Question 10.3.3: For the beam and loading shown in Figure 1 draw the shear fo...
BEAM WITH DISTRIBUTED LOAD
For the beam and loading shown in Figure 1 draw the shear force and bending moment diagrams.
Goal Draw the shear force and bending moment diagrams for beam ABC.
Given Information about the dimensions and loading of the beam, which allow us to model AC as a planar beam.
Assume The weight of the beam is negligible and the fixed support at C is ideal.
Draw We require two free-body diagrams to develop the equations that describe the shear and moment diagrams for the beam. First we cut the beam where the distributed load is applied (Figure 2a). The free-body diagram for this isolated beam segment is valid for a cut anywhere between A and B—that is, for 0 ≤ x ≤ 2 m. Then we cut between B and C , which is beyond the applied distributed load (Figure 2b), and draw a free-body diagram that is valid for a cut such that 2 m ≤ x ≤ 6 m.
Formulate Equations and Solve Analyzing equilibrium of the beam segments, we create equations that describe the shear and moment as a function of x.
For 0 ≤ x ≤2 m (Figure 2a):
In our equilibrium analysis, we must include only the portion of the distributed load that is acting on the segment of the beam we are considering. The length of the beam segment we have isolated to the left of the cut at P is x. Therefore the load is (10 kN/m)x, and:
\sum{F_{y}\left(\uparrow + \right) } = – 10 \frac{KN}{m} x – V_{y} =0
V_{y} = – 10\frac{KN}{m} x (0 ≤ x ≤ 2 m) (1)
This equation describes a line from 0 kN at x = 0 m to −20 kN at x = 2 m.
We choose P as our moment center in the moment equilibrium equation. This means that we measure the moment arm from P to the centroid of the distributed load. For free-body diagram in Figure 2a that distance is x/2.
\sum{M_{x @ P} }\left(\curvearrowleft + \right) = \left\lgroup 10 \frac{KN}{m} x\right\rgroup\frac{x}{2} + M_{bz} =0
M_{bz} = – \left\lgroup 5 \frac{KN}{m} \right\rgroup x^{2} (0 ≤ x ≤ 2 m) (2)
This equation describes a curve that starts from 0 kN.m at x = 0 m and bends parabolically down to −20 kN.m at x = 2 m.
For 2 m ≤ x≤ 6 m (Figure 2b):
The entire distributed load is acting on the isolated portion of the beam shown in Figure 2b. Therefore the load is (10 kN/m) (2 m) = 20 kN , and:
\sum{F_{y}\left(\uparrow + \right) } = – 20 kN – V_{y} =0
V_{y} = – 20 kN (2 m ≤ x ≤ 6 m) (3)
For the moment equilibrium equation we again measure the moment arm from P to the centroid of the distributed load, which is 1 m to the right of A. Therefore the length of the moment arm is (x − 1 m).
\sum{M_{x @ P} }\left(\curvearrowleft +\right) = 20kN (x – 1 m)+ M_{bz} =0
M_{bz} =20 kN.m – (20 kN)x (2 m ≤x ≤ 6 m) (4)
For this portion of the beam, the bending moment diagram is a straight line extending from −20 kN.m to −100 kN.m.
using (1)–(4) we create the shear force and bending moment diagrams for the beam (Figures 2c and 2d).
Check In addition to checking the value of the diagrams at several points, we 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 where no loads are applied. We illustrate another method for checking the shear and bending moment relationships in the next section.
