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Good morning. As you know already, we have been looking at influence lines for statically
determinate structures. To begin with, we introduced the concept of the M¸llerñBreslau
principle and then, we have spent the last two lectures looking at influence lines for
statically indeterminate. Again, to reiterate, we said the last time we spoke of qualitative
influence lines, M¸llerñBreslau principle, quantitative only the direct approach and
in the direct approach, we saw that the moment distribution method gives us the quickest
way of computing.
Let us now continue looking at influence lines. Hopefully, we will try to look at one or two
examples in this particular lecture. I am going to be moving a little bit faster because
by now, I do not have to reiterate every small step that I take ñ why am I releasing, why
am I doing this; by now, you should be able to be fairly comfortable with that. Let us
take an example.
Now we are going into real examples. The load only moves
in a, b and c ñ vertical loads only move between a, b and c. The flexural rigidity
for this is EI, the flexural rigidity for this is EI and the flexural rigidity of this
is 2 EI. We are given the fact that bending moment positive is this way
and this
is shear force positive ñ shear force and this is bending moment. This span is 4 meters,
this is 4 meters and this is 8 meters height.
We have to find out influence line for Ra, for Md, for Rd and for the bending moment
and shear force ñ bending moment at e and shear force at e; this is the center point, it is
2 meters. This is the problem: both qualitatively find out the influence lines for this, so
for these are qualitative and quantitative, find out Ra and bending, only quantitative
is Ra and BM; qualitative ñ all of the others. Let us go through these step by step. We can
do this reasonably quickly. Qualitatively, let us look at what happens.
For Ra, what we have to do isÖ. This is the system and for Ra this goes up, this point
goes nowhere and so, what we get is.Ö Note that this goes and this is fixed so this has
to be like this, goes like this, fixed goes; and this is going to be less because this
also has to go. However, the influence line is only over here. This is the influence line
for Ra. If we want to do it for bending moment Mba, this is for Mba, let me draw it this
way. For Mba, what happens is that this goes this way and this goes this way where this
and this has to be 1. That is the bending moment ñ this is for Mba. Then, we need to
find it out for Rd. For Rd, what do we do? Think about it.
We release this and push this up by 1, push this up by 1, this goes up by 1. What is going
to happen is that this is going to go
and this is going to be 1 and this is a, b, c. Note that alwaysÖ a, b, c, we do not care
what happens to this part because the load does not travel on that. Is that clear? Therefore,
even if you have a frame, the influence line is only to be drawn the shape that the part
on which the load travels, that is the only part that comes into the influence line ñ
remember that, never forget that.
Even if I have shown a frame here, I am only drawing the influence line for abc, because
that is the only position, that is the only kind of extent the load travels ñ we have
been given that. Then finally for Md; for Md, this has to go, this is going to be this
way and so, our thing will be this way and the influence line, this has to be equal to
1 but note that the influence line is this; even though we have drawn this part, it has
no role to play. This is my influence line ñ remember that. This is for Md although
I have given this. I am not going to draw anymore; okay, let me just draw the bending
moment at e for your benefit.
Bending moment and shear force ñ let me draw that. If I have bending moment at a, then
this is going to go straight and then this is going to go in this fashion, where this is a straight line and
this is going to be equal to 1. That is going to be bending moment at e. For the shear force
at e, it is going to be this and the shear force, positive isÖ this being pulled down
and that being pulled up, so it is going to be this way; of course, this is going to go
this way. Again, this is the only part that matters; this is going to go this way, it
is going to be less thanÖ if this did not exist, it is because this stiffness reduces
that, that is all. This is for the shear force at e. Qualitative influence lines is fine.
Now, we need to go into computations of this for this thing. The thing that we need to
know is from which position does this go? Let us look at that. Quantitatively, we need
to know this goes ab, bc and therefore, when it goes ab, (FEM)ba, when it goes bc, it is
going to be fixed bc and cb, but cb is anyway fixed, so it does not matter whether cb is
fixed or not. Therefore, we do not apply a fixed end moment ñ that fixed end moment
is going to go directly and at ab, we do not provide because Mab is always going to be
equal to 0. Therefore, the only ones we need to provide are bc and cb.
I am just going to show you notionally, I am not going to do it precisely. I am applying
and remember: this one, there is nothing, this one this and here, I have this, for this,
we have this side and this, this is 1.0. Let me go back and have a look. This is EI and
4 meters. This is going to be I upon l, three-fourths of I upon l, so this side, we have three-fourths
of I by 4; this side, we have I upon l, so we have I upon 4; this side, we have 2 I upon
8; so this is going to be 3 by 16 I plus I upon 4 and I upon 4. What we have then is
4, 4, 11 by 16, 11 by 16, 3 upon 16 divided by 11 upon 16 is 3 by 11 and this is 4 by
11, 4 by 11. What we have is 0.272, this is going to be 0.364 and 0.364, this is going
to be 0.728, yes 0.364 ñ these are the distribution factors that you have over here.
