OK.
So the first thing that we're gonna do at this question is work get these stiff Misses of the two different
souled members within the structure.
So we're starting as we always do with stiffness as well.
So the stiffness of a baby is going to be Thor.
We are over L because it's fixed at a and it is continuous over the support to be.
So that's got to be for over 12 years.
And you'll note that we're assuming he it's constant.
So that's equal to not point three three.
We are okay.
The stiffness of B C J P C which is going to be 3 E over L which is going to be 3 over 10.
We are.
And that's all we see because the structure is continuous.
Always support a B and it appears support at c.
So that's gonna be not Point 3 3.
T I O k we can jump straight on to the distribution factors.
Distribution.
Factor.
Thor B A is the stiffness of P and stiffness of ABC which is not Point 3 tree e.
I divide away the sum of this difference is what the memories meeting at the joint which is just two
stiffness as we've worked it previously with not point 6 3 E.
The E I said cancel and relief within four point five.
That means Did oestrogen factor for B C is one minus the distribution factor would be a.
So we have not worked for h k.
No we can go straight down to look at the fixed and moments.
So we look at a B first of all and we can see that a be subjected to point loads added third span positions.
And so the easiest way for us to analyze that is by using superposition because remember we once again
are fixed and moments from our Xand more tables and you are not going to find in most fixed and moment
tables an arrangement that has Juif stands and 2. point loads of different magnitudes.
So what we're going to do is as I said use superposition.
So a steratore off by sketching out the structure that we're interested in the substructure.
So Earth extend moment here is going to be so.
So so we have no that analyze these two individual structures and then add the fixed and moments at
each end.
So you will see from your fixed and moment tables that we can determine the fixed and moment.
We'll call this fixed and moment from a to b.
This will be one that's going to equal minus P where p is 100 Clinton's times a where the 4 years a
and the H is be so a B squared over L squared.
Which is equal to minus 1 7 7 point eight.
Killen meters answer the pharcyde.
We have the fix and moments be a one will be equal to.
So the fixed and moment B.
One is going to equal P A's squared B or L squared so p n squared.
B more L squared which will equal eighty eight point nine colon Majoras mulkey and we do the same thing
for below.
Down below we will change the subscripts to a B.
Number two a rattler were to B minus 44 point for Q and that means years and f e m the two is equal
to eighty eight point nine.
And it measures are right so why basic superposition adding up the moments we can work so that our fixed
end moment
for a B is equal to minus forty four point four minus one seven seven point eight which is gonna give
me minus two to two coins to kill anouk meters.
And he fixed and moments the opposite side.
Well equal worn seven seven point eight.
Tony majors.
Now we can look at our fixed and moments for ABC and remember the joint at sea ad supported C is a roller
support and that can average generate a moment.
So think stand a moment for C is going to be minus W L squared or 8 which really plugs and in his minus
one hundred and twenty five.
It shows and are pinned down and has obviously got a fixed and moment of zero.
Okay so we have a fixed and momentary redistribution factors and I see differences.
So now it's time to go in and look at our distribution.
Okay so we're ready to process through the distribution so we can see that the only joint that we have
here that's out of balance is joint B and the out of balance moment is a positive fifty two point eight.
Okay.
And so we know when we unlock joint B we need to add in negative fifty two point eight in order to balance
out that joint.
It's got to be added to both members are to both sides of the joint in proportion to these differences
of the members framing in the joint so that 50 2.8 52 percent of it will go in to be a and 48 percent
of us.
I'm looking here at the distribution factors were go into B C.
So when we work out God's relative there presented us we're going to get a negative twenty seven point
four six and a negative twenty five point three four.
OK.
Now we have to think of it carry over moments.
Well we're gonna carry 50 percent over to support a.
So that will be a negative.
Thirteen point seven three thirteen point.
But at the rate thirteen point seven three and we know that we carry no moment all rich were pinned
support so they carry or to opin support is zero.
And at that stage redondo is no all our joint B is now balanced and there is no further balancing required
so we can sum up the moments in each column.
So we heard it was 2 3 5 3 worn Plus one of fifty point three for balance way and minus one fifty point
three four.
And as you would expect zero at a pin support.
So that's the end of the distribution.
Now you actually conclude the question and to call the structure one fully analyzed.
We need to work out what are the reactions.
What is the sheer force diagram and what is the bendy model diagram.
So we have the moments that our developer support but we need to know what the span moments out to flesh
out before bending moment diagram.
