Centroid-of-Plane-Figures

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) Copyrighted material

ECE&EM

4-2

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

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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4-3


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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4-5


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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4-6


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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4-7 Ibhy

i.e.

^ J 2b

ill


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 )


4-8

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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


4-9

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Apply

(Z

A

)y


= I>-y)

.•. Distance from apex will be 161.17 mm. 4R )>!•• Example 4.2 : For a semicircular lamina, obtain y = -=— with usual notations. 3n
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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4-10


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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4-11


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


ECE&EM Example 4.4 : Prove that x =

4-12


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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4-19

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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4-15

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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4-16


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

Copyrighted m i . lal

ECE&EM

4-19 2ab 2 30

y =


\

/

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 :.

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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4-20


)>!•• 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


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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4-21


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


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


ECE&EM »»•

4-22


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


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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4-23 *3

= 150 - ^ ^

and

y3

= 128.779 mm

3 r.

y 2 = 150 -


= 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


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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4 - 24

3

- ^ ( 3 0 X 8 0 ) = - 1200

4

— rt(30) 2 = - 2827 43

V /

21572 57 mm2

30 IT

«„


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


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


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


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)(*)+

(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)


ECE&EM

4-27


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)


ECE&EM

4-28


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>


ECE&EM

4-29


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


ECE&EM

4-30


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.


«

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


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.


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.


ECE&EM

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


ECE&EM »

4-34 x = 5.27

•


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.


ECE&EM

4-35


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


ECE&EM

4-36

»*•• Example 4.26 :


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


ECE&EM


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.


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 Copyrighted material

ECE&EM

4-41


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

••• Copyrighted material


Centroid-of-Plane-Figures

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