---
title: "Ritz solution of beam under uniform load"
subject: "Finite Element Method"
chapter: 4
key: "semester_8/finite_element_method:ritz-simply-supported-beam"
source: https://www.bctengineer.com/q/semester_8/finite_element_method/ritz-simply-supported-beam
site: BCT Engineering
---

# Ritz solution of beam under uniform load

Finite Element Method · Chapter 4: Continuum Problems

- Asked in: year not given, 8 marks
- Video: [Rayleigh Ritz Method | Simply Supported Beam (SSB) with UDL | Finite Element Analysis (FEA)](https://www.youtube.com/watch?v=K4ND6YWSciw) — 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/ritz-simply-supported-beam
- Whole chapter: https://www.bctengineer.com/md/semester_8/finite_element_method/chapter-4

## Question

A simply supported beam of span $L = 4$ m and flexural rigidity $EI = 1.6\times10^6\ \text{N m}^2$ carries a uniformly distributed load $q = 10$ kN/m. Using the Rayleigh-Ritz method with the trial function $w(x) = a\sin\dfrac{\pi x}{L}$, find the coefficient $a$, the mid-span deflection and the mid-span bending moment. Compare with the exact values.

## Short answer

With trial w = a sin(πx/L), stationarity of Π gives a = 4qL⁴/(π⁵EI) = 20.9 mm mid-span deflection. Bending moment is 20.6 kN·m. Deflection matches exact (20.83 mm) within 0.4%; moment error is 3.2% because differentiation reduces accuracy.

## Answer

#### Choice of trial function

$w = a\sin(\pi x/L)$ gives $w(0) = w(L) = 0$, which satisfies the essential boundary conditions of a simply supported beam.

#### Potential energy

$$
\Pi = \frac{EI}{2}\int_0^L \left(\frac{d^2w}{dx^2}\right)^2dx - q\int_0^L w\,dx
$$

With $w'' = -a\left(\dfrac{\pi}{L}\right)^2\sin\dfrac{\pi x}{L}$, $\displaystyle\int_0^L\sin^2\dfrac{\pi x}{L}\,dx = \dfrac{L}{2}$ and $\displaystyle\int_0^L\sin\dfrac{\pi x}{L}\,dx = \dfrac{2L}{\pi}$:
$$
\Pi = \frac{EI}{2}\,a^2\left(\frac{\pi}{L}\right)^4\frac{L}{2} - qa\,\frac{2L}{\pi}
$$

#### Stationary condition

$$
\frac{d\Pi}{da} = EI\,a\left(\frac{\pi}{L}\right)^4\frac{L}{2} - \frac{2qL}{\pi} = 0 \Rightarrow a = \frac{4qL^4}{\pi^5EI}
$$

Substituting values:
$$
a = \frac{4(10\,000)(4)^4}{\pi^5(1.6\times10^6)} = \frac{1.024\times10^7}{306.02\times1.6\times10^6} = 0.02091\ \text{m}
$$

#### Mid-span values

- Deflection: $w(L/2) = a = 20.91$ mm.
- Bending moment: at $x = L/2$, the magnitude of $M = EI\,w''$ is $M = EI\,a\left(\dfrac{\pi}{L}\right)^2 = 1.6\times10^6(0.02091)(0.6169) = 20\,641$ N m.

#### Comparison

| Quantity | Ritz | Exact | Error |
|---|---|---|---|
| Mid-span deflection $\dfrac{5qL^4}{384EI}$ | 20.91 mm | 20.83 mm | 0.4 % |
| Mid-span moment $\dfrac{qL^2}{8}$ | 20.64 kN m | 20.00 kN m | 3.2 % |

The deflection (the primary variable) is accurate; the moment (a second derivative) is less accurate because differentiation reduces accuracy.

**Answer:** $a = 20.9$ mm; $w_{max} = 20.9$ mm; $M_{max} = 20.6$ kN m.

## Check yourself

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

1. Why is w = a sin(pi x / L) suitable for a simply supported beam?
   A. It satisfies w(0) = w(L) = 0, the essential boundary conditions
   B. It gives constant moment
   C. It has maximum at the ends
   D. It is linear
   Answer: A. The sine vanishes at both supports, satisfying the essential conditions. (Key point: Trial function)

2. What is the formula for a from dPi/da = 0?
   A. a = qL^4/(8EI)
   B. a = 4qL^4/(pi^5 EI)
   C. a = 5qL^4/(384 EI)
   D. a = qL^2/(8EI)
   Answer: B. Setting dPi/da = 0 gives a = 4qL^4/(pi^5 EI) = 0.02091 m. (Key point: Coefficient)

3. What is the Ritz mid-span deflection?
   A. 209.1 mm
   B. 2.091 mm
   C. 20.91 mm
   D. 20.83 mm
   Answer: C. At x = L/2, sin(pi/2) = 1, so w = a = 20.91 mm. (Key point: Mid-span deflection)

4. Why is the bending-moment error (3.2%) larger than the deflection error (0.4%)?
   A. The load is not uniform
   B. EI changes
   C. Moment is a primary variable
   D. The moment depends on the second derivative, and differentiation reduces accuracy
   Answer: D. Moment is EI w'', and differentiating an approximate function worsens accuracy; deflection is the primary variable. (Key point: Accuracy)

## Related in Finite Element Method

- [Simply supported beam with two elements](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Asimply-supported-beam-two-elements) — Finite Element Method, chapter 3
- [Ritz method for bar with distributed load](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aritz-bar-quadratic-trial) — Finite Element Method, chapter 4
- [Rayleigh-Ritz method and its steps](https://www.bctengineer.com/question.md?key=semester_8%2Ffinite_element_method%3Aritz-method-steps) — Finite Element Method, chapter 4

## Related in other subjects

- [Moment area: mid-span deflection under UDL](https://www.bctengineer.com/question.md?key=semester_4%2Ftheory_of_structures_i%3Amoment-area-ss-udl-midspan) — Theory of Structures I, chapter 4
- [Flexural rigidity; T-beam bending stresses (60 kN loads)](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aflexural-rigidity-t-beam-60kn) — Strength of Materials, chapter 7
- [Mid-span deflection of beam, 10 kN load](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Ass-beam-deflection-mid-span-4m) — Strength of Materials, chapter 7
- [Simply supported beam 6 m, 60 kN/m](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-60knm-udl) — Design of Steel and Timber Structure, chapter 7