Centroid of Plane Figures
Objective:
To understand significance of centroid ofplane area (lamina) and determine
its location for a given composite plane figure.
4.1 Introduction
We use and come across various types of shapes, laminae, grills/composite bars/wires
in the constructions, machines and mechanisms for different purposes.
Specific shape is chosen for cross section of a beam, column and other structural
member. One of the most important geometric property of the shapes is center of gravity
or centroid. Determination of C.G. of a composite figure or body is very essential for
strength of materials, fluid mechanics studies.
4.2 Basic Concepts
A given shape is considered to be lying in x-y plane. The reference axes and origin are
shown either as per our convenience or as given.
The element' or component' is length or area and is treated to be a force or vector (as
each and every particle is attracted by earth, i.e. force of gravity). The integral J x dA is
known as 'first moment of the area about (or with respect to) y axis'. Similarly J y dL is
known as 'first moment of the length about x axis'. Gravitational forces on particles is a
system of 'distributed forces' over the body under consideration.
4.3 Centroid and Center of Gravity
Center of gravity is the point where weight of the body can be assumed to be acting,
i.e. point of application of earth's gravitational force. Projection of center of gravity on
vertical plane for two dimensional bodies is termed as centroid of the body or figure. Thus
term centroid is often used for linear segments and plane figures or laminae, and term
center of gravity is often related to volumes (three dimensional bodies).
Varignon's theorem of moments is applied for determining centroids of composite
figures and linear segments.
(4-1)
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Centroid of Plane Figures
If weight of body W comprises of elements of weights W j , VvS,... etc. we have relation
W =
Wi + W 2 + ...
For obtaining co-ordinates x and y of centroid (these are measured from Y axis and X
axis respectively), consider moments of weights about Y and X axes respectively, i.e. apply
V.T.M. about Y and X axes respectively.
xW
= xi Wi +X2W2 + ....
If density of material p ( k g / m 3 ) , gravitational acceleration g ( m / s 2 ) and thickness of
body V are same, [W can be expressed as W = (pgb) (area) since W = pg x volume]
We get,
x-A
= X1A1+X2A2 +
_
J(x-dA)
x =
=
{ (dA)
Hence
£(xdA)
I(dA)
Here the product (x • A) is known as 'first moment of area'.
c.
..
.
_
J <y • dA >
I ( y d A)
For taking first moments of areas some basic figures should be known, (of course,
these can be obtained by integration as illustrated through the solved problems). Some
important points should be remembered while obtaining co-ordinates of centroids of areas.
4.4 Important Points to be Remembered
Few hints and basics about centroid should be remembered while determining position
of centroid of composite figures (areas).
1) Centroid always lies on the axis or axes of symmetry, if any.
2) It is not necessary that centroid must be on the composite figure (it may lie in
hollow portion).
3) Area portion is 'subtracted' when it is removed or cut.
4) If freely suspended, equilibrium position is achieved when centroid is vertically
below or above the point of suspension. Refer Fig. 4.1.
5) If area or length is on opposite side of origin, centroidal distance is negative.
(Do not get confused between 'negative area' and 'negative centroidal distance'.)
6) Divide the given figure in the most convenient
distances/lengths etc. clearly to avoid mistakes.
manner
and
show
all
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Centroid of Plane Figures
G_0
Hinge
rh^r Hinge or pin
(a) G lies below O
(b) G coincides with O
(c) G lies above O
Fig. 4.1 Freely suspended objects
4.5 Centroids of Common Plane Figures
Triangle, rectangle and circle (or its part) are regular basic shapes for which centroids
are given in the table 4.1 below.
Sr.
No.
1.
Description and area
Figure
-
Rectangle
A=b.d
x
X
y
b
2
d
2
b
h
3
~r
4k G
d
J1 _
-a
2
2.
H
!
Triangle
5
A= ^ b .h
•
*
-i
/
i
r
\
h
G
/
b/2
iiv
-'
\
b/2
:
H1
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4-4
Sr.
No.
Description and area
3.
Right angled triangle
Figure
b
H
A = ^ bh
T
i
H
4.
2( Rsina
Circular sector
A = aR2
H
a
Zero
)
O
Quarter circle
u=
6.
4R
3n
4R
Zero
4R
3a
3*
7t
4
Semicircle
2
2
J
Table 4.1 Basic Plane Figures
4.6 Method of Integration
By applying Varignon's theorem of moments (V.T.M.) and considering very small
elemental area, we will obtain formulas for the basic shapes given in the tables 4.1. This
mathematical procedure followed right from very basic or fundamental equation or law is
known as 'derivation from the first principles.'
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4.6.1 Rectangular Area
Consider a rectangle of size b x d as shown in the
Fig. 4.2. Though we know directly that area = b.d, the
area also can be obtained by considering a vertical strip
of thickness dx.
