---
title: "Derive bar element stiffness matrix"
subject: "Finite Element Method"
chapter: 3
key: "semester_8/finite_element_method:bar-element-stiffness-derivation"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/bar-element-stiffness-derivation
site: BCT Engineering
---

# Derive bar element stiffness matrix

Finite Element Method · Chapter 3: Direct Stiffness Method: Discrete Finite Elements

- Asked in: year not given, 5 marks
- Video: [Derivation of Shape Functions and Stiffness Matrix for a 1D (2-Node) Bar Element | FEA](https://www.youtube.com/watch?v=kO0GxWI9_-o) — The Mechanical Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/bar-element-stiffness-derivation
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-3

## Question

Derive the stiffness matrix of a two-node uniform bar element of length $L$, area $A$ and Young's modulus $E$ carrying only axial load. Hence write the element equation.

## Short answer

The stiffness matrix of a uniform bar element of length L, area A, modulus E is [k] = (AE/L)[[1,-1],[-1,1]]. Using direct spring analogy with stiffness AE/L, the element equation is {f1; f2} = (AE/L)[1 -1; -1 1]{u1; u2}. It is symmetric, singular, and each row sums to zero.

## Answer

A bar element carries only axial force. Take nodes 1 and 2 with axial displacements $u_1$ and $u_2$ and nodal forces $f_1$ and $f_2$.

```text
  f1 -->  o===================o  --> f2
         node 1     L       node 2
         (u1)               (u2)
```

#### Direct method

The bar behaves like a spring of stiffness $k = \dfrac{AE}{L}$. The elongation is $\delta = u_2 - u_1$ and the axial force is $F = k\delta$.

Equilibrium of the element requires $f_1 = -F$ and $f_2 = +F$:
$$
\begin{aligned}
f_1 &= \frac{AE}{L}(u_1 - u_2) \\
f_2 &= \frac{AE}{L}(-u_1 + u_2)
\end{aligned}
$$

In matrix form:
$$
\begin{Bmatrix} f_1 \\ f_2 \end{Bmatrix} = \frac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}\begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix}
$$

#### Check by energy method

With linear shape functions $N_1 = 1 - x/L$, $N_2 = x/L$, the strain is $\varepsilon = \dfrac{du}{dx} = \dfrac{u_2 - u_1}{L}$, so $[B] = \dfrac{1}{L}[-1\ \ 1]$. Then
$$
[k] = \int_0^L [B]^T AE [B]\,dx = AE L\cdot\frac{1}{L^2}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} = \frac{AE}{L}\begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}
$$

Both methods give the same result.

#### Properties

- Symmetric, since $k_{12} = k_{21}$.
- Singular ($\det = 0$): the element can move as a rigid body until supports are applied.
- Each row sums to zero, which expresses equilibrium for rigid-body motion.

## Check yourself

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

1. In the direct method, to what is the bar element analogous?
   A. A spring of stiffness AE/L
   B. A beam of stiffness EI/L
   C. A rigid link
   D. A spring of stiffness L/AE
   Answer: A. The bar acts like a spring with k = AE/L, elongation u2 - u1 and force F = k(u2 - u1). (Key point: Direct method)

2. Which equation gives the nodal force f1 of the bar element?
   A. f1 = (AE/L)(u2 - u1)
   B. f1 = (AE/L)(u1 - u2)
   C. f1 = (AE/L)(u1 + u2)
   D. f1 = (L/AE)(u1 - u2)
   Answer: B. Equilibrium gives f1 = -F, so f1 = (AE/L)(u1 - u2) and f2 = (AE/L)(-u1 + u2). (Key point: Nodal forces)

3. With linear shape functions, what is [B] for the bar?
   A. [1 1]/L
   B. [1 -1]/L^2
   C. [-1 1]/L
   D. [x/L 1-x/L]
   Answer: C. epsilon = (u2 - u1)/L, so [B] = (1/L)[-1 1], and [k] = integral of B^T AE B dx. (Key point: Energy method)

4. Why is the element stiffness matrix singular?
   A. Its terms are unequal
   B. The area is zero
   C. It is not symmetric
   D. The element can move as a rigid body until supports are applied
   Answer: D. det = 0 because the unsupported element can translate without deforming; each row sums to zero. (Key point: Matrix properties)

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