Moment of inertia for an isosceles Triangle-Ix,Iy.
Moment of inertia for an isosceles Triangle-Ix,Iy video.
The moment of inertia for an Isosceles triangle-Ix,Iy is our main subject. The triangle ABC, For which the two sides of the triangle AC and CB are equal. And we’ll find that the perpendicular From Point C To side AB will bisect the base, so the base length b will become (b/2) From the previous expression for the triangle.
We have estimated the moment of inertia Ix for the triangle ABC the value of Ix will be equal to b*h^3/12, and the area of the triangle will be equal to (1/2*b*h).
We can get the radius of Gyration, by using the expression K^2x=Inertia (b*h^3/12 )/ the area, which is (1/2*b*h).
We get the square value of the radius of gyration K^2x will be equal to h^2/6, as for the moment of inertia at the x-direction. For the CG the square of the radius of gyration is equal to b*h^3/36, So K^2x is the radius of Gyration k^2x at the Cg=Ix cg/Area. We have b*h^3/36/(1/2b*h). 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.
Moment of inertia for an isosceles triangle-Ix, Ixg.
1-The moment of inertia for an isosceles triangle Ix is obtained by considering the moment of inertia Ix for a Triangle, which we have obtained earlier an Ix= bh^3/12 and radius of gyration Kx^2 as Ix/area:b*h^3/12/(0.50*b*h)=h^2/6.
2-To get the moment of inertia at the Cg of the isosceles, which is termed Ix CG at the CG of the isosceles triangle we will deduct the product of area by the square of the distance from the Cg, which equals (A*y^2-bar) from the value of Ix, which equals bh^3/12, so we get the value of IxCg=b*h^3/36.
3-The square of the radius of Gyration, k^2cg can be estimated by dividing the moment of inertia about the Cg by the area, the value will be equal to Ixcg/A=h^2/18. For full data on the steps used, please refer to the next image.
Moment of inertia for an Isosceles Triangle Iy & Iyg.
A- The moment of inertia for an isosceles Iy can be obtained after adjusting the terms of the Iy of the triangle, where the y-axis is an external axis passing by point a.
The moment of inertia Iy for the triangle which we have obtained earlier will substitute the value of (a )in the relation as 1/2*b, where b is the base length, after adjustment.
We will get the value for the moment of inertia Iy for an isosceles triangle as Iy=(7/48)*h*b^3, where b is the base while h is the height of the isosceles triangle.
B- for the radius of gyration Ky for a triangle as k^2y=Iy/A=(7/48)*h*b^3/(0.50*b*h)= (7/24)b^2.
C-The moment of inertia Iy at the Cg of the isosceles triangle can be obtained by subtracting (A*x-bar ^2) from the estimated value of the moment of inertia Iy for a triangle as Iy=(7/48)*h*b^3, The distance from the Cg to the vertical y-axis which is x bar value =b/2.
Iy Cg=(h*b^3)/48.
D- for the radius of gyration Ky for an isosceles triangle as k^2y at the Cg =Iy cg/A=(1/48)*h*b^3/(0.50*b*h)= (13/24)b^2.
For more information, please refer to the next slide image.
This is the pdf file used in the illustration of this post.
For an external resource, the definition of the moment of inertia with solved problems is 2nd moment of inertia.
This is the next post, Product of inertia Ixy for an isosceles triangle.
