1
00:00:59,579 --> 00:01:08,290
So, we have talked about bending moment in
the last class. We can continue with that
2
00:01:08,290 --> 00:01:15,080
and switch over to other thing like shear
force. So, it is, basically, some component
3
00:01:15,080 --> 00:01:25,240
related to bending of beam. For the same problem
if we draw here again. So, we took a
4
00:01:25,240 --> 00:01:43,270
beam like that; put a load at some intermediate
point P; we got some reaction Rb Pa by l;
5
00:01:43,270 --> 00:01:58,979
and at a it was Ra Pb divided by l.
Now, one of the component, we have defined
6
00:01:58,979 --> 00:02:06,450
as bending moment. And we wrote some
expression for bending moment. And we have
7
00:02:06,450 --> 00:02:10,860
seen that quantity is not fixed within the
member; it is varying; and that was function
8
00:02:10,860 --> 00:02:14,370
of x; we have tried to plot that. In that
case,
9
00:02:14,370 --> 00:02:20,280
it was varying in a linear manner.
Similarly, there is a quantity called shear
10
00:02:20,280 --> 00:02:31,780
force. Basically, at this force, if we try
to take
11
00:02:31,780 --> 00:02:41,440
any section, so it will give some shearing
type of stress. So, if you take any section
12
00:02:41,440 --> 00:02:45,640
in
between and that shear stress will be that
13
00:02:45,640 --> 00:02:48,560
force divided by area, more or less it is
shear
14
00:02:48,560 --> 00:02:51,620
stress, and the total force is shear force.
15
00:02:51,620 --> 00:03:02,040
Now, here also if we define the total beam
segment into two regions: one a to this load
16
00:03:02,040 --> 00:03:16,120
point, and load point to the b. So, x, one
part it is a 0; here another is x a. So, shear
17
00:03:16,120 --> 00:03:20,990
force
sometimes we define in the form of b. It will
18
00:03:20,990 --> 00:03:28,480
be say Ra. So, here there is no other force.
Any section if you pick up, the shear force
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00:03:28,480 --> 00:03:33,320
is only Ra. If you cross that load, so b will
be
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00:03:33,320 --> 00:03:51,990
your Ra minus P or this Ra will be your Pb
by l or here it is Pb by l minus P; Pb by
21
00:03:51,990 --> 00:03:54,430
l into
P.
22
00:03:54,430 --> 00:04:09,090
Or if you write P it will be b minus L by
l or you can write P into a by l, because
23
00:04:09,090 --> 00:04:16,599
l minus
b is basically a. Now here ,there is one quantity
24
00:04:16,599 --> 00:04:25,970
Pb by l; another quantity Pa by l; and
they are practically matching with Ra and
25
00:04:25,970 --> 00:04:32,919
Rb. Pa by l is basically your Rb and Pb by
l
26
00:04:32,919 --> 00:04:44,100
equal to Ra, but here it is a plus sign, it
is a minus sign. And the bending moment
27
00:04:44,100 --> 00:04:49,610
expressions, what we have written earlier,
it was positive in both the cases, and we
28
00:04:49,610 --> 00:04:54,629
want
a positive moment bending diagram. So here,
29
00:04:54,629 --> 00:05:01,770
I will first draw it then we will try to define
what is the meaning of this plus, minus. So,
30
00:05:01,770 --> 00:05:06,300
we have to define some sign convention;
based on that, we have to draw the bending
31
00:05:06,300 --> 00:05:12,840
moment diagram and shear force diagram.
So, here, if we try to draw the shear force
32
00:05:12,840 --> 00:05:19,560
diagram, so we draw a line. So, this region
is
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00:05:19,560 --> 00:05:40,419
basically your Ra or Pb by l, and here it
is negative; this part will be Pa by l. So,
34
00:05:40,419 --> 00:05:49,939
this is a
positive part; this is a negative part. If
35
00:05:49,939 --> 00:05:58,139
we take that is a 0 line, this is above is
positive,
36
00:05:58,139 --> 00:06:08,830
below is negative. So, it is plus and minus.
37
00:06:08,830 --> 00:06:18,800
Now what is happening? Shear force is changing;
up to certain point, it is uniform, then
38
00:06:18,800 --> 00:06:26,610
it is changing; going to the negative side;
then again, it is remaining constant; negative.
39
00:06:26,610 --> 00:06:32,860
So, at this level, I should introduce the
concept of sign convention.
40
00:06:32,860 --> 00:06:59,879
Now, if I take shear force. Now, if we take
a shear force like this, we can define this
41
00:06:59,879 --> 00:07:12,259
is
positive; if we get shear force in that manner,
42
00:07:12,259 --> 00:07:22,309
it is negative. Here shear force and bending
moment, it is not a single force; a reaction
43
00:07:22,309 --> 00:07:33,389
is a single force. So, when there is a single
force, we follow some vector direction, say,
44
00:07:33,389 --> 00:07:40,129
it is in the positive direction of x, positive
direction of y, or positive direction of z,
45
00:07:40,129 --> 00:07:45,729
with that we define the plus or minus.
But when it is shear force… shear force
46
00:07:45,729 --> 00:07:51,270
is basically, there are… it is a system
of two
47
00:07:51,270 --> 00:07:56,879
forces; it is a balance system of force; bending
moment there are two forces. So, it is it
48
00:07:56,879 --> 00:08:00,069
is
trying to bend like this. So, this side there
49
00:08:00,069 --> 00:08:06,819
is a moment; this side there is a moment or
shear force. So, there is a counter force.
50
00:08:06,819 --> 00:08:11,229
So, if it is one of the force is along the
positive
51
00:08:11,229 --> 00:08:17,419
direction of x, other thing will be negative
direction of x. So, normally it is not defined
52
00:08:17,419 --> 00:08:24,129
according to the direction of vector; it is
defined in terms of the type of the deformation
53
00:08:24,129 --> 00:08:30,530
of the structure.
Now, here this is upward, this is downward;
54
00:08:30,530 --> 00:08:39,770
but as a whole, this sense is defined as
positive or this sense it is defined as negative.
55
00:08:39,770 --> 00:08:48,680
Similarly, if we talk about moment - this
is
56
00:08:48,680 --> 00:08:58,550
clockwise; this is anticlockwise. So, we will
not define as clockwise, anticlockwise. If
57
00:08:58,550 --> 00:09:01,350
it
bends like that, if the bending phenomenon
58
00:09:01,350 --> 00:09:05,540
is like that, we say it is positive; and if
it
59
00:09:05,540 --> 00:09:17,060
bends like this, we say it is negative. Not
necessarily, this negative, this is this positive,
60
00:09:17,060 --> 00:09:23,670
this is negative; it is a very universal thing.
In some cases, in some books, you may find
61
00:09:23,670 --> 00:09:28,980
it is positive, it is negative; in some cases,
it is positive. Some have written it is positive;
62
00:09:28,980 --> 00:09:36,370
it is negative. But whatever sign you will
follow, you have to follow consistently
63
00:09:36,370 --> 00:09:42,560
throughout the problem. Any problem if you
handle with this type of sign convention,
64
00:09:42,560 --> 00:09:47,160
you will follow up to the end of the problem.
Next problem you can change your sign
65
00:09:47,160 --> 00:09:54,800
convention, but in between if you change,
it will be total mix up.
66
00:09:54,800 --> 00:10:03,880
Now, bending form, we say it is… sometimes
that is defined as sagging type of moment;
67
00:10:03,880 --> 00:10:09,760
if the member is horizontal, if the loading
is vertical, gravity is downward, if the
68
00:10:09,760 --> 00:10:16,620
tendency is like that, so we say it is a sagging
moment. But if it is horizontal plane, there
69
00:10:16,620 --> 00:10:22,280
is no gravity. We cannot define it is sagging,
it is a very general case, but anyway in a
70
00:10:22,280 --> 00:10:29,470
particular case we can say it is sagging moment;
that is a hogging moment. So, plus and
71
00:10:29,470 --> 00:10:45,660
minus, and shear, this is plus, this is minus.
Now, this type of, say here, up to from the
72
00:10:45,660 --> 00:10:50,790
left end to the application of the load, if
you
73
00:10:50,790 --> 00:11:02,430
cut any section, say, if you cut here and
remove the left part, so this reaction Ra,
74
00:11:02,430 --> 00:11:07,880
it will
be acting on the upward side; or this left
75
00:11:07,880 --> 00:11:11,510
part if you take separately, you will find
to
76
00:11:11,510 --> 00:11:16,320
balance that load, that load will be acting
downward. So, anywhere if you cut, so there
77
00:11:16,320 --> 00:11:24,910
will be upward and downward forces. Now when
you will split, the right side face it is
78
00:11:24,910 --> 00:11:34,450
downward, and left side face it is upward;
or I can draw, say, this part little bit I
79
00:11:34,450 --> 00:11:37,700
can put
it in a bigger manner.
80
00:11:37,700 --> 00:11:46,390
So, this is your Ra; if you cut here, so this
part it will try to give one balancing force.
81
00:11:46,390 --> 00:11:50,900
So,
as a reaction, it will get some opposite force.
