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CentroidsandCenterof

Gravity

Lecture8

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Centerofgravitypointofapplica6onoftheweightofthebody

Centroidgeometriccenterofabody

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!

M! : summation of moments

My :! xW= xi"Wi!Mx :! yW= yi"Wi!

Wi

yi

xi

Several Wis

x

y

x

y

EquivalentForceSystems

F! : W = "Wi!

=G

W

x

y

(X,Y)

xW= xdW!yW= ydW!

(x, y )

Coordinatesof

thecentroid

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dW= !gadL

=density(masspervolume)

g=accelera6onduetogravity

a=cross-sec6onalarea

dL=lengthofdifferen6alelement

xW= xdW!yW= ydW!

W= ! g a L

W=weightoflineelement

x!gaL = x!gadL!y!gaL = y!gadL!

xL = x dL!yL = ydL!

Forahomogeneouswirewithuniformcrosssec6on

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5 - 6

x

y

)3,4(kg101 =m

)6,2(kg302 =m

)2,8(kg153

=m

Find the center of gravity (center of mass) of the three particles.

( ) ( )

27.3153010

)2)(15()6)(30()3)(10(

:

=++

++=

==

=

y

m

ym

m

ymy

ygmgmy

M

i

ii

T

ii

iiT

x

( ) ( )

0.4153010

)8)(15()2)(30()4)(10(

:

=

++

++

=

==

=

x

m

xm

m

xmx

xgmgmx

M

i

ii

T

ii

iiT

y

x

y

xy

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x

dAx

y

y

QX= yA

Qy = xA

= AY dAxQ

First moment of area about

First moment of area about

QX = ydAA!

Units? Length3(m3ormm3orin3)

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x

dA

y

y

x

dA

yy

x

yQx>0

Qx

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9

Centroidal axis any line passingthrough the centroid of the area.

Axis of symmetry an axis wherein

for every area on one side of the

axis, there exists a correspondingarea on the other side

x-axis: axis of symmetryQx= 0

y-axis: axis of symmetryQy= 0

Note: Axis of symmetry is also centroidal axis but not vice versa.

0'=XQ

'x0' =YQ

0=XQ

x

y

C

'y

0=YQ

x-axis: not axis of symmetryy-axis: not axis of symmetry

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5 - 10

Wherecanyoufindthecentroidofthefigure?Why?

BBisanaxisofsymmetry.Thefirstmomentoftheareawithrespecttotheaxisofsymmetryis

zero.Therefore,ifanareapossessesanaxisof

symmetry,itscentroidliesonthisaxis.


Ifanareacontainstwoaxesofsymmetry,thecentroidliesontheintersec6onofthetwoaxes.


AnareaissymmetricwithrespecttoacenterOifforeveryelementdAat(x,y)thereexistsanarea

dAofequalareaat(-x,-y).Thecentroidofthe

areacoincideswiththecenterofsymmetry.

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11

yAdAyQA

X == xAdAxQ AY ==where and are the coordinates of the centroidx y

x

A

x

y

y

yAQX =A

Qy x=

xAQY = A

Qx Y=

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12

Using a similar derivation for line elements:

yLdLyQL

X ==

xLdLxQL

Y ==

where and are the coordinates of the centroid of the linex y

yLQX=

L

Q

yx

=

xLQY =L

Qx Y=

x

dL

x

y

y

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x

dxy

2(x)

yy

1(x)

===

===

dAydydxydAyAy

dAxdydxxdAxAx

el

el Doubleintegra6ontofindthefirstmomentmaybeavoidedbydefiningdAas

athinrectangleorstrip.

x1 x2

xA = xel dA!= x y

1(x)" y

2(x)[ ]

x1

x2! dx

yA = yel dA!

