Centroids and Moments of Inertia

Centroids of Areas

The centroid is the geometric center of an area — the point where the area would balance on a pin and the location of the resultant of a distributed load or self-weight. By integration, x̄ = ∫x dA / ∫dA. In practice you almost never integrate on the FE exam; you look up standard shapes and combine them.

Composite-Area Centroids

Split the shape into simple parts, then take the area-weighted average of the part centroids:

$\bar{x} = \frac{\sum A_i \bar{x}_i}{\sum A_i}, \qquad \bar{y} = \frac{\sum A_i \bar{y}_i}{\sum A_i}$

Treat any hole as a negative area so it subtracts from both the numerator and denominator. A shape with an axis of symmetry has its centroid on that axis — use symmetry to halve your work.

Worked Example — L-Section Centroid

An L-shaped cross-section is made of two rectangles:

• Part 1 (horizontal flange): 100 mm × 20 mm, area A₁ = 2000 mm², centroid at (x̄₁, ȳ₁) = (50, 10) mm.
• Part 2 (vertical web): 20 mm × 80 mm, area A₂ = 1600 mm², centroid at (10, 60) mm.

Total area = 3600 mm².

$\bar{x} = \frac{2000(50) + 1600(10)}{3600} = \frac{116{,}000}{3600} = 32.2\text{ mm}$ $\bar{y} = \frac{2000(10) + 1600(60)}{3600} = \frac{116{,}000}{3600} = 32.2\text{ mm}$

The centroid lies at (32.2, 32.2) mm — inside the re-entrant corner, as expected for an L.

Moment of Inertia (Second Moment of Area)

The moment of inertia quantifies how an area is distributed about an axis and therefore how stiff a beam cross-section is in bending:

$I_x = \int y^2\,dA, \qquad I_y = \int x^2\,dA$

Key centroidal formulas:

Notice the cube on h: doubling a beam's depth multiplies its bending stiffness by eight. That is why I-beams place flanges far from the neutral axis.

Parallel-Axis Theorem

To move a moment of inertia from the centroidal axis to any parallel axis a distance d away:

$I = \bar{I} + A d^2$

This is the workhorse of composite-section analysis. Procedure for a built-up section: (1) locate the overall centroid, (2) compute each part's centroidal Ī, (3) transfer each to the composite centroid with +Aᵢdᵢ², (4) sum: I_total = Σ(Īᵢ + Aᵢdᵢ²).

Worked Example — Parallel-Axis Transfer

A 40 mm × 20 mm rectangle (b = 40, h = 20) has its centroid 50 mm above a reference axis. Find Ix about that axis.

Ī = bh³/12 = 40(20)³/12 = 40(8000)/12 = 26{,}667 mm⁴. A = 40 × 20 = 800 mm²; d = 50 mm. Ix = Ī + Ad² = 26{,}667 + 800(50)² = 26{,}667 + 2{,}000{,}000 = 2,026,667 mm⁴.

The Ad² term dominates — when material is far from the axis, the transfer term, not the local Ī, controls. Trap: d must be measured to the part's own centroid, and you may transfer only between an axis through the centroid and a parallel one — never directly between two arbitrary non-centroidal axes.

Related Quantities

• Polar moment of inertia: J = Ix + Iy (perpendicular-axis theorem). For a solid circle, J = πr⁴/2; it governs shaft torsion.
• Radius of gyration: k = √(I/A). It collapses I into an equivalent distance and appears directly in the column slenderness ratio L/k for buckling.
• Product of inertia: Ixy = ∫xy dA, which is zero whenever the area is symmetric about either axis; it is used to find principal axes.

Worked Example — Composite I of a Built-Up Section

Find Ix about the centroidal axis of a symmetric I-shape: two flanges 100 mm × 20 mm and a web 20 mm × 160 mm, total depth 200 mm. By symmetry the centroid is at mid-depth, so each flange centroid sits 90 mm from the neutral axis (100 mm to the flange center minus 10 mm half-thickness).

Web: Ī = bh³/12 = 20(160)³/12 = 20(4.096×10⁶)/12 = 6.83×10⁶ mm⁴ (web centroid is on the axis, so no transfer).

Each flange: Ī = 100(20)³/12 = 0.067×10⁶ mm⁴; transfer Ad² = (100×20)(90)² = 2000(8100) = 16.2×10⁶ mm⁴, giving 16.27×10⁶ mm⁴ per flange.

Total: Ix = 6.83 + 2(16.27) = 39.4×10⁶ mm⁴. Note the flanges contribute the bulk of the stiffness — the Ad² transfer terms dwarf the small local Ī values, confirming why flanges are placed far from the neutral axis.

Why These Quantities Matter on the Exam

Centroids and moments of inertia are the bridge from Statics into Strength of Materials. The bending-stress formula σ = Mc/I needs both the centroid (to find c, the distance to the extreme fiber) and I. The column-buckling formula uses the radius of gyration k = √(I/A). Torsion uses the polar moment J. Expect at least one or two FE questions that ask only for a composite centroid or a parallel-axis transfer, plus several downstream questions that depend on getting these geometric properties right first.