How V and M Change Along a Beam
In B3 you learned the bending stress σ = MyI. But that M, the bending moment, is different at every position along the beam. The middle of a bridge and the spot near a support cannot carry the same burden. So to design a beam you first need to know how the internal forces are distributed along its whole length. There are two internal forces: the shear V that tries to slide one face past the other, and the moment M that bends it. Plotting these two against the length of the beam gives the shear-force diagram and the bending-moment diagram, the V and M diagrams for short. The striking thing is that the two are not independent: the slope of M is exactly V, and the slope of V is the distributed load. And the single point on these diagrams where M is largest is where the beam will break.
First, set the stage. Take a simply supported beam, resting on a support at each end, and place a single point load on it. To hold the load up, the two supports must push upward, and those forces are the reactions. The equilibrium conditions fix them: the up-and-down forces sum to zero, and the rotation about any point sums to zero. That gives the intuitive result that the support nearer the load carries more. Drag the load position. Move the load left and the left reaction grows while the right shrinks. Precisely, R1 = P(L−a)L and R2 = P·aL. Finding these reactions first is the starting point of every beam analysis.
Now bring back the imaginary cut from A1. Cut the beam in your mind at some position and take only the left piece. For this piece to stay in equilibrium, the cut face must carry two internal forces: the shear V that balances the up-and-down forces, and the moment M that prevents rotation. Gather the forces on the left piece, the reaction and any load up to the cut, write equilibrium, and out come the V and M at that spot. Drag the cut position. Every time you move the cut, V and M change. This is the key point: the internal forces are functions of position. Each cut section has its own V(x) and M(x).
Instead of cutting at one place, slide the cut from one end of the beam to the other and record V; that graph is the shear-force diagram. For a beam with only point loads the shape is very simple. Between loads V stays constant, then the instant it passes a load it drops by that load's size. A staircase. Drag the load position. The left segment sits at height +R1, the right at −R2, and you see the jump of P at the load point. The area under this staircase actually connects to the moment, but that comes in the next scene. For now, remember the V diagram as a staircase made by the supports and the loads.
Now the moment diagram. For a simply supported beam under a point load, M is zero at both ends and rises to a sharp peak at the load: a triangle. But set it beside the V diagram and a hidden rule appears: the slope of M is exactly V. dM/dx = V. In the left segment where V is positive, M climbs at a constant slope; in the right where V is negative, it descends. And at the point where V hits zero, here the load point, M reaches its peak. Drag the load position. As the V staircase moves, the apex of the M triangle moves with it. Thanks to this relation, knowing V alone lets you draw M by integration: the area under the V diagram is the change in M.
Now for why all of this matters. In B3 the bending stress was σ = MS, proportional to the moment. So the most dangerous place in a beam is where M is largest. There the stress is maximum, so it yields or breaks first. This is the real reason for drawing V and M diagrams: to find that one maximum value of M. Knowing it alone lets you judge at once, through σmax = MmaxS, whether the section will hold. Drag the load position. The peak moves with it, and Mmax is largest when the load sits exactly at the center. So a designer assumes the worst load position, takes that Mmax, and sizes the section to match. That single point on the diagram decides the fate of the whole beam.