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Good morning. This is lecture 30, module 5 - we continuing with matrix analysis of beams
and grid. So, we will complete the reduce stiffness method, which we started in the
last class. For the beam element, and in the next class, we will take up flexibility method
for the beam element, and after that, we will do again reduced stiffness method for the
grid element and the conventional stiffness method. This is covered in the chapter on
beams and grids, in the book on advanced structural analysis.
Of the three methods, we are now looking a reduced element stiffness method.
If you recall we learnt how to derive the stiffness matrix using various techniques,
the easiest is the physical approach; that matrix is very easy to remember, 4 EI by L
2 EI by L 4 EI by L 2 EI by L. The diagonal elements are 4 EI by L; the half diagonal
elements are 2 EI by L; they are all positive and you physically know the meaning of each
coefficient in that matrix.
We also learnt how to setup the T D matrix. And the real problem is, when you have to
convert deflections in the main structure in the beam, to a reduced element stiffness
coordinate system at the local level. And we said, this can be done by converting the
chord rotations to equivalent flexural end rotations.
So, if you have a chord rotation of 1 by L corresponding a unit displacement, you can
treat that as an equivalent end rotation, but remember you have to reverse the sign;
that means, if you have a clockwise chord rotation, you should have an equivalent anti-clockwise
chord rotation. This is how you convert from typical global coordinates like m l m n p
to the two coordinates: 1 star, 2 star; we have done this in the last class. But if you
found this little difficult, there is an alternative way of arriving at the same derivation.
How do you do that?
No, not the, you have to bring in the transpose, you have to invoke the contra gradient principle
and some of you might find it easier to do it this way.
So, here, we do the reverse - we move from the local coordinates to the global coordinates;
and we invoke the T D i transpose matrix, which means you apply a unit, moment - end
moment - two of them, one at a time like this. You apply F 1 star equal to 1 satisfy equilibrium
and figure out what are the corresponding values that you get in the global coordinate
systems; next, you apply F 2 star equal to 1. And you can see that, after two end moments,
only one is non-zero and it has value of plus 1, and you get two reactions, you can see
very clearly - it is 1 by L.
If it is pointing upward, it is positive; if it is pointing downward, it is negative;
and you can very easily generate this matrix, you get it. So, this is another powerful way
of generating the same transformation matrix, but we are getting it from the transpose.
But my suggestion is, do it in the classical way - the T D approach using the displacement
transformation. Now, you recall we did this problem in the last class. And we will now
look at more problem, we will try to do three quick problems in this session; with that,
will close the reduced element stiffness method. We need to worry about how to deal with moment
releases. There are two kinds of moment releases that you can get: one is when the far end
is a hinged or roller support; the second is you have an internal hinge. Either way
the bending moment in the beam at that hinge location is known, usually it is 0, unless
you have an externally applied known value of a constant moment.
So, we have done this, we have taken advantage in terms of reduced kinematic indeterminacy,
when we applied the slope deflection method and moment distribution method, remember.
So, we can do the same thing; that means, here we just ignore that global degree of
freedom all together; we do not even give it a number. If you have a hinge, whether
an external hinge or an internal hinge, do not assign a global coordinate active or restrain
at all, just keep it; we will see how to do that.
And modify the element stiffness matrix, you know that, 4 EI by L will now become... 3
EI by L, when the far end is hinge; so, that advantage we will take care of. And you also
need to modify the fixed end forces, because normally you have the formulas for a cantilever
beam which is fix. If one of the ends has a moment release, it becomes a prop cantilever
and you know what to do, we have studied that. And the big change that you get is, in the
conventional stiffness method, how many degrees of freedom did you have? 4, you dealt with
the 4 by 4 element stiffness matrix; that reduced to 2, when we did the reduced element
stiffness method, where we did not take any shortcuts, that will now further reduce to
1. Remember, we had 4 EI by L and 2 EI by L, and then we had this 2 by 2 matrix.
Now, we do not worry about rotations at the other end - at the hinged end - and so you
have only 1 by 1 matrix, just 1 coordinate; the other end is hinged. So, that is a major
advantage we are trying to take in this method. You reduced a degree of kinematic indeterminacy
and you know that the element stiffness matrix is 1 by 1, and the coefficient is 3 EI by
L, is it clear? We have done exactly this in slope deflection and moment distribution
method, when we took this simplification. So, we will see how to do this.