The first one is going to be plus 100 here, 0 here, 0 here and of course, this is 0 and
so is this. We need to distribute it and we distribute it. This becomes minus 27.2, minus
36.4, minus 36.4, this one goes here, minus 18.2, this one goes here, so this becomes
this, this becomes minus 18.2 and that is it. Now, we go on to the next one. It is the
same thing that we have. It is going to be 0.364, 0.364, 0.272 and this time, this is
plus 100, this is 0, this is 0, this is 0, and so is this. When we do this, this becomes
minus 27.2, minus 36.4, minus 36.4 and this becomes minus 18.2, close, close, close. Nothing
goes here and here, we get minus 18.2.
If we look at this, we see that Mba is equal to 0.728 of mba and minus 0.272 of mbc ñ
this is the fixed end moments. Then, Mbc is equal to
minus 0.364 mba plus 0.636 mbc. Mcb ñ this is the moment at that end is going to be equal
to minus 0.182 mba minus 0.18 mbc plus mcb. Note that I have not done this moment distribution.
Why? Because at a fixed end moment if I apply a fixed end moment, what is going to be the
moment? The fixed end moments, so that is going to be directly this, it is not going
to get distributed anywhere else. These are the expressions that I get from my moment
distribution because if you look at this, if this is 100, then this is 72.8, this is
minus 36.4.
By the way I have not done the others because those do notÖ In this particular case, I
have only asked you to find out Ra and bending moment at e. If I had asked you to find out
Md, then you would of course have got it in terms of minus... let me do that, let me do
that, let me find out Md so that I would need to find out Mdc would be equal to minus 0.182
mba minus 0.182 mbc ñ that is what Md would be, so we will find that out. I have got these.
Once I have got these, then the next step is of course the same procedure that we have
already gone through and without much ado. We have already done this in the previous
case.
We know that if 0 is less than x is less than L, then (FEM)ba is going to be equal to minus
x squared into (L minus x) upon l squared minus half (L minus x) square into x upon
L squared and (FEM)bc is equal to 0. If L is less than x is less than 2 L, (FEM)ba is
equal to 0, (FEM)bc is equal to (x minus L) into (2 L minus x) squared upon L squared
plus half into (x minus L) the whole squared into (2 L minus x) upon L squared. Note that
the only thing that we have been asked to find out are three quantities.
Those three quantities if you look at it is Md. Md at all times for 0 less than x is less
than L by 2 is equal to Mdc. What are the other ones? The other two things that I have
been asked to find out are Ra ñ I need to find out Ra, so let us see what Ra is going
to be. I have over here Mba and I am only interested in finding out Ra and the bending
moment at e. When a load is between 0 and x by L, Ra is equal to (1 minus x by L) plus
Mba by L ñ we have already done this, I am not going to go into that.
When x is between L and 2 L, Ra is equal to Mba by L. That gives me the expressions for
Ra. Now, for the bending moment at a, when load is between 0 x and L by 2, bending moment
at E, let us look at what bending moment at E would be like. You would have a bending
moment in this fashion. If we look at it, Mba and the load is over here and so x is
between and you have Ra here and this is the bending moment. If it is between this and
this and I take moments about this point, what do we get? We get Ra into L by 2 ñ that
is being this thing, so it is Ra into L by 2 and here, we have this one going the opposite
way, so this is going to be minus 1 into L upon L minus x. This is the bending moment
and you can see that this is also equal to Mba minus x by L into L by 2, x by L into
L by 2 is minus x by 2, so we can actually put it in this fashion. If you substitute
into this, you will see that this is exactly the same as this, so there is nothing new
in it. We will probably use this ñ this is the easier one to use. What happens when x is between L by 2 to 2 L? Bending moment
is just equal to Ra into l upon 2. Once I have this, then I can write it down.
