WEBVTT
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Today, we shall start design of slabs. So,
we have a few more lecturers, so today we
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shall start part 1. And this is your lecture
number 14, design of slabs part 1.
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Before we start the design of slabs, let us
see what our words says related to your effective
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span.
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We have, say a masonry wall. This is our ground
level; it may be a beam or a slab. We can
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take as if it is supported by 2 masonry walls.
So, we have different spans. This is your
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ground level, we have
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foundation but we are not interested right
now. This may be a beam or a slab, we have
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this is your clear span like clear cover,
we can say clear span. And let us see the
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centre line of the supports; here
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this is masonry wall or brick wall. This is
called that centre to centre distance, but
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this is not the effective span. Our interest
here; we have to find out that effective span,
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like say effective depth,
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here also we can consider that effective span.
Where we have to use this effective span?
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We have to use this effective span, to calculate
moment.
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So, you can say for simply supported case,
let us say WL square by 8. You have find out
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the other way also; WL by 8, w is the total
load or w is the uniformly distributed load.
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Similarly, we can take say,
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for let us be specific simply supported. We
can specify 1 of the ways; whether you are
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specifying w udl or w the total load, whatever
way we specify. Similarly, for shear; WL by
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2 or simply we can write down
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w by 2 because, you will find out in codes
either it is in a total load. Total load is
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nothing but, w I can say, in this case which
is nothing, but w times l; total load.
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So, our interest here, what we have to find
out; what is the effective span? Or in the
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other way what is the value of l? For this
case, if you take the simply supported case
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what is the value of l that
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effective span, that we have to find out.
Now, let us see that what our quote says is
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456 what our quote says.
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We can take the effective span and we shall
get it in clause 22.2 is 456 2000. In design
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classes, we most of the time we Refer that
code because, wherever you have any dispute
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or another thing, that is the
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code it will help us to solve that problem.
So, if it is simply supported beam or slab,
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the effective span of a member that is not
built integrally with its supports that means:
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there is no movement
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developed in the support, shall be taken the
lesser of the following 2. That means: you
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are having 2, out of that which 1 will be
the lesser, that we you have to take.
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So, that means: you have to calculate that
clear span plus effective depth of slab or
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beam. That means: which are the case you are
considering. In our case, in few next few
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days, we shall design slabs, so we
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shall take here slab only. So, clear span
plus effective depth of slab or beam. So,
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that we have to take, that is 1 case that
is our effective span shall get. The other
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1 is centre to centre of supports. Out
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of these 2, which 1 will be minimum; that
we shall take it as effective span, for that
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problem.
So, here if we take, say 250 is the width
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of the wall brick wall, then we can find out
centre to centre distance, if the slab gets
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124 millimeter is the overall depth. So, you
can calculate, we can assume
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certain depth and on the basis of that, we
can go for calculation or replacement. If,
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it is 125 millimeter or 150 millimeter that
is, the thickness of the slab then, you can
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calculate effective depth and you
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can find out the effective span.
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But, if it is continuous beam or slab, so
we have the code says; it is in the same clause.
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If, the width of the support is less than
one- twelfth of the clear span, then it is
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dependent on the support width.
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The effective span shall be as per simply
supported case, the 1 which I have already
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told, so it will be according to that. Then
it is dependent on the width of that wall
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or the support but, it is not we do
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not stop it here, there are further clarification.
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Continuous beam if it is the other way, for
end span with 1 end fixed and the other continuous,
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what is continuous? We can start in this way
that means: the slab, we are specifying the
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slab or beam whatever
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you specify, if we take it as a say simply
supported, it can move like this. So, you
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can go like this let us say stop it here.
Let us say there is a column here, we can
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put it like this just to make it
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there is a column here, a column this side
so that means: I can say this side is fixed.
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So, our case in this case we can say for end
span with 1 end fixed and the other continuous
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or for the intermediate spans, the effective
span shall be the clear span between supports.
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So this is the case, so
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you will simply take that clear span between
supports.
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So, we can say for this case; this is the
clear span and also it is told that for intermediate
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supports, the intermediate span. These are
the intermediate span but, so we shall take
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the clear span. What
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about the other case? For end span with 1
end free and the other continuous, the effective
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span shall be the lesser of the following
2; clear span plus half the effective width,
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note half the effective width
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depth, clear span plus half the width of the
discontinuous support. That means: here in
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this case we are talking this problem.
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So, what should be the effective span? Because,
this side is free discontinuous support and
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we shall get for this case, your effective
span will be different. And that 1 will contribute,
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in calculating your
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moment shear other things; that is why it
is so important. Because, if you consider
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moment, so it becomes a l square. So obviously;
the effective span it will little bit crucial.
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Now, we have 1 more case: continuous beam
or slab. In the case of spans with roller
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or bearings, the effective span shall always
be the distance between the centers of bearings,
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so this is very simple. So,
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when you are talking say bearings, so then
it will be just you take the centre to centre
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distance between 2 bearings. So, that all
of them you will get in the is 456, but I
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think this is required to know
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with examples.
