Hello, everyone, we are coming to New lecture.
Unit number 5, we're going to talk about the ...
product moment of inertia.
For
rectangular section.
We want to prove .
that
the product moment of inertia for rectangular section =0.
We draw first.
a rectangle
The Base is b and the height is h,
and we introduce the external two.
axes.
X and Y.
Then we come at the cg and we draw another two axes, which are
X' and y'
We divide this rectangle into 4 areas.
namely A1, A2,
A3,
A4 , and for each area, we are going to locate.
the cg.
of every area.
for example, A1 this is the area the width =
= (b/2) the height is (h/2).
For area no. 2, the c g.
Lies to the right.
of y' axis by distance =b/4 and
Above the x' axis by.
h/4 and similarly , we can
Draw A3.
in the third quarter
And A4 in the fourth quarter.
For every.
Area ,we are going to estimate,the product
moment of inertia about x' and y'.
We're going to proceed as follows, Ixg
I x y  at g, where g is passing by the center of
the area
For the first area will be =A1*
Horizontal distance *y distance,
the horizontal distance for the cg of A1=(-b/4)
this the horizontal distance , we call x1,
y1 =(h/4) with positive sign
So (A1*-b/4*h/4)
Will you give us, A1.
*(-b*h/16)
While for Ixy g for area A2.
The x
value, which is x2 =+b/4
While
y2= +h/4.
Our Ixy at g
for A2= A2*b*h/16
For the area A3, we have horizontal
x distance =-
b/4 and we have y distance to the CG =-h/4,
that's why the product moment of inertia.
at cg for the area A3. will be A3*
(-)*(-), it will be (+) of the same ...
value (b/h/16), finally for the last area A4
The product moment of inertia .
of inertia for A4=A 4 * +.
(b/4*-h/4)
(+)*(-)s will give us (-), sign (-b*h/16) knowing that ...
all these areas are equal.
A1=A2=A3=A4.
each area of these areas is equal to b/2*h/2,
that's why this every area is b*h / 4.
Instead of A1,A2,A3,A4, we're going to write only one ...
area which is A1.
If you add minus and plus the other same value .
will turn them to zero, and we have (+)and(-) of the ...
same value which is turn them to zero, that is why A1
will be multiplied by zero, at the end
summation of Ixy about G
The center of gravity of the.
rectangle
We are going to use summation from A =1 to A=4 ...
will be turns to zero.
That's why for symmetrical.
section, the Ixy value willl be=0,
that is why , we will call this x'and y' principal axes
The principal axes , these are the axes for which
The product moment of inertia will be equal to zero.
But some sections they  don't exhibit such.
property that's why we are going to readjust these axes ...
but this will be discussed
Later.
The same result can be obtained for the product of ...
inertia, by introducing this rectangle ...
for the bottom left corner with X and Y.
And here we are going to draw this our strip dA
With breadth =b and with height .
=dy.
But in that regard, our product will be =Area of strip * x
value from the cg  to the Y axis.
Multiplied by y distance  from the CG.
to the x axis.
Since, this is rectangular strip
It's horizontal distance will be x/2, which is
=b/2
So Ixy  y equal to the integration our starting from.
y=0 and our  end is y =h.
And inside  the integration will contain.
(b*dy), the area ,
multiplied by (x/2)
(*y), while our x/2=  b / 2 .
So we have b/2*b which will give us
(b^2/2) * y dy.
We get (b^2/2)  outside.
after making integration for y dy , it will become .
y^2/2, we substitute.
Starting from h minus.
the value
When y=0, So we have b^2*/4
open one bracket
(h^2-0) at the end we have a b^2*h^2/4
Since the area the big A =b*h,
so we can write.
This expression Ixy is equal to A*b*h/4
This is the product moment of inertia about external two axes
at the corner X and Y, if you want.
to Estimate the product moment of inertia to CG,
we have to deduct.
Area(x*y)
x bar *y bar, our x bar
Is b/2 and our y bar is h/2.
This is
Ixy at G
And this is our.
Ixy
our Ixy g= (A*b*h/4)- area *(b/2*h/2),
will give us the same result.
(A*b/h/4) which is the same result.
At the end it will give us 0 value .
Thanks a lot. See you next lecture, Goodbye.
