Graphing the internal forces
So what do the formulas tell us?
The shearing force formulas:
f = x - 15 Left half of the beam
f = x + 5 Right half of the beam
The shearing force can be thought of as the vertical stress in the beam at any point along its length. The material must be able to take this stress without shearing (or breaking laterally), so the points of maximum shearing force are critical. They show where this type of material failure is most likely to occur.
From the graph of f = x - 15, for 0 < x < 5 (we're only dealing with the left half of the beam at the moment), we can see that this is where x = 0, i.e. at A itself.
This is always the case for beams - maximum shearing forces act at the points at which the external 'point' or 'knife-edge' forces act, be they at the ends or somewhere along the beam.
It might seem strange, at first, to have different formulas for the right and left halves of the beam. In fact, in this particular case, since the weight of the beam and that of the load are both centrally placed, we find that the shearing forces in the right half of the beam will indeed be symmetrical with those in the left half.
Though identical in magnitude, we will find them to be the negative of those in the corresponding position on the left, (for the 'equal-and-opposite' reasons that Mr Newton would be sure to remind us of were he still around).
Where the downward force of the 'point' load acts, at the very middle of the beam, there is a sudden change in the shearing force (a 'discontinuity') of a magnitude equal to that load.
So the complete shearing force graph would look like this:
At the mid-point, the shearing force appears to jump from -10 to +10.
In practice, we talk about the 'magnitude' of the shearing force, regardless of whether it is negative or positive. If the beam is going to snap, it doesn't much matter in which sense it does so, and indeed the initial diagram shows us clearly enough in which sense the failure will be.
Another way of looking at this is to imagine ourselves considering the beam from the other side: all the forces would then change signs, illustrating the arbitrary nature of our signing convention.
The bending moment formula:
M = 0.5x2 - 15x Left half of the beam
M = 0.5x2 + 5x – 100 Right half of the beam
The bending moment can be thought of as exactly that - the tendency of the beam to bend at that point. In simple cases, it will usually correspond to the actual physical deformation of the beam.
This can be in a downwards direction ('sagging'), usually due to the loads, or an upward direction ('hogging'), usually due to the support forces. These are illustrated below.
As you can see from the formula, the bending moment starts at zero when x = 0. Then it traces a quadratic parabola in a downward direction, until x = 5 in the centre of the beam, at which juncture the formula changes, and the right-hand side is the reflection of the left-hand side in the central line (x = 5).
There is no need for the left and right to obey the equal-and-opposite effect we saw in the shearing force graph, as they are both in 'sag', rather than one in 'sag' and one in 'hog'. The beam will clearly sag most in the middle. No surprise there. The fact that M = 0 at both ends shows that there is no tendency to bend at the ends of the beam itself, as there is no force constraining it beyond the end.
This is the result of our initial assumption that the beam does not extend beyond the support forces at the ends. As we noted at the time, this is not a realistic assumption as to the nature of a real bridge, but the latter can be modelled comprehensively as a combination of beam-like sections, each one behaving in the way we are describing.
Finally, we usually place the two graphs beneath each other, so we can see the relationships between them.
You may notice that the shearing force graph actually gives the gradient (or differential) of the bending moment graph. Looked at another way, the bending moment is the integral of the shearing force.
This gives us a quick way to sketch one graph from the other.
Now we have the ability to analyse the forces and stresses within a beam, as well as the forces that support the beam externally.
Putting these two aspects together brings us to the point where we can predict the effect on any type of beam (uniform or not) of any load (wherever it is positioned), whatever the magnitude of their respective weights.
This is the desired model for a train, or other significant load, moving across a bridge.
The shearing force and bending moment diagrams allow us to see what happens within the bridge as this takes place, and therefore give us insight into the points of maximum stress, and the conditions under which they are likely to be felt.