1
00:01:01,579 --> 00:01:08,790
So, we shall continue whatever we are trying
to cover in the last class, that is deflection
2
00:01:08,790 --> 00:01:15,830
of beam, and we are trying to solve the problem
with differential equation technique. We
3
00:01:15,830 --> 00:01:22,530
have taken a simple beam problem. It was a
cantilever, with some point loaded at the
4
00:01:22,530 --> 00:01:29,060
end, with that I have tried to explain you
how to find out the final equations, how to
5
00:01:29,060 --> 00:01:32,220
find
out the constants through boundary conditions,
6
00:01:32,220 --> 00:01:38,560
and from there, we try to obtain some
standard values of maximum deflection, maximum
7
00:01:38,560 --> 00:01:49,390
slope at one of the end.
Now, we will try some other standard cases,
8
00:01:49,390 --> 00:01:54,909
and try to get some important relationship.
Technique is same; problem will be little
9
00:01:54,909 --> 00:02:14,690
different. Now, I can take, again, cantilever
problem with a little different type of loader.
10
00:02:14,690 --> 00:02:47,170
So, structure is same; it has a length l,
EI, support identical, it is fixed, that is
11
00:02:47,170 --> 00:02:56,769
free, only
the loading is different. The idea is to handle
12
00:02:56,769 --> 00:03:03,920
a different type of loader. Also, I think,
before changing the type of structure; next
13
00:03:03,920 --> 00:03:07,010
step we can change the type of structure.
So,
14
00:03:07,010 --> 00:03:17,370
our first job is we have to find out the expression
for the bending moment. Now, at any
15
00:03:17,370 --> 00:03:26,510
distance x, so there is one cross section.
What will be the expression of bending
16
00:03:26,510 --> 00:03:33,210
moment? We can calculate from the right end,
because that is a free end it is easier to
17
00:03:33,210 --> 00:03:38,750
calculate. So, beyond that section right side,
the loading is first of all the length of
18
00:03:38,750 --> 00:03:46,750
that
will be L minus x. And the loading is say
19
00:03:46,750 --> 00:03:55,880
w omega; that is intensity of the distributed
load; the load per unit length. So, length
20
00:03:55,880 --> 00:04:00,620
is l minus x and w is per unit length - that
is the
21
00:04:00,620 --> 00:04:08,060
load. So, total load will be W into l minus
x and that is any distributed manner.
22
00:04:08,060 --> 00:04:15,670
So, moment will be, it is a uniform uniformly
distributed load, means we can assume that
23
00:04:15,670 --> 00:04:21,970
load is acting at the centroid of that part
again. So, it is it will be acting half distance
24
00:04:21,970 --> 00:04:28,090
of
lyx. So, we can write omega is the intensity,
25
00:04:28,090 --> 00:04:32,639
l minus x is the length, that multiplied by
l
26
00:04:32,639 --> 00:04:43,270
minus x divided by 2. So, that is one of the
very familiar type of expression. If we start
27
00:04:43,270 --> 00:04:49,879
from this side, well at omega into x that
will be the load into x by 2, or straightaway
28
00:04:49,879 --> 00:04:52,069
we
will write omega x square by 2; in that case
29
00:04:52,069 --> 00:05:06,270
x is not x, it is l minus x. So, it will be
omega l minus x square divided by 2.
30
00:05:06,270 --> 00:05:19,839
Now, we have to put some sign here again.
At this moment, we have to decide which
31
00:05:19,839 --> 00:05:24,189
sign convention you will take positively.
If you take sign as positive, it should be
32
00:05:24,189 --> 00:05:28,999
negative, because it will give a falling type
of moment. So, if we follow the same thing,
33
00:05:28,999 --> 00:05:37,940
it will be minus and minus.
I think this part is clear - bending moment
34
00:05:37,940 --> 00:05:44,400
expression. Once that expression is ready,
we
35
00:05:44,400 --> 00:05:52,531
can use that moment-curvature relationship;
that directly we will put, the moment
36
00:05:52,531 --> 00:06:06,479
expression. So, there it will be EI d 2 y
dx square equal to minus M. So, it will be,
37
00:06:06,479 --> 00:06:10,110
we
can say it is minus M, and we can put here
38
00:06:10,110 --> 00:06:25,069
omega l minus x square divided by 2. So,
minus, minus; it will be plus. We are sagging;
39
00:06:25,069 --> 00:06:32,060
sagging we have taken positive, but here
moment is hogging, so we have put minus. And
40
00:06:32,060 --> 00:06:36,719
if it sagging as the positive moment,
expression will be with a minus M. So, this
41
00:06:36,719 --> 00:06:43,110
minus, minus it will be plus.
So, in a similar manner, we will integrate
42
00:06:43,110 --> 00:06:49,559
as we did in our previous problem. Iit will
be
43
00:06:49,559 --> 00:07:05,379
omega l minus x cube divided by 6 minus C
1. And EIY equal to omega l minus x 4. So,
44
00:07:05,379 --> 00:07:22,430
4 into 6 24, and minus and minus it will be
plus, C 1 x plus C 2. Absolutely same. Only
45
00:07:22,430 --> 00:07:44,210
the right side part is different. Once you
have that expression, our job is to determine
46
00:07:44,210 --> 00:07:46,779
C 1
and C 2 - these unknowns - through boundary
47
00:07:46,779 --> 00:07:52,660
condition. So, it is in a same manner at x
is
48
00:07:52,660 --> 00:08:01,240
equal to 0, so theta or dy by dx is equal
to 0. So, this is the first expression or
49
00:08:01,240 --> 00:08:05,779
first
condition. So, if we put here, I think. So,
50
00:08:05,779 --> 00:08:11,639
it will be EI into 0. It will be minus omega
l
51
00:08:11,639 --> 00:08:29,610
cube divided by 6 plus C 1 or we can say C
1 equal to omega l cube divided by 6. So,
52
00:08:29,610 --> 00:08:44,740
your C 1 part is available.
