Solution to Problem 224 Triaxial Deformation

$\varepsilon_z = \dfrac{1}{E} [ \, \sigma_z - \nu (\sigma_x + \sigma_y) \, ]$
 

Where
σ_x = 6.0 ksi (tension)
σ_y = 5.0 ksi (compression)
σ_z = 9.0 ksi (tension)
 

$\varepsilon_z = \dfrac{1}{29 \times 10^6} [ \, 9000 - 0.3(6000 - 5000) \, ]$

$\varepsilon_z = 3 \times 10^{-4}$
 

ε_z is positive, thus positive stress is needed in the x-direction to eliminate deformation in z-direction.
 

The application of loads is still simultaneous:
(No deformation means zero strain)
$\varepsilon_z = \dfrac{1}{E} [ \, \sigma_z - \nu (\sigma_x + \sigma_y) \, ] = 0$

$\sigma_z = \nu(\sigma_x + \sigma_y)$
 

Where
σ_y = 5.0 ksi (compression)
σσ_z = 9.0 ksi (tension)

 

$9000 = 0.30(\sigma_x - 5000)$

$\sigma_x = 35\,000 \, \text{psi}$
 

$\sigma_{added} + 6000 = 35\,000$

$\sigma_{added} = 29\,000 \, \text{psi}$

$\dfrac{P_{added}}{2(4)} = 29\,000$

$P_{added} = 232\,000 \, \text{lb}$

$P_{added} = 232 \, \text{ kips}$           answer