---
title: "Limiting plastic moment is 1.5 times elastic moment"
subject: "Strength of Materials"
chapter: 7
key: "semester_3/strength_of_materials:plastic-moment-rectangular-1-5"
source: https://www.bctengineer.com/q/semester_3/strength_of_materials/plastic-moment-rectangular-1-5
site: BCT Engineering
---

# Limiting plastic moment is 1.5 times elastic moment

Strength of Materials · Chapter 7: Theory of Flexure

- Asked in: 2067 Asar (old course), 6 marks
- Video: [Example 1 - Plastic Section Properties](https://www.youtube.com/watch?v=A7Ip9Qf5mdY) — Luis Santos
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/plastic-moment-rectangular-1-5
- Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-7

## Question

Prove that the limiting plastic moment of a rectangular beam made of elastoplastic material is 1.5 times the maximum elastic moment.

## Short answer

For a rectangular section, the elastic moment capacity is σ_y bd²/6 and the fully plastic moment is σ_y bd²/4, giving a ratio of 1.5. Thus the plastic moment is 1.5 times the maximum elastic moment.

## Answer

Consider a rectangular section of width $b$ and depth $d$ made of an elastic-perfectly plastic material with yield stress $\sigma_y$.

```text
   Elastic limit        Fully plastic
   +--+                  +------+
   | /|                  |      |   sigma_y
 --|/-|-- NA          ---|------|--- NA
   |\ |                  |      |   sigma_y
   +--+                  +------+
```

#### Maximum elastic moment
Stress is linear and just reaches $\sigma_y$ at the extreme fibres:

$$
M_y = \sigma_y Z = \sigma_y\cdot\frac{bd^2}{6}
$$

#### Fully plastic moment
Every fibre has reached $\sigma_y$. The compression and tension zones are the two halves of the section, each of area $bd/2$, and each resultant acts at the centroid of its half, $d/4$ from the neutral axis:

$$
T = C = \sigma_y\frac{bd}{2}
$$

The lever arm between the two forces is $d/2$:

$$
M_p = T\times\frac{d}{2} = \sigma_y\frac{bd}{2}\times\frac{d}{2} = \sigma_y\frac{bd^2}{4}
$$

(The plastic section modulus is $S = bd^2/4$.)

#### Ratio
$$
\frac{M_p}{M_y} = \frac{\sigma_y\,bd^2/4}{\sigma_y\,bd^2/6} = \frac{6}{4} = 1.5
$$

**Hence $M_p = 1.5\,M_y$.** The shape factor of a rectangular section is 1.5.

## Check yourself

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

1. What is the maximum elastic moment M_y of a rectangular section?
   A. sigma_y x bd^2/4
   B. sigma_y x bd^2/6, when the extreme fibres just reach yield
   C. sigma_y x bd^2/12
   D. sigma_y x bd^3/6
   Answer: B. M_y = sigma_y Z with Z = bd^2/6. (Key point: Maximum elastic moment)

2. What is the stress state at the fully plastic moment?
   A. Stress is linear with sigma_y at the extremes only
   B. Only the middle fibres yield
   C. Every fibre has reached sigma_y, tension above and compression below the NA (or vice versa)
   D. Stress is zero at the extremes
   Answer: C. At M_p the whole section has yielded in a rectangular block. (Key point: Stress block at full plasticity)

3. How is M_p found?
   A. T = sigma_y bd times lever arm d/2
   B. T = sigma_y bd/2 times lever arm d
   C. T = sigma_y bd/2 times lever arm d/4
   D. T = C = sigma_y bd/2 times lever arm d/2, giving sigma_y bd^2/4
   Answer: D. Each half carries sigma_y bd/2 and the centroids are d/2 apart. (Key point: Plastic moment)

4. What does M_p/M_y equal for a rectangle?
   A. 1.5
   B. 1.12
   C. 1.7
   D. 2.0
   Answer: A. (bd^2/4)/(bd^2/6) = 6/4 = 1.5, the shape factor of a rectangle. (Key point: Shape factor)

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