Moment of inertia for an isosceles Triangle-Ix, Iy.
Brief content of the video.
The moment of inertia for an Isosceles triangle 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).
In the end, we get the Expression of h^2/18.Regarding the Iy estimation, again from the previous triangle calculation for the moment of inertia for y-direction, for the triangle, we have that Iy final =(b*h/12)*(a^2+a*b+b^2). This is a part of the video, which has a closed caption in English.
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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 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 (A*y^2-bar) from the value of Ix, which is 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 Ixcg/A=h^2/18. For full data of 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 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.
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.
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