Planetary Gearboxes available from jbj Techniques Limited

Planetary Gearboxes Gear Selection Selection Example 1: Concrete Pump Driven by Hydraulic Motor Known data: Power input via hydraulic motor (radial piston crankshaft design). » Pressure drop 175 bar » Motor displacement 154 cm³. » Total motor efficiency 0.92 (92%). » Input speed n = 150 rpm. 1 » Output speed n = 30 rpm. 2 » Power output via chain drive. » Sprocket eff. diameter = 127 mm » Sprocket mounted at 40 mm from shaft shoulder. » Duty cycle: continuous operation for 30 minutes up to 4 times a day. » Ambient temperature 40°C, horizontal mounting position. » Required life: 400 hours/year for 10 years = 4000 hours » Possible static (external) overload: 250% of nominal load. » OK OK OK OK 39 www.jbj.co.uk/planetary-gearboxes.html #DriveLineHarmony OK OK OK 1.1 Ratio n 150 1 u = = = 5 n 30 2 1.2 Transmitted torque: T = 175 bar x 154 cm³ x 0.92 / 20p = 394.6 Nm 1 T = T x u = 394.6 Nm x 5 = 1973 Nm 2 1 1.3 Application factor: K = 1.5 (from table 1) A 1.4 Radial load on output shaft: F = 1973 Nm x 2000 / 127 mm = 31071 N = 31 kN R 1.5 Shock factor: K = 1.5 (estimated, same as K ) S A 5 1.6 Service life: n x h = 30 x 4000 = 120000 = 1.2 x 10 2 5 1.7 Life factor gears: f = 0.71 (from graph 1, page 41, for n x h = 1.2 x 10 ) G 2 5 1.8 Life factor bearings: f = 0.47 (from graph 2, page 41, for n x h = 1.2 x 10 ) B 2 1.9 Required torque rating: T req. = T x K x 1 = 1973 x 1.5 x 1 = 4168 Nm 2 2 A f 0.71 G 1.10 Adjusted radial shaft load: F req. = F x K x 1 = 31 x 1.5 x 1 = 99 kN R R S (at 40 mm from shaft shoulder) f 0.47 B 1.11 Selected gearbox: PS 42 - L1 - X16 - 5.1 - G2 ratio 5.1:1; T = 4200 Nm, T = 4300 Nm, T = 5500 Nm nom act peak F adm = 108 kN (from bearing chart at 40 mm load distance) R P = 7.5 kW, P = 26 kW th max 1.12 Verification: T req = 4168 < T act = 4300 Nm 2 2 T peak = 2.5 x 1973 Nm = 4933 Nm < 5500 Nm 2 P = T x n = 394.6 x 150 = 6.2 kW < P = 26 Kw 1 1 max 9550 9550 P < P cor = 7.5 kW x 1.01 x 2 = 15.15 kW th (with f = 1.01 from table 3, page 40, f = 2 from table 4, page 40). t v F req = 99kN < F adm = 108 kN R R F peak = 2.5 x 31 kN = 78 kN < F adm = 108 kN R R Shaft fatigue: 31 kN x 1973 Nm = 0.13 < 0.5 108 kN 4200 Nm

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