Calculating the Centroid of Compound Shapes Using the Method of Geometric Decomposition
Centroids of simple geometric shapes
John Ray Cuevas
What Is a Centroid?
A centroid is the central point of a figure and is also called the geometric center. It is the point that matches the center of gravity of a particular shape. It is the point which corresponds to the mean position of all the points in a figure. The centroid is the term for 2-dimensional shapes. The center of mass is the term for 3-dimensional shapes. For instance, the centroid of a circle and a rectangle is in the middle. The centroid of a right triangle is 1/3 from the bottom and the right angle. But how about the centroid of compound shapes?
What Is Geometric Decomposition?
Geometric Decomposition is one of the techniques used in obtaining the centroid of a compound shape. It is a widely used method because the computations are simple, and requires only basic mathematical principles. It is called geometric decomposition because the calculation comprises decomposing the figure into simple geometric figures. In geometric decomposition, dividing the complex figure Z is the fundamental step in calculating the centroid. Given a figure Z, obtain the centroid Ci and area Ai of each Zn part wherein all holes that extend outside the compound shape are to be treated as negative values. Lastly, compute the centroid given the formula:
Cx = ∑Cix Aix / ∑Aix
Cy = ∑Ciy Aiy / ∑Aiy
Step-By-Step Procedure in Solving for the Centroid of Compound Shapes
Here are the series of steps in solving for the centroid of any compound shape.
1. Divide the given compound shape into various primary figures. These basic figures include rectangles, circles, semicircles, triangles and many more. In dividing the compound figure, include parts with holes. These holes are to treat as solid components yet with negative values. Make sure that you break down every part of the compound shape before proceeding to the next step.
2. Solve for the area of each divided figure. Table 1-2 below shows the formula for different basic geometric figures. After determining the area, designate a name (Area one, area two, area three, etc.) for each area. Make the area negative for designated areas that act as holes.
3. The given figure should have an x-axis and y-axis. If x and y-axes are missing, draw the axes in the most convenient means. Remember that x-axis is the horizontal axis while the y-axis is the vertical axis. You can position your axes in the middle, left, or right.
4. Get the distance of the centroid of each divided primary figure from the x-axis and y-axis. Table 1-2 below shows the centroid for different basic shapes.
Centroid for Common Shapes
| Shape | Area | X-bar | Y-bar |
|---|---|---|---|
Rectangle | bh | b / 2 | d / 2 |
Triangle | (bh) / 2 | - | h / 3 |
Right triangle | (bh) / 2 | h / 3 | h / 3 |
Semicircle | (pi (r^2)) / 2 | 0 | (4r) / (3(pi)) |
Quarter circle | (pi (r^2)) / 4 | (4r) / (3(pi)) | (4r) / (3(pi)) |
Circular sector | (r^2) (alpha) | (2rsin(alpha)) / 3(alpha) | 0 |
Segment of arc | 2r(alpha) | (rsin(alpha)) / alpha | 0 |
Semicircular arc | (pi) (r) | (2r) / pi | 0 |
Area under spandrel | (bh) / (n + 1) | b / (n+2) | (hn + h) / (4n + 2) |
5. Creating a table always makes computations easier. Plot a table like the one below.
| Area Name | Area (A) | x | y | Ax | Ay |
|---|---|---|---|---|---|
Area 1 | - | - | - | Ax1 | Ay1 |
Area 2 | - | - | - | Ax2 | Ay2 |
Area n | - | - | - | Axn | Ayn |
Total | (Total Area) | - | - | (Summation of Ax) | (Summation of Ay) |
6. Multiply the area 'A' of each basic shape by the distance of the centroids 'x' from the y-axis. Then get the summation ΣAx. Refer to the table format above.
7. Multiply the area 'A' of each basic shape by the distance of the centroids 'y' from the x-axis. Then get the summation ΣAy. Refer to the table format above.
8. Solve for the total area ΣA of the whole figure.
Recommended
9. Solve for the centroid Cxof the whole figure by dividing the summation ΣAx by the total area of the figure ΣA. The resulting answer is the distance of the entire figure's centroid from the y-axis.
10. Solve for the centroid Cy of the whole figure by dividing the summation ΣAy by the total area of the figure ΣA. The resulting answer is the distance of the entire figure's centroid from the x-axis.
