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So, last class we talked about energy method
and I was talking there is a method called
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unit load method. It is basically derivative
of your unit energy method. So, energy
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method, if we look in a separate angle, we
will get this method, which we are trying
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to
define as unit load method.
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So, it is not a separate method. It is basically
energy method, but looking - the whole
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thing is looked at in a different angle. The
idea is to make the treatment much more
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systematic or much more mechanical. So, we
need not start from the very basic thing.
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So, energy principle is basically its background,
but in a different form we want to apply;
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rather I was trying to explain you here.
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This problem we have solved. M square part
we have put there, after taking derivative,
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we are getting this plus another component.
So, this part was basically the moment and
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this part is the derivative of that moment
about that P. Now, if we write here, in general,
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U equal to half integral M square EI dx. So,
delta we are trying to tell, this is du, dU
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by
dP; it will be this half 2 M dM dP divided
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by EI dx.
So, U equal to half M square EI dx, where
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this M is… M was same function of P. So,
P I
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am writing it is not the P what we have considered
earlier; P may be P, MB, or any other
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quantity. So, it is a variable. So, if P is
some function of… if M is function of P.
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So, M
square by EI dx, this part if we take derivative,
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so it will be dU by dP. So, here, this M
square will be 2 M, 2 it will cancel, it will
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be M, and this part will be dM by dx EI into
dx. Now, this part, we can write as this is
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capital M, and this derivative part if we
write
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small m divided by EI into dx. You try to
go through that expression. So, delta, finally,
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we are getting integral M m EI dx capital
M small m EI dx.
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Now, you can… You may have… take this
problem. So, I am keeping both the sheets
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here, or this I got, the earlier sheet I am
putting it here. So, delta. So, this is basically
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M
and this part is small m. What is small m?
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Derivative of that with respect to P.
Now, for that problem, this is basically the
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M and this part is derivative of that with
respect to P. So, P here it is MB. So, if
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we will generalize. So, M is what? M is the
bending moment generated by the actual loading;
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say here, at this level, this MB we can
drop. So, this is the actual bending moment
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expression due to the applied load, or here
also, this part we can drop, because x already
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we have obtained. So, this is due to the
actual applied load. So, this capital M is
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basically the bending moment due to the actual
applied load. So, we need not bother about
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taking some artificial load and all those
thing.
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But what is small m? Small m is derivative
of the moment expression with that artificial
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load and taking derivative about this artificial
load. If there is no load there actually,
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we
are talking in terms of artificial load or
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load something. Now, try to see the expression
of
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that. Same way it was P and we are taking
expression about that. So, for P we are getting
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Px and its derivative about P we are getting
x x. So, what is x? Because when you are
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taking about P, this part will not participate;
only of it is derivative about P. So, other
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force will not be appearing. So, it is continuation
of the P and we are taking about P. So,
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for moment generated by P, if you take derivative
about P, means it is basically if you
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put a unit load here. So, 1 into x will be
the derivative. So, P there, some moment will
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P
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into x; if you take the derivate about P,
so Px will be x. So, what is x? It is basically
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if
you put a unit load there, whatever moment
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we are supposed to get, that will come.
Now from this concept, we are coming to unit
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load method. So, it is basically derivative
of the energy principle; nothing else. The
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first component is N. It will be generated
by
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actual load. And second part - small m; this
is the moment expression, generated by unit
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load. And unit load where we will get? Where
you want to get the deflection, there we
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have to put 1 unit load. Say, here, this cantilever
problem, we are interested for finding
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out the deflection here; there is no load
at this point. So, just capital M part omega
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x
square by 2 due to the actual load; this is
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the capital M, because P is 0 here. So, it
is
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omega x square by 2 is the bending moment
expression for the actual load.
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And x is what? If you put a unit load here,
1 into x that should be the small m. So, this
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small m is due to the unit load applied at
the free end where we want to get the
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deflection. Or in that case, the slope calculation,
if you come at that level, so this part is
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the bending moment where MB is equal to 0.
So, this is basically the expression for
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capital M. Capital M is moment generated by
actual load. And this is what? This is the
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moment, if you put a 1 unit moment. So, anywhere
moment will be 1 unit. So, it will be
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1. So, this is the small m; small m is the
moment generated by 1 unit moment at the free
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end where we want to get the rotation.
So, it is just 1 unit.
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One unit, because we are putting some value.
If you take the derivative, so it is P into
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something, if you take the derivative of that,
so P part will be going there, So, that will
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be defined in terms of 1 unit. From there,
we are trying to define it is a unit load
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method,
because it is coming just like putting a unit
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load there, unit load or unit moment. If you
are interested in terms of slope, we have
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to put 1 unit moment; if you are interested
for
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deflection, we have to put 1 unit - 1 unit
of force there.
