In the generalized Hooke's Law equations, let ${\epsilon }_x={\epsilon }_{p1}$ and ${\epsilon }_y={\epsilon }_{p2}$ and compute the principal stress ${\sigma }_{p1}$:

${\mathrm{\sigma }}_{\mathrm{x}}=\frac{\mathrm{E}}{(1-{\mathrm{\nu }}^2)}({\mathrm{\epsilon }}_{\mathrm{x}}+{\mathrm{\nu \epsilon }}_{\mathrm{y}})$ $=\frac{70,000 \mathrm{MPa}}{{(1-0.33)}^2}[0.001330+(0.33\left)\right(-0.000910\left)\right]$ $=80.89 \mathrm{MPa}$

Note: The strain values must be converted from $\mathrm{\mu \epsilon }$ to dimensionless values (i.e., mm/mm) before performing the calculation.

Let ${\epsilon }_x={\epsilon }_{p1}$ and ${\epsilon }_y={\epsilon }_{p2}$ and compute the principal stress ${\sigma }_{p2}$:

${\sigma }_y=\frac{E}{(1-{\nu }^2)}({\epsilon }_y+\nu {\epsilon }_x)$
$=\frac{70,000 \mathrm{MPa}}{{(1-0.33)}^2}[-0.000910+(0.33\left)\right(-0.001330\left)\right]$ $=-37.01 \mathrm{MPa}$