1
00:00:59,320 > 00:01:05,320
So, last day we were talking about deformation
of the structure or more specifically it is
2
00:01:05,320 > 00:01:14,050
deflection of beam. The method we used it
is basically integration of differential
3
00:01:14,050 > 00:01:19,130
equation. We tried to handle some cantilever
type of problem and then switched over to
4
00:01:19,130 > 00:01:22,490
a
simply supported case with uniformly distributed
5
00:01:22,490 > 00:01:29,750
load. Point loaded case we have taken,
very standard type of caseload is acting at
6
00:01:29,750 > 00:01:34,530
the center of the beam up to certain extent
we
7
00:01:34,530 > 00:01:42,000
solve it. So, remaining part we can continue.
So, last class we got more or less something
8
00:01:42,000 > 00:01:45,140
like this.
9
00:01:45,140 > 00:01:57,640
So, we took a beam like this; this side some
support; here also some support and center
10
00:01:57,640 > 00:02:11,510
we put P. Reaction we got P by 2; reaction
P by 2; if we say this side is x, that side
11
00:02:11,510 > 00:02:18,950
this is
y. Now, the main difficulty, what we were
12
00:02:18,950 > 00:02:28,330
discussing in the last class, it has two regions,
if we say it is A, and if it is B, and if
13
00:02:28,330 > 00:02:32,989
it is C. So, A to B there is one segment or
a part
14
00:02:32,989 > 00:02:40,820
where your moment as some expression and the
second part B to C it has a different
15
00:02:40,820 > 00:02:47,840
expression for the bending moment expression.
So, these two parts more or less we have
16
00:02:47,840 > 00:02:57,230
tried to express in this manner, say, M that
we have written P by 2 into x for the left
17
00:02:57,230 > 00:03:05,549
part;
for the right part it was P by 2 into x minus
18
00:03:05,549 > 00:03:13,519
Pxl by 2. So, that we have written in the
last
19
00:03:13,519 > 00:03:21,070
class, because here only one load will generate
some moment. So, P by 2 into the
20
00:03:21,070 > 00:03:29,699
distance. So, if x is beyond that limit, here
the total length is l, and we have divided
21
00:03:29,699 > 00:03:30,940
in
two parts l by 2, l by 2. So, the distance
22
00:03:30,940 > 00:03:44,720
we have written earlier; it was say l by 2,
l by 2.
23
00:03:44,720 > 00:03:52,200
Now, once we cross l by 2, the centre load
P will contribute something in bending
24
00:03:52,200 > 00:04:00,620
moment. So, that was P into l minus x minus
l by 2.
25
00:04:00,620 > 00:04:09,130
Now, after integration, we got two sets of
constant here and here left side C 1 C 2;
26
00:04:09,130 > 00:04:14,600
right
side D 1 D 2. And we try to establish the
27
00:04:14,600 > 00:04:18,290
slope and deflection continuity. From there
we
28
00:04:18,290 > 00:04:26,420
found C 1 is nothing but D 1 and C 2 is nothing,
but D 2. So, from that information, we
29
00:04:26,420 > 00:04:35,949
have tried to combine both the expressions.
So, we can say, if we integrate that, we will
30
00:04:35,949 > 00:04:40,140
get two integration constant, here also two
integration constant, and both integration
31
00:04:40,140 > 00:04:50,040
concept are nothing but identical values,
because that common point, from both the side
32
00:04:50,040 > 00:04:59,260
if we compare, the deflection slope, we have
checked it; it will be equal. Now, from
33
00:04:59,260 > 00:05:07,150
there we have tried to write a common expression
for M equal to P by 2 into x minus Px
34
00:05:07,150 > 00:05:18,420
minus l by 2 multiplied by a function xl by
2.
35
00:05:18,420 > 00:05:26,761
Now, this function we have defined as unit
step function, and that step function, if
36
00:05:26,761 > 00:05:31,600
we
just draw below the beam, it will be something
37
00:05:31,600 > 00:05:43,730
like this. Say, it is the load point, this
side
38
00:05:43,730 > 00:05:52,900
we have mentioned it; it will be 1; that side
it will be equal to 0. So here, the x is starting
39
00:05:52,900 > 00:06:00,810
from here, and total length we are telling
it is l by 2 l by 2.
40
00:06:00,810 > 00:06:11,230
Now, the property of H is such, when within
the bracket, the quantity x minus l by 2,
41
00:06:11,230 > 00:06:16,480
if
that is a negative quantity and it will yield
42
00:06:16,480 > 00:06:20,510
a value 0. And if it is a positive quantity,
it
43
00:06:20,510 > 00:06:34,920
will yield a value 1. So, it is a function;
we have defined it is a unit step function.
44
00:06:34,920 > 00:06:45,931
Now,
idea here is, if you see the equations here,
45
00:06:45,931 > 00:06:49,990
here P by 2 into x, here also P by 2 into
x. So,
46
00:06:49,990 > 00:06:57,480
this part is common; only there will be additional
part on the right span. So, this part will
47
00:06:57,480 > 00:07:02,480
be the additional part. So, this is the additional
component. This part should not be here
48
00:07:02,480 > 00:07:08,450
when we will be treating the left part, but
that part will be there when we cross the
49
00:07:08,450 > 00:07:13,380
concentrate force at the center.
Now, we are talking all in language: it will
50
00:07:13,380 > 00:07:15,700
be there; it will not be there; where it will
be
51
00:07:15,700 > 00:07:20,750
there; where it will not be there. So, that
part if we try to put it in a mathematical
52
00:07:20,750 > 00:07:25,370
manner,
so in this function H what we are defining
53
00:07:25,370 > 00:07:32,461
as unit step function, so basically that
understanding in a physical sense, it is trying
54
00:07:32,461 > 00:07:43,190
to express in a mathematical format,
because if it is before allowing the mid span,
55
00:07:43,190 > 00:07:49,750
before reaching the appearance of the load,
your internal value within the bracket it
56
00:07:49,750 > 00:07:52,090
will be negative, and whole quantity will
be 0,
57
00:07:52,090 > 00:07:59,010
automatically this component will not be there.
