---
title: "Collapse load of frame with UDL on column"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-frame-20kn-10knm"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-frame-20kn-10knm
site: BCT Engineering
---

# Collapse load of frame with UDL on column

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

- Asked in: 2079 Baishakh, 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-20kn-10knm
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Evaluate the collapse load for the given frame. [Figure: portal frame ABCD, height 5 m, A hinged, D fixed; beam BC of 6 m with 20 kN at 2 m and 20 kN at 4 m from B; plastic moments: AB = $M_p$, beam BC = $2M_p$, column CD = $2M_p$; 10 kN/m UDL on column CD.]

## Short answer

Using kinematik theorem: beam mechanism gives λ=M_p/12≈0.0833M_p; sway mechanism λ=M_p/25=0.04M_p. The lowest is from combined mechanism (hinges at B, under the 4m load, D) yielding λ=13M_p/370≈0.0351M_p. Thus collapse load factor is 13M_p/370; all loads proportionally increase.

## Answer

#### Assumptions
- Portal frame: A (0,0) hinged, B (0,5), C (6,5), D (6,0) fixed. Column AB has plastic moment $M_p$, beam BC and column CD have $2M_p$.
- The loads (two 20 kN point loads at 2 m and 4 m from B, and 10 kN/m on column CD) are applied proportionally with a load factor $\lambda$; the 10 kN/m acts horizontally towards the left on CD. The collapse load factor $\lambda_c$ is found in terms of $M_p$ (in kN·m).
- Method: kinematic (upper-bound) theorem with virtual work, $\sum M_p\theta = \lambda \sum P\delta$.

#### Possible hinge positions
B (in the weaker member AB, $M_p$), under each 20 kN load, C and D. Degree of indeterminacy = 5 - 3 = 2 (A hinged gives 2 reactions, D fixed gives 3), so with 5 critical sections the number of independent mechanisms is 5 - 2 = 3: two beam mechanisms (one with the hinge under each load) and the sway mechanism. Combining them gives further mechanisms.

**Mechanism 1: beam mechanism**

Joints B and C do not move. The beam BC forms hinges under the second load P2 (4 m from B) and at C; at B the hinge forms at the top of the weaker column AB ($M_p$). With rotation $\theta$ at P2, the left portion (4 m) rotates $\theta/3$ and the right portion (2 m) rotates $2\theta/3$ (rigid portions); the deflection under the second load is $4\times\theta/3$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| under the second 20 kN | $1\theta$ | $2M_p$ | $2\,M_p\theta$ |
| C | $\frac{2}{3}\theta$ | $2M_p$ | $\frac{4}{3}\,M_p\theta$ |
| B (top of column AB) | $\frac{1}{3}\theta$ | $M_p$ | $\frac{1}{3}\,M_p\theta$ |

Internal work $= \frac{11}{3}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| 20 kN at 2 m | $\frac{2}{3}\theta$ | $\frac{40}{3}\,\text{kN·m}\,\theta$ |
| 20 kN at 4 m | $\frac{4}{3}\theta$ | $\frac{80}{3}\,\text{kN·m}\,\theta$ |
| 10 kN/m UDL on CD | - | $0\,\text{kN·m}\,\theta$ |

**Equation:** $\lambda\times40\,\theta = \frac{11}{3}\,M_p\theta$, so $\lambda = \frac{11}{120}M_p = 0.0917\,M_p$.

With the hinge under the first load instead (2 m from B), the portions rotate $2\theta/3$ and $\theta/3$: internal work $= 2M_p\theta + M_p\cdot\frac{2}{3}\theta\ (\text{B}) + 2M_p\cdot\frac{1}{3}\theta\ (\text{C}) = \frac{10}{3}M_p\theta$, and the external work is again $40\theta$, so $\lambda = \frac{1}{12}M_p = 0.0833\,M_p$ (the lower beam value).

**Mechanism 2: sway mechanism**

Columns rotate through $\theta$ about A and D, and the beam translates horizontally by $5\theta$. Hinges form at B (the weaker section is AB, $M_p$), C and D. The vertical loads do no work. (The column CD is moved to the left by the UDL.)

