Hello, Everyone in this new lecture,
there is a third method to estimate the Ixy g,
as a third proof.
Again, we have our rectangle with base b and height h,
and we have external.
Two axes x and y at the left.
edge left corner of the rectangle, and we have x cg
and y cg , but this time our strip will be small strip
with  width =dx and height =dy
We are going to estimate  the horizontal distance  as X to y cg and y  .
to the xcg.
We're going to write directly as Ixy the product moment of inertia
at the cg
but this time we are going to make double integration ...
because we have dx and dy
So we have the area.
dx *dy*x, horizontal distance to y cg *y
we are going to perform.
First integration, then second integration,
first integration
We are going to make integration for .
(x*dx)
starting from .
-b / 2.
to +b/2.
And the integration of  x*dx will be x^2/2
from (-b/2) to (+b/2).
This will give us.
(b^2)/4.
b^2/ 4/2-
b^2/(4)
While.
This value will  cancel each other.
So the 1/2 will  come out and this * (b^2/4-
b^2/4)
This will cancel each other, then we don't need to ...
estimate the integration of y/dA,
at the end the Ixy for
The center of gravity at the center of gravity equal to zero.
Our last item for the rectangular section is the ...
polar moment of inertia for the rectangular section the J0.
if we are going to estimate at the intersection of X and Y will be equal to Ix+Iy,
Knowing  that the moment of inertia Ix.
for the external  X coordinate.
is=(+b*h^3/3), while for the y .
Iy for the y axis
will be b^3*h/3, add them together will give ...
us the polar moment of inertia.
Which we will be equal to( b h /
3) and we open one brackets ,*(h^2+b^2)
And if we would like to get the expression for the polar ...
moment of inertia.
At intersection of x cg and y cg.
we recall that Ix cg=b*h^3/12
+
For Iy passing by the cg=h*b^3/12
We have common, (b*h/12).
And then we have one bracket, we will put (h^2+b^2).
Inside the bracket, that's why we we can ...
estimate and we have got the expression for the Polar ...
moment of inertia for the external two axes ,as well as.
the X & Y axis
passing by the cg, thanks a lot and see you
In the next lecture , Goodbye
