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

plane of loading is a plane of symmetry. In fact, the loading plane is nothing but, the

y x plane and that plane is a plane of symmetry.

So, plane of loading is a plane of symmetry. Another important assumption that was made

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

and the loading is in the plane symmetry. So,

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

is equal to phi under no loading. As you apply

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

in compression. Plane of the loading is a plane

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.

Topics.

* Bending of curved bars

* Applications

* T-Section

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

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..

Nice video sir

Free course n u hv Monitize this video

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

Bs bhi kro ab kamai

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