Question 3.8: A solid steel shaft ABC with a 50 mm diameter (Fig. 3-35a) i......
A solid steel shaft ABC with a 50 mm diameter (Fig. 3-35a) is driven at A by a motor that transmits 50 kW to the shaft at 10 Hz. The gears at B and C drive machinery requiring power equal to 35 kW and 15 kW, respectively.
Compute the maximum shear stress \tau _{max} in the shaft and the angle of twist \phi _{AC} between the motor at A and the gear at C. (Use G = 80 GPa.)
Use a four-step problem-solving approach.
1. Conceptualize:
Torques acting on the shaft: Begin the analysis by determining the torques applied to the shaft by the motor and the two gears. Since the motor supplies 50 kW at 10 Hz, it creates a torque T _{A} at end A of the shaft (Fig. 3-35b) calculated from Eq. (3-40):
P=2\pi fT \ \ \ \ \ (f=Hz=s^{-1}) (3-40)
T _{A}=\frac{P}{2\pi f}=\frac{50 \ kW}{2\pi (10 \ Hz)}=796 \ N\cdot mIn a similar manner, calculate the torques T _{B} and T _{C} applied by the gears to the shaft:
T _{B}=\frac{P}{2\pi f}=\frac{35 \ kW}{2\pi (10 \ Hz)}=557 \ N\cdot mT _{C}=\frac{P}{2\pi f}=\frac{15 \ kW}{2\pi (10 \ Hz)}=239 \ N\cdot m
These torques are shown in the free-body diagram of the shaft (Fig. 3-35b). Note that the torques applied by the gears are opposite in direction to the torque applied by the motor. (Think of T _{A} as the “load” applied to the shaft by the motor, then the torques T _{B} and T _{C} are the “reactions” of the gears.) The internal torques in the two segments of the shaft are now found (by inspection) from the free-body diagram of Fig. 3-35b:
T _{AB}=796 \ N\cdot m T _{BC}=239 \ N\cdot m
The TMD is shown in Fig. 3-35c.
2. Categorize: Both internal torques act in the same direction; therefore, the angles of twist in segments AB and BC are additive when finding the total angle of twist. (To be specific, both torques are positive according to the sign convention adopted in Section 3.4.)
3. Analyze:
Shear stresses and angles of twist: The shear stress and angle of twist in segment AB of the shaft are found in the usual manner from Eqs. (3-14)
\tau _{max}=\frac{16T}{\pi d^3} (3-14)
and (3-17):
\phi =\frac{TL}{GI_p} (3-17)
\tau _{AB}=\frac{16T_{AB}}{\pi d^3}=\frac{16(796 \ N\cdot m)}{\pi (50 \ mm)^3}=32.4 \ MPa\phi _{AB}=\frac{T_{AB}L_{AB}}{GI_p}=\frac{(796 \ N\cdot m)(1.0 \ m)}{(80 \ GPa)\left\lgroup\frac{\pi}{32} \right\rgroup (50 \ mm)^4}=0.0162 \ rad
The corresponding quantities for segment BC are
\tau _{BC}=\frac{16T_{BC}}{\pi d^3}=\frac{16(239 \ N\cdot m)}{\pi (50 \ mm)^3}=9.7 \ MPa\phi _{BC}=\frac{T_{BC}L_{BC}}{GI_p}=\frac{(239 \ N\cdot m)(1.2 \ m)}{(80 \ GPa)\left\lgroup\frac{\pi}{32} \right\rgroup (50 \ mm)^4}=0.0058 \ rad
4. Finalize: Thus, the maximum shear stress in the shaft occurs in segment AB and is
\tau _{max}=32.4 \ MPa \hookleftarrow
Also, the total angle of twist between the motor at A and the gear at C is
\phi _{AC}=\phi _{AB}+\phi _{BC}=0.0162 \ rad+0.0058 \ rad=0.0220\ rad=1.26^\circ \hookleftarrow
The TDD is shown in Fig. 3-35d. As explained previously, both parts of the shaft twist in the same direction; therefore, the angles of twist are added.