WEBVTT
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Good afternoon this is lecture 15 on course
in advanced structural analysis.
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We are about to complete this topic of review
of basic structural analysis two. We were
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looking at displacement methods and this is
covered in part five in the book on basic
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structural analysis.
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So you will see that the problems that you
need to solve fall in four broad categories
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we finished the first two. In this session
we look at how to solve problems by the displacement
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method when you have multiple unknown rotations
but, no unknown translations that is the fourth
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type, which we will see later.
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If you recall we had done this problem in
the last session. This is a continuous beam
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subjected to relative settlement of two supports,
we had solved this problem by slope deflection
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method. Let us see how to solve such problems
by the moment distribution method.
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So, the first step is to convert the relative
settlements to chord rotations there are three
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members you can work out those chord rotations
we have done this in the earlier session you
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will notice that the member C D is the one
which undergoes an anticlockwise rotations,
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so you give a negative sign to that.
What is a degree of kinematic indeterminacy
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in this problem?
Two, I mean we can reduce it to two and the
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unknown rotations are at B and C, so at B
and C we need to do moment distribution up
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till now we have done problems which needed
moment distribution only at one location and
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you need to do it in one step and you get
an exact solution but, now when you have more
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than one location where you do distribution
it gets to be a little tricky.
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So, let us see how to do that your first job
is to find fixed end moments. Are there any
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fixed end moments in this beam subject to
support settlements?, not due to external
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loads but, you need to calculate the additional
fixed end moments you get due to this chord
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rotations and I hope you remember the formulas
these are easy to work out. In the slope deflection
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method, you do not call them fixed end moments
they come in the slope deflection equations
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under chord rotation but, they are known quantities.
In the moment distribution method, you have
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to explicitly calculate these quantities and
we call them additional fixed end moments
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right? and what is the next step in moment
distribution method after you get the fixed
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end moments?, up to this step we have done
in the last class.
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What do we do next? distribution factors carry
over factors shall we do that?
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You have to find out distribution factors
at B and C. Take B it is 3 E i by L and the
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i's are different for the three elements and
4 E i by L for the middle element right?,
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so this is straightforward you do not get
clean integers here so you can find them to
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as high and order of accuracy as you wish
so here I have given four significant figures,
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so I hope you know how to calculate the distribution
factors at joints B and C, which is the ratio
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of K B A to K B C and K C B to K C D there
is the support at D and as far as carry over
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factors are concerned you do not carry over
to the simple supports, so you have only a
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carryover from B to C and from C to B and
its equal to half.
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So, this is the distribution table, as you
can see from the number of lines I have drawn
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you do not get it in one cycle. So, how do
we do this? first you write down the element
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numbers, write down the distribution factors
please note at any joint the sum of all the
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distribution factors should add up to one
then you write down the carry over factors
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and you write down the fixed end moments in
this case the fixed end moments are caused
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by the support settlements right?
Now, let us also give with physical meaning.
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You have this beam in which you have arrested
the joints B and C you got those fixed end
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moments and at the joint B wherever you arrested
it you got a you got a moment, which you need
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to release and when you release it, it is
like balancing that moment which you get by
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distributing the moments to the two ends.
So, now you have take minus 100.74 minus 221.87
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add it up to get minus some quantity and that
total you have to distribute to the two ends,
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so how do you do that? and when you do that
you also have to do a carryover then you go
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to the next joint and do the same thing.
Now, there are many ways it can be done and
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the solution procedure follows pattern technique
called relaxation method you know it is a
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numerical way of solving simultaneous equations.
If you notice the superiority of the moment
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distribution method over the slope deflection
method is that we do not calculate theta B
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and theta C at all, we directly get the moments
where as in the slope deflections methods
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you calculates theta B theta C and plug those
values back into the slope deflection methods.
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So, here there are many ways to do it, the
one that we recommend is you first do at this
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joint if you have your calculators you can
work out it is pretty straight forward you
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add up the total and distribute 0.4497 to
the left end and to the right end, so at this
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stage if you add up all the moments at that
joint it should turn out to be 0 right?
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Now, the suggestion is do not do the carry
over now, we will do the carry over later
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for all the joints in one go it really does
not matter because the end you have to balance,
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so this is what we call the first cycle of
balancing and then you move over to the next
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joint and do exactly the same thing you will
get something right? It is interesting to
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note in the second joint you seem to have
smaller numbers to distribute in this particular
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instance it need not to be so for all loading
now, is a time to do the carry over.
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How do you do the carry over?
You can put some arrow marks to help you 50
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percent of the moment gets carried over plus
half, so is this self explanatory everything
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is clear?
Now, after you carried over at the joint B
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what is the unbalanced moment? It is plus
10.43, so what should you do now? again balance
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it, so that is what you do and you have to
balance 88.77 at joint C, you can do all that
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in one step, is it clear? it is a simple thing
you can do it manually very fast, now what
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do you do? so this is balance number two carry
over again but, as you go down, the quantities
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which you carry over are becoming smaller
and smaller, so when do you stop?
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What?
Less than some percent.
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Yea, when you done enough number of cycles
then you got tired. In practice you see the
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power of the moment distribution method is
lost if you have to put in lot of effort to
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it, so usually in the olden days they stop
to two cycles and they said we live with the
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error but, today we do not do that we say
that stop till you can tolerate you accept
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as an error that unbalanced moment and that
should it is left to you, it is very good
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if you say if it is less than 1 percent of
the maximum moment some people say 5 percent.
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So, you can keep on doing this and we have
done four cycles here and you can you should
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end with the balancing because if you leave
an unbalanced moment and then you would not
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satisfy equilibrium, so we leave we stop with
the fifth cycle of balancing and then we add
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up the total and when you add up one way to
know whether you got this right to some extent
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is that when you add up the two moments M
B A and M B C they should add up to 0, because
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there is no net moment there, same at C B
and C D, so you got the answers but, there
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is a little error but, that is a negligible
error right? The error comes from also in
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carry over you will have a little rounding
of problem because as you know you have to
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round off to the decimal places, so but, we
will leave with that.
