In the calculations of the moment of inertia of the area about an inclined axis, we have some transformation done. They are: (a)Iu = Ix cos2θ + Iysin2θ – 2Ixycosθsinθ

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In the calculations of the moment of inertia of the area about an inclined axis, we have some transformation done. They are:

(a) Iu = Ix cos2θ + Iysin2θ – 2Ixycosθsinθ

(b) Iv = Ixcos2θ + Iysin2θ – 2Ixycosθsinθ

(c) Iu = Ixcos2θ + Iysin2θ – 2Ixycosθsinθ

(d) Iv  = Ixcos2θ + Iysin2θ + 2Ixycosθsinθ

This interesting question is from Moments of Inertia for an Area About Inclined Axis in division Moments of Inertia of Engineering Mechanics

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The correct option is (d) Iv  = Ixcos2θ + Iysin2θ + 2Ixycosθsinθ

The explanation is: In the moment of inertia calculations we see that the net force acts at the centroid of the loading body. That is if the loading system is in the form of the triangle then at the distance 2 by 3 of the base the net force of the loading will act. And the load will be half the area of the loading. Thus when there is the inclination of the axis, we use transformation.

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