Strength of materials (expanded) - Mechanical Engineering Video Lecture - Mechanical Engineering | Best Video for Mechanical Engineering

Hi, this is Dr. S. P. Harsha from Mechanical
and Industrial Department, IIT Roorkee. I
am going to deliver my lecture 28 of the course
of The Strength of Materials, and this course
is developed under the National Program on
Technological Enhanced Learning, NPTEL. Prior
to start this lecture 28, actually I would
like to refresh you know like the concept
of previous lecture in which we discussed
mainly about if we have a beam and like if
it is under the consideration of bending moment,
then what will happen, you see you know like
means when the bending moment is applied,
then what exactly the relations were there
that part we discussed. So, I would like to
refresh those things because you see in that
lecture, we are going to discuss mainly on
the numerical problems based on the previous
formula which we derived, and also you see
we would like to discuss what the shear stresses
in this lecture.
So, you see in the previous lecture if you
just remind, then you will find that we had
a beam is there, and when this under you know
like the consideration of or under the application
of the bending moment M, only pure bending
moment, there is no other forces are there.
Then you see you know like simple bending
equations which you derived that sigma by
y equals to M by I equals to E by R in which
sigma was the bending stress which induce
due to the application of this bending moment.
So, sigma by y - y is the distance you see
on which we just want to find it out the bending
stresses, and that fiber is having a distance
from the neutral axis y. So, this y is the
distance from the neutral axis of that particular
fiber.
So, sigma by y equals to M by I. M was you
see you know like the bending moment which
is the applied bending moment, and you know
like due to that the bending stresses are
coming sigma, and then M by I was there. So,
I was you see you know like this second moment
of area was there which we defined the new
term which was nothing but equals to summation
of y square dA. So, there were you see the
two main theories which we described to calculate
this I. That was you say the second moment
of area because it has you know like the summation
of y square dA. The y square term is there
that is what you see we defined that as the
second moment of area.
So, you see by integrating, if you have a
regular structure, then you know like this
tapered bar or the soft or that kind of thing,
then it is pretty easy to find it out by integration
for y square dA or if we have different kind
of selections, then you see summation of y
square y square dA will give you this second
moment of area. Then, you see the third term
is sigma by y equals to M by I equals to this
E by R. So, E is the Young’s modulus of
elasticity and R is the curvature radius.
So, you see this is a simple bending like
the moment equation is there, and if you see
this equation, then you will find that only
the pure bending is there.
There is no shearing part is there, there
is no other stress components. Only these
stress components due to the bending is there
in this particular equation and you see due
to that we can also find it out the maximum
bending stresses. The sigma maximum which
is nothing but equals to you see you know
like we discussed that this M by I into you
like the y maximum distance. So, M and I are
you know the applied part. So, this is maximum
bending. This bending moment is applied and
I is the second moment of area is there. So,
it can be easily calculated with the using
of the dimensional one and the applied one,
but the key feature is that actually what
is the location is there at which this maximum
bending stresses are coming. That part we
discussed on that, and also we discussed that
actually M is equal to Z times the sigma maximum.
So, Z was the section modulus and it can be
easily calculated based on you see M by I
because you see the section modulus absolutely
depends on that. What is the exact structure
is there of that kind in which we want to
find it out the maximum bending moment with
corresponding to the maximum bending stresses.
So, that part you see we discussed in the
first section of those things and then, you
see you know like other section which came
up with that if we have a bending moment in
that or if we have a bending you know like
these, this second moment of area is there
on the neutral axis, then also we can calculate
these second moment of area at any section,
or any axes using perpendicular theory and
using parallel theory.
So, in perpendicular theory we described that
we are using the perpendicular axis then like
a polar moment of section. That means, you
see if you are taking area and you see these
two axes on which we want to find it out the
section area IX and IY, we can simply you
know like make the relations. The polar section
of this inertia with this because the polar
section of inertia, or we can say this is
J, J is perpendicular to the area concerned
and this IX and IY are the two corresponding
axes to this moment of inertia. So, with that
you see you know like if we want to relate
those things using this perpendicular axes,
we calculate that Z is equal to IX plus IY.
So, we can calculate that one, and the another
relation which we describe using one theorem
that was the parallel axis theorem.
So, that was you see you know like it is pretty
easy that if you want to find out at any axis,
let us say Z axis. So, IZ is equal to IX plus
integration of this you know this Y into dA,
or we can say that this we described that
H distance was there. So, H square into A.
So, meaning is pretty simple that we can calculate
you know like this second moment of area or
the moment of inertia for any section. Once
this moment of inertia for you know like the
axis which is passing from the centroid or
we can say the neutral axis, so these theorems
which we discussed and also, we find it out
that actually if we have a circular section
or if we have a rectangular section, then
also we can calculate those things. So, for
rectangular section which has this height
is D and breadth is B.
So, for that you see the calculated INA. That
means, you see the moment of inertia at this
neutral axis is nothing but equals to cube
by 12, or also you see if we have you know
like the uniform diameter from 0 to d, then
we have you see is equal to bd cube by 3.
