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Chapter 9 Center of Gravity and Centroid

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CENTER OF GRAVITY AND CENTROID (Chapter 9)

Objective :

Students will:

a) Understand the concepts of center of

gravity, center of mass, and centroid.

b) Be able to determine the location of

these points for a system of particles

or a body.

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APPLICATIONS

To design the structure for supporting awater tank, we will need to know the

weights of the tank and water as well as the

locations where the resultant forcesrepresenting these distributed loads are

acting---center of gravity (CG).

.

One of the important factors indetermining SUVs stability is

its center of mass (CM).

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CONCEPT OF CG and CM

The center of gravity (G) is a point which

locates the resultant weight of a system of

particles or body.

From the definition of a resultant force, the sum of moments due to

individual particle weight about any point is the same as the moment

due to the resultant weight located at G. For the figure above, try taking

moments about A and B.

Also, note that the sum of moments due to the individual particles

weights about point G is equal to zero.

Similarly, the center of mass is a point which locates the resultant

mass of a system of particles or body. Generally, its location is the

same as that of G.

3m

4N

1m

A B1 N

3 N

G

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CG / CM FOR A SYSTEM OF PARTICLES

Consider a system of n particles as shown in

the figure. The net or the resultant weight is

given as WR = W.

Similarly, we can sum moments about the x and z-axes to find the

coordinates of G.

By replacing the W with a M in these equations, the coordinatesof the center of mass can be found.

