centre of gravity

Centroid and Moment of Inertia

CHAPTER NO. 5

CENTROID AND MOMENT OF INERTIA

Centre of gravity:

Centre of gravity of a body is the point through which the whole weight of the body acts. A

body will have only one centre of gravity for all positions of the body. It is generally

represented as C.G. or simply G.

Centroid:

It is the point at which the total area of a plane figure is assumed to be concentrated. It is also

represented as C.G. or simply G. The centroid and centre of gravity are at the same point.

(They don’t have size or have a very small dimension as compared to other two dimensions).

Following table shows the centroid of some known shapes:

Name of Shape Figure

Rectangle

Triangle


Semicircle

T-section

I-section

Quarter Circle


L-section

Sector of Circle

Note: Small black dot indicate the centroid of the shapes.

Difference between centroid and centre of gravity:

Centroid Centre of gravity

1) Applicable to plane areas 1) Applicable to bodies with mass and

the weight.

2) It is a point in a plane area at which

the whole area concentrated.

2) It is the point of a body through

whole weight of the body is acting for

any orientation of the body.

Properties of Centre of gravity or Centroid:

1) It lies on axis of symmetry i.e. it is a meeting point of two or more axes of symmetry.

2) If the body is suspended freely from point P, then the C.G. lies on the vertical line of

suspension through P.

3) If some portion is added to the body, centroid or C.G. gets shifted towards addition

but if some portion of body is removed, it gets shifted away from deduction.

4) Centroid and C.G. are coincident if the body is homogeneous i.e. if its unit weight of

the material is same throughout the body.


Centre of gravity of plane figures by the method of Moments:

(Moment: It is the product of force and the perpendicular distance of the line of

action of that force to the axis of rotation.

In same way in case of plane figures or shapes we have take moment of area of plane

figure about the reference axis.)

Following Figure 1 shows a plane figure of total area ‘A’ whose C.G. is to be determined.

Let area ‘A’ is composed of a number of small areas a1, a2, a3, a4,…… etc.

∴ A = a1 + a2 + a3 + a4 + …..

Let x1 = the distance of the C.G. of the area a1 from axis OY

x2 = the distance of the C.G. of the area a2 from axis OY

x3 = the distance of the C.G. of the area a3 from axis OY

x4 = the distance of the C.G. of the area a4 from axis OY and so on.

The moments of all small areas about the axis OY

= a1 x1 + a2 x2+ a3 x3 + a4 x4 + ……

Let G be the C.G. of the total area whose distance from the axis OY is x̅

Figure 1

Then the moment of total area about OY = Ax̅

The moment of all small areas about the axis OY must be equal to the moment of total area

about the same axis.

a1 x1 + a2 x2+ a3 x3 + a4 x4 + …… = Ax̅


x̅ = a1 x1+ a2 x2+ a3 x3 + a4x4 + ….

A

Where, A = a1 + a2 + a3 + a4 + …..

If we take the moments of the small areas about the axis OX and also the moment of total

area about the axis OX, then

y̅ = a1 y1 + a2 y2 + a3y3 + a4 y4+ ….

A

Problems:

1) Find the centre of gravity of the T-section as shown in Figure 1.1 below.

2) Find the centre of gravity of symmetrical I-section as shown in Figure 1.2 below.

3) Find the centre of gravity of the I-section as shown in Figure 1.3 below.

4) Find the centre of gravity of the L-section as shown in Figure 1.4 below.

Figure 1.1 Figure 1.2

Figure 1.3 Figure 1.4


5) For rectangular lamina ABCD 10 cm x 14 cm a rectangular hole of 3 cm x 5 cm is cut

as shown in figure below. Find the centre of gravity of the reminder lamina.

Moment of Inertia:

Consider a thin lamina or a plane Figure of any shape.

Let,

x = distance of the C.G. of area A from the axis OY

y = distance of the C.G. of area A from the axis OX

Then the moment of area about the axis OY

= (Area) x (perpendicular distance of C.G. of area about axis OY)

= Ax -------------- (1)

Equation (1) is known as first moment of area about the axis OY

If the moment of area given in equation (1) is again multiplied by a perpendicular distance of

C.G. of area about axis OY (i.e. distance x), then Ax2 is known as moment of area or second

moment of inertia about the axis OY.