Next part is the C 2. So, next requirement
53
00:08:44,740 --> 00:08:51,240
is to find out C 2. And that we can find out
with the second boundary condition. That is
54
00:08:51,240 --> 00:08:53,930
deflection at this point - at x is equal to
0, it
55
00:08:53,930 --> 00:08:56,910
will be equal to 0.
56
00:08:56,910 --> 00:09:08,560
So, substitute that boundary condition - at
x equal to 0, y equal to 0. So, we will get
57
00:09:08,560 --> 00:09:20,690
EI
into 0. That will be omega l 4 24 plus C 1
58
00:09:20,690 --> 00:09:25,100
into 0 plus C 2 or we can get from here C
2
59
00:09:25,100 --> 00:09:38,110
equal to minus omega l 4 24 or these values
we can put in the expression of slope and
60
00:09:38,110 --> 00:10:45,830
deflection. So it will be 1 by EI minus.
So, C 1 and C 2 we have substituted the values
61
00:10:45,830 --> 00:10:57,010
in the expressions we obtained earlier for
slope and deflection, and it will be something
62
00:10:57,010 --> 00:11:04,730
like this - that is the expression of theta;
that is the expression of your deflection.
63
00:11:04,730 --> 00:11:16,750
Now, we will get the maximum deflection,
maximum slope. So, you here at x equal to
64
00:11:16,750 --> 00:11:24,790
l, theta will be theta max, and Y will be
Y
65
00:11:24,790 --> 00:11:36,269
max. So that part if we find out, so theta
x it will be 1 by EI, and here, this term
66
00:11:36,269 --> 00:11:43,200
will be 0;
it will be omega l cube by 6 or we can say
67
00:11:43,200 --> 00:12:02,220
omega l cube divided by 6 EI.
Now, if I find out Y max, 1 by EI; this part
68
00:12:02,220 --> 00:12:05,360
- it will be 0; and this part will be omega
l to
69
00:12:05,360 --> 00:12:16,560
the power 4 divided by 6, omega l to the power
4 divided by 24. So it is omega l to the
70
00:12:16,560 --> 00:12:23,110
power 4. So, 1 by 6 minus 1 by 24. So, if
you operate, it will be 24; this side it will
71
00:12:23,110 --> 00:12:29,680
be 4
minus 1. So 3. So, 3 24 means it will be 8.
72
00:12:29,680 --> 00:12:42,170
So, it will be omega l to the power 4 divided
by 8 EI. So, this is Y max; this is theta
73
00:12:42,170 --> 00:12:54,860
max. So, this is one of the quantity; that
is another
74
00:12:54,860 --> 00:13:15,790
important quantity. So, same problem with
a distributed load; we have tried to repeat
75
00:13:15,790 --> 00:13:19,640
that
process with a different moment expression,
76
00:13:19,640 --> 00:13:32,560
and got the values theta max, Y max.
Now, I will take a problem, where your type
77
00:13:32,560 --> 00:13:36,120
of structure will be different. So, beam will
not be a cantilever beam, say, we will take
78
00:13:36,120 --> 00:13:39,400
a simply supported beam. So, two supports
at
79
00:13:39,400 --> 00:13:48,210
the two end, and take some load, say, we can
take uniformly distributed load, whatever
80
00:13:48,210 --> 00:13:59,290
we have taken, at least load will be same
like this, but support condition will be different.
81
00:13:59,290 --> 00:14:19,090
So, this is a distributed load of intensity
omega or w; and the member is some EI; the
82
00:14:19,090 --> 00:14:53,880
flexible rigidity; this is the length; this
is x this is Y. So, our first job is to determine
83
00:14:53,880 --> 00:14:57,190
the
expression of bending moment. Now, if we start
84
00:14:57,190 --> 00:15:04,760
from this end or the other end, both the
sides supports are there. So, we have to know
85
00:15:04,760 --> 00:15:12,010
the value of those reactions.
Earlier case, support was on the left side
86
00:15:12,010 --> 00:15:14,540
and one free end was there. We were trying
to
87
00:15:14,540 --> 00:15:19,760
proceed from the free end; we need not bother
how much will be the reaction at the
88
00:15:19,760 --> 00:15:24,970
supports, but here we cannot start from this
end or other end; both end are supported.
89
00:15:24,970 --> 00:15:28,630
So,
first job is we have to find out the reaction.
90
00:15:28,630 --> 00:15:31,730
And that will be utilized for finding out
the
91
00:15:31,730 --> 00:15:40,120
bending moment expression. And that will go
to the differential equation.
92
00:15:40,120 --> 00:16:00,930
Now, here there will be one reaction and here.
So, this part we can say y; and this part
93
00:16:00,930 --> 00:16:20,790
we can say Ry; it is say Rx; let us say A,
that A; this is B; say left end if we say
94
00:16:20,790 --> 00:16:28,020
A, right
end if we say B; it is R along x at A; Ry
95
00:16:28,020 --> 00:16:41,910
y at A; this is RB.
So, it is a beam problem and loading here
96
00:16:41,910 --> 00:16:48,339
are all in vertical directions. So, RxA
automatically it will be 0. Still we have
97
00:16:48,339 --> 00:16:54,330
to put, because there might be some inclination
in the loading, and that will make the structure
98
00:16:54,330 --> 00:17:02,370
unstable. So, we have to keep the
provision of some resistance along the horizontal
99
00:17:02,370 --> 00:17:03,370
direction.
