In this world example the first thing that we've got to do is think about whether or not this is a sway
or enhance way frame.
Now obviously this is in the nuance way section of the course so we know it announced way around logically
but we should confirm for ourselves why that is not to look for too long to see that these support at
E is in fact a pin support.
And so that is going to stop this freight this this frame from sway that a writer left.
And so that's why it's in John's way Fred.
Okay so it dallow the way we can start thinking about the moment distribution so this can be tackled
in exactly the same way that we tackled our beam structure.
So it's a straight forward process.
So what we want to do is after doing our moment attribution we want and then move on to work at our
reactions.
Joshua forestieri armature bending moment diagram and something that's new far as is we wanna use the
bending moment diagram to help us determine what's called a qualitative deflected shape.
So are not interested in knowing the magnitude of the deflection at any point.
Bright light.
Now roughly speaking how does this frame deflect or what shape will it take.
OK.
Under loading.
So that's the last thing that we'll do in this particular question and we're going to do that for all
of our frame while reframe questions.
So first things first we're going to move straight on and look at the stiffness.
So the stiffness of a B Well it's got a piano here and it's a continuous structure B and so therefore
it will be the lesser of two stiffness.
So K A B is going to be 3.
E r y over L which is going to be 3 over 9.
E r e e r is constant shrout the structure so there at the very end of the day when we work at our distribution
of factors those e values will cancel out.
But for the time being that's not Point 3 3.
E e r n o k so the stiffness for a b e again a P in support only and a continuous structure at B.
So again the lesser of two stiffness is.
So k b e e is gonna be 3 E R over L..
Which again is tree or 8 in this case.
Times the I.
Which is not.
Point 3 7 5.
E I.
And the stiffness for a b c.
Well it's it's not a pin support it's actually a fixed support at c and so there's moment resistance
at c or rotational resistance rather at c and the structure continues at B and so we have the Graser
of the two stiffness options in this case.
So K B C is equal to 4.
E r e o her L which is gonna be for over 5 eai which is not point easy.
Very briefly before moving on faustman all those differences we know is that we haven't more jealous
stiffness for D B and that is because we're not going to work any distribution factor for idb because
it is a cantilevering which we've seen before.
We know that we will not transmit and bending moments while we will distribute rodder transmissions
from work.
We will distribute Annie bending moment into that cantilever throughout the distribution process so
something of the stiffness is that we do care of for the members meeting at joint B.
The song.
Of the K.
I'll use this guy in equal one point 5 0 5.
K now we can work at our distribution factors.
There are 4 members meeting at joint B.
We are going to only distribute moment in 2 3 of those members so the distributions factor for B is
d stiffness of be a k.
B a time see some of this stiffness is the members at that point into which we will ultimately distribute
moment not point 3 3 over 1 point 5 0 5 which is not pointed to 1 9 D distribution factor for a B E
is equal to no point 3 7 5.
The stiffness of B or one point five zero or five which is not point to 4 and 9.
And the distribution factor for b c we know the distribution factors most add up to 1 so we can simply
say one minus the previous two with 2 1 9 wins.
Now point to 4 9 which is equal to not point 5 3 2.
Excellent.
Now let's just for completeness sake we will just say the distribution factor for a b d is equal to
zero and that is because it has cantilevering.
So if you're a little bit fuzzy on the details there called back to the video area in the course where
we discussed exactly what's going on for kansanshi restructures.
Okay the next step is the fixed end moments the fixed and moments Effie and remember the whole process
here is we go through our structure and we lock our joints.
In this case we only have one internal choice or on a locked at joint and work out why are these fixed
end moments that would develop at that point.
So we will simply just as a little refresher reminder just say we are locking locking.
Joint.
B because it's got to reach these members that meet at point B.
So the first one will be Member B.
De.
That is our counselee run at sketcher though.
