337

Section 10Design Data

TopicFluid Power Troubleshooting Fluid power graphic symbols Hydraulic troubleshooting Cylinder and valve testing Replacing a pump or motor Power steering calculationsFluid Power Design Data Formulas - Fluid power formulas - Fluid power formulas, metric - English/metric conversions - Fluid power equivalents and abbreviations - Vehicle drive calculations Cylinders - Operating principles - Cylinder types - Hydraulic cylinder force and speed calculations - Hydraulic cylinder force and speed tables - Pneumatic cylinder air consumption - Pneumatic cylinder force table - Internal fluid PSI on tubing - Piston rod column strength - Calculating tie rod stretch - Designing with cylinders - Square head cylinder sizes - Regenerative circuit design - Regenerative circuit examples - Synchronizing several cylinders - Cylinder speed control - Counterbalance valve technical information Pumps and motors - Horsepower to drive a pump - Hydraulic pump unloading - High-low hydraulic circuits

Page

340342345346347

348349350351352

353354

355356360361362363364365369371372373374375

376378380

(continued on next page)

338

Topic Pumps and motors (cont.) - Torque/HP/RPM table - Pump and motor torque table - Mechanical transmission efficiency - Hydraulic motor circuits - Hydraulic motor speed control - Pump/motor shafts and flanges Valves - Directional valve patterns - Cv factors for hydraulic valves - Cv factors for pneumatic valves - Calculating required flow for operating air and hydraulic cylinders HP to compress air Vacuum - Vacuum applications - Tank pump-down time - Hydraulic oil filtering Heat exchangers - Cooling in hydraulic systems - Adding cooling capacity Closed loop hydrostatic transmissions Air over oil applications Accumulators - Accumulator sizing - Accumulator circuits Oil - Viscosity rating systems - SSU viscosity variation - Seal compatibility with common fluids Plumbing - Hydraulic pipe table - Oil flow capacity of pipes - Oil pressure loss through pipes - Carbon steel tubing data - Copper tubing data - Stainless steel tubing data - Oil flow capacity of tubing - Air line pipe size - Air pressure loss - Air flow through orifices and vacuum flow - Oil flow through orifices - SAE flange dimensional data - Straight thread fitting sizes - Equivalent pipe and tubing sizes - ISO standardization effort - Thread forms of fluid connectors - Common O-ring sizes

Page

381381382383385386

387388389

390391

392393394

395397398400

401403

405406407

408409410411412413414415415416417418419419419420421

(continued on next page)

339

TopicDrilling, Tapping and Punching U.S. twist drill diameters Tap drill sizes for U.S. threads Tap drill sizes, pipe tap drill sizes, SAE pads Drilling and tapping factors Tonnage for hole punching Drilling speed, HP and thrust Metric tap drill sizes Metric machine screw threads Suggested cutting speedsElectrical Control and Automation Design Data Electrical diagram schematic symbols, etc. Three phase motor data Motor installation data Three-phase AC motor windings, etc. Control transformer sizing and motor starter ratings Wire selection guide Color codes resistor, thermocouple and wiring Electrical enclosures NEMA ratings Electrical enclosures, IP rating Intrinsically safe Motion control formulasMiscellaneous Formulas and Conversions Circumference and area of circles Calculating areas of plane figures Table of equivalents Decimal and metric equivalents Conversion between English and (SI) Standard units Interchange between units Temperature conversion chart Table of standard wire gauges Densities, specific gravities and mechanical properties of common materials National Electric Code Article 505 International Electrotechnical Commission (IEC) ATEX Directive The New Conformity Mark North American classification system for hazardous locations European standards Classification group and category Temperature rating

Page

422423424425425426428428429

430432434435436437438439440441442

447448450451452453455456

457458458459459

460461462462

340

Double-Acting Cylinder

Double-Acting Cylinder with Double End Rod

Fixed Displ.

Unidirect.

Fixed Displ.

Bidirect.

Variable Displ.

Unidirect.

Variable Displ.

Over Cntr.

Variable Displ., Pres.

Compensator

Over Center

w/Comp.

Fixed Displ.

Unidirect.

Fixed Displ.

Bidirect.

Variable Displ.

Unidirect.

Bidirect. w/Press.

Partial Revolution Oscillator

Electric Motor

HYDRAULIC PUMPS

HYDRAULIC & ELECTRIC MOTORS

2-WAY & 3-WAY VALVES

4-WAY VALVES

2-Position Double Actuator

3-Position Spring Centered

*SPOOL CENTERS FOR 3-POSITION VALVES

Tandem Center

Float Center

Open Center

Proportional Solenoid Valve

Single-Acting Cylinder

3-Way Selector

2-Position Single Actuator

Closed Center

=

M

2-Way, N.C. 2-Position

3-Way, Direc- tional Control

2-Way, N.O. 2-Position

Graphic SymbolsFor Use on Fluid Power Drawings

These are the more common ANSI graphic symbols from the American National Standards Institute for use on fluid power circuit drawings. A more complete list

can be obtained from the National Fluid Power Association, 3333 N. Mayfair Rd., Milwaukee, WI 53222. Write for listing and prices.

341

GeneralPurpose

ACTUATORS FOR VALVES

ManualLever

FootOperated

CamOperated

PilotOperated

ButtonBleeder

SolenoidOperated

SpringReturn

PressureCompensated

Pilot andSolenoid

3-PositionDetent

PalmButton

Air TrioUnit

Air or OilFilter

Air LineLubricator

Air LineRegulator

Air Filterw/Drain

ComponentOutline

DrainLines

PilotLines Lines

CrossingLines

Connecting

PressureGauge

AirMuffler

ShuttleValve

ManualShut-Off

LiquidFlow

Air orGas Flow

Flow ControlValve with

Bypass

PressureCompensatedFlow Control

w/Bypass

HeatExchanger

4-WayServo Valve

Accum-ulator

Pilot toOpen Check

Valve

CheckValve

Pilot toClose Check

Valve

FixedOrifice

PressureCompensatedFlow Control

Valve

NeedleValve

ReliefValve

Relief Valvewith vent

SequenceValve

PressureReducing

Valve

Graphic Symbols (continued)

342

Hydraulic TroubleshootingMany of the failures in a hydraulic system show similar symptoms: a gradual or sudden loss

of pressure, resulting in loss of power or speed in the cylinders or motors. In fact, the cylinders may stall under light loads or may not move at all. Often the loss of power is accompanied by an increase in pump noise, especially as the pump tries to build pressure.

Any major component (pump, relief valve, directional valve, or cylinder) may be at fault. In a sophisticated system, other components could be at fault, but this would require the services of an experienced technician.

By following an organized step-by-step testing procedure in the order given here, the problem can be traced to a general area, then, each component in that area can be tested or replaced. All testing of a hydraulic system should be performed with the fluid in the system at normal operating temperature, about 100 degrees F.

PUMP

PRESSUREGAUGE

RELIEFVALVE

4W AYVALVESUCTION

STRAINER

WORKCYLINDER

C J

H H

KD FE

A

B

STEP 1 Pump Suction Strainer Hydraulic system problem encountered most often is cavitation of the hydraulic pump inlet

caused by restriction due to a dirt build-up on the suction strainer. This can happen on a new as well as an older system. It generally produces the following symptoms: increased pump noise, loss of high pressure and/or speed. A good indication of a clogged suction strainer is when a pump delivers low flow at both low and high pressure.

If the strainer is not located in the pump suction line it will be found immersed below the oil level in the reservoir (point A). Some operators of hydraulic equipment never give the equipment any attention or maintenance until it fails. Under these conditions, sooner or later, the suction strainer will probably become sufficiently restricted to cause a breakdown of the whole system and damage to the pump.

The Suction strainer should be removed for inspection and should be cleaned before re-installation. Wire mesh strainers can be washed in a solvent that is compatible with the systems fluid. Do not use gasoline or other explosive or flammable solvents. The strainer should be cleaned even though it may not appear to be dirty. Some clogging materials cannot be seen except by close inspection. If there are holes in the mesh or if there is mechanical damage, the strainer should be replaced. When reinstalling the strainer, inspect all joints for possible air leaks, particularly at union joints (points, E, G, H, J, and K). There must be no air leaks in the suction line. Check the reservoir oil level to be sure it covers the top of the strainer by at least 3" inches minimum oil level, with all cylinders extended. If it does not cover to this depth there is danger of a vortex forming above the suction strainer, which may allow air to enter the system when the pump is running.

STEP 2 Pump and Relief Valve If cleaning the pump suction strainer does not correct the trouble, isolate the pump and

relief valve from the rest of the circuit by disconnecting at (point E) so that only the pump, relief valve and pressure gauge remain in the pump circuit. Cap or plug both ends of the plumbing, which was disconnected. The pump is now deadheaded into the relief valve. Before proceeding, determine the normal system operating pressure. Adjust the relief valve to zero restriction by unscrewing the adjusting screw. Start the pump and begin screwing in the relief valve adjusting screw, watch for pressure build-up on the pressure gauge while tightening the adjustment on the relief valve. Care should be taken not to exceed the system operating pressure plus 300 psi. If full system pressure can be developed, the pump and relief valve are operating correctly and the trouble is to be found further down the line. At this point set the relief valve to the system pressure plus 300 psi. If full pressure cannot be developed in this test, continue with STEP 3.

343

Hydraulic Troubleshooting (continued)

STEP 3 Pump or Relief Valve If high pressure cannot be obtained in STEP 2 by running the pump against the relief valve,

further testing must be conducted to see whether the fault lies in the pump or in the relief valve. Proceed as follows:

If possible, disconnect the reservoir return line from the relief valve at point H and insert a flow meter in the line. Start the pump and run the relief valve adjustment up and down while observing the flow through the meter. If the pump is bad there will probably be a flow of oil through the meter when the relief is backed out but the flow of oil will diminish or stop as the adjustment is increased. The flow should be compared with the pump catalog rating. If the pump flow is more than 20% less than the rated flow it should be repaired or replaced.

If the gauge pressure does not rise above a low value, say 100 PSI, and if the volume of flow does not substantially decrease as the relief valve adjustment is tightened, the relief valve is probably at fault and should be cleaned, repaired, or replaced as instructed in STEP 5. If the oil substantially decreases as the relief valve adjustment is tightened, and if only a low or moderate pressure can be developed, this indicates trouble in the pump Proceed to STEP 4.

STEP 4 Pump If a full flow of oil is not obtained in STEP 3, or if the stream diminishes as the relief valve

adjustment is tightened, the pump is probably at fault. Assuming that the suction strainer has already been cleaned and the inlet plumbing has been examined for air leaks, as in STEP 1, the oil is slipping across the pumping elements inside the pump. This can mean a worn-out pump, or too high an oil temperature. High slippage in the pump will cause the pump to run considerably hotter than the oil reservoir temperature. In normal operation, with a good pump, the pump case will generally run about 20 degrees F above the reservoir temperature. If the temperature is greater than this, there is excess slippage, generally caused by wear.

Check for slipping belts, sheared shaft pin or key, broken shaft, broken coupling, bad coupling insert, or loosened set screw.

STEP 5 Relief Valve If the test in STEP 3 has indicated the trouble to be in the relief valve, point D, the quickest

remedy is to replace the valve with one known to be good. The faulty valve may later be disassembled for inspection and cleaning. A pilot-operated relief valve has small orifices that may be become blocked with accumulations of dirt. This problem can be remedied by simply cleaning the orifices inside the valve. The valve spool should be checked for free movement as well. In a relief valve with pipe thread connections in the body, the spool may bind if a pipe fitting is over tightened during installation. If possible, test the spool for binding before unscrewing threaded connections from the body, or screw in fitting tightly during inspection of the valve.

STEP 6 Cylinder If the pump will deliver full pressure when operating across the relief valve in STEP 2, both

pump and relief valve can be considered good, and the trouble is further downstream. The cylinder should be first tested for worn-out or defective packing using the method described on page 345.

STEP 7 Motor If it has been determined that the system flow and pressure is correct yet a motor is not

operating at the proper speed, there may be a problem with slippage inside the motor. For piston motors install a flow meter in the case drain to determine if there is excessive leakage inside the motor. Alternatively check the output RPM of the motor to determine if the speed matches the manufacturers published output for the flow being provided to the motor.

STEP 8 Directional Control Valve If the cylinder has been tested (STEP 6) and found to have reasonably tight piston seals,

the 4-way valve should be checked next. Although it does not often happen, an excessively worn valve spool can slip enough oil to prevent build up of maximum pressure. Symptoms of this condition are a loss of cylinder speed together with difficulty in building up to full pressure even with the relief valve adjusted to a high setting. This condition would be more likely to occur with high pressure pumps of low volume output, and would develop gradually over a long period of time. Four-way valves may be tested by the method described on page 345.

344

Hydraulic Troubleshooting (continued)

Other Components Check other components such as bypass flow controls, hydraulic motors, etc. Consider

solenoid 4-way valves or pilot-operated valves with tandem or open center spools that may not have sufficient pilot pressure to shift the spool.

Other Typical Hydraulic System Problems

Excessive heat in the system1. Worn out or improperly performing components in the system.2. Improper heat exchanger selection or heat exchanger not operating properly.3. Improper use of relief valve in the system.4. Improper component selection.5. Fluid viscosity too low causing leakage past components. Improper fluid selection.6. High velocity through piping components causing pressure drop resulting in heat.7. Improper installation of the reservoir. Air should circulate around and if possible below the reservoir.8. Ambient air temperature too high for system design.9. Improper flow of oil in reservoir causing channeling of the oil. Add baffles.10. Reservoir volume too low to dissipate the heat returning from the system.

Excessive noise1. Cavitation in the pump suction.2. Pump speed too high on engine applications.3. System operating over relief valve4. Improper mounting of components.5. Velocity too high in system.6. Improper valve operation.7. Improper plumbing of the system

Dirt in System Fluid1. Filters not being replaced at the proper intervals for the application.2. Air breather not installed or clogged.3. System not sealed from the atmosphere.4. Contamination introduced into the system while being fabricated or being serviced.

Air in System Fluid1. Pump Suction line leak.2. Return line not discharging below the level of the oil in the reservoir.3. Insufficient baffles in the reservoir.4. Insufficient oil in the reservoir.

Water in the System Fluid1. Condensation from heating and cooling of the reservoir.2. Water in containers being used to fill the reservoir.3. Leak in-water cooled heat exchanger.

345

Cylinder and Valve TestingIn an air system, if air is detected

escaping from a 4-way valve exhaust while the cylinder is stopped, this air is either blowing by worn out piston seals or is leaking across the spool in the 4-way valve. These two leakage paths are shown in figure to the right.

Most air cylinders and valves have soft seals and should be leak-tight. Those air valves having a metal-to-metal seal between the spool and body may be expected to have a small amount of leakage.

If leakage is noted, it is more likely to be coming through the cylinder than across the valve spool, and the cylinder should be tested first.

Cylinder Testing Run the piston to one end of its

stroke where the cylinder is stalled. Remove the line to the cylinder on the same end as the piston is stalled. Install a pressure gauge in this port and plug the open line. With the line plug and gauge installed, re-pressure the cylinder. If fluid is leaking past the seals inside the cylinder you should see a rise in pressure on the pressure gauge.

After checking for leakage, re-install the original plumbing and run the cylinder to the opposite end and repeat the test. Occasionally a cylinder will leak at one point in its stroke due to a scratch or dent in the barrel. Check suspected positions in mid stroke by installing a positive stoop at the suspected position and run the piston rod against it for testing as stated above. Once in a while a piston seal may leak intermittently. This usually caused by a soft packing or O-ring moving slightly or rolling into different positions on the piston, and is more likely to happen on cylinders of large bore.

Cylinder pistons with metal ring seals can be expected to have a small amount of leakage across the rings. Even leak-tight soft seals may have a small bypass during new seal break-in or after the seals are well worn.

4-Way Valve Testing For testing 4-way valves, either air or hydraulic, it is necessary to obtain access to the

exhaust or tank return ports so that the amount of leakage can be observed. To make the test, disconnect both cylinder lines and plug these ports on the valve. Install a flow-measuring device on the exhaust or tank line. Start up the system and shift the valve to one working position. Any flow out the exhaust or tank return line while the valve is under pressure is the amount of leakage. The leakage recorded should be compared to the manufacturers specifications for the valve. Repeat the test in all other working positions of the valve. It may be necessary to remove the valve and return it to the manufacturer for testing if it is not possible to measure the leakage.

Safe Pump Inlet Vacuum

The suction strainer should be cleaned or replaced when inlet vacuum on a hydraulic pump reaches these values. Sustained operation at these vacuums may damage the pump. When the suction strainer is clean, the inlet vacuum should not be more than 1/3 of these values.

ValveExhausts

PressurizedFluid In

ValveCylinderPorts

Cylinder Piston Leakage

ValveSpoolLeakage

Gear Pumps Vane Pumps Piston PumpsMax. Safe Inlet Vacuum, PSI 3 to 5 2 to 3 2Max. Safe Inlet Vacuum, In. Hg 6 to 10 4 to 6 4

2 Leak Paths

346

Replacement of Pump or MotorCalculating the Theoretical GPM of a Pump by Measuring Its Internal Parts.

To select a replacement for a broken or worn out hydraulic pump or motor which has no nameplate or has no rating marked on its case, use the formulas below after making internal physical measurements.

When replacing a pump, catalog ratings will usually be shown in GPM at a specified shaft speed. On a motor, catalog ratings will usually be in C.I.R. (cubic inches displacement per shaft revolution). Formulas are given for calculating either GPM at 1800 RPM or calculating C.I.R. Use the formula which is appropriate. Make all measurements in inches, as accurately as possible. Convert fractional dimensions into decimal equivalents for use in the formulas.

Make sure the catalog pressure rating is adequate for your application, and in the case of a pump, be sure direction of shaft rotation is correct.

D

W

W

DD

LL

LL

Gear Pumps and Motors1. Measure gear width, W.2. Measure bore diameter of one of the gear chambers: this is D.3. Measure distance across both gear chambers; this is L.

GPM @ 1800 RPM =

A speed of 1800 RPM is used in the formula. At other speeds, GPM is proportional to RPM.

C.I.R. Displacement =

Vane Pumps and Motors(Balanced type, not variable displ.)

1. Measure width of rotor. This is W.2. Measure shortest distance across bore; this is D.3. Measure longest distance across bore: this is L.

GPM @ 1800 RPM =

A speed of 1800 RPM is used in the formula. At other speeds, GPM is proportional to RPM.

C.I.R. Displacement =

(L - D)2

47 s W s (2D - L) s

(L - D)2

6 s W s (2D - L) s

(L - D)2

94 s W s s(L + D)4

(L - D)2

12 s W s s(L + D)4

Gear Pump

Vane Pump(Balanced Type Only)

Piston Pumps and Motors1. Find piston area from piston diameter; this is A in formula.2. Measure length of stroke; this is L in formula.3. Count number of pistons; this is N in formula.

GPM @ 1800 RPM = A s L s N s 1800 231

A speed of 1800 RPM is used in formula. At other speeds, GPM is proportional to RPM.

C.I.R. Displacement = A s L s N

If a pump of higher GPM has to be used, it will require more HP at the same pressure and cylinders in the system will move faster. If one with lower GPM is used, the system will have plenty of power but cylinders will move more slowly than originally.

If a motor with greater displacement is used, it will deliver more torque at a reduced RPM, but will require no more fluid HP from the pump. If it has less displacement it will rotate faster with less torque.