Area of strip dA = (dx) d
b
For complete rectangle, A = J d ( d x )
-X
0
= d[x]J
... d is a constant
= b.d
Fig. 4.2
Now first moment of dA about Y axis = x.dA
= d (x.dx)
Apply V.T.M. about V axis.
b
A x = d J x dx
0
•
•
bdx
= d
(b^
2
/
Now refer Fig. 4.3 showing horizontal strip
of thickness dy so that area of strip = b-dy
I
7y
. I0
b
H
d
Ay
-
b j y- dy
0
Fig. 4.3
(by applying V.T.M. about X axis.)
b*d*y = b
d^
/
Note : Same vertical strip (used for x) can also be used for y.
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Centroid of Plane Figures
4.6.2 Right Angled Triangular Area
For the triangle of base V and height 'h'
as shown in Fig. 4.4, consider vertical strip of
thickness dx and height 'hi' at distance x
from origin O.
dA =
•
A
•
h]dx
= | hi-dx
0
From similar triangles, we have
=
hx
Hence dA = -p- dx
b
Fig. 4.4
A
• / G)
x dx
o
=
-
y
hfb2
b|T
A = I
bh
Now consider first moment of the strip about Y axis and apply V.T.M
b
A x = J xdA
0
^ bh x
" J
^ 1 x 2 dx
0
h
b
^ bh x
3
Now consider first moment of the same strip about X axis. (Alternatively, a new
horizontal strip also can be considered). Apply V.T.M. also.
Ibhy
=
fnUldA
0
Y £ i I dx
- f t Ueb A b
0
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Ibhy
i.e.
^ J
2b
ill
Centroid of Plane Figures
x2 dx
0
( u3
2b2
Note : Depending on orientation of the triangle or measurement, understand the terms
'height' and 'base' carefully.
4.6.3 Sector of a Circle
Consider sector of a circle of radius R and angle
2 a as shown in Fig. 4.5. (Note that y = 0).
Area of very small elemental sector
= ^ R (R dO)
... treated as A
Centroidal distance for elemental sector from Y
axis
~>x
COS0
a
^
_
x =
2J
R2
de
sine
a
(!M ]
0
o
r
2j
0
•
cosG
a
R2
iLdO
Fig. 4.5
m
Note : Again
rad.
_
substituted, to obtain x
4R
for quartercircle and
rad.
f°r
a
semicircle can be
(measured along axis of symmetry from the center).
4.7 Composite Figures
For any non-standard or composite figure, either mathematical method (integration)
can be used or given shape can be divided into basic/standard shapes and formulas given
in the tables can be applied. (This is incidently, the 'principle of superposition')
Formulas,
x = Z
I
(a-x)
(a)
and
y
"
1 ( a )
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Centroid of Plane Figures
can be used for composite figure/lamina or plate of uniform thickness and density.
Remember to measure (calculate) distances at right angles to reference axes X and Y
which are either given in the problem or assumed by us as per our convenience.
Study the solved examples carefully
4.8 Solved Examples
)>!•• Example 4.1 : Determine centroid of the shaded area with reference apex.
( V T U , July - 2006)
Fig. 4.6
Solution : For the given Fig. 4.6, there is a vertical axis of symmetry. Hence we have to
find y only. [Question is not very clear due to wording 'with reference apex']. Given
shaded area = Triangle - Circle - Rectangle - Semicircle.
Assuming center of semicircle as origin, measure the centroidal y distances as shown
in the table below.
2
Component area a (mm )
^ (160) 240
- -J(40)
£A=
2
= 19200
= - 1256.64
-(40) (60)
= - 2400
_ .1(40)2
= - 2513 27
13030.09
Vertical (y) centroidal
distances (mm)
^ (240)
160+ ~
80+^
4
<
40>
3
Product a.y (mm )
= 80
+1536000
= 180
- 226195.20
= 100
- 240000
= 1 6 98
—
- 42675.325
1027129.5
Table 4.2
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Apply
(Z
A
)y
Centroid of Plane Figures
= I>-y)
.•. Distance from apex will be 161.17 mm.
4R
)>!•• Example 4.2 : For a semicircular lamina, obtain y = -=— with usual notations.
3n
<VTU, July - 2003)
Solution :
Consider a semicircle of radius R (Refer Fig. 4.7).
Width of elemental strip as shown = 2 ^ R 2 - y 2
R
Area of strip = ^2-J R 2 - y
R
•-X
2
) dy
.'. First moment of strip about diameter
(2 / R
2
^ ) ydy
Now use
Fig. 4.7
v dA
J dA
J (2A/R2-y2)ydy
•• y = 0
J
( i J W ^ y y
0
R
Numerator =
-J
(R2-y2)V2<-2y)dy
0
3/2 R
(
R 2
-y2)
3/2
0
•
Denominator
-
! [ ° - (
r
2
)
3
/
2
R
= 2 J { j R 2 - y 2 ^ dy
0
Put y = R sin 0. Hence dy = R cos 0 d 0 and limits change from 0 to ^
*/2
Denominator = 2 J R cos 0 (R cos 6 dG)
0
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n/2
2R 2 J
cos2ede
0
n/2
=
2R
2U
0
=
R2
cos 26 + 1
d9
v
+ sin 2 6
TCR2
+ 0
0
y
»»•• Example 4.3 :
=
R2
TZ-zr-
Determine co-ordinates of centroid of shaded portion (the spandrel).
