Given the generalized Hooke’s law for linear isotropic materials (on the equation sheet), derive the simplified equations for plane stress and plane strain conditions. Material Property Relations Poisson’s ratio v= Elat Elong Generalized Hooke’s Law -=[σ-ν(σ +σ)] y = [-1(x+2)] € =[2-(x+y)] 1 1 Yxy=xyYyzyzYzx=x where E 2(1+1)
The Correct Answer and Explanation is :
Derivation of Simplified Equations for Plane Stress and Plane Strain Conditions
Generalized Hooke’s Law:
For a linear isotropic material under 3D stress, the strain components are expressed as:
[
\varepsilon_x = \frac{1}{E} \left[\sigma_x – \nu (\sigma_y + \sigma_z) \right]
]
[
\varepsilon_y = \frac{1}{E} \left[\sigma_y – \nu (\sigma_x + \sigma_z) \right]
]
[
\varepsilon_z = \frac{1}{E} \left[\sigma_z – \nu (\sigma_x + \sigma_y) \right]
]
[
\gamma_{xy} = \frac{\tau_{xy}}{G}, \quad \gamma_{yz} = \frac{\tau_{yz}}{G}, \quad \gamma_{zx} = \frac{\tau_{zx}}{G}
]
where ( E ) is Young’s modulus, ( \nu ) is Poisson’s ratio, and ( G = \frac{E}{2(1 + \nu)} ) is the shear modulus.
Plane Stress Conditions:
Plane stress assumes that out-of-plane stresses are zero (( \sigma_z = \tau_{yz} = \tau_{zx} = 0 )). Substituting ( \sigma_z = 0 ) into the generalized equations:
[
\varepsilon_x = \frac{1}{E} \left[\sigma_x – \nu \sigma_y \right]
]
[
\varepsilon_y = \frac{1}{E} \left[\sigma_y – \nu \sigma_x \right]
]
[
\varepsilon_z = \frac{-\nu}{E} (\sigma_x + \sigma_y)
]
Here, ( \varepsilon_z ) is a secondary strain caused by the in-plane stresses due to Poisson’s effect.
Plane Strain Conditions:
Plane strain assumes that out-of-plane strains are zero (( \varepsilon_z = \gamma_{yz} = \gamma_{zx} = 0 )). From the strain equation for ( \varepsilon_z ):
[
\varepsilon_z = \frac{1}{E} \left[\sigma_z – \nu (\sigma_x + \sigma_y) \right] = 0
]
Solving for ( \sigma_z ):
[
\sigma_z = \nu (\sigma_x + \sigma_y)
]
Substitute ( \sigma_z = \nu (\sigma_x + \sigma_y) ) into the in-plane strain equations:
[
\varepsilon_x = \frac{1}{E} \left[\sigma_x – \nu (\sigma_y + \nu (\sigma_x + \sigma_y)) \right]
]
[
\varepsilon_y = \frac{1}{E} \left[\sigma_y – \nu (\sigma_x + \nu (\sigma_x + \sigma_y)) \right]
]
Simplify:
[
\varepsilon_x = \frac{1 – \nu^2}{E} \sigma_x – \frac{\nu (1 + \nu)}{E} \sigma_y
]
[
\varepsilon_y = \frac{1 – \nu^2}{E} \sigma_y – \frac{\nu (1 + \nu)}{E} \sigma_x
]
Explanation:
The derivation shows how 3D stress-strain relations reduce to simplified equations under specific assumptions. In plane stress, the absence of out-of-plane stress simplifies the strain directly, while in plane strain, out-of-plane constraints induce an equivalent stress (( \sigma_z )). These relations are critical for structural analysis where one dimension is negligible (plane stress) or constrained (plane strain).