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Now, what about the cantilever? For cantilever,
the effective length of a cantilever shall
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be taken as 1, its length to the face of the
support plus half the effective depth, note
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in few cases you are taking
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effective depth and few cases you are taking
half of the effective depth, so that you please
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note. Or the length to the centre of support,
where it forms the end of a discontinuous
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beam, then this is the
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case where it is a discontinuous; it is the
end of a continuous beam.
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So, when it is continuous beam that means:
here if we take like this this problem, it
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goes. So, in this case what will happen, then
we shall take that these portion we shall
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take it as say your, we are
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talking this, as I say continuous beam. This
is a continuous beam we are considering say
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and at the end we are talking say you say
as slab, so there we shall take the second
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case, the length of the centre of
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support while it forms the end of a continuous
beam. That we shall take for that problem.
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So, according to this effective span, we shall
decide which is the case? What is the support
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condition? And on the basis of that we shall
find out the effective span.
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But, what about frames because we have to
analysis say your problem, we need this 1
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for analysis purpose. If, it is a frame structure,
that which you can analysis it with some say
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1 say method different
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methods by say any program is still now a
days we do it.
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So, in this case what will happen? In the
analysis of a continuous frame, centre to
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centre distance shall be used. So, you shall
not go of any experiment on anything on the
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effective depth or anything. So,
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we shall do the analysis for frames; multiple
frames for say seismic analysis, dead load
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live load, different load cases, we have to
take and accordingly we have to solve it.
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In that case, we shall take that
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effective span, simply the centre to centre
distance.
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We do that, our code says; that moment and
shear coefficients for continuous beams, which
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we can get it in clause 22.5 I think because,
I always Referring this clauses, so at least
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you should note down so
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that, you can refer those clauses. So, where
shall we use this? Our code gives few coefficients
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so substantially uniformally distributed loads
over 3 or more spans. This is for continuous
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beams or slabs when
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you shall take, substantially uniform, it
is not exactly but, it will be take 10 percent
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or 15 percent that way we say.
So, if it is a more than over 3 or more spans,
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which do not differ by more than 15 percent
of the longest. There is a longest span out
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of that and if you find that, not within more
than 15 percent, then we
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can use these coefficients.
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So, 1 table 12 that simply, I have copied
that 1 that is; bending moment coefficients.
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But, we have to use the different coefficients,
we have to use. So, for span moments and for
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support moments; there is 1
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span moment and there is support moment; so
near middle of end span, for span moment.
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And we are talking this 1 say dead load and
imposed load fixed, dead load and imposed
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load fixed.
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So, loading which are say you shears or any
machine that is fixed in its position, then
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we take it as a fixed imposed load. So in
that case, near middle of end span what does
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it mean? It means: so you can
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take this problem. So, near middle of end
span, why near middle of end span? Because,
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when in diagram may be sometimes, it will
be say something like that, for uniformity
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distributed load, so it will go like
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this and here it thins because, at this end
we do not have any end 1.
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So, it will be near middle of not exactly
in the middle, so to the near middle of end
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span and that 1 we shall get and have to take
that 1 by 12 that coefficient. So, that coefficient
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what is that
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coefficient? It means: the movement equal
to, if I takes a total load, so 1 by 12, w
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is the total load and l.
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So, let us write down w total load on that
span, where which 1 you are considering and
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l effective span. So, M equal to 1 by 12 WL
here that is, the coefficient we are taking.
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At the middle of interior span,
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at middle of interior span that means; this
is the interior spans, these are called interior
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spans, this is called interior span. So, we
shall get in the middle of the interior span;
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here, here and that how
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much you shall get? That is 1 by 16 times
that WL.
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Now, what about your supports? At support
next to the end support because, we are at
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the end support there is no movement, so at
support next to the end support, that 1 will
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be even minus 1 by 10 and at
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other interior support we shall get minus
1 by 12. So, here we shall get; here we shall
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get that you say support next to the end support
and this is your interior support, in this
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particular problem. So,
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this is your case, we are considering dead
load and imposed load fixed.
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Now, this 1onethat imposed loads not fixed,
which is movable that means; today at this
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time you are say, you are in the bedroom itself,
you can say that you say dressing table then,
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you are say cot and so
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many other things almirah, all those things
you just move around . So, that way we can
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say this is not fixed. So in that case, we
have to take these coefficients.
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Now, what about your other part? There is
1 shear force coefficients. This shear force
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coefficients here, again the similar fashion,
that dead load and imposed load that is fixed,
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at end support and at
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support next to the end support and at all
other interior supports. So, you shall get
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and we are having that, next to the end support
we have 2 positions; 1 is the outer side other
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1 in the inner side. What
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does it mean? It means: this is your outer
site and this is your inner site. And that
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v, that 1 will be equal to say 0.6W, W is
the total load.