82
00:11:50,900 --> 00:11:55,690
Here, again, if we cut then there will be
a
83
00:11:55,690 --> 00:12:03,250
force, and if you continue, it will be like
this. So, if you combine, so it is just like
84
00:12:03,250 --> 00:12:06,210
electric
plus minus charger. So, there is here one
85
00:12:06,210 --> 00:12:10,680
negative charge, one positive charge; if you
combine they will be balanced. So, there is
86
00:12:10,680 --> 00:12:13,850
no external force there, but internally it
is
87
00:12:13,850 --> 00:12:21,681
there. It is just like fluid pressure; some
disturbance is there inside, it will be transmitted
88
00:12:21,681 --> 00:12:27,230
through interaction between the particle.
So, this is the force. So, if you cut here,
89
00:12:27,230 --> 00:12:30,300
only
one force; if you think that face, it will
90
00:12:30,300 --> 00:12:36,390
be upward. So, in that process, if you take
a small
91
00:12:36,390 --> 00:12:43,020
element, so it is more or less like this.
So, that type of phenomena, we say it is a
92
00:12:43,020 --> 00:12:46,991
positive
type of shear force. If it is opposite, if
93
00:12:46,991 --> 00:12:53,560
you come from the right side, it is your opposite
phenomena. So, though both the reactions are
94
00:12:53,560 --> 00:12:59,230
upward - Ra and Rb - but this side shear
force is negative; if you take shear force
95
00:12:59,230 --> 00:13:14,700
on the left side, is positive.
Now, with that idea, we can take a beam problem,
96
00:13:14,700 --> 00:13:18,440
with some loading, having some
support condition. That support we can remove
97
00:13:18,440 --> 00:13:23,270
with some appropriate support reaction.
We can apply the equation of statics; get
98
00:13:23,270 --> 00:13:25,620
the support reaction; once that will be known,
it
99
00:13:25,620 --> 00:13:32,250
will be member with some loading. So, from
one end, we can go on adding the load. So,
100
00:13:32,250 --> 00:13:38,460
summation of the vertical forces will give
the shear force.
101
00:13:38,460 --> 00:13:42,630
Say this is a member, some loading are there,
so at this station, from this end, if you
102
00:13:42,630 --> 00:13:45,930
go
on adding the forces here. So, summation of
103
00:13:45,930 --> 00:13:50,640
all the forces will be the shear force here.
And incidentally, if you calculate from this
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00:13:50,640 --> 00:13:58,660
side and that side, it should be equal, because
whole object is in equilibrium. So, if you
105
00:13:58,660 --> 00:14:03,720
cut here, the entire force, and entire force
this
106
00:14:03,720 --> 00:14:07,630
side, this would balance each other. So, you
can proceed from this side or you can
107
00:14:07,630 --> 00:14:13,670
proceed from that side; both the cases you
are supposed to get the same shear force.
108
00:14:13,670 --> 00:14:22,530
Normally, we try to adapt one procedure.
If we want to calculate shear force here,
109
00:14:22,530 --> 00:14:28,760
we will see which end is narrow. So, our
calculation will be little easier, if the
110
00:14:28,760 --> 00:14:32,460
less number of forces we have to count. So,
we can
111
00:14:32,460 --> 00:14:37,060
calculate shear force.
Similarly, bending moment is we have to take
112
00:14:37,060 --> 00:14:41,110
moment of all the forces acting on one
side above that section; that is the bending
113
00:14:41,110 --> 00:14:44,029
moment. So, if you count from left end, if
you
114
00:14:44,029 --> 00:14:48,730
count from the right end, you will get the
same bending moment. So, this way, it may
115
00:14:48,730 --> 00:14:53,030
be
clockwise, this end it will be anticlockwise;
116
00:14:53,030 --> 00:14:57,010
but clockwise anti clockwise; they will give
a sagging type of moment, so it will be positive;
117
00:14:57,010 --> 00:15:01,460
or it will give a hogging type of
moment, it will be negative.
118
00:15:01,460 --> 00:15:06,200
So, that is the basic idea for finding out
bending moment, shear force at any station
119
00:15:06,200 --> 00:15:11,960
within that beam, but you have to start with
finding out the reaction. Then we have to
120
00:15:11,960 --> 00:15:15,950
proceed inside for getting the expression
for the bending moment, shear force; then
121
00:15:15,950 --> 00:15:20,050
drawing out the bending moment diagram. Once
you will draw the bending moment and
122
00:15:20,050 --> 00:15:27,540
shear force diagram, from the diagram itself,
here say your Ra normally it will be more,
123
00:15:27,540 --> 00:15:36,680
because a is less than b. So, this reaction
should be more. So, it is nearer to this support.
