Okay so we have a lot of the heavy lifting done in terms of explaining what's new or what should be
new to you and this course which is the moment distribution procedure stiffness is fixed in moments
etc..
So there was a lot of there was a lot to cover in those first two lecturers.
So in this third lecture it's gonna be a little bit more easygoing because we're gonna be relying on
stuff that you should already know coming into this course.
So that basic and equally in we're going to be reverting back to equally we here.
Now what the moment distribution method basically does for us is it allows us to work out moments at
joints what we still have a lot of blanks to fill in before we can say aronne hours of destructionist
complete because we need to work out what reactions are we need to work out what the shear forces are.
We talk at the bending moment in the span.
But the good news is that we can use static equal theory for all of this.
So it's gone in the arm not known you if you like techniques or methods.
It's all the good old fashioned equal area.
So what we're going to do is we're going to try and flesh out our knowledge of the example structure.
We've been using up until now.
So we we do this by basically considering each of our subsections each of our so we have two subsections
in our class structure A to B and B to C we consider a free body diagram of each of those subsections
we draw on all of our unknowns and we just use equilibrium to identify what those nodes are.
So let's start off by looking at mamber a B.
OK.
So what are we done here.
I've drawn a freeway diagram.
Remember free body diagram.
We don't bother driving on sportswriting like that.
We literally just draw on the forces and moments styer affecting our structure so we can see when we
do that we have if the way actually that we think about this is everybody diagram we've got a Kosh here
and we have koosh here.
Now those courts have revealed internal bending moments which our Mormon distribution analysis has determined
far less.
And they've also revealed to shear forces the sheer force that a V8 which will also be the reaction
at A and we have the sheer force.
VB L where l signifies it's the sheer force to the left of the support.
I'm going to start off by taking this on at the moment about eight that would give me v.b. l..
So if I was to take some of the moments at that point.
Well I know that one of the moms were at a zero I assume clockwise are positive.
What I wake up minus taking some of the moments.
Mine sixty nine point seven plus one sixty point seven plus.
Let me see four times the one hundred killiney an external load when it's 8.
The B and that all equals zero.
Solving G B L is equal to sixty one point three 8.
And who is excellence.
Now what have we got no V.
So that is your point 3h point upwards.
And that would induce that that sheer force according to you I defamations sign convention would.
And again we went into that detail about that site and mentioned in a previous course that would induce
that of birds shear if light if that apply to an element like this and elements here that would cause
an element to rotate back which would be an anticlockwise rotation which would mean that is a negative
shear according to a defamation sine convention.
So if I share if the chute that's applied to the end of an element would tend to cause that element
to rotate clockwise.
Well then that would be a posit shear and that in a nutshell is all there is to that defamation sine
convention for sure.
Let me just make a quick note on that defamation.
I won't state the convention but I'll just say that it is based on the defamation signed convention
and it causes in anticlockwise are counter-clockwise.
Rotation.
Elements.
And that's why it's negative.
Now let's take a soul or these forces in the y direction.
The A minus 100 plus sixty one point three eights all equal to zero 0 solve gravity and its thirty eight
point six to kill Newton's Hulk.
What would that be.
That is a.
That is a sheer force that is pointing up in the direction assumed it's applied to the left hand of
an element.
If we assume there was some sort of a hinge at the right hand and that it rotates upwards like this
should be a clockwise rotation and it is therefore a positive shear.
All right then let's get rid of all that.
It's not a positive share.
Excellent so we can.
Now let me see.
What do we not know.
We know the moment's issues on each hand.
We don't know the bending moment the internal bending moment or underneath the point loss.
That's the next thing we need to do.
So to do that we're going to cut a deal.
Minus D minus sixty nine point seven plus four times thirty eight point six to all equal to zero.
And so energy is equal to eighty four point seven eight kill Nugent meters.
Okay.
And it is in the direction assumed which would induce tension on the bottom of the structure.
OK.
So we can sketch out what what our sheer force and bellemont eyebrow look like.
So the shear under left and side starts as a positive shear.
Oh.
Again I'm I'm I'm following the loads across the structure so my book was old thirty eight point six
to of the sheer force here draws a structure up thirty eight point six two there's nothing happen.
This is me.
There is nothing happening on the structure comes across comes across from grass.
Then we hit that point Lord.
One hundred Kill anythings down which calls sheer force down a total value of 100.
That brings us down to minus sixty one point three eight and then nothing is happening all the way across
until we get the B L which was plus sixty one point creator operates of sixty one point three days.
Which would bring us back to zero.
So that our sheer force diagrams knowing what we know about the relationship between shear force and
bending moment what can we say.
Well the sheer force diagram is made up of two horizontal lines which means that any one would diagram
will be two incline straight lines and the slope of all straight lines will be equal to the magnitude
of the shear.
What's that mean.
