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00:00:55,370 --> 00:01:05,680
So, we were discussing this problem in our
previous class. The starting of the problem
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00:01:05,680 --> 00:01:16,360
was like this. So, fixed beam, load placed
little bit away from the center. So, a load,
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00:01:16,360 --> 00:01:21,740
bending moment, at the two ends force, bending
moment separately we have applied.
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00:01:21,740 --> 00:01:30,040
And due to the load, the slopes we have written,
and due to your MA, MB, you wrote the
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00:01:30,040 --> 00:01:37,010
expression of slope at the two ends. So, next
part is just we have to merge the slope,
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00:01:37,010 --> 00:01:44,170
because slope here and slope here; it should
compensate, because it is a fixed support;
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00:01:44,170 --> 00:01:45,929
and here also same thing.
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00:01:45,929 --> 00:02:04,140
So, we can write theta A due to P should be
theta A due to moment. And theta A due to
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00:02:04,140 --> 00:02:07,590
P
and theta A due to moment, the expressions
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00:02:07,590 --> 00:02:18,260
whatever we have written, so if we just write
it on the left side, it will be your MAl 3
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00:02:18,260 --> 00:02:39,819
EI MB 6 EI; that was Pab a 2 b divided by
6 lEI.
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00:02:39,819 --> 00:02:48,980
So, definitely, this part on the left side,
the other part on the left part here, this
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00:02:48,980 --> 00:02:52,349
is on the
right side.
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00:02:52,349 --> 00:03:04,910
So, similarly, at B, due to the load, and
theta B due to the moment, if we equate, if
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00:03:04,910 --> 00:03:07,890
we
write here, so it will be MAl divided by 6
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00:03:07,890 --> 00:03:14,629
EI MBl by 3 EI; that is equal to Pab 2 a b
6
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00:03:14,629 --> 00:03:41,310
lEI. So, here it is a plus 2 b; here 2 b plus
a; here it is 3; it is 6; it is 6; it is 3.
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00:03:41,310 --> 00:03:48,950
So, these are
the two equations and two unknowns: MA, MB;
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00:03:48,950 --> 00:03:57,299
MA, MB. Now we have to just process
these two equations and find out MA and MB
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00:03:57,299 --> 00:04:07,340
Now the lower equation if we multiply
with 2, so this 6 will be 3, and you may part
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00:04:07,340 --> 00:04:17,989
it, we can cancel. So, this equation we can
write, we can straightaway multiply or we
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00:04:17,989 --> 00:04:28,330
can write in a separate line. So, it will
be MAl
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00:04:28,330 --> 00:04:50,550
by 3 EI; 2 MBl by 3 EI; it will be 2 Pab 2
ab divided by 6 lEI. Now, if we just subtract
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00:04:50,550 --> 00:05:00,060
that equation or from this equation, this
equation if we make it minus. So, your MAl
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00:05:00,060 --> 00:05:06,980
by 3
EI part will be cancelled and here MB part
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00:05:06,980 --> 00:05:11,380
will be remaining.
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00:05:11,380 --> 00:05:22,830
So, if I come to the next page, say, this
is your, if I say this is your equation one;
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00:05:22,830 --> 00:05:29,360
if we
say it is equation two. So, if we make equation
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00:05:29,360 --> 00:05:37,349
two minus equation one. So, it is equation
two minus equation one, if we write. So, this
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00:05:37,349 --> 00:05:52,750
MAl by 3 EI MAl by 3 EI that part will be
cancelled. And here it will be 2 Mbl by 3
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00:05:52,750 --> 00:06:08,200
EI minus Mbl by 6 EI, and here, your 2 Pab
2 a
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00:06:08,200 --> 00:06:34,349
plus b divided by 6 lEI minus Pab a plus twice
b 6 lEI. So, straightaway we just put the
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00:06:34,349 --> 00:06:45,669
expressions. Now, there are some common quantities
that we will consider in the next
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00:06:45,669 --> 00:06:52,720
step. So, here it is 3 EI 6 EI, we can just
take 6 EI below; and it will be how much?
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00:06:52,720 --> 00:07:00,030
So,
here it will be 4, 4 minus this side will
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00:07:00,030 --> 00:07:06,330
be 1. So, 4 minus it will be it will be 3
and 3 and 3
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00:07:06,330 --> 00:07:15,770
denominator will be 6. So, it will be half.
So, we can write, say we are writing MB MBl
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00:07:15,770 --> 00:07:27,000
divided by 2 EI. We take common 6. So, here
it will be 4 minus 1, 3, 3 6.
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00:07:27,000 --> 00:07:37,610
So, if you simplify you should get that. and
here denominator 6 lEI 6 lEI common; Pab,
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00:07:37,610 --> 00:07:46,050
Pab common. So, here the internal part you
have to adjust. So, it will be Pab divided
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00:07:46,050 --> 00:07:53,220
by 6
lEI. So, 2 if you put inside, it will be 4
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00:07:53,220 --> 00:07:59,900
a, here minus a; so, it will be 3 a; and it
will be 2
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00:07:59,900 --> 00:08:12,310
b; minus 2 b. So, it will cancel. So, it will
be 3. It will be 3 a. So, we can put it a
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00:08:12,310 --> 00:08:16,840
into a, it
will be a square; and this 3 we can cancel,
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00:08:16,840 --> 00:08:32,849
we can make it 2. So, your MB, it will be
Pa
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00:08:32,849 --> 00:08:42,430
square b; 2 2 cancel; El will be cancel; there
will be one l. So, it will be l square. So,
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00:08:42,430 --> 00:08:47,780
it
will be MB will be Pa square b by l square.
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00:08:47,780 --> 00:08:50,110
Unit also you can check, because a square
a
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00:08:50,110 --> 00:08:56,820
square unit will more or less cancel, and
P into b, so force, distance. So, it should
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00:08:56,820 --> 00:08:58,510
be MB.
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00:08:58,510 --> 00:09:06,000
Now, this MB if you put in any one of the
equation, equation two or equation one, or
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00:09:06,000 --> 00:09:15,111
again you just rearrange equation one and
two, cancel MB; you will get MA. And MA if
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00:09:15,111 --> 00:09:24,990
you calculate, you can check it; so it will
be simply MA; it will be Pab l square; the
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00:09:24,990 --> 00:09:34,839
square of a square will be shifted to b square.
So, that part you agree with me and if you
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00:09:34,839 --> 00:09:41,029
have some doubt, you can check it; rather
you should check it; it will be a good exercise.
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00:09:41,029 --> 00:09:47,500
So, MA will be Pab square by l square. So,
square will be also there.
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00:09:47,500 --> 00:09:57,490
Now, if the a and b, they are identical it
should be equal to l by 2. So, if a equal
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00:09:57,490 --> 00:10:00,810
to b
equal to l by 2. In that case, your MA will
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00:10:00,810 --> 00:10:08,930
be MB. And it will be ab; it will be equal.
So,
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it will be a cube, you can say, or you can
say it is b cube. So, both the side it will
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be a
cube, b cube or it is simply l cube divided
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by 8. So, l cube denominator l, it will cancel.
So, it will be pl square divided by 8; ab
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will be equal; so, a square b will be a cube;
a
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cube will be your l by 2. So, l cube 2 2 2
will be 8. So, cube this; this will not be
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your l
square ;it will be just l Pl by 8. Because
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it will be l cube l square, l square will
cancel; it
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00:11:00,430 --> 00:11:08,570
will be l. So, it should not be… it will
be pl by 8. So, power will not be 2, it will
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00:11:08,570 --> 00:11:20,269
be 1 or
you can write again your MA MB Pl divided
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by 8. So, force, length, moment divided by
8; this is one of the very standard expressions.
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If there is a beam; both end are fixed; we
sometimes defined that as a fixed beam; that
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is
a problem of fixed beam, means both the ends
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00:11:42,790 --> 00:11:52,529
are fixed and there is a load. At the center
if it is acting, so both the moment generate
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the support, we say fixed end moments. And
this fixed end moment is Pl by 8, under this
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condition. Now, if you take it fully
distributed load, in a similar way, we can
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00:12:05,670 --> 00:12:09,399
go ahead, and the value you will get omega
l
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00:12:09,399 --> 00:12:18,870
square divided by 12; fixed end moments.
So, some of the statically indeterminate a
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00:12:18,870 --> 00:12:29,010
problem having one degree of indeterminacy,
two degrees of indeterminacy - we have handle.
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00:12:29,010 --> 00:12:33,829
Now these problems are basically your
beam type of problems.
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00:12:33,829 --> 00:12:44,220
Now, let us come to a problem, which is little
bit different in look, not a beam. So, it
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00:12:44,220 --> 00:12:50,120
is
little bit frame type. So, beam, normally,
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00:12:50,120 --> 00:12:55,910
it is a straight member. So, frame definitely
it
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will have some change in orientation. Now,
the next step, if we take that type of
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problem, say, one of the most simple type
of frame problem is like this.
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So, there is a vertical member, and there
is one horizontal member. See, there is a
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00:13:49,170 --> 00:14:06,880
load
here; it is acting; it is P; say this length
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is l; for simplicity, say, this is also l;
and the
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00:14:23,050 --> 00:14:29,630
material is identical; it is E; cross section
of your section properties are identical;
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it is
also I; this is also I; it is l; it is l;
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00:14:34,540 --> 00:14:35,770
this part is fixed; there is a roller support;
this side there
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00:14:35,770 --> 00:14:51,149
is a load, it is l, and that is l.
So, you just think in that manner; this is
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a bar. So, we have tried to change its orientation,
just. So, it was straight; now it is like
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that. So, it is moving like this; now changing
its
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00:15:07,350 --> 00:15:14,230
orientation. So, it is not just connected
by means of a pin. So, it is same member.
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So,
here the connection is a fixed type of connection.
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We say it is a rigid type of connection.
There the idea is the angle initially it is
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90 degree; if there is a rotation of this
member vertical member, the horizontal member
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will rotate by the same angle. So, the relative
angle it will be unaffected or unchanged.
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So, whole thing will rotate by same amount.
We say it is a rigid joint.
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00:15:46,899 --> 00:15:54,020
Normally, any structural joint, we get some
plate, some welding, some riveting. So, it
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is
a solid type of connections. And as a whole
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it will try to rotate in a similar manner.
So,
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we define it is a rigid type of joint. So,
here there is a member. So, there is no
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discontinuity is there; there is no a pin
type of connection, so that it can easily
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rotate. So,
it cannot fly and change their relative angle.
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So, the angle would be identical. And that
type of problem we say it is a typical frame
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problem. So, frame and beam, basic
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difference is beam members are all in line;
frame, the orientation may be different. So,
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00:16:37,589 --> 00:16:41,100
frame we can say it is a three-dimensional
frame, two-dimensional frame.
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00:16:41,100 --> 00:16:46,360
So, if we take a frame, three-dimensional,
means member may be vertical, horizontal;
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it
may go to this side; it may go to inclined
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direction. So, it will be a total threedimensional
general type of frame. So, here at least members
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00:16:53,350 --> 00:16:58,949
are in the plane. We can
say it is a plane frame or two-dimensional
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frame. And this joint is a rigid type of joint.
And there is a load, we want to find out the
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00:17:07,030 --> 00:17:18,220
stresses, bending moment, shear force everything
of that particular frame.
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Now, first of all, this is a fixed support.
So, we have three reaction components, plus
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there is a roller, and this roller will have
one reaction. So, basically one extra unknown
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00:17:32,140 --> 00:17:36,450
is
there. If the roller is not there, so due
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to this load, it should be a rather a cantilever.
And
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this end was just extra. So, it is a fixed
one; something is connected; there is a load;
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00:17:48,210 --> 00:17:50,170
it
should undergo some bending, the vertical
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member; horizontal member should be just
there as an attachment. But here, it will
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try to bend, and when it will try to bend,
horizontal member will try to follow that;
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follow that means, it will it will try to
move in
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the right side, that is allowed. But here,
there will be some slope; this member when
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00:18:11,830 --> 00:18:15,100
it
will bend, there will be a slope at the end;
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00:18:15,100 --> 00:18:19,400
and this member, it will follow that slope,
means this end will try to come down, for
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00:18:19,400 --> 00:18:24,500
this member.
