Oh God.
In this letter I was introduced the idea of having Peter Downs to our to our structure.
Now the thing about a downed is.
So whether that's a role or a supporter of his support we know the final moment that will exist at that
support will be zero on Lester was an externally applied moment.
But there's no externally applied moment here in the moment there will be zero.
So we have to take that knowledge into account when we are proceeding shude the moment distribution
process.
So we're gonna we're gonna top this true by considering the same structure as the previous example.
But we're just gonna change and a and see into you appeal and a roller amidon a process should structure
it in the same manner as we did before.
So let's get to the structure.
Okay so this is the structure we're dealing with now exactly as before except n days now a pain and
and C is now a roller.
OK.
Well we've done is reduce or remove these rotational resistance at A and C..
So let's think about the stiffness is perko what the stiffness of a b is and the stiffness of a C.
So what I'd suggest is go ahead and work those two differences out now before watching me do it.
OK.
So we'll assume that you've done that so now it's considered to stiff masses of bolt members solar I
think would be the first word we got.
Well that's a pin no rotational resistance and B we have the structure is continuous over the support
so it continues on.
So it continues to see.
So we have rotational resistance provided a big but no rotational resistance provided that a.
So based on our last lecture or rodoreda electric for the last one you should be able to work at that
K A B is equal to 3 E or L OK.
It's the less stiff of the two options because there is no rotational resistance at one of the supports.
So that's just got to be equal to 3.
E over Asia again will soon ease concentrate.
This structure might give us not point 3 7 5.
Excellent stuff the exact same applies to B C and so K B C is gonna be 3.
E r e l l.
Also because we've done the same thing with remove rotational resistance.
One of this court's.
So what does eye-teeth me 3 E over it.
Which is not points re e r r c k now distribution factors distribution factors.
So the distribution factor for B.
Remember distribution factors was only our only coming into play at a joint where we have more than
one member coming into joint and that only happens here at Joint Base.
And so the distribution factor for B is the stiffness at a not point 3 75.
To some of the stiff misses of the members meeting at the joint which is going to be not point 6 7 5
and that gives us no point five six consecrates cries only and the distribution factor for B C is going
to be not three or four point six seven five which is not wait for FOUR exactly the same as we had when
we had to fix a court.
Why is that.
Well the only thing we have changed which is removing rotational resistance we've done the same to both
sides so we've applied the same change to both sides and so we've reduced the relative stiffness is
by the same amount and so the distribution factors remain the exact same.
Okay so same as before and that's logical that makes sense because you've altered both ends said.
Okay so now what we have to do is look up the fixed and momence Fixed and moments right.
What do we have.
So because a and c are pins right they can never develop a moment so we don't lock these joints.
OK so when we finish this question at the very very and the bending moment diagram will have zero moment
that a and zero moment at c because their pins and so we use that knowledge when we're working our flixton
moments and we say Right we're not gonna fix it and we're not going to fix it.
Makes no sense to fix them because they cannot develop a moment.
So we're going to say the cars.
And see our pins.
Okay.
And can not develop
moments.
We do not lock these joints.
We do not as a lark.
These joints.
Okay.
So this is what we end up with.
So what we're dealing with now instead of two beams of both ends fixed.
We're dealing with two crocked cantilevers.
Now the fixed moment for a B.
Dan from your table of 610 moments.
S e.
B is going to be plus three P L over 16 and the fixed end moments.
A B.
That's a maths B is equal to zero because it's a pin.
So we saw the numbers in we get f e m b a being equal to plus 150 and then we have a cross counselee
remodeller for B C.
The reaction moments that would develop but that fixed support is fixed and moments B C is going to
be w l squared over 8 so so the numbers in get f e n b c equal to minus three one two point five.
And because we have a PNC fixed and moment b c the fixed end moment for C B is zero.
Okay so that is one of the big big differences.
In fact you've seen both.
Big differences that are introduced by having pin downs the first difference was in our stiffness.
So the stiffness was 3i over L because with the and our substructure.
One of the supports had no rotational resistance.
That's the first difference.
The second difference is when we're determining our fixed and moments we don't lock the joint.
That is akin not clear where no moment can develop we just don't lock it off.
So our fixed end moments are based on the proft counselee remodel Rotterdam the fixed fixed the model
so we jump straight into our table at this point.
We have a plus one 15 on the bay side.
We have a minus three one two point five on the B C side and so what do we need to add read add a plus
we're are at 150 160 suites our plus 1 6 2.5 but point 5 6 all of that 160 2.5 is going to go into be
a and that works out to be Plus 91 and point four for all of that balancing moment of a plus 1 6 2.5
is going to go into b c that's going to be plus seventy one point five.
Now at this point usually what we would do is we would carry over 50 percent of that moment.
But remember the 50 percent only gets carried over when the adjacent support is a fixed support.
In both cases our adjacent support is a paean support and so that means the carry over moment is zero.
This is crucial that we remember this that we recognize this.
So the carry over moment here is zero because A B or rather support day is it.
Pin and the carry over moment to support c is also 0 because it appears.
So we do not carry over a moments to appear in support.
So I'll summarize all of these little quirks at the at the end of this lecture.
What for now.
Get it very clear in your mind that there is no carry over moment.
No carry over moments.
To pain support.
No carry over a moment to appear support.
All right.
So we then have that that's the end of the end of the distribution.
Joint B is no balanced.
We can solve the final final moments at joint B and we would end up wit to forty one and minus to 41
but as expected zero in a zero at c.
Excellent.
So let's summarize those key differences so the key differences key difference is.
Where's pinned ends.
When I have a P and am done with structure Premeds our item we're gonna get plenty of practice of analyzing
these when we get to a worked examples before now.
What are our key differences.
The first want that fixed end moments and appear in zero or fixed end moment AB pain is zero.
What does that mean.
That would mean that we determine the order fixed and moment at the order end of our particular substructure.
We're looking at the fixed end moment at the other end is determined.
As a prompt cancel lever and that's all we saw here in this example as a product cantlie.
That is the first thing.
All right.
So we're looking at a proper counselee now.
The second beagle is but there is no carry over.
There is no carry over moments.
To appeared.
Okay.
The final mole.
It must be zero.
That final moments most PS zero.
And I'll say the only caveat on less an external moment is applied then obviously the final moment in
the bending moment.
I'm am more busier when less an external bending moment was applied.
Otherwise the moment at that point is going to be zero.
So the rest of this question precedes exactly the same as before.
So you really want to give yourself a little bit of practice and work through the rest of this Questions
that.
That's gonna involve looking at free body diagrams of a b and then of B C and working out your reactions.
Your shear forces and your bending moments so what other do is I will sketch out the end solution and
leave it to you to work true and confirm that solutions.
So make sure you can get this solution out of that sketch.
That is the solution that you should be able to work your way down towards Key difference here is zero
moment at a zero moment at c.
So as I said the take aways from this lecture at these two key differences extend moment at a PIMS zero
and therefore the other fixed and moment is determined based on the proft cantilever model.
And we do not carry over any moment to appear in the final moment is always going to be zero what opinion
on Lester as an external moment applied there.
So that is all we have to remember.
For structures that have p and Ms.