WEBVTT
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Well, so today we shall learn how to calculate
deflection of rc beams reinforced concrete
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beams. If we take any the formula for calculation
of deflection is very easy. If we take a simply
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supported beam, if we take a simply supported
beam of say span L uniformly distributed load
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W. The deflection the central deflection say delta max that is equal to 5 by 384 W L to the power 4 by EI.
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We do not have any confusion regarding W the
uniformly distributed load over the span span
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L. But, we are having problem that E and I
because for reinforced concrete when you are
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talking say reinforced concrete we have to
calculate say E and i. So far, we have designed
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for the cracked section. That means, the only
complacent part that is uncracked but, bottom
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wherever we have reinforcement that crack
may appear.
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So what is the value of E and what is the
value of I that is the major difficulty here
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so for the formula concept. Formula looks
very simple but, providing that proper value
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of E and I we will give result that 1 differs
significantly; if we could not provide E and
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I. So we shall find out so what is the problem
let us find out. So what are the difficulties?
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Difficulties in calculation, number 1: let
me write down, variation in flexural stiffness
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that means E I along the span owing to but,
for owing to the uncertain nature and extent of cracking.
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So, we are faced difficulties because variation
in flexural stiffness E and I that EI that
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is the flexural stiffness along the span along
this span owing to the uncertain nature and
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extent of cracking. Because we do not know
cracking cracking also not uniform and also
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that uncertain wherever the crack will; there
is no uniformity. So, this is 1 of the cases.
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Number 2: variation in shrinkage and creep
these are the 2 properties we have told in
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the very beginning of our class. Owing to
what for it happens?
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Number 1: it may happen for curing that how
you have cured, then humidity, exposure, composition
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of concrete and also age at loading. That when you are
loading this structure it depends on that
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also. So, these are the difficulties that
we have faced due to these two cases that
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flexural stiffness you do not know the proper
ei it is not uniform like steel. And number
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2: due to shrinkage and creep and it may happen
due to say curing condition. That means, after
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the casting how it is cured humidity exposure
condition whether mild or that different exposure
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conditions.
Then composition of concrete what grade of
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concrete you are using that is it also matters
and also age at loading. When you are loading
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that what is the age of the concrete depending
on that. So, this formula we shall use the
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same formula only we have to provide that
proper E and I.
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We also write down the same formula say delta.
Let us write down it as 5 by 384 then W L
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to the power 4 EI. And we can make it as I
say F by EI F means here in this case what
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is f? F equal to you can write down 5 by 384
wl to the power 4 and it is this 1 it means
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we can say due to load span arrangement. So,
these value we are getting it here that f
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equal to 5 by that is if it is uniformly distributed
load over a simply supported beam, then we
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shall get these factor. So, EI is that value
F we are getting almost say we can say determinate
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the other part only we are facing difficulty.
So we can write down one more example we can
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show. Let us say, uniformly loaded by moment
bending moment that is uniform moment over
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the whole section. So, if we take a beam uniformly
distributed load say simply supported beam
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and moment is M that is uniform bending moment.
So, we can find out delta max delta max will
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be equal to, for this case the central deflection
delta max that one we shall get it as ML square
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by 8 EI this is the one deflection we shall
get it.
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We can find out also here that for these from
I do not know whether you know the moment
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area theorem. So ,according to that also you
can find out deflection otherwise you can
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find out the other way also but, any way delta
max is equal to ML square by 8 EI. So we can
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write down it as 1 by 8 times M by EI times
L square which comes as 0.125 M by EI times
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L square. In other way, we write down this
equation as beta times I times L square.
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Beta means, this coefficient this coefficient
is equal to 0.125 and sae, that is that curvature
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because psi equal to m by EI or 1 by o that
1 we can say and L is the span. So, we can
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write down in this fashion also that formula
we can write down. Infact what we do actually
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these value we use it for this shrinkage.
Indirectly we shall supply these value beta
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psi all those value we shall supply and on
the basis of that we can find out what is
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the deflection due to shrinkage.
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So, before going to that let me give the reference
that is your annex C. Annex C, that is calculation
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of deflection. And it is in page 88 IS 456
2000. There is one term is called that is
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you say c 2 clause c 2 that is your short
term deflection short term deflection. And
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let me give you the formula that whatever
given that I effective equal to Ir please
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note there are so many unknown things we are
writing. Ir minus Mr by M times z by d times
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1 minus x by d times bw by b.
