Find out Moment of Inertia:

Find out Moment of Inertia of an I shaped area around its centroidal axis as illustrated in Figure.

 Solution

 (i)        The area of I section may be divided into three parts namely, A₁, A₂ and A₃.

                  A =  A₁ +  A₂  +  A₃  = (400 × 150 × 2 + 300 × 200) = 18 × 10⁴

The centroid G is at mid-depth as illustrated in Figure (a).

Now, I _{A ( x)} = I _{A ( x)}  +  I _{A  ( x)}  +  I _{A ( x)}

where,

 I A ( x)  = I A ( x)  = (1 /12) × 400 × 150³ + 400 × 150 × (225)²

or,

I _{A 3 ( x)}  = I _{A 1 ( x)}  = 10⁶ [112.5 + 3037.5] = 3150 × 10⁶  mm⁴

I _{A2 ( x)}  = (1/12) × 200 × 300³ = 450 × 10⁶   mm⁴

∴ I _{A ( x)}  = [(2 × 3150) + 450] × 10⁶

= 6750 × 10⁶  mm⁴

 (ii)       On the other hand, the M. I. of I-section about x axis may be attained by subtracting the M. I. of area [2 × ( A₂′ )] from the M. I. of area ( A₁′ ) as illustrated in Figure (b).

Here,

A₁′ = 400 × 600 mm²      and  A₂′ = 100 × 300 mm²

∴ A =  A₁′ - 2 A2′

= 24 × 10⁴  - 6 × 10⁴  = 18 × 10⁴  mm²

Then,

 I A ( x)  = I( A₁′ )  - I ( A₂′ ) × 2

=400 × 600³/12 - 2 × 100 × 300³ / 12

=   (1/12)  × 10⁶ [86400 - 5400] = 6750 ×10⁶  mm⁴