124
00:15:36,680 --> 00:15:42,380
So, at this side shear force will be more
compared to shear force of this side. Naturally,
125
00:15:42,380 --> 00:15:48,050
the shear stress produced by the shear force
here, that will be more here compared to
126
00:15:48,050 --> 00:15:53,370
that.
Say bending moment. We have got some diagram.
127
00:15:53,370 --> 00:15:59,560
From there you can get the maximum
bending moment at some point. If we take a
128
00:15:59,560 --> 00:16:05,140
very complex problem, your bending
moment, some where you may get positive, some
129
00:16:05,140 --> 00:16:08,270
where you may get negative. Among
the positive part, there will be a maximum
130
00:16:08,270 --> 00:16:11,279
value; within the negative part, you will
get
131
00:16:11,279 --> 00:16:17,550
some maximum value or minimum value, whatever
you can say. So, those stations will
132
00:16:17,550 --> 00:16:22,930
be the most severe stations. So, we have to
investigate the stresses there; what will
133
00:16:22,930 --> 00:16:25,810
be the
bending stress generated due to that bending
134
00:16:25,810 --> 00:16:32,010
moment or whatever the shear stress
generated due to the shear force. And that
135
00:16:32,010 --> 00:16:35,390
will be our guideline for the design or you
will
136
00:16:35,390 --> 00:16:40,420
get the idea, your stress will be within certain
limit, that is your allowable stress. If that
137
00:16:40,420 --> 00:16:43,250
is
there, your safety will be assured; otherwise,
138
00:16:43,250 --> 00:16:51,190
you have to change your design; you have
to strengthen your structure, or basically,
139
00:16:51,190 --> 00:16:52,550
you have to make some alternative measure,
so
140
00:16:52,550 --> 00:16:59,450
that your structure will be within your safe
region; that is the basic purpose. So, what
141
00:16:59,450 --> 00:17:06,690
should be our next job? Next job will be once
bending moment, shear force we can find
142
00:17:06,690 --> 00:17:25,000
out, we can go for finding out the bending
stress and shear stress calculation.
143
00:17:25,000 --> 00:17:38,880
Now, say, I am drawing only a part of the
beam. Say some moments are here; some
144
00:17:38,880 --> 00:17:48,840
moments are here; say, this is the M; M it
is a typical station; and that is one of the
145
00:17:48,840 --> 00:17:56,970
maximum bending moment. Now, when moment will
be acting on that, it will generate
146
00:17:56,970 --> 00:18:02,150
some stress, and those stresses we say bending
stresses, because bending moment - these
147
00:18:02,150 --> 00:18:06,350
are moments - and that moment is generating
bending, we say it is bending moment. And
148
00:18:06,350 --> 00:18:10,379
that bending moment will generate some stress,
we will say it is bending stress.
149
00:18:10,379 --> 00:18:15,889
And which way it will generate bending stress?
So, bending stress, its distribution is
150
00:18:15,889 --> 00:18:43,440
something like this. So, if we go from the
centre line, a little away, our bending stress
151
00:18:43,440 --> 00:18:49,980
value will increase and it will be maximum
at the extreme (( )); at the top it will be
152
00:18:49,980 --> 00:18:54,779
maximum; at the bottom it will be maximum.
Maximum means one side it will be
153
00:18:54,779 --> 00:19:00,960
tensile; another side it will be compressive.
And in between there will be a whichever,
154
00:19:00,960 --> 00:19:08,010
that part, we say it is the neutral line or
neutral axis. So, neutral axis there will
155
00:19:08,010 --> 00:19:10,580
be no
stress, there will be no strain; it will be
156
00:19:10,580 --> 00:19:12,440
under neutral condition; so we say it is a
neutral
157
00:19:12,440 --> 00:19:17,840
axis.
So, if you go beyond that, more you will be
158
00:19:17,840 --> 00:19:26,440
away from the neutral axis, more will be the
stress; and the stress it carries a linear
159
00:19:26,440 --> 00:19:31,799
path. So, it follows absolutely the linear
path. So,
160
00:19:31,799 --> 00:19:38,409
here linearly it will increase. So, extreme
point, extreme stress, this side extreme point,
161
00:19:38,409 --> 00:19:45,210
extreme stress; and you can visualize which
side will be the tensile stress, because any
162
00:19:45,210 --> 00:19:52,240
member, if we apply moment or if we take our
finger apply moments, so this side will
163
00:19:52,240 --> 00:20:00,029
get stretched; stretched means tensile stress
will be developed. So, this will be tensile
164
00:20:00,029 --> 00:20:06,870
side and this will be your compressive side.
So, this side will try to compress. So,
165
00:20:06,870 --> 00:20:12,230
compressive stress will be generated, and
this side tensile stress will be generated.
