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Plane stress and plane strain differences

Short answer

Plane stress assumes sigma_z=0 in a thin body, giving epsilon_z=-nu/E(sigma_x+sigma_y); plane strain assumes epsilon_z=0 in a long body, giving sigma_z=nu(sigma_x+sigma_y). Their constitutive matrices differ: plane stress uses E/(1-nu^2) with 1, nu, 0; plane strain uses E/[(1+nu)(1-2nu)] with 1-nu, nu, 0. Examples: thin plate vs dam.

Differentiate between plane stress and plane strain conditions with practical examples. Write the stress-strain relations for each.

Both are two-dimensional idealisations of a 3D elasticity problem, which reduce the unknowns from six stress components to three.

Comparison

PointPlane stressPlane strain
DefinitionStress components in the z-direction are zero: σz=τxz=τyz=0\sigma_z = \tau_{xz} = \tau_{yz} = 0Strain components in the z-direction are zero: εz=γxz=γyz=0\varepsilon_z = \gamma_{xz} = \gamma_{yz} = 0
GeometryThin body (thickness much smaller than other dimensions)Very long body, with constant section and load along its length
LoadingIn-plane loads onlyLoads perpendicular to the length, not varying along it
Out-of-plane valueεz=−νE(σx+σy)≠0\varepsilon_z = -\dfrac{\nu}{E}(\sigma_x + \sigma_y) \neq 0σz=ν(σx+σy)≠0\sigma_z = \nu(\sigma_x + \sigma_y) \neq 0
ExamplesThin plate with a hole, gusset plate, deep beam webDam section, retaining wall, long cylinder or tunnel under pressure, rolling process
Unknownsσx,σy,τxy\sigma_x, \sigma_y, \tau_{xy}σx,σy,τxy\sigma_x, \sigma_y, \tau_{xy}

Constitutive matrices

Plane stress

{σxσyτxy}=E1−ν2[1ν0ν10001−ν2]{εxεyγxy}\begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = \frac{E}{1 - \nu^2}\begin{bmatrix} 1 & \nu & 0 \\ \nu & 1 & 0 \\ 0 & 0 & \frac{1 - \nu}{2} \end{bmatrix}\begin{Bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{Bmatrix}

Plane strain

{σxσyτxy}=E(1+ν)(1−2ν)[1−νν0ν1−ν0001−2ν2]{εxεyγxy}\begin{Bmatrix} \sigma_x \\ \sigma_y \\ \tau_{xy} \end{Bmatrix} = \frac{E}{(1 + \nu)(1 - 2\nu)}\begin{bmatrix} 1 - \nu & \nu & 0 \\ \nu & 1 - \nu & 0 \\ 0 & 0 & \frac{1 - 2\nu}{2} \end{bmatrix}\begin{Bmatrix} \varepsilon_x \\ \varepsilon_y \\ \gamma_{xy} \end{Bmatrix}

The plane strain matrix is obtained from the plane stress matrix by replacing E→E1−ν2E \to \dfrac{E}{1 - \nu^2} and ν→ν1−ν\nu \to \dfrac{\nu}{1 - \nu}. Plane strain is stiffer for the same EE and ν\nu, because the z-direction is constrained.

✓ Answer checked by Claude Opus 5.5 (high effort)

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