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STATICS - EXAMPLE


Area Between Curve and x and y-axis

  Example 1

 

Find moment of inertia of the shaded area about
a) x axis
b) y axis

 

   
  Solution (a)



 

Recall, the moment of inertia is the second moment of the area about a given axis or line.

     

For part a) of this problem, the moment of inertia is about the x-axis. The differential element, dA, is usually broken into two parts, dx and dy (dA = dx dy), which makes integration easier. This also requires the integral be split into integration along the x direction (dx) and along the y direction (dy). The order of integration, dx or dy, is optional, but usually there is an easy way, and a more difficult way.

   

Cross-section Area
 

For this problem, the integration will be done first along the y direction, and then along the x direction. This order is easier since the curve function is given as y is equal to a function of x. The diagram at the left shows the dy going from 0 to the curve, or just y. Thus the limits of integration is 0 to y. The next integration along the x direction goes from 0 to 4. The final integration from is

     

Expanding the bracket by using the formula,
     (a-b)3 = a3 - 3 a2 b + 3 a b2 - b3

     

   
    Solution (b)

   

Similar to the previous solution is part a), the moment of inertia is the second moment of the area about a given axis or line. But in this case, it is about the y-axis, or

     

   

Cross-section Area
 

The integral is still split into integration along the x direction (dx) and along the y direction (dy). Again, the integration will be done first along the y direction, and then along the x direction. The diagram at the left shows the dy going from 0 to the curve, or just y. Thus the limits of integration is 0 to y. The next integration along the x direction goes from 0 to 4. The final integration from is

     

     
    Comment
    The area is more closely distributed about the y-axis than x-axis. Thus, the moment of inertia of the shaded region is less about the y-axis as compared to x-axis.
     

  Example 2

 

Determine the moment of inertia of y = 2 - 2x2 about the x axis. Calculate the moment of inertia in two different ways. First, (a) by taking a differential element, having a thickness dx and second, (b) by using a horizontal element with a thickness, dy.

   
  Solution

     

 

a) The area of the differential element parallel to y axis is dA = ydx. The distance from x axis to the center of the element is named y.

     y = y/2

Using the parallel axis theorem, the moment of inertia of this element about x axis is

     

For a rectangular shape, I is bh3/12. Substituting Ix, dA, and y gives,

     

Performing the integration, gives,

     

     

 

 

(b) First, the function should be rewritten in terms of y as the independent variable. Due to the x2 term, there is a positive and negative form and it can be expressed as two similar functions mirrored about y axis. The function on the right side of the axis can be expressed as

     

The area of the differential element parallel to x axis is

     

Performing the integration gives,

     

Performing a numerical integration on calculator or by taking t = 2(2 - y) the above integration can be found as,

    

As expected, both methods (a) and (b) provide the same answer.