Fluid power cylinders

Approximately 85% of fluid power circuits incorporate some form of cylinder (or linear actuator). The cylinder converts pneumatic or hydraulic pressure into thrust to perform useful work. Both air and hydraulic cylinders come in ram, telescoping, single-acting/spring-return, double-acting, double rod end, rodless, and tandem types. Figure 6-1 shows the symbols for several of these types. Each machine has specific requirements that challenge the designer to determine which type of actuator to use.








Throughout this manual, many circuits show cylinders in a variety of applications. An explanation accompanies each example – noting the pumps, valves, and peripheral hardware used to do the work. Every design description also attempts to cover the limitations of a particular circuit and show other ways to perform the same task. This section covers several types of cylinder applications that do not fall under a particular heading.

Normally air cylinder circuits are less expensive than hydraulic circuits because there is no need for a power unit. An air compressor usually is part of the plant facility and compressed air is a commodity similar to electrical power. However, the cost of operating an air-powered machine may be four to seven times more than a hydraulically operated one.

Another disadvantage of air is the fluid’s compressibility. Hydraulic circuits are very rigid, while air circuits are quite spongy. This lack of control makes it almost impossible to accurately stop and hold an air cylinder in mid-stroke with standard air valves alone. After an air cylinder stops, it may start creeping or be forced out of position almost immediately.

When it comes to brute force, air cylinders fall far behind hydraulic cylinders because they normally operate only at 80 to 100 psi. Getting high force from low pressure requires large areas . . . with attendant large valves and piping. A general rule might be to look at hydraulics when an operation requires a 5-in. bore or larger air cylinder to develop the required force. However, another factor is how often the cylinder must cycle. Air circuits with very low cycle rates and long holding times could be more economical than hydraulics, but the faster the cycle time, the more it costs to operate an air cylinder. Another consideration is the operating environment. Around food or medicine, potential contamination from hydraulic oil could be a serious problem. Look at each application to see which fluid system best suits it.

Sizing hydraulic cylinders

Chart 6-1 provides an exercise in sizing a simple, single-cylinder hydraulic circuit with straightforward parameters. The example covers the basic requirements for sizing a hydraulic cylinder to power a specific machine.























Of course, in the real world of circuit design, experience, knowing the process, the environment, the skill of the user, how long will the machine be in service, and other items will affect cylinder and power unit choices.

Before designing any cylinder circuit it is necessary to know several things. The first is the required force. Usually, the force to do the work is figured with a formula. In instances where there is no known mathematical way to calculate force, use a mock-up part on a shop press or on a prototype machine to estimate the force requirements. If all else fails, an educated guess may suffice. (The sample problem in the chart requires a force of 50,000 Llb.)

The second requirement is the total cylinder stroke. Stroke length is part of machine function, but it is needed to figure pump size. Use a stroke of 42 in. in this problem.

Third, how much of the stroke requires full tonnage? If only a small portion of the stroke needs full force, a hi-lo pump circuit and/or a regeneration circuit could reduce first cost and operating cost. This cylinder requires full tonnage for the complete 42-in. stroke.

Fourth, what is the total cylinder cycle time? Make sure the time used is only for cycling the cylinder. While load, unload, and dwell are part of the overall cycle time, they should not be included in the cylinder cycle time when figuring pump flow. Use a cylinder cycle time of 10 seconds for this problem.

Finally, choose maximum system pressure. This is often a matter of preference of the circuit designer. Standard hydraulic components operate at 3000 psi maximum, so choose a system pressure at or below this pressure. If the company that will operate the machine has operating and maximum pressure specifications, adhere to them. Remember that lower working pressures require larger pumps and valves at high flow to get the desired speed.

In the example in Chart 6-1, the square root of the maximum thrust, divided by the maximum system pressure, divided by 0.7854 gives a minimum cylinder bore of 5.641 in. Obviously, a standard 6-in. bore cylinder should suit this system.

To figure pump capacity, take the cylinder piston area in square inches, times the cylinder stroke in inches, times 60 seconds, divided by the cycle time in seconds, times 231 cubic inches per gallon. This indicates a minimum pump flow of 61.7 gpm. A 65-gpm pump is the closest standard flow available. Never undersize the pump because this formula figures the cylinder is going at maximum speed the whole stroke. In the real world, the cylinder must accelerate and decelerate for smooth operation, so travel speed after acceleration and before deceleration should actually be higher than this formula indicates.

Figure horsepower by multiplying flow in gpm by pressure in psi by a constant of 0.000583. This comes out to 75.79 hp . . . and is close to a standard 75-hp motor. This should provide sufficient horsepower because the system pressure does not have to go to 2000 psi with the cylinder size used.

The tank size should be at least two to three times pump flow. For the example, 3 X 63 equals 195 gallons. A 200-gal tank should be satisfactory. When using single-acting cylinders or unusually large piston rods, size the tank for enough oil to satisfy cylinder volume without starving the pump.