Question

5 kW of power will be transmitted via a chain-sprocket system. The rotational speed of input shaft is 20 rpm. The required approximate rotational speed of the input shaft is 20 rpm. Required approximate rotational speed for the out shaft is 7 rpm. The approximate distance between center lines of input and output shafts should be 900 mm. This chain-sprocket system is used for an elevator and it is asked to use a single or multiple DIN 8187 chain with a 31.75 mm pitch.

a) Calculate the number of teeth on sprocket, which is on the output shaft, is the number of teeth on the sprocket on the input shaft is 19.

b) Calculate input power for the selection of chain and select the appropriate chain type.

c) Calculate pitch circle diameters of driving and driven sprockets, the linear velocity of the chain, length of chain, (number of chain links), and exact centerline distance between the driving and driven sprocket.

d) Calculate the factor of safety of the selected chain

Answer 1

Sel. Given P= 31.75mm 2,=19 n. - 20 rpm = 900 mm and 2-7 pm -> k2= lll (Power ) from mitted = 2.2kw 510 a. No. of teeth on sproket (22): - Z2 x02 = n, xZi Z₂x7 = 20x19 Z2 = 54.28 b. input power & type of chain :- (KW)rated = (kw)transmitted oks kixkz under given condition from table kg = 1.4 k = 1 .. (w)rated 02x1.4 2274-hew 6-3063 low TX1.11 from table at 20rpm under 2.774 kw suitable chain is 16B. c. Pitch circle diameter /in (180%) = 31.75 / sin (18/19) 193 mm D2 = P/in (18942) = 31-45/srin (1865428) = 548.87 mm Mo. of link tn = 2x ( 9 ) + 21+22 + ()a) = 2* ( 9 ) + 20+S4 28 13175, (14 320) *(70%) = 94.88 ~ 95 Total length of chain L = L x P = 95x31.75 = 30 16.25 mm Corrected centre distance 20+54-28 Ln-(2,422) = 95- 4:5627 +1-7227-8-1 31.75 $57.86+ / 57.862 – 8x (spes-20jey - 901.89 mm linear relocity . factor of safety. For 166 Chain Porculing load = 42300 N MP2/60x103 = 20x19x31.75/ (oxid) = 0.209 m/sec pe low x1100 4x100 10.201 : 3807 81343 Fos 3801 de 35 31343 2 + 2 =57.86 2 x 42300