# Lecture – 22 Advanced Strength of Materials

Today, we are going to study bending of curved
bars. I am sure in your earlier course you have studied the bending of straight bars.
You must have considered a component like this. It was subjected to bending moment,
because of the bending the bar gets into the shape. I am just showing a portion of
the bar here. And we select the coordinates that,
this is the direction aligned with the centroidal axis of the bar. And this is y axis, which
is pointed towards the center of curvature and z is out of the plane.
So, it is really trying to bend above the z x plane or rather the z x plane is trying
to get curved. You have seen that, for such cases,
if you just consider a portion of the bar. Let
us say this portion. This portion after the application of the load takes a step like
this. What you have considered? That this section
rotates by some angle. So, also this section rotates and all the
fibers of the bar gets or becomes curved. And they
are going to have common center of curvature. And you have considered, that the radius
of curvature of the neutral axis, which pass through the center of gravity of the cross section is equal to R. In this case, you are
going to have the bending stresses developed under the action of the bending moment.
And the distribution of the bending stress is going to be linear. It is going to give
rise to compressive stress at the top fiber and tensile
stress at the bottom fiber. You might recollect the flexure formula that was derived
in connection with this. That sigma x divided by the distance of the fiber from
the neutral axis is equal to M by I is equal to E
by R, where sigma x is the stress. So, we are plotting the sigma x stress in this direction
and y is in this direction. So, therefore, this is the variation of this
stress. And this sigma x by y, M is the bending moment, I is nothing but, the moment of inertia
of this section of the beam about the z axis. So in fact, this I is nothing but, moment
of inertia about the z axis I z z. But, it is
simply written as I, E is the modulus of elasticity of the material. And R is the radius of
curvature of the fiber, which passes through the centre of gravity of the cross section
or that is the neutral fiber.
Now, in this case you must note that the fiber all the fibers initially are of length. If
you consider that this distance is equal to l
0. Then, all the fibers initially are of the same
length and after getting built, they are going to have different lengths. So, the strengths
are going to be different. Just to refresh, what were the assumptions that you made in
connection with the derivation of this formula. You assumed, that the modulus of elasticity
of the material in tension E t is equal to the
modulus of elasticity of the material in comparison. Similarly, you also assume that the
y x plane and that plane is a plane of symmetry.
that after the bending the section does not get distorted. This section, which as initially
plane it is also plane after it is bent. So, plane sections remain plane. The other assumption is that, the radius of curvature
that we get out of this bending is going to be very
large compared to the cross sectional dimensions. Like the dimensions in the horizontal
direction or in the vertical directions. So, R is very large compared to cross sectional
dimensions. So, these are the major assumptions best of that, we derived this
relationship. Now, let us consider the situation, that the bar is not straight initially. But,
it is curved with some radius of curvature. Let us
say R bar for the centroidal axis. And it is reasonable to assume, that this R bar is
again very large compared to cross sectional dimensions.
Such that all the fibers can be considered to have the same center of curvature. So,
if the beam is initially curved and it is subjected
to bending, what will be the difference in the
stress distributions? How can we calculate them. And you must also understand that this
type of components do arise in practice. So, you must have come across the chain links,
which are carved to apply the load it is going to have bending action developed. So, just
I would like to show you certain components,
which are going to have configurations that are initially curved. Look at some of the
components which are shown here. Think of a circular ring. It is subjected
to diametrical compression. Think of the hook of
a crane, which is curved here. And it is subjected to loading and under the action of the
loading, there is going to be bending, developed. Similarly, you comes across pipelines
like this, which are subjected to loading. You can also have the press frame c press
frame, which is going to look like this. And it is subjected to loading and that is going
to cause bending in the portion here.
Similarly, you also have come across chain links. Think of the chain links and it is