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So, we came to much more systematic formula
energy method. So, what we have to do?
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We have to take the structure; we have to
add the expression for the bending moment,
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for
the different zone due to the load that is
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capital M.
What is small m? Where we want to get the
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deflection, there we have to put 1 unit load
or 1 unit moment, and find out the expression
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for the bending moment for the different
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segment. Then capital M, small m we have to
multiply, divided by EI, integrate for the
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different zone, and go on adding, we will
get the corresponding deflection of slope
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there.
Small m is the expression of moment generated
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by 1 unit force or 1 unit moment, where
we want to get the deflection or slope. Say,
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for a structure, at any point, say point A,
we
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want to get the deflection; so, we have to
give 1 unit displacement at A. If you want
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to
get the rotation about A, we have to give
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1 unit moment at A. That is the only force
on
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the structure; no other force; due to that
will get the moment expression; that is basically
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small m. Because small m it is derivative
of dm by dp; dm by dp m involves all the
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loading plus the applied load - what we are
trying to say pseudo load or some artificial
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load something. So, if you take the derivative,
all the effect of pseudo artificial load will
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be reflected. Now we are taking derivative
about this means, that times it will come
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down. So, if it is P, P will create the moment
P into x, if you take derivative it will be
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x
or if it is Px square by 2 it will be… just
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x square by 2. So, it is just becoming, we
are
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taking derivative about P; automatically it
is becoming the effect of 1 unit load there.
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So, at this level, if you want to forget what
we have done and try to follow the steps,
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what we have to do? Take the structure, due
to the actual load different segment we will
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find, write down the moment expression - capital
M
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Now, next question will be - what we want
to do? We want to get the deflection at point
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C. So, we have to apply 1 unit load at point
C; only 1 unit load; no other load; and
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different segment we will get the moment expression,
that will be small m. And segment
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wise capital M small m divided by EI, we have
to integrate. And all the segment if we
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take the contribution, we will get the deflection
at C.
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Now next step. We may be intersect for finding
out slope at B. So, we have to take the
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structure, just put 1 unit moment at B. We
will get another set of expression for small
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m
due to getting slope at B. Now, the capital
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M part is fixed; that is the expression of
moment due to actual loading. So, capital
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M, and the new set of m, due to 1 unit moment
at B go on integrating in that manner. So,
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we should multiply divided by E and just
integrate. So, we will get the slope at B.
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Now you can say at this timing, thing is becoming
much more systematic or mechanical.
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So, if you just follow the steps, we are in
a position for finding out the deflection
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and
slope.
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Small m it will be for the different, different
cases; cases means, if I want to find out
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deflection at two places, slope at three places,
so there will be five set of small m. Then,
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we have to calculate separately.
Now, this problem can be… we can take one
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problem to explain how we can utilize
that? Main part is it is dependent on the
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expression of m, and it may be the different
from
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segment to segment; it may vary; in one segment
there will be one expression; another
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segment it may be different expression. Apart
from that EI is also there. If EI varies,
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automatically segment number will be more.
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So, I am taking one problem intentionally
of that type. Say, total length of the beam
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is l,
and its central part this l by 4, l by 4,
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that is l by 2; it is bit heavier. We know
the bending
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moment will be more at the mid span. Accordingly,
say, the designer thought the middle
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part will have a heavier type of section.
So, the moment of inertia there it is 2 I,
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and that
quarter span, left and right, it it is half
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the value of I. So, it is simply I - second
moment
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of area.
So, middle part is double line and at the
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both the end it is I and I. And the whole
thing is
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subjected to a load W. And we are interested
for finding out the slope at one of the
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support. So, if it is A, if it is B; say here,
if I say C, if it is D, and if it is E. So,
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beam has
a symmetrical type of load and geometry. So,
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load is at the centre; though the cross
section is different, but it is symmetrically
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placed. So, slope at A and slope at B will
be
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identical. We are interested for finding out
the slope. We may be interested for slope
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at A
or we may be interested at deflection at the
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middle point D.
Now, say, we are interested for finding out
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the slope at A. The slope is little bit different
in idea, because slope means we have to put
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some moment and there is no load here. So,
we have taken in that manner. Now, moment
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expression, first of all the reaction will
get
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here, it is W by 2; here also W by 2. So,
A to point B, there will be a moment expression
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of W by 2 into x. And here also, here to here,
there will be another expression; it may be
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W by 2 into x minus W of x minus l by 2 something.
So, it will have a distribution like
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this, bending moment, central beam with a
central load; it will be linearly go to the
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peak;
come down. So, we can say there are two segment
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A to D and D to B; but we are
calculating capital M small m divided by EI.
