Rectangular Section Inertia

Rectangular section moment of inertia: I = b·h³/12

Take a rectangle and measure its second moment of area about the centroidal x-axis (the neutral axis when it bends): you land on I = b·h³/12, where b is the width and h the height. What makes the formula tick is that h shows up cubed. A 200×400 mm section gives I = 200·400³/12 ≈ 1.067·10⁹ mm⁴; lay it on its side as 400×200 and I collapses to 200·200³, roughly a quarter of the value. Stiffness rides on h³, and that is the whole reason beams are built tall. I governs two things at once: how much a beam deflects (δ ∝ 1/I) and how much bending it can take, through the section modulus W = I/(h/2). When the axis isn't centroidal, the parallel-axis theorem fills the gap: I' = I + A·d².

Applications

It shows up when you size reinforced-concrete and timber beams, when you work out the properties of I-beam and W-shape steel profiles, in FEM modal analysis where the natural frequency hangs on E·I, in buckling checks through Euler's column load (P_cr ∝ I), and in lightweight design, where the trick is pushing h as high as you can while keeping the mass down.

FAQ

Why divide by 12 and not by 3? You get the 1/12 by integrating y² across the cross-section about the centroid. Shift to an edge axis and it turns into b·h³/3 instead; the parallel-axis theorem is what ties those two results together.

Is I the same as polar moment J? No. Bending is the job of I, while torsion answers to the polar moment J = I_x + I_y. In a rectangle the two are connected, yet they don't come out equal.

Why are I-beams so efficient? Most of their material sits well away from the neutral axis, out where y is large, which lifts the I you get per unit of mass. A solid rectangle does the opposite, burying material near the centroid where the stress barely matters.