shear and moment diagrams
The last lesson met the bending stress σ = MyI. But that moment is not the same at every place on the beam. The middle of a bridge and the neighborhood of a support do not carry the same burden. To design a beam you first need how the internal forces spread over the whole beam. There are two. The shear V that tries to slide along a face, and the moment M that tries to curve it. Drawn along the length of the beam, they are the shear diagram and the bending-moment diagram, the V·M diagrams for short. They do not live apart. The slope of M is exactly V, and the slope of V is the distributed load. The one place where M is largest is where the beam will break. The stress of B3 becomes, here, the story of the whole beam. The same beam carries a different burden at every place, and without that map the section cannot be chosen. The diagram points at the place.
First set the stage. A simply supported beam, held at both ends, carries one point load. To hold the load the two supports must push upward, and those forces are the reactions. Equilibrium fixes them. The sum of the vertical forces is 0, and the sum of the rotations about any point is 0 too. Then the support closer to the load carries more, which is the intuitive result. Move the load to the left and the left reaction grows while the right one shrinks. Exactly, R1 = P(L−a)L and R2 = P·aL. Finding these reactions first is the start of every beam analysis. The nearer support carrying more is what equilibrium requires, and that is the start.
Now bring back the imagined cut from before. Cut the beam in your mind at some place and look only at the left piece. For that piece to stay in equilibrium, the cut face needs two internal forces. The shear V that balances the vertical forces, and the moment M that stops the rotation. Gather the forces on the left piece, the reaction and the loads up to there, write equilibrium, and the V and M at that place come out. Move the cut and V and M change every time the place changes. The internal forces are functions of position, so each cut section has its own V(x) and M(x). The cut of A1 walks along the length when the member is a beam.
Record V while you walk the cut from the start of the beam to the end, and that graph is the shear diagram. With only point loads the shape is simple. Between loads V stays constant, and the moment you pass a load it drops by the size of that load, a stair. Move the load and the left stretch sits at height +R1, the right stretch at height −R2, jumping by P at the load. The area of this stair is tied to the moment, and that comes in the next scene. For now, remember the shear diagram as the stair made by the supports and the loads. The jump in V is where the load sits. The jump marks the load, and the flat stretches between loads are the quiet parts of the stair.
Now the moment diagram. For a simple beam with a point load, M is 0 at both ends and reaches a sharp peak at the load, a triangle. Set it beside the shear diagram and a hidden rule shows. The slope of M is exactly V. dM/dx = V. On the left, where V is positive, M climbs at a steady slope, and on the right, where V is negative, it comes down. And where V becomes 0, here at the load, M reaches its peak. Move the load and the V stair moves, and the top of the M triangle follows. Because of this relation, knowing only V lets you draw M by integration. The area of the V diagram is the change in M. When the slope changes sign, a peak appears, and that peak is what the designer is hunting.
All of this matters because of the bending of the earlier lesson. In B3 the bending stress was σ = MS, proportional to the moment. So the most dangerous place on the beam is where M is largest. The stress is greatest there, and that is where it yields or breaks first. That is the real reason to draw the V·M diagrams. You are looking for that one largest M. Know only that value and σmax = MmaxS tells at once whether the section holds. Move the load and the peak follows, and Mmax is largest when the load sits exactly at midspan. A designer assumes the worst place for the load, takes that Mmax, and chooses the section to match. That one point on the diagram decides the fate of the whole beam.