WEBVTT
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So, welcome today's lecture; today we shall
continue the same design of slabs and that
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is part 2. So, and this is our lecture number
15, design of slabs part 2. So, let me first
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remind you the problem that which we have
started solving in the last class, that design
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of 1 way slab we have started with the design
of 1 way slab. Design a simply supported RCC
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slab for a roof of a hall; 3.5 meter into
8 meter inside dimensions with 250 millimeter
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walls all around.
Assume a live load of 4.0 kilo Newton per
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square meter and finish 1 kilo Newton per
square meter concrete grid M 20 and steel
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Fe 415. Just to show you what we have done.
so, far how far we have done. So, we have
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done the calculation of factored loads step
one and that, we have done on the basis of
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step span by depth ratio.
02:17.680 --> 02:25.280
The next 1 we have calculated dead loads,
on the basis of assumed thickness of the slab
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and finish already you have taken on kilo
Newton per square meter, live load of 4 kilo
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Newton per square meter.
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We have to calculate effective span, that
also we have calculated and effective span
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we have got out of these 2 the lesser 1 that
is 3.64 meter and total load per meter width
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of the slab that is 49.14 kilo Newton.
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And finally, we are getting the ultimate movement
as shear which, we have to do it for design
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and that we have got it mu equal to 22.3587
and Vu the shear force 24 .5 7.
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So, let us start the check number 3; check,
depths for binding we have to we are considering
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binding first. So, you have to consider binding
here. We shall use the same beam formula of
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rectangular beam only difference we are taking
1 meter; that means, 1000 millimeter and this
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is your overall depth which is 160 in our
case. We have to provide the reinforcement
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in the slab that we have to provide as reinforcement
of the slab. So, we are taking the rectangular
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slab the rectangular beam, 1 meter width and
depth we have assumed 160 millimeter and that,
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we have to check whether that is sufficient.
So, mu equal to 0. 138 fck bd square; d equal
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to root of Mu by 0.138 fck times b; movement
equal to the movement we have calculated that
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is, 22.3587 kilo Newton meter. So, movement
we have calculated 22.3587 kilo Newton per
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meter; so, 22.3587 into 10 to the power 6
Newton millimeter divided by 0.138 times 20
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times 1000 equals 90 millimeter and which
is, less than 140 millimeter; the effective
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depth that we have got it. So, d is equal
to 160 minus 15 minus 10 by 2 which is coming
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as one-forty millimeter; this is the 1 that
we have we have providing.
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So, d computed to make it clear; d computed
equal to 90 millimeter less than d provided
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equal to 140 millimeter this quite high of
course, but anyway we shall come back again.
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So, this you that check; so, which is all
right, but not economic.
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Number 4; check for shear. So, what about
tov v equal to Vu by bd equals just once more
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let me show you that, Vu equal to 24.57 kilo
Newton. So, Vu 24.57 into 10 to the power
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3 Newton divided d 1000 times d 140 which
comes as 0.1755 Newton per square millimeter
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Table 19 is 456 :2000 it gives for less than
equal to 0.15 that is equal that this is what
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is this 1 this is Ast 100 Ast by bd. For this
0.15 percent we get tov c equal to 0.28; this
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1 we will get it in table 19 is 456 tov c
0. 2.
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So, tov v less than tov c. So, we do not need
any that shear reinforcement and that we do
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not need any stirrup. So, that why we do not
provide the stirrup, but for other purposes
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we may need it.
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Number 5 calculation of steel areas. So, directly
we can calculate this or we can go from the
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mu that formula or directly we can remember
this formula 1 point 2 minus root bar 1.2
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whole squares minus 6.6 Mu by fck bd square.
So, this formula we can directly we can use
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it if you can remember otherwise, you can
start from the first principle that using
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the same formula. So, 6 .6 mu by fck bd square
equals 6.6 times 22.3587 into 10 to the power
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6 Newton millimeter divided by 20 times 1000
times 140 square which comes as 0.376. Therefore
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x by d equal to 1.2 minus root bar 1.44 minus
0.376 equals 1.2 minus 1.0315 equals 0.1685.
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So, what about your Zx by d this is less than
0.48; x u yd this is less than, 0.48; as I
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said write down this is less than 0.48.
