Solid Mechanics-materials - Mechanical Engineering Video Lecture - Mechanical Engineering | Best Video for Mechanical Engineering

Hi, this is Dr. S. P. Harsha from Mechanical
and Industrial Engineering Department, IIT
Roorkee. I am going to deliver my lecture
27 on the course of Strength of Materials
which is developed under the National Programs
on Technological Enhanced Learning, NPTEL.
Prior to start this lecture, I would like
to refresh you know like the previous concept
which we discussed in the previous lecture.
We discussed you see you know like that if
a beam is under the combined loads like you
see if we have the point load and the U dl
plus, if we have you see you know like that
we have the U dl plus the triangular element,
then how we can analyze those things.
So, in the previous lecture, you know like
that at which point the maximum bending moment
is there, at which point the minimum bending
moment is there, and what is the relation
between shear force and the bending moment.
That kind of discussion we made in the previous
lecture, and we found that there is a point
where there is a change in the bending moment
diagram from positive to negative. That means,
you see when there is a change from the sagging
to hogging of the bending, the bending moment
diagram, then this point is known as the point
of contra flexure. That means, you see you
know like we have the drastic change is there
from the positive sign of bending moment to
the negative sign of bending moment, so that
you see we define and even actually we found
that at those points, where the drastic changes
are there in shear force diagram, we found
the maximum bending moment.
So, this kind of relations even we derived
in the previous lectures that when F is equal
to dM by dX. Then what exactly you see whatever
the changes are there in the bending moment
with respect to the domain x, it gives you
shear force and if we are keeping shear force
0, then the bending moment will remain constant.
M will be constant because dM by dX is equal
to 0. So, that discussion you see we made
in the first part of our lecture in the previous
class, and then you see we derive a simple
bending theory that with certain assumptions
that you see like when the beam has to be
in straight form and the plane should be symmetrical
towards the longitudinal axis, and also you
see the plane should remain plane before and
after the bending. And whatever the material
which we are using for the bending, the beam
material, it should be homogeneous isotropic
material.
So, this kind of assumptions we made and then
you see we concluded that actually you know
like that if we have the straight beam and
if we have the curved beam, there are certain
portions which are affected by the stresses.
That means you see when there is a straight
beam, no bending moment is there. Then you
see all the fibers of that particular beam
is you know like just existed in the symmetrical
way, but when you see you know like a bending
moment is applied at the two extreme corners.
Then the upper fiber is always you know like
experienced by the tensile forces, and the
tensile stresses are occurring on those fibers
while the lower fibers or we can say the bottom
fibers because it is just like the curved
part.
So, the bottom fibers you see the experienced
by these compressive forces, or we can say
we have the compressive stresses, but within
those tensile and compressive, there is intermediate
layer on which there is no impact of this
shear stresses, and if we will focus those
points you know like if we just find those
trajectory of those particular points, these
trajectory or these locus is known as the
neutral surfaces. We can say actually whatever
the axis which is forming on these particular
surfaces is known as the neutral axis. So,
neutral axis is that axis you see which you
can say that whatever the impact is there
on the bending moment, it has no impact on
that particular thing because there is no
shear, there is no deformation and there is
no strain component all along this particular
line.
That is what you see we define the neutral
axis and since, it is a curvature part, you
see we define that actually if you want to
define the radius, then always starting from
the locus, this radius of curvature of the
locus to this curvature path. So, whatever
the distance will come that distance is known
as the R radius of curvature. So, you see
now our reference point is the neutral axis,
and now you see whatever the things will be
going to happen with respect of the bending
moment and all those things, our main focus
will be there on the neutral axis to define
the neutral axis and then correspondingly
you see you know like whatever the changes
are there, we need to define. So, that kind
of discussion we made in the previous lecture.
Now, you see in this lecture, we are going
to again analyze those things that actually
what will happen you see if we have a different
cross section of the beam irrespective of
you see if we have the rectangular cross section
or I section or whatever, then how we can
find out the neutral axis for those sections.
First, we will also go for the numerical exercises
and then you see the key feature is that actually
what will be the bending stresses in that
because still now you see we discussed only
about the bending moment that what will happen
to the bending moment, what will be the relation
between the bending moments and shear force
diagrams.
So, all those bla bla, these bla bla discussions
which we made only with that shear force and
bending moment diagram, still we did not discuss
about what will happen when the shear forces
are existing within the beam element, and
then due to that what will happen with the
bending stresses, or how they induce the bending
stresses within those components, and then
how these neutral axis will react when the
bending stresses are there, and what are the
potential area within the beam when it is
under subjected by various forces through
which we can say that, yeah this is the potential
area which the maximum bending stresses are
there. This is the minimum bending stresses
are there.
So, if you want to design the beam irrespective
whether cantilever beam or simply supported
beam, you can simply you know like focus on
those potential areas and use the factor safely
higher recording to the safety reasons. So,
this kind of you know like discussions which
we are going to discuss in this particular
first lecture. So, first of all as we discuss
the bending stresses in the beams or the derivation
of the elastic flexure formulas because you
see first of all we need to see that how the
bending stresses will form and then how it
will be distributed all along the entire span
of this beam.
So, in order to compare the value of bending
stresses developed in a loaded beam, first
of all we need to consider the two cross-sections
of the beam like HE and GF originally parallel
because there is no bending moment and then
they are absolutely parallel to each other.