Let us take this problem, we have done this problem earlier? Yes, I think so, in matrix
methods or in moment distributions slope deflections, one of them, anyway. So, how do you or did
we do it in the... I think we did the conventional stiffness method, remember. What was the degree
of indeterminacy? We did this, yes. 3
Not 3, it was 5, because theta B was D 1, theta C was D 2, theta D was D 3, then delta
E - we did not take any shortcuts - was D 4, and theta E was D 5, remember. They were
five active degrees of freedom and they were also five restrain degrees of freedom, we
had a 10 by 10 structure stiffness matrix; the k A was 5 by 5, remember that?
Now, we are going to have a massive reduction in kinematic indeterminacy. So, what should
we do first? Because this is aimed at manual analysis, so we do not want to solve 5 by
5 matrices. The first thing you can note is, you have an overhang, which you can just get
rid of it, because that reason is statically determinate.
So, that is the first thing you do; you separate it out and you bring it back, at the end when
you need to draw the bending moment and shear force diagrams. So, when you remove it, you
know that, if you have a clockwise moment of 50 kilo Newton meter at the free end E,
you have an anti-clockwise moment at D. And Newton's third law says, the same moment will
act in the opposite direction in the segment A B C D at D.
So, you are now applying this as a nodal moment, except that it is not really going to a global
coordinate, we will see, we will explain that shortly. Now, tell me we have got a reduced
problem now - just a 3 span continuous beam. What is the degree of kinematic indeterminacy
in this? Is just 2, if you want to take advantage of the hinge at D, right.
So, from 5, it is a massive plunge to 2; so, that is a clever thing and we should do it
correctly without making any mistakes; so, please pay attention. By the way, we did this
in slope deflection method, we did this in moment distribution method, we will now do
it in matrix method.
So, the first thing is, the procedure is quite clear. You have to do the coordinate transformation,
write down your displacement transformation; matrices - you have to figure out the fixed
end force vector, you have to do this manually, because you have a problem in the reduced
element stiffness method, because you do not have a constant fixed end moment, it can change.
So, you have to do it manually; you have to get the net resultant force vector F A by
F fA. Then, of course, you generate the stiffness
matrices in the usual way and you solve the equilibrium equations, find the unknown displacements,
use the disclosed active displacements to find the support reactions, at the end you
find the fixed end forces. And if you recall all previous discussion that last equation
what is it really represent in terms of what we did earlier, they are nothing but slope
deflection equations; remember, I said m a b is m f a b plus 4 EI by L.
So, all that, it is a matrix formulation of slope deflection equation. So, let us proceed,
the first thing you need to do is find the coordinate transformation. So, you got global
coordinates here; we agree that, we will have only two active degrees of freedom: 1 and
2. At the end D, what should you do? Leave it in piece, do not a put curled arrow there
and give it a number, because we are not interested in that rotation; we do not want to know theta
D, because we know the bending moment at D is, in this case 50 kilo Newton meter.
We will bring all that later, but the first step is, if you want to work with reduce kinematic
indeterminacy, do not include that as an active degree of freedom, but you have restrained
degrees of freedom 3 4 5 6 7 - the usual convention will do that.
Why are we putting the restrain degrees of freedom? Because we want to find support reaction,
that is a real . Also if you have, in this case we also have that, what do we have? We
have support settlements in this problem; so, you have to bring that in the picture.
And let us look at that we have two support settlements, remember, B and C go down by
5 and 10 mm respectively. So, you will label that as D 5 equal to minus
0.005 meter and D 6 as minus 0.010 meter. Next question, are there any nodal loads in
this structure F A? You will find that there were only distributed loads and there was
some constant load in the middle of that beam. You had a nodal moment at D, but it does not
have a number, so how do you deal with that? You understand the moment at D, that becomes
in a load which you have to account for in the fixed end force vector for the element
three, that is the only catch; you have to do it carefully and then you can solve the
problem. And we did it in slope deflection moment distribution
method, so let see how to proceed. These are the element level local coordinates, we have
three elements; and for each element, we usually have two degrees of freedom expect for the
third element. The third element has one degree of freedom; please pay attention, 1 degree
of freedom and you have to be careful about that.
