mechanics - 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 Wj, VvS,... etc. we have relation

W = Wi + W2 + ...

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.

x W = xi Wi +X2W2 + ....

If density of material p (kg/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 d A ) Hence x = =

{ (dA) I ( d A )

Here the product (x • A) is known as 'first moment of area'.

c . .. . _ J <y •d A> 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 manner and show all distances/lengths etc. clearly to avoid mistakes.



(a) G lies below O

G_0 rh^r Hinge or pin

(b) G coincides with O

Hinge

(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. Description and area Figure X y

1. Rectangle A = b . d

-a

-

x b 2

d 2

1. Rectangle A = b . d

-a

4

- !

k G

2 H

~r d

J1 _

b 2

d 2

2. Triangle

A = ^ b . h • * - i

/ i \

/ G i i v \

- H b/2 ' b/2 1

r h

:

b 5

h 3



Sr. No. Description and area Figure

3. Right angled triangle

A = ^ bh H

T

i H

b

4. Circular sector A = aR2

O

H a ) 2( Rsina Zero

Quarter circle

7t u = 4

4R 3 n

4R 3*

6. Semicircle

2

2 J

Zero 4R 3 a

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.'

ed material

ECE&EM 4 - 5

I

Centroid of Plane Figures

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(dx) 0

= d[x]J

... d is a constant = b.d

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

• • b d x = d (b^ 2

/

Now refer Fig. 4.3 showing horizontal strip of thickness dy so that area of strip = b-dy

d Ay - b j y- dy

0

(by applying V.T.M. about X axis.)

b*d*y = b d ^

- X

Fig. 4.2

I I 7

y

. I 0 b H

Fig. 4.3

/

Note : Same vertical strip (used for x) can also be used for y.



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 = h ] d x

A = | h i - d x 0

• •

From similar triangles, we have =

hx Hence dA = -p- dx

b

A x dx

Fig. 4.4

• / G) o y

= h f b 2 - b | T

A = I b h

Now consider first moment of the strip about Y axis and apply V.T.M b

A x = J x d A 0

^ bh x " J

^ bh x

0

h b

^ 1 x 2 dx

3

Now consider first moment of the same strip about horizontal strip also can be considered). Apply V.T.M. also.

I b h y = f n U l d A

X axis. (Alternatively, a new

0

- f t e Y £ i I dx

0 U b A b



i.e. I b h y ^ J x2 dx 2 b

i l l 2b2

0 ( u3

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 COS0

a 2J R2

de ^ _ o

cosG

x = (!Msine ] a

0

r R 2 2 j i L d O

~>x

a

0 Fig. 4.5

• m

Note : Again rad. for quartercircle and r a d. f° r a semicircle can be _ 4R

substituted, to obtain x (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 ( a - x )

I ( a ) and

y " 1 ( a )



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. (VTU, 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 ) Vertical (y) centroidal

distances (mm)

3 Product a.y (mm )

^ (160) 240 = 19200 ^ (240) = 80 +1536000

- -J(40)2 = - 1256.64 160+ ~ = 180 - 226195.20

-(40) (60) = - 2400 8 0 + ^ = 100 - 240000

_ .1(40)2 = - 2513 27 4<40> = 1 6 98 - 42675.325

£ A = 13030.09 — 1027129.5

Table 4.2



Apply ( Z A ) y = I > - y )

.•. Distance from apex will be 161.17 mm. 4 R )>!•• 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 R Area of strip = ^2-J R 2 - y 2 ) dy

•-X .'. First moment of strip about diameter

( 2 / R 2 ^ ) ydy

Now use v dA

J dA

Fig. 4.7

•• y = J ( 2 A / R 2 - y 2 ) y d y 0

J ( i J W ^ y y 0

R Numerator = - J ( R 2 - y 2 ) V 2 < - 2 y ) d y

0

( R 2 - y 2 ) 3 /2

3/2 R

0

• - ! [ ° - ( r 2 ) 3 / 2

R Denominator = 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


ECE&EM 4 - 1 0 Centroid of Plane Figures n/2

2R2 J cos 2 ede 0

n/2 = 2R 2U 0 v

cos 26 + 1 d9

= R2 + sin 2 6

+ 0

y = R2 TZ-zr-

0

TCR2

»»•• Example 4.3 : Determine co-ordinates of centroid of shaded portion (the spandrel). (VTU, July - 2006)

