## Moment of inertia Iy for parallelogram.

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### Divide into areas and estimate inertia for each individual area about the y-axis.

The post includes how to estimate the moment of inertia Iy for a parallelogram. The axis y is passing by the left corner of the parallelogram and intersecting with the base of the Parallelogramhe at Point a.

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 dimension of the parallelogram is b*a, whereas the side length and the height is=h, h can be considered as=a*sinθ.

To get the expression for Iy, we will divide the parallelogram into two triangles and a rectangle.

We will use the previous data for the moment of inertia about the y-axis obtained for inertia for the right-angle triangle and rectangle.

Iy for right-angle case-1, Iy for right-angle triangle case-2, and Iy for the rectangle. The sum of inertia for the two triangles will be deducted from the inertia of the big triangle.

### The inertia of the left triangle about the y-axis.

The left triangle has an upper base of (a* cos θ) and height h, the inertia for the y-axis will be estimated about an axis passing by the left corner point of the base of the Parallelogram Can be considered as the sum of Iy about the CG plus the Inertia value from the product of(A* x bar^2).

The expression for the inertia Iy for the left triangle can be simplified as shown in the next slide image.

### The inertia of the right triangle about the y-axis.

As for the right triangle, it has a bottom base of (a* cos θ) and height h, the inertia for the y-axis will be estimated about an axis passing by the CG point then add the product of the area*x cg^2, where xcg is the distance from the Cg to the external Y-axis. Iy2 is the sum value, which is shown in the next slide image.

More simplification is done for Iy_{2} is shown in the next slide image

The detailed calculations for Iy for parallelogram

We will add the inertia of the two triangles to get their sum, which will be later subtracted from the inertia Iy of the big triangle.

### The steps for inertia Iy for parallelogram.

The inertia of the big rectangle can be considered as the height* base^3/3, the left side of that rectangle coincides with Y-axis.

The final value of the inertia Iy for parallelogram is to be obtained by deducting the inertia of the two triangles from the Iy of the rectangle.

The steps of estimation are shown in the following slide images. The common items can be cleared.

The final expression inertia Iy for parallelogram is shown in a similar form to the expression shown in the NCEES Handbook.

For the value of Iy for the rectangle, we have θ=90 degrees, when substituted in the equation of inertia Iy for the parallelogram, we get the same expression for the rectangle inertia about the y-axis.

### The steps for inertia Iy for parallelogram at the Cg.

The inertia Iy for the parallelogram at the Cg can be estimated by subtracting the product of the parallelogram Area by the Xcg^2.

The Xcg is the horizontal distance from the parallelogram Cg to the external Y-axis from the inertia Iy for the parallelogram. The matching items are to be cleared.

The steps of the calculations of the inertia Iy for parallelogram at the Cg are shown in detail in the next slide images.

The final expression for the moment of inertia for the parallelogram about the Y-axis.

The slide image contains the values of the moment of inertia for the parallelogram, at the Y-axis and the radius of gyration and matches with the previous calculation as shown in this post.

This is the end of our post for the Moment of inertia Iy for the parallelogram.

This is the pdf file used in the illustration of this post.

This is the next post, Iy for the trapezium.

For the use of a calculator for various shapes, please find Moments of Inertia – Reference Table.