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Unformatted text preview: ugh + 1 revolution (i.e. 1
revolution anticlockwise) 0 2. Gear B fixedGear C rotated
through + x revolutions 0 +x + x× 3. Add + y r evolutions to all
elements +y +y Total motion +y x+y y +x× – TC TE
×
= –1
TE TD +y 4. Gear D 1. +1 (3TC = TD )
TC
TE –x
+y TC
TE y –x Since the speed of gear B is 200 r.p.m., therefore from the fourth row of the table,
y = 200 ...(i) Also, the speed of road wheel attached to axle Q or the speed of gear D is 210 r.p.m., therefore from the fourth row of the table,
y – x = 210 or x = y – 210 = 200 – 210 = – 10 ∴ Speed of road wheel attached to axle P
= Speed of gear C = x + y
= – 10 + 200 = 190 r.p.m. Ans. 13.11. Torques in Epicyclic Gear Trains Fig. 13.25. Torques in epicyclic gear trains. When the rotating parts of an epicyclic gear train, as shown in Fig. 13.25, have no angular
acceleration, the gear train is kept in equilibrium by the three externally applied torques, v iz.
1. Input torque on the driving member (T 1),
2. Output torque or resisting or load torque on the driven member (T 2),
3. Holding or braking or fixing torque on the fixed member (T 3). Chapter 13 : Gear Trains l 463 The net torque applied to the gear train must be zero. In other words,
T 1 + T2 + T 3 = 0
∴ ...(i) F1.r1 + F2.r2 + F3.r3 = 0 ...(ii) where F1, F2 and F3 are the corresponding externally applied forces at radii r1, r2 and r3.
Further, if ω1, ω2 and ω3 are the angular speeds of the driving, driven and fixed members
respectively, and the friction be neglected, then the net kinetic energy dissipated by the gear train
must be zero, i.e.
T 1.ω1 + T 2.ω2 + T 3.ω3 = 0 ...(iii) But, for a fixed member, ω3 = 0
∴ T 1.ω1 + T 2.ω2 = 0 ...(iv) Notes : 1. From equations (i) and (iv), the holding or braking torque T 3 may be obtained as follows : ω1
ω2 ...[From equation (iv)] T 3 = – (T 1+ T 2 ) ...[From equation (i)] T2 = – T1 × and ω N = T1 1 – 1 = T1 1 – 1
ω2
N2 2. When input shaft (or driving shaft) and output shaft (or driven shaft) rotate in the s...
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 Fall '08
 CELGHORN

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