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Now, EI part is different; E is same; I is
different. So, ultimately, the problem will
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be
reduced to four segments. So, one segment,
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two segment; moment wise it is two
segment, but that EI part is different. So,
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earlier, we are taking the break when the
moment expression was different; here at least
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it is same, but EI part is different. So,
here
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to here - one segment; here to here another
segment; here to this point one segment. So,
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four segments are there.
Now, this problem can be solved if we write
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in a tabular form; it will be much more
systematic. Say, we write segments. So, first
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is segment AB; not AB; B is at that end; it
is AC; then CD; then DE; and your EB. Now,
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origin, limits, I, and say, M. Now, AC,
CD, DE, and AB. So, there are four segments.
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So, we have to just put for the different
segment.
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Now, say, I part we can put; say AC I is I;
CD I equal to 2 twice I; DE is also twice
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I;
and ED is I. E, you can also vary. Now you
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could write EI. So, it will be EI 2, EI 2,
EI I.
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Now, the moment expression, we have to write,
and here, the moment expression will be
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in the form of some x, at this one we can
write as - the first one - it will be W divided
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by
2 multiplied by x. Second one, we can give
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some expression. So, it will be W divided
by
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2 x. Which one? Say W by 2 is the reaction;
reaction into x, reaction into x; expression
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of moment is W by 2 into x, which one?
So, we are writing there. Now, here we have
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to put the limits. Now, if we take AC, its
origin is A; we are taking x from here; and
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limit will be 0 to l by 4. Now if we take
CD,
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origin again it is A, because x is starting
from A, but limit is your l by 4 to l by 2.
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Now,
this limit we could change 0 to l by 4. In
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that case, origin should be C; in that case,
moment expression we could write W by 2 into
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x plus l by 2. x, we could start from here.
So, we could make the origin C, from C. So,
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if it is x. So, x plus l by 2 should be the
moment. So, moment expression we could write
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W by 2 x plus l by 2, and this limit we
could change 0 to l by 4, and the origin we
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could make it C. You will get the same
expression.
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So, which one will be convenient, that we
normally follow I think. So, if the, one of
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the
limit is 0, most of the cases that part will
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not contribute, I think. Now, here this
expression will be more. So, somehow, we have
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to compromise.
Now, the much more interesting part will come,
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it will come to the right side. Right side
if I come here or here, so expression will
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be this or W by 2 into that distance; this
into
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that distance. The other alternative, we can
start calculating from this side and that
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will
be much more easier. So, here, so we will
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take the last span. So, last span origin will
be
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your B; not A. And limit will be, so x we
will take counting from that side. So, it
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will be
your 0 to l by 4 and expression will be your
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again W by 2 into x. So, I have tried to filled
up this origin limit little later, I think,
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because you write the expression of M, so
which
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will be your convenient, accordingly we have
to pick up the limit and origin.
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Now, here, we can again follow it is B; if
we follow in a similar manner. So, it will
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be
your l by 4 to l by 2 and expression will
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be W by 2 into x. So, entire structure, we
can
194
00:26:02,490 --> 00:26:08,900
start the origin from this side or that side
or if we take much more complicated case,
195
00:26:08,900 --> 00:26:15,190
say,
if we take a frame, so we can start from here;
196
00:26:15,190 --> 00:26:19,230
next stage we can start from there; next
stage we can start from there. So, at different
197
00:26:19,230 --> 00:26:25,940
level we can start, and accordingly our
limit will be adjusted, and all moment expression
198
00:26:25,940 --> 00:26:31,670
will be also little bit different
depending on the origin. So, origin is very
199
00:26:31,670 --> 00:26:36,030
important; limit is very important; and
depending on that, your moment expression
200
00:26:36,030 --> 00:26:45,461
will be different.
Now, once that part is there, a small m, in
201
00:26:45,461 --> 00:26:51,530
a similar manner we have to write. Small m
here, we are interested for finding out the
202
00:26:51,530 --> 00:26:59,580
slope at A, means we have to take the beam,
and put a moment here, and we will get the
203
00:26:59,580 --> 00:27:08,070
moment expression. So, there entire beam
will get one moment expression, but as M has
204
00:27:08,070 --> 00:27:17,600
these four segments, there also M we have
to just take in four segments. Now, we can
205
00:27:17,600 --> 00:27:22,070
put it, we can put the small m here. So, if
we
206
00:27:22,070 --> 00:27:27,640
are interested for finding out only one deformation,
we can just put the expression of
207
00:27:27,640 --> 00:27:33,940
small m here, but normally in a problem, it
will be required to find out slope here, slope
208
00:27:33,940 --> 00:27:36,820
here, deflection here, deflection here, many
cases.
