OK.
This structure is a continuous beam with varying different spans lengths.
So we're going to start off by looking at the stiffness of a B.
This is a continuous beam with fixed support at A and E and it's continuous over rori support and we
get two different magnitude of uniform we distribute load.
And we notice as well that the spans are quite uneven so we consider dockerty stiffness of mamber a
year and we know that it's gone to be for E I over L. because the structure is built in or fixed at
a and it's continuous or with support at B.
So K A B is equal to four E I am assuming here is constant or L which is Asian.
So not point five.
E A B C C continuous or a bolt support and so it's gone again only before E R E R L and L in this case
is 4.
So we have worn the E.
Hi saeedi.
Same again.
Or e.
I.
Over 6.
Which is no point 6 7.
E e r e e.
And no one is going to be.
D e.
Which again is sorry I overall where l is H.
So knock point five.
Okay we can use these differences to work at the distribution factors.
Distribution factor.
War is going to be the stiffness of a b or stiffness of baby plus the stiffness of B C.
Which is simply not point five feet high or one point five.
And because his eyes concentrate the structure in all cases are your ideal will counsel was soul will
spin not point three 3 which means by well that means of a distribution factor for B C as one line is
not point three 3 has only two members meeting at that point of not went 6 7.
And we can go down and process through the rest of the distribution.
Factors that you see in l.a. times now so I'll just list them out.
Now with you can't look at it fixed and moments.
Look at member baby first.
So join day.
It started as being fixed and we have locked joint B against rotation and so that's why we have a fixed
support now shown to be.
Thus just the locked joint.
So that was 5 killer Newtons promesa in acts over a span of a meter is now.
The fixed and moments at the reaction moments that would develop at those supports and so weak in C
and again we would get the formulas for these if you didn't really remember them after the topic ahead.
Foremans would be obtained from a table of fixed moments at the end.
Maybe his ego is huge minus W L squared or 12 bookstand moment for B is clockwise and will therefore
be a positive.
W L squared over 12 and we plugged the numbers in.
We have twenty six point seven million food in these years.
And my rather bottom line is and then this will be a plus twenty six point seven killer it needs was
okay.
And the.
In each case the structure.
Sorry to respond.
A B B C C D D E E.
We've got we've got boats of course fixed em and we're dealing with new details saraa and redrawing
every structure.
We're just going to be changing the W value on the L value so I'll just list out the fixed extent marmot's.
Moving on to a distribution.
Okay so now we're ready to go and start balancing out each of our joints as the first joint we'll balance
out is joint bape now joined B has the positive malmedy of 26 27 and a negative of six point seven.
So to balance all joint B we would need to add a negative 20 continued measures distribution between
being a and b c and so under be a side will add 20 or rather without any native 20 times not point 3
tree which is the distribution factor for B.
So that would give me minus six point six and undecide minus thirteen point four so together they adopted
a negative 22 that's needed to balance out that joint.
Moving on and doing the same thing to join C will end up with a needing to add a positive twenty three
point tree and 60 percent of that will be 14 and 40 percent of it will be 9 points 3 2.
No moving on to joint D.
We have what we need to add.
We need to add a positive 23 points rate balance at joint and so 57 percent of that will be thirteen
point to 8.
And 10 points year old to the first site.
Now we can carry over all the joints are fixed in this case so we've got no pin joints and here in the
structure.
And so that means we're going to carry over to every every adjacent joint and so carrying 50 percent
or to joint A gives us a negative three point three.
And essentially that's what we're going to it's why just carry all of these over.
OK.
And our joints our regain balance because of these carry over moments and so we go through a second
round of balancing.
So.
Joint B.
We got to add a negative seven.
And so that's going to be a minus two point three warn us that be a side a minus four point six nine
on the B C side joint C by coincidence is more or less balanced.
And so we are going to do anything to joints see will leave that as it is adding no additional moments
moving onto joints.
We're going to balance that we're at minus four point six six so that will be minus two point six six
and minus two.
Excellent.
We can move on and do our carry overs again.
So now the only joint that is ever balance is joint see.
And so we can bounce up joints see it's it needs to be read as a three point six eight to balance of
joints.
