Figure 3–20 shows a crank in which it is necessary to visualize the three-dimensional arrangement. The 60-lb downward force tends to rotate the shaft $ABC$ around the x-axis. The reaction torque acts only at the end of the shaft outboard of the bearing support at $A$ . Bearings $A$ and $C$ provide simple supports.
Draw the complete free-body diagram for the shaft $ABC$ , along with the shearing force and bending moment diagrams.

Question

Figure 3–20 shows a crank in which it is necessary to visualize the three-dimensional arrangement. The 60-lb downward force tends to rotate the shaft [latex]ABC[/latex] around the x-axis. The reaction torque acts only at the end of the shaft outboard of the bearing support at [latex]A[/latex]. Bearings [latex]A[/latex] and [latex]C[/latex] provide simple supports.
Draw the complete free-body diagram for the shaft [latex]ABC[/latex], along with the shearing force and bending moment diagrams.

Accepted Answer

Objective: Draw the free-body diagram of the shaft [latex]ABC[/latex] in Figure 3–20. Draw the shearing force and bending moment diagrams for that part.
Given: The layout from Figure 3–20.
Analysis: The analysis will take the following steps:
1. Determine the magnitude of the torque in the shaft between the left end and point [latex]B[/latex] where the crank arm is attached.
2. Analyze the connection of the crank at point [latex]B[/latex] to determine the force and moment transferred to the shaft [latex]ABC[/latex] by the crank.
3. Compute the vertical reactions at supports [latex]A[/latex] and [latex]C[/latex].
4. Draw the shearing force and bending moment diagrams considering the concentrated moment applied at point [latex]B[/latex], along with the familiar relationships between shearing force and bending moments.

Results: The free-body diagram is shown as viewed looking at the x–z plane. Note that the free body must be in equilibrium in all force and moment directions. Considering first the torque (rotating moment) about the x-axis, note that the crank force of 60 lb acts 5.0 in from the axis. The torque, then, is

[latex]T=(60 \mathrm{lb})(5.0 \mathrm{in})=300 \mathrm{lb} \cdot \mathrm{in}[/latex]

This level of torque acts from the left end of the shaft to section [latex]B[/latex], where the crank is attached to the shaft.
Now the loading at [latex]B[/latex] should be described. One way to do so is to visualize that the crank itself is separated from the shaft and is replaced with a force and moment caused by the crank. First, the downward force of 60 lb pulls down at [latex]B[/latex]. Also, because the 60-lb applied force acts 3.0 in to the left of [latex]B[/latex], it causes a concentrated moment in the x–z plane of 180 lb. in to be applied at [latex]B[/latex].
Both the downward force and the moment at [latex]B[/latex] affect the magnitude and direction of the reaction forces at [latex]A[/latex] and [latex]C[/latex]. First, summing moments about [latex]A[/latex],

[latex]\begin{gathered}(60 \mathrm{lb})(6.0 \mathrm{in})-180 \mathrm{lb} \cdot \mathrm{in}-R_{d}(10.0 \mathrm{in})=0\R_{C}=[(360-180) \mathrm{lb} \cdot \mathrm{in}] /(10.0 \mathrm{in})=18.0 \mathrm{lb} ext { upward }\end{gathered}[/latex]

Now, summing moments about [latex]C[/latex],

[latex]\begin{gathered}(60 \mathrm{lb})(4.0 \mathrm{in})+180 \mathrm{lb} \cdot \mathrm{in}-R_{A}(10.0 \mathrm{in})=0 \R_{A}=[(240+180) \mathrm{lb} \cdot \mathrm{in}] /(10.0\mathrm{in})=42.0 \mathrm{lb} ext { upward }\end{gathered}[/latex]

Now the shear and bending moment diagrams can be completed. The moment starts at zero at the simple support at [latex]A[/latex], rises to 252 lb . in at [latex]B[/latex] under the influence of the 42-lb shear force, then drops by 180 lb . in due to the counterclockwise concentrated moment at [latex]B[/latex], and finally returns to zero at the simple support at [latex]C[/latex].

Comments: In summary, shaft [latex]ABC[/latex] carries a torque of 300 lb . in from point [latex]B[/latex] to its left end. The maximum bending moment of 252 lb . in occurs at point [latex]B[/latex] where the crank is attached. The bending moment then suddenly drops to 72 lb . in under the influence of the concentrated moment of 180 lb . in applied by the crank.

Question 3.15: Figure 3–20 shows a crank in which it is necessary to visual...

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