OK.
So in this video we're going to pick up where we left off from the last one and we're going to finish
this question so we have the hard work so to speak.
Don so we are Stage One and Two hours is done and we have our bending moments.
Now we're just gonna work through everybody diagrams to fleshing out the internal shear forces and span
moments that are different required locations within the structure.
OK.
So at this stage I'm cycling through the shropshire and I'm looking to see for each of the individual
members within the structure.
What information do I not yet know and when I first identified something that I need to work out that
I don't yet know.
That's when I know I need to go and construct a free body diagram for that particular member within
the structure.
So I'm on the go and first of all look at Member B.
Because I don't yet know the sheer force at the right hand side of that mamber which we will call.
VB L which is the shear to the left of joint.
B.
Okay so.
VB L. Well fairly easily I can just look at this by inspection say VB out most balanced a downward effect
of that 10 Killian's trimeter acting over three meters and so straight away.
We can say v.b. l most equal 30 killed Newtons acting both in the direction shown because as soon as
the force is in the vertical direction most equals zero.
Right now I can move on to member B.
Okay so I have the shears there right of member or rather the shear to the right of point B within the
structure is or not.
And the virtual reaction at E which recalling here V E is also not just the sum of the moment.
Point B and work out with E.
Okay so each vertical reaction and a roller is thirteen point five three Killen heathen's acting in
the direction shown and the sum of the force in the wider action hero is equal to zero.
Words are positive.
And so v.b. or is equal to twenty six point four seven kill any heathens again in the direction child.
Our rights are were we the eudi l uniform we distribute alone acting on this mamber.
And so we're going to want to work out what is the maximum moment between points.
Be any.
So what is the maximum span moment.
Now we know at this stage that that occurs at a location where the sheer force is zero.
And so I can work out that because another sheer force at each end I can say the maximum moments occurs
at take the shear on the left hand side.
Twenty six point four seven reducing at a rate of five.
Killen isn't perimeter which is equal to five to nine.
So dividing by the the rate at which the shear is reducing will give me the point at which it is zero.
And that's 5.2 nine measures to the right of B.
OK.
So now in a way to make my court in order to evaluate the maximum span moment.
I'll equals zero.
The only unknown is Max.
And so Max
is eighteen point three to kill any in these years causing tension on the bottom.
All right let's go and look at member of the sea.
Taken some of them I was there a point to see.
And so the we observe is equal to thirteen point zero of seven colonnades
in the direction sharne and by inspection hate.
C is gonna be thirteen point zero seven the other direction.
All right let's look at member of a.
Okay so let's start with take in the some of the moments about Point B.
So hey Jay's eleven point eight nine killed Newton's horizontal reaction and a taking taken some of
the forces in the horizontal or extraction equals zero.
The be lower by inspection very easily is equal to thirteen point one one kill and needs to balance
that 25 acting to the right.
Okay again what's the what's the bending moment at the centre of that member at the planeload.
And we can say pretty easily that by inspection the moment at the centre which lets us an made if I
were still of course I would very easily see that and Mitt was just going to be for example the horizontal
reaction times the lever of 4.5 metres we'd end up with eleven point eight nine times that lever of
four point five which is fifty three point five one Killen hidden meters.
Okay so we have considered each of our free body diagrams we've identified all of the unknown information
that's required in order for us to draw our sheer force bending moment diagram and deflect the shape.
So now we can just crack on and start summarizing all of our analysis with does diagrams size the sheer
force diagram first.
Okay so let me see the sheer force.
This was thirteen point zero seven.
Again I'm not worrying about driving aren't signs for the sheer force or indeed the bending moment because
the diagrams themselves indicate what sign the sheer force or bending moment is.
OK.
So we can use the information embedded in that sheer force stick around to help us construct the bending
moment diagram.
So I can start off by noting arm the values that I know am at known locations.
So for example we had a value causing tension on the left hand side of forty seven point seven killed
it metres appear at the other end of that mamber the internal bending moment caused tension on the right
hand side and had a value of seventeen point six six.
So seventeen point six six now.
The bending moment or rather the sheer force stick around for member B C is a line parallel to the structure.
And so I know obviously that the bending moment diagram must be inclined between those two points.
And we use the same logic Senshi to work our way shred structure.
OK so the last diagram that we want to do to complete the picture is a qualitative and deflected shape.
So this is just a diagram that gives us a feel for how this structure is going to deflect.
Now to build this diagram we are relying on the bending moment diagram to tell us are three each member
which which side of the member is in tension and that tells us they give us information on which way
the structure is curving.
We also know that this is the sway structure that's gone sway to the right.
And so we can use that knowledge as well while building out our qualitative deflected shape does a good
diagram to summarize our reaction forces on.
Now I haven't distributed the vertical reaction between member between joints a and C..
I'm simply saying with talked in a previous that read how you could distribute that vertical reaction
based on the actual differences of those two members were facilities simplicity's sake.
I'm just going to say here that the reaction that passes true the reaction force that passes through
a and c is going to be 30 plus twenty six point four sevens So the sum of the forces along that line
of action.
With fifty six point four seven kill and it happens and again we discussed in a little bit more detail
previously how you could go about treating that vertical reaction force.
So I'm just going to summarize it.
There can be a good idea to draw on our
apply loads.
Nix the process of building to deflect that shape a little bit more.
Sure.
Right so it sways to the right swing and a look at member of b.c first.
And I know it's that attention on the left flicking detention on the right so it's gonna do something
like this.
We have a point of Fletcher here.
So tension on the left then tension on the right we have tension on the right hand side about tension
for a very short amount of very short distance at the start of mamber be a weight very quickly.
We know from the bending moment diagram changes the tension on the other side.
The right hand side of the way to the pin support so we can have another point account reflection there
and we really only know that because depending on what diagram is telling us there is a point of zero
moment there.
The horizontal mamber is pretty straight forward.
Remember the whole thing has shifted over to the right.
So we're going to have something like this frac Hans lever bending down and then we have tension on
the top here.
Point to counter flexor and then tension on the bottom so tension and somewhere along here.
We flicked attention on the bottom.
Noting that the entire structure has shifted over to the right by an amount equal to Delta.
Okay so that is a deflected shape for our structure.
So in those tree diagrams we have summarized our complete analysis for that structure.