WEBVTT
01:03.540 --> 01:11.320
So, let us start with continue with the design
of columns. Last class we have done that,
01:11.320 --> 01:19.210
axially loaded column and we have axial uniaxial
bending as well biaxial bending. But today
01:19.210 --> 01:26.630
we shall with that one problem, example problem.
We shall start one example problem on axial
01:26.630 --> 01:27.509
loaded column.
01:27.509 --> 01:44.779
So, let us take one example design
01:44.779 --> 02:09.319
and axially loaded tied column, pinned at
both ends. So, this will give the effective
02:09.319 --> 02:59.890
length with an unsupported length of 3.5 meter for carrying characteristics load of 1500 kilo newton and
02:59.890 --> 03:25.170
concrete grade M20, steel grade Fe 415, please note, the characteristics load of 500 Kilo Newton.
03:26.360 --> 03:33.540
So, what about the design load? And this is
axially loaded column, design load which is
03:33.549 --> 03:45.150
Pu, we shall take it as a 1 5 times. The characteristics
load unless otherwise, specified. So, it will
03:45.150 --> 03:49.780
be different for different cases sismic load,
wind load, like that but in way even if it
03:49.780 --> 03:59.870
is not specified anything. So, we shall assume
1.5 which comes as 2250 Kilo newton. What
03:59.870 --> 04:14.560
about the Effective length? Effective length,
say le will be equal to same as the unsupported
04:14.560 --> 04:26.020
length. Because, it is pinned so, it is pinned
condition. So, for this case it will have
04:26.020 --> 04:41.140
3.5 metre what we have to do, we have to assume
a section we can and how shall what is the
04:41.140 --> 04:46.520
guideline? That guideline is that, l by d
less than 12. Because, we are considering
04:46.520 --> 04:54.190
it as a short column, since it is a short
column so, l by d less than 12.
04:54.190 --> 05:14.760
So, if l by d equal to 12 D will be equal
to l by 12 equal to 3500 divided by 12 which
05:14.760 --> 05:25.860
comes as 291.6 millimetre. So, we can assume
side of the column 291.6 why shall I got 325
05:25.860 --> 05:46.650
or 350 for the time being let us, start that
assume size of column 300 by 300 whatever
05:46.650 --> 06:02.940
the e mean as per the code eccentrically minimum
l by 500 plus D by 30 which comes as 3500
06:02.940 --> 06:19.450
divided by 500 plus 300 by 30 which comes
as a 17 less than here 20 millimetre.
06:19.450 --> 06:34.700
Let us, check the slenderness obviously; it
will be less than 12. So, you can check it
06:34.700 --> 06:47.020
le by d equal to 3500 by 300 which comes as
11.66 less than 12. So, this is your short
06:47.020 --> 06:53.170
column was also less than 20 and we have so
far we have assumed that 300 by 300. But we
06:53.170 --> 06:58.000
do not know, that whether 300 by 300 sufficient
from the strength point of view. We are talking
06:58.000 --> 07:03.370
this is the preliminary design, that we are
getting these dimensions 300 by 300 on the
07:03.370 --> 07:06.980
basis of that we would like to make it a short
column.
07:06.980 --> 07:13.710
So, for that we have taken 300 by 300, but
so far we do not know whether that your 2250
07:13.710 --> 07:21.050
kilo newton the design load which is applied
here; whether, this section is sufficient
07:21.050 --> 07:33.840
that we do know. So, let us continue we can
find out the design load compute area of steel.
07:33.840 --> 07:46.530
List of that, how shall we do compute area
of steel. So, Pu as per the code Pu equal
07:46.530 --> 08:07.500
to 0.4 fck Ac plus 0.67 fy area of steel.
Let us, write down Pu design load, Ac area
08:07.500 --> 08:34.849
of concrete, As area of steel, fck that I
can say characteristics value of concrete,
08:34.849 --> 09:02.610
fy for steel. So, we can now find out what
about the Ac? Ac equal to Ac we can find out.
