Problem 2: Uniformly distributed load, simply supported beam...An 8 m long beam is simply supported at the ends and carries a distributed load of 2 kN/m. Draw the shear and bending moment diagrams.

Solution: The total load is (2 kN/m)(8 m) = 16 kN. From the symmetry, the reactions are equal and are 8 kN each. To determine the shear and bending moment at any point within the beam, I draw a free body diagram using a cut between A and B:
 

V = 8 - 2x kN M = 8x - x2 kN-m At x = 0, V = 8 kN. At x = 8 m, V = -8 kN. These agree with free body diagrams (x almost zero from either end) that you can do mentally. Note that, at x = 0 and at x = 8, M = 0 as it should be. You can show that the maximum moment occurs at x = 4 m (by taking the derivative of M and equating to zero). At 4 m, M has its maximum value of 16 kN-m.

By convention, V is positive down and M is positive counterclockwise.

The plots are:

Sorry for the poor drawing. I can't draw curves well, but the maximum should be at x = 4 m.