Question 10.2.1: A simply supported beam is loaded as shown in Figure 1. Dete...
INTERNAL LOADS IN A PLANAR SIMPLY SUPPORTED BEAM
A simply supported beam is loaded as shown in Figure 1. Determine the axial force, shear force, and bending moment at cross sections at B and D.
Goal Find the axial force, shear force, and bending moment at two locations (B and D).
Given Information about the dimensions and loading of the beam. because all loads act in the xy plane, this is treated as a planar beam.
Assume The weight of the beam is negligible.
Draw First we need to find the loads at supports A and G. To accomplish this we draw a free-body diagram (Figure 2) of the entire beam. Later we will cut the beam at B and D and draw free-body diagrams of portions of the beam to find the internal loads at B and D.
Formulate Equations and Solve First we use the equations for planar equilibrium to find the loads at supports A and G ( F_{Ax} = 0, F_{Ay} = 3 kN, and F_{Gy} = 3 kN). (Calculations not shown.)
Internal Loads at B: We make a cut at B and draw a free-body diagram of the left portion of the beam (Figure 3). When drawing the diagram, we assume that the unknown internal loads are positive (using the internal load sign convention). We then apply the planar equilibrium equations (using the equilibrium sign convention):
\sum{F_{x} \left(\rightarrow + \right) } =0 \Rightarrow N_{x}=0
We sum moments about the cut at B, allowing us to eliminate V_{y} from the equilibrium equation so that we have only one unknown. As we apply (5.5C), with the moment center at B, each position vector is measured from the cut to the load. For the distributed load, we measure the position vector from the cut to the centroid of the load.
\sum{M_{z }}=0 (5.5C)
\sum{M_{z @ B} }\left(\curvearrowleft + \right) = – 3 kN(1 m)+1\frac{kN }{m} (1 m)(0.5 m)+M_{bz}=0
M_{bz}=2.5 kN.m
Finally we solve for V_{y}:
\sum{F_{y}\left(\uparrow + \right) } = -V_{y} +3 kN -1\frac{kN }{m} (1 m)=0
V_{y}=2 kN
Internal Loads at D: using the free-body diagram in Figure 4 as a reference, we calculate the internal forces at D using the same approach we used for cross section B.
\sum{F_{x} \left(\rightarrow + \right) } =0 \Rightarrow N_{x}=0
We determine M_{D} by summing moments with the moment center at D:
\sum{M_{z @ D} }\left(\curvearrowleft + \right) =- 3 kN(3 m)+1\frac{kN }{m}(2 m)(2 m)+M_{bz}=0
M_{bz}=5 kN.m
Finally we calculate V_{y}:
\sum{F_{y} \left(\uparrow + \right) } = – V_{y}+3 kN – 1\frac{kN }{m}(2 m)=0
V_{y}= 1 kN
Comment: While it might be tempting to simplify the beam loading and calculations by replacing the distributed load by a 2-kN point load at B, this will lead to incorrect results. Prove this to yourself by drawing a free-body diagram of beam portion AB using the 2-kN replacement load and comparing it with Figure 3. Will you get the same M_{bz}?
Check An alternative method of solving for the internal loads, and also a check on our solution, is to analyze the right portion of the beam (Figure 5).
\sum{F_{x} \left(\rightarrow + \right) } =0 \Rightarrow N_{x}=0
based on Figure 5 with the moment center at B:
\sum{M_{z @ B} }\left(\curvearrowleft + \right) =-1\frac{kN }{m} (1 m)(0.5 m) – 4 kN(3 m) + 3 kN(5 m) -M_{bz} = 0
M_{bz} = 2.5 kN.m
Summing vertical forces we get:
\sum{F_{y}\left(\uparrow + \right) } =V_{y}-1\frac{kN }{m} (1 m) – 4 kN+ 3 kN = 0
V_{y}=2 kN
These are the same results we obtained earlier.
We can use a similar approach to check our results at D by analyzing the portion of the beam to the right of the cut at D.
