1
00:00:57,190 --> 00:01:04,559
So, we are talking about moment-area theorem,
and two cases - simple beam problem,
2
00:01:04,559 --> 00:01:12,150
cantilever beam - with some deep load and
maximum deflection slope, we have
3
00:01:12,150 --> 00:01:20,400
calculated and tried to compare our values
with those obtained by your direct integration
4
00:01:20,400 --> 00:01:23,680
of the differential equation.
5
00:01:23,680 --> 00:01:26,039
And you must have that feeling, that it is
much more easier
6
00:01:26,039 --> 00:01:28,570
compared to the earlier approach.
7
00:01:28,570 --> 00:01:32,470
So, it is not a very general statement, for
that problem,
8
00:01:32,470 --> 00:01:33,470
it might be easier, I think.
9
00:01:33,470 --> 00:01:39,020
So, it depends; problem to problem, you have
to choose the
10
00:01:39,020 --> 00:01:41,310
method in a appropriate manner.
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00:01:41,310 --> 00:01:44,149
Sometimes the direct method may be much more
easier
12
00:01:44,149 --> 00:01:46,750
or convenient to get the solution.
13
00:01:46,750 --> 00:02:01,300
Now, I will take another problem of a cantilever
type of beam, because the values are
14
00:02:01,300 --> 00:02:07,530
quite important here.
15
00:02:07,530 --> 00:02:18,330
So, length is l, and again it is EI.
16
00:02:18,330 --> 00:02:33,250
Here, there is a moment acting at
the end, say, this moment is a M. Now, for
17
00:02:33,250 --> 00:02:46,950
that beam, you can draw the bending moment
diagram.
18
00:02:46,950 --> 00:02:59,250
So, it is M. Throughout the beam, bending
moment will be M or if we divide by
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00:02:59,250 --> 00:03:04,800
EI, we will get the M by EI diagram on the
beam.
20
00:03:04,800 --> 00:03:17,350
Now, if I draw the deflected shape of
the cantilever.
21
00:03:17,350 --> 00:03:36,390
So, here, there is one advantage - this point
is fixed.
22
00:03:36,390 --> 00:03:40,030
So, tangent at that
point is the initial line.
23
00:03:40,030 --> 00:03:45,140
So, that we have already mentioned.
24
00:03:45,140 --> 00:03:53,150
So, at the free end, if we
draw a tangent, so both that tangent will
25
00:03:53,150 --> 00:03:56,650
have a difference in slope, that will be theta.
26
00:03:56,650 --> 00:04:02,410
And incidentally, that value will be the actual
slope of the point at the free end.
27
00:04:02,410 --> 00:04:04,990
So, it is what?
28
00:04:04,990 --> 00:04:11,040
It is basically area of that curve.
29
00:04:11,040 --> 00:04:14,450
So, it is l, it is M by EI.
30
00:04:14,450 --> 00:04:20,600
So, it will be
simply Ml by EI.
31
00:04:20,600 --> 00:04:31,340
And delta is, from this point, if we move
in a direction perpendicular to
32
00:04:31,340 --> 00:04:37,180
the original line, it is meeting that tangent
drawn at that point here.
33
00:04:37,180 --> 00:04:40,500
So, that is delta.
34
00:04:40,500 --> 00:04:42,590
So,
we are moving along this line.
35
00:04:42,590 --> 00:04:45,560
So, about this point we have to take the first
moment of
36
00:04:45,560 --> 00:04:48,040
area of this M by EI diagram.
37
00:04:48,040 --> 00:04:52,669
So, it will be ranging from here to here,
moment about
38
00:04:52,669 --> 00:04:53,879
that.
39
00:04:53,879 --> 00:05:01,460
So, area already we have written Ml by EI.
40
00:05:01,460 --> 00:05:02,661
And its centroid will be at the midpoint.
41
00:05:02,661 --> 00:05:06,030
So,
that will be l by 2.
42
00:05:06,030 --> 00:05:15,230
So, it will be Ml square by 2 EI.
43
00:05:15,230 --> 00:05:24,440
Now these values are quite
significant values.
44
00:05:24,440 --> 00:05:33,030
Like the case of a cantilever subjected to
a load and the free end, we
45
00:05:33,030 --> 00:05:38,130
have written Pl square divided by 2 l or Plq
divided by 3 EI.
46
00:05:38,130 --> 00:05:44,210
Similarly, if there is a
moment, it will be Ml by EI or Ml square by
47
00:05:44,210 --> 00:05:50,389
2 EI for deflection and slope.
48
00:05:50,389 --> 00:05:55,520
Now, cantilever type of problem it is very
easier to handle, because one of the end is
49
00:05:55,520 --> 00:06:00,680
fixed, and there we can draw the tangent,
and that will be identical to the initial
50
00:06:00,680 --> 00:06:01,680
line.
