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Today, we will discuss statically indeterminate
problem. Already we have discussed
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what is statically indeterminate problem,
earlier. Now, here our job is how to handle
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that
type of problem. Now, if I write structure
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in two forms, though we have written it earlier,
still we write here a structural problem.
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So, it is these two groups: statically determinate;
statically indeterminate. The division is
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made basically on the idea of equation of
statics. Any problem, there are some equations
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of statics. If we take a three-dimension problem,
we will have six equations of statics. If
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we take a plane problem, it will have three
equations of statics. In the space, summation
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of all the forces along xyz that will be 0;
plus moment about xyz, it will be 0. In plane,
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summation of force along x will be 0, along
y will be 0, plus moment about z means
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moment perpendicular to that plane - so, that
will be 0. So, these are the three basic
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equations.
Now, if we have a problem, at the basic unknowns,
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which normally we find in the form
of support reaction. If it is exactly equal
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to three, it is statically determinate; means
that
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problem or those unknowns can be determined
with the equation of statics. So, we define
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it as statically determinate. If the number
of unknowns, it is more that number of
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equations of statics, so that problem cannot
be solved. So, this problem is statically
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indeterminate. So, it cannot be determined
with the equation of statics.
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Say, if we take a beam problem like that;
here there is a hinge support; here there
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is a
roller support; there might be any load on
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that. Now, at the hinge there will be two
reaction components; so, one vertical and
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another horizontal. In the roller there will
be
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only one vertical reaction component. Now,
here we can say the horizontal component,
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it
will be 0. Normally, in beam problem, we take
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a loading all perpendicular to the axis of
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the beam; usually, we do not get any component
of force along horizontal direction along
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x. So, most of the cases we get it 0. But
we cannot provide that hinge support as a
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roller support, because if we put roller on
both the sides, loading, usually it is vertical.
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So, it will take care of both the problem,
but if there is a small component of horizontal
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force due to accidental reason, so structure
will not be stable in space; it will go on
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rolling on the two roller; so, it will not
be placed, where you want to keep it. So,
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for any
unknown or undesirable horizontal force, we
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have to prevent that, at least left hand or
right hand. So, it may be 0 or any small value
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depending on the type of load. So, it
should be there.
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So, theoretically we can say for a wind problem,
usually, we are not going to take any
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vertical load; all the time it will be 0.
So, that horizontal component, all the force,
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that
condition is automatically satisfied. So,
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from the three equations, normally, one equation
is automatically satisfied. So, other two
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equations are there; from there we can find
out
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the two reactions.
Now, here the two equations are: we can take
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summation of vertical force along y
direction; and we can take moment about this
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point a; or we can take moment about the
other point b. Now, if we have some third
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support here, definitely there will be three
reaction components plus one horizontal component.
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Normally out of three already one
equation we have taken care of, in order to
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handle the horizontal part. So, vertical and
moment part - two reactions we can determine.
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So, if we have three unknowns and two
equations we cannot find out. Normally the
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number of equations should be equal to the
number of unknowns; if it is more than that,
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it cannot be determined.
Now, I was talking about taking moment. Now,
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there are three forces: one option is we
can take moment at one point; and other option
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is, other part is we can take all the forces
in the y direction equal to 0; and alternative
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of that, say, if I write summation of y equal
to 0 and summation of moment at any point,
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say, if it is A, if it is B, if it is C, we
can
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take MA equal to 0.
Other option we can take we can take MA equal
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to 0; MB equal to 0; that also you can
do. Say, initially that support is not there.
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So, we can, if we take moment about this,
that
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will get; if we take moment about that, the
other reaction we will get. Now, this may
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give
some idea, that instead of taking force - all
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the forces - along y direction, we can take
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moment about MA and MB, we were getting two
equations. Now, we have three
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unknowns, but two equations we cannot solve.
So, there might be a temptation that let us
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take moment about C or there might be a number
of support, we can go on taking
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moment at different point and go on generating
a number of equations.
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So, it can be theoretically solved, but really
if you try to generate that equation, you
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will
find all the equations are not independent
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equations. Independent means, you will find
one equation is identical to the other equation.
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It may not be exactly identical, but if we
multiply with the factor minus 1 plus 1 plus
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2 minus 2, you will get the other equations
or two equations if you combine, you will
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get the third equation. Just I can give you
one
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simple example. Say 5 x plus 7 y equal to
10. If I have minus 10 x minus 14 y equal
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to,
say 3, anything actually; right side not necessary
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it will be equal. Now, the first equation
if you multiply with 2 and add, so x will
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cancel; y will also cancel. So, in that process,
you cannot find out neither x nor y; none
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of the quantities you can find out from this
system.
