Solved Example & Practice Problems: Centroid of Composite Areas

Centroid of Composite Areas

Solved Examples for Understanding

Example 5.3.1 

Locate the centroid of the area shown in Fig. 5.3.1. The dimensions are in mm.

Solution: 

The given area can be obtained by subtracting triangle and semicircle from rectangle as shown in Fig. 5.3.1 (a).

The calculations are tabulated as follows:

Example 5.3.2 

A square area is removed from a thin circular lamina of radius r as shown in Fig. 5.3.2. The diagonal of the square is equal to the radius of the circle. Locate the centroid of the remaining lamina.

Solution: 

Due to symmetry, = 0

For the circle,

Α1 = πr2

Centroid of the circle is at its centre.

⸫ x1 = r

Length of diagonal of square = r

⸫ Length of each side = r/√2

For square,

Centroid of square is at the centre of its diagonal.

Example 5.3.3 

Locate the centroid of the plane area shown in Fig. 5.3.3.

Solution: 

Divide the given area as shown in Fig. 5.3.3 (a).

The calculations are tabulated as follows:

Example 5.3.4 

Determine the centroid of the unsymmetrical I-section shown in Fig. 5.3.4

Solution : 

Take co-ordinate axes and origin as shown in Fig. 5.3.4 (a) and divide the given section into three rectangles.

The given section is symmetric about a vertical line passing through the centre.

Example 5.3.5 

Determine the centroid of the channel section shown in Fig. 5.3.5

Solution: 

The given section is symmetric about a horizontal line passing through the centre.

To find take the leftmost vertical line to be reference and divide the area into three rectangles as shown in Fig. 5.3.5 (a).

Example 5.3.6 

Find the centre of gravity of the L-section shown in Fig. 5.3.6.

Solution: 

Take G as the origin and divide the L section into two rectangles GDEF and ABCD.

For rectangle GDEF,

A1 = 8 × 2 = 16 cm2

x1 = 4 cm, y1 = 1 cm

For rectangle ABCD,

A2 = 10 × 2 = 20 cm2

x2 = 1 cm, y2 = 2 + 5 = 7 cm

Example 5.3.7 

Determine the position of centroid for the lamina with a circular cutout shown in Fig. 5.3.7

Solution: 

Divide the given area into a triangle, a rectangle and a semicircle from which a circle is removed as shown in Fig. 5.3.7 (a).

The calculations are tabulated as follows:

Example 5.3.8 

Locate the centroids of the following area. Refer Fig. 5.3.8.

Solution: 

The shaded portion in the Fig. 5.3.8 can be obtained by subtracting the area of quarter circle (2) from the square (1) in second quadrant and then adding the area of quarter circle (3) in first quadrant.

The calculations are tabulated as follows:

Example 5.3.9

Determine centroid of the shaded area with reference apex.

Solution: 

The shaded area can be obtained by removing the circle, rectangle and semi-circle from the triangle. As the area is symmetric about a vertical line passing through the centre, centroid lies on this line, of symmetry. We take base of the triangle as reference for writing Y-co-ordinates of centroid.

The distance from apex = 240 - 82.02 = 157.98 mm

Example 5.3.10 

Determine the position of centroid with respect to 'O'

Solution: 

As the given area is symmetric about a vertical line passing through the centre,

The given area can be divided into three rectangles.

A1 = 200 × 10 mm2 ; y1 = 5 mm

A2 = 10 × 300 mm2; y2 = 160 mm

A3 = 100 × 10 mm2; y3 = 315 mm

Example 5.3.11

Determine the position of the centroid for the shaded area with respect to the axes as shown in Fig. 5.3.11.

Solution: 

The shaded region can be obtained by adding areas of square, rectangle, quarter circle and then subtracting a quarter circle as shown in Fig. 5.3.11 (a).

The calculations are tabulated as follows:

Example 5.3.12 

Find the co-ordinates of centroid of the lamina, shown in Fig. 5.3.12 with Fig. 5.312 with respect to point A.

Solution: 

The given area can be divided into a square, triangle and semi-circle as shown in Fig. 5.3.12 (a).

The calculations are tabulated as follows :

Example 5.3.13 

Determine location of centroid of shaded portion of lamina with respect to origin O. Refer Fig. 5.3.13.

Solution: 

The shaded area can be obtained by subtracting area of circle (3) from total area of rectangle ADEH, triangle BCD and semicircle EFG as shown in Fig. 5.3.13 (a).

The calculations are tabulated as follows:

Example 5.3.14 

A metal piece of uniform thickness is placed on horizontal surface as shown in Fig. 5.3.14. Find distance 'd' so that the piece will just be prevented from tipping. Diameter of the hole is 50 mm.

Solution :

Tipping will take place about O.

Divide the area as shown in Fig. 5.3.14 (a). The triangle (4) and circle (3) have to be removed from the total area of rectangles (1) and (2). Take moment of area about O, treating moment of area to the left of O as positive and to the right as negative.