100
00:17:03,370 --> 00:17:10,072
So, any way this Ry A and R B we have to determine
here. So how we will determine it?
101
00:17:10,072 --> 00:17:20,410
You can take moment about any one of the end.
So it is better to take moment about A,
102
00:17:20,410 --> 00:17:26,150
at this both the forces will be eliminated.
So, first step we can take summation of
103
00:17:26,150 --> 00:17:45,000
moment about A equal to 0. Now, the reaction
- RB into l. So, that is one force into the
104
00:17:45,000 --> 00:17:51,840
length; that will be balanced by the moment
generated by the externally applied load omega.
105
00:17:51,840 --> 00:17:56,220
So, that is will be equal to… Because this
is the this way moment and load will
106
00:17:56,220 --> 00:18:00,480
be giving in a different way.
So, we can put minus equal to 0 or we can
107
00:18:00,480 --> 00:18:05,049
directly put on the other side. So, we can
put
108
00:18:05,049 --> 00:18:16,049
here minus, and the omega is the load per
unit length into total length l, and its resultant
109
00:18:16,049 --> 00:18:23,360
will be acting at the centre of the beam,
because it is throughout uniform. So, into
110
00:18:23,360 --> 00:18:27,799
your l
by 2, and total moment will be equal to 0.
111
00:18:27,799 --> 00:18:43,249
So, RB you will get from here it will be omega
l by 2, because one l will cancel. So, omega
112
00:18:43,249 --> 00:18:51,179
l by 2.
Now, the other equation you can apply. It
113
00:18:51,179 --> 00:18:56,580
will be say summation of vertical force or
the
114
00:18:56,580 --> 00:19:03,100
all the forces along Y is equal to 0. Somewhere,
we will say summation of B equal to 0;
115
00:19:03,100 --> 00:19:08,570
somewhere, Y equal to 0; sometimes, YI; it
is a matter of convention. So, it will be
116
00:19:08,570 --> 00:19:16,530
RYA
upward, RB upward minus omega into l, it will
117
00:19:16,530 --> 00:19:20,179
be downward, whole quantity will be
equal to 0. Now, RB already you have obtained.
118
00:19:20,179 --> 00:19:33,250
So, RYE plus omega l by 2 minus
omega l equal to 0. So, RYA it will be omega
119
00:19:33,250 --> 00:19:42,289
l by 2. It will be omega l, and it will be
omega l by 2; this is minus plus, right side
120
00:19:42,289 --> 00:19:48,590
if you take it will be omega l by 2.
So, we have tried to use equation of statics,
121
00:19:48,590 --> 00:19:55,970
tried to write so many equations, ultimately
we got both the reactions are omega l by 2.
122
00:19:55,970 --> 00:20:00,880
From our basic idea also you know it will
be
123
00:20:00,880 --> 00:20:09,110
omega l by 2. So, it is, you can say, verification
of that or from this experience later on
124
00:20:09,110 --> 00:20:15,049
we can do one thing. There are some problems
where physically we can find out, we
125
00:20:15,049 --> 00:20:20,759
need not go for that calculation. So, straightaway
we will write it is omega l by 2, omega
126
00:20:20,759 --> 00:20:27,080
l by 2, because it is quite symmetrical structure;
structure fully loaded. uniform. So, half
127
00:20:27,080 --> 00:20:34,320
way, half way it will be distributed. So,
omega l, omega l by 2, omega l by 2. Now,
128
00:20:34,320 --> 00:20:37,850
once
these reactions are available we can write
129
00:20:37,850 --> 00:20:49,590
the expression at any point or rather any
section.
130
00:20:49,590 --> 00:20:59,920
So, we can write the beam in this manner;
here it will be omega l by 2; there it will
131
00:20:59,920 --> 00:21:06,850
be l
by 2; omega l by 2; and this part is omega;
132
00:21:06,850 --> 00:21:26,649
and here it is x; and that side y. So, moment
will be, so at any station, so at any distance
133
00:21:26,649 --> 00:21:40,289
x is omega l by 2, that reactive force
multiplied by x, that will give a moment like
134
00:21:40,289 --> 00:21:43,019
this – clockwise, and this load it will
give a
135
00:21:43,019 --> 00:21:54,730
anti clockwise moment. So that will be omega
x square by 2, because up to here the
136
00:21:54,730 --> 00:22:05,289
length is x omega into x into x by 2, omega
x square by 2. So, this is x and per unit
137
00:22:05,289 --> 00:22:09,640
length it is omega. So, omega into x will
be the load - total load - here that will
138
00:22:09,640 --> 00:22:14,160
be acting
at the middle of x. So, it will be x by 2.
139
00:22:14,160 --> 00:22:19,440
So, omega x into x by 2, omega x square by
2.
140
00:22:19,440 --> 00:22:29,419
And this moment will be a sagging moment,
because this is reaction is giving a moment
141
00:22:29,419 --> 00:22:35,990
that will generate tensile stress at the bottom.
So, here there is a reaction. This reaction
142
00:22:35,990 --> 00:22:43,120
will give a tensile type and this load it
will give… tensile here means hogging type.
143
00:22:43,120 --> 00:22:51,140
So, it
is we put minus. So according to sign convention
144
00:22:51,140 --> 00:22:54,009
we have written the bending moment
expression.
