ch07-distributed-forces-centroids-and-centers-of-gravity.ppt

7/30/2019 ch07-distributed-forces-centroids-and-centers-of-gravity.ppt

1/57

2007 The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: StaticsEighth

Edition

5 - 1

CE 102 Statics

Chapter 7Distributed Forces:

Centroids and Centers of Gravity


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Edition

5 - 2

Contents

Introduction

Center of Gravity of a 2D Body

Centroids and First Moments of Areas

and Lines

Centroids of Common Shapes of AreasCentroids of Common Shapes of Lines

Composite Plates and Areas

Sample Problem 7.1

Determination of Centroids by

Integration

Sample Problem 7.2

Theorems of Pappus-Guldinus

Sample Problem 7.3

Distributed Loads on Beams

Sample Problem 7.4

Center of Gravity of a 3D Body:Centroid of a Volume

Centroids of Common 3D Shapes

Composite 3D Bodies

Sample Problem 7.5


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Edition

5 - 3

Introduction

The earth exerts a gravitational force on each of the particles

forming a body. These forces can be replace by a single

equivalent force equal to the weight of the body and applied

at the center of gravity for the body.

The centroid of an area is analogous to the center of

gravity of a body. The concept of thefirst moment of anarea is used to locate the centroid.

Determination of the area of asurface of revolution and

the volume of a body of revolution are accomplished

with the Theorems of Pappus-Guldinus.


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Edition

5 - 4

Center of Gravity of a 2D Body

Center of gravity of a plate

dWy

WyWyM

dWx

WxWxM

y

y

Center of gravity of a wire

EE


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Edition

5 - 5

Centroids and First Moments of Areas and Lines

x

QdAyAy

yQdAxAx

dAtxAtx

dWxWx

x

y

respect tohmoment witfirst

respect tohmoment witfirst

Centroid of an area

dLyLydLxLx

dLaxLax

dWxWx

Centroid of a line

EE


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Edition

5 - 6

First Moments of Areas and Lines

An area is symmetric with respect to an axis BB

if for every pointPthere exists a pointP

suchthat PP is perpendicular to BBand is divided

into two equal parts by BB.

The first moment of an area with respect to a

line of symmetry is zero.

If an area possesses a line of symmetry, its

centroid lies on that axis

If an area possesses two lines of symmetry, its

centroid lies at their intersection.

An area is symmetric with respect to a centerO

if for every element dA at (x,y) there exists an

area dAof equal area at (-x,-y).

The centroid of the area coincides with the

center of symmetry.

EE


7/57 2007 The McGraw-Hill Companies, Inc. All rights reserved.


Edition

5 - 7

Centroids of Common Shapes of Areas

EE




Edition

5 - 8

Centroids of Common Shapes of Lines

V M h i f E i S iEE




Edition

5 - 9

Composite Plates and Areas

Composite plates

WyWY

WxWX

Composite area

AyAY

AxAX

V t M h i f E i St tiEE




Edition

5 - 10

Sample Problem 7.1

For the plane area shown, determine

the first moments with respect to the

x andy axes and the location of the

centroid.

SOLUTION:

Divide the area into a triangle, rectangle,

and semicircle with a circular cutout.

Compute the coordinates of the area

centroid by dividing the first moments by

the total area.

Find the total area and first moments ofthe triangle, rectangle, and semicircle.

Subtract the area and first moment of the

circular cutout.

Calculate the first moments of each area

with respect to the axes.





Edition

5 - 11

Sample Problem 7.1

33

33

mm107.757

mm102.506

y

x

Q

Q Find the total area and first moments of the

triangle, rectangle, and semicircle. Subtract the

area and first moment of the circular cutout.





Edition

5 - 12

Sample Problem 7.1

23

33

mm1013.828mm107.757

A

AxX

mm8.54X

23

33

mm1013.828

mm102.506

A

AyY

mm6.36Y

Compute the coordinates of the area

centroid by dividing the first moments bythe total area.