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If, it is simply supported beam say then,
W by 2 that is 0.5 instead were, we are having
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0.4 0.55 0.5 0.6 like that. And that we have
to take it, the moment and shear force you
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shall get it, so that means;
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if the spans are uniform, that we say almost
constant, all the spans same and load also
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does not differ much then, we can use the
table 12 and table 13. These 2 tables we can
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immediately we can use it and we
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can find out the moment and shear force.
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Now, our quote says that for moments at supports,
where 2 unequal spans meet or in case where
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the spans are not equally loaded. It may be
the either case, that means: the spans are
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equal and but loading is
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different, so that is why you will get for
in 1 side you will get 1 moment, considering
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the other side you will get the other movement,
in the same point same junction, same support.
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If the spans are
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different then also, you shall get the different
movement. In that case, our quote says that,
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the average of the 2 values for the negative
moment because, we are talking say support
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at the support, may be
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taken for design. In that case we simply,
we shall take the average value. Because,
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at each and every level, there should not
be any confusion or dispute and that is dissolved
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in this quotes. That is why
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this quote is so important.
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So, this is another clause where, a member
is built into a masonry wall. So far, we have
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taken that masonry wall that mean; simply
supported, but it may happened that you are
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say beam or slab that is
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interiorly, that is inserted in the embedded
in the masonry wall itself. In that case,
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certain partial fixity may occur and our quote
says; where a member is built into a masonry
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wall, which develops only
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partial restraint, not full that you, not
WL squared by 12.
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The member shall be designed to resist a negative
moment, at the face of the support, at the
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face of the support of WL by 24. Instead we
are having to say, WL by 12 w is the total
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load l in this case and l
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is the effective span. So WL by 12 24 that
mean: we are taking half of that. So, per
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shelf fixed effected you can say half that
means, it is semi rigid you can say in 1 since
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or such other restraining moment
20:54.859 --> 21:00.220
as may be shown to be applicable or you can
argue, this is a very determinant problem.
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So, you have personally fix it you know, from
other some other sources or measurement. So,
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you can take that 1 also, so
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our quote says; that is what you can take,
but there should be only taken if the client
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of both sides that agree.
21:17.560 --> 21:21.810
Beams or slabs over free end supports, where
a member is built into a masonry wall, which
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develops only partial restraint, the shear
coefficients at the end support, that is which
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is given in table 13 that
21:29.139 --> 21:33.830
is a shear coefficients, may be increased
by 0.05 that means: you can add 0.05 that,
21:33.830 --> 21:40.609
if it is 0.4 then, you add 0.45. If, it is
0.55 then, you add 0.6. So that way you can
21:40.609 --> 21:42.970
take, that is for shear and the
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other 1 for moment.
21:45.060 --> 21:50.820
Now, let us come to our problem that, so far
we know that what is the support condition,
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what should be the effective span, how to
calculate shear and bending movement, how
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to calculate shear force and that
21:56.950 --> 22:04.869
we shall follow. So but, what are the different
kinds of slab? If we take, there are reinforced
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concrete solid slabs. Mainly, we shall take
say 1 way slabs, 2 way slabs, then flat slabs
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and flat plates.
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These are the 4 different we shall use it.
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If, we say, let us say this is 1 hall what
will happen this 1? You will find out that
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bending, you can take a piece of paper or
you can take say your any type of say, thin
22:45.909 --> 22:47.909
membrane type of thing, if it is
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just a plastic also, you keep it over this
type say arrangement, what will find out?
22:54.139 --> 23:10.159
You see that it is bending like this. If,
we make it says almost a square then, we can
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find over that you will see that
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both sides; this side as well as this side.
23:18.730 --> 23:32.929
Now, depending on these let us say Lx and
this is say Ly; the span in both sides. Depending
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on the situation Ly by Lx, we find out that
whether if it is 1 way or 2 ways. How to design
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that means; whether we
23:44.550 --> 23:50.730
shall consider only 1 side movement or whether
we shall consider the movement both sides.
23:50.730 --> 24:00.659
So, in our case 1 way; if Ly by Lx greater
than 2, then we shall take it as say 1 way,
24:00.659 --> 24:01.770
if it is say within less than
24:01.770 --> 24:07.770
equal to 2, then we shall take that 1 in this
problem we shall take 2 ways. We shall count
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the other 2 flat slabs and flat plates that,
we shall not design but that 1, we shall come
24:13.119 --> 24:14.409
later stage. At least, I
24:14.409 --> 24:19.090
shall show 1 or 2 examples how to design that
1.
24:19.090 --> 24:32.139
Here what happens, we are talking say frames
structures; if you take a grid, if you take
24:32.139 --> 24:46.570
a grid say like this say something like this;
which is coming, these are all beams these.
24:46.570 --> 24:47.599
And over that, we have the
24:47.599 --> 24:56.300
slab. So these are all slabs and that you
have to design. Now, for this problem if I
24:56.300 --> 25:03.619
discontinue the beam here, then we can get
this is 1 panel and here we can design it,
25:03.619 --> 25:07.419
as if it will beamed like this, this
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side.
