Question #208127

An epicyclic gear consists of bevel wheels as shown in Fig. 13.49. The driving pinion A has 20 teeth

and meshes with the wheel B which has 25 teeth. The wheels B and C are fixed together and turn freely

on the shaft F. The shaft F can rotate freely about the main axis X X. The wheel C has 50 teeth and

meshes with wheels D and E, each of which has 60 teeth. Find the speed and direction of E when A

rotates at 200 r.p.m., if

1. D is fixed, and 2. D rotates at 100 r.p.m., in the same direction as A.

In both the cases, find the ratio of the torques transmitted by the shafts of the wheels A and E, the

friction being neglected


1
Expert's answer
2021-06-18T07:54:33-0400


The speed of A

x+y=200    y=200xx+y=200 \implies y=200-x

Part a

D is fixed

y+xTATCTBTD=0    200x+x20502560=0y+x \frac{T_A*T_C}{T_B*T_D}=0 \implies 200-x+x \frac{20*50}{25*60}=0

x=600;y=200600=400x=600 ; y= 200-600= -400

Speed of E, yxTATCTBTE=0    40060020502560=800rpmy-x \frac{T_A*T_C}{T_B*T_E}=0 \implies -400-600 \frac{20*50}{25*60}=-800 rpm

E rotates at 800 rpm in the direction opposite to A

Part b

D rotates at 100 rpm in the same direction of A

y+xTATCTBTD=100    x+23x=100rpm    x=300y+x \frac{T_A*T_C}{T_B*T_D}=-100 \implies -x+ \frac{2}{3}x=-100 rpm \implies x= 300

y=200x=100y=200-x=-100

Speed of E

yxTATCTBTE=10030023=300y-x \frac{T_A*T_C}{T_B*T_E}=-100 -300* \frac{2}{3}= 300

E rotates at 300 rpm in the direction to A

T1ωA=T0ωE    T1T0=ωEωAT_1 \omega _A=T_0\omega _E \implies \frac{T_1}{T_0}=\frac{\omega _E}{\omega _A}

Case A

T1T0=800200=4\frac{T_1}{T_0}=\frac{800}{200}=4

The ratio of torque transmitted by the shaft of wheel A and E is 4


Case B

T1T0=300200=1.5\frac{T_1}{T_0}=\frac{300}{200}=1.5

The ratio of torque transmitted by the shaft of wheel A and E is 1.5


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