subjected to loading in the axial direction. Under the action of this loading you are going to have bending in this portion. So, there
are certainly applications and we would like to
see, how the stresses in these components can be calculated.
You know that to derive the relationship for the bending stresses,
in response to some bending moment applied on the component. Let us concentrate on a
portion, which is uniform in cross section. Let us also assume that the cross section
is like this. It does have a plane of symmetry
we make the same assumption, as you deal in the case of straight beams.
Let us now consider that the c g of the cross section is here. And therefore, this is the
y axis. And let us consider that this axis passing
through c g is now z bar. Just I would like to differentiate with z I will come to why
it is different little later. Let us concentrate on a
portion of the beam, which is included between this section and a section, which is here.
So, let us just consider a portion which is included between these two sections. And let
us say, that this portion is included it makes an angle at the center of curvature, which
the bending moment, this cross section is going to rotate. Let us say that it is going
to rotate like this. And this rotation, that is very
small compared to phi and let us represent this by delta phi.
Now, you can see that the fibers, if you consider this fiber it has some length initially. Or
if you consider a fiber here, this is the length of the fiber initially. And it got
changed by this distance, after the application of the
bending moment. And you see here, that the fiber got compressed when it is located at
this position. On the other hand, if you consider a fiber, which is somewhere located
there this fiber is of length this much. But, it got extended by this much after the
application of the bending moment. So, we see the that phi bar here is getting stressed.
And the phi bar here it is getting compressed. The same picture you have noted in the case
of straight beam, the fibers which are located towards the center are getting compressed.
And the fibers which are located they are getting extended.
Obviously, if the fibers are getting stretched, it is getting compressed here and stretched
here. Obviously, there will be some location in between, where there is no change in the
length of the fiber. And that is the fiber, which is considered to be neutral fiber. So,
in this case let us we do not know, what is the
location of that fiber? Let us consider that the neutral fiber is going to be located somewhere
at this location. And let us show that this is the neutral phi bar.
So, this is the neutral axis of the beam after bending. The corresponding point on the
cross section is here. And let us indicate the axis, which passes through this point
by z. So, we are now trying to have the origin located
at the point, where there is no strain. I would like to emphasize, that we will like
to have the derivation again. Considering that
the modulus of elasticity of the material in tension is equal to the modulus of elasticity
of symmetry. In this case it is the y x plane, which is
the plane of symmetry. And we also make assumptions, that there is no distortion of
the cross section, after the load is applied. And
we are also assuming, that the radius of curvature. This radius of curvature, now we
would like to indicate that radius of curvature of the neutral axis or neutral fiber, that
is the fiber which is neutral and let us indicate
the radius of curvature of this fiber as R. So,
this r is now very large compared to cross sectional dimensions.
In fact, in this case we can also consider that this R bar, which is the radius of the
centroidal axis, this is also large compared to the cross sectional dimension. So, with
these assumptions, we would now like to proceed for the derivation of the formula. So, I
would like to now segregate that portion, which we considered which is making an angle
of phi at the center of curvature. So, that is the portion and this section gets
rotated by an angle, this much which we have indicated by delta phi. So, this is the this
axis is centroidal axis and this is the neutral axis. So, this one is neutral axis. If we
now concentrate on a fiber, which is located here.
Please note that we have indicated y axis here and our x axis is located in this direction.
So, therefore, this is really the direction of x axis.
Now, we are considering a fiber, which is at a distance y from the neutral axis. So,
we are concentrating on a fiber, which is a distance
of y from the neutral axis. If R is the radius of curvature of the neutral fiber,
then the initial length of this fiber l 0 is going to
be given by R minus y multiplied by phi, because phi was the angle made by the sector at
the center of curvature. After the loading has been applied, the section
gets rotated. And therefore, there is a change in the length here. And we can approximate
that length by this cord, which is delta l let us say. This delta l is equal
