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, A_{1}, A_{2} and A_{3}.
A = A_{1} + A_{2} + A_{3} = (400 × 150 × 2 + 300 × 200) = 18 × 10^{4}
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 × 150^{3} + 400 × 150 × (225)^{2}
or,
I _{A 3 ( x)} = I _{A 1 ( x)} = 10^{6} [112.5 + 3037.5] = 3150 × 10^{6} mm^{4}
I _{A2 ( x)} = (1/12) × 200 × 300^{3} = 450 × 10^{6 } mm^{4}
∴ I _{A ( x)} = [(2 × 3150) + 450] × 10^{6}
= 6750 × 10^{6 } mm^{4}
(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 × ( A_{2}′ )] from the M. I. of area ( A_{1}′ ) as illustrated in Figure (b).
Here,
A_{1}′ = 400 × 600 mm^{2} and A_{2}′ = 100 × 300 mm^{2}
∴ A = A_{1}′ - 2 A2′
= 24 × 10^{4} - 6 × 10^{4} = 18 × 10^{4} mm^{2}
Then,
I A ( x) = I( A_{1}′ ) - I ( A_{2}′ ) × 2
=400 × 600^{3}/12 - 2 × 100 × 300^{3} / 12
= (1/12) × 10^{6} [86400 - 5400] = 6750 ×10^{6} mm^{4}