Hello everyone.
In this segment we're going to talk about the moment of inertia for the triangle.
But this time we are going to estimate Iy
or the moment of inertia about y direction.
Again we have the triangle ABC.
the base=b and the height=h,  we have two perpendicular  axes X and y  they are intersecting
at point B of the triangle
we can consider that this triangle is composed of two right-angle triangles one is number-1
the other one is called number two and for the triangle Number one we will find that the x and y axes
are intersecting at point a, while for the other one the y-axis and x-axis  intersection are apart  from
the line cd by a distance =a
before we have estimated that for the right angle, we have the moment of inertia for the y direction is
Iy=(h*a^3/4)  and this is about the y-axis 
While our calculation for  The Moment of Inertia at the y-direction based on 
that  ydc, coincides with the opposite side of the triangle.
But at that regard Iycd=(base)^3*
height/12
So we have in our case where it's the base=(b-a)
Our expression can be written Iy for the line Cd =(b-a)^3
*h/12.
While our Iycg
=(b-a)^3*h/36 
we can not evaluate the Iy, the moment of inertia at the Y-axis  from using the value of the moment of inertia at the ycd
but again we have to proceed always, in order to get the moment inertia, for any location we
have to proceed from the moment of inertia at the C G which is considered the minimum value for any
place of  any shape we have to proceed from the Cg and follow the principle of parallel axes 
theory, so our procedure in order to get the moment of inertia Iy finalو we have to get the moment of
inertia at the Y=Iy1 from  triangle #1plus  a moment of inertia from triangle no#2
from y
from case number one, we have our Iy, 
Iyc.g
ِA2*
multiplied by the square of the horizontal distance
from the CG to our y axis  .
Here we have the CG distance is (1/3*base), which will be=(1/3)*(b-a)
minus H.
We are going to add the distance a
Then we raise to the power of 2, we have here.
Iy
Which was (b-a)
sorry
Iydc= (b-a)^3*h
Iycg=(b-a)^3/36 + A2, which is (1/2*(b-a)
*h
And here we have (a+1/3(b-a)^2
So add a, then ((a+1/3(b-a))^2
We will adjust these terms.
We have (a+
Then we have (-a)/3+b/3), so we have 
(1/3(2a+b))^2
So and we call this one as x2.
We will evaluate the x2^2
As =(4*a^2+4*a*b+ b^2)/9 
so our A2*x2^2= (1/2)*( b-a)*h *
((4*a^2+4*a*b+ b^2)/9 )
We are going to add to Iycg ,which is
=(b-a)
raised to the power of 3,^3
*h/36
Add them together and readjusting the terms by taking (b-a) /36.
as a common factor *h.
Then we are left with (b-a)^2
plus we have the 36 and we have denominator as 18.
We are going to multiply by two both open another bracke,t in our bracket
(4a^2+4ab+b^2)*2
We move to the next slide
then we are going to add the condition of Iy=(h*a^3/4)  for the triangle number one in order
to get the Iy total.
So from the previous expression, we have ((b-a)/36)*h
And that was our factor inside the bracket,  which we can more readjust by introducing (b^2-2*a*b+a^2
we are going to add 
(8a^2+8*A*B+2B^2)
(b-a)*h/36*(we have b^2+2b^2),so =*(3b^2)
and we have 8*a*b-2*a*b=6ab.
then we have a^2+8a^2= 9a^2, inside bracket.
and then we are going to add the moment of inertia  Iy for the right angle  for the first triangle
which is (h*a^3/4)
again we are going to adjust the terms.
Still we have (b-a)*h/36.
We are going to introduce it inside.
So we have (3b^2+6a*b+9a^2)*b, we have
(3b^2*b=3b^3,6a*b*b,6a*b^2,9a^2*b,9a^2*b , then again for negative sign
(-a)*( 3b^2), 
(- 3*a*b^2) and we have (-a)*(6*a*b^2),-6a^2*b^2 
this term and finally we have (-a)*(9a^2)=-9a^3 
again once we are making the denominator as 36,  it means that we have to multiply here by 9
and we take it out so we left with
9a^3 and (-9a^3) will  cancel each other and then we have (1/36)*h*
(3*b^3+(6ab^2-3a*b^2) 
gives(3a^2*b) and we have (9a^2*b-6a^2*b) 
gives(3a^2*b) we have a common b, we get it out.
So we have (b*h/36) and also we have(3), we get out and we are left with a bracket 
(b^2+a*b+a^2), we continue on the next slide, then we can find out that the Iy final value
will  be ( b*h/12)* because we have here 3/36 then
we have b*h/12*(b^2+a*b+a^2)
back this is the final expression (b*h/12)*(a^2+a*b)b^2)
This is a moment of inertia at the y-direction for the triangle .
ABC, then we can get the radius of gyration by using the k^2y=Iy /Area  
so we have the ( (b*h/12)*(a^2+a*b)b^2)/area 
which is (1/2*b*h), then k^2y=
(a^2+a*b+b^2)/6, next lecture we are going to estimate the moment of inertia  for the y-direction at the Cg
thanks a lot  and see.