51
00:06:01,680 --> 00:06:06,460
So,
at any point, if we draw a tangent, difference
52
00:06:06,460 --> 00:06:12,620
between the slope of this tangent and this
second tangent is basically the actual slope
53
00:06:12,620 --> 00:06:16,660
of that point; deflection also directly you
are
54
00:06:16,660 --> 00:06:18,160
getting in a similar manner.
55
00:06:18,160 --> 00:06:24,770
But if you take a simply supported case, problem
is little
56
00:06:24,770 --> 00:06:26,050
different.
57
00:06:26,050 --> 00:06:34,340
Now, how will we handle a case of a simply
supported type of problem?
58
00:06:34,340 --> 00:06:41,350
Now, it will
undergo some deformation like this.
59
00:06:41,350 --> 00:06:46,660
It may be some loading.
60
00:06:46,660 --> 00:06:56,470
Now, if you are
interested
61
00:06:56,470 --> 00:07:05,130
to find the slope at the supported end or
deflection at the middle of the span or some
62
00:07:05,130 --> 00:07:13,820
intermediate point, just like cantilever,
we cannot get it; but here, more or less we
63
00:07:13,820 --> 00:07:15,320
will
follow that type of procedure.
64
00:07:15,320 --> 00:07:23,230
About this point, suppose if you draw a tangent,
it will go like this.
65
00:07:23,230 --> 00:07:26,050
Now, here to here,
we can calculate.
66
00:07:26,050 --> 00:07:28,620
What we are doing?
67
00:07:28,620 --> 00:07:39,360
This is the deflected line and from this deflected
line this is a typical point on the elastic
68
00:07:39,360 --> 00:07:40,360
curve.
69
00:07:40,360 --> 00:07:48,570
From this point, we are moving along a
path normal to the initial line and it is
70
00:07:48,570 --> 00:07:50,919
meeting the tangent drawn at that point.
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00:07:50,919 --> 00:07:56,750
So, here to
here, it is basically, it may have some M
72
00:07:56,750 --> 00:07:58,960
by EI diagram over it.
73
00:07:58,960 --> 00:08:04,540
So, we have to take the M by EI diagram from
here to here, about this point we have to
74
00:08:04,540 --> 00:08:09,920
take the amount, that will give this value.
75
00:08:09,920 --> 00:08:15,720
So, if the area is this one, and it has a
centroid,
76
00:08:15,720 --> 00:08:22,150
a distance is this, so this area multiplied
this by the centroidal distance from the right
77
00:08:22,150 --> 00:08:29,260
end, that will be the value of this.
78
00:08:29,260 --> 00:08:32,070
And once that is there, that divided by l
will be the
79
00:08:32,070 --> 00:08:35,570
slope here.
80
00:08:35,570 --> 00:08:42,630
So, we can say this is your some value delta.
81
00:08:42,630 --> 00:09:05,029
So, delta is your area of M by EI
diagram, from A to B. Say it is A; it is B.
82
00:09:05,029 --> 00:09:13,470
So, that multiplied by, say, x bar.
83
00:09:13,470 --> 00:09:24,110
So, x bar will
be this; this will be x bar; say, we say it
84
00:09:24,110 --> 00:09:32,600
is x bar B. So, x bar B is centroid of the
M by EI
85
00:09:32,600 --> 00:09:42,680
diagram from your point B and that will be
the delta according to our second part of
86
00:09:42,680 --> 00:09:44,510
the
moment-area theorem.
87
00:09:44,510 --> 00:09:50,270
Now, if delta is there say theta a.
88
00:09:50,270 --> 00:09:56,800
So, theta a will be straightaway delta by
l; l is the
89
00:09:56,800 --> 00:10:05,170
length of the total beam.
90
00:10:05,170 --> 00:10:13,180
Now, at some intermediate point, if we want
to find out the
91
00:10:13,180 --> 00:10:21,050
deflection, it can be calculated like this:
theta a we can calculate, similarly, that
92
00:10:21,050 --> 00:10:25,950
side we
can draw a line, and this side we can find
93
00:10:25,950 --> 00:10:30,800
out something, that divided by l will be theta
b.
94
00:10:30,800 --> 00:10:37,839
So, end rotation we can find out.
95
00:10:37,839 --> 00:10:42,470
Say at some point intermediate point here
we want to get the deflection; this deflection
96
00:10:42,470 --> 00:10:45,800
we are interested in.
97
00:10:45,800 --> 00:10:50,000
Now, this is a point.
98
00:10:50,000 --> 00:10:55,740
So, from this point we can move vertically
downward, perpendicular to the initial line.
