Hydraulic cylinder calculator with piston rod buckling check
Size a double-acting hydraulic cylinder with a single piston rod: bore, rod diameter and operating pressure give the piston force when extending and the smaller pull force when retracting; the flow rate gives piston speeds, stroke times, the oil demand per stroke and cycle as well as the required pump drive power. The difference is the buckling check: the extended piston rod is a compression member, and that is exactly where long cylinders fail. The calculator performs it per Euler and Tetmajer, rates the buckling safety with a traffic light and reports the maximum permissible stroke.
Calculation
Double-acting cylinder with a single piston rod
Piston forces, speeds, stroke times, oil demand and drive power are calculated for the steady state. The buckling verification of the extended piston rod follows Euler and Tetmajer, with the mounting case providing the buckling length factor.
Model: double-acting cylinder with a single piston rod, static analysis without acceleration, friction or back-pressure terms. The efficiency η covers cylinder and drive losses together. The buckling check treats the piston rod conservatively as a straight bar of rod diameter over the full free length, centrically loaded, without initial deflection and without guidance forces from the cylinder tube.
Results
Calculating …
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Formulas and fundamentals
Piston force when extending and retracting
When extending, the operating pressure acts on the full piston area A_1 = π/4·D²; when retracting it acts only on the annulus area A_2 = π/4·(D² − d²), because the rod occupies part of the area. The effective piston force is pressure times area times efficiency:
The efficiency η covers seal and guide friction; typical values are 0.85 to 0.95. The ratio of the two forces is the area ratio φ = A_1/A_2: a Ø63 cylinder with a 36 mm rod pulls with only about two thirds of its pushing force. Rule of thumb: force in kN equals pressure in bar times area in cm² divided by 100.
Piston speed, stroke time and oil demand
The piston speed follows from the supplied flow rate and the area being pressurised:
Because the annulus area is smaller than the piston area, the cylinder retracts faster than it extends by the area ratio φ at the same flow rate. The oil demand per stroke is V = A·s and per cycle the sum of the extend and retract strokes - the basis for tank size, line cross sections and valve selection.
Hydraulic power and drive power
The hydraulic power is the product of flow rate and pressure; the drive power at the pump shaft is higher by the efficiency:
The numerical equation with the factor 600 comes from Tabellenbuch Metall and converts l/min and bar directly into kW.
Buckling verification of the piston rod
The extended piston rod is a slender, centrically loaded compression member. The governing state is the worst case: full extension at full operating pressure. The free buckling length is the distance between the mounting points with the rod extended, i.e. l = l_0 + s from the retracted mounting dimension and the stroke. The mounting case gives the buckling length factor β (fixed with a free end: β = 2; pinned/pinned: β = 1, the standard case for trunnion and rod eye mountings; fixed with a pinned end: β = 0.699; fixed at both ends: β = 0.5, to be avoided because misalignment introduces constraint forces). With the buckling length l_k = β·l and the radius of gyration i = d/4, the slenderness ratio is:
Above the limit slenderness λ_g the rod buckles elastically and Euler applies; below it the material yields first and Tetmajer applies:
The buckling stress gives the buckling force F_k = σ_k·A and the available safety S = F_k/F. The verification is run conservatively against the theoretical compressive force without efficiency, because the efficiency does not relieve the rod while the pressure is applied. Fluid power practice requires S = 3.5 to 5. The calculator additionally reports the maximum permissible stroke at which the required safety is just met and checks the compressive stress σ_d = F/A against the yield strength - for short, thick rods it is not buckling but crushing that governs.
Worked example
A double-acting hydraulic cylinder Ø63/d36 per ISO 6020-2 operates at 160 bar with an efficiency of 0.9. The piston area is A_1 = π/4·63² = 3117 mm², the annulus area A_2 = π/4·(63² − 36²) = 2099 mm², the area ratio φ = 1.48. With 160 bar = 16 N/mm² the theoretical pushing force is 16 N/mm²·3117 mm² = 49.9 kN, effectively 44.9 kN when extending and 30.2 kN when retracting with η = 0.9.
At a flow rate of 20 l/min (= 333,333 mm³/s) the piston extends at v_1 = 333,333/3117 = 107 mm/s and retracts at v_2 = 333,333/2099 = 159 mm/s. For a 600 mm stroke that is 5.6 s to extend and 3.8 s to retract, 9.4 s per cycle. The displaced volume is 1.87 litres extending and 1.26 litres retracting, i.e. 3.13 litres per cycle. The hydraulic power is 20·160/600 = 5.33 kW, the required drive power 5.33/0.9 = 5.93 kW.
Buckling check: the cylinder is mounted on a trunnion with a rod eye (pinned/pinned, β = 1). With a retracted mounting dimension of 250 mm and a 600 mm stroke, the free length is l = 850 mm and the buckling length the same. The E335 rod has i = d/4 = 9 mm, giving λ = 94.4 - above the limit slenderness λ_g = 89, hence the Euler range: σ_k = π²·210,000/94.4² = 232 N/mm². With A = 1018 mm² this gives F_k = 237 kN and, against the theoretical compressive force of 49.9 kN, a safety of S = 4.7, above the required 3.5. The verification holds up to a stroke of 739 mm; beyond that the rod becomes critical and the next larger rod diameter is chosen.
Frequently asked questions
How do you calculate the piston force of a hydraulic cylinder?
The piston force is pressure times effective area times efficiency: F = p_e·A·η. When extending the effective area is the piston area A_1 = π/4·D², when retracting the annulus area A_2 = π/4·(D² − d²). With 1 bar = 0.1 N/mm², a Ø63 cylinder at 160 bar delivers 49.9 kN in theory and about 44.9 kN effectively with η = 0.9.
Why is the pull force when retracting smaller than the push force?
When retracting, the pressure only acts on the annulus area because the piston rod occupies part of the piston area. The ratio of the two forces is the area ratio φ = A_1/A_2. For the standard pairings per ISO 6020-2 it ranges from roughly 1.2 to 2.0 depending on rod diameter; the cylinder retracts faster by exactly the same factor.
What is the free buckling length and how do I find the mounting dimension?
What counts for the buckling check is the distance between the two mounting points with the rod fully extended, not the cylinder length. It is the retracted mounting dimension l_0 (the catalogue distance between the mounting points) plus the stroke s. That is why every additional millimetre of stroke degrades the buckling check quadratically as long as the Euler range applies.
Which buckling safety factor is usual for hydraulic cylinders?
Fluid power practice uses S = 3.5 to 5, and many manufacturers base their charts on 3.5. Choose the upper value for shock loading, off-centre loads or uncertain guidance. The safety refers to the buckling force, not to the yield strength; the compressive check of the rod is performed in addition.
When does Euler apply and when does Tetmajer?
Euler is valid only in the elastic range, i.e. above the limit slenderness λ_g (89 to 105 for structural steels depending on grade). Below it the rod reaches the yield point before buckling elastically - there the Euler formula returns values that are too high and Tetmajer is used instead (straight line σ_k = a − b·λ). Short, thick piston rods are almost always in the Tetmajer range; the calculator selects the range automatically and reports it.
How large does the pump for the cylinder have to be?
The flow rate follows from the desired piston speed: Q = v·A_1 when extending. The hydraulic power is P = Q·p_e (with Q in l/min and p in bar: P in kW = Q·p/600) and the drive power at the pump shaft P/η. Note that when retracting the cylinder pushes the larger piston-side volume back to the tank - size the return line accordingly.
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