So, these kinds of discussion you know like
a derivation which we described. So, in this
lecture you see we are going to use again
the similar kind of formula for that, and
then the last section of the previous lecture
which we described that if we have you know
like big blocks are there in which the bricks
are to be settled, and there is you know like
the small gaps are there in between those
things for setting up any pillars for that
kind of structure you know like we can calculate
this INA for the bigger portion and the shaded
portion.
So, I is the total you know like INA is nothing
but equals to I for total rectangle section
minus I is the shaded portion. So, while comparing
those things, we can calculate bd cube by
12 minus two times because you see the two
different bricks were there if you remember
two times of bd cube by 12 or we can say bd
cube by 6. So, you see here these kinds of
discussions which we made, but in those discussions
you see there was no stress component due
to the shear. So, in this lecture you see
first we will discuss about you know like
what will happen when you know like all these
shear stress component is there. That is a
second part, but first part is focused on
you know like whatever we discussed in the
previous lecture, the formulae for the bending
stresses and the bending moment we are going
to use you know like for the numerical problem.
So, here you see the first numerical problem
is that we have I section girder. So, we have
you see the girder for you know like the basis
of any building, you see you will find that
on the top of roof you see the girder is there
which has the dimension of 200 millimeter
wide by 300 millimeter depth flange. So, one
is the flange which has this diameter and
the web in between. The web, it has the thickness
is 20 millimeters is used to simply support
beam for a span of 7 meters. So, you see here
for flanges, both of the flanges you see the
dimensions are 200 millimeter wide by 300
millimeter depth, and there you see the web
thickness is 20 millimeter is given for the
total span of beam is 7 meters.
The girder carries a distributed load of 5
kilo Newton. So, you see per meter here the
load distribution is given on that particular
girder which has the intensity of 5 kilo Newton
per meter, and the concentrated load. That
means, you see we have a combined load here.
So, one is the UDL in which 5 kilo Newton
per meter is there, and distributed load is
there and other one is the concentrated load
or we can say the point load that is 20 kilo
Newton at the mid span. So, this is the configuration
of the problem. Now, our main intension is
to find out the second moment of area of the
cross-section of the girder because I cross-section
is there. So, for that we need to find out
and also, we want to find out the maximum
shear stress, maximum stresses which are being
set up due to this load application, the combined
UDL as well as the concentrated load.
So, for that you see here the second moment
of area of the cross-section can be easily
described when you see we know the total regular
structures. So, for that we have taken the
section with the symmetricity about the neutral
axis can be easily made if we have I section.
So, for that you see you know like since it
is I, so centroid is absolutely passing from
the rectangular section of the mid-portion
of this I which has the total this I is nothing
but equals to bd cube by 12 which we have
described for you know like the rectangular
section, and here you see you know like it
is this middle portion of the web, I should
say is rectangular portion.
So, we can simply get those things bd cube
by 12, and the section can be divided into
convenient rectangles because you see these
are all the rectangles for each of which the
neutral axis passes through the centroid because
it is well balanced structure. So, all the
neutral axes just passing from the centroid
of those I section. For example, you know
like in the case enclosing by girder is just
by rectangular. So, you see what are the description
there of this kind of problem, it is absolutely
based on I section. In I section, you see
this neutral axis is just passing from this
particular centroid and you see you know like
it has symmetricity all along, both of the
side. So, we can say that whatever the things
we will be analyzing for one section, it has
symmetricity on the other section also. So,
for that you can see this particular diagram.
So, in that what is that you see here?
This is I section you see and this is the
web part, and this is I section. So, you know
like just go with my cursor, you will find
I structure here. So, this is I structure
here you see these two, this is one flange
and this is another flange and this is the
web part. So, you see here you know like as
it is being discussed here that this is 300
millimeter of this, and this is 200 millimeter
and 20 millimeter, this web is there. So,
this distance is 90 and 90. So, total is 180
and this is 20. So, this is 200. So, just
keep this thing in your mind that this is
I section and the neutral axis is just passing
from this. As I told you, it is symmetricity
on both the side.
So, for that we can simply calculate I girder
which is nothing but equals to I rectangular.
The total rectangular part you know like minus
I shaded area. If you remove this shaded area,
then you have I section. So, for that you
see you know like if you are talking about
the total rectangular area of this, then we
have 200 into 300. So, 200 into 300 cube divided
bd cube by 12, so b into d cube by 12. So,
this is there into 10 to the power minus 12
because it is in the millimeter. So, we just
want to calculate in the meter side. So, we
have this 1 minus two times because these
two shaded areas are there. So, two times
into 90 is this as I told you because this
is 20. So, 90-90 and 12, the total is 200.
