Centroids and Moments of Inertia
Key Takeaways
- The centroid is the geometric center of an area — the point where the area would balance on a pin.
- Composite area centroids: x̄ = Σ(Aᵢxᵢ) / ΣAᵢ, ȳ = Σ(Aᵢyᵢ) / ΣAᵢ.
- Moment of inertia (second moment of area) I measures resistance to bending; larger I = stiffer.
- Parallel Axis Theorem: I = Ī + Ad² transfers moment of inertia to a non-centroidal axis.
- For composite shapes: find I of each part about the desired axis, then add (for solids) or subtract (for holes).
- The radius of gyration k = √(I/A) is a measure of how far the area is distributed from the axis.
Centroids and Moments of Inertia
Centroids of Common Shapes
| Shape | Area | x̄ (from left) | ȳ (from bottom) |
|---|---|---|---|
| Rectangle (b × h) | bh | b/2 | h/2 |
| Triangle (base b, height h) | bh/2 | b/3 (from vertex) | h/3 |
| Circle (radius r) | πr² | r | r |
| Semicircle (radius r) | πr²/2 | r | 4r/(3π) from flat edge |
| Quarter circle (radius r) | πr²/4 | 4r/(3π) | 4r/(3π) |
Composite Area Centroids
For a composite area made of n simple shapes:
For areas with holes: Subtract the area and first moment of the hole.
Example: An L-shaped cross-section consists of:
- Rectangle 1: 100 mm × 20 mm (horizontal flange), centroid at (50, 10)
- Rectangle 2: 20 mm × 80 mm (vertical web), centroid at (10, 60)
A₁ = 2,000 mm², A₂ = 1,600 mm², A_total = 3,600 mm²
x̄ = (2,000 × 50 + 1,600 × 10) / 3,600 = 116,000/3,600 = 32.2 mm
ȳ = (2,000 × 10 + 1,600 × 60) / 3,600 = 116,000/3,600 = 32.2 mm
Moment of Inertia (Second Moment of Area)
The moment of inertia measures how the area is distributed relative to an axis.
Moments of Inertia for Common Shapes (about centroidal axes)
| Shape | Īx (about centroidal x) | Īy (about centroidal y) |
|---|---|---|
| Rectangle (b × h) | bh³/12 | b³h/12 |
| Triangle (base b, height h) | bh³/36 | b³h/36 (about vertex side) |
| Circle (radius r) | πr⁴/4 | πr⁴/4 |
| Semicircle (radius r) | 0.1098r⁴ | πr⁴/8 |
Parallel Axis Theorem
To find the moment of inertia about any axis parallel to the centroidal axis:
where:
- Ī = moment of inertia about the centroidal axis
- A = area of the shape
- d = distance between the centroidal axis and the desired axis
This is the most important formula in this section. Nearly every composite cross-section problem requires the parallel axis theorem.
Composite Cross-Section Moment of Inertia
Steps:
- Find the overall centroid of the composite section
- For each part, calculate Ī (centroidal moment of inertia)
- Use the parallel axis theorem to transfer each Ī to the composite centroid
- Sum all transferred moments of inertia
Radius of Gyration
The radius of gyration represents the distance from the axis at which all the area could be concentrated to produce the same moment of inertia. It is important in column buckling analysis.
Product of Inertia
- Ixy = 0 for shapes symmetric about either axis
- Used to find principal axes and maximum/minimum moments of inertia
- Parallel axis theorem: Ixy = Īxy + A·dx·dy
What is the centroidal moment of inertia Ix of a rectangle with base 6 cm and height 4 cm?
Using the parallel axis theorem, what is Ix of a 4 cm × 2 cm rectangle about an axis 5 cm below its centroid?
A composite section is made by removing a circular hole (r = 2 cm) from a 10 × 10 cm square. The centroid of the composite section is: