Hello everyone, in this lecture, we are going ...
to talk about the moment of inertia for the isosceles triangle.
The isosceles triangle ABC.
For which the two sides of the triangle AC and CB are equal.
And we'll find that
The perpendicular.
From the.
Point C.
To side AB will  bisect.
The base, so the b will become (b/2)
From the previous.
Expression that for the triangle we have IX for the ...
triangle ABC would be=b*h^3/12
And the area = (1/2*b*h)
We can get the radius of Gyration
by using the expression K^2x=b*h^3/12
/divided by the area, which is  (1/2*b*h).
We  get K^2x= h^2/6
As for the moment of inertia at the x- direction for the CG.
It will be= b*h^3/36 .
So K^2x is the radius of Gyration squared.
k^x at the Cg=Ix cg/Area
So we have b*h^3/36
/(1/2b*h)
At the end we get.
the Expression of
h^2/18
Regarding the Iy estimation.
again from 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).
And we have our a
=b/2, we are going to substitute.
for a=b/2 again we are going to adjust the Iy.
So it will become =b*h/12*(for the (a)^2,
it will be (b/2)^2.
+(a*b), which will be (b^2/2).
+(b^2).
We're going to sum.
The similar items we have b^2+ b^2/2
+(b/2)^2, so we have +b^2/4 + (b^2/2+b^2)
We need to make denominator as 4 .
so we have the (b*h) / 12.
*(b^2+2b^2+4b^2)
all / 4.
So Iy at the end willl be b*h/48*(7b^2),this is the ...
expression for the moment of inertia of the Y- direction.
and our y will pass by point C.
of the triangle
So K^2y  radius of gyration.
Iy/Area
So we have.
7*b^2.
*(b*h)/48.
so we have at the end K^y= 7b^2 / 24.
Regarding the expression for the CG we are going.
to estimate the moment of inertia at the CG.
From the triangle =(b*h/36)*(b^2+a^2-a*b)
And we are going to substitute for a =b/2.
then
We have Iycg=(b*h/36)*(b^2
+ b^2/4-a*b, which is -b*b/2
Then Iycg=b*h/36*( b^2.
+ b^2/4-b^2/2), we make the denominator as 4.
So we have here, (4b^2+b^2-b^2) ,
so we left with.
3*b^2.
/4.
(b*h.
/36.
So we have at the end Iycg=(1/48)*(h*b^3) then k^2y ...
at Cg=IyCG / area ,(b^2
*b*h/48.
Over the area.
Which is (1/2 b*h).
So
We have.
(b*h) will go with (b*h)
Then we are left with b^2/24 as Ky^CG,
thanks a lot and see you.
