We were discussing the “Theorem of perpendicular axis” and “Radius of gyration” and also we have seen the “Difference between centroid and centre of gravity” with the help of our previous posts.

Now we will determine the moment of inertia of small strip area dA about the AB axis.

Therefore we can brief here the theorem of parallel axis as the moment of inertia for a lamina about an axis parallel to the centroidal axis (axis passing through the center of gravity of lamina) will be equal to the sum of the moment of inertia of lamina about centroidal axis and product of area and square of distance between both axis.

Now we are going further to start our discussion to understand the theorem of parallel axis with the help of this post.

### Let us see now theorem of parallel axis

According to the theorem of parallel axis, the moment of inertia for a lamina about an axis parallel to the centroidal axis (axis passing through the center of gravity of lamina) will be equal to the sum of the moment of inertia of lamina about centroidal axis and product of area and square of distance between both axis.

IAB = IG + Ah2

Where,

IG = Moment of inertia of entire area A about the centroidal axis,

IAB = Moment of inertia of entire area A about the axis AB which is parallel to the centroidal axis

A = Entire area of the lamina

h= Perpendicular distance between both axis i.e. centroidal axis and Axis AB and it is displayed in following figure.

Let us see the above figure which indicates one lamina and area of the lamina is A. Let us assume that axis X-X is passing through centre of gravity of lamina and axis AB is parallel to the axis X-X or centroidal axis as displayed in above figure.

Let us assume one small strip, in given lamina, parallel to the centroidal axis and at a disthance Y from the centroidal axis. Let us assume that area of small strip is dA.

Moment of inertia of small strip area dA about the X-X axis = dA.Y2

Hence, moment of inertia of entire area A about the X-X axis, IXX will be determined as mentioned here.

IXX =Ʃ dA.Y2

Moment of inertia of entire area A about the X-X axis could also be written as moment of inertia of entire area about the centroidal axis because axis X-X is passing through the centre of gravity of the lamina i.e. we can write IG in above equation instead of IXX.

IG =Ʃ dA.Y2

Now we will determine the moment of inertia of small strip area dA about the AB axis.

IAB= dA. (h +Y) 2

Therefore moment of inertia of entire area of lamina about the AB axis will be concluded as

IAB= Ʃ dA. (h +Y) 2

IAB= Ʃ dA. (h2 +Y2+2hY)

IAB= Ʃ dA. h2 + Ʃ dA Y2 + Ʃ dA.2hY

We can easily observe here that Ʃ dA = A, because distance between axis i.e. h will be constant. Therefore we can write above equation as mentioned here

IAB= A. h2 + IG + 2h. ƩdA Y

As we can observe here that dA.Y represents the moment of area of small strip dA about the XX axis and Ʃ dA.Y represents the moments of entire area about the XX axis or centroidal axis.

As we know that the moments of entire area about the XX axis will be equal to the product of entire area and distance between C.G of entire area and XX axis. But as axis XX passes through the C.G of entire area and therefore distance between C.G of entire area and XX axis will be zero and therefore Ʃ dA.Y will be zero.

Hence, we can write above equation as mentioned here

IAB= A. h2 + IG

Therefore we can brief here the theorem of parallel axis as the moment of inertia for a lamina about an axis parallel to the centroidal axis (axis passing through the center of gravity of lamina) will be equal to the sum of the moment of inertia of lamina about centroidal axis and product of area and square of distance between both axis.

Do you have any suggestions or any amendment required in this post? Please write in comment box.

### Reference:

Strength of material, By R. K. Bansal

Image Courtesy: Google

We will see another important topic i.e., in the category of strength of material, in our next post.

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