to it is reducing in length, which is nothing but y
multiplied by delta phi. So, the strain in the direction x is equal to delta l by the
initial length l 0. And therefore, this is equal to
y by R minus y into delta phi by phi. So, the strain is going to have a non-linear
variation. You remember that in the case of straight beam, we had the strain beam varying
in the proportion y or the exact expression was minus y by R. But, in this case it is
minus y by R minus y. So, therefore, it is a nonlinear variation. Now, if you would like
to plot this variation, let us consider that the
section is here. This is our z axis, this is y axis.
Now, if I plot the strain in this direction Epsilon x and this is y. Then, this strain
variation you will find that, it is going to have a
variation like this. For y equal to 0, the strain is 0
and it is going to have comparison value here. And it is going to be opposite nature in the
other direction. When, y is negative which is true in this direction, we are going to
have positive strain.
So, therefore, this strain is tensile in the bottom fibers and it is compressive in the
top fiber. So, this strain variation is now non-linear.
In fact, you see that the length of this radius of curvature, the initial length of
the fibers are different depending on this y. And
therefore, you are going to have different starting length. And hence, the strains are
going to be different and it is not going to vary linearly as in the case of straight
beam. You can now also consider this point. That
is the fiber is here, which is what the center of curvature at a distance of y. Let us consider
exactly the fiber, which is located on the other side of the neutral axis at a distance
of y. Now, if you consider this fiber, which is
located again at a distance of minus y from the neutral axis.
Now, the extension of this is exactly the same as this. But, since the length of this
fiber is larger to begin with the strain is going to
be less. So, therefore, the strain what we are
saying is that, we have the fiber here which is a distance of y. Another fiber exactly
located on the other side of the neutral axis at the same distance. The strain in this fiber
is going to be more than the strain in this fiber.
So, therefore, you will find that the strains are less in the tensile zone.
If we want to calculate the stresses, stresses are going to be given by directly sigma x
is equal to modulus of elasticity E multiplied
by strain e x. And therefore, this is equal to E
delta phi by phi into y by R minus y. So, this is the stress variation and these are
constants for the sector. And therefore, they are not varying the strain, the stresses are
going to vary along similar line as in the case of strain.
So, if would like to plot the variation of space, you will get a variation of this type,
same type or to a some scale, this variation gives
the stress variation. So, we would like to draw it separately. So, we are plotting sigma
x stress here and the variation is going to be
given by again this y. So, here the stress is compressive and here it is tensile.
So note that, this is due to the fact, we have applied the bending moment in this fashion.
Therefore, we get compression and tension in this direction. So, this is due to the
direction of the moment, that we get compression there tension there. If you try to
reverse this directions, you are going to get tension there and compression on the other
side, again a phi bar which is located at the same distance from the neutral axis.
You will find that, stress in the phi bar which is located towards the center of curvature
is going to be more than the stress, which is
located away from the center of curvature. So,
we have got the variation of stress. But, we are not yet in a position to say, how this
stress is dependent on the applied bending moment. And what is this radius of curvature
R, we may be given the value of the radius of curvature of the centroidal axis. But, we are not so far clear about the value
of this radius of curvature R. So, therefore, we really have to dissolve the two issues.
How these constants delta phi by phi are related to the bending moment, and also how
we can calculate this radius. So, we will proceed to determine these things in terms
of the bending moment. How do you do that? You remember in the case of straight beam,
what we did is that we considered a fiber at a
distance y from the neutral axis. Therefore, the force acting on this fiber
is nothing but, sigma x. If you consider that this
area is nothing but, d A. Then, sigma x into d A is the force acting on that element. And
now if we integrate the total force acting on the cross section. That is going to be
sigma x into d A. This particular cross section is
only subjected to bending moment in and there is no axial force acting on the cross section.
And therefore, we must get the sum of the axial stresses to be equal to 0.
