00:01
Okay, so for the load diagram and free body diagram, they're kind of the same thing.
00:05
So we've got the beam and we draw on the forces.
00:12
So on the left hand side, which i'll call point a, we've just got a roller support.
00:18
And on the right hand side, which i call point b, we've got a pin support.
00:24
And so at point a, we've got a reaction force acting upwards.
00:30
And at point b, we've got one acting upwards and to the side.
00:35
So i'll call this rbiy, the upwards one, and rbx, the across one.
00:42
We've then got a force here of 279 kiloons, and then a distributed load here are 15 .75 kiloons per meter, and we can draw on the distances here.
01:08
This is 2 meters, and this is 1 .25 meters, and this is 0 .75 meters.
01:19
And so we want to then determine the reaction forces.
01:22
So this is kind of both the free body and load diagram.
01:26
And then the reaction forces, we can see that there's no other horizontal force, so rbx has to be zero.
01:32
And then we know that ra plus rbiy has to equal the sum of the downwards forces, which is 5 .75 times 2 plus 279, and that's equal to 310 .5.
01:49
Then doing the moment about point a, we get that the downward moment is going to be 15 .75 times 2.
02:00
That's the total weight of the distributed load times the distance of the centre of mass from the pivot point, which is just 1 metre, plus 279 times its distance from the pivot, which is 3 .25, and then minus 4rby, and that has to equal 0.
02:22
And so we can find from this that r .b .y is equal to 269 .4375, and therefore r .a.
02:32
Is equal to 41 .625.
02:37
And these are in kiloons.
02:43
For question three, then we want to draw the internal forces diagrams, but these are just the sheer force and bending moment diagrams.
02:51
So this is also question four as well, so they're kind of coupled into one.
02:54
So we first see the axial force, which is the horizontal ones, and that's just flat the whole way...