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Lecture 16 Engineering Mechanics
Center of Gravity & Centroid¢ Center of gravity
ÄThe center of gravity is a point which locates the resultant weight of a system of particles or body.
¢Determining the center of gravity
ÄApplying the principle of moments to the parallel system of gravitational force
xdwx
w=
ò ydwy
w=
ò zdwz
w=
ò
xdmx
m=
ò ydmy
m=
ò zdmz
m=
ò
Lecture 16 Engineering Mechanics
Center of Gravity & Centroid¢ Centroid
ÄThe centroid C is a point which defines the geometric center of an object when the density r of a body is uniform throughout.
¢ Centroid of a line
ÄFor a slender rod or wire of length L, cross-sectional area A, & density r, the mass of an element becomes
Ø If r & A are constant over the entire length of the rod, then;
xdLx
L=
ò ydLy
L=
ò zdLz
L=
ò
dm AdLr=
Lecture 16 Engineering Mechanics
Center of Gravity & Centroid¢ Centroid of area
ÄWhen a body of density r has a small but constant thickness t, the mass of an element becomes
Ø If r & t are constant over the entire area, then the centroid C of the surface area may be written
¢ Centroid of volume ÄFor a general body of volume V & density r, the
element has a mass
Ø If r is constant over the entire volume, then the centroid C of the volume may be written
xdAx
A=
ò ydAy
A=
ò zdAz
A=
ò
dm AdLr=
xdVx
V=
ò ydVy
V=
ò zdVz
V=
ò
dm AdLr=
Lecture 16 Engineering Mechanics
Example 1¢ Locate the centroid of a circular arc as shown
ÄChoosing the axis of symmetry
ÄThe centroid of the arc line can be located by
xdLx
L=
ò
0.ydL
yL
= =ò
dL r dq= ×
2L ra=
cosx r q=
x at any point on the arc can be found from;
( )( )
2coscos
2 2
r rd rx d
r r
a
aa
a
q qq q
a a-
-= =
òò
[ ]
2 sinsin sin( )
2 2
r rx
aa a
a a= - - =
sinrx
a
a\ =
Lecture 16 Engineering Mechanics
Example 2¢Determine the distance y from the base of a triangle of altitude h
to the centroid of its area
ÄThe area of the strip dA = x.dy
0( )
2
h by h y dyydA hy
bhA
-= =
òò
( )x b b
x h yh y h h
= ® = --
2
bhA =
( )b
dA h y dyh
\ = -
Triangle area
Applying the centroid of area equation
2
1
6
3
2
bhh
ybh
= =
Lecture 16 Engineering Mechanics
Example 3¢ Locate the centroid of the area of a circular sector
with respect to its vertex
ÄTechnique I
The area of the strip
2 2
2
( )
2A r r
ap a
p= =
sinoC
rx
a
a=
2 .o odA r dra=
Centroid of the strip from example 1
Total area A
Applying the centroid of area equation
( )
0
2
sin
2 .r
oo
rr drxdA
xA r
aa
a
a
æ öç ÷è ø= =
òò
Lecture 16 Engineering Mechanics
Example 3
2 2
2
( )
2A r r
ap a
p= =
2
cos
3Cx r q=
2 2
1
( )
2 2
ddA r r d
qp q
p= =
Technique II
Total area A
Applying the centroid of area equation
3
2
2sin 2 sin
3 3
r rx
r
a a
a a= =
The area of the strip
Centroid of the strip
2
2
2 1
cos .
3 2r r dxdA
xA r
a
aq q
a
-
æ öæ öç ÷ç ÷è øè ø= =
òò
[ ]
3
2
1
sin sin( )
3
rx
ra a
a= - -
2 sin
3
rx
a
a=
Lecture 16 Engineering Mechanics
Example 3
For a semicircular area 2a = p
2 sin / 2 4
3 / 2 3
r rx
p
p p= =
Lecture 16 Engineering Mechanics
Example 4¢ Locate the centroid of the area shown
ÄThe x
The area of the strip
1
2
0
1
3A x dx= =ò
Cx x=
2.dA y dx x dx= =
Centroid of the strip
Total area A
Applying the centroid of area equation
1
2
0( )
1 / 3
x x dxxdAx
A= =
òò
1
3
0
3
1 / 3 4
x dxx m= =
ò
Lecture 16 Engineering Mechanics
Example 4ÄThe y
The area of the strip
1
2
0
1
3A x dx= =ò
Cy y=
(1 ).dA x dy= -
Centroid of the strip
Total area A
Applying the centroid of area equation
( )
1
0
(1 )
1 / 3
y x dyydAy
A
-= =
òò
1
3/2
0
3
1 / 3 10
y y dyy m
-= =
ò
Lecture 16 Engineering Mechanics
Example 5¢ Locate the centroid of the area shown
The area of the strip
100 100
2
0 0
. 100 .A z dy y dyp p= =ò ò
Cy y=
2.dV z dyp=
Centroid of the strip
Total area A
500000A p=
Applying the centroid of Volume equation
100 100
2
0 0
( 100 . )
500000 5000
y y dy y dyydVy
V
p
p= = =
ò òò
66.67y mm=