac{200\times10^3}{1.3(0.4)} = 384\,615\ \text{MPa}$$ $$[D] = 384\,615\begin{bmatrix} 0.7 & 0.3 & 0 \\ 0.3 & 0.7 & 0 \\ 0 & 0 & 0.2 \end{bmatrix} = \begin{bmatrix} 269\,231 & 115\,385 & 0 \\ 115\,385 & 269\,231 & 0 \\ 0 & 0 & 76\,923 \end{bmatrix}\ \text{MPa}$$

Stresses for $\varepsilon_x = 0.001$

The shear modulus term is the same in both cases. Plane strain gives higher stresses because $\varepsilon_z = 0$ is enforced.

Answer: plane stress $\sigma_x = 219.8$ MPa, $\sigma_y = 65.9$ MPa, $\sigma_z = 0$; plane strain $\sigma_x = 269.2$ MPa, $\sigma_y = 115.4$ MPa, $\sigma_z = 115.4$ MPa.