Calculating the Centroid of Compound Shapes Using the Method of Geometric Decomposition
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 to 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 2dimensional shapes. The center of mass is the term for 3dimensional shapes. For instance, the centroid of a circle and a rectangle is at 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 C_{i} and area A_{i} of each Z_{n} part wherein all holes that extend outside the compound shape are to be treated as negative values. Lastly, compute the centroid given the formula:
C_{x} = ∑C_{ix} A_{ix} / ∑A_{ix}
C_{y} = ∑C_{iy} A_{iy} / ∑A_{iy}
StepByStep 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 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 12 below shows the formula for different basic geometric figures. After determining the area, designate a name (Area one, area two, area three, etc.) to each area. Make the area negative for designated areas that act as holes.
3. The given figure should have an xaxis and yaxis. If x and yaxes are missing, draw the axes in the most convenient means. Remember that xaxis is the horizontal axis while the yaxis 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 xaxis and yaxis. Table 12 below shows the centroid for different basic shapes.
Centroid for Common Shapes
Shape
 Area
 Xbar
 Ybar


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 yaxis. 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 xaxis. Then get the summation ΣAy. Refer to the table format above.
8. Solve for the total area ΣA of the whole figure.
9. Solve for the centroid C_{x}of 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 yaxis.
10. Solve for the centroid C_{y} 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 xaxis.
Here are some examples of obtaining a centroid.
Problem 1: Centroid of CShapes
Solution 1
a. Divide the compound shape into basic shapes. In this case, the Cshape 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 millimeters
c. X and Y distances of each area. X distances are the distances of each area's centroid from the yaxis, and Y distances are the distances of each area's centroid from the xaxis.
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 millimeters
d. Solve for the Ax values. Multiply the area of each region by the distances from the yaxis.
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 millimeters
e. Solve for the Ay values. Multiply the area of each region by the distances from the xaxis.
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 (C_{x}, C_{y}) 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
The centroid of the complex figure is at 66.90 millimeters from the yaxis and 65.00 millimeters from the xaxis.
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 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 millimeters
c. X and Y distances of each area. X distances are the distances of each area's centroid from the yaxis, and y distances are the distances of each area's centroid from the xaxis. Consider the orientation of x and yaxes. For Quadrant I, x and y are positive. For Quadrant II, x is negative while y is positive.
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 millimeters
d. Solve for the Ax values. Multiply the area of each region by the distances from the yaxis.
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 millimeters
e. Solve for the Ay values. Multiply the area of each region by the distances from the xaxis.
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 (C_{x}, C_{y}) 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 millimeters
The centroid of the complex figure is at 17.23 millimeters from the yaxis and 110.24 millimeters from the xaxis.
Did you learn from the examples?
Moment of Inertia of Irregular or Compound Shapes
 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
Is there any alternative method for solving for the centroid except this geometric decomposition?
Yes, there is a technique using your scientific calculator in solving for the centroid.
Helpful 106in area two of triangle in problem 2...how 210mm of y bar has obtained?
It is the ydistance of the centroid of the right triangle from the xaxis.
y = 130 mm + (2/3) (120) mm
y = 210 mm
Helpful 20How do you calculate wbeam centroid?
Wbeams are H/I beams. You can start solving the centroid of a Wbeam by dividing the whole crosssectional area of the beam into three rectangular areas  top, middle, and bottom. Then, you can start following the steps discussed above.
Helpful 11How did the ybar for area 3 become 135 millimeters?
I am very sorry for the confusion with the computation of the ybar. 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.
Helpful 10In problem 2, why is the quadrant positioned at the middle and the quadrant in problem 1 is not?
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 yaxis at the middle will yield to an easier and short solution.
Helpful 4
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