So, please write down the transformation matrix for... what will be the size of that matrix?
It is going to look like this, at least get it right. So, for the first element, what
is the size of the T D matrix? How many rows will there? Two rows for the first element.
How many columns will there? There will be seven, because there are seven coordinates,
of which two belong to T DA and five belong to T DR. Can you fill up? Similarly, you will
do it for the second element; and for the third element, you can do it all in one go,
the third element, it is a just one row. So, you give it a shot very easy; if you apply
D 1 equal to 1 in the main structure, what do you think will happen? Most of those numbers
will be 0, except D 2 star for element 1, and D 1 star for element 2, that is it; so,
I have given the solution here. Let us go through it carefully, if I apply
D 1 equal to 1 here, you agree that the second element in the... but we have 2 elements here:
one is element in the matrix and you have the actual beam element. So, when I see element,
I am referring to the element in the matrix; the second element in the first beam element
will be 1 and the first beam first element in the second beam element will be 1 and the
rest will be 0, clear. Now, you apply D 2, that means, this one D
2 equal to 1; you agree this will get effected and this will get effected, which will show
up as 0 0 0 1 1, clear. Next, we apply D 3 equal to 1, which means we are lifting up
the left end A and restraining all the other coordinates; this will result in a clockwise
chord rotation in the element 1 alone. Clockwise chord rotation of magnitude 1 by
8, 1 by 8 is 0.125. And if you have a clockwise chord rotation, you end up with equivalent
anticlockwise positive end rotations. And that is why, you write plus 0.125 plus 0.125
for the first element, and you write 0 0 0 4 for the other element and so on; so, I am
sure you can derive for all the others. Let us look at the last one; let us say D
7 equal to 1, if you lift up the last one, only element 3 will be affected, agree? Only
element three will be affected; you have a clockwise or anti-clockwise chord rotation?
You have a anticlockwise chord rotation and the equivalent flexural rotation will be clockwise,
which means negative and it will be 1 divided by 6; that is why, you have in the last row,
you have minus 0.1667, but all the others are 0, is it clear?
Can you all do this correctly without making any mistake? Do you have any doubts on this?
This is the first step, if you get it right, you will get everything right.
Global coordinate 4 Global coordinate 4, yes.
Global coordinate 4 is, ya, yes. You have a unit rotation, it will affect only beam
element 1.
You have to be careful when you read these rows. This is 1 0 0 0 0; it is my fault I
did not align all the rows correctly all the columns correctly, is it clear? No mistake,
everything is fine. We will proceed. Now, we have to find the
fixed end forces; we have done this problem earlier, so can I go fast over this? You can
work out the fixed end forces. Remember, you also need to find the vertical reactions.
Why do we need to do that? Not only the fixed end moments, also the vertical reactions,
because in the global coordinate, they pile on to those vertical forces in the coordinates
you identified. So, you need to do that little extra exercise, but that is easy to do; we
have done that earlier. Take the second element, we have done this
before. Remember, here the only thing to worry about is the formula, when half the span is
loaded with the u d l, for which you may have to look up a table, because you may not remember
it. Remember, it is 5 by 192 W L square, and 11 by 192, that is the only form, but we have
done this earlier. What about the third one? The fourth element,
you do not worry, because you have already dealt with it. In the third element, you have
to be careful, because the end D is now roller, and you are going to apply a 50 kilo Newton
meter at that end D clockwise as a load; please tell me, what is the fixed end moment you
get at C? Please calculate and tell me, because this is where you can go wrong, this value;
that is F, for the element 3 1 star; calculate it and tell me, you have done this, when you
did slope deflection method moment displacement; do it once more.
Well, due to the 150 kilo Newton load, for a fix-fix beam, you know the formula - W A
B square by L square, for the left end, and for the right end, it is W A square B by L
square; and you need to release that, so half of it gets carried over to this side; so,
that part you can deal with. What about that 50 kilo Newton meter? Half of it gets carried
over with the correct sign, so that is all you need to do.
