Okay so now we're gonna consider each section of the structure individually and consider its free body
diagram and that will allow us to identify the or known shear forces on either end of each segment within
our structure.
So we're going to consider member B C first.
So where unknowns are the right of B B or and the shear to the left of C V C else taken some of the
moments may be first of all that will reveal vcl.
So we have minus forty four point five three plus thirty one point five.
Nine plus two times seventy five minus four the ecl equal to zero.
And we have the C L equal to thirty four point two seven.
Killen is taking these sewn all of these forces vertically equal to zero.
His force is positive.
We have v.b. or aroney on or left minus seventy five plus thirty four point two seven are equal to zero.
And so the VB or is equal to forty point seven three.
Killen is and will boxtrolls off so we can move straight on to consider at the span moment so I would
like to know what the moment is in the span.
Underneath the point logs.
That's very easy to work out we just do a simple court at the point loud and consider it the left hand
side of that koosh.
So taking the song at the moment about the Kosh we would see that M made very easily is equal to two
times forty point seven three minus forty four point five three and so an NID is equal to thirty six
point nine three killiney meters.
Okay we can consider member C D.
So again let's take some moments about point C..
Sol minus eighty six point four minus six times v. de plus three times 75 all equal to zero.
And so video is equal to twenty three point six.
Killen is taking the sum of the forces in the y direction.
VELSHI Or.
Mine's seventy five plus twenty three point one six equals zero.
And so with e.s. or is fifty one point eight for clinicians.
Those are both in the directions shown.
All right.
Let's work at the moment underneath the point load in this case.
And we have made would equal.
Very simply.
Fifty one point eight four times leverage of three minus the end moment of eighty six point zero four.
And so am made is sixty nine point four eight kill and mutant meters and we can stall the for later
on sock member B E.
Right.
Let me see.
Mine's two points sworn.
Mine's five point four six plus for the b l equals zero and so the B L is equal to two point zero to
kill and he wins in the direction shown and therefore he only had a horizontal force acting is Haiti
which must balance at two point zero to killiney and also in the direction shown on the diagram.
Have straight down to member C G.
So mine's three point three.
Three plus three.
Hey H.
G.
Equal to zero and so hate G is one client 1 kill in Newtons.
Therefore v c upper is equal to one point one killiney hands again the positive numbers are indicating
here that the initial direction I have assumed for those forces is correct so they are in the directions
sharne right.
Let's consider remember C.F..
We've got a fifty seven point seven eight minus sixty one point zero six plus 20 times a LEO acting
over six Misa is leveraged with three meters minus six vcl all equal to zero.
And so the C L is equal to fifty nine point four five kill any of us.
Again it's nine that's in the direction shown right that some of the forces horizontally must equal
zero.
Take forces to the right as positive and we're going against her.
You'd yell of twenty six meters minus the fifty nine point four five with just words.
Mine is the only unknown.
Hey Jeff was equal 0 and so hey Jeff is equal to 60 point five.
Five killed and it was again in the direction shown.
Okay we are going to want to work out where in the span between C and F does the maximum moment occur
and so we're gonna use the same method that we've used previously.
The maximum moment we know occurs at the point of zero sheer force.
Now I know the sheer force has a value of sixty point five five at position F and it reduces at a rate
of 20 kg promesa.
Well then it will be value of 0 0 at.
Let me just set this OK.
Maximum moments at sixteen point five.
Five.
Divide that by reducing rate of 20 promesa will give me three point zero or three measurers Bob frahm
F for a ball of F so that's with a maximum old curse.
That's that's where I'm going to make my cross.
We are the minus Max Wayne is sixty one point zero six plus three point zero three times sixty point
five five minus 20 times three point zero three squares times a half.
That's the moment effect of the yugio that are equals zero.
And so we can work out that Max is equal to thirty point six million in meters.
All right and so we have finished considering freeway diagrams No what I want to do is work.
What is the what is the vertical reaction force that must act true.
Point C are rather what is the vertical reaction that must be generated at point death and at point
G in case a line of action of that reaction will also pass through C obviously.
So we're just going to set it up here by saying consider a vertical force equilibrium of joint C..
So based on what you've seen in previous example videos you should deftly try and tackle this yourself
first.
So one's gonna just draw a joint.
See here first of all.
And I am going to isolate it from the member to the right which is C D E C S car that C D and we've
got.
C v arm the left right what's happening.
So when we looked at some member of B c we had determined that the the sheer force at this location
would be twenty four point two 7 acting up which meant on the opposite side of a court.
It was going to be acting down.
So that's the effect on joint see twenty four point two or three four point two seven down.
OK.
Now when I look at member C D and I work at the shear at c we worked out that it was when we we're looking
at everybody diagrams it was fifty one point eight four which means on this side of that koosh it will
act down the value of fifty one point eight four.
So we can see that the net effect is actually to have fifty one point four plus thirty four point two
seven polling join see down.
And so the vertical reaction through see.
Is equal to fifty one point eight four plus thirty four point to seven which is eighty six point one
killer Noon's.
Okay.
So we know that that is the vertical force reaction that is acting vertically up through F through C
or trujillo's.
So we as we've seen in our previous fedia we have a number of different ways we can handle this rich
reaction.
And I have chosen to say here that there will be loud sharing between mamber C.F. and mamber C G.
OK.
So what I'm going to say is the vertical reaction is going to get distributed or it's going to get developed
at support and support death.
And that's going to develop in proportion to the actual stiffness of member acg and members C.F..
So member C F is gonna go into compression and get shorter and member siji is gonna go into tension
and get longer.
K C G.
Now the actual stiffness is Young's modulus times a cross-sectional area a or out now where I'm going
to assume that E and A are constant and so the only variable here between members is going to be the
length.
So for siji is going to be one or three which is not.
Point 3 3 and we said we were going to give that in all rubar to signify a difference between the flexural
stiffness and the actual stiffness doing the same from Embassy F..
And so we can work out some distribution factors for axial load.
If I again look at joined C.
A biopsy in member G is going to be the distribution factor of acg which is not point six six times
the total vertical reaction force that must be developed.
So that's going to be fifty six point eight three.
And the reaction force below.
So in other words the compression developed in member C F is going to be NARTH point three four times.
Eighty six point one one which is twenty nine point two.
Heat kill and events.
So we map that to the adjoining member on the opposite side of the cloche a tavern a member drawn in
here.
So remember that of course that's of course.
And so that really means that the member here is in compression.
And again mapping to the opposite side of the court here.
We would say that this member is in tension.
Okay.
And again with we sort of flesh out this idea in a previous previous example video.
Okay so that's all of our free body diagrams completed we've worked up all of these shear forces and
corresponding reaction forces.
Now all that remains is to work out our diagrams or summary diagrams.