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On a hydraulic guillotine shear, the pump motor is selected not for the peak power at the hardest moment of the cut, but for the equivalent (thermal) load over the whole cutting cycle; the peak power is checked separately to confirm that it stays below the motor's breakdown torque. Peak power results from the flow the pump delivers at maximum pressure; the equivalent load depends on the number of cuts per minute, the duration of each phase and how long the motor runs idle. This article works through the guillotine shear's cycle step by step to show how the motor power is verified and what to watch when replacing the motor.
Hydraulic and mechanical guillotines: the motor's role differs
Guillotine shears have two drive designs, and motor selection differs accordingly:
- Mechanical (flywheel) guillotine: The motor turns a heavy flywheel via a belt; at the moment of cutting, energy is transferred from the flywheel to the blade through a clutch. Between cuts the motor brings the flywheel back up to speed. In this design, flywheel inertia and the starting condition are decisive.
- Hydraulic guillotine: The motor turns a hydraulic pump; the pump feeds the cylinders, first pressing the hold-downs onto the sheet and then lowering the blade. The motor load changes within seconds together with pump pressure.
This article covers the hydraulic guillotine. The blade beam, hold-downs, safety valves and hand guards are the machine manufacturer's responsibility; motor selection does not replace or change these arrangements.
What does the motor go through in a cutting cycle?
A typical hydraulic guillotine cycle has four phases. In each phase the pump pressure, and therefore the motor load, is different:
| Phase | Hydraulic condition | Motor load |
|---|---|---|
| Idle waiting | Pump discharges to tank, pressure is low | Low; only friction and circulation losses |
| Hold-downs descend and blade approaches | High flow, medium pressure | Medium |
| Cutting | Pressure rises quickly to its maximum | Highest; short duration |
| Blade return | High flow to the return cylinder, low to medium pressure | Medium |
Cutting pressure depends on sheet thickness, material strength, cutting length and blade rake angle. On a machine cutting thin sheet, the pressure may never reach its maximum. So how often you cut the thickest sheet in the machine's capacity table directly affects the motor load.
Peak power and equivalent power: two separate checks
Peak power is the power the pump draws at maximum pressure and at the flow at that moment. An approximate calculation divides the product of pressure and flow by the pump efficiency; the value is taken from the pump manufacturer's table or the machine design. An induction motor can exceed its rated power for short loads, but this peak load must not approach the motor's breakdown torque without a safe margin. The motor's breakdown torque ratio is obtained from the motor manufacturer.
Equivalent power is the value that determines the motor's heating. A widely used method for intermittent loads is the root mean square: the square of each phase's power is multiplied by the duration of that phase, divided by the total cycle time and the square root is taken. The motor's rated power should not fall below this equivalent value. The method is an approximation; if there are long high-pressure phases or very frequent cuts, it is assessed together with the motor manufacturer.
The two checks complement each other. A motor selected only for peak power is larger than necessary and runs at low load in operation; a motor selected only for the average loses speed during the cut and slows down, in the extreme case stalls.
Hypothetical calculation example: a medium-sized hydraulic guillotine
The values below are hypothetical figures chosen only to illustrate the method; they are not data from a real machine. Let each cutting cycle of a hydraulic guillotine last 6 seconds:
- Cutting phase: 2 seconds, pump power 14 kW
- Approach and return: 2 seconds in total, 6 kW
- Idle waiting: 2 seconds, 2 kW
Root mean square: (14² × 2 + 6² × 2 + 2² × 2) / 6 ≈ 78.7; its square root is about 8.9 kW. Since the equivalent load in this cycle is about 8.9 kW, the next standard rating, an 11 kW motor, is a candidate. The peak load of 14 kW is about 1.3 times the motor's rated power; whether this ratio stays safely below the motor's breakdown torque is confirmed with the motor data. If the operator waits a long time between cuts, the equivalent load falls further; if the machine cuts continuously in series, the idle phase becomes shorter and the equivalent load rises.