25:08.470 --> 25:16.240
So, we have to design that 1, as I say 1 way.
Whereas this 1, it will bend this way as well
25:16.240 --> 25:21.070
as this way. So you have to design this 1
as I said 2 slabs that, I should will show
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you, that in the next few
25:23.859 --> 25:30.460
classes, how to design that 1. Now, what about
you say but that means it is supported by
25:30.460 --> 25:34.970
the beams or may be walls.
25:34.970 --> 25:47.300
Whereas, it may happen just to show you that
flex slabs. It may happen that it is like
25:47.300 --> 26:12.649
this. This is your beam column, so this is
your, we can design it as I said flat slab.
26:12.649 --> 26:15.080
We can also take it as, simply
26:15.080 --> 26:27.030
say flat plate as if it is supported by that
columns only. There is no beam, it is just
26:27.030 --> 26:32.030
supported by as if that is there are different
say prop, there are different supports and
26:32.030 --> 26:34.040
over that, you are keeping 1
26:34.040 --> 26:38.700
plate, whereas here you are just making little
bit bigger and then what that you have keeping
26:38.700 --> 26:43.669
the slab or plate. So this is your flex slab,
this one is talking that say here that flat
26:43.669 --> 26:43.859
plate.
26:43.859 --> 26:48.990
So, these are the 4 different types of say
your slab part, we generally design. But in
26:48.990 --> 26:53.450
our case, we shall only concentrated in this
class that is; 1 way slab and 2 way slabs,
26:53.450 --> 26:57.730
which we mainly generally use.
26:57.730 --> 27:04.629
The other portion, at least we should remember,
that is, loading on slabs for buildings. Let
27:04.629 --> 27:15.760
us say, at least we should remember this 1
and what our is 875 1987 that 1 says, obviously;
27:15.760 --> 27:16.990
we have to take self
27:16.990 --> 27:20.929
weight for reinforced concrete, already we
have used this 1 25 kilometer per cubic meter,
27:20.929 --> 27:25.270
finishes and partitions. We use finishes and
partitions generally at 1.5 kilo Newton per
27:25.270 --> 27:26.800
square meter. Note use cubic
27:26.800 --> 27:30.250
meter and here we are using 1.5 kilo Newton
per square meter.
27:30.250 --> 27:36.889
So, you finish and partition also that we
say, partition wall that 1 it is distributed,
27:36.889 --> 27:42.109
say may be say 125 millimeter thick, say your
brick wall. So, you can place it anywhere
27:42.109 --> 27:43.480
so that, we shall distribute
27:43.480 --> 27:49.589
that load and that is your coming on that
1.5 kilometer per square meter. The other
27:49.589 --> 27:58.300
1 that imposed loads for roofs, this is 1.5
kilo Newton per square meter with access,
27:58.300 --> 28:00.450
that means: if you access the roof,
28:00.450 --> 28:04.409
if you have access then, we take it 1.5 kilo
Newton per square meter.
28:04.409 --> 28:12.280
If, it is not accessible, possibly your hall
possibly that your roofs are not accessible,
28:12.280 --> 28:18.730
so in that case, we have to take say 0.75
kilo Newton on per square meter. For floors,
28:18.730 --> 28:20.580
we are having 2 different 1
28:20.580 --> 28:24.129
that, if it is say residential buildings,
so you take 2 kilo Newton on per square meter.
28:24.129 --> 28:29.740
If, it is office floors then, you take 3 kilo
Newton on per square meter. So, we should
28:29.740 --> 28:32.649
remember these few and if it
28:32.649 --> 28:36.800
is not mentioned then, we shall assume on
the basis of this that means: if it is a office
28:36.800 --> 28:42.139
building, if it is not mentioned the load,
then we have to take that 3 kilo Newton on
28:42.139 --> 28:43.909
per square meter.
28:43.909 --> 28:49.820
What about the concrete cover? That you will
find out in table 16. We have already used
28:49.820 --> 29:00.639
clear cover, that is, 25 millimeter for beam.
So, you will find out in table 16 clause 26.4.2
29:00.639 --> 29:02.639
is 456 2000. And I have
29:02.639 --> 29:07.629
just given say mild exposure because, severe
there are so many other exposure environmental
29:07.629 --> 29:13.560
1. So, nominal concrete cover should not be
less than 20 millimeter. It says it should
29:13.560 --> 29:15.450
not be less than 20
29:15.450 --> 29:15.830
millimeter.
29:15.830 --> 29:22.389
However, for main reinforcement up to 12 millimeter
diameter bar, for mild exposure, the nominal
29:22.389 --> 29:26.710
cover may be reduced by 5 millimeter. That
means: if it is say mild exposure and the
29:26.710 --> 29:27.790
body you are using for
29:27.790 --> 29:34.589
slabs; if it is less than 12 millimeter or
upto 12 millimeter then, you can use we can
29:34.589 --> 29:39.280
reduce that 20 millimeter to 15 millimeter,
that means: we can use 15 millimeter.