OK so the fixed end moment at that size B is going to be the 10 kalnins promesa acting over 3 in measurers
terms a resultant of one point five meters and so the fixed end moment for B which will remain at this
value and will not change schreck distribution is 45 kill and hidden meters and K moving it straight
on to member B E.
So we have a proper hansley round for a cross counselee rather fixed and moment is W L squared over
8.
Now this is a anticlockwise moment and therefore it's minus W L squared over 8 and so the fixed end
moment or B is equal to minus 40.
Kill any major cities just box these off so we can easily identify them.
All right let's move on to member MBA.
So the fixed end moment for B is equal to 3 p l o 16 positive and so will will.
It's clockwise will maintain a positive number so that would be that its quick numbers in three times
25 times 9 or 16 and so f g n B is equal to forty two point one 9 kill new meters.
Let's now think about what's happening at joint B consider.
Consider moment equilibrium.
Joint B.
Okay.
So let's just zoom in and have a look at point B.
So I have I am essentially isolating joined B by cutting off the members that frame into which I am
cutting and making my caution infinitesimally small distance from the joint.
In each case and Kate and we were all born according to virtual mamber rising a ball joint B because
there is no moments out was no external loading on that joint right.
What we're saying here is fixed any moment to the right hand side the joint was 40 quileute measures
in k..
If we just isolate again just to make it even clearer we have a 40.
Golly.
That way we has forty five going that way and we have forty two point one nine going again anticlockwise
sob.
The question now is w the out of balance moment at that point.
How out of balance is the joint so very easily.
We can say the out of balance the oh or be moments is 40 minus forty two point one 9 minus 45 which
is equal to minus forty seven point one nine and we all know now at this point at this stage in the
course that that means therefore we need to add with a balance that joint we cant.
We can have an error bars joint and so therefore we add a plus forty seven point one nine kill and it
measures to the joint.
To balance to balance to balance the joint.
And we also know that we added in proportion to the stiff masses of the members meeting at the joint.
So what does that mean.
Well if I come down here again and look at my joy.
I'm going to add a positive and therefore clockwise moment up here of forty seven point one nine times
the distribution factor for this member which is not Point 5 3 2.
That gives me twenty five point one one.
We're going to do the same here.
It's gonna be forty seven point one nine.
Why not.
Point two 4 9.
Distribution factor which is eleven point seven five kilometres that was also killed in new meters and
another clockwise moment here of forty seven point one ninths were balancing up the joint.
But we're doing it by adding moments in proportion to the stiffness of the members meeting at the joint
which is 10.
In this case and am 23 three killed in metres and very very importantly zero moments added to the cantilever
member because that moment was established at the outset and will not change.
So the final moments at the joint at the balanced joint at the superposition already addition of the
balancing moment that we have just added All right here plus the original fixed end moments in case
so we can see what the final member rather the final moments at the joint.
So the final moments.
The balanced joint.
Dark dark dark.
What are we got.
Okay.
And if you were to take a song of the moments at joint B.
Sure enough it equals zero.
And so Joy B is now balance.
Obviously you notice that there's no table I've not got a moment distribution table.
And that's basically because if I'm dealing with structure and instruction it's case framed structure
that has only one internal joins.
Well then I don't really need to resort with table to keep track of things.
The table is really only beneficial when I'm dealing with multiple internal joints and I have a lot
of carry over mom as a Dane now have to be rebalanced.
So dealing using a table to keep track of things in that situation is invaluable.
Well in this situation I only have one internal member I can handle it.
Like I don't hear very easily without working or when I'm going through a full table.
So let's just make a quick note from that very very quickly just putting into words what I've just said.
So analyzing the structure analyzing a structure.
With a single internal.
Joint.
B Don.
Without a table.
A moment distribution table to keep.
Track of the distribution.
That's all is haber's really for helps us keep on top of what's happening.
But we can sort of logic our way through.
If we only have one internal joint.
Okay so that wraps up the distribution part of this question.