9.1

Chapter 9: Center of Gravity and Centroid

Chapter Objectives

• To discuss the concept of the center of gravity, center of mass, and the centroid.

• To show how to determine the location of the centroid for a body of arbitrary

shape.

• To use the Theorems of Pappus and Guldinus for finding the surface area and

volume for a body having axial symmetry.

• To represent the method for finding the resultant of a general distributed loading.

9.1 Center of Gravity, Center of Mass, and the Centroid of a Body

Center of Gravity

Contrast the difference between the “center of gravity” and “centroid” using a

composite consisting of Styrofoam and lead.

Center of gravity Centroid

• “Center of gravity” is a function of specific weight.

• “Centroid” is a function of dimensions; that is, it is a geometric property only.

The earth (that is, gravity) exerts a force on each particle forming a rigid body.

• The summation of these forces adds up to the total weight of the rigid body W.

∑Fz : W = ∆W 1 + ∆W2 + ∆W3 + • • • + ∆Wn

9.2

• The point of application (i.e. the center of gravity) of the total weight of the

rigid body W may be found using the “principle of moments.”

∑My : x W = x 1 ∆W1 + x 2 ∆W2 + x 3 ∆W3 + • • • + x n ∆Wn

∑Mx : y W = y 1 ∆W1 + y 2 ∆W2 + y 3 ∆W3 + • • • + y n ∆Wn

x = x W/W y = y W/W

If we increase the number of elements into which the “plate” (or rigid body) is

divided and decrease the size of each element, we obtain the following equations.

W = ∫dW x W = ∫x dW y W = ∫ y dW

• The coordinates (x and y ) define the center of gravity of the plate (or of the

rigid body).

Center of Mass of a Body

Center of mass is a function of density.

Centroid of a Volume

The centroid defines the geometric center of an object.

Centroid of an Area

• In the case of a homogeneous plate of uniform thickness, the magnitude ∆W is

∆W = ∂ t ∆A

where ∂ = specific weight (weight per unit volume)

t = thickness of the plate

∆A = area of the element

• The weight of the entire plate is W = ∂ t A

where A = total area of the plate

• If we substitute these into the previous equations and divide through by the

constant “ ∂ t “:

∑My : x A = x 1 ∆A1 + x 2 ∆A2 + x 3 ∆A3 + • • • + x n ∆An

∑Mx : y A = y 1 ∆A1 + y 2 ∆A2 + y 3 ∆A3 + • • • + y n ∆An

x = x A/A y = y A/A

9.3

If we increase the number of elements

into which the “plate” (or rigid body) is

divided and decrease the size of each

element, we obtain the following

equations.

A = ∫dA

x A = ∫x dA

y A = ∫ y dA

If the plate is not homogeneous, these equations cannot be used to determine the

“center of gravity” of the plate.

• However, the equations still define the “centroid” of the area.

Centroid of a Line

For a wire of uniform cross section, the magnitude ∆W of the weight of an

element of wire may be expressed as follows.

∆W = ∂ a ∆L

where

∂ = specific weight of the material

a = cross sectional area of the wire

∆L = length of the element

and

L = ∫ dL

x L = ∫ x dL

y L = ∫ y dL

For a wire of uniform cross section and with

a constant specific weight, the center of

gravity of the wire coincides with the

centroid C.

Symmetry

• When an area or line possesses an axis of symmetry (that is, an axis that causes

a mirror image on either side of the axis) the centroid of the area or line must

be located on that axis.

9.4

• If an area or line possesses two axes of symmetry, then the centroid of that

area or line is located at the intersection of the two axes of symmetry, and the

following is true.

x = y = 0

First Moments of Areas and Lines

• The integral ∫ x dA is known as the “first moment of the area A with respect

to the y-axis” and is denoted by Qy.

Qy = ∫ x dA

• Similarly, the integral ∫ y dA is known as the “first moment of the area A with

respect to the x-axis” and is denoted by Qx.

Qx = ∫ y dA

• Using a finite summation, the first moments of the area A with respect to the

x-axis (i.e. Qx) and with respect to the y-axis (i.e. Qy) may be determined as

follows.

Qx = y A = y 1 ∆A1 + y 2 ∆A2 + y 3 ∆A3 + • • • + y n ∆An

Qy = x A = x 1 ∆A1 + x 2 ∆A2 + x 3 ∆A3 + • • • + x n ∆An

Procedure for Analysis

The centroid of an object or shape can be determined by integration using the

following equations.