29:39.280 --> 29:45.240
So, for slabs we shall use 15 millimeter.
If, it is not mentioned any other say in your
29:45.240 --> 29:50.080
environmental condition and your exposure
condition or any you are using that within
29:50.080 --> 29:51.460
12 millimeter, then we shall use
29:51.460 --> 29:55.030
this 15 millimeter.
29:55.030 --> 30:00.389
Then one more part that is; very important
because, this is the one that which will be
30:00.389 --> 30:05.500
giving the serviceability. I have already
told this is the 1 that, so far whatever I
30:05.500 --> 30:07.089
am talking, that is called actually
30:07.089 --> 30:13.599
you limit state of collapse. We are talking
flexure here; for beams as well as for slabs.
30:13.599 --> 30:19.800
But, there is a 1 more part, that is you say
serviceability limit state so that, the users
30:19.800 --> 30:22.129
will not feel any
30:22.129 --> 30:27.429
discomfort. And that 1 obviously; that you
say deflection, that if you have say; deflection
30:27.429 --> 30:33.690
that if you say that you have slab or the
1 where your are supporting structure, if
30:33.690 --> 30:35.520
it is vibrating too much, if
30:35.520 --> 30:39.820
you are having deflective more, then obviously;
that you will feel it will be scary.
30:39.820 --> 30:44.369
So, at least it should be serviceable and
that 1 obviously 1 part that is you say control
30:44.369 --> 30:50.669
of deflection, that even say crack section.
The crack section if you find out cracks,
30:50.669 --> 30:52.139
then what will happen, that
30:52.139 --> 30:57.700
also it may be scary. But, even if you give
guarantee but unfortunately or fortunately
30:57.700 --> 31:02.220
that our assumption is that, it is based on
crack section. The design you are doing that
31:02.220 --> 31:04.790
is you would actually crack
31:04.790 --> 31:10.679
section design that only the assumption. So,
that means even if you find cracks then you
31:10.679 --> 31:18.300
can say, even then it is safe but, the users
who are using that 1 residence of that building,
31:18.300 --> 31:18.710
obviously; that it may
31:18.710 --> 31:23.859
be scary. So, it is little bit irony also
you can say and that is true for all of us
31:23.859 --> 31:25.300
also, even if you design.
31:25.300 --> 31:33.810
So, this control of deflection, you will find
out in clause 23.2. The final deflection due
31:33.810 --> 31:38.530
to over loads, but how shall we find out?
Because, you are not going to calculate that
31:38.530 --> 31:41.320
deflection, so we do it on
31:41.320 --> 31:46.730
the basis of something say span. So the final
deflection due to all loads, including the
31:46.730 --> 31:54.329
effects of temperature, due to temperature
also the deflection may occur, creep it occurs
31:54.329 --> 31:57.200
and shrinkage, this will be
31:57.200 --> 32:02.700
are measured from the as-cast level of the
supports.
32:02.700 --> 32:12.109
So, from one, you can take 1 say bench mark
point, that is, the support of floors or roofs
32:12.109 --> 32:18.490
and all other horizontal members, should not
normally exceeds span by 250, it should not
32:18.490 --> 32:20.490
exceed span by 250. That
32:20.490 --> 32:25.950
means: if we know span, so effective span
say you are say 3000, so you can take say
32:25.950 --> 32:32.960
divide by 3000 divided by 250, that you should
take it as I say you, say maximum deflection
32:32.960 --> 32:35.530
allowed.
32:35.530 --> 32:41.829
The deflection, including the effects of temperature
creep and shrinkage occurring, occurring after
32:41.829 --> 32:47.470
erection of partitions and the application
of finishers, that means: partition or any
32:47.470 --> 32:49.329
other that final 1
32:49.329 --> 33:11.490
and finishers, should not, so normally exceeds
span by 350 or 20 millimeter whichever is
33:11.490 --> 33:16.270
less. So, you should remember that span by
350 that is, the maximum allowed that may
33:16.270 --> 33:17.780
deflection or 20 millimeter that
33:17.780 --> 33:26.490
is, it specified that in our case it should
not be more than 20 millimeter.
33:26.490 --> 33:37.260
Now, let us see the vertical deflection limits,
may generally be assumed to be satisfied,
33:37.260 --> 33:43.099
provided that the span to depth ratios are
not greater than the values obtained as below.