347

Step 1. Calculate torque required at the kingpin for steering with this formula:

T = Kingpin torque in inch lbs. This is total for both wheels. If these wheels are power driven, double the value calculated to allow for approximate additional dynamic loads.

W = Vehicle weight on the steered axle. Note that this is not total vehicle weight, only that part of the weight which is on the steered axle.

u = Coefficient of friction between tire and road. This can be assumed to be 0.7 for most applications, but for narrow tires this would be less, and can be taken from the graph.

B = Nominal width of tire.E = Kingpin offset. This is the distance, in

inches, measured on the road between the tire centerline and the kingpin projection onto the road.

Step 2. Calculate the bore diameter of the hydraulic cylinder(s) as follows: Take the torque found in Step 1 and divide it by the length of the radius arm(s), in inches. This gives the cylinder force. Use the chart on pages 19 and 20 to find the cylinder bore. Select a bore which will produce the required force at about 75% of the maximum PSI, or relief valve setting of the system. If only one cylinder is used in the system, be sure to use rod area when figuring force and bore diameter. If two single-end-rod cylinders are used, figure with the rod area of one combined with the blind end area of the other.Step 3. Calculate or measure the cylinder stroke length according to the vehicle geometry. A scale layout on paper may be used to measure the stroke needed to turn from hard left to hard right. This can also be calculated by the use of trigonometry.Step 4. Calculate the volume in cubic inches of the oil required to move the cylinder(s) through their entire stroke. Volume = square inch area on piston s length of stroke in inches.Step 5. Before proceeding further a decision must be made as to the number of steering wheel turns desired to move the road wheels from hard left to hard right. American vehicles vary from 256O to 556O turns of the steering wheel with 4 turns being a good average working number.Step 6. Select a fluid steering with the correct C.I.R. (cubic inch per revolution) displacement to operate the power cylinder through its full stroke with the selected number of steering wheel turns. For example, if the cylinder oil volume calculated in Step 4 was 30 cubic inches, and the number of steering wheel turns chosen in Step 5 was 4 turns, the fluid steering displacement would have to be 30 4 = 7.5 C.I.R..Step 7. Pump GPM is determined by how fast the system must follow the movement of the steering wheel. A typical figure is 2 seconds from hard left to hard right. Pump GPM = Cyl. Vol. (cubic inches) s 0.26 steering time in seconds. The factor 0.26 takes care of converting seconds to minutes and cubic inches to gallons.

Power Steering Calculations

From Steering

Valve

WE

IGH

TW

EIG

HT

KINGPIN

0.2 0.4 0.6 0.8 1.0 1.2

0.7

0.6

0.5

0.4

0.3

0.2

0.1

u (C

OE

FF

ICIE

NT

OF

FR

ICT

ION

)

RATIO E w B

EB

Graph for Estimating Coefficient of Road Friction

!T Wu + E2B28

348

Torque and horsepower Relations:T = HP s5252 RPMHP = T sRPM 5252RPM = HP s5252 T

Torque values are in foot pounds.

Hydraulic (fluid power) horsepower:HP = PSI sGPM 1714

PSI is gauge pressure in pounds per square inch; GPM is oil flow in gallons

per minute.

Velocity of oil flow in pipe:V = GPM s0.3208 AV is oil velocity in feet per second;GPM is flow in gallons per minute;A is inside area of pipe in square inches.

Charles Law for behavior of gases:T1V2 = T2V1, or T1P2 = T2P1T1, P1, and V1 are initial temperature, pressure, and volume, and T2, P2, and V2 are final conditions.

Boyles Law for behavior of gases:P1V1 = P2V2P1 and V1 are initial pressure and volume; P2 and V2 are final conditions.

Circle formulas:Area = !r 2, or !D 2 4Circumference = 2!r , or !Dr is radius; D is diameter, inches.

Heat equivalent of fluid power:BTU per hour = PSI sGPM s156O

Hydraulic cyl. piston travel speed:S = CIM AS is piston travel speed, inchesper minute;CIM is oil flow into cylinder, cubic inches per minute;A is piston area in square inches.

Force or thrust of any cylinder:F = A sPSIF is thrust or force, in pounds;A is piston net area in square inches; PSI is gauge pressure.

Force for piercing or shearing sheet metal:F = P sT sPSIF is force required, in pounds;P is perimeter around area to be sheared, in inches;T is sheet thickness in inches;PSI is the shear strength rating of the material in pounds per square inch.

Side load on pump or motor shaft:F = (HP s63024) ( RPM sR)F is the side load, in pounds, against shaft;R is the pitch radius, in inches, of sheave on pump shaft;HP is driving power applied to shaft.

Fluid Power FormulasEffective force of a cylinder working at an angle to direction of the load travel:

F = T ssin AT is the total cylinder force, in pounds;F is the part of the force which is effective, in pounds;A is the least angle, in degrees, between cylinder axis and load direction.

Heat radiating capacity of a steel reservoir:HP = 0.001 sA sTDHP is the power radiating capacity expressed in horsepower;A is surface area, in square feet;TD is temperature difference in degrees F between oil and surrounding air.

Burst pressure of pipe or tubing:P = 2t sS OP is burst pressure in PSI;t is wall thickness, in inches;S is tensile strength of material in PSI;O is outside diameter, in inches.

Relationship between displacement and torque of a hydraulic motor:

T = D sPSI 24!T is torque in foot-lbs.;D is displacement in cubic inches per revolution;PSI is pressure difference across motor; ! = 3.14.

Rules-of-Thumb

Horsepower for driving a pump: For every 1 HP of drive, the equivalent

of 1 GPM @ 1500 PSI can be produced.

Horsepower for idling a pump: To idle a pump when it is unloaded will require about 5% of its full rated

horsepower.

Compressibility of hydraulic oil: Volume reduction is approximately 1/2% for every 1000 PSI of fluid pressure.

Compressibility of water: Volume reduction is about 1/3% for every 1000 PSI pressure.

Wattage for heating hydraulic oil:Each watt will raise the temperature of 1

gallon of oil by 1F per hour.

Flow velocity in hydraulic lines:Pump suction lines 2 to 4 feet per

second; pressure lines up to 500 PSI, 10 to 15 feet per sec; pressure lines 500

to 3000 PSI, 15 to 20 feet per sec.; pressure lines over 3000 PSI, 25 feet

per sec.; all oil lines in air-over-oil system, 4 feet per sec.

349

Fluid Power Formulas inSI Metric Units

Fluid power formulas in English units are shown in the left column. SI (International) unit equivalents of these formulas are shown in the right column.

T = HP s5252 RPMHP = T sRPM 5252RPM = HP s5252 T

T = Torque, foot-lbs.RPM = Speed, revs/minuteHP = Horsepower

T = Kw s9543 RPMKw = T sRPM 9543RPM = Kw s9543 T

T = Torque, Nm (Newton-meters)RPM = Speed, revs/minuteKw = Power in kilowatts

HP = PSI sGPM 1714

HP = HorsepowerPSI = Gauge pressure, lbs/sq. inchGPM = Flow, gallons per minute

Kw = Bar sdm3/min 600

Kw = Power in kilowattsBar = System pressuredm3/min = Flow, cu. dm/minute

N = A sBar s10

N = Cylinder force in NewtonsA = Piston area, sq. centimetersBar = Gauge pressure

F = A sPSI

F = Force or thrust, in pounds.A = Piston area, square inchesPSI = Gauge pressure, lbs/sq. inch

S = V A

S = Travel Speed, inches/minuteV = Volume of oil to cyl., cu. in/min.A = Piston area, square inches

S = V A

S = Travel Speed, meters/sec.V = Oil flow, dm3/minuteA = Piston area, square centimeters

P = 2t sS O

P = Burst pressure, PSIt = Pipe wall thickness, inchesS = Tensile strength, pipe matl, PSIO = Outside diameter of pipe, inches

P = 2t sS O

P = Burst pressure, Bart = Pipe wall thickness, mmS = Tensile strength, pipe matl, BarO = Outside diameter of pipe, mm

V = GPM s0.3208 A

V = Velocity, feet per secondGPM = Oil flow, gallons/minuteA = Inside area of pipe, sq. inches

V = dm3/min 6A

V = Oil velocity, meters/seconddm3/min = Oil flow, cu. dm/minuteA = Inside area of pipe, sq. cm

fps = feet per second

Pump suction lines 2 to 4 fpsPres. lines to 500 PSI 10 to 15 fpsPres. lines to 3000 PSI 15 to 20 fpsPres. lines over 3000 PSI 25 fpsOil lines in air/oil system 4 fps

mps = Meters per second

Pump suction lines 0.6 to 1.2 mpsPres. lines to 35 bar 3 to 4 56O mpsPres. lines to 200 bar 456O to 6 mpsPres. lines over 200 bar 756O mpsOil lines in air/oil system 156M mps

English Units Metric (SI) Units

Hydraulic Power Flowing through the Pipes

Force Developed by an Air or Hydraulic Cylinder

Travel Speed of a Hydraulic Cylinder Piston

Barlows Formula Burst Pressure of Pipe & Tubing

Velocity of Oil Flow in Hydraulic Lines

Recommended Maximum Oil Velocity in Hydraulic Lines

Torque, HP, Speed Relations in Hydraulic Pumps & Motors

350

English/Metric Conversions

2030405060708090

100200300400500600700800900

1000

1.3792.0692.7593.4484.1384.8285.5176.2076.89713.7920.6927.5934.4841.3848.2855.1762.0768.97

110012001300140015001600170018001900200022502500275030003500400045005000

75.8682.7689.6696.55103.5110.3117.2124.1131.0137.9155.2172.4189.7206.9241.4275.9310.3344.8

