So now what we want to do is go through each of the members within this structure joffre body diagram
and work out what these shear forces and where necessary what are the span moments and that we need
to determine our reactions and to flesh out a sheer force and bending diagrams.
So looking at MEMBER Maybe first.
Right.
So I know the moments at the top and bottom.
From weight distribution the unknowns in this free body diagram.
The horizontal reaction at A and the sheer force just below joint B solve the L.
So let's take the saw the moment by point day which will yield VB.
So at ten point five serbin POST twenty one point five three minus six.
The L. Arly with a zero or solving of the b l equal to five point three five kearneys and the positive
sign.
Neza is in the direction shown as that would be a negative shear because it is cars it was tending to
cars anticlockwise rotation of that member by inspection by taking the sum of the horizontal forces
we would see that hate J is equal to five point three five keelen evens and has to balance that that
sheer force and again at odds with the direction schantz we've assumed the right directions are to start
moving on to look at mamber ABC.
Case where unknowns in this case add these shears on the right and left of the member taking some of
the moments to point B.
Mine is twenty one point four three plus thirty five point zero six.
Plus 25.
Times five times.
Two point five minus five.
Easy.
C. L. are equal to zero.
And so the ecl is equal to sixty five point two three kill and he wins taking the sum of the forces
in the y direction equal to zero.
Or is that's positive.
What do we get the VB or plus the sixty five point two three four vcl Minus the Udy out has twenty five
kms promoter acting or five meters all equal to zero and solve the unknown the B or is equal to fifty
nine point seven seven killiney was excellent.
That we see while we're here let's look and see what is the span.
Moment in the sea.
So we'd like to know the location and magnitude of the maximum moment Sol Previously we've done this
by working out an expression for the bending moment as a function of x differentiating that.
But now we we will take another angle.
We will simply.
We know that the maximum bending moment occurs where the sheer force is zero.
So looking at the span.
Moment.
The point p-t the point of zero shear.
We know at this stage is equal to the point of Max moment.
So where is the point of zero.
Sure.
Well the sheer force on the left is VB or is fifty nine point seven seven and that reduces it at a rate
of 25 killens promesa.
So if I'm minus 25 Killian's cremator at times some distance x when all of that is equal to zero.
The sheer force is zero.
And so if I solve for the distance x at which that equation is true I get X equals two point treat nine
means was that is really the sheer force zero and that is also where the bending moment is a maximum.
So I now know where I need to make my koosh in order to evaluate the maximum bending modes.
Okay so taking the sort of the moment both.
We got mine's twenty one point four three miners M. Max.
Plus fifty nine point seven seven times two point three nine minus the twenty five times two point three
nines squeers over to an all equal zero.
Therefore Max was equal to fifty point zero to kill Newton miseries.
So I now know the maximum moments are the last thing to consider is the last substructure which is member
at saeedi.
So my unknowns in this situation and a horizontal reaction are deep and the sheer force below sea.
So let's just take some of the moments about de.
We end up with minus 1721 three straight forward.
In this case there is no external loads minus 35 points 0 6 plus 3 v.c. or all equal to zero.
Therefore the C or is equal to seventeen point four islands and then by inspection.
Hey Judy is also seventeen point four killiney ones and both of those are in the directions assumed
Heard's directions shown in the diagram.