This time, I will do it with L by 4 so that we can actually find out more details. This
is going to be
x and the overall distance that it travels is from 0 to 8. I will have 0, 1, 2, 3, 4,
5, 6, 7, 8. The first ones we have to find out are mba and mbc ñ these are the fixed
end moments and those are the first things that you have to evaluate. Once you evaluate
that, if we look at it, we need look at this so all we need to do is to evaluate Mdc and
Mba because these are the only ones that we need. We are going to do that. I am going
to evaluate Mdc and Mba. I do not need to evaluate any of the others. That will give
me directly Md, Ra and the bending moment at e, the bending moment at e. Let me now
do this. We have done this now. We know that mbc is going to be 0 while the member goes
between 0 and 4. We also know that mba is going to be 0 when it travels between 4 and
8 and of course, when it is at the end, these are going to be 0. All we have to do really
is evaluate these and this one we know, it is 3 by 16 and this is negative and this is
going to be positive 3 by 16, here plus 6, this is not correct, this is plus 3 by 16 at the center
span. Let us find out at three-quarter span.
I am going to plug in the value of x equal to one-fourth in this. When I plug in one-fourth,
I get 1 upon 16, this is going to become 1 minus, so this is going to be three-fourths
and this is going to be 9 upon 16 and this is going to be one-fourth. What we have over
here is minus 3 upon 64 and then we have minus half of 9 upon 128, so this is going to be
equal to 6, so minus 15 upon 128. The other one with three-fourths is going to be just
the opposite, so this is going to become minus 9 upon 64 and the other will become 3 upon
128 and so when we do this, 18, we get minus 21.
What you have here is that since we are doing this thing, this is going to be 18 and 3,
plus 21, this is going to be plus 21 upon 128 and this is going to be plus 15 upon 128.
If you look at this, this turns out to beÖ as the load comes in, the bending moment increases
and this is more and this is less. Now, let us look at directly Mdc. This one if you really
look at it, this into L, so this is going to be 60, so this is going to be 4. I will
have to do this again, this is going to be 32 ñ this is into L, L is 4 here, this is
3 upon 4, this is going to be 21 upon 32, 21 upon 32, 3 upon 4, 21 upon 32.
If we write those down in numbers, you will see that you get 0.4 on 28, 12, 12, so that
is going to be 4.44, this is going to be 0.75, this is going to be equal to 0.6, 6 will make
it 192, that is going to give me 12, so it is going to be 0.64; this is going to be minus,
minus, minus, this is going to be minus 0.64, this is going to be minus 0.75, this is going
to be minus 0.44. When we plug those values in here in Mdc, you get minus 0.182 into 0.44
and if you look at that, that basically becomes minus, so it is going to be plus, so plus
0.072, so this is going to be 0.08.
Then if you do this into 3 by 4, it is going to be 0.75, this is going to be 3 by 4, 0.55,
0.55, 3, so this going to be 0.55 divided by 4, that is going to be 0.14 and in 0.64,
so this is going to be 0.11, this is going to be equal to 0 and this is going to be minus,
note that these are plus, plus, plus, so this Mdc is going to be minus, it is going to be
minus of 0.11, this way, it is going to be minus 0.14 and minus 0.08 and 0. Let us look
at for Md. This is exactly how it looks. This is Md and the point that I am trying to make
is this way you can find out mba, plug it in and you will get mba and then, you can
plug in these. Once you get Mba, you can find out Ra values and you can find out moments.
Once you know this, Mdc is directly this. Now the point that we are trying to make over
here is that once you put this factor in, you can get Ra and bending moment, you can
get Mba. I am not going to do all of this. The overall procedure boils down to three
steps. The first step is to determine where exactly your loads are. In this particular
case, for example, load is going from a to c.
Now, a is hinged, it cannot have a fixed end moment. Therefore, when the load is between
a and b, you have a fixed end moment at ba. Therefore, you need to do one moment distribution
with ba. Then when it is between b and c, you need to do the moment distribution. There
is going to be a (FEM)bc as well as cb, so you need to do a moment distribution with
bc and cb. But note that the there is no need to do Mbc because that is a fixed end and
you do not release that point, so there is no fixed end moment, so it only comes into
M into M, the moment at cb ñ these are important points. Therefore, once you know the levels
of moment distribution that you have to do, the next step is once you have done the moment
distribution and you have to now write down what are the member end moments in terms of
the fixed end moments ñ those are the coefficients you have found out from the moment distribution.
Once you have written those down, then you need to do the equilibrium of each member
and get it from there. Is that clear? This is the step and now I am going to quickly
look at another problem so that I can illustrateÖ one other type of problem to you. Right now,
we have set up the entire step and now all we need to do is essentially put down all
the values that you are going to get so that we do not need to do anything anymore. We
were done with this, this is it. Just let me put my papers together and then we will
quickly move on to the next one. The next problem is, let me take a particular example,
let us take this example.