So, therefore, sigma x d A is equal to 0. So, that is the equilibrium equation, that
is one of the equilibrium equations. Now, if we try
to consider the stresses and their moment about the z axis. So, the moment of the force
sigma d A, which is at a distance of y. So, that is the moment. Now, if we try to integrate
this moment over the whole cross sectional area.
Obviously, this is the positive x direction. The stress which is considered to be positive
is going to be directed like this. And it is
moment is going to be clockwise, but the external moment is applied anti clockwise. So, therefore,
we must have the equality like this, that this integral is equal to minus m, which is
the applied bending moment in the anti clockwise direction.
So, this is the second equations of equilibrium. And from these two equilibrium
equations, we can resolve the constants like R, and also this delta phi which is also
unknown. Let us proceed with the first equation. So, starting from 1 we have sigma x d
A is equal to 0 I think I would like to point out at this stage. That, this neutral axis
which you have arbitrarily taken, we do not know
it is location. So, therefore, that is also an unknown. What
is this y or what is the reference for this y is
really unknown for us, so far. So, therefore, we have to in that and in fact, you are going
to see that. This is going to be obtained from this relationship. So, if you now substitute
the value of sigma x, we will have E delta phi by phi y by R minus y d A is equal to
0. Or we can take this for the element, that
we have considered are all constants. So, therefore, what we really have is integration
of this quantity R minus y d A is equal to 0, we can these are constants. So, therefore,
we can now consider this integration is nothing but, y by R minus y d A equal to 0.
And if we do the manipulation, we can write now, we can add R subtract R. And therefore,
will have R by R minus y d A minus integration of A over d area is 0. So, what
we are doing is that, we have written y equal to R minus R minus y.
So, this gives us this R is a constant and this will be area of the cross section. So,
therefore, we will have R is equal to A by integration d A by R minus… So, that is
really the value of the radius of curvature. So,
we are in a position to get this radius of curvature, which is going to come up after
the application of the bending moment. So, this is a property of the cross section, note
that this is a property of the cross section, it
does not depend on the amount of load which is applied. We will consider now the 2nd equation which
is nothing but, sigma x multiplied by y d A over the whole cross sectional area is equal
to minus M. Or you can write minus E delta phi by phi y square by R minus y d A
is equal to minus M or E delta phi by phi y
square R minus y d A is equal to M. We can now consider one simplification here or
geometries shown . In this, we are considering the typical fiber,
which is at a distance of y. And let us represent, that the radius of curvature of
this fiber is molar. So, what I am saying, that the
radius of curvature of this is equal to… So, if you make it represent it here, that
this radius of curvature is equal to R. Therefore,
that is nothing but, the radius of curvature is
R here and this is the distance y. So, therefore, R minus y is nothing but, this molar. So,
making use of that fact, we can write here that y is equal to R minus R. And therefore,
we can write E delta phi by phi R minus R whole square divided by R d A is equal to
M. This will give us E delta phi by phi, you
will find that these are going to have very simple form. So, it is A R square by R plus
this is going to be R. Then, we are going to
have minus 2 R. So, that is d A. So, R square by R plus R, this is equal to M E delta phi
by phi. I would like you to consider, the point we had already derived that R is equal
to A by integration d A by R minus y which is
nothing but R. So, that was the relationship. So, therefore, here what we can do, that we
can write integration of d A by R is equal to
A by R. So, we can now substitute here for this we can write R square into A by R. And
this is nothing but, integration of R d A the whole cross section minus 2 R into A and
that is equal to M. So, you can now simplify E delta phi by phi is equal to twice I beg
your pardon, this is A into R plus what is this?
Please note this point . This quantity, the radius of curvature of this
fiber radius of curvature of this phi bar or it is at a distance of R from the center
of curvature. So, therefore, R into d A is nothing
but, the moment of this area about the center of curvature. And therefore, integration
of this whole quantity would be nothing but, A into R bar.
So, this quantity is nothing but, whole cross sectional area multiplied by R bar. So,
therefore, we will have A R bar minus twice R A is equal to M. Finally you will find that
E delta phi by phi into A and we will have now R bar minus R equal to M. And if we