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So, we can calculate now the lever arm we
can now calculate the lever arm. So, what
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about your lever arm lever arm Z equal to
d1 minus 0.416 x by d equals d equal to 140;
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1 minus 0.416 times 0 point 1 6 8 5 equals
1 thirty point 1 8 6 millimeter
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What about then area of steel area of steel
equal to mu by 0.87 fy times Z equals 22.358
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7 into 10 to the power 6 divided by 0.87 times
fu 415 times 130.186, which comes as 475.6
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millimeter square. What about the spacing,
spacing required because here we do not provide
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the number of bars here we provide that how
many bars at what spacing we will provide
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the reinforcement. Unlike say, your beam where
we provide the number of bars 3 bars 4 bars
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like, but here we provide that 10 millimeter
at the rate of that spacing we provide the
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where we providing stirrup.
So, spacing required 1000 times; so, how many
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bars we require 475.6 that, is the area of
the steel individual bar. Let us take say
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10 millimeter; so, we can get as bar if i
consider. So, we can get here 78.5 equals
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78.5 and which comes as, we are assuming that
we shall provide 10 millimeter bar and we
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are getting here 165.05 millimeter. So, we
can provide 10 at the rate of we can provide
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160 also, but let us provide 160.So, 150 millimeter
center to center and the area of steel provided,
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that is coming as 523.
So, this is your 10 at the rate of 150 millimeter
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center to center area of steel this is provided
and area of steel this is computed; 1 is computed
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and another 1 is provided. So, you have to
check that how much actually we are providing.
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So, we are providing here 523.
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So, now we can check let us see area of steel
500 or let us say make it very clear area
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of steel provided, equal to 523 square millimeters.
So, percentage of steel equals 523 into 100
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divided by 1000 times 140, 1000 is the width
and 140 is the depth effective depth which
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comes as 0.373 percent. There is 1 which is
called fs this factor, point because we are
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interested we would like to modify we have
assumed by l by d 25. Now, we would like to
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check we have that l by d we have started
with l by d that is 25.
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Now, I would like to check whether that is
sufficient or not because we have to control
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the deflection also. L by d 25 we have assumed
if it is less than, 25 which here comes then
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we have to change the depth, but if it is
more if we do not consider the economic consideration
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then, we can stop there otherwise if we consider
it is too much then; obviously, again we have
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to reduce the section.
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So, whatever the way our code says let us
see. So, we shall use this figure 4 modification
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factor for tension reinforcement. We have
given that tension reinforcement that, which
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Ast that, we calculated and that here we are
getting 0.713 percent. So, for 0.3734.373
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percent what shall we do? There is so, many
stress level fs 120, 145 this is 145; 192
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400 190, but it is not related with directly
not related with the fe 415; we have to do
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certain kind of modification.
What is the modification? this line if we
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do not if you could not get it let me write
down here that formula fs equal to 0.58 fy;
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fy is 415 fy is point fs is equal to 0.58
times fy divided by, area of cross section
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of steel required; this 1 will be required.
Area of cross section of steel required and
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area of cross section of steel provided. So,
we have to directly we cannot calculate not
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exactly 0.5 8 fy. So, here we have to do certain
kind of modification. So, area of cross section
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of steel required, in our case that is, 475.6
and we are providing 523.
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So, let me write down here clearly Ast required
equal to 475.6 square millimeters, Ast provided
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equal to 523 square millimeters. So, fs equal
0.58 times fe 415. So, 415 divided by 475.6
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divided by 523 which come as, 218 Newton per
square millimeter. So, we are getting here
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218 Newton per square millimeter. So, this
is the 1 that value through which we can find
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out, the corresponding that modification factor.
So, now if you come to this figure 4 of is
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4562, 1000 we know this is 218 and percentage
of steel we are getting here 0.373. So, 0.373
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means; somewhere it will be somewhere here
it will be somewhere here.
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So, you can write down. So, if we write down
here i think if we go something this way.
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So, we can get somewhere here we shall get
it and what is the value 218; so, 218 you
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can see that fs it is dependent on the value
fs 120 fs 145, 190; it means that, if you
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provide more reinforcement. So, you have that
you say modification factor also will be more
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that way also you can little bit you can manipulate.
