WEBVTT
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So far we have done the design for beams,
to resist bending moment and shear forces
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which is also applicable for footing as well
as slab and the other one we have done that
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axial, axial loaded column. Now, we shall
learn how to design for torsion where it is
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applicable. So, for that this lecture concerned
this is our lecture number 27, design for
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torsion.
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And now let us come, that where the torsion
is really applicable, there were x if you
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take a cantilever slab say we would like to
say cantilever slab supported by a beam. A
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cross section I am drawing it could be other
way also beam so, this is your beam that this
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beam is supporting this cantilever slab. So
obviously, you will find out that if you take
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1 beam say like this may be simply supported
also possible.
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In that case, what will happen then you will
find out not only the bending moment or forces
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acting here; you will have that torsion due
to this moment. So that means here, you have
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to design it for torsion also and this beam
you have to design for bending moment for
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shear forces, as well as for torsion developed
due to this slab.
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Similarly, it is possible if you draw it in plan this is a cross section if we might draw say beam plan, we are
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drawing the plan here. So, if we draw this is plan
please note it is not the elevation. So, what
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will happen? Due to this beam, torsion will
be developed at this point.
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Because, this beam will develop that apply
certain kind of torsion like this and for
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that also it will design. But that torsion
only applicable at this point or may be you
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can say what the width of this beam whereas,
in this case this torsion is applicable over
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the like this over the whole beam from one
end to other end. So, this is the so you can
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find out that there are so many other cases,
where it is possible that where torsion will
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be developed. So, how shall you design that?
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What we generally do, there there we can say
that direct that means when you are talking
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say bending moment
and the other 1 shear force. So, we can directly
we can compute the effective depth that 0
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138 fck b d square equal to say Mu as per
the limit state design. So you can find out
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the depth directly, then Mu is the ultimate
moment developed due to this bending moment
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that you are whatever load applied. And shear
force also we have that shear force, that
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is your say b or b u.
So, from that also you can find out the shear
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stress developed say tau v and we can find
out so tau v times say bd equal to say Vu.
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So, you can find out the shear stress and
critical stress also you can find out, so
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tau v minus tau c from there we can provide
the shear reinforcement.
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But regarding torsion, we do not go directly
what we do; we find out the equivalent bending
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moment developed due to torsion say Tu. Tu
the torsion applied after say factored 1 that
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means, after we multiply with 1 5 or whatever
the appropriate factor. And then we can find
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out the Tu and this Tu that there is no such
scheme or no such formula from where we shall
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directly get the d.
Because, here we are having that you can say
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3 parameters or 3 things are being applied:
one is bending moment, another one is shear
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force and another 1 is torsion. So, what we
shall do it here? We shall find out the Tu
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that how it is actually applicable? what is
the equivalent 1 that for bending moment as
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well as for shear force. And this one we shall
find out in clause for your reference clause
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4, page 74 , IS 456: 2000.
So, this is your that clause for limit state
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of collapse and that is torsion. So, we shall do this
particular one here so, what is our objective
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then ? If say this is your Mu, this is your
Vu. Then we shall find out, which is a function
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of Tu and Mu also that we shall find out Me that equivalent bending moment and Ve that equivalent shear
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force.
So, which is a function of that your say Tu
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as well as Mu, Vu like that. And what we have
to do? What we have to provide? We have to
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provide the reinforcement so obviously, the
reinforcement will be here; for bending moment
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the reinforcement will be here. The longitudinal
reinforcement and for this Ve that you have
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to provide the shear or web reinforcement
or the stirrups. So, these two things we have
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to provide.
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So, what is the idea behind that ? If we take
it say any section say this is your say torsion
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then shear stresses developed like this, about
the surface which is true for circular one
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also, for the noncircular one also. So that
means, shear force you will find out the shear
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stresses at different fibre that is developed.
So, if we can now if the shear stress developed
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here; so, if we can resist the shear stress
providing say reinforcement longitudinal which
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is coming equivalent to your say bending moment
from where you are getting. And the web reinforcement
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or stirrups you are providing so, which is
your say shear reinforcement. And that we
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shall do it here; that is the thing that is
the philosophy here; how we are taking care
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of the torsion.