99
00:10:55,740 --> 00:10:59,970
So, it will meet the tangent drawn on a.
100
00:10:59,970 --> 00:11:05,280
So,
this part is basically about this point we
101
00:11:05,280 --> 00:11:09,430
have to calculate the first moment of area
of this
102
00:11:09,430 --> 00:11:16,100
diagram; so, M by EI diagram; here to here,
about this point if you calculate the first
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00:11:16,100 --> 00:11:20,320
moment of area, you will get this part.
104
00:11:20,320 --> 00:11:22,970
And this slope into that length you will get
the full
105
00:11:22,970 --> 00:11:28,490
part; this theta into this one will be the
total length.
106
00:11:28,490 --> 00:11:34,860
So, total length minus this length will
be the actual deflection.
107
00:11:34,860 --> 00:11:49,110
So, if we are interested to find out deflection
at a distance of x, so this will be the actual
108
00:11:49,110 --> 00:11:50,110
deflection.
109
00:11:50,110 --> 00:12:01,860
So, actual deflection that will be your theta
a into x; theta a into x will be the
110
00:12:01,860 --> 00:12:13,339
here to here; theta a into x, minus this part
we have to put there, so that we will get
111
00:12:13,339 --> 00:12:14,980
the
actual deflection.
112
00:12:14,980 --> 00:12:17,990
So, here to here it will be the actual deflection.
113
00:12:17,990 --> 00:12:37,541
So, it will be minus
your area of M by EI diagram, between A and
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00:12:37,541 --> 00:12:43,410
say it is C, this point is C, distance is
x,
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00:12:43,410 --> 00:12:54,930
into, say, here it will have some centroid,
and that centroid, that distance, there might
116
00:12:54,930 --> 00:13:05,540
be
something say C bar; so, it will be C bar.
117
00:13:05,540 --> 00:13:11,080
So, that part is only the small part and this
part
118
00:13:11,080 --> 00:13:12,080
is the total component.
119
00:13:12,080 --> 00:13:17,529
So, if you make it minus, the effective value
what you are
120
00:13:17,529 --> 00:13:20,339
interested at any point you can find out.
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00:13:20,339 --> 00:13:30,930
So, moment-area theorem in cantilever form
we have tried to explain, in a simply
122
00:13:30,930 --> 00:13:37,339
supported condition, how we will get deflection
at some intermediate point; it may be
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00:13:37,339 --> 00:13:44,300
mid span or some other point; you can calculate
in this manner.
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00:13:44,300 --> 00:13:47,750
But initially you have to
find out the slope at the two ends.
125
00:13:47,750 --> 00:13:51,529
Say, you have to calculate some small delta
in this
126
00:13:51,529 --> 00:13:55,790
form, that divided by l will give the slope
at the end.
127
00:13:55,790 --> 00:14:04,800
So, we will get the initial slope;
from there if we follow that procedure, deflection
128
00:14:04,800 --> 00:14:07,329
at any point you can find out.
129
00:14:07,329 --> 00:14:11,089
We have
already, I mentioned you, it is not the only
130
00:14:11,089 --> 00:14:16,910
method, there are different methods; you have
to apply your method according to the type
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00:14:16,910 --> 00:14:20,040
of problem.
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00:14:20,040 --> 00:14:31,440
Now, we can take one example
and try to utilize some of the standard values
133
00:14:31,440 --> 00:15:36,401
to get the deformation of a structure.
134
00:15:36,401 --> 00:15:42,380
So, the beam is like this ABC; at A there
is one support - simple support; B another
135
00:15:42,380 --> 00:15:43,959
simple support; C is free.
136
00:15:43,959 --> 00:15:49,579
So, free end, there is a load P. A to B it
is fully loaded with
137
00:15:49,579 --> 00:15:53,300
some uniformly distributed load; intensity
is omega.
138
00:15:53,300 --> 00:15:59,040
AB length say it is defined as l 1;
BC it is defined as l 2.
139
00:15:59,040 --> 00:16:08,149
For simplicity, entire beam having a identical
cross section EI.
140
00:16:08,149 --> 00:16:24,360
Now, for that beam we are interested to find
out the form, say, for the deflected shape
141
00:16:24,360 --> 00:16:25,890
of
the beam.
142
00:16:25,890 --> 00:16:31,130
Now, here beam is a statically determined
beam, because you can find out the support
143
00:16:31,130 --> 00:16:32,630
reaction.
144
00:16:32,630 --> 00:16:38,089
Here there are two components of reaction,
because it is hinge, one is vertical,
145
00:16:38,089 --> 00:16:44,730
and another horizontal; here one vertical;
and horizontal part will be 0, because the
146
00:16:44,730 --> 00:16:46,829
external loads are all 0.
