## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, tutorials, and more.

No credit card required

(93)

Upon solution, Eqs. (92), (93) are written as

$\left\{\begin{array}{l}{\sigma }_{x}^{a}=\frac{{E}_{a}}{1-{\nu }_{a}^{2}}\left({\epsilon }_{x}^{a}+{\nu }_{a}{\epsilon }_{y}^{a}\right)-\frac{{E}_{a}\left(1+{\nu }_{a}\right)}{1-{\nu }_{a}^{2}}{\epsilon }_{ISA}\\ {\sigma }_{y}^{a}=\frac{{E}_{a}}{1-{\nu }_{a}^{2}}\left({\nu }_{a}{\epsilon }_{x}^{a}+{\epsilon }_{y}^{a}\right)-\frac{{E}_{a}\left(1+{\nu }_{a}\right)}{1-{\nu }_{a}^{2}}{\epsilon }_{ISA}\end{array}\left(\mathrm{PWAS}\right)$ (94)

(94)

$\left\{\begin{array}{l}{\sigma }_{x}=\frac{E}{1-{\nu }^{2}}\left({\epsilon }_{x}+\nu {\epsilon }_{y}\right)\\ {\sigma }_{y}=\frac{E}{1-{\nu }^{2}}\left(\nu {\epsilon }_{x}+{\epsilon }_{y}\right)\end{array}\left(\mathrm{structure}\right)$ (95)

(95)

Recall the strain-displacement relations

$\left\{\begin{array}{l}{\epsilon }_{x}^{a}=\frac{\partial {u}_{x}^{a}}{\partial x}\\ {\epsilon }_{y}^{a}=\frac{\partial {u}_{y}^{a}}{\partial y}\end{array}\left(\mathrm{PWAS}\right)$ (96)

(96)

$\left\{\begin{array}{l}{\epsilon }_{x}=\frac{\partial {u}_{x}}{\partial x}\\ {\epsilon }_{y}=\frac{\partial {u}_{y}}{\partial y}\end{array}\left(\mathrm{structure}\right)$ (97)

(97)

Substitution of ...

## With Safari, you learn the way you learn best. Get unlimited access to videos, live online training, learning paths, books, interactive tutorials, and more.

No credit card required