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STATICS - EXAMPLE
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Area Between Curve and x and y-axis
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Example 1
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Find moment of inertia of the shaded area about
a) x axis
b)
y axis
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Solution (a)
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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. |
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Cross-section Area |
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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
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Solution (b)
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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
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Cross-section Area |
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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
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Comment |
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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. |
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Example 2
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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. |
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Solution
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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 namedy.
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,
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(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. |
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