Edition

5 - 13

Determination of Centroids by Integration

ydxy

dAyAy

ydxx

dAxAx

el

el

2

dxxay

dAyAy

dxxaxa

dAxAx

el

el

2

drr

dAyAy

drr

dAxAx

el

el

2

2

2

1sin

3

2

2

1cos

3

2

dAydydxydAyAy

dAxdydxxdAxAx

el

el Double integration to find the first moment

may be avoided by defining dA as a thin

rectangle or strip.





Edition

5 - 14

Sample Problem 7.2

Determine by direct integration the

location of the centroid of a parabolic

spandrel.

SOLUTION: Determine the constant k.

Evaluate the total area.

Using either vertical or horizontal

strips, perform a single integration tofind the first moments.

Evaluate the centroid coordinates.

V t M h i f E i St tiE

E




Edition

5 - 15

Sample Problem 7.2

SOLUTION:

Determine the constant k.

2121

22

2

2

2

yb

axorxaby

a

bkakb

xky

Evaluate the total area.

3

30

3

20

2

2

ab

x

a

bdxx

a

bdxy

dAA

aa

V t M h i f E i St tiEi

Ed




Edition

5 - 16

Sample Problem 7.2

Using vertical strips, perform a single integration

to find the first moments.

1052

2

1

2

44

2

0

5

4

2

0

22

2

2

0

4

2

0

2

2

abx

a

b

dxxa

bdxy

ydAyQ

bax

a

b

dxxa

bxdxxydAxQ

a

a

elx

a

a

ely

V t M h i f E i St tiEi

Ed




Edition

5 - 17

Sample Problem 7.2

Or, using horizontal strips, perform a single

integration to find the first moments.

10

42

1

22

2

0

23

21

21

21

2

0

22

0

22

abdyy

b

aay

dyyb

aaydyxaydAyQ

badyyb

aa

dyxa

dyxaxa

dAxQ

b

elx

b

b

ely

Vector Mechanics for Engineers: StaticsEi

Ed



Vector Mechanics for Engineers: Staticsighth

dition

5 - 18

Sample Problem 7.2

Evaluate the centroid coordinates.

43

2baabx

QAx y

ax4

3

103

2abab

y

QAy x

by10

3


Ed



Vector Mechanics for Engineers: Staticsghth

dition

5 - 19


Surface of revolution is generated by rotating a

plane curve about a fixed axis.

Area of a surface of revolution is

equal to the length of the generatingcurve times the distance traveled by

the centroid through the rotation.

LyA 2


Ed




dition

5 - 20


Body of revolution is generated by rotating a planearea about a fixed axis.

Volume of a body of revolution is

equal to the generating area times

the distance traveled by the centroid

through the rotation.

AyV 2

Vector Mechanics for Engineers: StaticsEig

Ed




dition

5 - 21

Sample Problem 7.3

The outside diameter of a pulley is 0.8

m, and the cross section of its rim is as

shown. Knowing that the pulley is

made of steel and that the density ofsteel is

determine the mass and weight of the

rim.

33 mkg1085.7

SOLUTION:

Apply the theorem of Pappus-Guldinusto evaluate the volumes or revolution

for the rectangular rim section and the

inner cutout section.

Multiply by density and accelerationto get the mass and acceleration.


Ed




dition

5 - 22

Sample Problem 7.3

SOLUTION:

Apply the theorem of Pappus-Guldinusto evaluate the volumes or revolution for

the rectangular rim section and the inner

cutout section.

3393633 mmm10mm1065.7mkg1085.7Vm kg0.60m

2

sm81.9kg0.60 mgW N589

W

Multiply by density and acceleration to

get the mass and acceleration.


Ed




dition

5 - 23

Distributed Loads on Beams

A distributed load is represented by plotting the loadper unit length, w (N/m) . The total load is equal to

the area under the load curve (dW = wdx).

AdAdxwWL

0

AxdAxAOP

dWxWOPL

0

A distributed load can be replace by a concentratedload with a magnitude equal to the area under the

load curve and a line of action passing through the

area centroid.