09:02.610 --> 09:17.750
So, we can write down here the 0.4 fck is
20 then Ac will be equal to 300 square minus
09:17.750 --> 09:30.080
As. Because, we are taking only concrete part,
300 square minus is plus 0.67 times 415 times
09:30.080 --> 09:38.280
As should be equal to 2250 into 10 to the
power 3 let us make it everything in newton.
09:38.280 --> 09:58.170
So, we can get this part minus 8 As 0.4 into
20 plus this part is coming as 278.05 as will
09:58.170 --> 10:08.940
be equal to 2250 into 10 to the power of 3
minus 8 into 300 square I am taking this side.
10:08.940 --> 10:31.270
Or 270.05 As equals 15300 so, this is the thing or As area of steel equal to 5665.61 millimetre. Let us, say
10:31.270 --> 10:42.440
what about the percentage of steel p? That we have
to check it equal to let us, ay 5665 only
10:42.440 --> 10:54.529
divided by 300 into 300 into 100 which comes
as 6.29 percent greater than 6 percent. So,
10:54.529 --> 11:03.820
this section is not we cannot provide this
section.
11:03.820 --> 11:09.720
Because, the area of steel is coming greater
than 6 percent so, we cannot provide this
11:09.720 --> 11:15.170
section. So that means, here that we have
started with the start column with the criteria
11:15.170 --> 11:21.240
that it should be less than 12 even then,
you have to come to that here that from the
11:21.240 --> 11:27.110
strength the load that, the load you have
to bear the column has to bear. So, with that
11:27.110 --> 11:39.320
300 by 300 that steel is coming more let us,
say 6.29 percent 5665 what is the what difficult
11:39.320 --> 11:46.340
we can face? Even we provide that 1 say.
11:46.340 --> 11:52.540
Let us, find out the area of steel As 5665
11:52.540 --> 12:16.170
millimetre square let us consider 25 tau bar.
So, area of steel of each bar or As bar I
12:16.170 --> 12:31.940
can say that is coming as which comes as 491
square millimetre. So, number of bars required
12:31.940 --> 13:00.430
5665 by 491 which comes as 11.53. So, we have
to provide 12 numbers 25 tau it means, this
13:00.430 --> 13:21.450
is a section this is 300 1 2 3 4 corner 5
6 7 8 9 10 11 12.
13:21.980 --> 13:41.240
So, each side 2 more so what is spacing here we are getting. The space we will get it here, the cover the clear
13:41.240 --> 13:48.270
cover for columns it is 40 millimeter. The
clear cover is 40 millimeter for columns so,
13:48.270 --> 14:02.220
from both sides 300 minus 40 minus 40 which
comes as 220 millimeter. So, 220 minus 25
14:02.220 --> 14:12.900
by 2 minus 25 by 2 it means, from both sides
from the center I am talking.
14:12.900 --> 14:25.720
So, what we are getting it here? This is the
Let us, see that how much gap we have. So,
14:25.730 --> 14:33.770
we will say 12 bars uniformly distributed
after all it is an axially loaded column so,
14:33.770 --> 14:48.920
we can distribute uniformly. So, this distance
centre to centre of outer bars equal to 300
14:48.920 --> 14:56.170
minus 40 is the clear cover minus 40 the clear
cover other side minus 25 by 2 minus 25 by
14:56.170 --> 15:06.960
2 which comes as.
So, 80 plus 25 105 so 195 there are how many
15:06.960 --> 15:20.779
gaps 1 2 3 so, 3 gaps are there let us find
out. So, 195 by 3 the center to center gap
15:20.779 --> 15:40.660
195 by 3 which comes as 65 the center to center
gap was between these 2 bars. So, clear gap
15:40.660 --> 15:53.450
between 2 bars will be equal to 65 minus 25
by 2 minus 25 by 2 which comes as 40 millimeter.
15:53.450 --> 16:01.240
Now, the question is here this is say 40 millimetre
let us say, that you would aggregate maximum
16:01.240 --> 16:05.209
aggregate dimension that is also 20 millimeter
core segregate. So, that means it should come.