(VTU, July - 2006)
Y
A
-•> X
IN
a
Fig. 4.8
ydx as shown in Fig. 4.9.
Solution : Consider vertical strip of area dA
n
A s J ydx
0
Now
y2
m kx
k
i i r o t w f n t n m :
*
HK
Fig.
4.9
>x
i
r
A =
dx
a
kJ x2dx
0
at point (a, b) gives
b_
a2
a*
a2
T
ab
J
Moment ef the elemental area about Y a*i§ = * (ydx),
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Centroid of Plane Figures
Moment of entire area = J x ( k x 2 ) d x
o
k | x 3 dx
0
\
/
' b
x =
V
ab
or
Similarly moment of entire area about X axis = J (y dx)
\
0
k2
fy
J x 4 dx
0
5^
2a4 { 5
ab 2
10
fab2N
»
t
t
y =
10
i
ap
"3
• ff
y
3b
10
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Example 4.4 : Prove that x =
4-12
Centroid of Plane Figures
and y =
/or the part of parabolic area bounded by X
axis, line x = a and parabola i / 2 = kx.
[p\jf May - 1992]
Y
A
ii
*
I*
i
ij
i
i
i
>x
x =a
Fig. 4.10
Solution : Consider vertical strip of thickness dx at distance x as shown in Fig. 4.11
Area of strip dA
•
«
= y dx
Complete shaded area = J y dx
0
Now y 2 = kx applied at y = b gives k =
b2
A
= J ( - / k x ) dx = / k j
0
0
A
=
L
^dx
3 2
/ Jo
Putting value of k and simplifying,
2
A = — (ab)
Now consider moment of elemental area (strip)
about X axis.
Moment of strip = (y dx)
^
Fig. 4.11
a
For entire area, moment = J
1
— y 2 dx
0
i
a
xdx
0
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Centroid of Plane Figures
b2
2a
Moment of area about X axis
y =
Total area
Now,
•
2
4
Similarly consider moment of strip about Y axis.
a
Moment for entire area = J (y dx) x
0
a
= yfkj
9
xVxdx
0
1x5/2
=
/ k
[5/2_
0
2a 2 b
2a 2 b
x =
5
or
2ab
3 J
)»•• Example : 4.5 :
x2
+y
2
a2
fc2
Determine x and y for quarter of an ellipse (in the first
quadrant)
_
=1.
Solution : Equation for the ellipse is
x2
y2
2
b2
For x = 0, y = b and y = 0, x = a.
Assuming a > b quarter of ellipse will be as
shown.
x = a sin 0 and y = b cos 0
Let
*
•
tWIM|N|IMI
• 1*114
M
a
Fig. 4.13
>x
dx = (a cos 0) d0
•
and
dy = ( - b s i n 0 ) d 0
At
n
x = a, 0 = 2 and
At
y = b, 0 = 0.
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4-14
Hence for a vertical strip, (Fig. 4.12)
dA = y dx = ab cos 2 0 dG
ir/2
A =
J ab
0
ab
cos 2
8+
A
=
A
nab
=
~4~
+ cos2G
0 d0 = ab
sin 2G -.n/2
dO
abf k
T { l
0
Moment of strip about Y axis = x (y dx) = (ab sin 6 cosG) (a cosG dG)
71/2
•
Moment of entire area =
«
J ( a 2 b s i n 0 cos 2 e ) de
o
n/2
= a 2 b j (sin© c o s 2 0) dG
0
n/2
a 2 b J C O S 2 6 ( - sinG) d0
0
= -a
2
b
cos3 G
n/2
0
= - a2b
2b
Moment of entire area about Y axis =
x =
/
V
xrab
HI
or
Moment of strip about X axis = (y dx) y
i 2
.*. Moment of entire area about X axis = J - y dx
a
0
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Centroid of Plane Figures
b2ar
J cos 2 0 (a cosG d0)
0
ab2 r
2 J
0
cos 3 0 d 0
ab2p-l
2 [ 3
... n = 3 (odd power)
ab2
Moment about X axis =
ab2
ab2^
y =
/
nab
)>»•• E x a m p l e 4 . 6 :
or
Obtain centroidal distances from reference axes for the shaded area shown
in the Fig. 4.23.
[PU, May - 1998 Old]
AY
1
i
(a.b)
y = kx
(•n»«»«mi a
au m i
aam i aw * m w i m t « ^ « n '
»«t«
—>X
Fig. 4.13
»
b
b
Solution : At point P, we get m = — and k = — . Let y . = m x and y 2 = k x 2 . Hence height
a
a2
of elemental strip = y 1 - y 2 at a distance x as shown, in Fig. 4.14.
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Centroid of Plane Figures
Y
A
A = J (yi-y2)dx=J
a
I
0
I•
!
