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According to parallel axis theorem, the moment of inertia of a section about an axis parallel to the axis through center of gravity (i.e. $${I_{\text{P}}}$$) is given by (where, A = Area of the section, $${I_{\text{G}}}$$ = Moment of inertia of the section about an axis passing through its C.G. and h = Distance between C.G. and the parallel axis.)

A. $${I_{\text{P}}} = {I_{\text{G}}} + {\text{A}}{{\text{h}}^2}$$

B. $${I_{\text{P}}} = {I_{\text{G}}} - {\text{A}}{{\text{h}}^2}$$

C. $${I_{\text{P}}} = \frac{{{I_{\text{G}}}}}{{{\text{A}}{{\text{h}}^2}}}$$

D. $${I_{\text{P}}} = \frac{{{\text{A}}{{\text{h}}^2}}}{{{I_{\text{G}}}}}$$

Answer: Option A


This Question Belongs to Mechanical Engineering >> Engineering Mechanics

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Comments ( 1 )

  1. Sagar Chavan
    Sagar Chavan :
    3 years ago

    MI for Square section is

    a) b/12

    b) bd/12 c) 6/12

    d None

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