147
00:16:46,829 --> 00:16:50,589
So, there are two quantities; two quantities
you can find out
148
00:16:50,589 --> 00:16:54,180
summation of moment, summation of vertical
forces.
149
00:16:54,180 --> 00:16:57,980
Now that part is basically our first
job work.
150
00:16:57,980 --> 00:17:09,780
So, we can say there will be a reaction; it
is RXA - it will be 0; and here RYA
151
00:17:09,780 --> 00:17:20,640
- that we have to find out; and here RYB - that
also we have to find out.
152
00:17:20,640 --> 00:17:26,589
Now this is a
problem, straightaway you cannot say RYA will
153
00:17:26,589 --> 00:17:32,659
be this, RYB will be something like
this.
154
00:17:32,659 --> 00:17:42,490
Now, you can take moment about A of all the
process, so it is moment about A, due
155
00:17:42,490 --> 00:17:46,270
to all the process summation it will be 0.
156
00:17:46,270 --> 00:17:47,929
So, A means both the process will cancel.
157
00:17:47,929 --> 00:17:56,960
So,
from here we can say RYB into your l 1 minus
158
00:17:56,960 --> 00:18:04,580
P into l 1 plus l 2 that will be equal to
0.
159
00:18:04,580 --> 00:18:24,850
So, RYB will be Pl 1 l 2 divided by l 1.
160
00:18:24,850 --> 00:18:31,879
The uniform load - that part - we have not
taken I think.
161
00:18:31,879 --> 00:18:35,390
So, it will be additional problem.
162
00:18:35,390 --> 00:18:40,019
So, here, that part also we have to consider.
163
00:18:40,019 --> 00:18:42,110
So, we cannot write here 0.
164
00:18:42,110 --> 00:18:52,320
So, that
additional part will be minus L 1 square divided
165
00:18:52,320 --> 00:18:56,860
by 2; that will be equal to 0.
166
00:18:56,860 --> 00:19:14,760
So, RYB it
will be Pl 1 plus l 2 plus omega l 1 square
167
00:19:14,760 --> 00:19:25,450
by 2 divided by l 1.
168
00:19:25,450 --> 00:19:36,539
So, if that is the value,
summation of RYA and RYB will be equal to
169
00:19:36,539 --> 00:19:37,600
the total force.
170
00:19:37,600 --> 00:19:42,990
So, from there also we can
bring or you can take another moment of about
171
00:19:42,990 --> 00:19:45,220
BS, whatever you like.
172
00:19:45,220 --> 00:19:52,049
So, I think if you
take moment about B, we will get the another
173
00:19:52,049 --> 00:19:53,049
force.
174
00:19:53,049 --> 00:19:58,980
So, here this will be 0; automatically
it is 0.
175
00:19:58,980 --> 00:20:09,001
So, here it will be RYA into your l 1 minus
omega l 1 square by 2 plus P into l 2
176
00:20:09,001 --> 00:20:15,249
that will be equal to 0.
177
00:20:15,249 --> 00:20:20,440
Say RYA if it is clockwise; P is also clockwise.
178
00:20:20,440 --> 00:20:23,679
So, both are
positive; this is negative.
179
00:20:23,679 --> 00:20:36,759
So, from here, RYA, we can write omega l 1
square by 2 minus
180
00:20:36,759 --> 00:20:43,879
Pl 2 divided by l 1.
181
00:20:43,879 --> 00:20:52,549
Now, the reactions we have determined or evaluated
in the form of your l 1, l 2, omega,
182
00:20:52,549 --> 00:20:57,799
and P. Now, it depends which method we will
apply.
183
00:20:57,799 --> 00:21:00,879
In some cases, it is necessary you
have to find out these reactions.
184
00:21:00,879 --> 00:21:06,999
In some cases, it may not be necessary.
185
00:21:06,999 --> 00:21:09,229
But how it can
be determined?
186
00:21:09,229 --> 00:21:13,639
That we have shown it.
187
00:21:13,639 --> 00:21:23,389
Now, here there are two loads: one is P; another
is omega.
188
00:21:23,389 --> 00:21:31,179
Omega is acting from here to here, fully distributed,
for this entire span and
189
00:21:31,179 --> 00:21:40,390
these two loads are acting on the entire structure,
and we want to find out the
190
00:21:40,390 --> 00:21:41,850
deformation of that.
191
00:21:41,850 --> 00:21:49,419
Now, once the reactions are available, due
to this P and omega, we can find out the
192
00:21:49,419 --> 00:21:52,470
expression for this span.
193
00:21:52,470 --> 00:21:57,340
Expression for the bending moment, for this
span, because RYA
194
00:21:57,340 --> 00:22:01,970
into x omega x square by 2 will be the bending
moment here.
