8.3 Find centroids of composite areas | University of Alberta ENGG 130: Engineering Mechanics: Statics

University of Alberta ENGG 130: Engineering Mechanics: Statics

Centroids and Centres of Gravity

ENGG 130 study topic 8.3

A plane area can be divided into small pieces. Its centroid is the area-weighted average location of those pieces. For a uniform, thin plate, this is the point at which the area would balance. To find the centroid of a composite area, represent its outline as simpler shapes, find each shape’s area and centroid, and add their contributions. A hole is treated as removed area, so its contribution is negative. This lesson concerns the geometry of areas; no forces or support reactions are involved.

What you will learn

Explain a centroid as the area-weighted average position of a plane area.
Find area and centroid coordinates for basic component shapes.
Combine component areas, including holes, to find a composite centroid.
Check results using units, symmetry, first moments, and geometric location.

1. Centroid and first moment of area

Choose perpendicular axes in the plane and a common origin. A small area element

dA

at coordinates

(x,y)

contributes in proportion to its area and its distance from the axes. The centroid coordinates are therefore area-weighted averages.

The first moment about the

y

-axis weights each area element by its horizontal coordinate. The first moment about the

x

-axis weights each element by its vertical coordinate. For components, multiply each signed area by the corresponding centroid coordinate and sum. These first moments have units of length cubed: for example, square millimetres multiplied by millimetres gives cubic millimetres.

For a composite area, use positive areas for included regions and negative areas for holes. The same sign must be used when calculating both first moments. Every component centroid must be expressed from the same origin.

\bar{x}=\frac{\sum A_i x_i}{\sum A_i},\quad \bar{y}=\frac{\sum A_i y_i}{\sum A_i}

Area has units of length squared; first moment of area has units of length cubed.
A centroid coordinate has units of length.
A hole is subtracted from both the total area and the first moments.

2. Choose axes and list the components

First describe the area and choose axes that make component dimensions and locations straightforward. A corner or line of symmetry is often a useful reference. Take positive

x

to the right and positive

y

upward, and use those directions consistently.

A rectangle with width

b

and height

h

has area

bh

and centroid halfway along both dimensions. A right triangle has area one-half its base times its height. Its centroid is one-third of each perpendicular leg measured from the right-angle corner. These locations provide the component data needed for many composite-area problems.

Make a list of the components before doing the sums. For each, record its signed area and centroid coordinates measured from the shared origin. If a component’s local coordinates start at a different corner, translate its centroid location into the shared coordinate system. Then sum the signed areas and the two first moments.

A=\sum A_i,\quad Q_y=\sum A_i x_i,\quad Q_x=\sum A_i y_i

Choose the origin and axes before recording coordinates.
Use one shared coordinate system for every component.
The total area is the net area when holes are present.

3. Calculate and check the result

Divide each first-moment sum by the net area to find the corresponding centroid coordinate. Keep units visible during the calculation: an area in

\mathrm{mm}^2

times a distance in

\mathrm{mm}

gives a first moment in

\mathrm{mm}^3

, and division by area returns a distance in millimetres.

Use geometry to check the result. If the area is symmetric about a vertical line, its centroid lies on that line; symmetry about a horizontal line similarly fixes the vertical coordinate. For an area made entirely of included regions, the centroid should lie within the overall boundary. When a hole is present, the centroid can lie in the empty region, so check the shape and signed sums rather than applying the boundary check blindly.

Finally, verify the defining first-moment relationships: multiplying the net area by each reported coordinate must reproduce its signed first-moment sum. This is an arithmetic and geometric check, not a force or moment equilibrium calculation.

Q_y=A\bar{x},\quad Q_x=A\bar{y}

Check units, symmetry, signs, and whether the location is reasonable.
A hole shifts the centroid away from the removed area.
The centroid must reproduce both signed first moments when multiplied by net area.

Worked example

Rectangle with a circular hole

A rectangular area spans

0\leq x\leq120\,\mathrm{mm}

and

0\leq y\leq80\,\mathrm{mm}

. A circular hole of radius

20\,\mathrm{mm}

has centre

(80,40)\,\mathrm{mm}

. Find the centroid of the remaining area.

Assign signed areas
Use the lower-left corner as the origin. The rectangle is included area and the circle is removed area, so assign the circle a negative sign. The rectangle centroid is at its midpoint; the hole centroid is its centre.
$A_1=9600\,\mathrm{mm}^2,\ (x_1,y_1)=(60,40)\,\mathrm{mm};\quad A_2=-400\pi\,\mathrm{mm}^2,\ (x_2,y_2)=(80,40)\,\mathrm{mm}$
Sum area and first moments
Multiply each signed area by its centroid coordinates. Both component centroids have the same vertical coordinate, so the combined centroid also has that vertical coordinate.
$A=9600-400\pi,\quad Q_y=576000-32000\pi,\quad Q_x=384000-16000\pi$
Find the coordinates
Divide each first-moment sum by the net area. Since the hole removes area to the right of the rectangle centre, the remaining centroid shifts left of $60\,\mathrm{mm}$ .
$(\bar{x},\bar{y})=\left(\frac{576000-32000\pi}{9600-400\pi},40\right)\,\mathrm{mm}\approx(56.99,40.00)\,\mathrm{mm}$

Answer: The remaining area’s centroid is approximately

(56.99,40.00)\,\mathrm{mm}

from the rectangle’s lower-left corner.

Check: The net area is approximately

8343.36\,\mathrm{mm}^2

. The symmetry about

y=40\,\mathrm{mm}

confirms the vertical coordinate. Multiplying the net area by the reported coordinates reproduces the signed first-moment sums, within rounding.

