Centroid and Centre of Gravity

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Centroid and Centre of Gravity

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.

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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

W = W1 + W2 + W3 ........+ Wi ......Wn =  Wi

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

W1 × X1 + W2 ×X2 +.... + Wi × Xi ......+ Wn × Xn

= W×X

B  Wi Xi = W × X

X=

W

i

Xi

W Similarly if the moment are taken about x axis Moments of individual weights

=

W X W i

……6.1 (a)

i

i

=

Moment of the total weight

about x axis

about x axis

W1 × Y1 + W2 ×Y2 +.... + Wi × Yi ......+ Wn × Yn = W×Y

W Y

= W×Y

W Y

=

i

Y =

i

W

i

i

W Y W i

i

i

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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,

  ρ×t×g  A X =  A X   ρ×t×g  A A   ρ×t×g  A y =  A y Y =   ρ×t×g  A A i

X=

i

i

i

Similarly,

i

i

i

i

……6.2 (a)

i

i

i

…….6.2 (b)

i





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 the centroid ‘G’ are with respect to the axis shown in the figure.

Sr.No. FIGURE AREA X

 X, Y  of 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

π r2 2

0

4r 3r

5

π r2 4

4r 3r

4r 3r

6

r 2 α*

2 r sin 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

CO-ORDINATES

Ai.Xi

Ai.Yi

Ai

Xi

Yi

1. RECTANGLE

A1

X1

Y1

A1.X1

A1.Y1

2. SEMI-CIRCLE

A2

X2

Y2

A2.X2

A2.Y2

3. RT. ANGLE

A3

X3

Y3

A3.X3

A3.Y3

TRIANGLE

A

 A .X

i

i

i

 A .Y i

i

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

A

i

b) Add up the product of area and x co-ordinate of different parts to get

 A .X i

i

c) Add up the product of area and y co-ordinate of different parts to get

 A .Y i

i

4. The co-ordinates of the centroid of the composite figure are obtained by using relations (a) and (b) viz.

X

 A .X A i

i

i

and Y

 A .Y A i

i

i

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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 P =

1 1 1  h × h   h 2 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 ď ą = 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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