You can see the first diagram here on the
screen that these two you know like we are
taking these two elements in which these two
lines are very, they are simply parallel to
this EH and GF. And you see you know like
these as we discussed already in the previous
chapter that the middle section which is passing
from you know like entirely just you know
like this neutral surfaces. This is nothing,
but our neutral axis and we just want to focus
that actually if we take element, let us say
from distance y from the neutral axis, this
AB, then what will happen is means how the
fibers will behave when it is under subjected
by bending moment. That is our prime focus.
So, for that again you see what we are doing
here since it is plane structure in which
this first figure 1 A plane structure in which
there is no bending moment, there is no shear
forcing is applied. A straight beam is there.
Well it means equilibrium manner and you see
for that we have simple structure. There is
no deformation you can see on the diagram,
but when a beam is to bend, that means if
your bending moment is applied, it is assumed
that this section remain parallel. That means,
you see H dash, E dash and G dash, F dash,
they have to be there in the parallel because
you know like we assume that even the plane
should remain plane even before and after
the bending.
So, with that particular assumptions, we can
assume this kind of theory and the final position
of the sections are still together you know
like the straight lines are there corresponding
the previous lines. And they also you know
like subtended by the angle theta, and you
can see this particular diagram figure 1B
as we discussed previously also that when
you apply the bending moment at the extreme
corners, it will bend. And then you see the
extreme top corner will be extended due to
the tensile forces, and the tensile stresses
occurring on this particular surface, but
you see this particular surface which is under
the compression, we have the compression stresses
on these particular surface, but we have the
neutral axis line, this A dash, this B dash,
this C dash and D dash. There is no strain
component on that. That means you see C dash,
D dash is equal to CD. Just keep this thing
in your mind because you see there is no deformation
occurring along this particular line. This
is the neutral axis is there.
So, C dash, D dash which we are denoting in
the figure 1B is exactly similar to this figure,
this point CD in figure 1A, but there is a
change in A dash, B dash to AB because you
see these certain deformations are there,
the extension is there. So, we have the tensile
deformation and this is due to you see you
know like we have a kind of you know like
some extension is there, and this extension
we can easily form because we know that this
distance is R, and you see this distance because
the neutral axis distance as we defined that
we can always you know like suppose to define
the radius from this locus point to the neutral
axis.
So, this distance is R and also, we define
in the previous section that actually this
AB distance, this AB section has a distance
from CD as y. So, this 1 is y. So, the total
distance if you want to find out for the curvature
part is nothing, but equals to R plus y and
then this theta is there. So, we can define
with the radius and R definition for that,
and then you see you know like we have one
more thing that we have this H dash and G
dash which is also in the extensive part.
So, we have the maximum deformation in tensile
way on these particular points, and this is
the minimum deformation at these points you
know like or we can say the 0 deformation
at C to D point, and you see like again some
compressiveness is there of this E dash, F
dash points in this particular figure.
So, these figures you simply justify that
actually whenever there is a kind of bending
moment, then how you know like the fibers
are playing in this particular way, and how
the tensile and the bending this compressive
stresses are coming due to the bending only.
So, that is why you see we are saying that
this normal stresses which are coming due
to the bending action is always the flexural
stresses.
Now, you see consider the fiber AB of the
material, you see as I told you in the intermediate
section, if you are just focused on the fibers
of that particular path of the AB line at
the particular distance y from the neutral
axis, then the bend will be stress to A dash
B dash. That means, you see now due to the
bending action, there is a deformation in
this beam right from AB to A dash B dash.
So, now our main intension is that actually
how the deformation can be measured, and how
we can you know relate this strain to the
stress component with the using off because
you see the important feature here is whatever
the deformation is there, due to these bending
moment is always the elastic deformation.
And that is what you see we can straight way
apply the Hooke’s law for that in which
the stress is proportional to strain, and
since here the bending moment is applied.
So, we have the bending stresses which is
correspondingly you know like proportional
to the strain component. So, here with that
particular theory, again what we are doing
here is the strain in the fiber AB which is
our main focus is on fiber AB only that actually
how the strain will be coming and what kind
of deformations are there along that particular
path. So, strain in the component or we can
say the fiber of AB which is nothing, but
equals to change in length divided by original
length, and the final distance or final length
of this fiber AB is A dash B dash earlier
distance, or we can say earlier length of
the fiber AB was AB. So, A dash B dash minus
AB divided by A AB.
So, that means, you see whatever the original
path is there means how much deformation is
there divided by how much original length
is there. This will give you the total component
means how much deformation is there, and this
deformation generally we are measuring in
terms of the strain. So, strain definition
is like that, but as far as the neutral axis
is concerned, we already assume that AB is
equal to CD because you see you know like
there is a straight part. When there is no
bending moment is there, the straight regions
are there. So, AB is just cutting exactly
as a CD.
So, there is no difference in that same distances
are there. So, because there is no bending
moment, there is no deformation in that particular
you know like either the neutral axis and
above. So, we have AB equals to CD, but CD
equals to C dash D dash because the main thing
is that CD and C dash D dash are the points,
which are always lying on the neutral axis,
and as you know like we know that there is
no strain or there is no deformation is coming
on this particular neutral axis. So, we can
straight way you know like conclude that whatever
the distance is there before bending moment
is applied and after bending moment is applied
on the neutral axis. They have the same distance
without any deformation or without any strain.
