Section modulus: rectangle versus circle

Section modulus

The section modulus is $Z = I/y_{max}$, where $I$ is the second moment of area about the neutral axis and $y_{max}$ the distance to the extreme fibre. From the flexure formula the moment of resistance is

$$M_r = \sigma_{all}\,Z$$

so for a given stress, the larger $Z$, the larger the moment the section can carry. Unit: mm$^3$ or m$^3$.

Comparison for equal area

Rectangle of width $b$ and depth $h = 2b$:

$$A = 2b^2, \qquad Z_r = \frac{bh^2}{6} = \frac{b(2b)^2}{6} = 0.667\,b^3$$

Circle with the same area:

$$\frac{\pi d^2}{4} = 2b^2 \;\Rightarrow\; d = 1.596\,b$$ $$Z_c = \frac{\pi d^3}{32} = \frac{\pi (1.596b)^3}{32} = 0.399\,b^3$$ $$\frac{M_{r,\text{rect}}}{M_{r,\text{circle}}} = \frac{Z_r}{Z_c} = \frac{0.667}{0.399} = 1.67$$

For example, with $b = 100$ mm: $Z_r = 666.7\times10^3$ mm$^3$ and $Z_c = 398.9\times10^3$ mm$^3$. With $\sigma_{all} = 10$ MPa they carry 6.67 kN m and 3.99 kN m respectively. The rectangle (on edge) is about 67 % stronger in bending for the same weight of material.

Why I-sections

In bending the material near the neutral axis carries little stress. An I-section puts most of the area in the flanges, far from the NA, so $I$ and $Z$ are large for a small area. Its web resists shear. It therefore gives the highest strength-to-weight ratio, which is why rolled I-sections are used for beams.