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So, what will be happening? There should be
a tendency, initially, we can take moment
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at any number of point, generate any number
of equation, and any number of unknown
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we can solve; but theoretically if you make
an attempt, we will more or less getting that
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type of equations. So, we will basically get
two independent equations here, I think, apart
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from the horizontal summation of the force.
So, two independent equations you will get,
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if you try to generate more, so those equations
will be basically coming out from these
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equations. May be, this plus this if you add
or this multiplied by 10 minus this multiplied
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by 5 add you will get the third equation or
the fourth equation. So, whole thing will
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be
not a independent system and equation cannot
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be solved.
So, our main idea is three equations and we
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may take all in the form of moment, but the
number of equations it will be just three,
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or horizontal part, if we just forget, it
will be
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only two equations in a moment form or one
moment, another just force form.
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Now, this problem, what I have? So, this problem,
what is drawn here? Here basically
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there three reactions and we can take equation
in this form or in that form. So, the third
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reaction, we cannot find out, because total
number is more.
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Now, here other problem can be solved. The
problem can be solved, in addition with that
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equation, we require some other equation,
say, three is the reaction, and equation of
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statics is two. So, we require another one
extra equation. And that equation will come
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from the deformation of the structure or deflection
of the structure. So, more or less we
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are trying to handle deflection of the structure
in last two, three classes. So, those ideas
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will be required here, and if we use those
equation in the form of structural deformation,
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structural deflection, we will get some information
in the form of additional equation.
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So, apart from these two, we will get the
third equation, from our structural deflection
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or
structural deformation. And that will help
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to find out the reaction of the structure.
Now,
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let us take one simple case, which is a very
popular type of problem.
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Say, this a cantilever, and here, this end
is supported. We take any load, say, it is
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distributer load throughout. This part is
l it is EI; you can take it is A; it is B.
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So, length
is l; this end is a clamp support; this is
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a roller support; fully distributed load;
uniform
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flexural rigidity EI. Now, how many components
of reactions are there? First of all, this
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clamp support will have horizontal component,
vertical component reaction; and here
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there is one reaction. So, 3 plus 1, 4; number
of equation is w; so, one unknown is more.
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So, we say degree of indeterminacy. So, degree
of indeterminacy is - how many extra?
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So, 3 is the normal number. So, here 1 extra
is there. So, degree of indeterminacy here
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it
is 1. Or if we just remove that support, it
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is a cantilever beam, There is a load and
that
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problem already we have solved, because it
was a determinate problem. Three reactions
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at one point; that is support A. Now due to
putting A support here, at the right hand
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B, a
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problem becomes statically indeterminate one.
And it cannot be solved with the equation
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of statics. We cannot determine the reaction.
And if we cannot determine the reaction
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anywhere, we cannot go and find out the shear
force and bending moment.
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So, there is one set of force; that is applied
load. Another set of force, it will come from
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your support, in the form of reactive force.
Once all those forces will be known, then
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only we can come to a particular place and
sum up all the force, we get shear force;
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or
we can take summation of all the force in
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the form of moment at the station, we will
get
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bending moment there. Now, the main job here
is to determine the reaction. Once that
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will be obtained, we can find out anything
inside.
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Now, how the problem can be handled? Now this
problem we can just write in this form.
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Say it is a cantilever problem, with that
load plus the roller here, it will give a
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reaction
RB. So, at B there is a roller; roller will
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give a reaction; reaction force is R at B;
RB is
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the reaction. And this force, normally, this
omega is the load per unit length intensity
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of
distributed load, that value will be a known
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value, some numerical value will be given,
because that is the external applied force.
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Now, this RB, this force will be supplied
by the support in the form of reaction, and
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that
force is unknown here. Now, this problem can
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be solved in two steps. So, this may be
divided as… Already we have talked about
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method of super position. If a structure is
subjected to more number of load, we can get
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the effect of the individual R component of
force, then we can combine their effect, through
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simple algebraic addition, that we say it
is method of super position, because whole
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thing is a linear system.
Now, here there are two loads; one is omega.
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So, first step we have put this distributer
load. And this RB we have removed; that we
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are putting in the second level. So, this
RB,
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that force, and this force, they are applied
separately. So, deformation of the beam
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subjected to w, and deformation of the beam
subjected to RB, if we combine we are
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supposed to get the deformation of the actual
structure. Now, due to omega, it will take
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a
shape like this; and due to RB it will take
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a shape like that.
