ch05-distributed forces (centroids and centers of gravity)

8/14/2019 Ch05-Distributed Forces (Centroids and Centers of Gravity)

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VECTOR MECHANICS FOR ENGINEERS:

STATICS

Ninth Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

2010 The McGraw-Hill Companies, Inc. All rights reserved.

5Distributed Forces:

Centroids and Centers

of Gravity


2/32

2010The McGraw-Hill Companies, Inc. All rights reserved.

Vector Mechanics for Engineers: StaticsNinth

Edition

Contents

5 - 2

Introduction

Center of Gravity of a 2D Body

Centroids and First Moments of

Areas and Lines

Centroids of Common Shapes ofAreas

Centroids of Common Shapes of

Lines

Composite Plates and AreasSample Problem 5.1

Determination of Centroids by

Integration

Sample Problem 5.4

Theorems of Pappus-Guldinus

Sample Problem 5.7

Distributed Loads on Beams

Sample Problem 5.9

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

Centroids of Common 3D

Shapes

Composite 3D BodiesSample Problem 5.12


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Edition

Introduction

5 - 3

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 gravityfor the body.

The centroid of an areais analogous to the center of

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

Determination of the area of asurface of revolutionand

the volume of a body of revolutionare accomplished

with the Theorems of Pappus-Guldinus.


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Edition

Center of Gravity of a 2D Body

5 - 4

Center of gravity of a plate

dWy

WyWyM

dWx

WxWxM

y

y

Center of gravity of a wire

NE


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Edition

Centroids and First Moments of Areas and Lines

5 - 5

x

QdAyAy

yQdAxAx

dAtxAtx

dWxWx

x

y

respect tohmoment witfirst

respect tohmoment witfirst

Centroid of an area

dLyLydLxLx

dLaxLax

dWxWx

Centroid of a line

NE


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Edition

First Moments of Areas and Lines

5 - 6

An area is symmetric with respect to an axis BB

if for every pointPthere exists a pointP

suchthat PPis 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 center O

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

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

The centroid of the area coincides with the

center of symmetry.

NE


7/32 2010The McGraw-Hill Companies, Inc. All rights reserved.


Edition

Centroids of Common Shapes of Areas

5 - 7 NE




Edition

Centroids of Common Shapes of Lines

5 - 8

f SNE




Edition

Composite Plates and Areas

5 - 9

Composite plates

WyWY

WxWX

Composite area

AyAY

AxAX

V t M h i f E i St tiNE




Edition

Sample Problem 5.1

5 - 10

For the plane area shown, determine

the first moments with respect to the

xandyaxes 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

Sample Problem 5.1

5 - 11

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

Sample Problem 5.1

5 - 12

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

Determination of Centroids by Integration

5 - 13

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 dAas a thin

rectangle or strip.





Edition

Sample Problem 5.4

5 - 14

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.





Edition

Sample Problem 5.4

5 - 15

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 tiN

Ed




Edition

Sample Problem 5.4

5 - 16

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 tiNi

Ed




Edition

Sample Problem 5.4

5 - 17

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 StaticsNi

Ed



Vector Mechanics for Engineers: Staticsinth

dition

Sample Problem 5.4

5 - 18

Evaluate the centroid coordinates.

43

2baabx

QAx y

ax4

3

103

2abab

y

QAy x

by10

3

Vector Mechanics for Engineers: StaticsNi

Ed



Vector Mechanics for Engineers: Staticsinth

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



Vector Mechanics for Engineers: Staticsnth

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


Ed




dition

Sample Problem 5.7

5 - 21

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.

Vector Mechanics for Engineers: StaticsNin

Ed




dition

Sample Problem 5.7

5 - 22

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

sm81.9kg0.60mgW N589

W

Multiply by density and acceleration to

get the mass and acceleration.


Ed




dition

Distributed Loads on Beams

5 - 23

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.

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

Sample Problem 5.9

5 - 24

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

Sample Problem 5.9

5 - 25

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

Sample Problem 5.9

5 - 26

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

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

5 - 27

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

Centroids of Common 3D Shapes

5 - 28


Edi


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tion

Composite 3D Bodies

5 - 29

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: Staticsthtion

Sample Problem 5.12

5 - 30

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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Sample Problem 5.12

5 - 31

Vector Mechanics for Engineers: StaticsNint

Edit


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Sample Problem 5.12

34 in.2865in08.3 VVxX

34 in.2865in5.047 VVyY

34 in.2865in.6181 VVzZ

in.577.0X

in.577.0Y

in5770

Z

ch05-distributed forces (centroids and centers of gravity)

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