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This is how you do moment distribution method
when you have more than one joint is it clear?
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Here I have got an example of two joints,
if you have three joints you do parallelly
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three but, then today we live in the world
of computers and we say it is not worth it,
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so it is really worth it if you have only
one member one rotation then you do it very
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fast and it is an instant solutions 100 percent
accurate this is not that accurate compare
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to slope deflection method and you get exactly
the same solution as you get in slope deflection
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method.
Now, again I want to go back historically,
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this is how they did multi storied frames
in the olden days I mean it is too much to
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analyze the full frame, so we have this concept
of substitute frame you take anyone floor
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and you assume that there is not much interaction
between one floor and the next floor and you
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can prove this using Mueller bridge Law's
principle and as long as there is no sway
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in the frame you can isolate one frame and
assume the columns should be fixed at the
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top and bottom, so that is what this is for
example, what you can do in a three storied
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a three bay multistoried frame under gravity
loads not under lateral load because under
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lateral loads it will sway.
Now I made our job easier, because I have
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deliberately chosen a symmetric loading condition
in which case you can take further advantage
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and cut it in the middle and what do you put
there?
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Guided fixed support
Guided fixed support okay, guided roller support
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and you know very well how to analyze this.
You have two unknown rotations there theta
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A and theta B and you can you need not worry
about that delta C, because we know the modified
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stiffness how do we proceed?
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Well, first we calculate the fixed end moments
for the column elements they are no fixed
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end moments because there is no lateral element
on those columns. For the beam elements A
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B and B C dash I hope by know now you know
how to do those calculations okay, so let
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us assume you can calculate these moments
correctly then you need to find the rotations
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of the distribution and carry over factors.
So, let us pause for a while and see how to
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do this, see at this joint A you have three
elements, so you have K A F, K A E and K A
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B right?, so for K A F it is 4 E i by the
height which is 3.5 for K A E likewise it
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is 4 E i by 3.5 and for K A B,it is 3 E I,
4 E i but, the E i is, 3 E i divided by the
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span which is 8 are you getting it?
So, you got three members meeting at that
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joint, so when you make the table you have
to arrange it in such a way that you can handle
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all of them simultaneously and the same thing
you do at joint B, at joint B the only point
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to remember here you have four elements is
that the stiffness of B C will be E i and
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that i is 2 i divided by i, is it clear? So,
you can work this out, I want to get the concept
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clear. So, let us see you got this then it
is easy you make your table carry over factors
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from B to A and A to B is plus half and from
B to C dash it is minus 1, clear? Make your
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table.
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So, let us not do this, let us just see how
it is done, it is exactly the same way you
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have to arrange your table in such a manner
that you can fit in most of the quantities
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nobody does this anymore, so it is a little
more of historical importance but, I just
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wanted you to know that this is how it can
be done then you can draw your free bodies
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and bending moment diagram. This is just a
demonstration of how moment distribution method
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was widely used to design buildings till about
the 1980s or 1985 very common.
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Today we do not do it because we got nice
software, which do it automatically for us.
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What is this structure that we see here?.
It is called a box culvert, where is it used?
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It has many purposes you can use it for drainage
but, let say you got ditch in front of your
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house and you want to bring your car across
instead of putting a slab with a separate
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foundation and there is a drain going below
you can put a neat box like this which is
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very efficient very stiff and it is called
a box culvert right? and if you had a concentrated
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load in that box culvert say due to your vehicular
load coming then it will be balanced by a
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soil pressure from below which is conveniently
assumed here to be uniformly distributed and
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you could have, what are those triangular
pressures you get on this side? Yeah, they
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come from the soil and may be if there is
water around there, so it is a self-equilibrating
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system. This is a self-equilibrating system
you do not talk about supports here okay?
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It is self equilibrating and you need to analyze
this and you can do it beautifully by either
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moment distribution or slope deflection method
but, there is symmetry here, is not it? So,
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why do not we take advantage of that. When
you do that it look like that divide that
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load 18 by 2 you get 40 right?
Now, how many what is a degree of indeterminacy
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here? Two, theta A and theta B so can you
find out the fixed end moment? yes you know
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all the formulas, so let us say you know how
to do all the fixed end moments of course,
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you are more used to beams with loads acting
downward, so when the loads are acting upward
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you have to stand on your head in your mind
then and get the directions correct it is
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plus clockwise is always positive and anticlockwise
is always negative, so do not that is the
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only place where you can make a mistake.
Okay, then the distribution factors are easy
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to calculate I think by now, you are taking
advantage of all the modified stiffnesss remember
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they are only three magic numbers remembers
4 E i by L is very normal 3 E i by L is if
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you have a hinge support at the extreme end
and E i by L is when you have a guided fixed
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support, is it clear?
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So, you can work out these and you can work
out your carry over factors do your table.
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Sorry.
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Soil pressure when it acts on the retaining
wall have you gone through a geotechnical
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course you are going through, so you will
know that. Take water if it is water is hydrostatic
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if it is water Pascal's law says that the
pressure is the same in all directions if
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it soil it is not the same, so the lateral
pressure it is called the active worth pressure
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is less than the vertical pressure how much
less depends on the soil properties, so if
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you have sandy soil it is roughly around one-third,
so whatever vertical load you applying one-third
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of that pressure goes, so it is a linear variation
it is conveniently assume this is ran kinds
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soil pressure profile but, do not in structural
analysis we say somebody is find out those
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loads and given it to you and you but, in
real life you need to work out those loads
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yourself, so that is why you have to ask those
question okay.
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Then you get a beautiful symmetric, symmetric
across to planes no across one plane diagram
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okay. It is very fast so the structure looks
little formidable when you saw it in the beginning
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but, now you see it is quite easy to handle
okay.
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So, the idea is for you to know that such
things can be done you done the assignments
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but, beyond that we are going to do everything
automatically using matrix methods, so it
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is it is important know that this is how many
structures were analyzed for many years.
