Question 10.3.1: The simply supported beam in Figure 1 is loaded with a point...
SHEAR, MOMENT, AND AXIAL FORCE DIAGRAM FOR A SIMPLY SUPPORTED BEAM
The simply supported beam in Figure 1 is loaded with a point load at a distance L/3 to the right of support A. Determine the axial force, shear force, and bending moment diagrams for the beam.
Goal For beam AB, find the axial force, shear force, and bending moment diagrams.
Given The dimensions, support conditions, and loading of the beam. The loads are all in one plane so we can treat beam as planar.
Assume The weight of the beam is negligible. We also assume A and B are ideal supports. This means the pin at A is frictionless and imparts no moment to the beam, and that the roller at B is frictionless and imparts no moment or axial force to the beam.
Draw by creating a free-body diagram of the entire beam and applying the conditions of equilibrium, we find the forces at supports A and B in terms of the applied force to be F_{Ax} = 0 , F_{Ay} = 2Q/3 , F_{By} = Q/3 (Figure 2a).
We follow the same procedure we demonstrated for the cantilever beam and create a free-body diagram of a portion of the beam. However, because a point load is applied to beam AB, we need to create two free-body diagrams, one with P to the left of the load (x < L/3) (Figure 2b) and the other with P to the right of the load at x > L/3 (Figure 2c).
Formulate Equations and Solve using the free-body diagram in Figure 2b (x < L/3), we write the planar equilibrium equations with the moment center at P:
\sum{F_{x} \left(\rightarrow + \right) } = 0 \Rightarrow F_{Ax} + N_{x} = 0
\sum{F_{y}\left(\uparrow + \right) } = 0 \Rightarrow \frac{2Q}{3} – V_{y}=0
\sum{M_{x @ P} }\left(\curvearrowleft + \right) =0 \Rightarrow – \frac{2Q}{3}x + M_{bz} =0
These equations are valid for any x such that 0 ≤ x ≤ L/3 and can be rearranged as
N_{x} = – F_{Ax} = 0 axial force (1A)
V_{y}= \frac{2Q}{3} shear force (1B)
M_{bz} = \frac{2Q}{3} x bending moment (1C)
Referring to the free-body diagram of the portion of the beam with the boundary at x > L/3 (Figure 2c), we write a second set of planar equilibrium equations with moment center at P (where P is now at L/3 ≤ x ≤ L):
\sum{F_{x} \left(\rightarrow + \right) } = 0 \Rightarrow F_{Ax} + N_{x} = 0
\sum{F_{y}\left(\uparrow + \right) } = 0 \Rightarrow \frac{2Q}{3} – Q – V_{y}=0
\sum{M_{x @ P} }\left(\curvearrowleft + \right) =0 \Rightarrow – \frac{2Q}{3}x +Q\left\lgroup x – \frac{L}{3} \right\rgroup + M_{bz} =0
The moment equilibrium equation can be rearranged as
\frac{Q}{3} \left(x – L\right) + M_{bz} =0
These equations are valid for any x such that L/3 ≤ x ≤ L and can b rearranged as
N_{x} = – F_{Ax} = 0 axial force (2A)
V_{y}= – \frac{Q}{3} shear force (2B)
M_{bz} = \frac{Q}{3} \left( L – x \right) bending moment (2C)
We now create axial force, shear force, and bending moment diagrams for the simply supported beam. We use (1A)–(1C) to draw the diagrams for 0 ≤ x ≤ L/3, and (2A)–(2C) for L/3 ≤ x ≤ L , as shown in Figure 3a, b, and c.
We notice that at x = L/3, the bending moment equations for the two portions intersect at M_{bz} = 2QL/9 (this is the value of bending moment given by both (1C) and (2C) at x = L/3). At the same point, the shear force changes from positive to negative, represented by the step change on the shear force diagram.
From Figure 3c the largest magnitude of bending moment occurs at x = L/3 ; is this consistent with where you thought it would be? Is this where you would expect the beam to fail?
Check In the next section where we explore the mathematical relationship between V and M diagrams, we will discover that there are some consistencies between the two diagrams that help us to check if they are correct. In the meantime we can draw free-body diagrams and calculate values of N_{x} , V_{y} , and M_{bz} at several points and compare these results with the diagrams as a check.
Also, we can recognize that the shear diagram must be constant between applied point loads, and that the “jump” in the shear diagram at the location of the point load is equal in magnitude to the load. For this example V_{y} jumps from 2Q/3 to −Q/3 at C. The total change is −Q , which is equal to the applied downward load.