So, we can say that CD is equal to C dash
D dash and if we apply those things here,
what we have? We have the strain component
is nothing, but equals to A dash B dash which
is the previous thing and AB is equal to CD
and CD is equal to C dash D dash. So, we have
A dash B dash minus C dash D dash divided
by C dash D dash. So, now you see here our
main conclusion for the strain component is
just focused on what will happen when the
strain component is there. That means, you
see A dash B dash is nothing, but the total
deformed path along the line A, B. C dash
D dash is the new location for the points
on the neutral axis, and you see you know
like when both are comparing, then we can
get the strain component.
So, since CD and C dash D dash as I told you
like they are on the neutral axis, and it
is assumed that the stress on the neutral
axis is always you see line 0 and since, there
is no stress component is there on that particular
part. So, there is no strain, no deformation
is there on that particular part. Therefore,
they would not have, they would not be having
any strain component on the neutral axis and
there is no deformation there. So, we cannot
measure anything. So, whatever the points
are there before bending moment and after
bending moment, they have the equal distances
on the neutral axis.
So, for that now our main intentions to find
out with the use of the curvature theory that
what will be the distance of A dash B dash
and C dash D dash, so that we can straightway
get the strain and once you have the strain,
then you can easily go with the stress component
with using of the modulus of elasticity. Because
whatever the deformation which we are assuming
for this particular kind of analysis is always
the elastic deformation, and all those Hooke’s
law is valid for that, and you see all those
elastic modulus of Young’s modulus of elasticity
or whatever the components which are valid
for the elastic deformation, they are also
valid for this kind of analysis. So, come
to the main point that the strain is nothing,
but equals to A dash B dash minus C dash D
dash divided by C dash D dash. So, if you
focus on the A dash B dash in the previous
diagram, then we will find that the total
distance for A dash B dash was R plus y.
Now, you see this is the curvature path was
there. So, if you multiply this arc into this
particular theta, then you have the total
distance. So, R plus y into theta, that is
A dash B dash minus C dash D dash, straightway
you see it is on the neutral axis where the
distance is R. So, R theta will give you the
total distance. So, R plus y into theta will
be A dash B dash minus R theta is the C dash
D dash. So, if we put these things in this
particular formula, the strain formula, then
we have R plus y into theta minus R theta
divided by R theta and if we generalize those
things, then we have the strain equals to
y by R, where y is the distance of the fiber
from the neutral axis.
So, you see here always it is above you know
like neutral axis, and you see we just want
to observe the effect of this bending moment
on those fibers from the neutral axis, they
have the distance of y. So, now you see as
I told you that it is the elastic deformation
where the Hooke’s law is valid. So, we can
again use the stress is proportional strain
or you see stress by strain will give you
Young’s modulus of elasticity E, and then
you see if we are keeping those values there,
then what we have is this stress. The sigma
is always denoting the stress. So, sigma by
E will give you the strain and you see in
the previous formula, we just derived that
the strain is nothing, but equals to y by
R.
So, we have now real formula which is the
basic formula for any bending stresses which
are occurring in any beam kind of you see
irrespective whether it is a cantilever beam
or simply supported beam. The bending stress
is sigma divided by E which is equal to y
by R, or we can say that this bending stress
sigma is nothing, but equals to y by R into
E, or we can say that actually E which is
Young’s modulus of elasticity R which is
the radius of curvature into if you multiply
y as, so as y is you know like just increasing,
definitely you have more bending stresses.
Obviously we have because you see if you compare
the realistic way this bending of beam, then
we will find that the upper surfaces because
as you are moving from upper direction from
the neutral axis, we have more and more stresses,
and this will you know like conclude from
this formula also that sigma is equal to E
by R into y.
So, as you increase y, again you have more
and more bending stresses in that particular
beam, and whatever the fibers are there on
the extreme corner, they are experienced with
the maximum bending stresses in that particular
way, and we will describe many more things
in the further lecture. So, you see here now
again if you focus on just distance at y,
but if we cut the section of dy at this particular
y, so you see we have just you know like unsymmetrical
section and this you see for that we have
a neutral axis for which is passing from this
particular way, and we have a segment for
which actually our main studies focused that
we have this particular small segment, and
it has you know like the total width is dy
which has a distance y from the neutral axis.
So, now you see what will happen when the
bending moment is applied on that particular
section, and how you see the area will affect
of that.
So, considering this particular arbitrary
cross section of beam, as you know like shown
in that main you see now what will be the
strain on the fiber at distance y. That means,
you see you know like what will happen when
the bending moment will apply, how these fibers
will play under the action of this bending
moment? Then you see you know like again we
would like to start from the same formula
that when it is under the action of bending
moment, then we have the bending stresses
sigma which is equal to E by R into y, and
you see if the shaded area which I just shown
you and that dA. Now, if our main focus is
on that, what will happen? You know like when
you apply the bending moment, then you know
how fibers will play their key role in set
up of the bending stresses in this particular
strip dy.
So, for that you see again our main focus
is that actually what will be the force on
that particular strip. So, the force is nothing,
but equals to whatever the stresses are there
into the area and area we consider dA. So,
sigma into dA will give you the force, or
we can say sigma since already we have described
that when it is under the action of bending
moment. The bending stresses are there. So,
sigma is the bending stress which is equal
to E by R into y into delta A. So, now you
see here we have the force. Once you have
the force, then you can easily get the moment
for those things.
