FORCE ANALYSIS OF GEARS Week-2 FORCES ON SPUR GEARS 3 T12 = r2 Ft32 O3 O3 3 B Ft23 r2 A O2 2 F13 Fr32 F23 Ft32 = P/ r2 2 Ft 2 12 F r t F 32 = F 32 tan 12 F32 = Ft32 / cos P = T12 2 = r2 Ft32 2 T13 Fr23 O2 Fr12 Ft32 T12 F32 PROBLEM y 2 • 1 2 18 t 3 4 20 t 36t F12 P= T122 20,000= T12 (900)(2/60 ) T12 = 212.2 N.m Fr12 2 O2 Ft12 Pressure Angle: = 20 ° Metric modul: m= 4 mm Speed : n = 900 rpm x Power: P = 20 kW z: Number of teeth T12 d = mz d2 = 72 mm d3 = 144 mm d4 = 80 mm MO2=0 T12 - Ft32 r2=0 Ft32 = T12 / r2 2 Ft32 = 212.2 / (0.072/2) Ft32 =5894N=5.894 kN Ft32 Fr32= Ft32 tan20 = 5.894 tan20 Fr32 = 2.145 kN F32= Ft32 / cos20 = 5.894 / cos20 F32= 6,27 kN Fr32 F32 F=0 F12= - F32 F23 Fr23 MO3=0 - Ft23 r3+ Ft43 r3=0 F13 Ft23 Ft23= Ft43 Fr23= Fr43 F23= F43 3 Fr43 F=0 F + F + F =0 23 43 13 3 O3 F34 Fr34 Ft43 Ft34 4 O4 4 T14 (5.894i -2.145j)+(-2.145i+5.894j)+ F13 =0 F43 F13 =-3.75i -3.75j kN = 5.3 /-135 kN F=0 F14 = -F34 MO4=0 T14 = Ft34 r4 =5.984 (0.04) T14= 235.76 N.m T14 > T12 (DISCUSS) F14 Otherwise, from kinematics 2/4= d4/d2 4= (900)(2/60 )(72/80)=84.82 r/s T14 = P /4 T14 = 20,000 /84.82 T14 = 235.79 N.m (DISCUSS) Helical Gears Ft=T/r is helix angle Straight Bevel Gears Ft=T/r
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