9. Center of Gravity, Centroid

and Moment of Inertia

Engineering Mechanics

Statics

Dr. Rami Zakaria

Mechanical Systems Engineering_2016

To design the structure for supporting a water

tank, we will need to know the weights of the

tank and water as well as the locations where

the resultant forces representing these

distributed loads act.

One of the important factors in

determining its stability is the

SUV’s center of mass.

One concern about a sport utility

vehicle (SUV) is that it might tip

over while taking a sharp turn.

Applications

CONCEPT OF CENTER OF GRAVITY (CG)

A body is consist of an infinite number of

Particles. So if the body is located within a

gravitational field, then each of these particles

will have a weight dW.

Therefore, the sum of all moments due to

individual particle weights about any point is

equal to the moment due to the resultant weight

located at G.

Also, note that the sum of moments due to the individual particle’s

weights about point G is equal to zero.

The center of gravity (CG) is a point (often

shown as G) which locates the resultant weight

of a system of particles or a solid body.

The location of the center of gravity, measured

from the y axis, is determined by equating the

moment of W about the y axis to the sum of the

moments of the weights of the particles about

this same axis. ~ ~ ~

x W = x dW ~ _ If dW is located at point (x, y, z), then

Similarly, y W = y dW ~ _ z W = z dW ~ _

Therefore, the location of the center of gravity G with respect to

the x, y,z axes becomes

Are the coordinates of the center of gravity (CG) zyx ,,

Center of Mass OF A BODY

By replacing the W (weight) with an m (mass) in these equations, the

coordinates of the center of mass can be found.

Similarly, the coordinates of the

centroid of area, or length can be

obtained by replacing V by A, or L,

respectively (only x-y).

Centeroid of a Volume

dV

dVzz

dV

dVyy

dV

dVxx

dVdm

Vddm

V

Vm

~

,

~

,

~

;

is density

is volume

but the density is constant for a homogeneous material

CONCEPT OF CENTROID

The centroid coincides with the center of

mass or the center of gravity only if the

material of the body is homogenous (density

is constant throughout the body).

If an object has an axis of symmetry, then

the centroid of object lies on that axis.

Note: In some cases, the centroid is not

located on the object.

The centroid, C, is a point which defines the

geometric center of an object.

STEPS TO DETERME THE CENTROID OF AN AREA

1. Choose an appropriate differential element dA at a general point

(x,y). Hint: Generally, if y is easily expressed in terms of x

(e.g., y = x2 + 1), use a vertical rectangular element. If the

converse is true, then use a horizontal rectangular element.

2. Express dA in terms of the differentiating element dx (or dy).

Note: Similar steps are used for determining the CG or CM.

These steps will become clearer by doing a few examples.

4. Express all the variables and integral limits in the formula

using either x or y depending on whether the differential

element is in terms of dx or dy, respectively, and integrate.

3. Determine coordinates ( x , y) of the centroid of the

rectangular element in terms of the general point (x,y).

~ ~

EXAMPLE

2. dA = y dx = x3 dx

3. x = x and y = y / 2 = x3 / 2 ~ ~

Solution

1. Since y is given in terms of x, choose

dA as a vertical rectangular strip.

Given: The area as shown.

Find: The centroid location (x , y)

4. x = ( A x dA ) / ( A dA ) ~

0 x (x3 ) d x 1/5 [ x5 ]1

0 (x3 ) d x 1/4 [ x4 ]1

= ( 1/5) / ( 1/4) = 0.8 m

1 = =

1

0

0

1 A y dA 0 (x

3 / 2) ( x3 ) dx 1/14[x7]1

A dA 0 x3 dx 1/4

1 = y =

~

=

= (1/14) / (1/4) = 0.2857 m

0

Review: we found that the location of the center of gravity G with

respect to the x, y, z axes is:

Are the coordinates of the center of gravity (CG) zyx ,,

Similarly we found the coordinates (position) of the center of mass:

Center of volume:

Center of area (Centroid):

dV

dVzz

dV

dVyy

dV

dVxx

~

,

~

,

~

dA

dAzz

dA

dAyy

dA

dAxx

~

,

~

,

~

Example: Find the centroid of this plate.

1- Select a random x,y point

2- make a small rectangular

3- calculate the area of that small rectangular

4- find the centroid of the small rectangular

5- Find the centroid of the entire plate by applying:

ydxdA 2

0~,~ yxx

dA

dAxx

~

2

0

2

0

)2(

)2(~

dxy

dxyx

xdA

dAxx ,but

5.02 )2(2 xyxy

m

x

x

dxx

dxx

dxx

dxxx

x

2.1)2(6.002

026.0

)(22

22

)2(2

))2(2(

5.15.1

5.25.2

2

0

5.15.1

1

2

0

5.25.2

1

2

0

5.0

2

0

5.1

2

0

5.0

2

0

5.0

0y because the plate is symmetric around x axis

COMPOSITE BODIES

A composite body in this section refers to a body made

of a number of simpler bodies (for example rectangular

triangular, circles, semi-circles, etc…)

The composite method for determining the location of

the center of gravity of a composite body requires

simple arithmetic instead of integration.

Instead of calculating infinite number of (differential )

weights, we only need a finite number of weights,

therefore:

Same way, if we need centre of mass we replace W by m.

