Question 8.SP.3: The solid shaft AB rotates at 480 rpm and transmits 30 kW fr......
The solid shaft AB rotates at 480 rpm and transmits 30 kW from the motor M to machine tools connected to gears G and H; 20 kW is taken off at gear G and 10 kW at gear H. Knowing that \tau_{\text {all }}=50 MPa , determine the smallest permissible diameter for shaft AB.
STRATEGY: After determining the forces and couples exerted on the shaft, you can obtain its bending-moment and torque diagrams. Using these diagrams to aid in identifying the critical transverse section, you can then determine the required shaft diameter.
MODELING: Draw the free-body diagram of the shaft and gears (Fig. 1).
Observing that f = 480 rpm = 8 Hz, the torque exerted on gear E is
T_E=\frac{P}{2 \pi f}=\frac{30 kW }{2 \pi(8 Hz )}=597 N \cdot m
The corresponding tangential force acting on the gear is
F_E=\frac{T_E}{r_E}=\frac{597 N \cdot m }{0.16 m }=3.73 kN
A similar analysis of gears C and D yields
\begin{array}{ll} T_C=\frac{20 kW }{2 \pi(8 Hz )}=398 N \cdot m & F_C=6.63 kN \\ \\ T_D=\frac{10 kW }{2 \pi(8 Hz )}=199 N \cdot m & F_D=2.49 kN \end{array}
Now replace the forces on the gears by equivalent force-couple systems as shown in Fig. 2.
ANALYSIS:
Bending-Moment and Torque Diagrams (Fig. 3)
Critical Transverse Section. By computing \sqrt{M_y^2+M_z^2+T^2} at all potentially critical sections (Fig. 4), the maximum value occurs just to the right of D:
\left(\sqrt{M_y^2+M_z^2+T^2}\right)_{\max }=\sqrt{(1160)^2+(373)^2+(597)^2}=1357 N \cdot m
Diameter of Shaft. For \tau_{\text {all }}=50 MPa, Eq. (7.32) yields
\sigma_1=2 \sigma_2 (7.32)
\frac{J}{c}=\frac{\left(\sqrt{M_y^2+M_z^2+T^2}\right)_{\max }}{\tau_{\text {all }}}=\frac{1357 N \cdot m }{50 MPa }=27.14 \times 10^{-6} m ^3
For a solid circular shaft of radius c,
\frac{J}{c}=\frac{\pi}{2} c^3=27.14 \times 10^{-6} \quad c=0.02585 m =25.85 mm
Diameter = 2c = 51.7 mm