I«
r
•
1!
o
3^
j:
/
m
/
/
\
\
\
\
\
(mx-kx2)dx
V1-V2
/
ba
A
ba
ab
=
Fig. 4.14
a
Moment of area about Y axis = J x ( y j - y 2 ) dx
0
a
J ( m x 2 - kx) dx
0
/
3 \
m
4>
-k
/
a^b
fa b
2
12
12
\
j
... by putting values of m and k
or
/
Moment of area about X axis = J
N
yi
(Y]")^)^*
+
y2
0
=
^J
0
= IJ
2
(yi -y
2
2
)
d x
(m2x2-k2x4)dx
0
1
2
2ab 2
30
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2ab 2
30
y =
Centroid of Plane Figures
\
/
ab
6
2b
•
y
»
t
=
»»•• Example 4.7 : Locate the centroid of the section shoum in Fig. 4.15.
[Anna Univ., May-2005]
Rad. 10 cm
Quarter
circle
30 cm
H—20 cm
Fig. 4.15
Solution :
Let
A i = (20 x 30) = 600 cm 2
2
A2 = -
= - 78.54 c m 2 and £ A = 521.46 c m 2 .
With respect to left bottom corner (Fig. 4.16).
20
*i
x,
=
= 10 cm,
= 20 -
3
y, = —0 = 15 cm
4(10)
3TC
Y
= 15.756 cm
4(10)
y 2 = 30 -
' M V A V . ' A
3TC
»*<•>
.•VmU
= 25.756 cm
_
x
—
h
r
—
i i v i
X v . v l
\ * > > v :<v>.
i.v.
A y X s v M v . i
S x
* •
A i xi + A 2 x?
=
>• < • > * • > A
"
. ^ i v ^ - M s ' X - f i '
r-x-s
*
"
i «iri
. • «•..•• A . t ,
r> • * i
«V. SV.S'i*.'. SV.\*»\VV»,.1tiV.'. ^ . - . ' i V . S V . W ^
.».» i . r t a • v.* i • • i i
•
• • • C u • «.tj.i.*.< J
.i •
W . ' . v ,
I
a
A i + A2
yi
V A S V . ' i V A S V . S ' . V . y ^ . ' . « , V , S V V. V i 1 . ' . V / ^ ' . V . V ,
A W . V / . S V . ' . W / A V SVA
V.".
V.'
v >•
x-fc
v.*
^ V A V i ' . V A V / . V i W . ' & i V M V . ' . V M y . ' .
v.* \ v
. \V.y .v.v.*
v . - M v • *.v. v .
(600)(10)+(-78.54)(15.756)
(521.46)
y2
x
o
i
H
Fig. 4.17
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4-18
_
_
y =
A i yl+A2
y2
A i + A2
( 6 0 0 ) ( 1 5 ) + ( - 78.54) (25.756)
(521.46)
))
Example 4.8 : Determine the centroid of area shown in Fig. 4.27 by taking moment of
[Anna Univ Dec.-2004]
area about the given aa-axis and bb-axis.
12 cm
6 cm
12 cm
Fig. 4.17
Solution : Given axes means the origin is at right bottom corner of trapezeum at O as
shown in Fig. 4.18. There are three component areas - rectangle, triangle and semicircle.
6 cm
• X
6 cm
U
1
-X
6 cm
1
cm
Fig. 4.18
A1
= ( 1 2 x 6 ) =72 c m 2
A2
= ^(12)(6) = 36cm2
A3
=
71(6)
= 56.55 c m 2
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4-19
£
A = 164.55 c m 2
xi
12
= t
*
6
=
cm,
x2 = 1 ( 1 2 )
a
Yi = o = 3 cm
6
... for 'x' measurement, 12 cm is "h" and 6 cm is 'b'.
*2 = 4 cm
y2
= 6 + ^(6)
y2
= 8 cm
*3 = —
=
- 2.546 cm
ys = 6 cm
v
A—
~
£
... now 'h' is 6 cm and 'b' is 12 cm
due to symmetry for semicircle.
(A x) _ ( 7 2 ) ( 6 ) + ( 3 6 ) ( 4 ) + ( 5 6 . 5 5 ) ( - 2 . 5 4 6 )
I
A
X == 2.625 cm
I
y =
... on opposite side of the origin.
164.55
i.e. to the left of the aa-axis
(A y)
I A
(72) ( 3 ) + ( 3 6 ) ( 8 ) + (56.55) ( 6 )
164.55
y == 5.125 cm
i.e. above the bb-axis
)>»•• Example 4.9 : A semicircular area having radius 100 mm is located in the xy plane such
that its diametral edge coincides with the Y-axis. Determine X-coordinate of its centroid.
[Anna Univ., May-2003]
Solution :
The area will be as shown in Fig. 4.19. Centroid of
the area lies on the horizontal axis of symmetry at a distance
4R
x =-=— from the Y-axis. (There is no relevance of X axis.)
371
i.e.
[Refer solved example 4.2 to get the same result from first
principles (by turning Fig.4.7 through 90° anticlockwise) if
required/asked.]
Fig. 4.19
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Centroid of Plane Figures
)>!•• Example 4.10 : Determine the centroid of the cross sectional area of an unequal I-section
shown in Fig. 4.20.