Application

Using this method we can find the

centre of gravity of a complex shapes,

such as this water tank

The I-beam or tee-beam shown are

commonly used in building various types

of structures.

When doing a stress or deflection

analysis for a beam, the location of the

centroid is very important.

Many industrial objects can be considered as composite bodies

made up of a series of connected “simple” shaped parts or holes,

like a rectangle, triangle, and semicircle.

Knowing the location of the centroid, C, or center of gravity, G,

of the simple shaped parts, we can easily determine the location

of the C or G for the more complex composite body.

Steps for Analysis

1. Divide the body into pieces that are known shapes.

Holes are considered as pieces with negative weight or size.

2. Make a table with the first column for segment number, the

second column for weight, mass, or size (depending on the

problem), the next set of columns for the moment arms, and,

finally, several columns for recording results of simple

intermediate calculations.

3. Fix the coordinate axes, determine the coordinates of the center

of gravity of centroid of each piece, and then fill in the table.

4. Sum the columns to get x, y, and z. Use formulas like

x = ( xi Ai ) / ( Ai ) or x = ( xi Wi ) / ( Wi )

This approach will become clear by doing examples!

Area Cx Cy

b.h b/2 h/2

b.h/2 (a +b)/3 h/3

π.a2 a a

π.a2 /2 a 4.a /3.π

For more shapes see: http://www.roymech.co.uk/Useful_Tables/Maths/M_of_Inertia_2.html

EXAMPLE

Solution:

Step 1. This body can be divided into the following pieces:

rectangle (a) + triangle (b) + quarter circular (c) – semicircular

area (d). Note the negative sign on the hole!

Given: The part shown.

Find: The centroid of the part.

Steps 2 & 3: Make up and fill the table using parts a, b, c, and d.

39.83 76.5 28.0

27

4.5

9

- 2/3

54

31.5

– 9

0

1.5

1

4(3) / (3 )

4(1) / (3 )

3

7

– 4(3) / (3 )

0

18

4.5

9 / 4

– / 2

Rectangle

Triangle

Q. Circle

Semi-Circle

y A

( in3)

x A

( in3)

y

(in)

x

(in)

Area A

(in2)

Segment

x = ( x A) / ( A ) = 76.5 in3/ 28.0 in2 = 2.73 in

y = ( y A) / ( A ) = 39.83 in3 / 28.0 in2 = 1.42 in

Step 4. Now use the table data results and the formulas to find the

coordinates of the centroid.

0x

Because the shape is symmetric around (y)

No. Area A (in2)

y (in)

A.y (in3)

1 6(1) 3 18

2 6(1) 3 18

3 6(1) 5.5 33

4 6(1) 9 54

Σ 24 123

inA

Ayy 13.5

24

123~

1 2

3

4

No. Area A (ft2)

x (ft)

y (ft)

A.x (ft3) A.y (ft3)

1 1.5 -2.23 5.3

2 3(3)=9 1.5 1.5 13.5 13.5

3 0.5 (3)(3)=4.5 4 1 18 4.5

4 0 1.5 0 -4.71

Σ 13.86 29.27 18.59

3

)5.1(45.3

2

)5.1( 2

14.3)1( 2

ftA

Axx 11.2

86.13

27.29~

ftA

Ayy 34.1

86.13

59.18~

Moment of Inertia

APPLICATIONS

Many structural members like beams and columns have cross sectional

shapes like an I, H, C, etc..

The design of the structural members require calculation of its cross-

sectional moment of inertia.

The moment of inertia about a given axis describes how difficult it is to change its angular

motion about that axis. Therefore, it shows how far each bit of mass is from this axis. The

farther out the object's mass is, the more inertia the object has, and the more force is required

to change its rotation rate

DEFINITION OF MOMENTS OF INERTIA FOR AREAS

The force on the area dA at that point is dF = p dA.

The moment about the x-axis due to this force is y (dF).

The total moment is A y dF = A y2 dA = A( y2 dA).

This sort of integral term is used not only in fluid mechanics but also appears

in solid mechanics when determining stresses and deflection.

This integral term is referred to as the moment of inertia of the area of the

plate about an axis.

Consider a plate submerged in a liquid. The

pressure of a liquid at a distance y below the

surface is given by p = y, where is the

specific weight of the liquid.

The moments of inertia for the entire area are obtained by

integration.

Ix = A y2 dA ; Iy = A x2 dA

JO = Ix + Iy = A ( x2 + y2 ) dA = A r2 dA

The MoI is also referred to as the second moment of an area and

has units of length to the fourth power (m4 or in4).

DEFINITION OF MOMENTS OF INERTIA FOR AREAS

For the differential area dA, shown in the figure:

d Ix = y2 dA ,

d Iy = x2 dA , and,

d JO = r2 dA , where JO is the polar

moment of inertia about the pole O or z axis.

Example: Moment of Inertia of a Rectangular Area around x and y.

Around x Around y

Questions:

1. The definition of the Moment of Inertia for an area involves an

integral of the form

A) x dA. B) x2 dA.

C) x2 dm. D) m dA.

2. Select the SI units for the Moment of Inertia for an area.

A) m3

B) m4

C) kg·m

D) kg·m3