1 PSI = 0.0689655 barPSI Bar PSI Bar

Pressure - PSI and Bar

123456789

101520253035404550

14.5029.0043.5058.0072.5087.00101.5116.0130.5145.0217.5290.0362.5435.0507.5580.0652.5725.0

556065707580859095

100150200250300350400450500

797.5870.0942.5101510881160123313051378145021752900362543505075580065857250

1 bar = 14.5 PSIBar PSI Bar PSI

12345

10152025303540455055606570

3.7857.57011.3615.1418.9337.8556.7875.7094.63113.6132.5151.4170.3189.3208.2227.1246.0265.0

7580859085

100125150175200225250275300325350375400

7580859085

100125150175200225250275300325350375400

GPM = 3.785 liters/minGPM l/min GPM l/min

Hydraulic Flow - GPM and Liters per Minute

5102030405060708090

100125150175200225250275

1.322.645.287.9310.613.215.918.521.123.826.433.039.646.252.859.466.172.7

300350400450500550600650700750800900

100011001200130014001500

79.392.5 106119132145159172185198211238264291317343370396

1 liter/min = 0.2642 GPMl/min GPM l/min GPM

2030405060708090

100200300400500600700800900

1000

1.3792.0692.7593.4484.1384.8285.5176.2076.89713.7920.6927.5934.4841.3848.2855.1762.0768.97

110012001300140015001600170018001900200022502500275030003500400045005000

75.8682.7689.6696.55103.5110.3117.2124.1131.0137.9155.2172.4189.7206.9241.4275.9310.3344.8

1 CFM = 0.47195 cu. dm/sec

CFM CFM

Air Flow CFM and Cubic Decimeters per Second

5101520253035404550607080

100125150175200

10.5921.1931.7842.3752.9763.5674.1584.7595.34105.9127.1148.3169.5211.9264.8317.8370.7423.7

225250300350400450500550600700800900

100011001200130014001500

476.7529.7635.6741.5847.5953.4105911651271148316951907211923312542275429663178

CFM CFM

(dm3/s)1 cu. dm/sec = 2.1187 CFM

(dm3/s)

351

1 U.S. gallon: = 231 cubic inches = 4 quarts or 8 pints = 128 ounces (Liquid) = 133.37 ounces (weight) = 8.3356 pounds = 3.785 liters 1 Imperial gallon = 1.2 U.S. gal.1 Liter = 0.2642 U.S. gallons1 Cubic foot: = 7.48 gallons = 1728 cubic inches = 62.4 pounds (water)1 Cu. ft. water weighs 62.4 lbs.1 Bar at sea level: = 14.504 PSI = 0.98692 atmosphere = 33.6 foot water column = 41 foot oil columnApprox. 1/2 PSI decrease each 1000 feet of elevation.1# Hg = 0.490 PSI = 1.131 ft. water1 Horsepower: = 33,000 ft. lbs. per minute = 550 ft. lbs. per second = 42.4 BTU per minute = 2545 BTU per hour = 746 watts or 0.746 kw

1 PSI = 2.0416# Hg = 27.71# water = 0.0689 bar1 Atmosphere: = 1.013 bar = 29.921# Hg = 14.696 PSI = 760 mm Hg1 Foot water column = 0.432.PSI1 Foot oil column = 0.354 PSI1 Barrel oil = 42 gallons1 Micro-meter (mm): = 0.000001 meter (micron) = 0.001 centimeter = 0.00004 inch25 Micro-meters = 0.001 inch

Approximate Equivalents1 Pint = 2 cups = 32 tablespoons = 96 teaspoons = 16 fl. oz. = 1 lb.1 Quart = 4 cups = 2 pints = 32 fluid ounces = 2 pounds.1 Gallon = 16 cups = 4 quarts = 8 pints = 128 fl. oz. = 231 cu. ins.1 Cup = 16 tablespoons = 48 tsp.1 Tablespoon = 3 tsp. = 1/2 fluid oz.1 Fluid oz. (volume) = 600 drops hydraulic oil.1 Cubic inch = 330 drops (oil).

Fluid Power EquivalentsExact Equivalents

abs absolute (as in psia)AC alternating currentBhn Brinell hardness numberBtu British thermal unitC degrees Centigrade (Celsius)cc closed centerccw counter clockwise cfm cubic feet per minutecfs cubic feet per secondcir cubic inches per revolutioncim cubic inches per minutecom Commoncpm cycles per minute cps cycles per second cu. in/rev cubic inches per revolutioncw clockwisecyl cylinderDC direct currentdia diameterext externalF degrees Fahrenheitfl fluidfpm feet per minuteft footft-lb foot poundgal gallongpm gallons per minuteHg MercuryHP horsepowerHz HertzID inside diameterin inchin-lb inch poundint internal

pm inches per minuteips inches per secondlb poundmax maximummin minimummtd mountedNC normally closedNO normally openNPT national pipe threadNPTF dryseal pipe threadsoc open centeroz. ounceP.O. pilot operatedpres pressurePSI pounds/square inchpsia psi absolutepsig psi gaugept pintqt quartr radiusrms root mean squarerpm revolutions per minuterps revolutions per secondscfm standard cu. ft. per minuteSmls seamlesssol solenoidSSU Saybolt seconds universalSUS Saybolt universal secondsm micro-meters or micronsT torquevac vacuumVI viscosity indexvisc viscosity

Fluid Power Abbreviations

352

The force to drive a vehicle is composed of the sum of (1) road resistance, (2) force necessary to climb a grade, (3) force needed to accelerate to final velocity in the allowable time, (4) force to overcome air resistance, on fast moving vehicles. Each of these forces can be calculated or estimated from the formulas on this page, then added together. In selecting an engine, allow enough extra power to make up for losses in the mechanical transmission system including gear boxes, clutches, differentials, chain or belt drives.

Travel Speed in MPH (miles per hour) is found by multiplying wheel RPM wheel circumference.MPH = RPM sd 336, orRPM = 336 sMPH dd is wheel diameter in inches.

Axle Torque for driving the vehicle is found by multiplying drawbar pull (or push) times wheel radius.T = F s r or, F = T rT is axle torque in inch pounds.F is drawbar pull in pounds.r is wheel radius in inches.

Drawbar Pull to keep the vehicle in steady motion on level ground depends on the road surface. The following figures are pounds of drawbar pull per 1000 lbs. of vehicle weight.Concrete . . . . . . . . . . . . . . . . . 10 to 20 lbs.Asphalt . . . . . . . . . . . . . . . . . . . 12 to 22 lbs.Macadam . . . . . . . . . . . . . . . . . 15 to 37 lbs.Cobbles . . . . . . . . . . . . . . . . . . 55 to 85 lbs.Snow . . . . . . . . . . . . . . . . . . . . 25 to 37 lbs.Dirt . . . . . . . . . . . . . . . . . . . . . 25 to 37 lbs.Mud . . . . . . . . . . . . . . . . . . . . 37 to 150 lbs.Sand . . . . . . . . . . . . . . . . . . . 60 to 300 lbs.

Horsepower required on vehicle wheels is torque times RPM:HP = T sRPM 63024T is wheel torque in inch pounds.NOTE: Additional HP is required at the engine to overcome transmission system losses.

Conversion Formula between torque, HP, and speed.T = HP s63024 RPMTorque values are in inch pounds.

Momentum of a vehicle is equivalent to that constant force which would bring it to rest in one second by resisting its movement.Momentum = Weight sV gWeight is in pounds.V is velocity in feet per second.g is gravity acceleration = 32.16

Acceleration of a vehicle is expressed in this formula involving weight, accelerating force, and time.F = (V sW) ( g sT) F is accelerating force in pounds.V is final velocity in feet per second.W is vehicle weight in pounds.g is gravity acceleration = 32.16T is time in seconds that force acts.Note : The gravity acceleration symbol, g,converts weight into mass.

Grade , in mobile work, is usually expressed in percentage rather than in degrees. For example, a 10% grade has a rise of 10 feet in a distance of 100 feet, etc.

Grade Resistance is the drawbar pull needed to keep the vehicle in constant motion up a grade. This is in addition to the drawbar pull to overcome road resistance as expressed by another formula.F = GR sWF is drawbar pull in pounds.GR is grade resistance in percent (20% is written as 0.20, etc.)W is gross vehicle weight in pounds.

Air Resistance will be important only on fast moving vehicles (over 20 to 30 MPH).

F = FA s0.0025 MPH2

F is additional drawbar pull needed to overcome air resistance. FA is frontal area of vehicle in square feet.MPH is vehicle speed, miles per hour.

Axles and drive shafts must have a diameter large enough to transmit the torque without excessive deflection. The angle of deflection for a solid round axle may be calculated from this formula:

A = 583.6 sT sL ( D4 sE) A is angle of deflection in degrees.T is applied torque in inch pounds.L is shaft length in inches.E is modulus of elasticity of material. (12,000,000 for steel)D is shaft diameter in inches.

Some authorities say that a steel shaft should be limited to an angular deflection of 0.08 degrees per foot of length to avoid failure.

Vehicle Drive Calculations

353

CylindersCylinders are used to convert fluid power

energy, air or hydraulic, into push-pull mechanical power.

They are available in a wide range of bore sizes, pressure ratings, and mounting styles to suit almost any application.

Operating PrinciplesThe operating principle of a cylinder is simple - fluid pressure is applied on one side of its

piston, and the opposite side of the piston is vented to a lower pressure or to atmosphere.Force developed on the piston and rod is easily calculated by multiplying gauge pressure, in PSI, times the square inch area on the piston which is exposed to the pressure. The square inch area on a circular piston can be calculated from this formula:

Area (sq.in. ) = ! sD2 4, in which D is diameter in inches

EXAMPLE : Find the force produced by an air cylinder with 6# diameter piston operating on an air pressure of 90 PSI.

SOLUTION: First, find the piston area: A = $ s 62 4 = 28.27 square inches. Force = 28.27 s 90 PSI = 2544 pounds. The answer can also be obtained from tables in this manual.