166
00:20:12,230 --> 00:20:18,059
So, one will be tensile; another will be compressive.
So, in between there will be a
167
00:20:18,059 --> 00:20:25,080
switch over smoothly from in a linear manner,
crossing the 0 point through neutral axis.
168
00:20:25,080 --> 00:20:31,869
Now, so here I have shown like this; I have
shown like that. So, stress at any point,
169
00:20:31,869 --> 00:20:37,159
it can
be obtained with that relation sigma is your
170
00:20:37,159 --> 00:20:49,619
My by I.
Now, this diagram is taken from side of the
171
00:20:49,619 --> 00:20:58,200
beam. If we take a diagram along this, means
if we take the cross section of that beam,
172
00:20:58,200 --> 00:21:11,100
so it may be any arbitrary shape. Say, that
is the
173
00:21:11,100 --> 00:21:32,980
neutral axis; now here, say, any point, that
distance we are defining as y. So, if y equal
174
00:21:32,980 --> 00:21:36,940
to
0, My by I it will be 0. So, at neutral level
175
00:21:36,940 --> 00:21:43,330
it will be 0. If y is maximum at the extreme
level, it will be maximum stress; here also
176
00:21:43,330 --> 00:21:46,990
it will be plus or minus. If we define tensile
as
177
00:21:46,990 --> 00:21:52,570
positive; compressive will be negative. Here
also there is a sign convention; we can
178
00:21:52,570 --> 00:21:57,370
follow anything; either tensile positive or
compressive positive; other thing will be
179
00:21:57,370 --> 00:22:02,710
negative.
Now, My by I is the sigma or this is the bending
180
00:22:02,710 --> 00:22:15,369
stress here. Now what is I? I is the
moment of inertia of the section above the
181
00:22:15,369 --> 00:22:24,740
neutral axis. So, I is moment of inertia or
sometimes we say it is second moment of area.
182
00:22:24,740 --> 00:22:40,070
So, above neutral axis if we calculate
moment of inertia of the cross section or
183
00:22:40,070 --> 00:22:48,879
sometimes we say it is second moment of area;
so, that will be I. So, that will give M is
184
00:22:48,879 --> 00:22:52,880
fixed for that station; I is fixed. So, it
will just
185
00:22:52,880 --> 00:22:57,529
vary on, it will dependent on y, which is
a variable quantity.
186
00:22:57,529 --> 00:23:05,220
So, y it is varying from 0 to some value;
0 to some value. So, accordingly our stress
187
00:23:05,220 --> 00:23:08,610
will
change. So, we will get maximum stress. So,
188
00:23:08,610 --> 00:23:23,110
sigma top, that will be M; say this is C 1;
this is C 2. So, it will be MC 1 by I and
189
00:23:23,110 --> 00:23:33,190
the bottom part will be MC 2 by I. I am not
putting plus and minus here; only numerical
190
00:23:33,190 --> 00:23:45,720
value I am putting here. So, sigma top and
sigma bottom will be MC 1 by I, MC 2 by I.
191
00:23:45,720 --> 00:23:52,480
Now it will depend which one is bigger - C
1 is bigger or C 2 is bigger - based on that
192
00:23:52,480 --> 00:23:57,999
sigma t or sigma b it will be maximum.
193
00:23:57,999 --> 00:24:11,779
Again, there is another aspect, another requirement
that one part it will be tensile,
194
00:24:11,779 --> 00:24:20,009
another part it will be compressive. Steel
normally tensile compressive part not much
195
00:24:20,009 --> 00:24:25,999
varying, unless the buckling part we consider
it; but there are some material where
196
00:24:25,999 --> 00:24:29,470
compressive strength is different compared
to tensile strength.
197
00:24:29,470 --> 00:24:38,789
So, might be one of the component is small
sigma t or sigma b, but in that mode, the
198
00:24:38,789 --> 00:24:44,149
capacity of the material may be less. So,
which ever will be guiding that depends on
199
00:24:44,149 --> 00:24:51,009
that
nature of stress, and its value; both will
200
00:24:51,009 --> 00:24:55,019
be important. But this extreme fiber stresses
are
201
00:24:55,019 --> 00:25:04,429
quite important. Now at this moment let us
introduce one term, say this one - this I
202
00:25:04,429 --> 00:25:10,169
MC 1
by I; we can sometimes write as Mt equal to
203
00:25:10,169 --> 00:25:26,779
MI by C 1 or we can write M by Z 1.
Similarly, sigma b it will be M by say Z 2.