So, due to this torsion what how the shear
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stresses developed here on the surface and
accordingly you have to provide the sufficient
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reinforcement. So that, we can resist shear
that cracking can be avoided and so, that
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longitudinal reinforcement and your web reinforcement.
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So, what is the that what our code says this
is very simple one we can say for equivalent
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shear force. So, we can say in a particular
section we can say Ve, where that torsion
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is acting. So, Vu the shear force from the
usual whatever way you have got it plus 1
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6 Tu by b. This one available, in clause just
for your reference clause 41 3 1, page 75,
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IS 456: 2000.
So, Ve equivalent shear, Vu that shear that
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developed due to your say applied load whatever
is coming in the usual way. It will come from
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the analysis also, that we shall do it in
the next few classes for a total frame we
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shall do it, Tu torsional moment and b breadth of beam. So, here one parameter is in your hand rather
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in our hand that is b Vu and Tu that is coming
from the analysis, due to applied load say.
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Now, the equivalent shear that we can reduce
it taking say proper width of the beam that
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means, since it is at the bottom denominator.
So, what we can do it here, denominator so,
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we can take that b say may be that generally
we take it say 250 which is governed by the
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wall thickness, we can change it to 300 also.
So that means, in unusual cases that where
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we are getting the shear force is quite high,
then we can at least we can do some kind of
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say your changes reduce it by taking proper
b though it is in linear parameter. But even
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then, we can change it. So, b that means there
is no such big thing only thing we have that,
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we have to change the using this equation
let us, find out the corresponding equivalent
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shear force. Then, we shall go as per the
standard procedure of design of shear.
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The other 1 that is for the equivalent bending moment. So, I can write down so let us write down here Me 1 equal
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to Mu plus say Mt this is for the applied load and
this is for the torsion. So, Mt equal to Tu
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divided by 1 plus d by b
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moment due to torsion we are getting. D is
the overall depth, b breadth, Tu torsional
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moment.
Now, we have one more thing Me1 equal to Mu
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plus Mt and this is available that clause
for your reference clause 41 4 to be more
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specific I can say 41 1 2, page 75, IS 456:
2000. The parameter 1 7 or the parameter 1
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6 that obviously, because we generally come
to this 1 finally, we come to say optimum
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one so that it is safe. But in actual case,
if one is interested one can go for the whole
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analysis, one can find out little different.
But finally, we come to thus 1 6 1 7 for design
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purpose so that, we can it is applicable for
the wide range of say your rectangular sections
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or for beams. That is the objective here,
that how we generally come to the that your
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say all the clauses, wherever you are having
the different parameters; one can say say
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sometimes we find out something say modulus
of stress say e that is say 5700 root over
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fck.
So, if one is interested for design say you
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are so from the analysis or from the experiment
one can find out the value, may be it is coming
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say 5200 or 5300 like that. But finally, we
are coming to that particular value so, that
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it is applicable to wide range that is the
basic concept in any design code and for any
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country.
So, even if you find out that in say American
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code they may vary little bit say UK European
code it may vary little bit. But almost you
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can say it is coming towards certain value,
which is closer to that. And there is a committee
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for each and every country also that for a
particular code say IS 456 that is a committee
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similarly, for say IS 800 for steel.
Similarly, say IS 875 for each and every code
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there is a committee who are looking after
that things and coming to a poin, that this
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value may be quite sufficient for wide range
and applicable to our say Indian standard
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or Indian condition rather.
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So, now I would also like to tell something
more that here also you can find out that
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Mt that 1 that d by b that means, the earlier
case only we have the parameter that b with
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by which we can manipulate or rather we can
reduce the value. Similarly, here also choosing
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say d by b and then we can find out t hat
Mt we can reduce, that Mt that also we can
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here.