Worked example

An L-shaped area from two rectangles

An L-shaped area consists of a bottom rectangle

100\,\mathrm{mm}

wide and

20\,\mathrm{mm}

high, spanning

x=0

to

100\,\mathrm{mm}

and

y=0

to

20\,\mathrm{mm}

, plus a

40\,\mathrm{mm}

by

50\,\mathrm{mm}

rectangle above its left end, spanning

x=0

to

40\,\mathrm{mm}

and

y=20

to

70\,\mathrm{mm}

. Find the centroid.

Locate component centroids
Take the lower-left corner of the whole area as the origin. Find each rectangle’s midpoint in this shared coordinate system.
$A_1=2000\,\mathrm{mm}^2,\ (x_1,y_1)=(50,10)\,\mathrm{mm};\quad A_2=2000\,\mathrm{mm}^2,\ (x_2,y_2)=(20,45)\,\mathrm{mm}$
Calculate the totals
Both rectangles are included regions, so their areas and first-moment contributions are positive. Add the two contributions for each total.
$A=4000\,\mathrm{mm}^2,\quad Q_y=140000\,\mathrm{mm}^3,\quad Q_x=110000\,\mathrm{mm}^3$
Divide by net area
Divide each first-moment sum by the total area. The result is between the component centroid coordinates, as expected for two included areas.
$(\bar{x},\bar{y})=(35.0,27.5)\,\mathrm{mm}$

Answer: The centroid is

(35.0,27.5)\,\mathrm{mm}

from the lower-left corner.

Check: The products

4000(35.0)=140000\,\mathrm{mm}^3

and

4000(27.5)=110000\,\mathrm{mm}^3

reproduce both first-moment sums. The coordinates are consistent with the L-shaped outline.

Worked example

Right triangle joined to a rectangle

A right triangle with vertices

(0,0)

,

(60,0)

, and

(60,40)\,\mathrm{mm}

is joined edge-to-edge to a

60\,\mathrm{mm}

by

40\,\mathrm{mm}

rectangle spanning

x=60

to

120\,\mathrm{mm}

and

y=0

to

40\,\mathrm{mm}

. Find the centroid of the combined area.

Find component areas and centroids
The triangle’s right-angle corner is at $(60,0)\,\mathrm{mm}$ . Its centroid lies one-third of each leg from that corner, toward the other vertices. The rectangle centroid is at its midpoint.
$A_1=1200\,\mathrm{mm}^2,\ (x_1,y_1)=(40,40/3)\,\mathrm{mm};\quad A_2=2400\,\mathrm{mm}^2,\ (x_2,y_2)=(90,20)\,\mathrm{mm}$
Sum first moments
Both regions are included in the combined area, so add their positive areas and first moments.
$A=3600\,\mathrm{mm}^2,\quad Q_y=264000\,\mathrm{mm}^3,\quad Q_x=64000\,\mathrm{mm}^3$
Calculate centroid
Divide the first moments by total area. The result lies between the component centroid coordinates and is closer to the larger rectangle’s centroid.
$(\bar{x},\bar{y})=(73.33,17.78)\,\mathrm{mm}$

Answer: The centroid is approximately

(73.33,17.78)\,\mathrm{mm}

from the origin at the triangle’s first vertex.

Check: Using the rounded coordinates gives

3600(73.33)\approx264000\,\mathrm{mm}^3

and

3600(17.78)\approx64000\,\mathrm{mm}^3

, consistent with both first-moment sums.

Common mistakes and how to avoid them

Treating a hole as positive area.

Correction: Assign the hole a negative area and subtract its first-moment contributions as well.

Using each component’s local corner as the coordinate origin.

Correction: Translate every component centroid to one common origin before adding contributions.

Taking an unweighted average of component coordinates.

Correction: Weight each coordinate by its signed area, add the first moments, then divide by net area.

Giving a centroid coordinate in square or cubic units.

Correction: First moments have cubic-length units, but dividing by area leaves length units.

Lesson summary

Choose a common origin and record each component’s area and centroid coordinates.
Use positive areas for included regions and negative areas for holes.
Divide each signed first-moment sum by the net area.
Check units, symmetry, and whether the centroid is geometrically reasonable.

Check your understanding

Question 1

A rectangle is

80\,\mathrm{mm}

wide and

30\,\mathrm{mm}

high, with its lower-left corner at the origin. What are its centroid coordinates?

$(40,15)\,\mathrm{mm}$
$(80,30)\,\mathrm{mm}$
$(30,40)\,\mathrm{mm}$
$(20,7.5)\,\mathrm{mm}$

Show answer and explanation

(40,15)\,\mathrm{mm}

The centroid is halfway along each dimension:

(80/2,30/2)=(40,15)\,\mathrm{mm}

.

Question 2

A circular hole is removed from the right side of a rectangle. Compared with the rectangle’s centroid, which way does the remaining area’s centroid shift horizontally?

Left, away from the removed area
Right, toward the removed area
It must remain at the rectangle’s centroid
Its horizontal coordinate becomes zero

Show answer and explanation

Left, away from the removed area

Removing area on the right reduces the right-side contribution to the first moment, shifting the remaining centroid away from the hole.

Key terms

Centroid: The area-weighted average position of a plane area.
First moment of area: The sum of area contributions weighted by distance from an axis.
Composite area: An area represented as a collection of simpler shapes.
Signed area: An area counted positively when included and negatively when removed as a hole.

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8.3 · Find centroids of composite areas

University of Alberta ENGG 130: Engineering Mechanics: Statics

ENGG 130 study topic 8.3

What you will learn

1. Centroid and first moment of area

2. Choose axes and list the components

3. Calculate and check the result

Rectangle with a circular hole

An L-shaped area from two rectangles

Right triangle joined to a rectangle

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Key terms

Continue through ENGG 130

About this lesson and its review