Y A

I N -•> X

a Fig. 4.8

Solution : Consider vertical strip of area dA ydx as shown in Fig. 4.9. n a

A s J ydx k J x 2 d x 0 0

Now y 2 m kx at point (a, b) gives b_ 2 k

i i r o t w f n t n m : > x H K

i r A = * dx

Fig. 4.9

a

a 2 a* T

ab J

Moment ef the elemental area about Y a*i§ = * (ydx),


ECE&EM 4-11 Centroid of Plane Figures

Moment of entire area = J x ( k x 2 ) d x o

x =

' b

V

ab or

k | x 3 dx 0

\ /

Similarly moment of entire area about X axis = J (y dx) 0

fy \

k2 J x4dx 0

» t t y =

2a 4 { 5

ab2

10 f a b 2 N

10 i

a p "3

5 ^

• ff y 3b 10


ECE&EM 4 -12 Centroid of Plane Figures

Example 4.4 : Prove that x = 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

*

I * i i i i i i j 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 = b 2

A =

A =

J ( - / k x ) dx = / k j ^ d x 0 0

Fig. 4.11

L 3 / 2 Jo 2 Putting value of k and simplifying, A = — (ab)

Now consider moment of elemental area (strip) about X axis.

Moment of strip = (y dx) ^

a 1 For entire area, moment = J — y 2 dx 0

i a

x d x 0



Now,

b 2

2a 2

y = Moment of area about X axis Total area

• 4

Similarly consider moment of strip about Y axis. a

Moment for entire area = J (y dx) x 0

9

a = y f k j x V x d x

0

= / k 1x5/2 [ 5 / 2 _

0

2a 2 b

x =

2a 2 b 5

2ab

3 J or

)»•• Example : 4.5 : Determine x and y for quarter of an ellipse (in the first quadrant) x2 +y2 _ a 2 fc2

=1.

Solution : Equation for the ellipse is x 2 y 2

2 b 2

*

• 1 * 1 1 4 t W I M | N | I M I > x M

a Fig. 4.13

For x = 0, y = b and y = 0, x = a. Assuming a > b quarter of ellipse will be as shown.

Let

• •

x = a sin 0 and y = b cos 0

dx = (a cos 0) d0

and dy = ( - b s i n 0 ) d 0

n At x = a, 0 = 2 and

At y = b, 0 = 0.



Hence for a vertical strip, (Fig. 4.12)

dA = y dx = ab cos2 0 dG ir/2

A = J ab cos2 0 d0 0

= ab + cos2G dO

A = ab 8 + sin 2G -.n/2

0 abf k T { l

A = nab ~4~

• «

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© cos 2 0) dG

0

n/2 a 2 b J COS 26(- sinG) d0

0

= - a 2 b cos 3 G n / 2

0

= - a 2 b HI Moment of entire area about Y axis =

2 b

x = V /

xrab or

Moment of strip about X axis = (y dx) y

a i .*. Moment of entire area about X axis = J - y 2 dx

0



b 2 a r J cos 20 (a cosG d0) 0

ab 2 r 2 J

0 cos 3 0 d 0

a b 2 p - l 2 [ 3

ab 2

... n = 3 (odd power)

Moment about X axis = ab 2

ab2^

y = /

nab or

)>»•• Example 4 .6 : 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 » « » « m i a a u m i a a m i a w * m w i m t « ^ « n ' »« t« —>X

Fig. 4.13 »

b b Solution : At point P, we get m = — and k = — . Let y . =mx and y 2 =kx 2 . Hence height a a 2

of elemental strip = y 1 - y 2 at a distance x as shown, in Fig. 4.14.



Y A

a

I I •

! I « r •

j : \ 1 / \ ! / \ / \

m \

V1-V2

Fig. 4.14

A = J ( y i - y 2 ) d x = J (mx-kx 2 )dx 0 o

/

A =

ba ba

ab

3 ^

a Moment of area about Y axis = J x ( y j - y 2 ) dx

0 a J ( m x 2 - kx) dx 0

m / 3 \

- k 4 >

/

a^b 12 ... by putting values of m and k

f a 2 b \

12 j or

Moment of area about X axis = J ( Y ] " ) ^ ) ^ * / N

y i + y 2

0

= ^ J ( y i 2 - y 22 ) d x

0

= I J ( m 2 x 2 - k 2 x 4 ) d x 0

1 2

2ab 2

30

Copyrighted m i . lal


2ab2

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

30 cm

Quarter circle

H—20 cm

Fig. 4.15

Solution : Let A i = (20 x 30) = 600 cm 2

2 A 2 = - = - 78.54 cm 2 and £ A = 521.46 cm 2 .

With respect to left bottom corner (Fig. 4.16).