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

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

| Load | Displacement along the load | External work |
|---|---|---|
| 20 kN at 2 m | $0\theta$ | $0\,\text{kN·m}\,\theta$ |
| 20 kN at 4 m | $0\theta$ | $0\,\text{kN·m}\,\theta$ |
| 10 kN/m UDL on CD | - | $125\,\text{kN·m}\,\theta$ |

**Equation:** $\lambda\times125\,\theta = 5\,M_p\theta$, so $\lambda = \frac{1}{25}M_p = 0.0400\,M_p$.

**Mechanism 3: combined mechanism**

Combine mechanisms 1 and 2 so that the hinge at C cancels. Both columns rotate $2\theta/3$ (sway to the left), the beam portion from B to the hinge rotates $\theta/3$ in the opposite sense (hinge at B $= 2\theta/3 + \theta/3 = \theta$), and the portion from the hinge to C rotates $2\theta/3$ with the column (hinge at the load $= \theta$, no hinge at C).

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| B (in column AB) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| under the second 20 kN | $1\theta$ | $2M_p$ | $2\,M_p\theta$ |
| D | $\frac{2}{3}\theta$ | $2M_p$ | $\frac{4}{3}\,M_p\theta$ |

Internal work $= \frac{13}{3}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| 20 kN at 2 m | $\frac{2}{3}\theta$ | $\frac{40}{3}\,\text{kN·m}\,\theta$ |
| 20 kN at 4 m | $\frac{4}{3}\theta$ | $\frac{80}{3}\,\text{kN·m}\,\theta$ |
| 10 kN/m UDL on CD | - | $\frac{250}{3}\,\text{kN·m}\,\theta$ |

**Equation:** $\lambda\times\frac{370}{3}\,\theta = \frac{13}{3}\,M_p\theta$, so $\lambda = \frac{13}{370}M_p = 0.03514\,M_p$.

(The same combination with the hinge under the first load gives $\lambda = \frac{11}{245}M_p = 0.0449\,M_p$, which is higher.)

#### Collapse load
The smallest value governs (combined mechanism, lower than both the beam value 0.0833 and the sway value 0.04):
$$
\lambda_c = \frac{13}{370}M_p = 0.0351\,M_p
$$
So the collapse loads are: point loads $20\lambda_c = 0.703\,M_p$ kN each and UDL $10\lambda_c = 0.351\,M_p$ kN/m (with $M_p$ in kN·m). A check with a linear-programming search over all possible hinge sets gave the same value $0.03514\,M_p$.

**Answer:** collapse load factor $\lambda_c = 13M_p/370 \approx 0.0351\,M_p$ (combined mechanism with hinges at B, under the 20 kN load at 4 m, and at D).

## 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 2, how many independent mechanisms exist?
   A. 2
   B. 3
   C. 5
   D. 7
   Answer: B. Independent mechanisms = critical sections - degree of indeterminacy = 5 - 2 = 3. (Key point: Mechanism count)

2. Which value of the load factor governs the collapse and why?
   A. The largest, because it is safest
   B. The smallest, as the kinematic theorem gives an upper bound and the lowest is true
   C. The average of all mechanisms
   D. The beam mechanism only
   Answer: B. Each mechanism gives an upper bound; the lowest is the true collapse load. (Key point: Governing choice)

3. In the sway mechanism, what does the external work come from?
   A. The two 20 kN vertical loads
   B. The 10 kN/m horizontal load on CD only, as vertical loads do no work
   C. All loads equally
   D. The weight of the beam
   Answer: B. The beam moves horizontally, so vertical loads do no work; the UDL on CD gives 125 kN.m x theta. (Key point: Sway mechanism)

4. What is the collapse load factor and the mechanism?
   A. 0.1 Mp, beam mechanism
   B. 0.04 Mp, sway mechanism
   C. 0.0351 Mp (13Mp/370), combined mechanism
   D. 0.0351 Mp, beam mechanism
   Answer: C. The combined mechanism gives 13Mp/370, lower than 0.1 and 0.04. (Key point: Result)

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