[Anna Univ., Dec.-2002]
h
20 cm
H
5 cm
5 cm
15 cm
5 cm
b
30 cm
H
Fig. 4.20
Solution :
As the reference axes are not given, we assume origin and X, Y axes as shown
in Fig. 4.21 i.e. x = 0 (or x = 15 cm from left or right bottom corner). Hence we have to
find only y where
y
A]
_ A i y 2 + A 2 y 2 + A 3 y3
Aj+A2+A3
= 3 0 x 5 =150
A 2 = 1 5 x 5 =75
h
cm2
20 cm
H
5 cm
cm2
A 3 = 2 0 x 5 = 100 c m 2
£
5 cm
A = 325 c m 2
yi = 2
y2
= 2 5
15 cm
c m
15
= 5 + — - 12.5 cm
-X
y 3 = 5 + 15 + ^ = 22.5
+X
30 cm
Fig. 4.21
(150)(2.5)+(75)(12.5)+(100)(22.5)
325
y = 10.962 cm
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Centroid of Plane Figures
Find the centroid of the shaded area shown in Fig. 4.22.
[Anna Univ., July-1999]
Example 4.11
Y
1
27 mm
35 mm
40 mm
13 mm
X
h
4*
47 mm
H
35 mm
Fig. 4.22
Solution :
Area x and y are already given.
A i = 40 x (47 + 35) = 3280 m m 2
A 2 = - 3 5 x 1 3 = - 4 5 5 mm2
A
= -
3
rc^
= -628.31 mm2
47 + 35
„
= — - — = 41 mm,
xi
and £
A = 2196.69 m m 2
40
„
y i = — = 20 mm
35
13
X2 = 47 + — = 64.5 mm, y2 = — = 6.5 mm
X3
= 2 7 mm,
X
A
t
—
X
•
And
=
y XT
—
„„
4 ( 2 0 ) = 31.51 mm
(A.x)_(3280)(41)+(-455)(64.5)+(-628.31)(27)
2196.69
L A
40.14 mm
I
(A y)
z*
( 3280) ( 2 0 ) + ( - 4 5 5 ) ( 6 . 5 ) + ( - 628.31)( 31.51)
2196.69
y = 19.50 mm
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»»•
4-22
Centroid of Plane Figures
Example 4.12 : Locate the centroid of the plane area shown in Fig. 4.23.
[Anna Univ., June-2001]
100 mm
25 mm
100 mm
50 mm
25 mm
150 mm
Fig. 4.23
Solution :
Reference axes are not given. Assume left bottom corner as origin O and X
axis horizontal, Y axis vertical through O. There are three basic shapes square, triangle and
semicircle. Refer Fig.4.24 (cross marks show individual centroids).
Y
^
50
100
100
Fig. 4.24
Let
Ai =
(150) 2 = 22500 m m 2 ,
A 2 = -i(50)(HX>)
- 2500 m m 2
and
A3
=
*
-7i (50) 2
-3927 mm2
Centroidal distance are :
xi = y i
= 75 mm
50
X2 = -=- = 16.667 mm
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ECE&EM
4-23
*3
= 150 - ^ ^
and
y3
= 128.779 mm
3 r.
y 2 = 150 -
Centroid of Plane Figures
= H6.667 mm
= 7 5 mm
(22500)(75)+(-2500)(16.667)+(-3927)(128.779)
x =
22500 - 2500 - 3927
X
Y
Y
= 70.934 mm
(22500)(75)- (-(- 2500)(116.667)+ ( - 3927)(75)
16073
= 68.519 mm
Example 4.13 : Find the centroid of the shaded area shown in Fig. 4.25.
[Anna Univ., Dec.-1997]
30 mm
140 mm
60 mm
h
H
200 mm
Fig. 4.25
Solution :
There are four basic figures involved. Hence it is better to prepare tabular
form for convenience as shown below.
Sr. No.
Area (mm 2 )
x (mm)
y (mm)
A.x (mm 3 )
A.y (mm 3 )
1
200x140= 28000
100
70
2800000
1960000
2
-(40)(60) = - 2400
40
2 0 0 - - ^ = 180
- 432000
- 264000
80 + ^
=110
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ECE&EM
4 - 24
3
- ^ ( 3 0 X 8 0 ) = - 1200
4
— rt(30) 2 = - 2827 43
V
/
21572 57 mm2
30
IT
«„
Centroid of Plane Figures
6 0 + 3 ( 8 0 ) = 113 333
- 12000
- 136000
70
- 282743.34
- 197920.34
—
2073256 7
1362079.7
100
—
Table 4.3
I
X =
Mow
(A x)
I
A
2073256.7
21572.57
x = 96.106 mm
Z
y =
And
(A y) 1362079.7
X A
" 21572.57
V = 63.139 mm
>»•• Example 4.14 : Locate the centroid of area slwam Fig. 4.26 ivith respect to the cartesian
coordinate system shown.