195
00:22:01,970 --> 00:22:07,789
Here also, you can find out
the expression for the bending moment.
196
00:22:07,789 --> 00:22:11,879
And you can apply your differential equation
technique.
197
00:22:11,879 --> 00:22:17,510
So, there will be one segment, another segment;
load is started means we have
198
00:22:17,510 --> 00:22:23,340
to apply throughout, then we have to add something,
in this range one equation, another
199
00:22:23,340 --> 00:22:28,179
equation step function, then two constant,
we have to put boundary condition.
200
00:22:28,179 --> 00:22:36,139
So, if you start from there, you are basically
try to solve the problem from your basic
201
00:22:36,139 --> 00:22:37,139
equations.
202
00:22:37,139 --> 00:22:43,609
You need not remember anything, start from
your initial equations; or other
203
00:22:43,609 --> 00:22:52,869
option is some of the standard parameters
already you have derived and those values,
204
00:22:52,869 --> 00:22:58,440
if it
is in our mind, so we can utilize that, and
205
00:22:58,440 --> 00:23:01,820
try to get the solution of the problem.
206
00:23:01,820 --> 00:23:07,399
Now,
that is utilizing some of the known parameters;
207
00:23:07,399 --> 00:23:10,019
some known information.
208
00:23:10,019 --> 00:23:15,440
Now, here there is a technique called method
of superposition.
209
00:23:15,440 --> 00:23:18,210
Method of superposition.
210
00:23:18,210 --> 00:23:25,220
The method of superposition is the effect
of this and effect of that, it can be combined
211
00:23:25,220 --> 00:23:30,899
here, because our analysis is a linear type
of analysis.
212
00:23:30,899 --> 00:23:34,549
So, we are assuming stress is
proportional to strength.
213
00:23:34,549 --> 00:23:37,669
So, stress strength-relationship is falling
in a linear manner.
214
00:23:37,669 --> 00:23:40,789
And
our deformations are very small.
215
00:23:40,789 --> 00:23:44,149
So, whole problem is a linear problem.
216
00:23:44,149 --> 00:23:49,590
So, if it linear
problem, means, load and deflection relationship
217
00:23:49,590 --> 00:23:50,909
will follow a linear curve.
218
00:23:50,909 --> 00:23:56,200
So, if the
load is 100, say, only this load is acting,
219
00:23:56,200 --> 00:23:59,330
whatever slope we will get, if we make the
load
220
00:23:59,330 --> 00:24:08,080
200 instead of 100, slope will be just double;
means 100 plus 100 effect will be just
221
00:24:08,080 --> 00:24:10,669
added; if it is 300 it will be 3 times.
222
00:24:10,669 --> 00:24:13,649
So, all the effect will be simply added with
this.
223
00:24:13,649 --> 00:24:18,950
So,
effect of w and effect of P, if we can study
224
00:24:18,950 --> 00:24:27,789
separately, put the deformation, we can
simply superpose or simply algebraically add
225
00:24:27,789 --> 00:24:29,340
to get the total effect.
226
00:24:29,340 --> 00:24:33,799
The reason is load
deflection curve is linear.
227
00:24:33,799 --> 00:24:41,759
So, this is called your method of superposition.
228
00:24:41,759 --> 00:24:56,570
So, method of superposition is valid for a
linear system and the type of problem we are
229
00:24:56,570 --> 00:24:58,460
handling it is a linear problem.
230
00:24:58,460 --> 00:25:03,140
So, if we have a number of loads or system
of loads, we
231
00:25:03,140 --> 00:25:08,820
can get the response of a load or a system
of load, and get the response of second load
232
00:25:08,820 --> 00:25:11,749
or
a second system of load.
233
00:25:11,749 --> 00:25:16,029
If we combine or algebraically add both the
responses, it is the
234
00:25:16,029 --> 00:25:21,739
response of the structure when it is subjected
to both system of load or both are loaded.
235
00:25:21,739 --> 00:25:28,320
So, that is one of the principle; we said
is principle of superposition, sometimes method
236
00:25:28,320 --> 00:25:35,169
of superposition, that we normally apply here.
237
00:25:35,169 --> 00:25:38,749
Now, we have started with this problem.
238
00:25:38,749 --> 00:25:46,500
Now will take the effect of w in one step;
second step we will take the effect of P;
239
00:25:46,500 --> 00:25:52,869
and we will combine, to get the total effect
of P
240
00:25:52,869 --> 00:25:58,090
as well as w, when they will act in a simultaneous
way.
241
00:25:58,090 --> 00:26:05,509
Now, let us take the first case; say it is
only w.