Ed




dition

5 - 24

Sample Problem 7.4

A beam supports a distributed load as

shown. Determine the equivalent

concentrated load and the reactions atthe supports.

SOLUTION:

The magnitude of the concentrated load

is equal to the total load or the area under

the curve.

The line of action of the concentrated

load passes through the centroid of thearea under the curve.

Determine the support reactions by

summing moments about the beam

ends.


Ed


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ition

5 - 25

Sample Problem 7.4

SOLUTION:

The magnitude of the concentrated load is equal to

the total load or the area under the curve.

kN0.18F

The line of action of the concentrated load passes

through the centroid of the area under the curve.

kN18

mkN63 X m5.3X


Ed


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ition

5 - 26

Sample Problem 7.4

Determine the support reactions by summing

moments about the beam ends.

0m.53kN18m6:0 yA BM

kN5.10yB

0m.53m6kN18m6:0 yB AM

kN5.7y

A


Edi


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ition

5 - 27

Center of Gravity of a 3D Body: Centroid of a Volume

Center of gravity G

jWjW

jWrjWr

jWrjWr

G

G

dWrWrdWWG

Results are independent of body orientation,

zdWWzydWWyxdWWx

zdVVzydVVyxdVVx

dVdWVW and

For homogeneous bodies,


Edi


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ition

5 - 28

Centroids of Common 3D Shapes


Edi


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tion

5 - 29

Composite 3D Bodies

Moment of the total weight concentrated at the

center of gravity G is equal to the sum of themoments of the weights of the component parts.

WzWZWyWYWxWX

For homogeneous bodies,

VzVZVyVYVxVX


Edi


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Vector Mechanics for Engineers: Staticshthtion

5 - 30

Sample Problem 7.5

Locate the center of gravity of thesteel machine element. The diameter

of each hole is 1 in.

SOLUTION:

Form the machine element from arectangular parallelepiped and a

quarter cylinder and then subtracting

two 1-in. diameter cylinders.


Edit


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5 - 31

Sample Problem 7.5

Vector Mechanics for Engineers: StaticsEigh

Edit


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5 - 32

Sample Problem 7.5

34 in.2865in08.3 VVxX

34 in.2865in5.047 VVyY

34 in.2865in.6181 VVzZ

in.577.0X

in.577.0Y

in.577.0

Z

P bl 7 6


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33

y

x

20 mm 30 mm

Problem 7.6

36 mm

24 mm

Locate the centroid of the planearea shown.

y

Problem 7 6


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34

Solving Problems on Your Own

Several points should be emphasized

when solving these types of problems.

Locate the centroid of the plane area

shown.

y

x

20 mm 30 mm

36 mm

24 mm

1. Decide how to construct the given area from common shapes.

2. It is strongly recommended that you construct a table

containing areas or length and the respective coordinates ofthe centroids.

3. When possible, use symmetry to help locate the centroid.

Problem 7.6

Problem 7 6 Solutiony


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35

Problem 7.6 Solutiony

x

24 + 12

20 + 10

10

30

Decide how to construct the given

area from common shapes.

C1

C2

Dimensions in mm



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36


x

24 + 12

20 + 10

10

30

C1

C2

Dimensions in mm

Construct a table containing areas and

respective coordinates of the

centroids.

A, mm2 x, mm y, mm xA, mm3 yA, mm3

1 20 x 60 =1200 10 30 12,000 36,0002 (1/2) x 30 x 36 =540 30 36 16,200 19,440

S 1740 28,200 55,440



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37


x

24 + 12

20 + 10

10

30

C1

C2

Dimensions in mm

XSA = SxAX (1740) = 28,200

Then

or X = 16.21 mm

and YSA = SyA

Y (1740) = 55,440or Y = 31.9 mm

A, mm2 x, mm y, mm xA, mm3 yA, mm3

1 20 x 60 =1200 10 30 12,000 36,000

2 (1/2) x 30 x 36 =540 30 36 16,200 19,440

S 1740 28,200 55,440

Problem 7 7


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38

Problem 7.7

The beamAB supports two

concentrated loads and

rests on soil which exerts a

linearly distributed upward

load as shown. Determine

(a) the distance a for whichw

A= 20 kN/m, (b) the

corresponding value wB.