Here are some examples of obtaining a centroid.
Centroid for complex figures: C-shapes
John Ray Cuevas
Problem 1: Centroid of C-Shapes
Solution 1
a. Divide the compound shape into basic shapes. In this case, the C-shape has three rectangles. Name the three divisions as Area 1, Area 2, and Area 3.
b. Solve for the area of each division. The rectangles have dimensions 120 x 40, 40 x 50, 120 x 40 for Area 1, Area 2, and Area 3 respectively.
Area 1 = b x h
Area 1 = 120.00 mm x 40.00 mm
Area 1 = 4800.00 square millimeters
Area 2 = b x h
Area 2 = 40.00 mm x 50.00 mm
Area 2 = 2000 square millimeters
Area 3 = b x h
Area 3 = 120.00 mm x 40.00 mm
Area 3 = 4800.00 square millimeters
∑A = 4800 + 2000 + 4800
∑A = 11600.00 square millimetersc. X and Y distances of each area. X distances are the distances of each area's centroid from the y-axis, and Y distances are the distances of each area's centroid from the x-axis.
Centroid for C-shapes
John Ray Cuevas
Area 1:
x = 60.00 millimeters
y = 20.00 millimeters
Area 2:
x = 100.00 millimeters
y = 65.00 millimeters
Area 3:
x = 60 millimeters
y = 110 millimetersd. Solve for the Ax values. Multiply the area of each region by the distances from the y-axis.
Ax1 = 4800.00 square mm x 60.00 mm
Ax1 = 288000 cubic millimeters
Ax2 = 2000.00 square mm x 100.00 mm
Ax2 = 200000 cubic millimeters
Ax3 = 4800.00 square mm x 60.00 mm
Ax3 = 288000 cubic millimeters
∑Ax = 776000 cubic millimeterse. Solve for the Ay values. Multiply the area of each region by the distances from the x-axis.
Ay1 = 4800.00 square mm x 20.00 mm
Ay1 = 96000 cubic millimeters
Ay2 = 2000.00 square mm x 65.00 mm
Ay2 = 130000 cubic millimeters
Ay3 = 4800.00 square mm x 110.00 mm
Ay3 = 528000 cubic millimeters
∑Ay = 754000 cubic millimeters| Area Name | Area (A) | x | y | Ax | Ay |
|---|---|---|---|---|---|
Area 1 | 4800 | 60 | 20 | 288000 | 96000 |
Area 2 | 2000 | 100 | 65 | 200000 | 130000 |
Area 3 | 4800 | 60 | 110 | 288000 | 528000 |
Total | 11600 |
|
| 776000 | 754000 |
f. Finally, solve for the centroid (Cx, Cy) by dividing ∑Ax by ∑A, and ∑Ay by ∑A.
Cx = ΣAx / ΣA
Cx = 776000 / 11600
Cx = 66.90 millimeters
Cy = ΣAy / ΣA
Cy = 754000 / 11600
Cy = 65.00 millimeters
Centroid for C-shape
John Ray Cuevas
The centroid of the complex figure is 66.90 millimeters from the y-axis and 65.00 millimeters from the x-axis.
Centroid for complex figures: Irregular figures
John Ray Cuevas
Problem 2: Centroid of Irregular Figures
Solution 2
a. Divide the compound shape into basic shapes. In this case, the irregular shape has a semicircle, rectangle, and right triangle. Name the three divisions as Area 1, Area 2, and Area 3.
b. Solve for the area of each division. The dimensions are 250 x 300 for the rectangle, 120 x 120 for the right triangle, and a radius of 100 for the semicircle. Make sure to negate the values for the right triangle and semicircle because they are holes.
Area 1 = b x h
Area 1 = 250.00 mm x 300.00 mm
Area 1 = 75000.00 square millimeters
Area 2 = 1/2 (bh)
Area 2 = 1/2 (120 mm) (120 mm)
Area 2 = - 7200 square millimeters
Area 3 = ((pi) r^2) / 2
Area 3 = ((pi) (100)^2) / 2
Area 3 = - 5000pi square millimeters
∑A = 75000.00 - 7200 - 5000pi
∑A = 52092.04 square millimetersc. X and Y distances of each area. X distances are the distances of each area's centroid from the y-axis, and y distances are the distances of each area's centroid from the x-axis. Consider the orientation of x and y-axes. For Quadrant I, x and y are positive. For Quadrant II, x is negative while y is positive.
