Centroid
The centroid is the geometric centre of an area or volume; the centre of gravity is where weight acts. Learn integral and composite-area methods, standard shape centroids, and why centroids matter for beams, stability and loads.
When a distributed quantity — weight, area, pressure — acts over a region, it can be replaced by a single resultant at a special point. For area that point is the centroid; for weight it's the centre of gravity (the same for uniform materials).
Centroids decide where the neutral axis of a beam lies, whether a vehicle or ship is stable, and where equivalent loads act.
An analogy
The centroid of a flat shape is the point where you could balance it on a pencil tip.
Symmetry first
If a shape has an axis of symmetry, the centroid lies on it. Rectangles and circles have centroids at their centres; I-sections at mid-height. Use symmetry to skip half the calculation.
Composite shapes
Split a complex section into simple rectangles, triangles and circles. Take moments of area about a reference axis: ΣA·y. Holes are treated as negative areas. Divide by total area.
Centre of gravity and mass
For bodies made of different materials, weight each part by its mass: x̄ = Σm_i x_i / Σm_i. Vehicle CG height decides rollover and braking weight transfer; aircraft CG must stay within limits for stability.
Pappus–Guldinus theorems
The surface area of a revolved curve equals its length times the distance travelled by its centroid; the volume of a revolved area equals the area times the centroid's path length. Quick for tanks, pulleys and toroidal parts.
Formulas
- Composite area:
x̄ = ΣA_i x_i / ΣA_i, ȳ = ΣA_i y_i / ΣA_i - Integral definition:
x̄ = ∫x dA / ∫dA - Triangle:
ȳ = h/3 from base - Semicircle:
ȳ = 4r/(3π) from diameter
Worked example: Centroid of a T-section
Given: Flange 100 × 20 mm on top; Web 20 × 80 mm below
Find: ȳ from the bottom.
- Flange: A = 2000 mm², y = 80 + 10 = 90 mm
- Web: A = 1600 mm², y = 40 mm
- ȳ = (2000 × 90 + 1600 × 40)/3600 = 67.8 mm
Answer: ȳ = 67.8 mm
Worked example: Centroid of an equal angle
Given: L 100 × 100 × 10 mm
Find: x̄ = ȳ.
- Vertical leg 100 × 10: A = 1000, x = 5
- Remaining horizontal 90 × 10: A = 900, x = 10 + 45 = 55
- x̄ = (1000 × 5 + 900 × 55)/1900 = 28.7 mm (= ȳ by symmetry)
Answer: 28.7 mm from each back face
Common mistakes
- Wrong: Measuring component centroids from different reference axes. — Right: Use one common reference axis for every y_i.
- Wrong: Adding holes as positive areas. — Right: Subtract holes (negative area and negative moment).
Applications
- Beam design: The neutral axis passes through the section centroid, so bending stresses are measured from it.
- Vehicle & ship stability: CG height versus track width (vehicles) or metacentre (ships) decides rollover and capsize margins.
- Lifting: Lifting points are placed above the CG so loads hang level.
Key takeaways
- x̄ = ΣA_i x_i / ΣA_i; use a single reference axis.
- Exploit symmetry; subtract holes.
- Centroid = balance point; CG uses mass weighting.
- Centroids locate neutral axes and stability margins.
Going further
- CAD mass properties: CAD tools compute centroid, mass and inertia tensors exactly from solid models — the standard source for CG in product design.
- Hydrostatics: The centre of pressure on a submerged plate lies below its centroid; dam gates and tank walls are designed from it.
- Shear centre: For open thin-walled sections (channels), loads must pass through the shear centre, not the centroid, to avoid twisting.