So now we move on and look at our sheer forces and our reaction.
So again I started off by thinking about the shear forces the shear forces and reactions.
So as they said there's nothing hugely different.
For iframe and that we haven't already Don for a beam structure the only slight difference is with a
frame.
We're also aware that actual forces will develop within some of the memories but in that frame we're
going to have to track those actual forces true the structure in order to work at reactions ravina Various
different situations.
So we'll we'll we'll talk more about that when we get to it.
So the way to handle this systematically is to look at my structure and break up every single part of
that structure.
I consider a free body diagram of these subsections already substructures within that structure.
So I'm basically just going to look at each straight segment of my structure and consider it equally
agrium of that segment.
So let's start off by looking at member Beadie.
Now why is why do I want to look at this.
This substructure on its own.
Well I want to know what is the sheer force that develops the G the L to the left of joint.
Be clear and it's not that shear forces.
That's why I'm considering this substructure.
Well it very easily irritates some of the forces in the y direction.
They know they have to equal zero.
I'm assuming opera's forces are positive.
Based on my static sign convention and we can very easily see that VB L. sheer force to the left of
B is going to be 30 killiney happens to balance that Udy L..
Now I note as well that that is in fact a negative a negative shear based on a defamation sine convention
because that sheer force would tend acting on that court.
Right hand face that sheer force would tend to rotate that element Beadie counterclockwise.
And based on a defamation sine convention I'm calling that a negative shear so negative.
Shear the car's it causes it causes.
A. clockwise rotation.
And this was this idea of a defamation sine convention was something that we covered in the mastering
shear force and bending modes course right and so now we can move on and look at member B joffre body
diagram.
See what we don't know and work out are user egli room equations to find out the values of the of the
forces that we don't know.
Okay so in this case we have an unknown shear to the right of point B which is v.b. or and we have what
is actually the vertical reaction at joint E which recalling the e in this diagram.
So the House to take some of the moments about B.
I would be able to determine straight away what v e e is.
So with minus 50 one point seventy five clockwise Mormans positive minus 8 deleveraged times the E plus
the moment effect of the d l which is 5.
Acting over 8 meters with a lever for meters.
All that equals zero and therefore the E is equal to thirteen point five three Killen is in meters.
Excellent.
Got a box around it make it stand.
Right now we can't take this song of the forces in the y direction.
They all have to equal 0 0 broods are positive and we end up with VB or.
VB or.
Plus thirteen point five three minors eight times five equals zero.
And so the B or is equal to twenty six point four seven kill any lines.
Okay so that allows us to look at the bigger picture now in terms of the vertical forces acting on this
structure.
Okay so what do we mean by that.
Let's just think about vertical force equilibrium at joint B.
So let's consider.
You'll see why I'm sort of doing this.
Once I gotten to the end of what I'm about to when I'm about to explain so I'll consider a vertical
force equilibrium.
Joint B.
And let me see that's isolated from the structure of and structure below.
Now we've looked at the cantilever to the left and we'd seen that in fact he shear on either side or
rather the shear on the right hand side was 30 which meant the shear on this side was an breeds of 30.
And when we add a sheer force that we've just worked out.
VB or we end up with the shear.
Twenty six point four seven acting operas that are right.
The Kosh which would mean on the left of the court.
It would act down twenty six point four seven.
Now I have not drawn the external forces arm they're just.
Just for clarity let's think of what we have here.
We have joined be subject to a combined fifty six point four serve and kill any fans acting down.
So if you like the influence of the effect of the structure to the right and left of joined be the effect
of those substructures is to pull down joint B and cake.
So the downward force that was break this was do the step by step so the downwards force.
Acting on B joint B is going to equal let me see it's gone in equal that thirty plus the twenty six
point four seven which is equal to fifty six point four seven kill any heads are right.
And that simply means that the vertical reaction through B must be fifty six so as jump down block structure
here.
Therefore the vertical.
Reaction.
Through B.
Must be fifty six point four.
7 killin it is OK.
Now we are at a three way crossroads here.
We know that there must be a vertical force acting operates here of fifty six point four seven.
Now the question now arises is where is that reaction.
Where does that reaction develop Let's go back up to the top of the structure.
Here are the original structure.
We know we must have fifty six point four seven acting Brits trivia.
Now the question is does that all act down here or does it act all up here or is it a combination of
an opera's force developed at joint's C and in auberry's force reaction force developed that joint.