So, that is what 90 into 260 as it is given
you see you know like this is 260 for this
description. So, 260 cubes of bd cube divided
by 12 into 10 to the power minus cube. So,
now if you analyze those things by that, then
you have 1.68 into 10 to the power minus 4
meter 4. So, now you see here since I section
is well symmetrical section all along this
two flanges and one web. So, it is pretty
easy to calculate you see with the use of
this. First you calculate for entire rectangle
and then, you know like reduce the shaded
portion out of that. So, you will be handicapped
with this kind of structure, and the maximum
stress may be found from you know like the
simple bending theory of equation. As you
see we discussed in the previous case that
sigma by Y is equal to M by I is equal to
E by R, or we can say that if you want to
find out the maximum bending stresses, then
it is nothing but equal to the maximum bending
moment M by I into Y maximum.
So, for that you see you know like it is pretty
easy to put those values and get the final
thing because you see here there are two different
kind of loadings. So, first of all our main
focus is to find out this bending moment.
So, for that you see here the two main loadings.
As I told you, one is the UDL which has you
know like the intensity of 5 kilo Newton per
meter, and then we have a concentrated load
you know like some magnitude. So, in order
to obtain the maximum bending moment, the
technique will be you know like considered
each loading on a beam separately and get
the bending. You know like we need to just
go with the separate kind of loading and get
the bending moment due to these loadings as
if no other forces acting on the structure,
and then you like superimpose of these two
results because you see first the UDL is there
and then, the concentrated load is there.
So, we need to calculate different segments
and then, combine it because our main focus
is that there is no shearing occurs on that.
Only this bending stresses are there due to
this bending moment. So, that is why you see
we need to compute these bending moments or
bending stresses very carefully along with
this particular kind of load.
That is why you see you know like the maximum
bending moment is nothing but equals to wL
by 4 plus wL square by 8. WL by 4 is coming
due to this point load, or we can say the
concentrated load plus you see whatever the
bending moment is coming due to the UDL. So,
this distributed load wL square by 8 is there.
So, if you are keeping those things, then
we have this w which is 20 kilo Newton of
20 into 10 to the power this cube. So, that
is in you know like the Newton, sorry in kilo
Newton. Then, we have L that is 7 meter divided
by 4 plus w which has the 5 kilo Newton per
meter. So, 5 into this 10 to the power cube
into 7 square because the total length of
the beam is 7. So, 7 square divided by 8.
So, if you compute, then we have the maximum
bending moment is 65.63 kilo Newton meter.
Once we have these things, then we can pretty
easily calculate what will be the maximum
bending stresses and for that the sigma maximum
is nothing but equals to M maximum divided
by I into y maximum or if you are keeping
those value 65.63 into 10 to the power cube.
So, this is you see the maximum bending which
we calculated on the above side into you know
like we want to find out at this 150. You
know like this millimeter from that 150 meter
from that side.
So, you see here we are keeping those things,
and you see we just calculated 1.06 into 10
to power minus 4. We are keeping for I this
value for this I section which we calculated.
So, if you are putting all those values, then
we have y maximum is 51.8 mega Newton per
meter square. So, you see this is the exact
processor to calculate first of all what is
I because you see the second area, second
moment of area is very important for any kind
of structure that what exactly the cross-section
is. Here, we have just chosen the cross-section
of the beam is I section. So, that is why
you see first of all we need to apply that
part and then, you see you know like what
will be the maximum bending moment is there.
So, it comes only due to that what kind of
loading is there. So, once you know the loading
or combination of loading, then again segregate
those parts and calculate that combined effect
of those things.
So, we have you see the maximum bending moment
here. Once you have the bending moment, once
you have you see the second moment of area,
then it is pretty easy to calculate the maximum
bending stresses by using the simple bending
equation theory. So, you see here these values
are again you see like all these things, it
can be easily predictable for these kinds
of examples.
So, you see here we have an example that if
we have you know like simply supported beam
on which the UDL is there, so one is UDL which
has the intensity of 5 kilo Newton meter as
we discussed in the previous case, and you
see this at the middle mid of this pan, what
we have. We have 20 kilo Newton of the point
load, and you see you know like this distance
because the total distance is 7 meters. So,
at the central point, we have the 3.5 meters.
So, with that now first of all our main intention
is to combine those loading because you see
here we have the UDL as well as we have the
point load.
So, with the combination of those things as
I told you, see the maximum bending moment
is coming by wL by 4 or wL square by 8. So,
with that you see here we simply combine those
loads here you know like with the UDL plus
these things and then, you see reaction forces
are coming out of those things, and then once
we have the reaction forces, we can go for
shear force diagram. And the bending moment
diagram also as well because here if you go
for these reactions here, all these you know
like the joint points, this reaction point
is just going on the top of side. This is
you see are just going on the bottom of side.
So, we have you see this positive direction
here, and if you go for this right hand direction,
this reaction is going on the top of side.
This is going on the bottom of side.