2m
(VTU, Jan.-2003)
2m
5m
1m
6m
10m
Fig. 4.26
1m
ECE&EM
4-25
Centroid of Plane Figures
Solution : Given figure (section of dam) comprises of rectangles and triangles
1.5 m
-X
2m
2m
1 m
5m
Fig. 4.27
Divide the section into triangle and rectangle as shown in Fig. 4.27
A,
= 1920(6) = 6 m 2
x,
=
3(6) = 2
Vi
O is origin
(2) = 1.333 m
m
A 2 = (2) (7.5) = 15 m 2
2
7.5
= 3.75 m
x2 = 2 + ^ = 3 m , y2 =
A 3 = i(5)(5) = 12.5 m 2
x 3 = 4+1 3 ! = 5.667 m
ya = 1 +[ 3 1 = 2.667 m
Aa = (1) (6) = 6 m 2
=
4
= 7 m, y4 = i - 0.5 m
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ECE&EM
4-26
SA
Hence
= Aj + A 2 + A 3 + A 4 = 39.5 m
A 1x1 + A2X2 + A 3 x 3 + A 4 X 4 _ 165.8333
A1 + A 2 + A 3 + A 4
39.5
X
or
X
And
= 4.198 m
A i y i + A 2Y2 + A 3 y 3 + A 4 y 4
A1+A2+A3+A4
Y
or
Centroid of Plane Figures
Y
104.583
39.5
= 2.648 m
Hence centroid will be at G (x, y) as shown approximately.
For a trapezium of parallel sides 'a' and 'b' with height 'h', show that
)>»• Example 4.15 :
y=
— xvhere a < ft.
A3 J
{a+h
Solution : Let the trapezium be as shown. Divide it into a rectangle and two right angled
triangles. Let PU = c. Therefore TS = b - a - c. Individual centroids are shown by small
cross marks. (Refer Fig. 4.28).
A i = — ch, A 2 = ah and
A 3 = - ( b - a - c ) h with A
H
u
Also
Fig. 4.28
[(i£)hl,
.
('ch)(*)+<ah>
(a + b) y =
y =
1
H
y i = 3 ^ y2 ="2 and y3
Consider moments
trapezium.
( b - a - c ) h" h
+
. m H.
U
about
A
the
base
1
h
U
of
3
+ 3a + b - a - c ]
2a
a
m
Example 4.16 : Determine the position of centroid for the lamina with a circular cutout
shown in Fig. 4.29.
(VTU, July - 2003)
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4-27
Centroid of Plane Figures
B
60 mm 60 mm
120 mm
Fig. 4.29
Solution : Let Ax = area of triangle = ^ (60) (120)
.\Aj = 3600 mm 2 , x, = ^ (120) = 40 mm and y, = 100 + ^ ( 6 0 ) = 120 mm
A 22 = Area of rectangle = (100) (120) = 12000 mm 2
x2
A,
= 60 mm and y 2 = 50 mm
"(50) 2
2
= Area of semicircle = — - — = 3926.991 mm
=
4(50)
120 + ~3TT = 141.221 mm and y 3 = 50 mm
A 4 = Area of circular hole = n(20) 2 = 1256.637 mm :
x
=
A i x i + A 2X2 + A 3 x 3 - A 4x4
I
_
_
y =
A
Aiyi+A2y2+A3y3~A4y4
I
= 73.517 mm
A
_
= 63.793 mm
Hence co-ordinates of centroid with respect to point A are (73.517, 63.793)
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ECE&EM
4-28
Centroid of Plane Figures
Example 4.17 : Locate the centroid of the shaded area shoivn in Fig. 4.30 with respect to
the axes shown
(VTU, Jan.-2006)
Lx
Fig. 4.30
Solution : For the isosceles triangular hole, b = 20 mm, h = 30 mm.
A 3 = ^(20)(30) = 300 mm 2 and X3 = 30 mm
Assume base of triangle at 40 mm level (from top) as appears in the Figure
1
y 3 = 50 +—(30) = 60 mm
J
Let A 2 = area of quarter circle (cut out) =
A2
x2
= - 1963.5 mm 2
= 90 - ^S^- = 68.78 mm
3K
and
4(50)
y2 = ~~3TT = 21.22 mm
Lastly,
A1
and
• •
= area of square = 8100 mm
= yi
I A
n(50)
= 45 mm
= A 1 - A 2 - A 3 = 5836.5 mm2
X = A1X1-A2X2-A3X3
2>
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ECE&EM
4-29
Centroid of Plane Figures
8100(45) - (1963.5)(68.78) - (300)(60)
5836.5
8100(45) - (1963.5X21.22) - (300) - (60)
y =
5836.5
Also
Example 4.18 :
A
metal plate having
uniform
thickness is shown
in
the
figure,
Determine the position of its centre of gravity with reference to point O.
T
500
350
Fig. 4.31
Solution : Given figure can be divided into components square OACE - triangle OAF
-
square BCDG + quarter circle.