And the answer to this question really lies in what assumptions we are making for our structure.
So we could do a couple of different things here.
So the question we're trying to wrestle with here is where does the reaction.
Fifty six of fifty six point four seven kill and holdens develop Well several options as I say.
The first one is we said you remember at the very start we said that we're neglecting one of the assumptions
is we neglect the axial change in length of our members change in length.
Right now what does that mean.
If we're saying that the vertical members there is no change in length.
Well the consequence of that is they must have infinite stiffness and k..
Well if they have infinite stiffness the top member is just as stiff as the bottom member are either
member above.
B is just as stiff as the member blowby and if they are equally as stiff well then that simply means
that the reactions will be shared.
The reaction will be shared 50 50 or so.
If I sketch in my structure here it was very roughly speaking it was something like this.
It's like that.
Well what that would mean is that the member a bull B would go in to tension because remember he's being
pulled downwards by the loading that's acting on this member and the loading It's acting on this member
has been pulled down.
So what that would mean is a bushby structure would go in to tension and below be a structure it would
go in to compression.
Okay.
And that makes sense.
And if there if there's no change in length and therefore they are infinitely stiff and therefore they
are equally as infinitely stiff.
Well then they got a shared reaction so you're going to have half of it develop here as a opera's reaction
force of fifty six point four or six over two and you're going to have half of it develop down here.
We get half of fifty six point four seven overtip.
So that's one option we could say and that's perfectly valid once you sort of lay out your logical reasoning
for why you assumed half and half.
And I've given you the sorrow that the logical development of that argument that's fine we can do that
we can say if we neglect the actual change in and which was one of our initial assumptions to our analysis
therefore we're dealing with infinite axial stiffness.
Infonet axial stiffness and the consequence of that would be that the load the reaction the load is
shared
50 50 between.
Structure.
Bar-lev and below the.
Right then capital B.
So that was one option.
And what's another option.
Well let's imagine for a minute that we take our sanction our starting assumption that we neglect actual
change in and then we put that on the shelf.
We press pause on that assumption and we say well for the purposes of working out the distribution to
vertica reaction between the support above and below.
Let's let's let's say that the structure of bauby can get longer and the structure blowby can get shorter.
So that would bring into play the axial stiffness of the structure a ball.
That's that's game that went over so we can say we're going to parse.
Assumption.
Ignoring.
Changes in lengths.
So that's the precursor to this discussion.
We're on a path that one of us say that that no longer applies well.
Well that would mean that member in the or short for a member be B it will go into compression.
Go in to compression and member B C
would go in to stage.
So there is a much more realistic scenario of load sharing between the two and a degree to which the
load is shared is gonna be proportional to the actual stiffness of the two members.
So it's very very analogous to how we've looked at moments being distributed according to the flexural
stiffness is within our moment distribution process.
Replying the same methodology to the same logic to distribute axial loading between the mamber in the
top and the member in the bottom so we could follow a very similar procedure and we could say the degree
in this model the degree to which
load is shared.
Between members is determined.
Is determined based on their axial stiffness.
My right hand and axial stiffness is simply E A R L R right and so we can work out an actual stiffness
for each member.
We could say the actual stiffness as that's put in a hat or a barrel that the actual stiffness ribisi
is going to be is.
Well let's zoom in a car since it will be 1 over L which is 1 or 5 does not point to and we could say
the actual stiffness for man rippy is worn over L. 1 or 9 which is not point 1 1 1.
We could sondos zob and we with get no point 3 1 1 and in exactly the same logic a manner that we distribute
moments we can work out a distribution factor for axial load.
Also we could say the distribution factor for B C is just not equate to the actual stiffness of b c
o with the sum of its differences not point 3 1 1 which is not.
Point 6 for 3 and we could say the distribution factor for B is equal to not point 1 1 1 0 and our point
3 1 1 which is not Point 3 5 7 are right let's go and first EP and then say that the tension in B C
would equal not points 6 for three times the fifty six point four 7 which would equal thirty six point
three one Q Newtons.
Excellent.
And then we could say that the compression in be a used zero and gone with this is equal to not point
three five seven times fifty six point four seven.
And that is equal to twenty point one six it is all right now that's the other option as a second option.
So a little bit more realistic you might say.
Certainly moralistic than saying 50 50 between the top and the bottom.