145
00:22:54,009 --> 00:23:01,749
So, next step it will be our moment-curvature
relationship. So, it will be EI d 2 by dx
146
00:23:01,749 --> 00:23:10,639
square will be the minus moment, because sagging
is positive. So, it will be minus
147
00:23:10,639 --> 00:23:26,899
omega lx by 2 plus omega x square divided
by 2. So, once that relation is available,
148
00:23:26,899 --> 00:23:31,040
we
have to integrate twice. So, it will be EI
149
00:23:31,040 --> 00:23:41,179
dy by dx minus omega lx square divided by
4
150
00:23:41,179 --> 00:23:52,309
plus omega x cube divided by 6 plus some constant
will come again. So, second level
151
00:23:52,309 --> 00:24:04,100
integration it will come minus omega lx cube
3 into 4 12, plus omega x to the power of
152
00:24:04,100 --> 00:24:32,919
4, 4 into 624 C 1 x plus C 2.
So, next job is to find out the value of C
153
00:24:32,919 --> 00:24:42,929
1 and C 2 with the help of boundary conditions.
In previous problem, we got at x equal to
154
00:24:42,929 --> 00:24:50,820
0, slope, deflection - both quantities become
0,
155
00:24:50,820 --> 00:24:58,609
but here it is a simply supported problem.
So, it will deform like this. So, slope is
156
00:24:58,609 --> 00:25:02,659
not 0
here; slope is not 0 here; only deflection
157
00:25:02,659 --> 00:25:13,049
is 0; and deflection is 0; these two information
we have. So, our boundary condition is at
158
00:25:13,049 --> 00:25:18,220
x equal to 0, y equal to 0 - this is one of
the
159
00:25:18,220 --> 00:25:29,509
boundary condition. So that part we can write
here - EI y is equal to this will be 0; this
160
00:25:29,509 --> 00:25:38,359
will be 0; this will be 0; and C 2; and that
will lead to C 2 equal to 0.
161
00:25:38,359 --> 00:25:51,750
Because everywhere it is in 2 x, x cube, x
4, x. So, x is equal to 0, and entire part
162
00:25:51,750 --> 00:25:55,440
will be
equal to 0; and y equal to 0, C 2 will be
163
00:25:55,440 --> 00:26:02,500
equal 2 - this is one of the boundary condition.
Now, the other boundary condition at x equal
164
00:26:02,500 --> 00:26:16,119
to l, y is equal to 0. Now that boundary
condition if we apply, we will have a relatively
165
00:26:16,119 --> 00:26:17,539
large expression.
166
00:26:17,539 --> 00:26:32,960
So, that expression will be, your EI deflection
is 0, it will be minus omega l 4 divided by
167
00:26:32,960 --> 00:26:43,399
12 plus omega l 4 divided by 24 plus C 1 l,
C 2 it becomes already 0, for C2 is 0, C 1
168
00:26:43,399 --> 00:26:48,480
l
omega l to the power 4 by 24 minus omega l
169
00:26:48,480 --> 00:26:59,159
2 the power by 12 that will be equal to 0.
So, C 1 we can find out from here. So, C 1
170
00:26:59,159 --> 00:27:04,320
will be, this part is 0, this part is 0, 1
l it will
171
00:27:04,320 --> 00:27:12,289
cancel, it will be omega l cube, omega l cube
divided by 24, because it is 12, it is 24,
172
00:27:12,289 --> 00:27:17,139
half
of that. And this is minus it will go the
173
00:27:17,139 --> 00:27:25,850
plus side. So, C 1 is omega l to the power
24. And
174
00:27:25,850 --> 00:27:33,790
C 2 is 0. And these two quantities we can
put in the expression of our slope and
175
00:27:33,790 --> 00:27:40,559
deflection. We will get the equation for the
deflection slope, of the deflected line.
176
00:27:40,559 --> 00:27:47,659
So, we can write here theta is equal to, because
from here already we have the
177
00:27:47,659 --> 00:28:04,259
expression, it is 1 by EI, and this part is
minus omega l x square by 4 plus omega x,
178
00:28:04,259 --> 00:28:14,669
x 3
divided by 6 plus, C 1 is omega l cube divided
179
00:28:14,669 --> 00:28:25,900
by 24. So, C 1 omega l cube divided by
24. And other part we have just put there.
180
00:28:25,900 --> 00:28:35,049
And y equal 1 by EI, omega lx cube divided
by
181
00:28:35,049 --> 00:28:52,019
12, omega x to the power 4 divided by 24,
and this part will be omega l cube 24 x C
182
00:28:52,019 --> 00:29:06,419
2 is
0. So, these are the equations. Definitely,
183
00:29:06,419 --> 00:29:10,749
you can rearrange in some other form, but
I
184
00:29:10,749 --> 00:29:16,070
want to find some important parameters.
So, what will be the maximum slope? What will
185
00:29:16,070 --> 00:29:23,833
be the maximum deflection? Now, here
the maximum slope we will get, where will
186
00:29:23,833 --> 00:29:31,840
get it? It will be at the two ends. Slope
will
187
00:29:31,840 --> 00:29:35,859
be maximum at the two ends. So, gradually,
so it is a simply supported beam problem.
188
00:29:35,859 --> 00:29:42,950
So, here it will be maximum, and here it will
be 0 slope; and again it will go the
189
00:29:42,950 --> 00:29:48,320
maximum, it will be minus side; in a minus
side maximum value will be this. It is
190
00:29:48,320 --> 00:29:52,859
symmetrical one. So, slope this side, slope
this side - it will be identical; only one
191
00:29:52,859 --> 00:29:56,039
will be
plus and another will be minus. And deflection
192
00:29:56,039 --> 00:30:02,330
maximum we will get at the center. So,
deflection maximum means slope 0. So, that
193
00:30:02,330 --> 00:30:12,320
way also we can get it.
So, say at x equal to 0, theta will be your
194
00:30:12,320 --> 00:30:19,269
theta max. So, theta max will be here. We
can
195
00:30:19,269 --> 00:30:29,200
put x equal to 0, in that equation. So, it
will be 1 by EI. And these two terms, it will
196
00:30:29,200 --> 00:30:38,200
be 0;
it will be omega LQ by 24 or sometimes we
197
00:30:38,200 --> 00:30:53,229
write omega LQ divided by 24 EI.
