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

# Collapse load P: portal frame with UDL 2P/l

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

- Asked in: 2068 Chaitra, 8 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-portal-p-p-udl
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

For the given portal frame with same plastic moment of resistance $M_p$ for all the members, calculate the value of P at collapse. [Figure: portal frame, both bases hinged, height l, span l divided into three parts of l/3; vertical loads P at the first and second third-points of the beam; UDL of 2P/l on the left column.]

## Short answer

Beam, sway, and combined mechanisms yield P = 6Mp/l, 2Mp/l and 2Mp/l. The critical mechanism has a plastic hinge inside the UDL-loaded column, giving collapse load Pc = (1 + √3/2)Mp/l ≈ 1.866 Mp/l.

## Answer

#### Assumptions
- Portal frame: A and F hinged, columns AB and EF of height $l$, beam BE of span $l$; the vertical loads $P$ act at C and D (the third points). A UDL of $2P/l$ acts horizontally to the right on column AB. All members have the same plastic moment $M_p$. Take $l = 1$ in the tables.
- Hinged bases: degree of indeterminacy $= 4 - 3 = 1$.

#### Mechanisms

**Mechanism 1: beam mechanism**

Taking $l = 1$. No sway; hinges at B, C and E. Part BC ($l/3$) rotates $2\theta/3$ and CE ($2l/3$) rotates $\theta/3$, so the hinge at C turns $\theta$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| B (beam end) | $\frac{2}{3}\theta$ | $M_p$ | $\frac{2}{3}\,M_p\theta$ |
| E (beam end) | $\frac{1}{3}\theta$ | $M_p$ | $\frac{1}{3}\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| P at C | $\frac{2}{9}\theta$ | $\frac{2}{9}P\,\theta$ |
| P at D | $\frac{1}{9}\theta$ | $\frac{1}{9}P\,\theta$ |
| 2P/l on AB | - | $0P\,\theta$ |

**Equation:** $P\times \frac{1}{3}\,\theta = 2\,M_p\theta$, so $P = 6\,M_p/l$.

**Mechanism 2: sway mechanism (hinges at B and E)**

The columns (height $l$) rotate $\theta$ about the hinged bases and the beam moves $l\theta$. The hinges are at B and E, at the beam ends.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| B (beam end) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| E (beam end) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| P at C | $0\theta$ | $0P\,\theta$ |
| P at D | $0\theta$ | $0P\,\theta$ |
| 2P/l on AB | - | $1P\,\theta$ |

**Equation:** $P\times 1\,\theta = 2\,M_p\theta$, so $P = 2\,M_p/l$.

**Mechanism 3: combined mechanism**

Combine 1 and 2 so that the hinge at B cancels; hinges at C and E.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| E (beam end) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| P at C | $\frac{2}{9}\theta$ | $\frac{2}{9}P\,\theta$ |
| P at D | $\frac{1}{9}\theta$ | $\frac{1}{9}P\,\theta$ |
| 2P/l on AB | - | $\frac{2}{3}P\,\theta$ |

**Equation:** $P\times 1\,\theta = 2\,M_p\theta$, so $P = 2\,M_p/l$.

**Mechanism 4: sway with the hinge inside the loaded column.** With a UDL on AB, the maximum moment in the column is not at B but within the column, at height $y$ above A. Let the column part AK rotate $\theta$ about A; the upper part of the column and the beam translate $y\theta$, and the right column rotates $\phi = y\theta/l = y\theta$ (with $l = 1$). Hinges: K ($\theta$) and E ($\phi$).
$$
\begin{aligned}
\text{Internal} &= M_p\theta(1 + y) \\
\text{External} &= \tfrac{2P}{l}\left[\tfrac{y^2\theta}{2} + (1-y)\,y\theta\right] = 2P\theta\,y\left(1 - \tfrac y2\right)
\end{aligned}
$$
$$
P = \frac{M_p(1+y)}{2y - y^2}
$$
Minimising: $y^2 + 2y - 2 = 0$, so $y = \sqrt3 - 1 = 0.732\,l$, and
$$
P = \frac{\sqrt3\,M_p}{4\sqrt3 - 6} = \left(1 + \frac{\sqrt3}{2}\right)\frac{M_p}{l} = 1.866\,\frac{M_p}{l}
$$
#### Collapse load
This is lower than the sway (2.0), combined (2.0) and beam (6) values, and a check of the other hinge positions (for example, adding a hinge at C to mechanism 4) gives nothing lower:

**Answer: $P_c = \left(1 + \dfrac{\sqrt3}{2}\right)\dfrac{M_p}{l} = 1.866\,\dfrac{M_p}{l}$.**

## Check yourself

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

1. In the beam mechanism (hinges at B, C, E), what is the internal work and the resulting P?
   A. Internal work 2Mp theta, P = 6Mp/l
   B. Internal work Mp theta, P = 3Mp/l
   C. Internal work 2Mp theta, P = 2Mp/l
   D. Internal work 3Mp theta, P = 9Mp/l
   Answer: A. Hinge rotations are theta, 2theta/3 and theta/3, total 2theta; external work is P(2/9+1/9)theta = P theta/3, so P = 6Mp. (Key point: Beam mechanism result)

2. In the sway mechanism, which load does external work and what P results?
   A. Both vertical loads P; P = 6Mp/l
   B. Only the UDL on the left column; P = 2Mp/l
   C. None; the frame cannot sway
   D. The UDL, giving P = 4Mp/l
   Answer: B. The beam translates horizontally, so vertical loads do no work; the UDL (2P/l over height l) does P theta of work, giving P = 2Mp/l. (Key point: Sway mechanism)

3. Why is an extra mechanism with a hinge inside the loaded column considered?
   A. The columns are heavier than the beam
   B. The base hinges cannot rotate
   C. Under the UDL the maximum column moment occurs within the column, not at joint B
   D. To increase the number of hinges to four
   Answer: C. A UDL gives a maximum moment at a point of zero shear inside the member, so the hinge forms there. (Key point: Hinge inside the loaded column)

4. What is the collapse load, and why?
   A. P = 6Mp/l, the beam mechanism
   B. P = 2Mp/l, sway and combined
   C. P = 1.866 Mp/l, the lowest value from the sway mechanism with the internal hinge
   D. P = 8Mp/l
   Answer: C. Collapse load is the lowest of the mechanism values; 1.866 Mp/l (y = 0.732 l) is below 2.0 and 6. (Key point: Collapse load)

## Related in Theory of Structures II

- [Collapse load P of portal frame](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-load-portal-p-p-p) — Theory of Structures II, chapter 5
- [Collapse moment of portal frame, load factor 1.75](https://www.bctengineer.com/question.md?key=semester_5%2Ftheory_of_structures_ii%3Acollapse-moment-portal-load-factor-1) — Theory of Structures II, chapter 5
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