And one more thing I shall tell you that,
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if we take say span by depth ratio say here
say for simply supported case say 20 then,
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we shall find out that it will not be with
that. If we start then it may happen that
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we are really going that that value that one
only very very less. Anyway so let us, find
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out what more thing we can do it let us find
out.
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There is one more thing that is your say for
transverse reinforcement, that is your say
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shear reinforcement. We find out here say
Asv that your say stirrup that area of stirrup
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or reinforcement it says Tu by Asv; Asv the
spacing of stirrups or web reinforcement b1
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d1 times 0 87 fy plus Vu Sv divided by 2 5
d1 times 0 87 fy. We can find out the area
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of steel using this formula in clause 41 4.3,
page 75, IS 456: 2000.
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What we shall do it here? That Asv we can
find out that means, I know Tu, I know Vu
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b1 what is b1? fy we know, say Fe 415 so,
450 newton per square millimeter. So, Tu Tu
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is the torsional moment, Vu shear the is not
the 1 other than that say Tu not we have not
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taken equivalent please note that Vu the shear
whatever it is coming due to applied load,
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Sv spacing of stirrups.
Let us say because finally, you have to provide
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that 1 as stirrups we call it web reinforcement.
So, we have to provide the stirrups Asv and
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this b1 and d1; b1 centre to
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centre distance of corner bars in breadth
direction. What is that? So, we have b1 and
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we have d1.
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If this is your beam so, we are having reinforcement
so b1 means there are many so many bars b1
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means this is your b1, along breadth direction
b1. So, what about d1? Along the depth direction.
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So, b1 will be equal to b if this is your
b so b minus that clear cover this side 25
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millimeter, clear cover this side 25 millimeter
minus that half of the diameter here; half
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of the diameter so that means, 1 diameter.
So, centre to centre distance of corner bars
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in breadth direction so let us, write down
for d1 also d1 will be equal to similar fashion
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centre to centre distance of corner bars in
depth direction. So, overall depth D minus
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that clear cover for beam 25, 25 for usual
cases minus that half of the top bar half
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of the bottom bar and we can find out d1.
So, what we can do it then? Asv that means
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we know Tu, we know b1 d1 fy everything we
know, only thing we do not know Sv.
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Now, what we in usual cases what we do in
general cases Sv we generally take it say
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may may be that say we have stirrup spacing
say 200 millimeter say. So, if we assume that
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we shall provide 200 millimeter as per our
that coded provision which should not be more
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than certain value, on the basis of that we
can choose say 200 millimeter 250 millimeter.
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So, like that if we choose it then we can
find out Asv and now if we know Asv. So let
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us, choose the the diameter of bar and if
it is whether it is 2 legged or 3 legged or
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4 legged depending on the situation we shall
provide. So that means, for each and every
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design there is a standard there is a specific
procedure.
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Here what we shall do? That we shall do it
that we provide Sv and then we can find. In
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other way also we can do it, that means from
this equation we can take Sv out and we can
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Asv we can take it say 2 legged or 3 legged
on the basis of that also you can find out.
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Or in other way also we can do it say Asv
by Sv also we can make it, that means no the
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thing is that whether we shall make it say
youâ€¦ I am telling, number 1 choose Sv as
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per the code that which should not be more
than certain value all those things.
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Find Asv and provide that bar dia this this
may be one case, second case could be choose
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bar diameter
in other way, that Asv find Sv and provide
as per the code. So that, way there it is
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coming within the less than that value within
the limiting one. The third could be the standard
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one that we have done it, for the beam in
the very in the beginning. That means, I can
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take it as Asv by Sv also, Asv by Sv also
we can find out and accordingly from this
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because it is not so difficult.
So, Asv by Sv we can take it we can find out
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Asv by Sv and then from there we can find
out that, what should be corresponding value
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either selecting Asv or selecting Sv we can
find out that. So, these are the three procedures
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that whatever way we we shall do it. Now,
let us start 1 example but, before that I
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shall tell you one more thing that is I have
told you in the very beginning that Me1 equal
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to Mu plus Mt that is there. And Method we
have calculated here.