33:43.099 --> 33:45.129
We can satisfy that 1 in
33:45.129 --> 33:50.859
other way; indirect way. We take that certain
span to depth ratios span by depth ratio l
33:50.859 --> 33:57.530
by d, l is the effective span and effective
depth d. So, if we take certain ratio which
33:57.530 --> 33:59.960
we can find out, the code
33:59.960 --> 34:04.320
says; if it is cantilever and then our code
says that means: here we can get cantilever
34:04.320 --> 34:11.510
l by d, we will getting that 1 say 7. And
here this is your basic values of span to
34:11.510 --> 34:13.369
effective depth ratios, for spans
34:13.369 --> 34:20.609
up to 10 meter. That means: if the span is
up to 10 meter then, we shall take, for cantilever
34:20.609 --> 34:27.690
span by depth ratio that is 7, because it
will give us that to choose the, it will also
34:27.690 --> 34:29.869
give us to choose the
34:29.869 --> 34:31.940
effective depth.
34:31.940 --> 34:39.510
In other way, we shall use these value 720
or 26, 7 for cantilever, 20 for simply supported
34:39.510 --> 34:45.109
and 26 for continuous beam. So, what we can
do, what is the when you are starting the
34:45.109 --> 34:48.149
design, then effective depth
34:48.149 --> 34:52.820
I have to find out. That we calculate on the
basis of because, I know span or effective
34:52.820 --> 34:58.450
span. So, effective span, that too I know,
so effective span divided by these value basic
34:58.450 --> 35:00.050
value, it will give me to
35:00.050 --> 35:06.120
select you are say effective depth and also
the overall depth, which we can use it for
35:06.120 --> 35:08.440
the calculation of load and other things also.
35:08.440 --> 35:13.780
So, this one also will help us to take that
you are say, clearly assumption of you say,
35:13.780 --> 35:19.370
effective depth as well as overall depth.
So, we are getting cantilever 7 simply supported
35:19.370 --> 35:21.660
20 continuous 26, this is
35:21.660 --> 35:24.750
basic values.
35:24.750 --> 35:34.840
But, one more that for spans above 10 meter,
the basic values may be multiplied by 10 by
35:34.840 --> 35:40.990
span. So, if it is more than 10 meters or
10 by span, so it will be further reduced
35:40.990 --> 35:42.710
in meters, except for cantilever
35:42.710 --> 35:48.660
in which case, the special calculation should
be made. That means: for cantilever beam for
35:48.660 --> 35:57.020
the slab, if it is more than 10 meter, then
in that case you have to calculate your deflection.
35:57.020 --> 35:59.040
Otherwise, you can
35:59.040 --> 36:04.170
use that coefficient or that value, we can
multiply with that 10 by span and then we
36:04.170 --> 36:06.620
can find out the new value and that we can
use it.
36:06.620 --> 36:11.000
So, you have started with some basic value,
then we are modifying according to the span
36:11.000 --> 36:15.550
that is 1, we can say that 1 say your q 1.
36:15.550 --> 36:24.190
Now, depending on the area and the stress
of steel for tension reinforcement, the basic
36:24.190 --> 36:28.330
values that means, the first 1 that which
are given or the second 1 depending on the
36:28.330 --> 36:31.290
that span by 10 by span
36:31.290 --> 36:38.620
multiplying that 1, shall be modified multiplying
with the modification factor obtained as per
36:38.620 --> 36:44.040
figure 4. There is 1 figure so that, we can
take it and that 1 what is the case? Area
36:44.040 --> 36:45.800
and the stress of steel,
36:45.800 --> 36:50.820
that from there we shall find out and we can
further modify the basic value and let us
36:50.820 --> 36:52.540
see that figure 4.
36:52.540 --> 36:59.880
So, this is your that figure 4, for different
grid different stress we can take it and this
36:59.880 --> 37:03.710
is 1 say service load in Newton per square
millimeter, from there that is the modification.
37:03.710 --> 37:06.330
We can use this figure
37:06.330 --> 37:11.980
and from there from the percentage of steel,
we can further we can modify. Now, how do
37:11.980 --> 37:16.510
I know that initially if we start with some
say simply supported beam 20?
37:16.510 --> 37:22.430
So, here we can have to modify the value,
but how do I know? That means; it will come
37:22.430 --> 37:27.350
from our experiments, that what is the percent
of steel generally we use for slabs. So, on
37:27.350 --> 37:28.630
the basis of that, we can find
37:28.630 --> 37:35.630
out certain factor. Or in other way, we have
to redesign it. We have to start with basic
37:35.630 --> 37:40.160
value then, you do the whole process again
you do it, but that we generally don’t do
37:40.160 --> 37:42.470
it because, if you are express if
37:42.470 --> 37:48.890
you know that 1 so the very first trial itself
we can go.
37:48.890 --> 37:55.560
Depending on the area of compression reinforcement,
the value of span to depth ratios be further
37:55.560 --> 37:59.760
modified by multiplying with the modification
factor, obtained as per figure 5. There is
37:59.760 --> 38:00.650
1 more figure in that
38:00.650 --> 38:04.260
is 456, from there we can get another modification.
38:04.260 --> 38:09.510
So, this is your depth depending on the compression
reinforcement. The first 1 figure 4 that is,
38:09.510 --> 38:14.520
on the tension reinforcement and the other
1 on the basis of the compression reinforcement.
