Tuesday, 21 February 2017

STATE THE PERPENDICULAR AXIS THEOREM WITH PROOF

STATE THE PERPENDICULAR AXIS THEOREM WITH PROOF

We were discussing “Area moment of inertia” and “Radius of gyration”,” and also we have seen the “Basic principle of complementary shear stresses” with the help of our previous posts.

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

 

Let us first see here theorem of perpendicular axis

According to the theorem of perpendicular axis, if IXX and IYY are the moments of inertia of an irregular lamina about tow mutually perpendicular axis X-X and Y-Y respectively in the plane of lamina, then there will be a moment of inertia IZZ of lamina around the Z-Z axis which will be perpendicular to the plane of lamina and passing through the intersection of X-X and Y-Y axis.

Moment of inertia IZZ of lamina around the Z-Z axis will be given by following equation as per the theorem of perpendicular axis.

IZZ = IXX + IYY

Let us see the following figure which indicates one irregular lamina in the plane of X-Y and area of the lamina is A. Let us assume one small elemental area dA in the plane of X-Y.
Where,
X= Distance of the C.G of the small elemental area dA from OY axis
Y= Distance of the C.G of the small elemental area dA from OX axis
R= Distance of the C.G of the small elemental area dA from OZ axis

We can easily note it here that distance of the C.G of the small elemental area dA from OZ axis could be written as mentioned here.
R2= X2+Y2
Moment of inertia of small elemental area dA about the OX axis = dA.Y2
Hence, moment of inertia of entire area A about the OX axis, IXX will be determined as mentioned here.
IXX =Ʃ dA.Y2

Moment of inertia of small elemental area dA about the OY axis = dA.X2
Hence, moment of inertia of entire area A about the OY axis, IYY will be determined as mentioned here.
IYY =Ʃ dA.X2

Moment of inertia of small elemental area dA about the OZ axis = dA.R2
 
Hence, moment of inertia of entire area A about the OZ axis, IZZ will be determined as mentioned here.

IZZ =Ʃ dA. R2
IZZ =Ʃ dA. [X2+ Y2]
IZZ =Ʃ dA.X2+ Ʃ dA.Y2
IZZ = IYY +IXX

IZZ = IXX + IYY

Therefore we can brief here the theorem of perpendicular axis as moment of inertia of an irregular lamina about an axis normal or perpendicular to it will be equal to the sum of the moment of inertia about any two mutually perpendicular axis in the plane of the lamina.

 
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

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