204
00:25:26,779 --> 00:25:36,419
So, what is Z 1? Z 1 is nothing but your I
by C 1. So, second moment of area divided
205
00:25:36,419 --> 00:25:39,700
by
the distance of the extreme fibre from the
206
00:25:39,700 --> 00:25:43,590
neutral axis. So, that quantity we define
as
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00:25:43,590 --> 00:26:11,080
section modulus. So, it is called section
modulus. So, if we think in terms of our design
208
00:26:11,080 --> 00:26:17,259
problem, we will be mostly interested about
the maximum stress.
209
00:26:17,259 --> 00:26:24,129
So, any section if it is not symmetrical,
it will have two section - two values of section
210
00:26:24,129 --> 00:26:31,741
modulus; one will be based on C 1, and another
will be based on C 2 - the top one and
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00:26:31,741 --> 00:26:36,129
bottom one. So, we will get Z 1 and Z 2, but
if it is a symmetrical section, say, if we
212
00:26:36,129 --> 00:26:41,890
take
a rectangular section, up on I section. So,
213
00:26:41,890 --> 00:26:47,009
here it is absolutely symmetrical about the
neutral axis. So, C 1, C 2 identical; it will
214
00:26:47,009 --> 00:26:53,409
have only one value of section modulus; only
nature is different one will be tensile and
215
00:26:53,409 --> 00:26:56,210
compressive, but the stress evaluation part
will
216
00:26:56,210 --> 00:27:01,460
be only one step.
But if the section is different if we take
217
00:27:01,460 --> 00:27:07,399
a t section, so it is not symmetrical about
the
218
00:27:07,399 --> 00:27:12,659
neutral axis. So, top flange will be very
nearer to the neutral axis; this part will
219
00:27:12,659 --> 00:27:16,779
be little
away from that. So, as well this numerical
220
00:27:16,779 --> 00:27:24,940
value of the stress will be more here compared
to the stress at the flange, but again it
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00:27:24,940 --> 00:27:34,200
may be dependent on the nature plus or minus.
Now, if we can get the bending moment, we
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00:27:34,200 --> 00:27:46,720
can find out the bending stresses. Now, the
next step is shear force. So, shear force
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00:27:46,720 --> 00:27:51,840
will give some stress and that stress we say
it is a
224
00:27:51,840 --> 00:27:57,480
shear stress. So, similarly, My by I, there
will be something for finding out the shear
225
00:27:57,480 --> 00:27:58,919
stresses.
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00:27:58,919 --> 00:28:23,259
Now, if we take say this side is V; this side
is V; and some cross section here. So, this
227
00:28:23,259 --> 00:28:43,210
cross section may be looking like this. Now,
this V will be responsible for generating
228
00:28:43,210 --> 00:28:48,970
the
shear stresses. So, shear stress it can be
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00:28:48,970 --> 00:29:03,739
calculated as VQ by Ib.
Now, V is the sheer force; I already have
230
00:29:03,739 --> 00:29:08,169
defined. So, b and Q these are the two terms
if
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00:29:08,169 --> 00:29:17,119
we know, we can get the shear stress at any
point. Now, b is, say at any level, and say
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00:29:17,119 --> 00:29:25,869
here we want to find out the stresses. So,
this is the b; b is the width of the section
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00:29:25,869 --> 00:29:44,850
at that
level. See here this is the b. And Q; Q is
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00:29:44,850 --> 00:29:53,429
basically the beyond that line whatever area
you
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00:29:53,429 --> 00:30:14,850
will get, you have to find out the first moment
of area of that about neutral axis. Now b
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00:30:14,850 --> 00:30:18,509
is
clear. Q is, say, here we are interested,
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00:30:18,509 --> 00:30:20,960
we can draw a line, we can get b. If it is
a
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00:30:20,960 --> 00:30:29,629
rectangular section all the time it will be
b, fixed. Now, Q is where we want. So, if
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00:30:29,629 --> 00:30:33,229
we
draw a line, above that some portion we will
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00:30:33,229 --> 00:30:39,580
get; and that portion if we consider, we
have to took the first moment of area of that
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00:30:39,580 --> 00:30:50,239
region about that neutral axis.
So, what is that? That is basically area of
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00:30:50,239 --> 00:30:54,059
that portion multiplied by the centroid of
this
243
00:30:54,059 --> 00:31:01,989
shaded part, say, it may be here. So, from
the neutral level whatever is the length that
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00:31:01,989 --> 00:31:07,460
we
have to multiply. So, Q part, say, I am trying
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00:31:07,460 --> 00:31:16,350
to define with a rectangular section; it will
be easier to calculate. So, it will give a
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00:31:16,350 --> 00:31:27,370
better understanding.