When it will try to come down, automatically
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00:18:24,500 --> 00:18:31,309
the reactive force will be generated. So,
in
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the form of physical deform, if you think
in a physical way, the deformation of the
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00:18:35,730 --> 00:18:41,570
structure. So, it will try to come down, some
reaction will be regenerated. Or other way
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00:18:41,570 --> 00:18:48,340
you can think, there is a force, there will
be a reaction force, and this reaction force
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00:18:48,340 --> 00:18:52,320
will
makes generate some moment, and that moment
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00:18:52,320 --> 00:18:59,750
some part will be taken care by the
support and some part, because there will
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be equal and opposite reactions, and that
part
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00:19:02,760 --> 00:19:10,080
will take care the remaining part of the moment.
So, whole problem is the force system is integrated
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00:19:10,080 --> 00:19:15,140
inside; you cannot say, this will be
this one or other part will be 0. So, everything
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00:19:15,140 --> 00:19:19,510
will be associated. Now, the main
difficulty here it is a statically indeterminate
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00:19:19,510 --> 00:19:25,650
problem, and any one of the component you
can take as a additional unknown, and that
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00:19:25,650 --> 00:19:27,429
part you can find out from the deformation
of
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00:19:27,429 --> 00:19:28,909
the structure.
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00:19:28,909 --> 00:19:34,700
Now, one of the simplest way is this roller
part you can remove and put a reactive force.
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00:19:34,700 --> 00:19:40,850
And that reactive force you can find out from
the deformation of that point, because it
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00:19:40,850 --> 00:19:42,620
is
a support, at least vertical deformation will
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00:19:42,620 --> 00:19:48,190
be 0. Now, if we say if it is A, and if it
is B, if
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00:19:48,190 --> 00:19:57,850
it is C. Now, that problem can be drawn in
that manner.
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00:19:57,850 --> 00:20:06,630
So, I am just putting with a line diagram;
not giving the thickness of that. So, there
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00:20:06,630 --> 00:20:16,900
will
be a P and there will be a reaction Rc. So,
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00:20:16,900 --> 00:20:21,540
this is just represented in that manner. So,
here
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00:20:21,540 --> 00:20:29,399
it is fixed; there is a P; there is a Rc.
And Rc, this force, is not known; if that
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00:20:29,399 --> 00:20:36,179
is known,
we can find out everything, because it is
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00:20:36,179 --> 00:20:38,169
this end is now free end, I think. So, it
is just a
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00:20:38,169 --> 00:20:52,210
cantilever; only it is changing its direction.
Now, this part of problem is basically two
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00:20:52,210 --> 00:20:55,280
forces are acting; one is P, another is Rc.
So,
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00:20:55,280 --> 00:21:03,400
effect of P and Rc we can handle in a different
manner, and we can try to compare the
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00:21:03,400 --> 00:21:10,580
deflection at this end, and it should be equal
to 0, because here there is a support. So,
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00:21:10,580 --> 00:21:21,310
these two parts if we take separately.
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00:21:21,310 --> 00:21:51,890
So, this is P, and here, if we add, now our
basic problem was like this: this is P and
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00:21:51,890 --> 00:21:56,510
Rc;
this is clamped; it is free end. Now, that
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00:21:56,510 --> 00:22:01,880
is only P and Rc, in two steps. Now, if we
apply
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00:22:01,880 --> 00:22:08,060
P, what will we have? Only this vertical member
will have a cantilever mode of
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00:22:08,060 --> 00:22:22,910
deformation. So, I can draw it in a different
colour. So, it will be like this, and this
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00:22:22,910 --> 00:22:39,980
member, it will come like that. So, here,
due to this P, there will be a horizontal
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00:22:39,980 --> 00:22:49,980
displacement plus a rotation, and that rotation
we know, that rotation is simply Pl by 2
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00:22:49,980 --> 00:22:58,789
EI. So, that slope will be this slope; this
slope and this slope will be identical, because
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00:22:58,789 --> 00:23:06,130
initially it was 90 degree, it will be 90
degree. So, whole thing will be rotated; vertical
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00:23:06,130 --> 00:23:14,370
member will have a rotation about vertical
axis and same angle will be produced by the
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00:23:14,370 --> 00:23:27,169
horizontal member. So, this angle and this
angle is basically your theta B. And it is
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00:23:27,169 --> 00:23:49,320
your
Pl by 2 EI. I think it is Pl square or Pl?
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00:23:49,320 --> 00:23:50,774
So, P into l into l half of that it will Pl
square by 2
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00:23:50,774 --> 00:24:01,630
EI, that will be the theta B, theta B. So,
this end we will undergo a vertical deformation;
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00:24:01,630 --> 00:24:16,909
that is your delta, or due to your P, this
theta B, it is basically due to P, and delta
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00:24:16,909 --> 00:24:21,540
P will
be how much? So, this angle multiplied by
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00:24:21,540 --> 00:24:28,480
length, because by horizontal, vertical - both
the members - are having the same length.
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00:24:28,480 --> 00:24:36,200
So, this part should be equal to your Pl cube
divided by 2 EI.
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00:24:36,200 --> 00:24:46,950
Pl square by 2 EI is the slope. So, slope
into length, slope into length, will be the
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00:24:46,950 --> 00:24:57,440
vertical
deformation. So, it is a vertical member,
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00:24:57,440 --> 00:25:04,010
there is a load. So, this end we have a rotation
of theta B due to P of defining is theta Bp
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00:25:04,010 --> 00:25:06,970
Pl square 2 EI and that horizontal member
it
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00:25:06,970 --> 00:25:18,140
will just follow that angle. So, slope into
l will be the vertical deformation. So, it
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00:25:18,140 --> 00:25:22,730
will be
Pl cube by 2 EI. Slope is 2; deflection is
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00:25:22,730 --> 00:25:27,010
3. So, here the deflection will be Pl cube
by 3 EI
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00:25:27,010 --> 00:25:36,090
- here to here - and this slope is Pl square
by 2 EI, the slope into that length.
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00:25:36,090 --> 00:25:46,399
Now the next case, next case this force it
will give a cantilever type of deformation
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00:25:46,399 --> 00:25:52,010
of
this member. If we assume this is rigid, means
185
00:25:52,010 --> 00:25:59,050
there is no deformability of the vertical
member, so there will be a cantilever deformation.
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00:25:59,050 --> 00:26:05,200
Now, second level if you take the
deformability of that, so this force will
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00:26:05,200 --> 00:26:07,299
be transmitted here plus there is a moment.
So,
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00:26:07,299 --> 00:26:10,929
there is a moment; that moment will give the
cantilever deformation of the vertical
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00:26:10,929 --> 00:26:20,090
member. So, both the deformation will be combined
together and we will get the final
190
00:26:20,090 --> 00:26:30,171
deformation here. So, here deformation will
be upward; here it will be downward; and
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00:26:30,171 --> 00:26:32,650
ultimately it will match.
Now, the cantilever deformation of the vertical
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00:26:32,650 --> 00:26:44,700
member it will be more or less like this.
So, if we take this is a rigid member, so
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00:26:44,700 --> 00:26:49,409
Rc if we cut here, so here we will get a force
Rc
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00:26:49,409 --> 00:26:57,070
upward plus there is a moment, and due to
that moment, there will be a cantilever of
195
00:26:57,070 --> 00:27:14,020
moment. Now, if we take this is, this part
is not under deformation. So, here there will
196
00:27:14,020 --> 00:27:19,662
be
a slope, or about this, there will be a slope;
197
00:27:19,662 --> 00:27:28,149
so, this slope will be equal to this slope.
And
198
00:27:28,149 --> 00:27:41,419
what will be that slope? That slope we can
say it is theta at B due to say that reaction.
199
00:27:41,419 --> 00:27:47,600
And that reaction will be this Rc into l that
will be the moment; that moment will
200
00:27:47,600 --> 00:27:59,990
generate some slope. So, moment is your…
it will be… let me rewrite here. This will
201
00:27:59,990 --> 00:28:00,990
be
202
00:28:00,990 --> 00:28:07,159
equal to your Rc into l; that will be the
moment. So, it will be, if there is a moment
203
00:28:07,159 --> 00:28:12,340
there,
so, it will be, slope will be how much? Ml
204
00:28:12,340 --> 00:28:15,730
by EI. If there is a cantilever, if there
is a
205
00:28:15,730 --> 00:28:19,700
moment, so entire beam will have a uniform
moment. So, M by EI diagram will be M by
206
00:28:19,700 --> 00:28:27,990
EI. So, area will be Ml by EI. So, it will
be Ml divided by EI; that will be the theta
207
00:28:27,990 --> 00:28:37,980
here;
this angle, this angle, and if that is the
208
00:28:37,980 --> 00:28:46,910
angle, so this upward displacement will be
how
209
00:28:46,910 --> 00:28:59,590
much? That into l. So, this value, say, it
will be, say, delta; we have written in terms
210
00:28:59,590 --> 00:29:05,289
of P,
that will be in terms of R, and say it is
211
00:29:05,289 --> 00:29:08,860
one part. We have not taken the deformation
of
212
00:29:08,860 --> 00:29:18,810
that member. So, this part will be this Rc
already l l’s, it will be l cube divided
213
00:29:18,810 --> 00:29:29,890
by EI.
Now, if we add the deformation of the member
214
00:29:29,890 --> 00:29:58,649
here. So, you try to get the two
deformation. First of all, say, this part
215
00:29:58,649 --> 00:30:03,250
is not under deformation, say, assume for
the time
216
00:30:03,250 --> 00:30:11,360
it is rigid. So, this R will be shifted here
in the form of axial force plus irritation;
217
00:30:11,360 --> 00:30:14,110
axial
force plus a moment. So, moment will be Rc
218
00:30:14,110 --> 00:30:21,660
into l. So, if there is a cantilever, subjected
to a moment, so slope produce will be Ml by
219
00:30:21,660 --> 00:30:26,169
EI.
So, Ml, it will be Rc into l or Rcl square
220
00:30:26,169 --> 00:30:33,210
by EI will be the slope here. That slope into
distance; that will be Rcl l into l l squared
221
00:30:33,210 --> 00:30:37,799
into l l cubed divided by EI; that will be
delta
222
00:30:37,799 --> 00:30:45,250
R for this moment component. Now, it if it
is rigid, it will be like this - a straight
223
00:30:45,250 --> 00:30:49,429
line, but
it is not rigid; it it has also flexibility.
224
00:30:49,429 --> 00:30:53,549
Now, here to here, this load, it will give
a
225
00:30:53,549 --> 00:31:01,340
cantilever type of deformation. So, that is
delta R dash and this delta R dash - this
226
00:31:01,340 --> 00:31:14,480
part will be just your Rc; I think it is l
cube divided by 3 EI. So, delta R will be
227
00:31:14,480 --> 00:31:20,330
delta R dash
plus delta R double dash. So, we can write
228
00:31:20,330 --> 00:31:32,890
here delta R.
Now, delta R dash already we have written
229
00:31:32,890 --> 00:32:01,409
it is Rc l cube divided by EI. And here, this
part, Rcl cube divided by 3 EI. Now, if we
230
00:32:01,409 --> 00:32:05,850
add that, this will give, so it is Rcl cube
by EI,
231
00:32:05,850 --> 00:32:24,350
so one-third. So, it will be 4 Rcl cube 3
EI. So, this delta is the total delta R. So,
232
00:32:24,350 --> 00:32:28,659
delta R
it is upward and delta P it is downward. So,
233
00:32:28,659 --> 00:32:34,230
delta P is due to P and delta R due to Rc.
So,
234
00:32:34,230 --> 00:32:40,419
both the deformation should be and it should
be, because it will be equal and opposite,
235
00:32:40,419 --> 00:32:50,230
cancel each other. So, this value and this
value, we can just compare, and from there,
236
00:32:50,230 --> 00:32:53,420
we
will get the value of reactive force.
237
00:32:53,420 --> 00:33:01,600
Now, in the next step, so if we take this
data and try to match your delta P and your
238
00:33:01,600 --> 00:33:08,419
delta
R. So, your delta P should be equal to delta
239
00:33:08,419 --> 00:33:13,200
R. And delta P we have written here; it is
Pl
240
00:33:13,200 --> 00:33:32,230
cube 2 EI; that is equal to 4 Rcl cube divided
by 3 EI. So, Rc part we can write here 3;
241
00:33:32,230 --> 00:33:37,940
it
will go there. So, it will be 3 divided by
242
00:33:37,940 --> 00:34:00,460
8, EI will cancel. So, it will be just P.
