Using Singularity Functions

Problem 3: Uniformly distributed load, simply supported beam, concentrated moment...An 8 m long beam is simply supported at B but attached at A, carries a distributed load of 2 kN/m, and has an external moment of 5 kN-m applied at its center. Write the shear and bending moment as functions of x.

Solution: The total load is (2 kN/m)(8 m) = 16 kN. Symmetry cannot be used to determine the reactions! Taking the usual directions as positive...

Next, I mentally draw a very short free body diagram around point A; since there is very little (zero) load in that diagram, the shear equals the reaction (but opposite), or 8.625 kN down. I then do the same around point B and get the 7.375 kN down. However, in this latter case I need the shear on a free body encircling A and very close to B, so I have to reverse the shear, getting 7.375 kN up (or negative, since shear down is positive). Can you see that the moments are zero at each end?

This gives the correct values for the shear at x = 0 and at x = 8 m.

This is zero at x = 0 and at x = 8, as it must be. Notice that I had to subtract the external moment?

The moment curve is discontinuous at its maximum (as so is the slope), so we would not expect to be able to use derivatives to determine where it reaches its maximum.

For your comparison, the plots are: