Area Moment of Inertia

What I actually means

For a cross-section in the y-z plane bending about the z-axis, the area moment of inertia is

      I_zz = ∫∫ y² dA

where y is the distance from the neutral (centroidal) axis and dA is a differential area element. Each scrap of cross-section contributes its area times its distance from the axis squared. A scrap 10 mm from the neutral axis contributes 100× as much as one 1 mm away — the quadratic scaling is the whole story.

The role of I in beam bending shows up in two formulas:

      Stress:    σ = M·y / I           (extreme fiber σ_max at y = c)
      Curvature: κ = M / (E·I)
      Deflection (centre point load on simply-supported span):
                 δ = W·L³ / (48·E·I)

Stiffness scales as EI; deflection scales as 1/I. Doubling I halves the deflection.

Worked example: a rectangular beam

For a rectangle of width b and depth h with the bending axis horizontal (z-axis through the centroid):

      I_zz = ∫_{-h/2}^{h/2} y² · b dy = b · [y³/3]_{-h/2}^{h/2} = b · h³ / 12

For a 100 mm × 200 mm timber joist:

      I = b·h³/12 = 100 × 200³ / 12 = 100 × 8,000,000 / 12 = 6.67 × 10⁷ mm⁴
        = 6.67 × 10⁻⁵ m⁴

If you flip the joist on its side (200 mm wide, 100 mm deep) the depth in the bending direction is now 100 mm instead of 200, so:

      I_flat = 200 × 100³ / 12 = 200 × 1,000,000 / 12 = 1.67 × 10⁷ mm⁴

The joist on edge is 4× stiffer than the joist flat — same volume of wood, same total area, just a different orientation. This is why floor joists are installed on edge.

I for standard cross-sections

Shape	I about centroidal axis	For 100 mm characteristic	Best feature
Rectangle (b × h)	bh³/12	I = 8.33×10⁶ mm⁴ (100×100)	Cheap, isotropic in plane
Solid circle (D)	πD⁴/64	I = 4.91×10⁶ mm⁴ (D=100)	Same I about every axis
I-beam (W14×90 equivalent)	bh³/12 − (b−tw)hw³/12	I ≈ 4.16×10⁸ mm⁴ (h=350)	Mass at extreme fibers
T-section	Mixed; centroid offset; ∑(I_i + A_i d_i²)	Computed by parts	Asymmetric, mixed I_x ≠ I_y
Hollow square (b, t)	(b⁴ − (b−2t)⁴)/12	I = 6.07×10⁶ mm⁴ (b=100, t=5)	Equal I about both axes
Circular tube (D_o, D_i)	π(D_o⁴ − D_i⁴)/64	I = 2.93×10⁶ mm⁴ (D_o=100, t=5)	High torsional + bending stiffness

Why I-beams dominate

An I-beam concentrates material in the flanges (top and bottom), connected by a thin web. Because I integrates y², the flanges — being far from the neutral axis — contribute most of the moment of inertia. The web's job is just to keep the flanges spaced apart and to carry shear; it can be thin without losing bending stiffness.

Consider a standard W14×90 wide-flange — a steel I-beam with 14-inch nominal depth weighing 90 lb/ft. Its actual dimensions are: depth 14.0 in, flange width 14.5 in, flange thickness 0.71 in, web thickness 0.44 in. Cross-section area: 26.5 in². Moment of inertia about the strong axis: I_xx = 999 in⁴ = 4.16 × 10⁸ mm⁴.

A solid square of the same 26.5 in² cross-sectional area would be 5.15 in × 5.15 in. Its moment of inertia is

      I_solid = (5.15)⁴ / 12 = 58.6 in⁴

The W14×90 has 999 in⁴ vs the solid square's 58.6 in⁴ — 17× more bending stiffness for the same weight of steel. That's the I-beam's advantage in one number.

          ▓▓▓▓▓▓▓▓▓▓▓▓▓     ← top flange (most of I)
                ║
                ║          ← web (carries shear)
                ║
                ║
          ▓▓▓▓▓▓▓▓▓▓▓▓▓     ← bottom flange (most of I)

The parallel axis theorem

To find I about an axis other than the centroid, use:

      I_axis = I_centroid + A · d²

where d is the distance between the centroidal axis and the new axis, and A is the cross-section's area. The Ad² term grows quadratically with offset — which is the engineering reason composite cross-sections (sandwich panels, hollow steel decks) work so well.

Example: built-up I-beam from rectangles. Suppose two 200 mm × 30 mm flanges sit at top and bottom of a 250 mm × 10 mm web. Each flange's centroid is 140 mm above and below the overall centroid. For one flange:

      I_flange_centroid = 200 × 30³ / 12 = 4.5 × 10⁵ mm⁴
      Area · d² = (200 × 30) × 140² = 6,000 × 19,600 = 1.176 × 10⁸ mm⁴
      Total per flange ≈ 1.18 × 10⁸ mm⁴ (dominated by Ad²)

Two flanges together: 2.36 × 10⁸ mm⁴. The web's I about its own centroid is 10 × 250³ / 12 = 1.30 × 10⁷ mm⁴. Sum: ≈ 2.49 × 10⁸ mm⁴. Notice that the flanges' bare centroidal I (9 × 10⁵ total) is negligible — almost all the bending stiffness comes from the Ad² offset of the flange material from the neutral axis.

I_xx, I_yy, I_polar, principal

A real cross-section has multiple moments of inertia depending on the axis you pick:

• I_xx — about the horizontal centroidal axis. Resists bending in the vertical plane. The "strong axis" for an I-beam.
• I_yy — about the vertical centroidal axis. Resists bending in the horizontal plane. The "weak axis" for an I-beam.
• I_xy (product of inertia) — non-zero for asymmetric sections. Couples bending in two directions.
• J = I_polar = I_xx + I_yy — about the axis perpendicular to the cross-section. Resists torsion. The perpendicular-axis theorem makes this exact only for thin plates; for solid 3D bodies it's an upper bound.
• I_principal — the I_xx and I_yy in the rotated frame where I_xy = 0. For symmetric sections these align with the geometric axes; for an L-angle they don't.

For an L-shaped angle iron (asymmetric section), I_xx and I_yy about the geometric axes are not the principal moments. The product of inertia I_xy is non-zero, meaning a vertical load on a horizontal angle bends it both vertically and sideways simultaneously — a coupling that surprises designers who haven't run a principal-axis analysis.

Real-world numbers

• 2x4 lumber (1.5 × 3.5 in actual): I_strong = 1.5 × 3.5³ / 12 = 5.36 in⁴. I_weak = 0.98 in⁴. The 5.5× ratio explains why studs go on edge in walls.
• Standard steel pipe NPS 4 (4.5 in OD, 4.026 in ID): I = π(4.5⁴ − 4.026⁴)/64 = 7.23 in⁴. About the same as a 2x6 solid wood (I = 20.8 in⁴) — but the steel pipe is hollow and dramatically stiffer per pound.
• Aircraft wing spar (carbon-fiber I-section, 250 mm tall): typical I ≈ 5×10⁹ mm⁴, achieved with under 200 g of cross-section per metre. The composite layup puts unidirectional fibers exactly where the y² integral pays.
• Bicycle frame tubing (aluminum, 50 mm OD, 1.5 mm wall): I = π(50⁴ − 47⁴)/64 = 6.4×10⁴ mm⁴. Doubling diameter to 100 mm at the same wall would give I ≈ 5.5×10⁵ mm⁴ — 8.5× stiffer for 2× the material — explaining the trend toward oversized tubes.

Common design errors

• Using strong-axis I for biaxial loads. A column laterally loaded both ways must check I about both axes. The weak axis usually governs. I-beams in wind-loaded conditions need bracing perpendicular to the strong axis or they buckle sideways.
• Ignoring the Ad² term. When summing I for built-up sections, beginners sometimes add only the centroidal I of each piece and forget the parallel-axis offset. The error can be 100× for sections where the offsets are large.
• Confusing I (area moment) with mass moment of inertia. They're different quantities — area moment has units of length⁴, mass moment has length²·mass. Mass moment governs angular acceleration in dynamics; area moment governs bending in statics.
• Using I_xx for an asymmetric section. An angle iron loaded vertically also bends horizontally because I_xy ≠ 0. Calculating only the geometric-axis I and assuming planar bending under-predicts deflection by 30–50% in L-shapes and channels.
• Forgetting shear flow in built-up sections. Splitting a solid beam into stacked planks doesn't preserve I — slip between layers reduces effective depth. Bonded laminates, bolted plates, or welded built-ups must transmit horizontal shear along the bond line; if they slip, the effective I drops sharply.
• Local buckling of thin-walled sections. A high I from thin flanges far from the neutral axis only delivers full stiffness if the flange doesn't buckle locally. Slenderness limits in steel codes (b/t ≤ ~16 for compact flanges) prevent this failure mode.