Now, here, if we say this is your delta due
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to w, and this is basically delta due to RB;
these two values will be equal and opposite.
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Because in a real structure, joint B is
connected to a support. So, it cannot move
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vertically upward or downward. So, under w,
whatever w - delta w – we will get, that
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should be compensative by RB in the form of
w
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RB. So, if you would take it as positive,
it should be negative; if it is 2 millimeter
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it
should be minus 2 millimeter. Ultimately,
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whole thing if we combine, we should get
back to the original situation, where deflection
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at B should be 0.
Now, this delta w delta RB, the deflection
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values already we have determined. If we take
a cantilever, at the end if we apply a load,
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whatever deflection we will get or if it is
fully
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distributed case whatever will be the standard
value that already we have determined.
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Now, we can write this value here again, delta
w. So, it is fully distributed load and that
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value was, already we have determined, it
is omega l to the power 4 divided by 8 EI.
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So,
it will be omega l to the power 4 8 EI.
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Now, this force will be how much? It will
be RB L cube divided by 3 EI. Now these two
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deflections should be equal and opposite.
So, if we just equate these two equations,
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we
can find out RB in the form of your W, l,
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EI; something we will get from there I think.
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Now, next step if we just equate that. Your
delta due to w should be equal to delta due
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to
your RB and this we have written as… So,
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from there RB we will get, it will be this
l this
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3 divided by 8. So, EI will cancel; lq will
cancel; it will be l remaining; omega will
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be
there; 3 will go to that side, it is below.
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So, 3 by 8 omega l will be the reactive force
there. Now, if I draw the beam again. So,
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actual structure, it can be represented by
a
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simple beam without the support at the right
end, in lieu of that, we will supply the
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reactive force supplied by the support. So,
the reactions supplied by the support is RB
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and the value we have calculate this 3 omega
l by 8. So, total load is omega l on the
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beam and it will be three-eighth of that,
it will be coming in the form of reactive
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force
there. Now, at any point, if we try to find
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out the shear force, it will be your 3 wl
by 8
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minus omega into x or moment if you want to
take, so this 3 wl by 8 multiplied by x
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minus omega x square by 2.
So, this plus or minus, depending on the sign
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,effect of that can be taken as plus, so effect
of this will be minus or it may be reverse.
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So, or you can draw the bending moment
diagram of a beam, you can draw the shear
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force bending movement of the beam.
Finally, we may be interested for finding
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out the maximum stress. We will be interested
for finding out the maximum shear stress.
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So, all this quantity we can find out from
the
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structure.
So, the main bottleneck was number of reactive
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force was more and in a normal way we
could not proceed; naturally we had to take
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the help of deformation of the structure.
So,
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we have tried to match the deformation. So,
this technique sometimes we say it is a
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consistent deformation technique; means structure
will have a deformation. It will
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maintain some consistency, because there is
a support. So, deformation should be equal
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to 0. Or somewhere there might be a support,
left side, right side slope should be equal,
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or deflection should be equal. So some form
of deformation compatibility we have to
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establish. So, basically your, this displacement
part, we have to take care and we have to
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get a additional equation. Now, in the next
case, if we take a problem like this: say,
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let us
take a simply supported problem.
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This is fully loaded omega and I am making
it quite simple. So, l is the span; l is the
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span; AB. So, it is ABC, and this side l;
and this is l; fully loaded omega. Now, already
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that problem was defining, because there are
three vertical reactions plus 1 horizontal
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reaction, though that will be a 0, but we
have to take. So, 4 is the total number of
199
00:25:32,200 --> 00:25:42,260
equations; 3 equation of statics. So our extra
equations we require is that extra number
200
00:25:42,260 --> 00:25:44,770
of
unknowns. So, extra number of unknowns, though
201
00:25:44,770 --> 00:25:51,740
that we were defining in terms of
degree of statical indeterminacy. So, degree
202
00:25:51,740 --> 00:25:54,270
of statical indeterminacy, here it is one.
So,
203
00:25:54,270 --> 00:25:59,710
at least one additional information, in the
form of some equation, we should get for
204
00:25:59,710 --> 00:26:12,760
finding about these reactions.
Now, this problem can be handled like this.
205
00:26:12,760 --> 00:26:37,090
Now say at this level we remove the central
support. Now due to that we will get some
206
00:26:37,090 --> 00:26:56,580
deformation here. Now, we are not supposed
to remove the support at the middle. To take
207
00:26:56,580 --> 00:27:08,520
care, we have to supply a reaction, say Rc.