wA wB

A B

a 0.3 m24 kN 30 kN

1.8 m

Problem 7 7


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39

The beamAB supports two

concentrated loads andrests on soil which exerts a

linearly distributed upward

load as shown. Determine

(a) the distance a for which

wA

= 20 kN/m, (b) the

corresponding value wB.

wA wB

A B

a 0.3 m24 kN 30 kN

1.8 m


1. Replace the distributed load by a single equivalent force.

The magnitude of this force is equal to the area under the

distributed load curve and its line of action passes throughthe centroid of the area.

2. When possible, complex distributed loads should be

divided into common shape areas.

Problem 7.7

Problem 7.7 Solution


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40

wB

A B

a 0.3 m24 kN 30 kN

20 kN/m

C

0.6 m 0.6 m

RI

RII

We have RI = (1.8 m)(20 kN/m) = 18 kN1

2

RII = (1.8 m)(wBkN/m) = 0.9 wBkN1

2

Replace the distributed

load by a pair of

equivalent forces.



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41

wB

A B

a 0.3 m24 kN 30 kN

C

0.6 m 0.6 m

RI = 18 kN RII = 0.9 wBkN

(a) SMC= 0: (1.2 - a)m x 24 kN - 0.6 m x 18 kN- 0.3m x 30 kN = 0

or a = 0.375 m

(b) SFy= 0: -24 kN + 18 kN + (0.9 wB) kN - 30 kN= 0or wB= 40 kN/m

+

+


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42

Problem 7.8

y

x

0.75 inz

1 in2 in2 in

3 in

2 in2 in

r= 1.25 in

r= 1.25 in

For the machine element

shown, locate thezcoordinateof the center of gravity.

y Problem 7.8


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43


y

x

0.75 inz

1 in2 in2 in

3 in

2 in 2 in

r= 1.25 in

r= 1.25 in

XS V = Sx V YS V = Sy V ZS V = Sz V

whereX, Y, Zandx, y, zare the coordinates of the centroid of the

body and the components, respectively.

For the machine element

shown, locate thezcoordinateof the center of gravity.

Determine the center of

gravity of composite body.

For a homogeneous body

the center of gravity coincides

with the centroid of its volume. For this case the center of gravity

can be determined by



44/57

44

y

x

0.75 inz

1 in2 in2 in

3 in

2 in 2 in

r= 1.25 in

r= 1.25 in

Determine the center of gravity

of composite body.

First assume that the machine

element is homogeneous so

that its center of gravity will

coincide with the centroid of

the corresponding volume.

y

x

z

I

IIIII

IVV

Divide the body into

five common shapes.

y y


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45

y

x

z

I

IIIII

IVV

y

x

0.75 inz

1 in2 in2 in

3 in

2 in2 in

r= 1.25 in

r= 1.25 in

V, in3 z, in. zV, in4

I (4)(0.75)(7) = 21 3.5 73.5

II (/2)(2)2 (0.75) = 4.7124 7+ [(4)(2)/(3)] = 7.8488 36.987

III -(11.25)2

(0.75)= -3.6816 7 -25.771IV (1)(2)(4) = 8 2 16

V -(/2)(1.25)2 (1) = -2.4533 2 -4.9088

S 27.576 95.807

ZS V = Sz V: Z(27.576 in3 ) = 95.807 in4 Z= 3.47 in

Problem 7 9


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46

Problem 7.9

Locate the centroid of the volumeobtained by rotating the shaded area

about thex axis.

h

a

x

yy = kx1/3

k 1/3y Problem 7.9


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47


1. When possible, use symmetry to help locate the centroid.

Locate the centroid of the volume

obtained by rotating the shaded areaabout thex axis.

h

a

y = kx1/3

x

y

The procedure for locating the

centroids of volumes by direct

integration can be simplified:

2. If possible, identify an element of volume dV which produces

a single or double integral, which are easier to compute.