When it will cross P, this component will
58
00:07:59,010 > 00:08:04,950
be greater than 0  a positive quantity automatically
this will appear into a equation. So, it is
59
00:08:04,950 > 00:08:11,669
a matter of automatic adjustment.
You are talking about Macaulay’s method.
60
00:08:11,669 > 00:08:15,880
Basically, it is same thing and there you
put
61
00:08:15,880 > 00:08:22,010
some form; it is basically in a same manner;
different places it is representing in a
62
00:08:22,010 > 00:08:27,700
different way. So, basic idea is different
equations are applicable at the different
63
00:08:27,700 > 00:08:32,829
zones,
but you will find some common part; once you
64
00:08:32,829 > 00:08:38,070
will cross a limit, you will find some
additional term. So, that additional term
65
00:08:38,070 > 00:08:41,000
will be there in some part; in some other
part it
66
00:08:41,000 > 00:08:45,230
will not be there. So, that will be controlled
by this function; this is one of the ways
67
00:08:45,230 > 00:08:46,230
of
representation.
68
00:08:46,230 > 00:08:55,110
Now, this is a simple case; it is P; there
might be another load here again. So, there
69
00:08:55,110 > 00:09:00,839
should be one segment, two segment, three
segments. So, there should be first is P by
70
00:09:00,839 > 00:09:04,199
2
into x; second part P by 2 into x minus P
71
00:09:04,199 > 00:09:11,030
x minus l by 2; and there should be a third
segment, so this plus something. So, it should
72
00:09:11,030 > 00:09:15,899
H something, H something, like this. So, a
number of unit step function may be there;
73
00:09:15,899 > 00:09:21,060
only x minus l by 2 might be x minus say 3
l
74
00:09:21,060 > 00:09:28,260
by 4 or something, where the load will be
appearing on the span of the beam.
75
00:09:28,260 > 00:09:38,839
Now, the whole problem we can treat as a beam,
where the moment expression is this
76
00:09:38,839 > 00:09:44,330
expression, valid for the entire span. If
that is the common expression valid for the
77
00:09:44,330 > 00:09:49,110
entire
span, we can write the momentcurvature relationship
78
00:09:49,110 > 00:09:50,520
in this manner.
79
00:09:50,520 > 00:10:05,080
Your EI d 2 y dx square that is equal to minus
M or just now we have written as minus P
80
00:10:05,080 > 00:10:21,870
by 2 x minus minus plus P x minus l by 2 and
Hl by 2. Now, if we integrate. So, in the
81
00:10:21,870 > 00:10:38,630
first step we will get say EI dy dx; that
will be minus P x square by 4 plus Pl by 2
82
00:10:38,630 > 00:10:46,730
square
divided by 2 H x divided by 2 plus C 1. So,
83
00:10:46,730 > 00:10:52,440
again if you take EI y Px cube divided by
12.
84
00:10:52,440 > 00:11:10,770
So, it will be 6; it will be C 1 x plus C
2.
85
00:11:10,770 > 00:11:21,120
So, it is nothing but writing two expressions
simultaneously, we are writing in a
86
00:11:21,120 > 00:11:28,920
combined fashion, because this constant C
1 and D 1 are identical, C 2 will be equal
87
00:11:28,920 > 00:11:33,630
to D
2. Only this unit step function will control
88
00:11:33,630 > 00:11:38,750
this part. We are interested automatically
this
89
00:11:38,750 > 00:11:44,490
part will be there or this part will be dropped
out. Now, that is a common expression.
90
00:11:44,490 > 00:11:49,360
Next step, normally, what we do we substitute
the boundary condition and try to find out
91
00:11:49,360 > 00:11:55,443
C 1 C 2.
So, let us try here and find out the value
92
00:11:55,443 > 00:12:02,399
of C 1 and C 2. So, we know at x equal to
0, y
93
00:12:02,399 > 00:12:11,540
equal to 0. So, if I put here, so EI into
y equal to 0 and this part depending on x.
94
00:12:11,540 > 00:12:19,330
So, this
part will be also 0; and this part, if we
95
00:12:19,330 > 00:12:23,880
are in the left side, that x equal to 0 means
the unit
96
00:12:23,880 > 00:12:29,960
step function within the bracket, the quantity
will be minus l by 2. So, automatically we
97
00:12:29,960 > 00:12:33,899
might be getting some value here, but that
should be multiplied with 0 means whole
98
00:12:33,899 > 00:12:42,490
quantity will be 0. So, that will be also
0 and here it will be C 1, 0, C 2. So, from
99
00:12:42,490 > 00:12:47,311
here we
will get C 2 equal to 0. If you take the write
100
00:12:47,311 > 00:12:57,730
end that x equal to l, your y is again 0.
Now, here it will be 0 equal to minus Pl cube
101
00:12:57,730 > 00:13:06,850
divided by 12 plus, in that situation this
term will be activated, because the unit step
102
00:13:06,850 > 00:13:15,580
function value will be 1. So, that is the
important point. You should remember that
103
00:13:15,580 > 00:13:18,070
this part will not be there in the earlier
case,
104
00:13:18,070 > 00:13:25,050
but when it is greater than l by 2 it will
be there. So, if I put x equal to l, it will
105
00:13:25,050 > 00:13:32,350
be l by 2
to the power 3. So, it will be Pl cube. So,
106
00:13:32,350 > 00:13:36,520
denominator it will come 8. So, 8 already
6 is
107
00:13:36,520 > 00:13:46,730
there plus your C 1 l, C 2 automatically we
obtain  it is 0. So, from here C 1 you can
108
00:13:46,730 > 00:13:58,839
find out. So, C 1 will be your Pl square by
12 minus Pl square, because this l part will
109
00:13:58,839 > 00:14:11,890
cancel this will be 48. So, what will be the
value? So, it will be 1 by 12 minus 1 by 48.
110
00:14:11,890 > 00:14:32,310
48 if you pick up, so it will be 4 minus 1.
So, 3 divide by 48. So, it will be 3 1 16.
111
00:14:32,310 > 00:14:41,190
It will
be Pl square divided by 16.
