---
title: "Collapse value of P: portal frame with horizontal P/2"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-p-portal-p-half-ml"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-p-portal-p-half-ml
site: BCT Engineering
---

# Collapse value of P: portal frame with horizontal P/2

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

- Asked in: 2067 Asar, 15 marks
- Video: [Plastic Analysis Problem No -  26 ( Finding Collapse Load in a Frame )](https://www.youtube.com/watch?v=-amh6vptYwc) — 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-p-portal-p-half-ml
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

For the frame shown, calculate the collapse value of 'P' assuming $M_P$ as the plastic moment of resistance for all the members. [Figure: portal frame, columns of height $\ell/2$, beam span $\ell/2 + \ell/2$, both bases fixed; P vertical at mid-span and P/2 horizontal at the top-left joint.]

## Short answer

Collapse load is governed by beam and combined mechanisms equal to Pc = 8Mp/l; sway gives 16Mp/l. Thus vertical collapse load is 8Mp/l with horizontal load 4Mp/l.

## Answer

#### Assumptions
- Portal frame with fixed bases A and E, columns of height $\ell/2$ and a beam of span $\ell$ ($\ell/2 + \ell/2$) with mid-point C. All members have the plastic moment $M_P$.
- Loads: $P$ vertical at mid-span C and $P/2$ horizontal (to the right) at the top-left joint B. Take $\ell = 1$ in the tables.
- Fixed bases: 6 reactions, so the degree of indeterminacy $= 6 - 3 = 3$. Critical sections: A, B, C, D, E (5), so there are $5 - 3 = 2$ independent mechanisms, beam and sway; their combination is the third mechanism.

#### Mechanisms

**Mechanism 1: beam mechanism**

Taking $\ell = 1$. No sway; hinges at B, C and D. The hinge at C rotates $\theta$ and the hinges at B and D rotate $\theta/2$ each (the two halves of the beam turn $\theta/2$ in opposite senses).

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (under P) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| B (beam end) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |
| D (beam end) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |

Internal work $= 2\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| P/2 horizontal at B | $0\theta$ | $0P\,\theta$ |
| P at C | $\frac{1}{4}\theta$ | $\frac{1}{4}P\,\theta$ |

**Equation:** $P\times \frac{1}{4}\,\theta = 2\,M_p\theta$, so $P = 8\,M_p/\ell$.

**Mechanism 2: sway mechanism**

The columns (height $\ell/2$) rotate $\theta$ and the beam moves $\ell\theta/2$. Hinges at A, B, D and E, each rotating $\theta$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| A (fixed base) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| B | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| D | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| E (fixed base) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

Internal work $= 4\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| P/2 horizontal at B | $\frac{1}{2}\theta$ | $\frac{1}{4}P\,\theta$ |
| P at C | $0\theta$ | $0P\,\theta$ |

**Equation:** $P\times \frac{1}{4}\,\theta = 4\,M_p\theta$, so $P = 16\,M_p/\ell$.

**Mechanism 3: combined mechanism**

Combine 1 and 2 so that the hinge at B cancels. Hinges at A, C, D and E.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (under P) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| D | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| A (fixed base) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |
| E (fixed base) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |

Internal work $= 3\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| P/2 horizontal at B | $\frac{1}{4}\theta$ | $\frac{1}{8}P\,\theta$ |
| P at C | $\frac{1}{4}\theta$ | $\frac{1}{4}P\,\theta$ |

**Equation:** $P\times \frac{3}{8}\,\theta = 3\,M_p\theta$, so $P = 8\,M_p/\ell$.

#### Collapse load
The beam and combined mechanisms give the same value, which is smaller than the sway value (16):

**Answer: $P_c = 8M_P/\ell$** (and the horizontal load at collapse is $P_c/2 = 4M_P/\ell$).

## Check yourself

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

1. For this fixed-base portal frame, how many reactions, indeterminacy and independent mechanisms are there?
   A. 6 reactions, indeterminacy 3, 2 independent mechanisms (beam and sway)
   B. 4 reactions, indeterminacy 1, 2 mechanisms
   C. 6 reactions, indeterminacy 3, 5 mechanisms
   D. 6 reactions, indeterminacy 3, 3 independent mechanisms
   Answer: A. Indeterminacy = 6 - 3 = 3; five critical sections (A to E) give 5 - 3 = 2 independent mechanisms; combination is the third. (Key point: Independent mechanisms)

2. Why does the sway mechanism need 4 Mp theta of internal work?
   A. Hinges form at A, B, D and E, each rotating theta
   B. Only two hinges form
   C. Hinges at A, C and E rotate 2theta
   D. Hinge at C rotates 4theta
   Answer: A. Fixed bases and beam ends all hinge with equal rotation theta, giving 4Mp theta; with external work P theta/4, P = 16Mp/l. (Key point: Sway mechanism)

3. In the combined mechanism, what is cancelled and what is P?
   A. Hinge at C cancels; P = 16Mp/l
   B. Hinge at B cancels; hinges at A, C, D, E give P = 8Mp/l
   C. Hinge at D cancels; P = 4Mp/l
   D. Nothing cancels; P = 24Mp/l
   Answer: B. Internal 3Mp theta; external (1/8 + 1/4)P theta = 3P theta/8, so P = 8Mp/l. (Key point: Combined mechanism)

4. Beam, sway and combined give 8, 16 and 8 Mp/l. What is the collapse load?
   A. 16Mp/l, the largest
   B. 12Mp/l, the average
   C. 8Mp/l, the lowest (so horizontal load is 4Mp/l)
   D. 32Mp/l, the sum
   Answer: C. The collapse load is the least value from the upper-bound mechanisms: Pc = 8Mp/l. (Key point: Collapse load)

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