Now, if I write the maximum deflection at
198
00:30:53,229 --> 00:30:59,299
x equal to l by 2, your Y will be equal to
Y
199
00:30:59,299 --> 00:31:13,239
max. So, it will be at the center of the beam.
So, your Y max. So, here we have to put 1
200
00:31:13,239 --> 00:31:28,669
by EI minus omega l. So, we have to put l
by 2 here. So it will be l to the power 4.
201
00:31:28,669 --> 00:31:34,080
So it
is l by 2, 2 to the power 3 it will be 8.
202
00:31:34,080 --> 00:31:37,600
So, 8 will come at the denominator; it will
be LQ
203
00:31:37,600 --> 00:31:52,399
and one l is there; it will be l 4 plus omega
l 4 divided by 24 is there and l by 2 4 times.
204
00:31:52,399 --> 00:32:06,950
So, it will be 16 here. And here omega l to
the power 4 plus omega l to the power 4, it
205
00:32:06,950 --> 00:32:16,220
will be 24 below and 1 2 you will get, because
here l by 2 half it will be 16, it will be
206
00:32:16,220 --> 00:32:22,519
8.
Now, all are omega l 4 and it will be EI.
207
00:32:22,519 --> 00:32:27,090
Only some factors are there; it is 12 into
8, 24
208
00:32:27,090 --> 00:32:35,679
into 16, 24 by 2. So, this calculation if
we carry out here again, so it will be something
209
00:32:35,679 --> 00:32:44,769
like this: it will be omega l to the power
4 divided by EI, and this factor we have to
210
00:32:44,769 --> 00:32:47,679
find
out; this factor we can calculate here in
211
00:32:47,679 --> 00:32:49,450
as a rough calculation.
212
00:32:49,450 --> 00:33:00,119
So, denominator is maximum 24 into 16. So,
24 12 means, here 2, here 2, it will be 2
213
00:33:00,119 --> 00:33:11,009
into
2 4, minus plus 1. And here it is 2, 2 into
214
00:33:11,009 --> 00:33:19,080
8 24, 24, it will be 8. So, it will be 8 plus
1 9, 9
215
00:33:19,080 --> 00:33:42,509
minus 4 it will be 5, and 24 into 16, 24 into
16 if you calculate. So, 4 into 6 24 2. 384.
216
00:33:42,509 --> 00:33:51,220
I
think it will be 384. So, 24 4. So, it will
217
00:33:51,220 --> 00:34:00,249
be 384. So, this is one of the standard formal
thing, but I want to check whether it is exactly
218
00:34:00,249 --> 00:34:04,859
coming out. Because this is one of the
important term; this is one of the important
219
00:34:04,859 --> 00:34:08,600
term. So, 5 by 3 into omega l to the power
of
220
00:34:08,600 --> 00:34:30,100
4 by EI and that is omega LQ by 24 EI.
So, this is one very, very standard case - a
221
00:34:30,100 --> 00:34:35,550
simply supported beam with uniformly
distributed load, maximum deflection, maximum
222
00:34:35,550 --> 00:34:46,109
slope. So, in the next case, I will take a
point load on a simply supported type of problem,
223
00:34:46,109 --> 00:34:50,180
because cantilever we have started
with a point load, then distributed load.
224
00:34:50,180 --> 00:34:55,639
Then we have kept the loading same, support
was different; now support is again simple
225
00:34:55,639 --> 00:35:04,440
support at the both end, but loading is point
loader. Now let us see what are the things
226
00:35:04,440 --> 00:35:40,640
we will get.
227
00:35:40,640 --> 00:36:09,790
So, the length is l; left end a it is in support.
So, Rxa Rya; this is roller Rb; xy is like
228
00:36:09,790 --> 00:36:17,000
that;
P is acting at the mid span. So, it is at
229
00:36:17,000 --> 00:36:32,140
the center of the support. Now, the first
job here is
230
00:36:32,140 --> 00:36:41,320
to determine the support reactions, because
both the end are supported; that we have
231
00:36:41,320 --> 00:36:49,900
found in the earlier problem. So, one option
is you can take moment at support a, then
232
00:36:49,900 --> 00:36:57,580
you can take summation of all the forces.
But you need not do it, because from the
233
00:36:57,580 --> 00:37:02,490
structure itself or the way load is put there,
directly you can calculate. So, what will
234
00:37:02,490 --> 00:37:03,490
be
235
00:37:03,490 --> 00:37:08,270
the reaction here? PY 4. It will be straight
way PY by 2; half, half load. So, Rb we can
236
00:37:08,270 --> 00:37:16,430
write straightaway P by 2, and Ra it is P
by 2, and Rxa equal to 0. There is no horizontal
237
00:37:16,430 --> 00:37:25,110
force. So, the reactions we obtain from the
look of the structure.
238
00:37:25,110 --> 00:37:31,330
Next part is we have to get the expression
of the bending moment, and that we will
239
00:37:31,330 --> 00:37:39,911
substitute in our moment curvature relationship.
Here there is one problem; say this a, it
240
00:37:39,911 --> 00:37:57,450
b, the center point is c, and total length
is l; and here to here, it is say ly 2. Now
241
00:37:57,450 --> 00:38:01,140
your
moment expression is not valid for the entire
242
00:38:01,140 --> 00:38:10,340
span. So, E to C, there will be one
expression for bending moment; if you cross
243
00:38:10,340 --> 00:38:20,820
C moment expression will be different.
Cantilever problem, point load case was much
244
00:38:20,820 --> 00:38:28,320
easier compared to the distributed load,
but here situation is just reversed.