If I show you the actual problem. So, P if
243
00:34:00,460 --> 00:34:10,090
we apply there will be a reaction of three-eight
of the P. Now, if there is three-eight of
244
00:34:10,090 --> 00:34:18,940
the P, there will be a force downward to balance
this. When these two will give a moment and
245
00:34:18,940 --> 00:34:24,890
P there will be a reactive force here.
Automatically, if you calculate all the moments,
246
00:34:24,890 --> 00:34:32,260
that will be the moment here or we can
say if you start from there, there is a reaction
247
00:34:32,260 --> 00:34:35,630
force you can calculate bending moment at
any point.
248
00:34:35,630 --> 00:34:43,510
So, Rc into l that will be the moment here;
that is this way; and if you follow that
249
00:34:43,510 --> 00:34:51,740
direction minus P into x, some reverse way
moment will be starting, generated from this
250
00:34:51,740 --> 00:34:57,210
end. And here there will be a moment, it should
match with the support moment
251
00:34:57,210 --> 00:35:03,130
generated at A. So, bending moment anywhere
you can find out; shear force also you can
252
00:35:03,130 --> 00:35:11,710
find out, because shear force is this force
Rc, Rc here. And here it will be only P, but
253
00:35:11,710 --> 00:35:17,060
there will be axial force also in the system,
because this reaction, here it will generate
254
00:35:17,060 --> 00:35:18,860
axial force, Rc part.
255
00:35:18,860 --> 00:35:25,660
Because there will be a reaction. So, Rc there
will be some member. This member there
256
00:35:25,660 --> 00:35:30,330
will be no axial force, because there is no
resistance in that end. So, anywhere if you
257
00:35:30,330 --> 00:35:34,800
cut,
there will be no axial force, but vertical
258
00:35:34,800 --> 00:35:40,280
member some axial force will be there
So, normally in a frame problem, we get bending
259
00:35:40,280 --> 00:35:47,110
moment, shear force, and axial force,
and deformation. Normal beam problem, we get
260
00:35:47,110 --> 00:35:53,230
shear force, bending moment, and we
get the deformation, deflection, and slope,
261
00:35:53,230 --> 00:35:56,870
but when it is a frame problem, in a three,
in a
262
00:35:56,870 --> 00:36:01,280
two-dimensional manner, so any point may undergo
horizontal moment, vertical
263
00:36:01,280 --> 00:36:06,620
moment, as well as a rotation. So, beam there
is no horizontal moment, but here there is
264
00:36:06,620 --> 00:36:10,270
a
horizontal moment; there is a vertical moment
265
00:36:10,270 --> 00:36:15,830
plus rotation. So, there are three
displacement components plus there are three
266
00:36:15,830 --> 00:36:21,080
force components - bending moment,
shear force, axial force. Beam it is only
267
00:36:21,080 --> 00:36:26,050
bending moment, shear force, and two
displacement component; displacement - vertical
268
00:36:26,050 --> 00:36:39,870
displacement plus rotation.
So, we have tried to change our problem a
269
00:36:39,870 --> 00:36:45,160
little bit from beam. Earlier, mostly we have
handled beam type of problem. In handling
270
00:36:45,160 --> 00:36:52,760
a determinate problem through differential
equation technique or moment area method,
271
00:36:52,760 --> 00:36:57,790
normally, beam type of problem we have
handled. Here also at the beginning we have
272
00:36:57,790 --> 00:37:02,530
taken all beam. So, this is one of the case,
where we are going to take a frame type of
273
00:37:02,530 --> 00:37:07,010
problem.
Now, in a real case, you will not get everything
274
00:37:07,010 --> 00:37:13,861
in beam. If we take any problem, say,
wave frame is also a frame; wave frame at
275
00:37:13,861 --> 00:37:20,900
least you have the idea; or any framing system
it will have some horizontal components, some
276
00:37:20,900 --> 00:37:27,350
vertical component; may be some
inclined component. So, it will be basically
277
00:37:27,350 --> 00:37:32,401
a structural system where there is no
guarantee that all the member will be like
278
00:37:32,401 --> 00:37:37,400
a beam. Beam we take, because that is a very
simple type of problem to explain the basic
279
00:37:37,400 --> 00:37:42,410
understanding. But actual system it will be
a
280
00:37:42,410 --> 00:37:50,120
frame; most simple case it will be plane frame;
much more complicated case it will be a
281
00:37:50,120 --> 00:38:00,910
space frame; three-dimensional frame.
Now, this period and last period, we are more
282
00:38:00,910 --> 00:38:06,990
or less trying to handle a problems
statically indeterminant type of structure.
283
00:38:06,990 --> 00:38:14,530
And we have seen the number of reactive force
are more and we are trying to utilize the
284
00:38:14,530 --> 00:38:20,790
deformation of the structure. And whatever
understanding we had, for the deformation
285
00:38:20,790 --> 00:38:23,471
of beam under simple cases - a cantilever,
a
286
00:38:23,471 --> 00:38:29,930
simply supported case, a load fully distributed
load, or a point load at the end, or at the
287
00:38:29,930 --> 00:38:37,930
middle - we are trying to get standard relations
to or get the overall deformation of the
288
00:38:37,930 --> 00:38:42,030
structure from there we are trying to manipulate
deformation due to this force,
289
00:38:42,030 --> 00:38:46,860
deformation due to the reaction, try to compare,
from there the reactive force we are
290
00:38:46,860 --> 00:38:51,010
trying to find out.
Now, the activity, whatever we have done here
291
00:38:51,010 --> 00:38:56,690
with two members, now you try to think,
if we add another member here, or if there
292
00:38:56,690 --> 00:39:03,450
is a inclined member, so whole problem will
be much more complicated, because the effect
293
00:39:03,450 --> 00:39:10,360
of reaction, so it is shifted in two manner.
So, deformation of that, plus it is shifted
294
00:39:10,360 --> 00:39:15,860
in the forward moment, then some slope, slope
will be reflected in the form of displacement.
295
00:39:15,860 --> 00:39:22,670
So, if you go on adding the members, it
will be very difficult to handle in that manner.
296
00:39:22,670 --> 00:39:32,390
So, we have to, theoretically speaking, we
are talking about different methods, and we
297
00:39:32,390 --> 00:39:37,530
can think with one method I can solve
everything - theoretically you can solve everything;
298
00:39:37,530 --> 00:39:44,670
but it will be very, very difficult job
to solve everything with a single method.