16:05.209 --> 16:11.160
Now, the question is coming whatever would
the lapping? Because, we shall get the bar
16:11.160 --> 16:16.930
from the bottom to the top we shall not get
that bar, that 1 single bar having that sufficient
16:16.930 --> 16:22.830
laid may be say 2 storied, 3 storied, or 5
storied we shall not get, that long bar. So,
16:22.830 --> 16:28.740
we will get may be 5 meter or maybe say 10
meter or whatever it is. So, if I get that
16:28.740 --> 16:37.230
we have to give lapping of bars required.
So, it means even I say that 50 percent of
16:37.230 --> 16:44.930
the bar will be staggered. That all we shall
such a way, that all the bars will not curtail
16:44.930 --> 16:54.279
in the same position that means, all are going
up and then at this level and then, again
16:54.279 --> 17:00.089
you are putting on that I do not want that
and that is not a good solution. So, it is
17:00.089 --> 17:06.709
not desirable then the section will be weak,
it will be that your it you can say that crowded
17:06.709 --> 17:10.699
with so many bars.
So, even you say that 6 bars will be there
17:10.699 --> 17:16.299
even then, increasing that percentage still
will be l you are getting 6 point certain
17:16.299 --> 17:21.939
percentage I m not talking with the code provisions.
So, 6.29 percent so, it will be 9 percent
17:21.939 --> 17:28.439
then, because 50 percent of that also will
be there that means, if there are 12 bars.
17:28.439 --> 17:39.139
So, I mean to say 12 into 491 divided by 300
into 300 times 10s 100 this is the percentage
17:39.139 --> 17:48.879
of the steel provided in this case. So, it
comes 12 into 491 divided by 300 divided 300
17:48.879 --> 17:57.600
into 1. So, even then we are getting 6.54
percent so, why we are providing the lap that
17:57.600 --> 18:04.090
means, it should be 12 bars plus 6 bars in
a particular section. Then, in the side by
18:04.090 --> 18:09.779
side if I take it.
18:09.779 --> 18:19.090
Because, it is very important otherwise, the
cracks lot of other problem will arise, that
18:19.090 --> 18:26.950
is why that detailing reinforcement all those
things are very important. So, what shall
18:26.950 --> 18:39.269
we do we can start if I take 50 percent at
a particular section. Let us, say we are having
18:39.269 --> 18:46.470
this bar and this bar say.
So, I can provide 1 bar here and I can provide
18:46.470 --> 18:54.970
1 bar similarly, I can have 1 bar here, I
can have 1 bar here, like that it will go
18:54.970 --> 18:59.909
similarly, I can have say 1 bar here and 1
bar here. That means, in a particular section
18:59.909 --> 19:06.239
because, these bars are left you are giving
the lapping these bars. So, that in particular
19:06.239 --> 19:19.409
section you are having 18 bars if you have
18 bars 18 into 491. So, the percentage of
19:19.409 --> 19:30.279
steel becomes at the lapped section where
you are providing the lapping for 50 percent.
19:30.279 --> 19:42.009
Let us, be specific 50 of the bars we are
getting here 9.82 percent.
19:42.009 --> 19:58.619
So, it will very difficult to do the concreting
and then, the if you if there is a in between
19:58.619 --> 20:04.940
the core segregate, if there is a achieve
the concrete strength. So, that is why it
20:04.940 --> 20:10.559
is very important to use that is why the code
says that, you should not use more than 6
20:10.559 --> 20:15.739
percent. But if you uses 6 percent because,
you have to provide the lapping so, that ways
20:15.739 --> 20:22.190
in that case I can say, it is better the designer
we should always prefer you should not be
20:22.190 --> 20:29.009
more that 3 percent.