So, anyway. So, if come here our case will
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be somewhere here our case will be somewhere
here. This line possibly somewhere almost
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in the middle zone, somewhere may be saying
some where here it will be.
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So, which is coming here this value that,
I have computed here that, is 1.55 because
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it is 1.4. So, I have taken 1.55.
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So, I can take that modification factor modification
factor that is equals 1.55; so, l by d equal
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to 1.55 this is a simply supported case times,
20 which comes as 31 we started that assumed,
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l by d that is 25. So; that means, it is from
here we can say; that means, we can further
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reduced the d. So, in this case that l by
or let us say d will be equal to l by 31.
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So, which may be? So, we can just come back
to the effective depth effective span.
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So, 36 43.64; so, 3640 divided by 31; so,
which comes as 117 point say 42 that means
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we can come to this, 117.42 and we have seen
also that d that is coming 90. So; that means,
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in this case what we can do? 125 that effective
depth; so, that way we can take it, but any
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way let us complete the problem here, but
it is the; that means, it comes the solution
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is not economic it is safe, but not economic.
So, in that case you can take 1 can again,
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further 1 can do the problem and he can again
do the next trial and he can modify the section.
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So, next 1 what about the secondary steel
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secondary steel will be Ast secondary means
the other way; that means, here whatever the
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longitudinal steel we were providing here
along the span say, this is along the span
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and then this is the other side that you say
secondary steel or we call it distribution
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steel. And this is the main steel, which you
are computing on the basis of your bending
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moment. So, Ast will be equal to 0.12 percentage
that is the minimum you have to provide times
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bd for mild steel it is point 1 5 for tod
steel it is 0.12 percent which comes as, 0.12
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times 1000 times.
This overall depth overall depth divided by
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100 equals 0.92 square millimeter. Let me
check it once more. So, we are getting here
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0.12 times 1000 times 160 divided by 100;
192. So, if you provide 8 millimeter dia bar
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as of bar area of bar, that is equal to say
which comes as 50 square millimeter. So, spacing
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required that is equal 1000 times 50 divided
by 192 equal to 260 millimeter center to center.
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So, we can get that area of steel of individual
bar, that is 50 square millimeter. So, on
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the basis of that we can find out the spacing
required that, is because we are calculating
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on the basis of everything per meter width
that is why you are taking this 1000; just
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to Refer this think.
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We should refer that, there is a clause 26.3.3
b this 1 related to the spacing of reinforcement;
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let us make it here. So, spacing of reinforcement;
there are 2 cases 3d or 300 millimeter, whichever
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less is and another 1 this is also including
against shrinkage and temperature. There is
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another 1number 2 this is also doing shrinkage
and temperature that is your 5d or 450 millimeter
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whichever is less. We shall in our case it
is coming 260 millimeter center to center.
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So, we can provide which is less than these
values. So, you can provide the distribution
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steel; that is, 8 at the rate of let us provide
260. So, let us provide 250 millimeter center
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to center.
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So, we can provide 8 at the rate of 250 millimeter
center to center. Now, finally, we have to
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show the detailing for each reinforcement
that, reinforcement concrete structures. We
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have to show the detailing; that means, for
beams we have to provide cross section also
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your 1 longitudinal section we have to provide
for beams and also the cross section we have
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to provide, in the longitudinal section.
We have to provide curtailment of the reinforcement
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and how is the shear spacing all those things
it will come and your cross section, will
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show a typical cross section. That means;
at which section you are taking along the
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longitudinal 1 that you have to provide. Here
also we have to provide thus for slab we have
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to provide the reinforcement. That detailing
you have to provide.
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Let us take; this is your 1 panel for which we are considering and as our problem says this is the slab
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supported over masonry wall in both sides, this is the
symbol of masonry. Our reinforcement, what
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we shall provide our reinforcement please
look, I think we can keep the dotted line
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here this is the support. What we shall do?
We shall give 1 reinforcement our main reinforcement.
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How much we have provided on main reinforcement
10 at the rate of 150 millimeter center to
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center we have provided on main reinforcement
10 at the rate of 1 fifty millimeter center
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to center.
And this is our span that 3640 and it is more
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than that, but effective span 3640 and this
1 how much this 1 is 3500 plus 250 plus 250.