112
00:14:41,190 > 00:14:48,901
Now, once we have the value C 1 is equal to
Pl square by 16 and C 2 is equal to 0, we
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00:14:48,901 > 00:14:53,870
have the expression of y we have the expression
for dy by dx we can substitute and get
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00:14:53,870 > 00:14:59,800
the important values, say maximum deflection
at the center of the beam, maximum slope
115
00:14:59,800 > 00:15:06,089
at the supports. So, from here we can calculate
all those quantities.
116
00:15:06,089 > 00:15:23,670
So, we can write the slopes theta 1 by EI,
that should be minus Px square divided by
117
00:15:23,670 > 00:15:36,970
4
plus P square divided by 2 and C 1 will be
118
00:15:36,970 > 00:15:44,920
there; it will be Pl square divided by 16;
y will
119
00:15:44,920 > 00:16:07,300
be 1 by EI Px cube divided by 12 Px minus
l by 2 is 6. So, whatever expression we have
120
00:16:07,300 > 00:16:15,839
derived, just the value of C 1 we have substituted
1 by EI part I have brought it here. So,
121
00:16:15,839 > 00:16:29,149
it appeared as 1 by EI. So, we will get at
x equal to 0, theta will be your theta max.
122
00:16:29,149 > 00:16:35,339
And
at x equal to l by 2, your y will be y max.
123
00:16:35,339 > 00:16:45,399
So, theta max it will be just we have substitute
the value of x equal to 0. So, this part will
124
00:16:45,399 > 00:16:54,410
be 0, this part will not be 0, but the step
function will be 0. So, in a automatic process
125
00:16:54,410 > 00:16:59,360
there is a tendency we calculate something,
but there is no point of calculating because
126
00:16:59,360 > 00:17:03,790
the step function will be 0 here. So, both
the
127
00:17:03,790 > 00:17:08,959
quantity will be 0 and ultimately the Pl square
by 16 will be remaining that will be
128
00:17:08,959 > 00:17:19,559
divided by, so, that will be the value.
Similarly, the maximum deflection, it will
129
00:17:19,559 > 00:17:23,000
not be all 0; we will get some value. So,
y will
130
00:17:23,000 > 00:17:35,080
be 1 by EI Pl cube divided by 12. And this
term will be 0, because x will be l by 2 minus
131
00:17:35,080 > 00:17:40,450
l by 2. So, this term will be 0, and there
is a ambiguity here, whether unit step function
132
00:17:40,450 > 00:17:46,159
will be 0 or 1, because that is the transition.
So, if it is just in the right side, it is
133
00:17:46,159 > 00:17:50,750
1; if it is
just in the left, it is 0; and at that point
134
00:17:50,750 > 00:17:56,450
what will be the value? There might be a deviate
here again. So, that deviate can be solved
135
00:17:56,450 > 00:18:01,180
by this term, because this term will be
automatically 0, because l by 2 minus l by
136
00:18:01,180 > 00:18:06,100
2. So, whole quantity will be 0. So, you need
not think about that confusion what will be
137
00:18:06,100 > 00:18:08,700
the value of H here, I think.
138
00:18:08,700 > 00:18:16,789
So, automatically that part will be 0, and
the remaining part it will be Pl, if I put
139
00:18:16,789 > 00:18:20,610
and if I
make l cube. So, here it will be half, it
140
00:18:20,610 > 00:18:25,879
will be 32. So, a little simplification we
have to
141
00:18:25,879 > 00:19:00,370
meant. So, it is 1 by 32 and it is 1 by 12.
So, it will be how much? Pl cube by 62. Here
142
00:19:00,370 > 00:19:02,989
if
we substitute, it is not pl cube divided by
143
00:19:02,989 > 00:19:06,259
12, this half term will be there, it will
be 2 into
144
00:19:06,259 > 00:19:13,149
2 into 2, 8 terms will be there; otherwise,
this term will be larger and we are getting
145
00:19:13,149 > 00:19:15,549
a
negative value; it should not be. So, it should
146
00:19:15,549 > 00:19:24,529
be 1 by 12 into 8 plus 1 by 32. So, here I
can calculate a little bit; it will 1 by 32
147
00:19:24,529 > 00:19:41,690
minus 1 by 12 into 8.
96. How much? It is 2 by 96.
148
00:19:41,690 > 00:19:45,159
That is 1 by 40.
1 by 40 it is the correct answer.
149
00:19:45,159 > 00:19:52,909
No, you calculate and check it from there
again.
150
00:19:52,909 > 00:19:55,059
Say 96 is the common, into how many times
will 32?
151
00:19:55,059 > 00:19:59,940
That is it will be 3 times.
3 times 1 by 24.
152
00:19:59,940 > 00:20:15,019
It will be 1. So, automatically it will be
2. So, it will be automatically 48.
153
00:20:15,019 > 00:20:23,830
Now that 48 the y max value.
Now, this part is one of the common, very
154
00:20:23,830 > 00:20:28,899
standard form. If Pl cube 48 EI, theta max
is
155
00:20:28,899 > 00:20:38,440
equal to Pl cube by 16 EI. So, by that time
we have to tried to find out some standard
156
00:20:38,440 > 00:20:44,489
expressions, say, beam fully loaded, what
whatever maximum deflection we are getting,
157
00:20:44,489 > 00:20:50,019
whatever maximum slope we are getting, a beam
with a central load, this value you are
158
00:20:50,019 > 00:20:55,700
getting, cantilever some form we got.
Apart from that, there are some other values,
159
00:20:55,700 > 00:21:06,659
but in this context, this unit step function,
which is required for a new load appearing
160
00:21:06,659 > 00:21:11,450
on the span, we are going to use, so if I
take a
161
00:21:11,450 > 00:21:24,090
very general type of case I want to show how
we will apply this unit step function to
162
00:21:24,090 > 00:21:26,960
represent a number of load.
