SSC JE Civil Engineering 2014 Set 2 Solved Paper

Question : 188 of 200

Marks: +1, -0

Moment of Inertia of rectangular section shown in Fig. about its base is :

Solution:

Moment of inertia of the cross - section given rectangular section about its base -
By parallel axis theorem About its base

I_\base =I_{X-X}+Ah^2

Now

I_{X-X}={bD^3}/12

∴I_{\base}={bD^3}/12+ (bd)(d/2)^2

={bd^3}/3

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