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So, clear span 3500 plus 250 plus 250 which
comes as 4000.And effective span we have calculated
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3640 3460. What we shall do; we do not provide
the reinforcement here we can just simply
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provide the reinforcement say just 1 horizontal
line, but we have to provide the reinforcement
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at the the bottom as well as at the top.
So, what we do here in the support also we
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shall provide certain nominal reinforcement
that we shall provide. If we are provide that,
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nominal reinforcement here then what we shall
do? We shall calculate let us start, this
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is the 1 we are starting here 1 bar and then,
it will be cranked and go up like this. We
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are starting from here, then go up and this
1; the other 1 what we shall do this is 1
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reinforcement. Similarly, what we can do here
the other 1 i can start go up and this one;
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that means, alternate. So, your reinforcement
will be same, the bars will be same only thing
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1 will be placed this side and another load
will be placed other side.
31:04.200 --> 31:11.840
So, it will be cranking 1 side only when it
will be at the side they will crank in 1 side,
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but they will place alternate way. So, I shall
get the other reinforcement, I shall get it
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like this this , is your reinforcement and
it will keep on going like that. So, here
31:29.669 --> 31:45.889
because we want simply supported case, we
want this 1 as 10 at the rate of 150 center
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to center. And if we take a section here;
let us say this is section, a. So, we can
32:03.720 --> 32:17.559
get we can say this is our section a and we
have to provide, we have to provide the where
32:17.559 --> 32:28.210
shall will crack what about this distance.
If it is say Lx center to center distance
32:28.210 --> 32:39.500
is lx then, this 1 will be 0.1 Lx that is
the 1 it will 0.1 Lx.
32:39.500 --> 32:57.769
So, in our case lx equal to 3500 plus 250
center to center, which comes as 3750 millimeter?
32:57.769 --> 33:05.669
So, point 1 lx equal to 3 hundred seventy
5 millimeters. So, from the end of the support
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that, you will crank at 375 millimeter and
on both sides you will do it like that, but
33:15.049 --> 33:23.909
we have to provide the distribution steel
also like this, where to provide the distribution
33:23.909 --> 33:30.080
steel also and if I provide the distribution
steel. So, since you are cutting this section.
33:30.080 --> 33:43.529
So, we shall get it like that we shall get it like that because that all the longitudinal that you say other
33:43.529 --> 33:48.330
direction that on got. So, you will get all say the
circular 1 will get that the respective to
33:48.330 --> 33:57.009
the bar diameter and similarly at the top
also, similarly at the top also please. Note,
33:57.009 --> 34:10.360
we have to provide at least 2 distribution
steel, at least 2 bars at the top to hold
34:10.360 --> 34:14.290
the 1 at least 2 bars you have to provide
even if it is small, but you have to provide
34:14.290 --> 34:22.990
at least 2 bars you have to provide.
Now, what about the other side this side what
34:22.990 --> 34:34.500
about the other side; that means, I can take
another section that, is bb and which we say
34:34.500 --> 34:47.960
provide here though it is not a good choice.
So, generally will provide somewhere here
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and this is your that bar. Please note the
dots in 2 different sections please, note
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the dots in 2 different sections this is your
section bb. If I say this is 1, this is bar
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type 1 and this is bar type 2; the distribution
steel. So, 1 and this 2 here this is 1 this
35:28.609 --> 35:47.430
bar 1 and the other 1 the dots are 2 whereas,
here this 1 is your 2 and this dots 1;
35:47.430 --> 35:53.760
So, because we are taking in the other section
that is, very important whenever you are showing
35:53.760 --> 35:58.740
the section. So, which side you are cutting.
So, that is very important and you can see
35:58.740 --> 36:06.869
the main reinforcement always the main reinforcement
will have the maximum effective depth main
36:06.869 --> 36:13.700
reinforcement here this is your main reinforcement
that should have maximum depth effective depth.
36:13.700 --> 36:18.819
So, that is why it is always in the below
of the distribution steel.
36:18.819 --> 36:31.960
So, we have to provide the reinforcement also
we provide the thickness of slab that is here
36:31.960 --> 36:42.920
160 millimeters. So, we should provide the
detailing including the summary of results
36:42.920 --> 36:50.069
1 detailing. So, that from there at least
a draftsman should be able to draw the figure,
36:50.069 --> 36:59.930
as per the convenient he can place it, but
we need this say 1 plan and 2 section .