And so that will mean adding a plus two points to one 60 percent of three point six aces to point to
one and 40 percent will be a plus worn point for seven.
And you minyon you know a b or c is the only joints that we have got to balance OK.
Because B is in balance because there was no carry over moment to take him to joint.
B it means it remained balanced and similarly there was no carry over moments taken to join d into it
remained balance which just left joint C on balanced.
After the carry over.
Phase.
Now we have rebalanced to look our own carry all over again.
Take it 50 percent.
We at one point worn worn and marked coin 74.
So we're we're out of balance is now is quite small.
But I'll just do one more.
Just to bring it down in order to watch Soul balancing on the Joy behar again.
We'll be left with mine as not quite three sabon.
And minus Narek weight 7 4 and then bouncing off joint deep with God.
Mine's not quite for two and a minus nine point three too.
And so now our joints are balanced.
And if we look and see what the carry of would be 50 percent of those very small moments is going to
be a very very small imbalance.
If we were to do a further carry over.
And so it's not really necessary or worth our while gong show another carry over because we're already
approximating very very closely the actual bending moments in the structure so we can sew more up each
column determine our final moments or our final support moments more specifically.
Okay so the Momi distribution process has done its job for us.
It has removed an off or known innsmouth structure that I can revert back to using the equations of
equal Librium to consider each segment of the structure and turn and work at what all of the unknowns
answered also knowns are going to be the shear forces at the ends of each substructure as well as span.
Moments that I need to know.
So I'm going to go through each of these substructures in turn now with a free body diagram and work
my remaining on notes that I can produce of final summary solution for the structure which will be all
the reactions to complete sheer force diagram and the complete blending diagram.
Considering.
Member.
A B.
First.
So I wanted termin the church girl share v.v. else to do that I'll take to some of the Mormons point
A.
Mine is 30 warned.
Point 1 6 plus seventeen point four to mine is eight times.
V p l o paulose 15.
Now 15 its five year five acting over eight.
Nature is really rare for Misa as that all equals zero solving for avp l..
We have eighteen point to eight kill in humans.
Okay we can take some of the forces in the y direction now
and we have the Year close.
Eighteen point two eight minus five times eight.
To zero.
So the A is equal to twenty one point seventy five conclusions.
So now we like to determine the point of maximal moment.
So the point is.
Max moment.
Well.
Equal the shear at the at a V.A. So that's gonna be twenty one point seventy five vialli the rate at
which it declines.
Which is fine.
Me are five Killian's perimeter which is going to give me four point three five metres to the right
of a.
So now I know where to make my cut from my maximal moment.
Okay so we know we want to make our cut four point three like metres to the right of POINT day the cut
We've just marked as x x the cut section and that has revealed the internal bending moment which I am
labelling here as I am worn one because were dealing with the first man in the structure.
So I take to some of the moments both cut to determine and one.
So minus 30.
We want six plus to twenty one point seventy five times four point three five minus fives acting over
four point three five metres were Levering all four which recoil to us they're saying full which required
squared time.
Now point five.
Minus morn aisle equals zero and solving and warned the maximum moment in the span between a and b is
equal to sixteen point one of five killed in metres.
Now that is in the orientation of Shonin which will induce tension on the bottom.
So that's the first of three spans are the first of four spans dawn.
The procedure for analysing span b c c d and d e is exactly the same which is no different.
Start off rotating almost bit one side taking the sum of the vertical forces to get the other on known
virtual sheer phonier where the maximum moment occurs or the rather to be more specific real goal maximum
or minimum occurs and then making a call at that location.
So if you haven't done so already at this point I will strongly urge you to pause the video and go ahead
and analyse the following three spans yourself with a practice.
And after you've done that combat if you've if you've had difficulty come back and wash the remainder
of this video.
So for anyone who's had any difficulty with doing that I'll carry on now and repeat this analysis procedure
for the remaining three spans.
And so will the vcl equal to nine point one to feel any offence taking the song with the forces in the
y direction equals zero.
And we therefore have with loebe or equal to ten point eight eight kill and it was our right then Max
moment.
And solving and two is equal to a negative five point five.
Eight Killen and meters What does the negative mean.
It means at the moment actually goes this way which will induce tension on the top of the structure.
