hello everyone for the objective of this lecture , we have the first one.
is the Theory of pure bending assumptions and the relation and second objective  why we estimate the product
of inertia?.
And the third objective we are going to talk about Mohr circle of inertia and
How can we use it to get the maximum and minimum values of the moment of inertia?
First, we have some assumptions when there is a beam acted upon by pure bending, it is going 
to be deformed.
But first let us examine the beam before deformation.
We have two axes X and Y and we consider the section of the beam as a rectangular section
at the first stage  before the deformation.
We have vertical
section perpendicular  to the neutral axis and in our assumption that the plain section.
perpendicular  to the neutral axis before the deformation remains perpendicular  to the neutral neutral
axis after deformation .
After deformation, the beam will take the form of the curved shape  and perpendicular  to
that it will not remain any more as vertical but   the section will be it will
remain perpendicular  to the neutral axis
after  deformation, the second assumption that deformation due to the bending is small.
The third assumption the beam is initially straight and all the longitudinal  filaments bend ito circular
arcs  with a common center of curvature.
This means.
All these perpendicular sections in this section and if we assume there is another section that would
be plain section perpendicular to neutral axis.
And third portion there will another section perpendicular to the neutral axis.
All these sections if you are going to extend they will meet at a common point too.
And the line constituted will be a form of circular arc number# 4 the deformation in 
the vertical direction i.e the transverse strain, epsilon εyy.
May be neglected in driving expression of the longitudinal strain when we have bending
moment there will be an extension in the lower chord accompanied by compressioninthe upper chord, for
the extension.
Always.
in a direction that will  be compression or transverse strain
This transverse  strain in our case will be represented by the εyy =dh/h and it will be negative, while
the longitudinal  deformation will  be the extension of length divided by the original length, extension
of length will be replaced by dl/ l, which will be positive in εyy.
we consider if it has a small value and it will be deleted.
Number five the radius of curvature is larger as compared to the dimension of the cross-section.
Number six the value of the young's modulus of elasticity  in both tension and compression will
be the same, next step.
We are going to talk about the pure bending.
What is the pure bending?, the pure bending.
it means that the beam  subjected to moments on both sides and they are both  equals, the shear
value will  be zero.
Why is the shear value will be zero?  because this bending moment  on the right-hand side will produce a couple
it is anti-clockwise so the couple will be upward here the shear here will be accompanied by downword
shear foce on the right hand side, while this moment will produce a resisting couple anti clockwise
so the shear force will  be downward and the other shear force it will be upward so downward
and upward opposite to each other and both , suppose will cancel each other.
That is why the shear force will be equal to zero, while for the bending moment diagram 
if we want to draw this is sagging.
It will  be positive and value will go  to the right support and that will be closed here.
So this, from left to right.
it will  be subjected to bending of equal value positive bending  , for the left hand
side moment  and the right hand side moment  they will cause the beam to bend in a form of Arc and
the enclosed angle would be α  as we can see in the diagram.
The second  case when we have a pure bending  in case if you have a simply
supported beam, hinged, simply supported beam acted upon by two shear forces a P and P are of equal.
Values and they are acting on the third point of the beam and L/3 and L/3 and, in that regard.
We have shear forces diagram we have P positive shear force coming up and go down until coming
to B and from the right-hand side we have negative shear will  continue until point d and closing.
while for the bending  moment diagram,  we have (p*L/3) 
(Pl/3) this is the positive bending moment which we have , so for this the intermediate section
will be having a bending moment while shear force between C and d =0
This is the second case of the pure bending moment the assumption number #3  that beam is initially
straight.
If we can see the that is a rectangular section and this is the x-axis  and y-axis and before the deformation
if we take, a section with  apart distance by (dx) and there is an element small element apart by distance y
before deformation and let us have a look after deformation will find that the beam will bend in
this form and the neutral axis which was horizontal becoming curved like that and we can see here
the x axis and y and perpendicular axis 3
If we assume that the arc radius will  be R and that is an enclosed arc angle will be considered
as the dα.
We are going to equate and we are going to find the strain value before deformation.
There was no strain, after deformation.
This segment with length the (dx) its length it will become are the Rdα
for the element at Y distance up, its length originally was  dx while after deformation the length will
become (R-Y) multiplied by (dα).
So the length for this red line=(R-y*dα), was originally the length 
it was dx.
If we are going to compare it actually it's becoming in the neutral axis.
There is no change so we can take the dx as =(Rdα) , the deformation is tension or compression.
For this case we are above the neutral axis.
So the deformation will  be coming negative.
We have (R-y)*dα-Rdα ,+Rdα and -Rdα
will cancel each other and we are left with (-y*dα)
And since it a strain is deformation over the origin length the deformation is (-ydα)
and the origin length was R*dα
so we left with(-y/R) and since due to Young 's modulus of elasticity = stress/strain 
so we have our strain equal to (-y)/R .
This is a strain, so the f= (e *E )the strain * young 's modulus  of elasticity.
It means that f=(-y/R)*(E)  and since we have the 
assumption of 4 that Young 's modulus of elasticity is   is the same for case of tension and compression
will continue into next topic.
We are going to get the relation for the stress and how can we get the Ix value
and Ixy at the end.
It has to become zero.
Thanks a lot and see you.
