Mechanical Engineering - Engineering Mechanics - Discussion

Discussion Forum : Engineering Mechanics - Section 4 (Q.No. 31)
31.
According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through centre of gravity (i.e. IP) is given by(where, A = Area of the section, IG = Moment of inertia of the section about an axis passing through its C.G., and h = Distance between C.G. and the parallel axis.)
IP = IG + Ah2
IP = IG - Ah2
IP = IG / Ah2
IP = Ah2 / IG
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
1 comments Page 1 of 1.

Manabjyoti Doley said:   4 years ago
Give me the explanation, please.

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