---
title: "Collapse moment of portal frame, load factor 1.75"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:collapse-moment-portal-load-factor-1"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/collapse-moment-portal-load-factor-1
site: BCT Engineering
---

# Collapse moment of portal frame, load factor 1.75

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

- Asked in: 2069 Asar, 8 marks
- Video: [Plastic analysis (part-3)-Collapse load in beam](https://www.youtube.com/watch?v=6035iMwKLmY) — TECHNICA ONLINE LECTURES
- 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-moment-portal-load-factor-1
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Calculate the collapse moment after establishing possible failure mechanisms for the portal frame shown in figure below. Use load factor 1.75. [Figure: portal frame, both bases hinged, height 4 m, span 6 m (3 m + 3 m); 40 kN horizontal at the top-left joint; 120 kN vertical at mid-span; all members $M_P$. Printed as the alternative (OR) to the continuous beam question above.]

## Short answer

Factored loads: 70 kN horizontal, 210 kN vertical. Beam, sway and combined mechanisms require Mp of 157.5, 140 and 227.5 kN·m. The combined mechanism governs; collapse moment is 227.5 kN·m.

## Answer

#### Given and factored loads
Portal frame with both bases hinged, height 4 m, span 6 m, all members $M_p$. Working loads: 40 kN horizontal at the top-left joint B and 120 kN vertical at mid-span C. With the load factor 1.75, the factored loads are $40\times1.75 = 70$ kN and $120\times1.75 = 210$ kN. The frame must just collapse under these loads, so each mechanism gives a required $M_p$ and the largest one governs.

Degree of indeterminacy $= 4 - 3 = 1$; the critical sections B, C and D give $3 - 1 = 2$ independent mechanisms (beam and sway), plus their combination.

#### Mechanisms

**Mechanism 1: beam mechanism**

No sway; hinges at B, C and D. Each half-span rotates $\theta/2$ (in opposite senses), so the hinge at C turns $\theta$ and C deflects $3\times\theta/2 = \frac{3}{2}\theta$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (mid-span) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| B | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |
| D | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| 70 kN horizontal at B | $0\theta$ | $0\,\text{kN·m}\,\theta$ |
| 210 kN at C | $\frac{3}{2}\theta$ | $315\,\text{kN·m}\,\theta$ |

**Equation:** $M_p\times 2\,\theta = 315\,\theta$ (kN·m), so $M_p = \frac{315}{2} = 157.50$ kN·m.

**Mechanism 2: sway mechanism**

The columns (4 m) rotate $\theta$ about the hinged bases and the beam moves $4\theta$; hinges at B and D only.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| B | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| D | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| 70 kN horizontal at B | $4\theta$ | $280\,\text{kN·m}\,\theta$ |
| 210 kN at C | $0\theta$ | $0\,\text{kN·m}\,\theta$ |

**Equation:** $M_p\times 2\,\theta = 280\,\theta$ (kN·m), so $M_p = 140$ kN·m.

**Mechanism 3: combined mechanism**

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

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (mid-span) | $1\theta$ | $M_p$ | $1\,M_p\theta$ |
| D | $1\theta$ | $M_p$ | $1\,M_p\theta$ |

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

| Load | Displacement along the load | External work |
|---|---|---|
| 70 kN horizontal at B | $2\theta$ | $140\,\text{kN·m}\,\theta$ |
| 210 kN at C | $\frac{3}{2}\theta$ | $315\,\text{kN·m}\,\theta$ |

**Equation:** $M_p\times 2\,\theta = 455\,\theta$ (kN·m), so $M_p = \frac{455}{2} = 227.50$ kN·m.

#### Collapse moment
The combined mechanism needs the largest $M_p$:

**Answer: $M_p = 227.5$ kN·m** (the collapse moment of the frame).

## Check yourself

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

1. With load factor 1.75, what loads are used in the mechanism equations for the 40 kN horizontal and 120 kN vertical loads?
   A. 70 kN and 210 kN
   B. 40 kN and 120 kN
   C. 70 kN and 120 kN
   D. 22.9 kN and 68.6 kN
   Answer: A. Working loads are multiplied by the load factor: 40x1.75 = 70 kN and 120x1.75 = 210 kN. (Key point: Factored loads)

2. For this hinged-base portal frame (indeterminacy 1, critical sections B, C, D), how many independent mechanisms are there?
   A. 3 (beam, sway, combined all independent)
   B. 2 (beam and sway); the combined one is derived from them
   C. 1 (sway only)
   D. 4
   Answer: B. Independent mechanisms = critical sections - indeterminacy = 3 - 1 = 2; the combined mechanism is a combination of these. (Key point: Number of independent mechanisms)

3. In the beam mechanism, why is the external work of the 70 kN horizontal load zero?
   A. Because the load is smaller than the vertical load
   B. Because the hinge at B cancels it
   C. Because there is no sway, so B has no horizontal displacement
   D. Because horizontal loads never do work
   Answer: C. The beam mechanism has no sway, so the point B does not move horizontally and the horizontal load does no work. (Key point: Beam mechanism work)

4. The mechanisms give Mp = 157.5, 140 and 227.5 kN·m. Which is taken as the collapse moment and why?
   A. 140, the smallest, as it fails first
   B. 157.5, the average
   C. 227.5, the combined mechanism, because the largest required Mp governs
   D. The sum of all three
   Answer: C. The frame must resist every mechanism, so the largest required Mp (combined, 455/2 = 227.5 kN·m) governs. (Key point: Governing collapse moment)

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