Centroid and centre of gravity

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P
INTRODUCTION
We know everybody is attracted to the centre of the earth by a force of
attraction, known as the weight of the body.
The weight being a force acts through a point known as the centre of gravity
of the body. In Chapter 2 we have emphasised in article 2.2 that the point of
application of the force is one of the necessary data to define a force. Hence
the location of centre of gravity becomes important while dealing with the
weight force.
In this chapter we will learn to find the centre of gravity of bodies, plane areas
and lines. We will also study the approach using integration method to find
the centre of gravity of figures bounded by curve. Finally we will study the
application of the location of centre of gravity to certain engineering problems.
CENTROIDS AND CENTRE OF GRAVITY DEFINE
Centre of Gravity
It is defined as a point through which the whole weight of the body is
assumed to act. It is a term used for all actual physical bodies of any size,
shape or dimensions e.g. book, cupboard, human beings, dam, car, etc.
Centroid
The significance of centroid is same as center of gravity. It is a term used for
center of gravity of all plane geometrical figures. For example, two dimensional
figures (Areas) like a triangle. Rectangle, circle, and trapezium or for one
dimensional figures (Lines) like circular arc, straight lines, bent up wires, etc.
Centroid and Centre of Gravity

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RELATION FOR CENTRE OF GRAVITY
Consider a body of weight whose centre of gravity is located at G  X,Y as
shown. If the body is split in n parts, each part will have its elemental weight
Wi acting through its centre of gravity located at Gi (xi yi). Refer Fig.
The individual weights W1, W2, W3, ……Wi……..Wn form a system of parallel
forces. The resultant weight of the body would then be
1 2 3 i n
i
W = W + W + W ........+W......W
= W
To locate the point of application of the resultant weight force W using
Varignon's theorem (discussed earlier in Chapter 2).
Taking moments about y axis
Moments of individual weights = Moment of the total weight
about y axis about y axis
1 1 2 2 i i n nW × X +W ×X +.... + W × X ......+W × X = W×X
B i iW X = W × X
i i i i
i
W X W X
X = =
W W
 

……6.1 (a)
Similarly if the moment are taken about x axis
Moments of individual weights = Moment of the total weight
about x axis about x axis
1 1 2 2 i i n nW × +W × +.... + W × ......+W × = W×Y Y Y Y Y
i iW = W ×Y Y
i i i i
i
W W
= =
W W
Y Y
Y
 


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Using the relations (a) and (b), the centre of gravity G of a body having co-
ordinates  X,Y can be located.
Relation for Centroid
We recall that weight = mass × g
= (Density × Volume) × g
= (Density × Area × Thickness) × g
W = ρ × A × t × g
= (ρ × t × g) A
For uniform bodies i.e. of same density and thickness throughout the body,
we get,
 
 
i i i i
i i
ρ×t×g A X A X
X = =
ρ×t×g A A
 
 
……6.2 (a)
Similarly,
 
 
i i i i
i i
ρ×t×g A A
= =
ρ×t×g A A
y y
Y
 
 
…….6.2 (b)
Using the relation (a) and (b), the centroid G having co-ordinates  X,Y of a
Plane area can be located.

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AXIS OF SYMMETRY (A. O. S.)
Axis Of Symmetry is defined as the line which divides the figure into two equal
parts such that each part is a mirror image of the other.
If the geometrical figure whose centroid has to be located is a symmetrical
figure, then the centroid will lie on the axis of symmetry (A.O.S.). If the figure
has more than one axis of symmetry, the centroid will lie on the intersection
of the axis of symmetry. Fig below shows the importance of identifying the
axis of symmetry.

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CENTROIDS OF REGULAR PLANE AREAS
Table shows the centroids of regular plane areas. The co-ordinates  X,Y of
the centroid ‘G’ are with respect to the axis shown in the figure.
Sr.No. FIGURE AREA X Y
1 b × d b
2
b
2
2 1
b × h
2