So, in actual case, this omega will be there
208
00:27:08,520 --> 00:27:12,441
plus there will be a reactive force that we
see
209
00:27:12,441 --> 00:27:20,170
due to this roller support. So, if we want
to remove that, we have to take the effect
210
00:27:20,170 --> 00:27:23,470
of that
in the form of some reactive force. So, that
211
00:27:23,470 --> 00:27:29,270
is Rc. So, that we are putting in the second
line. So, if we combine we are supposed to
212
00:27:29,270 --> 00:27:43,929
get the actual structure.
Now due to Rc, the beam should deform like
213
00:27:43,929 --> 00:27:57,540
this. Now, whatever deformation you will
get here, if we say this is delta for omega,
214
00:27:57,540 --> 00:28:01,990
and this is the deformation whatever we will
get delta for your Rc. Similar to your earlier
215
00:28:01,990 --> 00:28:11,290
case, this value and that value should match.
216
00:28:11,290 --> 00:28:31,690
So, if we write your delta…so this value
should be equal and opposite, and that opposite
217
00:28:31,690 --> 00:28:37,980
part that we have already drawn here. So,
we need not bother about the sign, because
218
00:28:37,980 --> 00:28:50,910
sign part is reflected through drawing. Now,
this delta omega, delta Rc, this values are…
219
00:28:50,910 --> 00:28:57,370
this is a simple determinant problem. A simply
supported beam, fully loaded, the centre
220
00:28:57,370 --> 00:29:02,820
deflection we know, already we have the expression.
Now, here also it is a simply
221
00:29:02,820 --> 00:29:09,490
supported beam with a centre load. And for
that problem, also we have the standard
222
00:29:09,490 --> 00:29:14,920
expressions. Some of the cases we have derived;
some of the cases, any time, we can
223
00:29:14,920 --> 00:29:21,059
derive with our basic understanding in the
form of differential equation technique or
224
00:29:21,059 --> 00:29:28,921
moment-area theorem or we can just consult
some standard table; from there we can
225
00:29:28,921 --> 00:29:40,690
pickup. So, what will be this value? Omega
l; omega l will be your 5 omega l to the your
226
00:29:40,690 --> 00:29:46,280
4 divided by 3 84 EI.
Only this l part here, it will be l and l,
227
00:29:46,280 --> 00:29:57,940
it will be 2 l. So, it will be 5 omega and
2 l 4
228
00:29:57,940 --> 00:30:30,260
divided by 384 EI. And that should be equal
to… this will be Rc 2 l cube divided by
229
00:30:30,260 --> 00:30:36,940
48
EI. So, it is EL by 48 here and this will
230
00:30:36,940 --> 00:30:41,680
be 5 omega l divided 384 EI; only l part in
this
231
00:30:41,680 --> 00:30:51,049
problem is equal to 2 l; just we have substituted
that. Now, from here, this Rc can be
232
00:30:51,049 --> 00:30:58,400
easily determined; just we have to put the
expressions. So, if we just simplify that
233
00:30:58,400 --> 00:30:59,400
equation.
234
00:30:59,400 --> 00:31:14,270
So, your Rc will be, it will be 5 omega, and
2 l part it will be q and 4 it will cancel;
235
00:31:14,270 --> 00:31:21,820
2 l;
and
236
00:31:21,820 --> 00:31:34,490
it will be 384, and this 48, that will go
off. So, it will be multiplied by 48. So,
237
00:31:34,490 --> 00:31:35,490
it will
238
00:31:35,490 --> 00:31:49,420
be 480 on the numerator, because 5 into 2
10 48, it will omega l divided by 384. I am
239
00:31:49,420 --> 00:32:12,590
sure it can be further. It will be 5 by 4.
So, that will be its reduced form. Now we
240
00:32:12,590 --> 00:32:33,380
can
draw the beam again. It will be 5 omega l
241
00:32:33,380 --> 00:32:42,940
by 4 and total length is equal to your l and
l.
242
00:32:42,940 --> 00:32:48,840
So, twice l.
So, the support, we have first replaced in
243
00:32:48,840 --> 00:32:56,620
the form of some reactive force unknown Rc.
And that Rc we got by 5 w by l divided by
244
00:32:56,620 --> 00:33:02,160
4. So, it is a simply supported beam, with
a
245
00:33:02,160 --> 00:33:06,419
vertically downward applied fully distributed
load, plus at the mid span there is a
246
00:33:06,419 --> 00:33:15,370
concentrated load, value is known, but that
value is acting in the upward direction. So,
247
00:33:15,370 --> 00:33:21,140
this problem can be easily solved. The remaining
reactions, and you can find out shear
248
00:33:21,140 --> 00:33:26,550
force, bending movement, shear stress, bending
stress; any important information you
249
00:33:26,550 --> 00:33:36,561
want, you can find out from there.