* i = 2 0 3 0 = 10 cm, y, = — = 15 cm

x, = 20 -4(10)

3TC

= 15.756 cm

y2 = 30 -4(10)

3TC

= 25.756 cm _ A i xi + A 2 x? x = — h r — a A i + A2

(600)(10)+(-78.54)(15.756) (521.46)

Y

yi

o

' M V A V . ' A

»*<•> >• < • > * • > A

.•VmU i i v i X v . v l i . v . A y X s v M v . i

\ * > > v : < v > . S x . ^ i v ^ - M s ' X - f i ' r - x - s

* • " * "

i « i r i . • « • . . • • A . t , r> • * i . i • « V . S V . S ' i * . ' . S V . \ * » \ V V » , . 1 t i V . ' . ^ . - . ' i V . S V . W ^ W . ' . v , .».» i . r t a • v.* i • • i i • • • • C u • « . t j . i . * .< J I

V A S V . ' i V A S V . S ' . V . y ^ . ' . « , V , S V V. V i1 . ' . V / ^ ' . V . V , A W . V / . S V . ' . W / A V S V A V . " . V . ' v >• x - f c 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 .

y2

i H x

Fig. 4.17


ECE&EM 4 - 1 8 Centroid of Plane Figures

_ _ A i yl+A2 y 2 y = A i + A2

(600) (15 )+( - 78.54) (25.756) (521.46)

)) Example 4.8 : Determine the centroid of area shown in Fig. 4.27 by taking moment of area about the given aa-axis and bb-axis. [Anna Univ Dec.-2004]

6 cm

12 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.

• X

6 cm

U 6 cm 1 6 cm 1

- X

cm

Fig. 4.18

A1 = ( 1 2 x 6 ) =72 cm 2

A 2 = ^ (12) (6 ) = 3 6 c m 2

A 3 = 71(6)

= 56.55 cm 2


£ A = 164.55 cm 2

1 2 * 6 a xi = t = 6 cm, Yi = o = 3 cm

x2 = 1 ( 1 2 ) ... for 'x' measurement, 12 cm is "h" and 6 cm is 'b'.

*2 = 4 cm

y2 = 6 + ^ ( 6 ) ... now 'h' is 6 cm and 'b' is 12 cm

y2 = 8 cm

*3 = — = - 2.546 cm ... on opposite side of the origin.

ys =

v —

6 cm due to symmetry for semicircle.

£ (A x) _ (72) (6 )+(36) (4 )+(56 .55) ( -2 .546) A ~

I A 164.55

X = = 2.625 cm i.e. to the left of the aa-axis

y = I (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

4 R 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



)>!•• 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]

20 cm h H

b 30 cm

5 cm

5 cm

15 cm

5 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

_ A i y 2 + A 2 y 2 + A 3 y 3 y A j + A 2 + A 3

A ] = 3 0 x 5 =150 cm 2

A 2 = 15x5 =75 cm 2

A 3 = 20x5 = 100 cm 2

£ A = 325 cm 2

yi = 2 = 2 5 c m

15 y2 = 5 + — - 12.5 cm

y3 = 5 + 15 + ̂ = 22.5 -X

h 20 cm

H

5 cm

5 cm 15 cm

+X

30 cm

Fig. 4.21

(150)(2.5)+(75)(12.5)+(100)(22.5) 325

y = 10.962 cm



Example 4.11

Y 1

Find the centroid of the shaded area shown in Fig. 4.22. [Anna Univ., July-1999]

27 mm 35 mm

40 mm

t •

And

13 mm

h 47 mm 4 * 35 mm H X

Fig. 4.22

Solution : Area x and y are already given.

