OK.
So we're going to start off with sheer force diagram.
Right.
And so we can say for a member be well we know the sheer force is going to come out.
A magnitude of two point zero two positive shear.
And there's no external forces acting on this member so it stays at a constant value all the way to
the end.
But Member B and k..
Now let's look at the horizontal member of B.
Well the shear for over a side B we worked from everybody diagram was forty point seven threes and that's
going to be an approach year of forty point seven three and that continues unchanged in effect until
we hit the point load in the centre of that span which brings it down a total value of 75 which will
be down to a minus 34 point to seven.
And again it's unchanged all the way to the end of the member.
And if we look at the share on member C D.
And referring back to our free body diagrams we have a value of fifty one point eight for let me see
that members probably a little bit longer.
So there are 75 brings it down to minus twenty three point one six.
One change until the end.
All right.
So now we can look at the virtual member let's go for CGP acg was pretty straightforward it was just
one point one colludes of positive share.
Okay.
So looking at member CEF denin death we are going to have the reaction causing the sheer force to step.
A value of sixteen point five five.
And that's going to reduce linearly at a rate of 20 Killian's from edar OP until it hits minus fifty
nine point four phys.
We can use the information in Baghdad and our sheer force diagram to help us build a bending moment
diagram.
So we have a moment an external moment applied at point A and that's immediately gonna give us a value
of 50 kilometres.
Point day and that's gonna continue on changed all the way until we get in to point B.
Now we looked at the moment at B and we worked out from our distribution of that was a value of forty
four point five three.
So then we just put in my 50 here.
The moment underneath the point load was thirty six point ninety three.
So that was going to be around here somewhere.
Now the sheer force diagram is a straight line running parallel to the axis of the member.
And that tells me immediately that my very moment I ran is an incline straight line.
So we're gonna come down to our point now the sheer force changes sign at that point from a positive
to a negative which means we're going to have at peak an hour bending moment diagram.
And that makes sense to us because we know the bending moment at sea is going to be thirty one point
five nights.
It's going to go up to thirty one point five nine.
Right now let's deal with the moment in B E.
So we have a value of.
Let me see five point four six from our distribution.
And that was a fixed moment at the base.
So we're going to have our fix support of the base which can sustain a moment of minus two point six
one and five point four six.
Right then.
Now let me see we if we did what member G where we know the bending moment at G is going to be zero.
And so that's going to come down to a value of three point three in a straight line.
So it's gonna be me see three point three.
In fact let me just pull that out there first.
Let's let's put on the moment from member C.F. Well we have a uniform We distributed load which gives
an inclined straight line for shear first round which gives a curve for our bending moment Ira.
So we know of any moment did we work at the moment at the top here at sea as being fifty seven point
seven eight it's gonna curve down.
It's gonna reach a maximum point somewhere here.
And on a curve back up.
To a value of sixty one point zero six.
Sixty one point zero six and fifty seven point seven eight was top value fifty seven point seven eight.
And we've got three.
And you see the maximum value we worked out was.
The maximum value was thirty point six.
We do at member C D.
We have a minus 60.
Eighty six point zero four.
So a hugging moment of mine is a six point zero for here.
Coming down to a peak value of sixty nine point four ace.
And we have a pina D.
So the moment at d got to be zero.
K. So that was minus 86 point 0 4.
And then a sixty nine point four eight.
That completes our bonding moment Ira.
Now we're going use the information in our bending moment diagram to help us build a quality of deflected
shape for our structure.
Okay.
So we have surmised our structure again.
We've put on our reaction forces noting our propping force at point deep of fifty eight point five three.
So now we can build equality out of any moment diagram and really our doing here is we're looking at
the bending moment diagram we're spotting what surfaces are what parts of the structure our intention.
And we're bending our curves for the structure based on that.
So for example the cantilever member from A to B has tension across the top of the mamber from A to
B.
So I know that that's curving down so the top face is the tension face the curve bends down like so.
So tension on the tough face.
Now we have let me see.
Tension on the top.
And then we come to a point of counter Fletcher another point to counter flexion and we start to bend
up again.
So what am I saying here.
We've got tension on the bottom indicated by the bending moment diagram the two points of counterfactual
to red dot to the point of zero moment where the curve changes from a concave curve to a complex curve
or from tension on the bottom to tension on the top.
So there's tension on the top here and tension on the top here.
Now the 90 degree angle is maintained at all joints and so the joint joints see here maintains a 90
degree angle which means that the member from cdg was to have rotated counterclockwise and therefore
at the member in its curved or in its deflected shape it heads off to the left but has to curve back
around because we know it has to go to support G.
And that has tension on the left hand face which is confirmed by depending on would diagram which also
shows that there is tension on the left hand side of that mamber from CTG or Kate.
Now we can look at C to D.
Right.
So again the joint has rotate counterclockwise which means the member starts off heading off like so.
Very quickly we get to a point of counter flexor where the curve starts to go down and it remains so
until it comes back up.
Solid point counter flexural somewhere here according to the bending moment diagram the attention on
the top here and tension on the bottom here.
Right.
And then the last member is member C to F and he again just drawing our curves in correspondence with
what the bending man would diagram is telling us we're going to have something like this counter flesher
curves not a point to counter Fletcher and the curve changes direction.
There's something there and something there.
So we got tension on the outside tension on the inside tension on the inside.
So that is a very very exaggerated example of deflected cheikh plass structure so that is the Those
are the faces that will be in tension.
And as a result of that that is the shape that structure it will take all underload.