Question 1.3: A stepped circular shaft is fixed at A and has three gears t...
A stepped circular shaft is fixed at A and has three gears that transmit the torques shown in Fig.1-13. Find the reaction torque M_{Ax} at A and then find the internal torsional moments in segments AB, BC, and CD. Use properly drawn free-body diagrams in your solution.
Use the following four-step problem-solving approach to find internal torsional moments T(x).
1. Conceptualize : The cantilever shaft structure is stable and statically determinate. The solution for the reaction moment at A(M_{Ax}) must begin with a proper drawing of the FBD of the overall structure (Fig.1-14). The FBD shows all applied and reactive torques. Separate FBDs showing internal torques T in each segment are obtained by cutting the shaft in regions AB, BC, and CD in succession and are given in Fig.1-15(a–c). Each cut produces a left-hand and a right-hand free-body diagram.
2. Categorize : The shaft is subjected to applied torques that act along the centroidal axis of the shaft, so only internal torsional moment T(x) is present at any section cut along the shaft. There is no distributed torque acting on this shaft, so the internal torsional moment T is constant within each segment.
3. Analyze :
Solution for external reaction moment M_{Ax}:
Sum the moments about the x-axis to find the reaction moment M_{Ax} . This structure is statically determinate because there is one available equation from statics \left(\Sigma M_{x}=0\right) and one reaction unknown (M_{Ax} ). A statics sign convention is used [i.e., right-hand rule or counterclockwise (CCW) is positive].
M_{A x}-17,000 \text { lb-in. }+9000 \text { lb-in. }+5000 \text { lb-in. }=0\begin{aligned}M_{A x} &=-(-17000 lb – in .+9000 lb – in .+5000 lb – in .) \\&=3000 lb – in .\end{aligned}
The computed result for M_{Ax} is positive, so the reaction moment vector is in the positive x direction as assumed.
Solution for internal torsional moments T in each shaft segment:
Start with segment AB and use either FBD in Fig.1-15a to find:
Left FBD: Right FBD:
T_{A B}=-M_{A x}=-3000 \text { lb-in. } \begin{aligned}T_{A B}=&-17,000 \text { lb-in. }+9000 \text { lb-in. } \\&+5000 \text { lb-in. }=-3000 \text { lb-in. }\end{aligned}
Next consider segment BC. Summing moments about the x axis in Fig.1-15b gives
Left FBD: Right FBD:
\begin{aligned}T_{B C} &=-M_{A x}+17,000 \text { lb-in. } \\&=14,000 \text { lb-in. }\end{aligned} \begin{aligned}T_{B C} &=9000 lb – in .+5000 lb \text {-in } \\&=14,000 lb – in .\end{aligned}
Finally, moment equilibrium about the x axis leads to a solution for the internal torsional moment in segment CD:
Left FBD: Right FBD:
\begin{aligned}T_{C D}=&-M_{A x}+17,000 lb – in \\&-9000 lb – in .=5000 lb – in\end{aligned} T_{C D}=5000 lb – in
In each segment, the internal torsional moments computed using either the left or right FBDs are the same.
4. Finalize : Segment BC has the maximum positive internal torsional moment, and segment AB has the maximum negative torsional moment. This is important information for the designer of the shaft. Properly drawn free-body diagrams are essential to a correct solution. Either the left or right free-body diagram can be used to find the internal torque at any section.
