In the epicyclic gear train shown in Fig.15.35, the arm A, carrying the compound wheels D and E, turns freely on the output shaft. The input speed is 1000 rpm in counter - clockwise direction when seen from the right. Input power is 7.5 kW. Calculate the holding torque to keep the wheel C fixed

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[latex] Z_{b}=20, z_{d}=60, z_{e}=30, z_{f}=32, z_{c}=80, n_{b}=1000 rpm ccw , P_{i}=7.5 kW [/latex].

[latex] ext { Let } n_{a}= rpm ext { of arm. Considering train B, D, C, we have } [/latex].

[latex] \left(n_{b}-n_{a} ight) /\left(n_{c}-n_{a} ight)=-z_{c} / z_{b}, n_{a}=0 ext { being fixed } [/latex].

[latex] (1000-n) /(0-n)=-80 / 20=-4 [/latex].

[latex] n_{a}=200 rpm [/latex].

Considering train B, D, E, F, we have

[latex] \left(n_{b}-n_{a} ight) /\left(n_{2}-n_{a} ight)=\left(-z_{f} / z_{e} ight) imes\left(z_{d} / z_{b} ight) [/latex].

[latex] (1000-200) /\left(n_{2}-200 ight)=(-32 / 30) imes(60 / 20)=-3.2 [/latex].

[latex] n_{2}=-50 rpm [/latex].

[latex] \omega_{2}=-2 \pi imes 50 / 60=-5.236 rad / s [/latex].

[latex] T_{1}=\left(7.5 imes 10^{3} imes 60 ight) /(2 \pi imes 1000)=71.62 ext { N.m } [/latex].

[latex] T_{1} \omega_{1}+T_{2} \omega_{2}=0 [/latex].

[latex] T_{2}=7.5 imes 10^{3} / 5.236=1432.4 ext { N.m } [/latex].

[latex] ext { Holding torque } T_{3}=-T_{1}-T_{2}=-71.62-1432.4=-1504.02 N ext {.m } [/latex].

Question 15.31: In the epicyclic gear train shown in Fig.15.35, the arm A, c...

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