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PRINCIPAL MOMENTS Of INERTIA

Fig. 4.40 (a) shows a body of area A with respect to old axes (x, y) and new axes (x1, y1), The new axes x1 and y1 have been rotated through an angle O in anticlockwise direction. Consider a small area dA. The co-ordinates of the small area with respect to old axes is (x, y) whereas with respect to new axes, the co-ordinates are x’ and y’. The new co-ordinates (x’,y’) are expressed in terms of old co-ordinates (x, y) and angle S as [Refer to Figs. 4.40 (b) and 4.40 (c)] 

These are the values of principal moment of inertia.

Problem 4.18. For the section shown in Fig. 4.41 (a) determine:

(i)                           Moment of inertia about its centroid along (x,y) axis.

(ii)                        Moment of inertia about new axes which is turned through an angle of 300 anticlockwise to the old axis.

(iii)                      Principal moments of inertia about its centroid.

All dimensions are in cm.

Sol. Given:

The Fig. 4.41 (a) shows the given section. It is symmetrical about x-axis. The C.G. of the section lies at O (origin of the axes). To find moment of inertia of the given section, it is divided into three rectangles as shown in Fig. 4.41 (b).

First the moment of inertia of each rectangle about its centroid is calculated. Then by using parallel axis theorem, the moment of inertia of the given section about its centroid is obtained.

PRINCIPAL MOMENTS Of INERTIA

 

(a) Consider rectangle ①

The C.G. of rectangle ① is at a distance of 20 cm from y-axis and at a distance of 25 cm from y-axis.

(Ixx)1 = (lG)1x + A1(k1x)2                                                                               … (i)

where (Ixx)1 = M.O.I. of rectangle ① about y-axis passing through the centroid of the given section.

(IG)1x = M.O.I. of rectangle ① about an axis passing through C.G. of rectangle ① and
parallel to y-axis

PRINCIPAL MOMENTS Of INERTIA

 

 

 

 

 

= 2.25 × 104 cm4

A1 = Area of rectangle ① = 10 × 30 = 300

(k1x) = Distance of C.G. of rectangle ① from x-axis

=20

Substituting the above values in equation (i), we get
(Ixx)1 = 2.25 × 104 + 300 × 202

= 2.25 × 104 + 12 × 104

= 14.25 × 104 cm4                                                                           … (A)

Similarly the M.O.I. of rectangle ① about x-axis passing through the centroid of the
given figure is given by

(Iyy)1 =(IG)1y + A1(k1y)2                                                  … (ii)

PRINCIPAL MOMENTS Of INERTIA

 

 

 

 

 

 

(k1y) = Distance of C.G. of rectangle ① from x-axis = 25

(Iyy)1 = 0.25 × 104 +300 × 25

= 0.25 × 104 + 18.75 × 104

= 19 × 104 cm4                                                                            … (B)

(b) Consider rectangle

The C.G. of this rectangle coincides with the C.G. of the given section. Hence

PRINCIPAL MOMENTS Of INERTIA