Example 5.3.15 

A thin lamina of uniform thickness is as shown in Fig. 5.3.15. If it is suspended from the point A by a wire, what angle will line AB make with vertical ?

Solution: 

As the given lamina is suspended from point A, choose A as the origin and find co-ordinates of centroid G. For convenience we take X-axis towards left and Y-axis downward. When the lamina is suspended from point A, the line AG rotates and X 1 becomes vertical. AG will rotate by angle which is the angle made by AG in anticlockwise sense with vertical as shown Fig. 5.3.15 (b). AB will also rotate through the same angle from its horizontal position. Hence angle made by AB with vertical will be 90 - θ.

We divide the given area into 4 components as shown in Fig. 5.3.15 (a). We subtract triangle (2), rectangle (3) and quarter circle (4) from the semicircle (1). The locations of centroids of the basic shapes are shown as G1, G2, G3, and G4.

The calculations are tabulated as follows:

⸫ Angle made by AB with vertical = 90 - 63.42 = 26.58°

Example 5.3.16 

Determine, the co-ordinates of centroid of the shaded portion shown in Fig. 5.3.16.

Solution: 

Divide the area into three components as shown in Fig. 5.3.16 (a). The calculations are tabulated as follows.

Example 5.3.17 

Locate the centroid of shaded portion of a lamina if AB = 90 mm is diameter of semicircle. Refer Fig. 5.3.17.

Solution: 

The shaded area can be obtained by subtracting the two right angled triangles from the semicircle.

From figure,

⸫ h = 42.426 mm

The centroids of the three components are as shown in Fig. 5.3.17 (a).

The calculations are tabulated as follows.

Example 5.3.18 

Locate the centroid of the shaded lamina as shown in Fig. 5.3.18.

Solution: 

The shaded area can be obtained by subtracting the area of quarter circle and triangle from the area of square. The centroids of components are as shown in Fig. 5.3.18 (a).

The calculations are tabulated as follows.

Example 5.3.19 

Locate the centre of gravity of the area shown in the following Fig. 5.3.19 with respect to the x and y axis. The dimensions shown are in cm.

Solution: 

Divide the area as shown in Fig. 5.3.19 (a) and mark the centre of gravity of each standard shape.

The calculations are tabulated as follows:

Example 5.3.20 

For the section shown in Fig. 5.3,20 below, locate the horizontal and vertical centroidal axes.

Solution: 

Take origin as shown in Fig. 5.3.20 (a) and divide the given area into three rectangles.

The calculations are tabulated as follows:

The vertical line passing through = 515.625 is the vertical centroidal axis.

The horizontal line passing through = 268.75 is the horizontal centroidal axis.

Example 5.3.21 

Locate the centroid of the plane lamina as shown in Fig. 5.3.21.

Solution: 

The calculated are tabulated as follows:

Example 5.3.22 

Locate centroid with respect to O. Refer Fig. 5.3.22.

Solution: 

The given lamina can be divided into 3 parts a rectangle, an equilateral triangle and a semicircle with centroids as shown in Fig. 5.3.22 (a). The calculations are tabulated as follows:

Example 5.3.23 

Find the centroid of the shaded area OPQ, shown in Fig. 5.3.23. The curve OQ is parabolic.

Solution :

At x = b, y = h

Consider the differential element to be a horizontal strip of length x and width dy as shown in Fig. 5.3.23(a).

Area of differential element is

dA = x dy

Only positive root is taken as the given area is entirely in first quadrant.

Taking moment of the area about y-axis,

Taking moment of the area about x-axis,

Example 5.3.24 

Determine the coordinates of the centroid of the shaded area Fig. 5.3.24.

Solution :

Divide the given area into a rectangle, a triangle and a circle as shown in Fig. 5.3.24 (a).

The calculations are tabulated as follows:

Example 5.3.25 

Determine the centroid of the area shown in Fig. 5.3.25.

Solution: 

The given area can be obtained by adding areas of rectangle and quarter circle and then subtracting area of quarter circle at the bottom as shown in Fig. 5:3.25 (a).

The calculations are tabulated as follows:

Example 5.3.26 

Find the centroid of the shaded area shown in Fig. 5.3.26 about X and Y axes.

Solution :

Example 5.3.27 

From a semicircular lamina of radius r, a circular lamina of radius (r/2) removed as shown in Fig. 5.3.27. Find the position of centre of gravity of the remainder.

Solution :

Given Fig. 5.3.27(a) is symmetric about y-axis, hence

Examples for Practice

Q.1

Locate the centroid of the shaded area as shown in Fig. 5.3.28.

Q.2

For a composite Fig. 5.3.29 shown, find 'a' if centroid of the figure is to coincide with the centroid of rectangle ABCD. 

[Ans. : a = 30.23 mm]

Q.3

Locate centroid with respect to O. Refer Fig. 5.3.30.

Q.4

Find distances of centroid of shaded area from EF and ED. Refer Fig. 5.3.31.