So, if it is 3 percent then, what happens
20:29.009 --> 20:35.169
even somebody uses the lapping all the lapping
in 1 section even then, also we shall not
20:35.169 --> 20:41.139
go beyond 6 percent. So, that we do not though
we write down specifically that only 50 percent
20:41.139 --> 20:44.929
of the lapping will be done at particular
section. But even then, we shall not take
20:44.929 --> 20:52.509
a chance so, it is preferable that you percent
of steel should within 3 percent that should
20:52.509 --> 20:57.489
be the or there are sometime we say, within
4 percent sometime we make it like that.
20:57.489 --> 21:17.850
So, let us take the section then, assume the
column section as 375 by 375 we shall take
21:17.850 --> 21:34.409
that column section as 375 by 375. So, what
we can do it hear emin that l by 500 plus
21:34.409 --> 21:45.159
375 by 30 which comes as 7 plus 12.5 equal
to 19.5 less than 20 millimeter whatever,
21:45.159 --> 22:03.789
the area of steel we can take it as same formula
0.4 fck Ac plus 0.67 fy As equal to 2250 into
22:03.789 --> 22:12.619
10 to the power of 3.
Or 0.4 times 20 times 375 square minus As
22:12.619 --> 22:27.779
plus 0.67 times 415 times As equal to 2550
into the 10 to the power of 3 or minus 8 As
22:27.779 --> 22:53.700
plus 278.05 As equal to 2250 into 10 to the
power of 3 minus or 270.05 As equal to 1125000
22:53.700 --> 23:07.749
therefore, As equal to 4165 square millimeter. So, you’re
As area of steel got it as 4165 square millimeter
23:07.749 --> 23:16.049
if we have that section 375 by 375. What about
the percentage of steel? We can find out the
23:16.049 --> 23:16.950
percentage of steel.
23:16.950 --> 23:30.489
So, p equal to 4165 into 100 by 375 into 375
equal to 2.96 percent. So, you can get 2.96
23:30.489 --> 23:50.820
percent we can get. Number of bars use 25
tor number of bars equal to 4165 by 491 that
23:50.820 --> 23:58.979
is the area of for that 25 millimetre bar
which comes 8.48 what we can do, we can provide
23:58.979 --> 24:25.509
10 numbers of 20.5 tor. So, area of steel
provided equal to 10 into 491 into 100 by
24:25.509 --> 24:33.190
375 into 375 which comes as 3.49 percent.
Even then, though it says that it should less
24:33.190 --> 24:39.820
than 4 percent because, 4 percent means 4
plus half of that 2 so, 6 percent that 50
24:39.820 --> 24:43.139
percent lapping.
So, in this case alright, but even then I
24:43.139 --> 24:51.359
prefer less than 3 percent, it is a designer
choice. So, you will find out the reinforcement
24:51.359 --> 24:56.629
design or steel design you will find out it
is something like that your say different
24:56.629 --> 25:01.340
school of thought different philosophy you
cannot say this is wrong, you cannot say that
25:01.340 --> 25:07.269
is wrong only thing I can say that, it is
an idea that it is the philosophy from where
25:07.269 --> 25:14.989
we have learnt. It is also sometimes, it happens
in the institute itself so, with whom we have
25:14.989 --> 25:17.340
learnt that 1 teacher so, that way also it
goes.
25:17.340 --> 25:21.570
Similarly, in the industry also with whom
you have worked so, that philosophy that actually,
25:21.570 --> 25:27.239
you follow that particular 1 that is the idea.
So, I prefer less than 3 percent so, that
25:27.239 --> 25:36.429
is the idea I usually do it. So, now we can
say 10 numbers of 25 tor. So, what about the
25:36.429 --> 25:47.059
that say cross section? Cross section the
I shall provide here always 4 corners, we
25:47.059 --> 25:56.450
can provide here 6 then, we can provide 2.
So, we are having though it is not symmetric,
25:56.450 --> 26:13.080
but anyway 10 numbers of bars. Now, we are
to provide the tie bars so, if we have to
26:13.080 --> 26:15.700
provide the tie bars.
