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 tall = 50 MPa, determine the smallest permissible diameter for shaft AB.

Question

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 [latex]t_{all}[/latex] = 50 MPa, determine the smallest permissible diameter for shaft AB.

Accepted Answer

Torques Exerted on Gears. Observing that f = 480 rpm = 8 Hz, we determine the torque exerted on gear E:

[latex]T_{E}=\frac{P}{2 p f}=\frac{30 kW }{2 p (8 Hz )}=597 N \cdot m[/latex]

The corresponding tangential force acting on the gear is

[latex]F_{E}=\frac{T_{E}}{r_{E}}=\frac{597 N \cdot m }{0.16 m }=3.73 kN[/latex]

A similar analysis of gears C and D yields

[latex]T_{C}=\frac{20 kW }{2 p (8 Hz )}=398 N \cdot m[/latex] [latex]F_{C}=6.63 kN[/latex]

[latex]T_{D}=\frac{10 kW }{2 p (8 Hz )}=199 N \cdot m[/latex] [latex]F_{D}=2.49 kN[/latex]

We now replace the forces on the gears by equivalent force-couple systems.

Bending-Moment and Torque Diagrams

Critical Transverse Section. By computing [latex]\sqrt{M_{y}^{2}+M_{z}^{2}+T^{2}}[/latex] at all potentially critical sections, we find that its maximum value occurs just to the right of D:

[latex]\sqrt{M_{y}^{2}+M_{z}^{2}+T^{2}}{ }_{\max }=\sqrt{(1160)^{2}+(373)^{2}+(597)^{2}}=1357 N \cdot m[/latex]

Diameter of Shaft. For [latex]t _{ ext {all }}=50 MPa[/latex], Eq. (7.32) yields

[latex]s_{1}=2 s _{2}[/latex] (7.32)

[latex]\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 imes 10^{-6} m ^{3}[/latex]

For a solid circular shaft of radius c, we have

[latex]\frac{J}{c}=\frac{ p }{2} c^{3}=27.14 imes 10^{-6}[/latex] c = 0.02585 m = 25.85 mm

Diameter = 2c = 51.7 mm

Question 8.S-P.3: The solid shaft AB rotates at 480 rpm and transmits 30 kW fr...

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