Find the centroid and second moment of area of a built-up section with the parallel axis theorem.
Why it matters
The shape of a cross-section matters as much as the material. A plank laid flat sags easily; the same plank on edge is many times stiffer. Steel I-beams put most of their material far from the middle for the same reason. Two section properties capture this: the centroid (the geometric centre of the area, where the bending axis passes) and the second moment of area I (often just called the moment of inertia of the section), which measures how far the area is spread from that axis. Bending stress and deflection, two lessons from now, both depend on I.
The centroid of a composite area
For a shape built from simple pieces (rectangles, circles, triangles), the centroid is the area-weighted average of the pieces' centroids:
ȳ = Σ(Ai · yi) / Σ Ai
where Ai is the area of piece i and yi is the distance from a reference line (say, the bottom of the section) to that piece's own centroid. Holes are included as pieces with negative area. The centroid of a rectangle is at its middle; of a triangle, one third of the height from its base.
The second moment of area
For a rectangle of width b and height h, about a horizontal axis through its own centroid:
I = b·h³ / 12
The h³ is why the plank on edge is so much stiffer. A 50 mm × 150 mm board on edge has I = 50 · 150³ / 12 = 14.1 × 10⁶ mm⁴. Laid flat (b = 150, h = 50) it has I = 150 · 50³ / 12 = 1.56 × 10⁶ mm⁴: nine times less, for the same amount of wood.
Other common results: a solid circle of diameter d has I = π·d⁴ / 64.
The parallel axis theorem
The I of each piece must be about the whole section's centroid, not its own. The parallel axis theorem shifts it:
I = I_own + A · d²
where d is the distance between the piece's own centroid and the section centroid. The A·d² term is usually the big one, which is why material far from the axis is so effective.
Worked example: a T-section
A T-section has a flange 100 mm wide and 20 mm thick sitting on top of a web 20 mm wide and 100 mm tall. Measure y from the bottom of the web.
| piece | b × h (mm) | A (mm²) | y (mm) | A·y (mm³) |
|---|---|---|---|---|
| web | 20 × 100 | 2,000 | 50 | 100,000 |
| flange | 100 × 20 | 2,000 | 110 | 220,000 |
| total | 4,000 | 320,000 |
ȳ = 320,000 / 4,000 = 80 mm from the bottom
Now the second moment of area about the horizontal axis through ȳ:
web: I_own = 20·100³/12 = 1,666,667 d = 80 − 50 = 30 A·d² = 2,000·900 = 1,800,000
flange: I_own = 100·20³/12 = 66,667 d = 110 − 80 = 30 A·d² = 1,800,000
I = 1,666,667 + 1,800,000 + 66,667 + 1,800,000 = 5,333,333 mm⁴ ≈ 5.33 × 10⁶ mm⁴
The flange's own I is tiny (it is thin), but its A·d² term is large because it sits 30 mm from the axis.
Units
Area is mm² or m²; the first moment A·y is mm³; the second moment I is mm⁴ or m⁴. Convert carefully: 1 m⁴ = 10¹² mm⁴. Steel section tables usually list I in mm⁴ × 10⁶ or in in⁴.
Common mistakes
- Forgetting the parallel axis term, or adding I_own values about different axes directly.
- Using d from the reference line instead of from the section centroid.
- Swapping b and h. h is always the dimension perpendicular to the bending axis, and it is the one that is cubed.
- Confusing I with mass moment of inertia. The second moment of area (mm⁴) is a geometric property used for bending; the mass moment of inertia (kg·m²) is used for rotation in dynamics.
Recap
- Centroid: ȳ = Σ(Ai·yi) / ΣAi.
- Rectangle: I = b·h³/12 about its own centroid.
- Parallel axis: I = I_own + A·d², with d measured to the section centroid.
- Putting area far from the axis increases I much faster than adding area near it.
Try it
On the Do tab, finish the parallel-axis step in a function that computes the centroid and I of any section made of rectangles, then check the lesson's T-section.