10 26 Planetary Gearbox Selection Process 2 selection examples including formula, and associated specification tables & graphs
Planetary Gearbox Selection: Beyond Power & Ratio A gearbox enquiry usually begins with a power figure and a required output speed. Those are useful starting points, which lead to further questions about the application. A suitable selection also depends on how the machine operates, how the gearbox connects to it and the required lifetime of the gearbox. Two concrete pumps can illustrate the difference. Both require a planetary gearbox, yet their drive arrangements, output loads and operating lives lead to substantially different selections. Understanding those differences is where application engineering contributes to the specification. Why planetary gearing? Planetary gearing uses planet gears arranged around a central sun gear, working within a surrounding ring gear. In a typical reduction arrangement, the sun gear provides the input, whilst the planet carrier delivers a lower output speed and higher output torque according to the reduction ratio. Sharing the transmitted load across several planet gears enables high torque capacity within a compact assembly. However, that capability must be assessed against the requirements of the complete installation. Start with the operating conditions The first worked example concerns a hydraulic motor driving a concrete pump through a gearbox and chain drive. The hydraulic motor runs at 150 rpm. With a geometric displacement of 154 cm³, a pressure drop across the motor of 175 bar and the example’ stated efficiency of 92%, the calculated motor output torque is approximately 394.6 Nm. The required gearbox output speed of 30 rpm gives a calculated reduction ratio of 5:1 and an output torque of approximately 1,973 Nm. The specified operating life is 4,000 hours, based on intermittent operation over ten years. The second example uses an electric motor with a variable frequency drive, or VFD. Its input speed is 1,430 rpm, while the required output speed is just 4 rpm. The calculated ratio is therefore 357.5:1, with an output torque of approximately 22,681 Nm. This calculation is based on a transmitted motor power of 9.5 kW, equivalent to approximately 63.4 Nm at the gearbox input. The installation is required to run for 6,000 hours per year over twenty years, giving a required life of 120,000 hours. The description “concrete pump” tells us the general application. It does not establish the gearbox specification.
Consider how the output is connected In the hydraulic example, the gearbox drives a chain sprocket. That connection imposes a radial load on the output shaft as well as transmitting torque. The calculation uses both the sprocket diameter and its position relative to the shaft shoulder. The initial calculated radial load is approximately 31 kN. After applying the example’ shock and bearing-life factors, the required radial-load capacity becomes approximately 99 kN. Selection therefore includes checks on output-bearing capacity, peak radial load and the combined effect of torque and radial loading on the shaft. The electric example uses a shrink-disc connection and a torque reaction arm. The arm restrains the gearbox casing against the reaction torque generated during power transmission, transferring that reaction to the supporting structure. This arrangement requires consideration of the mounting and reaction forces, together with verification that the shrink disc can transmit the required peak torque. How the gearbox connects to the machine is part of selection from the outset. Establish normal and peak torque Normal operating torque is only one part of the assessment. The hydraulic example includes a possible external static overload of 250% of nominal load. The selected gearbox must be checked against that peak torque requirement as well as the calculated operating requirement. In the electric example, the VFD limits the motor peak factor to 1.6. That specified limit is used when checking peak torque against the capacities of both the gearbox and shrink disc. Knowing how a drive starts, operates and responds to overload helps establish the loads its components must withstand. Account for operating life The required life also influences selection. The examples use output speed and operating hours to obtain gear-life factors. Where external radial loading is assessed, the hydraulic example also applies a bearing-life factor. A gearbox intended for occasional operation over 4,000 hours faces a different requirement from one expected to accumulate 120,000 hours. The required torque rating must take that operating life into consideration.
Check thermal capacity The selection checks use the relevant temperature, application and speed factors. Both examples also verify thermal capacity. The electric example transmits 9.5 kW during operation. Its corrected thermal capacity is approximately 10.3 kW, so this check forms a meaningful part of the selection. Mechanical torque capacity and thermal capacity answer different questions. Both must suit the intended duty. Define the application before choosing the gearbox A useful enquiry brings together: » Driver type, power and operating speed. » Required output speed and torque. » Starting conditions and possible overloads. » Operating hours, duty pattern and required life. » Output connection and external shaft loads. » Mounting arrangement, ambient conditions and available space. Drawings and a discussion of the machine’ operation help establish how those details fit together. At jbj Techniques, planetary gearbox selection begins with that understanding. It allows the gearbox, input connection, output arrangement and supporting components to be considered as one installation. That is Engineering Beyond Product Supply: using the application requirements to arrive at a specification that can be explained and checked. The following pages present the two worked selection examples, including the calculations, reference tables and graphs.