163
00:21:26,960 > 00:21:31,119
e
Say there might some support; there might
164
00:21:31,119 > 00:21:41,341
be some support here. Now, there might be
a
165
00:21:41,341 > 00:21:53,440
load here, say, P 1; there might be another
load P 2; there might be a distributed load
166
00:21:53,440 > 00:22:05,200
like
this, say it is W that intensity of the distributed
167
00:22:05,200 > 00:22:22,750
load per meter length or per unit length;
say, I am writing here it is l 1, and this
168
00:22:22,750 > 00:22:35,369
part is say l 2, here it is l 3; say it is
extended up
169
00:22:35,369 > 00:22:47,059
to this, it is l 4. You can add some other
load as a concentrated load or some form
170
00:22:47,059 > 00:22:54,460
distributed load. I am trying to show you
a very typical case. You can increase the
171
00:22:54,460 > 00:22:58,979
number.
Now what will happen? Here we will get some
172
00:22:58,979 > 00:23:10,609
reaction. So, some reactive force; here
also we will get some reactive force. Say
173
00:23:10,609 > 00:23:15,090
if we say this is support A, it is support
B, that
174
00:23:15,090 > 00:23:25,999
reactive force; I can use a different color,
I do not know how much it will be prominent
175
00:23:25,999 > 00:23:41,529
here. So, this is say Ra and this Rb. This
side is x; that side is y. Now our objective
176
00:23:41,529 > 00:23:46,629
is to
get a common expression of bending moment.
177
00:23:46,629 > 00:23:52,900
In that case, we will use the unit step
function to combine different segments.
178
00:23:52,900 > 00:24:04,460
Now, if it is Ra 0 to l 1, your bending moment
should be your Ra into x, or let me write
179
00:24:04,460 > 00:24:15,950
the expression of M. M will be Ra into x.
So, far we are within l 1; x is 0 to l 1;
180
00:24:15,950 > 00:24:21,200
this will
be valid, if you cross that, you will have
181
00:24:21,200 > 00:24:32,490
say P 1 x minus l 1, but that part will be
after l
182
00:24:32,490 > 00:24:43,619
1. So, here you have to put some step function.
So, this part will be 0, when x will be less
183
00:24:43,619 > 00:24:49,099
than l 1; when it will be more than that,
so that part will be 1 or you will get some
184
00:24:49,099 > 00:24:51,580
contribution from here.
185
00:24:51,580 > 00:25:00,499
Similarly, if I take the second load; it is
basically P 1 not P. So, it will be P 2 x
186
00:25:00,499 > 00:25:18,850
minus l 2
Hx minus l 2. So, P 2 x minus l 2. So, there
187
00:25:18,850 > 00:25:24,450
is another unit step function; that is x minus
l
188
00:25:24,450 > 00:25:31,929
2. So, if it is less than l 2, this part will
be 0, and this part will not be there, but
189
00:25:31,929 > 00:25:38,129
if you
cross that it will be there. Now, this part
190
00:25:38,129 > 00:25:44,090
is also coming, if you cross your l 3. So,
it will
191
00:25:44,090 > 00:26:05,999
be minus omega x minus l 3 square divided
by 2 x minus l 3. So, if I am here, we cross
192
00:26:05,999 > 00:26:13,950
this l 3. So, if we reach here, I think effect
of that load will come, and intensity of the
193
00:26:13,950 > 00:26:20,899
load is this one. So, if I am here, so here
to here it is x. So, x minus l 3 will be only
194
00:26:20,899 > 00:26:25,779
this
small part. So, omega into xl minus 3 will
195
00:26:25,779 > 00:26:30,070
be the load into half of the distance. So,
square
196
00:26:30,070 > 00:26:38,330
divided by 2, that will be the effect. And
that part is valid when you are here, I think,
197
00:26:38,330 > 00:26:41,330
not
before this. So, another step function it
198
00:26:41,330 > 00:26:53,320
will be Hx minus l 3.
Now, here there is a problem. R 1, P 1, P
199
00:26:53,320 > 00:26:59,010
2 at if you cross R 1 anywhere, it is R 1
into x;
200
00:26:59,010 > 00:27:09,080
if you cross P 1, it is P 1 into x minus l
1; or beyond this, it is P 2 x minus l 2.
201
00:27:09,080 > 00:27:12,590
But if you
cross here, up to this it is, but if you come
202
00:27:12,590 > 00:27:22,129
here this expression is not valid. If you
are
203
00:27:22,129 > 00:27:29,609
here, because it is x minus l minus 3. So,
if your x here, l minus 3 is this one. So,
204
00:27:29,609 > 00:27:34,390
your
loading is not there. The load is already
205
00:27:34,390 > 00:27:38,960
over here. So, this part will give a message
that it
206
00:27:38,960 > 00:27:45,940
is a extended up to the end, but it is not
applicable. In that case, the procedure is,
207
00:27:45,940 > 00:27:50,970
if there
is a distributor load starting from some part,
208
00:27:50,970 > 00:27:55,109
if you write that type of form, so we have
to
209
00:27:55,109 > 00:28:04,859
follow that equation and add this load, and
that load is not in the system . So, we have
210
00:28:04,859 > 00:28:08,470
to
make it minus here.
211
00:28:08,470 > 00:28:19,830
So, upper one it can be tackle by the equation
whatever we have written, and the lower
212
00:28:19,830 > 00:28:26,090
part is the correcting force, for that we
have to add another term, and that will be
213
00:28:26,090 > 00:28:28,710
in a
reverse manner, it will be minus into plus.
214
00:28:28,710 > 00:28:33,129
So, it will be omega and it will be x minus
l 4
215
00:28:33,129 > 00:28:43,419
divided by 2 Hx minus l 4.
So, this is one of the very tricky type of
216
00:28:43,419 > 00:28:49,059
idea. You should remember when there is
distributor load, starting from any point,
217
00:28:49,059 > 00:28:52,070
if it is valid up to the end of that member,
it is
218
00:28:52,070 > 00:28:59,490
fine, but if it is terminating at some point,
then there is a problem, because the way we
219
00:28:59,490 > 00:29:05,779
will take the equation, it will give the message
 load is extended up to the end of the
220
00:29:05,779 > 00:29:13,679
member. So, in that case, we have to really
extend the force and we have to put some
221
00:29:13,679 > 00:29:20,789
correcting force there. So, that correcting
term will come here and this term will take
222
00:29:20,789 > 00:29:25,210
care with this black and ring  both the load.
