---
title: "Plastic moment and shape factor of a beam"
subject: "Mechanics of Solids"
chapter: 12
key: "semester_5/mechanics_of_solids:shape-factor-plastic-moment-rectangle"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/shape-factor-plastic-moment-rectangle
site: BCT Engineering
---

# Plastic moment and shape factor of a beam

Mechanics of Solids · Chapter 12: Introduction to Plastic Range Stress and Strain

- Asked in: year not given, 8 marks
- Video: [How to calculate plastic, elastic section modulus and Shape Factor of a cross-section?](https://www.youtube.com/watch?v=T2XMORhJ49Q) — Bold Learning
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/shape-factor-plastic-moment-rectangle
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-12

## Question

Define plastic moment and shape factor. For a rectangular beam of width $b$ and depth $h$ of an elastic-perfectly plastic material, derive the yield moment, the fully plastic moment and the shape factor. A beam of rectangular section 60 mm wide and 100 mm deep has $\sigma_y = 250$ MPa. Find the yield moment, the plastic moment, the moment when the elastic core is 60 mm deep, and the collapse load of the beam if simply supported on a span of 3 m with a central point load.

## Short answer

Plastic moment is the fully plastic moment capacity; shape factor S = Mp/My. For a rectangle, My = sigma_y b h^2/6, Mp = sigma_y b h^2/4, S = 1.5. Given beam: My = 25 kN m, Mp = 37.5 kN m, moment at 60 mm elastic core = 33 kN m, collapse load = 50 kN.

## Answer

#### Definitions

- **Yield moment** $M_y$: the moment at which the extreme fibre first reaches $\sigma_y$.
- **Plastic moment** $M_p$: the moment when the whole section has yielded (a plastic hinge forms).
- **Shape factor** $S = M_p/M_y = Z_p/Z$: depends only on the section shape.

Assumptions: elastic-perfectly plastic stress-strain curve, same behaviour in tension and compression, plane sections remain plane, and no axial force.

#### Derivation (rectangular section)

**Yield moment**

$$
M_y = \sigma_y Z = \sigma_y\,\frac{bh^2}{6}
$$

**Elastic-plastic stage.** Let the elastic core have half-depth $y_0$ ($y_0 \le h/2$). The stress is $\sigma_y y/y_0$ in the core and $\sigma_y$ in the outer zones.

$$
M = \underbrace{\sigma_y\,\frac{b\,(2y_0)^2}{6}}_{\text{core}} + \underbrace{2\sigma_y\,b\,\left(\frac h2 - y_0\right)\left(\frac{h/2 + y_0}{2}\right)}_{\text{outer zones}} = \sigma_y\,b\left(\frac{h^2}{4} - \frac{y_0^2}{3}\right)
$$

**Fully plastic** ($y_0 \to 0$):

$$
M_p = \sigma_y\,\frac{bh^2}{4} = \sigma_y Z_p,\qquad Z_p = \frac{bh^2}{4}
$$

**Shape factor**

$$
S = \frac{M_p}{M_y} = \frac{bh^2/4}{bh^2/6} = 1.5
$$

#### Numerical values ($b = 60$, $h = 100$, $\sigma_y = 250$ MPa)

$$
\begin{aligned}
M_y &= 250\times\frac{60\times100^2}{6} = 250\times10^5 = 25.0\ \text{kN m} \\
M_p &= 250\times\frac{60\times100^2}{4} = 250\times1.5\times10^5 = 37.5\ \text{kN m} \\
\end{aligned}
$$

Elastic core 60 mm deep, so $y_0 = 30$ mm:

$$
M = 250\times60\left(\frac{100^2}{4} - \frac{30^2}{3}\right) = 15000\times(2500 - 300) = 33\times10^6\ \text{N mm} = 33.0\ \text{kN m}
$$

(A value between $M_y$ and $M_p$, as expected.)

#### Collapse load (simply supported, central load $W$, span $L = 3$ m)

At collapse a plastic hinge forms under the load, with $M_{max} = WL/4 = M_p$:

$$
W_p = \frac{4M_p}{L} = \frac{4\times37.5}{3} = 50.0\ \text{kN},\qquad W_y = \frac{4M_y}{L} = 33.3\ \text{kN}
$$

**Answer:** $M_y = 25$ kN m, $M_p = 37.5$ kN m, shape factor $= 1.5$, $M = 33$ kN m for a 60 mm elastic core, and the collapse load is $50$ kN (first yield at $33.3$ kN).

## Check yourself

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

1. What is the shape factor of a rectangular section and what does it mean?
   A. 1.0; no reserve strength
   B. 1.5; the plastic moment is 1.5 times the yield moment
   C. 1.7; the yield moment is 1.7 times the plastic moment
   D. 2.0; the section carries twice the load
   Answer: B. S = Mp/My = (bh^2/4)/(bh^2/6) = 1.5. (Key point: Shape factor)

2. What is the yield moment of the 60 x 100 mm section with sigma_y = 250 MPa?
   A. 37.5 kN m
   B. 50 kN m
   C. 25 kN m
   D. 15 kN m
   Answer: C. My = sigma_y b h^2/6 = 250 x 1e5 N mm = 25 kN m. (Key point: Yield moment)

3. What is the moment with a 60 mm deep elastic core (y0 = 30 mm)?
   A. 33 kN m, between My and Mp
   B. 25 kN m
   C. 37.5 kN m
   D. 28 kN m
   Answer: A. M = sigma_y b (h^2/4 - y0^2/3) = 15000 x 2200 = 33 kN m. (Key point: Elastic-plastic moment)

4. How is the collapse load of the simply supported beam with a central point load found?
   A. Equate WL/4 to Mp, giving 50 kN
   B. Equate WL/4 to My, giving 50 kN
   C. Equate WL/8 to Mp, giving 100 kN
   D. Equate WL/2 to Mp, giving 25 kN
   Answer: A. A plastic hinge forms under the load when WL/4 = Mp, so W = 4 x 37.5/3 = 50 kN. (Key point: Collapse load)

## Related in Mechanics of Solids

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