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 at A, then find the internal torsional moments in segments AB, BC, and CD. Use properly drawn free-body diagrams in your solution.
(1) Draw the FBD of the overall shaft structure. The cantilever shaft struc-ture is statically determinate. The solution for the reaction moment at A must begin with a proper drawing of the FBD of the overall structure (Fig. 1-14). The FBD shows all applied and reactive torques .
(2) Sum the moments about the x axis to find the reaction moment M_{A x}.
This structure is statically determinate because there is one available equation from statics (ΣM_{ x}= 0) and one reaction unknown (M_{A x}). A stat-ics sign convention is used (i.e., right-hand rule or CCW is positive).
M_{A x} – 1900 N m + 1000 N m + 550 N m = 0
M_{A x} = -(-1900 N m + 1000 N m + 550 N m)= 350 N m
The computed result for M_{A x} is positive, so the reaction moment vector is in the positive x direction as assumed.
(3) Find the internal torsional moments in each segment of the shaft. Now that reaction moment M_{A x} is known, we can cut a section through the shaft in each segment creating left and right FBDs (Fig. 1-15). Internal torsional moments then may be computed using statics. Either FBD may be used; the computed internal torsional moment will be the same.
- Find the internal torque T_{A B} (Fig. 1-15a).
Left FBD:
T_{A B}=-M_{A x}=-350 \mathrm{~N} \cdot mRight FBD:
\begin{aligned}T_{A B}=&-1900 {~N} \cdot {m}+1000 {N} \cdot {m} \\&+550 {N} \cdot {m}=-350 {~N} \cdot {m}\end{aligned}- Find the internal torque T_{B C} (Fig. 1-15b).
Left FBD:
\begin{aligned}T_{B C} &=-M_{A x}+1900 {N} \cdot {m} \\&=1550 {~N} \cdot {m}\end{aligned}Right FBD:
\begin{aligned}T_{B C} &=1000 {N} \cdot {m}+550 {N} \cdot {m} \\&=1550 {N} \cdot {m}\end{aligned}- Find internal torque T_{C D} (Fig. 1-15c).
Left FBD:
\begin{aligned}T_{C D}=&-M_{A x}+1900 {N} \cdot {m} \\&-1000 {N} \cdot {m}=550 {N} \cdot {m}\end{aligned}Right FBD:
T_{C D}=550 {~N} \cdot {m}In each segment, the internal torsional moments computed using either the left or right FBDs are the same.
