---
title: "Axisymmetric stress analysis formulation"
subject: "Finite Element Method"
chapter: 6
key: "semester_8/finite_element_method:axisymmetric-analysis-strain-displacement"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/axisymmetric-analysis-strain-displacement
site: BCT Engineering
---

# Axisymmetric stress analysis formulation

Finite Element Method · Chapter 6: Applications in Solid Mechanics

- Asked in: year not given, 8 marks
- Video: [Axi-symmetric Problems in FEM I finite elements analysis I Axisymmetric Problems in FEM Explained](https://www.youtube.com/watch?v=8lImesNTLm4) — Mahesh Gadwantikar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/axisymmetric-analysis-strain-displacement
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-6

## Question

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.

## 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.

## Answer

#### 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 $r$-$z$. All quantities are independent of the circumferential angle $\theta$. Examples: pressure vessels, pipes, flywheels, rotors, pistons and shrink fits.

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

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

#### Unknowns and strains

Displacements: radial $u$ and axial $w$. Because $u$ displaces a ring, the circumference changes, producing a hoop strain.
$$
\{\varepsilon\} = \begin{Bmatrix} \varepsilon_r \\ \varepsilon_\theta \\ \varepsilon_z \\ \gamma_{rz} \end{Bmatrix} = \begin{Bmatrix} \partial u/\partial r \\ u/r \\ \partial w/\partial z \\ \partial u/\partial z + \partial w/\partial r \end{Bmatrix}
$$

For the triangular element with $u = \sum N_iu_i$:
$$
[B] = \frac{1}{2A}\begin{bmatrix} \beta_1 & 0 & \beta_2 & 0 & \beta_3 & 0 \\ \frac{\alpha_1}{r} + \beta_1 + \frac{\gamma_1z}{r} & 0 & \frac{\alpha_2}{r} + \beta_2 + \frac{\gamma_2z}{r} & 0 & \frac{\alpha_3}{r} + \beta_3 + \frac{\gamma_3z}{r} & 0 \\ 0 & \gamma_1 & 0 & \gamma_2 & 0 & \gamma_3 \\ \gamma_1 & \beta_1 & \gamma_2 & \beta_2 & \gamma_3 & \beta_3 \end{bmatrix}
$$
with $\alpha_i$, $\beta_i = z_j - z_k$ and $\gamma_i = r_k - r_j$ defined as for the CST using $(r, z)$.

#### Stress-strain matrix

$$
[D] = \frac{E}{(1 + \nu)(1 - 2\nu)}\begin{bmatrix} 1 - \nu & \nu & \nu & 0 \\ \nu & 1 - \nu & \nu & 0 \\ \nu & \nu & 1 - \nu & 0 \\ 0 & 0 & 0 & \frac{1 - 2\nu}{2} \end{bmatrix}
$$

#### Stiffness matrix

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

The hoop-strain row of $[B]$ contains $1/r$ and $z/r$ terms, so $[B]$ varies with position, unlike the CST in the plane problem. The integral also contains the factor $r$. Therefore $[k]$ is not constant. In practice it is approximated by evaluating $[B]$ at the centroid $(\bar r, \bar z)$:
$$
[k] \approx 2\pi\,\bar r\,A\,[\bar B]^T[D][\bar B]
$$
This is accurate when elements are small compared with their distance from the axis. Loads are applied as total forces on the full ring, $2\pi r$ times the load per unit circumference.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. When can a 3D body be analysed as an axisymmetric 2D problem in r-z?
   A. When loads vary with theta
   B. Only when strain in z is zero
   C. Geometry, material, loads and supports are all symmetric about the axis, so nothing depends on theta
   D. Only when the body is thin
   Answer: C. All quantities are independent of the circumferential angle, so one plane section represents the body; each element is a ring. (Key point: Axisymmetric idealisation)

2. Which relation gives the hoop strain?
   A. eps_theta = dw/dz
   B. eps_theta = u/r
   C. eps_theta = du/dz + dw/dr
   D. eps_theta = du/dr
   Answer: B. A radial displacement u changes the ring circumference, giving hoop strain u/r. (Key point: Strain relations)

3. How many strain components appear and what is the shear term in [D]?
   A. 3 components; (1 - nu)/2
   B. 4 components; (1 + nu)/2
   C. 6 components; 1
   D. 4 components; (1 - 2nu)/2 in the shear position, with E/[(1+nu)(1-2nu)]
   Answer: D. Axisymmetric analysis uses eps_r, eps_theta, eps_z and gamma_rz, with the same [D] form as plane strain extended to a 4x4. (Key point: Constitutive matrix)

4. Why is the axisymmetric triangle stiffness matrix not constant?
   A. The element has too many nodes
   B. [B] contains 1/r and z/r terms and the integral has a factor r
   C. Thickness varies
   D. The shape functions are quadratic
   Answer: B. The hoop-strain row depends on position and the volume element includes r, so it is usually approximated by evaluating [B] at the centroid: k = 2 pi r_bar A B^T D B. (Key point: Non-constant stiffness)

## Related in Finite Element Method

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- [Axisymmetric triangle strains and stresses at centroid](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aaxisymmetric-triangle-centroid-strains) — Finite Element Method, chapter 6

## Related in other subjects

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