So, ir atleast let me write down few things
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ir that is second moment of area of the cracked
section. Ir this is ir that is the one second
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moment of area of the cracked section. Mr
that is cracking moment mr that is cracking
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moment and which will be equal to fcr times Igr 1
more term divided by yt. Fcr modulus of rupture
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and that we shall take it as 0.7 root bar
fck. Fck again you know that is characteristic
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strength 0.7 root fck 0.7 root fck we shall
get it that clause that reference I should
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give IS 456 that is clause 6.2 0.2.
You will get it as flexural strength also
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modulus of rupture of page 16 IS 456 2000.
So, fcr you ll get it this formula you may
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refer it to clause 6.2 0.2. Then Fcr we know
what about Igr.
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So, our main difficulty here to find out this
I effective Ir we have to find out that we
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shall show you one example. Mr that is the
cracking moment so from there Fcr. Fcr we
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can directly we can get if it is m twenty
grade of concrete. So, I can get it fcr directly.
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What about Igr? Igr that is gross second moment of area. That means, overall section we are not taking any
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say craked section or anything that overall section or whatever we shall take it that is the 1 Igr. And yt
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distance from centroidal axis of gross section
and here we shall take it in the tensile 1.
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If we take a section if this is your section,
so yt will be we are assuming that tensile
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1 in the bottom. So, this 1 we shall take
it as yt and bw obviously bw this is bw and
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this one we shall take it as b or bf whatever
we call it b.
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So, bw is the width of the web and then b
that 1 say flange and d and overall depth
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that is say D. So, for cross section we shall
take it overall depth d. X I think that 1
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that is neutral axis. So, from the top we
can take the neutral axis depth and Z lever
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arm. And what about m? One more term is there
M maximum moment under service loads. So,
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if we now come back to the formula again once
more I hope I have defined everything. So,
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let us find out whether anything left or not.
So this is our formula we have to find out
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I effective and here also there is no end
to it. So I effective equal to Ir. Ir we have
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to find out for the cracked section Mr cracking
moment. We shall find out from here where
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igr is the for the cross section it also we
can calculate for the gross section. Z lever
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arm d effective depth x the neutral axis depth
bw width of the web and b width of the flange.
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So, everything we know; so we have to find
out. And here it doesnot end here there is
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1 restriction and that I should say that Ir
is should be that I effective and Igr. Then
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whatever we are getting that I effective that
whatever we are getting if it comes less than
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Ir then you have to take it Ir that value
we have to take it.
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So, I think we can take one example that will
be easier. Let us take one t section and let
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us assume that we have designed it due to
say your applied load. Now, you have to check
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the deflection. So, this is 2200 let us say
that the slab depth which is the depth of
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the flange also 125. Let us say, we have got 325 bar
and these depth that is say 575 overall depth
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700 concrete M 2. We are talking say service
load steel Fe 4 1 5 dead load span 7.5 meter
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and simply supported.
Then d equal to dead load fifteen kilo newton
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per meter this is the service load not the
factored 1. When we talking the deflection
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that 1 we are talking say service load. Live
load 10 kilo Newton per meter. D equal to
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700 overall depth 700 minus cover minus 25
by 2 it comes as 662. 5 millimeter. Area of
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steel that is 3 into pie by 4 into 25 square
so 3 into 491 1 4 7 3 square millimeter. M
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that is the moment we are getting it here.
So, WD plus WD that is the maximum moment.
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For the simply supported case WL square by 8 which comes as equals 175.78 kilo Newton meter. So, that
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Mr by M so this m that is the maximum moment
under the service load that we have got one
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component so far. So, let us find out so M
we have got it 175.78 kilo Newton meter. So,
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let us find out that Ec.
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Ec equal to 5000 root fck in the new code
in IS 456 2000 it is 5000 in the earlier code
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of 1978 it was 5 7 0 0. So, moment we have
got 175.78 kilo Newton meter and area of steel
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you are getting 1 4 7 3 square millimetre.
Effective depth 662.5 with clear cover 225
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millimeter and that bar we have used say 25
dia so which we have got it. These one you
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can say that you have got this information
everything from your say either by working
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stress method or by limit state method. That
you have got these values.
