Question 3.16: Figure 3-23 shows a crank in which it is necessary to visual...

Objective    Draw the free-body diagram of the shaft ABC in Figure 3-23. Draw the shearing force and bending moment diagrams for that part.
Given    The layout from Figure 3-23.
Analysis        The analysis will take the following steps:

1. Determine the magnitude of the torque in the shaft between the left end and point B where the crank arm is attached.
2. Analyze the connection of the crank at point B to determine the force and moment transferred to the shaft ABC by the crank.
3. Compute the vertical reactions at supports A and C.
4. Draw the shearing force and bending moment diagrams considering the concentrated moment applied at point B, 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 A-axis, note that the crank force of 60 lb acts 5.0 in from the axis. The torque, then, is

F = (60 lb)(5.0in) = 300 lb.in

This level of torque acts from the left end of the shaft to section B, where the crank is attached to the shaft.
Now the loading at B 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 B. Also, because the 60-lb applied force acts 3.0 in to the left of B, it causes a concentrated moment in the .x-z plane of 180 lb.in to be applied at B, Both the downward force and the moment at B affect the magnitude and direction of the reaction forces at A and C. First, summing moments about A

(60 \ lb)(6.0 \ in) – 180 \ lb.in – R_c(10.0 \ in) = 0

R_c = [(360 – 180)lb-in]/(10.0 \ in) = 18.0 \ lb upward

Now, summing moments about C.

(60 \ lb)(4.0 \ in) + 180 \ lb.in – R_A(10.0 \ in) = 0
R_A = [(240 + 180) lb-in]/(10.0 \ in) = 42.0 lb upward

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

Comments        In summary, shaft ABC carries a torque of 300 lb.in from point B to its left end. The maximum bending moment of 252 lb.in occurs at point B 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.