38:14.520 --> 38:15.890
So, our basic
38:15.890 --> 38:22.270
value will be modified with the tension reinforcement
also on the basis of compression reinforcement.
38:22.270 --> 38:31.210
Now, for flanged beams, the basic values be
modified as per figure 6 and the reinforcement
38:31.210 --> 38:35.600
percentage for use in figure 4 and figure
5 should be based on area of section b and
38:35.600 --> 38:38.280
d, note that this is b and d
38:38.280 --> 38:40.390
that way we have to take.
38:40.390 --> 38:49.430
And this is that figure 6. So for slab design
or also for beam design, we have to take all
38:49.430 --> 38:53.280
those things that which, I have not given
previously, because I thought at least we
38:53.280 --> 38:55.030
should concentrate on that
38:55.030 --> 39:00.820
formulas, that which we are using say nv equal
to 0.138 fck bd square like that and now we
39:00.820 --> 39:06.240
are slowly we are coming to that codes and
other things that how to how to design it.
39:06.240 --> 39:12.940
So, the beam slab design is nothing but you
can say, beam design itself. Only thing that,
39:12.940 --> 39:19.160
in the slab design, we take that width of
the slab that is 1000 millimeter, that is
39:19.160 --> 39:26.850
we take 1 meter width and that is so we are
doing the same beam design. Only thing that,
39:26.850 --> 39:29.100
here we are taking that 1 meter
39:29.100 --> 39:34.450
and here also that, another by simple thing
that we are taking only rectangular 1. So,
39:34.450 --> 39:39.840
now let us at least start the problem, in
which we shall continue in the next class.
39:39.840 --> 39:45.620
So, our problem let us start with 1
39:45.620 --> 39:59.450
example, design of 1 way slab.
39:59.450 --> 40:35.660
Design a simply supported RCC slab for a roof
of a hall 3.5meter into 8 meter. And this
40:35.660 --> 41:08.030
is a inside dimensions with 250 millimeter
walls all around, which is supported by 250
41:08.030 --> 41:12.740
millimeter walls around. Let us
41:12.740 --> 41:33.750
assume a live load of 4 kilo Newton per square
meter and finish, because there is no partition
41:33.750 --> 41:56.340
finish 1.0 kilo Newton square meter. Use concrete
M20 and steel Fe 415.
41:56.340 --> 42:06.380
So, we have to design of 1 way slab, design
a simply supported RCC slab for a roof of
42:06.380 --> 42:12.950
the hall 3.5 meter into 8 meter that is the
inside dimension. So, clear span here 3.5
42:12.950 --> 42:15.160
meter in 1 side and another side 8
42:15.160 --> 42:22.840
meter, as increase 1 way, so we have to take
3.5 meter with 250 millimeter wall around.
42:22.840 --> 42:28.660
Assume a live load of 4 kilo Newton per square
meter, though it is quite high but any way
42:28.660 --> 42:30.020
just for problem check let us
42:30.020 --> 42:34.720
take it and finish 1 kilo Newton per square
meter. I have given 1.5, but here we are not
42:34.720 --> 42:42.800
using any partition. Use concrete M20 and
steel Fe 415. So this is the problem.
42:42.800 --> 43:08.350
So, let us do the first step, that is, the
calculation of factored loads. Let us take
43:08.350 --> 43:23.270
we shall assume span by d is equal to say 25, simply supported here it is 20, but constantly say the
43:23.270 --> 43:24.480
reinforcement other
43:24.480 --> 43:39.080
things you have taking say you say 25. So,
you can find out d equal to span by 25. And
43:39.090 --> 43:43.750
we have not yet calculate the effective span
because, this is after all some preliminary,
43:43.750 --> 43:45.230
that estimate to find out your
43:45.230 --> 43:54.230
loading other things. So, you can take it
here directly say 3500, that we are taking
43:54.230 --> 44:08.490
this much and on the basis of that, we shall
get 140 millimeter. Total depth equal to 140
44:08.490 --> 44:13.770
plus clear cover let us write
44:13.770 --> 44:24.320
down clear cover plus phi by 2. I think, I
should write down here d, d plus clear cover
44:24.320 --> 44:31.560
plus phi by 2, d equal to here 140, clear
cover how much we have take? Let us take assume
44:31.560 --> 44:40.970
10 mm dia bar. For slab, we
44:40.970 --> 44:46.640
shall generally use that 8 mm 10 mm or 12
mm, generally we use.
44:46.640 --> 44:53.210
So, in our case we can use the clear cover,
even for mild exposure, we are getting say
44:53.210 --> 44:57.800
15. So, for mild exposure we are getting 15
because, we are using less than or equal to
44:57.800 --> 45:02.560
12 millimeter plus 15 by 2 and
45:02.560 --> 45:06.550
we get 160 millimeter.
45:06.550 --> 45:34.750
So, we can now we can calculate the dead load.
