Problem 5.114

The free body diagram is

Taking mental free body diagrams about each end, I find that there is no shear or bending moment at end A. At end C, and remembering to reverse the signs resulting from a free body from the right, the shear is

and the bending moment is

. These will provide checks on the results obtained using the singularity functions.

To write the load as a singularity function, it cannot end in the middle of the beam. So I redraw the beam with equivalent loads :

These distributed loads may be expressed as:

Check: function #1 - At x = 0, it yields as it should.

At x = 2a, it yields

as it should.

At x = a, it yields zero.

Check: function #2 - At x = 0, the singularity function is identically zero (by definition)

At x = 2a, it yields

as it should.

At x = a, it yields zero.

Now I use the relationships:

Again, I check this. At x = 0, V = 0 (as it should be). And at x = 2a, V = (which is also correct). Now I am ready to compute the bending moment:

Check: At x = 0, the moment should be zero, and it is! At x = 2a, the moment should be , and it is!

To determine the bending moment just to the right of C (not required) , I substitute x = a to get .