Okay so in this leasure we're going to talk about this stiffness and distribution factors now.
We skipped over that in the last lecture so we're gonna drill down here in this lecture book before
I jump into that.
There is a lot of little time saver that I think we should talk about and that is this practice of capturing
the mommy distribution analysis in a table or tracking it through a table.
And this is a really important and it's even more important when we get into more complex structures.
So what I'll do is I'll lay out in the standard table everything we did in the last lecture and you'll
see hopefully the benefits of using using one of these tables.
So this table for a mom distribution always looks the same or that the sections of the table always
at the same.
It starts off the top row table will list all of the joints in our structure.
So aren't we had joint a we had joined B and we had joint C..
OK.
Then below that we're gonna look at what are the members at each of those joints.
The man was meeting each of those joints so our member our joint day we just had MEMBER A B coming in
now at joint B.
We had both one of my down at point B.
We had the A coming in on the left hand side and we had.
B c coming in on the right hand side and then rejoined C we just had a CB Okay so that's going to be
the members.
All right.
So the next one is the distribution factors wider distribution factors and distribution factors are
gonna come into play.
Joint's where there are more than one member coming in.
So where we're focussing in on joint be here for the minute.
So we said the distribution factor was nine point six six.
Remember this only came from made ops differences.
We're gonna address this in this lecture.
And that was not point 3 3.
Then the next thing that we like to lay out in this table is Why are the fixed end moments fixed and
moments sort of we have mines 100 plus 100 and minus two or eight point three plus two or eight point
three.
Okay.
Now this is where the table really comes into its all because we go to each joint where there is more
than one member coming in and we look at the moment in balance and we can very clearly see that the
moment in balance here at a is such that I need to add plus one hundred eight point three to join joint
being in or for joint beat balance and I do that in proportion to the stimuluses as we said so I add.
Seventy two point two here and thirty six point one here.
So that was my Distribution Committee us a bit more space.
That was the distribution.
So thats just throw a notion here.
We had a balancing moment.
Balancing moment.
Of plus 1 0 8 3 divided in two ratio of stiffness is or rather divided in ratio of stiffness.
OK.
Great.
Then we had the next layer.
It's going to be what's called carry all Relayer the carry over role and because joint a is a fixed
joint.
We carry over.
50 percent which would be thirty six point one.
And we carry over half to here which is to join to see which would make that eighteen point one.
No joint D balance so we can sum up our final moments.
Okay so this summing up should the columns so we have minus 100 plus thirty six point one would give
us minus sixty three point nine.
So it's just what we did previously with Don in a table.
So again for b.a. we have 100 plus 72 which was a worn seventy two point two.
And the process continues minus one seventy two point two and two to six point four.
And those are our final moments.
So again as Sean Aldrin here we had carry the carry over all 50 percent carry over 50 percent to adjacent
to adjacent fixed and fixed is excellence.
And then the model are the North for here is our final moments.
Well we just laming final moments really.
OK.
So that is a process laid out on the table now it seems fairly trivial for such a straightforward structure
such a trivial structure at home.
But this process or this practice of going into a table and tracking everything to a table becomes invaluable.
And it will be invaluable for us as we proceed to more complex structures later on in this course so
we'll.
And this is the method we're going to use from here on an economy using this table c should become very
very familiar with it.
So we'll draw a line on that one and say that's that little nugget finished and we're going to move
on to what we really want to talk there.
In this letter which is the stiffness of our members.
Okay.
Let's make our way through this.
So in the last letter we said that member maybe was twice as stiff as member ABC and because it was
twice as stiff sort of fundamental a basic structure as are basic mechanics would tell us that if it's
twice as stiff it receives or attracts sumps and people think abella is attracting twice as much moment
twice as much Maung.
So the stiffer I make a structure the more moment that it will resist whereas the more flexibility is
less mohmed it can resist.
So actually just true though those figures you said a ratio two to one right.
Let's just show that in very light allows us to crack ahmadis ample when actually we need to be more
careful when we're assessing that.
The memories stiffness is so we're going to say here is that the flexural rigidity or stiffness so the
flexural rigidity.
Brigid Ishi also known as the stiffness depends on a number of factors.
Right.
So it depends on what's the first thing it depends on.
It depends on the Yon's much this.
The Young's modulus usually donal's it with a capital Yon's much less.
This is taken into account the material that it did.
The structure is made of.
Okay so what's then what how is the material contributing to this structure the ability to resist bending
the next it depends on is the second moment of area set and moment of area.
