Okay.
Here we have in on symmetrical structure so straight away.
That should tell us.
That should set off an alarm bells first at this structure is going to want to sway now.
Fortunately for us we have a roller support at sea that is providing a propping force reaction force.
So what we'll see is that the horizontal reaction that develops at sea is a is in fact what's gonna
stop this thing from swaying right or left.
Now this is pretty interesting to note about this particular question is that Eli is for the first time.
Not constant.
So what we're saying here is try to replicate reality a little bit more and we're saying that these
stiffness the flesh would stiffness or rigidity of the horizontal beam between the bnc is twice that
of the columns on either side supporting it.
So this is much more realistic and will see here that it's actually very very easy to build in this
kind of them.
This kind of variable stiffness into our analysis.
So we're in Nigel straight into it here.
We're going to work out these differences.
First of all.
Southridge member a.b Well it's fixed at 8.
It's a continuous structure at B and so it is gonna be for at times.
E over overwhelmed.
Now what have we got so that's gonna be for over six EE.
And so that is equal to not point six seven.
A Member B C.
We're going to be for orell remembered this stiffness of E of B C is gonna be twice that of the older
two vertical members.
So that's gone a beat too e of her L which is going give me a five.
We are.
One point six yahiye are right.
Excellent.
And then finally for the stiffness is kcet with a much shorter member which will tell us that it's stiffer
fixed at dare thickset deep it continues at c and so we're gonna have four eai overwell which is for
all three.
We are in which is one point.
Three three.
E I.
Excellent.
So the distribution factor for a B is simply wannabe point six to seven the stiffness of that member.
What was some of the stimuluses meeting at that joint point six of them plus one quite six years.
Nine point two 9 5.
Distribution factor for ABC is going to be one minus Nage point 2 9 5 which is not win 7 5 makes sense
and half lot more Mormons is going to be distributed in that stiffer horizontal mamber distribution
factor for CB is one point six olver one point six plus one point three three which is Naka 5 4 6 and
so distribution factor for c d is one minus NARTH point five for six point four five four excellent
arcangel extraneous year fixed and moments only one member to consider.
So they fixed and moments from B to see.
It's anticlockwise and so are going to be minus W L squirrel which Well 25 times fives five's weird
over 12.
So the fixed end moment is warps the fixed and moment for B to C is minus fifty two point.
Oh eight killiney Media is excellent.
And so we can see very easy that the fixed and moments from C to B.
This is now clockwise and so therefore positive.
So it will be a plus 52 Quinto eight.
Q Any in meters.
OK.
Now unlike the last example that we looked up that I only have one internal joint and I said that we
could analyze that structure without resorting to a table.
This structure has two internal joints and so like our beam structures with multiple internal joints
we're gonna go straight into using a table to keep track of our distribution.
Okay so we're set up to do a distribution.
We have our distribution factor as we are fixed and Mormans.
Now it's time to balance each joint So we're going to balance joint B by adding a positive fifty two
point zero weight and we're obviously by distribute that between the two members in proportion to the
esteve misses which will work out to be a positive fifteen point three six and a positive thirty six
point seven two.
Okay.
And we do the same over our joints.
Balancing it out with a minus 28 point four or four.
Mine's twenty eight point four or four.
Minus twenty three point six four.
Okay now remembering to carry over.
So we are going to carry over a moment to join day because as a fixed support and so they carry over
the moment will be half of my fifteen point three six which is going to be a positive seven point six
eight.
So that's that carry over.
We're going to carry over from b or to see which will be eighteen point three six.
Carrying from C back over to be 50 percent which will be mines fourteen point two two and carrying over
to Duke from C over to D will be minus 11 point eight 2.
Okay now.
So we bounced the joints.
We then carried over.
Moments.
The process of carrying over 50 percent of each balancing moment the opposing the opposite points meant
that we have no On balance those joints all over again.
But we've unbalanced them with smaller magnitude.
Moments are gonna go through the same process and win on iterate through this process until the balance
moment at the joints are sufficiently small so as to be insignificant.
So rather don't talk my way through doubt down through all of these distribution and carry all resist
the same process repeated.
So I will just work my way through it here.
Encourage you to stop at this point and actually work your way through all the ways till the end and
when you've done that.
Just watch me rattle through the rest of this distribution table.
Okay so remember you always want to finish on a distribution.
So what I'm saying is you always want to finish good balanced joint So we now have balance joints to
carry over a moments that would subsequently unbalanced those joints sufficiently small that we can
stop our distribution now.
And so more up the bending moments in each of the columns to work at the final moments and each of these
joints.
Okay so we see that there is.
The moments are balancing a joints bnc which is exactly what we would expect a minor discrepancy here.
Reaches down to rounding errors.
So that is the endable of the distribution for this particular question.