---
title: "Collapse load of propped cantilever, midspan load"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-propped-cantilever-p-midspan"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-propped-cantilever-p-midspan
site: BCT Engineering
---

# Collapse load of propped cantilever, midspan load

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

- Asked in: 2072 Chaitra, 4 marks
- Video: [Collapse Load in Propped Cantilever Beam with Point Load on the centre ( static & Kinematic Method)](https://www.youtube.com/watch?v=3kZOFU4LFMo) — 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-propped-cantilever-p-midspan
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

A propped cantilever beam of uniform $M_p$ is loaded as shown in the figure below. Find the collapse load. [Figure: beam AC, A fixed, C roller, span L/2 + L/2, point load P at B (mid-span).]

## Short answer

Two hinges at fixed end and under load; virtual work yields collapse load Pc = 6Mp/L.

## Answer

#### Set-up
Propped cantilever AC of span $L$: A fixed, C a roller, uniform $M_p$, point load $P$ at mid-span B. The beam is indeterminate to the first degree, so $1 + 1 = 2$ hinges cause collapse. The largest moments occur at the fixed end A and under the load B, so the plastic hinges form at A and B.

#### Mechanism and virtual work
Part AB rotates $\theta$ about A. Part BC rotates about C so that B has the same deflection $\delta = \dfrac{L}{2}\theta$. Since BC has the same length, it rotates $\theta$ the other way.
- Rotation at A $= \theta$; rotation at B $= \theta + \theta = 2\theta$.
- Internal work $= M_p\theta + M_p(2\theta) = 3M_p\theta$.
- External work $= P\delta = P\,\dfrac{L}{2}\theta$.

$$
P_c\,\frac{L}{2}\theta = 3M_p\theta \;\Rightarrow\; P_c = \frac{6M_p}{L}
$$

#### Check by equilibrium
With $M_A = -M_p$ (hogging) and $M_C = 0$, the moment under the load is $M_B = \dfrac{PL}{4} + \dfrac{M_A + M_C}{2} = \dfrac{PL}{4} - \dfrac{M_p}{2}$. Setting $M_B = M_p$ gives $P = \dfrac{6M_p}{L}$. The moment diagram is nowhere above $M_p$, so the static and kinematic values agree and this is the true collapse load.

**Answer: $P_c = 6M_p/L$.**

## Check yourself

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

1. A propped cantilever is indeterminate to the 1st degree. How many hinges cause collapse?
   A. 1
   B. 2
   C. 3
   D. 4
   Answer: B. Hinges = degree + 1 = 2. (Key point: Number of hinges)

2. Where do the hinges form for a mid-span point load?
   A. At the roller and mid-span
   B. At the fixed end A and under the load B
   C. At A and at the roller C
   D. Only at mid-span
   Answer: B. Largest moments occur at the fixed end and under the load. (Key point: Hinge locations)

3. If AB rotates theta about A and BC rotates theta the other way, what is the rotation at B?
   A. theta
   B. 3 theta
   C. 0
   D. 2 theta
   Answer: D. The relative rotation is theta + theta = 2 theta. (Key point: Hinge rotation under the load)

4. What is the collapse load P_c?
   A. 4 Mp/L
   B. 6 Mp/L
   C. 8 Mp/L
   D. 3 Mp/L
   Answer: B. P (L/2) theta = Mp(theta + 2theta) = 3 Mp theta gives P = 6 Mp/L. (Key point: Collapse load)

## Related in Theory of Structures II

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## Related in other subjects

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