So, for moment about the neutral axis, this
will give you force into the distance because
you see what we are doing here is, we are
simply taking strip which has distance y from
the neutral axis. So, we can easily compute
the moment you know like along with this particular
neutral axis because we have a distance. So,
when we apply the force and the distance is
there, this will give you the moment. So,
F into y will be the moment or you see once
you have the force, you can easily replace
that force E which equals to E by R y into
del A into you see this distance is y. So,
what we have? We have the moment which is
equal to E by R y square del A.
So, now you see if I just want to compute
this moment for whole of the section because
till now you see we discussed about a small
strip which has you know like the area elemental
area dA which has the total you know like
the width of the strip is dy. So, for that
you see this moment is there which is equal
to E by R y square into del A, but now it
is sum up all those small element which has
a distance you see certain distance from the
neutral axis, and if we sum up with the summation,
these thing we have the total moment which
is there you know like moment applied on that
particular section, and due to that we have
the bending stresses. So, for that this total
bending moment over the cross section is summation
of E by A y square dA or since, thus total
summation is always focused on y square into
dA.
So, E and R are the constant one. As we discussed
E is the modulus of elasticity which is a
property of the material which you are using.
So, Young’s modulus of elasticity will come
and the curvature is always focus that actually
how what the dimensional parameter which we
are using for that. So, if you just take it
out once, then we have the summation of y
square dA, E is the total you know like will
give you when you sum up those things, the
small element, it will give you a real feeling
about the moment.
So, once you do that, then you have the total
moment for that section and you see in this
there are two components. One component E
by R which is as I told you is subjected to
like what kind of material and the dimension
which you are using, but another component
in that is the summation of y square dA. This
is a specific property and this property you
know like of a material is known as the second
moment of area. That means, you see you know
like it is a very different thing because
it has the square term, y square and it has
area term. So, you see here we are saying
at different way. You see first area, it is
second moment of area of the cross section
and generally, we are denoting I. So, I is
equal to summation of y square dA.
So, now if you are keeping those things, then
we have the total moment y M is nothing, but
equal to E by R. That was the first component.
Second component was the summation of y square
dA which we replaced by I. So, we have new
equation of bending for you know like any
kind of cross section which is equal to M
by I equal to E by R, and you know like E
by R, we already concluded that sigma by y
is there in the first equation.
So, the total you know like the bending equation
for this generalized bending equation, I should
say nothing but equals to Sigma by y which
is equal to M by I which is equal to E by
R in which sigma is the bending stresses which
is coming due to the bending moment, and this
bending moment is M you know like, which is
always applied at the extreme end or you see
because due to the couple also, it can be
formed y is the distance which you know like
the fiber which we just want to find out this
bending stress, and the distance of that fiber
from the neutral axis is y. So, this y is
there, I is IE. I just told you that it is
the second moment of area which is nothing
but equals to summation of y square dA.
So, one can easily calculate those things
and E is the Young’s modulus of elasticity.
It is specifically you know like the property
of a material which you are using for that
elastic property is there and R is the radius
of curvature. So, with all this six points
or all the six properties, we can easily describe
generalized theory of the bending. Sigma by
y is equal to M by I which is equal to E by
R, and it is a very good you know like equation
for solving many of the numerical problems,
and this equation is known as the bending
theory equation, generalized bending theory.
So, whatever the structure which we are using,
if it is under the action of bending moment,
then this equation is valid if we are considering
the deformation up to the elastic one.
So, the above proof you know like has involved
the assumption. That pure bending without
any shear force being present because you
see like we have sigma, it is the bending
stress M which is the bending moment and you
see other sections are the dimensional parameters
like y I E and R and E is nothing, but the
property of material. So, all those components
if you conclude, then you will find that there
is no other stress component is presenting.
Only we have bending moment and the bending
stresses, and that is why you see we can say
that when the pure bending is there, there
is no shear force occurring within that.
Therefore, you see this term as the pure bending
equation as we discussed, and this equation
gives the description you know like about
the stresses which are normal to the cross
section. That means in the x direction. So,
that is why you see all these normal stresses
are there as we discussed that whenever bending
moment is there, they have the tensile and
the compressive stresses. They are the normal
stresses or we can say that they are the flexural
stresses. So, this theory which is known as
the pure bending theory has only the component
related to the bending moment, bending stresses.
There is no shear component is occurring in
that. We have the normal stress component
along with the bending stresses. So, now you
see in that we found that there are some special
components within this particular structure.
So, we would like to first focus on those
components.
First is the section modulus. In that you
see from the simple bending theory, whatever
we discussed in the previous equation sigma
by y is equal to M by I is equal to E by R,
from that equation you see our main intention
is that when there is you know like the bending
moment is applied on this particular beam,
then which area is influenced by the maximum
bending stresses or the minimum bending stresses.
So, for that you see here the simple cases
just give you the maximum stresses which can
be obtained at any cross section sigma maximum.
That is the bending stress maximum is equal
to M by I, whatever the bending moment which
you are applying I is nothing, but the summation
of y square dA.
So, these are the constant one, but the key
feature is this distance. So, y maximum means
you see you know like where this maximum bending
stresses are occurring means which fiber is
experiencing this maximum bending stresses
which can be easily calculated. Once you know
the maximum value or once you know the location,
you can get the maximum value. So, sigma maximum
is basically depending on y maximum. That
is the distance from the neutral axis. So,
sigma maximum is equal to M by I y maximum.