So, you see here we have the magnitude directions,
and you see in between there is a point load
which will give you abrupt change in shear
force. So, just with that concept you see
we can simply draw shear force diagram. So,
this is you see at this particular point,
we have the maximum shear force due to the
reaction, and then it is just going to reduce
here up to this point and then, there is 20
kilo Newton you know like the force is there
which is on the downward direction. So, there
is you know like the change is there in shear
force and then, there is a negative direction
because of the sign convention. So, we can
calculate those things pretty easily. Now,
you see you know like because as I told you
this is pretty simple to calculate. We can
again segregate into two parts. As we discussed
in the previous case, you see here that first
just go with the UDL.
So, this is you see the first case with the
UDL only. So, in this 5 kilo Newton per meter
is there and the 7 meter is the total length
for that. This CFD is there as we discussed
in the previous cases you see and this is
the bending moment diagram. Then, you see
here this is one part another part is that
that this beam is influenced by also this
twenty kilo Newton. So, this is a simple simply
supported beam point load is there at the
mid of this span. So, for that you see this
w is coming and then, you know like since
it is segregating two exact equal parts. So,
we have one positive, one negative part in
shear force diagram. So, this is the positive
and the negative part and since, there is
no UDL, so simple triangle is coming because
of the point load.
So, now you see if we combine those things,
then you see it is very important thing that
once you combine both of the things, then
we are ending up with those things. So, you
see here this is the combination of maximum
these things are there, then it is decreasing
and you see up to that point, it is decreasing.
So, you see up to this it is bisecting here,
it is also bisecting here. So, this you see
the changing area is there. So, it is very
interesting you know like if you want to calculate
those things, then you know like just on the
top of that you can see that this is the real
problem. Our problem equating to one problem
by this UDL plus one problem by point load.
So, you see here we can simply analyze those
things or in the bending moment if you go
for the previous slide. Then, you will find
that the maximum bending moment was wL by
4 due to this point load, 20 kilo Newton plus
wL square by 8 due to this UDL of 5 kilo Newton
per meter, and we can calculate those things.
So, you see here we have shear force diagram
for both things and we have the bending moment
diagram for both the cases.
So, you see here in these cases, we only focused
on the bending moment as well as the bending
stresses which are coming due to you know
like this applied moments at this particular
beam in irrespective whether it is cantilever
beam or simply supported beam. Now, you see
here main the main key feature of this particular
lecture is to focus on shearing stresses in
the beam, and all of the theory, which has
been discussed earlier. As I told you just
on the bending stresses in the beam you know
like for any kind of pure bending is there.
That means, constant bending moment is just
acting along the entire length of the beam.
Now, you see here we would like to focus that
we have a beam you know like just let us say
AB beam is there which is transversely loaded,
and then you see you know like with that we
have kind of forces which is just creating
the shearing.
So, together with the shearing force and the
bending moment, we can simply note that the
middle portion CD of the beam is free from
shear force, and that you see the bending
moment always is loaded into whatever the
distance is there at which the load is acting
P into a is always uniform between these two
particular sections. That is why you see we
are always saying that whenever the pure bending
is there, it always gives you the bending
moment, which is applied to both the sections.
So, you see here since we know that this point
loads are coming along with this particular
section in which I just told you that you
see here the constant shear forces are there
for which there is no bending. The dM by dX
is 0, since there is no shearing. So, for
that the bending moment is constant. So, you
see here when we are talking about those things,
this is pretty simple diagram on which the
two point loads are there. Both are just going
downward direction. The reactions are coming
from this which is on upward direction.
So, if you are going for the bending moment,
then you will find that this and this direction
you see on the left hand force, so we have
the positive shear force. So, this is the
positive shear force and this is the negative
shear force, but there is no load application
in between. So, there is no shearing or there
is no resultants shear forces existing in
between this action. So, when we have you
know like more shearing force, so F is equal
to dM by dX. If you are keeping F equals to
0, dM by dX is equal to 0. That means M is
constant. When M is constant, then that means
only bending moment is applied on a beam and
this is the main condition of pure bending.
When you see only you know like the bending
moment is applied like the total stresses
which are inducing in this particular beam
is just the bending stresses due to this M.
So, you can visualize this thing in this bending
moment diagram. So, this bending moment is
P into a. As we discussed M is equal to P
into a, which has constant value of this particular
section and these are there due to the shear
forces. So, this is you see one of the example
in which we can say that there is a condition
in which the pure bending is there, but it
is also computing with certain kinds of this
shearing stresses. So, our main focus is just
to focus on that actually. If we have this
kind of loading, then how we can conclude
you know like the shearing altogether with
the bending stresses. So, conclusion is hence
one of the conclusion which is coming from
the pure bending theory was shear force at
each cross-section is 0 because you see you
know like all the pure bending is coming,
no shearing is there. And the normal stresses
are due to you know like this bending are
there only which are producing on the top
of the fiber, we have the tensile stresses
on the bottom of the fiber, we have the compressive
stresses.
