Question

Shaft Rotates @ 500 rpm f = 48f 500 rpm = \frac{500}{60} Hz = 8\frac{1}{3} Hz. 10kw 20kw A 400 400 400 rc = 70 mm = 0.07 m \Gamma_D = 90 mm = 0.09 m \Gamma_E = 200 mm = 0.2 m \tau_{all} = 50 MPa \phi_{AB} min ? So: T_E = \frac{30}{2\pi(8\frac{1}{3})} = 573.2 N.m Corresponding Forces: F_E = \frac{573.2}{0.2} = 2866 N Gears C \frac{1}{3} D : T_C = \frac{20 kw}{2\pi(8\frac{1}{3})} = 381.99 N.m F_C = \frac{381.99}{0.07} = 5456.96 N

          Shaft Rotates @ 500 rpm
f = 48f 500 rpm = \frac{500}{60} Hz
= 8\frac{1}{3} Hz.
10kw
20kw
A
400 400 400
rc = 70 mm = 0.07 m
\Gamma_D = 90 mm = 0.09 m
\Gamma_E = 200 mm = 0.2 m
\tau_{all} = 50 MPa
\phi_{AB} min ?
So: T_E = \frac{30}{2\pi(8\frac{1}{3})} = 573.2 N.m
Corresponding Forces:
F_E = \frac{573.2}{0.2} = 2866 N
Gears C \frac{1}{3} D :
T_C = \frac{20 kw}{2\pi(8\frac{1}{3})} = 381.99 N.m
F_C = \frac{381.99}{0.07} = 5456.96 N

Shaft Rotates @ 500 rpm
f = 48f 500 rpm = (500)/(60) Hz
= 8(1)/(3) Hz.
10kw
20kw
A
400 400 400
rc = 70 mm = 0.07 m
= 90 mm = 0.09 m
= 200 mm = 0.2 m
τall = 50 MPa
ϕAB min ?
So: TE = (30)/(2π(8(1)/(3))) = 573.2 N.m
Corresponding Forces:
FE = (573.2)/(0.2) = 2866 N
Gears C (1)/(3) D :
TC = (20 kw)/(2π(8(1)/(3))) = 381.99 N.m
FC = (381.99)/(0.07) = 5456.96 N

Added by Jack G.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Supreeta N

11/24/2023

Step 1

14159). T = 20,000 / (2 * 3.14159 * 500) = 63.66 N.m Show more…

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Thanks for your feedback!

Please explain how you found the shear force and draw a diagram. Rotates 500 rpm. Given: D = 48 mm 500 rpm H = 2 m B = 30.1 m rc = 70 mm c = 0.07 m 20 kW = 90 mm c = 0.09 m D = 200 mm O = 2 m 400 rpm 500 rpm AB Solution: Torque: T = 573.3 N.m Corresponding Force: 573.2 N Gears: 20 Nm Tc WN bbl88 %0.07 = 545696