Now we're going to move to stage two.
So the logic here is that Grace.
I know what the proffering force is and it stops this thing swaying and I've got the bending moment
for the proper frame now I need to work at the bending moments for a frame where I apply that propping
force in the opposite direction.
So what I'm trying to do and we just label this as stage two.
Ultimately what I'm trying to do here is stage two is do this take a frame that looks like that.
Apply a noun propping force of P eagle to fourteen point one nine killed and.
And that will generate this way which is absolutely fine we'll call that sway Delta the actual sway.
And we want to work at the bending moment resulting from that sway.
But as we've seen already we cannot work out fixed and moments without knowing the magnitude of the
sway and we don't know what magnitude that sway is right.
So let's put that into words here.
So we want to apply P eagle to fourteen point one 9 killens as shown because remember we're building
this whole thing around the idea of superposition and we need to apply this this way force to counteract
that additional support that we've put in there for our stage for analysis.
So applied our force our shard to get my stage two moments before the spanner in the works is that I
don't know.
I don't know.
Delta the magnitude delta of the Sway caused by P eagle to fourteen point one.
Night therefore we can calculate can't calculate fixed and moments in order to do a distribution and
that's where the cleverness of this analysis technique comes in.
So we will instead apply an arbitrary.
Treasury Square a Delta prime and back calculate
p prime.
That would cause would cause Delta prime and this whole tank works because we know that there is a linear
relationship between the sway force that I apply to the structure and the bending moment that is generated
within the structure.
And that's why this whole little technique works.
So now we can determine our arbitrary fixed end moment so we can look at each of our elements of the
structure that sway and determine what with the fixed end modes be.
So remember a B.
That has undergone in our obituary sway Delta prime and that induces fixed and molins.
So the arbitrary fixed and moment.
Hey it is going to be minus because it's counterclockwise.
6.
E e r.
Delta prime over L squared the length of this member is 3 meters or L squared is gonna be 9 which is
going to give me minus 2 over 3.
He.
Delta prime and we're going to get the exact same for the fixed end moment.
On the other side.
So the fixed and.
Moment from a to b is also going to be minus two thirds.
E r e at a prime.
We can now look at member C D.
So the magnitude of the Sway is exactly the same.
And so f e m from C to D.
It's going to be minus 6 e Delta prime over L squared.
That's gonna be 6 sixes is 36 which is gonna be minus 1 or 6.
E a Delta prime and we have the same for the opposite end.
Okay so again it's the relative magnitude of these these fixed end moments that is of importance to
us.
We're not interested in their absolute values we just want to work.
We're really just concerned about the values of them in relation to each other so we can condense those
down into more manageable numbers are numbers that are easy to handle in our distribution so we can
do that next.
So let's start by stating what they were worked.
From our free body diagrams.
So because it's only their relative magnitudes.
We care about it straightaway.
We can't ignore or get rid of the E..
Delta primes out there.
So we just have numbers and if we multiplied across by 6 we'd end up with a minus 4 and minus 4 and
minus one minus one.
And that's what I'm gonna use in my moment distribution next.
Okay.
So now again it's this label these up these error obituary stage to them almost the moment is caused
by our obituary propping force last thing to do here is to determine what was the propping force that
would generate those moments.
And again we just look at the necessary free body diagrams in order to determine that propping force.
Taking some of the moments of point A.
So the lower is going to be one point five six colon heads in the direction shonn go straight on to
look at saeedi.
Taking some of the moments that point Dee will reveal the C lower.
P.S. lower is equal to not.
Point three one hill and events in the direction sharne.
Therefore p prime which would cause C A U S E to cause Delta prime is equal to one point eight seven.
Kill any evidence and an axe to the right place at that is p prime.
That is the force that is required to cause Delta prime.
So
is the logic here is if p prime equal to one point eight seven cars as my two prime.
Moments above.
Well then the logic simply is to scale.
We want to scale and to prime.
Moments when my sway factor by Lambda ego to p over p prime to get em two moments and M The two moments
are just a stage two moments are those moments generated by the actual propping force.
That is that is debt is generated in our structure.
So we can work out our lambda factor.
Our sway factor is equal to fourteen point one 9.
The actual propping force divider by the Arab literary propping force.
Do you mean a linear relationship which is the Lambda value of seven point five nine excellence audacity.
The the scale value descaling value that we scale at our retreat stage to moments by to get our actual
stage moments.
So again the best thing to do here is summarize all this in a little table.
Okay.
So it's a surprise what each one of these were these were at the top.
And one moment where might prompt structurer moments.
These were my arbitrary sway moments.
These were my actual sway moments.
And then these were my final moments obtained using superposition.
So now we're in a good place.
We are able to do the bending what we diagram around the sheer force diagram under deflect and shape
simply by considering free body diagrams of the different parts of the structure in order to determine
the relevant internal shear forces and various different span moments.
So all I'm going to focus on here for the rest of this video lecturer is developing the bending moment
diagram.
And so in order for me to do that I need to work get this span moment in member a b and also the span
moment in member.
So that's what I'll do next.
By just considering some free body diagrams.
Okay so we need to know what is v.b. Lower and what is he.
A.
B.
Cars.