209
00:27:36,820 --> 00:27:43,680
So, there will be a separate table just for
small m. So, there you need not write origin
210
00:27:43,680 --> 00:27:48,760
and
all those, because that part are common. Only
211
00:27:48,760 --> 00:27:55,250
small m expressions you can put in a
separate table or if the small m number is
212
00:27:55,250 --> 00:28:00,990
small, if you have sufficient space, you can
write 1 M 1 for this, M 2 is for this like
213
00:28:00,990 --> 00:28:05,820
this. So, we can put MA here, small a here,
small
214
00:28:05,820 --> 00:28:13,170
c or small d; something you can go on adding.
So, it is, there is no rule here; rather we
215
00:28:13,170 --> 00:28:20,300
will follow certain norm, which will help
us to solve the problem. So, we can make in
216
00:28:20,300 --> 00:28:28,890
that manner. So, small m part we can put it
here, but rather draw the beam, and put the
217
00:28:28,890 --> 00:28:29,890
load.
218
00:28:29,890 --> 00:28:54,350
So, it will be... So, here, I am not putting
all the thickness; just try to indicate this
219
00:28:54,350 --> 00:28:58,870
part is
heavy compared to this. Now, here we have
220
00:28:58,870 --> 00:29:08,260
to put just 1 unit moment; that is all. So,
it
221
00:29:08,260 --> 00:29:21,360
was A, it was B, it was C, it was D, it was
E. So, you can put all this l by 4, l by 4,
222
00:29:21,360 --> 00:29:25,110
l by 4,
l 4, EI, everything we can put; it is identical;
223
00:29:25,110 --> 00:29:33,670
already we have written. Now, due to the
moment, we have to find out the expression
224
00:29:33,670 --> 00:29:40,910
of bending moment, and those are nothing
but the small m. How we will get it? First,
225
00:29:40,910 --> 00:29:47,559
we have to find out the reactions. So, what
will be the reaction here? A simply supported
226
00:29:47,559 --> 00:30:05,960
beam, there is a moment M, total length is
l.
227
00:30:05,960 --> 00:30:19,290
Say, there is a…. the beam is subjected
to only one moment - unit moment. So, that
228
00:30:19,290 --> 00:30:22,930
unit
moment will try to rotate that member; who
229
00:30:22,930 --> 00:30:28,170
will prevent that? The reactions. So,
reactions will be… there is no other force.
230
00:30:28,170 --> 00:30:30,770
So, this reaction and that reaction should
be
231
00:30:30,770 --> 00:30:39,140
equal and opposite; otherwise force equilibrium
will not be balanced. So, this will be one
232
00:30:39,140 --> 00:30:43,700
force, just opposite force, and they will
give a couple, and that value of that couple
233
00:30:43,700 --> 00:30:49,240
should be equal to the applied couple, applied
couple, applied moment. So, one is there.
234
00:30:49,240 --> 00:30:53,400
So, these two forces will give in a reverse
manner. So, this is clock wise; it will be
235
00:30:53,400 --> 00:30:58,240
anticlockwise. So, this side reaction will
be this side; this will be that side; and
236
00:30:58,240 --> 00:31:02,380
they will
produce a moment of 1. So, this force will
237
00:31:02,380 --> 00:31:05,650
be total length is l. So, it should be 1 by
l; it
238
00:31:05,650 --> 00:31:14,490
should be 1 by l.
The
239
00:31:14,490 --> 00:31:21,170
moment part is little bit confusing; sometimes,
force we can handle much more
240
00:31:21,170 --> 00:31:26,980
conveniently; moment sometimes creates problem,
because this moment is applied here,
241
00:31:26,980 --> 00:31:35,210
and reaction we will get in that manner. So,
there is a applied moment; if you take the
242
00:31:35,210 --> 00:31:41,930
free body of the beam, so it is a simple bar;
there is a physically applied moment; and
243
00:31:41,930 --> 00:31:46,700
there might be some reaction; there might
be some reaction. So, if I take moment about
244
00:31:46,700 --> 00:31:54,720
this, so there is a applied moment; so, it
is 1; and the force into that distance, say,
245
00:31:54,720 --> 00:31:59,360
R into l
it should be equal to 1. So, R will be 1 by
246
00:31:59,360 --> 00:32:03,830
l. In that way also you can know. Or physically
you can think there is a moment, it should
247
00:32:03,830 --> 00:32:07,510
be balance by this; this moment, if I put
it
248
00:32:07,510 --> 00:32:15,480
here, then also you will get the same reaction.
If the moment is applied in the middle, anywhere
249
00:32:15,480 --> 00:32:22,780
if you apply, because moment is…
basically. this moment will create same reaction.