223
00:29:25,210 > 00:29:34,500
So, with that type of understanding, you can
handle a problem with any number of point
224
00:29:34,500 > 00:29:44,320
load and any form of uniformly distributed
load; not only uniformly distributed load,
225
00:29:44,320 > 00:29:47,889
if
the variation is little different only that
226
00:29:47,889 > 00:29:53,570
equation we have to take and you cannot
terminate. So, you have to extend some of
227
00:29:53,570 > 00:29:56,539
the portion and you have to put some
correcting force.
228
00:29:56,539 > 00:30:00,979
So, finally, you will get a common expression
for bending moment, which will be valid
229
00:30:00,979 > 00:30:07,749
for the entire span with a number of unit
step functions, and they will start from here,
230
00:30:07,749 > 00:30:14,450
here, here, here, different levels; before
that it will be 0, after that it will be invoked,
231
00:30:14,450 > 00:30:21,489
means it will give a value of 1, means it
is now 1; it is just behave as a switching
232
00:30:21,489 > 00:30:26,630
function. So, it is now on; it is now off;
something like this.
233
00:30:26,630 > 00:30:33,919
Now, once you get M, you can write M is equal
to minus 82 by dx square; integrate
234
00:30:33,919 > 00:30:40,549
twice, you will get C 1 x C 2, and at the
two end you can substitute the boundary
235
00:30:40,549 > 00:30:46,269
condition; you get C 1 and C 2. Finally, you
will get the expression of y or you can say
236
00:30:46,269 > 00:30:52,570
dy dx. So, any point if you substitute x equal
to 0 or l l by 2 l 1, l 2, l 3 you will get
237
00:30:52,570 > 00:30:57,500
the
values of y, that is deflection D 2 by dx
238
00:30:57,500 > 00:31:04,369
square slope. So, you can find out all those
quantities for very unknown type of problem.
239
00:31:04,369 > 00:31:19,659
Sir, there we reached to a conclusion. We
knew that at this point slope is 0. We have
240
00:31:19,659 > 00:31:28,169
taken the deflection in that y equal to 0;
in that at the end we have taking.
241
00:31:28,169 > 00:31:34,889
Because we starting with the seconddegree
equation. So, moment minus EI d 2 dx
242
00:31:34,889 > 00:31:39,730
square; we will get two constants. So, two
boundary conditions are unnecessary. And the
243
00:31:39,730 > 00:31:47,330
boundary conditions will be in the form of
deflection and slope. So, if it a cantilever
244
00:31:47,330 > 00:31:52,129
problem, in one point we are getting the condition
deflection as well as slope are 0 at the
245
00:31:52,129 > 00:31:58,429
fixed end; but simply supported case, it is
slope is a unknown quantity here; slope is
246
00:31:58,429 > 00:32:01,179
also
unknown quantity here. So, slope we cannot
247
00:32:01,179 > 00:32:06,349
utilize. We can only take the help of
deflection of A and B, means at the two supported
248
00:32:06,349 > 00:32:13,690
points.
Now, this is more or less dealing with deflection
249
00:32:13,690 > 00:32:22,289
of beam with the help of differential
equation technique. We can take some numerical
250
00:32:22,289 > 00:32:31,830
example in our tutorial classes; there
we can check different situations. So, my
251
00:32:31,830 > 00:32:40,659
next target here is move on to a little different
method, which is called moment area method.
252
00:32:40,659 > 00:32:57,940
So, this method is called moment area method.
So, it is another method like your
253
00:32:57,940 > 00:33:01,799
differential equation technique; very simple
and elegant method, typically applicable for
254
00:33:01,799 > 00:33:16,269
a small size problem. Now, the idea is coming
again from your momentcurvature
255
00:33:16,269 > 00:33:31,149
relationship. We have the understanding that
EI d 2 by dx square equal to your minus M,
256
00:33:31,149 > 00:33:41,149
minus or plus whatever M is there. So, this
we can write as d 2 by dx square equal to
257
00:33:41,149 > 00:33:46,719
M
by EI or that equation we can write in this
258
00:33:46,719 > 00:34:01,379
form d theta by dx minus M by EI, because
theta is basically dy by dx. So, if we put
259
00:34:01,379 > 00:34:04,389
theta equal to dy by dx we are supposed to
get d
260
00:34:04,389 > 00:34:16,850
2 y by dx square. Now this is one equation.
So, d theta will be your M by EI dx.
261
00:34:16,850 > 00:34:34,070
Now, let us draw a beam with some deformation.
It may have some load, some support.
262
00:34:34,070 > 00:34:57,200
So, it will undergo some deformation here.
So, from our momentcurvature relationship,
263
00:34:57,200 > 00:35:06,070
we have written d theta equal to minus M by
EI dx. Now if I take two adjacent points;
264
00:35:06,070 > 00:35:15,150
so,
there is a point here; if we draw the tangent
265
00:35:15,150 > 00:35:18,130
on the elastic curve at that point, and if
I take
266
00:35:18,130 > 00:35:31,350
from some adjacent point here, if we draw
tangent here, the angle between the two lines
267
00:35:31,350 > 00:35:40,710
is the d theta, because this line with x,
it is basically theta, and here this line
268
00:35:40,710 > 00:35:45,910
with x, it is d
theta plus some increment of theta. So this
269
00:35:45,910 > 00:35:51,480
value is nothing but d theta. And what is
the d
270
00:35:51,480 > 00:36:01,740
theta? d theta is M by EI into dx. So, what
is dx? dx is… basically this is dx.
271
00:36:01,740 > 00:36:13,310
So, this is a point say at x; this is point,
it is x plus dx; and here at x it is theta;
272
00:36:13,310 > 00:36:18,420
here x plus
dx it is theta plus d theta. So, difference
273
00:36:18,420 > 00:36:23,380
in their slope is d theta, and that is equal
dx M
274
00:36:23,380 > 00:36:31,640
by EI. Now what that means is some loading
are there you find out the reactions; from
275
00:36:31,640 > 00:36:35,560
the reactions, you can draw the bending moment
diagram of the beam. If you can draw
276
00:36:35,560 > 00:36:39,370
the bending of moment diagram of the beam,
that you can divided by EI of the beam;
277
00:36:39,370 > 00:36:47,130
you will get some diagram; it is the M by
EI diagram. Now, here say a bending moment
278
00:36:47,130 > 00:37:00,570
is something, and that we have divided by
EI. So, that if we say M by EI diagram. So,
279
00:37:00,570 > 00:37:04,700
M
by EI into dx is basically this part.