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Now, that could be another possibility that
Mt that we have got it that is greater than
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Mu. The case first of all that is the we are
adding up that Mu plus Mt and where we shall
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provide that one say in the longitudinal reinforcement
that Mu in most of the cases Mu which is actually
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in the bottom the reinforcement we have to
provide that means, Mu plus Mt.
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Now, in another case it may be it is possible
that Mt is greater than Mu, when Mt is greater
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than Mu then we have to provide the reinforcement
at the top for and what is that value, how
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what is the moment, that will be Mt minus
Mu that we have to take it. So that means,
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it is possible Me 2 also that is another case
which is equal to Mt minus Mu.
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So Mt minus Mu that means, say let us say
I have told Me1 equal to Mu plus Mt, where
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Mt equal to Tu 1 plus d by b divided by 1
7 and this is one case. The other case is
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that, if Mt is greater than Mu then I shall
take Me2 will be equal to Mt minus Mu. So,
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this is other case and for that we have to
provide the reinforcement at the top, we shall
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provide the reinforcement taking Me2 equal
to Mt minus Mu.
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Here also I have seen in this one of our code
says and that you will find out, that is in
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clause 41 4 2 1, page 75, IS 456. Now but,
I have seen in few other books at least 1
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book I have seen and that is where that professor
ex professor of IIT Khanpur he is having 1
29:03.010 --> 29:09.100
book of that limited design of reinforced
concrete structures, where I have seen papers
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that even if it is say not less than say Mt
and Mu.
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Let us not check that one whether it is greater
than or less than even if it less than then
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let us, provide the top reinforcement taking
that Mt only Mt. So that, that means say here
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what I would like to say that in design even
that our code says specifically, even then
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also there is some kind of say practice also.
And obviously, that it depends that those
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thing mainly we are following the Professor
Vargheese book but, here also you will find
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out so, there is a difference of opinion also.
So anyway, it does not matter that way but,
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what we can do actually after all we are providing
that say top reinforcement for the hanging
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bars. So, that also we can check and we can
find out that whether it is sufficient or
30:00.810 --> 30:05.960
not considering that Mt value only; that also
we can check, if it is sufficient or we can
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provide at least that value. There is nothing
wrong that way at least because, after all
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we are providing the top reinforcement that
1 say for hanging bars even if we do not require.
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So, that also we can check but, strictly speaking
that as per this clause 41 4.2.1 that we do
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not need it if Mt is less than Mu.
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So, let us start one example problem. Let
us take, simply supported beam span 5 meter,
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live load 25 kilonewton per meter, torsion
25 kilo newton meter please note 25 kilonewton
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meter from end to end. That means, all along
the beam this torsional moment is acting;
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concrete M20, steel Fe 415.
So, data given fck 20 newton per square millimeter,
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fy 415 newton per square millimeter, l effective
span, say here we can take it say 5 meter
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live load or wll say live load 25 kilonewton
per meter, T unfactored that not the design
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1, T given as 25 kilonewton meter. Let us,
calculate the load.
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So, step 2 say load calculation we shall try
say few trials. Because, instead of directly
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going to the value it is worth that going
say all the calculation because; what are
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the process, how we repeat the whole calculation.
let us start like that. Because, I do not
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want to make the final value and see that
as if it is a magic way we have got it but,
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it is not like that when we are designing
then we do lot of say you are repeating steps.
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So, let us say assume width of the beam 250
millimeter. If we take the span by depth let
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u, take say span by depth say let us take
say 20 because for simply supported beam.
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So what about the depth then? 5000 by 20 which
is coming as 250. If we start using this value
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then you will find out that it will be very
very less this value. So, let us start at
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least say because that is why I do not want
to go to that computation. Because, if we
34:44.520 --> 34:51.040
start say you have to calculate the self weight
of the beam and then your say that live load
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given.
Then, I have to calculate that Mu that bending
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moment due to that applied load and then,
I have to find out that equivalent bending
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moment due to torsion. So, let us because
if I start with 250 millimeter then it will
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be very very less. So, let us take for calculation
of the self weight the overall depth let us
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take say 500 millimeter.