Second equation is that for any given allowable
stresses means you see we have these allowable
stresses for the bending of beam, and for
that you see we just want to find out the
maximum moment which is acceptable to any
particular shape of the cross-section we can
easily get because now our beam can be you
know like absorb this much amount of bending
stresses. So, we can get once we know that
you know like this is the allowable stresses
for this kind of beam.
So, for that we can easily get the maximum
bending moment and the formula for maximum
bending moment is nothing, but equals to I
by y maximum into sigma maximum. That means,
you see I which is you know like having all
the summation of the cross-sectional like
summation of y square dA, but you see what
will be the distance of the fiber at which
this maximum bending moment is coming. So,
first of all, this y maximum which will influence
this M, and second is how much bending stresses
are occurring on those fibers, the value of
that. So, that will also you know like depending
parameter is there for calculating the maximum
bending moment.
So, you see M is nothing, but equals to I
by y maximum into sigma maximum. These formulas
are very important to calculate the remaining
parameter because you see in the question
they can ask you that now calculate the maximum
bending stresses. So, for that first of all
you need to obtain that what will be y maximum
is there. Once you know the y maximum because
other parameters are very constant M by y
M by I. So, you see first of all you need
to locate the location y maximum and then
you can calculate the sigma maximum. That
means the maximum bending stresses. Similarly,
you see you can also calculate the maximum
bending moment with that formula as I shown
you I by y maximum into sigma maximum.
In the last one whatever you see you know
like these compressions are coming in that
particular curved beam of the strength, this
particular beam of any cross section we can
simply give with some relation and that relation
is coming with the section modulus Z. As we
defined earlier also that the section modulus
is always you know computing the geometrical
parameters like you see you know like if we
know that, yeah this is the location at which
the fiber is there which is experiencing the
maximum bending stresses.
So, once you calculate that one and once you
know that what will be I is there, you can
simply compute this section modulus, and then
you see whatever the bending moment is coming
which is always maximum focused on the section
modulus into whatever the maximum stresses
are coming. So, if you conclude all that part,
then you will be having M equals to Z into
sigma maximum, where Z is the section modulus
which is equal to I by y maximum. I is nothing,
but you know like the second moment of area
in which we can calculate the summation of
y square dA. Y maximum is the distance from
the neutral axis. So, we can calculate all
those things with the corresponding way.
So, you see here we have three different ways
to calculate, either the maximum bending stress
or maximum bending moment or also, you see
with the using of section modulus, we can
calculate you see how we can relate the maximum
bending moment and the maximum bending stresses.
Those issues are discussed here because of
you know like once you have the basic theory
of beam, you know that actually because of
the bending moment, we have the bending stresses
at different locations, but to get the maximum
value or the minimum value. We should first
identify the location that which fiber is
experiencing because in that particular if
you know the concept of the bending, then
you know that we have tensile stresses, we
have the compressive stresses, but we have
some of the section, which does not exhibit
any kind of stresses.
So, for within those limits, you see our main
intension is to get that which fiber is experiencing
maximum bending stresses, and what will happen
you see once you locate that part, how we
can relate you know like with the maximum
stresses or bending moment or bending stresses
with the section modulus Z. So, all these
you know like relations, one has to remember
and these relations are there in front of
you.
So, the higher value of Z, now you see we
have M is equal to Z into sigma maximum. So,
M is straight way depending on Z section modulus.
So, higher the value of Z for a particular
cross section, higher the bending moment is
there because it is directly proportional
to that which it can withstand for given maximum
stresses. That means you see if you want to
find out that what will be the bearing capacity
of the beam is.
First of all, our main intention is to find
out that how much you know like the maximum
value will come for the section modulus, and
once you know the section modulus, maximum
value you can say that we can go up to this
much maximum. The maximum value of bending
moment and the corresponding change is there
in the bending stresses. So, that is why you
see Z which is the section modulus is a very
important parameter to design you know like
this particular beam for loading conditions.
And now you see we would like to put one theorem
to determine the second moment of area because
you see in previous formula, you will find
that I is a very important parameter.
So, our main intention is that how to find
out those things because practically you see
if you are going for individual sections and
then sum up is not an easy thing because you
see we have simply linearly varying this uniform
structure or any kind of you know like this
kind of structure is there in which we can
simply put the integration from 0 to some
value. Then, we can get here. We have a different
uniform section in which the different segments
are there. So, it is not easy to get all the
small segment area, this y square dA and then
sum up those things.
So, here you see like we need to go with certain
theorems. So, here there are two different
methods. One method which I just discussed
that you know like there are two theorems
which are helpful to determine the value of
the second moment of area, and which is required
to be you know like used while solving simple
bending theory of the equation, the integration
of y square dA. So, first is summation of
y square dA, and second is integration of
y square dA which is helpful in which condition
this is the prime you know like the matter
of concern.
So, first of all we first focus on the second
moment of area in which taking an analogy
from the mass moment of inertia as we discussed
in the previous case, the second moment of
area is also defined as the summation of area
times distance squared from the fixed axis
because you see here what we have is y square
into dA. So, first we need to calculate if
we are taking any small segment.
So, what will be the area of that segment?