250
00:32:22,780 --> 00:32:29,770
So, here, here, anywhere if we put, it
will be balanced by this reaction; it will
251
00:32:29,770 --> 00:32:37,240
generate a moment; and that moment and this
moment should cancel. Only moment expression
252
00:32:37,240 --> 00:32:41,630
will be different; if we keep the
moment here, means applied coupled here, and
253
00:32:41,630 --> 00:32:55,490
applied coupled here.
So, let us draw the bending moment diagram.
254
00:32:55,490 --> 00:33:19,380
Here, moment will be how much? 1. And
here moment will be 0; it will come like this.
255
00:33:19,380 --> 00:33:28,750
Anywhere, if you pick up, it will be 1
minus 1 by l into x. So, moment expression
256
00:33:28,750 --> 00:33:33,260
will be 1 minus 1 by l into x; when x will
be
257
00:33:33,260 --> 00:33:37,400
l, so 1 by l into l, 1 one cancel, it will
be 0. And it should be 0, because it is a
258
00:33:37,400 --> 00:33:40,290
free end;
there is no applied moment; it is also free
259
00:33:40,290 --> 00:33:46,840
end, but there is a applied moment.
You can just think in this form: there is
260
00:33:46,840 --> 00:33:50,730
a cantilever beam with a load, so there is
a load.
261
00:33:50,730 --> 00:33:55,270
So, shear force will be that force. So, it
is a free end; there is a moment. So, moment
262
00:33:55,270 --> 00:33:56,270
will
263
00:33:56,270 --> 00:34:06,860
be that applied moment. And this moment, and
inside there is no load, so it will vary in
264
00:34:06,860 --> 00:34:09,236
a
linear manner; it will come to 0. If there
265
00:34:09,236 --> 00:34:16,419
is a moment here, we should get in a different
form. So, it should be like that. If we put
266
00:34:16,419 --> 00:34:24,700
a moment here, we should get up to this, then
there should be a change over by 1 unit, and
267
00:34:24,700 --> 00:34:47,980
go like this. So, here also, if we just write
segment… it is the moment 1.
268
00:34:47,980 --> 00:35:09,099
Now, one option is we can put a column M,
or here, say, space is not there, again, we
269
00:35:09,099 --> 00:35:15,730
have to write segment and M, we can put origin
limit I, or already it is there, we can just
270
00:35:15,730 --> 00:35:34,690
follow it. So, small m… AC will be how much?
So, A to C it will be 1 minus x by l. So,
271
00:35:34,690 --> 00:35:44,089
1 minus this reaction 1 by l into x. So, 1
by l into x. So, here also same origin. So,
272
00:35:44,089 --> 00:35:53,000
1
minus x by l, because both the segment we
273
00:35:53,000 --> 00:35:56,440
have taken origin here. So, this line will
be
274
00:35:56,440 --> 00:36:04,599
valid. So, it is basically what? This is 1,
minus there is a linear curve x by l. So,
275
00:36:04,599 --> 00:36:08,809
if x
equal to l, it will cancel, it will be 0.
276
00:36:08,809 --> 00:36:12,180
So, if we make a box – rectangle - it is
basically the
277
00:36:12,180 --> 00:36:25,010
diagonal, but this one EB, we have taken origin
from this side. So, it will be just 1 by l
278
00:36:25,010 --> 00:36:33,319
into x. So, it will be x by l and it will
be x by l.
279
00:36:33,319 --> 00:36:43,950
Now, earlier case, if the origin was only
A, we could retain the same expression, but
280
00:36:43,950 --> 00:36:49,440
whatever we have decided, that will more or
less follow. So, for capital M some origin,
281
00:36:49,440 --> 00:36:57,500
some limit, we have fixed up. So, for small
m also same origin, same limit will be valid.
282
00:36:57,500 --> 00:37:06,420
Now this capital M, say capital M, small m,
the sets of expressions we obtained. This
283
00:37:06,420 --> 00:37:14,250
small m is due to the applied moment at A;
means, we are going to calculate the slope
284
00:37:14,250 --> 00:37:19,809
at
A. So, with these two tables - this table
285
00:37:19,809 --> 00:37:23,210
and this table - basically these expressions
and
286
00:37:23,210 --> 00:37:34,660
these expressions plus this I value, with
that limit, we can write our expression, what
287
00:37:34,660 --> 00:37:38,180
we
have derived for unit load; that is M small
288
00:37:38,180 --> 00:37:42,569
m EI dx for different segments.
289
00:37:42,569 --> 00:38:05,109
So, we can write theta A; it should be integral
0 to l by 4; capital M was Wx by 2; small
290
00:38:05,109 --> 00:38:33,839
m was 1 minus x by l divided by EI dx. This
plus it was l by 2 l by 4 to l by 2 Wx by
291
00:38:33,839 --> 00:38:47,739
2 1
minus x by l 2 EI dx, plus it is your l by
292
00:38:47,739 --> 00:39:04,200
4 to l by 2 Wx by 2. And here x by l twice
EI dx
293
00:39:04,200 --> 00:39:18,940
plus 0 l by 4.