The way to do this is to consider each of our Sobe members.
So a to b and then B to C separately and look at free body diagrams for those members.
So the next section is reactions were just call it reactions.
And S F T sure of sheer force diagram and bending or diagram be empty.
Okay so considering member a B first.
Okay.
Where unknowns.
Our V A and the b l the sheer force at a and a B.
Now we'll.
We're also going to want to know what it's the bending moment underneath each of those quite loads.
And so we'll do clutz when it comes to mccombs at that point.
So let's start off.
We're trying to work out what VB L is we can do that simply by taking the sum of the moments point A.
So we have a negative two three five point nine three plus one fifty point three four so taking care
of two moments first.
Plus the moment the fact of the one hundred Krsna load plus the embalmed effect of the 50 bullet in
C minus 12 lever times the B L.
So the on your 9 that equation is v v L So it's trivial to solve that and determine if e e the air is
equal to fifty nine point five three kill and it is ok to solve each positive number indicates at the
direction that I assume which is operative that sheer force is the correct direction.
And what does that mean.
Well if I was to apply that sheer force to the end of the mamber and as soon at the pin at the other
end of the mamber what would that do.
That would cause that to rotate anticlockwise and then anticlockwise rotation tells me they somewhat
defamation sign convention that I'm dealing with a negative share.
I'm gonna go through this each time with finer or I determine its sheer force but just to give you an
idea of where this sign is coming from.
That's a negative shear.
K we can take the sum of the forces in the y direction to determine the only one known vertical force
which is V A.
So here minus 100 minus 50 plus fifty nine point five three which is aeb l..
All equal zerox solving this is equal to ninety point four.
Seven.
Kill any of us and that is a positive share based on the.
Based on the same sign convention let's work out what the moment D is so we can take a quash Abdi and
determined the internal bending moment.
If we marker.
Course as Section X x.
Minus M. de minus two three five point nine.
Three plus four times ninety point four seven equals zero and solving M. de is equal to one to five
point ninety five.
Killen it means yes.
Now we can take the moments are we can at the moment at e we do the very same thing making a course.
Courting and considering the right answer side this time.
So we have many.
Plus one fifty point three four minus four times.
Fifty nine point five.
Three mile equals zero and solving three M E equals eighty seven point seven eight.
Kill any meters.
Okay so that's the first amp substructure dealt with.
We had this stage we like summarized as sheer force and bending moment for this part of the structure.
That's not a bad idea to capture all of the information that we have determined at this stage.
Somewhere I'll look into a into a sheer force and brandywell diagram.
So now we can do the exact same thing for a member of B C.
OK so we'd like to determine v c and to do that we can take moments about B.
Minus 150.
Point three four minus 10 v c plus 10 kilohertz Rumi her acting over 10 meters with a lever arm or five
measures equals zero.
And solving these C is equal to 35 kg and taken to some of the forces in the y direction.
It's very easy to work out VB or plus 30 flies which was easy we just worked minus 10 Killian's perimeter
acting or written in wheelchairs all equal zero nonso VB or the shear on the right hand side of me is
guys equal 65 kill and humans.
Okay so we want to know where those the sheer force becomes zero because it is going to be a positive
sheer VB or as a positive share and vcl Negative share so it changes from positive to negative and we
must know where that is because at that point the bending moment diagram will experience a log of maximal
minimum so we can very easily do that by working out the.
Let me see if we take the sheer force VB or which was 65.
Let's just say in point of zero shear which in this case is also going to be maximum moment.
So that point we'll call it X k where X is gonna be measured from point B reinhardt's B.
So X is gone egal whatever the shear forces at that point which is 65 killiney means divide away the
rate at which it declines.
Which is 10 Clinton's perimeter.
That's gonna give me X equal to six point five meters.
That is where the sheer force crosses through zero or is equal to zero and nattering on a want to make
my Kosh to identify what the maximum bending moment is in that span.
So we can do our quote at the point of maximum Merks moment.
In making this cult we revealed the internal bending moment which we'll just call an axe and we will
take the song or the moment about that koosh.
Some of the moments most kursch Equals zero.
So we have minus M x plus.
Sixty five times six point five measurers minus 10 acting all were six point five meters.
Leverage what.
Six point five or two.
I'll equal to zero unseals solving for x we have IMX equal to sixty point nine worn in the orientation
assumed which would induce tension on the bottom which we choose which is what we would expect intuitively
killiney metres.