280
00:37:04,700 > 00:37:14,390
Say some bending moment diagram divided by
EI means we are basically representing
281
00:37:14,390 > 00:37:21,600
the same diagram with some scaling factor.
This scaling is 1 by EI. So, that diagram
282
00:37:21,600 > 00:37:25,880
is M
by EI diagram, say, M by… it is not M by
283
00:37:25,880 > 00:37:30,410
EI, say it is say y  function of x. So, M
by EI
284
00:37:30,410 > 00:37:37,970
is a variable quantity; it is a function x
into dx. So, function of x dx; this small
285
00:37:37,970 > 00:37:43,670
incremental; this is function of x into dx;
and dx is very, very small; means within that
286
00:37:43,670 > 00:37:51,160
range we can assume this is more or less uniform
value. Now, if I take some other point,
287
00:37:51,160 > 00:37:59,300
little bit away, it is not adjacent point,
it is little bit away. So, this tangent will
288
00:37:59,300 > 00:38:01,454
move like
this; there e is a point here. So, say from
289
00:38:01,454 > 00:38:09,630
here if I draw a tangent. So, that was the
slope;
290
00:38:09,630 > 00:38:16,840
that is the lope; the difference between the
slope is basically your theta. This was d
291
00:38:16,840 > 00:38:22,760
theta.
Now, if you can go on taking many points.
292
00:38:22,760 > 00:38:28,150
So, d theta, d theta, d theta, if you go on
accumulate, you will get difference between
293
00:38:28,150 > 00:38:31,700
the two tangent line theta. So, theta will
be,
294
00:38:31,700 > 00:38:40,340
basically, our integration of d theta, and
integration will be found here to here. So,
295
00:38:40,340 > 00:38:58,390
if it is
point C, and if it is point B. So it will
296
00:38:58,390 > 00:39:02,550
be point C to point B and this is nothing
but your
297
00:39:02,550 > 00:39:08,580
integral M by EI dx. So, we obtain integral
M by EI dx, means from C to D, if we
298
00:39:08,580 > 00:39:19,990
calculate the area of the curve, M by EI,
ranging from here to here. So, the total area
299
00:39:19,990 > 00:39:28,480
from C to D if we integrate, that is shown
here, because this is M by EI and dx, and
300
00:39:28,480 > 00:39:33,420
we
are integrating from C to dx. And that integration
301
00:39:33,420 > 00:39:40,090
will give, if we draw slope here, and
slope here, the difference between their angle
302
00:39:40,090 > 00:39:46,660
is theta.
So, if you have a deflected curve, if you
303
00:39:46,660 > 00:39:49,420
take two points, if you draw a slope here,
if you
304
00:39:49,420 > 00:39:55,460
draw a slope, both the slope will meet, and
this slope is change. So, the direction is
305
00:39:55,460 > 00:40:02,980
changing. So, this amount of change is basically
here to there; whatever area under M by
306
00:40:02,980 > 00:40:10,400
EI diagram. So, this is one of the component
of your moment area theorem, in terms of
307
00:40:10,400 > 00:40:11,791
calculation of slope.
308
00:40:11,791 > 00:40:25,740
So, next step is the deflection part. So,
if we draw the beam again, and try to investigate
309
00:40:25,740 > 00:40:43,650
what is happening with the deflection, it
will be as I told, assume any loading. Now,
310
00:40:43,650 > 00:40:52,200
this
point we have drawn some tangent. So, that
311
00:40:52,200 > 00:41:01,110
we have defined as C. Now, we took some
adjacent point; there we have drawn another
312
00:41:01,110 > 00:41:13,380
slope. Now, there is point D. So, C to D we
want to see what is happening here.
313
00:41:13,380 > 00:41:24,210
Now this angle is D theta. Now if I draw a
line perpendicular from D, perpendicular to
314
00:41:24,210 > 00:41:31,950
the initial line if you drop the line so this
value will be how much? This value will be
315
00:41:31,950 > 00:41:40,110
D
theta into the distance; say this angle is
316
00:41:40,110 > 00:41:46,070
your D theta. And d theta into here to here
this
317
00:41:46,070 > 00:41:55,600
distance if you multiply, so you will get
this the small distance. Now if we take another
318
00:41:55,600 > 00:42:01,860
point, draw it, there will be another D theta.
So, here to here you have to multiply. If
319
00:42:01,860 > 00:42:06,280
you
take another tangent, so it is D theta multiplied
320
00:42:06,280 > 00:42:12,570
by this one. So, like that if you go on
taking small, small value, so here to here
321
00:42:12,570 > 00:42:26,480

this part  it will be how much? It will be
322
00:42:26,480 > 00:42:30,540
your
D theta, already we have written has D theta
323
00:42:30,540 > 00:42:44,030
you have written minus M by EI dx.
Now, I am trying to tell from here, D theta,
324
00:42:44,030 > 00:42:52,270
D theta into, say, this is some distance,
say, z
325
00:42:52,270 > 00:43:02,721
or something, and that part if you go on accumulating,
integrate, so if that is delta, you
326
00:43:02,721 > 00:43:19,320
should get this delta. And this is nothing
but your minus M by EI into your z into your
327
00:43:19,320 > 00:43:32,420
dx. Now, here this z and x, we will take in
a similar fashion, because you have to
328
00:43:32,420 > 00:43:47,020
integrate dx, there is no limit actually.
So, this part will be your M x M by EI; you
329
00:43:47,020 > 00:44:01,640
can
say it is xdx, say here, we are trying to
330
00:44:01,640 > 00:44:06,300
tell it is z. So, D theta into z.
331
00:44:06,300 > 00:44:17,500
Now if you take the next one it will be different
z. So, you have to go on accumulating;
332
00:44:17,500 > 00:44:19,790
definitely, you will get this value, this
value, this value, this value; finally, we
333
00:44:19,790 > 00:44:24,740
will get
the total value, so it is… and D theta from
334
00:44:24,740 > 00:44:34,040
earlier expression we have M by EI into dx.