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Now, we go to the fourth category, the toughest
that is when you have sway which you cannot
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get rid of it is unknown it is not like a
support settlement which is known the chord
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rotations are unknown in some cases you can
take advantage of them. We will look at those
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cases in tomorrow's class for the time being,
let us look at how to deal with this.
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So, if you have a symmetric portal frame with
the symmetric loading what is a deflected
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shape? will it be also symmetric? Yes, it
will be symmetric. If the loading is slightly
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eccentric you have an a symmetric loading
then the question I want to ask you is, is
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it going to sway to the lefts or is it going
to sway to the right, what is your hunch?
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To the right side.
Did we discussed this earlier?
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Okay, now what you need to do to get it intuitively
right is keep pushing the load to the extreme
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length and press it down hard and your mind
will tell you which way is going to move,
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but you can also give a very beautiful analytical
way of explaining why it is going sway to
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the right. Let us look at that, let us say
you do moment distribution of this frame okay,
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you have two joints where you have unknown
rotations you can do it right?
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Will you get the correct answer? will you
get the correct answer you will get some moments
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but, you will have a problem with the answers
you get. What is the problem that you get?
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that is the problem you get in moment distribution
method not with slope deflection method. What
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is a degree of indeterminacy of such a frame?
Two.
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But, there is a sway, so it is 3, so you cannot
wish a way that sway degree of freedom, so
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the moment distribution method the way we
have learnt till now where you have unknown
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rotations is not able to handle sway, so if
you do by moment distribution method you are
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actually doing it for a braced frame which
means without your knowledge you are not leading
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it sway that means there is a restraints there,
right? So, you get something from the distribution
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table but, the frame you are analyzing is
not the original frame but, a frame in which
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someone has arrested the sway degree of freedom
right? and then your results will be correct
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for that problem.
Now, where do you think the moments will be
21:54.009 --> 22:00.029
more, in the left column or the right column?
Let us take the beam. If it is a fixed beam
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where will the fixed end moments be more,
left or right? Left, so do not you agree that
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the it is not fully fixed it is partially
fixed, so those moments get affected and whatever
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moment you get in the beam end will be passed
on equal and opposite to the column end, so
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do not you realize that the column on the
left side is likely to have more bending moment
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at the top than on the right side, right?
and the moment at the bottom is 0, because
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it is a hinge at the bottom.
Now, if you will you have a horizontal force
22:42.340 --> 22:49.340
at the bottom? Yes. What will be the value
of that force? it will be the moment at the
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top of the column divided by the height of
the column, so you will get two horizontal
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forces, which one will be more or will they
be equal?
23:02.769 --> 23:09.769
Sir they are equal.
Why do they have to be equal?
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In opposite
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Yes, right.
So to satisfy equilibrium where will that
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force have to go?
Hinge that hinge. That artificial.
23:24.330 --> 23:31.330
Artificial support.
That is a key to it you are right. So, you
23:31.789 --> 23:35.950
are if you do moment distribution method you
would not be able to satisfy equilibrium because
23:35.950 --> 23:41.389
your horizontal forces are you will find you
need a net horizontal force the way I have
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shown you and then you have to do another
analysis, which is called a pure sway analysis
23:47.220 --> 23:51.259
where you have a horizontal load acting, so
this is a problem with moment distribution
23:51.259 --> 23:58.259
method that is you need to do two distributions,
one with the sway arrested and the other you
23:58.940 --> 24:03.859
have to figure out how much that sway then
do that analysis, so it is not worth it and
24:03.859 --> 24:08.169
so, we would not follow moment distribution
method beyond this point. We will say if you
24:08.169 --> 24:15.169
have such problems you can do it but, we grown
up and we do not do it that way okay and here
24:16.070 --> 24:20.419
you can clearly see why it is going to sway
to the right because the net load is going
24:20.419 --> 24:24.700
to act in that direction you have to reverse
the load acting on your artificial restrain
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this is a beautiful powerful left brain understanding
of why the sway is always to the right okay.
24:34.509 --> 24:41.509
Now, let us say how to deal with sway. I have
suggested that do not deal directly with translation
24:43.179 --> 24:49.950
always deal with chord rotation, because they
are much easier to deal with. So, you will
24:49.950 --> 24:55.320
have sway degrees of freedom in a frame like
this to figure out how many sway degrees of
24:55.320 --> 25:02.059
freedom you need to have you? You conveniently
put hinges at all the junctions and let it
25:02.059 --> 25:08.159
sway like a mechanism then you will find this
there is one unknown degree of freedom here
25:08.159 --> 25:15.159
that is delta, is it clear?
So, the other way to do it is at every joint
25:17.539 --> 25:23.149
in a plane frame you have two translation
degrees of freedom any two orthogonal directions
25:23.149 --> 25:30.149
you can take vertical and horizontal, so you
have at A and D both are arrested, so you
25:30.159 --> 25:36.009
have four potential degrees of freedom at
B and C agreed? four translation degrees of
25:36.009 --> 25:43.009
freedom but, if you assume members to be axially
rigid you are bringing in as many constraints
25:44.539 --> 25:51.539
as there are members. So, you have three constraints
because three lengths remain unchanged the
25:52.639 --> 25:58.119
length of A B, the length of B C and the length
of C D, so that is a clever way of figuring
25:58.119 --> 26:04.129
out how many independent sway degrees of freedom
you have four potential translations degrees
26:04.129 --> 26:10.009
of freedom three constraints constraining
equations because of the length 4 minus 3
26:10.009 --> 26:17.009
is equal to 1, got it? and you can choose
anyone in this case delta. And then you work
26:19.499 --> 26:24.909
out your chord rotations if delta is to the
right positive then the chord rotation phi
26:24.909 --> 26:30.100
A B is delta by H and C D is also delta by
H.
26:30.100 --> 26:37.100
What happens if you if your roof is tilted?
