---
title: "Plastic moment capacity Mp of frame"
subject: "Theory of Structures II"
chapter: 5
key: "semester_5/theory_of_structures_ii:plastic-moment-capacity-frame-3mp-2mp"
source: https://www.bctengineer.com/q/semester_5/theory_of_structures_ii/plastic-moment-capacity-frame-3mp-2mp
site: BCT Engineering
---

# Plastic moment capacity Mp of frame

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

- Asked in: 2066 Jestha, 8 marks
- Video: [Plastic Analysis Problem No -  23 ( Finding Plastic Moment Capacity in a Frame )](https://www.youtube.com/watch?v=HsnNrS1vR3M) — 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/plastic-moment-capacity-frame-3mp-2mp
- Whole chapter: https://www.bctengineer.com/md/semester_5/theory_of_structures_ii/chapter-5

## Question

Determine the value of plastic moment capacity Mp for the frame loaded as follows. [Figure: frame; beam B-C (3Mp) of 4 m (2 m from B to the 20 kN load) with a 10 kN load/moment at the right end; left column BA (2Mp) with A fixed; column from the beam to a hinged support D (Mp); heights 3 m and 1 m marked.]

## Short answer

The combined mechanism governs: internal work 6Mp theta equals external work 35 theta, giving Mp = 35/6 = 5.83 kNm. Hence member capacities are AB 11.67, BC 17.5, CD 5.83 kNm.

## Answer

#### Assumptions (the figure is not fully clear)
- Frame: A (0,0) fixed, B (0,3), C (4,3), D hinged 1 m below C. Beam BC is 4 m with the 20 kN at mid-span J (2 m from B); the 10 kN at the right end C is taken as a horizontal force to the right.
- Plastic moments: column AB $2M_p$, beam BC $3M_p$, column CD $M_p$. The loads are ultimate loads and the required $M_p$ is the largest value given by the mechanisms.
- Degree of indeterminacy $= 5 - 3 = 2$; critical sections A, B, J, C: $4 - 2 = 2$ independent mechanisms (beam and sway) plus the combination.

#### Mechanisms

**Mechanism 1: beam mechanism**

No sway; hinges at B (in the column BA, $2M_p$ is weaker than the beam $3M_p$), J (2 m from B, mid-span) and C (in the column CD, $M_p$). B and C rotate $\theta/2$ and J rotates $\theta$.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| J (under 20 kN) | $1\theta$ | $3M_p$ | $3\,M_p\theta$ |
| B (column BA) | $\frac{1}{2}\theta$ | $2M_p$ | $1\,M_p\theta$ |
| C (column CD) | $\frac{1}{2}\theta$ | $M_p$ | $\frac{1}{2}\,M_p\theta$ |

Internal work $= \frac{9}{2}\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| 20 kN at J | $1\theta$ | $20\,\text{kN·m}\,\theta$ |
| 10 kN at C | $0\theta$ | $0\,\text{kN·m}\,\theta$ |

**Equation:** $M_p\times \frac{9}{2}\,\theta = 20\,\theta$ (kN·m), so $M_p = \frac{40}{9} = 4.44$ kN·m.

**Mechanism 2: sway mechanism**

Column AB (3 m) rotates $\theta$ about the fixed base A and the beam moves $3\theta$; column CD (1 m high) rotates $3\theta$ about the hinge D. Hinges at A, B (in BA) and C (in CD).

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (column CD) | $3\theta$ | $M_p$ | $3\,M_p\theta$ |
| A (fixed base) | $1\theta$ | $2M_p$ | $2\,M_p\theta$ |
| B (column BA) | $1\theta$ | $2M_p$ | $2\,M_p\theta$ |

Internal work $= 7\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| 20 kN at J | $0\theta$ | $0\,\text{kN·m}\,\theta$ |
| 10 kN at C | $3\theta$ | $30\,\text{kN·m}\,\theta$ |

**Equation:** $M_p\times 7\,\theta = 30\,\theta$ (kN·m), so $M_p = \frac{30}{7} = 4.29$ kN·m.

**Mechanism 3: combined mechanism**

Combine 1 and 2 so that the hinge at B cancels; hinges at A, J and C.

| Hinge | Rotation | Plastic moment | Internal work |
|---|---|---|---|
| C (column CD) | $2\theta$ | $M_p$ | $2\,M_p\theta$ |
| J (under 20 kN) | $1\theta$ | $3M_p$ | $3\,M_p\theta$ |
| A (fixed base) | $\frac{1}{2}\theta$ | $2M_p$ | $1\,M_p\theta$ |

Internal work $= 6\,M_p\,\theta$.

| Load | Displacement along the load | External work |
|---|---|---|
| 20 kN at J | $1\theta$ | $20\,\text{kN·m}\,\theta$ |
| 10 kN at C | $\frac{3}{2}\theta$ | $15\,\text{kN·m}\,\theta$ |

**Equation:** $M_p\times 6\,\theta = 35\,\theta$ (kN·m), so $M_p = \frac{35}{6} = 5.83$ kN·m.

#### Required plastic moment
The combined mechanism requires the largest $M_p$:

**Answer: $M_p = \dfrac{35}{6} = 5.83$ kN·m** (so AB $= 11.67$, BC $= 17.5$ and CD $= 5.83$ kN·m).

## Check yourself

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

1. At a hinge at joint B, which plastic moment is used for the internal work in the beam mechanism?
   A. 3Mp of the beam
   B. The weaker member meeting there: 2Mp of column BA
   C. The average 2.5Mp
   D. Mp of column CD
   Answer: B. The hinge forms in the weaker member at the joint, so B uses the column BA value 2Mp. (Key point: Different Mp members)

2. In the sway mechanism, why does the hinge at C in column CD rotate 3theta?
   A. Because CD is 1 m high while AB is 3 m, so rotating about hinge D it must turn 3theta for the same beam translation 3theta
   B. Because C is a fixed base
   C. Because the load at C is 10 kN
   D. Because CD has 3Mp
   Answer: A. The beam moves 3theta horizontally; CD, 1 m high, rotates 3theta about D. (Key point: Sway mechanism with unequal columns)

3. In the combined mechanism, what is Mp and what is the internal work?
   A. Internal 6Mp theta, external 35 theta, Mp = 5.83 kN·m
   B. Internal 7Mp theta, Mp = 4.29
   C. Internal 4.5Mp theta, Mp = 4.44
   D. Internal 6Mp theta, external 20 theta, Mp = 3.33
   Answer: A. Internal 2+3+1 = 6Mp theta; external 20 + 15 = 35 theta, so Mp = 35/6 = 5.83 kN·m. (Key point: Combined mechanism)

4. Mechanisms give 4.44, 4.29 and 5.83 kN·m. Which governs?
   A. 4.29, the smallest
   B. 4.44, the beam mechanism
   C. 5.83, the largest, from the combined mechanism
   D. Their average, 4.85
   Answer: C. The largest required Mp is adopted so the frame is safe against all mechanisms. (Key point: Governing mechanism)

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