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Axisymmetric: constitutive, strain-displacement

Short answer

Axisymmetric problems describe bodies of revolution with symmetric loading, reduced to r-z plane. Displacements u (radial) and w (axial) give ε_r = ∂u/∂r, ε_z = ∂w/∂z, ε_θ = u/r, γ_rz = ∂u/∂z+∂w/∂r. The constitutive matrix [D] = (E(1-ν)/[(1+ν)(1-2ν)]) [[1, ν/(1-ν), ν/(1-ν), 0]; [ν/(1-ν), 1, ν/(1-ν), 0]; [ν/(1-ν), ν/(1-ν), 1, 0]; [0,0,0,(1-2ν)/(2(1-ν))]].

What do you understand by axisymmetric problem? Explain with examples and write the constitutive relations and strain-displacement relations for axisymmetric condition.

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}
        z
        ^   |p|
        |   v v v      axis of symmetry
   -----+---------
   |    |   r ->  |   Only the r-z half-plane
   |    |         |   is analysed.
   -----+---------

✓ Answer checked by Claude Opus 5.5 (high effort)

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