I hope you have understood, do not go wrong here? Because we are taking shortcuts, shortcuts
can be taken only by intelligent people, otherwise you get lost in the woods. But the work load
reduces enormously, we have reduced the indeterminacy. Any doubts on this? We have done similar problems
earlier; keep the science intact correctly. What do we do next?
We write down the fixed end force vector, you can put it all in 1 box or you can keep
them separately as I shown here; that is, these are only the end moments corresponding
to that degrees of freedom. Typically, two degrees of freedom per element except for
element number 3, where you have only 1 degree of freedom.
From here, manually, you have to build up the fixed end forces, that act on the main
structure. Finally, we want that, we want F fA and F fR.So, we can get that we need...
so, this is the fixed end force vector in the global coordinates, F fA correspond to
the 2 coordinates - 1 and 2, remember 1 and 2, 1 and 2 are here these two moments and
these two moments. So, you just algebraically add up those two
moments, you will get F 1 f F 2 f. The net moment that you get corresponding to the fixed
end moments at B and C; that is, F 1 f and F 2 f. And what is F 3 f, F 4 f? You can pick
up from here, remember, this was F 3 f.
So, that is plus 120, F 4 f is plus 160, F 5 f is a summation of 120 kilo Newton and
61.25, 181.25, F 6 f is likewise summation of these two and F 7 f is 34.72. It is easy
to do, but you have to do it with your eyes open manually. Now, let us proceed; well,
not really manually, you can also program it, if you do it skillfully.
So, you can find the net load vector, which is the nodal force vector; in this case, it
is a null vector, there are no nodal forces in this problem, and F fA which we got in
the last step. You have to put a minus sign, because you have to satisfy equilibrium. And
you will find that, you are actually getting you have reduce that complicated problem;
it was simple problem where you apply only two nodal moments in the direction shown.
And this is equivalent to the original problem, in the sense, you get exactly the same theta
B and theta C and that is the beauty of the displacement method; such a complicated problem,
you reduce to a simple problem, where you have only nodal forces.
Next, generate the element and structure stiffness matrices; this is easy. Remember 4 EI by L
2 EI by L, so put the right EI, in this case EI is 4 EI; in this problem, L is... so easy
to write. Next element, 3 EI and L is 6, so we see is very easy; please write down for
the third element, because third element is different; there is only one, it is a 1 by
1 matrix; the EI here is 2 EI and the L is 6, but you have to it is 3 EI by L EI into
1, it is easy.
Now, can you assemble? Well, you first do this product; we have done this before, let
us not waste time, you have got the T D matrix in the beginning; you understood how to do
it, just pre multiply that with the k i star matrix, get this, then what do you do again?
Pre- multiply this with T D transpose and they just add up everything, add up mechanically,
no slotting you required here.
That is a beauty of this T D matrix; not to reduce, when you use T D even in the conventional,
you get this advantage. And the top upper corner, on the left side, ease of k AA and
that is your real; you do not worry about the size of the whole matrix, it is 7 by 7;
it looks scary, but we want to only invert 2 by 2. That is k AA, you can pick it out
and inverting it is easy and you are ready to take any load, but the loads are also known,
these equations are well known, it is always good to find out k AA D A directly. In this
case, you have support settlements. So, D R is given as, you know we did this
earlier, minus 0.005 meter minus 0.010 meter at 5 and 6 coordinates, but EI is 80000. You
have an option either to plug in 80000 in the stiffness matrix or you try to get rid
of EI by converting these deflections in millimeters to equivalent EI terms, one side; in other
words, if you multiply these two with 80000, you can write this as 1 by EI into minus 400
minus 800. In short, if you take this product, you will get minus 0.005 meter.
It is a clever trick to do, so that you do not you leave your stiffness matrix in piece,
that EI gets canceled out. And there are five elements here, because there are five restrained
coordinates; you know that there is no rotational slip or settlement anywhere else, so it is
0 0 minus 400 minus 800 0.
Now, you are all set to apply and solve; you first take care of the D R k R, you can do
this step wise or you can do it one shot; I leave it you, finally, you get the answer
- you get the unknown displacements. And we have solve this problem earlier by the hard
conventional stiffness method, you get if you check up you get exactly the same solutions.