Okay, it is called a bent portal frame then
26:38.559 --> 26:45.559
also you have three members four potential
degrees of freedom 4 minus 3 1 then also you
26:47.669 --> 26:54.149
have only one unknown and it is delta both
will move both will move parallelly both the
26:54.149 --> 26:59.609
column will move parallelly as shown here
and what is interesting is the beam does not
26:59.609 --> 27:06.100
have any chord rotation whether the beam is
horizontal or inclined there is no chord rotation,
27:06.100 --> 27:11.720
so you will have sway you have chord rotation
only in the vertical elements and the chord
27:11.720 --> 27:18.720
rotation A B is delta by H 1 the chord rotation
in C D is delta by H 4, got it? So, we are
27:21.239 --> 27:23.649
going to increase the complexity now.
27:23.649 --> 27:29.029
What do you do when you have frame like this?
when we will solve this problem? Tomorrow.
27:29.029 --> 27:36.029
When do you encounter frames like this? Well
you got a house in a hilly sites banjara hills
27:37.679 --> 27:42.690
in Hyderabad may be and you are went off from
the first floor here and the ground floor
27:42.690 --> 27:49.539
there and your architect wants a fancy design
with incline walls, so you can have a frame
27:49.539 --> 27:56.539
like that, you know your foundation are at
two different levels, right? Then also you
27:57.559 --> 28:02.739
should remember even though it look complicate
and this is a very interesting problem you
28:02.739 --> 28:06.940
do not need to solve it but, at least conceptually
you should know how to solve it.
28:06.940 --> 28:13.940
You still have only one sway degree of freedom,
why? Because four potential translations at
28:15.119 --> 28:22.119
B and C, three members, three constraining
equations 4 minus 3 1. You can still choose
28:22.450 --> 28:28.509
delta but, now your beam is going to have
a chord rotation because if you play it with
28:28.509 --> 28:35.090
it like as you play with the MeCcano's set
it is a mechanism, if you play with it if
28:35.090 --> 28:42.090
the if these are rigid members it will take
a shape like this, agreed? And chord rotations
28:43.869 --> 28:50.869
for A B is simple it is phi A B, phi A B is
delta by H 1 for C D also it simple it is
28:53.279 --> 28:58.090
delta by H 2 but, it is a little tricky when
you calculate you can use trigonometry and
28:58.090 --> 29:01.149
calculate and that is the formula that you
get.
29:01.149 --> 29:06.279
I would like you to you know this these book
are available for you but, that is how you
29:06.279 --> 29:10.149
do it, this is how it was done in the olden
days in slope deflection method you do not
29:10.149 --> 29:17.149
have to worry at all about these things you
can do it directly. In the matrix method also
29:18.720 --> 29:24.429
its done directly.
If you have a pitch portal frame you can similarly,
29:24.429 --> 29:28.249
workout those equations do not worry too much
about it.
29:28.249 --> 29:35.249
Normally, our buildings do not look that funny,
you do not have inclined column you do not
29:35.519 --> 29:41.669
have inclined beam you have frames which are
called reticulated frames made up of rectangles
29:41.669 --> 29:47.210
where your life is made easier and if you
say axial deformations are negligible, then
29:47.210 --> 29:54.210
you have only one rotational dig unknown at
every joint and one sway degree in every floor,
29:56.220 --> 29:59.659
right.
Why only one sway degree in every floor? Because
29:59.659 --> 30:06.659
all the joints in that floor will move like
a rigid body okay, and there are serviceability
30:07.779 --> 30:13.320
limitations in a tall building the maximum
drift in the building should be limited so
30:13.320 --> 30:19.169
usually it is height by 500 under lateral
loads, okay?
30:19.169 --> 30:22.330
Can you analyze the frame like this exact
solution?
30:22.330 --> 30:29.330
Yes, that is the idea we want to be able to
handle any structure any skeletal structure
30:29.619 --> 30:36.619
of any shape subject to any loading including
indirect loading that is the objective of
30:37.639 --> 30:44.210
structural analysis.
Okay now, in slope deflection equations till
30:44.210 --> 30:49.049
now we were conveniently having only rotational
unknowns and the moment equilibrium equations
30:49.049 --> 30:56.049
were M B A plus M B C is equal to 0 kind of
equations. Now, you have a translational unknown,
30:59.570 --> 31:06.279
so moment equation is not going to work here
so you need a force equilibrium equation it
31:06.279 --> 31:10.820
is called a shear equilibrium equation. Let
us demonstrate that.
31:10.820 --> 31:16.879
Let us take this example this example okay,
where you have a portal frame with an a symmetric
31:16.879 --> 31:23.879
load un-symmetric load, right?. First identify
what is the what are your unknown displacements,
31:27.659 --> 31:34.659
theta B, theta C and you can call it just
delta you can call it delta B C because every
31:38.659 --> 31:45.440
point in that beam is going to translate by
the same delta, so theta B, theta C delta
31:45.440 --> 31:52.440
B C fixed end moments well your columns do
not have any fixed end moments but, your beam
31:54.129 --> 31:59.590
does and you can easily calculate that, right?,
you can calculate a fixed end moments at the
31:59.590 --> 32:06.590
beam, right? Minus W A B squared by L squared
plus W A squared b by L square fine.
32:06.799 --> 32:13.799
Now, what do we do next? Slope deflection
method. What do we do next? you have to write
32:19.649 --> 32:25.509
down the slope deflection equations chord
okay, rotations are clear here, right, so
32:25.509 --> 32:32.080
what will they look like? will you try, write
down the slope deflection equations for the
32:32.080 --> 32:39.080
three members A B, B C and C D, so how many
equations to you need to write down six equations
32:39.700 --> 32:46.700
six equations and the you can write them mechanically
of the six equations, in the six equations
32:47.190 --> 32:52.969
you have to put them in terms of three unknown.
What are the three unknowns?
32:52.969 --> 32:59.969
Well, instead of phi you say delta, delta
by 4 is phi, right?
