---
title: "Strain-displacement [B] matrix for CST"
subject: "Computational Techniques in Civil Engineering"
chapter: 4
key: "semester_8/computational_techniques_in_civil_engineering:cst-b-matrix"
source: https://www.bctengineer.com/q/semester_8/computational_techniques_in_civil_engineering/cst-b-matrix
site: BCT Engineering
---

# Strain-displacement [B] matrix for CST

Computational Techniques in Civil Engineering · Chapter 4: Finite element method

- Asked in: 2073 Magh, 6 marks
- Video: [2D Strain/Stress Analysis in FEA | FEA Done Right – Section 5](https://www.youtube.com/watch?v=qXwTII0ko90) — MechCADemy
- 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/cst-b-matrix
- Whole chapter: https://www.bctengineer.com/md/semester_8/computational_techniques_in_civil_engineering/chapter-4

## Question

Derive the relation of the strain-displacement [B] matrix for the constant strain triangle.

## Short answer

For the CST, displacements are linear and shape functions are Ni=(ai+bi*x+ci*y)/(2A). Its strain-displacement matrix is B=(1/(2A))*[b1 0 b2 0 b3 0; 0 c1 0 c2 0 c3; c1 b1 c2 b2 c3 b3], with bi=yj-yk and ci=xk-xj. B is constant, so strains and stresses are constant in the element.

## Answer

#### Constant strain triangle (CST)
The CST has three nodes with coordinates $(x_i,y_i)$, $i=1,2,3$ (anticlockwise) and two DOFs per node: $\{d\}=\{u_1,v_1,u_2,v_2,u_3,v_3\}^T$.

#### Displacement field
A linear polynomial is assumed (3 constants per component, one for each node):
$$
u=\alpha_1+\alpha_2x+\alpha_3y,\qquad v=\alpha_4+\alpha_5x+\alpha_6y
$$
Fitting the three nodal values of $u$ and solving for $\alpha_1,\alpha_2,\alpha_3$ gives
$$
u=N_1u_1+N_2u_2+N_3u_3,\qquad N_i=\frac{1}{2A}(a_i+b_ix+c_iy)
$$
where $A=\tfrac12\left|\begin{matrix}1&x_1&y_1\\1&x_2&y_2\\1&x_3&y_3\end{matrix}\right|$ is the area and
$$
\begin{aligned}
a_1&=x_2y_3-x_3y_2,& b_1&=y_2-y_3,& c_1&=x_3-x_2\\
a_2&=x_3y_1-x_1y_3,& b_2&=y_3-y_1,& c_2&=x_1-x_3\\
a_3&=x_1y_2-x_2y_1,& b_3&=y_1-y_2,& c_3&=x_2-x_1
\end{aligned}
$$
The same $N_i$ are used for $v$. In matrix form $\{u\ v\}^T=[N]\{d\}$ with $[N]=\begin{bmatrix}N_1&0&N_2&0&N_3&0\\0&N_1&0&N_2&0&N_3\end{bmatrix}$.

#### Strain-displacement relation
For plane problems
$$
\{\varepsilon\}=\begin{Bmatrix}\varepsilon_x\\\varepsilon_y\\\gamma_{xy}\end{Bmatrix}=\begin{Bmatrix}\partial u/\partial x\\\partial v/\partial y\\\partial u/\partial y+\partial v/\partial x\end{Bmatrix}
$$
Differentiating, $\dfrac{\partial N_i}{\partial x}=\dfrac{b_i}{2A}$ and $\dfrac{\partial N_i}{\partial y}=\dfrac{c_i}{2A}$ (the constants $a_i$ drop out). So
$$
\varepsilon_x=\frac{1}{2A}(b_1u_1+b_2u_2+b_3u_3),\quad \varepsilon_y=\frac{1}{2A}(c_1v_1+c_2v_2+c_3v_3),
$$
$$
\gamma_{xy}=\frac{1}{2A}(c_1u_1+b_1v_1+c_2u_2+b_2v_2+c_3u_3+b_3v_3)
$$
Hence $\{\varepsilon\}=[B]\{d\}$ with
$$
\boxed{[B]=\frac{1}{2A}\begin{bmatrix}b_1&0&b_2&0&b_3&0\\0&c_1&0&c_2&0&c_3\\c_1&b_1&c_2&b_2&c_3&b_3\end{bmatrix}}
$$

#### Remarks
- $[B]$ contains only the nodal coordinates, so it is **constant** over the element. Strain and stress (from $\{\sigma\}=[D][B]\{d\}$) are therefore constant, hence the name "constant strain" triangle.
- The element stiffness follows as $[k]=t\,A\,[B]^T[D][B]$ for an element of thickness $t$.

## Check yourself

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

1. What displacement field is assumed in a CST?
   A. Quadratic in x and y
   B. Linear in x and y, u = a1 + a2 x + a3 y (and similarly for v)
   C. Constant displacement
   D. Cubic in x only
   Answer: B. Three nodes give three constants per component, so a linear polynomial is used. (Key point: Assumed displacement)

2. With N_i = (a_i + b_i x + c_i y)/2A, what are dN_i/dx and dN_i/dy?
   A. b_i/(2A) and c_i/(2A); a_i drops out
   B. a_i/(2A) and b_i/(2A)
   C. c_i/(2A) and b_i/(2A)
   D. Both zero
   Answer: A. Differentiation removes the constant a_i and leaves b_i/2A and c_i/2A. (Key point: Derivatives of shape functions)

3. Why is the triangle called a constant strain triangle?
   A. Its thickness is constant
   B. [B] contains only nodal coordinates, so strain (and stress) are uniform over the element
   C. Loads are applied uniformly
   D. Its area is constant
   Answer: B. [B] depends on b_i, c_i and A, not on x or y, so e = B d is constant. (Key point: Why constant strain)

## Related in Computational Techniques in Civil Engineering

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