Chapter eight Laith Batarseh - Philadelphia University · Chapter eight Laith Batarseh Home ......

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Chapter eight

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Center of Mass and Center of Gravity

In this lecture we will

Define the concepts of center of mass and center of gravity and the centroid

Learn how to locate the center of mass and center of gravity and the centroid

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Center of gravity

x

y

z dw

x

y

z

x

y

z W

'x

'y

CG

'z


Center of gravity

dwW

Weight

The weight of the body is found by integrating the small portions of dw.

The weight of a body acts at a point in the body called the center ofgravity

The position of the center of gravity is found by applying the equilibriumof moment equations where the summation of moment at C.G equalzero

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Center of gravity

The location of C.G of a certain body is found as:

dw

dwxx'

dw

dwyy'

dw

dwzz'


Center of mass

As in the center of gravity, center of mass represent the point where wecan assume that all the mass is concentrated in it. Mathematically, thelocation of the center of mass can be found by:

dm

dmxx'

dm

dmyy'

dm

dmzz'

dW = g dm

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Center of volume

If the body is made from a constant density material, then the center ofvolume can be found from the relation between the mass and thedensity:

dV

dVxx'

dV

dVyy'

dV

dVzz'

dm = r dV


Centroid of an Area

dA

dAxx'

dA

dAyy'

Centroid of an Line

dL

dLxx'

dL

dLyy'

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Analysis procedures

Differential Element

1. Define the coordinate system.

2. Chose a suitable differential element for the integration

a. For line: line segment dL

b. For area: rectangular element (dA = L dW; L is the length and W is

the width)

c. For volume: circular disk ( dV = πr2.dt; r is the radius and t is the

thickness)

3. Be sure that the chosen element touch an arbitrary point P(x,y,z) located

at the curve that define the boundary of the body


Analysis procedures

Size and Moment Arms

1. Express the chosen element in terms of the pre described coordinates

2. Express the moment arms ( , , ) in terms of the pre described

coordinates (x, y, z)

x y z

Integration

1. Substitute the formulas of moment arms and the chosen elements in

the appropriate equation.

2. Rearrange the equation in the integration to be in the same variable

3. Define the limits of the integration

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Example: problem 9-18

The plate is made of steel having a density of 7850 kg/m3. If the thickness of the plate is 10 mm, determine the horizontal and vertical components of reaction at the pin A and the tension in cable BC.



Solution

Differential element and coordinate system

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Solution

F.B.D



Solution

Centroid

dxxxdydA .26.1 31Area element

Centroid location xx 2/yy

Area2

4

0

31 6.26.1 mdxxdAA

Weight (w) NAtgmgw 462181.901.067850

Centroid location

m

dxxx

dA

dAxx 3.2

6

.26.1

'

4

0

31

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Solution

Equilibrium conditions

NAAFy yy 19640462126570

NFFM CBBCA 265703.2462140

00 xAFx

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Centroid

Centroid of an Area

dA

dAxx'

dA

dAyy'

Centroid of an Line

dL

dLxx'

dL

dLyy'

Centroid

Analysis procedures

Differential Element

1. Define the coordinate system.

2. Chose a suitable differential element for the integration

a. For line: line segment dL

b. For area: rectangular element (dA = L dW; L is the length and W is

the width)

c. For volume: circular disk ( dV = πr2.dt; r is the radius and t is the

thickness)

3. Be sure that the chosen element touch an arbitrary point P(x,y,z) located

at the curve that define the boundary of the body

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Centroid

Analysis procedures

Size and Moment Arms

1. Express the chosen element in terms of the pre described coordinates

2. Express the moment arms ( , , ) in terms of the pre described

coordinates (x, y, z)

x y z

Integration

1. Substitute the formulas of moment arms and the chosen elements in

the appropriate equation.

2. Rearrange the equation in the integration to be in the same variable

3. Define the limits of the integration

Centroid

Example: centroid of triangle

Locate the location of the centroid from the base for thetriangle given below

h

b

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Centroid


Solution


h

b

y

x

dy

x

(x,y)( , )

y

x y

Centroid


Solution

Area and integration formulas

xdydA Area element

Relate x to y using triangles symmetry dyyhh

bdA

3

.

.