This is A, this is B, this is C and this is D; for this, I have I, for this, I have I,
for this, I have I; this is 4 meters, this is 3 meters, from here to here is 4 meters
and from here to here, it is 4 meters. The vertical load only goes between B and C and
D ñ this is important; it does not go on AB. Is that clear? The vertical load is only
in this zone. What does that mean? When we draw the influence line, we draw the deflected,
use the M¸llerñBreslau principle and get the deflected shape. We only need to draw
the deflected shape of BCD for the influence line.
Let me now say that the influence lines for Ma, for Ra, then at E, which is the center
point, I need for the shear force, shear force, the reaction at C andÖ Ra, Ma, VE, Rc and
MD. We have five for which we need to find outÖÖ I leave this as an exercise for you
ñ you are going to do it yourself, qualitatively over this entire scope at 1 meter intervals,
you are going to find out VE. I leave this to you as an exercise ñ I will give you the
answer but I will not go and solve the details. However, what I will do is I shall use the
M¸llerñBreslau principle and show you how this thing can be done.
First, Ra: I am going to make it a question that is going to become very very important
and that is the reason why I have given you this particular problem, nothing else. Here,
note that now going to become a fixed roller which is going to move in this direction,
so this way it has to be equal to 1. What does that mean? Ideally, this would go, 1.
However, this cannot move, so therefore this has to now move perpendicular and how much
will the perpendicular be in turn, think about it. If this is 1 then the perpendicular is
going to be 3 by 4 so that this then becomes 5 by 4. this is going to go like this and
I am not going to draw this, this is going to go like this. Therefore, this is going
to rotate in this direction if it has to move up; the tangent is not going to be this way,
the tangent is going to be this way, so it is going to go like this, go here and then
go this way and this is going to be 5 by 4. Note that this is for Ra.
Note this very importantly and that is the whole reason why I have given this particular
problem and that is why I have given Ra in this direction instead of vertical. Vertical
would have been very easy. But I gave it in this direction for you to understand that
when this has to go this way, you have to satisfy the boundary conditions. Remember
that the M¸llerñBreslau principle has to satisfy all the boundary conditions that you
have in your specific problem ñ this is the point, this was the whole point, everything
else is now trivial. However, I am going to go through with it. Let us look at the next
one which is Ma. I am going to Ma.
Ma means I rotate it so this becomes now a hinge and rotate it here. How will it look?
It will look like this. When this goes this way, this would tend to do a little bit and
so this is where it becomes interesting. We knew that this is 1and this is the influence
line. However, qualitatively we know nothing over here. This is for Ma. Now, let me do
it for Rc.
For Rc this is going to remain this, this is going to remain this, we are going to remove
this and move it up and up by 1. How will this thing look? You will see that when this
goes up, this is going to go like this, so this is going to go a little bit like this
and so this is going to get a little bit like this, probably this will have an angle this
way but this in essence gives you for Rc.
Now, Md, let us make it for Md. For Md, now what do I do? I release this, make it into
a hinge, this is anyway a hinge, this is fixed, make it into a hinge and then rotate it. What
is going to happen here? You are going to see that this is going to go in this fashion
and this is going to go this way and this way, so this is going to be where I know that
this slope is equal to 1. This is for Md. Finally, for the shear force also I am going
to draw this because this is important. This is quantitative.
Then we have to actually get the quantitative also. This is fixed, this is hinged, this
is fixed. Therefore, what we need to do isÖ here, get it up and here, get this to go down;
when you make it go down, this this will go in this fashion and go this way, this is going to go in this fashion ñ this
and this have to be the same, so that this is going to be something like this; this does
not matter. It is this, this and this and this total is going to be equal to 1. This
is for Ve. Finally, as I said, I am not going to show you how I have done it but I am going
to put down for x, so x is moving from b to c to d at 1 meter intervals. It is 1 meter
here, here, here, here, then here, here and here. At 1 meter intervals, I am plotting
it starting from here ñ my x starts from here.
For 0, 1, 2, 3, 4, 5, 6, 7 and 8 I am putting down this as x, I am putting down the derived
values of Ve for you ñ you are going to actually calculate it. For 0 this is 0, for 1 it is
plus 0.2235, for 2 there is going to be a 2 minus and then there is going to be a 2
plus; for 2 minus this is plus 0.5112 and for 2 minus it is minus 0.4888 ñ this is
for 2; for 3, it is minus 0.2068, for 4, it is 0, for 5, it is plus 0.0803, for this,
it is plus 0.0733, this is 0.0273 and for 8 it is 0.
I am going to leave this with you; you have to solve this problem and get these Ve ñ
use this. Next time, I am going to do it in detail and show it to you. Do not wait for
that ñ try to get these values using the procedure that I have developed for you. Thank
you. See you next time.