Question 10.3.2: A moment is applied at point B on the simply supported beam ...

Goal Find the shear and bending moment diagrams for beam AC.
Given Information about the geometry and loading of the beam.
Assume The weight of the beam is negligible and supports are ideal.
Draw We draw a free-body diagram of the entire beam and solve for loads at the supports (Figure 2a). Next, we create a coordinate system with its origin at A and make two different cuts in the beam: the first to the left of M_{o} , and the second to the right of M_{o} . We draw free-body diagrams of isolated portions AP when x ≤ B (Figure 2b), and when x ≥ B (Figure 2c).
Formulate Equations and Solve We analyze equilibrium of the beam portions and create equations that describe the shear and moment as a function of x.
For 0 ≤ x ≤ B (Figure 2b):

\sum{F_{y}\left(\uparrow + \right) } =  –  \frac{M_{o}}{L} –  V_{y}= 0
V_{y}= –  \frac{M_{o}}{L}              (1)
\sum{M_{x @ P} }\left(\curvearrowleft + \right) = –  \frac{M_{o}}{L} x +M_{bz} =0
M_{bz} =  –  \frac{M_{o}}{L} x              (0 ≤ x ≤ B)      (2)

For B ≤ x ≤ L (Figure 2c)

\sum{F_{y}\left(\uparrow + \right) } =  –  \frac{M_{o}}{L} –  V_{y}= 0
V_{y}= –  \frac{M_{o}}{L}        (3)
\sum{M_{x @ P} }\left(\curvearrowleft + \right) =  \frac{M_{o}}{L} x +  – M_{o} + M_{bz} =0
M_{bz} = M_{o}\left\lgroup 1 -\frac{x}{L}\right\rgroup        (B ≤ x ≤ L)                  (4)

using (1)–(4) we create the V and M diagrams (Figures 2d and 2e).
Check We can check the value of the diagrams at several points, and also look at the diagram shapes. The shear force is constant between A and C, because no forces are applied between the supports. The bending moment diagram exhibits a discontinuity of magnitude M_{o} at the location of the applied moment.