36:59.930 --> 37:13.150
There is 1 more thing that is called; so,
far we have done the udl, but if we have say
37:13.150 --> 37:31.400
slabs carrying concentrated load. If the slabs
are carrying concentrated load then, what
37:31.400 --> 37:48.980
will happen in that case we have 1 is called
effective width method. We have to calculate
37:48.980 --> 38:00.510
the effective width, it happens where we need
that we can experience the concentrated load.
38:00.510 --> 38:08.540
We can experience the concentrated load in
bridges in bridges we can that, because wheel
38:08.540 --> 38:14.839
load whatever we have that is you say concentrated
load. So, if we take generally; I think it
38:14.839 --> 38:24.010
is what it mentioning before going to that
the there are so, many kinds of bridges that
38:24.010 --> 38:30.430
are different types 1 is called say highway
bridges another 1 the railway bridges. So,
38:30.430 --> 38:37.250
these are the 2 different purpose; highway
bridges there we had having say wheel load
38:37.250 --> 38:46.940
and another 1 we have say that is called tracked
load. The loading whatever we get, the 1 say
38:46.940 --> 38:56.690
tyre, this is you the tyre or it can have
say some track load that having say very little
38:56.690 --> 38:58.290
bit longer length.
38:58.290 --> 39:07.000
So, while you have that impression on the
surface of the bridge, how the bridge comes
39:07.000 --> 39:13.920
there is 1 bridge that is called, a popularly
used for reinforced concrete that is pre-stressed
39:13.920 --> 39:24.609
girder, I shall, if I get opportunity I shall
tell little bit on pre stressed girder. So,
39:24.609 --> 39:32.599
this pre stressed grider, which we can construct
with which we can make it outside pre stressing
39:32.599 --> 39:41.050
it means; I can show you 1 example, let us
say that we have stake of books, which has
39:41.050 --> 39:47.470
simply 1 after another all the books not like
this, you are keeping the books 1 after another
39:47.470 --> 39:52.200
in this way.
Now, if would you like to carry it what you
39:52.200 --> 39:59.290
will do; you have to press it otherwise it
will fall. If what I am trying to say, the
39:59.290 --> 40:05.940
books are not placed like this 1 after another
instead of that books are placed like this.
40:05.940 --> 40:08.740
And you are pressing it and then you are carrying
it
40:08.740 --> 40:15.640
That means you have to compress it to that
way and then you have to do it. Here what
40:15.640 --> 40:23.510
happened, here it does the same way there
is 1 that cable the steel cables are there
40:23.510 --> 40:30.589
high tension cables are there and those are
pulled. If they are pulled and if they are
40:30.589 --> 40:34.660
restrained on that 2 ends then it will try
to compress.
40:34.660 --> 40:41.480
You are pulling the there is once a hole,
there are 2 there is once a hole through that
40:41.480 --> 40:48.910
we have done the casting of the beam say and
you have kept 1 hole through the that longitudinal
40:48.910 --> 40:55.710
section longitudinal way effective through
that, after 28 days after that, what you are
40:55.710 --> 41:02.670
doing when it is hardened fully hardened then
you are passing once a pre stressing cable.
41:02.670 --> 41:10.700
Now, while you are putting that 1 then what
you are doing you are pulling it in 2 sides
41:10.700 --> 41:16.140
and then, you are restraining something resting
somehow and then when you have rested then,
41:16.140 --> 41:21.720
it will try to compress when it will try to
compress. So, what will happen then initially
41:21.720 --> 41:29.329
the beam was like this; I want to say something
like this and keeping say here, I am not going
41:29.329 --> 41:34.890
in detail, I am keeping here, I am giving
certain kind of say eccentricity E and this
41:34.890 --> 41:42.920
is your force P which is put like this.
So, it will be like that due to bending due
41:42.920 --> 41:48.109
to that moment developed there is a moment
developed here. So, initially it will be like
41:48.109 --> 41:54.059
this not that much, but anyway it will try
to bend in this fashion because you are applying
41:54.059 --> 41:58.710
that I am not directly, I am not putting that
center instead of that, I am giving intentionally,
41:58.710 --> 42:04.099
I am giving this 1 at a certain eccentricity.
