chapter 5: distributed forces: centroids and centers of...

Chapter 5: Distributed Forces: Centroidsand Centers of Gravity

School of Mechanical Engineering

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Contents

Introduction

Center of Gravity of a 2D Body

Centroids and First Moments of Areas and Lines

Centroids of Common Shapes of Areas

Centroids of Common Shapes of Lines

Composite Plates and Areas

Sample 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 Bodies

Sample Problem 5.12

School of Mechanical Engineering 5 - 2

Introduction


Centroids and First Moments of Areas and Lines




Sample Problem 5.1


Sample Problem 5.4


Sample Problem 5.7


Sample Problem 5.9



Composite 3D Bodies

Sample Problem 5.12

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 the first moment of an area is used to locate the centroid.


• The centroid of an area is analogous to the center of gravity of a body. The concept of the first moment of an area is used to locate the centroid.

• Determination of the area of a surface of revolution and the volume of a body of revolution are accomplished with the Theorems of Pappus-Guldinus.

• Determination of the area of a surface of revolution simplifies the analysis of beams subjected to distributed loads


집중력은 없다 : 어떤 물체에 기계적으로 가한 모든 외력은 아무리 작아도 유한한

접촉면적에 분포되기 때문에 분포력으로 볼 수 있다.

※ “concentrated” forces does not exist in the exact sense

※ internal forces

R R : Resultant of distributed forces


◎ 물체의 중심(Center of gravity ) : 물체의 각 부분에 작용하는 중력의 합력인작용점

※ internal forces

C

CC

C

Stress : internal distributedforces in solids

5 - 4

Center of Gravity of a 2D Body• Center of gravity of a plate • Center of gravity of a wire

nz WWWWF D++D+D=Ûå L21


òååòåå

=Û

D++D+D=D=Û

=Û

D++D+D=D=Û

dWyWy

WyWyWyWyWyM

dWxWx

WxWxWxWxWxM

nniix

nniiy

L

L

2211

2211

nz WWWWF D++D+D=Ûå L21

모멘트 원리 : 임의의 축에 대한 중력의 합력 W의 모멘트는

물체의 미소 요소로 생각한 각 질점에 작용한 중력 dW의 모멘트의 합과 같다.

좌표 중심의 무게 물체의:),( yx

5 - 5

Centroids and First Moment of Areas and Lines• Centroid of an area • Centroid of a line

specific weight(비중량) )N/morlb/ft( 33grg = 323 N/m))(m/seckg/m()( ==gr


( ) ( )

모멘트차의면적대한축에

모멘트차의면적대한축에

1 A x

1 A y

Þ

==

Þ

==

=Û=

ò

òòò

x

y

QdAyAy

QdAxAx

dAtxAtxdWxWx gg ( ) ( )

òò

òò

=

=

=Þ=

dLyLy

dLxLx

dLaxLaxdWxWx gg

specific weight(비중량) )N/morlb/ft( 33grg = 323 N/m))(m/seckg/m()( ==gr

좌표의 C 도심 의A 면적:),( yx

좌표의선 C 도심 의L :),( yx

òò ==\ ydLL

yxdLL

x 1,1

òò ==\ ydAA

yxdAA

x 1,1

5 - 6

First Moments of Areas and Lines

• An area is symmetric with respect to an axis BB’if for every point P there exists a point P’ such that PP’ is perpendicular to BB’ and 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 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 Oif for every element dA at (x,y) there exists an area dA’ of equal area at (-x,-y).

• The centroid of the area coincides with the center of symmetry.


• Composite plates

åååå

=

=

WyWYWxWX


• Composite area

åååå

=

=

AyAYAxAX

Sample Problem 5.1

SOLUTION:

• Divide the area into a triangle, rectangle, and semicircle with a circular cutout.

• Find the total area and first moments of the 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.


For the plane area shown, determine the first moments with respect to the x and y axes and the location of the centroid.

• Compute the coordinates of the area centroid by dividing the first moments by the total area.

• Find the total area and first moments of the triangle, rectangle, and semicircle. Subtract the area and first moment of the circular cutout.

Sample Problem 5.1


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.

Sample Problem 5.1

23

33

mm1013.828mm107.757

´

´+==

åå

AAxX

• Compute the coordinates of the area centroid by dividing the first moments by the total area.


mm 8.54=X

23

33

mm1013.828mm102.506

´

´+==

åå

AAyY

mm 6.36=Y연습문제: 5.3, 5.13, 5.20


òòòòòòòò

===

===

dAydydxydAyAy

dAxdydxxdAxAx

el

el • Double integration to find the first moment may be avoided by defining dA as a thin rectangle or strip.


( )

( )ydxydAyAy

ydxx

dAxAx

el

el

ò

òòò

=

=

=

=

2

( )[ ]

( )[ ]dxxay

dAyAy

dxxaxadAxAx

el

el

-=

=

-+

=

=

òò

ò

ò

2

÷øö

çèæ=

=

÷øö

çèæ=

=

ò

ò

ò

ò

qq

qq

drr

dAyAy

drr

dAxAx

el

el

2

2

21sin

32

21cos

32

Sample Problem 5.4

SOLUTION:

• Determine the constant k.

• Evaluate the total area.

• Using either vertical or horizontal strips, perform a single integration to find the first moments.


Determine by direct integration the location of the centroid of a parabolic spandrel.

• Using either vertical or horizontal strips, perform a single integration to find the first moments.

• Evaluate the centroid coordinates.

Sample Problem 5.4

SOLUTION:

• Determine the constant k.

2121

22

22

2

yb

axorxaby

abkakb

xky

==

=Þ=

=


2121

22

22

2

yb

axorxaby

abkakb

xky

==

=Þ=

=

• Evaluate the total area.

