---
title: "Collapse load of fixed beam with UDL"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-fixed-beam-udl"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-fixed-beam-udl
site: BCT Engineering
---

# Collapse load of fixed beam with UDL

Theory of Structures II · Chapter 5: Introduction to plastic analysis

- Asked in: 2074 Chaitra, 4 marks
- Video: [Collapse Load in Fixed Beam ( Standard Loading Cases )](https://www.youtube.com/watch?v=bfnfGkGRKi8) — Stan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-fixed-beam-udl
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Determine collapse load for the following beam. [Figure: beam of span L m fixed at both ends carrying W kN/m UDL.]

## Short answer

For a fixed-ended beam with UDL, collapse needs hinges at both ends and midspan; virtual work gives internal work 4Mpθ and external work WL²θ/4. Hence Wc = 16Mp/L² kN/m (total load 16Mp/L).

## Answer

#### Set-up
A beam of span $L$ fixed at both ends carries a UDL of $W$ kN/m. The plastic moment of the section is $M_p$. The beam is indeterminate to the second degree, so $2+1 = 3$ hinges are needed for collapse. By symmetry the hinges form at the two fixed ends A and B and at mid-span C (where the free moment is largest).

#### Mechanism and virtual work
Let the mid-span deflection be $\delta = \dfrac{L}{2}\theta$.
- Rotation at A: $\theta$; at B: $\theta$; at C: $2\theta$.
- Internal work $= M_p\theta + 2M_p\theta + M_p\theta = 4M_p\theta$.
- External work $=$ load $\times$ average deflection $= W L \times \dfrac{\delta}{2} = W L\times\dfrac{L\theta}{4} = \dfrac{WL^2}{4}\theta$.

Equating:
$$
\frac{W_cL^2}{4}\theta = 4M_p\theta \;\Rightarrow\; W_c = \frac{16M_p}{L^2}
$$

#### Check by equilibrium (static method)
At collapse the end moments are $-M_p$ (hogging) and the mid-span moment is $+M_p$. The free bending moment at mid-span is $WL^2/8$, and the moment diagram is lowered by $M_p$, so $\dfrac{WL^2}{8} = M_p + M_p = 2M_p$, giving $W = \dfrac{16M_p}{L^2}$, the same value. Hence the upper and lower bounds coincide and this is the true collapse load.

**Answer: $W_c = 16M_p/L^2$ kN/m (total load $16M_p/L$).**

## Check yourself

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

1. A beam fixed at both ends is indeterminate to the 2nd degree. How many plastic hinges are needed for collapse?
   A. 2
   B. 4
   C. 1
   D. 3
   Answer: D. Hinges needed = degree of indeterminacy + 1 = 3. (Key point: Number of hinges)

2. Where do the three hinges form for a fixed beam under a full UDL?
   A. At the two ends and at mid-span
   B. At the quarter points and mid-span
   C. At mid-span only
   D. At the ends and the third points
   Answer: A. By symmetry, the ends carry the largest hogging moment and mid-span the largest sagging moment. (Key point: Hinge locations)

3. With rotations theta at A, 2 theta at C and theta at B, what is the internal work?
   A. 2 Mp theta
   B. 3 Mp theta
   C. 4 Mp theta
   D. 8 Mp theta
   Answer: C. Mp(theta + 2theta + theta) = 4 Mp theta. (Key point: Internal work)

4. What is the collapse UDL for the fixed beam?
   A. 8 Mp/L^2
   B. 16 Mp/L^2
   C. 12 Mp/L^2
   D. 4 Mp/L^2
   Answer: B. Wc L^2/4 = 4 Mp gives Wc = 16 Mp/L^2, confirmed by statics WL^2/8 = 2Mp. (Key point: Collapse load result)

## Related in Theory of Structures II

- [Collapse load of two-span beam](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-two-span-beam-20kn) — Theory of Structures II, chapter 5
- [Slope deflection: beam, 360 kN load, fixed end](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aslope-deflection-beam-10kn-20knm-360kn) — Theory of Structures II, chapter 3
- [Collapse load of portal frame, 8 m span](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-frame-w-2mp-8m) — Theory of Structures II, chapter 5
- [Collapse load of frame with UDL on column](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-frame-20kn-10knm) — Theory of Structures II, chapter 5
- [Collapse load P: portal frame with UDL 2P/l](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-portal-p-p-udl) — Theory of Structures II, chapter 5
- [Section modulus of fixed beam at collapse](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Afixed-beam-section-modulus-collapse-60kn) — Theory of Structures II, chapter 5

## Related in other subjects

- [Bar fixed at both ends: load shares, displacements](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Acircular-bar-fixed-both-ends-50kn) — Strength of Materials, chapter 3
- [Beam 6 m span with 100 kN secondary beam load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abeam-6m-20knm-100kn-midspan) — Design of Steel and Timber Structure, chapter 7
- [Euler's formula: both ends fixed](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-fixed) — Strength of Materials, chapter 8, asked 2×
- [Euler's formula both ends fixed, with limitations](https://www.bctengineer.com/question.md?key=semester_3%2Fstrength_of_materials%3Aeuler-both-ends-fixed-limitation) — Strength of Materials, chapter 8