Even all the 50 50 approach was much more in line with our initial assumptions through our wider analysis
of frames w a turd show just a turd option.
The third option would be to say.
Real stink real is.
Re is Dick Lee.
Model the support the supports realistically models supports.
Right so our structure did so.
Think like this.
MRS.
We have pin here.
I think we said we had a pin here and we had a fixed support up here.
Now while we would have to ask ourselves to follow down this third route is is that realistic.
Is that actually what's happening up there.
Because what that's saying is that whatever the structure joins on to or at support.
See here that is capable of providing a vertical reaction it's capable of developing a vertical reaction
that might not be what we actually have because if we have it this is a column does vertical columns
supporting a larger structure.
And we're only seeing part of the structure in this diagram.
All of the vertical support may be developed at the foundation here.
And a better way to draw this a moralistic or SRU real-life ways you draw that would be as follows.
It would be to say I actually have something that looks like this or the frame geometry has the same
frame geometry is the same.
And again we we may have a pin down here.
But what I actually might have take a dashed line across here signified top of the structure.
What I might have.
Is something like this.
Now bear with me.
What is that.
That is a support that resists moments.
I rather resist rotation so there is no rotation.
No lateral movement.
What does it allow.
Vertical movement.
I sold of like if I just zoom in.
What does that actually look like.
Well I have my structure here coming up.
Let's imagine it's some form of steel beam and what I'm saying is I have some clamping roller structure.
Right.
So.
So that clamping roller structure stops any rotation will appear so weak and if it stops rotation it's
capable of developing a moment.
And it also stops and he lateral movement.
And so therefore it's capable of generating a lateral reaction force.
What it doesn't stop so it's doesn't stop vertical.
Okay so the paint that the structure is capable of moving operates and downwards with these rollers
here.
And so we can't develop a vertical reaction and that might be much more realistic to what we have in
our real structure.
And so in that case all of the reaction will be generated at the base of the structure and cake.
And there will be no load sharing because it will be no capability of developing a vertical reaction
or a Join Sea.
Okay.
So there are the two three possible options that we could have and each one of them is perfectly valid.
It's really don't you as the engineers to decide what most realistically represents your structure for
the remainder of this course.
I think what we'll do is I like option 2 because I feel like it's the most nuanced of the different
options.
But the key thing for you really is to understand the difference between the tree different options
of the tree different treatments of the vertical reaction that gets developed that gets shared between
two possible supports.
All right so now we can get back to the question in hand.
So we've dealt with virtual force.
Now what we're going to do is look at the horizontal forces.
And in doing that we're gonna consider the virtual members where the sheer force is on those and therefore
horizontal forces.
So looking at mamber ABC.
Okay so we have no external loads applied at aren't a member ABC B.C. But we do have moments in juist
and therefore we're going to have reactions on the end.
So the first thing to say is what is the moment at sea.
Well it was a fixed support and so it was subject to a carry over a mortgage which wasn't immediately
obvious because I didn't perform the distribution in a table.
But we should be comfortable with the fact that if I have a moment at B and I have a fixed support and
see 50 percent at that moment will be carried over as my carry over mode.
Sorry got not point five times twenty five point one one which gives me twelve point five six Killen
had meters as my carry olver moments.
OK.
So we can no work.
What is hate chc is the horizontal reaction at joined C and hate.
Be You is the horizontal force are just a bully.
B which essentially is a sheer force ram just calling it hate here because it's horizontal.
So if I was to take the saumur the moment about C some of the moments about C equal to zero clockwise
positive.
What do we end up with.
Minus Twenty five point one minus twelve point five six plus five.
HAGE be all equals zero.
And solving for hate b are rather h.b. Opper each b Oprah is equal to seven point five.
Three.
Kill any instance and that is the sheer force and immediately a ball joint beat.
Taking the same the horizontal force being equal to zero we can see by inspection very obviously that
Hage C is also going to be seven point five three kill and humans acting in the direction chahal right.
So that was quick and easy.
We can now move on and look at member a B.
Okay so we have hate B.
Lower sheer force blowby And with hallé the horizontal reaction in a.
So again taking the sum of the moments below point B.
Thirty one point eight six minus four point five times twenty five plus nine.
Hey they legal 0 or solving for.
Hey Jay.
We have eight point nine six kuehl and who.