245
00:38:28,320 --> 00:38:42,750
So, if we divide into two regions, say x,
it is l by 2 0, this is say one part; another
246
00:38:42,750 --> 00:38:55,940
part l
by 2 l; here your moment is P divided by 2
247
00:38:55,940 --> 00:39:00,990
into x; that is, because P into the distance;
P
248
00:39:00,990 --> 00:39:06,460
into distance that is giving a sagging type
of moment, positive moment. So, this P by
249
00:39:06,460 --> 00:39:11,750
2
into x. Now, if we cross point C, your moment
250
00:39:11,750 --> 00:39:20,020
will be it is P by 2 into x minus
something will be there, that will be P into
251
00:39:20,020 --> 00:39:31,940
x minus l by 2. So, additional force will
involve additional term in the expression.
252
00:39:31,940 --> 00:39:39,610
So, if you have some other loads, say, if
you put another load. So, we will have one
253
00:39:39,610 --> 00:39:44,090
segment, two segments, three segments, or
from here there will be a distributed load.
254
00:39:44,090 --> 00:39:46,860
So,
once a different type of load will come, it
255
00:39:46,860 --> 00:39:50,150
will go on inviting one, one additional term,
so
256
00:39:50,150 --> 00:39:57,770
expression will no longer valid there. So,
it will be, with any additional term means
257
00:39:57,770 --> 00:40:03,361
it
will be different. So, for that region we
258
00:40:03,361 --> 00:40:10,410
have to apply your moment curvaturerelationship
with this moment, and for the other region,
259
00:40:10,410 --> 00:40:15,990
you have to put this moment in
the moment-curvature relationship. So, simultaneously
260
00:40:15,990 --> 00:40:18,320
for different segment you have to
run it.
261
00:40:18,320 --> 00:40:28,570
Now, let us put the expression and see what
is happening. First of all your EI d 2 y dx
262
00:40:28,570 --> 00:40:36,760
square equal to minus M. So, it will be minus
Px by 2. And this side it will be minus EI
263
00:40:36,760 --> 00:40:44,510
d
2 y dx square equal to minus Px by 2 plus
264
00:40:44,510 --> 00:40:53,680
Px minus l by 2. So, this side is more or
less
265
00:40:53,680 --> 00:41:08,510
same thing, only this part is additional term.
Now, if we integrate, so it will be EI Px
266
00:41:08,510 --> 00:41:21,780
square by 4 plus C 1, and this side will be
EI dy by dx equal to minus Px square by 4
267
00:41:21,780 --> 00:41:29,200
plus Px minus ly 2 square divided by 2 plus
say d 1.
268
00:41:29,200 --> 00:41:43,120
Again if we integrate, so it will be EI into
y minus Px cube divided by 12 C 1 x plus C
269
00:41:43,120 --> 00:41:58,480
2
and here it will be Px cube 12 plus P. 6.
270
00:41:58,480 --> 00:42:22,350
So, it is d 1 x plus d 2. Can you read d 2?
Or it is
271
00:42:22,350 --> 00:42:27,460
going beyond limit. Now, it is more or less
similar type of expression. Here it is up
272
00:42:27,460 --> 00:42:32,500
to
that; here this plus some quantity. And this
273
00:42:32,500 --> 00:42:40,260
region we are getting C 1 x C 2 and here we
are getting d 1 x d 2.
274
00:42:40,260 --> 00:42:50,020
There might be another segment; it should
be, say, C, D, E, F, G, some sets of constants.
275
00:42:50,020 --> 00:43:02,680
So, how we will get all these constants? Now,
here the deformed shape will be
276
00:43:02,680 --> 00:43:14,900
something like this. And if you take the load
point, under the load, the deflection if you
277
00:43:14,900 --> 00:43:19,410
calculate from the left span, and the deflection
if you calculate from the right span, they
278
00:43:19,410 --> 00:43:23,330
are suppose to be equal; it cannot be different
deflection, because it is a continuous
279
00:43:23,330 --> 00:43:31,240
structure. Similarly, the slope, if you calculate
from the left equation - means equation
280
00:43:31,240 --> 00:43:36,270
for the left part - and the slope if you calculate
from the equation for the right part, it
281
00:43:36,270 --> 00:43:42,070
should be identical, because slope should
be identical; there should be not be any ((kink))
282
00:43:42,070 --> 00:43:52,290
here.
So, there might be a number of span, but in
283
00:43:52,290 --> 00:44:00,450
between all those interface point, where there
is any new load, your expression will change,
284
00:44:00,450 --> 00:44:08,560
your continuity of slope and deflection, we
can use; those will give the additional information
285
00:44:08,560 --> 00:44:20,130
like boundary conditions.
Now, if we utilize those say at x equal to
286
00:44:20,130 --> 00:44:25,380
l by 2, your theta from this side and theta
from
287
00:44:25,380 --> 00:44:39,920
that side should be equal. So, we can say
theta at C from left, it should be equal to
288
00:44:39,920 --> 00:44:52,870
theta
at C from right. So, if you calculate from
289
00:44:52,870 --> 00:44:58,710
the left side equation, if you calculate theta
from the right side equation, at C it should
290
00:44:58,710 --> 00:45:11,530
be equal. So, that is the equation.
So, if you write that expression, it will
291
00:45:11,530 --> 00:45:28,140
be, say here, it will be minus pl square divided
by
292
00:45:28,140 --> 00:45:44,750
4 into 4 into 4. So, it will l by 2, means
2 2 4 plus C 1, that should be equal to. Here,
293
00:45:44,750 --> 00:45:48,960
it
will be again pl square divided by 4 into
294
00:45:48,960 --> 00:45:53,930
4 plus… This quantity will be how much?