b
3
h
3
3 1
b × h
2

_ h
3
4 2
π r
2
0 4 r
3 r
5 2
π r
4
4r
3r
4 r
3 r
6 2
r α*
#
2 rsin α
3 a
0

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CENTROID OF COMPOSITE AREA
An area made up of number of regular plane areas is known as a Composite
Area.
To locate the centroid of a composite area, adopt the following procedure.
PART AREA
Ai
CO-ORDINATES Ai.Xi Ai.Yi
Xi Yi
1. RECTANGLE A1 X1 Y1 A1.X1 A1.Y1
2. SEMI-CIRCLE A2 X2 Y2 A2.X2 A2.Y2
3. RT. ANGLE
TRIANGLE
A3 X3 Y3 A3.X3 A3.Y3
iA i iA .X i iA .Y
Table 2
1. Divide the composite area into regular areas as in Fig. (b)
2. Mark the centroids G1, G2, G3, .....on the composite figure as shown in Fig.
(b) and find their co-ordinates w. r. t. the given axis. Let the area of a
regular part be Ai and the co-ordinates be Xi and Yi.
3. Prepare a table as shown (Table 2)
a) Add up the areas of the different parts to get iA
b) Add up the product of area and x co-ordinate of different parts to get
i iA .X
c) Add up the product of area and y co-ordinate of different parts to get
i iA .Y
4. The co-ordinates of the centroid of the composite figure are obtained by
using relations (a) and (b) viz.
i i
i
A .X
X
A


and
i i
i
A .Y
Y
A



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APPLICATION OF CENTRE OF GRAVITY
For solution of certain engineering problems, we require the location of centre
of gravity. For example,
The centre of gravity for vehicles should be at minimum distance from the
ground, so that tipping of the vehicles is avoided while negotiating curves at
high speed.
Also in case of dam, which is a structure built
across a river to store water, the centre of gravity
of the dam should lie within the middle one-third
of the base of the dam. If the C. G. goes beyond
the middle one-third, the dam may loose its
stability. Refer Fig.
Another area where location of C.G. becomes
important is when the load acting on a
structure is distributed as per a given
relation and one is required to find the
resultant load and its location.
For example, refer figure. The water pressure
exerted on the face of the dam varies along
the depth being zero at the water surface and
having a maximum intensity  h at the base (where )  is the unit weight of
water). This load diagram is triangular in shape and therefore the resultant
pressure due to water acting on the dam would be the area of the triangle.
 Resultant water pressure 21 1 1
P = h × h h
2 2 2
    
This pressure shall act at the C.G. of the triangle i. e. at h/3 from the base of
the dam.

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Another example is of a simply supported beam (refer Fig.) which has a
distributed load having zero intensity at the ends and variation is as per the
relation y = f(x) along the length of the beam. The resultant load would lie at
the C.G. of the load diagram and value of the resultant load would be the area
under the load diagram.

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EXCERCISE
1. Locate the centroiod coordinates of the given figures and fill them in the
blanks. All dimensions are in cm units.
2. Locate the centroid of the composite figure shown.

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3. Find the centroid coordinates of the plane lamina shown w.r.to O.
4. Determine the centroid of the shaded area shown.
5. Locate the centroid of the shaded area shown in figure.
6. Determine centroid of plane area ABCDE w.r.t. A.
7. Find centroid of the shaded area.

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8. G is the centroid of an equilateral triangle ABC with base BC of side 60 cm.
If GBC is cut from the lamina, find the centroid of the remaining area.
9. Locate the centroid of the section.
10. Determine the co-ordinates of centroid of the shaded portion.
11. Find the centroid of the shaded area shown in figure. Note that OAF is a
quarter part of a circle of radius 750 mm.
12. Determine the co-ordinates of the centroid of the lamina shown with
respect to origin. Note that the circle of diameter 5O mm is cut out from
the plane lamina, with centre at (15O, 25).

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13. Determine the centroid of the shaded portion shown.
14. Find centroid of shaded plane area.
15. Determine the centroid of the shaded area shown.
16. Determine centroid of the plane lamina shown. Shaded portion is removed.

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UNIVERSITY QUESTIONS
1. A thin homogeneous wire of uniform section is
built into a shape as shown in figure. Determine
the position of c.g. of a wire. Take  = 30° and r =
15 cm. (5 Marks)
2. Determine forces in all the members of the plane truss as shown in figure.
(10 Marks)
3. Locate the centroid of the shaded area. (8 Marks)
4. Find centroid of the shaded area. (8 Marks)
5. Find Centroid of shaded area. (8 Marks)

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