Remember, when calculating force developed on the retraction stroke, the pressure does not act on the piston area which is covered by the piston rod; therefore to find the effective (or net) area, the rod area must be subtracted from the full piston area.

For convenience in selecting cylinder size, various charts are included on the following pages for calculation of cylinder force, speed, air consumption, and minimum rod size to prevent rod buckling.

Direction of Travel. A double-acting air or hydraulic cylinder is usually controlled with a 4-way directional control valve connected as shown. The valve may have only two working positions as in this illustration, or it may have a center neutral position for stopping the cylinder in mid stroke.

Direction of travel of a single-acting (spring or gravity returned) cylinder is usually controlled with a 3-way directional valve; or with a 4-way valve in which the unused cylinder port is plugged. For a single-acting cylinder which requires 3 functions, such as raise-hold-lower, the control valve must have a center neutral position.

Travel Speed. The speed of an air cylinder may be regulated with flow control valves placed in the lines between 4-way valve and cylinder. Listings in this manual show both needle-type and pressure compensated flow control valves covering a full range of sizes.

The exact travel speed of an air cylinder cannot be predicted because of many variables such as amount of load, piping size and restrictions, valve size, etc. The only way to be certain of sufficient speed is to select a cylinder which has at least 25% to 50% more force than required by the load, or one with double the force required if travel speed must be rapid.

DDPP

SquareHead IndustrialCylinder

V1

354

The speed of a hydraulic cylinder can be accurately predicted if the volume of oil flow is known. The formula for calculating speed in inches per minute is: Speed = oil flow (cubic inches per minute) square inch area of the piston. If oil flow is gi ven in GPM (gallons per minute) convert to cubic inches per minute by multiplying by 231. Forward and return speed of a hydraulic cylinder are shown in charts on the next few pages.

Cushions : Cylinders can be ordered with cushions on either or both ends of the stroke. The purpose of a cushion is to decelerate the piston at a point about 1# from the end of the stroke. Cushions cannot be added in the field; they must be specified at the time the cylinder is purchased.

A cushion consists of a closed chamber in the cylinder end cap which traps fluid when the cushion nose, mounted on the moving piston, enters the chamber. The rate of fluid escape, or the rate of deceleration, is controlled from this point through the remainder of the stroke, by the external adjustment of a built-in needle valve. A check valve, also built in, allows the cylinder to start up in the opposite direction at full speed and force.

Cushions are highly effective in decelerating a hydraulic cylinder, but are effective on air cylinders only when the load being stopped consists principally of friction with very little mass which creates a momentum energy which has to be dissipated while the cylinder is being decelerated. For decelerating a high momentum load on an air cylinder, a limit switch or cam valve should be used to reduce the air flow into or out of the cylinder. This switch or valve should be located far enough ahead of the final position to give the load time to decelerate properly.

Cylinder TypesStandard Double-Acting . Provides a

power stroke in both directions. This is the type used in the majority of applications, both air and hydraulic.

Single-Acting Cylinder . Where force is needed in only one direction, a standard double-acting type may be used, with the inactive end vented to atmosphere through a filter/breather in the case of an air cylinder, or to tank above oil level in the case of a hydraulic cylinder.

Double-End-Rod Cylinders . Available from most manufacturers. They are used when an equal displacement on both sides of the piston is desirable, or where the additional rod is needed for mechanical support. Sometimes the extra rod is used to move a second device on the opposite end from the main load, or is used to actuate limit switches or cam valves.

Spring Return, Single-Acting . This type is usually limited to very small, short stroke applications such as clamping or holding. The extra barrel length required to contain the return spring makes them impractical where a long stroke is required.

Ram-Type Single-Acting . These cylinders have only one fluid chamber. They are quite often mounted vertically working upward, and are retracted by load weight. They are practical for very long strokes. Sometimes they are called displacement cylinders and are used for applications like house jacks or filling station lifts.

Telescopic Cylinders . Although expensive as compared to standard cylinders, they are used where collapsed length must be shorter than could be obtained with a standard cylinder. They are available with up to 4 or 5 telescoping sleeves, and either in single-acting or double-acting types.

355

Calculation of Hydraulic Cylinder Force

EXAMPLE : A certain application requires a cylinder force of 25 tons. What should be the cylinder bore diameter used and at what gauge pressure?

SOLUTION: The required force is 25 tons s 2000 = 50,000 pounds. Refer to the Hydraulic Cylinder Force table on pages 356 and 357 which shows several combinations of piston diameter and PSI pressure which will produce 50,000 pounds of force or more. For example, a 6 inch piston will produce 56,550 pounds at 2000 PSI; a 7 inch piston will produce 57,725 lbs at 1500 PSI; an 8 inch piston will produce 50,265 lbs at 1000 PSI, a 10 inch piston will produce 58,900 lbs. at 750 PSI, etc. So there are many combinations which could be used, and the final choice is a matter of preference or of matching the pressure and flow capability of other components, particularly the pump.

In practice, choose a combination which will produce from 10% to 25% more than actually required by the load alone. This will provide a safety allowance which will take care of pressure losses in valves and piping, and mechanical losses in the cylinder.

EXAMPLE : How many pounds of force will be developed on the extension stroke of a 356M !bore cylinder operating at 1500 PSI? If this cylinder has a 16M !diameter!piston!rod,!how!much!force will be developed on the retraction stroke?

SOLUTION: Refer to the Hydraulic Cylinder Force table on pages 356 and 357. The chart shows 12,444 lbs. A solution can also be obtained by using the piston area (8.296 square inches) and multiplying by the pressure (1500 PSI); 8.296 square inches s 1500 PSI = 12,444 lbs.

On the retraction stroke the amount of force developed on the 2.41 square inch rod area must be subtracted: 12,444 3608 = 8836 lbs.

EXAMPLE : What!PSI!gauge!pressure!is!required!for!retraction!of!a!50,000!lb.!load!with!an!8!inch bore cylinder having a 4 inch diameter rod?

SOLUTION: The net piston area must be found which is the full piston area minus the rod area. 50.27 (piston area) 12.57 (rod area) = 37.7 square inches. PSI = 50,000 37.7 = 1326 PSI. The actual pressure will be slightly greater due to friction of the piston in the barrel.

Calculation of Hydraulic Cylinder Speed

EXAMPLE : At what speed would the piston of a 4 inch bore cylinder extend on an oil flow of 12 GPM?

SOLUTION: The table of Hydraulic Cylinder Speeds on pages 358 and 359 may be used or the speed figured with the formula which says that speed is equal to the incoming flow of oil in cubic inches per minute, divided by the square inch area of the piston. The speed will be in inches per minute.

A flow of 12 GPM is 231 s 12 = 2772 cubic inches per minute. The speed is 2772 (flow rate) 12.57 (piston area) = 220.5 inches per minute. This checks very closely with the value shown in the table on page 358.

EXAMPLE : Find the GPM flow necessary to cause a 5 inch bore cylinder to travel at a rate of 175 inches per minute while extending.

How fast would this cylinder retract on the same oil flow if it had a 2 inch diameter piston rod?

SOLUTION: Flow is determined by multiplying the piston area in square inches times the travel rate in inches per minute. This gives flow in cubic inches per minute. Divide by 231 to convert to GPM: 19.64 (piston area) s 175 = 3437 cubic inches per minute. 3437 231 = 14.88 GPM. This checks very closely with 15 GPM at 174 inches per minute shown on the chart on page 358.

To find the retraction speed on 14.88 GPM, the net piston area must be found. This is the full piston area minus the rod area: 19.64 (piston area) 6.5 (rod area) = 16.5 square inches. The flow rate is 3437 cubic inches per minute (equivalent to 14.88 GPM) 16.5 (net area) = 208 inches per minute. Note that this is faster than the extension speed on the same oil flow.

Hydraulic CylinderForce & Speed Calculations

356

Low Pressure Range 500 to 1500 PSI 1 56O" to 14" BoresCylinder forces, both extension and retraction, are shown in pounds. The chart on this page

covers cylinder operation in the pressure range of 500 to 1500 PSI, and the chart on the next page covers the 2000 to 5000 PSI range. Lines in bold type show extension force, using the full piston area. Lines in italic type show retraction force with various size piston rods.

Remember that force values are theoretical, derived by calculation. Experience has shown that probably 5%, but certainly no more than 10% additional pressure will be required to make up cylinder losses. For pressures not shown, the effective piston areas in the third column can be used as power factors. Multiply effective area times pressure to obtain cylinder force produced. (continued on next page)

Hydraulic Cylinder Force

BoreDia.,Ins.

156O

2

256O

3

356M

4

5

6

7

8

10

12

14

RodDia.,Ins.