So, it will try to make 1 moment that developed
42:04.099 --> 42:08.819
and due to that, and also excel force. So,
what will happen it will make like this.
42:08.819 --> 42:15.569
So, what will happen if you would like to
for failure if we apply the load like this
42:15.569 --> 42:25.460
then what will happen, it has to fast neutralize
and then moves. So, automatically we can understand
42:25.460 --> 42:34.940
that, it will take more loads. The other 1,
I can give you example that is, you say that
42:34.940 --> 42:43.420
cycle that 1 that ring, there also the spokes
that, whatever you are using there also if
42:43.420 --> 42:50.240
you have noticed you will see there also they
are doing certain kind of pre stressing. Because
42:50.240 --> 42:56.050
all the say your spokes they always rotate
like that and there are doing the that is
42:56.050 --> 43:00.650
why it does not bend; otherwise what will
happen, it will just make certain kind of
43:00.650 --> 43:06.160
say bending of the your ring
So, there also they are doing the pre stressing;
43:06.160 --> 43:12.130
so, this is the thing what they are doing,
but anyway we are going little away, but anyway
43:12.130 --> 43:18.119
just to conclude this pre stressing. This
is called pre stressed girder and so, we can
43:18.119 --> 43:25.510
go say 20 meter 25 meter 40 meter which we
shall not be able to do it by reinforced concrete
43:25.510 --> 43:31.900
reinforced concrete beams that 1 we cannot
go say 5 meter 6 meter 7 meter maybe we can
43:31.900 --> 43:38.040
go say 10 meter, but your depth will be enormous
that depth you have to because mu equal to
43:38.040 --> 43:42.819
0.138 fck bd square. So, from that you can
find out that, what will be corresponding
43:42.819 --> 43:49.450
depth, if we have the span more with the same
load the depth will be more. But here what
43:49.450 --> 43:56.589
we do we use that, pre stressing girder pre
stressed girder we use it and that can go
43:56.589 --> 44:03.400
say 40 meter also we can go and depth will
be say you 1 meter 1.5 something like that,
44:03.400 --> 44:08.730
it can happen and we use number of cables
and in bridges we use it
44:08.730 --> 44:17.480
So, while you have 2 girders say like this
there are so, many may be say the bridge having
44:17.480 --> 44:23.500
say 6 girders, if you notice now a day’s
say, bridges when if you go by train itself
44:23.500 --> 44:30.609
always there will be always 1 another highway
bridge at least. So, if you notice that you
44:30.609 --> 44:37.670
can see that, depth slab and the depth of
girder that you can see there is a difference.
44:37.670 --> 44:44.750
So, that is now here, we are having that wheel
load these wheel load is coming as I say concentrate
44:44.750 --> 44:49.800
load. So, now, if I have the concentrated
load here then, what will how I shall take,
44:49.800 --> 44:54.079
the effective width effective width of the
loading.
44:54.079 --> 44:59.130
These loading directly that, whatever the
impression we do not take it we take little
44:59.130 --> 45:06.050
more. So, how much more we can take it that
we have to find out. So, what we shall do
45:06.050 --> 45:14.930
it here, in this case there is always 1 wearing
coat; now, come to the problem here there
45:14.930 --> 45:32.500
is always 1; this is
45:32.500 --> 45:47.440
called wearing coat this is not reinforced
and then we are having that slab let us say,
45:47.440 --> 45:54.309
this is up to the this is the reinforcement
and this is the loading that, concentrate
45:54.309 --> 46:07.720
load having certain area we take 45 degree
dispersion.
46:07.720 --> 46:25.720
If we take this 1 as h and we have to find
out what should be the this length a and let
46:25.720 --> 46:36.990
us take this 1, say your x1 is the impression
say in 1 way other way say y1. So, x 1 times
46:36.990 --> 46:46.010
y1 that is, at the top that is the impression
x1 by y1 and we can take this is you say d
46:46.010 --> 46:57.079
is effective depth this is the effective depth.
There are 2 concept, 1 concept is that particularly
46:57.079 --> 47:04.349
this portion you go up, to the top of the
slab other alternative that 1 because this
47:04.349 --> 47:09.490
1; if you take this much then that this we
divide this 1; that means, your concentrated
47:09.490 --> 47:15.309
load and so, you are taking actually moment
if you compute moment that, will be more because
47:15.309 --> 47:19.670
this 1 this length is less. So, distributed
less.