'

0

0 h

dyyhh

b

dyyhh

by

dA

dAyy

h

h

Centroid location

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Centroid

Common cases

Arc segment

y

x

0'y

sin

3

2'

rx

Ө

r

Centroid

Common cases

Quarter circle

y

x

3

4''

ryx

Semicircle

y

x

r r

0'y3

4'

rx

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Centroid

Common cases

Rectangle Circle

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Centroid

Centroid of an Area

dA

dAxx'

dA

dAyy'

Centroid of an Line

dL

dLxx'

dL

dLyy'

Centroid


Determine the area and the centroid of the area.

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Centroid


Solution


Centroid


Solution

Centroid

dyy

xdydA4

2

Area element

Centroid location

82

4/2/

22 yyxx yy

Area 2

4

0

2

33.5.4

mdyy

dAA

Centroid location

m

dyyy

dA

dAxx 2.1

33.5

.48

'

4

0

22

m

dyy

y

dA

dAyy 3

33.5

.4

'

4

0

2

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Centroid


Determine the area and the centroid of the area from the y-axis.

Centroid


Solution


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Centroid


Solution

Centroid

Centroid location xx

Area 2

0

2

3

2.

22. kadx

a

xxkdxydAA

a

Centroid location

aka

ka

ka

dxa

xxkx

dA

dAxx

8

5

3/2

12/5

3/2

.2

2

'2

3

2

2

0

2

Centroid


Determine the area and the centroid of the area from the y-axis.

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Centroid


Solution


Centroid


Solution

Centroid

dxxxdxyydAxyxy .., 22/1

21

2

2

2/1

1 Area element

Centroid location xx yy

Area 2

1

0

22/1

3

1. mdxxxdAA

Centroid location

m

dxxxx

dA

dAxx 45.0

20

9

3/1

20/3

20

.

'

1

0

22/1

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Composite Bodies

Basic concept

Composite bodies are a series of simple shaped

The centroid of the composite body is found by finding the centroid for each shape and then apply superposition principle.

W

Wxx'

W

Wzz'

W

Wyy'

Where:

W is the individual weight of each shape

∑W is the total weight of the body

Composite Bodies

For length

L

Lxx'

L

Lzz'

L

Lyy'

For area

A

Axx'

A

Azz'

A

Ayy'

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Composite Bodies


Each of the three members of the framehas a mass per unit length of 6 kg/m.Locate the position of the center of mass.Neglect the size of the pins at the jointsand the thickness of the members. Also,calculate the reactions at the pin A androller E.

Composite Bodies


Solution

Centroid location

Seg L

C-D-E 8 4 13 32 104

B-D 7.2 2 10 14.4 72

A-B-C 13 0 6.5 0

∑ 28.21 46.4 260.6

x y Lx Ly

mL

Lxx 65.1

21.28

4.46'

mL

Lyy 24.9

21.28

60.260'

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Composite Bodies


Locate the centroid of the cross-sectional area of the channel.

Composite Bodies


Solution

Centroid location

Seg A

A-B (9)(1) =9 5.5 0.5 49.5 4.5

B-C (1)(24)=24 0.5 12 12 288

C-D (9)(1) =9 5.5 23.5 49.5 211.5

∑ 42 111 504

x y Ax Ay

''64.242

111'

A

Axx ''12

42

504'

A

Ayy

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Chapter eight

Laith Batarseh

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Composite Bodies

In this lecture

we will solve some examples on how to find the centroid of compositebodies

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Composite Bodies


Locate the centroid of the composite area shown in the Fig

Composite Bodies


Solution

Decompose the area

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Composite Bodies


Solution

Decompose the area

Composite Bodies


Solution

Centroid location

Area A

1 9 1.5 1.5 13.5 13.5

2 4.5 2 4 9 18

3 42 6.5 3 273 126

4 -(π32)/4 10-(4/π) 6-(4/π) -61.7 -33.4

∑ 48.4 233.8 124.1

x y Ax Ay

''83.44.48

8.233'

A

Axx ''56.2

4.48

1.124'

A

Ayy

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Composite Bodies


Locate the centroid of the composite area shown in the Fig

Composite Bodies


Solution

Decompose the area

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Composite Bodies


Solution

Decompose the area

Composite Bodies


Solution

Centroid location

x y Ax AyArea A

1 (9/4)π (2/π) 1.5 4.5 (27/8)π

2 9 1.5 1.5 13.5 13.5

3 4.5 4 1 18 4.5

4 -π 0 1.5 -0 -1.5π

∑ 17.43 36 23.9

'06.243.17

36'

A

Axx '37.1

43.17

9.23'

A

Ayy

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Chapter eight Laith Batarseh - Philadelphia University · Chapter eight Laith Batarseh Home ......

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