Cylinder force per ISO 15552 with buckling check
Size double-acting pneumatic and hydraulic cylinders: bore, rod diameter, operating pressure and efficiency yield the theoretical and effective extend and retract forces. The calculator also determines the pressure required for a target force together with a suitable standard bore, performs the Euler buckling check of the piston rod for four mounting styles, and computes the air consumption per double stroke in standard litres for pneumatics. The standard series per ISO 15552 (pneumatics) and ISO 6020-2/6022 (hydraulics) are built in.
Cylinder calculator (pneumatics/hydraulics)
Model: double-acting cylinder with single-sided piston rod, static analysis. Back pressure on the exhaust/return side is included as a lump sum in the efficiency. Buckling check per Euler with a global safety factor, conservatively against the theoretical push force; bending due to dead weight of long horizontal cylinders is not covered. Dynamics and end-position cushioning are outside the scope of this calculator.
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Formulas and fundamentals
On extension the operating pressure acts on the full piston area A_K = π/4·D²; on retraction only on the annulus area A_R = π/4·(D² − d²), because the piston rod occupies part of the area. The theoretical force is pressure times area: F_D,th = p·A_K when pushing and F_Z,th = p·A_R when pulling, with 1 bar = 0.1 N/mm². A handy rule: F in N equals p in bar times A in cm² times 10. Seal and guide friction is covered by the efficiency: F_eff = F_th·η, with typical values of 0.85 to 0.95 for pneumatics and 0.90 to 0.95 for hydraulics.
If a target force is specified, rearranging the force equation gives the required pressure p_erf = F_target/(A_K·η). In addition, the calculator suggests the smallest standard bore whose effective push force reaches the target force at the set operating pressure. For pneumatics a traffic light rates the load ratio F_target/F_D,th: up to about 0.7 there is enough force reserve for a reliable, brisk motion; above that the cylinder becomes sluggish – in practice the bore is therefore chosen 25 to 50 percent larger than theoretically necessary.
The extended piston rod is a slender compression member and is checked for buckling per Euler. Conservatively, the calculator treats it as a strut with the rod diameter over the entire free length l between the mounting points. The mounting style sets the effective length factor β (pivot/pivot: β = 1, the standard case; rigid with free rod end: β = 2; rigid with pivoted load guidance: β = 0.707; rigid at both ends: β = 0.5, not recommended due to constraint forces). With L_k = β·l, the second moment of area I = π·d⁴/64 and E = 210,000 N/mm², the buckling force is F_K = π²·E·I/L_k². The safety factor S = F_K/F is evaluated against the theoretical push force without efficiency and should typically reach 3.5 to 5. This also yields the permissible free length l_zul and – given the dead length l_tot – the maximum permissible stroke. For slenderness ratios below about 90, Euler no longer strictly applies (Tetmajer range); the calculator flags this.
The air consumption of a double-acting pneumatic cylinder follows from the volume displaced per double stroke, V_DH = (A_K + A_R)·s. It is converted to the standard state via the absolute pressure ratio ε = (p + 1.013)/1.013 (isothermal compression, Boyle-Mariotte): q_DH = V_DH·ε in standard litres. With the cycle rate n and a dead-volume allowance of typically 5 to 10 percent for end-cap chambers, tubing and fittings, the consumption per minute is Q = q_DH·n·f_S – the basis for sizing valves and the compressor.
Worked example
A pneumatic cylinder of size Ø63/d20 per ISO 15552 operates at 6 bar gauge pressure with an efficiency of 0.9. The piston area is A_K = π/4·63² = 3117 mm², the annulus area A_R = π/4·(63² − 20²) = 2803 mm². This gives the theoretical push force F_D,th = 0.6 N/mm²·3117 mm² = 1870 N and the pull force F_Z,th = 1682 N – exactly the manufacturer's catalogue values. With η = 0.9, about 1683 N are effectively available on extension and 1514 N on retraction.
Buckling check: the cylinder is mounted via a rear pivot eye and a rod clevis (standard case pivot/pivot, β = 1); the free length with the rod fully extended is 500 mm. With I = π·20⁴/64 = 7854 mm⁴, the Euler buckling force is F_K = π²·210,000·7854/500² ≈ 65.1 kN. Against the theoretical push force of 1870 N the safety factor is S = 34.8 – far above the required buckling safety of 3.5. Conversely, at S = 3.5 the free length could be up to 1577 mm; only beyond that does the rod become critical.
Air consumption: at 100 mm stroke the cylinder displaces V_DH = (3117 + 2803) mm²·100 mm = 0.59 litres per double stroke. With the compression ratio ε = (6 + 1.013)/1.013 = 6.92 this amounts to q_DH = 4.10 standard litres per double stroke. At 10 double strokes per minute and a 5 percent dead-volume allowance, the consumption is about 43 sl/min.
Frequently asked questions
Why is the retract force smaller than the extend force?
On retraction the pressure acts only on the annulus area A_R = π/4·(D² − d²), because the piston rod occupies part of the piston area. The pull force is therefore smaller than the push force by the area ratio φ = A_K/A_R – about 5 to 20 percent for common standard pairs, considerably more for oversized hydraulic rods.
Which efficiency should I use?
Typical values from manufacturer literature: pneumatics 0.85 to 0.95 (default 0.9), hydraulics 0.90 to 0.95 on the piston side (default 0.95), rather 0.80 to 0.90 on the rod side; some manufacturers conservatively use 0.85. The efficiency covers seal and guide friction; small cylinders sit at the lower end, large ones at the upper end.
Which mounting style corresponds to which effective length factor?
A rear pivot eye or trunnion with a pivoted load is the standard case in practice (β = 1). A rigidly mounted cylinder with a free rod end is the worst case (β = 2). Rigid mounting with pivoted load guidance gives β = 0.707. Rigid at both ends (β = 0.5) is not recommended: alignment errors create constraint forces on rod and guide – the calculator shows a warning for this.
Why does the buckling check ignore the efficiency?
The check is conservatively performed with the theoretical push force. The friction described by the efficiency does not relieve the rod when full pressure is applied – for instance when driving against a hard stop. As an additional conservative assumption, the entire free length is treated as a strut with the rod diameter, although the cylinder tube is stiffer.
What does air consumption in standard litres mean?
The compressed air displaced in the cylinder is converted back to ambient conditions so that it can be compared with the compressor output. The conversion uses the absolute pressure ratio ε = (p + 1.013)/1.013. Some catalogues simplify with p + 1 bar – at 6 bar the difference is only about one percent.
Is this calculator sufficient for selecting a cylinder?
For the static sizing of force, bore, rod buckling and air consumption, yes. Not covered are dynamics and end-position cushioning, back pressure on the exhaust or return side (included as a lump sum in the efficiency), bending of long horizontal cylinders due to dead weight, and special designs such as single-acting or telescopic cylinders. For pneumatics the load-ratio traffic light helps: keep the load ratio below about 0.7 for brisk motion.
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