So, there are four segments; first segment
294
00:39:18,940 --> 00:39:24,150
0 to l by 4; second one is l by 4 to l by
2; third
295
00:39:24,150 --> 00:39:31,859
one is l by 4 to l by 2; fourth one 0 l by
4. Wx by 2, this capital M expression, is
296
00:39:31,859 --> 00:39:41,369
identical. For the left, small m first two
span is 1 minus x by l, 1 minus x by l; right
297
00:39:41,369 --> 00:39:47,710
part
is x by l x by l. Here it is EI 2 EI 2 EI
298
00:39:47,710 --> 00:39:51,319
EI. Now, this part if we just integrate and
put all
299
00:39:51,319 --> 00:40:03,660
limit, we will get the slope at theta A. Now,
if you are interested for theta B, we have
300
00:40:03,660 --> 00:40:09,069
to
generate another set of m - small m. So, we
301
00:40:09,069 --> 00:40:16,380
have to put one unit of moment here, and we
will get some reactive force, and find out
302
00:40:16,380 --> 00:40:19,809
the expression for small m. And this small
m
303
00:40:19,809 --> 00:40:25,750
we have to put here, here, here, and here.
So, if you calculate in that manner, so you
304
00:40:25,750 --> 00:40:43,539
will
get theta B. Now, which one?
305
00:40:43,539 --> 00:40:57,690
Now, this problem we can solve, and the value,
whatever you will get, I can supply you
306
00:40:57,690 --> 00:41:32,880
what should be the answer for that. It will
be 5 Wl square 128 EI. So, because
307
00:41:32,880 --> 00:41:38,520
everywhere EI will come out, W will come out,
and here x into 1 x square by 2, it will be
308
00:41:38,520 --> 00:41:43,230
some l square some factor, so for those factors
if you go on accumulating. after
309
00:41:43,230 --> 00:41:51,569
calculation it will come 5 W l square 128
EI. So, we can check this value, later on,
310
00:41:51,569 --> 00:41:52,569
after
integration.
311
00:41:52,569 --> 00:42:09,799
Now, the method I have shown with a beam,
where there are different segments and the
312
00:42:09,799 --> 00:42:17,009
loading is acting at the middle, we are trying
to get the slope. Now, you will find much
313
00:42:17,009 --> 00:42:25,190
more use of that method, if we handle a little
difficult type of problem, we talked about
314
00:42:25,190 --> 00:42:31,249
frame. So, one vertical member, one horizontal
member we have taken. If we take a
315
00:42:31,249 --> 00:42:40,210
much more generalized form of a frame, there
is one example here, if we pick up.
316
00:42:40,210 --> 00:44:06,710
You can nicely analyze that problem with this
unit load method. See there is a frame like
317
00:44:06,710 --> 00:44:13,220
this; there are two vertical members; not
only that their lengths are not identical,
318
00:44:13,220 --> 00:44:18,180
but
there is a horizontal member. Horizontal member
319
00:44:18,180 --> 00:44:23,579
is subjected to a load 96-kilo Newton.
On horizontal members, there is a load 48-kilo
320
00:44:23,579 --> 00:44:31,329
Newton. This is a roller type of support;
this is a hinge type of support. Now, the
321
00:44:31,329 --> 00:45:15,739
length here, it is 5 meter, and here there
is 4.5
322
00:45:15,739 --> 00:45:23,729
meter; that is 3 meter. So, this is 4 and
half meters; this is 3 meter; this is 3 meter;
323
00:45:23,729 --> 00:45:30,749
this is
3 meter; this is 5 meter; these are the loads.
324
00:45:30,749 --> 00:45:41,160
And say, this is I, this is I, and this is
2 I.
325
00:45:41,160 --> 00:45:48,759
So, horizontal member has a heavier section;
I will twice I. And vertical members are
326
00:45:48,759 --> 00:45:57,239
both I and I, but height is different; here
it is 4.535 meter, 3 meter, 3 meter, 96 kilo
327
00:45:57,239 --> 00:46:05,519
Newton, and 48 kilo Newton. So, this is hinge;
this is roller. Now, this structure if we
328
00:46:05,519 --> 00:46:08,589
try
to tell you, if you try to solve by moment-area
329
00:46:08,589 --> 00:46:14,579
theorem of differential equation
technique. The problem theoretically it can
330
00:46:14,579 --> 00:46:17,309
be solved, but, it will very, very difficult
to
331
00:46:17,309 --> 00:46:27,800
solve. So, many steps we have to carry out
to get the solutions of that.