So, this flexural stresses or we can say this
normal stresses are always coming, but there
is no shear stresses are there. In that case
you see the non-uniform bending of a beam,
where the bending moment varies from one cross-section
to another, there is a shearing force on each
of the cross-section and shearing stresses
are also inducing in the material. That means,
you see whenever a non-uniform you know like
the bending is there if it is a uniform bending,
there is no point to go with any kind of shearing
any kind of twisting or any kind of couple
is there. But whenever non-uniform bending
is there, always it gives for different cross-section
at different kind of shearing forces are there,
and corresponding shearing stresses are there
in the material.
The deformation associated with those shearing
stresses causes warping because you see like
whatever deformations are coming, they always
cause the warping of the cross-section, so
that the assumption which we assume while
deriving the relation for this plane cross-section
after bending remains plane will be violated
because you see after once it is warping,
once it is deviating from the original position,
then we cannot say that you see once you remove
the load, it will be in the main form of all
those fibers. It would not be possible.
So, for that you see this assumption is violated.
That means, this condition, whatever the condition
sigma by sigma y equals to M by I equals to
E by R is absolutely violated for this kind
of relations. Now, due to the warping, the
plane cross-section before bending does not
remain. The plane after the bending as we
told you and this come you know like this
complicates the problem, but more we can say
elaborated analysis shows that normal stress
due to the bending as calculated by a simple
equation like sigma by y is equal to M by
I equal to E by R.
This equation you know like gives the distribution
of the stresses which are normal to the cross-section.
That is in the cross-section or we can say
along the span of a beam are not greatly alerted
because by the presence of these shearing
stresses. Thus, you see you know it is just
to use the theory of pure bending in case
of non-uniform bending and it is accepted
in practices also to do these things. That
means you see sometimes when we know that
non-uniform bending stresses are also there
in those cross-sections, then also we can
apply these normal stress concept that these
normal stresses are there along with the particular
fibers. On top of the fibers we have the tensile
stresses, and on the bottom of the fibers
we have this kind of compressive stresses
and we have a neutral axis.
So, we can apply all these perpendicular axis
theorems or this parallel axis theorem for
those things and also, we can apply this equation
sigma by y equals to M by I equals to E by
R, so that this is justified case. Therefore,
this kind of theory in which you see the non-uniform
bending is there, and we can say that whenever
pure bending is there, it is just due to this
particular M concept. Now, you see here in
that you see we just focused on the uniform,
on non-uniform bending or now, our focus is
that shear stresses are there in the beam.
So, by the earlier discussion which we had,
we found that all in the bending moment represent
the resultant of certain linear distribution
of the normal stresses sigma x or the cross-section
that you see. The upper fibers are there in
the tensile stresses which is one form of
the normal stress, and lower fibers are there
of the compressive stresses which is also
one of the form of the normal stresses. Similarly,
the shear forces Fx over any cross-section
must be resultant of a certain distribution
of shear stresses because you see whenever
shearing is there, whenever the kind of twisting
is there, always induces some sort of the
shearing stresses, and these stresses are
always coming along with this particular bending
stresses and our main focus is on that only.
So, you can see here this particular diagram
as we discussed, we have simple structure
of a particular kind of strain being in which
this cross-section is like that, and on that
particular we have this particular section
influenced by shearing force and on top of
that this cross-section. After this you know
like the delta x segment, we have the cross-section
or at this particular shearing forces are
F plus delta F, and then you see we have the
bending moment which in the sagging form.
So, M and this is at this particular side
after this delta x, we have M plus dM.
So, with that our main focus is that you know
like that will happen along with the neutral
axis and on above. So, this is the neutral
axis which you see you know like our main
plane of the matter of this mean plan is concerning,
or which our main focus is coming on this
particular way. So, that is what will happen,
and how these fibers will react due to this
F, and this F plus delta F along with this
way or we can say in sagging form that M on
this particular way and M plus dM will act.
So, if you see those things, then you will
find that the resultant stresses are because
you see here not only the bending stresses
are there, but along with this we have the
shearing stresses. So, that means, you see
whatever these things are coming, these are
coming with the resultant part. So, we have
the resultant stresses of these particular
things. So, resultant stresses are there,
and you can see here they are coming in the
triangular form at this particular point and
at this particular point, the triangular form.
So, we can say that the resultant stresses
the sigma at this point, and here it is sigma
plus delta sigma along with this point. And
this side you know like just to be compared
with the other side in the sense that this
small element is in the equilibrium position
under the action of the shearing force as
well as the bending moment.
So, we need to concern that you see when it
is coming along with this x and y axis, so
what are the stress components along the x
axis, what are the stress components along
the y axis, and how this twisting will take
place? Due to this twisting and shearing forces,
what kind of interactions are there? So, now
if you focus on this element, let us say the
y distance is there from this neutral axis,
and we have you see the AB element. So, now
you see of this fiber our main focus is that
now if you are taking a small element dA,
you can see this particular part. So, we have
a small element on which the stresses are
acting.
So, when the stresses are acting or just the
clear cut you know like the description is
there, it has the area dA is there which has
a distance from you know like the neutral
axis y, and the section modulus is Z for this
kind of thing. So, now with this particular
width of the cross-section and width is dA
area of this small element, and with these
stress components from the left hand side,
the sigma from right hand side, the sigma
plus delta sigma, and under all these forces,
the shearing forces and the sagging moments,
these objectives in the equilibrium portion.