26:22.560 --> 26:49.620
The diameter 25 by 4 so, we can provide use
8 mm dia bar what about the spacing? This
26:49.629 --> 27:03.889
will not be less than spacing number 1: the
dimension of column say 375 millimeter in
27:03.889 --> 27:16.749
this case. Number 2: 16 times the longitudinal
bar 400 millimeter. Number 3: our code says
27:16.749 --> 27:32.479
300 millimeter. So, we can provide 8 tor at
the rate of say 300 millimeter centre to centre.
27:32.479 --> 27:42.759
So, this is the 1 we can provide for tie bars.
So, this is your that column design for fca
27:42.759 --> 27:49.470
loaded column. But here this is not the end
of the column design because, it is not so
27:49.470 --> 27:56.169
simple that we can have eccentricity or we
can have that moment developed. So, there
27:56.169 --> 28:04.809
are 2 possible cases so, we would like to
do.
28:05.889 --> 28:33.329
Design of columns having axial load and moment;
the moment would be uniaxial or biaxial. If
28:34.460 --> 28:44.789
this is your beam say and here is an column
if the span is equal loading is equal I can
28:44.789 --> 28:55.210
say this side there is no moment, about this
side there is no moment. Whereas, about this
28:55.210 --> 29:04.409
axis there is a moment of in addition, to
that axial load the moment is also here. In
29:04.409 --> 29:25.429
this context, I think I can tell you that
if there is a plane here the plan I am drawing
29:25.429 --> 29:32.570
the plan here.
Then, if the columns are rectangular, this
29:32.570 --> 29:40.799
is your column what is the orientation of
the column then, with respect to these the
29:40.799 --> 29:49.839
orientation of that 1. Because, if we take
the whole building say rectangular building
29:50.000 --> 29:58.300
type say flanged rectangular. So in that case
obviously, we can say it is easier to bend
29:58.840 --> 30:07.760
let us say this is a building it is rectangular
1 that cross section plan there are so many
30:07.769 --> 30:11.109
rooms.
So obviously, I can say it is easier to bend
30:11.109 --> 30:20.389
along this about bend this way, but where
as I cannot bend this way I do but it is difficult.
30:20.389 --> 30:29.299
If we say wind load or earthquake load due
to ground motion it moves like that. So obviously,
30:29.299 --> 30:33.989
it will vibrate you can see that it is vibrating
like this rather, it is vibrating like this.
30:33.989 --> 30:41.590
So, if note this 1 if you observe this so,
if rectangular comes if the column is square
30:41.590 --> 30:47.049
there is no problem.
Then, if the column is your say circular than
30:47.049 --> 30:51.210
here is no problem, there is no orientation
problem. But if the column is rectangular
30:51.210 --> 30:56.629
the question is whether, shall we provide
the column this way or shall we provide the
30:56.629 --> 31:07.340
column this way obviously, we shall provide
the column this way. Because, the moment of
31:07.340 --> 31:18.090
second moment very less to resist that 1 we
shall provide the column in this direction,
31:19.400 --> 31:27.640
not the other way because, the bending not
only the axial load, but bending also there.
31:27.649 --> 31:37.820
So, sometimes it happens that we can find
out there are few cases for example here
31:40.520 --> 31:47.980
this corner columns that ends here. That means,
here the moment would be about both sides,
31:47.989 --> 31:55.080
both axis this column. The moment will be
about 1 axis whether, this column moment about
31:55.080 --> 32:02.969
the axis whereas, this middle columns you
say almost axially loaded you may have due
32:02.969 --> 32:09.549
to say your changes span or loading may happen
that, there is a the moment.
32:09.549 --> 32:14.450
But in most of the cases, you can find out
that these columns are auxiliary loaded whereas,
32:14.450 --> 32:23.879
these columns are auxiliary loaded as well
as, bending about both axis. This bending
32:23.879 --> 32:32.859
about 1 axis whereas, these and these bending
about to say the axis so that means, we have
32:32.859 --> 32:44.399
2 more cases: 1 is called uniaxial bending
and the other is called biaxial bending. Now,
32:44.399 --> 32:51.789
if we consider here that uniaxial bending
and we have biaxial bending.
