Hello,Everyone, in this course, we are going to talk about the moment of inertia,
The moment of inertia, it falls into two categories the first category is the area
moment of inertia, or sometimes called second moment of area.
This term is used for determining the stresses due to  shear and moment and torsion
acting on a given section. This is the moment of inertia, Under this category
is dealing with the statics.
Problems and how to solve these Static Problems by use of the Moment of inertia .
The second category is about the mass moment of inertia, this  represents the resistance  of a body to angular
acceleration due to torque and this topic is very important in the study of Dynamics.
In this course ,we are going  to talk about the area moment of inertia,
And for the mass moment of inertia it will  be later on discussed.
How are going to perform?
our estimation for the second moment of inertia, we are going to draw the x, and y axes.
and for example, if we have an area in this shape we're going to introduce.
Strip and we call that strip area dA, if we are going to
evaluate the inertia about the X axis, we need to know the y dimension or the normal dimension to x axis.
We are going to say that, the moment of inertia about x axis ,
is equal to the integration of d A * y^2
While,  for the moment of inertia about the y-axis, as we can see, is equal to  the integration of d A *x^ 2, where is the x is normal
Distance to the y axis , in this regard we are going to use Vertical strip, this strip,
area we call it d A, the integration itself.
The integration symbol is the same like summation of individual areas and.
In that regard each area will  be multiplied by its X component, but  symbol
integration,  it means that,
We are going. Move this is strip multiplication by x^2
Starting from one point Ending to another point.
Next item will be the product of inertia, suppose
we have big area like this, where is the area we call it A
we will  introduce only small strip.
of area =dA
from that strip, we are going to estimate the cg distance.
for both the  x, y coordinates, for x  is a normal horizontal distance perpendicular to Y axis, while y,
Normal distance perpendicular to the x axis.
The expression for I xy = the integration of dA* (x*y), for that
the strip, this = the summation
for example A*x1*y1+A2*x2 y2
etc, where  i=1 till i=n
At the end, instead of performing this task we are going to introduce
only one small strip and we start.
integration from our start point  until our end .
What are the steps to estimate  thee moment of inertia?
First we draw the element,
to know what is the shape of that element, either  a rectangle.
triangle, Trapezoidal, etc.....
Number 2 for  an expression for the dA, by introducing the strip and we
And we will  find out this is a horizontal or vertical strip.
Number 2 for  an expression for the dA, by introducing the strip and we
And we will  find out this is a horizontal or vertical strip.
what is the relation between x and y, this is the third step, write the
expression for both X and Y and making interrelation between them.
We need y if we are going to estimate the inertia
At x axis, and  the x  distance if you are going to calculate the I Y
and of course, we need both if we are going to calculate  the product moment of inertia
Number 4 , we are going to set the limits  moving the
the strip, what point of start?
until what point it will end,and in some cases
Especially in the triangle,
We are taking the integration in two steps.
As we are going to see, number 5 make the integration.
we Estimate the  integration values  and the end  we're going to evaluate the moment of inertia.
there is a very important,
theory, which is called parallel axis theory, we are going to prove for Ix, suppose we have two
intersecting axes, namely x and y
this  is the area we are going to estimate.
The moment of inertia
by introducing small area d A
we are going to discuss the first point, first question to ask ...
where are you going to estimate?
the moment of inertia, you are going to estimate regarding the x axis
or to the y axis
from the sketch shown
let x, y two external axes, while
x' and y' are two axes passing by the cg.
we are going to estimate the moment of inertia, for example about the x axis
we need the vertical distance from the CG of the dA
small infinitesimal area d A
This y actually is can be estimated as,
The sum of two components, the first component is called y1
y1 is the vertical distance between the CG of the dA
to the x', which is passing by the c.g.
+ the y bar
Which we know, that the vertical cg distance from the external
chosen x y,
If you're going to.
write Ix= to the integration of dA*
big distance Y^2, it can be equated to (y1
+y bar)^2
And if we're going to estimate for the I y for the moment of inertia
About the y axis, we are going to integrate d A x *x^2,as we are going to. see
In the first case, Continuation of the Ix estimation, we we need the integration
of d A *. y^2, and this y^2 can be put as (y1+y bar)^2
we continue, Ix =  the integration of dA and , we are going to
to get the product of the square, this is
dA*(y1^2+2*y1*y bar +y bar^2)
all ,we will put inside the bracket
And will  be multiplied by dA , then Ix= .
the integration of dA*y^2+the integration
of dA*2* y1*y bar+
the integration of d A*y bar ^2
This first item, the multiplication of area *y1^2
will resemble the moment of inertia about the x ', which is the axis passing by the cg
The moment of inertia about the  x'
will resemble the moment of inertia about the x ', which is the axis passing by the cg
will resemble the moment of inertia about the x ', which is the axis passing by the cg
Which is a X'passing by the cg, we are going to write as x '+
will resemble the moment of inertia about the x ', which is the axis passing by the cg
Which is a X'passing by the cg, we are going to write as x '+
2 *( y bar ,being constant will come outside the bracket*, we left with
dA*y1 , the dA*y1
is the first moment of area, and since this is the axis about the cg
as we know the first moment of area =0
this term will become 0 +A*Y bar^2
So, finally the moment of inertia = Ix', about the axis passing by the cg, +A*y bar^2, in this regard, some times it is easy
to estimate moment of inertia about. external axis
subtracting the multiplication of the area by y^2, we can get the expression
of the moment of inertia about the cg, thanks a lot , we will continue  in the next lecture
See you and Goodbye.