47:19.670 --> 47:24.369
So, moment will be more computed moment will
be more. Whereas if i take this 1 then the
47:24.369 --> 47:29.319
moment will be less because it distributed
along this 1 if you take this length then
47:29.319 --> 47:36.809
it will be further distributed. So, your moment
also will be further less. Now, the designers
47:36.809 --> 47:42.349
1 who are a 1 type of designer, 1 class of
designers they take it say up to this another
47:42.349 --> 47:49.319
1 they take it up to this obviously; I can
say they are more conservative the designers
47:49.319 --> 47:54.089
who are using this 1 upto this; that means,
they are more conservative then, the designer
47:54.089 --> 48:02.229
who are taking this 1; because here your moment
all this things will be calculated less compared
48:02.229 --> 48:03.380
to this 1.
48:05.000 --> 48:41.880
So, I can further write down let us say this
is continuation. So, if we take this length
48:41.880 --> 49:01.470
say B and this is your span
let us say, this is your le that effective
span and we have certain impression here,
49:01.470 --> 49:13.420
this is a impression either from the tyre
or any other concentrated load and after dispersion
49:13.420 --> 49:24.270
we are getting, say like this is our a is
the 1; this is the support span means this
49:24.270 --> 49:29.380
1 span. So, we can assume that there are 2
girders 1 girder say this is 1 girder this
49:29.380 --> 49:35.040
is another girder and this is the over the
there is bridge depth may be say 250 millimeter
49:35.040 --> 49:40.410
or say 200 millimeter in that range it comes,
but whereas the girder if may be say 1 meter
49:40.410 --> 49:44.650
1 point 5 meter. So, that 1 will be considered
as support.
49:46.460 --> 49:56.040
So, a is the 1 parallel to this girder and
from the nearest support, this distance is
49:56.049 --> 50:04.960
x from the nearest support this distance is
x. So, what we can do? This is your dispersed
50:04.960 --> 50:12.920
area after the dispersed area means after
the 45 degree the dispersion this is dispersed
50:12.920 --> 50:24.349
area and we have to find out, the effective
width be. Let us say, B is the effective width
50:24.349 --> 50:29.260
if we have to find out and that impression
that, 1 having say at a distance say x from
50:29.260 --> 50:40.240
the nearest support. So, how shall we find
out that, we can say a equal to either we
50:40.240 --> 51:01.000
can write down say x1 plus twice the thickness of wearing coat or we can write down a equal to x1 plus
51:01.000 --> 51:08.230
twice say, let us say; thickness of wearing coat
plus d. There are 2 options I have told, I
51:08.230 --> 51:11.869
have told either it could be up to this or
it can go after to the effective depth.
51:11.869 --> 51:15.950
So, you can write down that x1 plus twice
thickness of wearing coat or x1 plus twice
51:15.950 --> 51:24.650
thickness of wearing coat plus d. So, that
we can find out and our code says our code
51:24.650 --> 51:40.290
says, be the effective width equal to k times
x1 minus x by le plus a. So, this is 1, a;
51:40.290 --> 51:50.260
I know x, I know, Le I know. So, I can find
out Kx Le effective span and a, I can find
51:50.260 --> 51:56.569
out the effective width this is the effective
width instead of taking this a part of part
51:56.569 --> 51:59.130
of meter width or anything. So, it could be
that effective width.
51:59.130 --> 52:04.349
So, I have got this 1 effective width and
what is the value of k this 1; I shall get
52:04.349 --> 52:17.049
it from table 14 is 456 2000. So, we have
to calculate that due to concentrated load
52:17.049 --> 52:21.589
we have to calculate that, effective width
instead of taking say 1 meter you have to
52:21.589 --> 52:25.730
calculate that effective width and that we
can find out from this formula and which is
52:25.730 --> 52:27.109
I have in the code.
52:27.109 --> 52:34.130
Similarly, for the cantilever beam also. So,
we shall stop it here then we shall continue
52:34.130 --> 52:38.250
in the next class the slab. I think we shall
take few more classes for slab we have 1 more
52:38.250 --> 52:43.970
that is you say, 2 way slab also this is called
1 way slab similarly we have 2 way slabs .