Engineering

A Hydraulic cylinder (also known as linear hydraulic motor) is a mechanical actuator used to give a unidirectional force through a unidirectional stroke. It has many applications, notably in construction equipment and engineering vehicles, manufacturing machinery and civil engineering.

Hidromec thanks to the expertise of his staff is able to invent, design, build, maintain and improve cylinders for your individual applications.

Engineering is a technique of execution and control of compliance: a difficult result to achieve. A proven and certificated ISO9001model which splits into several phases: it starts from the verification and control of the materials of each single component to the analysis of the surface treatments (coatings) up to Reverse Engineering of subcomponents made in order to reach a products and subcomponents quality standard and create a conformity database.


  • TECHNICAL STAFF IN A REVIEW MEETING
  • TECHNICAL DEPARTMENT
  • DIMENSIONAL INSPECTION & COMPLIANCE CHECK
  • SURFACE ROUGHNESS & CHROME PLATING THICKNESS TESTING
  • TECHNICAL STAFF IN A REVIEW MEETING
  • TECHNICAL DEPARTMENT
  • DIMENSIONAL INSPECTION & COMPLIANCE CHECK
  • SURFACE ROUGHNESS & CHROME PLATING THICKNESS TESTING


Buckling
Buckling without side loading

Special problems to do with stability occur when cylinders with long stroke lengths are used.
For the purpose of calculation these cases are divided into areas:

 Non elastic buckling loads (Tetmajer’s calculation) andn
 Elastic or Hook’s buckling loads (its critical limiting load is determined by Euler’s equation)
n

In hydraulic cylinder Euler’s calculation is basically the calculation used, as the piston rod may usually be considered to be a slender strut (negligible diameter).

Buckling load and operation load are then calculated as follows:

Buckling load K= ………In N (1)
i.e. the rod buckles under this load
 
Max. operating load F = K/S In N (2)
Sk = free buckling length in mm
E = modulus of elasticity (2.1 x 10 * for steel ) in N/mm

J = moment of inertia for circular cross-section in mm* …..
S = safety factor (3.5)

The length to be used as the free buckling length may be determined from the Euler loading cases (see table 4). In order to simplify the calculation the stiffening due to the cylinder tube is ignored. This provides the required safety margin in standard cylinders, the installation position of which is usually not known, in order to cater for any superimposed bending loads..

Table 4: Euler’s loading cases

Euler’s loading case

Case 1 One end free, one end rigidly connected.

Case 2 (Basis case) Two ends pivoted.

Case 3 One end pivoted, one end rigidly connected.

Case 4 Two ends rigidly connected.

Illustration

Free buckling length

Installation position for cylinder









Note


Mounting type FA,FB,LA,LB, MF3, ME7,MF4,ME8,MS2

Mounting type TA,TC,CA,CB,MP3,MP1,
MT4


Mounting typeFA,FB,LA,LB,MF3,ME7,
MF4,ME8,MS2

Load must be carefully
guided, or else possible bracing.


Mounting type FA,FB,LA,LB,MF3,ME7,
MF4,ME8,MS2

Not suitable, as
 
bracing is to be
expected.

 

 

Hydraulic force - area formulas and calculator

Hydraulic force - area formulas and calculator

hydraulic cylinder force

The force produced by a double acting hydraulic piston on the rod side can be expressed as

F1 = (π (d22 - d12) / 4) P1         (1)

where 

F1 = rod pull force (lb, N)

d1 = rod diameter (in, m)

d2 = piston diameter (in, m)

P1 = pressure in the cylinder (rod side) (lff/in2 (psi), N/m2 (Pa))

The force produced opposite the rod can be expressed as

F2 = (π d22 / 4) P2         (2)

where 

F2 = rod push force (lb, N)

P2 = pressure in the cylinder (opposite rod) (lff/in2 (psi), N/m2 (Pa))

Push Diagram

Rod pushing force for hydraulic cylinders are indicated below: 

hydraulic cylinder push force diagram

  • 1 psi (lb/in2) = 144 psf (lbf/ft2) = 6,894.8 Pa (N/m2) = 6.895x10-3 N/mm2 = 6.895x10-2 bar
  • 1 N/m2 = 1 Pa = 1.4504x10-4 lb/in2 = 1x10-5 bar = 4.03x10-3 in water = 0.336x10-3 ft water = 0.1024 mm water = 0.295x10-3 in mercury = 7.55x10-3 mm mercury = 0.1024 kg/m2 = 0.993x10-5 atm
  • 1 lbf (Pound force) = 4.44822 N = 0.4536 kp
  • 1 N (Newton) = 0.1020 kp = 7.233 pdl = 7.233/32.174 lbf = 0.2248 lbf = 1 (kg m)/s2 = 105 dyne = 1/9.80665 kgf
  • 1 in (inch) = 25.4 mm
  • 1 m (meter) = 39.37 in = 100 cm = 1000 mm