32:53.280 --> 33:04.080
So, how many cases we have that loading depending
upon the loading, number 1 what are the loading
33:05.500 --> 33:14.460
Pu and Mu say, Mu is the moment ultimate moment,
Pu is the design axial load. So, if we have
33:14.469 --> 33:24.679
to design this for this, we have to provide
the section. 1 case: we can have Mu equal
33:24.679 --> 33:41.059
to 0 that means, axially loaded. Number 2:
it can happen that Pu equal to 0 that means,
33:41.059 --> 33:49.169
it is nothing but purring bending that means
it is nothing, but purring bending that bending
33:49.169 --> 33:54.590
is your beam problem you can say that it is
nothing, but doubly reinforce section.
33:54.590 --> 34:14.610
The number 3 both Pu and Mu are present our
code says if it is axially loaded, the strain
34:14.610 --> 34:27.700
is 0.002. If it say bending in that case what
we shall say, that our code says that in the
34:27.710 --> 34:35.929
maximum stressed compression side. They are
the strain will be equal to epsilon c will
34:35.929 --> 34:47.720
be equal to point 0.0035 minus 0.75 times
epsilon dash c I mean to say, if this is the
34:47.720 --> 34:58.430
section then, we can find out this 0.0035
this is D somewhere here we have the pivot
34:58.430 --> 35:09.970
that is 0.002.
So, I can say the these this 0.0035 is a total
35:09.970 --> 35:18.809
0.0035 minus this is epsilon dash c minus
three-fourth of epsilon dash c will given
35:18.809 --> 35:27.890
me this 1. So, this value will be 0.0035 minus
three-fourth of epsilon dash c or nothing,
35:27.890 --> 35:34.829
but 0.75 dash c. This garbage required to
find out the strain as well as to find out
35:34.829 --> 35:42.169
the stress. So, we can find out the strain
and then, we can find out stress what we do
35:42.460 --> 35:48.220
here. Because, it is very we say for example,
Pu and Mu is given, now what should be percentage
35:48.230 --> 35:53.020
of steel that we do not know.
So, what value of Pu and what value of Mu;
35:53.020 --> 35:58.770
that means, if Pu moved that is not mean that
Pu and Mu there is no linear relationship.
35:58.770 --> 36:03.079
That means, you should certain kind of trial
and error and we have to find out at what
36:03.079 --> 36:10.569
percentage of steel both Pu and Mu satisfied.
For what percentage of steel and section say
36:10.569 --> 36:15.599
if I take, certain section 300 by 300.
So, for that section if we start then, we
36:15.599 --> 36:20.549
have to find out and then let us take say
1 percent percentage of steel. For that, we
36:20.549 --> 36:24.819
can find out we can use this formula we can
find out the strain and all those things.
36:24.819 --> 36:30.490
And then, we can get the Pu and Mu and then,
we can say it is perfectly alright that is
36:30.490 --> 36:33.079
the thing we generally do it.
36:33.079 --> 36:46.139
But here what we generally do, that is called
interaction curve P verses M interaction curve.
36:47.600 --> 37:06.260
So, that 1 you can get it I shall show you
anyway, just to tell you.
37:07.780 --> 37:20.599
Actually, I have already referred this 1 that
design aids for reinforced to concrete IS:456-1978
37:20.599 --> 37:24.069
of course, this is SP 16 the special publication
SP 16.
37:24.069 --> 37:30.049
Where you, find out lots of charts and tables
all those things you can find out even nowadays.
37:30.049 --> 37:36.250
one can write a simple small program and can
generate all those things. For each table
37:36.250 --> 37:41.349
all those things 1 can generate that 1 you
say. 1 small program it is not very big 1.
37:41.349 --> 37:53.650
So, what we shall do it here what I like to
show I shall come in detail.
37:53.650 --> 38:12.430
Let us, take I think we can show you these
1 anyone of them . So, we have like this chart
38:12.430 --> 38:18.940
what we can do let me, see whether available
here 1 second.
38:18.940 --> 38:44.890
I can do it I think it may be better right.