Again that's usually a capital secon moment of area is taken into account the cross-sectional shape
of my nambour.
Okay how is mass distributed throughout the trek the member.
Or typically how is mass distribution trekked cross action.
Because again fundamental structures will tell you that the more mass of the more area I have away further
away from the neutral axis the more stiff my structure will be.
That's why steel beam is steel IBM has those two horizontal flanges at the extremities of the member
of the top bottom cos were concentrating mass or area structural mass as far away from the neutral axis
of the structure as we can.
As it makes for a stiffer structure so that that's where he comes in.
The next thing that affects the stiffness of a structure or its ability to resist bending is the length.
The length of the member how long is it.
How long is the member between restraints.
Length of member between.
Restraints.
And this is it.
This is a very intuitive one that say between rotational restraints.
So no surprise to anybody that the longer a mamber is the less stiff it is or the more flexible it is.
Whatever everywhere you look at it the shorter remember is the stiffies smartass.
That's L..
And then the final one which is what we're going to spend our time focussing on.
Number four is the support conditions OK.
So when I get down to those different supports they pinned are they fix or support conditions.
Right.
What kind of support condition they got it fixed.
Is it pinned I mean in reality support is rarely ever fully fixed fully when it's going to be somewhere
in between.
To facilitate or to allow us to modeller structures we are we approximate them as either fixed complete
rotational resistance or pinned zero rotation resistance.
And again this is where we're going to spend our time looking at support conditions.
So focussing on let's focus on the influence of supporting conditions
right.
And you'll be happy to hear we have only two cases to consider.
There are only two different scenarios right.
So two cases aswarm first cases where I have rotational fixity provided at the adjacent support.
Now the word adjacent here is gone it is going to make more sense to you when I start trying to diagram
when s just in words trot out these two different cases case one rotational fixity OK a rotational fixity
provided.
At the adjacent.
Support.
That's case 1.
And then case you you can probably guess where I'm going with this.
Row tension.
Let me see if we can see if we see rotational fixity nosh provide it.
Adjacent.
Support.
Ok.
Now I'll tell you the end result here we're on where I've got this result but I'll tell you now that
for a case one situation where we have rotational fixity provided to Jason's support stiffness for that
member K is going to be for e r o r l and for the opposite case stiffness is gonna be three E I O L.
Now if you're purely interested in the moment distribution method that's all you need to know for stiffness
is this guy and this guy right now what I'm gonna do is show you give you a bit more of an idea of where
these two equations confronter what we're going to arrive back at this equation of this equation and
that's all we're really going to use for a home distribution method.
What to flesher understanding let's consider these a little bit more teater so we're gonna go in here
and look at case 1.
What do we have.
Simple.
Let's start off with a simple structure.
We have a simple frocked cantilever now at joint B.
I'm gonna apply a moment NDA.
Now this is red that word adjacent is gonna come in the adjacent support.
When I say adjacent they mean the support that is adjacent to where I'm applying the moments in applying
the moment that support B the adjacent supporters support a.
And this what makes this case 1 is that support Day has a fixed OK so there's rotational fixie provided
the adjacent support our right.
So this moment and the well cause this structure to bend and the bending can be quantified by where
the ways it quantified is by this angle.
D be a case a thief should be a is just the angle between the horizontal axis which is the axis of the
structure and the tangent to the deflected shape at point B.
OK so that's how much the structure is rotating at point B is the other way.
Think about that.
So what relates and the date it.
These are the is the stiffness.
OK so the stiffness we have n the a that's going to equal the stiffness of structure B.
The rotational stiffness times things at B.
Ah I.
This is what I'm hunting for.
This is what I'm after Kaybee.
Right so what's the stiffness reisch.
So what we're going to do at this stage is to get us to that equation.
I'm going to introduce two other equations which were not going to drive but I'm going to introduce
them here to facilitate the process.
So were going to introduce introduce what are called.
You may have come across as before.
What I call the slope deflection.
Relationships.
Right.
We're focussed here a moment distribution.
So we knock on a branch of time going through the derivation of these slope to electoral it's really
only just introduced a minute without derivation.
We're essentially going to state them the slope deflection relationships.
This B is equal to l over 6 e times to n be a net that should be to be a and b a minus an A B right.
What is that telling me that is some relationship that is some equation that relates this angle to be
a.
To the moment at either ends of the structure and the other parameters we talked at which were at length.
The ones martius and second of area.
So we're gonna just state that and assume that it is true without driving it.
