00:01
So in this problem we have our beam and we need to draw a free body diagram.
00:05
So there's going to be a reaction at a.
00:07
Now there's also going to be an x reaction, but we can see that's simply zero, so we don't need to bother with that.
00:13
Our reaction at b, a downward force here, it's six point something kilonewtons, but we don't know what that is, so i'm just going to use six and whichever x and y is, you can just plug that in in the arithmetic, but the process will be the same.
00:27
And then we have our applied force, which is again two point something kilonewtons per meter, but i'm just using two kilonewtons per meter here.
00:36
Now we have all our distances between a and six is one meter, and then our point c here that we want to solve for, 0 .5, or actually no, that side is one meter again, and then 0 .5 here, so that means the rest of this is 1 .5 meters given the whole beam is four meters long.
01:02
Now let's set our axis at b, x, and y.
01:09
So we want to find the shared moment diagram, which means we're going to need to know some or all of the reactions a and b.
01:16
So i think the quickest way to do it is to start with b.
01:19
The beam is in equilibrium, so the moments are also in equilibrium, so sum of moments should equal zero.
01:25
We'll take counterclockwise positive by convention, and the moment is a force times a perpendicular distance.
01:34
So these two around b are positive moments, and then a is a negative moment by the direction we picked.
01:43
Now our applied, our distributed force, we can represent that as a concentrated force.
01:50
Now it's uniform, so it's going to be halfway in between.
01:52
It's 1 .5 meters long, so we have 0 .75 meters distance here, and then the whole length is two meters, so the magnitude of this force is two kilonewtons per meter times 1 .5 meters.
02:15
And so we can do that.
02:16
We start, we said a is a negative moment, has a magnitude of a, and a distance four meters away.
02:22
Then the six kilonewtons is three meters away, and our distributed force, the two times 1 .5, which is a force component, and then the distance 0 .75.
02:35
This is all equal to zero.
02:36
Force b acts right on the axis, zero distance, zero moment.
02:40
So that gives us our force a to be 5 .063 kilonewtons.
02:51
And so that's all we need in terms of math, because, well, some basic arithmetic.
02:55
But we can start our shear moment diagrams from the a end and then work to get to c.
03:03
So starting with the shear diagram, and we can use that and integrate to the moment diagram.
03:08
The shear diagram follows the applied forces, so we have a at 5 .063 kilonewtons upwards...