A = ∫dA x A = ∫ x dA y A = ∫ y dA

• Denoting x el and y el as the coordinates of the element of dA, the last two

equations may be written as follows.

x A = ∫ x el dA and y A = ∫ y el dA

The centroid of an object or shape can be determined by finite summations using

the following equations.

A = ΣAi x A = Σx i Ai y A = Σ y i Ai

Likewise, the centroid of a line can be determined by finite summations using the

following equations.

L = Σ Li x L = Σx i Li y L = Σ y i Li

L = ∫dL x L = ∫x dL y L = ∫y dL

9.5

Examples - Centroids of Areas

Given: The triangular area shown.

Find: Centroid (x , y ), using a

horizontal element.

Applicable equations:

A = ∫ dA dA = x dy

Qy = x A = ∫ x el dA x el = x/2

Qx = y A = ∫ y el dA y el = y

To define values of “x” in terms of “y” to allow the integration with respect to y,

write an equation for the boundary of the area (i.e. line AB).

• The general form for the equation of a line is y = m x + c

• The slope “m” of the line AB is m = - h/b

• Inserting the slope, the equation may be written y = (-h/b) x + c

• To solve for the y-intercept “c”, insert the coordinates (b, 0), a point on the

line, into the equation and solve for “c”.

0 = (-h/b) (b) + c c = h

• The equation for the line is then y = (-h/b) x + h

• Rewrite the equation to define “x” in terms of “y”. x = (b/h) (h – y)

• Now define dA and x el in terms of “y”.

dA = x dy = (b/h) (h – y) dy

x el = x/2 = ½ (b/h) (h – y)

Now solve for area A and the first moments of area Qx and Qy.

A = ∫ dA = ∫ x dy = h

0(b/h ) (h – y) dy = (b/h) (h y – y2/2)|

h

0

= (b/h) [(h2 – h2/2) – 0] = (b/h) (h2/2)

A = bh/2

9.6

x A = ∫x el dA = ∫(x/2) x dy = ½ ∫x2 dy = ½ h

0(b/h)2 (h-y)2 dy

= ½ (b/h)2 h

0(h – y)2 dy = ½ (b/h)2 (1/3) (h – y)3 (-1) |

h

0

= - ½ (b/h)2 (1/3) (0 – h3)

x A = b2 h/6

y A = ∫ y el dA = ∫ y x dy = h

0y (b/h) (h – y) dy

= (b/h) h

0y (h – y) dy = (b/h)

h

0(h y – y2) dy

= (b/h) [(h y2/2 – y3/3)] |h

0 = (b/h) (h3/2 – h3/3)

y A = bh2/6

Finally, determine the location of the centroid.

x = x A/A = (b2 h/6)/(bh/2) = b/3

y = y A/A = (bh2/6)/(bh/2) = h/3

9.7

Given: Area shown.

Find: Centroid (x , y ), using a

vertical element.

Applicable equations:

A = ∫ dA dA = (y2 – y1) dx

Qy = x A = ∫ x el dA x el = x

Qx = y A = ∫ y el dA y el = ½ (y2 + y1)

Define values of “m” and “k” in terms of “a” and “b”.

For the line, when x = a and y = b: y = mx b = ma m = b/a

For the parabola, when x = a and y = b: y = kx2 b = ka2 k = b/a2

To define values of “y” in terms of “x” to allow the integration with respect to x,

write equations for the boundaries of the area (i.e. the line and the parabola).

• The equation for the line is y = (b/a) x

• The equation for the parabola is y = (b/a2) x2

• Now define dA and y el in terms of “x”.

dA = (y2 – y1) dx = [(b/a) x - (b/a2) x2] dx

y el = ½ (y2 + y1) = ½ [(b/a) x + (b/a2) x2]

Now solve for area A and the first moments of area Qx and Qy.

