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Plane stress, strain, axisymmetric with constitutive

Short answer

Plane stress: thin plate loaded in its plane, sigma_z=0, D=E/(1-nu^2)[[1,nu,0],[nu,1,0],[0,0,(1-nu)/2]]. Plane strain: long prism, eps_z=0, D=E/((1+nu)(1-2nu))[[1-nu,nu,0],[nu,1-nu,0],[0,0,(1-2nu)/2]]. Axisymmetric: solid of revolution with symmetric loads, fields in r,z, hoop strain u/r; D has four components with nu/(1-nu) coupling. Examples: thin plate with hole, dam, footing.

Describe the plane stress, plane strain and axisymmetric problems with their examples and constitutive relations to be used for stress analysis problems.

Plane stress

Plane stress: a thin plate loaded only by forces in its own plane, so the stresses through the thickness are zero: σz=τxz=τyz=0\sigma_z=\tau_{xz}=\tau_{yz}=0. The stress state is two-dimensional, {σ}={σx,σy,τxy}T\{\sigma\}=\{\sigma_x,\sigma_y,\tau_{xy}\}^T, although the strain εz=−νE(σx+σy)≠0\varepsilon_z=-\dfrac{\nu}{E}(\sigma_x+\sigma_y)\neq0. Examples: thin plate with a hole, gusset plate, deep beam, shear wall. Constitutive relation:

{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

Plane strain: a long prism (length much larger than cross-section) with uniform cross-section and load that does not vary along its length and is perpendicular to it. Movement along the axis is prevented, so εz=γxz=γyz=0\varepsilon_z=\gamma_{xz}=\gamma_{yz}=0, but σz=ν(σx+σy)≠0\sigma_z=\nu(\sigma_x+\sigma_y)\neq0. Examples: gravity dam, retaining wall, long tunnel, strip footing. Constitutive relation:

{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}

Axisymmetric

Axisymmetric problem: a body of revolution about an axis (the zz-axis) whose loading and supports are also symmetric about that axis. All quantities are independent of the circumferential angle θ\theta, so the 3D problem reduces to a 2D problem in the rr-zz plane. The circumferential displacement is zero (vθ=0v_\theta=0), the shear strains γrθ=γzθ=0\gamma_{r\theta}=\gamma_{z\theta}=0, and the non-zero components are εr,εz,εθ,γrz\varepsilon_r,\varepsilon_z,\varepsilon_\theta,\gamma_{rz}.

Examples: circular footing, pile or well under axial load, pressure vessel and cylindrical tank, water tank, shaft, tunnel/borehole under uniform pressure, thick-walled cylinder under internal pressure, soil sample in a triaxial test.

Strain-displacement relations (displacements uu along rr and ww along zz):

{ε}={εrεzεθγrz}={∂u∂r∂w∂zur∂u∂z+∂w∂r}\{\varepsilon\}=\begin{Bmatrix}\varepsilon_r\\\varepsilon_z\\\varepsilon_\theta\\\gamma_{rz}\end{Bmatrix}=\begin{Bmatrix}\dfrac{\partial u}{\partial r}\\[2mm]\dfrac{\partial w}{\partial z}\\[2mm]\dfrac{u}{r}\\[2mm]\dfrac{\partial u}{\partial z}+\dfrac{\partial w}{\partial r}\end{Bmatrix}

The hoop strain εθ=u/r\varepsilon_\theta=u/r arises because a ring at radius rr stretches from circumference 2πr2\pi r to 2π(r+u)2\pi(r+u).

Constitutive relation {σ}=[D]{ε}\{\sigma\}=[D]\{\varepsilon\} with {σ}={σr,σz,σθ,τrz}T\{\sigma\}=\{\sigma_r,\sigma_z,\sigma_\theta,\tau_{rz}\}^T:

[D]=E(1−ν)(1+ν)(1−2ν)[1ν1−νν1−ν0ν1−ν1ν1−ν0ν1−νν1−ν100001−2ν2(1−ν)][D]=\frac{E(1-\nu)}{(1+\nu)(1-2\nu)}\begin{bmatrix}1&\dfrac{\nu}{1-\nu}&\dfrac{\nu}{1-\nu}&0\\[2mm]\dfrac{\nu}{1-\nu}&1&\dfrac{\nu}{1-\nu}&0\\[2mm]\dfrac{\nu}{1-\nu}&\dfrac{\nu}{1-\nu}&1&0\\[2mm]0&0&0&\dfrac{1-2\nu}{2(1-\nu)}\end{bmatrix}

✓ Answer checked by Claude Opus 5.5 (high effort)

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