Cylinders and Actuators

Force and speed, the annulus side, regeneration, cushioning, buckling of long rods and sizing a cylinder.

Cylinder types

TypeDescriptionUse
Double-actingoil pushes in both directions; the rod side has less areamost machines
Single-actingoil pushes one way; a spring or the load returns itjacks, clamps, lifts
Telescopicnested stages for a long stroke from a short bodydump trucks, tippers
Tie-rod and weldedtie-rod: serviceable industrial (ISO 6020/6022); welded: strong, compact, mobile and heavy dutyindustrial and mobile
Ram (plunger)rod the same diameter as the piston, single acting, high thrustpresses, lifts
Rotary actuatorrack-and-pinion or vane; oscillating rotationturntables, valves, tilts

Force and speed

On the extend stroke oil acts on the full piston area; on the retract stroke only on the annular area Aann = Acap − Arod, so the pull is less and the speed, for the same flow, is higher:

Fout = p Acap    Fin = p Aann    v = Q ÷ A

Worked example: the press cylinder

Worked example: 80/45 mm at 160 bar and 30.2 L/min

QuantityExtendRetract
Area (cm²)50.2734.36
Force at 160 bar80.4 kN55.0 kN
Speed at 30.2 L/min0.100 m/s0.146 m/s
Time for 400 mm4.0 s2.7 s
  • The area ratio is 1.46; the return is 1.46 times faster, and the return pull is 68 percent of the push.
  • With a back pressure of 20 bar on the rod side (from a counterbalance valve or a restrictive return line) the net push falls to 80.4 − 6.9 = 73.6 kN.
  • A vertical load of 1 000 kg hanging on the rod needs only 28.5 bar on the rod side to hold it; a counterbalance valve stops the load running away.

Regenerative extension

To extend faster, the rod-side oil can be routed into the cap side along with the pump flow, so the pump flow is added to the returning oil. The net area that needs oil is just the rod area, so the speed rises (and the force falls) accordingly:

vregen = Q ÷ Arod    Fregen = p Arod

For the press: Arod = 15.90 cm², so v = Q ÷ Arod = 0.316 m/s (3.2 times faster) with a push of only 25.4 kN: a fast approach followed by a slow, strong pressing stroke by switching the regenerative valve off. Many presses use a separate small cylinder for rapid advance.

Cushioning and end-of-stroke loads

A moving mass carries kinetic energy that the cylinder must absorb at the end of stroke. Cushions throttle the exhausting oil over the last 20 to 40 mm to slow the piston. A 500 kg load at 0.3 m/s has KE = ½ m v² = 22.5 J; stopping it in 30 mm needs an average force of 22.5 ÷ 0.030 = 750 N, and a pressure spike on the cushion side if the cushion is badly adjusted. Over the maker’s energy rating, use an external shock absorber or decelerate with the valve (Valves and Flow Control).

Rod buckling

A long, slender rod in compression behaves like a column and can buckle under a load far below its yield. The Euler critical load is Pcr = π² E I ÷ (K L)², with I = π d⁴ ÷ 64, E = 210 GPa for steel and an end-fixity factor K (0.5 fixed-fixed, 0.7 fixed-pinned, 1 pinned-pinned, 2 fixed-free). The length L is the full extended length between the supports, not the stroke.

Piston rod buckling limit against stroke0.51.01.52.02.50100200300400500600700unsupported rod length (m)critical load (kN)cylinder force × safety factor 3 = 241 kNlimit 0.66 mP cr = π² E I ÷ (K L)², E = 210 GPa, rod 45 mm, K = 2 (one end free). Pinned ends (K = 1) allow twice thelength: 1.31 m here.
Critical buckling load of a 45 mm rod against its unsupported length.

Worked example: is the press rod safe?

  • Rod stress in compression: 80.4 kN ÷ 15.90 cm² = 51 MPa, well under a chrome rod’s yield of about 500 MPa.
  • Buckling, free at one end (K = 2), L = 0.4 m: Pcr = 652 kN, 8.1 times the 80 kN load: safe on a 400 mm stroke.
  • With a safety factor of 3 the load limit is reached at 0.66 m of unsupported length. A longer stroke needs a larger rod, a guided or pinned end (twice the length for K = 1), or a stop tube.

Sizing a cylinder

Worked example: choosing a bore

  • The job needs 50 kN at a working pressure of 140 bar (leave headroom below the 160 bar relief setting): A = 50 000 ÷ 14 000 000 = 35.71 cm², so D = 67.4 mm.
  • Round up to a standard bore (40, 50, 63, 80, 100, 125, 160 mm): 80 mm. At 160 bar it can push 80.4 kN, which leaves a margin for friction, back pressure and acceleration.
  • Choose the rod from the buckling check; for a push cylinder a ratio of rod to bore of 0.5 to 0.7 is common (the press’s 45/80 is 0.56).
  • Check the flow for the speed (Q = A v), the cushions for the load and the seal type for the temperature and the speed.

Related

FAQ

Why does a cylinder retract faster than it extends?

The rod takes up area on the rod side, so the same flow fills the smaller volume in less time.

What is stroke versus length?

Stroke is the travel; the extended length (stroke + closed length) is what matters for buckling.

What bore ratio is best?

A 2:1 area ratio (rod about 0.7 of the bore) gives equal speed in both directions with regeneration; most push cylinders use a thinner rod.

Do I need to cushion every cylinder?

Any cylinder that stops a load at speed needs a cushion or an external stop. Check the kinetic energy against the maker’s rating.

References

  • Esposito, A., Fluid Power with Applications, Pearson (cylinders and their sizing); Parker Hannifin, Hydraulic Cylinder Engineering Data.
  • ISO 6020 and ISO 6022 (mounting dimensions, 160 and 250 bar), ISO 3320 (bore and rod sizes) and ISO 4393 (stroke).
  • Timoshenko, S. P., Gere, J. M., Theory of Elastic Stability (the Euler formula).

Last reviewed: October 2026. Safety factors and friction are typical.