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Question 15.25: An epicyclic gear train shown in Fig.15.29, consists of two ...

An epicyclic gear train shown in Fig.15.29, consists of two sun wheels A and D with 28 and 24 teeth respectively, engaged with a compound planet wheels B and C with 22 and 26 teeth. The sun wheel D is keyed to the driven shaft and the sun wheel A is a fixed wheel coaxial with the driven shaft. The planet wheels are carried on an arm E from the driving shaft which is coaxial with the driven shaft.
Find the velocity ratio of gear train. If 1.2 kW is transmitted and input speed is 120 rpm, determine the torque required to hold the sun wheel A.

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\text { Given: } \quad z_{a}=28, z_{d}=24, z_{b}=22, z_{c}=26 \text {, power transmitted }=1.2 kW , N=120 rpm .

Table 15.23 is used to find the speed of gears.

y=+120 rpm .

n_{d}=y+\frac{91 x}{66}=120+\frac{91 x}{66}=0 .

x = – 87 rpm.

n_{a}=x+y=-87=120=33 rpm .

Table 15.23
Revolutions of Operation
Gear D, 24 Gear B, C, 22, 26 Gear A, 28 Arm, E

+\frac{z_{a}}{z_{b}} \times \frac{z_{c}}{z_{d}}

=\frac{28}{22} \times \frac{26}{24}=\frac{91 x}{66}

 -\frac{z_{a}}{z_{b}}=\frac{-28}{22}=\frac{-14}{11} +1 0 1. Arm E fixed, + 1
revolutions given
to gear A, ccw
\frac{91 x}{66} \frac{-14 x}{11} +x 0 2. Multiply by x
y+\frac{91 x}{66} y-\frac{14 x}{11} y+x y 3. Add y

Speed of driven shaft (gear A) = 33 rpm ccw
Speed ratio          =\frac{n_{d}}{n_{a}}=\frac{87}{33}=2.637 .

M_{1}=\frac{1.2 \times 10^{3} \times 60}{(2 \pi \times 120)}=+95.5 Nm .

\omega_{1}=\frac{2 \pi \times 120}{60}=12.57 rad / s .

\omega_{2}=\frac{2 \pi \times 33}{60}=3.456 rad / s .

M_{2}=M_{1} \times \frac{\omega_{1}}{\omega_{2}}=95.5 \times \frac{12.7}{3.456}=347.35 Nm .

\text { Holding torque, } M_{3}=-M_{1}-M_{2}=-95.5-347.35=-442.85 Nm .

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