Once you know the segment area, then how much
distance is there of that particular segment
from the neutral axis square term. So, y square
into dA will give you this property arises
while we were deriving bending theory equation
and this is also known as the moment of inertia
because you see here we are using the same
concept of the mass moment of inertia in which
we are taking individual sections and sum
up those things. So, here also this moment
of inertia which we are calculating, the same
concept, alternate name is given to the second
moment of area because the first moment of
area is different than second moment of area.
It is the sum of the area times there distances
from the given axis.
So, if we are you know like if we are summating
the summation y into dA, then that is your
first this moment of area, but we are saying
that the second moment of area because here
we are multiplying this first moment of area
into distance, or we can say that we are doing
the summation of y square into dA. That is
why we can say it is the second moment of
this area or we can say that is a moment of
inertia. Both have the same meaning. Only
our focus is that actually what we are doing,
how we are dealing this problem and what are
the influencing parameters which are coming
within that particular you know like things.
So, that is what you see we need to cover
those things.
So, for that to calculate the second moment
of area, again we are coming with the same
structure that we have. Any structure in which
you know like just the O is the origin which
is coming from both of the axis x, x to x
dash y to y dash. Then, you see we have a
small segment which we are taking for which
there is a small element, dA is the area is
there and for that we have the distance R.
So, R can be you know like the polar coordinate
is there which can be again related with x
to y Cartesian coordinate.
So, it has a distance x and y. So, you say
that R is nothing, but equals to x plus y.
Sorry, square root of this x square plus y
square and then you see you know like once
you focus on this small element of area, then
once you know like we just have the bending
moment and then there is a fiber which is
influencing by the bending moment. So, for
this particular element area, how these fibers
will you know like deviating from their original
shape. Under the action of this bending moment,
it is a prime matter of concern. So, considering
any cross-section having the small elemental
area dA, we need to define what will be the
mass moment of inertia is there, because what
we are doing here is the same as we discussed
in the previous slide that because we are
summing up all those things, so it can also
be known as the second moment of area, and
is also be known as the mass moment of inertia
So, Ix equals to integration of y square dA.
So, in that again there is a term y square
is coming because it is a second moment of
inertia and dA is the total area which is
our focused area in that.
So, if we focus on this, the second moment
of area, we would also always see that it
is a kind of integration because we have a
regular structure. Now, the moment of inertia
about the axis which is you know like passing
through the O, as you know like I just told
you that x to x dash, it is passing or y to
y dash, it is passing from O and perpendicular
to the plane which you know like of the figure
like this one perpendicular to that is also
called the polar moment of inertia, or polar
moment of inertia is also you know like known
as the area moment of inertia. So, this is
the new term which is you know like the concern
of area is just perpendicular to this plane
of figure.
So, for that matter you see always whatever
the second moment of area is coming, the new
term is there for that which is the J which
is also known as the polar moment of inertia,
and it can be you know like calculated because
it is just focused on this. We have the R
of the distance for that which I have shown
you this R distance from O to our elemental
area. So, this integration of R square dA
will be equal to J or since, R is nothing,
but was you see x and y was there.
So, R square is nothing, but equals to x square
plus y square into dA, or you see if we segregate
those matters integration of x square dA plus
integration of y square dA will give you the
new terms of the second area moment. So, polar
moment of area is nothing, but equals to this
Ix plus Iy or we can say J J is equal to Ix
plus Iy. So, you see all three, one is along
the x axis, along the y axis and the third
one is just perpendicular to the plane of
figure. All three you know like this second
area of the second moment of area is coming
together, and this you know like relation
is coming you know like due to that is known
as the perpendicular axis theorem, and it
is just stated like that some of the moment
of inertia about two axis is x or y.
Whatever you see you know like which we are
concerning here is the plane is equal to the
moment of inertia about an axis perpendicular
to plane. That is the polar moment of inertia
that is J, but condition is that all three
axis must being you know concurrent. That
means, they must exist together in the particular
plane, so that we can say that Z is equal
to Ix plus Iy. So, this theorem is only valid
when they are you know like concurrent or
we can say they are just existing together.
So, in that we can say that Ix plus Iy will
give you J value that is just perpendicular
to these two planes.
So, you see here these two relations are there
with the polar moment of area for that kind
of section, but now you see here if we are
focused on the circular section, so you can
see the figure for a circular section. The
polar moment of inertia can be easily you
know like computed in just similar manner
that again we need to just concern on the
circular portion of that. So, what we have?
We have you know like a circular portion having
the diameter d, and you see our main focus
is with the small strip which has a thickness
dr and which is located at the distance of
r from the O.
So, with that you see you know like we can
simply calculate the circular strip area dA
which is nothing, but equals to 2 into pi
into r into this thickness dr. So, 2 pi r
will give you the total length, whatever is
there into dr will be the area. So, dA is
equal to 2 pi r dr for the circular section
and with that now we know that the polar moment
of inertia is nothing, but equals to integration
of r square dr.
Now, you have these you know like the total
symmetrical sections. For that we can simply
integrate those things within those definite
integrals 0 to d by 2 or 0 to r, I should
say. So, 0 to d by 2 r square into 2 pi r
dr because of the small segment, our main
focus is there in which the thickness of dr,
small dr. So, this one you see that is what
dA can be easily replaced by as we discussed
in the previous section. 2 pi r dr r square
is there and since, we are defining from 0
2 r, so 0 2 d by 2 pi is the constant one.
So, it can be simply taken out from that.