Because, from 250 to 500 to going to that
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level obviously, it is better you can try
checking with that value at least you try
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250 and check that how it is coming instead
of directly going to that 500. So, self weight
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0 25 is the width, times 0 5 is the overall
depth, times 25 kilonewton per meter cube,
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which gives us 3 125 kilonewton per meter;
design load 1 5 times 3 125 self weight approximately
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you can say 300 k g per square meter.
So that means, in few cases what we can do
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instead of going directly we can take say
400 k g per square meter also we can take
36:29.650 --> 36:34.700
it or 4 kilometer per square meter. In that
case also,sometimes we take it that way also.
36:34.700 --> 36:48.970
So, 1.5 3 125 plus 25 that is the live load
and which comes as 42 1875 kilonewton per
36:48.970 --> 36:51.250
meter.
36:51.250 --> 37:17.809
Whatever the bending moment, Mu equal to Wu
say ultimate times l square by 8 42 1875 times
37:17.809 --> 37:36.099
5 square divided by 8 equal to 131 83594 kilonewton
meter we can take it at say 131 84 kilonewton
37:36.099 --> 37:43.170
meter. Few designer paper say, okey we shall
go since it is coming in the calculator so
37:43.170 --> 37:48.869
as much as possible we can go. The other alternative
also we can take it say as a simple say after
37:48.869 --> 37:54.000
say up to 2 decimal that also we can do it.
So, what I am trying to say, that when you
37:54.000 --> 37:59.460
design it sometimes it happens that you have
to submit the design also, design report also,
37:59.460 --> 38:05.140
that how we have come not only that our final
objective though to make the drawings, which
38:05.140 --> 38:10.319
will go to the site. But at the same time
sometimes for proof checking or for other
38:10.319 --> 38:13.650
purpose for record also, we have to submit
the design also.
38:13.650 --> 38:19.619
Now in that case, it happens that we can go
to this say as much as possible but, there
38:19.619 --> 38:25.049
we can taking the very beginning we can write
that what whatever whether we are gone up
38:25.049 --> 38:29.750
to say 2 decimal or 3 decimal. Because, after
all that after that it will only unnecessary
38:29.750 --> 38:35.349
it will take time, but we can round up to
that say 2 decimal for when you are talking
38:35.349 --> 38:40.130
say kilonewton meter.
So similarly, for meter we can go up to say
38:40.130 --> 38:44.940
only say after meter we can go say may be
say 2 or 3 decimal because, up to millimeter
38:44.940 --> 38:50.079
we can go. If it is say millimeter then we
can go only 1 decimal if it is or we can keep
38:50.079 --> 38:54.630
it in millimeter only. So, those things we
can write down in the very beginning of your
38:54.630 --> 38:59.650
say design note that these are the things
we are followed.
38:59.650 --> 39:17.910
Now, what about the torsion? That is equal
to 1 5 times 25 so 37 5 kilonewton meter.
39:17.910 --> 39:35.150
So, we can find out the equivalent
39:35.150 --> 40:01.009
moment due to torsion. So, equal to Tu 1 plus
d by b divided by 1 7 equal to 37 5 times
40:01.009 --> 40:13.460
1 plus d we have taken 500 b, we have taken
250, you can take it 0 525 also but, after
40:13.460 --> 40:22.940
all we are taking this only millimeter millimeter.
So, we can here and 1 7 which comes as let
40:22.940 --> 40:34.920
us see 37 5 plus 3 by multiplied by 3 by 1
7 what I mean to say that.
40:34.920 --> 40:39.789
Let us see that, how what is the factor I
can get it directly I can get the whole value,
40:39.789 --> 40:44.750
but that I am not interested I am interested
in usual cases it may happen that, sometimes
40:44.750 --> 40:51.349
it happens that your checking that what should
be the Tu times how much. So that, let us
40:51.349 --> 41:01.390
say that that means here we can get get it
as say 37 5 into 1 7647 that means, 76 percent
41:01.390 --> 41:06.079
extra the 1 76 that is what I am interested
to know.