Solution for irregular shape
John Ray Cuevas
Area 1:
x = 0
y = 125.00 millimeters
Area 2:
x = 110.00 millimeters
y = 210.00 millimeters
Area 3:
x = - 107.56 millimeters
y = 135 millimetersd. Solve for the Ax values. Multiply the area of each region by the distances from the y-axis.
Ax1 = 75000.00 square mm x 0.00 mm
Ax1 = 0
Ax2 = - 7200.00 square mm x 110.00 mm
Ax2 = - 792000 cubic millimeters
Ax3 = - 5000pi square mm x - 107.56 mm
Ax3 = 1689548.529 cubic millimeters
∑Ax = 897548.529 cubic millimeterse. Solve for the Ay values. Multiply the area of each region by the distances from the x-axis.
Ay1 = 75000.00 square mm x 125.00 mm
Ay1 = 9375000 cubic millimeters
Ay2 = - 7200.00 square mm x 210.00 mm
Ay2 = - 1512000 cubic millimeters
Ay3 = - 5000pi square mm x 135.00 mm
Ay3 = - 2120575.041 cubic millimeters
∑Ay = 5742424.959 cubic millimeters| Area Name | Area (A) | x | y | Ax | Ay |
|---|---|---|---|---|---|
Area 1 | 75000 | 0 | 125 | 0 | 9375000 |
Area 2 | - 7200 | 110 | 210 | -792000 | -1512000 |
Area 3 | - 5000pi | - 107.56 | 135 | 1689548.529 | -2120575.041 |
Total | 52092.04 |
|
| 897548.529 | 5742424.959 |
f. Finally, solve for the centroid (Cx, Cy) by dividing ∑Ax by ∑A, and ∑Ay by ∑A.
Cx = ΣAx / ΣA
Cx = 897548.529 / 52092.04
Cx = 17.23 millimeters
Cy = ΣAy / ΣA
Cy = 5742424.959 / 52092.04
Cy = 110.24 millimetersThe centroid of the complex figure is 17.23 millimeters from the y-axis and 110.24 millimeters from the x-axis.
Final answer to irregular shape
John Ray Cuevas
More Math Tutorials
- How to Calculate the Area of a Composite or Compound Shape (Rectangles, Triangles, Circles)
Learn how to calculate the area of a composite or compound shape. - Calculator Techniques for Polygons in Plane Geometry
Problems related to plane geometry, especially polygons, can be easily solved using a calculator. Here is a comprehensive set of problems about polygons solved using calculators. - How to Solve for the Moment of Inertia of Irregular or Compound Shapes
This is a complete guide in solving for the moment of inertia of compound or irregular shapes. Know the basic steps and formulas needed and master solving moment of inertia.
This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.
Questions & Answers
Question: Is there any alternative method for solving for the centroid except this geometric decomposition?
Answer: Yes, there is a technique using your scientific calculator in solving for the centroid.
Question: in area two of triangle in problem 2...how 210mm of y bar has obtained?
Answer: It is the y-distance of the centroid of the right triangle from the x-axis.
y = 130 mm + (2/3) (120) mm
y = 210 mm
Question: How did the y-bar for area 3 become 135 millimeters?
Answer: I am very sorry for the confusion with the computation of the y-bar. There must be some dimensions lacking in the figure. But as long as you understand the process of solving problems about centroid, then there's nothing to worry about.
Question: How do you calculate w-beam centroid?
Answer: W-beams are H/I beams. You can start solving the centroid of a W-beam by dividing the whole cross-sectional area of the beam into three rectangular areas - top, middle, and bottom. Then, you can start following the steps discussed above.
Question: In problem 2, why is the quadrant positioned at the middle and the quadrant in problem 1 is not?
Answer: Most of the time, the position of the quadrants is given in the given figure. But in case that you are asked to do it yourself, then you should place the axis to a position where you can solve the problem in the most easy way. In problem number two's case, placing the y-axis at the middle will yield to an easier and short solution.