Solved Problems:1. Two gears in mesh have 100 and 25 teeth respectively. The module of the gears is 4mm. Find out the distance between the centers of the two gears. (Set 1-Nov:2012 Regular)

Sol: T1 = 100, T2 = 25, m = 4mm, We know that module equal to, m = d/T i.e. m = d1/T1 = d1/100 = 4 d1 = 400mm m = d2/T2 = d2/25 = 4 d2 = 100mm Distance between the centers of the two gears (C) C = d1 + d2/2 = 400 + 100/2 = 250mm.2. A wheel has 50 teeth and circular pitch 20 mm. Find (a)pitch circle diameter (b) module.Sol:T = 50,Pc = 20 mm

(a) Pitch circle diameter (PCD):

Therefore , Pc = d/T d = PcX T/ = 20 X 50/ = 318.30mm(b) Module (m) = d/T = 318.30/50 (m) = 6.36 mm /tooth. 3. A toothed wheel module 5mm and 50 teeth rotates at 100 rpm. Find the peripheral speed of gear wheel.Sol: m = 5mm T = 50 N = 100 rpm Peripheral speed of the wheel V = dN/60 m/sec Module (m) = d/T = 5 = d/50 d = 250mm = 0.25meters. Therefore, V= X.25X100/0 = 1.308 m/sec.4. A gear of 45 teeth pitch circle diameter of 350 mm. What is its module, circular pitch, dedendum and addendum.

Sol: T = 45,d = 350mm, m = d/T = 350/45 = 7.7 mmDedendum = 1.25 X module = 1.25 X 7.7 = 9.62 mmAddendum = 1Xmodule = 1X7.7 = 7.7 mm5. Two axes of two parallel shafts are approximately 2 m apart. The motion is transmitted from one shaft to another by spur gears, whose module is 20mm. One shaft is to rotate 4.5 times as fast as the other. Find the number of teeth on each and exact centre distance.Sol:

Fig: (1)Centre distance (C) = 2 m = 2000mmN2/N1 = 4.5 = N2 = N1 X 4.5N2/N1 = d1/d2 = T1/T2 = 4.5 (1) d1 = 4.5 d2 C = d1 + d2/2 = 2000,Therefore, d1 + d2 = 4000 (2) From the equations (1) & (2)4.5 d2 + d2 = 4000 d2 =727.27mm, d1 = 3272.72mm, m= d1/T1 = T1 = 3272.72/20 T1 = 163.6m= d2/T2 = T2 = 727.27/20T2 = 36.36