41:06.079 --> 41:10.660
Because, only from this one you will not get
that idea you will not consider that what
41:10.660 --> 41:16.749
should be the value that means here, even
37 5 in worst case I can take it as say just
41:16.749 --> 41:23.079
simply I can multiply with 2 also, the twice
of that. So, what I mean to say that in that
41:23.079 --> 41:31.990
way those values factors are coming. So I
can say that 37 5 into 1 7647 and which is
41:31.990 --> 41:41.740
equal to 66 176 equal to say 66 18 kilonewton
meter.
41:41.740 --> 41:51.519
So, I can find out 66 18 kilonewton meter.
This one thing I would like to make it here
41:51.519 --> 41:59.900
comment that generally it happens these values
that whether you can for any design purpose
41:59.900 --> 42:05.309
if you have to feel it. In a sense that whenever
you are getting any numerical value, the numerical
42:05.309 --> 42:10.470
value whether it is coming all right or not.
So that, you have to check at each and every
42:10.470 --> 42:14.910
step whenever you are doing the design that
each and every step it would take that whether
42:14.910 --> 42:18.839
that value is coming all right or not.
Because, if you do at least few may be say
42:18.839 --> 42:24.299
4 or 5 problems like these then you will have
an idea that that value at in each step if
42:24.299 --> 42:28.910
you find out any unusual value. Then, immediately
you can find out you can understand, okay
42:28.910 --> 42:34.619
there is something wrong. So, thats why these
few steps that you all each and every step
42:34.619 --> 42:38.349
you have to do it and accordingly you can
find out. So that means, say it if it whether
42:38.349 --> 42:43.220
you are doing any calculation mistake or not.
So, just simply putting all those value in
42:43.220 --> 42:49.390
your calculator you can find out this value
and that what we do it. But since we are getting
42:49.390 --> 42:52.940
this idea the twice or something that it should
not be more than twice or something like that
42:52.940 --> 42:57.579
what is the range. So, immediately you can
find out whether you are getting that value
42:57.579 --> 43:05.880
correct or not, that is the standard procedure
of design that you have to do.
43:05.880 --> 43:26.019
So, what about the equivalent bending moment?
equal to 1 31 84 that is the Mu we have got
43:26.019 --> 43:33.390
it due to the applied load plus equivalent
moment, which is coming as 66 18 and which
43:33.390 --> 43:45.930
is coming as 198 02 kilonewton meter. So,
what about the effective depth we can compute
43:45.930 --> 44:01.819
we can check effective depth let us compute
effective depth. So, we have the same formula
44:01.819 --> 44:20.259
so 0 138 fck bd square equal to Mu.
Let us, write down here Me this is our Me
44:20.259 --> 44:29.660
so Mu means here, as per our that say due
to torsion so it should be applied here Me.
44:29.660 --> 44:56.799
So, we can find out d equal to 198 02 into
10 to the power 6 divided by 0 138 times 20
44:56.799 --> 45:15.740
times 250 equals 535 71 millimeter overall
depth d equal to 535 571 plus 25 is the clear
45:15.740 --> 45:22.960
cover plus let us take that bar diameter little
higher side 25 by 2 that is 25 millimeter
45:22.960 --> 45:26.529
dia bar.
Let us, assume so we are taking little higher
45:26.529 --> 45:37.640
side may be 20 millimeter may be sufficient.
So, which is coming as 573 21 millimeter so
45:37.640 --> 45:55.480
we can provide overall depth say 575 millimeter.
We have come to a problem, the problem is
45:55.480 --> 46:02.630
that we have selected we have taken the self
weight as say 500 millimeter, breadth width
46:02.630 --> 46:10.380
of the beam that is 250 millimeter, 573 21
millimeter that is the overall depth to be
46:10.380 --> 46:16.730
provided.
So, just we are really in a marginal situation
46:16.730 --> 46:24.359
575 millimeter. So that means, we do not know
that due to the self weight, whether that
46:24.359 --> 46:34.999
we shall steel will be in the safer side.