A i = 40 x (47 + 35) = 3280 mm 2

A 2 = - 3 5 x 1 3 = - 4 5 5 mm 2

A 3 = - rc^ = -628.31 m m 2 and £ A = 2196.69 m m 2

xi = 47 + 35 „ 40 „ — - — = 41 mm, y i = — = 20 mm

35 13 X2 = 47 + — = 64.5 mm, y2 = — = 6.5 mm

X3 = 2 7 mm, „„ 4(20) = 31.51 mm

X ( A . x ) _ ( 3 2 8 0 ) ( 4 1 ) + ( - 4 5 5 ) ( 6 4 . 5 ) + ( - 6 2 8 . 3 1 ) ( 2 7 ) A —

L A 2196.69

X = 40.14 mm

XT — I (A y) ( 3280) ( 2 0 ) + ( - 455 ) (6 .5 )+( - 628.31)( 31.51)

y - z* 2196.69

y = 19.50 mm



» » • Example 4.12 : Locate the centroid of the plane area shown in Fig. 4.23. [Anna Univ., June-2001]

100 mm

50 mm

25 mm

100 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

Let

Fig. 4.24

A i = (150)2 = 22500 mm 2 ,

A 2 = -i(50)(HX>)

and A 3 =

- 2500 mm 2

-7i (50) 2

* - 3 9 2 7 mm 2

Centroidal distance are : xi = y i = 75 mm

50 X2 = -=- = 16.667 mm



and

* 3 =

y 2 =

150 - ^ ^ = 128.779 mm 3 r.

150 - = H6.667 mm

y 3 = 7 5 mm

(22500)(75)+(-2500)(16.667)+(-3927)(128.779) x = 22500 - 2500 - 3927

X = 70.934 mm

Y (22500)(75)- (-(- 2500)(116.667)+ ( - 3927)(75)

Y 16073

Y = 68.519 mm

Example 4.13 : Find the centroid of the shaded area shown in Fig. 4.25. [Anna Univ., Dec.-1997]

60 mm

h 200 mm

30 mm

140 mm

H

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 (mm2) x (mm) y (mm) A.x (mm3) A.y (mm3)

1 200x140= 28000 100 70 2800000 1960000

2 -(40)(60) = - 2400 40 2 0 0 - - ^ = 180 80 + ̂ =110 - 432000 - 264000



3 -^(30X80) = - 1200 30 «„ IT 6 0 + 3 ( 8 0 ) = 113 333 - 12000 - 136000

4 — rt(30)2= - 2827 43 100 70 - 282743.34 - 197920.34

V / 21572 57 mm2 — — 2073256 7 1362079.7

Table 4.3

Mow

And

X = I (A x) I A

x = 96.106 mm

y = Z (A y) X A "

V = 63.139 mm

2073256.7 21572.57

1362079.7 21572.57

>»•• Example 4.14 : Locate the centroid of area slwam Fig. 4.26 ivith respect to the cartesian coordinate system shown. (VTU, Jan.-2003)

2 m

1 m

6m

2m 5 m 1 m

10m

Fig. 4.26


Solution : Given figure (section of dam) comprises of rectangles and triangles

1.5 m

- X

2 m 2m 5 m 1 m

Fig. 4.27

Divide the section into triangle and rectangle as shown in Fig. 4.27

A, = 1920(6) = 6 m2

x, =

Vi

(2) = 1.333 m

3 ( 6 ) = 2 m

O is origin

A2 = (2) (7.5) = 15 m2

2

x2 = 2 + ^ = 3 m , y2 =

A3 = i(5)(5) = 12.5 m2

7.5 = 3.75 m

x3 = 4+1 3 ! = 5.667 m

ya = 1 +[ 3 1 = 2.667 m

Aa = (1) (6) = 6 m2

= 4 = 7 m, y4 = i - 0.5 m



Hence S A

X

= Aj + A2 + A3 + A4 = 39.5 m

A 1x1 + A2X2 + A 3 x 3 + A 4 X 4 _ A1 + A2 + A 3 + A 4

165.8333 39.5

or X = 4.198 m

And Y A i y i + A 2Y2 + A 3y3 + A 4 y 4 104.583

And Y A 1 + A 2 + A 3 + A 4 39.5

or Y = 2.648 m

Hence centroid will be at G (x, y) as shown approximately.

)>»• Example 4.15 : For a trapezium of parallel sides 'a' and 'b' with height 'h', show that

y = — xvhere a < ft. {a+h A 3 J

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 . m H. H u

Also 1 U H A 1 U y i = 3 ^ y2 ="2 and y3 h

Fig. 4.28

[ ( i £ ) h l , . ( ' c h ) ( * ) + < a h >

Consider moments about the base of trapezium.

