Okay this structure similar to example too with the exception that we have a cantilever and an E and
we have a moment applied to that cantilever.
We also have an open pointing point load between bnc at point F..
So that's going to mix things up a little bit in terms of the distribution.
That's absolutely fine.
We just have to follow the same rules and procedures that we've established sulpher.
So we'll go straight on and work out our stiffness.
So if you're walking along with this question if you have any tried ash go ahead and try and work out
your differences yourself first before you watch me do.
So assuming that you've done that we can say that the stiffness k a B from embryo B is going to be a
4 eai Overall because we are we are fixed point and continuous support at B.
So B is for E R.
Or the length of eight meters which is not quite.
And k b c will be the same it's going to be for me and I over the span in this case six meters and that's
going to be not quite six.
Then he I.
Now the stiffness for c d is going to be three E or it out because at c the structure is continuous
or support giving it a rotational resistance when at D.
It does fly over this support but it's just a cantilevered slut connecting the structures not connecting
to anything and E.
And so there's no rotational resistance provided inherently in the structure.
And so we've got three overall which is going to be three or four which is not a point seven five.
We are at the difference of D E is not required because there will be no moment transmitted intuity
we can determine what the moment at is at the very outset of the question and that won't change.
So moving forward to look at our distribution factors this is the usual procedure now the distribution
factor for be a they won't cross the line.
And now we'll just say that it is not.
Point 4 3.
And the distribution factor for DC.
The left hand side of that cantilevers is going to be worn.
And that's always the case would a cantilever because the distribution factor for the actual account
leveritt s d e will always be zero because as I said we never distribute moment into the cantaloupe.
So now we can look at our fixed and moments.
Well we have a eudi l. acting on a fixed fixed structure.
It is after we have fixed the joint that be until it is fixed fixed structure and we know at this stage
we've seen enough example that the fixed and moment.
In that case for maybe it is going to be minus W L squared or 12 which we can work it to be minus fifty
three point three Kule meters.
Similarly he's fixed in Longwood on the first side for B and will be the positive version.
So these are the reaction warmers that would develop at those fixed supports.
Now we can lock in a little bit more in depth at the fixed end moment for mamber APC.
So in this case the fixed end moments will be the opposite sense to what you are used to seeing because
in this case the load is pointing OK.
The seventy five killed on it and zagging open it and remember these tricks and moments are the reaction
moms that would develop but those support B and C in response to the loading that's all in this case
those reaction moments would in fact go the opposite to the way that we're used to seeing them and so
we can say that the fixed end moment for B C is in fact clockwise now and so what it closes.
And we've seen the formula for this before.
It's gonna be p times.
We see a KNOW WHERE THAT IS A and those B times X so it's going to be 2 times be squared which is 4
squared all or elsewhere were a total length which is 6.
So it's going to be sixty six point seven kill and hutin meters and the fixed end moment for CB is going
to be seventy five times B which was four times square which is 2 squared or 6 squared which is minus
thirty three point three.
Kill Newton meters.
Okay and then the plastron we'll consider it again we'll look at this in a bit of detail is see.
So this is going to be a difficult and moment for ecd gomi Bay style props Carnally remodel and that
is always the case when we have a cantilever let's say it right here so that's Dee and we know it for
destruction as a cantilever there that is a cantilever there.
The kickstand moment on the adjacent spam are always going to be determined based on a proper cantilever
model which is what you're seeing we drill here.
It's a fixed and moaned for proceed is a minus W L squared over 8 foot.
Can't lever which will be minus 10.
Okay and obviously fixed any moment for the first zero because it's a pin.
D C is zero.
Okay we can quickly consider the cantilevering at the cancellara moment currently removement.
The fixed mortgage is going to be the reaction moment.
F e m g e e equal to minus 50.
Okay so we can now go ahead and work on our distribution.
OK so balancing a joint B first the error balance moment is a hundred twenty and so I need to balance
that with a minus 120.
So Forty three percent of that will be minus fifty one point six.
And on the far side minus sixty eight point four join our balance and join C..
We have a forty three point three a positive forty three point three required so that will mean that
will mean twenty point three five on the left.
Twenty two point ninety five on the right.
And then balancing oppty cantilever Well the negative 50 at deee must be balanced by a positive 50.
The distribution factor is one first for D.C. and so forth 50 goes in to balancing the cantilever on
the left hand side of course that's your plus 50.
Now we can carry over our carry over moments.
Now balancing a joint began.
We have to have a negative four point three h a negative five point eight on this side we have four
point three two and four point eight a room and the cancellation balance that the stack and then it's
left well alone carrying over.