So, now our main analysis will come that what
will be the resultant bending stresses are
there and what will be the resultant shearing
stresses are there to calculate those you
see and to put the relation in between those
things, certain assumptions are always needed
because to avoid the complication in the system.
So, for that you see first of all the first
assumption is that stress is uniform all across
the width of those thing. That means, you
see you know like it should be parallel to
the neutral axis. If it is not parallel, then
you see you know like the fibers are playing
in the different way and then, you see whatever
the stresses are coming on the fibers due
to these combinations of the shear stresses
and the bending moment all together, it is
different. So, that is why you see first of
all the first assumption is very important
that whatever the stresses are occurring or
inducing in this particular beam, it has to
be uniform all across the width. That means
you see they have to be you know like parallel,
just parallel to that neutral axis. This is
one.
The second is the presence of the shear stress
does not affect the distribution of the normal
stresses. That means, you see you know like
since we have shear stress presence, and along
with that there is also this bending moment
and due to that the bending stresses are there,
but whatever the presence of shear stresses
are there, they will not affect in any way
to the distribution of the normal bending
stresses. So, normal bending stresses are
always coming as per own concept. That is
what we can again apply the same theory, the
sigma by y which equals to M by I equals to
E by R in this equation also, and it may be
noted that the assumption number two cannot
be rigidly true. Sometimes you see as an existence
of shear stresses will cause distortion of
the transverse plane, and that is what you
see you know like which will not be exactly
there with the same plane. That means, you
see once we apply, once we remove the load,
it will not be exactly in the same plane before
and after bending.
So, sometimes this assumption two will not
be valid for certain cases, but in general
cases, we can you know like agreed upon those
things that these are valid for certain things.
And in the above you see you know like all
those figure which we described in the previous
one that we have simple segment, and we cut
one portion or which one portion is affected.
One side is affected by sigma and one side
is affected by sigma plus delta sigma.
So, for those you see you know like for the
two main transverse sections are there which
we concern which has a distance of delta x
apart from one point to another point and
the shearing forces and the bending moment
as we discussed, they are at one point. It
is F and in this direction and another point,
we have F plus dF in this particular direction.
Then, you see we have the bending moment also.
At one point, it is M which is going in this
direction and M plus dM which is going in
this direction. So, we you know like both
of the sides are being influenced by both
shear forces as well as the bending moment
in that.
Now, due to the shear stresses on the transverse
plane, there will be complementary shear stresses
are always be there on the longitudinal planes
parallel to the neutral axis just to balance
the equation. Otherwise, you see what will
happen if you know like because of this shear
forces. And because of these things, the bending
moment, the sagging bending moment, we cannot
you know like gear to exact equilibrium position
because if for any analyses, this is a must
condition that the equilibrium position must
be coming within those things, and then only
we can apply all those equations. So, for
that you see there is complimentary shear
stresses are there just to balance these shear
stresses acting. So, to counter the shear
stresses or we can say the complimentary shear
stresses are always there on the longitudinal
planes, they are which are just parallel to
the neutral axis and they will form always
you see just to balance these equations.
If I am saying that tau is the value of the
complimentary shear stresses which are coming
you know like in the transverse shear stresses
at any distance of Y 0 from the neutral axis
Z, then from the neutral axis, then we can
say that it is one more component which has
to be defined which is Z which is nothing
but equals to the width of the cross-section
at any position. So, now you see we have two-dimensional
parameters. One is where the tau is acting.
That means, you see where the complimentary
shear stresses are acting and that is now
at the distance of Y 0, and another one is
from the neutral axis and another one is the
width of the cross. This width is there of
any cross-section which we are defining by
the Z like it is absolutely based on what
kind of cross-section which we are using.
So, the Z value is varying.
A is the area as I told you like in the cross-section
cut off by the line parallel to the neutral
axis, and Y bar is the distance of the centroid
of the area from the neutral axis. So, we
found that these are some of the geometrical
parameters which can be analyzed for any kind
of numerical problems while we are dealing
with those things. So, here you know like
this tau, which we found that the complimentary
shear stresses are there which are always
inducing in that kind of structure, whenever
even the shear forces are there or we can
say that these bending moments are there for
that kind of thing. So, here you know like
with those components as we discussed for
this left hand side and right hand side for
this small element, again we found that sigma
and sigma plus delta sigma which is you know
like the resultant stresses are there due
to shear force.
Bending moment are just normal stresses on
the element on the area dA at the two transverse
sections as we have shown in the previous
diagram, and then you see if we want to find
out the difference of the longitudinal forces.
On one side we have the force sigma into dA;
on another side we have the force of sigma
plus delta sigma into dA. So, if you want
to find out the differences of the longitudinal
forces which is always equal to delta sigma
into delta dA for the small element, and you
see here we just want to find out this quantity
is d sigma into dA is the quantity which is
sum up over the entire area. This area A is
in equilibrium with the transverse shear stress
tau on a longitudinal plane of area which
has Z into dx. That means you see you know
like our main focus is how these stresses
are being setup, and one side you see we have
you know like this because our total thickness
which we have assumed is delta x.
