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Axisymmetric stress analysis formulation

Short answer

Axisymmetric analysis treats a body of revolution with symmetric loads as a 2D r-z problem. Strains are {epsilon_r, epsilon_theta, epsilon_z, gamma_rz} = {u_r, u/r, w_z, u_z + w_r}; D is the 4x4 isotropic matrix with E/((1+nu)(1-2nu)) and diagonal entries 1-nu,1-nu,1-nu,(1-2nu)/2. Stiffness is not constant because [B] contains 1/r and z/r terms and the volume integral includes r dr dz.

Explain axisymmetric stress analysis. Write the strain-displacement relations and the stress-strain matrix for an axisymmetric solid, and explain why the stiffness matrix of an axisymmetric triangular element is not constant.

Axisymmetric problems

A body of revolution (geometry and material symmetric about an axis) with loads and supports also symmetric about the same axis can be analysed on a single plane section, as a 2D problem in rr-zz. All quantities are independent of the circumferential angle θ\theta. Examples: pressure vessels, pipes, flywheels, rotors, pistons and shrink fits.

        z (axis)
        |    +------+
        |    | mesh |  <- triangular ring
        |    +------+     elements
        +------------- r

Each element is really a ring of material of volume 2πrˉA2\pi\bar r A.

Unknowns and strains

Displacements: radial uu and axial ww. Because uu displaces a ring, the circumference changes, producing a hoop strain.

{ε}={εrεθεzγrz}={∂u/∂ru/r∂w/∂z∂u/∂z+∂w/∂r}\{\varepsilon\} = \begin{Bmatrix} \varepsilon_r \\ \varepsilon_\theta \\ \varepsilon_z \\ \gamma_{rz} \end{Bmatrix} = \begin{Bmatrix} \partial u/\partial r \\ u/r \\ \partial w/\partial z \\ \partial u/\partial z + \partial w/\partial r \end{Bmatrix}

For the triangular element with u=∑Niuiu = \sum N_iu_i:

[B]=12A[β10β20β30α1r+β1+γ1zr0α2r+β2+γ2zr0α3r+β3+γ3zr00γ10γ20γ3γ1β1γ2β2γ3β3][B] = \frac{1}{2A}\begin{bmatrix} \beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0 \\ \frac{\alpha_1}{r} + \beta_1 + \frac{\gamma_1z}{r} & 0 & \frac{\alpha_2}{r} + \beta_2 + \frac{\gamma_2z}{r} & 0 & \frac{\alpha_3}{r} + \beta_3 + \frac{\gamma_3z}{r} & 0 \\ 0 & \gamma_1 & 0 & \gamma_2 & 0 & \gamma_3 \\ \gamma_1 & \beta_1 & \gamma_2 & \beta_2 & \gamma_3 & \beta_3 \end{bmatrix}

with αi\alpha_i, βi=zj−zk\beta_i = z_j - z_k and γi=rk−rj\gamma_i = r_k - r_j defined as for the CST using (r,z)(r, z).

Stress-strain matrix

[D]=E(1+ν)(1−2ν)[1−ννν0ν1−νν0νν1−ν00001−2ν2][D] = \frac{E}{(1 + \nu)(1 - 2\nu)}\begin{bmatrix} 1 - \nu & \nu & \nu & 0 \\ \nu & 1 - \nu & \nu & 0 \\ \nu & \nu & 1 - \nu & 0 \\ 0 & 0 & 0 & \frac{1 - 2\nu}{2} \end{bmatrix}

Stiffness matrix

[k]=2π∫A[B]T[D][B] r dr dz[k] = 2\pi\int_A [B]^T[D][B]\,r\,dr\,dz

The hoop-strain row of [B][B] contains 1/r1/r and z/rz/r terms, so [B][B] varies with position, unlike the CST in the plane problem. The integral also contains the factor rr. Therefore [k][k] is not constant. In practice it is approximated by evaluating [B][B] at the centroid (rˉ,zˉ)(\bar r, \bar z):

[k]≈2π rˉ A [Bˉ]T[D][Bˉ][k] \approx 2\pi\,\bar r\,A\,[\bar B]^T[D][\bar B]

This is accurate when elements are small compared with their distance from the axis. Loads are applied as total forces on the full ring, 2πr2\pi r times the load per unit circumference.

✓ Answer checked by Claude Opus 5.5 (high effort)

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