+ ( b - a - c ) h" h

3

(a + b) y = + 3a + b - a - c ]

y = 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)



60 mm 60 mm

120 mm

Fig. 4.29

B

Solution : Let Ax = area of triangle = ^ (60) (120)

.\Aj = 3600 mm2, x, = ^ (120) = 40 mm and y, = 100 + ^(60) = 120 mm

A2 = Area of rectangle = (100) (120) = 12000 mm2 2

x 2 = 60 mm and y2 = 50 mm

A, = "(50)2 2

Area of semicircle = — - — = 3926.991 mm

= 120 + 4(50) ~3TT

= 141.221 mm and y3 = 50 mm

A4 = Area of circular hole = n(20)2 = 1256.637 mm:

A ixi + A 2X2 + A 3 x 3 - A 4x4 x = I A

= 73.517 mm

_ _ A i y i + A 2 y 2 + A 3 y 3 ~ A 4 y 4 _ y =

I A = 63.793 mm

Hence co-ordinates of centroid with respect to point A are (73.517, 63.793)



Example 4.17 : Locate the centroid of the shaded area shoivn in Fig. 4.30 with respect to the axes shown (VTU, Jan.-2006)

L x

Fig. 4.30

Solution : For the isosceles triangular hole, b = 20 mm, h = 30 mm.

A 3 = ^(20)(30) = 300 mm2 and X3 = 30 mm

Assume base of triangle at 40 mm level (from top) as appears in the Figure 1

and

Lastly,

and

• •

y 3 = 50 +—(30) = 60 mm J

Let A 2 = area of quarter circle (cut out) = n(50)

A 2 = - 1963.5 mm2

x 2 = 90 - ^S^- = 68.78 mm

y2 = 4(50) ~~3TT

3K

= 21.22 mm

A1 = area of square = 8100 mm

= yi = 45 mm

I A =

X =

A 1 - A 2 - A 3

A 1 X 1 - A 2 X 2 - A 3 X 3

2 >

= 5836.5 mm2



Also

8100(45) - (1963.5)(68.78) - (300)(60) 5836.5

y = 8100(45) - (1963.5X21.22) - (300) - (60) 5836.5

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 BCDG + quarter circle.

A B C

square OACE - triangle OAF -

D

A i = (1000) (1000) = 1 x 10 6 mm 2

A 2 = \ (650) (1000) = 0.325 x 10 6 mm 2

A 3 = (500)2 = 0 .250x10 6 mm 2

A 4 = |(500)2 = 0 . 1 9 6 x l 0 6 mm 2

. O Fig. 4.32



Hence table 4.4 shown below.

Area A (mm2) x (mm) y (mm) A.x(mm3) A. y(mm3)

1*106 500 500 500x106 500 x 106

- 0.325 x 10® I (650) ^(1000) - 70.417 x10® - 108.333 x106

- 0.25 x106 500+ 500 . 5

2°° - 187.5 x10® - 187.5 x106

0.196 x106 500 500+ 4 <f °> <3 I t

139.592 x106 139.592 x106

0.621 x106 — — 381.675 x106 343.759 x10® i

Table 4.4

gives

I > y y = ^ T T g i v e s

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

50 mm

JL

Fig. 4.33

R = 25 mm

X axis

R = 50 mm

Solution : Shaded area comprises of : rectangle + triangle -f semicircle - circle.


«


Area A (mm2) x (mm) y (mm) A.x(mm 3 ) A.y(mm 3 )

200x50 = 10000 100 25 1*10® 0.25 *106

^ (100) (50)=2500 1 0 0 + ^ 9 = 133.333 SO 5 0 + 6 6 . 6 6 7 4

0.333 xlO6 0.167 x106

^(50)2 = 3926.99 150 « . 2 1 . 2 2 1 0.589 x10® - 0.083 x106

-JI(25)2 = - 1963.5 150 25 - 0.295 x106 - 0.049 x106

14463.49 — — 1.628 x106 0.284 x106 I

Table 4.5 Note that 'y' for semicircle being below X axis, is negative

Z A x 1.628 xlO6 x =

I > 14463.49

• •

• •

• •

x = 112.55 mm

y = l A y

2 >

0 .284xl0 6

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

400 mm

150 mm

»X

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.



z

Area A (mm2) x (mm) y (mm) A.x (mm3) A. y(mm3 )

Rectangle 400 x 300 = 12000

200 150 24x106 18 X106

Triangle

-^(200) (150)

= - 15000

200 + 1 (200)

= 333.333 150+ ^(150) =250 - 5 xlO6 - 3 75 x106

Circle -^(200)2

= - 31415.927 X Y - 31415.927 X - 31415.927 y

73584.07 — — 19 x106 -

31415.927 x 14.25 x 10® — 31415.927 y

Now „ . i a X =

Z A

Table 4.6 _ 1 9 x l 0 6 - 3 1 4 1 5 . 9 2 7 x X = 7358407

Also _ _ Z A y _ _ 1 4 . 2 5 x l 0 6 - 3 1 4 1 5 . 9 2 7 y y ~ g l V C S y 73584.07

» 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.