Because, that 535 71 another 1 we have taken
46:34.999 --> 46:47.740
that your say Tu we have taken here Tu that
d by b so d is that say your say 575 that
46:47.740 --> 46:53.099
means, Tu due to this torsion also that your
bending moment will change.
46:53.099 --> 46:59.559
So, if we really provide that 1 say 1 alternative
could be that we can go say 600 millimeter
46:59.559 --> 47:05.200
and we can check all of them we can do it,
instead of that what we can do it we can do
47:05.200 --> 47:19.609
it the other way. Let us try this way that
we shall change the width. So, let us change
47:19.609 --> 47:22.900
the width.
47:22.900 --> 47:43.390
So, assume width of beam say 300 millimeter,
overall depth 575 millimeter here since I
47:43.390 --> 47:52.589
have changed the 300 I could reduce it or
anyway let us, do not change this. Because,
47:52.589 --> 47:58.930
it may happen that that value will come less
but, anyway let us do not reduce it at this
47:58.930 --> 48:20.670
stage. So, Tu 1 plus d by b by 1 7 equal to
37 5 1 plus 575 by 300 divided by 1 7 equal
48:20.670 --> 48:37.930
to 64 33 kilonewton meter, self weight equal
to 0 3 width of the beam times 0 575 overall
48:37.930 --> 48:46.880
overall depth of the beam, times 25 kilonewton
per meter cube comes as 4 3125 kilonewton
48:46.880 --> 48:59.769
per meter, design load equal to 1 5 times
4 3125 plus 25 equals 43 96875 kilonewton
48:59.769 --> 49:01.329
per meter.
49:01.329 --> 49:29.890
So, we have got the design load we can calculate
that Mu equal to 43 96875 times 5 square by
49:29.890 --> 49:56.660
8 equals 137 40234 kilonewton meter. Let us,
say 1 hundred thirty 7 point 4 zero kilonewton
49:56.660 --> 50:18.089
meter. So, we have got Mu so equivalent bending
moment
50:18.089 --> 50:29.239
equal to 137 40 plus 64 33 equals 201 73 kilonewton
meter.
50:29.239 --> 51:01.369
D equal to let us calculate d. So let us you
see that 493 millimeter only, overall depth
51:01.369 --> 51:21.690
4930 59 plus 25 let us still take 25 millimeter
dia bar we shall provide. So, 531 09 millimeter
51:21.690 --> 51:38.809
provide overall depth D equal to say 550 millimeter
540 also we can provide but, anyway let us
51:38.809 --> 51:44.619
say 550 millimeter.
So, we can now we can still we can change
51:44.619 --> 51:48.559
that value because, self weight also that
means if we provide a 550 that means here;
51:48.559 --> 51:55.559
so, we can provide that self weight let us
check the self weight again.
51:55.559 --> 52:08.489
So, 0 3 times 0 55 times 550 times 25 comes
as 4 125 kilonewton per meter design load
52:08.489 --> 52:30.160
equal to 1 5 times 4 125 plus 25 equals 43
6875 kilonewton per meter Mu equal to 43 6875
52:30.160 --> 52:48.690
times 5 square divided by 8 equals 136 52
kilonewton meter equivalent moment so, 37
52:48.690 --> 53:12.819
5 1 plus 550 by 300 by 1 7 equals 62 5 kilonewton
meter. So, this is due to torsion only this
53:12.819 --> 53:16.419
is due to torsion only.
53:23.060 --> 53:43.880
So, equivalent bending moment equal to 136 52 plus 62 5 equals 199 02 kilonewton meter. So that means finally,
53:43.880 --> 53:53.880
we can set to this value that 3 by 550 millimeter and then, we can take it 199 02 kilonewton meter for
53:53.880 --> 54:00.369
the bending moment and we can provide the
reinforcement for this bending moment. I think
54:00.369 --> 54:05.019
we shall stop it here, we shall continue in
the next class.