Pull Diagram

Rod pulling force for hydraulic cylinders are indicated below:

hydraulic cylinder pull force diagram

Piston Rod Size Selection

Piston Rod Size Selection

The selection of a piston rod for thrust (push) conditions requires the following steps to be carried out:

1. Determine the type of cylinder mounting style and rod end connection to be used. Consult the Stroke Factor table and determine which factor corresponds to the application.
2. Using this stroke factor, determine basic length from the equation:
Basic Length = Net Stroke x Stroke Factor
( The Piston Rod Selection Chart, below, is prepared for standard rod extensions beyond the face of the gland retainer. For rod extensions greater than standard’ add the increase to the stroke to arrive at the basic length’)
3. Find the load imposed for the thrust application by multiplying the full bore area fo the cylinder by the system pressure, or by referring to the Push and Pull Force charts.
4. Using the Piston Rod Selection Chart below, look along the values of ‘ basic length ’ and ‘ thrust ‘ as found in 2. and 3. above, and note the point of intersection.

The correct piston rod size is read from the diagonally curved line labeled ‘ Rod Diameter ‘ above the point of intersection.


Long Stroke and Stop Tubes

When considering the use of long stroke cylinders, the piston rod should be of sufficient diameter to provide the necessary column strength.
For tensile(pull)loads, the rod size is selected by specifying standard cylinders with standard rod diameters and using them at or below the rated pressure.


For long stroke cylinders under compressive loads, the use of stop tubes should be considered, to reduce bearing stress.
Selection of a stop tube is described.

 

CALCULATION OF CYLINDER DIAMETER

CALCULATION OF CYLINDER DIAMETER

Given that the load and operating pressure of the system are known, and that a piston rod size has been estimated taking account of whether the rod is in tension(pull) or compression (push), then the cylinder bore can be selected.
If the piston rod is in compression, use the ‘Push Force’ table below.

1. Identify the operating pressure closest to that required.
2. In the same column, identify the force required to move the load (always rounding up).
3. In the same row, lock along to the cylinder bore required.

If the cylinder envelope dimensions are too large for your application increase the operating pressure, if possible, and repeat the exercise.

Push Force

Bore φ

Cylinder Bore Area

Cylinder Push Force in kN

10 Bar

40 Bar

63 Bar

100 Bar

125 Bar

160 Bar

40

1257

1.3

5.0

7.9

12.6

15.7

20.1

50

1964

2.0

7.9

12.4

19.6

24.6

31.4

63

3118

3.1

12.5

19.6

31.2

39.0

49.9

80

5027

5.0

20.1

31.7

50.3

62.8

80.4

100

7855

7.9

31.4

49.5

78.6

98.2

126

125

12272

12.3

49.1

77.3

123

15.3

196

160

20106

20.1

80.4

127

201

251

322

200

31426

31.4

126

198

314

393

503

250

49087

49.1

196

309

491

614

785

320

80425

80.4

322

507

804

1005

1287

If the piston rod is in tension, use the ‘Deduction for Pull Force’ table. The procedure is the same but, due to the reduced piston surface area resulting from the piston rod, the force available on the ‘pull’ stroke will be smaller, To determine the pull force.

1. Follow the procedure given for ‘push’ applications as described above.
2. Using the ‘Deduction for Pull Force’ table below, establish the force indicated according to the rod diameter and pressure selected.
3. Deduct this from the original ‘Push’ force. The resultant is the net force available to move the load.


If this force is not large enough, go through the process again but increase the system operating pressure or cylinder diameter if possible. If in doubt, Our design engineers will be pleased to assist.

Deduction for Pull Force

Piston Rod φ

Piston Rod Area

Reduction in Force

10 Bar

40 Bar

63 Bar

100 Bar

125 Bar

160 Bar

22

380

0.4

1.5

2.4

3.8

4.8

6.1

28

616

0.6

2.5

3.9

6.2

7.7

9.9

36

1018

1.0

4.1

6.4

10.2

12.7

16.3

45

1590

1.6

6.4

10.0

15.9

19.9

25.5

56

2463

2.5

9.9

15.6

24.6

30.8

39.4

70

3848

3.8

15.4

24.2

38.5

48.1

61.6

90

6362

6.4

25.5

40.1

63.6

79.6

102

110

9503

9.5

38.0

59.9

95.1

119

152

140

15394

15.4

61.6

97.0

154

193

246

180

25447

25.4

102

160

254

318

407

220

38013

38.0

152

240

380

475

608

 

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