242
00:26:05,509 --> 00:26:41,289
So, this is the actual problem.
243
00:26:41,289 --> 00:26:47,669
That can be
split into two forms: one with a load; another
244
00:26:47,669 --> 00:26:50,340
with a distributed load.
245
00:26:50,340 --> 00:26:53,509
So, response of that
plus response of that will be the response
246
00:26:53,509 --> 00:26:55,399
of this structure.
247
00:26:55,399 --> 00:27:04,659
So, that we are trying to
define as a principle of superposition or
248
00:27:04,659 --> 00:27:06,059
method of superposition.
249
00:27:06,059 --> 00:27:13,840
So, we will utilize that
to find out the separate effect and combine.
250
00:27:13,840 --> 00:27:20,440
Now, if we take this w, what will be the type
of deformation?
251
00:27:20,440 --> 00:27:25,309
So, here to here it is a
simply supported beam, under distributed load,
252
00:27:25,309 --> 00:27:27,400
and decide there is no load.
253
00:27:27,400 --> 00:27:33,909
So, it will be
a deformation like this.
254
00:27:33,909 --> 00:27:36,049
And we have that information; already we have
solved that
255
00:27:36,049 --> 00:27:42,719
problem; and this part will be, it will just
follow that line; there will be no bending
256
00:27:42,719 --> 00:27:44,129
of this
member.
257
00:27:44,129 --> 00:27:51,799
So, here to here it is a simply supported
case with full load.
258
00:27:51,799 --> 00:27:55,870
We have the idea what will
be the maximum deflection and what will be
259
00:27:55,870 --> 00:27:57,460
the slope here.
260
00:27:57,460 --> 00:28:00,029
And here also it will be the
same slope.
261
00:28:00,029 --> 00:28:01,239
And there is no bending.
262
00:28:01,239 --> 00:28:05,639
So, this slope will be continued; there will
be no
263
00:28:05,639 --> 00:28:09,739
deformation of this member.
264
00:28:09,739 --> 00:28:10,769
Straight.
265
00:28:10,769 --> 00:28:16,429
So, this information we have already in the
form
266
00:28:16,429 --> 00:28:18,369
of maximum deflection and maximum slope.
267
00:28:18,369 --> 00:28:24,909
So, slope into that l 2 will be the deflection
here and slope here.
268
00:28:24,909 --> 00:28:26,999
Now, this part.
269
00:28:26,999 --> 00:28:33,580
This part there is a simply supported force
and without any load.
270
00:28:33,580 --> 00:28:36,729
And
there is a part extending beyond the support
271
00:28:36,729 --> 00:28:39,089
and there is a load here.
272
00:28:39,089 --> 00:28:45,640
Now, this load, we
can shift from here to here.
273
00:28:45,640 --> 00:28:49,679
So, it will come as a force and a moment.
274
00:28:49,679 --> 00:28:59,399
So, if I shift the
load , so its equivalent part will be P into
275
00:28:59,399 --> 00:29:03,200
P into l 2 is a moment.
276
00:29:03,200 --> 00:29:10,269
So, instead of putting
the load, we can shift the load here as P
277
00:29:10,269 --> 00:29:12,789
plus a moment.
278
00:29:12,789 --> 00:29:15,649
Now, this P will be consumed by
279
00:29:15,649 --> 00:29:18,950
the support, because it is on the support.
280
00:29:18,950 --> 00:29:26,990
And this moment is a moment acting at the
hinge of a simply supported beam.
281
00:29:26,990 --> 00:29:31,340
So, it will be a simply supported beam problem
with
282
00:29:31,340 --> 00:29:34,679
some moment.
283
00:29:34,679 --> 00:29:44,929
Now, this problem if we just redefine or if
we draw the deformed shape,
284
00:29:44,929 --> 00:29:57,799
so due to this P and the moment Pl 2.
285
00:29:57,799 --> 00:30:00,659
So, P, it will not contribute anything, because
it
286
00:30:00,659 --> 00:30:06,690
will be just possible by the supporter, only
this moment part.
287
00:30:06,690 --> 00:30:16,539
So, there is a moment
means, it will try to rotate like this.
288
00:30:16,539 --> 00:30:28,499
So, this P if I shift here, P into moment,
P will not do anything, only the moment part
289
00:30:28,499 --> 00:30:31,809
will give a slope here.
290
00:30:31,809 --> 00:30:34,149
And this part will be simply extended.
291
00:30:34,149 --> 00:30:40,729
Now, I have done only one
violation, the P part we have shifted from
292
00:30:40,729 --> 00:30:42,989
here to here.
293
00:30:42,989 --> 00:30:50,399
Now, regarding this part, you can
put the load here or here with a moment is
294
00:30:50,399 --> 00:30:54,169
a same thing, because it is a theory hierarchy.