So, we have integration 0 2 d by 2 r square
dr, or we can say that 2 pi r you know like
since it is a cubic term is there from these
multiplications, so we have r to the power
4 by 4 0 to d by 2.
So, we have the polar moment of inertia for
circular section is pi d 4 by 32 and it is
just for you know like since it is perpendicular
axis theorem, it is coming. So, we can say
that J is nothing, but equals to Ix plus Iy
because we are simply concerning about the
rth value, but in the previous section, we
found that when you know like all three axes
are existing together and you see they have
the relation Ix plus Iy is equals to J. So,
by applying those things, what we have? Since,
it is a symmetrical section of both thing,
Ix and Ir always must be equal to the moment
of inertia about this particular diameter
because we are simply concerning about a symmetrical
geometry. So, for that we can say that whatever
Ix and Ir Iy is coming, they have the equal
value.
So, you know like I diameter because we are
considering about the diameter. So, I diameter
equals to J by 2 or we can say that the real
I is equal to pi d 4 by 64, and this for you
know like this is for the solid circulars
section in which we have the radius r. We
can say the diameter is d. So, for that we
can simply calculate the value. The second
moment of area that is I is equal to pi d
4 by 64, but if we have a hollow section in
which there are two diameters, the capital
diameter, the outer diameter and the small
d is the inner diameter. So, if we have this
kind of thing, then we can calculate this
J that is the polar moment of inertia is nothing,
but equals to pi d 4 D to the power capital
D to the power 4 minus small d to the power
4 divided by 32, Or we can say I which is
second moment of area is nothing, but equals
to pi D to the power minus D 2, the power
4 small d to the power 4 divided by 64.
So, you see here these two formulas are very
important for solid as well as the circular
shaft because once you know the value of J
and I, you can easily calculate the different
property described for the beam. Because you
have the circular cross-section of the beam
and you see you know like all those property,
the bending stresses, the bending moment,
they are absolutely focused on what the values
of I and J are. So, for that you see we need
to calculate those things and for calculation
of I and J, the parallel axis theorem was
very important to the perpendicular axis.
Now, you see we have the next theorem that
is the parallel axis theorem So, in that the
moment of inertia about any axis is equal
to the moment inertia about a parallel axis
through you know like which is passing through
the centroid plus the area times square. You
see what the integration of area into distance
whatever is there between the axes. So, you
see this kind of you know like theorems are
always useful if we want to find out the moment
of inertia at a particular section of the
total object because you see here we know
the distance. Once you know the distance,
multiply the distance by the area and you
have one component. Then, you see you know
like the moment of inertia which is for the
axis which is passing through from the centroid
or we can say the neutral axis.
So, now you have I plus you know like integration
of y into d. So, this will give you the new
moment of inertia about the particular axis.
So, if in this particular diagram which you
can see here, what we have? We have a particular
diagram in which G is you know like the center
of gravity through which we just want to find
out. If we are taking a small element, this
one which has a diameter dA and which has
area dA and the distance is y is there from
this axis. So, now our main focus is to find
out you know like that at the ZZ dash with
this particular section, what will be the
moment of inertia because we know this XX
dash this moment of inertia because it is
passing from the centroid. So, it can be easily
computed once you know these things, you know
the distance you can get the moment of inertia
by this particular formula.
So, here you see we have Ix is equal to integration
of y plus h, this square term into del A.
So, del A is the area because our main focus
is that one. So, with that here we are defining
the moment of inertia about the axis ZZ dash
because we know that XX dash, this moment
of inertia can be easily calculated because
it is passing through the centroid. Now, what
will be the distance there? The distance we
are assuming is h. So, with that y plus h
into whole square, this y plus h whole square
into dA will give you the Iz. So, once you
have the Iz, again you can describe those
things y square plus h square plus 2 y h into
dA and since, we know that y into dA because
you see y is nothing, but you know like it
is passing from this centroid.
So, y into dA integration will give you 0
because it is always you see there is no deformation
within that particular element because it
is passing from the centroid. So, with that
particular assumptions, now we have Iz equals
to integration y square dA plus h square integration
dA. So, integration dA is nothing, but you
see the total area which will give you, so
we have A and here you see integration y dA
is always we are saying that since it is passing
from this square, it is a second area moment
for the neutral axis, or we can say the centroid.
So, it is Ix plus A times h square.
So, you see here we have with the using of
this parallel axis theorem, we can calculate
this moment of inertia for any section which
has a parallel relation from this neutral
axis or we can say centroidal axis. So, for
that Iz is nothing, but equals to Ix plus
area which is the total area into h square
and Ix is nothing, but equals to Iz as I told
you which is passing from the centroid, and
A is the total area of the selection. So,
this was therefore circular section. Now,
if we have a rectangular section for this
one you know like if you want to calculate,
so for that rectangular cross section of beam,
the second moment of area can be easily calculated
for this.
So, for that the diagram is pretty simple.
We have cross-section here. You see the rectangular
cross-section. This is the total distance
is D in the vertical way and B is the width
is there for this, and the neutral axis is
passing from this or we can say the centroidal
axis is passing from this. It is showing by
the dotted line and this we are taking strip
on which our main focus is there which has
total width is dy in this particular direction
and the total distance of this strip from
neutral axis is y as we are using generally.
So, with this configuration, now our main
focus is to calculate that what will be the
second moment of inertia is there for that.