None*5/81None*116None*11616MNone*11616MNone*1616M2None*16M2256ONone*2256O3356ONone*256O3356O4None*3356O4456O5None*356O4456O5556ONone*456O5556O7None*556ONone*7

Effec.Area,Sq.In

1.76721.46040.98183.14162.35621.65674.90874.12333.42382.50347.06866.28325.58374.66338.29586.81095.89055.154212.56710.1629.42547.658319.63516.49314.72612.56610.01428.27423.36521.20518.65315.70738.48531.41628.86425.91822.58118.85050.26640.64537.69934.36230.63126.50878.54062.63658.90554.78240.055113.1089.339153.94115.46

500 PSI

884730491

15711178828

245420621712125235343142279223324148340529452577628450814713382998188247736362835007

14,13711,68310,603

93277854

19,24315,70814,43212,95911,291

942525,13320,32318,85017,18115,31613,25439,27031,31829,45327,39120,02856,55044,67076,97057,730

750 PSI

13251095

7362356176712433682309225681878530147124188349862225108441838669425762270695744

14,72612,37011,045

94257511

21,20617,52415,90413,99011,78028,86423,56221,64819,43916,93614,13837,70030,48428,27425,77222,97319,88158,90546,97744,17941,08730,04184,82567,004

115,45586,595

1000 PSI

17671460

982314223561657490941233424250370696283558446648296681158915154

12,56710,162

94257658

19,63516,49314,72612,56610,01428,27423,36521,20518,65315,70738,48531,41628,86425,91822,58118,85050,26640,64537,69934,36230,63126,50878,54062,63658,90554,78240,055

113,10089,339

153,940115,460

1250 PSI

22091826122739272945207161365154428031298836785469805830

10,370851473636443

15,70912,70311,782

957324,54420,61618,40815,70812,51835,34329,20626,50623,31619,63448,10639,27036,08032,39828,22623,56362,83350,80647,12442,95338,28933,13598,17578,29573,63168,47850,069

141,375111,673192,425144,325

1500 PSI

2651219114734712353424857363618551363755

10,603942583766996

12,44410,216

88367731

18,85115,24314,13811,48729,45324,74022,08918,84915,02142,41135,04831,80827,98023,56157,72847,12443,29638,87733,87228,27575,39960,96856,54951,54345,94739,762

117,81093,95488,35882,17360,082

169,650134,009230,910173,190

Pressure Differential Across Cylinder Ports

*These figures are for extension force. No piston rod diameter is involved.

357

High Pressure Range 2500 to 5000 PSI 1 56O" to 14" Bores(continued from previous page) Values in two or more columns can be added for a pressure

not listed, or, force values can be obtained by interpolating between the next higher and the next lower pressure columns.

Pressure values along the top, of each chart are differential pressures across the two cylinder ports. This is the pressure to just balance the load, and not the pressure which must be produced by the system pump. There will be circuit flow losses in pressure and return lines due to oil flow, and these will require extra pressure. When designing a system, be sure to allow sufficient pump pressure, probably an extra 25% to 30% on the average, to supply both the cylinder and to satisfy system flow losses.

Hydraulic Cylinder Force

BoreDia.,Ins.

156O

2

256O

3

356M

4

5

6

7

8

10

12

14

RodDia.,Ins.

None*5/81None*116None*11616MNone*11616MNone*1616M2None*16M2256ONone*2256O3356ONone*256O3356O4None*3356O4456O5None*356O4456O5556ONone*456O5556O7None*556ONone*7

Effec.Area,Sq. In

1.76721.46040.98183.14162.35621.65674.90874.12333.42382.50347.06866.28325.58374.66338.29586.81095.89055.154212.56710.1629.42547.658319.63516.49314.72612.56610.01428.27423.36521.20518.65315.70738.48531.41628.86425.91822.58118.85050.26640.64537.69934.36230.63126.50878.54062.63658.90554.78240.055113.1089.339153.94115.46

2000 PSI

3534292119646283471233139817824768485007

14,13712,56611,167

932916,59213,62211,78110,30825,13420,32418,85115,31739,27032,98629,45025,13220,02856,54846,73042,41037,30631,41476,97062,83257,72851,83645,16237,700

100,53281,29075,39868,72461,26253,016

157,080125,272117,810109,564

80,110226,200178,678307,880230,920

2500 PSI

441836512455785458914142

12,27210,308

85606259

17,67215,70813,95911,66120,74017,02714,72612,88631,41825,40523,56419,14649,08841,23336,81531,41525,03570,68558,41353,01346,63339,26896,21378,54072,16064,79556,45347,125

125,665101,613

94,24885,90576,57866,270

196,350156,590147,263136,955100,138282,750223,348384,850288,650

3000 PSI

530243812945942570694970

14,72612,37010,271

751021,20618,85016,75113,99224,88720,43317,67215,46337,70130,48628,26622,97558,90549,47944,17837,69830,04284,82270,09563,61555,95947,121

115,45594,24886,59277,75467,74356,550

150,798121,935113,097103,086

91,89379,524

235,620187,908176,715164,346120,165339,300268,017461,820346,380

4000 PSI

706958423927

12,56694256627

19,63516,49313,69510,01428,27425,13322,33518,65733,18327,24423,56220,61750,26840,64837,70230,63378,54065,97258,90450,26440,056

113,09693.46084,82074,61262,828

153,940125,664115,456103,672

90,32475,400

201,064162,580150,796137,448122,524106,032314,160250,544235,620219,128160,220452,400357,356615,760461,840

5000 PSI

883673024909

15,70811,781

828424,54420,61717,11912,51735,34331,41627,91923,32241,47934,05529,45325,77162,83550,81047,12738,29298,17582,46573,63062,83050,070

141,370116,825106,025

93,26578,535

192,425157,080144,320129,590112,90594,260

251,330203,225188,495171,810153,155132,540392,700313,180294,525273,910200,275565,500446,695769,700577,300

Pressure Differential Across Cylinder Ports

*These figures are for extension force. No piston rod diameter is involved.

358

Figures in body of chart are cylinder piston speeds in inches per minute. Piston and rod diameters are in inches. Horizontal lines in bold type are extension speeds. Lines in italic type are retraction speeds for the bore and rod diameters shown in the first two columns, using net piston area.

For Fluid Flows from 1 to 20 GPM

Hydraulic Cylinder Speeds

PistonDiam.

156O

2

256O

3

356M

356O

4

5

6

7

8

10

RodDiam.

None*5/81

None*3/41

16None*

11616M

None*1

156O2

None*1616M2

None*156M16M2

26None*

156M16M2

256O26M

None*156O2

256O3

356ONone*

16M256O3

356O4

None*3

356O4

456O5

None*356O4

456O5

556ONone*

456O5

556O7

1 GPM

131158236748698

13947566792333744592834394524283236441820232530351213141618238.28.9101112156.07.48.08.910124.65.76.16.77.58.72.93.73.94.25.8

3GPM

3924757062212572944181411682022779811013117684

102118134728396

1071335561687490

1053539424755692527303337441822242731371417182023268.811121317

5GPM

654791

- - - -36842849069723528033746116318421829413917019622412013816017822292

10211412315117459657078921154145495462743037404551612328313438441518202129

8GPM

- - - -- - - -- - - -588684784

- - - -37644854073826129434947122327131435919222025628535614716318219624127994

10311212514718665717987991184859647182983745495460702430313446

12GPM

- - - -- - - -- - - -882

- - - -- - - -- - - -565672810

- - - -39244152370633440747153828833038442853422124427329436241814115516818822127798

107119131149176728896

1071231475568748190

1053544475169

15GPM

- - - -- - - -- - - -- - - -- - - -- - - -- - - -706840

- - - -- - - -49055165488241850958867236041348053566727630634136845252317619421023527634612313414816318622190110120134153184698592

1011131314455596387

20GPM

- - - -- - - -- - - -- - - -- - - -- - - -- - - -941

- - - -- - - -- - - -654735871

- - - -5576787848964805506407138903684074554906036972352592803143684611631791982182482941201471601782052459211412313415117459747884115

*These figures are for extension force. No piston rod diameter is involved.

359

Cylinder Speeds for Fluid Flows From 25 to 100 GPM

PistonDiam.

3

356M

356O

4

5

6

7

8

10

12

14

RodDiam.

None*1

None*1616M

None*156M16M2

None*156M16M2

256O26M

None*156O2

256O3

356ONone*

16M256O3

356O4

None*3

356O4

456O5

None*356O4

456O5

556ONone*

456O5

556O7

None*556O7

856ONone*

7856O10

25GPM

817919696848980600688800891460509568613754871294323350392460577204223247272310368150184200223256306115142153168189218749298

105144516577

10238505977

30GPM

980- - - -835

- - - -- - - -720825

- - - -- - - -551611682735905

- - - -35338842047155169224526829732737244118022124026730736813817018420222626188111118127173617893

12345607192

40GPM

- - - -- - - -- - - -- - - -- - - -960

- - - -- - - -- - - -735815909980

- - - -- - - -47151756062773592332735739543649558824029432035740949018422724526930234911814815716923182

103124164608095

123

50GPM

- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -919

- - - -- - - -- - - -- - - -- - - -588646700784919

- - - -40944649454561973530036840044651161323028430633637743614718419621128810212915520575

100119153

60GPM

- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -706776840941

- - - -- - - -49053659365474388236044148053561473527034136840345252317622123525334612315518624690

120143184

75GPM

- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -882970

- - - -- - - -- - - -- - - -613670741817929

- - - -450551600668767919345420460504566654221277294316433153194232307113150178230

100GPM

- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -- - - -817893989

- - - -- - - -- - - -600735800891

- - - -- - - -460568613672754871294369392422577204259310410150200238306

Interpolation of Cylinder Speed ChartsCylinder speed is directly proportional to GPM. To find speed at a flow not shown in charts,

add speeds in two columns. Example : Speed with 35 GPM is the sum of speeds in the 5 GPMand 30 GPM columns.

Calculation of Cylinder SpeedThese charts were calculated from the formula: S = CIM A, in which S is piston speed in

inches per minute; CIM is flow in cubic inches per minute; and A is cross sectional area of cavity being filled, in square inches. GPM flows must be converted to cubic inches per minute. Multiply GPM times 231.

Extension speeds are calculated with full piston area, retraction speeds with net area which is piston area minus rod area. (See Hydraulic Cylinder Force)

*These figures are for extension force. No piston rod diameter is involved.

360

The purpose of estimating air consumption of a cylinder is usually to find the HP capacity which must be available from the air compressor to operate the cylinder on a continuous cycling application.

Air consumption can be estimated from the table below. The consumption can then be converted into compressor HP.

Using the Table to Calculate Air ConsumptionFigures in the body of the table are air consumptions for cylinders with standard diameter

piston rods. The saving of air for cylinders with larger diameter rods is negligible for most calculations.

