Interactive Lecture
Centroids, Second Moments of Area, and the Parallel Axis Theorem
by Dr Arnaldo Delli Carri
The centroid of a shape is the geometric centre of the shape - the balancing point if it were made of a uniform material.
For a composite shape made up of multiple simple shapes, the centroid can be found using:
\[ y_c = \frac{\sum A_i y_i}{\sum A_i} \]Where \(y_c\) is the y-coordinate of the centroid, \(A_i\) is the area of each component, and \(y_i\) is the centroid of each component.
The Second Moment of Area (also called Moment of Inertia) measures how far a shape's area is distributed from an axis. It determines a beam's resistance to bending.
For a simple rectangle about its centroidal x-axis:
\[ I_x = \frac{bh^3}{12} \]Where \(b\) is the width and \(h\) is the height perpendicular to the x-axis.
For composite shapes, we use the Parallel Axis Theorem:
\[ I = I_c + Ad^2 \]Where \(I_c\) is the moment about the centroidal axis, \(A\) is the area, and \(d\) is the distance between axes.
The following formulas give the second moment of area about the centroidal axes (x and y) for common shapes:
\[ I_x = \frac{bh^3}{12} \quad \text{and} \quad I_y = \frac{hb^3}{12} \]
\[ I_x = I_y = \frac{\pi D^4}{64} \]
\[ I_x = \frac{bh^3}{36} \]
Use the parallel axis theorem to calculate the total second moment by summing contributions from flanges and web. See the interactive example below!
Select a section type from the dropdown menu and adjust the dimensions using the sliders. Observe how changing the dimensions affects the centroid location and second moment of area.
Key observations to note:
The second moment of area shows up wherever bending matters:
These principles are used in the design of structural elements in buildings, bridges, mechanical components, and more.
| Concept | Equation | Description |
|---|---|---|
| Rectangle | \(I = \frac{bh^3}{12}\) | Centroidal horizontal axis |
| Circle | \(I = \frac{\pi D^4}{64}\) | Diameter \(D\) |
| Parallel Axis Theorem | \(I = I_c + Ad^2\) | Any parallel axis |
| Bending Stress | \(\sigma = \frac{My}{I}\) | Stress in bending |
| Centroid | \(y_c = \frac{\sum A_i y_i}{\sum A_i}\) | Composite shapes |
A T-section beam has a flange width of 200 mm, flange thickness of 25 mm, web width of 20 mm, and web height of 175 mm. Calculate:
Show Solution
Given: Flange: \(b_f = 200 \text{ mm}, t_f = 25 \text{ mm}\); Web: \(b_w = 20 \text{ mm}, h_w = 175 \text{ mm}\)
Step 1: Calculate areas and individual centroids
Flange area: \(A_1 = 200 \times 25 = 5000 \text{ mm}^2\), \(y_1 = 12.5 \text{ mm}\) from top
Web area: \(A_2 = 20 \times 175 = 3500 \text{ mm}^2\), \(y_2 = 25 + 87.5 = 112.5 \text{ mm}\) from top
Step 2: Calculate composite centroid
\(y_c = \frac{A_1 y_1 + A_2 y_2}{A_1 + A_2} = \frac{5000 \times 12.5 + 3500 \times 112.5}{5000 + 3500} = \frac{456,250}{8500} = 53.68 \text{ mm}\) from top
Step 3: Calculate second moment of area using parallel axis theorem
Flange: \(I_1 = \frac{200 \times 25^3}{12} + 5000 \times (53.68 - 12.5)^2 = 260,417 + 8,477,509 = 8,737,926 \text{ mm}^4\)
Web: \(I_2 = \frac{20 \times 175^3}{12} + 3500 \times (112.5 - 53.68)^2 = 8,932,292 + 12,110,727 = 21,043,019 \text{ mm}^4\)
Total: \(I_{xx} = 8,737,926 + 21,043,019 \approx 29,780,944 \text{ mm}^4 = 29.78 \times 10^6 \text{ mm}^4\)
Answer: Centroid \(y_c = 53.68\) mm from the top; \(I_{xx} \approx 29.78 \times 10^6\) mm⁴
Two rectangular beams have the same cross-sectional area of 12,000 mm². Beam A is 100 mm wide and 120 mm deep. Beam B is 60 mm wide and 200 mm deep. Which beam has greater resistance to bending about its horizontal axis?
Show Solution
Beam A: \(b = 100 \text{ mm}, h = 120 \text{ mm}\)
\(I_A = \frac{bh^3}{12} = \frac{100 \times 120^3}{12} = 14.4 \times 10^6 \text{ mm}^4\)
Beam B: \(b = 60 \text{ mm}, h = 200 \text{ mm}\)
\(I_B = \frac{bh^3}{12} = \frac{60 \times 200^3}{12} = 40.0 \times 10^6 \text{ mm}^4\)
Beam B has a second moment of area 2.78 times larger than Beam A, making it much more resistant to bending despite having the same cross-sectional area. Distributing material farther from the neutral axis significantly increases bending resistance.
Answer: Beam B (\(I_B = 40.0 \times 10^6\) mm⁴ vs \(I_A = 14.4 \times 10^6\) mm⁴, 2.78× greater) resists bending better.
An I-beam has the following dimensions: flange width = 150 mm, flange thickness = 20 mm (top and bottom), web thickness = 15 mm, total height = 250 mm. Calculate the second moment of area about the centroidal x-axis and determine the maximum bending stress if a moment of 50 kN·m is applied.
Show Solution
Step 1: Calculate second moment of area
For a symmetric I-beam, the centroid is at mid-height: \(y_c = 125 \text{ mm}\)
Top flange: \(I_1 = \frac{150 \times 20^3}{12} + 150 \times 20 \times (125 - 10)^2 = 100,000 + 39,675,000 = 39,775,000 \text{ mm}^4\)
Web: \(I_2 = \frac{15 \times 210^3}{12} = 11,576,250 \text{ mm}^4\)
Bottom flange: \(I_3 = 39,775,000 \text{ mm}^4\) (by symmetry)
Total: \(I_{xx} = 39,775,000 + 11,576,250 + 39,775,000 = 91,126,250 \text{ mm}^4 = 91.13 \times 10^6 \text{ mm}^4\)
Step 2: Calculate maximum bending stress
Convert the moment to consistent units: \(M = 50 \text{ kN·m} = 50 \times 10^3 \text{ N} \times 10^3 \text{ mm} = 50 \times 10^6 \text{ N·mm}\)
\(\sigma_{max} = \frac{My_{max}}{I} = \frac{50 \times 10^6 \times 125}{91,126,250} = 68.6 \text{ MPa}\)
Answer: \(I_{xx} \approx 91.13 \times 10^6\) mm⁴; \(\sigma_{max} \approx 68.6\) MPa