Question 10.3.4: Given the beam and loading shown in Figure 1a, draw the shea...

Goal Draw the shear force and bending moment diagrams and locate the maximum bending moment.
Given Information about the dimensions and loading of the beam (indicating a planar beam model).
Assume The weight of the beam is negligible and the supports are ideal. This means the pin at B is frictionless and imparts no moment to the beam, and that the roller at D is frictionless and imparts no moment or axial force to the beam.
Draw We draw a free-body diagram of the beam and solve for the loads at B and D (Figure 1b). We then create three free-body diagrams: one for a cut between A and B, one for a cut between B and C, and one for a cut between C and D (Figures 1c, 1d, and 1e).
Formulate Equations and Solve We perform an equilibrium analysis of each of the three free-body diagrams to create equations that describe the shear and bending moment as a function of x.
For 0 ≤ x ≤ 2 ft (Figure 1c), equilibrium in the y direction is:

\sum{F_{y}\left(\uparrow + \right) } =  –  2 \frac{kip}{ft} x  –  V_{y} =0
V_{y} =   –  2 \frac{kip}{ft} x         (0 ≤ x ≤ 2 ft)        (1)

This equation describes a line from 0 kip  at  x = 0 ft  to  − 4 kip at x = 2 ft. The moment equilibrium equation is:

\sum{M_{x @ P} }\left(\curvearrowleft +\right) = \left\lgroup 2 \frac{kip}{ft} x\right\rgroup \frac{x}{2}  + M_{bz} =0
M_{bz} = –  \left\lgroup 1 \frac{kip}{ft} \right\rgroup x^{2}                      (0 ≤ x  ≤ 2 ft)                (2)

This equation describes a parabolic curve from 0 kip.ft at x = 0 ft to −4 kip.ft at x = 2 ft.
For 2 ft ≤ x ≤ 6 ft (Figure 1d), equilibrium in the y direction gives:

\sum{F_{y}\left(\uparrow + \right) } =  –   \left\lgroup 2 \frac{kip}{ft} \right\rgroup (2   ft) + 7  kip  –  V_{y} =0
V_{y} = 3  kip                (2 ≤ x ≤ 6 ft)               (3)

Summing the moments about a z axis at the cut gives:

\sum{M_{x @ P} }\left(\curvearrowleft + \right) =0=2 \frac{kip}{ft}(2  ft) (x – 1  ft)  –  7  kip(x – 2  ft) + M_{bz} =0

With rearranging and simplifying this becomes

M_{bz} = –  10  kip.ft (3  kip)x          (2 ft ≤ x ≤ 6 ft)          (4)

For this portion of the beam, the bending moment diagram is a straight line extending from −4 kip.ft at x = 2 ft to 8 kip.ft at x = 6 ft.
For 6 ft ≤ x ≤ 8 ft (Figure 1e), we get:

\sum{F_{y}\left(\uparrow + \right) } =  –   \left\lgroup 2 \frac{kip}{ft} \right\rgroup (2   ft) + 7   kip  – 7   kip   –  V_{y} =0
V_{y} = –  4  kip               (6 ft ≤ x ≤ 8 ft)               (5)

and:

\sum{M_{x @ P} } = 0\left(\curvearrowleft + \right)
2 \frac{kip}{ft} (2   ft) (x – 1  ft)  –  7  kip(x – 2  ft) + 7  kip (x – 6  ft) + M_{bz} =0

With rearranging and simplifying this becomes

M_{bz} =32  kip.ft (4  kip)x               (6 ft ≤ x ≤ 8 ft)           (6)

The bending moment diagram is a straight line extending from 8 kip.ft at x = 6 ft to 0 kip.ft at x = 8 ft.
using (1)–(6) we create the shear force and bending moment diagrams for the beam (Figures 2b and 2c). We see from the bending moment diagram that the maximum moment is 8 kip.ft and it occurs at point C.
Check We can calculate V and M at several points and compare the values with the diagrams. We also can look at the shapes of the diagrams. The shear force is decreasing linearly between A and B due to the uniformly distributed load, and is constant between B and C, and between C and D where no loads are applied. We’ll learn more about the shape of the moment diagram in the next section. We will check the results for this example in the next section using the soon-to-be-developed relationships between ω, V_{y} and M_{bz}.