33:01.009 --> 33:08.009
So, let us demonstrate for A B, do you agree
with this? M A B is M F A B plus 4 E i by
33:11.509 --> 33:18.509
L of theta A is gone because theta A is, so
you are left with 2 E i by L of theta B minus
33:20.049 --> 33:27.049
6 E i by L into phi, phi is delta by 4, is
it clear? Likewise you can write for M B A,
33:32.749 --> 33:39.749
M B A will be M F B A plus 4 E i by L theta
B, minus 6 E i by L into phi, right?
33:41.419 --> 33:48.419
Theta C, where is it? It is outside A B, so
do not worry, right? Theta c is not same thing
33:58.879 --> 34:03.549
in a continuous beam, right? you had theta
B and theta C when you are dealing with this
34:03.549 --> 34:09.869
beam, C was far away, so you do not have to
worry about it, is it clear?
34:09.869 --> 34:14.290
If you have open them out it will look like
continuous beam you are writing down same
34:14.290 --> 34:21.290
equations. Is there any doubt on this? In
the same manner you can write for B C in B
34:21.690 --> 34:26.270
C will there be any sway any chord rotation?,
No.
34:26.270 --> 34:33.270
So, for B C it is a straightforward thing
fixed end moment you can write down in simple
34:33.270 --> 34:40.270
4 E i by L theta B in 2 E i by l theta C,
is it clear? For M B A, for M B C, and for
34:41.450 --> 34:48.450
M C B straightforward and for M C D and D
C, they will look similar to M A B and M B
34:52.470 --> 34:57.160
A, right?
So, actually you have to just copy those equations
34:57.160 --> 35:02.180
only thing your theta B gets replaced now
by theta C but, you have to put the right
35:02.180 --> 35:08.010
theta at the right place. I hope you know
now understand that slope deflection equations
35:08.010 --> 35:13.000
are a blind method of doing it you do not
have to worry about anything you get this.
35:13.000 --> 35:17.210
Now, what is the next step?
35:17.210 --> 35:24.210
Write down the. So, you need three equations,
two of them are very easy, so corresponding
35:24.309 --> 35:31.309
to okay, I can summarize a nice rightly in
writed nicely in this matrix form, so you
35:34.020 --> 35:41.020
look carefully here these are my six moments,
right? three pairs three elements these are
35:42.230 --> 35:46.160
my fixed end moments only the beam has fixed
end moments the columns do not have fixed
35:46.160 --> 35:53.160
end moments these are my coefficients related
to E i theta B, E i delta C and E i delta
35:53.880 --> 36:00.880
B C which comes from either 4 E i L or 2 E
i by L or 6 E i by L square, right? And if
36:04.170 --> 36:10.980
somehow I get the answers for theta E i theta
B E i theta C E i delta then I just plug it
36:10.980 --> 36:14.000
in I get the final answers, very simple, is
it clear?
36:14.000 --> 36:20.260
Now the big question is how do I get those
three equations? Of two of them are straight
36:20.260 --> 36:24.970
forward we have done it in continuous beams.
There is no net moment at B, no net moment
36:24.970 --> 36:31.059
at C you pick up the second and third rows
and just add it up you will get two equations.
36:31.059 --> 36:38.059
Where you get third equation?
Please tell me how to write, how to write
36:38.170 --> 36:42.329
the third equation?
The clue is you have to write it in terms
36:42.329 --> 36:49.180
of these moments. How will you write?
36:49.180 --> 36:56.180
No, no.
M A equal to M A B equal to.
36:57.150 --> 36:59.960
You tell me yeah this is your structure. Tell
me what to do?
36:59.960 --> 37:05.369
Vertical reactions.
You will be able to write the reaction in
37:05.369 --> 37:07.250
terms of M B C and M C B.
Sir M A B is equal to M D C
37:07.250 --> 37:14.079
How does that help you?
I will give a clue; no I will give a clue.
37:14.079 --> 37:18.099
Sir M A B is equal to M D C, M A B is equal
to M D C D C who told?
37:18.099 --> 37:23.869
Horizontal force at B and C those moments.
Horizontal force is different from moment.
37:23.869 --> 37:29.630
That that into this.
I will give you a clue. See you wrote M B
37:29.630 --> 37:36.339
A plus M B C equals 0, because there is a
B common there, because theta B was the corresponding
37:36.339 --> 37:40.559
unknown rotation.
Now, you have delta B C you have a sway degree
37:40.559 --> 37:45.039
of freedom, you have to write a force equilibrium
equation, right? not a moment equilibrium
37:45.039 --> 37:52.039
equation, which force what equilibrium?
The shear force in column A B.
37:52.839 --> 37:59.839
The shear force in column A B, is it a constant
shear force? is it a constant shear force
38:04.349 --> 38:09.520
is it a constant shear force? Yes.
Yes.
38:09.520 --> 38:13.510
Because you have a column, if you take the
free body of the column you have a moment
38:13.510 --> 38:18.660
only at the top and the bottom there is no
intermediate load and let us say they are
38:18.660 --> 38:25.660
clockwise positive you have M B A and you
have M A B, so your shear force is?
38:26.700 --> 38:32.130
Well, there are both positives, so where is
a difference?
38:32.130 --> 38:38.720
2 M B A by the 4.
You add them M A B plus M B A divide by H
38:38.720 --> 38:43.369
will give you a shear force well to make your
life easier, why do not you look at it as
38:43.369 --> 38:50.369
horizontal reaction at A and D? What should
those two reactions add up to? see easiest
38:52.180 --> 38:53.279
thing you can do.
38:53.279 --> 38:57.839
So, you do not break your head too much over
it that is all you need to do. Horizontal
38:57.839 --> 39:04.789
reaction at A and D should add up to 0 and
for the sign convention we have assumed M
39:04.789 --> 39:10.599
A B plus M B A dived by 4 clockwise, where
is that reaction pointing from left to right
39:10.599 --> 39:17.599
or right to left?
39:21.579 --> 39:28.579
This is, so for equilibrium this will point
this way and this will point this way. This
39:34.140 --> 39:41.140
is what we called H A and
it will be, right? And this will be equal
to this end wherever you cut a section the
39:49.329 --> 39:56.329
shear force is constant. What happens if you
have instead of A B you have D C and C D?