The magnitude of the shear here is thirty eight point sixty two which means and it starts numerically
that is a smaller magnitude than sixty one point three agents of the slope will the shallower.
We going to have a shallow slow down to my span moment.
Well it doesn't start at zero It starts off at a moment value here.
What do we say as port moment was sixty nine point seven sixty nine point seven and that's gonna come
down to attention on the bottom.
The value of eighty four point seven eight.
Easy for seven eight.
And then the support.
Moment.
B was 160.
Negative.
Because it's a hugging moment.
One sixty point seven.
OK.
And that would also be a negative.
So we see that comes up up up up up.
Right so as we have as you would expect looking at the sheer force died down the slope of the inclined
line on the left the downward slope is shallower than the more steep slope on the right hand side and
atwood's that's predicted by the larger value of 61 quints create versus thirty eight point sixty two
so that our benchwarmer diagram for that half of the structure.
Now we go and repeat the same process for the other each other substructure So we take another free
body diagram.
OK so we now have the shear on the right hand side of the support.
VB or as unknown and the shear also the reaction at the seals is unknown.
So taking some of the moments about B again we're going to basically follow the same procedure that
we did for the left hand substructure.
So we're gonna assume clockwise are positive and we end up with minus 160 points plus the moaned on
the right hand side to treat two point one plus 25 acting orbit 10 meters with a lever five meters minus
10.
See equal to zero.
And so v.c. is one three two point one for kill and it is our right.
And so that is a negative shear negative shear.
Then we take a swarm of the forces in the y direction to get my only other vertical unknown force.
And we end up with VB or.
Plus one three two point one for minus 25 times 10 all equal to zero V.
B or is one one seven point eight six.
Okay.
Now what does the sheer force diagram look like.
Well we have a yellow understructure which means that the sheer force would be an incline straight line.
Anybody with any experience of bending moment and shear forces will look at that point at which it crosses
zero and will instantly say that there is the location of the the peak in the bending moment Kyra.
OK.
So that is the point.
Well maximum moment.
OK.
And that all comes from the equation.
D n d x equal to V.
Again you take my previous course should be very very familiar with that.
So now what we have is what is that distance what distance is really easy to work out.
It's just whatever the sheer force is on the left divided by the rate at which it declines which is
25 Killian's per meter.
Now that will be four point seven worn meters.
So I know if I want to know wear them I want to know the magnitude of the maximum moment.
I now know where I need to structure.
So let's cut.
Point of maximum moment.
Then we have minus Amex minus one 60 point seven plus one more seven point eight six times four point
seven one.
Mine is the mohmand effect of the eudi l..
She's twenty five times four point seven one four point seven one or two which is just the same as saying
four point seven one squared or two.
And so then we have X equal to one 1 7 point one killer Newton metres and that would be inducing tension
on the bottom of the structure.
All right so what would my bending moment diagram look like.
We have a moment here.
One sixty point seven.
It's a hogging moment.
It's negative.
And we have a moment on the first site.
Minus two three two point one and then.
Four point seven one meter is in the moment.
Peaks on the bottom and a value.
One in seven point one.
And we know it curves between those three points.
So we can just sketch in sketch in a curve.
That is all there is to it.
That's that's you're shear force and bending moment diagram out for the complete structure.
It's good practice at this stage to just sketch out the structure from the beginning and to sketch out
the sheer force and bending moment diagrams So that's what I'll do now.
Oh Clay right.
That was a lot of work.
So we have this structure.
This is when I would say the analysis is complete when we have the final summary of my structure my
reactions my sheer force diagram and my bending moment diagram.
So again if you.
I'm assuming I'm yeah I am assuming that you have a good understanding of the relationship between loading
and sheer force and shear force and bending moment so we can summarize here by saying we have a point
load here which means my sheer force diagram is composed of two horizontal lines and there's a logical
development of that which says that the bending moment diagram will therefore be to incline straight
lines.
The slopes of those straight lines is equal to the magnitude of the threat of sheer force.
And what I tend to do is say right well if that's a small number means the slope of this line will be
less than the slope of this line because the slope of this line is equal to this larger number.
We know that the change and side between the shear forces means a change and the sign of the slope which
means we're going to have a peak.
And at any moment diagram the used between B and C tells me I will have an incline straight line for
my sheer force diagram that in turn will tells me that I have a slow precursory we have a curve lying
like bending moment diagram.
The point at which the sheer force is zero is the local maximum in the bending moment diagram.
If you're unsure about what's going on here in this particular video free body diagrams bending moment
shear forces you really need to take a step back and look at the more basic more basic material on shear
forces and bending walls.
So now what we're going to do is start talking about some of the different things that we need to keep
in mind when we're analyzing different types of structures with different kinds of conditions.
So we'll draw a line on this lecturer and will pick it up in the next round.