Tag: Inertia Ix for Parallelogram.

  • 21- How to find Moment of Inertia Iy for Parallelogram?

    21- How to find Moment of Inertia Iy for Parallelogram?

    Moment of inertia Iy for parallelogram.

    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 intersects with the base of the Parallelogram 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.

    How to derive the expression for inertia Iy for parallelogram?

    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).

    inertia Iy for parallelogram by diving into shapes.

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

    Iy for the left triangle.

    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 a height of 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.

    Iy for the right angle portion.

    More simplification is done for Iy2 is shown in the next slide image. The detailed calculations for Iy for parallelogram is shown in the next slide image.

    Iy calculation for the right triangle.

    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.

    Iy calculation for the two triangles.

    The steps for inertia Iy for the 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.

    Steps to estimate inertia Iy for parallelogram
    The calculation for Inertia Iy for Parallelogram.

    The final value of the inertia Iy for the 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.

    Rearrangement of terms for inertia.
    The calculations for Inertia Iy for Parallelogram

    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.

    Final Iy value for inertia Iy for parallelogram.
    Inertia Iy for Parallelogram-case of rectangle checked.

    The value of the radius of gyration at the external corner for the parallelogram can be obtained by dividing Iy at the left edge by the area of the parallelogram. The expression for r^2y for the parallelogram is shown in the next slide image.

    The radius of gyration about y axis for the parallelogram
    Inertia Iy for Parallelogram-radius of gyration at Y

    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.

    Iy value for inertia for parallelogram at CG.

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

    Iyg calculation for the parallelogram.

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

    The final expression for Iy g for the parallelogram.
    Inertia Iy for Parallelogram at the Cg

    The value of the radius of gyration at the Cg for the parallelogram can be obtained by dividing Iy at the Cg by the area of the parallelogram. The expression for r^2g for the parallelogram is shown in the next slide image.

    The radius of gyration for the parallelogram
    Inertia Iy for Parallelogram-radius of gyration.

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

    List of inertia for trapezoid and parallelogram

    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.

  • 20-How to get a moment of Inertia Ix for Parallelogram?

    20-How to get a moment of Inertia Ix for Parallelogram?

    Moment of inertia Ix for parallelogram.

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

    The post includes how to estimate the moment of inertia Ix for a parallelogram. The axis x 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 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 Ix, we will divide the parallelogram into two triangles and a rectangle, and we will use the previous data for the moment of inertia about the x-axis that was obtained for inertia for the right-angle triangle and rectangle.

    Here are the relevant data for Ix for the right-angle-case 1, Ix for the right-angle triangle case-2, and Ix for the rectangle.

    How to estimate inertia Ix for parallelogram?

    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 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.

    Divide parallelogram into Two triangles and a rectangle for Ix estimate.

    The common items will be grouped and adjusted. The final expression for inertia Ix for parallelogram can be estimated as Ix=(1/3)bh^3. To get an expression in terms of a, the value of inertia Ix for parallelogram will be=ba^3(sin θ)^3(1/3).

    The final expression for inertia ix for a parallelogram.

    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 radius of gyration about y-axis.

    The radius of gyration (rx)^2 for the parallelogram about CG.

    The expression for the moment of inertia about Cg can be derived by using the parallel Axes theorem, we consider Ix for 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 next slide image.

    Calculation of Ixg , inertia about Cg for a parallelogram.

    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).

    Calculation of Ixg , inertia about Cg for a parallelogram.

    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 value of the radius of gyration at the Cg about y' axis.

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

    List of inertia for trapezoid and parallelogram

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

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

    This is the next post: Moment of Inertia Iy for Parallelogram.