Remember if I pretended that that was fixed and I was looking at applying my morning to kansanshi ribband
the cantilevered down inducing tension on the TOC.
So we will just stick a box around that and say tension harp.
Now moving it straight onto member saeedi.
And solve vitiello is equal to thirty five point to eight kill incidents.
Some of the forces in the y direction equals zero.
And we can see very easily into the sea or is equal to twenty four point seven.
To cure an illness that's simply 60 miles thirty five point to eight.
So the maximum moment.
And solving and three is equal to sixteen point six.
Five kill anything meters.
Okay.
And that's gonna induce tension on the bottom.
So moving straight on to member d e.
And the he therefore is equal to forty one point four six killiney.
Are right some of the forces in the wider section equals zero or birds being passed of.
And again very easily.
We have 80 minus the forty one point four six which gives me V.T. or equal to thirty eight point five
for kill any of us in any direction sound.
So the maximum.
Moment.
And so m 4 is equal to twenty eight point six seven killiney and meters excellence.
Tension on the bottom.
Okay so that was quite a.
That the procedure that follows after the actual moment distribution in that particular question is
quite quite a laborious procedure.
But it is a really useful example structure to be able to solve.
Because anyone who has had any experience yet of designing concrete structures concrete beams the continuous
concrete beams are very very common structure.
And in many cases there are tables lookup tables that you can go to to determine bending moments in
such structures or continuous beam structures but in many cases those tables start to apply because
the spans of your beam are very uneven.
And that was the case in this particular example that beams spans are very uneven and so some form of
manual analysis to determine the bending moment within the structure would be required.
And if you didn't have recourse to a computer analysis and solid this is although it's quite a laborious
structure to work way through it's a very very good structure to be able to analyze because it crops
up in practical engineering very very often so we can summarize our whole solution.
Now in our story diagrams at the end so I'm going to sketch out the structure with its reaction forces.
Then below the sheer force and bending moment diagram as a way of summarizing everything that we've
done so far.
So the sheer force diagram is constructed by simply following the loads across the structure.
So the reaction and a draws a sheer force diagram up and value of twenty one point seventy five the
uniformly distributed log between a and b reduces it linearly at a rate of five killens promesa Sol
five times eight is forty so on the far side would have stepped down 40 in total so it will be twenty
one point seventy five.
Minus 40 would bring us to minus eighteen point twenty five.
And then the reaction of twenty nine point one six were bring this up to a value of ten point nine one.
Now four times five is twenty soldiers stepped down a total 20 ringing loss to nine point zero nine.
Then we had step walked thirty three point eight for bringing us to twenty four point seventy five.
So whatever you're up let's put on the moment that we know for a start.
Well this is going to be a hugging moment of thirty one point one six.
So it has tension it's inducing tension on the top with the shropshire and that's what we call a hugging
moment.
Same is true down here a moment inducing tension on top of a fifty seven point three one.
We have a moment of be the moment of seventeen point four to minus minus seventeen point four two.
Here we have a minus thirteen point nine and year we have a minus.
Or C Five.
Point six.
Now we span moment at a B was a sixteen point one five inducing tension on the bottom.
That's going to happen where the shear forces zero.
So if I draw a curved close it helps.
So it's out there at sixteen point four right now for b c the sheer force is zero here which means that
were a local maximum or minimum occurred and it turns out that our value here was in fact a local minimum
rather than a lock of maximum because we had tension on the top where a magnitude of five point five
eight and saw shifts are bending one diagram never drops below.
So we have a minus five point five eight retention at the top of the structure.
Now for Cedi the sheer force is zero there.
And we worked out the landing moment at that location induce tension on the bottom where a magnitude
of sixteen point six five.
So it does drop below.
In that case.
So that's sixteen point six five.
And then finally our sheer force was zero there and the bending moment at that location induced tension
on the bottom at a magnitude of twenty eight point sixty seven.
It's a curved down and all the way up.
Renard's twenty eight point six seven all case love fully defines our bending moment diagram in Clinton
meters.
So we have a sheer force diagram the bending moment diagram and all of our reaction shots.
That is the best way to wrap up a question like that and summarize everything at the very end.
So we've completely analyzed that continuous beams Russia.