Ari in this lecture what we want to do is take everything we talked about in the last lecture and apply
it to a very simple way frame structure.
So we're gonna take destruction.
This is a simple portal frame which has a canopy which What role what we could call a canopy to the
right hand side which is basically just an overhanging cantilevering mamber with a point not applied
to the end.
So the fact that this frame is unsymmetrical and also has an on symmetrical loading arrangement means
that it's going to sway.
So let's apply our principle of superposition to this framed structure so I can take that structure
and say that Dad is using the principle of superposition that will be the same as a structure like so.
So the first one would be the structure.
In a proper state.
So were adding support to point B on the structure which is essentially stopping it from swayne's or
propping the structure so that will be stage worn.
Analysis.
Now we have to take into account the sway moments at the moment generated by sway.
And so if we just focus on the stage what structure of four seconds what we're going to get is a some
force developing at support B.
And that force we will call P A propping force.
We now need to counteract that propping force by applying the reverse of that force.
Aren't you a sway stroke.
And so just to note that this is opposition to the propping force because we're essentially counteracting
the effect of that propping force.
Opposite to propping force or whatever I would gather propping force to be from my statement analysis.
I apply the same magnitude but opposite direction in the stage to analysis.
And notice again at the Stage 2 analysis does not have any other external loads on the frame.
And the propping force.
So stage one analysis is what we've looked at in a previous section on this course and so that should
be no problems now in the interests of completeness.
I'm gonna go and complete the stage when ours is here in this video but I'm not a rattle to it pretty
quickly OK.
So Stage 1 and hours as was the first thing we got to do was to determine what is the propping force.
So a standard prop frame analysis will work its still.
Distribution factors fixed n moments.
Then do a distribution table and use free body diagrams to work out.
P.
Okay so we've reached the end of our distribution.
All that is fairly standard stuff which hopefully should have no trouble with after completing the last
section of the course.
Now we have to think about what is the objective here that we're trying to achieve is and it is principally
to determine that probing force P So we only want to consider the free body diagrams at this point which
are going to be helpful in allowing us to determine that propping force p off we look at the structure
again.
Really what I want to find out is what is the sheer force at the top of member a B.
So that's a B C and D.
I want to know what is the sheer force acting at the top of nambour a.b. and one is sheer force acting
at the top of member Cedi because they're essentially gonna be how I determine the cropping force P
So I don't need to consider any more than that.
The only free by diagrams I consider at this point are the ones that will allow me to determine the
propping force p.
So the first free body diagram I consider is a B.
Oh and we should also note that these are M worn moments.
We're going to need these later on.
So we've got as much information out of member a.b as we need at this stage and we can move on to members
saeedi.
So let's zoom in and think about what exactly is happening at the top of each of those members.
If I drop a member a B.
Well I've worked out that the sheer acting of the top of member a.b So that's be an It's A is a sheer
force in that direction.
So thirteen point zero two.
And and if I drop mamber seedy What have I got.
I had the sheer force acting that way and that was going to have a magnitude of fifty three point one
three hour right now on the other side of the kulisz Cause I remember we had to have taken a cut through
the structure to reveal those internal shear forces on the other side of that court.
We have the member.
Which was member B C.
And if I look at the sheer force acting on membered D.C. It's going to be opposite to the sheer force
acting on the old the opposite side of the court.
So what we're going to end up with is the thirteen point zero to acting this way.
And the fifty three point one tree acting that way.
So what we can say from that is the net force that's acting through through member B C is going to equal
taking forces to the right as positive.
It's gonna be fifty three point one three minus thirteen point zero two.
So the net force is a 40 point one 1 killens acting to the right.
All right.
Now what does that mean.
That means that the force of the propping force that's restraining that heart stopping that structure
from moving after that arise that restraining our reaction force which is our propping force is just
going to be forty point one one can Ethans acting to the left.
Okay so it's just a reaction.
So the propping.
Or restraining
force is.
Forty point one won't kill is acting to the left so that our propping force are so.
What would we say P is equal to if we wanted to dispense with the arrow and only have a number we would
put in at minus forty point one one that allows us to get rid of the the arrow that are propping force
so that essentially is our stage one and ours is complete.
So what do we got from our stage 1 analysis Well we have determined a stage one moments so propped frame
moments.
And we've also determined the magnitude of our propping force.
So that is where we move on to stage two.