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Now, you are checking the deflection and which
you will do it by say under working stress
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condition that means under service load. So
Ec we are getting here say 5000 root fck and
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which comes as 2 2 3 6 1 newton per square
millimeter. Modulus ratio that modular ratio
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there is 1 more formula but, that we can use
directly this formula also Es by Ec. For steel
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it is this newtom per square millimeter and
which we got it for concrete from the flexural
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strength we are getting that is from characteristic
strength we can find out ec and which comes
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as 8.94. So modulus ratio 4 that we have got
it 8.94.
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Now, what about fcr fcr equal to 0.7 root
fck already I have told you that is in clause
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6.2 0.2 page 16. These value also you will
get it in the same page page 16 the next clause
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of IS 456 and which comes as 3.13 Newton per
square millimetre. You will get it as the
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flexural strength also modulus of rupture
which is defined in the list of symbols in
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IS 456. Let us assume, x is less than 125
millimeter. I mean to say that your neutral
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axis will be within your flange then our problem
will be easier.
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So, this your 125 that is the slab also we
can say or depth of the flange. So, neutral
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axis may be somewhere here. So this is your
x. So, if that be the case so you can use
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the transformed section it means half this
is your b times x square that means b times
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x into x by 2. B times x is the area times
x by 2 that is the 1 we are taking with respect
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to the neutral axis. That 1 should be equal
to modulus ratio times ast this one will change
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the transformed section this one will change
the steel to the equivalent concrete section.
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Times d minus x so I am getting this one with
respect to this neutral axis and taking the
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moment and from there we can get these values.
So, in our case in our case it will be half
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b is 2200 thats what we have given that in
the problem please I would like to refer it
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back, this is 2200. So, b and x I do not know
so far so we have got it. So, 1 4 7 3 whatever
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depth, depth is 662.5 minus x. So, from this
equation I can find out the value x. So, from
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this equation I can get that value x.
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So, let us just simply write down the equation
x square plus 1.05 equal to 0. And x comes
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as 83.27 millimeter less than 125 millimeter
the assumption whatever we have done that
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is correct; that means, it is coming within
flange. Now, we can calculate the lever arm
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lever arm working stress method. So 662.5
minus 83.27 by 3 and it comes as 634.75 millimeter.
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What about that Ir? Ir is the for the cracked
section second moment of area of cracked section.
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to let me write down the formula first b times
x square by 3. Let me draw a figure.
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So, this is your b somewhere here we are having
. So, b that means we are taking about this
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neutral axis, So, b x square by 3 we are talking
this portion so this is uncracked one. Plus
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m times ast where it is ast here times d minus
x whole square that means this 1 your d so
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d minus x. So, this is your d minus x.
So, this is your that formula so I can write
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down it 2200 into 83.27 whole cube by 3 plus
m is 8.94 which we have computed times 1 4
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7 3 the area of steel times 662.5. That effective
depth minus neutral axis depth 3.27 whole
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square which comes as 4.8414 into 10 to the
power 9 millimeter to the power 4. So, this
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is your ir for the cracked section we are
getting this value.
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Now other part distance of centroid from tension
fibre. So, this is 2200, this 1 your 125,
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this is 575 overall depth, this is overall
depth 700 the section which we are using this
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is 250. And we are interested to find out
yt may be somewhere here. And yt will be equal
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to because from the tension fibre and we are
for the simply supported case it is at the
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bottom. So, yt will be equal to 250 times
700 times 700 by 2 plus ... Let me write down
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here 2200 minus 250 times 125 times 575 plus
125 by 2.
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We are taking this portion this is 1 250 into
700 as if we are talking rectangle this rectangle
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we are taking. So, 250 into 700 into 700 by
2 plus the remaining portion which is 2200
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minus 250. So, that means this portion times
125 that is the area remaining portion times
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we are we shall get somewhere here. So, these
depth that is your 575 plus 125 by 2 thats
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what I have written. So we have made it 2
parts: 1 part is that the rectangular portion
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overall and the another 1 the remaining portion.
So, 2200 minus 250.
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Then we shall get it here divided by 250 into
700 plus 2200 minus 250 times 125 and which
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comes as 517 millimeter. So, we have got yt
and let us find out that Igr the gross moment
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second moment of area. Gross second moment
of area we can find out because I know now
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that 517 millimeter this 1. So 250 times 517
whole cube by 3 that means, I am talking about
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these portion this portion I am talking so
with respect to this neutral axis.
