Plastic Section Modulus⚠ unverified
Mechanical / Beams · Compute the plastic section modulus of a rectangular section
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| b | b | m | 1.0 | Width of the section |
| h | h | m | 1.0 | Height of the section |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | Z | m**3 | Plastic section modulus, in metres^3 (m**3) |
The science & history
Understanding the Parameters
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Width $b$ — enters linearly, exactly as in the elastic modulus: it scales the area carrying stress but not the lever arm.
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Height $h$ — enters as $h^2$. Compare the elastic section modulus of a rectangle, $S = bh^2/6$: both go as $bh^2$, but the plastic value has the larger coefficient $1/4$ versus $1/6$.
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Shape factor. The ratio $Z/S = (bh^2/4)/(bh^2/6) = 1.5$ for a rectangle. This shape factor is the extra moment capacity available between first yield and full plasticity. It is a property of the cross-section shape: ~1.5 for a rectangle, ~1.7 for a solid circle, but only ~1.12–1.18 for an I-beam (whose material is already near the extreme fibres, leaving little reserve). The I-beam's low shape factor is the price of its high elastic efficiency.
Derivation (Approaching a Proof)
At the fully plastic condition, the material is idealised as elastic–perfectly-plastic: every fibre carries exactly the yield stress $\sigma_y$, in tension below the neutral axis and compression above (or vice versa). Two conditions define the plastic moment:
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Zero net axial force. The tension and compression forces must balance. Since the stress magnitude is uniform ($\sigma_y$), the tension and compression areas must be equal. For a symmetric section this plastic neutral axis is the axis that bisects the area — here, mid-height.
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Plastic moment = sum of force couples. Each half-area $A/2$ carries a force $\sigma_y (A/2)$, acting at the centroid of that half-area. The two forces form a couple with lever arm equal to the distance between the two half-centroids, $\bar{y}_1 + \bar{y}_2$: $$M_p = \sigma_y \left(\frac{A}{2}\right)(\bar{y}_1 + \bar{y}_2) \equiv \sigma_y Z.$$
For the rectangle: each half has area $A/2 = bh/2$, and the centroid of each half-rectangle lies at $h/4$ from the mid-height axis, so $\bar{y}_1 + \bar{y}_2 = h/4 + h/4 = h/2$. Therefore
$$Z = \frac{A}{2}(\bar{y}_1 + \bar{y}_2) = \frac{bh}{2}\cdot\frac{h}{2} = \frac{b h^2}{4}.$$
So $Z$ is the first moment of area of the two halves about the plastic neutral axis — a different (larger) quantity than the elastic $S = I/c$, which weights fibres by their distance rather than counting them all at $\sigma_y$.
Dimensional check. $[Z] = \text{m}\cdot\text{m}^2 = \text{m}^3$. ✓
History and Development
Plastic (limit-state) analysis grew from the work of Kazinczy (1914) and was developed into a design theory by J. F. Baker, Horne, and the Cambridge school in the 1930s–50s, and by van den Broek in the US. Recognising that ductile steel structures redistribute moment and form plastic hinges before collapse led to more economical designs and underlies modern Load and Resistance Factor Design (LRFD): steel-beam capacity is quoted as $M_p = \sigma_y Z$, with $Z$ tabulated for every rolled shape in the AISC manual.
Related Concepts: Section Modulus, Rectangular Moment of Inertia, Beam Bending Stress, Moment Of Inertia I Beam, Von Mises Stress
Notes: Formula is for a solid rectangle; the shape factor $Z/S = 1.5$ applies to rectangles only. For other sections compute $Z$ as the first moment of the two equal areas about the plastic neutral axis (the area-bisecting axis, which is not the centroid for unsymmetric sections). Assumes a ductile, elastic–perfectly-plastic material.