Question

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

(a) Iuv = Ixcosθsinθ – Iysinθcosθ + Ixy (cos^2θ-sin^2θ)

(b) Iuv = Ixcosθsinθ + Iysinθcosθ – Ixy (cos^2θ-sin^2θ)

(c) Iuv = Ixcosθsinθ – Iysinθcosθ – Ixy (cos^2θ-sin^2θ)

(d) Iuv = Ixcosθsinθ + Iysinθcosθ + Ixy(cos^2θ-sin^2θ)

This question was posed to me in final exam.

Question is from Moments of Inertia for an Area About Inclined Axis topic in section Moments of Inertia of Engineering Mechanics

Answer 1

The correct choice is (a) Iuv = Ixcosθsinθ – Iysinθcosθ + Ixy (cos^2θ-sin^2θ)

The explanation: 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.