Summing the moments about the y-axis, we get

x WR = x1W1 + x2W2 + .. + xnWn

where x1 represents x coordinate of W1, etc..

~~~

~

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CG / CM / CENTROID OF A BODY (Section 9.2)

A rigid body can be considered as madeup of an infinite number of particles.

Hence, using the same principles as in

the previous slide, we get thecoordinates of G by simply replacing the

discrete summation sign (

) by the

continuous summation sign (

) and W

by dW.

Similarly, the coordinates of the center of mass and the centroid

of volume, area, or length can be obtained by replacing W by m,

V, A, or L, respectively.

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CONCEPT OF CENTROID

The centroid C is a point which defines the

geometric center of an object.

The centroid coincides with the center

of mass or the center of gravity only if

the material of the body is homogenous

(density or specific weight is constant

throughout the body).

If an object has an axis of symmetry, then

the centroid of object lies on that axis.

In some cases, the centroid is not

located on the object.

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Centroids of volumes

If the density (mass per unit volume) of

the material that makes up a body isconstant, then

dVV,dmm ==

=

V

dVV

V

dVx

xdVxdMVx VCV

yzC === ,

V

dVy

ydVydMVyV

C

V

xzC

=== ,

V

dVz

zdVzdMVz VCV

xyC

=== ,

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Centroids of Areas

If the body consists of a homogeneous, thin

plate of uniform thickness t and surface areaA, then tdAdmtAm == ,

=A

dAAA

dAx

xdAxdMAx ACA

yzC

=== ,

A

dAy

ydAydMAy

A

CA

xzC

=== ,

A

dAz

zdAzdMAz ACA

xyC

=== ,

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Centroid of Lines

If the body consists of a homogeneous

curved wire with a small uniform cross-sectional area A and length L, then AdLdmLAm == ,

=

L

dLL

L

dLx

xdLxdMLx LCL

yzC

=== ,

L

dLy

ydLydMLyL

CL

xzC

=== ,

L

dLz

zdLzdMLz LCL

xyC

=== ,

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STEPS FOR DETERMING AREA CENTROID

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 CG, CM, etc.. 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 interms of dx or dy, respectively, and integrate.

3. Determine coordinates (x , y ) of the centroid of the rectangularelement in terms of the general point (x,y).

~ ~

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EXAMPLE

Solution

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

dA as a vertical rectangular strip.

x,y

x , y~~2. dA = y dx = (9 x2) dx

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

Given: The area as shown.

Find: The centroid location (x , y)

Plan: Follow the steps.

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EXAMPLE (continued)

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

0

0

0 x ( 9 x2) d x [ 9 (x2)/2 (x4) / 4] 30

( 9 x2) d x [ 9 x (x3) / 3 ] 3

= ( 9 ( 9 ) / 2 81 / 4 ) / ( 9 ( 3 ) ( 27 / 3 ) )

= 1.13 ft

3= =

3

33.60 ftA y dA 0 ( 9 x

2) ( 9 x2) dx

A dA 0

( 9 x2) d x

3

=y = =~

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CONCEPT QUIZ

1. The steel plate with known weight and non-uniform thickness and density is supported

as shown. Of the three parameters (CG, CM,

and centroid), which one is needed for

determining the support reactions? Are allthree parameters located at the same point?

A) (center of gravity, no)

B) (center of gravity, yes)C) (centroid, yes)

D) (centroid, no)

2. When determining the centroid of the area above, which type ofdifferential area element requires the least computational work?

A) Vertical B) Horizontal

C) Polar D) Any one of the above.

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EXAMPLE

Given: The area as shown.

Find: The x of the centroid.

Plan: Follow the steps.

Solution

1. Choose dA as a horizontal rectangular

strip.(x1,,y) (x2,y)

2. dA = ( x2

x1) dy

= ((2 y) y2) dy

3. x = ( x1 + x2) / 2

= 0.5 (( 2 y) + y2 )

~

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EXAMPLE(continued)

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

A dA = 0

( 2 y y2) dy

[ 2 y y2 / 2 y3/ 3] 1 = 1.167 m2

1

0

A x dA = 0

0.5 ( 2 y + y2 ) ( 2 y y2 ) dy

= 0.5 0

( 4 4 y + y2 y4 ) dy

= 0.5 [ 4 y 4 y2 / 2 + y3 / 3 y5/ 5 ] 1

= 1.067 m3

0

1

1

~

x = 1.067 / 1.167 = 0.914 m

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ATTENTION QUIZ

1. If a vertical rectangular strip is chosen as the

differential element, then all the variables,

including the integral limit, should be in

terms of _____ .

A) x B) y

C) z D) Any of the above.

2. If a vertical rectangular strip is chosen, then what are the values of

x and y?

A) (x , y) B) (x / 2 , y / 2)

C) (x , 0) D) (x , y / 2)

~ ~

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homework

9-7, 9-9, 9-13, 9-34

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COMPOSITE BODIES (Section 9.3)

Objective:

Students will be able to determine

the location of the center of

gravity, center of mass, or

centroid using the method of

composite bodies.

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CONCEPT OF A COMPOSITE BODY

Many industrial objects can be considered as composite bodies

made up of a series of connected simpler shaped parts or

holes, like a rectangle, triangle, and semicircle.

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

of the simpler shaped parts, we can easily determine the

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

a

be

d

ab

e

d

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This can be done by considering each part as a particle andfollowing the procedure as described in Section 9.1. This is a

simple, effective, and practical method of determining the

location of the centroid or center of gravity.

CONCEPT OF A COMPOSITE BODY (continued)

a

be

d

ab

e

d

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Centroids of common shapes of areas and volumes

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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.

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EXAMPLE

Given: The part shown.

Find: The centroid of the part.

Plan: Follow the steps for analysis.

Solution:

1. This body can be divided into the following pieces

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

area (d)

a

bc

d

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EXAMPLE (continued)

39.8376.528.0

27

4.5

9- 2/3

54

31.5

90

1.5

1

4(3) / (3 )4(1) / (3 )

3

7

4(3) / (3 )0

18

4.5

9 / 4 / 2

Rectangle

Triangle

Q. CircleSemi-Circle

A y

( in3)

A x

( in3)

y

(in)

x

(in)

Area A

(in2)

Segment

Steps 2 & 3: Make up and fill

the table using parts a,

b, c, and d.

a

bc

d

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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

4. Now use the table data and these formulas to find the

coordinates of the centroid.

EXAMPLE (continued)

C

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EXAMPLE - 2

Given: Two blocks of differentmaterials are assembled as shown.

The specific weights of the

materials are

A = 150 lb / ft3 andB = 400 lb / ft3.

Find: The center of gravity of thisassembly.

Plan: Follow the steps for analysis

Solution

1. In this problem, the blocks A and B can be considered as two

segments.

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EXAMPLE 2 (continued)

Weight = w =

(Volume in ft3)

wA = 150 (0.5) (6) (6) (2) / (12)3 = 3.125 lb

wB = 450 (6) (6) (2) / (12)3 = 18.75 lb

62.559.3831.2521.88

6.25

56.25

3.125

56.25

12.5

18.75

2

3

1

3

4

1

3.125

18.75

A

B

wz(lbin)

w y(lbin)

w x(lbin)

z (in)y (in)x (in)w (lb)Segment

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EXAMPLE 2(continued)

~x = ( x w) / ( w ) = 31.25/21.88 = 1.47 iny = ( y w) / ( w ) = 59.38/21.88 = 2.68 in

z = (

z w) / ( w ) = 62.5 /21.88 = 2.82 in~~

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homework

9-44, 9-55, 9-61, 9-83

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