And again some of the forces in the x direction by the horizontal direction would very easily give me
Hage be lower Eagle that sixteen point O for accuracy.
Again the direction show.
So now we have considered all of the vertical members which have given me all of the horizontal shear
forces which allows me to determine what is the propping force of the propping reaction horizontal reaction
developed on the right hand the right hand support.
Right so let's consider horizontal force egri of being with joint B.
And again I'm not going to draw on the external externally applied loads just for clarity.
So what do we have.
We draw the shear forces that we got for the opera member of bushby.
We had so we had a reactional period seven point five three a sheer force of gold be a seven point five
three which meant that the shear on the opposite side of that court acted in the opposite direction
which was seven point five three doing the exact same logic for every member of the lobby.
What do we have.
Well we worked out that this was a shear of sixteen point four which meant on the opposite side of the
Joint we had sixteen point four nets for completeness draw on the horizontal reaction a eight point
nine six races so now we can see we can isolate joined B we can look jointly and see that we've got
a combined.
We have a combined seven point five.
Three plus sixteen point four which is gonna give me twenty three point five seven.
Dan is this force pushing jointly to the right which means that we have to have a reaction opposing
it to the left which is our group which is going to act a true member of B where E is after right here.
And that means on the opposite side of that court it's gonna act that way which means that this member
is in compression which means that the reaction force develops E is to the left of twenty three point
five seven kilotons.
So that are propping force right there and that's all developed by just tracing the loads.
True our courts so our free body diagram gave us that and that's the fact that we caught and isolated
the structure above and below.
This meant that we could tell on the opposite side of those courts the sheer force acted to the right.
In that case and answered a right in that case.
And that meant that we had to have twenty three point five seven opposing that those two forces and
was.
We traced those that reaction force.
True the quotes that we made to isolate the joint we were able to work out that the reaction force on
the right hand side.
He was in fact the magnitude of twenty three point five seven acting to the left and what that tells
us is that fray wants to sway to the right.
And if that reaction at e was a roller instead of a pin it would sway to the right with the fact that
it's a pin it's able to develop that horizontal reaction of twenty three point five seven.
Now in order to flesh out a benchmark diagrams are the last thing that we have to do is work out our
Span moments.
OK.
We have the moments at all of the different junctions.
Now we need to work out where the moments in the span's an answer very very easy.
OK.
We know the reaction was eight point nine six at the point at which we make this caution there is 25
killens coming on.
And the moment we want to know is the internal moment reveals we're making that Kosh up four point five
metres from a.
Were calling an arm span by inspection.
Span is equal to eight point nine six times four point five meters which is forty point three to Killen.
OK.
So if you talk to some of the moments about that koosh you would arrive at this equation now that you
are a span moment we want is a Span moment in memory be we.
We have a udrs acting as a member of the year and so as engineers were gonna want to know wherein that
beam is the maximum moment occurring and what is the magnitude of that maximum moment.
So this is a process that we've we've seen several times before.
Right.
What I'm gonna do is I'm gonna make a khush at a distance x to the left of joint B.
So I made a cloche assistance exit left of joint e that has revealed an internal bending moment which
is a function of x and if I take the song of the moments about the chord that equals zero sume clockwise
or positive I end up with m a function of Max minus thirteen point five three x plus two point five
x squared equal to zero.
OK.
So I essentially have an equation for the bending moment in that member as a function of x.
So I have an equation that describes a curve that curve just happens to be the bending moment diagram.
So although I want to find out the location of a local maximum in an he in and he curved us the scribe
by a polynomial.
I just need to differentiate that polynomial that gives me the slope and once I have the slow by can
let the slow equal to zero because a slog will equal zero on a local maximum in a.
And then that essentially will be the point of maximum moment that I could saw that back in to my equations
to get the magnitude of the mode allocation right.
Sounds an awful lot more complicated and is I simply want to take Didi's X of M of x which is going
to be thirteen point five three minus 2.5 X let that equal to 0 0 because that is the slope of my bending
moment diagram so letting it equal to zero to thirteen point five three wins two point five.
The XY was a zero and therefore X is equal to two point seven zero or six meters.
And that is my point of maximum moment.
And I can so now value back in to that equation to get m at x equal to two point seven zero or six is
equal to.
So that's going to be eighteen point three one kill and it means I was right.
Now in order allocation and magnitude of the maximum bending moment in member B.