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Sunday, 5 March 2017

MOMENT OF INERTIA OF AREA UNDER CURVE


Today we will see here the method to determine the moment of inertia of an area under a curve of given equation with the help of this post.

Let us see here the moment of inertia of an area under a curve of given equation

Let us consider one curve which equation is parabolic as displayed in following figure and let us consider that equation of this parabolic curve is as mentioned here.

x = ky2

Let us determine the moment of inertia of this area about the YY axis. Let us consider one small strip of thickness dx and at a distance x from the YY axis. We can also observe here from above figure that, x= a and y= b

Let us first determine the area of this small strip, dA = y.dx
Area of this small strip, dA = (x/k)1/2.dx

As we have considered already that equation of above curve is x = ky2

Now we can easily find here that value of constant k and it could be written as mentioned here
k= x/y2=a/b2
k=a/b2

Now area of this small strip, dA = (x/k)1/2.dx
Now area of this small strip, dA = [x/ (a/b2)]1/2.dx
Now area of this small strip, dA = [x1/2b/ (a1/2)].dx  

Now we will determine here the moment of inertia of the small strip area about the axis YY and we can write here as

Moment of inertia of the small strip area about the axis YY = x2.dA
 
Moment of inertia of the small strip area about the axis YY = x2. [x1/2b/ (a1/2)].dx
Moment of inertia of the small strip area about the axis YY = (b/a1/2) x5/2.dx

Now let us integrate the above equation from 0 to a in order to secure the moment of inertia of this entire area about the axis YY and it is displayed here in following figure.
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.Moment of inertia for the triangular section about its base line, in the category of strength of material, in our next post.

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