So, at one side we have this sigma, at one
side we have sigma plus delta sigma and then,
you see what will be the total you know like
the resultant longitudinal forces are there
which is being setup due to the this shear
forces as well as the bending moment. So,
that that is why you see it is coming like
that and then, after focusing on those things
what we have is this tau z into delta x which
is the resultant part is there the longitudinal
forces.
So, tau which is coming due to the complimentary
shear stresses is the width, the Z is the
width there and dx which is the total thickness
which we are assuming. So, this tau which
is stress into this area will give you the
force which equals to the integration of d
delta into dA because you see these are the
longitudinal transverse forces are coming
due to the resultant forces, difference is
there. So, if you integrate for entire region,
then you will be ending up with the integration
of d delta into dA. So, from that particular
bending theory, again we can simply apply
the sigma by y which is equal to M by I. So,
from that we have the bending stresses sigma
which is nothing but equals to M into y divided
by I or we can say that since this is on the
first section.
On the second section, you see we have sigma
plus d sigma. This is the total stresses on
the other side and due to that we have the
bending which is coming on the sagging way
M plus delta M into y divided by I. So, from
those above two equations, we can conclude
that d sigma is nothing but equals to delta
M into y divided by I. So, here you see we
have shear stresses tau into Z into delta
x equals to integration of d sigma into dy
dA, and we have you see the total another
stress formula. This d sigma which is nothing
but equals to delta M into y divided by I.
So, if you compare both of the equations,
then you will find that these stresses are
the resultant stresses coming due to the shear
part of the shear forces, that is tau is there
in one figure and due to the bending moment,
delta M is there. That means you see you know
like whatever normal and shear stresses are
coming, they are the resultant part due to
these two portions.
So, you can again visualize those equations
from this particular figure, and this figure
says that we have a small rectangular section
like this cross-section, and the total length
is delta x. And for that you see this neutral
axis passing from this and this is our main
strip of you know like which our main focus
is that how the stresses will play. So, out
of one side of that, we have sigma into dA
and on another side, we have sigma plus del
sigma into del A.
So, you see here how they will balance. So,
exactly you see we discussed this particular
section that this is there and then, we have
the distances from you know like the neutral
axis of this particular area that is the y
distance is there, and this y 0 is the initial
distance, or we can say y bar is there that
we define for which you see this Ix axis is
there is nothing but equals to integration
of y square dA. So, for that you see you know
like along with those neutral axis and along
with this particular axis, since both are
coming altogether you know like in the same
way, and there is a combined effect of shear
stresses as well as you know like the shear
forces as well as the bending moment.
So, we can simply visualize that actually
when these fibers are being influenced by
both of the way, then the resultant is coming
from this side sigma into dA. That is the
force is there and on this side, sigma plus
delta sigma into dA from both the side. So,
the resultant since we are saying that this
and this small element is in equilibrium position
along with this particular you know like whatever
the forces are coming. So, they will you know
like cancel each other, and that is why you
see complimentary shear stresses also come
in tau which are there at all across the parallel
to you know like this cross-section, or we
can say that the parallel to the fibers, and
that is what you see we know like we assume
in the previous section.
So, here if you substitute those things which
we got d sigma is delta M into y by I, and
from the second equation by you know like
the basic bending, this equation and the second
part you see the first thing was there due
to this tau into z into delta x which is equal
to you know like integration of this d sigma
into dA. So, now if I replace this d sigma,
then what we have? We have this tau into z
into dx equals to integration of delta M into
y divided by A into delta A. So, now you see
here you know like if you consider this particular
equation, then you will find that we have
both of the components, we have shear stress
component, we have the bending component and
due to that you see you know like the result
is coming.
So, if you analyze those things, then dM by
I simply taken out, and we have you see this
tau into z into dx equals to delta M by I
integration of y into dA, or you see you know
like just go to the main relation between
shear force and the bending moment. We have
shear force F equals to dM by dx. So, with
that you see you know like again we can simply
put those things here. So, by keeping those
values here, this F is equal to dM by dx.
We are ending up with tau which is equal to
F. Shear force is there divided by I into
z because z will come, this delta M and delta
x will be giving you the F. The F will come
here I into z integration of y into dA, but
from the definition of integration of y into
dA as we discussed in the previous section,
which is equal to A times of y bar.
Again you see with this y integration of y
into dA is nothing but the first moment of
area because you see in the previous section,
we discussed about the second moment of inertia
that is the integration of y square into dA,
and the second moment of inertia will give
you the real bending feeling. Here you see
you know like we have the first moment of
area that is the integration of y into dA
of the particular shaded portion which is
there, and y dash is as I just told you in
the previous diagram that it is nothing but
just the location of the centroid from the
neutral axis.