Now, you need the support reactions; this is where you have to look at the big k RA
and k RR matrices, which we generated in... but we done this in the conventional stiffness
method, so this step is same to that, similar; there is no difference. So, you should get
exactly the same answers, because only D A was unknown here, the rest we had derived
earlier in the conventional stiffness method and you will find you are getting the same
reactions.
Last step, member end forces; so, remember those slope deflection equations. So, you
note all the displacements, both the active and the restrain. and for the first element,
plug in those values; second element, third element, you get the same answers as we did
earlier. Finally, you have to draw the same bending
moment and shear force diagram, is this clear? It is a powerful shortcut method; instead
of solving for five unknown displacements, we solved for only two; our interest was to
get the final solution to design this beam. So, we need to know the bending moment diagram
and shear force diagram and we have got it. We will do the same problem tomorrow using
the flexibility method and you can decide which is the easiest to do.
We have not done this in moment distribution or slope deflection method. How to deal with
an internal hinge in the reduced element stiffness formulation? You will find that it is easier
to do this here than in the conventional stiffness method, because there you have to artificially
create a clamp and all that; here, life is easy, this is all you need to do.
First of fall, whenever issue a hinge avoid putting any global coordinate their active
or restrain, because we want to take a shortcut; so, remember, earlier we had a coordinate
2 here and we had a clamp there. Now, we just leave it in piece, there is a hinge there.
So, that is what we have to note; there is no coordinate corresponding to rotation at
B, because there is no unique value for rotation there; both the ends can move and you do not
care with how much they move, because you just want the bending moment and shear force
diagram. You are not looking for the absolute value of the rotations there, that is a real
idea. Ideally, we should show a hinge also at B to remind us.
The other coordinates are clear? Two; so, the coordinate 2 now shifts here, 2 3 4 5,
so one active degree of freedom, 4 restrain degrees of freedom - same methodology. This
is beautiful, because you have only one degree of freedom for each of the two elements and
the stiffness is 3 EI by L. Can you write down the T D matrix for this? It is very easy,
one row for each. Fill up this matrix, so it is going to look
like this. Let us check it out, if I lift up B by unity and do not allow any other moments,
then I have an anticlockwise chord rotation 1 by 10 for this element, which will give
me... should it be minus or plus? Minus, because anti-clockwise chord rotation is clockwise
equivalent flexural rotation; so, it is minus, and for this, I get plus, because it is the
opposite, so that is how I fill up the first column.
Second column, how do I fill up? Second column is simple; I will lift up these, if I lift
up these, I have a clockwise chord rotation for this; and nothing for the second one,
so I get anticlockwise positive 0.10. Third one, this will give me 1 here, and 0 here
and so on, and very clear, simple, easy to do, ones you get the hang of it.
We will move ahead, fixed end forces. Here, remember, in there was a nodal moment 30 kilo
Newton meter to the left of the hinge, so can you.. so you have to find out the fixed
end forces and we have done this before, is this clear? Due to this 100 kilo Newton load,
you get a fixed end moment here, which is W L by 8 into one and half times, because
of the prop cantilever effect; I hope you remember that.
And it is anti-clockwise, so it is positive, but you have a clockwise end moment of 30
kilo Newton meter given as a load in the problem, but you are taking it at the element level
and so half of it gets carried over with the same sign and it is clockwise, so you should
put a minus sign here, is it clear? Once you calculate this, it is easy to calculate your
vertical reactions. There is a reaction here, there is an arrow that should be here, you
get the reactions; we did this earlier also. Similarly, you do it for the next beam. There
is a hinge, but there is no nodal moment at B; this is a straight forward problem, W L
squared by 12 into one and half 3 by 2, find the vertical reactions.
Now, can you assemble the fixed end force vector? Yes. can you find out F fA and F fR?
Well, it is easy, do it by inspection. Corresponding to this coordinate 1, you have to add up this
reaction and this reaction, right; it adds up to 73.25. And similarly, you can fill all
the other fixed end coordinates: 1 active and 4 restrain, is it clear? Actually the
procedure is similar to the earlier procedure. So, we do not have time to actually sit and
write down all the numbers, but that is something you need to do, when you go back home and
refresh your understanding, prepare for the exam but the procedure, is it clear?
