Hydraulic Cylinder Sizing: How to Select Bore, Rod and Stroke

Start with the force at the worst point of the stroke. For a vertical lift it is the weight, F = m × g; for a sliding load, F = μ × m × g; on an incline, F = m × g × (sin α + μ × cos α). Add the force to accelerate the load (F = m × a) and any process forces.

Through a lever, the rod force is the load force times the lever-arm ratio, divided by the sine of the angle between cylinder and lever; check both ends of the stroke.

Then add a margin — typically a factor of 1.2–1.5, towards the upper end for shock, uncertain friction or long hose runs — and divide by the mechanical efficiency explained below.

The working pressure is the pressure at the cylinder port while the load moves: the relief valve setting minus the losses in valves, hoses and fittings. Size on that figure, not on the pump’s maximum.

A higher pressure gives a smaller bore, less oil and a lighter cylinder; a lower pressure gives gentler control and longer seal life. Fenitsa FDH cylinders are rated for 160 bar (2,321 psi) and tested at 240 bar, with a 250 bar (3,626 psi) option when space is tight.

With the force in newtons and the pressure in N/mm² (1 bar = 0.1 N/mm²), the piston area is A = F / p and the bore D = √(4A / π). Round up to the next standard bore — never down — and recalculate the real force.

Forces are in metric tonnes-force (1 t = 9.81 kN). Bores above Ø200 mm are built to order. For pull force or other pressures, use the hydraulic cylinder calculator.

In tension the rod is rarely critical; in compression at full extension a long, slender rod behaves like a column. The stroke and the mounting — not the pressure — decide the rod size.

Use E = 210,000 N/mm² for steel. L is the free length at full extension — pin to pin for a cylinder pinned at both ends — and K the mounting factor below. Divide F_k by a safety factor (3.5 is common engineering practice); the result must be at least the maximum push force on the rod.

Real mountings are never perfectly rigid, so many engineers use case 2 (K = 1) for pinned cylinders and take no credit for case 4; a centre trunnion (CTM) lies between cases 1 and 2.

Stroke comes from the machine geometry: the travel the load needs, plus a small margin so the piston does not act as the machine’s end stop. Fenitsa FDH standard strokes are 100, 200, 300, 400 and 500 mm, with a maximum of 5,000 mm.

Speed follows from flow. Retraction is faster than extension, and more oil leaves the cap side than the pump delivers, so size the return line for it. Keep piston speed within 0.5 m/s, the FDH rating.

Example: a Ø100 mm cylinder extending at 0.1 m/s needs 47.1 L/min, or 12.6 kW of hydraulic power at 160 bar. The hydraulic flow rate calculator converts flow and speed for any size.

A hydraulic cylinder turns most, but not all, of its pressure into force. Seal friction typically leaves a mechanical efficiency of about 85–95 %. Losses are proportionally larger at low pressure, with tight new seals and at very low speed, where breakaway friction exceeds running friction and can cause stick-slip.

Volumetric losses in a healthy cylinder are close to zero. The bigger losses are pressure drops in valves and hoses — hence step 2 uses the pressure at the cylinder.

Example: a Ø100 mm cylinder at 160 bar has a theoretical push force of 125.7 kN; at a mechanical efficiency of 0.9 it delivers about 113.1 kN.

Task. A platform and its load weigh 5,000 kg and must be lifted 900 mm in 9 s. The cylinder is vertical and pinned at both ends (rear clevis and rod eye), with an extended pin-to-pin length of 2,000 mm. The power unit gives 160 bar at the cylinder. Safety factor 1.25, mechanical efficiency 0.9.

1 — Force. Load force = 5,000 × 9.81 = 49.03 kN; with the safety factor 61.29 kN; design force 61.29 / 0.9 = 68.10 kN.

2 and 3 — Pressure and bore. 160 bar = 16 N/mm², so A = 68,102 / 16 = 4,256 mm² and D = 73.6 mm. The next standard bore is Ø80 mm: 80.4 kN (8.20 t) at 160 bar.

4 — Rod. FDH rods for Ø80 mm are Ø45 and Ø56 mm; pinned at both ends, K = 1 and L_k = 2,000 mm. Ø45 mm: F_k = 104.3 kN, permissible 29.8 kN — less than the 61.3 kN factored load. Ø56 mm: F_k = 250.1 kN, permissible 71.5 kN — enough for the 61.3 kN factored load. The Ø56 mm rod is chosen.

5 — Speed and flow. v = 900 / 9 = 100 mm/s, so Q = A × v = 30.2 L/min. Lifting the load actually needs F_load / (A × η) = 108 bar, well inside 160 bar, and 108 × 30.2 / 600 = 5.4 kW of hydraulic power. Retracting with the same flow on the 25.6 cm² annulus gives 196 mm/s, within 0.5 m/s, while 59.1 L/min leaves the cap side.

Try your own numbers in the hydraulic cylinder calculator, then check the rod with the buckling table above.