38:49.940 --> 38:57.660
So, the same thing it will be now, available
in our library also online.
38:58.450 --> 39:08.000
So, you can see the same thing now, available
here this is for this is called interaction
39:08.000 --> 39:10.900
curve.
39:12.060 --> 39:21.840
What we do, this is say there is 1 chart just
to tell you in the later class. So, here what
39:21.849 --> 39:31.880
happen here this particular 1 this is for
fy say 415 what will we do it, we make it
39:31.880 --> 39:33.300
here.
39:44.620 --> 40:04.160
Here, write down by pu by fck bD Pu is th
design load axial load fck bD and Mu by fck
40:04.420 --> 40:12.640
bD square this for a certain percentage of
the steel. I can get different curve for different
40:12.650 --> 40:20.710
percentage of steel. What about this point?
B ecause, there are few salient points this
40:20.710 --> 40:39.380
point axially loaded, that is certain point
here that is balanced, this point no axial
40:39.380 --> 40:54.369
load, this side is compression zone and this
portion is tension zone.
40:54.369 --> 41:00.990
So, depending on the neutral axis position
we can find out there is 1 case say you are
41:00.990 --> 41:06.880
axially loaded that means, it is uniformly
distributed that 0.002 the strain is 0.002.
41:06.880 --> 41:15.970
Then, we have 1 case where there is no axial
load that means, it is simple beam problem
41:15.970 --> 41:21.900
and neutral axis is the within the section
somewhere, we are having the balanced section
41:21.900 --> 41:29.150
then, we are having say tension zone.
Finally, the neutral axis is going out the
41:29.150 --> 41:37.359
section and where you are getting purely compression
zone. That means, that even the distribution
41:37.359 --> 41:44.690
of say that 1 here it may be something like
this, the strain this it can go like this
41:44.690 --> 41:49.730
something like this. So, what we can do we
are starting somewhere here like this 0.002
41:49.730 --> 41:58.750
is the strain. Then, somewhere you are getting
0.0035 minus three-fourth of this side so,
41:58.750 --> 42:03.839
like that it will go.
So, this is your interaction curve this SP
42:03.839 --> 42:12.730
16 provide you this so many curves. So, what
we have to do depending on these charts depending
42:12.730 --> 42:20.789
on the d dash by d means that clear that effective cover.
42:23.360 --> 42:51.800
So, what we do here so, it says what is does
d dash means equal to 40 millimetre plus pie
42:51.800 --> 43:01.560
by 2 diameter of the bar pie by 2. So, d dash
by d depending on different d dash by d, we
43:01.569 --> 43:11.359
can get different strain that why we are having
different d dash by d it starts with 0.05,
43:11.359 --> 43:17.240
0.1, 0.15, 0.2.
So, we can have different charts see chart
43:17.240 --> 43:31.319
31 for Fe 415 d dash by d like that. Then,
we can have the other 1 d dash by d 0.1 this
43:31.319 --> 43:50.619
is for 0.15 and this is for 0.2 what we can
do? Let me, show you what we shall do.
43:50.619 --> 43:59.559
So, let us take a particular chart it depends
on that is for mile steel this for fy means
43:59.559 --> 44:07.390
mile steel. Since, we are using say Fe 415
so, we have to use a Fe 415 what we shall
44:07.390 --> 44:35.039
do it here I think I can yes, what we do it
here, 1 sec this is the case I can show you
44:35.039 --> 44:46.519
1 example then, it will clear what we are
going to do, what we shall do.
44:46.519 --> 45:02.970
Let us, take Pu say 1500 Kilonewton and Mu
say 100 Kilonewton Meter we are getting this
45:02.970 --> 45:22.010
value. And let us, assume a section from a
section say 350 by 350 fck 20 newton square
45:22.010 --> 45:38.529
millimetre fy 415 newton per square millimetre
d dash let us, take d dash here 40 plus 20
45:38.529 --> 45:50.609
by 2 20 millimetre slab so 50 . So, we are
getting this 1 say 50 so, what about d dash
45:50.609 --> 46:14.789
by d? d dash by d will be 3, 50 by 350.