=

y1(x)+y

2(x)

2

#

$%&

'(y1(x)" y

2(x)[ ]

x1

x2! dx

x

dy

x2(y)

y

x1(y)

y1

y2

xA=

xel dA!=

x1(y)+x

2(y)

2

"

#$%

&'x2 (y)( x1(y)[ ]

y1

y2! dy

yA = yel dA!y x

2(y)( x

1(y)[ ]

x1

x2! dy

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( )

( )ydxy

dAyAy

ydxx

dAxAx

el

el

=

=

=

=

2

( )[ ]

( )[ ]dyxay

dAyAy

dyxaxa

dAxAx

el

el

=

=

+

=

=

2

Checkifthevariablesinintegrandareconstantor

arevaryingalongxory

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Find the coordinates of the centroid of the righttriangle whose base is band whose altitude is h.

x

b

y

h

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1

=

AX dAyQ

Since we need Qx, we can use

horizontal strips (parallel to x-axis).

==

AAX yedydAyQ

Since eis not totally independent

of y, express ein terms of y (using

similar triangles).

edydA =

x

dy

y

y

e

( )hyhbe

=

( )6

2

0

bhdyyhy

h

bQ

h

X == 3

h

A

Qy x ==

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= AY dAxQ

==A

AY xldxdAxQ

Using similar triangles,

ldxdA =

x

dx

y

x

l

( )b

xbhl

=

( )6

2

0

hbdxxbx

b

hQ

b

Y ==

3

b

A

Qx Y ==

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18

x

m0.4

2

5

1xy =

y

Findthexandycoordinatesofthespandrelshown.

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QY = xel dAA

!The expression fordA is dxdyin

the definition of first moment.

Using dxdy, we will be faced with

double integral in solving forQ.

To avoid this, we get strips as our

dA.

hdxdA =

x

dx

2

5

1xy =

y

x

h

QY =

xeldA

A! = xhdx

A

!

Since his varying along

x, express hin terms of

x.2

5

1xyh ==

Some6mes,itismoreconvenientto

orientthestripsparalleltotheaxis

wherethefirstmomentisrequired.

SinceweneedQy,wewilluse

ver6calstrips(paralleltoy-axis).

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hdxdA =

x

dx

2

5

1xy =

y

x

h

QY = xel dAA! = xhdx

A!

Note: The advantage of using strips that are parallel to the axis is that thecoordinate of the centroid is the distance of the strip to the axis. (x in this

example)

=4

0

2

5

1dxxxQY

3

4

0

4

8.1220

mx

QY ==

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==A

A

hdxdAA

=4

0

2

5

1dxxA

2

4

0

3

27.415

mx

A ==

For xA

Qx Y= We need to solve for the area first to get x

mm

m

A

Qx Y 0.3

27.4

8.12

2

3

===

x

y

mx 0.3=

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QX = y eldAA! If we use horizontal strips:

QX = y eldAA!= ybdy

A!

Since bis varying along y,

express bin terms ofy.

yxb 544 ==

bdydA =

x

dy

2

5

1 xy =

y

y

byx 5=

y

5

16=

y

QX = y 4! 5y( )1/2( )dy

0

165" = 4.096 y =

Qx

A= 0.96 m

Seatwork,setuptheintegralforfindingQxusinghorizontalstrips.

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QX = y eldAA! Alternatively, we could still usethe vertical strips:

QX = y eldAA!= h / 2( )hdx

A!

QX =1

2

1

5x2

!

"#

$

%&1

5x2

!

"#

$

%&dy

0

4

' = 4.096 y =Qx

A= 0.96 m

hdxdA =

x

dx

2

5

1xy =

y

x

h 2

5

1xyh ==

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24

Determine the location of the centroid of a semicircular arc ofradius r.

Since the curve is symmetrical

about the y-axis, 0=x

x

dld

y

rddL = sinry =

r

rL

Qy X

222

===

2

0

22sin

=== drydLQL

X

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x y

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x y

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Usually,flateplatescanbedividedintocommonshapes.Tofindthecenterofgravity,recall:

My :! xW= xi"Wi!

Mx

:! yW= yi"Wi!

Iftheplateishomogeneouswithuniformthickness,thecenterofgravitycoincideswiththecentroid.

Thefirstmomentcanbeexpressedasasumofeachelementaryarea.(Similartointegra6on).