This
295
00:45:53,930 --> 00:46:01,850
quantity, incidentally it will be equal to
0, because x equal to l by 2 l by 2 l by 2
296
00:46:01,850 --> 00:46:10,270
0 plus d
1. But that should be divided by your EI,
297
00:46:10,270 --> 00:46:13,960
because this is slope into EI; both the side
it
298
00:46:13,960 --> 00:46:20,150
will be divided by EI. So, you can put it,
again cancel it. So, basically this quantity
299
00:46:20,150 --> 00:46:22,930
will
be equal to this quantity. Now this pl square
300
00:46:22,930 --> 00:46:29,870
by 4 into 4 and pl square 4 by 4, it is
identical. And it is C 1; this is 0 D 1. So,
301
00:46:29,870 --> 00:46:41,950
from here we are getting your D 1 equal to
C 1.
302
00:46:41,950 --> 00:46:58,270
So, our equation is different, and we are
getting different sets of unknowns, but the
303
00:46:58,270 --> 00:47:03,500
slope
continuity at that point where we are changing
304
00:47:03,500 --> 00:47:06,880
our equation, that will give the D 1 is
305
00:47:06,880 --> 00:47:13,080
equal to C 1. So, first constant in this equation
will be identical to that equation on the
306
00:47:13,080 --> 00:47:17,430
second equation.
In a similar manner, if we just equate the
307
00:47:17,430 --> 00:47:23,890
y, so at x equal to l by 2, so this quantity
and
308
00:47:23,890 --> 00:47:32,210
this quantity if we equate, we will get D
2 will be equal to C 2. So, you can put and
309
00:47:32,210 --> 00:47:37,660
see
here, I think.
310
00:47:37,660 --> 00:47:55,490
Y at c, it is from left YC from right. So,
here we have the equation it will be minus
311
00:47:55,490 --> 00:48:07,650
pl
cube divided by 12 into 8 plus C 1 l by 2
312
00:48:07,650 --> 00:48:14,550
plus C 2; it will be equal to minus pl cube
12
313
00:48:14,550 --> 00:48:21,680
into 8. Incidentally, that term will be equal
to 0; the additional term what you will get
314
00:48:21,680 --> 00:48:28,860
at
that point plus D 1 already we have obtained
315
00:48:28,860 --> 00:48:36,950
it is equal to your C 1 it is l by 2 plus
D 2.
316
00:48:36,950 --> 00:48:42,280
Now, this C 1 and D 1 already we have established.
So, this term should be equal to this
317
00:48:42,280 --> 00:48:48,080
term; and this term is equal to this term;
it is 0. So, these two terms should be equal
318
00:48:48,080 --> 00:48:55,020
to 0.
So, we are getting D 2 equal to C 2.
319
00:48:55,020 --> 00:49:05,310
The main reason is I shall show you again
this page - previous page - the additional
320
00:49:05,310 --> 00:49:11,400
term
x minus l by 2 - this term - will be after
321
00:49:11,400 --> 00:49:18,360
l by 2, but at l by 2 exactly the value is
0. This
322
00:49:18,360 --> 00:49:25,210
moment, this load will give some moment on
the right side, but it will start from 0 value.
323
00:49:25,210 --> 00:49:30,260
So, when we are putting the boundary condition,
that part is becoming 0 whatever
324
00:49:30,260 --> 00:49:37,700
additional term we are getting it. So, ultimately
it is becoming identical constant within
325
00:49:37,700 --> 00:49:46,210
the equation. So, there might be say two segments
like this type of problem; there might
326
00:49:46,210 --> 00:49:51,520
be three segments or four segments; only first
segment will have some expression;
327
00:49:51,520 --> 00:49:57,610
second segment will have expression plus some
additional; term third one is the
328
00:49:57,610 --> 00:50:04,160
expression for the previous zone plus some
additional term. So, same thing will be there
329
00:50:04,160 --> 00:50:14,240
plus some one-one additional component plus
some constants C 1 plus C 1 x plus C 2 D
330
00:50:14,240 --> 00:50:20,340
1 x plus D 2 like that it will continue, but
from the continuity condition all the constants
331
00:50:20,340 --> 00:50:25,910
are identical.
So, it is this C 1, D 1, E 1 it will be all
332
00:50:25,910 --> 00:50:30,980
identical; similarly, C 2, D 2, E 2, F 2 it
will be all
333
00:50:30,980 --> 00:50:37,280
identical. Now, from there we can think in
that manner, we can take the expression of
334
00:50:37,280 --> 00:50:41,810
the
last span. So, last span it will retain all
335
00:50:41,810 --> 00:50:48,040
the quantity of the previous one with some
additional, additional component. And the
336
00:50:48,040 --> 00:50:49,890
constant whatever we will get, because it
will
337
00:50:49,890 --> 00:50:58,720
be identical part all the thing, so it can
be rewritten in this form; say, this particular
338
00:50:58,720 --> 00:51:04,700
beam
problem, we can write, say, M we can write
339
00:51:04,700 --> 00:51:09,510
in this manner. So, for the left span, we
got
340
00:51:09,510 --> 00:51:27,110
it was M equal to Px by 2, and the right span
we got
341
00:51:27,110 --> 00:51:51,460
it was minus Px minus l by 2. So, it
is minus Pxl minus 2 and Px by 2.
342
00:51:51,460 --> 00:51:59,520
Which one?
Student: Extreme reaction.
343
00:51:59,520 --> 00:52:03,620
Extreme reaction it will not come in the second
part.
