Solution to Problem 223 Triaxial Deformation

 

 

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,
$\varepsilon_y = \dfrac{1}{29 \times 10^6} [ \, -5000 - 0.30(6000 + 9000) \, ]$

$\varepsilon_y = -3.276 \times 10^{-4}$
 

ε_y 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