Last Updated on January 2, 2026 by Maged kamel
Moment of inertia Ix for parallelogram.
Divide the area into subareas and estimate the moment of inertia of each subarea about the x-axis.
The post includes how to estimate the moment of inertia Ix for a parallelogram. The x-axis is located at the base of the Parallelogram. A parallelogram is a skewed rectangle with an angle=θ between the base and the left side. When θ = 90, the shape becomes a rectangle.
The area of the parallelogram is b*a, whereas the side length and the height are h and h = a*sinθ.
To get the expression for Ix, we will divide the parallelogram into two triangles and a rectangle, and use the previously obtained moment of inertia about the x-axis for the right-angle triangle and the rectangle.
Here are the relevant data for Ix for the right-angle case 1, the right-angle triangle case 2, and the rectangle.
The left triangle has a base of (a* cos θ) and height h, the inertia for the x-axis will be estimated about an axis passing by the left corner point of the base of the Parallelogram. Ix for that triangle=base*height^3/12, or (a* cos θ)*(h^3/12).
As for the rectangle about the x-axis, Ix=base*height^3/3=(b-(a* cos θ)*(h^3/3).
For the last triangle inertia, Ix will be deducted, it is case-1 of the right-angle triangle, but this angle is inverted, we will get Ix about the Cg and add the multiplication of area by the ycg ^2, which is ((2/3)*h)^2=(4/9) h^2.
The matching items will be grouped and adjusted. The final expression for inertia Ix for a parallelogram can be estimated as Ix=(1/3)bh^3. To get an expression in terms of a, the value of inertia Ix for a parallelogram will be=ba^3(sin θ)^3(1/3).
The square of the radius of gyration (rx)^2 for the parallelogram about the x-axis.
Since the expression for rx^2=Ix/Area, we will divide the inertia Ix for a parallelogram by the value of the area, the final expression is shown in the next slide image and compared by the value obtained from the FE Reference Handbook.
The square of the radius of gyration (rx)^2 for the parallelogramabout CG.
The expression for the moment of inertia about Cg can be derived by using the parallel Axes theorem. We consider Ix for a parallelogram about the X-axis and subtract the product of the area*ycg^2 from the Ix value.
We have the Cg distance y cg=1/2(a*sin θ) from the X-axis that passes by the base, Area is =b*h. The final value and the relevant calculation can be derived from the image on the next slide.
A modification of similar terms is to be conducted, h value is to be replaced by the equation h=aasin θ. The final term of Ix cg=(1/12)ba^3*(sin^3 θ).
The square value of the radius of gyration for the parallelogram about the Cg in the x-direction can be estimated by dividing the inertia Ix for the parallelogram at the CG over the area, Ixg/A, we have Ixg=(1/12)*b*a^3*(sin^3 θ), while the area of the parallelogram=b*h, rg^2=(1/12)*b*a^3*(sin^3 θ)/(b*h).
The square value of the radius of gyration about the Cg in the x-direction can be found to be =(1/12)*a^2*(sin^2 θ).
The rectangle inertia Ix is similar to the inertia Ix for a parallelogram, for which the angle θ=90 degrees, then the square value of the radius of gyration about the Cg in the x-direction for the rectangle is =(1/12)a^2(1), a is the height h, finally =(1/12)*h^2, please refer to the next slide image for more details.
The slide image shows the moment of inertia for the parallelogram and its radius of gyration, and matches the previous calculation shown in this post.
You can download and review the content of this post through the following pdf file.
For a link to a calculator for various shapes, please find Moments of Inertia – Reference Table.
This is the following post: Moment of Inertia Iy for Parallelogram.