295
00:30:54,169 --> 00:31:01,200
If we just cut here, so it will be a portion,
there is a P and reaction will be P, and there
296
00:31:01,200 --> 00:31:02,279
is
a moment.
297
00:31:02,279 --> 00:31:05,999
And that basically we are taking it here.
298
00:31:05,999 --> 00:31:15,419
And due to this part there will be a
additional deformation as a cantilever.
299
00:31:15,419 --> 00:31:23,739
So, there is a load about this point, there
will be a
300
00:31:23,739 --> 00:31:28,359
bending like a cantilever.
301
00:31:28,359 --> 00:31:35,700
If I cut here will get a reaction and a moment,
so this extended part is just like a
302
00:31:35,700 --> 00:31:36,700
cantilever.
303
00:31:36,700 --> 00:31:40,149
So, it will have a cantilever type of deformation,
but this is not fixed.
304
00:31:40,149 --> 00:31:45,100
So,
whole thing is rotating, say, a cantileverâ€™s
305
00:31:45,100 --> 00:31:47,539
support is there; support is yielding.
306
00:31:47,539 --> 00:31:51,129
So, there
will be a normal deformation of the beam plus
307
00:31:51,129 --> 00:31:52,469
the support will yield.
308
00:31:52,469 --> 00:31:56,419
So, whole thing
will be yielded by certain extent.
309
00:31:56,419 --> 00:32:02,389
So, that part is basically the yielding of
this part.
310
00:32:02,389 --> 00:32:06,820
Is it clear this part?
311
00:32:06,820 --> 00:32:08,179
This part.
312
00:32:08,179 --> 00:32:10,230
No second part.
313
00:32:10,230 --> 00:32:11,599
Left part.
314
00:32:11,599 --> 00:32:26,340
Because it is a simply supported; it is not
fixed.
315
00:32:26,340 --> 00:32:30,360
So, this is a continuous member, it is just
supported.
316
00:32:30,360 --> 00:32:32,659
So, it is resting on the top.
317
00:32:32,659 --> 00:32:35,359
Yes, it is just resting at the two points.
318
00:32:35,359 --> 00:32:39,459
So, if you try to put the load automatically
there
319
00:32:39,459 --> 00:32:41,710
will be a bending of that.
320
00:32:41,710 --> 00:32:46,759
So, right part if we consider, separately,
so here I have drawn
321
00:32:46,759 --> 00:32:47,759
it.
322
00:32:47,759 --> 00:32:54,100
So, there is a load P. So, if we cut here
to balance that load there will be a P and
323
00:32:54,100 --> 00:32:58,479
there
will be a moment M. So, if we cut that part
324
00:32:58,479 --> 00:33:01,759
as a reaction there will be a force here;
this is
325
00:33:01,759 --> 00:33:04,710
upward and this will be downward; and that
will be that.
326
00:33:04,710 --> 00:33:12,119
So, this part I have basically
drawn here; or you can think - this part is
327
00:33:12,119 --> 00:33:13,709
strong part, it is a rigid - what will happen?
328
00:33:13,709 --> 00:33:17,159
There is a load; this part will be rigid one.
329
00:33:17,159 --> 00:33:20,299
So, this load will be simply shifted P into
this
330
00:33:20,299 --> 00:33:23,709
one; this part will undergo deformation or
this part is rigid.
331
00:33:23,709 --> 00:33:29,320
So, this part will not deform;
under this load, this part will be only bend.
332
00:33:29,320 --> 00:33:31,919
So, when it is rigid, this part will deform;
this
333
00:33:31,919 --> 00:33:33,570
is this part.
334
00:33:33,570 --> 00:33:36,609
When this is rigid, this part will be this
one.
335
00:33:36,609 --> 00:33:41,120
So, both the part are flexible; it
is not rigid; this part or that part.
336
00:33:41,120 --> 00:33:44,659
So, if you combine, you will get a combined
deformed
337
00:33:44,659 --> 00:33:50,599
shape of the structure.
338
00:33:50,599 --> 00:33:57,450
Now all these information are known; most
of the information; say, this one already
339
00:33:57,450 --> 00:33:58,669
we
have calculated.
340
00:33:58,669 --> 00:34:06,119
If you go through your earlier note, you will
get a beam with a fully
341
00:34:06,119 --> 00:34:16,200
distributed load, what will be the slope?
342
00:34:16,200 --> 00:34:24,940
It is already there in a note or I can supply
the
343
00:34:24,940 --> 00:34:28,520
value from here.
344
00:34:28,520 --> 00:34:33,090
It is beam fully loaded; it will be omega
l cube divided by 24 EI.