A = ∫ dA = ∫ x dy = a

0[(b/a) x - (b/a2) x2] dx

= [(b/a) (x2/2) – (b/a2) (x3/3)] |a

0 = (b/a)(a2/2) – (b/a2)(a3/3)

= ba/2 – ba/3

A = ba/6

9.8

x A = ∫x el dA = a

0x [(b/a) x - (b/a2) x2] dx

= a

0[(b/a) x2 – (b/a2) x3 ] dx = [(b/a)(x3/3) – (b/a2)(x4/4)] |

a

0

= (b/a)(a3/3) – (b/a2)(a4/4) = ba2/3 – ba2/4

x A = ba2/12

y A = ∫ y el dA = a

0½ [(b/a) x + (b/a2) x2] [(b/a) x - (b/a2) x2] dx

= ½ a

0[(b/a)2 x2 – (b/a2)2 x4 ] dx = ½ [(b2/a2)(x3/3) – (b2/a4)(x5/5)] |

a

0

= ½ [(b2/a2)(a3/3) – (b2/a4)(a5/5)] = ½ (b2a/3 – b2a/5)

y A = b2a/15

Finally, determine the location of the centroid.

x = x A/A = (ba2/12)/(ba/6) = a/2

y = y A/A = (b2a/15)/(ba/6) = 2b/5

9.9

Given: Area shown.

Find: Centroid (x , y ), using a horizontal

element.

Applicable equations:

A = ∫ dA dA = (x2 – x1) dy

Qy = x A = ∫ x el dA x el = ½ (x2 + x1)

Qx = y A = ∫ y el dA y el = y

To define values of “x” in terms of “y” to allow the integration with respect to y,

write an equation for the boundary of the area (i.e. the line).

• The general form for the equation of a line is y = m x + c

• The slope “m” of the line is m = (4 – 0)/(4 – 3) = 4/1 m = 4

• Inserting the slope, the equation may be written y = 4 x + c

• To solve for the y-intercept “c”, insert the coordinates (4, 4), a point on the

line, into the equation and solve for “c”.

4 = 4 (4) + c c = - 12

• The equation for the line is then y = 4 x - 12

• Rewrite the equation to define “x” in terms of “y”. x = (y + 12)/4

• Now define dA and x el in terms of “y”.

dA = (x2 – x1) dy = [(y + 12)/4 – y2/4] dy

x el = ½ (x2 + x1) = ½ [(y + 12)/4 + y2/4]

9.10

Now solve for area A and the first moments of area Qx and Qy.

A = ∫ dA = ∫ (x2 – x1) dy = 4

0[(y + 12)/4 – y2/4] dy

= ¼ 4

0(y + 12 – y2) dy = ¼ (y2/2 + 12y – y3/3) |

4

0 = ¼ (8 + 48 – 64/3)

A = 8.67 in2

x A = ∫x el dA = 4

0½ [(y + 12)/4 + y2/4] [(y + 12)/4 – y2/4] dy

= ½ 4

0{[(y + 12)/4] 2 – (y2/4)2 } dy = (1/32)

4

0[(y + 12) 2 – (y2)2] dy

= (1/32) 4

0(y2 + 24 y + 144 – y4) dy = (1/32) (y3/3 + 12 y2 + 144 y – y5/5) |

4

0

= (1/32) (64/3 + 192 + 576 – 1024/5)

x A = 18.27 in3

y A = ∫ y el dA = 4

0y [(y + 12)/4 – y2/4] dy =

4

0y {[(y + 12)/4] – y2/4} dy

= ¼ 4

0(y2 + 12y – y3) dy = ¼ [(y3/3 + 6y2 – y4/4] |

4

0 = ¼ (64/3 + 96 – 64)

y A = 13.33 in3

x = x A/A = 18.27/8.67 = 2.11”

y = y A/A = 13.33/8.67 = 1.537”

9.11

9.2 Composite Bodies

In many instances, an area may be divided into familiar shapes (rectangles,

triangles, squares, and circles).

• The centroidal distances are found by equating the first moments of area.

∑Mx : y 1 ∆A1 + y 2 ∆A2 + • • • + y n ∆An = y A

∑My : x 1 ∆A1 + x 2 ∆A2 + • • • + x n ∆An = x A

A tabular solution is often a convenient method to determine the location of the

centroid for composite areas.

9.12

Examples – Composite Areas

Given: Area shown.

Find: Centroid (x , y ).

Part Area, Ai x i y i x i Ai y i Ai

1 10,800 45.0 120.0 486,000 1,296,000

2 2,700 30.0 40.0 81,000 108,000

3 - 2,510 73.0 120.0 - 183,000 - 301,000

Totals 10,990 384,000 1,103,000

A3 = π r2/2 = π (40)2/2 = 2,510

x3 = 90 – 4r/3 π = 90 – 4(40)/3π = 73.0

x = Σx i Ai/ΣAi = 384,000/10,990

x = 34.9 mm

y = Σ y i Ai/ΣAi = 1,103,000/10,990

y = 100.4 mm

9.13

Given: The cover-plated beam shown.

Find: Neutral axis.

