---
title: "Collapse load of frame with W and W/2 loads"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-load-frame-wkn-1-5mp"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-load-frame-wkn-1-5mp
site: BCT Engineering
---

# Collapse load of frame with W and W/2 loads

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

- Asked in: 2078 Kartik, 10 marks
- Video: [Steel Structure- Plastic Analysis of Frames numerical-1 / Collapse Load of Frames (In Hindi).](https://www.youtube.com/watch?v=iZQ8JcfVek8) — Yogendra Ram Meena
- 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-wkn-1-5mp
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Determine the collapse load in the frame shown in the figure. [Figure: frame ABCDE; A hinged, E fixed; column AB (2Mp) of height 6 m with UDL (W/2) kN/m; beam BCD with 1.5Mp, load W kN at C, BC = CD = 4 m; column DE (Mp) with heights 6 m and 3 m as marked.]

## Short answer

Using virtual work on beam, sway, and combined mechanisms, the sway mechanism yields the lowest collapse load: Wc = 11Mp/18.

## Answer

#### Assumptions
- Frame ABCDE: A (0,0) hinged, B (0,6), C (4,6), D (8,6), E (8,3) fixed. Column AB $2M_p$ (6 m high), beam BCD $1.5M_p$ (BC = CD = 4 m), column DE $M_p$ (3 m high; the right base is 3 m above A's level, so the columns have different heights).
- Loads: W at C, and W/2 per metre on column AB acting horizontally to the right. $W_c$ is expressed in terms of $M_p$ (in kN·m, lengths in m).
- Method: virtual work, $\sum M_p\theta = \sum P\delta$.

#### Mechanisms

**Mechanism 1: beam mechanism**

Hinges at B, C and D (the weaker section at D is the column DE, $M_p$). Beam BC rotates $\theta/2$ at B, the two beam halves rotate $\theta/2$ each in opposite senses, so C deflects $4\times\theta/2 = 2\theta$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (under W) | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ |
| B (beam side) | $\frac{1}{2}\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{4}\,M_p\theta$ |
| D (column DE side) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |

Internal work $= \frac{11}{4}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| W at C | $2\theta$ | $2W\,\theta$ |
| W/2 per m on AB | - | $0W\,\theta$ |

**Equation:** $W_c\times 2\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{8}\,M_p = 1.3750\,M_p$.

**Mechanism 2: sway mechanism**

The beam BD translates horizontally by $3\theta$ (column DE, 3 m high, rotates $\theta$ about E); column AB rotates $\theta/2$ about A. Hinges at B (beam side, $1.5M_p$), D (in DE) and E. W does no work.

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

Internal work $= \frac{11}{4}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| W at C | $0\theta$ | $0W\,\theta$ |
| W/2 per m on AB | - | $\frac{9}{2}W\,\theta$ |

**Equation:** $W_c\times \frac{9}{2}\,\theta = \frac{11}{4}\,M_p\theta$, so $W_c = \frac{11}{18}\,M_p = 0.6111\,M_p$.

**Mechanism 3: combined mechanism**

Combination of 1 and 2 removing the hinge at B: hinges at C, D and E.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| D (column DE side) | $\frac{3}{2}\theta$ | $M_p$ | $\frac{3}{2}\,M_p\theta$ |
| C (under W) | $1\theta$ | $\frac{3}{2}M_p$ | $\frac{3}{2}\,M_p\theta$ |
| E (fixed base) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| W at C | $2\theta$ | $2W\,\theta$ |
| W/2 per m on AB | - | $\frac{9}{2}W\,\theta$ |

**Equation:** $W_c\times \frac{13}{2}\,\theta = 4\,M_p\theta$, so $W_c = \frac{8}{13}\,M_p = 0.6154\,M_p$.

**Refinement: sway with the hinge inside the column AB.** For a UDL the plastic hinge in the loaded column does not form at B but at the section of maximum moment, at a height $y$ above A. Let the beam translate by $\Delta = 3\phi$ ($\phi$ = rotation of DE). The part AK rotates $\theta$ about A, so $y\theta = \Delta$, and the part KB translates with the beam. Hinges: K (rotation $\theta$, $2M_p$), D ($\phi$, $M_p$), E ($\phi$, $M_p$), where $\phi = y\theta/3$.
$$
\begin{aligned}
\text{Internal} &= M_p\theta\,(2 + \tfrac{2y}{3}) \\
\text{External} &= \tfrac W2\left[\tfrac{y^2\theta}{2} + (6-y)\,y\theta\right] = \tfrac W2\,\theta\,y\left(6 - \tfrac y2\right)
\end{aligned}
$$
$$
W = \frac{M_p\,(4 + \tfrac{4y}{3})}{y\,(6 - \tfrac y2)}
$$
Setting $dW/dy = 0$ gives $y^2 + 6y - 36 = 0$, so $y = 3(\sqrt5 - 1) = 3.708$ m, and
$$
W_c = 0.5818\,M_p
$$
(At $y = 6$, a hinge at the top of the column, this expression gives 0.667; mechanism 2, with the hinge in the beam, gives 0.611.)

#### Collapse load
The smallest value is the sway mechanism with the hinge in the column AB, and it is lower than the beam (1.375) and combined (0.615) mechanisms:

**Answer: $W_c \approx 0.582\,M_p$ (in kN when $M_p$ is in kN·m and lengths are in m).**

## Check yourself

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

1. In the beam mechanism, what is the external work and result?
   A. W x 2 theta, so W_c = 11/8 Mp
   B. W x theta, so W_c = 11/4 Mp
   C. W x 4 theta, so W_c = 11/16 Mp
   D. zero, so W_c is infinite
   Answer: A. C deflects 2 theta so work = 2W theta; internal work 11/4 Mp theta gives W_c = 11/8 Mp. (Key point: Beam mechanism)

2. In the sway mechanism, W at C does no work because:
   A. the beam is rigid
   B. the beam translates horizontally and W is vertical
   C. W is below the hinge
   D. the hinge at C is absent
   Answer: B. Horizontal translation produces no vertical displacement of W; only the UDL on AB does work. (Key point: Sway mechanism)

3. Why is the sway mechanism refined with a hinge inside column AB?
   A. Because under a UDL the hinge forms at the section of maximum moment, not at joint B
   B. Because the beam is weaker
   C. Because the hinge at D cancels
   D. Because it gives a higher load
   Answer: A. For a UDL the hinge forms at the maximum-moment section at height y = 3.708 m. (Key point: UDL refinement)

4. What is the collapse load?
   A. 1.375 Mp
   B. 0.615 Mp
   C. 0.611 Mp
   D. 0.582 Mp, from sway with hinge in AB
   Answer: D. The refined sway gives 0.5818 Mp, lowest of all. (Key point: Result)

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