Question

1. A two-stage compound gear is driven by a 20-hp motor that rotates the shaft 'a' in CCW as shown in the figure at a speed of 100 rpm. Entire gear set has a diametral pitch of 4 teeth/in and a pressure angle of 20°. The tooth numbers are: N2 = 48, N3 = 72, N4 = 48, and Ns = 120. 5 С a 3 a. Draw the free body diagram of the compound gear 3 and 4 mounted on idler shaft b. Calculate all the force components in the free body diagram you drew. Use the template given below. 3 4 b. Calculate the RPM of gear 5 c. Calculate the pitchline velocity of gear 4.

Answer 1

teeth (t) & rpm (N) 1/3 power = 20 hp a 14.914 kW 2 Ft V N2 Irodlute profile 2 (drining gear) power = 14.914 = Torque x angular relocity speed of a3 24 will be same as compounded and Nq & CCW SON z 2 Nu will be counter clock wise. 14.914 kW = FAXR3) X 27X Hold 60 of oft 48 p= t = 4 4 - ty = 4 DU Dq = 12eich R₂ =binch dring gear 1 R₂ = 15:24 cm 100 60 14.914x10 = FtX 15 24 X 24X100 for gear or gear 2] If = 9.345kN FR = 34 01:31 N[ Radical force) F = Ft - 9944.76 N cos2o net force on gear

Now assume there is no pouver loss. 2 FAX Ru 14.914x103 60 * 2* (NU) [for gear t { law of gear N3=tz 100x 48 = 66.6666pm N3 = N2X 48 72 72 Ny = N4 = 66:666.rpm (clockwire) » D = 12 Rabin= 15.24cm p=4 teeth e t=4=48 Inch D D X 2AX 66.666 14.914x 103 = FtX 15.24x102 60 F = 14.017 kN FR 5.lola Normal = 14.9 16 KN ray of goor gear law 6 rpm of gear 5 [apply gear N = N4 = 66:6662km (Cw) tz » N- A 666 666 - 72 48 0 44. 4uy som m = Ny Nu ny Ny=th NG = 44 444 X 48 Ng -17. 77 76 mus {ccw) 10

③ pitch point hele relocity of gear 4 V= WR = 27X 66 666 x 15.24x10² 6o 60 Tup 1:063 m/s NOTE In enternal meshing of tooth direction will change. EN FNR FR I EN DF oft ft FB:1 of 3 F.Brd of 4