---
title: "Linear shape functions for one-dimensional element"
subject: "Finite Element Method"
chapter: 5
key: "semester_8/finite_element_method:linear-bar-shape-functions"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/linear-bar-shape-functions
site: BCT Engineering
---

# Linear shape functions for one-dimensional element

Finite Element Method · Chapter 5: Interpolation Functions

- Asked in: year not given, 6 marks
- Video: [Shape Functions - 1D Element](https://www.youtube.com/watch?v=g7ralyIjotI) — Postcard Professor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/linear-bar-shape-functions
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-5

## Question

Derive the shape functions for a two-node one-dimensional (linear) element of length $L$. State their properties and use them to write the strain-displacement matrix.

## Short answer

For a two-node bar element N1=1-x/L and N2=x/L. They satisfy Kronecker delta, sum to unity, and give C0 displacement with constant strain. Strain-displacement matrix is [B]=(1/L)[-1 1].

## Answer

#### Displacement function

Two nodal values $u_1$ (at $x = 0$) and $u_2$ (at $x = L$) allow a linear polynomial:
$$
u(x) = a_1 + a_2x
$$

```text
   u1                u2
   o-----------------o
  x=0               x=L
```

Apply the nodal conditions:
$$
u(0) = a_1 = u_1,\qquad u(L) = a_1 + a_2L = u_2 \Rightarrow a_2 = \frac{u_2 - u_1}{L}
$$

Substitute back:
$$
u = u_1 + \frac{u_2 - u_1}{L}x = \left(1 - \frac{x}{L}\right)u_1 + \frac{x}{L}\,u_2
$$

So $u = N_1u_1 + N_2u_2$ with
$$
N_1 = 1 - \frac{x}{L},\qquad N_2 = \frac{x}{L}
$$

#### Properties

- $N_i = 1$ at node $i$ and $0$ at the other node (Kronecker delta property).
- $N_1 + N_2 = 1$ everywhere (partition of unity); this allows rigid-body motion.
- Each $N_i$ is linear, so the displacement is continuous between elements ($C^0$), but the strain is constant per element.
- In natural coordinate $\xi = 2x/L - 1$: $N_1 = \dfrac{1 - \xi}{2}$ and $N_2 = \dfrac{1 + \xi}{2}$.

#### Strain-displacement matrix

$$
\varepsilon = \frac{du}{dx} = \frac{dN_1}{dx}u_1 + \frac{dN_2}{dx}u_2 = \frac{1}{L}\left[-1\ \ 1\right]\begin{Bmatrix} u_1 \\ u_2 \end{Bmatrix}
$$
$$
[B] = \frac{1}{L}\left[-1\ \ 1\right]
$$

## Check yourself

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

1. Why is a linear polynomial used for a two-node element?
   A. Two nodal values determine two coefficients
   B. It has three nodes
   C. The strain is zero
   D. The displacement is constant
   Answer: A. Two nodal values u1 and u2 fix the two coefficients of u = a1 + a2 x. (Key point: Displacement function)

2. What are N1 and N2?
   A. x/L and 1 - x/L
   B. 1 - x/L and x/L
   C. 1 and x
   D. x and x^2
   Answer: B. Substituting a1 = u1 and a2 = (u2 - u1)/L gives N1 = 1 - x/L and N2 = x/L. (Key point: Shape functions)

3. Which statement about shape functions is correct?
   A. N1 + N2 = 0
   B. N1 = 1 at both nodes
   C. N1 + N2 = 1 everywhere (partition of unity)
   D. N1 is constant
   Answer: C. Each N_i is 1 at its node and 0 at the other, and they sum to 1, allowing rigid-body motion. (Key point: Properties)

4. What is the strain in the element?
   A. Linear in x
   B. Zero
   C. Quadratic
   D. Constant, (u2 - u1)/L
   Answer: D. The derivative of a linear function is constant, so strain is constant per element. (Key point: Strain)

## Related in Finite Element Method

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## Related in other subjects

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