Okay all right so that is more or less everything that we need in order to fully define the complete
shear force and bending moment diagram and indeed to determine the reaction forces which is everything
we were asked for starrett but before going on to sketch the sheer force and then he one would diagram.
I'll just lay out an alternative a slightly different way of arriving at the point of maximum moment
so we can we can essentially do it using calculus.
So you're you're probably already familiar with the method.
But for anybody who isn't I'll just lay it out here quite quickly so I'll just say.
Alternatively alternatively.
He can analyze the bending moment as a function of a variable x analyse the bending moment as a function
of variable x.
Why do women will latch ranil who hasn't seen this before.
Well all I really wanna say is instead of quartering six point five metres to the right of the support
I'm gonna call at a distance x ray X is a variable.
So what does that look like.
So our cloche is here and in making the court I've revealed an internal bending moment.
Now the bending moment now is a function of x where x is the distance here to the koosh.
OK X X is now a variable that's the subtle difference so I can take some of the moments to court and
I'll end up with an equation which essentially is a quadratic equation in x.
So if I take these som.
Of the moments though the course equals zero.
What do I have minus 150.
Point three for mine is a function of x plus 65 times it's lehren which is now just the x variable minus
10 acting over X measure is would it leave rural X over 2 meters all equal to zero.
Simplifying that.
We have an axe bending moment is equal to.
Let me see minus 5 x squared plus 65 X minus one fifty point three four right.
That is an equation which defines the bending moment diagram as a function of x for this portion of
the structure between B and C. So that's quite useful because that allows me to do a couple different
things.
I can determine the position and alternate a way of determining the position whether it is bending on
and a maximum and I can also using this equation determine where the bending moment is zero and from
a design perspective does a Quelle useful thing to know.
Might like to know where is the banding moment zero because that's where the bending stresses or the
normal stresses due to bending will also be zero.
Okay so the first thing I wanna do is use that equation to determine the point of maximum moments and
what I'm looking for here.
Remember if I was just doing a rough manual diagram it looks something like that.
If that's the bending moment diagram myths just draw a horizontal axis in here.
The point of maximum moment is the point at which the slope to that curve is completely horizontal.
And if the slope of the curve is horizontal It means the slope is zero.
So I want to find an equation for the slope of that curve that bending moment and conveniently in off
calculus helps us out here because we can differentiate that formation of x with respect x and what
I will give us is the slope of that line.
But remember that equation is just the equation or that line.
And if I differentiated I get the slope.
So carrying out that differentiation were going to be left with minus 10 x plus 65.
So this is now the slope.
If I let up slope equal to zero on Sol for X I'll be finding the point at which the bending moment is
a maximum.
Okay so we can do that Prezi So we have minus 10 x plus 65.
The slope is equal to zero.
Solving for X and we have X equals a six point five and that's exactly what we got previously.
We're looking at where the sheer force is zero.
So this is not heard of a way of arriving at the point of maximum bending moment.
The benefit of doing this sorace are determining the any mommie's function of x is.
I'm also now able to work out where the bending moment is zero and all that is is determining what the
roots of that quadratic equation are.
So hopefully you are going ahead to find the roots of a quadratic equation.
So we'll just say let me see the ending moment bending moment now is zero.
At the roots of the m function of x equations so headway find the root to that equation.
It's your standard formula.
X is even a minus B plus minus the square root of B's squared minus for a c all over to a K.
So x is equal the minus Putin wins 65 plus or minus the square it should be squared which will be sixty
five squared minus for a C so that's minus four times the first term minus five.
The last term.
And when it's one fifty point three four square while that all of it divided by two times the first
term which is two times minus five.
Well.
Well I won't go through the longhand here.
But needless to say the x is equal to zero.
Three measures and X is eaglet zero at 10 meters.
Note the 10 meters is exactly what we would expect.
Because what that saying is the bending moment is 0 at point C which is what we know all the cases pin
support.
So this is essentially two different ways of handling at the bending moment in this region of the structure.
Were at this stage were able to sketch our shear force and bending moment diagram and essentially provide
our summary solution to the problem.
Okay so that is the summary of our solution.
It's always nice when you get to the end of a long long solution or when you have reams and reams of
pages of calculations it's nice to wrap it all up at the very end in a summary diagram and this is a
really you know bending moment and sheer force.
Questions lend themselves quite well to this so we can capture a reading with doyne in this question.
In these three diagrams that's how I like to wrap things up and I suggest you do the same.