---
title: "Plane stress constitutive; axisymmetric problems"
subject: "Computational Techniques in Civil Engineering"
chapter: 3
key: "semester_8/computational_techniques_in_civil_engineering:plane-stress-constitutive-axisym"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/plane-stress-constitutive-axisym
site: BCT Engineering
---

# Plane stress constitutive; axisymmetric problems

Computational Techniques in Civil Engineering · Chapter 3: Elasticity in solids

- Asked in: 2075 Bhadra, 5+3 marks
- Video: [Lec 14 : Mathematical formulation plane stress plane strain](https://www.youtube.com/watch?v=59RxQtSsnD0) — NPTEL IIT Guwahati
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/plane-stress-constitutive-axisym
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-3

## Question

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

## 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.

## Answer

#### Constitutive relation for plane stress
Start from Hooke's law for an isotropic material in 3D:
$$
\varepsilon_x=\frac{1}{E}[\sigma_x-\nu(\sigma_y+\sigma_z)],\quad \varepsilon_y=\frac{1}{E}[\sigma_y-\nu(\sigma_x+\sigma_z)],\quad \gamma_{xy}=\frac{\tau_{xy}}{G},\ G=\frac{E}{2(1+\nu)}
$$
For plane stress put $\sigma_z=0$:
$$
\varepsilon_x=\frac{\sigma_x-\nu\sigma_y}{E},\quad \varepsilon_y=\frac{\sigma_y-\nu\sigma_x}{E},\quad \gamma_{xy}=\frac{2(1+\nu)}{E}\tau_{xy}
$$
and $\varepsilon_z=-\nu(\sigma_x+\sigma_y)/E$. In matrix form $\{\varepsilon\}=[C]\{\sigma\}$:
$$
\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}=\frac{1}{E}\begin{bmatrix}1&-\nu&0\\-\nu&1&0\\0&0&2(1+\nu)\end{bmatrix}\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}
$$
Inverting the first two equations: from $E\varepsilon_x=\sigma_x-\nu\sigma_y$ and $E\varepsilon_y=\sigma_y-\nu\sigma_x$, solve for $\sigma_x=\dfrac{E}{1-\nu^2}(\varepsilon_x+\nu\varepsilon_y)$ and $\sigma_y=\dfrac{E}{1-\nu^2}(\varepsilon_y+\nu\varepsilon_x)$. Also $\tau_{xy}=G\gamma_{xy}=\dfrac{E}{2(1+\nu)}\gamma_{xy}$. Hence
$$
\begin{Bmatrix}\sigma_x\\\sigma_y\\\tau_{xy}\end{Bmatrix}=\frac{E}{1-\nu^2}\begin{bmatrix}1&\nu&0\\\nu&1&0\\0&0&\dfrac{1-\nu}{2}\end{bmatrix}\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}
$$
i.e. $\{\sigma\}=[D]\{\varepsilon\}$ with the plane stress matrix $[D]$ above.

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

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 $\{\varepsilon_r,\varepsilon_z,\varepsilon_\theta,\gamma_{rz}\}=\{\partial u/\partial r,\ \partial w/\partial z,\ u/r,\ \partial u/\partial z+\partial w/\partial r\}$ and the 3D problem is analysed on a 2D mesh in the $r$-$z$ plane (ring elements).

## Check yourself

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

1. Solving E eps_x = sigma_x - nu sigma_y and E eps_y = sigma_y - nu sigma_x for sigma_x gives what?
   A. sigma_x = E(eps_x + nu eps_y)
   B. sigma_x = E(eps_x - nu eps_y)/(1 - nu^2)
   C. sigma_x = E(eps_x + nu eps_y)/(1 - nu^2)
   D. sigma_x = E eps_x/(1 + nu)
   Answer: C. Eliminating sigma_y gives sigma_x = E(eps_x + nu eps_y)/(1 - nu^2). (Key point: Inversion of plane stress law)

2. How does the shear term (1 - nu)/2 in the plane stress [D] arise?
   A. From tau = G gamma with G = E/(2(1 + nu)) = E/(1 - nu^2) x (1 - nu)/2
   B. From nu/2
   C. From sigma_z = 0
   D. From epsilon_theta = u/r
   Answer: A. E/(2(1 + nu)) can be written as [E/(1 - nu^2)] x (1 - nu)/2. (Key point: Shear term)

3. Which set is the axisymmetric strain-displacement relation?
   A. {du/dx, dv/dy, du/dy + dv/dx}
   B. {du/dr, dw/dz, u/r, du/dz + dw/dr}
   C. {u/r, w/z, 0, 0}
   D. {du/dr, dw/dz, w/r, du/dr}
   Answer: B. epsilon_r = du/dr, epsilon_z = dw/dz, epsilon_theta = u/r, gamma_rz = du/dz + dw/dr. (Key point: Axisymmetric strains)

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