299
00:39:44,670 --> 00:39:51,490
So, depending on the problem, some method
may be much more suitable, and we are
300
00:39:51,490 --> 00:39:55,660
supposed to take that type of method, because
it will be much more convenient to handle
301
00:39:55,660 --> 00:40:04,140
that type of problem, with this typical method
of technique.
302
00:40:04,140 --> 00:40:10,950
Now, in that respect, if we try to handle
little complicated problem, a frame type of
303
00:40:10,950 --> 00:40:16,770
problem or beam little bit more, and if you
do not want to remember all this - Pl cube
304
00:40:16,770 --> 00:40:23,400
divided by 3 EI and all those, there is one
technique called energy method. And energy
305
00:40:23,400 --> 00:40:30,660
method is a very, very powerful method. In
many cases, we try to use that energy
306
00:40:30,660 --> 00:40:34,910
method, and try to get the deformation of
the structure, because statically indeterminate
307
00:40:34,910 --> 00:40:39,910
problem, means you have to handle the deformation
of the structure.
308
00:40:39,910 --> 00:40:46,950
So, this part, I have clarified earlier also.
In some cases we want to know what is the
309
00:40:46,950 --> 00:40:53,690
deformation at that point, but that type of
requirement is not very frequent; rather,
310
00:40:53,690 --> 00:40:57,940
we are
much more interested for finding out the bending
311
00:40:57,940 --> 00:41:03,430
moments, shear force stresses. But
statically indeterminate problem, we cannot
312
00:41:03,430 --> 00:41:09,630
find out the bending moment, shear force,
unless we get the reactions. To handle that
313
00:41:09,630 --> 00:41:14,630
we have to apply the knowledge of
deformation of the structure. So, that was
314
00:41:14,630 --> 00:41:21,780
the basic reason for going through the
deformation of the structure. Indeterminate
315
00:41:21,780 --> 00:41:24,770
form, then we just utilize that for finding
out
316
00:41:24,770 --> 00:41:32,970
the reaction for indeterminate type of case.
Now, I want to take the energy method. Energy
317
00:41:32,970 --> 00:41:42,350
method is much more useful, much more
rather suitable for handling little different
318
00:41:42,350 --> 00:41:44,310
type of problem - little complex type of
319
00:41:44,310 --> 00:41:51,620
problem. And there one thing it may be beneficial
for you, that you need not remember
320
00:41:51,620 --> 00:41:57,850
all this standard expressions. So, you have
to just write down the expression of the
321
00:41:57,850 --> 00:42:01,590
bending moment for the different segment.
You have to perform some integration there.
322
00:42:01,590 --> 00:42:07,170
You have to get the energy expression, you
have to take the derivative; from there, you
323
00:42:07,170 --> 00:42:11,200
can get the deflection slope at different
places.
324
00:42:11,200 --> 00:42:20,320
So, our next job we can start energy method,
but first we will try to apply with a
325
00:42:20,320 --> 00:42:26,340
statically determinate determinant type of
structures, but we have some understanding
326
00:42:26,340 --> 00:42:31,100
which we can handle easily. So, once that
understanding will be there, that problem
327
00:42:31,100 --> 00:42:33,220
can
be extended to a statically indeterminate
328
00:42:33,220 --> 00:42:35,320
type of case.
329
00:42:35,320 --> 00:43:03,890
Now, this energy method is our next objective.
Now, what is this energy? Basically, here
330
00:43:03,890 --> 00:43:12,580
we will be handling strain energy. Now what
is strain energy? So, let us discuss little
331
00:43:12,580 --> 00:43:18,860
bit,
very basic component of energy. Now, I can
332
00:43:18,860 --> 00:43:31,430
take a simple bar. I am sure you have the
idea of that. If you apply P, if the length
333
00:43:31,430 --> 00:43:36,220
is l, area is A, material property is E, so
there
334
00:43:36,220 --> 00:43:47,230
will be a elongation of that member; that
elongation will be Pl by AE. So, this is P;
335
00:43:47,230 --> 00:43:53,550
this is
l by AE; that will be the elongation.
336
00:43:53,550 --> 00:44:05,190
Now, this load P, if we apply from 0 and gradually
if we increase. So, it is starting from
337
00:44:05,190 --> 00:44:10,540
0, 1, 2, 3, 4 like that if we increase. So,
delta initially it will be 0; gradually it
338
00:44:10,540 --> 00:44:13,890
will
increase. So, if we plot the relationship
339
00:44:13,890 --> 00:44:17,941
between P and delta, so P is load; delta is
the
340
00:44:17,941 --> 00:44:24,530
deflection. Load-deflection curve for a linear
system, we are supposed to get a
341
00:44:24,530 --> 00:44:37,180
relationship like this. So, this side is P
and this side is delta. So, if P equal to
342
00:44:37,180 --> 00:44:41,470
0; delta will
be equal to 0. So, you increase the load,
343
00:44:41,470 --> 00:44:44,750
delta will increase, because lAE - these are
fixed
344
00:44:44,750 --> 00:44:51,140
parameters. So, you go on increasing the load,
it will follow a line. So, at any instant,
345
00:44:51,140 --> 00:45:02,340
say, that is the P, and under this load P,
the deflection is delta. Now, if I slightly
346
00:45:02,340 --> 00:45:12,530
increase
that load from P to delta P. So, delta will
347
00:45:12,530 --> 00:45:23,950
have a little change of d delta. So, here,
say, it
348
00:45:23,950 --> 00:45:36,180
is d delta and here also it is dp. So, here
to here it is P, it is dp; and here delta,
349
00:45:36,180 --> 00:45:38,420
it is d
delta.
350
00:45:38,420 --> 00:45:48,740
So, under P deflection was delta. So, if I
just increase little bit. So, there will be
351
00:45:48,740 --> 00:45:52,780
a slight
change in deflection. So, there P will do
352
00:45:52,780 --> 00:45:59,401
some work. So, at the end there is a P. So,
if you
353
00:45:59,401 --> 00:46:06,880
slightly increase this end, we will slightly
shift by d delta. So, what will be the small
354
00:46:06,880 --> 00:46:17,160
work? That d work, we can say, that small
incremental work, it will be P into your d
355
00:46:17,160 --> 00:46:27,480
delta. P is already there; delta is already
there. So, there is a further moment. So,
356
00:46:27,480 --> 00:46:32,930
this
further moment is P is the force and d delta
357
00:46:32,930 --> 00:46:36,970
is the moment. So, that will be the work.