344
00:52:03,620 --> 00:52:14,420
Say here; here to here M equal to P by 2 into
x; and here to here it is, say, this force
345
00:52:14,420 --> 00:52:19,620
into
x that part is there, minus P into x minus
346
00:52:19,620 --> 00:52:23,670
l by 2, because here if it is x, so l by 2
minus
347
00:52:23,670 --> 00:52:29,290
that is the distance P l minus x P; that part
it will give a hogging moment, and this part
348
00:52:29,290 --> 00:52:39,490
will give some sagging moment.
Yes, because these two force will give some
349
00:52:39,490 --> 00:52:45,610
moment. That moment will be balanced by
Rbh. So, if you come from Rb, whatever moment
350
00:52:45,610 --> 00:52:49,250
you will get, if you come from this
side you will get the same moment. So, either
351
00:52:49,250 --> 00:52:54,510
you have to take from this side or that side,
I think; both the thing you cannot take that,
352
00:52:54,510 --> 00:52:58,900
then it will be cancelling it. So, the main
part
353
00:52:58,900 --> 00:53:02,980
is, this is the expression for the left part;
this is the expression for the right part,
354
00:53:02,980 --> 00:53:07,340
where
right part is involving one additional component.
355
00:53:07,340 --> 00:53:15,700
Now, this expression, whatever we have written,
so we can say this is for left part; and if
356
00:53:15,700 --> 00:53:20,540
we take the full component - this plus something,
it will be the right part. Now, this
357
00:53:20,540 --> 00:53:26,500
equation can be written for the entire structure
in that manner. So, whatever we are
358
00:53:26,500 --> 00:53:32,800
talking in language, it can be written in
some equation form.
359
00:53:32,800 --> 00:53:48,070
So, same expression we are telling when it
is left span it is only this one, and right
360
00:53:48,070 --> 00:53:59,060
part, it
is some additional part. Now, this H l minus
361
00:53:59,060 --> 00:54:10,550
x - this is called unit step function, and
its
362
00:54:10,550 --> 00:54:27,240
value is something like this. Here it is 1
unit; it is l by 2; it is l by 2. So, up to
363
00:54:27,240 --> 00:54:40,030
it is 0, then
it will be 1. The reason is very simple. The
364
00:54:40,030 --> 00:54:51,230
function is 0, when it is less than l by 2.
So,
365
00:54:51,230 --> 00:54:56,100
when it is beyond the load, it is 0; when
it is beyond the load, means it is before
366
00:54:56,100 --> 00:55:00,860
the load.
When it is after the load the value is 1.
367
00:55:00,860 --> 00:55:06,480
So, here also when it is before the load,
only this
368
00:55:06,480 --> 00:55:12,180
term; when it is after that load, this plus
this quantity. So, that physical logic we
369
00:55:12,180 --> 00:55:15,120
are
trying to put in a mathematical form with
370
00:55:15,120 --> 00:55:24,820
some function called unit step function.
So, it will give a general expression for
371
00:55:24,820 --> 00:55:31,630
bending moment for the entire span, because
both the section here constant, will be identical.
372
00:55:31,630 --> 00:55:38,750
I think we can stop here at this class;
next class we can continue on.
373
00:55:38,750 --> 00:55:52,410
Preview of the Next Lecture
Most specifically it is deflection of beam
374
00:55:52,410 --> 00:55:55,320
the method we used it is basically integration
of
375
00:55:55,320 --> 00:56:03,160
differential equation we try to handle some
cantilever type of problem, then switch over
376
00:56:03,160 --> 00:56:09,630
to a simply supported case with uniformly
distributed load point loaded case we have
377
00:56:09,630 --> 00:56:18,200
taken very standard type of case load is acting
at the centre of the beam up to the certain
378
00:56:18,200 --> 00:56:25,490
extend we solve it. So, remaining part we
can continue. So, last class we got more or
379
00:56:25,490 --> 00:56:28,510
less
something like this.
380
00:56:28,510 --> 00:56:42,600
So, we took a beam like this this side some
support here also,me support and centre we
381
00:56:42,600 --> 00:56:55,650
put P reaction we got P by 2 reaction P by
2 if we say this side is x that side this
382
00:56:55,650 --> 00:57:03,630
is y.
Now, the main difficulty what we were discussing
383
00:57:03,630 --> 00:57:13,180
in the last class it has 2 regions if we
say it is A and if it is B and if it is C.
384
00:57:13,180 --> 00:57:17,550
So, A to B are there is 1 segment or a part
where
385
00:57:17,550 --> 00:57:24,900
your moment has some expression and the second
part B to C it has a different
386
00:57:24,900 --> 00:57:32,560
expression for the bending moment expression.
So, these two parts more or less we have
387
00:57:32,560 --> 00:57:41,320
tried to express in this manner say M that
we have written P by 2 into x for the left
388
00:57:41,320 --> 00:57:49,620
part
for the right part it was P by 2 into x minus
389
00:57:49,620 --> 00:57:58,680
P xl by 2 right.
So, that we have written in the last class
390
00:57:58,680 --> 00:58:05,900
because here only 1 load will generate some
moment. So, P by 2 into that distance. So,
391
00:58:05,900 --> 00:58:11,750
if x is beyond that limit here the total length
is
392
00:58:11,750 --> 00:58:21,190
l and we have divided into 2 part l by 2 l
by 2, I think. So, that distance we have written
393
00:58:21,190 --> 00:58:35,680
earlier it was say l by 2 l by 2 now once
we cross l by 2 the centre load P will contribute
394
00:58:35,680 --> 00:58:45,160
something in bending moment. So, that was
P into l minus we got x minus l by 2. Now,
395
00:58:45,160 --> 00:59:10,160
after integration we got 2 sets of constant
here and here left side say C 1 C 2 right
396
00:59:10,160 --> 00:59:25,510
side d
1 d 2.