---
title: "Strong and weak form for an axial bar"
subject: "Finite Element Method"
chapter: 4
key: "semester_8/finite_element_method:strong-weak-form-bar"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/strong-weak-form-bar
site: BCT Engineering
---

# Strong and weak form for an axial bar

Finite Element Method · Chapter 4: Continuum Problems

- Asked in: year not given, 8 marks
- Video: [The Strong and Weak Forms — Lesson 1](https://www.youtube.com/watch?v=GAOGcSnwQPk) — Ansys Learning
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/finite_element_method/strong-weak-form-bar
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-4

## Question

For an axially loaded bar of length $L$ with variable $A(x)$, modulus $E$ and body force $f(x)$ per unit length, fixed at $x = 0$ and loaded by an axial force $P$ at $x = L$: (a) state the strong form, (b) derive the weak form, (c) state why the weak form is preferred in FEM.

## Short answer

Strong form: d/dx(AE du/dx)+f=0 with u(0)=0 and AE du/dx=P at L. Multiplying by w, integrating, and integrating by parts gives ∫w'Ae u' dx = ∫wf dx + w(L)P. It needs only C0 functions and includes natural BCs.

## Answer

#### (a) Strong form

Find $u(x)$ such that
$$
\frac{d}{dx}\left(AE\frac{du}{dx}\right) + f = 0,\quad 0 < x < L
$$
with $u(0) = 0$ (essential BC) and $AE\dfrac{du}{dx}\Big|_{x=L} = P$ (natural BC).

It requires $u$ to be twice differentiable, and the equation must hold at every point.

#### (b) Weak form (Galerkin)

Multiply the equation by an arbitrary weight (test) function $w(x)$ with $w(0) = 0$, and integrate over the length:
$$
\int_0^L w\left[\frac{d}{dx}\left(AE\frac{du}{dx}\right) + f\right]dx = 0
$$

Integrate the first term by parts:
$$
\int_0^L w\,\frac{d}{dx}\left(AE\frac{du}{dx}\right)dx = \left[w\,AE\frac{du}{dx}\right]_0^L - \int_0^L \frac{dw}{dx}AE\frac{du}{dx}\,dx
$$

At $x = 0$, $w = 0$. At $x = L$, $AE\,u' = P$. So the boundary term is $w(L)P$. The weak form is:
$$
\int_0^L \frac{dw}{dx}\,AE\,\frac{du}{dx}\,dx = \int_0^L w f\,dx + w(L)\,P
$$
for all admissible $w$.

Note that the natural BC enters through the boundary term, while the essential BC is imposed on $u$ and $w$.

#### (c) Why the weak form is preferred

- Only first derivatives appear, so the trial and test functions need only be $C^0$ continuous; simple linear shape functions can be used.
- Natural boundary conditions are included automatically.
- It leads to a symmetric stiffness matrix, $K_{ij} = \int AE\,\phi_i'\phi_j'\,dx$.
- It is the physical statement of the principle of virtual work, and it still works when $A$, $E$ or the load are discontinuous.

## Check yourself

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

1. What does the strong form require of u?
   A. To be twice differentiable and satisfy the equation at every point
   B. Only to be continuous
   C. To be zero
   D. To be linear
   Answer: A. The strong form holds pointwise and so needs second derivatives of u. (Key point: Strong form)

2. What condition is placed on the weight function w(x) in the derivation?
   A. w(L) = 0
   B. w(0) = 0, where u is fixed
   C. w = u
   D. w = 1
   Answer: B. w must vanish where the essential BC is imposed, so the boundary term at x = 0 drops out. (Key point: Test function)

3. What does the boundary term from integration by parts become?
   A. w(0) P
   B. zero always
   C. w(L) P, introducing the natural BC
   D. u(L) P
   Answer: C. At x = L, AE u' = P, giving w(L)P; at x = 0, w = 0. (Key point: Boundary term)

4. Which is the weak form?
   A. Integral of w'' AE u dx = P
   B. Integral of w u dx = f
   C. AE u'' = f
   D. Integral of w' AE u' dx = integral of w f dx + w(L)P
   Answer: D. After integration by parts only first derivatives appear. (Key point: Weak form)

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