---
title: "Collapse load of two-span beam"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-two-span-beam-20kn"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-two-span-beam-20kn
site: BCT Engineering
---

# Collapse load of two-span beam

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

- Asked in: 2070 Asar, 8 marks
- Video: [Plastic Analysis Problem No - 9 ( Finding Collapse Load - in a Two Span Continuous Beam )](https://www.youtube.com/watch?v=WyQSYmpcfRA) — 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-two-span-beam-20kn
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Determine the collapse load for the two span beam shown in figure below if the plastic moment capacity is $M_P$. [Figure: beam ABC, A fixed, B and C supports; AB = 6 m (3 m + 3 m) with 20 kN at mid-span; BC = 6 m with 10 kN/m UDL.]

## Short answer

Two-span beam with fixed end A. Span AB mechanism gives λ = 0.0667M_p. Span BC mechanism (hinges at B and 3.52m from B) gives minimum λ_c = 0.0324M_p. Collapse loads: 0.648M_p kN at midspan AB, 0.324M_p kN/m on BC.

## Answer

#### Set-up
Beam ABC: A fixed, B and C supports (C is a simple end support). AB = 6 m with a 20 kN load at mid-span J; BC = 6 m with a UDL of 10 kN/m. The loads are increased proportionally by a load factor $\lambda$ (the collapse loads are $20\lambda$ kN and $10\lambda$ kN/m); $M_p$ is the same for both spans. Reactions: A (3) + B (1) + C (1) = 5, so the degree of indeterminacy is $5 - 3 = 2$. A single-span mechanism needs 3 hinges for a fixed-ended span, or 2 hinges for a span with a simple end.

#### Mechanism 1: span AB (hinges at A, J and B)
AB behaves as a fixed-ended beam. Mid-span deflection $\delta = 3\theta$; rotations are $\theta$ at A, $2\theta$ at J and $\theta$ at B.
- Internal work $= M_p(\theta + 2\theta + \theta) = 4M_p\theta$
- External work $= 20\lambda\times 3\theta = 60\lambda\theta$

$$
60\lambda\theta = 4M_p\theta \;\Rightarrow\; \lambda = \frac{4}{60}M_p = 0.0667\,M_p
$$

#### Mechanism 2: span BC (hinges at B and K in the span)
BC has a hinge at B and a hinge K at distance $x$ from the simple end C. Part BK rotates $\theta$ about B and part KC rotates $\phi = \dfrac{(6-x)}{x}\theta$ about C (both give the same deflection at K, $\delta_K = (6-x)\theta$).
- Rotation at B $= \theta$; at K $= \theta + \phi = \dfrac{6\theta}{x}$.
- Internal work $= M_p\theta\left(1 + \dfrac{6}{x}\right)$.
- External work $=$ UDL $\times$ area of the deflected triangle $= 10\lambda\times\dfrac{6\delta_K}{2} = 30\lambda(6-x)\theta$.

$$
\lambda = \frac{M_p\,(x+6)}{30\,x\,(6-x)}
$$
Minimising with respect to $x$: $\dfrac{d\lambda}{dx} = 0 \Rightarrow x^2 + 12x - 36 = 0 \Rightarrow x = 6(\sqrt2 - 1) = 2.485$ m from C (3.515 m from B). Then
$$
\lambda = \frac{M_p\times8.485}{30\times2.485\times3.515} = 0.03238\,M_p
$$

#### Mechanism 3: combined (hinges at A, J and in BC)
With no hinge at B, the parts J-B and B-K form one rigid piece that rotates about support B, so one of the two loads always moves against its own direction. If J moves down, the part of BC next to B lifts, the UDL does negative work and $\lambda > M_p/15$ (higher than mechanism 1). If the BC side moves down, J lifts and the 20 kN load does negative work; the best hinge position (about 4.5 m from B) gives $\lambda \approx 0.093M_p$. Neither is critical.

#### Collapse load
The smallest value governs, which is the failure of span BC:
$$
\lambda_c = 0.0324\,M_p
$$
So the collapse loads are $20\lambda_c = 0.648\,M_p$ kN at J and $10\lambda_c = 0.324\,M_p$ kN/m on BC ($M_p$ in kN·m).

**Answer: collapse load factor $\lambda_c = 0.0324\,M_p$ (collapse of span BC, with hinges at B and 3.52 m from B).**

## Check yourself

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

1. What does the load factor lambda represent?
   A. The factor by which loads are raised proportionally until collapse
   B. The ratio of Mp to My
   C. The span ratio
   D. The number of hinges
   Answer: A. Collapse loads are 20 lambda and 10 lambda, so lambda scales all loads together. (Key point: Load factor idea)

2. Span AB (fixed-ended behaviour) needs hinges at:
   A. A and B only
   B. A, J and B
   C. B and C
   D. J only
   Answer: B. A fixed-ended span needs three hinges: both ends and under the load. (Key point: Span AB mechanism)

3. In span BC, how is the position x of the sagging hinge K found?
   A. Assumed at mid-span
   B. By minimising lambda with respect to x
   C. Where shear is maximum
   D. At the support C
   Answer: B. Under a UDL the hinge position is unknown; set d(lambda)/dx = 0, giving x = 2.485 m from C. (Key point: Hinge position in BC)

4. Lambda for AB = 0.0667 Mp and for BC = 0.0324 Mp. Which governs?
   A. AB
   B. BC, the lower one
   C. Both equally
   D. Neither
   Answer: B. The smaller load factor governs, so span BC fails first. (Key point: Governing span)

## Related in Theory of Structures II

- [Plastic moment capacity of frame, 60 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aplastic-moment-capacity-frame-60kn-40kn) — Theory of Structures II, chapter 5
- [Collapse load of fixed beam with UDL](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-fixed-beam-udl) — Theory of Structures II, chapter 5
- [Plastic moment capacity of frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Aplastic-moment-capacity-frame-80kn-20knm) — 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
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