For triaxial deformation (tensile triaxial stresses):
(compressive stresses are negative stresses)
$\varepsilon_y = \dfrac{1}{E} [ \, \sigma_y - \nu (\sigma_x + \sigma_z) \, ]$
$\sigma_x = \dfrac{P_x}{A_{yz}} = \dfrac{48}{4(2)} = 6.0 \, \text{ ksi}$ (tension)
$\sigma_y = \dfrac{P_y}{A_{xz}} = \dfrac{60}{4(3)} = 5.0 \, \text{ ksi}$ (compression)
$\sigma_z = \dfrac{P_z}{A_{xy}} = \dfrac{54}{2(3)} = 9.0 \, \text{ksi}$ (tension)
Thus,
$\varepsilon_y = \dfrac{1}{29 \times 10^6} [ \, -5000 - 0.30(6000 + 9000) \, ]$
$\varepsilon_y = -3.276 \times 10^{-4}$
εy is negative, thus, tensile force is required in the x-direction to produce the same deformation in the y-direction as the original forces.
For equivalent single force in the x-direction:
(uniaxial stress)
$\nu = -\dfrac{\varepsilon_y}{\varepsilon_x}$
$-\nu \varepsilon_x = \varepsilon_y$
$-\nu \dfrac{\sigma_x}{E} = \varepsilon_y$
$-0.30 \left( \dfrac{\sigma_x}{29 \times 10^6} \right) = -3.276 \times 10^{-4}$
$\sigma_x = 31\,666.67 \, \text{psi}$
$\sigma_x = \dfrac{P_x}{4(2)} = 31\,666.67$
$P_x = 253\,333.33 \, \text{lb}$ (tension)
$P_x = 253.33 \, \text{ kips}$ (tension) answer
