Rectangular Moment of Inertia⚠ unverified

Understanding the Parameters

• Width $b$ — the dimension parallel to the neutral axis. It enters linearly: every horizontal strip of area contributes in proportion to $b$, but all strips at a given height are at the same distance $y$, so width adds area without adding lever arm. Width is "cheap" bending material.

• Height $h$ — the dimension perpendicular to the neutral axis, i.e. the depth in the bending direction. It enters as $h^3$ because the outermost fibres sit at $y = \pm h/2$ and their leverage in the $\int y^2\,\mathrm{d}A$ integral compounds with the extra area they represent. Height is where bending performance is won.

• The factor 1/12 — a geometric constant that falls out of integrating $y^2$ over a rectangle from $-h/2$ to $+h/2$. (About the edge rather than the centroid the constant becomes 1/3; the difference is the parallel-axis shift — see Parallel Axis Theorem.)

Units are length$^4$ (m⁴), reflecting area (length²) times distance-squared (length²). $I$ feeds directly into bending stress (Beam Bending Stress), deflection ($\propto 1/EI$), and buckling (Beam Buckling Load).

Derivation (Approaching a Proof)

By definition, the second moment of area about the neutral (here, centroidal $x$-) axis is

$$I = \int_A y^2\, \mathrm{d}A.$$

For the rectangle, take a horizontal strip of height $\mathrm{d}y$ at position $y$; its area is $\mathrm{d}A = b\,\mathrm{d}y$. The centroid lies at mid-height, so $y$ runs from $-h/2$ to $+h/2$:

$$I = \int_{-h/2}^{+h/2} y^2\, (b\,\mathrm{d}y) = b\left[\frac{y^3}{3}\right]_{-h/2}^{+h/2} = \frac{b}{3}\left(\frac{h^3}{8} + \frac{h^3}{8}\right) = \frac{b}{3}\cdot\frac{h^3}{4} = \frac{b h^3}{12}.$$

The two equal contributions from the tension and compression halves make the section symmetric about its centroid, confirming the neutral axis is at mid-height. The same integral about an axis through the base gives $bh^3/3$; the parallel-axis theorem relates the two by $I_{\text{base}} = I_{\text{cent}} + A(h/2)^2$.

Dimensional check. $[I] = \text{m} \cdot \text{m}^3 = \text{m}^4$. ✓

History and Development

The concept of the second moment of area emerged with the development of beam theory by Euler, Bernoulli, and Navier in the 18th and early 19th centuries, as engineers sought to predict beam stiffness and strength for the railways and iron bridges of the industrial era. Tabulated $I$ values for standard shapes fill references such as Roark's Formulas for Stress and Strain and the AISC Steel Construction Manual, all built on the same $\int y^2\,\mathrm{d}A$ definition.

Related Concepts: Beam Bending Stress, Section Modulus, Parallel Axis Theorem, Moment Of Inertia I Beam, Plastic Section Modulus, Beam Buckling Load

Notes: This is the strong-axis value (bending about the axis parallel to $b$). For bending about the other axis, swap $b$ and $h$: $I = h b^3/12$. For built-up or non-rectangular sections, sum component inertias with the Parallel Axis Theorem.