Using Singularity Functions

Problem 2: Uniformly distributed load, simply supported beam...This is similar to the problem that was presented in the last two classes, but with an extra force...An 8 m long beam is simply supported at the ends and carries a distributed load of 2 kN/m, with a force of 4 kN at the center. Write the shear and bending moment as functions of x.

Solution: The total load is (2 kN/m)(8 m) = 16 kN, plus 4 kN. From the symmetry, the reactions are equal and are 10 kN each. 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 10 kN down. I then do the same around point B and get the same result. However, in this latter case I need the shear on a free body encircling A and very close to B (on the left hand section), so I have to reverse the shear, getting 8 kN up (or negative, since shear down is positive). The moments are zero at each end.

The rate of change of the shear is the negative of the (distributed) load:

But this is wrong, since the shear changes where the 4 kN force acts.  We need to add in a singularity function:

Check it! At x = 0, V = 10 kN (the singularity function is identically zero for x = 0) as it should be. At x = 8 m, the singularity function becomes normal parentheses and V = - 10 kN, so that's also OK.  Also, at x = 4 m, the external force of 4 kN drives the shear down by 4 kN, from 2 kN to -2 kN.

Integrating again for the bending moment...

This is zero at x = 0 and at x = 8, so it checks!

Let's see if we can find the maximum value for the moment by setting the derivative of the moment (the shear) to zero.

Notice how both of the results contradict the conditions (I got 5 m assuming x is less than 4 m, for example).  This is because the shear is discontinuous (slope of the moment) so the calculus does not apply across the discontinuity.

The plots are (the areas under the shear are shown in parentheses):