What is the moment of inertia of a rectangle?

About the centroidal horizontal axis Ix = b·h³/12, and about the centroidal vertical axis Iy = h·b³/12, where b is the width and h the height of the section.

How do I get the moment of inertia about the base?

Use the parallel-axis theorem: I = Ix + A·d², with A = b·h and d = h/2 the distance from the centroid to the base. That gives I = b·h³/3.

Why does Ix depend on the cube of the height?

The integrand carries the square of the distance to the axis, and integrating along the height contributes one more power. Doubling the height therefore multiplies Ix by eight.

Moment of inertia of a rectangle: deriving Ix = bh³/12

The moment of inertia of a rectangular section

The moment of inertia is one of the key properties of a cross-section: it describes how the area is distributed about a chosen axis. For a rectangle — the most common shape in engineering calculations — the formula is easy to derive analytically, yet it rests on a fundamental integral definition.

The definition

For a plane figure the second moments of area about the x and y axes are defined as area integrals:

I_{x} = \int_{A} y^{2} d A, I_{y} = \int_{A} x^{2} d A

Here:

A — the area of the figure
x, y — the coordinates of a point inside that area
Ix — the moment of inertia about the x axis
Iy — the moment of inertia about the y axis

Setting up the problem

Take a rectangle of width b and height h placed symmetrically about the origin: its centre is at (0, 0) and its sides are parallel to the axes. The integration limits are then x from −b/2 to b/2 and y from −h/2 to h/2.

A rectangle of width b and height h with the centroidal x and y axes

Deriving Ix

I_{x} = \int_{- \frac{b}{2}}^{\frac{b}{2}} \int_{- \frac{h}{2}}^{\frac{h}{2}} y^{2} d y d x

First evaluate the inner integral over y:

\int_{- \frac{h}{2}}^{\frac{h}{2}} y^{2} d y = \frac{h^{3}}{12}

Then integrate over x, where the integrand no longer depends on x:

I_{x} = \frac{h^{3}}{12} \int_{- \frac{b}{2}}^{\frac{b}{2}} d x = \frac{b h^{3}}{12}

Deriving Iy

The argument is perfectly symmetric: the square is now taken of the x coordinate, so the width enters as a cube.

I_{y} = \int_{- \frac{h}{2}}^{\frac{h}{2}} \int_{- \frac{b}{2}}^{\frac{b}{2}} x^{2} d x d y = \frac{h b^{3}}{12}

What the result means

The formulas show that Ix leans on the height h, which is cubed, while Iy leans on the width b. That matches the intuition: the farther the area lies from an axis, the harder the figure is to rotate about it. A tall, narrow section therefore has a large moment of inertia about the horizontal axis and a small one about the vertical — which is exactly why a joist is installed on edge.

Where it is used

mechanics — resistance to bending and torsion
structural engineering — beams, trusses, bridges
machine design — shafts, frames, brackets
simulation — rigid body behaviour in physics engines

For arbitrary shapes the moment of inertia is computed numerically: the outline is split into triangles and their contributions are summed — the approach used by CAD and FEM packages, and by the calculator on this site. For simple shapes such as a rectangle, though, the closed-form result is instant and exact.