So, that is the… let us draw in a little
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00:31:27,370 --> 00:31:34,009
bigger manner. So, that will be the neutral
axis.
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00:31:34,009 --> 00:31:45,390
Now here if we are interested. So, we have
to take a line. Now this part can define as
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00:31:45,390 --> 00:31:46,570
say
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00:31:46,570 --> 00:31:55,169
it is d by 2; it is d by 2; d is the depth
of the beam; and b is the width that is uniform
251
00:31:55,169 --> 00:32:03,039
throughout. Now at this level we are interested
to find out the stress.
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00:32:03,039 --> 00:32:15,722
Now, this part we can take and we say it is
y. So, this part will be d by 2 minus y. So,
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00:32:15,722 --> 00:32:31,769
what will be the area of that? So Q. So b
is the width. G minus 2 minus y, that will
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00:32:31,769 --> 00:32:40,269
be the
height. And that multiplied by the centroid
255
00:32:40,269 --> 00:32:46,110
of the shaded part from the neutral axis.
We
256
00:32:46,110 --> 00:33:04,529
get this. So it will be y plus d by 2 minus
y divided by 2. It may be simplified. It is
257
00:33:04,529 --> 00:33:11,240
just
for our understanding we are writing all these
258
00:33:11,240 --> 00:33:17,191
terms. So this part will give Q and b here
it
259
00:33:17,191 --> 00:33:52,299
is uniform. Now, you can follow that expression
that tau equal to VQ by Ib. We can get
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00:33:52,299 --> 00:34:02,549
the shear stress at any point.
Now, theoretically we can say we can calculate.
261
00:34:02,549 --> 00:34:08,159
But we should have some idea: which
way it will vary; where it will be maximum.
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00:34:08,159 --> 00:34:15,849
Bending stress, it was clearer; it was varying
in a linear manner; and it was maximum at
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00:34:15,849 --> 00:34:22,540
the top and bottom. And it was 0 at the mid
level. Now b for this section it is some constant
264
00:34:22,540 --> 00:34:32,300
value Ib, I will be a constant, but b it
may change. For a rectangular type of case,
265
00:34:32,300 --> 00:34:40,570
it will be constant, but it will take a circular
one. b will be changing; at the centre it
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00:34:40,570 --> 00:34:43,090
will be maximum, gradually it will be less.
Or if
267
00:34:43,090 --> 00:34:44,440
you take a triangular type of section, it
will vary in the V part.
268
00:34:44,440 --> 00:34:53,280
The Q part is quite cumbersome. This b d minus
2 minus y, there is one y; here there is a
269
00:34:53,280 --> 00:35:03,480
y; here there is a y; it is depending on y.
Now here, this part is y and this part is
270
00:35:03,480 --> 00:35:06,250
y. So it
will be multiplied. So we will get some term.
271
00:35:06,250 --> 00:35:14,220
It will be Y squared. Say we are taking this
case only; very simple case - rectangular
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00:35:14,220 --> 00:35:21,580
case. Here in b d minus 2 minus y, there is
one
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00:35:21,580 --> 00:35:28,620
y; and this part we are getting y. If we multiply,
we get some constant term; some term
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00:35:28,620 --> 00:35:37,130
related to y; some term related to y square.
So here, if we put the expression of Q, in
275
00:35:37,130 --> 00:35:41,820
b Q
by I b, we will get one expression. There
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00:35:41,820 --> 00:35:50,710
maximum power of y you will get y squared;
then you can get y; then we can get some constant
277
00:35:50,710 --> 00:35:54,390
value.
So, that will indicate our shear force will
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00:35:54,390 --> 00:35:59,410
be not varying in a linear manner; it will
not be
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00:35:59,410 --> 00:36:03,700
constant throughout. So, it will be constant,
then linear, then it will be quadratic; means
280
00:36:03,700 --> 00:36:06,320
it
will follow a parabolic manner. So, it will
281
00:36:06,320 --> 00:36:17,010
follow a second order curve; second order
curve means it is a parabola. Now, here the
282
00:36:17,010 --> 00:36:24,820
maximum value will be interestingly at the
centroid, and it will be minimum, and it will
283
00:36:24,820 --> 00:36:32,170
be 0 at the end. The reason is very simple,
because what is Q ? Q is the first moment
284
00:36:32,170 --> 00:36:37,750
of area of this region. So, if you put y is
equal
285
00:36:37,750 --> 00:36:47,990
to d by 2. So, area will be 0. The shaded
part will gradually reduce; at the top, it
286
00:36:47,990 --> 00:36:48,990
will be
287
00:36:48,990 --> 00:36:57,000
0. So, Q part will be 0. So, at the extreme
level shear force will be 0, and if you come
288
00:36:57,000 --> 00:37:02,780
nearer to that, area will be more, and your
first moment area also will be more, and it
289
00:37:02,780 --> 00:37:07,910
will
maximum at the neutral level.