So far, we are handling a problem where the
250
00:33:36,561 --> 00:33:45,590
degree of your indeterminacy is 1. So, extra
unknown is 1. So, cantilever beam, end reaction
251
00:33:45,590 --> 00:33:52,430
we have removed; here the central part
we have removed. Here I want to mention, the
252
00:33:52,430 --> 00:33:56,050
central reaction we have removed, so not
necessarily that is the only alternative.
253
00:33:56,050 --> 00:34:02,490
So, if we come to the original problem. So,
here this was removed by a force; we could
254
00:34:02,490 --> 00:34:11,980
keep the support; remove this one; and put
a reaction here. That can be also done or
255
00:34:11,980 --> 00:34:14,559
we
can keep the support, remove that, put a reaction
256
00:34:14,559 --> 00:34:15,559
there.
257
00:34:15,559 --> 00:34:30,419
Or if I take the cantilever problem; cantilever
problem we have removed that. Here, the
258
00:34:30,419 --> 00:34:35,829
restrain against rotation we can release.
So, we can make it simply supported. So, we
259
00:34:35,829 --> 00:34:39,669
can
make it a hinge support, with a applied load,
260
00:34:39,669 --> 00:34:43,710
and that load will be in the form of your
bending movement. So, one bending movement
261
00:34:43,710 --> 00:34:47,649
will be acting at the end and that
bending movement is a unknown quantity. So,
262
00:34:47,649 --> 00:34:55,679
it will be simply supported beam, fully
loaded, plus a end moment. So, one extra quantity
263
00:34:55,679 --> 00:35:00,030
that.
So, there, due to the fully distributed load,
264
00:35:00,030 --> 00:35:07,000
there should be a slope. And due to that
moment, slope will be in the reverse side.
265
00:35:07,000 --> 00:35:12,440
So, the vertical load, it will try to make
a slope
266
00:35:12,440 --> 00:35:18,980
like this, and this moment it will give a
slope in a reverse manner, and it should be
267
00:35:18,980 --> 00:35:21,650
equal
and opposite, because it is a fixed support,
268
00:35:21,650 --> 00:35:30,880
so slope should be equal to 0. So, in that
manner not necessarily, this is the only place
269
00:35:30,880 --> 00:35:35,249
where we can remove and put the load.
There might be some other possibilities. So,
270
00:35:35,249 --> 00:35:39,150
same problem can be solved in different
manner.
271
00:35:39,150 --> 00:35:46,619
For all the cases, our extra-unknown degree
of indeterminacy is one. So, we are writing
272
00:35:46,619 --> 00:35:51,970
only one equation; we are trying to match
one deflection; and from there we are getting
273
00:35:51,970 --> 00:36:03,099
the result. Now, once we have solved two problems
- two different types of problems.
274
00:36:03,099 --> 00:36:12,190
Now, I want to take a beam, where at least
the number of unknowns will be more than
275
00:36:12,190 --> 00:36:21,059
one. So, your degree of indeterminacy will
be two. We can take a very simple type of
276
00:36:21,059 --> 00:36:22,309
problem.
277
00:36:22,309 --> 00:36:47,200
Say there is a beam; it is fixed at both the
ends. Now, there is a load, and this load
278
00:36:47,200 --> 00:37:05,069
is, say,
acting here. And this part, we can say, here
279
00:37:05,069 --> 00:37:12,569
to here is a; here to here it is b. It will
have
280
00:37:12,569 --> 00:37:21,970
vi, l, everything and we can say l is nothing
but a plus b. And it has some flexural
281
00:37:21,970 --> 00:37:41,759
rigidity EI. No, it will not stretch. So,
the right end is fixed along with the left
282
00:37:41,759 --> 00:37:47,369
end. So, if
we remove the right end, it will be a normal
283
00:37:47,369 --> 00:37:51,489
cantilever indeterminate. So, if we remove
the right end, how many reactions you will
284
00:37:51,489 --> 00:37:56,220
just eliminate or you have to take in the
form
285
00:37:56,220 --> 00:37:59,700
of some reactive force.
First of all, it will give a vertical reaction
286
00:37:59,700 --> 00:38:03,839
force; plus there will be a rotational part.
Now,
287
00:38:03,839 --> 00:38:09,410
horizontally part we are not considering,
because there is no horizontal force, and
288
00:38:09,410 --> 00:38:12,490
we say
this horizontal force here and here, it will
289
00:38:12,490 --> 00:38:18,009
cancel. So, basically we will have two
equations apart from the horizontal equation.
