Question 10.2.2: The cantilever beam in Figure 1 is loaded with a point force...
INTERNAL LOADS IN A PLANAR CANTILEVER BEAM
The cantilever beam in Figure 1 is loaded with a point force at the free end and a concentrated moment at a distance of L/2 from the free end. 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 cross sections at B and D.
Given Information about the dimensions and loading of the beam. because all of the applied loads cause the beam to bend in the xy plane, we can treat the system as planar.
Assume The weight of the beam is negligible.
Draw If we use a free-body diagram of the portion of the beam between the free end and the cross section of interest, we can determine the internal loads without calculating the loads at support E; therefore we do not need to start with a free-body diagram of the entire beam. Instead we cut the beam at B and analyze the portion of the beam to the left of the cut. We do the same for the cut at D. The free- body diagrams of the left portions of the beam for each cut are shown in Figure 2 and Figure 3. In each diagram the unknown internal forces are assumed to be positive according to the internal load sign convention.
Formulate Equations and Solve using the free -body diagram of the left portion of the beam (Figure 2) as a reference we first sum moments about the cut at B:
\sum{M_{z @ B} }\left(\curvearrowleft + \right) =P\left\lgroup\frac{L}{4} \right\rgroup +M_{bz}=0
M_{bz}= – \frac{PL}{4}
The minus sign indicates that the bending moment is in the opposite direction from what we assumed. Since we assumed a positive bending moment, the bending moment is actually negative. This means that the beam is curving downward with the bottom surface of the beam in compression and the top in tension.
We use the other two planar equilibrium equations to solve for N_{x} and V_{y} :
\sum{F_{x} \left(\rightarrow + \right) } =0 \Rightarrow N_{x}=0
\sum{F_{y}\left(\uparrow + \right) } = – P – V_{y}=0 \Rightarrow V_{y}=- P
The minus sign means that the shear force is acting in the opposite direction from that which was assumed and therefore is acting upward. Before going on to calculate the internal loads at D, we reflect on the answers we obtained for the loads at B. Figure 4 shows free-body diagrams of the left and right portions of the beam when it is cut at B and includes our calculated values from above. At the cut we see the equal and opposite force pairs and moment pairs that we expect because of Newton’s third law. Looking at the portion of the beam to the left of B, we notice that the applied load at A and the shear force at B form a counterclockwise couple with a magnitude of PL/4. The clockwise internal bending moment of magnitude PL/4 at B maintains equilibrium for this portion of the beam.
Now we find the internal loads at the cross section at D. Using Figure 3 as a reference, we calculate the internal forces at D using the same approach we used for the cross section at B.
\sum{M_{z @ D} }\left(\curvearrowleft + \right) =0=P\left\lgroup\frac{3L}{4} \right\rgroup +M_{c} + M_{bz}
Substituting for M_{c} (Figure 1 indicates that M_{c} = PL), we get
P\left\lgroup\frac{3L}{4} \right\rgroup +PL+M_{bz}=0 \Rightarrow M_{bz}= -\frac{7PL}{4}
based on vertical and horizontal equilibrium:
\sum{F_{x} \left(\rightarrow + \right) } =0 \Rightarrow N_{x}=0
\sum{F_{y}\left(\uparrow + \right) } = – P – V_{y}=0 \Rightarrow V_{y}=- P
There are no axial loads applied to the beam, and therefore the internal axial force is zero throughout the beam.
Check To check the results, we can analyze the portions of the beam to the right of the cuts at B and D. This requires us to first calculate the loads at support E.