So balancing.
Again we are going to have a negative not point nine three and a minus one point to three plus one point
three six and a plus worn loss worn point five four.
Okay.
Now at this point we evaluate and we say well joints are all balanced now.
If you not around carry overs we're going to win.
On balance the joint put the amount by which were on balancing them is really negligibly small.
So we want bother gone to do another carry on.
We reduced things sufficiently at this point so we can some the moments in each column to arrive at
the final moments for each joint so at a negative eighty one point to nine a negative three point six
warn a positive three point six warn negative forty four point three seven and a positive forty four
point three seven and then a casually read balanced those plus 50 minus 50 and plus 50.
Okay worth at this point just noting down in what way.
Whether there's tension on the top or bottom of each of those joints so the negative 81 point to 9 that's
a anticlockwise moment applied to the left hand end of a structure.
So that's like saying.
Negative moment negative anticlockwise would apply to at left an end at the structure at which would
induce tension on top.
A negative three point six one applied to the right hand side of a structure is a moment like this which
will induce tension on the bottom and do not Same meant lexus's each way accra's we would work at that
that's also tension on the bottom and that tension on the top.
Okay so we are back at our process now of analysing free body diagrams.
So once again like I said in the last example that that the important thing for you to be able to do
here is to carry this true frahm distribution table all the way to your fine of sheer force and bending
moment diagrams.
So as I said before I strongly urge you to go and do that for yourself.
First of all if you have any problems combat and washed rest this video.
So now I'm gonna process the freeway diagrams in turn and flesh out the remainder of the analysis.
Okay so looking at these are these are my free bloody diagrams and am looking at member a.b first.
So VB L is equal to twenty nine point three 9 killer Newtons.
Okay.
I mean the some of the forces in the right direction.
This is equal to fifty point six one kill and it means again in the direction of silt.
So following exactly what we've seen previously the max moment.
So one is equal to forty six point seven eight kill 11 meters.
And that is in the orientation that I Drano which would induce tension on the bottom of the structure.
All right.
Moving on to B C.
And so vcl is equal to thirty one point seventy nine.
Again that's the direction of sumed and I assumed that it was going to act downwards.
Which is correct.
And that is a positive would be a positive.
Shear.
According to a defamation signed convention.
Now we can take some of the forces in the wider section.
And so VB or is equal to forty three point two one killer nuisance again in the direction assumed which
was down which arm this side of the structure a downward acting sheer force would be a negative sheer
force.
Okay let's cut underneath the point load so called out f to determine what the bending moment is there.
And so m f is equal to eighty two point a one killiney meters.
Okay.
And again that would induce in that orientation tension on the top of the structure.
So we think go ahead and not mamber C D.
And so the v D L is equal to thirty three point five nine colon events.
Taking some of the forces in the y direction.
Lecy or is equal to thirteen point five nine kiren humans.
Get acting down.
Right.
So we can determine our maximal moment location.
And so m three is equal too.
It's actually equal to 11 weight 0 9 killiney meters causing tension on the bottom.
All right now we know enough to be able to summarize our solution with our earth's structural diagram
indicating our reactions and our shear force and bending moment I agree.
Okay so now we have the complete the complete summary of our solution.
So you will at this stage I certainly hope you have worked out how to go from the end of your distribution
table down to this complete summary solution.
So you should understand the relationship between the loading on the structure and the shape of the
of the sheer force diagram.
Because we're only determining shear forces in these last three examples we've done.
We've only been determining shear forces at discrete locations and so that relies on you having the
knowledge of how the sheer force diagram varies between those discrete locations and not how it changes
or has sheer force diagram varies depending on the type of loading on the structure.
And then similarly when we know the shape of our sheer force diagram we can use that to help us determine
the shape of the bending moment diagram between the discrete locations that we've already determined
the bending moment so we make our cuts and our structure.
And we determine span moments and the distribution has given us the moment that supports.
But we need to have some knowledge about how the bending moment will vary between those discrete locations
and that's back to more fundamental knowledge shear force and bending moment diagrams which I've done
I've also done a course on which if you're if you're fuzzy on you should maybe consider taking a look
at that course.
Okay what is.
As far as using the moment distribution method to analyze statically indeterminate beam structures.
I think we've more or less corret of everything that we can.
You've seen these different types of lodin that we can apply.
How that plays over We've seen how to deal with cantilever and how to deal with P and ends on what they
introduce into the distribution process.
So you should at this stage if you've managed to do those tree work examples you should be more or less
well equipped to tackle and he statically indeterminate been.
And so now we are going to draw a line under beams and move on to applying this method to structural
frames.