We got fixed end moments at the element level, yes. You are going back to... you cannot do
that in reduced element stiffness method, got it? You are right, you can normally do
it and you should it in conventional stiffness method, but do not ever think of doing it
in reduced element stiffness method, because reduced element stiffness method has reduced
coordinates, it cannot handle this complexity; you have to do it manually.
So, you have to look through inspection, is it clear? Please remember, because this is
a common mistake people might make, you cannot do it.
So, you have an equivalent joint load and this is what we have to analyze and that is
easy. Let us generate the element stiffness matrices, we have 2 elements; single degree
of freedom EI 3 EI by L, easy to do, you got those numbers, then you know the T D matrices.
So, T D k star T D, you can work out. And then, do the summation, you will get two contributions:
the first contribution comes from the first element, second contribution comes from the
second element; and you just add up, you will get the final stiffness matrix k A is 1 by
1; very easy to do.
You can find the inverse which is just the reciprocal; same formulas, these are now looking
very familiar, there is nothing great in all this; solve for the unknown single unknown
displacement D 1 in the middle; remember, we did this problem, we got 101 mm in the
conventional stiffness method. You will be also noted that, when the hinge
was not there, the deflection was very small; the hinge made the whole system very flexible,
the deflection shot up. By a large amount, find out the support reactions, they match
with what you solved earlier using the clamp, but this is much easier.
One unknown here and there you had to bring in the clamp. So, next member forces, same
procedure, same member end forces, same shear force diagram, same bending moment. So, we
are really doing the same problem by all the method, so that you get a clear idea and it
is an easy check. And the check is even better, when we do by flexibility method which is
the completely different approach and you are getting exactly the same solution.
Last problem and this has to do with guided fix supports. And this is a little, it requires
little understanding and please listen carefully. So, here, you have what can be called a shear
release, it is not a moment release, because in a guided fix support; there is no shear
transfer from the beam to the support. And like in the previous case, you ignore the
degree of freedom associated with the member end release; that means, the translation you
ignore we did this in moment distribution method, slope deflection method, do the same
thing. Modify the element stiffness, what will be
the element stiffness for a guided fix support case? remember those three magic numbers;
your 4 EI by L, 3 EI by L, which one you will use?
3 EI by L. That is all. Only one number to remember,
you remember it very well. You can prove it, it is a cantilever behavior and that is the
element that you can assume. This is here, we will take Ramesh's suggestion, remember,
in the beginning, he said, why do you always take a simply supported beam as the element,
why cannot you take a cantilever. Here it is ideal to take a cantilever, because
the behavior is exactly like a cantilever. So, my suggestion is you can take this element,
but it looks so complicated, right. Why do not we take a much simpler element? And that
simpler element is this, simple cantilever fixed at the other end and you know that the
stiffness is the same.
There are two ways of getting EI by L. So, we will use this in our symbols, so that our
understanding is simplified, is it clear? So, let us demonstrate this with this same
example which we did earlier. So, we can take advantage of the symmetry, cut it into two
and you get a guided fix support there, what is the degree of indeterminacy here - kinematic
indeterminacy? Two normally, which two...
Do not say A B; tell me theta, delta, which one?
Theta A and delta B. Theta A, theta B.
Theta B, delta C.
Well, strictly speaking, theta A, theta B, delta C, but the whole idea of demonstrating
this is to take advantage of... so, there are two advantages you can take: one is delta
C you do not worry about, the other is theta A also you do not worry about, what do the
previous examples? Far ended hinge. So, you guys should wake up; whenever you
get a nice shortcut, take it; from three, we have reduced to one, single unknown, what
is that single unknown? Theta B. Do you want to do a 3 by 3 matrix inversion or 1 by 1
matrix 1 by 1? So, be intelligent; do one, the procedure as the same.
Take a look here; this is the reduced beam problem, these are the global coordinate,
just 1 is it ok? Just 1, and 2 3 and 4 are restrained in the global coordinates. Local
coordinates just 1, but simply supported I have shown here and cantilever I have shown
here, is it ok? 1 star, 1 star, you know the stiffness; for the left one, it is 3 EI by
L; for the right one, it is EI by L, that is it.
Write down the T D matrix. What will be the size of the T D matrix for each of those elements?
