ES11-08 Stud Ver
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Transcript of ES11-08 Stud Ver
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1/30/12 1
CentroidsandCenterof
Gravity
Lecture8
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Centerofgravitypointofapplica6onoftheweightofthebody
Centroidgeometriccenterofabody
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5 - 3
!
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.
Wherecanyoufindthecentroidofthefigure?Why?
Ifanareacontainstwoaxesofsymmetry,thecentroidliesontheintersec6onofthetwoaxes.
Wherecanyoufindthecentroidofthefigure?Why?
AnareaissymmetricwithrespecttoacenterOifforeveryelementdAat(x,y)thereexistsanarea
dAofequalareaat(-x,-y).Thecentroidofthe
areacoincideswiththecenterofsymmetry.
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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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5 - 14
( )
( )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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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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5 - 26
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5 - 27
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5 - 28
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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44
+=YZQ
+=XYQ
=XZQ
The perpendicular distances
from the xy-plane (zs) are all
positive (above xy-plane)
thus yielding a positive Qxy.
The perpendicular distances
from the xz-plane (ys) are all
negative (left of xz-plane)
thus yielding a positive Qxz.
The perpendicular distances
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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48
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