290
00:38:18,009 --> 00:38:24,450
So, two equations, we can take one vertical
reaction; one moment. So, one vertical; another
291
00:38:24,450 --> 00:38:31,769
rotation, that part will be extra. So, two
additional unknowns are there. So, your degree
292
00:38:31,769 --> 00:38:41,839
of indeterminacy is equal to 2.
There is another possibility; we can just
293
00:38:41,839 --> 00:38:51,009
make both the clamped support as in support.
Take some reactive force and allow the rotation.
294
00:38:51,009 --> 00:38:57,609
So, it will be a simply supported
problem with moment. So, there, if we say
295
00:38:57,609 --> 00:39:02,760
it is A and if we say it is B, so there will
be a
296
00:39:02,760 --> 00:39:05,900
moment MA; there will be a moment MB - one
option.
297
00:39:05,900 --> 00:39:16,201
Another option is we straightaway remove that
support, and take a vertical reaction, and
298
00:39:16,201 --> 00:39:21,690
a moment. So, in two ways we can handle the
problem.
299
00:39:21,690 --> 00:40:04,779
Now this problem we can, say, if I idealize
in that manner. So, we can release the
300
00:40:04,779 --> 00:40:15,670
restraint imposed by support A and support
B against rotation; and put some moment in
301
00:40:15,670 --> 00:40:21,380
lieu of that resistance or that restraint.
So, restraint will basically give some moment.
302
00:40:21,380 --> 00:40:23,460
So,
that moment, say, we have given in the form
303
00:40:23,460 --> 00:40:34,829
of MA, and moment here is MB. And just it
was converted with in support, but our requirement
304
00:40:34,829 --> 00:40:49,140
is we have to find out ME and MB.
Now, this problem can be divided into component.
305
00:40:49,140 --> 00:41:34,569
So, if we put the load here P,
plus…Now, due to that load, there will be
306
00:41:34,569 --> 00:41:41,910
a deformation like this. And due to the applied
moment MA and MB, it will try to bend in the
307
00:41:41,910 --> 00:41:52,339
upward direction.
Now, the second part, we can put in two different
308
00:41:52,339 --> 00:42:02,450
steps also. We can put P, second step
MA, third step MB or here MA, MB second, third
309
00:42:02,450 --> 00:42:10,450
we have more or less combined. So,
more our maturity will come, we will try to
310
00:42:10,450 --> 00:42:15,089
combine the effect. We will just imagine this
is happening, so we need not break the different
311
00:42:15,089 --> 00:42:23,299
steps. So, it is better to break in some
steps at the beginning, to get a feeling what
312
00:42:23,299 --> 00:42:30,870
is happening.
Now, here due to that load P at A, we will
313
00:42:30,870 --> 00:42:39,910
get some slope; at B we will get some slope.
Now this slope at A will be produced by the
314
00:42:39,910 --> 00:42:49,509
load P only. Now, due to MA and MB, we
will get some slope here. So, this slope should
315
00:42:49,509 --> 00:42:56,359
be balanced by this slope, because in
actual structure there will be a clamp support,
316
00:42:56,359 --> 00:43:04,430
slope will be equal to 0. Similarly, here
slope and slope it should be balanced together.
317
00:43:04,430 --> 00:43:10,999
Now two additional unknowns we have taken
in the form of MA and MB. And these two
318
00:43:10,999 --> 00:43:21,680
unknowns can be evaluated from the compatibility
of slope at A, matching of the slope at
319
00:43:21,680 --> 00:43:29,859
A, and similar way we just equate the slope
at B, obtained from here and here. So,
320
00:43:29,859 --> 00:43:36,680
basically slope computability at A and B,
we will try to impose. And from there, we
321
00:43:36,680 --> 00:43:41,069
will
get two information in the form of two equations,
322
00:43:41,069 --> 00:43:46,719
and these two equations will help to
evaluate MA and MB. So, two unknowns, at two
323
00:43:46,719 --> 00:43:51,430
places we have to look after the
deformation of the structure.
324
00:43:51,430 --> 00:43:59,589
Now, intentionally, I have put the load in
a very arbitrary place. So, distance is A
325
00:43:59,589 --> 00:44:03,170
and it
is B. In a special case, if I put B is equal
326
00:44:03,170 --> 00:44:07,690
to at the center of the beam, say, it is at
l by 2.
