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Shear and Bending Moment Diagrams using the Direct Method

*Example: (continued)

Strategy: Next construct a FBD of the portion of the beam from 0 ≤ x ≤ L/2 as shown

below and use equilibrium to determine the shear force, V, bending moment, M, and

axial force, N in this section of the beam. wx is the equipollent force equivalent to the

distributed load in this section of the beam.

From equilibrium for 0 ≤ x ≤ L/2: → ΣF_x = 0 N = 0

CCW ΣM_C = 0 wx(x/2) + M – (3wL/8)x = 0 M(x) = 3wLx/8 – wx²/2

↑ ΣFy = 0 3wL/8 – V – wx = 0 V(x) = 3wL/8 – wx

Strategy: Next construct a FBD of the portion of the beam from L/2 ≤ x ≤ L as shown

below and use equilibrium to determine the shear force, V, bending moment, M, and

axial force, N in this section of the beam. wL/2 is the equipollent force equivalent to the

distributed load in this section of the beam.

From equilibrium for L/2 ≤ x ≤ L: → ΣF_x = 0 N = 0

CCW ΣM_C = 0 (wL/2)(x – L/4) + M – (3wL/8)x = 0 M(x) = – wLx/8 + wL²/8

↑ ΣFy = 0 3wL/8 – V – wL/2 = 0 V(x) = – wL/8

The final step is to plot these values of shear, V(x), and bending moment, M(x).

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