So, considering this rectangular cross-section
which has the area dA, and the thickness is
dy as we have taken and this breadth is B
which is located at distance of y from the
neutral axis, for that the second moment of
area is I. Generally, it can be easily defined
that this integration of y square d.
So, with that you see you know like the moment
of inertia at the neutral axis from minus
d by 2 d 2 because you see the neutral axis
is this centroidal part. So, it is coming
from minus d by 2 to d by 2 like y square
into B into dy because this is the area, the
width into the thickness of the strip. So,
B into dy is the area y square you see over
because it is a second moment of inertia is
there. So, with that minus D by 2 to D by
2 or we can say that since the B is the constant
one, there we are not changing any breadth
of this particular rectangular one. So, B
is taken out. So, minus D by 2 to D by 2 y
square dy or we can say that y cube y 3 and
when we are keeping those things, we are ending
up with B by 3 D cube by 8 plus D cube by
8, or we can say that when we are calculating
those things, we have D cube by 4 and outside
is 3.
So, we have BD cube by 12. So, here what we
have? We have the second moment of area about
the neutral axis is BD cube by 12 when you
know like the neutral axis is just passing
through the centroidal part and distance is
this D. And this is the width and if you are
changing those things because if you are taking
the neutral axis which is passing from B and
D is seemed parallel. This is a very special
case. In this case what we assume is, we are
taking you know like the dimension that the
diameter is this, and the neutral axis is
just cutting the diameter in this way that
B width is parallel. For that you see the
second moment of area about this neutral axis
is BD cube by 12, but in the other case, if
I am taking the same rectangular cross-section
and the neutral axis passing from the B which
is parallel to D, so in that case we have
I neutral axis is DB cube by 12. So, you see
here we need to be very conscious about those
things that what we are taking and how it
will come. So, you see again you know like
what we discuss that actually if it is passing
from those things.
The second moment of area for this particular
rectangular section which has the axis passing
from the lower edge of the section, it can
be easily found with use of the same procedure
which we are used from you know like 0 to
D, or we can say minus D by 2 to D in the
previous case. So, for that we have I which
is equal to B y cube by 3, 0 to D because
we are concerning about the whole diameter
from 0 to D 1. So, we have BD cube by D 3
because that kind of cross-section is symmetrical
along 0 to diameter.
These standard formulas prove very convenient
you know like in determination of this second
moment of area about the neutral axis I NA
to build up the section which can conveniently
divide into rectangular because you see in
the real structure, what we are doing here
is, we have a different cross-section. So,
we can easily divide into the rectangular
shape and then we can find out this second
moment of area along the neutral axis, and
once you know these things, then you can apply
the parallel and the perpendicular axis theorem
to get the other, this second moment of area
at the particular section.
For instance, you know like if we just want
to find out the moment of inertia about the
I section, then we can use the above relation
very easily because I section is having a
first two rectangular section and then there
is a middle rectangular section. So, it is
you see two horizontal and one is middle.
So, how the compatibility is there is based
on that we can easily get those values of
I for those kind of sections.
So, now here you see we have a different structure
all together if we have the neutral axis which
is passing from the D. There are two different
sections, which have same breadth, small B
and the total B is this one. So, as I told
you see you know like if you want to find
out for really structure that what will be
I NA is there. So, for that kind of thing
you know like it is just giving you the small
kind of gap.
So, the total, the brick we can say if we
are using the brick or any kind of the rectangular
structure, the breadth is B. So, for that
and the total breadth is capital B, the small
diameter is d and the capital diameter is
D for the whole section for that we can calculate.
You see I NA is nothing, but equals to I of
the total dotted rectangular minus I of the
standard area. For this you see there are
two different standard areas, and for dotted
one this capital B and the capital D are the
two main parameters. So, BD cube by 12 minus
two times of BD cube by 12 for a small section
standard area, or we can calculate the overall
result. From these two different kinds of
section which is coming in this moment of
second moment of inertia for this I NA is
equal to BD cube by 12 minus BD cube by 6.
So, this is you see the resultant like the
mass moment of inertia for this rectangular
cross-section.
So, in this you know like lecture we discussed
mainly about you know like what will happen
when beam is there under various kind of forces,
then what exactly the relations are there
when we are only talking about the bending
you know like only there is no other forces,
only bending moment is there. Then, you see
what will be the relation like sigma by y
is equal to M by I is equal to E by r. So,
this is the pure bending you know like the
relation is there. The equation is there through
which we can easily calculate the different
parameters once we know the bending stresses
of moment etcetera.
In second section, we discussed about the
section modulus and also we discussed about
the second moment of area. That means, you
see integration of y square dA. So, in these
sections, we discussed about the basic concept
of the bending stresses, but still we did
not discuss anything about when you see the
bending moment or any bending stresses are
there along with that always there is shear
stress.
So, in the next section, our main focus is
first is to solve some of the numerical problems
based on the geometrical parameters, and all
those you know like the attempted values of
these bending moment and bending stresses.
So, we just want to again use those you know
like formula just to refresh our concept.
So, in the next lecture our main focus is
to solve first the numerical problem and then
the second section is like dealing with the
shearing stresses that when you see we have
the bending moment with the shearing forces
and then how these shearing stresses are coming
all together with this particular bending
stresses. Then, what exactly the relations
are there, and how the interaction is there
in between these bending and the shearing
stresses, and what exactly the relations are
there in those two stresses. That is our main
focus. So, in the next lecture, we are going
to discuss all these things.Thank you.