HJD Institute Of Technical Education And Research - Kera

Presenting by,Shah Parth L. (120850106009)Vaghela Suyagn D. (120850106010)Saradhara Divyesh. (120850106016)Patel Jay H. (120850106029)Bhatti Bhishma J. (120850106046)

CENTROID AND CENTRE OF GRAVITY

Centroid word is used to represent centre of line like,

-Arc -Rectangle

-Square -Triangle, Circle etc.

Centre of Gravity word is used to represent centre of solid bodies like,

-Cube -Cone, Sphere etc.

CENTER OF GRAVITY

Definition

“ It is defined as a Point about which the entire weight of the body is assumed to be concentrated. ”

“ The point where the weight of the body acts. ”

Fig 14.1

CENTER OF GRAVITY

The location of the CG remains fixed as long as the body does not change shape.

If an object’s shape or position changes, the location of the CG changes.

Fig 14.3

“ The moment of the resultant gravitational force(weight) W about any axis is equal to sum of the moments of individual weights about the same axis. ”

CENTER OF GRAVITY

Consider small elements of equal size and shape whose weights are dw1, dw2, dw3, …..dwn.

here,dw1 = dwdw2 = dwdw3 = dwdwn = dw

Total wt. of the body = sum of all individual wt.

So, W = dw1 + dw2 + dw3 + ….dwn

W = = total wt. = resultant gravitational force.

∑𝑖=1

𝑛

𝑑𝑤

Now, sum of moments of all individual weights @ y-axis

= dw1 x1 + dw2 x2 + dw3 x3 + …..dwn xn

= i xi

= dw … (1)

Moment of resultant weight @ y-axis= W … (2)

Equating (1) and (2)

· ·· ·

·

·

∑𝑖=1

𝑛

𝑑𝑤 ·

∫𝑥

𝑥

W = dw

= .

= . … (A)

Similarly, = . … (B)

· ·

·

𝑥 ∫𝑥

𝑥 ∫𝑥 dw

∫𝒅𝒘

W

𝒙 ∫𝒙·dw

∫𝒚· dw𝒚

∫𝒅𝒘

Centre of Mass :

= . = .

Centre of Area : (Centroid of 2D Geometric Figure)

= . = .

Centre of Volume : (Centroid of 3D Geometric Figure)

= . = .

𝑥

𝑥

𝑥

∫𝑥

∫𝑥

∫𝑥

𝑦

𝑦

𝑦

∫ 𝑦

∫ 𝑦

∫ 𝑦dm dm

dvdv

dAdA

∫𝑑𝑚∫𝑑𝑚

∫𝑑𝐴∫𝑑𝐴

∫𝑑𝑣∫𝑑𝑣

· ·

· ·

··

Centroid/Centre of Gravity of Composites : 1) Composites linear elements : (Line, arc etc.)

Part 1 = line AB

Part 2 = line BC Part 3 = line CD

l1, l2, l3 = length of part 1, 2, 3 respectively x1, x2, x3 = dist. of centroid of part 1, 2, 3 from vertical reference axis y1, y2, y3 = dist. of centroid of part 1, 2, 3 from horizontal reference axis

_ X = l1 x1 + l2 x2 + l3 x3

l1 + l2 + l3  _ Y = l1 y1 + l2 y2 + l3 y3

l1 + l2 + l3

2) Composite lamina : (square, rectangle, circle etc.)

Part 1 = Rectangle Part 2 = Triangle Part 3 = Semicircle

a1, a2, a3 = area of part 1, 2, 3 respectively x1, x2, x3 = dist. of centroid of part 1, 2, 3 from vertical ` reference axis y1, y2, y3 = dist. of centroid of part 1, 2, 3 from horizontal reference axis

_ X = a1 x1 + a2 x2 + a3 x3

a1 + a2 + a3  _ Y = a1 y1 + a2 y2 + a3 y3

a1 + a2 + a3

3) Composite volume : (cone, cylinder, cube etc.)

_ X = v1 x1 + v2 x2 + v3 x3

v1 + v2 + v3  _ Y = v1 y1 + v2 y2 + v3 y3

v1 + v2 + v3

CENTROIDS OF STANDARD SHAPES :

Table – A. One dimensional (wires)

Table – B. Two dimensional (lamina)

C. THREE DIMENSIONAL FIGURES (SOLIDS)

Ex. : Using basic principle show that centroid of a semicircular line arc is 2r/ from the centre line at base.

Solution :Polar co-ordinates : dL = R = R = R

= .

EXAMPLES

𝛑

𝑦 ∗𝑥∗ sin θ

d θ

cos θ

𝑦 ∫ 𝑦∗∫𝑑𝐿

dL

= . = .

= . = .

∫0

𝜋2

2 (𝑅 sin 𝜃 ) 𝑅d 𝜃

∫0

𝜋2

2¿¿

∫0

𝜋2

sin θd 𝜃

∫0

𝜋2

d 𝜃

R

𝑅 [− cos𝜃 ]𝜋20

[𝜃 ]𝜋20

𝑅 [𝑐𝑜𝑠 𝜋2 −𝑐𝑜𝑠0 ][ 𝜋2 −0]

= .

=

So,,

due to symmetry of arc about y-y axis

-R𝜋2

2𝑅𝜋

𝒚=𝟐𝑹𝝅

.

Ex. : Find centroid of wire ABCD shown in figure.

Solution :

Wire is a one-dimensional matter.

Therefore, length of wire isto be considered.

Part-1, AB : l1 = 20 cm x1 = 10 cm y1 = 0 Part-2, BC : l2 = 15 cm x2 = 20 cm y2 = 7.5 cm Part-3, CD : l3 = 10 cm x3 = 20 + 5 = 25 cm y3 = 15 cm

=

= 20*10 + 15*20 + 10*25

= 750

= 16.67 cm

= = 20*0 + 15*7.5 + 10*15 = 5.83 cm So,

G( , ) = (16.67 cm, 5.83 cm)

l1 x1 + l2 x2 + l3 x3 l1 + l2 + l3 

𝑥

20 + 15 + 10

45

𝑦 l1 y1 + l2 y2 + l3 y3 l1 + l2 + l3  20 + 15 + 10

𝒙𝒚

Ex. : Find centroid of the section shown in figure.

Solution :

The section isSymmetrical aboutx-x axis.

So,

is to be calculated.

𝑦¿152

¿𝟕 .𝟓𝒄𝒎𝒚

𝒙

Part-1 :a1 = 10*2.5 = 25 x1 = 5 cm

Part-2 :a2 = 10*2.5 = 25 x2 = 2.5/2 = 1.25 cm

Part-3 :a3 = 10*2.5 = 25 x3 = 10/2 = 5 cm

=

= 25*5 + 25*1.25 + 25*5

= 281.25

= 3.75 cm

So,G( , ) = (3.75 cm, 7.5 cm)

a1 x1 + a2 x2 + a3 x3 a1 + a2 + a3 

𝑥

25 + 25 + 25

75

𝒙

𝑥𝑦

THANK YOU