1 by 4. So, you have to give me 8 numbers. I have be a covering all types of problems,
all types of complexities, all types of shortcuts. So, let us check it out; if you lift up, if
you put this rotation 1, it will effect this and this, right. So, you get 1 1, both anticlockwise
positive. You put D 2 equal to one; that means, you lift up this end, which is simply supported;
you get a clockwise rotation in element 1, which means anti-clockwise equivalent flexural
rotation. 1 by 4 is how much? 0.5
0.5, plus 0.25, but nothing happens to the second element, so this is 0. If you lift
up the middle here 3, left will undergo anticlockwise chord rotation of 1 divided by 4, which means
minus, because it is a clockwise equivalent end moment 0.25 whereas nothing; and for the
right end, now you have to be careful. In a cantilever, translations do not create
any problem. In a cantilever, remember, the it was guided fix here and support here. If
you have support settlement, it is not going to create any problems; you have to visualize
it. So, this is where you do not make mistakes; your T D value will be 0, because this is
a different ball game, right, because D is meant to deflect D; B is free, nobody stopping
that deflect; let it deflect, it is a rigid body moment, the whole thing will go up.
So, you have to be careful; this is where you recited from memory, it did not work here.
You have to always use your brain, that is the only thing. The last one is item number
4, anticlockwise moment , here you have to be careful. If you give an anti-clockwise
here, at the other end you will have an equivalent clockwise opposite moment.
So, this is the only thing you have to be careful about; may be, you might like it in
the old style. Let us try to generate the T D matrix for this guided fix support case
from first principles.
So, you have guided fix support here, prop here and we first take the case where you
have a movement here. If you have a moment here, it will just move like that, the rigid
body movement. So, in terms of rotations, you do not have any rotations, so that is
why the rotation value is 0. In the second case, we are dealing with a rotation an anti-clockwise
rotation at the right end, which means this rotation is now unity; and we need to draw the deflected shape and figure out, what
is the rotation at the left end. We will take the same boundary conditions
and you must recognize that there is a possibility for the beam to deflect at this end, because
you have a guided roller support here. So, this is little tricky, because you might be
tempted to draw it like this, in which case this angle will be half this angle, which
is not necessarily true, because this can deflect.
So, let us take some help from the contra gradient principle and let us see what happens.
If in a beam element to satisfy equilibrium, you have a moment - the unit moment - acting
here. To satisfy equilibrium, you also need an equal and opposite moment acting here one.
So, the T D value should be minus 1 and the minus is quite clear, because you have an
anti-clockwise rotation here, you have a clockwise rotation here, so it is certainly minus, but
it is not minus half, its minus 1, which means this has to deflect some more and that is
not a problem. So, let say this goes down to a point there, you can get the desired
deflection, still maintaining this change in angle is 1 here; and from half this, this
increases to 1, so that is a physical explanation. And the answer is absolutely crystal clear,
it has to be minus 1. So, there are two explanations, why you need
to put. Minus 1, when you are dealing with this case; and you have to put 0, when you
are dealing with this case, because there is no rotation happening in this case. Fixed
end forces, can you find out?
Let us look at fixed end forces; fixed end forces for the first one is 30 kilo Newton
meter, and for the second one, we have done this problem earlier, you get 5 kilo Newton
meter. You can add up and get the resultant force vector, which is 5 kilo Newton meter;
and you have this matrix, 5 kilo Newton meter and substitute...
You get 3 EI by L for the first element, EI by L for the second element; and plug in those
values, you will get the stiffness matrix. And find the inverse, same procedure, find
the unknown displacements, its minus 4 EI 4 by EI. Find the support reactions, you will
get exactly what we got earlier; member forces, that is it.
So, be careful about this particular shortcut, but you can cross check with your previous
examples. When you have doubts about T D, check with T D transpose, because you are
very comfortable with equilibrium, right. So, but we have covered all the possibilities
that you can get. So, we have now reached a stage where we have finished conventional
stiffness method, we did four examples; we have finished reduced element stiffness method,
we did the same four examples. Now, what is left is flexibility method. We
will do only two examples; we will finish it tomorrow. And the last topic left is, very
interesting topic grids; we are going into spatial structures, we will take some examples
and do it; you have an assignment to do three problems, one of them is a grid.
Thank you