So, we can get 0.1428 d dash by d 0.1428 what
46:14.789 --> 46:20.400
we can do though it is not correct we can
interpolate of course, but we can take the
46:20.400 --> 46:29.039
table per chart this chart. Because, this
is it says d dash y d equal to 0.15 what we
46:29.039 --> 46:35.240
should do we can take d dash by d for 0.1
and also we can take d dash by d for 0.15.
46:35.240 --> 46:41.289
So, here we are getting 0.1428 which is coming
closer to 0.15. So, we can let us assume that
46:41.289 --> 46:48.000
we shall take this 1, but actually we should
interpolate this 2 charts, what shall we do
46:48.000 --> 46:56.589
this here. So, d dash by d 0.15 this is the
section it is unixial about these line these
46:56.589 --> 47:03.160
axis the moment applied since, the moment
is applied about this axis that so, that’s
47:03.160 --> 47:05.779
why I am providing the reinforcement about
this.
47:05.779 --> 47:16.010
The thing that here, this is your section
moment is applied about these axis. So, we
47:16.010 --> 47:25.329
have to provide the reinforcement along this
then only we will get the maximum benefits.
47:25.329 --> 47:30.579
Because, it should be as far as possible it
will be far away from the neutral axis. So,
47:30.579 --> 47:35.809
instead of providing here we provide like
this for uniaxial bending we are talking uniaxial
47:35.809 --> 47:45.140
bending. That means, the bars can be provided
along this 2 lines is it clear?
47:45.140 --> 47:52.440
Then, we are having say moment Pu and Mu when
you are having Pu and Mu, if we do not want
47:52.440 --> 47:58.160
to provide the reinforcement uniformly distributed
in all 4 sides, instead of that I would like
47:58.160 --> 48:04.950
to get the maximum benefit here what I am
interested here. We can provide the reinforcement
48:04.950 --> 48:12.779
along these 2 axis only 2 sides only. Because,
this 1 whatever moment will be produced due
48:12.779 --> 48:18.470
to these wherever, you provide this reinforcement
it does not matter if it is axially loaded,
48:18.470 --> 48:22.980
but when you are talking say moment.
So, these bars if you provide say far any
48:22.980 --> 48:27.599
from this 1 then, it will take the maximum
moment. And that is why we are providing that
48:27.599 --> 48:34.519
in 2 sides only, we are providing only reinforcement.
But it is biaxial bending then, there we have
48:34.519 --> 48:42.000
to provide in all 4 sides. So, what we can
do if you have d dash by d equal to 0.1428
48:42.000 --> 48:53.579
I can take this section and I can take the
I think what I can do just to going to computation.
48:53.579 --> 49:00.039
So, what I shall do here say this is if the
reinforcement think it is not over anyway,
49:00.039 --> 49:39.099
what I can show you if the this is your Mu
by fck bD square what I shall do, I shall
49:39.099 --> 49:47.710
take along this line this is 1 axis which
I shall get it Mu by fck by bD square I shall
49:47.710 --> 49:54.630
get along this. And Pu by fck by bD I shall
get along this.
49:54.630 --> 50:00.700
So, shall get a point so, I shall get 1 line
and from there what are these are for different
50:00.700 --> 50:06.910
value of p by fck this is for different value
of p by fck and then, we can find out the
50:06.910 --> 50:11.740
w can calculate the percent of steel. So,
I think I shall do it in the next class I
50:11.740 --> 50:17.599
shall do it that will example the specific
example, how we can do it that 1 and if it
50:17.599 --> 50:23.940
possible if the time permits.
Then, I shall show how to make in the computer
50:23.940 --> 50:31.220
because, that 1 it is very simple 1 it is
not a very difficult problem 1 can do it.
50:31.220 --> 50:36.279
Because, all the charts everything 1 can simply
make it on his own all those things you can
50:36.279 --> 50:40.069
make it let us, finish it today.
Thank you