The “neutral axis” is an axis in the cross

section of a beam (a member resisting

bending) along which there are no

longitudinal stresses or strains.

• If the section is symmetric, then the

neutral axis is located at the geometric

centroid.

A vertical axis through the center of the web forms an axis of symmetry.

• Only the y distance is required.

Use the bottom of the bottom flange as the reference axis.

y = Σ y i Ai/ΣAi = [22.3 (18.2/2) + 1(12)(18.2 + 1.0/2)]/[22.3 + 1(12)]

= (202.93 + 224.40)/34.3

y = 12.46”

9.14

Given: Bent wire shown.

Find: Centroid (x , y ).

Segment Length, Li x i y i x i Li y i Li

1 4.00 0 5.00 0 20.00

2 6.00 3.00 7.00 18.00 42.00

3 7.00 6.00 3.50 42.00 24.50

4 6.71 3.00 1.50 20.13 10.07

Totals 23.71 80.13 96.57

x = Σx i Li/ΣLi = 80.13 / 23.71

x = 3.38”

y = Σ y i Li/ΣLi = 96.57 / 23.71

y = 4.07”

9.15

9.3 Theorems of Pappus-Guldinus

The two theorems of Pappus and Guldinus are used to find the surface area and

volume of any body of revolution.

• The theorems were first developed by Pappus of Alexandria during the 4th

century A.D.

• The theorems were restated later by the Swiss mathematician Paul Guldin (or

Guldinus) (1577 – 1643).

Surface Area

A “surface of revolution” is a surface that may be generated by rotating a plane

curve about a fixed axis.

Theorem 1: “The area of a surface of revolution is equal to the length of the

generating curve times the distance traveled by the centroid of the curve while

the surface is being generated.” (The curve must be non-intersecting.)

dL

2 π y el

A = ∫2π y el dL

= 2π ∫ y el dL

A = 2π y L

where 2π y is the distance

traveled by the centroid

of the line L

Volume

A “body of revolution” is a body that may be generated by rotating a plane area

about a fixed axis.

Theorem 2: “The volume of a body of revolution is equal to the generating area

times the distance traveled by the centroid of the area while the body is being

generated.” (The area must be non-intersecting.)

9.16

dA

2 π y el

A = ∫ 2 π y el dA

= 2 π ∫ y el dA

A = 2 π ( y A)

where 2π y is the distance

traveled by the centroid

of the area A

9.17

Examples – Theorems of Pappus and Guldinus

Given: The cone shown.

Find: The surface area and volume of

the cone.

Surface area using the Theorem

A = 2 π x L

= 2 π (1.5) 5.0 = 15.0 π

A = 47.12 in2

Surface area from formula

A = π r (r2 + h2)½

= π (3) (32 + 42)½ = 15.0 π

A = 47.12 in2

Volume using the Theorem

V = 2 π x A

= 2 π [(1/3) 3.0][½ (4.0) 3.0] = 12.0 π

V = 37.70 in3

Volume using formula

V = (π/3) r2 h

= (π/3) (32) 4 = 12.0 π

V = 37.70 in3

9.18

Given: A sphere with a radius of 12”.

Find: The surface area and volume of

the sphere.

Surface area using the Theorem

A = 2 π y L

y = 2r/π = 2(12)/π = 24/π

L = 2 π r/2 = π r = π (12) = 12 π

A = 2 π (24/π) (12 π) = 576 π

A = 1809.6 in2

Surface area from formula

A = 4 π r2

= 4 π (122) = 576 π

A = 1809.6 in2

Volume using the Theorem

V = 2 π y A

y = 4r/3π = 4 (12)/3 π = 16/π

A = π r2/2 = π (12)2/2 = 72π

V = 2π (16/π)(72 π) = 2304 π

V = 7238.2 in3

Volume from formula

V = (4π/3) r3 = (4π/3) (123) = 2304 π

V = 7238.2 in3

9.19

9.4 Resultant of a General Distributed Loading

The concept of the centroid of an area may be used to solve problems with beams

supporting a “distributed load.”

Magnitude of Resultant Force

A “distributed load” on a beam may be

replaced by an equivalent concentrated load.

• The magnitude of this single equivalent

concentrated load is equal to the area

under the load curve.

Location of the Resultant

The line of action of the resultant force

passes through the centroid of the

distributed loading.

9.5 Fluid Pressure

Pressure Distribution over a Surface

The concept of the centroid of an area is useful for this problem, whose solution is

discussed in “Fluid Mechanics.”