Similarly the moment of area about the axis OX = Ay

And second moment area about the axis OX = Ay2


If, instead of area, the mass (m), of the body is taken into account then the second moment is

known as moment of inertia.

Hence moment of inertia when mass is taken into account about the axis OY = mx2 and about

the axis OX = my2

Radius of Gyration

Radius of Gyration of a body is defined as the distance from an axis of reference where the

whole mass (or area) of a body is assumed to be concentrated so as not to alter the moment of

inertia about the given axis.

From above Figure the moment of inertia about the given axis is given by the equation as

I = a1 r12 + a2 r2

2 + a3 r32 + …..


Or I = ∑ 𝑎𝑟2

Theorem of Perpendicular Axis

Theorem of perpendicular axis states that if Ixx and Iyy be the moment of inertia of a plane

section about two mutually perpendicular axis X-X and Y-Y in the plane of the section then

the moment of inertia of the section Izz about the axis Z-Z perpendicular to the plane and

passing through the intersection of X-X and Y-Y is given by,

Izz = Ixx + Iyy

Proof:

A plane section of area ‘A’ and lying in plane XY as shown in Figure below. Let OX and OY

be the two mutually perpendicular axes and OZ be the perpendicular axis.

Consider a small area say dA.

x = distance of dA from axis OY

y = distance of dA from axis OX

r = distance of dA from axis OZ

Then,

R2 = x2 + y2

Now moment of inertia of dA about x-axis

= dA x (distance of dA from x-axis)2


= dA x y2

∴ Moment of inertia of total area ‘A’ about y-axis

Ixx = Σ dA x2

And moment of inertia of total area ‘A’ about z-axis

Izz = ΣdA r2

= ΣdA [ x2 + y2]

= ΣdA x2 + ΣdA y2

Izz = Ixx + Iyy

Theorem of Parallel Axis:

It states that, if the moment of inertia of a plane area about an axis in the plane of area

through the C.G. of the plane area be represented by IG, then the moment of inertia of the

given plane area about the parallel axis AB in the plane of area at a distance ‘h’ from the C.G.

of the area is given by

IAB = IG + Ah2

Where,

IAA = M.I. of the given area about AB

IG = M.I. of the given area about CG

A = area of the section and

h = distance between the C.G. of the section and the axis AB

Proof:

A lamina of plane area ‘A’ is shown in Figure below


Let,

x-x = the axis in the plane of area A and passing through the C.G. of the area

AB = The axis in the plane of area ‘A’ and parallel to the axis x-x

H = distance between AB and x-x

Consider a strip parallel to x-x axis at a distance ‘y’ from the axis.

Moment of inertia of area dA about x-x axis = dA y2

∴ Moment of inertia of the total area about x-x axis

Ixx or IG = Σ dA y2 --------------- (1)

Moment of inertia of the area dA about AB

= dA (h+y)2

= dA (h2 + 2hy + y2)

∴ Moment of inertia of the total area A about AB

IAB = Σ dA [h2 + 2hy + y2]

= Σ dA h2 + Σ dA y2 + 2 Σ dA hy


As h or h2 is constant and hence they can be taken outside the summation side

IAB = h2 Σ dA + Σ dA y2 + 2h Σ dA y

But Σ dA = A and also from (1) Σ dA y2 = IG

Substituting these values in the above equation, we get

IAB = h2 A + IG + 2h Σ dA y ----------- (2)

But dA y represents the moment of area of the strip about x-axis and ΣdA y represents the

moments of the total area about X-X axis. But the moments of the area about x-x axis is equal

to the product of the total area ‘A’ and the distance of the C.G. of the total area from x-x axis.

As the distance of the C.G. of the total area from x-x axis is zero. Hence ΣdA y will

be equal to zero.

From (2) we get,

IAB = h2 A + IG

∴ IAB = IG + Ah2 ---------(3)

Questions:

1) Determine the moment of inertia of the section about the horizontal and vertical axes,

passing through the C.G. of the section. (see Figure 1.1)

Figure 1


2) Find the moment of inertia of the shaded lamina as shown in Figure 1.5 below about

the horizontal axis passing through its centre of gravity.

3) A plane lamina ABCDE is shown in Figure below. Find the moment of inertia of

lamina about an axis passing through the centroid of the lamina and parallel to base

AB.

centre of gravity

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