327
00:44:07,690 --> 00:44:15,640
In that case, MA should be equal to MB. So,
we could calculate only one quantity, only
328
00:44:15,640 --> 00:44:22,549
matching with the slope at A or B. But if
we make it unsymmetrical, definitely your
329
00:44:22,549 --> 00:44:29,470
slope will not match this side and that side
due to P. And your MA and MB will be
330
00:44:29,470 --> 00:44:35,180
different. But anyway, if we handle a general
case, we can make A is equal to l by 2 for
331
00:44:35,180 --> 00:44:36,180
a
332
00:44:36,180 --> 00:44:41,650
specific case can be determined from there.
And there should be a checking, MA should
333
00:44:41,650 --> 00:44:56,289
be equal to MB. Now, these two steps, let
us find out. So, just draw it here again with
334
00:44:56,289 --> 00:44:58,950
some notations.
335
00:44:58,950 --> 00:45:23,700
So, the first part… and there is a P. Now
it will deform like this. So, we are talking
336
00:45:23,700 --> 00:45:27,989
about
the this is A, this is B. So, this angle will
337
00:45:27,989 --> 00:45:38,309
be your, say, A due to say P. And here, it
will
338
00:45:38,309 --> 00:45:50,289
be theta B due to P. So, there is a super
script P to identify this is responsible for
339
00:45:50,289 --> 00:46:01,490
P only.
Now, this theta A and theta B; that you can
340
00:46:01,490 --> 00:46:10,079
derive from with your earlier concept of
moment area theorem or differential equation
341
00:46:10,079 --> 00:46:16,859
technique. Or here, the theta A and theta
B
342
00:46:16,859 --> 00:46:34,890
it is available. So, theta A, if I put here,
theta AP, it is given Pb l square minus b
343
00:46:34,890 --> 00:46:40,200
square
divided by 6 l EI.
344
00:46:40,200 --> 00:46:50,969
So, 6 l EI is the denominator; Pb l square
minus b square is the rotation of the left
345
00:46:50,969 --> 00:46:56,260
end.
This part is A; this is B; total length is
346
00:46:56,260 --> 00:47:09,690
L; already we have written that. Now, this
equation, and theta B due to that P, that
347
00:47:09,690 --> 00:47:29,829
part is Pab 2 l minus b divided by 6 l EI.
Now,
348
00:47:29,829 --> 00:47:40,190
this is the rotation, at this end. I want
to change this expression little bit, because
349
00:47:40,190 --> 00:47:44,150
a plus b
is equal to l.
350
00:47:44,150 --> 00:47:51,380
Now, the numerator part, I want to keep entirely
in terms of ab; denominator it will be l,
351
00:47:51,380 --> 00:47:56,660
because there some part is a, some part is
b. And, if we express that, you will find
352
00:47:56,660 --> 00:48:02,009
expression will be more or less similar type.
So, it will be identical expression. So, here
353
00:48:02,009 --> 00:48:12,700
we can write P. So, l square minus b square
you can write as l plus b l minus b. So, l
354
00:48:12,700 --> 00:48:20,549
minus b is a; l minus b will be equal to a.
So, this a if we take common, so it will be
355
00:48:20,549 --> 00:48:26,109
ab.
So, here ab is there; I want to bring it there
356
00:48:26,109 --> 00:48:32,529
actually. So, l minus b already we have taken
in the form of a and l plus b; l plus b; l
357
00:48:32,529 --> 00:48:43,460
is what? A plus b. A plus b. So, it will be
a plus 2
358
00:48:43,460 --> 00:49:07,989
b divided by 6 l EI. Here also, this l part
will be ab; l is your 2 a plus 2 b. So, it
359
00:49:07,989 --> 00:49:17,410
will be, 2
a plus 2 b minus b will be b divided by 6
360
00:49:17,410 --> 00:49:23,309
l EI.
So, what is happening? E ab divided by 6 l
361
00:49:23,309 --> 00:49:27,099
EI, that part is entirely identical; only
this end
362
00:49:27,099 --> 00:49:34,910
it is a plus 2 b; and there it is 2 a plus
b. So, a plus b, there is a factor 2. So,
363
00:49:34,910 --> 00:49:40,509
here this end
load is away from a. If we take a is more
364
00:49:40,509 --> 00:49:45,519
compared to b. So, this slope will be little
less.
365
00:49:45,519 --> 00:49:51,400
So, here it is twice b. So, b is small; it
is twice b; and this side slope will be more,
366
00:49:51,400 --> 00:49:52,400
it is
twice a.
367
00:49:52,400 --> 00:50:01,160
So, if it is Pab a plus b; some of the quantity
will be twice; which one will be twice, that
368
00:50:01,160 --> 00:50:10,960
also from the sense we can find out; plus
the denominator will be 6 l EI that component.