332
00:46:27,800 --> 00:46:37,380
Now, this can be nicely handled by your unit
loop method. Now, I can write the different
333
00:46:37,380 --> 00:46:57,160
points set: if is A, if it is B, if it is
C, if it is D. So, E we have to take. F we
334
00:46:57,160 --> 00:47:03,609
have to take.
So, this is A, this is B, this is C, this
335
00:47:03,609 --> 00:47:06,510
is D. So, there is a load here. So, we are
taking E;
336
00:47:06,510 --> 00:47:22,210
there is load it is F. Now, we may be interested
for finding out - what will be the
337
00:47:22,210 --> 00:47:29,190
horizontal displacement at point B. Due the
load, whole thing may undergo some
338
00:47:29,190 --> 00:47:35,589
moment. So, the horizontal moment of B and
C will be identical, because they are
339
00:47:35,589 --> 00:47:39,320
connected by horizontal member, and we are
not going to take any extension of this
340
00:47:39,320 --> 00:47:46,150
member; only bending part we are considering.
So, horizontal moment of BC will be
341
00:47:46,150 --> 00:47:49,009
identical.
We may be interested for finding out the horizontal
342
00:47:49,009 --> 00:47:55,519
displacement of D; vertically it will
not come down, because extension we are not
343
00:47:55,519 --> 00:48:02,319
taking. So, B and C, it will have a
common displacement horizontally. D can move
344
00:48:02,319 --> 00:48:10,130
horizontally. Now, rotation we may get
here, here, here, here, under this small load,
345
00:48:10,130 --> 00:48:17,910
under this load, at any point we may get
rotation. So, there are so many possibilities,
346
00:48:17,910 --> 00:48:22,539
I mean, so many displacement components
we can find out from there. We can say this
347
00:48:22,539 --> 00:48:26,779
is a structure, you find out horizontal
displacement at C, horizontal displacement
348
00:48:26,779 --> 00:48:30,420
at D, rotation at A, rotation at B, rotation
at C
349
00:48:30,420 --> 00:48:36,650
rotation at D, rotation at E, rotation at
F. So, there will be a long list. So, small
350
00:48:36,650 --> 00:48:42,170
m you
will have a series, and capital M, one. So,
351
00:48:42,170 --> 00:48:45,619
capital M and one set of small m, you will
get
352
00:48:45,619 --> 00:48:50,049
one displacement component. Capital M and
another small m, you will get another
353
00:48:50,049 --> 00:48:59,869
displacement; like that, you can go on calculating.
Now, I have taken this problem, because this
354
00:48:59,869 --> 00:49:04,869
problem is little complex compared to the
earlier problem. Unit load we have just applied
355
00:49:04,869 --> 00:49:10,660
for a beam problem. So, this is a frame
with at least three members with some loading
356
00:49:10,660 --> 00:49:14,141
this side, that side. So, that structure,
if
357
00:49:14,141 --> 00:49:21,690
you pick up, first of all we have to define
the different segments, their origin, their
358
00:49:21,690 --> 00:49:24,869
limit,
plus due to the external load we have to find
359
00:49:24,869 --> 00:49:27,019
out the expression of capital M. So, that
part
360
00:49:27,019 --> 00:49:34,509
we can write.
Now, the number of segment: we can say A to
361
00:49:34,509 --> 00:49:40,469
E there will be a segment; E to B there
will be a segment; B to C there will be a
362
00:49:40,469 --> 00:49:43,420
segment; F to C there will be a segment; and
F
363
00:49:43,420 --> 00:49:53,420
to D there will be another segment. So, one,
two, three, four, five segments will be there.
364
00:49:53,420 --> 00:50:02,680
Now we can write the segments. So, next page
if we write the segments.
365
00:50:02,680 --> 00:50:39,969
Say if it is AE, and here say EB, BF, FC,
and CD. So, one will be your origin, limits,
366
00:50:39,969 --> 00:50:52,920
EI
you can write, capital M you can write. Now,
367
00:50:52,920 --> 00:51:07,109
our main job is we have to find out the
expression for capital M, and it should be
368
00:51:07,109 --> 00:51:13,319
taken in much more convenient form. So, that
limit will be more or less in reasonable order,
369
00:51:13,319 --> 00:51:20,739
moment expression will be little bit small,
and for that purpose we can change our limits.
370
00:51:20,739 --> 00:51:27,960
Now, for getting that moment expression,
let us return back to our earlier problem.
371
00:51:27,960 --> 00:51:34,640
Here, our first job is to find out the reaction
and
372
00:51:34,640 --> 00:51:40,049
that will help to find out the expression
for the bending moment.