The maximum bending moment occurs where the sheer force is zero.
So if I know the sheer force uneed around and I know the rate at which it changes I can more go where
it's equal to zero.
So let's take some of them moments about point a first to determine VB lower.
Now we can look at the soul of the forces in the horizontal direction to determine hate J.
Okay so I know the maximum moment occur is that a sheer force is zero and so the maximum the maximum
is at.
Let's take the sheer force at the bottom of twelve point six three and we know it's reducing at a rate
of six kms trimeter So let's divide up by six and we get two point one metres.
That is where the sheer force is zero.
So it's two point one meter is a ball of a.
So I know if I may get caught at that location and determine the internal bending moment it'll be at
its maximum value.
So Max is equal to minus five point six killer meters.
Okay.
The negative indicates at the moment is actually going the opposite orientation so the internal moment
location is a clockwise moment which will give me tension on the outside of the structure so tension
on the outside.
Because if I was to consider my structure as a cantilever are rather if I was to consider my internal
moment acting at the end of a cantilever.
We know it now goes clockwise and that would cause a counter lever structure to do this which would
give me tension on the outside and that's why I am able to say tension outside.
Okay all we need to do now is determine the moment in mamber b c c span.
Mormons are first of all I need to know what is the sheer force on either end.
Solidstate the soul of the forces are rather to some of the moments ago Point B.
So vcl is twenty seven point eighty three colonnades acting up.
And therefore we know that some of the forces in the vertical direction must equal zero.
So I can say that the B or is going to be my 50 kilotons minus twenty seven point eight three which
will be twenty two point one seven killens also acting up.
So to consider the span moment I'm just gonna make gay caught in the structure at point load of 50 kilometres
in the centre of the span.
So taken some of the moments about the court I can very easily say that Max by inspection is going to
equal the twenty two point one seven times to me a reliever minus 8 canoed metre moment.
So Max is going to be thirty six point three for kalinich metres and that would cause tension on the
bottom.
The positive still positive number here indicates that the moment is indeed anticlockwise and so off
I applied that moment on the end of cantilever it would do that to the counter lever which would cause
tension on the underside of that deflected shape.
All right so we are now able to go ahead and sketch out the bending moment diagram for this structure.
Okay so we have a base moment of magnitude at eighteen point nine.
We know the Udy L generates linearly varying sheer force diagram which means that the slope with the
bending moment diagrams will vary linearly which also means that the bending diagram would be a curve.
We know the magnitude at the top is going to have a value of 8.
And we know to what you say it was two point one meters above.
We're gonna reach a local max or min which actually turns out in this case is a local minimum so it
will appear at about two point one metres five point six.
So acre has to pass through those tree points so it looks something like this.
Where that minimum is five point six killed it in metres.
There must be moment equally among the joint which means that is also eight.
We have reached our point load.
We worked out a value of thirty six point three four thirty six point three four so straight line.
The sheer force changes sign at that location which means the slogan the bending moment changes sides
which means it has to.
There's essentially a peak in at any moment diagrams where starts going straight up again.
We know the value at joint C is 19.
Nineteen point three two as all the straight line between those two points.
Excellent.
Again momeni global joint dictates that this value out here is going to be also nineteen point three
too.
We worked at the value at the bottom was 13 points to 4 from our moment distribution and again a straight
line between those two points.
We can go ahead and sketch a deflected shape.
A classic sign of a frame that is swaying.
When you look at the bending moment diagram is when you have one of the vertical members that has tension.
Only on one side and we can see that would mamber a B so member a.b has tension only on the outside
which means the whole member must be bending over to the right like this.
And if the whole member bends over to the right it means our frame has swayed so that our sway Delta
and we can then flesh out the rest of the deflected shape according to the bending moment diagram drying.
Tension are where are rather where our benchwarmer diagram is telling us what face is our intention.
So with tension on the outside of the deflected shape that they feel linked that member we have tension
on the TARP or point of counter flexor pretty quickly detent moving to tension on the bottom.
Again another point of counter flexor and we end up with tension on the Tuck.
And again we have the same magnitude of sway Delta and then we have tension.
So let me see as just tension on the bottom.
That's tension on the top and we have tension therefore on the outside here and we passed sure a point
of counter flexer which means at tension flips the inside as we move back down into the fixed support
and we know that bending moments most Xed are rather deflected.
Shape most extant vertically out of that support.
So there we have our deflected shape for the structure so obviously the only diagram that's missing
here is the sheer force diagram.
So for practice you should go ahead and work at the sheer force diagram for this structure.
What really the main focus of these last two lectures has been in trying to understand the logical flow
of the analysis procedure.
What stage 1 analysis means.
What stage 2 analysis means.
Why are we going through this process of applying an arbitrary sway and how that linear relationship
between sway force and bending moment allows us to scale up those arbitrary moments or scale down as
the case may be to find our final stage two moments and then a final superposition between stage 1 and
Stage 2 gives us our final moments for our structure.
So there is a lot to dissect in there but hopefully over the last three lectures the picture has become
a little bit clearer for you.
Now it's really over to you to work out.
Some of these analysis for yourself.
Basically So that's what we're focussed on in the next few lectures.