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. » The blue highlighted areas link to the relevant graph or table required. 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 Application Factor Table) 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 Gear Life Factor Graph for n x h = 1.2 x 10 ) G 2 5 1.8 Life factor bearings: f = 0.47 (from Bearing Life Factor Graph 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 Temperature Factor Table and f = 2 from t v Speed Factor Table 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
Selection Example 2: Concrete Pump driven by Electric Motor with Variable Frequency Drive. Known data: Power input via electric motor (4 pole - IEC 160 B5). » Installed power 11 kW. » Actually transmitted power 9.5 kW. » Motor peak factor = 1.6 limited by VFD. » Input speed n = 1430 rpm. 1 » Output speed n = 4 rpm. 2 » Power output via shrink disc, torque arm length = 1 m. » Duty cycle: continuous operation for 24 hours, 3 start-ups per day. » Ambient temperature 30°C, horizontal mounting position. » Required life: 6000 hours/year for 20 years = 120000 hours. » The blue highlighted areas link to the relevant graph or table required. 2.1 Ratio: n 1430 1 u = = = 357.5 n 4 2 2.2 Transmitted torque: T = P x 9550 = 9.5 x 9550 = 22681 Nm 2 n2 4 2.3 Application factor: K = 1.2 (from Application Factor Table) A 2.4 Radial load on output shaft: Not applicable due to adequate length of torque reaction arm). 2.5 Shock factor: Not applicable, as above. 5 2.6 Service life: n x h = 4 x 120000 = 480000 = 4.8 x 10 2 5 2.7 Life factor gears: f = 0.59 (from Gear Life Factor Graph for n x h = 4.8 x 10 ) G 2 2.8 Life factor bearings: Not applicable, as 2.4 above. 2.9 Required torque rating: T req. = T x K x 1 = 22681 x 1.2 x 1 = 46131 Nm 2 2 A f 0.59 G 2.10 Adjusted radial shaft load: Not applicable, as 2.4 above. 2.11 Selected gearbox: PS 420 - L4 - P24 - 360 - E16 ratio 360:1; T = 42000 Nm, T = 51000 Nm, T = 58000 Nm nom act peak P = 12 kW, P = 30 kW th max 2.12 Verification: T req = 46131 < T act = 51000 Nm 2 2 Gearbox: T peak = 1.6 x 11 x 9550 = 42020 Nm < 58000 Nm 2 4 Shrink disc: T peak = 42020 Nm < 55000 Nm 2 P = 11 kW < P = 30 kW peak max P = 9.5 kW < P cor = 12 kW x 0.86 x 1 = 10.3 kW th (with f = 0.86 from Temperature Factor Table and t f = 1 from Speed Factor Table). v
3 10 4 10 5 10 6 10 7 10 Example 2 Example 1 n x h 2 fg 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 3 10 4 10 5 10 6 10 7 10 n x h 2 fB 0.6 0.8 1.0 0.4 0.2 Gear Life Factor Graph Bearing Life Factor Graph Selection Example 1 Selection Example 2 Links back to
Speed Factor Table f (--). Is used to adjust the table value of P for different input speeds than the rated value of 1500 rpm. v Speed Factor fv 3000 Input Speed n (rpm) 1 2500 2000 1750 1500 1250 1000 750 500 0.50 0.62 0.75 0.88 1.00 1.25 1.50 1.75 2.00 Horizontal Mounting 0.43 0.56 0.71 0.82 0.95 1.18 1.41 1.68 1.90 Vertical Mounting Uniform Agitators and mixers for uniform density, conveyors, generators, small winches. Moderate shock Crane slewing, travel and hoist, calenders, concrete mixers and pumps, piston pumps. Heavy shock Crushing machines, excavators, foundry equipment, mills, presses, rolling mills, rotary boring tools. Start/stop frequency Driven machine Type of loading Light shock Multi cyl. combustion engine 1.0 - 1.3 1.7 - 2.0 2.0 - 2.5 Moderate shock Single cyl. combustion engine 1.0 - 1.3 2.0 - 2.5 2.5 - 3.0 Type of loading Application Factor Table Uniform Electric and hydraulic motor 1.0 - 1.3 1.2 - 1.7 1.7 - 2.0 Fewer More Fewer More Fewer More Selection Example 1 Selection Example 2 Links back to Temperature Factor Table 100% 75% 67% 50% 33% 25% 20% 10% 1.29 1.48 1.57 1.82 2.24 2.57 2.87 3.00 0°C 1.14 1.32 1.40 1.62 1.99 2.29 2.56 3.00 10°C 1.00 1.15 1.22 1.41 1.74 2.00 2.24 3.00 20°C 0.86 0.99 1.05 1.21 1.49 1.71 1.92 2.71 30°C 0.71 0.82 0.87 1.01 1.24 1.43 1.60 2.26 40°C 0.57 0.66 0.70 0.81 0.99 1.14 1.28 1.81 50°C Ambient Temperature: Temperature Factor ft Percentage of Utilization at Equal Distribution over 1 Hour ft (--). This is used to adjust the table value of P for a choice of ambient temperatures and percentages of usage based on the principle: 1 hour continuous duty = 100% usage
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