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So 250 into 517 whole cube by 3 plus let us
take the whole so which is 2200 into 700 minus
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517 whole cube by 3. I am taking this portion
the whole minus 2200 minus 250 times these
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portion this portion. That means, I would
like to remove this portion and let me write
33:10.149 --> 33:26.549
down that 1 here as 700 minus 517 minus 125
that whole cube by 3. And which comes as 1.588238
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into 10 to the power 10 millimeter to the
power 4.
33:31.859 --> 33:41.999
So, you can get Igr this value that means
I am taking the only this portion in the bottom
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of the neutral axis. Then I am taking the
top portion minus this portion this 2 thats
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what I have done it here. You can do it other
way also there is no such but, we generally
33:55.549 --> 33:57.629
do it in this fashion.
33:57.629 --> 34:20.230
So, what about your cracking moment mr the
cracking moment equal to igr by yt times fcr
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which equals 1 0.588238 into 10 to the power
10 into Fcr we have got it calculated 3 point
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1 3 newton per square meter 0.7 root fck.
Yt is your 517 and everything if we divide
34:42.220 --> 34:48.810
it by 10 to the power 6 then that 1 I shall
get it in kilo Newton meter. So, everything
34:48.810 --> 34:54.560
was in newton millimeter so divided by 10
to the power 6 will give me kilo Newton meter
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and which comes as ninty six point 1 5 kilo
newton meter. So, this is your cracking moment
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we have got it and we know that m also.
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So, now we shall calculate now we shall calculate
our that I effective. So, effective that formula
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we can write now once more so ieffective equals
to Ir let us produce the formula once more
35:33.490 --> 35:49.320
minus Mr divided by M z by d 1 minus x by
d times bw by b. Let us find out seperately
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1.2 minus Mr by Mz by d 1 minus x by d bw
by b. Which comes as 1.2 Mr that just now
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we have computed 96.15. M we know 175.78 W
L square by 8 that 1 the maximum moment under
36:17.090 --> 36:31.520
the service load. Z we have got it as 634.75
z by d 662.5 1 minus x 83.27 that also we
36:31.520 --> 36:45.680
have computed d 662.5 times bw 250 by 2200.
So, everything we have got it and which comes
36:45.680 --> 36:56.600
as 1.2 minus 0.052 equal to 1.148. So, this
is your we are getting this formula. So, I
36:56.600 --> 37:14.160
effective will be equal to Ir by 1.148 equals
0.871 Ir. But, in our case ir the formula
37:14.160 --> 37:26.600
is like this so it will be in between. So,
here I effective will be equal to Ir only
37:26.600 --> 37:31.230
we shall take the cracked one only. In our
case, we shall take I effective equal to Ir
37:31.230 --> 37:38.400
because we have to it should be greater than
Ir and less than I gross. So in our case it
37:38.400 --> 37:46.310
is coming less so we have to take I effective
equal to Ir.
37:46.310 --> 38:05.650
So now we can calculate different cases. Number
1: deflection due to dead load that is the
38:05.650 --> 38:13.100
permanent load deflection due to permanent
load that dead load here. So, delta DL we
38:13.100 --> 38:24.510
can write down as 5 by 384 into that 15 kilo
Newton meter that means 15 Newton per millimeter
38:24.510 --> 38:44.790
also you can write down times 8. It will be
7000 500 whole cube by E 2 22361 times that
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iI or I effective that is equal to 4.8414
into 10 to the power 9 thats what we have
38:53.330 --> 38:57.490
computed.
So, we can find out this formula using this
38:57.490 --> 39:31.320
we can find out and which comes as 5 into
153 divided by 3 it will be 4 it will be 4.
39:31.320 --> 39:41.980
5 into 15 into 7500 4 divided by 384 divided
by 22361 divided by 4.84149 which comes as
39:41.980 --> 39:52.150
9 which comes as 5.7 millimeter it comes as
5.7 millimeter. This is your exact 5 by 384
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wl to the power 4 by L is 75007.5 meter 15
kilo Newton per meter is nothing but, 15 newton
39:59.290 --> 40:07.840
per millimeter E 22361 which we are getting
from 5000 root fck and this is you said I
40:07.840 --> 40:17.290
effective. So, you can get 5.7 millimeter.
