Calculating the Centroid of Compound Shapes Using the Method of Geometric Decomposition
Ray is a licensed engineer in the Philippines. He loves to write about mathematics and civil engineering.
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.
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.
d. Solve for the Ax values. Multiply the area of each region by the distances from the yaxis.
e. Solve for the Ay values. Multiply the area of each region by the distances from the xaxis.
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.
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.
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.
d. Solve for the Ax values. Multiply the area of each region by the distances from the yaxis.
e. Solve for the Ay values. Multiply the area of each region by the distances from the xaxis.
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.
The centroid of the complex figure is at 17.23 millimeters from the yaxis and 110.24 millimeters from the xaxis.
Recommended for You
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
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 ydistance of the centroid of the right triangle from the xaxis.
y = 130 mm + (2/3) (120) mm
y = 210 mm
Question: How did the ybar for area 3 become 135 millimeters?
Answer: 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.
Question: How do you calculate wbeam centroid?
Answer: 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.
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 yaxis at the middle will yield to an easier and short solution.
Question: Regarding Q1, there are graphical methods that can be used in many simple cases. Have you seen the game app, Pythagorean?
Answer: It looks interesting. It says that Pythagorea is a collection of geometric puzzles of different kind that can be solved without complex constructions or calculations. All objects are drawn on a grid whose cells are squares. A lot of levels can be solved using just your geometric intuition or by finding natural laws, regularity, and symmetry. This could really be helpful.
© 2018 Ray
Comments
JR Cuevas on August 23, 2020:
Hi, Luke and Dexter! Unfortunately, there is a lacking dimension on the figure. But you can follow this:
Given radius of 100 mm, assume 35 mm from the bottom up to the first point of the semicircle.
100 mm + 35 mm = 135 mm from the bottom of the figure up to the centroid of the semicircle.
I am very sorry for the inconvenience.
luke on August 23, 2020:
how did you get you y value for area 3 ?
Dexter on August 23, 2020:
How did you get the ybar for area 3
JackieZhang on June 30, 2020:
Thank you! Very very easy to understand!
Trying boy on June 11, 2020:
Nice. Easy to understand. Good explaination!
Ray (author) from Philippines on March 08, 2020:
Please, check on the "Questions and Answers" portion of the article to look for similar questions. Thanks!
Ray (author) from Philippines on January 23, 2020:
Hi, Good day Srikar! H/3 is the distance of the centroid of the triangle from the base of the triangle while 2H/3 is the distance of the centroid of the triangle from the vertex or tip of the triangle.
Srikar on January 23, 2020:
When to use b/3 and 2b/3 for x of triangle
Ray (author) from Philippines on December 17, 2019:
Hi, Mousa. 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.
Mousa Alzahrani on December 17, 2019:
in area three in problem 2... how 135 mm of y bar has obtained ?
K on November 01, 2019:
Sooper
Faiz on January 11, 2019:
Why centroid is out of fig in c shpae??