Air consumption was calculated assuming the cylinder piston will be allowed to stall, at least momentarily, at each end of its stroke, giving it time to fill up with air to the pressure regulator setting. If reversed at either end of its stroke before full stall occurs, air consumption will be less than shown in the table.

The first step in the calculation is to be sure that the bore size of the selected cylinder will just balance the load at a pressure of 75% or less of the maximum pressure available to the system. This leaves about 25% of available pressure which can be used to overcome flow losses through piping and valving. This surplus pressure must be available or the cylinder cannot travel at normal speed.

Determine the exact air pressure needed to just balance the load resistance. Add about 25% for flow losses and set the system regulator to this pressure. This is also the pressure figure which should be used when going into the table.

After determining the regulator pressure, go into the proper column of the table. The figure shown in the table is the air consumption for a 1-inch stroke, forward and return. Take this figure and multiply times the number of inches of stroke and by the number of complete cycles, forward and back which the cylinder is expected to make in one minute. This gives the SCFM for the application.

Cylinder Air Consumption per 1-inch Stroke, Forward and ReturnRegulator Outlet PSI (At Least 25% Above Load Balance PSI)

Pneumatic Cylinder Air Consumption

Cyl.Bore1.502.002.503.003.254.005.006.008.0010.012.014.0

60PSI.009.018.028.039.046.072.113.162.291.455.656.890

70PSI.010.020.032.044.053.081.128.184.330.516.7441.01

80PSI.012.022.035.050.059.091.143.205.369.576.8311.13

90PSI.013.025.039.055.065.100.159.227.408.637.9191.25

100PSI.015.027.043.060.071.110.174.249.447.6981.011.37

110PSI.016.029.047.066.078.119.189.270.486.7591.091.49

120PSI.017.032.050.070.084.129.204.292.525.8201.181.61

130PSI.018.034.054.076.090.139.219.314.564.8811.271.72

140PSI.020.036.058.081.096.148.234.335.602.9401.361.84

150PSI.021.039.062.087.102.158.249.357.6421.001.451.96

Converting SCFM Into Compressor HPCompression of air is an inefficient process because part of the energy is lost as heat of

compression and can never be recovered. By over-compressing the air and then reducing it to a lower pressure through a regulator, the system losses are increased. The amount of this loss is nearly impossible to calculate, but on the average system may amount to 5 or 10%. Also, there is a small loss due to flow resistance through the regulator.

After finding the SCFM to operate the cylinder, refer to the tables on page 391. Convert into HP according to the kind of compressor used. Add 5 to 10% for the miscellaneous losses described above. This should be very close to the actual HP capacity needed.

EXAMPLE : Find compressor HP needed to cycle an air cylinder through a 28-inch stroke, 11 times a minute, against a load resistance of 1000 lbs.

SOLUTION: A 4# bore cylinder working at 80 PSI would balance the 1000 lb. load. Add 25% more pressure (20 PSI) and set the pressure regulator at 100 PSI. From the above table, air consumption would be:

0.110 s28 (stroke) s11 (times a minute) = 33.88 SCFM

Refer to page 391. Assume a 2-stage compressor. At 100 PSI, 0.164 HP is required for each 1 SCFM. Total HP = 0.164 s 33.88 = 5.57 HP. Add 5% (or 0.278 HP) for miscellaneous losses. Total compressor HP = 5.57 + 0.278 = 5.848 HP.

361

Extension and Retraction 60 to 130 PSI Pressure RangeCylinder forces are shown in pounds for both extension and retraction. Lines in bold type

show extension forces, using the full piston area. Lines in italic type show retraction forces with various size piston rods. Remember that force values are theoretical, derived by calculation.

Pressures along the top of the chart do not represent air supply pressure; they are differential pressures across the two cylinder ports. In practice, the air supply line must supply another 5% of pressure to make up for cylinder loss, and must supply an estimated 25 to 50% additional pressure to make up for flow losses in lines and valving so the cylinder will have sufficient travel speed.

For good design and highest circuit efficiency, open the cylinder speed control valves as wide as practical and reduce the pressure regulator setting to as low a pressure as will give satisfactory cylinder force and speed.

For pressures not shown, use the effective areas in the third column as power factors, Multiply effective area times differential pressure to obtain theoretical cylinder force.

Pneumatic Cylinder Force

PistonDia.,In.

156O

16M

2

256O

3

356M

356O

4

5

6

7

8

10

12

14

RodDia.,In.

None5/81None5/8156MNone5/81None5/8116None116MNone11616MNone1None11616MNone116None1616MNone16None1616MNone16M2None2256ONone256O3

Effec.Area,Sq.in.

1.771.46.9852.412.101.183.142.832.354.914.604.123.437.076.284.668.307.516.825.899.628.84

12.5711.7811.0910.1619.6418.8518.1628.2726.7925.9038.4937.0150.2748.7947.9078.5476.1475.40113.1110.0108.2153.9149.0146.8

60 PSI

1068859

14412671

188170141295276247206424377280498451409354577530754707665610

11781131108916961607155223092220301629272872471245684524678665986491923489398810

70PSI

124102

69168147

83220198165344322289240495440326581526477413674618880825776712

13751320127119791875181126942590351934153351549853295278791776977573

107731042910278

80PSI

14211779

192168

95251227188393368330274565503373664601545472770707

1006943887813

15711508145222622143206930792960402239033829628360916032904887978655

123121191911747

90PSI

159132

89216189106283255212442414371308636565420747676613531866795

11311061

998915

17681697163425442411232834643331452443914308706968526786

1017998969737

138511340913215

100PSI

177146

98241210118314283235491460412343707628466830751681589962884

125711781109101619641885181628272679258638493701502748794786785476147540

113101099610819153901489914683

110PSI

195161108265231130345312259540506454377778691513913827750648

1058972

138212961219111821602074199731102946284542344071553053665265863983758294

124411209511901169291638916151

120PSI

212176118289252142377340283589552495411848754560996902818707

11551060150814151330122023572263217933923214310446194441603258545744942591369048

135721319512983184681787917620

130PSI

230190128313273154408368306638598536445919817606

1079977886766

12511149163415321441132125532451236036753482336250044811653563426222

1021098989802

147031429514075200071936919088

362

Internal Fluid PSI on TubingThis table is for use in selecting wall thickness of tubing. Figures in the body of the table

are internal fluid pressures in PSI that will produce a fiber stress of 10,000 PSI along the circumference, tending to rupture the tubing. If a tube is made of steel with an ultimate strength of 40,000 PSI, the safety factor would be 4.

TubeO.D.

156O

16M

2

256M

256O

26M

3

356M

356O

36M

4

456M

456O

46M

5

556O

6

656O

7

756O

8

856O

9

956O

10

1056O

The above table is calculated by Barlows formula: P = 2t sS O in which P is the internal pressure in PSI, t is wall thickness of tubing in inches, S is the fiber stress allowable in the tubing in PSI (a value of 10,000 PSI is used in the table), O is the outside diameter of the tubing in inches.

.120

1600

1371

1200

1067

960

873

800

738

686

640

600

565

533

505

480

436

400

369

.156

2080

1783

1560

1387

1248

1135

1040

960

891

832

780

734

693

657

624

567

520

480

446

416

.187

2493

2137

1870

1662

1496

1360

1247

1151

1069

997

935

880

831

787

748

680

623

575

534

499

468

.219

2920

2503

2190

1947

1752

1593

1460

1348

1251

1168

1095

1031

973

922

876

796

730

674

626

584

548

515

.250

3333

2857

2500

2222

2000

1818

1667

1538

1429

1333

1250

1176

1111

1053

1000

909

833

769

714

667

625

588

556

526

500

476

.313

4173

3577

3130

2782

2504

2276

2087

1926

1789

1669

1565

1473

1391

1318

1252

1138

1043

963

894

835

783

736

696

659

626

596

.375

5000

4286

3750

3333

3000

2727

2500

2308

2143

2000

1875

1765

1667

1579

1500

1364

1250

1154

1071

1000

938

882

833

789

750

714

.500

6667

5714

5000

4444

4000

3636

3333

3077

2857

2667

2500

2353

2222

2105

2000

1818

1667

1538

1429

1333

1250

1176

1111

1053

1000

952

.625

8333

7143

6250

5556

5000

4545

4167

3846

3571

3333

3125

2941

2778

2632

2500

2273

2083

1923

1786

1667

1563

1471

1389

1316

1250

1190

.750

7500

6667

6000

5455

5000

4615

4286

4000

3750

3529

3333

3158

3000

2727

2500

2308

2143

2000

1875

1765

1667

1579

1500

1429

.875

7000

6364

5833

5385

5000

4667

4375

4118

3889

3684

3500

3182

2917

2692

2500

2333

2188

2059

1944

1842

1750

1667

1.000

7273

6667

6154

5714

5333

5000

4706

4444

4211

4000

3636

3333

3077

2857

2667

2500

2353

2222

2105

2000

1905

Thickness of Tubing Wall, Inches

363

Piston Rod Column StrengthLong, slim piston rods may buckle if subjected to too heavy a push load. The table below

suggests the minimum diameter piston rod to use under various conditions of load and unsupported rod length, and is to be used in accordance with the instructions in the next paragraph. There must be no side load or bending stress at any point along the rod.

How to Use the Table . Exposed rod length is shown along the top of the table. This is usually somewhat longer than the actual stroke of the cylinder. The vertical scale, column 1, shows the load on the cylinder, and is expressed in English tons (1 ton = 2000 lbs.). If both the end of the rod and the FRONT end of the cylinder barrel are rigidly supported, a smaller rod will have sufficient column strength, and you may use, as Exposed Length of Piston Rod , one-half of the actual rod length.

For example, if the actual length is 80#, and if the cylinder barrel and rod end are supported as described, you could enter the table in the column marked 40. On the other hand if hinge mounting is used on both cylinder and rod (pin-to-pin), you may not be safe in using actual exposed rod length, and should use about twice the actual length. For example, if the actual length is 20#, you should enter the table in the 40# column.