represent this R bar minus R by E which is considered to be eccentricity. Finally what we have is that E delta phi by
phi is equal to M by A into E. So, the quantity delta phi is now related to the applied bending
moment and the cross sectional dimension, and also the eccentricity of the
cross section. So, this is relation number 4.
So, we have obtained that if delta phi by phi is equal to M by A earlier we have already
derived that the stress sigma x is equal to E delta phi by phi into R
minus y. So, if we now substitute the value in that
stress relationship, we now find that sigma x is
equal to minus M by A e y by R minus y, which is same thing as M y A e into R. The
radius of curvature of the fiber. So, what we are considering, that this is the fiber
we are concentrating and this fiber has a radius
R. So, the stress in this phi bar is given by M
into y, where it is located at a distance of y from the neutral axis.
So, this distance is y M y A e into R, where eccentricity between the centroidal axis and
the neutral axis is e. We can now see the value of this stresses, that is occurring
in the cross section. If you consider the fiber which
is at the extreme interior, it is at a distance of let us say h 1 from the neutral axis. Then,
the stress is going to be the highest and that
sigma x at A is equal to minus M h 1 A e. If you consider that the radius of curvature
of fiber is equal to r 1. Then, it is A e into r 1
and note that this stress is compressive in response to the bending moment, that we have
applied. Similarly, if you consider the fiber B, which is at a distance of minus h 2 from the neutral axis this distance is h 2. So,
it is minus h 2 y axis is positive towards the
center of curvature. And therefore, sigma x at B is going to be… So, minus h 2 will
cancel the negative sign. And therefore, it is going to be M h 2 A e
and if consider that the phi bar radius of this
fiber is r 2, then it is a e into r 2. So, this is a tensile stress. So, we expected
that the compressive should be there. And the tensile
stress should come up at this point. And in fact, we have also got the values accordingly.
So, these are the extreme stresses in this section. I would like you to again look back
into the relationship for the straight beam. So, if you consider the straight beam sigma
x for straight beam, it was nothing but, M into y by I. That is for straight beam. So,
the difference is quite evident, that here in it is
non-linear, whereas, in this case it is linear relation. So, you not to solve the problem
of curved beam, what is necessary? You will be
given bending moment applied. You will probably given the value of the radius of
curvature of the centroidal axis. And you will be required to find out the stresses
in the cross section to extreme stresses in the cross section and to solve the problem.
Since, the relationship is sigma x is equal to minus M into y by A e into R. What is necessary
is that, we have to first locate the neutral axis, which means that you have to
find out the radius of curvature of the neutral axis, which is capital R. And the difference
between the radius of curvature R bar and this capital R will give you the eccentricity
of the cross section. Then, you are in a position to find out exactly,
the radius of curvature of the extreme fibers also. And then you get back to the
relationship sigma x is equal to M into y by A e
R. That will give you the stresses at any layer of the cross section. So, we will now
look into what is the eccentricity, it is a property
of the cross section. So, it will be nice idea to
look into how we can calculate this eccentricity for number of cross sections. Let us consider the rectangular section. For
the rectangular section, we have the radius of
curvature of the neutral axes R. And the radius of curvature of the centroidal axis is r bar.
Let us now represent that the height of the inner fiber, from the neutral axis is h 1
and the height of the extreme outer fiber is h 2 from
the neutral axis. So, this it will be the convention that will make use of. And total
height of the cross section is h 1 plus h 2,
that is h. What is r bar that is what we have to determine.
And we have the formula that R is equal to cross sectional
area divided by d A by R minus y integrated over the whole cross section. So, if you consider
this section to have width equal to b. Let us say that width of the cross section
is b. And for any fiber that is arbitrary fiber, we
have this radius of curvature equal to r. And this fiber is at a distance of y from
the neutral axis.
So, we can write this integration that is there is nothing but, write one more step
d A by r. Since, r is equal to capital R minus y.
We can write also d A given by, if you consider that thickness of this layer is equal to d
y. Then, this d A is nothing but, b into d y. And
these limits of integrations are going to be minus h 2 to h 1.