What about delta LL? The same way we can find
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out 5 by 384 or we can take the ratio also
10 Newton per millimeter 7500 whole to the
40:27.510 --> 40:41.760
power 4 divided by 22361 into 4.8414 into
10 to the power 9 and which will be here equal
40:41.760 --> 40:59.000
to so 3.8 say. So, this is your that we are getting
that deflection so permanent due to the permanent
40:59.000 --> 41:08.370
load and your say service load say live load.
And now we have two more things, one is thatâ€¦
41:08.370 --> 41:14.420
means here there is no difficulty we can find
out that Ec 5000 root bar fck and I effective
41:14.420 --> 41:20.020
also we can get it from the formula the 1
which is given in annex. That annex C and
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from there we can find out.
41:21.690 --> 41:28.910
Now, the next part is coming that is your
say due to shrinkage and due to creep. So,
41:28.910 --> 41:52.820
what we shall do it here that we do it that
is in clause C 3. This is deflection due to
41:52.820 --> 42:06.280
shrinkage what we shall do it here say delta
cs say let us say that 1 equal to let us write
42:06.280 --> 42:22.710
K 3 psi cs L square. That is the formula which
is given in clause c 3.1 page 88 IS 456. You
42:22.710 --> 42:27.860
remember that I have told you that formula
that beta psi L square.
42:27.860 --> 42:33.390
So, for simply supported case here also it
says for simply supported case K 3. So, K
42:33.390 --> 43:10.050
3 has diffferent values so 0.5 for cantilevers
0.125 for simply supported. 0.086 for members
43:10.050 --> 43:32.760
continuous at 1 end and 0.063 for fully continuous
members. In our case k 3 will be equal to
43:32.760 --> 43:54.380
0.125. Psi cs will be equal to that is called
shrinkage curvature and that is equal to K
43:54.380 --> 44:05.010
4 epsilon cs by d. So, here whatever actually
you were assuming that there is a beam and
44:05.010 --> 44:10.120
there is a due to shrinkage there is a deflection.
And that deflection we have to find out this
44:10.120 --> 44:13.420
deflection we have to find out.
So, what we are doing that means indirectly
44:13.420 --> 44:17.920
what we are doing actually basically we are
calculating what is the strength. For shrinkage
44:17.920 --> 44:23.880
strength and from there we can find ou. So,
what about the value of K 4?
44:23.880 --> 44:38.560
K 4 let me write down here that K 4 here seperately.
So K 4 equal to 0.72 pt minus pc that percentage
44:38.560 --> 44:46.850
steel root over pt. That percentage steel
tensile 1 this is compression steel percentage
44:46.850 --> 45:03.800
square root of that tensile steel. And less
than equal to 1.0 for 0. 25 pt minus pc less
45:03.800 --> 45:11.320
than 1.0. If this is the case, that means
we shall use this formula if we find if we
45:11.320 --> 45:15.910
find pt minus pc difference of the tensile
steel and compression steel percentage. If
45:15.910 --> 45:23.870
it comes within this range then we shall use
this formula. Otherwise, we shall use the
45:23.870 --> 45:29.800
seperate formula that means the change value
here.
45:29.800 --> 45:42.100
Then again it should be less than equal to
1.0 for pt minus pc greater than equal to
45:42.100 --> 45:47.830
1.0. So, let us find out we do not have any
compression steel we have only given tensile
45:47.830 --> 45:57.040
1. So pt equal to how much let us find out
pt equal to so let us find out pt.
45:57.040 --> 46:13.000
What about the area of steel? Area of steel
here area of steel 1473 times 100 divided
46:13.000 --> 46:26.950
by b bw we shall use it 250 we shall use bw
times d 6625. So, let me write down here pt
46:26.950 --> 46:43.500
will be equal to 100 ast by bwd and equals
0.889; it comes in this range. So, k 4 equal
46:43.500 --> 47:01.080
to 0.72 then, 0.889 minus 0 there is no pc
0.889 which comes as 0.6788. So, K 4 we shall
47:01.080 --> 47:17.300
get it as 0.6788. So, pt that means so from
this pt we can get K 4 0.6788.
47:17.300 --> 47:28.410
Now, what about epsilon shrinkage strain unless
otherwise specified we shall take shrinkage
47:28.410 --> 47:37.980
strain as 0.0003. We shall take shrinkage
strain as 0.0003 we shall take it if it is
47:37.980 --> 47:50.030
not specified. And so, we can find out that
psi cs will be equal to K 4 epsilon cs by
47:50.030 --> 48:07.800
d and will be equal to 0.6788 divided by 0.0003
divided by 700 and let us write down this
48:07.800 --> 48:18.200
1.
So, 0.6788 into 0.0003 divided by 700 which
48:18.200 --> 48:32.270
comes as 2.9091429 into 10 to the power minus
7. And then, delta cs that deflection due
48:32.270 --> 48:44.920
to shrinkage K 3 times psi cs times L square
which equals to K 3 for simply supported case.
48:44.920 --> 48:57.680
For simply supported case K 3 0.125, so 0.125
times 2.9091429 into 10 to the power 7 times
48:57.680 --> 49:19.060
7500 whole square. So, we can get it which comes as 2.04 millimeter. So, due to shrinkage we shall get 2.04
49:19.060 --> 49:40.240
millimeter. Then next one we have that is your creep and that you will get it in c 4 deflection due
49:40.240 --> 49:57.020
to creep deflection due to creep. And that
is in page 89 IS 456 2000. Let me little bit
49:57.100 --> 50:09.820
faster ecc that is given here Ecc equal to
that is your creep equal to Ec by 1 plus theta.
50:09.830 --> 50:16.010
And theta that is creep coefficient, there
is no such direct method that we are doing
50:16.010 --> 50:22.000
it in indirect way and that one we are getting
it from here say that experimental values
50:22.000 --> 50:31.170
on the basis of that we get it so coefficient.
Creep coefficient and generally we take it
50:31.170 --> 50:37.010
theta equal to 1.5 generally we take it theta
equal to 1.5.
50:37.010 --> 50:51.750
So, we can find out say delta c that is the
say permanent equal to the same formula we
50:51.750 --> 51:01.340
are using the same way Ecc ieffective minus
f by Ec this is due to load the short term
51:01.340 --> 51:09.170
load I effective. And Ecc equal to Ec by 1
plus theta. So, we can write down this 1 as
51:09.170 --> 51:21.460
f by I effective times 1 plus theta by Ec
minus 1 by Ec because I am just simply writing
51:21.460 --> 51:33.780
this Ecc here. And which comes as f by Ec
I effective times 1 plus theta minus 1 this
51:33.780 --> 51:43.810
1 equals f by Ec I effective times theta.
That means, only due to creep only due to
51:43.810 --> 51:47.570
creep we can get this value.
This one actually we can get it only due to
51:47.570 --> 51:53.550
creep we can get the value as this is due
to short term load this is due to short term
51:53.550 --> 51:59.020
load times the theta theta is 1.5. I t is
not mentioned and nothing so we have to take
51:59.020 --> 52:08.240
so f by Ec times I effective times theta.
So, what we can do it then we can find out
52:08.240 --> 52:15.700
let us go back to the whatever the results
we have got it.
52:15.700 --> 52:28.630
So delta DL due to dead load 5.7 millimeter
delta live load 3.8 millimeter, delta shrinkage
52:28.630 --> 52:43.590
2.04 millimeter. And delta creep that one
we shall get it as that delta DL times theta.
52:43.590 --> 52:51.200
That means, here it will be 5.7 this is due
to permanent load 5.7 times 1.5 and which
52:51.200 --> 53:02.850
will be equal to 8.55 millimeter. So, we can
get it here the total deflection that means
53:02.850 --> 53:07.400
that is due to load short term load permanent
load this is due to live load what we have
53:07.400 --> 53:12.840
got it. This is due to shrinkage we have got
it and we have checked we have got that 1
53:12.840 --> 53:18.930
this due to permanent load dead load times
theta will give me due to creep. So, we shall
53:18.930 --> 53:27.450
get it 8.55.
So, total deflection, so we can get it as
53:27.450 --> 53:43.870
say 5.7 plus 3.8 plus 2.04 plus 8.55 equals
5.7 plus 3.8 plus 2 .04 plus 8.55. Which comes
53:43.870 --> 53:51.310
as 20.09 millimeter almost equal to 20. That
means, 20 millimeter is the maximum deflection
53:51.310 --> 53:57.260
that we can allow so it is coming almost this
1. So, this is your that whole calculation
53:57.260 --> 54:03.050
regarding the deflection. The procedure that
whatever I have given that 1 according to
54:03.050 --> 54:04.869
IS 456.
Thank you.