CENTROID CENTRE OF GRAVITY Centre of gravity : of a body is the point at which the whole weight of...

CENTROID

CENTRE OF GRAVITY

Centre of gravity : of a body is the point at which the whole weight of the body may be assumed to be concentrated.

It is represented by CG. or simply G or C.

A body is having only one center of gravity for all positions of the body.

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CENTRE OF GRAVITY

Consider a three dimensional body of any size and shape, having a mass m.

If we suspend the body as shown in figure, from any point such as A, the body will be in equilibrium under the action of the tension in the cord and the resultant W of the gravitational forces acting on all particles of the body.

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Resultant W is collinear with the Cord

Assume that we mark its position by drilling a hypothetical hole of negligible size along its line of action

Cord

Resultant

CENTRE OF GRAVITY

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To determine mathematically the location of the centre of gravity of any body,

Centre of gravity is that point about which the summation of the first moments of the weights of the elements of the body is zero.

we apply the principle of moments to the parallel system of gravitational forces.

CENTRE OF GRAVITY

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We repeat the experiment by suspending the body from other points such as B and C, and in each instant we mark the line of action of the resultant force.

For all practical purposes these lines of action will be concurrent at a single point G, which is called the

centre of gravity of the body.

CENTRE OF GRAVITY

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w

A

A

w

B

AG

B

A

w

AB

CGB

A

C

C

B

Example:CENTRE OF GRAVITY


if, we apply principle of moments, (Varignon’s Theorem) about y-axis, for example,

The moment of the resultant gravitational force W, about any axis

=the algebraic sum of the moments about the same axis of the gravitational forces dW acting on all infinitesimal elements of the body.

dWxWhere W =

dW

Wx

The moment of the resultant about y-axis =

The sum of moments of its components about y-axis

CENTRE OF GRAVITY


where = x- coordinate of centre of gravity

x

x

x

W

dWxx

Similarly, y and z coordinates of the centre of gravity are

W

dWyy

W

dWzz and ----(1)

CENTRE OF GRAVITY


x

W

dWxx

W

dWyy

W

dWzz

With the substitution of W= m g and dW = g dm

m

dmxx

m

dmyy

m

dmzz

----(1)

----(2),,

,,

(if ‘g’ is assumed constant for all particles, then )

the expression for the coordinates of centre of gravity become

CENTRE OF MASS

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dV

dVxx

dV

dVyy

dV

dVzz

and ----(3)

If ρ is not constant throughout the body, then we may write the expression as

,

CENTRE OF MASS

The density ρ of a body is mass per unit volume. Thus, the mass of a differential element of volume dV becomes dm = ρ dV .

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m

dmxx

m

dmyy

m

dmzz ----(2),,

This point is called the centre of mass and clearly coincides with the centre of gravity as long as the gravity field is treated as uniform and parallel.

CENTRE OF MASS

Equation 2 is independent of g and therefore define a unique point in the body which is a function solely of the distribution of mass.


When the density ρ of a body is uniform throughout, it will be a constant factor in both the numerators and denominators of equation (3) and will therefore cancel.The remaining expression defines a purely geometrical property of the body.

dV

dVxx

dV

dVyy

dV

dVzz

and, ----(3)

CENTROID


When speaking of an actual physical body, we use the

term “centre of mass”.

Calculation of centroid falls within three distinct

categories, depending on whether we can model the

shape of the body involved as a line, an area or a

volume.

The term centroid is used when the calculation concerns

a geometrical shape only.

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LINES: for a slender rod or a wire of length L, cross-sectional area A, and density ρ, the body approximates a line segment, and dm = ρA dL. If ρ and A are constant over the length of the rod, the coordinates of the centre of mass also becomes the coordinates of the centroid, C of the line segment, which may be written as

L

dLxx

L

dLyy

L

dLzz

The centroid “C” of the line segment,

,,

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AREAS: when the density ρ, is constant and the body has a small constant thickness t, the body can be modeled as a surface area. The mass of an element becomes dm = ρ t dA.

If ρ and t are constant over entire area, the coordinates of the ‘centre of mass’ also becomes the coordinates of the centroid, C of the surface area and which may be written as

A

dAxx

A

dAyy

A

dAzz

The centroid “C” of the Area segment,

,,

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VOLUMES: for a general body of volume V and density ρ, the element has a mass dm = ρ dV .

If the density is constant the coordinates of the centre of mass also becomes the coordinates of the centroid, C of the volume and which may be written as

V

dVxx

V

dVyy

V

dVzz

The centroid “C” of the Volume segment,

, ,

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Centroid of Simple figures: using method of moment ( First moment of area)

Centroid of an area may or may not lie on the area in question.

It is a unique point for a given area regardless of the choice of the origin and the orientation of the axes about which we take the moment.


Moment of Total area ‘A’ about y-axis

=Algebraic Sum of moment of elemental ‘dA’ about the same axis

where (A = a1 + a2 + a3 + a4 + ……..+ an)

(A) x = (a1) x1 + (a2) x2 + (a3) x3 +

……….+(an) xn

= First moment of area

The coordinates of the centroid of the surface area about any axis can be calculated by using the equn.


If an area has an axis of symmetry, then the centroid must lie on that axis.If an area has two axes of symmetry, then the centroid must lie at the point of intersection of these axes.

AXIS of SYMMETRY:

It is an axis w.r.t. which for an elementary area on one side of the axis , there is a corresponding elementary area on the other side of the axis (the first moment of these elementary areas about the axis balance each other)

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For example:

The rectangular shown in the figure has two axis of symmetry, X-X and Y-Y. Therefore intersection of these two axes gives the centroid of the rectangle.

B

DD/2

D/2

B/2 B/2

X X

Y

Y

xx

dada

da × x = da × x

Moment of areas,da about y-axis cancel each other

da × x + da × x = 020


AXIS of SYMMETYRY

‘C’ must lie at the intersectionof the axes of symmetry

‘C’ must lie on the axis

of symmetry

‘C’ must lie on the axis of symmetry


EXERCISE PROBLEMS

Locate the centroid of the shaded area shown

Problem No.1:

Ans: x=12.5, y=17.5

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Locate the centroid of the shaded area shown

Problem No.2:

D=600

Ans: x=474mm, y=474mm23

EXERCISE PROBLEMS


Locate the centroid of the shaded area w.r.t. to the axes shown

Problem No.3:

Ans: x=34.4, y=40.324

EXERCISE PROBLEMS



Problem No.4:

Ans: x= -5mm, y=282mm

10

25

EXERCISE PROBLEMS



Problem No.5

Ans:x =38.94, y=31.46

30

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EXERCISE PROBLEMS



Problem No.6

x

y

Ans: x=0.817, y=0.2427

EXERCISE PROBLEMS


Problem No.7


Ans: x= -30.43, y= +9.5828

EXERCISE PROBLEMS


Problem No.8

Locate the centroid of the shaded area.

Ans: x= 0, y= 67.22(about base)

20

29

EXERCISE PROBLEMS


Problem No.9

Locate the centroid of the shaded area w.r.t. to the base line.

Ans: x=5.9, y= 8.17

2

30

EXERCISE PROBLEMS


Problem No.10


Ans: x=21.11, y= 21.1131

EXERCISE PROBLEMS


Problem No.11


Ans: x= y= 22.2232

EXERCISE PROBLEMS


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CENTROID CENTRE OF GRAVITY Centre of gravity : of a body is the point at which the whole weight of...

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