Minimum Piston Rod Diameter

Figures in body of chart are suggested minimum rod diameters, in inches.

CAUTION: Hinge mounted cylinders, when mounted horizontally or at any angle other than vertical, create a bending stress on the rod when extended, due to cylinder weight. On large bore and/or long stroke hinge mounted cylinders, the trunnion mounting rather than tang or clevis mounting should be used, and the trunnion should be located in a position which will balance the cylinder weight when extended.

Tons

1/2

3/4

1

156O

2

3

4

5

756O

10

15

20

30

40

50

75

100

150

10

13/16

15/16

1

16QE

16

1556QE

2

26

26M

356

36M

46

56

20

5/8

11/16

3/4

7/8

1

156

156M

1(6QE

16M

2

2(6QE

26M

356

36M

46

56

40

3/4

13/16

7/8

15/16

1

156

16QE

16QE

1(6QE

16

1(6

256

256O

2(6

356M

3(6

46

56

60

1

156QE

156

16QE

156M

16

156O

16QE

16M

1(6

256

26

26M

3

36

4

456O

556O

70

156M

16

1(6QE

16QE

16

16M

1(6

2

256M

256O

26M

3

356O

4

46M

556O

80

16

156O

16QE

16

16M

1(6

2

256

26

26

2(6

356M

356O

45R

46M

556O

100

16QE

1(6

2

256

256M

2(6QE

2556QE

2(6

356M

356O

36M

46

4(6

56M

120

256M

26

256O

26M

3

356M

356O

36M

4

456O

5

6

Exposed Length of Piston Rod, Inches

364

Calculating Tie-Rod StretchThe amount of stretch (elongation) in

mechanical members under tension can be calculated very easily. The method described here can be applied to any structural member under stress. In fluid power design this might include the stretch in cylinder tie rods when the cylinder is mounted as in this illustration and subjected to internal fluid pressure, or when severe momentary shocks pass through hydraulic lines. It can be used to find the expected stretch in bolts holding a component to a manifold or holding SAE flange unions together or holding a pipe flange to the body of a hydraulic component. It can be used to estimate whether the tie rods or barrel of a foot mounted cylinder may distort under pressure, causing the piston to bind.

Gathering Data Collect the following data and you will be

ready to make a calculation:q , Length of stressed member, in inches.

w Cross sectional area of the stressed member, in sq. inches. e Total applied force, in lbs. r Modulus of elasticity of stressed member material from chart or from a machine handbook. There are two moduli of elasticity. One for torsion, the other for tension. Be sure to use the one for tension. Since the modulus of elasticity is only a ratio, it has no units.

Formula for Calculating Elongation

Elongation = (L s F) ( A s E)

In which L is length in inches; F is applied force in pounds; A is the cross section in square inches; and E is modulus of elasticity (from chart). Elongation will come out in inches.

Calculation Example for Cylinder Tie Rods PROBLEM : Find the stretch in the tie rods of a 6# bore cylinder operating at 3000 PSI; it has

4 steel tie rods of 1# diameter and they are 27# long.SOLUTION: First, get all the facts assembled. The piston area of a 6# bore cylinder is 28.27

sq. in. The tie rods are the stressed members and each rod has a sectional area of 0.785 sq. in, or a total of 3.14 sq. in. for all rods combined. The stressing force is 3000 PSI s 28.27 = 84,810 lbs. Modulus of elasticity is 30,000,000 (from chart). Now apply the formula:

Elongation = (27 s 84,810) ( 3.14 s 30,000,000) = 0.0243 inches.

How to Check for Over StressingThe stressed member will return to its normal length after the applied force has been removed

provided it was not stressed beyond its elastic limit. To determine if a given force will over stress a structural member, take the force value, in lbs., and divide by the cross sectional area, in sq. inches, of the member. This will give the stress value in PSI (lbs. per sq. inch). Compare this with the elastic limit of the material as given in machine handbooks. As in all mechanical designs, a suitable factor of safety should be allowed.

Pressure surges which pass through the system, even momentarily, may over stress and damage cylinders or other components. These surges may be generated when a 4-way valve shifts, when a hydraulic motor is suddenly stopped, when a hydraulic cylinder is allowed to decompress too rapidly, and by mechanical shocks or impacts against the cylinder piston transmitted from the load. The peak pressure of these surges can be determined by instrumentation. Sometimes they can be measured by a pressure gauge connected in series with a tight sealing check valve. The shock should be allowed to recur a dozen or so times before the gauge is read. The member can then be checked for over stressing as described above.

Tie rods stretch from fluid pressure on cylinders

MaterialAluminumBrassCast iron, grayCopperMagnesiumMonelNickelPhosphor bronzeSteel, carbonSteel, stainless

Modulus of Elasticity, E

10,000,00015,000,00012,000,00017,000,000

6,500,00026,000,00030,000,00016,000,00030,000,00028,000,000

Modulus of Elasticity forCommon Metals (in Tension)

365

Designing with CylindersStandard catalog cylinder models are not designed to take any appreciable side load on the

piston rod. They must be mounted so the rod is not placed in a bind at any part of the stroke. If the direction of the load changes during the stroke, hinge mounting must be used on both the rod end and the rear end. Use guides on the mechanism, if necessary, to assure that no side load is transmitted to the cylinder rod or piston.

Rod BucklingColumn failure or buckling of the rod may

occur if the cylinder stroke is too long relative to the rod diameter. The exact ratio of rod length to rod diameter at which column failure will occur cannot be accurately calculated, but the Piston Rod Column Strength table in this manual shows suggested safe ratios for normal applications.

Tension and Compression FailuresAll standard cylinders have been designed

with sufficiently large piston rods so failure will never occur either in tension

Rod Bearing FailureRod bearing failures usually occur when

the cylinder is at maximum extension, and occur more often on hinge or trunnion mount cylinders, in which the rear support point is located considerably behind the rod bearing. If space permits, it is wise to order cylinders with longer stroke than actually required, and not permit the piston to approach close to the front end while under full load.

Stop CollarOn those applications where it is necessary

to let the piston bottom out on the front end, the cylinder may be ordered with a stop collar. The stop collar should be especially considered on long strokes if the distance between supports exceeds 10 times the rod diameter, if the maximum thrust is required at full extension, and if the cylinder has rear flange, clevis, tang, or trunnion mounting.

15 F

EET

5 FE

ET

F2=2500LBS.

F1=7500 LBS.

F=15,000 LBS.

30o

45o 45o

45o

3536 LBS.

Cranes and BeamsSince the working angles on a

crane are constantly changing, it may be necessary to construct a rough model out of a sheet of paper, to exact scale, with cardboard arms and thumbtack hinge pins. This will show the point at which the greatest cylinder thrust is needed. An exact calculation can then be made for this condition.

Only that part of the cylinder thrust which is at right angles to the beam axis is effective for turning the beam. This can be calculated by the method shown elsewhere in this manual. For heavy beams, the beam weight will have to be considered in the calculations.

or compression, provided the cylinder is operated within the manufacturers pressure rating.

366

EXAMPLE : (Crane calculation using the figure on preceding page). Starting with a cylinder thrust, F, of 15,000 lbs., find the maximum load that can be lifted by the crane when the angles are as shown. First, translate the 15,000 lbs. into F1, 7500 lbs. at right angles to the beam using a power factor of 0.500 from the table on the next page, for a 30& angle. Next, translate this to F2, 2500 lbs. at the end of the beam where the weight is suspended. This is done with simple proportion by the length of each arm from the base pivot point. F2 is a third of F1 since the lever arm is 3 times as long. Next, find the maximum hanging load that can be lifted, at a 45& angle between beam and load weight, using a power factor for 450:

2500 lbs 0.707 = 3535 lbs .

Calculations for a Heavy Beam

5 FEET

21/2 FT71/2 FEET

9 FEET

15 FEET

750 LBS BEAM

WEIGHT

2250 LBS BEAM

WEIGHT

150 LBS. PER FOOT

FULCRUM

500 LBS

CYLINDER

CG2CG1

On heavy beams it is necessary to calculate not only for concentrated loads such as the suspended weights and cylinder thrust, but to figure in the weight of the beam itself. If the beam is uniform, so many pounds per foot of length, the calculation is relatively easy. In the example above the beam has a uniform weight of 150 lbs. per foot, is partially counterbalanced by a weight of 500 lbs. on the left side of the fulcrum, and must be raised by the force of a cylinder applied at a point 9 feet from the right side of the fulcrum.

The best method of solution is to use the principle of moments. A moment is a torque force consisting of so many pounds applied at a lever distance of so many feet or inches. The solution here is to find how much cylinder thrust is needed to just balance the beam. Then, by increasing the hydraulic cylinder thrust about 5 to 10% to take care of friction losses, the cylinder would be able to raise the beam.

Using the principle of moments, it is necessary to calculate all of the moment forces which are trying to turn the beam clockwise, then calculate all the moment forces trying to turn the beam counter-clockwise, then subtract the two. In this case they must be equal to balance the beam.

Clockwise moment due to the 15 feet of beam on the right side of the fulcrum: This can be considered as a concentrated weight acting at its center of gravity 756O feet from the fulcrum.

Moment = 150 (lbs. per foot) s15 feet s756O feet = 16,875 foot pounds .Counter-clockwise moment due to the 5 feet of beam on the left side of the fulcrum:

160 (lbs. per foot) s5 feet s256O feet (CG distance) = 1875 foot pounds .Counter-clockwise moment due to hanging weight of 500 pounds:

500 s5 feet = 2500 foot lbs .Subtracting counter-clockwise from clockwise moments:

16,875 1875 2500 = 12,500 foot pounds that mustbe supplied by the cylinder for balance condition .

To find cylinder thrust: 12,500 foot pounds 9 f