Question 8.S-P.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 t_{all} = 50 MPa, determine the smallest permissible diameter for shaft AB.
Torques Exerted on Gears. Observing that f = 480 rpm = 8 Hz, we determine the torque exerted on gear E:
T_{E}=\frac{P}{2 p f}=\frac{30 kW }{2 p (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
T_{C}=\frac{20 kW }{2 p (8 Hz )}=398 N \cdot m F_{C}=6.63 kN
T_{D}=\frac{10 kW }{2 p (8 Hz )}=199 N \cdot m F_{D}=2.49 kN
We now replace the forces on the gears by equivalent force-couple systems.
Bending-Moment and Torque Diagrams
Critical Transverse Section. By computing \sqrt{M_{y}^{2}+M_{z}^{2}+T^{2}} at all potentially critical sections, we find that its maximum value occurs just to the right of D:
\sqrt{M_{y}^{2}+M_{z}^{2}+T^{2}}{ }_{\max }=\sqrt{(1160)^{2}+(373)^{2}+(597)^{2}}=1357 N \cdot m
Diameter of Shaft. For t _{\text {all }}=50 MPa, Eq. (7.32) yields
s_{1}=2 s _{2} (7.32)
\frac{J}{c}=\frac{\sqrt{M_{y}^{2}+M_{z}^{2}+T^{2}}_{\max }}{ t _{ all }}=\frac{1357 N \cdot m }{50 MPa }=27.14 \times 10^{-6} m ^{3}
For a solid circular shaft of radius c, we have
\frac{J}{c}=\frac{ p }{2} c^{3}=27.14 \times 10^{-6} c = 0.02585 m = 25.85 mm
Diameter = 2c = 51.7 mm
