Lecture 16 Engineering Mechanics

Engineering Mechanics

Lecture 16

Center of Gravity & Centroid

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=

Lecture 16 Engineering Mechanics

Discussions

Any Question?