Product of inertia Ixy-for the right-angle triangle-Case-1.
Product of inertia Ixy-for the right-angle triangle-Case-1- video.
We are going to talk about the product moment of inertia for the right-angle triangle case-1 by using a horizontal strip. The same procedure we have used for the determination of the moment of inertia about the x-axis.
If we investigate the strip we will find out that the x distance from the CG will be (x/2).
The width will be considered as X and the Y distance will be considered as Y. For the product of inertia it is equal to the ∫ of our strip, dA*(x/2)*y, because we are dealing with a CG of the strip, the dA itself is=x*dy, so we have a product of inertia= ∫x*dy*(x/2)*y.
But since we have already an expression of y = (-h*x/b+h), which is the equation of the line Bc, we can utilize the x value as=(h-y)/(h/b), At the end X will be=(b*h-b*y)/h.
We will have to arrange our terms IX Y =∫ from y=0 to y= h.∫x^2*y/2*dy.
Again will introduce the X relation with Y as follows. Ixy =∫ from zero to h ∫ (b*h-b*y)^2/h^2*(y/2)*(dy). This is the expression of x^ 2, which is = (b*h-by)^2/h^2, then Ixy =1/2 h^2, being constant *∫ from 0 to h for ∫(b^2*h^2-2*b^2*h*y+b^2*y^2) *y*dy.
Our next step is to estimate the integration from 0 to h, we have already (b^2*y*dy-*b^2*h*y^2*dy+b^2*y^3*dy, all multiplied by (1/2*h^2). This is a part of the video, which has a closed caption in English. If you wish to see the video on U-tube, here is the link.
Step-by-step guide for the calculation of the product of inertia.
1-For the product of inertia Ixy estimation, a horizontal strip will be used, the strip width is dy and its height is x from the external axis X, which is passing the triangle base.
Since the strip height starts from the base and intersects with line Bc, y value of the strip is the same y value for the equation of line BC. y=-h/b*x+h.
We can check the equation by substituting the values of x =0 and x=b and check whether this y value is satisfied or not. 2- the moment of inertia due to that strip=dA*xcg*ycg, our xcg= x/2, while ycg=y.
3- We will substitute the value of dA, try to make all our items a function of y, since we will integrate from y=0 to y=h, for the Ixy value.
We have x^2=(b^2/h^2)(y^2-2yh+h^2), we can substitute this value and readjust the terms. dA=xdy, dIxy=xdy(x/2)y=(1/2)x^2y dy.
4-After performing the integration, the summing of items will be continued for the matched items.
Finally, we get the value of the product of inertia- as Ixy=h^2b^2/24 about the x and y axes intersecting at the left corner A. The product of inertia is written as Ixy=(Height^2base^2/24).
Product of inertia-for-right-angle triangle-Case-1 at the Cg.
5-for the product of inertia for-right-angle triangle-Case-1at the Cg, we will use the parallel axes theorem and deduct the product of A*(x-bar)*y bar from the product of inertia Ixy at the external axis, the final value of Ixy, for x bar it is=b/3, while y bar it is =h/3.
6- Finally, we get Ixyg=b^2h^2/24-(1/2)*(b)*(h)(b*3*h/3)=(-)b^2*h^2/72.The product of inertia for the right-angle triangle case-1 at the CG has a negative sign. Refer to the next image for more details. the moment of inertia about the CG of a right-angle triangle case-1=(-)*(base)^2*height^2/72.
There is a negative sign in the expression for the product of inertia Ixy at the Cg of the triangle.
As we can see the final result matches the values included in the table of inertia for plane shapes.
This is the pdf file used in the illustration of this post.
For an external resource, the Second Moment of Area for Standard shapes.
For the next post10: Ix-the moment of inertia for right-angle-Ix-Case-2.
