# Calculating the Centroid of Compound Shapes Using the Method of Geometric Decomposition

Updated on December 31, 2019 Ray is a Licensed Civil Engineer and specializes in Structural Engineering. He loves to write anything about education.

## 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 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 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 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 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.) to 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)
Table 1-2: Centroid for Common Shapes. X-bar is the distance of the centroid from the y-axis. Y-bar is the distance of the centroid from the x-axis.

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)
Table 1-1: Table Format. X is the distance of the centroid from the y-axis. Y is the distance of the centroid from the x-axis.

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.

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.

## 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 millimeters```

c. 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.

```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 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 millimeters```

e. 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
All units are in terms of millimeters.

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```

The centroid of the complex figure is at 66.90 millimeters from the y-axis and 65.00 millimeters from the x-axis.

## 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 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.

```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 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 millimeters```

e. 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
All units are in terms of millimeters.

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 millimeters```

The centroid of the complex figure is at 17.23 millimeters from the y-axis and 110.24 millimeters from the x-axis.

## Did you learn from the examples?

See results

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.

• 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.

• in area two of triangle in problem 2...how 210mm of y bar has obtained?

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

• How do you calculate w-beam centroid?

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.

## Related

0 of 8192 characters used
• AUTHOR

Ray

4 days ago from Philippines

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

5 days ago

When to use b/3 and 2b/3 for x of triangle

• AUTHOR

Ray

5 weeks ago from Philippines

Hi, Mousa. 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.

• Mousa Alzahrani

5 weeks ago

in area three in problem 2... how 135 mm of y bar has obtained ?

• 2 months ago

Sooper

• Faiz

12 months ago

Why centroid is out of fig in c shpae??

working