Qy = X!A = !xA

Qx =Y!A = !yA

x

y

mm50mm25

mm100

Composite Area

x

y

mm75

mm125

mm200

O30

Composite Line

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x

y

TA

( )TT yx ,

1A

3A

2A

x

y

( )33

, yx

( )22

, yx

( )11, yx

areasubithofareaAi =

)(. compositecentroidofcoordxxT =

)(. compositecentroidofcoordyyT = )(. areassubcentroidofcoordyyi =

)(. areassubcentroidofcoordxxi =

areatotalAAAAT

=++=321

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1A

3A

2A

x

y

2XQ

1YQ

1XQ

2YQ

3XQ

3YQ

321 XXXXT QQQQ ++= 321 YYYYT QQQQ ++=

332211yAyAyAyA TT ++=

332211xAxAxAxA

TT++=

=

i

ii

TA

yAy

=

i

ii

T

A

xAx

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( ) ( ) ( )mm

A

yAy

i

ii

T 0.375491

61.1049133.332500502500=

++==

( ) ( ) ( ) mmA

xA

xi

ii

T 945.05491

61.1049167.1625005.122500=

++==

1A

3A

2A

x

y

( )33, yx

( )22, yx

( )11

, yx

mm100

mm50mm25

Determinethexandycoordinateofthecentroid.

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ATyT = A1x1 +A2x2!A3x3

332211yAyAyAyA TT +=

=T

A

321AAAA

T +=

321 XXXXT QQQQ

+=

A1A2

A3

QYT=Q

Y1+Q

Y2!Q

Y3

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=T

A A1A2

A3

A(mm2) x y xA(mm3) yA(mm3)

A1 37500 125 75 487500 2812500

A2 525 275 50 154875 281250

A3 -(8835.7293) 75 118.190 -279 -1044109

34289 557195 204940

x =162.49mm

y=

59.77mm

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linesubithoflengthLi =

)(. compositecentroidofcoordxxT =

)(. compositecentroidofcoordyyT = )(. areassubcentroidofcoordyyi =

)(. areassubcentroidofcoordxxi =

lengthtotalLLLLT =++= 321

Acompositelinecanbebrokendownintosub-lineswherein

thegeometricproper6es(lengthsandindividualcentroids)of

sub-linesareavailable.

TL

( )TT yx ,

x

y

( )33

, yx( )

22, yx

( )11, yx

1L

3L2L

x

y

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321 XXXXT QQQQ++=

321 YYYYT QQQQ++=

332211yLyLyLyL TT ++= 332211 xLxLxLxL TT ++=

=

i

ii

TL

yLy

=

i

ii

T

L

xLx

XTQ

YTQ

x

y

TL

1L

3L

2L

3XQ 3YQ

x

y

2XQ

2YQ

1YQ

1XQ

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( ) ( ) ( )mm

L

xL

xi

ii

T 0.276.560

6.16120006.2355.137125=

++==

( ) ( ) ( )mm

L

yLy

i

ii

T 9.376.560

5020075.476.2350125=

++

==

1L

3L

2L

mm75

mm125

mm200

O30 x

y

321LLLL

T++=

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5 - 41

To find the center of gravity:

= jWjW

( ) ( ) ( )jWrjWrjWrjWr

G

G

=

=

== dWrWrdWW G

Resultsareindependentofbodyorienta6on,

=== zdWWzydWWyxdWWx

=== zdVVzydVVyxdVVx

dVdWVW == and

Forhomogeneousbodies,

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First Moment of a Volume w.r.t. a plane

x

y

dV

z

y

dxdydzdV =

z

x

= VXY dVzQ

= VXZ dVyQ

= VYZ dVxQ

Unit: length to the

fourth power e.g.444

,, ftcmm

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The first moment of a volume, Qcan have a positive, negative or

zero value (+, - , or 0) depending upon the location of the volume

relative to the axis where moment is required. Here are some

examples

+=XYQ

+=YZQ

The perpendicular distances from thexy-plane (zs) are all positive (above

xy-plane) thus yielding a positive Qxy.

