Last Updated on September 8, 2026 by Maged kamel
Product of inertia Ixy-for the right-angle triangle-Case-1.
Step-by-step guide for the calculation of the product of inertia for the right-angle triangle-Case-1.
1-To estimate the product of inertia Ixy, a horizontal strip will be used. The strip width is dy, and its height is x from the external axis X, which passes through the triangle base.
Since the strip height starts from the base and intersects with line Bc, the 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 x = 0 and x = b and checking whether the resulting y values satisfy the equation.
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, which is (x*dy), and 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 Expression and rearrange 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. The following slide illustrates the guide for obtaining the final values of the product of inertia.
Finally, we get the value of the product of inertia- as Ixy=h^2*b^2/24 about the x and y axes intersecting at the left corner A. The product of inertia is written as Ixy=(Height^2 by base^2/24). The product of inertia is positive at the corner point.
Product of inertia for a right-angle triangle- Case 1 at the Cg-by using the parallel axes theorem.
5-For the product of inertia for a right-angle triangle- Case 1 at the Cg, we will use the parallel axes theorem and deduct the product of A*(x-bar)*(y bar) from the product of inertia at the external axis.
The final value of the product of inertia can be estimated by selecting the appropriate value of x-bar, which equals +b/3. At the same Time, the vertical distance to y-bar is ±h/3. The product of A*xbar*ybar=1/2*(b*h)*1/3*b*1/3h=1/18*(b^2*h^2); this value is to be deducted from Ixy.
6- Finally, we get Ixyg=b^2*h^2/24-1/18*(b^2*h^2).=(-)b^2*h^2/72.
The product of inertia for case 1 of the right-angle triangle at the CG is negative. 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.
We can get any value for the product of inertia for any point by adding the product of inertia value at the CG to the product of the Area by the x*y values. Still, for points below the Cg and to the right side, the product of inertia value will be (+ve)*(-ve), which will produce a final negative value. This will apply, for instance, for the line AB at point B.
Point C is above and to the left of the Cg, with x distance= -b/3, while y distance for point C is + 2/3*h; so we expect the product of inertia to have a negative value.
There is a negative sign in the Expression for the product of inertia Ixy at the triangle’s centroid (Cg). Please refer to the following slide image for the Ixy values for the right-angle triangle case 1.
As we can see, the final result matches the values included in the table of inertia for plane shapes.
You can download and review the content of this post through the following PDF file.
This is a link to the Second Moment of Area for Standard shapes.
For the next post: the Moment of inertia for right-angle-Ix-Case-2.