Question 3.7: A motor driving a solid circular steel shaft transmits 30 kW......
A motor driving a solid circular steel shaft transmits 30 kW to a gear at B (Fig. 3-34).
The allowable shear stress in the steel is 42 MPa.
(a) What is the required diameter d of the shaft if it is operated at 500 rpm?
(b) What is the required diameter d if it is operated at 3000 rpm?
Use a four-step problem-solving approach. Combine steps as needed for an efficient solution.
Part (a): Motor operating at 500 rpm.
1, 2. Conceptualize, Categorize: Knowing the horsepower and the speed of rotation, the torque T acting on the shaft is found from Eq. (3-43).
H=\frac{2\pi nT}{60(550)}=\frac{2\pi nT}{33,000} (n = rpm, T = 1b-ft, H = hp) (3-43)
Solve that equation for T to get
T=\frac{60 \ P}{2\pi n} =\frac{60(30 \ kW)}{2\pi (500 \ rpm)}=573 \ N\cdot mThis torque is transmitted by the shaft from the motor to the gear.
The maximum shear stress in the shaft can be obtained from the modified torsion formula [Eq. (3-14)]:
3. Analyze: Solve that equation for the diameter d, and also substitute \tau _{allow} for \tau _{max} to get
d^{3}=\frac{16T}{\pi\tau _{allow}}=\frac{16(573 \ N\cdot m)}{\pi (42 \ MPa)}=6.948\times 10^{-5} m^3from which
d = 41.1 mm \hookleftarrow
4. Finalize: The diameter of the shaft must be at least this large if the allowable shear stress is not to be exceeded.
Part (b): Motor operating at 3000 rpm.
1, 2. Conceptualize, Categorize: Follow the same procedure as in part (a).
3. Analyze: Torque T and diameter d are now
d^{3}=\frac{16T}{\pi\tau _{allow}}=\frac{16(95.49 \ N\cdot m)}{\pi (42 \ MPa)}=1.158\times 10^{-5} m^3
d = 22.6 mm \hookleftarrow
which is less than the diameter found in part (a).
4. Finalize: This example illustrates that the higher the speed of rotation, the smaller the required size of the shaft (for the same power and the same allowable stress).