What to call that one number one word on a state the next one which was foetor A B C minus L over six
times n B A minus two and a P right.
Let's call that one number two.
So the first thing we can say remember dealing with case one here so because joint A is fixed we can
say since because it's fixed t's are a B is zero since D A B is zero because it's a fixed support from
equation 2.
We can say if you if you subthemes to a V equal to zero you would end up and you look at equations to
simplify it.
You would end up with a B equal to a half and B.
Okay that you carry your gramont.
We've already seen we've already used this result but this is where it comes from.
Carry over.
Moment.
So we use this result in lecture what now from equation 1.
So frahm no one seems to be a ego the L over six times to be a minus.
A b.
Remember.
Maybe we just worked out to be a half and b A minus a half.
B A who subbing in our previous results.
And let's simplify that guy out.
What would we end up with.
We'd end up with these to be equal to.
A hemsi BE IN TIMES OVER for E.
Let's rearrange that simply rearranging.
We'd end up with M The eagle too for I over all times these are B.
What is this.
This year this equation looks very much like this equation where the central piece in here cabi A is
equal to for I or L O K.
That's basically telling us that that there k be a is equal to 4 e.
Or else that's the result I told you that we were going to get it.
OK I guess that's case one dealt with the next case that we'll deal with is case number two.
So we look at Case 2.
Again we draw a simple structure.
This is case 2.
It is.
We're applying a mom with an MBA and the adjacent support which is support a has no rotational fixie
OK.
So we have no carry over mohmed The first thing to say is there is no carry over moment to a because
it's a spin.
So no carry over moment because we know we can't have a moment that the pin was now.
We can't have a mom develop hoping we can apply a moment at a pin externally but we can develop a momentary
reaction mode a pin does not carry over a moment to a because it's him.
OK now again going back to the equation number two.
Since an A B is zero if we so it may be equal to zero.
We would get these to be equal to over 6 e times to be a because any B.
That should say 0.
Let me see.
And then if we simplify and rearrange that we end up with n be a equal to 3 E R or L times he should
be.
And that is k e a place too.
So what have we done in these last.
In looking at case 1 and case 2 What we basically try to do is clarify a little bit more wary.
This stiffness comes from the stiffness is what relates the moment applied at a joint to the level of
rotation at that joint and that level of rotation at the joints that we're applying the moment is strongly
dependent on what is happening at the adjacent at the opposite point.
Whether that's a fixed point where a lot of pin joint and we've looked at those two cases and we've
seen that the stiffness of the member is different.
The rotational stiffness of the memory is different depending on what's going on at an ordered joint
and these are the two differences that we said we would arrive.
So what does all that was kind of a little bit of a detour if you like and sort of get back to how do
I apply that to the moment distribution method What does that mean to me as someone who is looking to
use that information within the moment distribution.
And we can summarize wondered at what it means.
This is where it's getting it's getting relevant again.
So the story is this.
I'm going to try at a couple of different structures a few different structural types and work on a
look and see what they have in common and how we use this information.
So let's think about the first take the structure.
Fixed both ends.
So that's one option.
We may end up with a structure before we like any joints that looks like this where the little Sobe
structure is got to is essentially to roller is going to roll.
The structure flies over the road and it's continuous all with roller so that that this dashed line
with the arrow is simply saying that the structure continues on past the roller arm to another support.
So we've got continuous olver both supports the structure continues past the support that we're interested
in are to some other or other support.
That's what we're showing with these dashed arrows here.
Then the little structure we may use we may have you rotate some structure will look like this.
Maybe it's a fixture built in support at one end and then at the other end it flies.
Let me see the other end.
It flies all over support.
And continues on to another support.
Okay.
So that's fixed and continuous.
Always support.
Now what all of these structures have in common is joint fixity or resistance to rotation at the joint.
So k a B in each case is for E R or L O K because we have rotational fixity here.
Here we have rotational fixity here and here not because of the rollers but because the structure flies
past a roller and connects to an order structure.
No other support rodder And so in doing so it resists the rotation at point B and a joint.
And here we've got rotational XY because it's built in.
And here because it flies off over to another structure.
What do we mean by that.
It's just so when I say it it continues always support and I'm saying here is if I was to take this
guy out and look at the full structure.
Imagine it does this.
We've got our built in support which we see a ball.
We have our Roller here.
Now this dashed arrow.
Here what that's saying is it's an instruction to something like this it continues on.
And maybe it continues and it stops at a rotor here.
