Moment of Inertia by Integration (From First Principles)

Moment of Inertia by Integration

(From First Principles)

• The moment of inertia can be obtained by choosing an appropriate differential element dA which can be either rectangular or a horizontal or vertical strip writing its moment of inertia and then integrating to cover the whole area. This procedure is illustrated for some standard figures.

1. Rectangle

• Consider a rectangle of dimensions b×d as shown in Fig. 6.5.1. Consider an elemental strip of length b and width dy at a distance y from the centroidal X-axis.

• The moment of inertia about the base AB can be obtained using parallel axes d theorem. The distance of AB from X-X axis is d/2.

• The polar moment of inertia can be obtained using perpendicular axes theorem.

• The radius of gyration about any axis can be obtained using Here, A = bd

2. Triangle

• Consider a triangle of base b and height h as shown in Fig. 6.5.2. Choose a horizontal strip of length b and width dy at distance. y from the base as the differential element.

By similarity of triangles,

• Moment of inertia of the strip about the base is,

• The centroid of triangle is at a height of h/3 from the base AB. Using parallel axes theorem, 

• The radius of gyration about centroidal axis is

3. Circle

• Consider a circle of radius R. Choose a differential element of angular width dθ at angle θ and radial width dr at distance r from the centre as shown in Fig. 6.5.3.

• The distance of this area element from centroidal X-axis, which is the diameter, is

y = rsin θ

• The polar moment of inertia is

4. Semicircle

• Consider a semicircle of radius R as shown in Fig. 6.5.4.

• Choose differential element of angular width dθ at angle θ and radial width dr at distance r from the centre.

• The area of the differential element will be

dA = (rdθ) (dr)

⸫ dA = rdr dθ

• The distance of this differential element from the diameter AB is

y = rsinθ

• M.I. of differential element about AB is,

• The centroid is at a distance of 4R/ Зл from AB.

• The moment of inertia about a centroidal axis perpendicular to diameter (Y-axis shown in Fig. 6.5.4) is

5. Quarter Circle

• Consider a quarter circle of radius R as shown in Fig. 6.5.5.

• Choose a differential element of angular width dθ at angle θ and radial width dr at distance r from the centre.

• The area of the differential element will be,

dA = (r dθ) (dr)

⸫ dA = r dr dθ

• The distance of this differential element from AB is

y = rsinθ

• M.I. of differential element about AB is