Dead load; so slab 0.16 times 25 4 kilo Newton
45:34.750 --> 45:49.690
per square meter, finish 1 kilo Newton per
square meter which already given, dead load
45:49.690 --> 45:54.370
5 kilo Newton per square
45:54.370 --> 46:14.670
meter, live load already specified 4 kilo
Newton per square meter. So, factored load
46:14.670 --> 46:23.510
or we call it as design load. So, we are talking
dead load live load only, so let us multiply
46:23.510 --> 46:28.380
with 1.5 only. So, it comes
46:28.380 --> 46:46.800
as equals 13.5 kilo Newton per square meter.
What about the effective span? So, we know
46:46.800 --> 46:53.470
factored load or design load that is 13.5
kilo Newton per square meter.
46:53.470 --> 47:24.640
Effective span, let us write down 1 3.5 plus
0.25, that width of wall 250 millimeter which
47:24.640 --> 47:40.910
comes as 3.75 meter. The other 1, 3.5 plus
effective depth, effective depth we have computed,
47:40.910 --> 47:41.200
effective depth we
47:41.200 --> 47:58.200
have computed 140 millimeter. So, you can
take it as 0.14 and this comes as 3.64 meter.
47:58.200 --> 48:37.240
Therefore, effective span
equal to 3.64 meter, total load per
48:37.240 --> 48:44.450
meter width equal 13.5 times 3.64 which comes
as 49.14
48:44.450 --> 48:52.950
kilo Newton. We are talking the total load;
so 3.64 is the span and 13.5 kilo Newton per
48:52.950 --> 49:00.540
square meter, that is effected load or design
load. So 13.5 times 3.64 which comes as 49.14
49:00.540 --> 49:05.270
kilo Newton.
49:05.270 --> 49:27.800
So, ultimate, now let us calculate ultimate
moment and shear. Mu equal to total load times
49:27.800 --> 49:46.130
effective span divided by 8 equals 49.14 times
3.64 divided by 8 equals to 22.3587 kilo Newton,
49:46.130 --> 49:48.910
Vu that is equal to
49:48.910 --> 50:06.350
W by 2 equals 49.14 divided by 2 equals 24.57
kilo Newton. This is your Mu and the other
50:06.350 --> 50:19.320
1 is Vu. Now, we know what we shall do it;
we shall take a beam, a rectangular beam and
50:19.320 --> 50:25.510
we have the width 1000
50:25.510 --> 50:31.680
millimeter, 1000 millimeter is the width and
if and then, we shall solve this problem taking
50:31.680 --> 50:34.100
that Mu and Vu.
50:34.100 --> 50:40.790
So, we shall find out and we have to check,
whether we have and how to provide the reinforcement
50:40.790 --> 50:46.460
that reinforcement, so all ready we have done
that, overall depth we have already computed.
50:46.460 --> 50:47.369
So, we have to
50:47.369 --> 50:55.020
find out that, so this is 160 little bit higher
side. So, you have to check it taking 160,
50:55.020 --> 51:01.570
whether this is sufficient whether it can
resist this moment. And we have to provide
51:01.570 --> 51:04.869
their reinforcement, so we have
51:04.869 --> 51:09.750
to provide their reinforcement. In this case
what we provide actually, we do not provide
51:09.750 --> 51:15.980
the reinforcement their number of bars. These
are the reinforcement this 1 normal to the
51:15.980 --> 51:17.610
paper I say, longitudinal
51:17.610 --> 51:20.010
reinforcement, but we do not provide it as
a number of bars.
51:20.010 --> 51:30.310
The one we have done it for say your beams,
that say 2 times 2 20 320 or 31 like that.
51:30.310 --> 51:39.720
But, here what we do, we take it say the diameter
of the bar say 10 millimeter at the rate of
51:39.720 --> 51:41.520
the spacing, we provide
51:41.520 --> 51:48.050
the spacing here. So that we have to find
out, that means: we have already taking that
51:48.050 --> 51:53.920
say overall depth and also know the effective
depth. So, you have to check that taking this
51:53.920 --> 51:54.860
moment; whether that
51:54.860 --> 52:01.630
effective depth is coming less than the computed
figure less than the provided effective depth.
52:01.630 --> 52:06.720
If it is okay, then we can go for that we
can accept that. And then, you have to find
52:06.720 --> 52:13.300
out the area of steel. Then we have to check
that, Vu generally we don’t check it, but
52:13.300 --> 52:14.670
we shall check for this problem;
52:14.670 --> 52:21.790
whether that V that shear stress we have computed
on the basis of this Vu, whether we are getting
52:21.790 --> 52:29.140
that 1 less than that c. Whether you are getting
less than the 0.15 percent as per the table
52:29.140 --> 52:29.650
19 of is
52:29.650 --> 52:34.890
physics, so that it less than. If it is less
than, then we shall not provide any reinforcement,
52:34.890 --> 52:42.160
then we do not need any shear and for slabs
we do not provide any stirrups. And that shall
52:42.160 --> 52:43.040
we continue in the
52:43.040 --> 52:44.980
next class.