---
title: "Collapse load of rectangular portal frame"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-rect-portal-2p-3p"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-rect-portal-2p-3p
site: BCT Engineering
---

# Collapse load of rectangular portal frame

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

- Asked in: 2074 Asoj, 6 marks
- Video: [Plastic Analysis Problem No -  24 ( Finding Collapse Load in a Frame )](https://www.youtube.com/watch?v=ume4u1BAVqI) — 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-rect-portal-2p-3p
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Determine the collapse load, $W_p$, for the rectangular portal frame loaded as shown in figure below. [Figure: portal frame, fixed bases; left column Mp of height 4 m, right column 2Mp of height 6 m, beam Mp with span 3 m + 3 m; 2P horizontal at the top-left joint and 3P vertical at mid-span.]

## Short answer

Virtual work gives beam 0.444Mp, sway 0.500Mp and combined 0.353Mp; the lowest is the combined mechanism with hinges at A, C, D and E. Therefore Pc = 6/17 Mp ≈ 0.353Mp, so the collapse loads are 2Pc = 0.706Mp horizontal and 3Pc = 1.059Mp vertical.

## Answer

#### Assumptions
- Portal frame with fixed bases. Beam B-C-D, span 6 m (3 m + 3 m), $M_p$. Left column AB, 4 m high, $M_p$; right column DE, 6 m high, $2M_p$. The beam is horizontal, so the left base A is 2 m above the right base E.
- Loads: $2P$ horizontal (to the right) at the top-left joint B and $3P$ vertical at mid-span C. $P$ is increased until collapse; the collapse load is the value $P_c$ (the actual loads are $2P_c$ and $3P_c$), given in terms of $M_p$ (kN·m, lengths in m).
- Degree of indeterminacy $= 6 - 3 = 3$, critical sections A, B, C, D, E (5): number of independent mechanisms $5 - 3 = 2$ (beam, sway), plus the combined mechanism.

#### Mechanisms

**Mechanism 1: beam mechanism**

No sway. Hinges at B, mid-span C and D, all in the beam ($M_p$ is less than the column $2M_p$ at D). The two beam halves rotate $\theta/2$ each, with $\theta$ under the load.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (under 3P) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| 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 |
|---|---|---|
| 2P horizontal at B | $0\theta$ | $0P\,\theta$ |
| 3P at mid-span | $\frac{3}{2}\theta$ | $\frac{9}{2}P\,\theta$ |

**Equation:** $P_c\times \frac{9}{2}\,\theta = 2\,M_p\theta$, so $P_c = \frac{4}{9}\,M_p = 0.4444\,M_p$.

**Mechanism 2: sway mechanism**

The left column (4 m) rotates $\theta$ about A, so the beam moves $4\theta$ to the right; the right column (6 m) rotates $2\theta/3$. Hinges at A, B, D (beam end) and E.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| A (fixed base) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| B (column) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| D (beam end) | $\frac{2}{3}\theta$ | $M_p$ | $\frac{2}{3}\,M_p\theta$ |
| E (fixed base) | $\frac{2}{3}\theta$ | $2M_p$ | $\frac{4}{3}\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| 2P horizontal at B | $4\theta$ | $8P\,\theta$ |
| 3P at mid-span | $0\theta$ | $0P\,\theta$ |

**Equation:** $P_c\times 8\,\theta = 4\,M_p\theta$, so $P_c = \frac{1}{2}\,M_p = 0.5000\,M_p$.

**Mechanism 3: combined mechanism**

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

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (under 3P) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| D (beam end) | $\frac{5}{6}\theta$ | $M_p$ | $\frac{5}{6}\,M_p\theta$ |
| A (fixed base) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |
| E (fixed base) | $\frac{1}{3}\theta$ | $2M_p$ | $\frac{2}{3}\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| 2P horizontal at B | $2\theta$ | $4P\,\theta$ |
| 3P at mid-span | $\frac{3}{2}\theta$ | $\frac{9}{2}P\,\theta$ |

**Equation:** $P_c\times \frac{17}{2}\,\theta = 3\,M_p\theta$, so $P_c = \frac{6}{17}\,M_p = 0.3529\,M_p$.

#### Collapse load
The lowest value is the combined mechanism (beam 0.444 and sway 0.5 are higher), and the LP check over all hinge combinations gave the same:

**Answer: $P_c = \dfrac{6}{17}M_p \approx 0.353\,M_p$**, i.e. collapse loads $W_p = 2P_c = 0.706\,M_p$ horizontal and $3P_c = 1.059\,M_p$ vertical (kN, with $M_p$ in kN·m).

## Check yourself

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

1. With 5 critical sections and degree of indeterminacy 3, how many independent mechanisms exist?
   A. 2
   B. 3
   C. 5
   D. 8
   Answer: A. 5 - 3 = 2 (beam and sway); the combined mechanism is formed from them. (Key point: Independent mechanisms)

2. In the beam mechanism the work equation is P x (9/2)theta = 2 Mp theta. What is Pc?
   A. 0.5 Mp
   B. 0.3529 Mp
   C. 2.25 Mp
   D. 0.4444 Mp
   Answer: D. Pc = 2/(9/2) Mp = 4/9 Mp = 0.4444 Mp. (Key point: Beam mechanism)

3. In the sway mechanism, hinge E (rotation 2theta/3) lies in the 2Mp column. What internal work does it contribute?
   A. (2/3) Mp theta
   B. (4/3) Mp theta
   C. 2 Mp theta
   D. (1/3) Mp theta
   Answer: B. Internal work = 2Mp x 2theta/3 = 4/3 Mp theta, since the stronger column has plastic moment 2Mp. (Key point: Role of the 2Mp column base)

4. Why does the combined mechanism (Pc = 6/17 Mp) govern?
   A. Because it has the most hinges
   B. Because it cancels the hinge at A
   C. Because it is the lowest of 0.444, 0.5 and 0.353 Mp
   D. Because it gives the highest load
   Answer: C. The collapse load is the minimum over mechanisms; combining beam and sway cancels the hinge at B and gives the lowest value. (Key point: Combined mechanism and governing load)

## Related in Theory of Structures II

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