Now, what is D theta? D theta M by EI into
335
00:44:34,040 > 00:44:45,430
dx. So, this dx part is this one. Now dx we
are we are measuring x from here. And dx is
336
00:44:45,430 > 00:44:50,720
this one. So, if we shift our origin, say,
dx
337
00:44:50,720 > 00:44:58,700
we can take as dz, no problem, because from
here it is z, and this part is dz, so dz
338
00:44:58,700 > 00:45:05,810
into…So, entire quantity we can write dz
dz or xdx in any form we can write . So, it
339
00:45:05,810 > 00:45:10,360
will
more or less same thing. So, here what are
340
00:45:10,360 > 00:45:15,940
we getting? Delta equal to M by EI is the
say
341
00:45:15,940 > 00:45:28,080
any diagram here; this is M by EI diagram;
that is x into dx. So, we say this is the
342
00:45:28,080 > 00:45:34,520
first
moment of area of this diagram, between your
343
00:45:34,520 > 00:45:43,820
C to D, but this x or z whatever we are
talking about, you have to take from here,
344
00:45:43,820 > 00:45:55,940
not from here.
So, ultimately, this area you have to consider,
345
00:45:55,940 > 00:46:05,780
and you have to take the first moment of
this area about this line, because z we have
346
00:46:05,780 > 00:46:15,040
taken this way, x we have took in that
manner; in order to unify x and z, we have
347
00:46:15,040 > 00:46:19,970
taken the expression as x, we could write
it is
348
00:46:19,970 > 00:46:26,952
z also; rather it is better to write z, and
z we are taking from here. So, it is… this
349
00:46:26,952 > 00:46:32,640
is the
ordinate; it is xdx; this is the typical expression
350
00:46:32,640 > 00:46:39,580
of taking this area  moment of that area
about this line.
351
00:46:39,580 > 00:46:48,990
So, if you take any two points, and if we
calculate the first moment of area under that
352
00:46:48,990 > 00:46:58,330
curve from C to D, about d, so we will get
this value. This value is what? From the
353
00:46:58,330 > 00:47:05,500
deflected curve D, if we come down perpendicular
to the original line, it will meet the
354
00:47:05,500 > 00:47:12,680
tangent drawn at C. So, whatever distance
you will get, that can be calculated from
355
00:47:12,680 > 00:47:16,860
here.
So, this the second step of our moment area
356
00:47:16,860 > 00:47:20,580
theorem.
So, it has two steps: first was related to
357
00:47:20,580 > 00:47:24,060
slope calculation; the beam will be given;
loading
358
00:47:24,060 > 00:47:28,390
will be given; reaction we can calculate;
we can draw the bending moment diagram; that
359
00:47:28,390 > 00:47:34,180
bending moment diagram can be divided by EI.
If EI is constant throughout, it will be
360
00:47:34,180 > 00:47:39,390
uniform scaling; it may be a beam having a
different type of EI, so it will be just divided
361
00:47:39,390 > 00:47:44,650
by EI at the corresponding range. So, you
will get M by EI diagram.
362
00:47:44,650 > 00:47:50,890
Now, any two points if you consider and draw
two slopes, that difference between their
363
00:47:50,890 > 00:47:56,580
slope, one will have theta 1 and another will
have theta 2. So, theta 1 minus theta 2 will
364
00:47:56,580 > 00:48:04,810
be the actual difference; that will be area
of M by EI diagram, from your C to D, or point
365
00:48:04,810 > 00:48:12,940
1 to point 2. And if you take first moment
of area about D or point 2 whatever you can
366
00:48:12,940 > 00:48:23,870
say, it will give the deflection measure from
D towards the tangent point drawn at A, but
367
00:48:23,870 > 00:48:28,820
your measurement should be perpendicular to
the initial line.
368
00:48:28,820 > 00:48:37,230
Now, if I take some example, cantilever is
a very good example in that case, because,
369
00:48:37,230 > 00:48:47,290
say, this is
a cantilever problem; and this problem you
370
00:48:47,290 > 00:48:51,080
have solved by you differential
equation technique; some of the information
371
00:48:51,080 > 00:48:56,870
it is already there. So, we can compare our
result at this level using moment area theorem.
372
00:48:56,870 > 00:49:05,820
Say, this is the beam, and there is a load
P. This case you have handle at the beginning.
373
00:49:05,820 > 00:49:19,980
Say total length is L, and this is EI, and
length is L.
374
00:49:19,980 > 00:49:33,890
So, our first job is we have to draw the bending
moment diagram. It is a cantilever
375
00:49:33,890 > 00:49:39,270
problem. We need not calculate the reactions,
because we have one free end; from that
376
00:49:39,270 > 00:49:44,490
end we can start. So, P into x that will be
the expression of the bending moment diagram
377
00:49:44,490 > 00:49:49,480
or if we draw the bending moment diagram it
will be something like this.
378
00:49:49,480 > 00:50:02,300
So, this value will be B into L. This is what
your bending moment diagram. And if you
379
00:50:02,300 > 00:50:23,680
divide by EI, it will be M by EI diagram.
Now, here I will draw the deflated shape of
380
00:50:23,680 > 00:50:31,480
the
beam. So, this is a initial line; after deformation
381
00:50:31,480 > 00:50:37,300
it will be like this. Now at this point, if
I
382
00:50:37,300 > 00:50:54,430
draw one tangent. So, this is one of the point,
say, starting point. This is the initial line
383
00:50:54,430 > 00:50:58,770
plus if you draw the tangent also, it will
be the same line, because it is a fixed end;
384
00:50:58,770 > 00:51:04,210
it will
not undergo any slope; it will be 0 slope
385
00:51:04,210 > 00:51:08,110
and 0 deflection. So, this is the initial
line is the
386
00:51:08,110 > 00:51:13,890
tangent to that point, and here tangent is
this one.
387
00:51:13,890 > 00:51:26,290
So, difference between their slope of the
two tangent is theta. And according to your
388
00:51:26,290 > 00:51:29,680
first
part of the momentarea theorem, difference
389
00:51:29,680 > 00:51:37,200
between the slope is area under the curve.
So, theta will be area of this curve. So,
390
00:51:37,200 > 00:51:41,780
what is the area? This is L; this is Pl by
2; and it
391
00:51:41,780 > 00:51:53,930
will be half of that. So, it will be Pl by
EI half of that into L. Now we can say Pl
392
00:51:53,930 > 00:51:58,980
square
divided by 2 EI. And that value we obtain
393
00:51:58,980 > 00:52:09,910
earlier also. And here one of the advantage
one of the support is fixed. So, as it is
394
00:52:09,910 > 00:52:15,930
fixed, one of the tangent line is the initial
line; that
395
00:52:15,930 > 00:52:22,170
is a great benefit here, I think. Earlier
we have taken a simply supported type of the
396
00:52:22,170 > 00:52:27,450
beam. The first slope, last slope, it is an
not the original line. Here the advantage
397
00:52:27,450 > 00:52:31,790
is left
end, the tangent is the initial line. So,
398
00:52:31,790 > 00:52:36,180
any point if you draw the tangent here, that
will be
399
00:52:36,180 > 00:52:42,100
the effective slope. So, straight away we
can calculate.
400
00:52:42,100 > 00:52:51,230
Similarly, if you want to find the deflection
delta. So, try to remember the second part.
401
00:52:51,230 > 00:52:58,810
Second part means, so from here  this is
the deflected line  from here if we move
402
00:52:58,810 > 00:53:06,810
perpendicular to the initial line, so we will
meet a tangent drawn at other point. So, that
403
00:53:06,810 > 00:53:14,360
is the deflection, means the area of curve,
we have to take first moment of area about
404
00:53:14,360 > 00:53:21,480
this, because we are moving about this point.
So, delta will be the area under the curve
405
00:53:21,480 > 00:53:27,930
we have already written  it is Pl square
by 2 EI  that is the area, and centroid will
406
00:53:27,930 > 00:53:33,510
be 2
by 3 l. So2 by 3 l. So, 2 by 3 l. So, it will
407
00:53:33,510 > 00:53:43,180
be Pl cube divided by 6 EI. That is also we
have derived. Earlier we have written some
408
00:53:43,180 > 00:53:48,410
equation and obtained C 1, C 2, calculated.
So, it is much more easier at least that type
409
00:53:48,410 > 00:53:59,930
of problem. Now, with that this class I am
trying to finish. Next class we will continue
410
00:53:59,930 > 00:54:16,880
with some other type of problems.
Preview of Next Lecture
411
00:54:16,880 > 00:54:27,010
So, we are talking about momentarea theorem,
and two cases  simple beam problem,
412
00:54:27,010 > 00:54:36,161
cantilever beam  with some deep load and
maximum deflection slope, we have
413
00:54:36,161 > 00:54:42,970
calculated and tried to compare our values
with those obtained by your direct integration
414
00:54:42,970 > 00:54:48,560
of the differential equation. And you must
have that feeling, that it is much more easier
415
00:54:48,560 > 00:54:54,710
compared to the earlier approach. So, it is
not a very general statement, for that problem,
416
00:54:54,710 > 00:55:01,560
it might be easier, I think. So, it depends
problem to problem, you have to choose the
417
00:55:01,560 > 00:55:06,670
method in a appropriate manner. Sometimes
the direct method may be much more easier
418
00:55:06,670 > 00:55:14,130
or convenient to get the solution. Now, I
will take another problem of a cantilever
419
00:55:14,130 > 00:55:17,330
type of
beam.
420
00:55:17,330 > 00:55:41,650
Because the values are quite important here.
So, length is l, and again it is EI here.
421
00:55:41,650 > 00:55:48,040
There
is a moment acting at the end, say, this moment
422
00:55:48,040 > 00:56:17,010
is a M. Now for that beam, you can draw
the bending moment diagram. So, it is M. Throughout
423
00:56:17,010 > 00:56:24,110
the beam, bending moment will be
M or if we divide by EI, we will get the M
424
00:56:24,110 > 00:56:48,300
by EI diagram on the beam. Now, if I draw
the deflected shape of the cantilever. So,
425
00:56:48,300 > 00:56:59,020
here, there is one advantage  this point
is fixed.
426
00:56:59,020 > 00:57:08,100
So, tangent at that point is the initial line.
So, that we have already mentioned. So, at
427
00:57:08,100 > 00:57:12,690
the
free end, if we draw a tangent, so both that
428
00:57:12,690 > 00:57:21,440
tangent will have a difference in slope, that
will be theta. And incidentally, that value
429
00:57:21,440 > 00:57:24,010
will be the actual slope of the point at the
free
430
00:57:24,010 > 00:57:32,720
end.
So, it is what? It is basically area of that
431
00:57:32,720 > 00:57:40,710
curve. So, it is l, it is M by EI. So, it
will be
432
00:57:40,710 > 00:57:54,850
simply Ml by EI. And delta is, from this point,
if we move in a direction perpendicular to
433
00:57:54,850 > 00:58:03,070
the original line, it is meeting that tangent
drawn at that point here. So, that is delta.
434
00:58:03,070 > 00:58:06,760
So,
we are moving along this line. So, about this
435
00:58:06,760 > 00:58:12,070
point we have to take the first moment of
area of this M by EI diagram. So, it will
436
00:58:12,070 > 00:58:17,220
be ranging from here to here, moment about
that.
437
00:58:17,220 > 00:58:26,210
So, area already we have written Ml by EI.
And its centroid will be at the midpoint.
438
00:58:26,210 > 00:58:34,830
So,
that will be l by 2. So, it will be Ml square
439
00:58:34,830 > 00:58:48,400
by 2 EI. Now these values are quite
significant values. Like the case of a cantilever
440
00:58:48,400 > 00:58:54,000
subjected to a load and the free end, we
441
00:58:54,000 > 00:58:58,010
have written Pl square divided by 2 l or Plq
divided by 3 EI. Similarly, if there is a
442
00:58:58,010 > 00:59:11,090
moment, it will be Ml by EI or Ml square by
2 EI for deflection and slope.