369
00:50:10,960 --> 00:50:21,329
So, this part we can keep in that form, and
the other part, we took in that manner. So,
370
00:50:21,329 --> 00:50:40,329
we
can draw it here, and there is a MA, there
371
00:50:40,329 --> 00:50:58,749
is one MB . So, it should with theta A due
to
372
00:50:58,749 --> 00:51:16,020
moment and theta B due to moment. So, theta
A plus moment; so this is theta A due to
373
00:51:16,020 --> 00:51:24,539
moment; this is theta B due to the moment.
Now this theta AM, theta BM - they are
374
00:51:24,539 --> 00:51:32,630
produced by MA and MB.
Now, if we take a beam, apply at a moment,
375
00:51:32,630 --> 00:51:40,849
the slope we will get - how much? It is Ml
by 3 EI. So, if you take a simply supported
376
00:51:40,849 --> 00:51:47,940
beam, apply a moment, so where we are
applying a moment it will be Ml by 3 EI; other
377
00:51:47,940 --> 00:51:57,700
end it will be half, Ml by 6 EI.
If we take a simply supported beam, apply
378
00:51:57,700 --> 00:52:01,049
a moment, so if it is not there, if we apply
a
379
00:52:01,049 --> 00:52:09,319
moment, this angle will be Ml by 3 EI, and
this angle will be Ml by 6 EI. Similarly,
380
00:52:09,319 --> 00:52:12,039
due
to this moment, if that is not there, this
381
00:52:12,039 --> 00:52:17,059
end slope will be MB l by 3 EI, and this end
slope
382
00:52:17,059 --> 00:52:28,660
will be MB l by 6 EI. So, both the actions
are there. So, in that case our theta AM,
383
00:52:28,660 --> 00:52:43,730
it will
be MA l divided by 3 EI plus MB l divided
384
00:52:43,730 --> 00:52:57,239
by 6 EI.
Similarly, theta BM; it will be MA l by 6
385
00:52:57,239 --> 00:53:14,369
EI plus MB l divided by 3 EI. Can you see
that? So, it will be MA l by 3 EI inherent
386
00:53:14,369 --> 00:53:20,349
and MB l by 6 EI. It is coming from the
moment applied at the other end, because this
387
00:53:20,349 --> 00:53:24,779
will be this way, this will be this way; both
side, it will be just additive.
388
00:53:24,779 --> 00:53:29,650
Now, next part is - this slope and this slope
should match; and this slope and this slope
389
00:53:29,650 --> 00:53:36,269
should match. So, if we match that, we will
get the reactive force.
390
00:53:36,269 --> 00:53:47,150
Now, up to this, I am keeping it for this
class. So, next class; in the next class,
391
00:53:47,150 --> 00:53:53,200
we will
continue from equating these slopes, and finally,
392
00:53:53,200 --> 00:54:02,700
get the reactive force. Gradually, we
will switch over to other type of problem.
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Preview of the Next Lecture
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So, we were discussing this problem in our
previous class. The starting of the problem
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was like this. So, fixed beam, load placed
little bit away from the center. So, a load,
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bending moment, at the two ends force, bending
moment separately we have applied.
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And due to the load, the slopes we have written,
and due to your MA, MB, you wrote the
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expression of slope at the two ends. So, next
part is just we have to merge the slope,
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because slope here and slope here; it should
compensate, because it is a fixed support;
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and here also same thing.
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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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P
and theta A due to moment, the expressions
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whatever we have written, so if we just write
it on the left side, it will be your MAl 3
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EI MB 6 EI; that was Pab a 2 b divided by
6 lEI.
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So, definitely, this part on the left side,
the other part on the left part here, this
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is on the
right side.
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So, similarly, at B, due to the load, and
theta B due to the moment, if we equate, if
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we
write here, so it will be MAl divided by 6
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EI MBl by 3 EI; that is equal to Pab 2 a b
6
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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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So, these are
the two equations and two unknowns: MA, MB;
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MA, MB. Now we have to just process
these two equations and find out MA and MB
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Now the lower equation if we multiply
with 2, so this 6 will be 3, and you may part
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it, we can cancel. So, this equation we can
write, we can straightaway multiply or we
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can write in a separate line. So, it will
be MAl
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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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that equation or from this equation, this
equation if we make it minus. So, your MAl
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by 3
EI part will be cancelled and here MB part
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will be remaining.
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So, if I come to the next page, say, this
is your, if I say this is your equation one;
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if we
say it is equation two. So, if we make equation
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two minus
equation one.