290
00:37:07,910 --> 00:37:15,440
Now, if we want to plot the sheer force here,
say this is the section, if we plot this,
291
00:37:15,440 --> 00:37:20,860
this
side, so here it will be maximum, and it will
292
00:37:20,860 --> 00:37:32,710
be like this. So, it will follow a parabolic
pattern having a maximum value at the neutral
293
00:37:32,710 --> 00:37:59,480
level and it will be 0 at the top and
bottom.
294
00:37:59,480 --> 00:38:19,642
Now, if I really put the value of Q here,
and we will get the expression of tau, and
295
00:38:19,642 --> 00:38:24,550
that
will give the equation of this curve, but
296
00:38:24,550 --> 00:38:28,000
you can more or less put it and simplify.
So, you
297
00:38:28,000 --> 00:38:33,930
will get the expression of this curve and
you can plot it. So, at this level I am not
298
00:38:33,930 --> 00:38:37,120
going to
find out the final formula, expression, any
299
00:38:37,120 --> 00:38:45,180
time you can determine, but this value at
the
300
00:38:45,180 --> 00:38:51,730
beginning of your last class we were talking
about shear stresses. So, we have taken two
301
00:38:51,730 --> 00:39:01,340
plates connected by a rivet trying to split
by two forces. So that force it was passed
302
00:39:01,340 --> 00:39:09,290
through shearing of that rivet. So, we told
P divided by area was the shear force, but
303
00:39:09,290 --> 00:39:19,270
actual case it will not vary in a uniform
manner. So, it will be starting from 0 here,
304
00:39:19,270 --> 00:39:21,370
and it
will diminish at 0, in between at the center
305
00:39:21,370 --> 00:39:25,690
it will be maximum, and it will go in a
parabolic manner.
306
00:39:25,690 --> 00:39:31,650
So, actual stress distribution will be little
different; it will be little complex compared
307
00:39:31,650 --> 00:39:34,320
to
the first side, we just… the entire load,
308
00:39:34,320 --> 00:39:36,060
entire area we have divided. So, that area,
that
309
00:39:36,060 --> 00:39:40,600
stress, we got rather some average stress,
and here we will get the actual stress
310
00:39:40,600 --> 00:39:48,620
distribution. So, in that case, if we calculate,
say V is the shear force, and b into d for
311
00:39:48,620 --> 00:39:51,990
the
rectangular case, it is the area, we will
312
00:39:51,990 --> 00:39:57,950
get tau average. And this tau average, this
value,
313
00:39:57,950 --> 00:40:02,600
and if we actually plot the shear stress distribution,
definitely this tau average is
314
00:40:02,600 --> 00:40:10,000
throughout uniform; that will be something
like this.
315
00:40:10,000 --> 00:40:23,750
So, some part it will be less; some part it
will be more. So, the maximum possible stress
316
00:40:23,750 --> 00:40:31,410
we will get, they are neutral axis and this
stress if we calculate, it is tau max; it
317
00:40:31,410 --> 00:40:41,880
can be
verified, the tau max, we can write here tau
318
00:40:41,880 --> 00:40:52,430
max equal to 15.5 times of the tau average
value, at least for the rectangular cross
319
00:40:52,430 --> 00:40:57,720
section.
So, if you really put the value of Q and simplify
320
00:40:57,720 --> 00:41:13,589
we will get the expression of tau, and at
y equal to 0, you will get the maximum stress
321
00:41:13,589 --> 00:41:59,840
- tau max - and if you calculate the average
shear stress - simply V by bd V by area - so
322
00:41:59,840 --> 00:46:25,860
we
323
00:46:25,860 --> 00:47:16,130
will
324
00:47:16,130 --> 00:49:00,500
find that maximum shear stress
325
00:49:00,500 --> 00:50:24,260
that
326
00:50:24,260 --> 00:51:45,470
will
327
00:51:45,470 --> 00:52:50,110
be 50 percent more than the average shear
stress. So, up to
328
00:52:50,110 --> 00:53:01,800
this
329
00:53:01,800 --> 00:56:09,441
we have the idea
about bending moment, shear force, bending
330
00:56:09,441 --> 00:56:54,170
stress, shear stresses. So initial part more
or
331
00:56:54,170 --> 00:57:07,260
less we
have covered today. So, in the next class,
332
00:57:07,260 --> 00:57:45,960
we
333
00:57:45,960 --> 00:59:28,440
will take some further problem,
which are little advance compared to this.