3

30

3

20

22

ab

xabdxx

abdxy

dAAaa

=

úúû

ù

êêë

é===

=

òò

ò

Sample Problem 5.4

• Using vertical strips, perform a single integration to find the first moments.

1052

21

2

44

2

0

5

4

2

0

22

2

2

0

4

2

0

22

abxa

b

dxxabdxyydAyQ

baxab

dxxabxdxxydAxQ

a

aelx

a

aely

=úúû

ù

êêë

é=

÷øö

çèæ===

=úúû

ù

êêë

é=

÷øö

çèæ===

òòò

òòò


1052

21

2

44

2

0

5

4

2

0

22

2

2

0

4

2

0

22

abxa

b

dxxabdxyydAyQ

baxab

dxxabxdxxydAxQ

a

aelx

a

aely

=úúû

ù

êêë

é=

÷øö

çèæ===

=úúû

ù

êêë

é=

÷øö

çèæ===

òòò

òòò

Sample Problem 5.4

• Or, using horizontal strips, perform a single integration to find the first moments.

( )

( )

10

421

22

2

0

2321

2121

2

0

22

0

22

abdyyb

aay

dyyb

aaydyxaydAyQ

badyyb

aa

dyxadyxaxadAxQ

b

elx

b

bely

=÷øö

çèæ -=

÷øö

çèæ -=-==

=÷÷ø

öççè

æ-=

-=-

+==

ò

òòò

ò

òòò


( )

( )

10

421

22

2

0

2321

2121

2

0

22

0

22

abdyyb

aay

dyyb

aaydyxaydAyQ

badyyb

aa

dyxadyxaxadAxQ

b

elx

b

bely

=÷øö

çèæ -=

÷øö

çèæ -=-==

=÷÷ø

öççè

æ-=

-=-

+==

ò

òòò

ò

òòò

Sample Problem 5.4

• Evaluate the centroid coordinates.

43

2baabx

QAx y

=

=

ax43

=

103

2ababy

QAy x

=

=


103

2ababy

QAy x

=

=

by103

=


• Surface of revolution is generated by rotating a plane curve about a fixed axis.


• 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 generating curve times the distance traveled by the centroid through the rotation.

LyA p2=


• Body of revolution is generated by rotating a plane area about a fixed axis.


• Body of revolution is generated by rotating a plane area 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 p2=

Sample Problem 5.7

SOLUTION:

• Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section.

• Multiply by density and acceleration to get the mass and acceleration.


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 of steel is determine the mass and weight of the rim.

33 mkg 1085.7 ´=r


Sample Problem 5.7

SOLUTION:

• Apply the theorem of Pappus-Guldinus to evaluate the volumes or revolution for the rectangular rim section and the inner cutout section.



( )( ) ÷øö

çèæ´´== - 3393633 mmm10mm1065.7mkg1085.7Vm r kg 0.60=m

( )( )2sm 81.9kg 0.60== mgW N 589=W

연습문제: 5.35, 5.42, 5.59


• A distributed load is represented by plotting the load per unit length, w (N/m) . The total load is equal to the area under the load curve.

òò === AdAdxwWL

0


• A distributed load is represented by plotting the load per unit length, w (N/m) . The total load is equal to the area under the load curve.

òò === AdAdxwWL

0

( )

( ) AxdAxAOP

dWxWOPL

==

=

ò

ò

0

• A distributed load can be replace by a concentrated load with a magnitude equal to the area under the load curve and a line of action passing through the area centroid.

Sample Problem 5.9

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 the area under the curve.


A beam supports a distributed load as shown. Determine the equivalent concentrated load and the reactions at the supports.


• Determine the support reactions by summing moments about the beam ends.

Sample Problem 5.9

SOLUTION:

• The magnitude of the concentrated load is equal to the total load or the area under the curve.

kN 0.18=F




kN 18mkN 63 ×

=X m5.3=X

Sample Problem 5.9

• Determine the support reactions by summing moments about the beam ends.

( ) ( )( ) 0m .53kN 18m 6:0 =-=å yA BM

kN 5.10=yB


kN 5.10=yB

( ) ( )( ) 0m .53m 6kN 18m 6:0 =-+-=å yB AM

kN 5.7=yA



• Center of gravity G

( )å D-=- jWjWrr

( ) ( )[ ]( ) ( ) ( )jWrjWr

jWrjWr

G

Grrrr

rrrr

-´D=-´

D-´=-´

åå

òò == dWrWrdWW Grr

• Results are independent of body orientation,

òòò === zdWWzydWWyxdWWx

òòò === zdVVzydVVyxdVVx

dVdWVW gg == and

• For homogeneous bodies,


School of Mechanical Engineering 5 - 30연습문제: 5.71, 5.77,5.113

Composite 3D Bodies

• Moment of the total weight concentrated at the center of gravity G is equal to the sum of the moments of the weights of the component parts.

åååååå === WzWZWyWYWxWX

• For homogeneous bodies,


åååååå === VzVZVyVYVxVX

Sample Problem 5.12

SOLUTION:

• Form the machine element from a rectangular parallelepiped and a quarter cylinder and then subtracting two 0.03-m- diameter cylinders.


Locate the center of gravity of the steel machine element. The diameter of each hole is 1 cm.

Sample Problem 5.12


Sample Problem 5.12

( ) ( )-3-7 10 3.347610.0245 ´´== åå VVxX


( ) ( )-3-7 10 3.347610 52.31 ´´-== åå VVyY

( ) ( )-3-7 103.34761022.25 ´´== åå VVzZ

chapter 5: distributed forces: centroids and centers of...

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