---
title: "Collapse load of portal frame, 8 m span"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-frame-w-2mp-8m"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-frame-w-2mp-8m
site: BCT Engineering
---

# Collapse load of portal frame, 8 m span

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

- Asked in: 2078 Bhadra, 10 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-load-frame-w-2mp-8m
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Find the collapse load of the following frame. [Figure: portal frame ABCD, A fixed, D hinged; column AB (2Mp) of height 4 m with horizontal W at B, beam BC (3Mp) of span 8 m with two vertical W loads each 2 m from the ends, column CD (Mp).]

## Short answer

Virtual work evaluates beam, sway, and combined mechanisms; the combined mechanism gives the minimum collapse load, Wc = 11Mp/10 = 1.1Mp.

## Answer

#### Assumptions
- Portal frame: A (0,0) fixed, B (0,4), C (8,4), D (8,0) hinged. Column AB $2M_p$, beam BC $3M_p$ (span 8 m), column CD $M_p$; both columns are 4 m high.
- Loads: W horizontal (to the right) at B, and two vertical loads W at 2 m and 6 m from B. Find $W_c$ in terms of $M_p$ (kN·m, lengths in m).
- Hinges may form at A, B, under the loads and at C (the weaker of the members meeting at a joint). The fixed base gives 3 and the hinge 2 reactions, so the degree of indeterminacy is 5 - 3 = 2. There are 5 critical sections (A, B, two load points, C), hence 5 - 2 = 3 independent mechanisms (two beam mechanisms, one for each load, and one sway); their combinations are also examined.

#### Mechanisms

**Mechanism 1: beam mechanism**

Joints do not rotate or sway. Hinges at B (in the weaker column AB), under the load 2 m from C and at C (in the weaker column CD). The rotations of the beam parts are $\theta/4$ (B to hinge, 6 m) and $3\theta/4$ (hinge to C, 2 m): the hinge under the load has $\theta$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| under W at 6 m from B (2 m from C) | $1\theta$ | $3M_p$ | $3\,M_p\theta$ |
| C (in column CD) | $\frac{3}{4}\theta$ | $M_p$ | $\frac{3}{4}\,M_p\theta$ |
| B (in column AB) | $\frac{1}{4}\theta$ | $2M_p$ | $\frac{1}{2}\,M_p\theta$ |

Internal work $= \frac{17}{4}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| W horizontal at B | $0\theta$ | $0W\,\theta$ |
| W at 2 m | $\frac{1}{2}\theta$ | $\frac{1}{2}W\,\theta$ |
| W at 6 m | $\frac{3}{2}\theta$ | $\frac{3}{2}W\,\theta$ |

**Equation:** $W_c\times 2\,\theta = \frac{17}{4}\,M_p\theta$, so $W_c = \frac{17}{8}\,M_p = 2.1250\,M_p$.

**Mechanism 2: sway mechanism**

The columns rotate $\theta$ about the bases and the beam moves horizontally by $4\theta$. Hinges at A, B (in AB) and C (in CD).

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| A (fixed base) | $1\theta$ | $2M_p$ | $2\,M_p\theta$ |
| B (in column AB) | $1\theta$ | $2M_p$ | $2\,M_p\theta$ |
| C (in column CD) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

Internal work $= 5\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| W horizontal at B | $4\theta$ | $4W\,\theta$ |
| W at 2 m | $0\theta$ | $0W\,\theta$ |
| W at 6 m | $0\theta$ | $0W\,\theta$ |

**Equation:** $W_c\times 4\,\theta = 5\,M_p\theta$, so $W_c = \frac{5}{4}\,M_p = 1.2500\,M_p$.

**Mechanism 3: combined mechanism**

Combining a beam mechanism (hinge under the load 2 m from B) with the sway mechanism removes the hinge at B. Hinges at A, under the first W, and at C (in CD).

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| under W at 2 m from B | $1\theta$ | $3M_p$ | $3\,M_p\theta$ |
| C (in column CD) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| A (fixed base) | $\frac{3}{4}\theta$ | $2M_p$ | $\frac{3}{2}\,M_p\theta$ |

Internal work $= \frac{11}{2}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| W horizontal at B | $3\theta$ | $3W\,\theta$ |
| W at 2 m | $\frac{3}{2}\theta$ | $\frac{3}{2}W\,\theta$ |
| W at 6 m | $\frac{1}{2}\theta$ | $\frac{1}{2}W\,\theta$ |

**Equation:** $W_c\times 5\,\theta = \frac{11}{2}\,M_p\theta$, so $W_c = \frac{11}{10}\,M_p = 1.1000\,M_p$.

A search of all other hinge combinations (including the hinge under the load at 6 m with the sway mechanism, 1.5 $M_p$, and sway with hinge in the beam, 1.5 $M_p$) gave higher values.

#### Collapse load
The lowest value is from the combined mechanism:

**Answer: $W_c = 1.1\,M_p$** (hinges at A, under the first W and at C; $M_p$ in kN·m, $W_c$ in kN with lengths in m).

## Check yourself

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

1. What is the sway mechanism result?
   A. W_c = 5/4 Mp (internal 5 Mp theta, external 4W theta)
   B. W_c = 5/2 Mp
   C. W_c = 17/8 Mp
   D. W_c = 4/5 Mp
   Answer: A. Hinges at A, B, C give 2+2+1 = 5 Mp theta; W at B moves 4 theta, so W_c = 1.25 Mp. (Key point: Sway mechanism)

2. At joints B and C, in which member does the hinge form?
   A. The stronger member
   B. The weaker member of those meeting at the joint
   C. The beam always
   D. At the centre of the joint
   Answer: B. The hinge forms in the weaker member, so C is in column CD (Mp) and B in column AB (2Mp) rather than the 3Mp beam. (Key point: Hinge at joints)

3. What is the aim in combining the beam and sway mechanisms?
   A. To increase the load
   B. To eliminate a hinge (at B) and lower the load factor
   C. To remove the fixed base
   D. To make the frame determinate
   Answer: B. Combining cancels the hinge rotation at B, giving a lower value 1.1 Mp. (Key point: Combined mechanism)

4. What is the collapse load?
   A. 2.125 Mp
   B. 1.25 Mp
   C. 1.5 Mp
   D. 1.1 Mp, with hinges at A, under first W, and C
   Answer: D. Combined mechanism: 11/2 Mp theta = 5W theta, so W_c = 1.1 Mp. (Key point: Result)

## Related in Theory of Structures II

- [Collapse load of portal frame, height 2L](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-portal-2w-3w-2l) — Theory of Structures II, chapter 5, asked 3×
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