Cylinder force per ISO 15552 and ISO 6020-2 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 buckling check of the piston rod per Euler and Tetmajer for four mounting styles along with the compression check against the yield strength, 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, live with every input.
Calculation
Model: double-acting cylinder with single-sided piston rod, static analysis. Back pressure on the rear side is included as a lump sum in the efficiency. Buckling check with a global safety factor per Euler and Tetmajer, capped at the yield strength, evaluated against the theoretical push force; plus the compression check σ_d = F/A against R_e. The piston rod is treated as a straight, centrically loaded strut without initial deflection and without guidance forces from the cylinder tube. Bending due to dead weight of long horizontal cylinders is not covered. Dynamics are outside the scope of this calculator. Air consumption is calculated isothermally (Boyle-Mariotte); end-position cushioning is outside the scope of this calculator.
Results
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Formulas and fundamentals
Theoretical and effective force
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:
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:
Required pressure for the target force
If a target force is specified, rearranging the force equation gives the required pressure:
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.
Buckling check per Euler and Tetmajer
The extended piston rod is a compression member and is checked for buckling. 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 and the radius of gyration i = d/4 the slenderness ratio is λ = L_k/i. Which formula applies is decided by the limiting slenderness λ_g of the selected rod material:
The switch does not happen at the tabulated λ_g but at the intersection of the two curves: for E295/E335/S355 it lies at λ = 85.8, for S235 at λ = 104.1. λ_g and the Tetmajer coefficients come from two different sources and do not match everywhere; between 85.8 and 89 the line would sit up to 6.9 % above the Euler hyperbola. No buckling stress can exceed it, because in the inelastic range σ_k = π²·T_K/λ² applies with the buckling modulus T_K < E. Within this narrow band the calculator is therefore up to 6.9 % on the safe side compared with a hand calculation that switches rigidly at λ_g = 89.
In both ranges the buckling stress is capped at the yield strength R_e - there the rod squashes instead of buckling. The buckling force is F_K = σ_k·A with A = π/4·d²; 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_allow and - given the dead length l_dead - the maximum permissible stroke. Because no closed-form inverse exists in the Tetmajer range, the calculator searches for l_allow on the very curve it also plots.
Compression check of the piston rod
A stocky rod yields long before buckling governs. A second criterion with its own traffic light therefore sits next to the buckling check:
How much this matters is shown by a case built from nothing but standard values: pair 32/14 per ISO 6020-2 at 400 bar with 150 mm free length gives λ = 43. Pure Euler reported a safety factor of 5.4 there and showed green, while the rod carries a compressive stress of 209 N/mm² - with E335 the buckling safety is 1.5 instead of the required 3.5.
Air consumption per double stroke
The air consumption of a double-acting pneumatic cylinder follows from the volume displaced per double stroke:
It is converted to the standard state via the absolute pressure ratio ε = (p + 1.013)/1.013 (isothermal compression, Boyle-Mariotte):
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 and the rod is made of E335. The buckling length is L_k = 500 mm, the slenderness ratio λ = 500/(20/4) = 100 and therefore above the limiting slenderness λ_g = 89 - Euler applies with σ_k = π²·210,000/100² = 207.3 N/mm². With the rod cross-section A = π/4·20² = 314 mm² the buckling force is F_K = 207.3·314 ≈ 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. The compression check is uncritical: σ_d = 1870/314 = 5.95 N/mm² against R_e = 335 N/mm² gives S_d = 56.
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.
Why do I have to specify the rod material?
Because it decides which formula applies at all. Euler describes elastic buckling only, above the limiting slenderness λ_g, and λ_g depends on the material: 105 for S235, 89 for E295/E335/S355. Below that the rod never reaches the Euler buckling force because the material yields first - the Tetmajer line applies there. The material also supplies the yield strength R_e for capping the buckling stress and for the compression check. Only materials with a documented yield strength to EN 10025-2 are offered here; the 16NiCr4 of the Tetmajer table is not among them because no source states a yield strength for it (it remains available in the buckling calculator, where R_e is entered freely). The default is E335: in practice a hard-chrome-plated piston rod is made of quenched and tempered steel (C45E, 20MnV6, 42CrMo4) with a considerably higher yield strength, so the calculation stays on the safe side.
When is compression governing rather than buckling?
For stocky rods, that is a short free length with a large diameter. An example using nothing but standard values: pair 32/14 per ISO 6020-2 at 400 bar with 150 mm free length gives λ = 43. A pure Euler check reported a safety factor of 5.4 there and showed green, although the rod carries a compressive stress of 209 N/mm². With Tetmajer and E335 the buckling safety is 1.5 instead of the required 3.5, and the compression check S_d = 335/209 = 1.6 is only just above the standard safety factor of 1.5. The calculator therefore runs both checks with their own traffic light and additionally reports which range was used.
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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