So, you see how much distance is there like
the centroid from this neutral axis. If they
are common, then you see you know like this
particular first moment of area, it is just
you know like coincide with each other. We
can calculate correspondingly, but here you
see whatever the distance is there of the
y bar which has you know like the main distance
from the centroid to neutral axis always coming
up in the co-operation of this first moment
of area. So, with that you see you know like
we can simply put this A into y bar here.
So, we are ending up with shear stress formula.
Shear stress is nothing but equals to shear
force F into the total area A into y bar which
has the distance of the centroid from the
neutral axis divided by I into z, where z
is the actual width of the section at which
the position you know like at any position
is there, and tau is you know like this complimentary
shear stress are there which can be easily
calculated, and I is the total moment of inertia
above the neutral axis.
So, now you see if you focus on this formula,
then the shear stresses mainly you know like
is coming due to the shear forces that what
are the shear forces are there in what direction
this shear forces are there, and you see what
is the affected area under which these you
know like the shear forces are acting and
then, you see again the key feature is that
what is the exact distance is there of the
centroid from the neutral axis. Then, you
know like this total moment of inertia I is
key feature because what kind of the cross-section
which we are taking, correspondingly you see
you know like the shear stresses are coming,
and then another part is you know like what
will be the total width is there of the cross-section.
So, the dimensional parameter of cross-section
is very important to visualize the shear stresses
in those things. So, now I would like to conclude
this chapter. So, you know like in the first
section of this chapter, we discussed about
when a beam is under the combined load, then
how we can you know like the bending moment
as well as you know like the bending stresses.
So, in that you see we found that if you know
UDL is there and the point load is there,
then how we can segregate both the issues
and then, we need to calculate you know like
the different sections. The maximum bending
moment like you see you know like in case
of point load, we found that wL by 4 is there
in case of you know like this UDL wL square
by 8 is there, and then we can calculate the
maximum like the shearing stresses that is
you know like this M by M maximum divided
by I into this Y maximum.
So, you know like this kind of situation,
it is pretty easy to analyze and then, in
the middle portion you see our main focus
was there that whenever a beam is under the
action of you know like shear force as well
as bending moment, then the main assumption
of this bending like once you remove the load
or once you remove this bending moment, then
the situation is exactly same as the previous.
That means, you see you know like the fibers
are having the same kind of plane of the section
as before and after the bending. So, whenever
this shear forces are acting, then this situation
will not remain exactly as the previous case.
So, that is what you see this theory is sometimes
uniform and non-uniform bending. In non uniform
bending, sometimes it is not valid, but you
know like it depends on what kind of structure
which we are using. Based on that you see
this sigma by y is equal to you know like
M by I equals to E by R. It can be valid for
uniform and the non-uniform bending part.
Then, in the final part you see we discussed
about the shearing stresses of a beam. In
this you know like some of the assumptions
which we made that whenever you see the forces
are applying on this particular you know like
the beam, then how we can you know like put
the cross-section of this particular beam
and how we can analyze for the shearing forces.
So, these two assumptions sometimes you know
like it is valid sometimes depends on what
kind of structures or cross-sections which
we are using. These assumptions are valid
and then, you know like in the last section
of this, we found that the shearing stresses
are always like they are the resultant stresses
coming due to shear force as well as the bending
moment, and you know like if you want to calculate
the shearing stresses, then it has you know
like the total component of shear force F
into area into y bar which is the distance
of the centroid from the neutral axis divided
by I which is like this total moment of inertia
into the total whatever the actual width is
there on the cross-section.
So, in that shearing force, the shearing stresses
if you want to calculate, it is highly sensitive
to first of all that what is the cross-section
of your like the objectives. That means, you
see whatever the beam which we are taking,
what exactly the cross-section is and then,
the second thing which is very important that
how you are you know like configurating those
things means actually if you are talking about
I section or if you are talking about the
rectangular section, then what is our affected
area on which our main study is. So, based
on that you see you know like we can simply
calculate the shearing stresses. We can calculate
the bending stresses. This is sigma which
is nothing but equals to sigma by y equals
to M by I or equals to E by R.
So, based on either that equation or this
equation tau which is F a y bar divided by
I z, we can simply calculate the bending stresses
and the shearing stresses for a beam which
is under the influence of shear force and
the bending moment. So, you see here this
in this section, we simply you know like go
with the simple assumptions and we simply
put certain shearing stress equation, and
the bending moment equations for bending stress
equations.
In the next chapter, our main focus is that
if you have different section as I discussed
like if you have a circular cross action,
if you have this I section or if I have a
rectangular cross-section, then how can one
can be calculated, you see this shearing stresses
and the bending stresses for that, and which
you see if I am just taking this I section
or any section, then which of the area is
having you know like the maximum shear stress
or the bending stresses. Then, how this interaction
is there you know like this bending stress
and shearing stresses, and what is the real
relation between those stresses. These are
our main matter of concern and that part you
see we are going to discuss about different
kind of cross-section of a beam. Thank you.