Engineering Mechanics
Lecture 8 : Properties of plane surfaces I: First moment and centroid of area
 



Centroid: Centroid of a bounded area is a point whose x-coordinates XC and y-coordinate YC are defined as

1

where A is its total area. We now solve some examples of calculating these quantities for some simple areas.

Example 1: We start with the simple example of the first moment and centroid of a triangle with the base along the x-axis. Let its base BC be of length b, and let the height of the triangle be h. (see figure 3)

2

To calculate the Mx, we take a strip of width dy at height y (see figure 3). Then

3

But by similarity of triangles 4. So

5

Thus

6

Let us now calculate the x-co-ordinates for the centroid. For this let the x-coordinate of A be a so that the coordinate of point A is (a, h).

7

Now

8

For dA let us now take a vertical strip (figure 4). Notice that 9. We will also perform the x-integration in two parts: one from x = 0 to x = a, and the other from x = a to x = b because in the two regions, dependent of y on x is different so

10

For the region x = 0 to x = a, we can write

11

and for the region x = a to x = b, we have

12

Thus

13

This gives

1415

Thus for a triangle 16.

Example2: As the second example, let us calculate the centroid of a semicircular disc of radius R . It would be quite easy to solve this problem if the centre D of the circle is kept at the origin but I want to do the problem with the disc positioned as drawn below to show you how to tackle the problem.

The equation of OBC (the circular boundary of the disc) is

18

where R is the radius of the circle. The total area of the plate is 19. To calculate XC , we take a vertical strip of width dx at x and calculate

20

With , we get

To evaluate this integral, we let 23 so that the limits of θ integration are from 24. Then

25

which gives

26

To calculate YC we need to calculate MX = 27, where dA represents as strip from x1 to x2 (see figure 6)

From the equation of the circle we get

29

This gives

30

and therefore

31

Substituting y = R sin θ , we get

32

This gives

33

Thus the centroid of the semicircle shown is at 34. Notice that the y coordinate of the centroid is less than 35 which is easily understood because more of the area is concentrated towards the x-axis.

We would no like to emphasize that the centroid (XC YC ) gives a point fixed in a given planar surface and no matter in which co-ordinate system we calculate this point, it will always come out to be the same point in the surface. Thus it is a property of a surface.