3. After setting up an expression fordV, integrate and determine

the centroid.



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48

x

y

z x

dx

y = kx1/3

r

Use symmetry to help locate the

centroid. Symmetry implies

y = 0 z = 0

Identify an element of volume dV

which produces a single ordouble integral.

Choose as the element of volume a disk or radius rand

thickness dx. Then

dV = r2 dx xel

= x



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49

x

y

z x

dx

y = kx1/3

r

Identify an element of volume dV

which produces a single or

double integral.

dV = r2 dx xel

= x

Now r= kx1/3 so that

dV = k2 x2/3dx

Atx= h, y= a : a = kh1/3 or k= a/h1/3

Then dV = x2/3dxa2

h2/3



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50

x

y

z x

dx

y = kx1/3

r

dV = x2/3dxa2

h2/3

Integrate and determine the centroid.

0

h

V = x2/3dxa2

h2/3

= x5/3a2

h2/335[ ]0

h

= a2h3

5

0h

xel dV = x ( x2/3 dx) = [ x8/3 ]a2

h2/3

a2h2/3

38

Also

= a2h23

8



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51

x

y

z x

dx

y = kx1/3

r

Integrate and determine the centroid.

V = a2h3

5

xel

dV = a2h23

8

Now xV= xdV: 38x ( a2h) = a2h23

5

x = h5

8

y = 0 z = 0

Problem 7.10


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52

The square gateAB is held in theposition shown by hinges along its

top edgeA and by a shear pin at B.

For a depth of waterd= 3.5 ft,

determine the force exerted on the

gate by the shear pin.

A

B

d

1.8 ft

30o

Problem 7.10


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53


The square gateAB is held in the

position shown by hinges along itstop edgeA and by a shear pin at B.




A

B

d

1.8 ft

30o

Assuming the submerged body has a width b, the load per unit

length is w= bgh, where h is the distance below the surface ofthe fluid.

1. First, determine the pressure distributionacting perpendicular

the surface of the submerged body. The pressure distribution

will be either triangular or trapezoidal.

Problem 7.10


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54


The square gateAB is held in the

position shown by hinges along itstop edgeA and by a shear pin at B.




A

B

d

1.8 ft

30o

2. Replace the pressure distribution with a resultant force, and

construct the free-body diagram.

3. Write the equations of static equilibrium for the problem, and

solve them.



55/57

55

A

B

Determine the pressure distribution

acting perpendicular the surface of the

submerged body.

1.7 ft

(1.8 ft) cos 30o

PA

PB

PA = 1.7 gPB= (1.7 + 1.8 cos 30

o

)g

Problem 7.10 SolutionAy


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56

A

B

(1.8 ft) cos 30o

Replace the pressure

distribution with a

resultant force, and

construct the free-bodydiagram.

y

Ax

FB

1.7 g

(1.7 + 1.8 cos 30o)g

LAB/3

LAB/3

LAB/3

P1

P2

The force of the water

on the gate is

P= Ap = A(gh)1

21

2

P1 = (1.8 ft)2(62.4 lb/ft3)(1.7 ft) = 171.85 lb

1

2

P2 = (1.8 ft)2(62.4 lb/ft3)(1.7 + 1.8 cos 30o)ft = 329.43 lb

1

2

Problem 7.10 SolutionAy


57/57

A

B

(1.8 ft) cos 30o

y

Ax

FB

1.7 g

(1.7 + 1.8 cos 30o)g

LAB/3

LAB/3

LAB/3

P1

P2

P1 = 171.85 lb P2 = 329.43 lb

Write the equations of

static equilibrium for the

problem, and solve them.

SMA = 0:

( LAB)P1 + ( LAB)P21

3

2

3

-LABFB = 0

+

(171.85 lb) + (329.43 lb) - FB = 0132

3 FB = 276.90 lb

30oFB = 277 lb

ch07-distributed-forces-centroids-and-centers-of-gravity.ppt

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