Shear and Bending Moment Diagrams using the Direct Method

*Example: The beam AB shown below is simply supported at A and by a roller at B. A

uniformly distributed load, w, acts on the first half of the beam. Plot the shear and

bending moment diagrams showing the distribution of shear and bending moment along

the axis of the beam.

Strategy: Use the direct method. Click here for a reminder of this method.

The FBD of the entire beam (where the equipollent load wL/2 acting L/4 from A

replaces the distributed load, w) is:

From equilibrium: → ΣF_x = 0 A_x = 0

CCW ΣM_A = 0 B_y L – (wL/2)(L/4) = 0 B_y = (wL/8)

↑ ΣFy = 0 A_y – wL/2 + wL/8 = 0 A_y = 3wL/8

Now the external loading changes at midspan. So break the beam into two sections, one

from 0 ≤ x ≤ L/2 and the other from L/2 ≤ x ≤ L. Construct a FBD of each section and

use equilibrium to determine the internal shear force and bending moment in each section.

Click here to continue with this example.

Return to Notes on Solid Mechanics

Copyright © 2019 Richard C. Coddington
All rights reserved.