345
00:34:33,090 --> 00:34:37,159
So,
this is a same slope.
346
00:34:37,159 --> 00:34:42,440
So, what will be this deflection?
347
00:34:42,440 --> 00:34:44,169
Omega.
348
00:34:44,169 --> 00:34:53,050
So, your slope is omega l
cube means l here is l 1 cube divided by 24
349
00:34:53,050 --> 00:34:54,050
EI.
350
00:34:54,050 --> 00:35:00,020
That will be the slope into l 2; that will
be
351
00:35:00,020 --> 00:35:06,830
the deflection.
352
00:35:06,830 --> 00:35:11,119
If we are interested for finding out the deflection
only, we can find out slope or other
353
00:35:11,119 --> 00:35:13,059
parameter in some other place.
354
00:35:13,059 --> 00:35:19,720
Now, here this slope, this slope or if I draw
a tangent, this
355
00:35:19,720 --> 00:35:21,309
slope will be how much?
356
00:35:21,309 --> 00:35:26,420
So, it is a simply supported beam; if we apply
moment at one
357
00:35:26,420 --> 00:35:36,000
end - this is one of the standard value - so,
here, the value is given this is ml by 3 EI.
358
00:35:36,000 --> 00:35:37,670
So,
it is, what is M?
359
00:35:37,670 --> 00:35:50,310
M is your Pl 2 into your l 1 3 EI.
360
00:35:50,310 --> 00:35:58,540
So, it is P Ml by 3 EI.
361
00:35:58,540 --> 00:36:05,319
So, M here is P
into l 2 and here l is l 1 divided by 3 EI.
362
00:36:05,319 --> 00:36:16,460
So, here to here, this value will be Pl 2
l 1
363
00:36:16,460 --> 00:36:24,720
divided by 3 EI multiplied by l 2, means another
l 2 will come here.
364
00:36:24,720 --> 00:36:31,720
And this part it is just a cantilever type
of case; there is a load P. So, it will be
365
00:36:31,720 --> 00:36:44,500
Pl 2
divided by 3 EI.
366
00:36:44,500 --> 00:36:50,100
Pl cube by 3 EI.
367
00:36:50,100 --> 00:36:59,700
So, l is, l is equal to this 3 EI, and this
part is this theta
368
00:36:59,700 --> 00:37:06,690
into l 2; and this part is theta into l 2.
369
00:37:06,690 --> 00:37:12,269
The idea is we have some information and we
try to
370
00:37:12,269 --> 00:37:24,960
explore all those standard values in different
places.
371
00:37:24,960 --> 00:37:39,289
Now, in a similar manner, if we can take some
other problem, and tick the different loads
372
00:37:39,289 --> 00:37:46,789
separately, and try to utilize the standard
values, and get the effect for, say, displacement
373
00:37:46,789 --> 00:37:49,019
we have tried to find out.
374
00:37:49,019 --> 00:37:51,910
Say, if we are interested for slope at that
end.
375
00:37:51,910 --> 00:37:57,329
So slope at that
end is same slope, and here slope is this
376
00:37:57,329 --> 00:38:02,109
slope plus here there will be some slope,
that we
377
00:38:02,109 --> 00:38:06,390
can calculate; it will be Pl 2 square divided
by 2 EI.
378
00:38:06,390 --> 00:38:13,829
So, in that manner, we can get the information
about the slope; go on adding; we will get
379
00:38:13,829 --> 00:38:18,460
the final deflection, final slope.
380
00:38:18,460 --> 00:38:21,339
It may be at the free end; it may be at this
point; it may
381
00:38:21,339 --> 00:38:26,319
be here; it may be central deflection; or
some other parameter.
382
00:38:26,319 --> 00:38:35,809
So, we have tried to show you with some simple,
simple problem and tried to find out
383
00:38:35,809 --> 00:38:37,160
some standard values.
384
00:38:37,160 --> 00:38:43,019
Those values can be utilized in this manner
to carry out a practical
385
00:38:43,019 --> 00:38:48,609
type of problem with some complex support
and load system.
386
00:38:48,609 --> 00:38:54,030
Now, we can take up some
numerical example in our tutorial class and
387
00:38:54,030 --> 00:38:59,920
go ahead with different type of situations
to
388
00:38:59,920 --> 00:40:59,609
have a better understanding of how we can
tackle different type of situations case
389
00:40:59,609 --> 00:41:02,230
to
case.
390
00:41:02,230 --> 00:44:43,390
I think with
391
00:44:43,390 --> 00:45:06,289
that
392
00:45:06,289 --> 00:45:36,859
we
393
00:45:36,859 --> 00:46:12,509
can
394
00:46:12,509 --> 00:59:23,400
conclude now.