4 m 2m

! d i

1.3 m

1.2 m

L.H.S. moment = (4x 0.3) (2) = 2.4 m 3

0.5 m R.H.S. moment = (2x 1.8) (1) -G* 8x1.3

. 2 + 1 x0.8 J x 0.5 x 0.5 j d

Equating the moments, we get

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 Hence OC = 10.392 cm

Area A (cm2) x (cm) y (cm) A x (cm3) A y(cm3)

^(10.392)x(6)

= 31.176

l(6) = 2 |(10.392) = 6928 62.352 215.987

*(12)2

6 = 75.398

MS) = 6.616

3Sin30x(12)x(sin3a>)

i = 3.82

498.833. P

288.020

106.574 — — 561.185 504.01

Table 4.7



» • x = 5.27 cm

and

Example 4.23 : For a general lamina in the form of triangle OAB as shown, show that

x - ^Ar1 where x = distance of centroid of the triangle from Y axis. (refer Fig. 4.38)

X axis

Fig. 4.38 1 1 Solution : Let height of two right angled triangles AD = h and A i = ah, A 2 ( L - a ) h

be their areas. Area of given triangle = i L h. Consider moments about Y axis.

L H ) ( X )

Lx =

( L - a ) h ( L - a ) 3 + nm

^ + ( L - a ) r 3 a + ( L - a )

2 a 2 - a (L+ 2a)

2 a z + L 2 + 2 a L - a L - 2 a 2

• « X =

L (a + L) 3

a + L

)) Example 4.24 : 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.



2a

Fig. 4.39

Solution : Moments of areas to the left and right of 'C' must be same

* ( a ) 2 (2 ab)

" 2 " r 2 or b = 0.816 a

Example 4.25 : 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) (120)2 [ m . n (60) 1 2 0 - 4x60 \ 3n J

20 a 2 = 864000 - 534584.01

a = 128.34 mm



»*•• Example 4.26 : A metal piece of uniform thickness is to he suspended in the position shown. Determine length L

/ / / / / / /

12 cm

O String

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,

i.e.

x + L = 12 + 30

L = 42 - x

Consider V.T.M. about Y axis.

^ + i (30) (24) V — "k(12)2_

A — 2

12 — + 3it )

(30) (24) 12 + 30

Solving, we get x = 16.176 cm

L = 25.824 cm



/

Fig. 4.44

Now _ _ Z A x _ 3280500 x = I A 32382.075

Also y =

x = 101.306 mm

_ _ £ A y _5013219.4 ~ 32382.075 I A

y = 154.815 mm

For hung position, (refer Fig. 4.44) we have

4 • tan 6 = 3 6 0 - y

tan B = 0.494

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.



50

I I I I i i i i

100

200 Fig. CP-4

100 mm

100 mm

- a [PU, May 19931 [Ans.: x = - 107.8 mm, y = 64.4 mm]

_ _ _ _ 2a 4. For shaded area, show that x = ̂ and y = —

|PU, May-1999]

i m i a m i i >»<•«« • i m i i i i i i n i u i m i i M i i M i i i

5. For shaded areas shown, find y

X ^

20 mm

100 mm

[PU, Dec-1998] [Ans.: 27.69 mm)

Fig. CP-5

30 mm

• X

30

30 mm

X

Hole 10 mm radius

X

60 60

[PU, Dec-2003] [Ans.: 16.72 mm]

Fig. CP-6



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

B r

20 mm D

40 mm

a

40 mm 20 mm

Fig. CP-7

7. Determine co-ordinates of centroid of plane as shown.

[Ans.: a = 30.228 mm]

[PU, Dec-1998 Old]

Y I

30

All dimensions in mm

— ^ X

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 _

a a [Ans. : 44.64°] [Ans. : 16°]

M N I I M *

Fig. CP-9

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mechanics - centroid of plane figures

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