SSC JE Civil Engineering 2018 Set 3 Solved Paper

Question : 153 of 200

Marks: +1, -0

If the moment of inertia of a section about its axis is I and its effective sectional area is A, its radius of gyration r about the axis is

Solution:

I = Moment of inertia
A = Cross-sectional Area
r = Radius of gyration
I = Ar

^2

r =√{I/A}

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