The perpendicular distances from theyz-plane (xs) are all positive (front of

yz-plane) thus yielding a positive Qyz.x

y

dV

z

y

dxdydzdV=

z

x

+=XZQ

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+=YZQ

+=XYQ

=XZQ

The perpendicular distances

from the xy-plane (zs) are all

positive (above xy-plane)

thus yielding a positive Qxy.


from the xz-plane (ys) are all

negative (left of xz-plane)

thus yielding a positive Qxz.


from the yz-plane (xs) are all

positive (front of yz-plane)

thus yielding a positive Qyz.

x

y

dV

z

y

z x

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Whenthesumofthefirst

momentsofthevolumesabove

thexy-plane(posi6vez-values)

equalsthesumofthefirstmomentsofthevolumesbelow

thexy-plane(nega6vez-values)

willresultto

Qxy=0

=XYQ

+=

XYQ

x

dV

z

yz

z

dV

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4

Centroidal plane any plane that

passes through the centroid of the

volume. The first moment of the

volume about any of these planes is

zero. Qxy= 0. xy-plane is a centroidalplane thus Qxy=0.

xy-plane: centroidal planeQxy= 0

yz-plane: centroidal planeQyz= 0

xz-plane: centroidal planeQxz= 0

xy-plane: centroidal planeQxy= 0

x

=XY

Q

+=XYQ

dV

z

yz

z

dV

C

x

z

yC

planexy '

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Plane of symmetry a plane whereinfor every point P there exists a point P

of the same volume, and the line PP is

perpendicular to the given plane and is

bisected by that plane.

Which are planes of symmetry in the

figure?

xy-plane: NOT plane of symmetry but

still Qxy= 0

Note: Plane of symmetry is also

centroidal plane but not vice versa.

x

=XYQ

+=XYQdVyz

z

dV

C

x

z

yC

planexy '

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The first moment of a volume can also be expressed as the product

of the volume and some value.

zVdVzQV

XY == yVdVyQ VXZ == xVdVxQ VYZ ==

where , and locate the centroid of the volume.x y z

V

Qy XZ=

V

Qx YZ=

V

Qz XY=x

y

x

y

z

z

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Thecentroidofavolumeboundedby

analy6calsurfacescanbedetermined

byevalua6ng:

dVusecubeofsizedx,dy,dzasdifferen6alvolumerequirestripleintegra6on

= dVxVx = dVyVy = dVzVz

x

dV

yz

dV

C

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WhenavolumeVpossessesaplaneofsymmetry,thefirstmomentofV

withrespecttothatplaneiszero

= xdVVx

0==zdVVz

First moment wrt x-z

First moment wrt x-y

First moment wrt y-z

x

y

z

0== ydVVy

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Single integration can be used by choosing a thin slab fordV

.

= dVxVx el

Illustration: For bodies of revolution, usecircular slabs for dV

0== zy

xxel=

dxrdV 2=

express xel and dV interms of x

x

y

z

r

xel

dx

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But eis varying along z

The relationship eand zcan be found

using similar triangles. x

z

yz

dz edzedV

2

=dzedV 2=

h

z

y

e

z

( )h

zhre

=

( )

===

h

VVxy

dzh

zhrzdzezzdVQ

0

2

2

12

22hrQxy

=

4

3

122

22

h

hr

hr

V

Qz

xy===

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SampleProblem5.13

Determinetheloca6onofthe

centroidofthehalfrightcircular

coneshown.

SOLUTION:

Volumeissymmetricaboutthex-yplane;zcoordinateofcentroidis0.

0=z

dVxVxel

=

dVyVy el=

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dxrdV2

2

1=

xxel =3

4ryel =

xh

ar

h

a

x

r==

First moment wrt y-z plane: dVxVxel

=

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First moment wrt x-z plane: dVyVy el=

dxrdV2

2

1=

xxel =3

4ryel =

xh

ar

h

a

x

r==

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Forhomogeneousbodies, === VzVZVyVYVxVX

MomentofthetotalweightconcentratedatthecenterofgravityGisequaltothesumofthemomentsoftheweights

ofthecomponentparts.

=== WzWZWyWYWxWX

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Forhomogeneousbodies,=== VzVZVyVYVxVX

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0

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34in.2865in08.3== VVxX

34 in.2865in5.047== VVyY

34in.2865in.6181== VVzZ

in.577.0=X

in.577.0=Y

in.577.0=Z

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ES11-08 Stud Ver