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Plane stress constitutive; axisymmetric problems

Short answer

Plane stress constitutive law: {σ}=[D]{ε} with [D]=E/(1-ν^2)[[1,ν,0],[ν,1,0],[0,0,(1-ν)/2]]. Axisymmetric problems: bodies of revolution with axisymmetric loading, e.g., circular footing, pressure vessel; strains ε_r=∂u/∂r, ε_z=∂w/∂z, ε_θ=u/r, γ_rz=∂u/∂z+∂w/∂r and a 4×4 constitutive matrix.

Derive the constitutive relation for plane stress problems. Explain axisymmetric problems with examples.

Constitutive relation for plane stress

Start from Hooke's law for an isotropic material in 3D:

εx=1E[σx−ν(σy+σz)],εy=1E[σy−ν(σx+σz)],γxy=τxyG, G=E2(1+ν)\varepsilon_x=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)],\quad \varepsilon_y=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)],\quad \gamma_{xy}=\frac{\tau_{xy}}{G},\ G=\frac{E}{2(1+\nu)}

For plane stress put σz=0\sigma_z=0:

εx=σx−νσyE,εy=σy−νσxE,γxy=2(1+ν)Eτxy\varepsilon_x=\frac{\sigma_x-\nu\sigma_y}{E},\quad \varepsilon_y=\frac{\sigma_y-\nu\sigma_x}{E},\quad \gamma_{xy}=\frac{2(1+\nu)}{E}\tau_{xy}

and εz=−ν(σx+σy)/E\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E. In matrix form {ε}=[C]{σ}\{\varepsilon\}=[C]\{\sigma\}:

{εxεyγxy}=1E[1−ν0−ν10002(1+ν)]{σxσyτxy}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}=\frac{1}{E}\begin{bmatrix}1&-\nu&0\\-\nu&1&0\\0&0&2(1+\nu)\end{bmatrix}\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}

Inverting the first two equations: from Eεx=σx−νσyE\varepsilon_x=\sigma_x-\nu\sigma_y and Eεy=σy−νσxE\varepsilon_y=\sigma_y-\nu\sigma_x, solve for σx=E1−ν2(εx+νεy)\sigma_x=\dfrac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y) and σy=E1−ν2(εy+νεx)\sigma_y=\dfrac{E}{1-\nu^2}(\varepsilon_y+\nu\varepsilon_x). Also τxy=Gγxy=E2(1+ν)γxy\tau_{xy}=G\gamma_{xy}=\dfrac{E}{2(1+\nu)}\gamma_{xy}. Hence

{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&\dfrac{1-\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}

i.e. {σ}=[D]{ε}\{\sigma\}=[D]\{\varepsilon\} with the plane stress matrix [D][D] above.

Axisymmetric problems

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.

Strains are {εr,εz,εθ,γrz}={∂u/∂r, ∂w/∂z, u/r, ∂u/∂z+∂w/∂r}\{\varepsilon_r,\varepsilon_z,\varepsilon_\theta,\gamma_{rz}\}=\{\partial u/\partial r,\ \partial w/\partial z,\ u/r,\ \partial u/\partial z+\partial w/\partial r\} and the 3D problem is analysed on a 2D mesh in the rr-zz plane (ring elements).

✓ Answer checked by Claude Opus 5.5 (high effort)

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