---
title: "Derive CST element stiffness matrix"
subject: "Finite Element Method"
chapter: 6
key: "semester_8/finite_element_method:cst-stiffness-matrix-derivation"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/cst-stiffness-matrix-derivation
site: BCT Engineering
---

# Derive CST element stiffness matrix

Finite Element Method · Chapter 6: Applications in Solid Mechanics

- Asked in: year not given, 8 marks
- Video: [Constant Strain Triangle Elements in FEM | CST in Finite Element Methods| Finite Element Analysis](https://www.youtube.com/watch?v=NXORiASKLcI) — 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/cst-stiffness-matrix-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-6

## Question

Derive the stiffness matrix of a constant strain triangular element for plane stress, starting from the principle of minimum potential energy. Write the form of the load vectors for body force and for a uniform traction on an edge.

## Short answer

Minimizing potential energy gives [k] = t A [B]^T [D] [B] for a CST, a symmetric 6x6 singular matrix. Body force load vector is (tA/3){X_x,X_y,X_x,X_y,X_x,X_y}^T; uniform traction on edge length ell gives (t ell/2)T_x and (t ell/2)T_y at each edge node, zero at the third.

## Answer

#### Displacement and strain

The element has nodes 1, 2, 3 with $\{d\} = \{u_1, v_1, u_2, v_2, u_3, v_3\}^T$. The displacement field is $\{u\} = [N]\{d\}$ with linear shape functions $N_i = \dfrac{1}{2A}(a_i + b_ix + c_iy)$.

Strains:
$$
\{\varepsilon\} = \begin{Bmatrix} \partial u/\partial x \\ \partial v/\partial y \\ \partial u/\partial y + \partial v/\partial x \end{Bmatrix} = [B]\{d\},\quad [B] = \frac{1}{2A}\begin{bmatrix} b_1 & 0 & b_2 & 0 & b_3 & 0 \\ 0 & c_1 & 0 & c_2 & 0 & c_3 \\ c_1 & b_1 & c_2 & b_2 & c_3 & b_3 \end{bmatrix}
$$

Stresses: $\{\sigma\} = [D]\{\varepsilon\} = [D][B]\{d\}$.

#### Potential energy

$$
\Pi = \underbrace{\frac{1}{2}\int_V \{\varepsilon\}^T[D]\{\varepsilon\}\,dV}_{U} - \underbrace{\left(\int_V \{u\}^T\{X\}\,dV + \int_S \{u\}^T\{T\}\,dS + \{d\}^T\{P\}\right)}_{W}
$$

Substituting $\{\varepsilon\} = [B]\{d\}$:
$$
\Pi = \frac{1}{2}\{d\}^T\left[\int_V [B]^T[D][B]\,dV\right]\{d\} - \{d\}^T\{f\}
$$

Minimising, $\dfrac{\partial\Pi}{\partial\{d\}} = 0$, gives $[k]\{d\} = \{f\}$ with
$$
[k] = \int_V [B]^T[D][B]\,dV
$$

#### Evaluation

For CST, $[B]$ and $[D]$ are constant, and the integral over the volume is just the element volume $tA$ ($t$ is the thickness):
$$
\boxed{[k] = t\,A\,[B]^T[D][B]}
$$

This is a symmetric $6\times6$ matrix, singular until supports are applied (three rigid-body modes: two translations and one rotation).

#### Load vectors

- **Body force** $\{X\} = \{X_x, X_y\}^T$ (per unit volume): $\{f_b\} = \displaystyle\int_V [N]^T\{X\}\,dV = \dfrac{tA}{3}\{X_x,\ X_y,\ X_x,\ X_y,\ X_x,\ X_y\}^T$, i.e. one-third of the total body force at each node.
- **Uniform traction** $\{T\} = \{T_x, T_y\}^T$ on an edge of length $\ell$ between nodes $i$ and $j$: $\{f_s\} = \displaystyle\int_S [N]^T\{T\}\,t\,dS$ gives $\dfrac{t\,\ell}{2}T_x$ and $\dfrac{t\,\ell}{2}T_y$ at each of the two edge nodes and zero at the third node.

## Check yourself

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

1. What does minimising the potential energy give for the element stiffness matrix?
   A. [k] = integral of [B][D][B]^T dV
   B. [k] = integral over V of [B]^T[D][B] dV
   C. [k] = integral of [N]^T[D][N] dV
   D. [k] = integral of [D][B] dV
   Answer: B. Substituting strain = [B]{d} in the strain energy and setting dPi/dd = 0 gives [B]^T[D][B] integrated over volume. (Key point: General stiffness)

2. Why does the integral reduce to t A [B]^T[D][B] for the CST?
   A. The shape functions are quadratic
   B. The thickness is zero
   C. The element is symmetric
   D. [B] and [D] are constant over the element, so the integral is just the volume tA
   Answer: D. Constant integrand means integration just multiplies by the element volume tA. (Key point: CST evaluation)

3. What is true of the 6x6 CST stiffness matrix before supports are applied?
   A. Symmetric and positive definite
   B. Diagonal
   C. Unsymmetric and singular
   D. Symmetric and singular, with three rigid-body modes
   Answer: D. Rigid-body motion (two translations and one rotation) causes no strain, so [k] is singular until supports are applied. (Key point: Matrix properties)

4. How are body force and edge traction distributed to the nodes?
   A. Body force: all at node 1; traction: all at one edge node
   B. Body force: one-third of the total at each node; traction: half of t*l*T at each of the two edge nodes
   C. Body force: half at each of two nodes; traction: one-third at each of three nodes
   D. Both: equal thirds at three nodes
   Answer: B. Integrating [N]^T times the load gives tA/3 X per node for body force, and t*l/2 T at each edge node with zero at the third node for uniform traction. (Key point: Load vectors)

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