A
B
C
A i = (1000) (1000) = 1 x 10 6 m m 2
A 2 = \ (650) (1000) = 0.325 x 10 6 m m 2
D
A 3 = (500) 2 = 0 . 2 5 0 x 1 0 6 m m 2
A4
= |(500) 2 = 0 . 1 9 6 x l 0 6 m m 2
. O
Fig. 4.32
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ECE&EM
4-30
Centroid of Plane Figures
Hence table 4.4 shown below.
Area A (mm2)
x (mm)
y (mm)
A.x(mm3)
A. y(mm3)
1*10 6
500
500
500x10 6
500 x 106
^(1000)
- 70.417 x10®
- 108.333 x106
- 187.5 x10®
- 187.5 x106
139.592 x106
139.592 x106
381.675 x106
343.759 x10®
- 0.325 x 10®
- 0.25 x106
500+
0.196 x106
i
I (650)
500 .
2 °°
4
500
0.621 x106
5
500+ < f ° >
<3 It
—
—
Table 4.4
gives
I>y
y = ^TT
gives
y = 553.6 mm
"Example. 4-19 : • Detom'ne location of centroid of shaded portion of lamina with respect
to origin O.
[PU, May - 1994]
axis
T
50
mm
R = 25 mm
50
mm
JL
X axis
R = 50 mm
Fig. 4.33
Solution : Shaded area comprises of : rectangle + triangle -f semicircle - circle.
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«
ECE&EM
4-31
Area A (mm 2 )
x (mm)
200x50 = 10000
y (mm)
A.x(mm3)
A.y(mm3)
25
1*10®
0.25 *10 6
0.333 xlO 6
0.167 x106
0.589 x10®
- 0.083 x106
- 0.295 x106
- 0.049 x106
1.628 x106
0.284 x106
100
^ (100) (50)=2500
SO
50+66.667
4
1 0 0 + ^ 9 = 133.333
2
« .
150
^(50) = 3926.99
2
-JI(25) = - 1963.5
I
Centroid of Plane Figures
150
14463.49
2 1
.
2 2 1
25
—
—
Table 4.5
Note that 'y' for semicircle being below X axis, is negative
x = Z
A x
I >
•
•
x = 112.55 mm
•
y =
•
•
1.628 xlO 6
14463.49
l A y
2 >
0.284xl06
14463.49
y = 19.65 mm
•
Example 4.20 :
Determine co-ordinates of center of circle to be cut from a plate such
that centroid of remaining plate ivill coincide with the center of circle itself.IPU, May-19951
200 mm
300
150 mm
»X
400 mm
Fig. 4.34
Solution : If (x, y) are co-ordinates of centroid of the shaded portion, center of circle will
also have same co-ordinates.
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ECE&EM
4-32
2
x (mm)
Area A (mm )
Rectangle
400 x 300 = 12000
Centroid of Plane Figures
3
3
y (mm)
A.x (mm )
A. y ( m m )
150
24x10 6
18 X106
150+ ^(150) =250
- 5 xlO 6
- 3 75 x106
Y
- 31415.927 X
- 31415.927 y
—
19 x106 31415.927 x
200
Triangle
200 + 1 (200)
-^(200) (150)
= 333.333
= - 15000
Circle -^(200) 2
X
= - 31415.927
z
73584.07
—
14.25 x 10® —
31415.927 y
Table 4.6
Now
„X .=
i
a
_
X =
Z A
Also
»
_
_
y
~
Z
A
y
__
glVCS
19xl06-31415.927 x
7358407
14.25xl06-31415.927y
73584.07
y
4
Example 4.21 :
A metal piece of uniform thickness is placed on horizontal surface as
shoivn. Find distance 'd' so that the piece will just be prevented from tipping. Diameter of
the hole is 0.5 m.
[PU, Dec. - 1995]
0.5 m
1.3 m
1.2 m
Fig. 4.35
Solution : Tipping may occur about point O as shown in Fig. 4.36. Therefore the moments
of portions to the left of O and to the right of O must be same.
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4-33
2m
4m
Centroid of Plane Figures
L.H.S. moment = (4x 0.3) (2) = 2.4 m
! d i
0.5 m
R.H.S. moment = (2x 1.8) (1)
1.3 m
. 2 + 1 x0.8 J
3
-G*
8x1.3
x 0.5 x 0.5 j d
Equating the moments, we get
1.2 m
Fig. 4.36
Example 4.22 :
Locate centroid of the lamina
R = 12 cm
I
Fig. 4.37
Solution : Prepare table for components.
OB = 12 cm
2
x (cm)
Area A (cm )
^(10.392)x(6)
Hence OC = 10.392 cm
l(6) = 2
y (cm)
|(10.392) = 6928
3
3
A x (cm ) A y(cm )
215.987
62.352
= 31.176
2
*(12)
6
= 75.398
106.574
3 Sin30 x(12)x(sin3a>)
MS)
i
498.833.