And
358
00:46:36,970 --> 00:46:43,210
this Pd delta is nothing but the area of this
particular state, because it is d delta and
359
00:46:43,210 --> 00:46:45,160
that
part is P.
360
00:46:45,160 --> 00:46:54,390
Now, you can say this point is P; this point
is not P; it is P plus dp. So, which one we
361
00:46:54,390 --> 00:47:03,350
should take? Someone may suggest we should
take the average one: P plus dp… P plus
362
00:47:03,350 --> 00:47:05,780
P
plus dp divided by 2. So, it will be P plus
363
00:47:05,780 --> 00:47:13,190
dp by 2; but this dp normally we take this
incremental concept in a very limiting manner,
364
00:47:13,190 --> 00:47:19,840
means they will tend to 0.
So, compared to P, this dp or dp plus dp by
365
00:47:19,840 --> 00:47:23,880
2 will be a very small quantity. So, if it
is
366
00:47:23,880 --> 00:47:34,210
take 100 and 100.00005 or 1, so that 00001
that part we can drop. So, it is in a
367
00:47:34,210 --> 00:47:41,150
incremental manner more or less is a area
under the curve. So, what we do? We take a
368
00:47:41,150 --> 00:47:45,510
strip. So y into dx. So, y; this end and that
end there is a difference, but within limiting
369
00:47:45,510 --> 00:47:51,320
manner that is tending to a point. So, it
is also like this; that is, there is a P and
370
00:47:51,320 --> 00:47:56,050
there is a
d delta. Now, if we just increase by another
371
00:47:56,050 --> 00:48:01,750
small amount, there will be another strip;
another strip like this. Or if we start working
372
00:48:01,750 --> 00:48:10,080
from the beginning, there will be a number
of strips like this. So, that will be, basically,
373
00:48:10,080 --> 00:48:17,500
our work. So, it is the total work; it will
be
374
00:48:17,500 --> 00:48:24,110
half P into delta, say, at any P deflection
is delta, it will be half P into delta, because
375
00:48:24,110 --> 00:48:26,470
this
is P, and this is delta, this is triangular
376
00:48:26,470 --> 00:48:34,160
part will be half of that.
Usually we can think the load is P and deflection
377
00:48:34,160 --> 00:48:40,000
is delta. So, what should be P into
delta? So, it is not like this, because P
378
00:48:40,000 --> 00:48:45,650
it is not applied from the beginning its full
value; it
379
00:48:45,650 --> 00:48:54,030
is starting from 0, say P is 10, and delta
is 2. So, 10 it is not applied at the beginning.
380
00:48:54,030 --> 00:48:57,980
So,
it is starting from 0 to 10. So, average value
381
00:48:57,980 --> 00:49:03,640
is 5; and that average value, that
displacement we are getting 5. So, we can
382
00:49:03,640 --> 00:49:06,830
say 10 into that value displacement divided
by
383
00:49:06,830 --> 00:49:15,630
2 half, means we are basically averaging it,
because it is following a line. So, here,
384
00:49:15,630 --> 00:49:18,560
it is a
line, if you take a non-linear curve, basically
385
00:49:18,560 --> 00:49:25,030
area under that curve will be that total
work, and that work will be stored in the
386
00:49:25,030 --> 00:49:30,550
form of energy, and that energy will stored
in
387
00:49:30,550 --> 00:49:35,340
straining the member. So, we say it is strain
energy. And if we release that load, that
388
00:49:35,340 --> 00:49:41,460
strain energy will be released. So, it will
help to bring back to the original level.
389
00:49:41,460 --> 00:49:56,530
So, this
part we say it is U or strain energy. So,
390
00:49:56,530 --> 00:50:03,210
U - this total work - we are defining as U
and this
391
00:50:03,210 --> 00:50:10,910
is called strain energy.
So, if there is a load, there is a deflection
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00:50:10,910 --> 00:50:16,120
d delta half will be the energy, because the
load
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00:50:16,120 --> 00:50:24,810
will act gradually from a 0 value to a certain
value. Now, we can make an argument that
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00:50:24,810 --> 00:50:31,130
- who is going to put the load in a very slow
manner, in a gradual manner? So, you just
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00:50:31,130 --> 00:50:35,980
put there; suddenly applied load. If there
is a beam, suddenly if you apply a load, so
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00:50:35,980 --> 00:50:42,990
whole problem will be a dynamic problem. So,
if there is a beam, put the load there. So,
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00:50:42,990 --> 00:50:50,570
what will be happening? Suddenly there will
be a force and there will be no deformation.
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00:50:50,570 --> 00:50:54,970
So, some equilibrium will not be maintained;
automatically some motion will be
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00:50:54,970 --> 00:50:59,450
initiated. So, it will come to the equilibrium.
Then some velocity will be there. So, it will
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00:50:59,450 --> 00:51:06,030
cross that limit; go beyond that. After that
it will come to rest, but at that level
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00:51:06,030 --> 00:51:11,580
equilibrium is violated. Again it will start
come back. So, some vibration will be there
402
00:51:11,580 --> 00:51:12,850
in
the system.
403
00:51:12,850 --> 00:51:19,090
Now, due to the damping and all those, ultimately
it will die out, come to the static stage,
404
00:51:19,090 --> 00:51:25,710
but external loading you may apply suddenly,
but internal stresses it will generate
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00:51:25,710 --> 00:51:31,880
gradually. Any structure if you put the load
suddenly, stress will not be generated from
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00:51:31,880 --> 00:51:35,113
0
to 10 suddenly. So, it will gradually take.
407
00:51:35,113 --> 00:51:38,330
If it is a static problem, dynamic problem,
we
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00:51:38,330 --> 00:51:44,680
are trying to put this energy, P is the load,
and delta is the elongation. Actually this
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00:51:44,680 --> 00:51:48,420
is the
external phenomena, that work will be stored
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00:51:48,420 --> 00:51:54,080
in the form of strain energy inside.