Alternatively we can write, that the radius of curvature in terms of the radius of curvature
of the extreme fibers. If you consider the radius of curvature of this extreme fiber
is equal to r 1. And the extreme bottom fiber
radius of curvature equal to r 2. Then, we can certainly write from this relationship d y
is equal to minus d r. So, we have now we will
have the limits r 2 to r 1 b h divided by d A is nothing but, b minus d r divided by
r. So, that is the we are moving from this fiber
to that fiber. So, d r is negative and now this
gives us b is canceling, this will give us h divided by we can now make the limits to
be interchanged. So, therefore, it will be d
r by r and this gives us h divided by logarithmic r
2 by r 1. So, that is the radius of curvature of the rectangular cross section. So, once
you have got this radius of curvature, it is given
from the three dimensions. Once you know the radius of curvature r bar, you know what
is the value of r 2 which is nothing but r bar
plus h 2. And the value of r bar is nothing but r bar minus I beg your pardon r 2 is going
to be equal to r bar plus h by 2. And r 1 is going to r bar minus h by 2. So, you do
have this value known. So, you can write in terms of the radius of
curvature of this centroidal axis. So, this is
nothing but, l n. This is r bar plus h by 2 divided by r bar minus h by 2. So, that
is the value for the radius of curvature. And the
eccentricity is nothing but, r bar minus R. So,
we have the eccentricity now given by h divided by l n r bar plus h by 2 divided by r bar
minus h by 2. Let us now consider a section which is different.
You will find that curved members are not always made up of rectangular cross section.
They could be made up of t section, they could be made up of trapezoidal section
for circular section. So, you would like to look into how to calculate the radius of curvature
for such cross sections. So, let us now consider a T section. So, the
section is shown here, that we have 1 t formed by a rectangular portion of dimension
4 t by t and another portion. So, the flange is of dimension 4 t by t and the is of dimension
3 t by t. The centroidal axis will pass through the centroid
of the cross section. Therefore, in this case it is not obvious, where the centroid is located.
So, it has got to be located first. And the problem is to find out, the eccentricity
of the cross section. So, if you consider that this is the axis or this is the neutral
axis or the z axis. Then, what is the difference between the locations of z and z bar. You
would like to consider again the top fiber. The
distance of the top fiber from the centroidal axis. Let us say, that is equal to h 1 bar
which is unknown to us. And you would also indicate, that the distance
of the top fiber from the neutral axis, as usual given by h 1. And the distance of the
bottom fiber from the neutral axis is h 2. So,
to solve this problem, we will have to first
of all find out the location of this centroid of
cross section. And then we have to find out the radius of curvature of the cross section.
So, we will get into solving this problem. If we take the datum, datum to be this lin.
Then, we can take the moment of this flange about that line. And moment of the ((Refer Time: 54:25)) above the same datum. And the
total moment would be equal to nothing but, total area multiplied by the distance
of this c g that is h 1 bar. So, we can write now h 1 bar into area of
the flange is 4 t square and the area of the is 3 t square. So, therefore, that is the
total area. So, moment of the total area about the datum is equal to the
moment of the individual area about the same datum. So, the flange area is 4 t square and
it is distance from the datum is t by 2. Similarly, the area is equal to 3 t square.
And it is distance is nothing but, this t plus half of 3 t. So,
therefore, it is going to be 5 t by 2. This gives us h
1 bar equal to 9.5 t cube by 7 t square which is nothing but, 9.5 t by 7 which is nothing
but 1.357 t. So, we have got the c g located with reference to the top fiber it is 1.357
t. We will consider the calculation of the radius
of curvature in our next presentation.

## 8 Replies to “Lecture – 22 Advanced Strength of Materials”

1. Diego Andres Alvarez Marin says:

Topics.
* Bending of curved bars
* Applications
* T-Section

2. Vijay Jadon says:

very poor explaination of the basics… not expected from a iit prof.

3. prashant mali says:

very nice explanation sir…
i wanted to know which reference book is to be reffered for the same notations.
please tell the book name sir..

4. RORBOYS kabaddi says:

Nice video sir

5. Mysterious File says:

Free course n u hv Monitize this video

6. Mysterious File says:

Shame on u nptel! Gareebo ke liye monitization hota naki paise wale iit institutions ke liye…kitna kamaoge tm log

7. Mysterious File says:

Bs bhi kro ab kamai

8. ketan tripathi says:

Why is there minus sign before bending stress in bending equation??