288.020
561.185
504.01
P
= 3.82
= 6.616
—
—
Table 4.7
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»
4-34
x = 5.27
•
Centroid of Plane Figures
cm
and
For a general lamina in the form of triangle OAB as shown, show that
Example 4.23 :
x - ^Ar1 where x = distance of centroid of the triangle from Y axis. (refer Fig. 4.38)
X axis
Fig. 4.38
Solution : Let height of two right angled triangles A D = h and A i =
1
ah, A 2
1
(L-a)h
be their areas. Area of given triangle = i L h. Consider moments about Y axis.
nm
(L-a) h
L H
) (X)
Lx
= ^
+
3 +
(L-a)
(L-a)r3a+(L-a)
2a2
-a
(L+ 2a)
2az+L2+2aL-aL-2a2
L (a + L)
3
X
• «
))
Example 4.24 :
=
a +L
A plate of uniform thickness is formed by attaching one rectangular and
one semicircular plates as shown. Determine 'b' in terms of V so that 'C will be centroid of
the plate.
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ECE&EM
4-35
Centroid of Plane Figures
2a
Fig. 4.39
Solution : Moments of areas to the left and right of 'C' must be same
*
"
2
(a)2
"
(2 ab)
r
2
Example 4.25 :
or
b = 0.816 a
A metal piece as shown must hove its center of gravity at point G.
Determine dimension 'a' for this purpose.
a
60 mm
60 mm
120 mm
Fig. 4.40
Solution : Just like previous problem, equate moments of portions to the left and right
of G.
\ (a) (120)
20 a 2
(120)2 [ m .
n (60)
120-
4x60 \
3n
J
= 864000 - 534584.01
a = 128.34 mm
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ECE&EM
4-36
»*•• Example 4.26 :
Centroid of Plane Figures
A metal piece of uniform thickness is to he suspended in the position
shown. Determine length L
/ / / / / / /
O
String
12 cm
30 cm
Fig. 4.41
Solution : Let the reference axes and origin be as shown in Fig. 4.42. As 'G' lies below O
i.e. on the line OA extended, from the geometry of figure,
Y
A
»X
Fig. 4.42
We get,
x + L = 12 + 30
L = 42 - x
i.e.
Consider V.T.M. about Y axis.
^
+
i(30)(24) V
A—
—
Solving, we get
"k(12) 2 _
2
12
—
3it
)
+
(30) (24)
12 +
30
x = 16.176 cm
L = 25.824 cm
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ECE&EM
Centroid of Plane Figures
4-38
/
Now
_ _ Z A x _ 3280500
x =
32382.075
I A
x = 101.306 mm
Also
_ _ £ A y
y =
I A
_5013219.4
~ 32382.075
y = 154.815 mm
For hung position, (refer Fig. 4.44) we
have
4
•
tan 6 =
360-y
tan B = 0.494
Fig. 4.44
4.9 Chapter Summary
•
Centroid is an important geometric property of a plane figure and its position
with respect to some reference lines (axes) can be obtained in the form of
coordinates (x,y). By considering elemental strip, area and coordinates can be
obtained by integration.
•
Varignon's theorem of moments is used to find these coordinates (x,y) by using
formulas.
A ) x = A\ x\ +A2 X2+
( 2 > ) y = A i -yi + A 2 -y2+
•
Centroid lies on axes or axis of symmetry. For freely suspended lamina, centroid
lies, vertically below point of suspension (for equilibrium position).
•
Composite plane figure should be divided into basic regular figures such as
rectangle, triangle, circle etc.
•
Area is taken negative for a hole or removed / cut portion while centroidal
distance is negative if it is measured on opposite side of reference X or Y axis.
Copyrighted material
ECE&EM
4-40
Centroid of Plane Figures
50
100 mm
100
I
I
I
I
100 mm
i
i
i
i
200
Fig. CP-4
[PU, May 19931 [Ans.: x = - 107.8 mm, y = 64.4 mm]
2a
a
4. For shaded area, show that x = ^ and y = —
__
__
-
i m i a m i i >»<•«« •
im
|PU, May-1999]
i i i i i i n i u i m i i M i i M i i
i
• X
Fig. CP-5
5. For shaded areas shown, find y
30 mm
Hole
10 mm radius
30
30 mm
X ^
X
X
60
20 mm
60
100 mm
[PU, Dec-1998]
[Ans.: 27.69 mm)
[PU, Dec-2003]
[Ans.: 16.72 mm]
Fig. CP-6
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ECE&EM
4-41
Centroid of Plane Figures
6. For a composite figure shown, find V if centroid of the figure is to coincide with centroid of
rectangle ABCD
[PU, Dec-2002]
t
r
B
40 mm
20 mm
D
a
40 mm
20 mm
Fig. CP-7
[Ans.: a = 30.228 mm]
7. Determine co-ordinates of centroid of plane as shown.
[PU, Dec-1998 Old]
Y
I
30
—^ X
All dimensions in mm
Fig. CP-8
[Ans.: x = 29.96 mm, y = 30.27 mm]
8.
Find angle made by line AB when lamina shown below is freely suspended from A.
Y
Y
4
4
i
.i
i
E
_
MNIIM*
a
a
[Ans. : 44.64°]
[Ans. : 16°]
Fig. CP-9
•••
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Copyrighted material