mech chapter 09

7/25/2019 Mech Chapter 09

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Center of Gravity and Centroid9

Engineering Mechanics:

Statics in SI Units, 12e

Copyright 2010 Pearson Education South Asia Pte Ltd


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Chapter Outline

1. Center of Gravity and Center of Mass for a System of

Particles

2. Composite Bodies

3. Theorems of Pappus and Guldinus 4. Resultants of a General Distributed Loading

5. Fluid Pressure


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Chapter Outline

1. Center of Gravity and Center of Mass for a System

of Particles

2. Composite Bodies

3. Theorems of Pappus and Guldinus4. Resultants of a General Distributed Loading

5. Fluid Pressure


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9.1 Center of Gravity and Center of Mass

for a System of Particles

Center of Gravity

Consider system of nparticles fixed within a region of

space

The weightsof the particles can be replaced by asingle (equivalent) resultant weight having defined

point G of application


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Center of Gravity

Resultant weight = total weight of nparticles

Sum of moments of weights of all the particles about x,

y, z axes = moment of resultant weight about these

axes

dWW


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Center of Gravity

dWzWz

dWyWy

dWxWx

~

~

~

dW

dWzz

dW

dWyy

dW

dWxx

~

~

~


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Center Mass

Provided acceleration due to gravity g for every

particle is constant, then W = mg

A rigid body is composed of an infinite number ofparticles

dm

dmzz

dm

dmyy

dm

dmxx

~

;

~

;

~





Centroid of a Volume

Consider an object subdivided into volume elements

dV, for location of the centroid,

V

V

V

V

V

V

dV

dVz

zdV

dVy

ydV

dVx

x

~

;

~

;

~





Centroid of an Area

For centroid for surface area of an object, such as

plate and shell, subdivide the area into differential

elements dA

A

A

A

A

dA

dAy

y

dA

dAx

x

~

~





Centroid of a Line

If the geometry of the object takes the form of a line,

the balance of moments of differential elements dL

about each of the coordinate system yields

L

L

L

L

dL

dLy

y

dL

dLx

x

~

~



Example (1)

Locate the centroid of the rod bent into the shape of a

parabolic arc.

L

L

L

L

dL

dLy

y

dL

dLx

x

~

~



Example

For differential length of the element dL

Since x = y2and then dx/dy = 2y

The centroid is located at

yyxx ~,~

dydy

dxdydxdL 1

2

22

dyydL 12 2


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Example

Integrations

m

dyy

dyyy

dL

dLy

y

m

dyy

dyyy

dyy

dyyx

dL

dLx

x

L

L

L

L

574.0479.1

8484.0

14

14~

410.0479.1

6063.0

14

14

14

14~

1

0

2

1

0

2

1

0

2

1

0

22

1

0

2

1

0

2

x = y2


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Example (2)

Locate the centroid of the area ()


A

A

A

A

dA

dAy

y

dA

dAx

x

~

~


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Example

For differential element dA


2

~

~

yy

xx

dxydA


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Example


m

dxx

dxxx

dxy

dxyy

dA

dAy

y

m

dxx

dxx

dxy

dxxy

dA

dAx

x

A

A

A

A

3.0333.0

100.0

)2

()2

(~

75.0333.0

250.0

~

1

0

2

1

0

22

1

0

1

0

1

0

2

1

0

3

1

0

1

0

y = x2

2

~

~

yy

xx

dxydA


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Test

Locate the y centroid for the paraboloid of revolution

(y)


V

V

dV

dVy

y

~


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Test

For differential element dV


yy~

dyzdV )( 2


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Test


mm

dyy

dyyy

dyz

dyzy

dV

dVy

y

V

V

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))100((

))100((

)(

)(

~

100

0

100

0

100

0

2

100

0

2


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Chapter Outline


Particles

2. Composite Bodies


5. Fluid Pressure


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9.2 Composite Bodies

Consistsof a series of connected simplershaped

bodies, which may be rectangular, triangular or

semicircular

A body can be divided into its composite parts Accounting for finite number of weights

W

Wzz

W

Wyy

W

Wxx

~~~


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9.2 Composite Bodies

Procedure for Analysis

1. Divide the body

2. Determine the coordinates of centroid of each part

3. Summations

W

Wzz

W

Wyy

W

Wxx

~~~


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Example

Locate the centroid of the plate area.


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Solution

Plate divided into 3 segments.

Area of small rectangle considered negative.


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Solution

Location of the centroid for each piece is determined and

indicated in the diagram.

A

Ayy

A

Axx

~~


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Solution

Summations

mmA

Ayy

mmA

Axx

22.15.11

14~

348.05.11

4~

A

Ayy

A

Axx

~~


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Test

Locate the centroid of the wire ()


L

Lz

z

L

Lyy

L

Lx

x

~

~

~

ry

2


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Test


ry

2


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Test



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Test



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Chapter Outline


Particles

2. Composite Bodies


5. Fluid Pressure


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9.3 Theorems of Pappus and Guldinus

The theorems of Pappus and Guldinus (

) are used to find the surfaces areaand volumeof

any object of revolution () provided the

generating curves and areas do not cross the axis

they are rotated


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Surface Area

A surface area of revolution is generated by

revolving a plane curveabout a non-intersecting

fixed axisin the plane of the curve


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Area of a surface of revolution = product of length of

the curve and distance traveled by the centroid in

generating the surface area

(A)(L)

LrA Note:


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Example

Show that the surface area of a sphereisA = 4R2.

(A = 4R2)


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Example -- Solution

Surface Area --

Generated by rotating semi-arcabout the x axis

For centroid,

For surface area,

/2Rr

242

2 RRR

A

LrA


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Volume

A volume of revolution is generated by revolving a

plane areabout a nonintersecting fixed axis in the

plane of area


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Volume of a body of revolution = product of

generating area and distance traveled by the centroid

in generating the volume

(V)(A)

ArV

Note:


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Example

Show that the volume of a sphereis V = 4/3 R3.

(V = 4/3 R3)


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Example -- Solution

Volume --

Generated by rotating semicircular area about the x axis

For centroid,

For volume,

32

3

4

2

1

3

42 RRR

V

ArV

3

4Rr


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Composite shapes

The total surface area or volume generatedis the

additionof the surface area or volumes generated by

eachof the composite parts

ArV ~

LrA

~


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Test

Determine the surface area and volume of the full solid.


ArV ~ LrA ~


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Test -- Answer


)(143)]1)(3()3)(5.3()11)(3()2)(5.2[(2

~2

~

222 m

LrA

LrA

(1) (3)

(4)

(2)

(1) (3) (4)(2)


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Test -- Answer


)(6.47)}12)(3()112

1)(1667.3{(2

~2

~

3m

ArV

ArV

(1)

(2)

(1) (2)


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Chapter Outline


Particles

2. Composite Bodies


5. Fluid Pressure


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9.4 Resultant of a General Distributed

Loading

Pressure Distribution over a Surface

The flat plate subjected to the loading function

p=p(x, y) [N/m2] Pa(Pascal)

The force dF acting on the differential area dA m2of

the plate, located at the differential point (x, y)

dF = p(x, y) [N/m2] dA[m2]

= p(x, y) dA [N]= dV


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Loading

Magnitude of resultant force (FR) = total volume under

the distributed loading diagram

VdVdAyxpFVA

R ),(


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Loading

Location of Resultant Forceis

V

V

A

A

V

V

A

A

dV

ydV

dAyxp

dAyxypy

dV

xdV

dAyxp

dAyxxpx

),(

),(

),(

),(


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Chapter Outline


Particles

2. Composite Bodies


5. Fluid Pressure


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9.5 Fluid Pressure

According to Pascals law, a fluid at rest creates a

pressurepat a point that is the same in all directions

Magnitude ofpdepends on the specific weight or

mass density of the fluid and the depth zof the point

from the fluid surface

p= z = gz

Valid for incompressible fluids

Gas are compressible fluids and the above equation

cannot be used


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9.5 Fluid Pressure



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9.5 Fluid Pressure

Flat Plate of Constant Width

Consider flat rectangular plate of constant width

submerged in a liquid having a specific weight

Plane of the plate makes an angle with the horizontal


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9.5 Fluid Pressure

Flat Plate of Constant Width

As pressure varies linearly with depth, the distribution

of pressure over the plates surface is represented by a

trapezoidal volume having an intensity of p1= z1 at

depth z1and p2= z2 at depth z2

Magnitude of the resultant force FR

= volume of this loading diagram


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Example

Determine the magnitude and location of the resultant

hydrostatic force acting on the submerged rectangular

plate AB. The plate has a width of 1.5m; w= 1000kg/m3.

(AB)


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Solution

The water pressures at depth A and B are

(AB)

kPamsmmkggzp

kPamsmmkggzp

BwB

AwA

05.49)5)(/81.9)(/1000(

62.19)2)(/81.9)(/1000(

23

23


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Solution

The intensities of the load at A and B

(AB()3D2D)

mkNkPambpwmkNkPambpw

BB

AA

/58.73)05.49)(5.1(/43.29)62.19)(5.1(


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Solution

The magnitude of the resultant force FR

This force acts through the centroid

of the area ()

measured upwards from B

N

trapezoidofareaFR

5.154)6.734.29)(3(2

1

mh 29.1)3(58.7343.29

58.73)43.29(2

3

1


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Test

Determine the magnitudeof the hydrostatic force acting

on gate AB, which has a width of 0.5m. The specific

weight of water is = 10 kN/m3. (AB

)



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Solution

The water pressures at depth A and B are

(AB)

2

2

/k30)2.18.1)(10(

/k18)8.1)(10(

mNzp

mNzp

BB

AA


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Solution

The intensities of the load at A and B

(AB()3D2D)

mNbpw

mNbpw

BB

AA

/k15)30)(5.0(

/k9)18)(5.0(


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Solution


KN

trapezoidofareaFR

18

)9.02.1)(159(21 22


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9.5 Fluid Pressure

Flat Plate of Variable Width

Since uniform pressure p = z (force/area) acts on dA,

the magnitude of the differential force dF

dF = p dA = (z) dA

)( AzF

AzzdA

zdAdFF

R

R


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Test

Determine the magnitudeof the hydrostatic force actingon gate AB, which has a width of 0.5m. The specific

weight of water is = 10K N/m3. (AB

)



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Solution


kN

AzFR

18

)5.0*9.02.1(*)22.18.1(*10

)(

22


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9.5 Fluid Pressure

**Curved Plate of Constant Width

When the submerged plate is curved, the pressure

acting normal to the plate continuously changes

direction

Integrationcan be used to determine FRand location of

center of centroid C or pressure P


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END OF CHAPTER 9

mech chapter 09

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