Problem 5.38 Revised

Problem: Draw the shear and bending moment diagrams for the beam and loading shown. Write the shear and bending moment using singularity functions. Determine the maximum normal stress due to bending.

Solution: First, I need the reactions:

The sum of the forces and moments over the entire hinged beam yields two equations and three unknowns:

To obtain another equation, I take a free body of the right portion (from the hinge at B to E):

Taking moments about B,

Subtracting equations (2) and (3), I get R_E = 11.2 kips and R_C = 49.6 kips. Substituting these into equation (1), I get R_A = 19.2 kips. From the sum of the forces (not shown), I get R_B = - 19.2 kips. This internal force at the hinge will be useful in checking the shear when I write its singularity function.

The distributed force given is unsuitable for expression as a singularity function. Any distributed forces to be expressed must continue all the way to the right end of the beam. Therefore I redraw the distributed load as shown below (I am not showing the other loads):

But this is incomplete because the other external forces (R_C = 49.6 kips and the 32 kip load) have to be included...

You might notice that R_A has been included as the initial value of the shear (the lower limit in the integration) but that R_E has not. Can you see why not? How would you write the singularity function to include R_E? Why doesn't it make any difference?

Before proceeding to determine the moment function, I check the shear for the known values:

Check at A (x = 0): V = 19.2 kips

Check at B (x = 8): V = - 19.2 kips

Check at E (x = 20): V = - 11.2 kips

Since each of these agree with the reactions found there, I integrate to produce the moment equation:

There are no externally applied moments to add in as additional singularity functions.

Check at A (x = 0): M = 0

Check at E (x = 20): M = 0

I use the shear and moment functions to compute values in order to plot them:

position (x - ft)	Shear (kip)	Moment (kip-ft)
0	+19.2	0
8	- 19.2	0
10	- 28.8	- 48
15	+20.8	+56
20	-11.2	0

It is clear from the linearity of the shear and the symmetry of the values at x = 0 and x = 8 ft that the shear reaches a zero value at x = 4 ft. Thus, the moment must reach a local maximum at x = 4 ft; M = 38.4 kip-ft. I now have enough information to plot the shear and bending moment:

Sorry about the parabolic arch - it's the best I could do with a freehand drawing.

To determine the maximum normal stress due to bending, I need to use the elastic flexure formula and the value of the section modulus from Appendix C for a W12 x 40: