Hello, everyone.
Now we are going to talk about the rectangular section.
the objective of this lecture is
to estimate the moment of inertia for the x axis  passing by cg.
Using parallel axes theorem, which means that, we are going to take
internal  X & Y axes.
we are going to estimate  the Ix about these two axes.
Then we are going to deduct the multiplication of the area.
by y̅^2.
In the case of Ix.
Second step is to estimate.
The moment of inertia about the y-axis Passing By the cg.
by
Using the parallel axis theory.
Third step, once we have
Estimated the moment of inertia about x.
and moment of inertia about y, we can get
easily estimate the radius of gyration for Ix and I y.
Number 4, is to 
Estimate the product moment of inertia for the rectangular section.
Number 5 is to estimate the polar moment of inertia.
J0
j0
Either at the cg or at external
parallel axes.
As we are going to see, first step,
we are going to
get the moment of inertia
Ix
We have two external axes
namely x and y.
We are going to draw our rectangular section,
the edge of the rectangular section bottom, left edge
coincides with the intersection of x and y
we have b as breadth and h as height.
then,
we have the cg, which we know that,
lies at a horizontal distance= b/2,
the vertical distance =h/2, since the h
is overall height of the rectangular section.
In order to estimate the Ix
We are going to
draw a rectangular strip, the width of that strip ...
=breadth of the
rectangle , which is =b
while , the
height will be = d y
and y is the vertical distance from the cg of that strip to our ...
external x axis.
This is the first step,
strip dA
which is horizontal strip for ix estimation
x and y are two axes at the corner of the rectangle.
while x' and y'
are two axes at the cg.
The x
distance from the cg to y is b/2, while the y cg =
h/2, while the area =
b*h, this is the area of the rectangle.
If we are going to estimate the area of the small strip, which we call dA, it will be=
b * dy
We need the y distance
from the cg to the x axis, we are going to square it
as we are going to see.
Ix, we can express,
is the integration
from 0 to h
of the multiplication of dA
*y^2
Recall that dA =b*dy
again *y^2, the integration performed from y=0
to y=h
The integration
the b will come out, the integration of y^2*dy from 0 to h
will be=y^3
/3
and the substitution from 0 to h
=b*(
we have h^3/3
-
0
0^3)=0
at
the end, we are going to
find out that the Ix=b*h^3
all/3
We know that this is the Ix for the external
x-axis
which is passing by the corner,
while our aim is to estimate the Ix'
which we are going to call it I x g,to denote
that this is at the cg.
from the parallel theorem, we have
The moment of inertia about the cg =the moment of inertia about any
external axis(-) Area of the section
by the square of the vertical distance, in our case
is
h/2 to be squared.
Our Ix already estimated as b*h^3/3
(-) the area ,which is b*h,*(h/2)^2
b*h^3/3- b*h^3/4
we are going to
put
denominator as 24
so
24/3=8,24/4=6
So we have
8(-) 6
inside bracket
*(b*h^3)/24
This 2/24=1/12
So I xg=b*h^3/12, this is the famous
equation of the
Moment of inertia for the rectangle,
which is = breadth* perpendicular^3/12
There is another way
if you want to estimate directly,
without going to the external
Parallel theory.
we are going to consider the x'
and y' and our strip,
will consider the vertical distance is y
from the x'
still
our strip
width is b and
the height is d y
Ix' or Ixg
=integration
from y=-
h/2 to +
h/2
The integration of dA*
by
y^2
this is y^2
=
the integration from -h/2 to +h/2
of b
b*dy ,as a term of dA
and y^2 , is the square of this vertical distance
Then Ixg=
The integration of y^2 dy =y^3/3
all multiplied by b
we get
b/3*( open one bracket)
inside the bracket is
h
(h/2)^3, which will give us, h^3/8-
(-h^3/8) afetr substituting y=-h/2
(-)*(-)=+, so we have b/3,
let h^3 outside, we have(1/8)+(1/8)
so
will come at the end, b*h^3/12,
which is the same result
obtained by using
the external parallel theory.
Thanks a lot, and we will continue
The next topic.
GoodBye.