373
00:51:40,049 --> 00:51:48,369
Now, here there will be two reaction components
or rather I can draw a member without
374
00:51:48,369 --> 00:52:14,410
this detail 3 meter or 5 meter or EI. So,
if I put it there, this is up to this. So,
375
00:52:14,410 --> 00:52:33,759
see I am
putting only 96, not kilo Newton, in order
376
00:52:33,759 --> 00:52:49,349
to make the diagram as much clear as possible.
Now, this point is hinge, so it can carry
377
00:52:49,349 --> 00:52:53,940
axial force, so it can carry horizontal as
well as
378
00:52:53,940 --> 00:53:02,280
vertical; this will carry only vertical. So,
all the horizontal force will be carried by
379
00:53:02,280 --> 00:53:04,170
this
support.
380
00:53:04,170 --> 00:53:13,739
Now, there will be two reaction components;
vertical here; you can take moment about
381
00:53:13,739 --> 00:53:20,489
this or moment about that. So, if we take
moment about this, it will be a 48 into 4.5
382
00:53:20,489 --> 00:53:28,530
plus
96 into 3 - that will be the moment; that
383
00:53:28,530 --> 00:53:33,089
will be balanced by this force. So, it will
be both
384
00:53:33,089 --> 00:53:44,839
clockwise. So, there will be a reaction. So,
it will be how much? It will , say, it we
385
00:53:44,839 --> 00:53:53,029
write,
should be 48 into 4.5 plus 96 into 3 divided
386
00:53:53,029 --> 00:54:01,349
by 6. If we take moment about A. So, 48 into
4.5, it will give a moment; 96 into 3 it will
387
00:54:01,349 --> 00:54:04,319
give a moment; that moment should be
388
00:54:04,319 --> 00:54:12,799
balance by your reactive force at D upward,
that in to the distance 6, it will be your
389
00:54:12,799 --> 00:54:28,410
reaction at D. So, it is just if you take
moment about A, that will give this value,
390
00:54:28,410 --> 00:54:33,460
and this
value will be how much? This value will be,
391
00:54:33,460 --> 00:54:53,209
this will be 84 kilo Newton.
Now, the
392
00:54:53,209 --> 00:55:01,020
next option is, if it is 84, but the vertical
force is 96, this is 48. So, there should
393
00:55:01,020 --> 00:55:12,670
be a another force here. So, it will be 12
kilo Newton; 12 plus 84 equal to 96; summation
394
00:55:12,670 --> 00:55:22,739
of all upward forces. Now, if you have the
reaction 48 12 84, we can proceed for finding
395
00:55:22,739 --> 00:55:28,390
out the moment at the different segments for
capital M.
396
00:55:28,390 --> 00:55:40,809
So, this is the actual structure. We have
removed the support and tried to find out
397
00:55:40,809 --> 00:55:44,359
the
reactions; the objective is we have to get
398
00:55:44,359 --> 00:55:49,430
the expression of capital M for the different
segment. So, reaction part we have determined.
399
00:55:49,430 --> 00:55:54,369
So, next job is for the different segment
we have to find out the expression of capital
400
00:55:54,369 --> 00:56:09,579
M, and we may vary our original limits.
I think with that we have to conclude today.
401
00:56:09,579 --> 00:56:21,609
We can try to solve it in the next class.
Answers.
402
00:56:21,609 --> 00:57:02,160
A number of answers are there: theta a, theta
b, theta c.
403
00:57:02,160 --> 00:57:21,630
Preview of Next Lecture
Particular type of energy method. So, the
404
00:57:21,630 --> 00:57:24,779
main concept behind that is, basically, in
the
405
00:57:24,779 --> 00:57:30,969
energy principle. So, that we have defined
in a particular fashion, and we defined that
406
00:57:30,969 --> 00:57:37,190
method as a unit-load method. That method,
we have tried to demonstrate with a beam
407
00:57:37,190 --> 00:57:41,569
problem.
And at the end, we have tried to apply to
408
00:57:41,569 --> 00:57:49,769
a much more complicated type of problem.It
was a portal frame, having absolutely arbitrary
409
00:57:49,769 --> 00:57:58,529
type condition, because the legs were
different. We can go ahead with that to get
410
00:57:58,529 --> 00:58:03,630
the feeling about the applicability of method
to this particular type of structure.
411
00:58:03,630 --> 00:58:29,459
So, the frame you must have remembered it
was something like that. So, this leg it was
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00:58:29,459 --> 00:58:43,529
quite long compared to the other leg. That
leg, the right side leg, it had a height of
413
00:58:43,529 --> 00:58:46,599
5
meter; and left side above the load, it was
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00:58:46,599 --> 00:58:52,680
3; below that it is 4.5. So, total 7.5 and
the
415
00:58:52,680 --> 00:58:53,359
horizontal.