Now, let us write in that manner, say, U is
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00:51:54,080 --> 00:52:03,640
equal to half P into delta. So, this part
it is
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00:52:03,640 --> 00:52:23,580
written in this manner, say, half P by A delta
by l A into l. So, it will be, we can write,
413
00:52:23,580 --> 00:52:36,100
half sigma epsilon into volume. P delta, P
we have divided by A, we are getting stress,
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00:52:36,100 --> 00:52:42,210
delta divided by l is the strength. So, A
and l unnecessarily we have taken at the
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00:52:42,210 --> 00:52:45,810
denominator; numerator we have to compensate
that. So, A into l is basically volume.
416
00:52:45,810 --> 00:52:53,120
So, half stress strain into volume; that will
be the total energy or this half into sigma
417
00:52:53,120 --> 00:52:59,090
into
epsilon that is called as strain energy density;
418
00:52:59,090 --> 00:53:04,070
means strain energy per unit volume. So,
multiplied by the volume will be the total
419
00:53:04,070 --> 00:53:05,860
energy.
420
00:53:05,860 --> 00:53:18,920
So, in a general case, we can write U is equal
to half sigma dv integral. So, this is a
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00:53:18,920 --> 00:53:24,670
problem where stress and strain are uniform
throughout. So, we are simply taking it is
422
00:53:24,670 --> 00:53:31,220
half sigma epsilon into v, but sigma epsilon
may vary point to point. So, in a general
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00:53:31,220 --> 00:53:37,550
case, we can say it is sigma epsilon dv. So,
it will be general integral, and this part
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00:53:37,550 --> 00:53:40,500
we
say it is intensity of the strain energy,
425
00:53:40,500 --> 00:53:49,210
and whole part is the strain energy. So, I
think we
426
00:53:49,210 --> 00:53:58,360
can drop it here. Just we have initiated the
strain energy in the form of simple extension
427
00:53:58,360 --> 00:54:05,450
of a bar. That idea will be much more generalized
and it will be applied to a structural
428
00:54:05,450 --> 00:54:26,110
problem in the subsequent classes.
Preview of the Next Lecture
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00:54:26,110 --> 00:54:40,540
Just we have started talking on this. The
technique is energy method; it is based on,
430
00:54:40,540 --> 00:54:46,600
basically, the concept of strain energy. The
strain energy we are trying to explain in
431
00:54:46,600 --> 00:54:50,520
the
form of a simple bar problem under tension.
432
00:54:50,520 --> 00:55:09,220
So, we took a bar in that form and applied
a load P. Now this is a case of a very simple
433
00:55:09,220 --> 00:55:15,590
stress problem, because entire bar will have
identical stress and strength. So, anywhere
434
00:55:15,590 --> 00:55:21,630
stress is P by A and strain will be your total
elongation divided by length of the member.
435
00:55:21,630 --> 00:55:27,270
So, it is a case of uniform strain, uniform
stress, whatever you can say.
436
00:55:27,270 --> 00:55:40,120
Now, the elongation we wrote in the last class,
it was Pl by AE, and we have drawn a
437
00:55:40,120 --> 00:55:49,440
curve like this, your load versus elongation
- straight line, because whole thing is a
438
00:55:49,440 --> 00:55:54,670
linear
system. And for any load, if you start applying
439
00:55:54,670 --> 00:56:01,520
load from 0 gradually increase go up to a
load P, that elongation will increase from
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00:56:01,520 --> 00:56:16,870
0 to delta, and the area under this load
deflection curve, we have found that it is,
441
00:56:16,870 --> 00:56:19,220
basically, the work carried out the force
P.
442
00:56:19,220 --> 00:56:25,390
Now, it is not straightaway P and delta, because
P is not constant throughout, because P
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00:56:25,390 --> 00:56:34,770
is starting from 0 to that P, and it is delta
is starting from 0 to delta. So, total work.
444
00:56:34,770 --> 00:56:41,360
So,
that we have put in the form of half P into
445
00:56:41,360 --> 00:56:50,540
delta. So, that work, it will be stored in
the
446
00:56:50,540 --> 00:56:55,790
form of some energy; we said strain energy;
that strain energy will be equal to this work
447
00:56:55,790 --> 00:57:02,890
done half P into delta. Now, here load, why
we are putting in a gradual manner? Because
448
00:57:02,890 --> 00:57:08,480
we want to maintain the static condition.
If we apply suddenly, a load whole thing will
449
00:57:08,480 --> 00:57:12,210
be a dynamic problem.
So, that will be much more difficult case.
450
00:57:12,210 --> 00:57:19,960
So, we want to understand with a very simple
type of problem. So, load in a static manner,
451
00:57:19,960 --> 00:57:24,220
if we want to apply, we have to apply
gradually. So, always there will be balance
452
00:57:24,220 --> 00:57:25,780
between internal external force system.
453
00:57:25,780 --> 00:57:31,320
Basically, this strain energy though it is
expressed in the form of P into delta, it
454
00:57:31,320 --> 00:57:36,370
is rather
the measure of some internal quantities; internal
455
00:57:36,370 --> 00:57:40,550
quantities, means there will be a
deformation that we have defined in the form
456
00:57:40,550 --> 00:57:48,350
of strain, and due to the strain, some stress
will be there. So, whole thing is a measure
457
00:57:48,350 --> 00:57:55,790
of the strain and stress inside. And you must
have remember we have written this one as
458
00:57:55,790 --> 00:58:00,240
this sigma epsilon into v, because that P
we
459
00:58:00,240 --> 00:58:06,430
have divided by area, epsilon divided by l.
So, it became stress, it became strain, and
460
00:58:06,430 --> 00:58:12,690
this
l and A it becomes v, that is the volume or
461
00:58:12,690 --> 00:58:16,670
sometimes we write it is U 0 v. So, U 0 is
this
462
00:58:16,670 --> 00:58:27,220
part. We sometimes say it is strain energy
density. So, half stress into strain is the
463
00:58:27,220 --> 00:58:31,910
strain
energy density multiplied by the volume will
464
00:58:31,910 --> 00:58:39,070
be the total energy. Now, this is a case of
uniform stress and strain. So, everywhere
465
00:58:39,070 --> 00:59:12,240
your sigma and epsilon is constant. For a
general case, we have written U equal to your
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00:59:12,240 --> 00:59:19,330
half.