Shear and Bending Moment Diagrams using Relations between Shear

and Bending Moment

*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 relations among shear and bending moment to expedite solution.

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

Solution: The shear force at x = 0 is 3wL/8 and the slope is dV/dx = ˗ w. So the shear force
distribution starts out at 3wL/8 and decreases with a slope of ˗ w. The change in slope from
0 ≤ x ≤ L/2 is V(x) = ʃ ˗ w dx = wL/2. Thus V(L/2) = ˗ wL/8. Since w(x) = 0 from
L/2 ≤ x ≤ L there is no change in shear force in this section of the beam. The figure on
the next screen shows the shear force diagram.

Click here to continue with this example.

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