39:59.059 --> 40:06.059
well the only change is this A becomes B and
you get, so this is the thinking you need
40:13.289 --> 40:20.289
to do and now you got expression of M A B,
M B A from your slope deflection equation
40:21.260 --> 40:26.460
plug it into that equation and you got three
equations.
40:26.460 --> 40:33.460
So, George Maney discovered this way back
100 years back powerful method you get actually
40:34.650 --> 40:41.650
the last equation you have to play a little
bit play bit with it to get the symmetry in
40:42.920 --> 40:43.859
that matrix, okay.
40:43.859 --> 40:48.799
Actually, they do not turned out to be the
way it is, I have shown you because there
40:48.799 --> 40:53.700
is a minus sign you have to multiply that
row with minus 1 you will get the symmetric,
40:53.700 --> 41:00.700
okay? Now, what is this matrix look like?
what is what is that matrix? Where displacement
41:02.859 --> 41:09.859
method or stiffness methods not flexibility
method that matrix is a property of the structure.
41:11.230 --> 41:17.740
What is that what is that matrix called?,
Well it is related to the stiffness matrix
41:17.740 --> 41:24.740
and the beautiful part about this is the vector
that you get here is caused by the loads that
41:26.789 --> 41:30.329
left side is completely a property of the
structure, you change the loads only this
41:30.329 --> 41:37.329
will change this will not change and, so we
will see how you can generate that stiffness
41:37.859 --> 41:43.630
matrix from first principles I am stepping
ead in to into matrix methods but, I want
41:43.630 --> 41:49.200
us to go through that exercise.
We got this blindly by just applying equilibrium
41:49.200 --> 41:55.180
equation, right? but, with our eyes open can
we generate this matrix. Let us do that than
41:55.180 --> 42:00.240
you have a real understanding of the physics
behind the mathematics okay.
42:00.240 --> 42:06.789
So, you can solve these equations and a now
days you calculators which do it you know
42:06.789 --> 42:12.950
just by pressing some button so I will not
waste time, you will get these answers okay?,
42:12.950 --> 42:18.390
and then what should you do? Plug them back
into the slope deflection equations you got
42:18.390 --> 42:25.390
the answers, so after you got the end moments,
what should you do? You draw free body diagrams,
42:27.410 --> 42:29.410
I hope you know how to do that.
42:29.410 --> 42:36.410
Draw the free body diagrams clockwise moment
put clockwise negative sign move put anticlockwise
42:38.720 --> 42:44.369
and then everything should match all the equilibrium
should be satisfied you can now find the vertical
42:44.369 --> 42:51.369
reactions in the beam and the horizontal reactions
in the column and you will find that this
42:52.190 --> 42:57.859
perfect equilibrium everything is balanced
you got the exact solution powerful method,
42:57.859 --> 43:02.680
agreed? Then you can draw the bending moment
diagram it will look like that. So, you not
43:02.680 --> 43:08.819
afraid of sway you know how to deal with sway
but, if you have multiple sway degrees of
43:08.819 --> 43:12.339
freedom you have to do it little more carefully.
43:12.339 --> 43:19.339
Now, I want ask to go through this small exercise
of generating the stiffness matrix of this
43:19.779 --> 43:23.619
frame from first principle.
Shall be give it a trail okay?
43:23.619 --> 43:30.619
You will love it, it is simple. How many degrees
of freedom are there?
43:30.650 --> 43:37.650
Three, shall we number them 1, 2, 3 Theta
B is our first degree of freedom theta C is
43:41.369 --> 43:47.279
our second degree of freedom and delta B C
is our third degree of freedom. Okay, to write
43:47.279 --> 43:51.990
to generate the stiffness matrix of this what
should I do, what is the size of the, what
43:51.990 --> 43:54.089
is the order of the matrix?
3 by 3
43:54.089 --> 44:01.089
3 by 3, what should I do? From first principle.
Give a unit displacement at that location.
44:01.190 --> 44:08.089
Give a unit displacement; let us say I want
to fill the first row. First column of the
44:08.089 --> 44:13.450
stiffness matrix what should I do?
First I should arrest everything.
44:13.450 --> 44:15.490
Yes sir.
Rotate
44:15.490 --> 44:18.319
And then I should allow only.
Unit displacement at.
44:18.319 --> 44:21.470
One displacement at a time, which one?.
First one.
44:21.470 --> 44:28.470
So, shall we call it D 1, D 1 should be 1,
D 2 should be 0. Can you sketch the deflected
44:29.420 --> 44:36.420
shape of such a frame? Do it. Sketch the deflected
shape we are trying to find the first column,
45:21.170 --> 45:28.170
so to fill this column we need to put
right draw a sketch, what will it look like?
it should look like this, right? It should
45:41.420 --> 45:42.339
look like this
45:42.339 --> 45:49.339
Only a unit rotation at B, no translation,
no rotation at C, right? Okay.
45:51.079 --> 45:58.079
Can you draw the next one can you draw the
sketch that you need to do for this one? What
45:59.010 --> 46:06.010
will it look like?
Okay, let us look at this. What is the physical
46:07.569 --> 46:12.019
meaning of K 1, 1.
46:12.019 --> 46:19.019
So, when you see when you hear of stiffness
you are reminded of a spring, spring constant,
46:22.190 --> 46:29.190
right? So, it is a stiffness, so spring stiffness
is force per unit deflection here you have
46:33.160 --> 46:40.160
two coordinates, so it is a force at the coordinate
one caused by unit displacement at the coordinate
46:40.210 --> 46:43.660
J with all other coordinates arrested
46:43.660 --> 46:50.660
So, here only theta one only D one is there
and, so one way to do it is you remove that
46:51.589 --> 46:56.890
restraint then you deliberately give an external
moment that moment you need to apply to make
46:56.890 --> 47:03.890
the joint rotate by a unit angle is K 1 and
the reaction that develops at the other artificially
47:07.269 --> 47:14.269
fixed end at B is K 2 1 and the reaction that
is developed in the direction of the coordinate
47:14.630 --> 47:20.329
you identified in the translation degree of
freedom is called K 3 1.
47:20.329 --> 47:24.160
Can you calculate these quantities?
Yes sir.
47:24.160 --> 47:30.630
Do it.
Maybe you let us draw this also the next case
47:30.630 --> 47:37.630
which is the similar your job now is to calculate
K 1 1. What is k 1 1?
47:38.890 --> 47:43.730
Yes.
8 E i by L.
47:43.730 --> 47:50.730
No, wrong.
Let us do it slowly. It is going to be 4 E
47:53.109 --> 48:00.109
i by, what is L? 4 E i 4 plus?
4 in to 2 6
48:01.859 --> 48:08.859
6 E i by L
These are the other notations, agreed? Right
48:11.190 --> 48:18.190
now, we need not do this is equal to what
from the other figure? K 1 1 will be equal
48:20.460 --> 48:27.460
to K 2 2, is it not? So we need not spend
more time on that K 1 1 and K 2 2 will be
48:28.859 --> 48:35.859
identical. Can you find out K 1 2 or K 2 1,
K 2 1 is how much? it is a carryover moment
48:39.509 --> 48:41.900
you get.
2 by 2 E i.
48:41.900 --> 48:44.299
2 by 2 E i.
Which one?
48:44.299 --> 48:48.339
2 by 2 E i.
Wait one minute; do you have a problem with
48:48.339 --> 48:50.650
this addition?
48:50.650 --> 48:56.420
1 plus 4 E i.
48:56.420 --> 49:03.420
7 by 3.
7 by 3 that is right.
49:04.839 --> 49:11.839
7 by 3 you are right well caught, 7 by 3 agreed,
7 by 3 okay.
49:15.160 --> 49:20.730
This one? This is correct?
Yes sir.
49:20.730 --> 49:27.730
Okay, it is 7 by 2 not 8 by 3 you are right,
good. This is equal to K 1 2, so there itself
49:30.390 --> 49:35.740
you see the symmetry in the matrix. What is
left over here? K 3 1, can you find out what
49:35.740 --> 49:42.740
k 3 1? That is a clue that is a clue that
is a clue,
49:46.589 --> 49:53.589
So, you had here if you take that column you
had how much? 4 E i 4 and what do you get
49:56.240 --> 50:03.240
here? 2 E i by 4 and this will give you shears
which are going to be equal to 6 E i by 4
50:16.490 --> 50:23.490
squared and so if you take that frame in which
you arrested this degree of freedom and arrested
50:27.799 --> 50:34.799
this degree of freedom and you applied K 1
1 equilibrium suggest that you will definitely
50:37.220 --> 50:44.220
get horizontal reaction here which is equal
to this quantity and you do not get anything
50:46.450 --> 50:53.309
here because this does not bend, so the reaction
you get here will be equal to how much?
50:53.309 --> 50:55.849
No.
Minus.
50:55.849 --> 51:00.549
Minus, because they should add up to 0, so
minus how much?
51:00.549 --> 51:05.279
6 E i by 16
6 E i by 16 what to?
51:05.279 --> 51:09.390
3 E i by .
That is it. This is the physical approach
51:09.390 --> 51:16.390
now it is easy now it is easy, right? This
will be equal to K 3 2 in the second figure.
51:17.170 --> 51:24.170
So, we have got the first two columns very
easily last column can you draw the sketch.
51:25.579 --> 51:28.910
Infinite rigidity of.
That is right.
51:28.910 --> 51:33.420
It is like a rigid beam this is how it is
going to move? Point of contra flexure will
51:33.420 --> 51:40.420
be in the mid height of every column, so what
will it look like? Can you calculate? can
51:44.049 --> 51:51.049
you calculate K 1 3 and K 2 3, it likes settlement
of supports it likes settlement of supports
51:52.390 --> 51:58.509
it is a chord rotation, so what is the moment
that you get? K 1 2 and K 2 3.
51:58.509 --> 52:02.109
K 3.
6 E i by 4.
52:02.109 --> 52:09.109
Sir K 3 into L by .
Into 1 by 4 because your chord rotation is
52:09.579 --> 52:16.470
1 by 4, so remember this remember this.
52:16.470 --> 52:23.470
If I have settlement of supports problem and
this is delta this is the shape that you get
52:29.990 --> 52:36.470
clockwise chord rotation will give you anticlockwise
end moments. What are these moments? well
52:36.470 --> 52:43.470
6 now do not talk about 5 it will come to
6 E i l square by delta, because delta by
52:43.930 --> 52:47.599
L is 5 do not make mistakes like that you
lose everything.
52:47.599 --> 52:54.599
Okay, so that is easy now, you got it? And
there is a minus sign clockwise chord rotation
52:56.730 --> 53:00.240
will give you anticlockwise end moments, got
it?
53:00.240 --> 53:06.990
So, you got K 1 3, you K 2 3, you got everything
in this matrix except this last fellow. What
53:06.990 --> 53:13.990
is k 3 3? will it be positive or negative?
Do you need to think to answer this all the
53:16.539 --> 53:23.539
diagonal elements have to be positive why?
Displacement caused it.
53:24.049 --> 53:27.049
You what it mean? I am pushing it is to the
right obviously it is going to deflect to
53:27.049 --> 53:33.970
the right, if kick a football to the front,
it is not going to come back, do you understand?,
53:33.970 --> 53:38.309
so that is what all the diagonal elements
are doing, I am pushing it in the in the same
53:38.309 --> 53:44.230
direction as identified positive degree of
freedom, so it has to be positive, do not
53:44.230 --> 53:47.900
break your head over it.
Okay, the half diagonal elements can be negative
53:47.900 --> 53:54.900
or positive diagonal elements will always
be positive. How much will it be? So, what
53:55.509 --> 54:01.109
are the shears you get here? This is 6 E i
by L square this is 6 E i by L squared the
54:01.109 --> 54:08.109
shear will be what? Tell me the value.
3 E i delta by L square L cube.
54:09.200 --> 54:12.410
3 E i delta by L.
54:12.410 --> 54:19.380
Where did you get three from?
3 E i delta by L.
54:19.380 --> 54:23.259
Come on how where you get 3 from?
By 2 sir.
54:23.259 --> 54:25.400
What by 2?
Moment plus.
54:25.400 --> 54:32.099
Take this free body, what will be this shear?
What did we do here? See this place is divided
54:32.099 --> 54:38.400
by that.
6 E i delta.
54:38.400 --> 54:43.970
If you are an accountant, your bank balance
will have reduced by one-fourth by now, if
54:43.970 --> 54:50.970
you play games like this, right? So it is
not 3, it is 12, one-fourth the answer you
54:53.390 --> 54:55.480
have given is one-fourth the correct answer,
is it not?
54:55.480 --> 55:02.480
This moment plus this moment divided by that
is span. 12 E i, it is right in front of you
55:05.640 --> 55:12.640
by L cube, now that is interesting, so what
is the answer? what is K 3 3? What is k 3
55:22.339 --> 55:29.339
3? Give me the answer final answer?
Is it 12 E i by 4 cubed?
55:35.259 --> 55:36.839
Yes sir.
No wrong, answer is wrong.
55:36.839 --> 55:38.950
No, no, no what a pity?
Twice.
55:38.950 --> 55:42.990
Twice, there are two columns. The two columns
at the bottom there you have two reactions
55:42.990 --> 55:47.650
they should all add up to that applied force,
so in to 2 do not make mistake like this,
55:47.650 --> 55:54.650
so physical approach is for bright students
who are alert who know how to satisfy equilibrium
55:55.180 --> 56:01.569
but, it is wonderful, is it not? So if you
do that.
56:01.569 --> 56:06.609
Apart from that small mistake of 8 by 3 which
should be 7 by 3 you get the same matrix which
56:06.609 --> 56:13.609
we got in the blind manner, right? We did
it the blind way and we got the same answer.
56:13.650 --> 56:20.650
As I said do not attempt this by moment distribution,
but if by chance you did it, then you get
56:20.700 --> 56:26.220
the same fixed end moments you get the distribution
factors you draw the table you get some values
56:26.220 --> 56:33.220
but, the problem is you will not satisfy equilibrium
and you can check it out you can check it
56:33.230 --> 56:39.069
out you will find that the horizontal reactions
do not match there is the balance and you
56:39.069 --> 56:45.640
need to analyze the frame for that lateral
load. So, let us not do moment distribution
56:45.640 --> 56:52.640
method when there is sway I will conclude
by dealing with one last problem of pure sway.
56:53.730 --> 56:59.980
You have this frame subject to pure sway okay.
How do you solve this problem, are there any
56:59.980 --> 57:06.980
fixed end moments? No fixed end moments deflected
shape will look like that. Remember we did
57:07.109 --> 57:12.079
an approximate analysis when we did portal
method cantilever method we do not do anything
57:12.079 --> 57:19.079
approximate now we do exact this is exact
solution, so same three unknown displacements
57:19.130 --> 57:23.980
I will not waste time write down slope deflection
equation because they are the same as in the
57:23.980 --> 57:29.670
previous problem with one additional advantage,
what is the advantage? No fixed end moments,
57:29.670 --> 57:36.660
right? The pure sway that that is going to
the node it is not going in between any member
57:36.660 --> 57:43.660
then what about your equilibrium equations?
Same same except, in that last equation when
57:46.259 --> 57:51.470
you put sigma F X equal to 0, you have to
add the lateral load which was not there in
57:51.470 --> 57:56.869
the earlier problem and do it properly put
plus 50 on the left hand side of the equation
57:56.869 --> 58:01.220
and not on the right hand side because you
will get totally wrong results otherwise,
58:01.220 --> 58:02.660
clear?
58:02.660 --> 58:07.799
Then you get the same stiffness may in fact
if you do stiffness method you can use this
58:07.799 --> 58:14.799
you do not have to do slope deflection method
F 1 is 0, F 2 is 0, F 3 is 50, right? So you
58:17.549 --> 58:19.299
can do it this way also.
58:19.299 --> 58:26.299
Now you see, this is the stiffness method
which we are going to do soon solve it, plug
58:29.410 --> 58:34.930
it into the back into the slope deflections
equations you get this answer which we.
58:34.930 --> 58:40.369
Now look at this problem I am going to raise
a question and will stop this class. There
58:40.369 --> 58:47.369
is something nice about this bending moment.
What do you call such figure? It is it is
58:47.999 --> 58:54.999
not symmetric it is anti-symmetric, so that
gives you a clue. It is anti-symmetric and
58:57.099 --> 59:03.740
the moment at the middle of the beam is 0,
so maybe we could have taken an advantage
59:03.740 --> 59:04.450
of this anti-symmetric.
59:04.450 --> 59:11.450
How do it take advantage of it? you can do
this you can do this symmetry plus anti-symmetry
59:16.910 --> 59:21.230
and when you cut it in the middle because
that is why your moment is 0 you can replace
59:21.230 --> 59:27.509
your original structure with this structure.
This structure does not have any bending pure
59:27.509 --> 59:33.029
axial compression in this member only this
will bend and now it is left you can either
59:33.029 --> 59:38.799
choose a left one or the right one, your problem
has become much less complicated. How many
59:38.799 --> 59:45.799
unknowns do you have there? Theta B why? You
can ignore sway, because it is a cantilever
59:48.349 --> 59:52.319
action we have done a problem earlier like
this, so life is easy you have to just take
59:52.319 --> 59:59.319
out that theta B is your unknown, right?
1:00:06.400 --> 1:00:13.400
We will look at this in the next class but,
this is a beautiful shortcut method this is
1:00:13.730 --> 1:00:18.980
what we will explore in the next class and
this is what you got in your assignment the
1:00:18.980 --> 1:00:22.769
last problem but, you got two floors.
Thank you.