In which case the fact that this part of the structure continues and connects to this and this is holding
it down is a reaction here.
If I try to rotate this joint here the structure this reaction will come into play and the structure
would be resisted at this point p.
It would be resisted in rotation so it wouldn't be free to rotate because this would have to bend.
Let me get rid of that.
That's rotate.
This would be dope.
Have to come back down to retention here.
So that's what I'm saying because the structure flies past to support it be conex on into or not of
support.
Maybe cozad conex onto another sport doesn't matter.
Point is it flies past B and connect to something else and that in connecting to that something else.
We're providing a resistance to rotation at point B.
So that's what these dashed arrows are all are signifying.
OK let's get that over the way.
Now I think with some more structures let's imagine we have a fixed support here and let's say this
was a pin a roll or even it stops at the end of the structure it doesn't continue off to the right hand
side of B.
It just stops there.
Well there is no rotational resistance provided B.
So this is a classic fixed point.
The other case would be let's imagine.
OK.
At this support a.
It's not fixed anymore but it's pinned.
And instead maybe doesn't cover our AP maybe it continues and flies over a roller and maybe continues
artemas structure here.
Key thing here is we don't have we now we don't have an rotational resistance at 8 so no is free to
rotate.
So in the opposite case there was three to rotate and a was fix had rotation resistance and then in
this structure down here was free to rotate but no B has rotational resistance because the structure
continues on.
So in both of these lower cases where this one is pinned and continuous in both of these lower places
we don't have a fix as he approached and we only have visagie at one end.
And so we can say that for those structures K A B as egot AT 3 E or L so we can think about how does
that apply to our regional structure.
So for our previous example.
Now we're assuming E is constant.
Troops are going to remove that as a variable right.
So between a B and B c e is constant.
Great.
Now does it reset the loading because the stiffness of the structure is totally independent of the loading
what Potter joined out on.
Now remember a.b. what is the stiffness of a.b.
Well if we think about it we have a fixed support at a and b the structure flies over a pin or a roller
but it continues on.
It continues on.
And we know it joins to see what we're saying that it is continuous continuous over support.
So you should be able to see that the stiffness maybe is equal to 4 e I o l which is going to be for
that length of eight times VII which is gonna be not quite five.
Eli that is the stiffness of that structure.
And then we can do a similar exercise for DC.
And that would mean four or a length of 10 times which is not quite force or sure enough has a lower
stiffness because it's longer so there any actual stiffness of our structure.
Now we want to divide the moments in the ratio of the stiffness and that is the redistribution factors
come in.
So there's no real mystery here we're just using this value and this value to divide up that balancing
moment in proportion to the two stiffness and we develop work all distribution factors for that.
So the distribution factor for a B S that the mom goes into is gonna be the stiffness.
Any stiffness of the structure on the left or the sum of all of the stiffness is of the members that
meet to that point which is simply say a B plus k. B C.
Thought B C is the stiffness of the member B C or solve this differences that meet at the joint.
Pay a B plus k b c which is simply mathematically the distribution factors.
The source of the distribution.
Fact is most stiegel when I write them.
So we'll just remind ourselves in our previous lecture or previously we just shoot two values from the
top four ahead and we said previously the distribution factor the made up distribution factors and weight
that were not point 6 6.
And let me see.
P.S. What was it.
It was not 3 3.
Remember that came from March 2 to 1 ratio.
Again totally made up totally fiction and what we can see now is the actual stiffness is already actual
differences that lead to these actual distribution factors.
So very very quickly what are able to do now is jump back to our table and actually repeat the analysis
with the right distribution factors and then we'll get the right bending moments for its structure.
So we have a right distribution factories so we can work the balancing moment was gonna be not point
five six times the one or eight point three which gave us sixty point six five and then on this side
we have not point four four times when I wait.
Point three which gives us forty seven point six five carry over 50 percent each fixed hand and we got
thirty point three.
Three twenty three point eighty three.
So what the moments minus sixty nine point seven worn sixty point six five minus one sixty point six
five and 2 3 2 point 1 3.
OK.
That's it.
That's all there is to it.
So let me see.
I think we've covered the not involved in these last few lectures of the entire process.
Am I at this stage it's just about identifying and a couple of exceptions.
But what I would suggest is spend the time on these first two lectures make sure you really grasp what's
gone on in these first few lectures.
And before we start thinking about and developing this further and going on to think about different
exceptions that the comment to play in more complex structures.
OK Sol we will leave it there.