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Fluid Power
Standard Equation

Hydraulic Cylinder Push & Pull Force Formula (F = P·A·η)

Calculate extension thrust force, retraction pull force, effective annulus area, and speed ratios for single-rod double-acting hydraulic cylinders.

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Primary Mathematical Expression

F = P * A * \eta

Nomenclature & Variables

SymbolVariable NameMetric UnitImperial UnitDescription
FCylinder output forcekNlbfThe mechanical thrust (push) or tension (pull) force generated by fluid pressure.
POperating pressurebarpsiThe working fluid pressure supplied by the hydraulic pump.
d_{bore}Cylinder bore diametermminThe inside diameter of the cylinder tube.
d_{rod}Piston rod diametermminThe diameter of the piston rod.
AEffective pressurized areamm²in²The cross-sectional area acting against hydraulic fluid.
\etaMechanical efficiency--Seal friction and mechanical transmission efficiency (typically 0.90 - 0.95).

Step-by-Step Derivation

  1. 1

    Hydraulic force originates from Pascal's Principle: fluid pressure applied to an enclosed surface exerts a uniform perpendicular force per unit area: F = P * A.

  2. 2

    For cylinder extension (Push Stroke), the entire bore surface area is exposed to pressurized fluid: A_{bore} = \pi * d_{bore}² / 4.

  3. 3

    The ideal push force is F_{push, ideal} = P * A_{bore}. Accounting for piston seal friction: F_{push} = P * (\pi * d_{bore}² / 4) * \eta.

  4. 4

    For cylinder retraction (Pull Stroke), the piston rod reduces the available pressurized area to an annular ring: A_{annulus} = A_{bore} - A_{rod} = \pi * (d_{bore}² - d_{rod}²) / 4.

  5. 5

    The actual pull force is F_{pull} = P * [\pi * (d_{bore}² - d_{rod}²) / 4] * \eta.

  6. 6

    Therefore, pull force is always strictly less than push force by the ratio (1 - d_{rod}² / d_{bore}²).

Worked Example Calculation

Problem Statement

Calculate the push force and pull force of a hydraulic cylinder with a bore diameter of 100 mm and a rod diameter of 50 mm operating at 200 bar with 95% mechanical efficiency.

Calculation Steps
  • •Calculate Piston Area: A_{bore} = \pi \times 100² / 4 = 7,853.98 mm².
  • •Calculate Annulus Area: A_{annulus} = \pi \times (100² - 50²) / 4 = 5,890.49 mm².
  • •Convert pressure: 200 bar = 20 MPa = 20 N/mm².
  • •Compute Push Force: F_{push} = (20 N/mm² \times 7,853.98 mm² \times 0.95) / 1,000 = 149.23 kN.
  • •Compute Pull Force: F_{pull} = (20 N/mm² \times 5,890.49 mm² \times 0.95) / 1,000 = 111.92 kN.
Final ResultPush Force = 149.23 kN, Pull Force = 111.92 kN

Engineering Assumptions

  • •Hydraulic fluid is virtually incompressible.
  • •Pressure is distributed uniformly across the entire active piston and annulus areas.
  • •Mechanical seal drag accounts for approximately 5% frictional loss (\eta = 0.95).

Design Limitations

  • •Does not account for dynamic pressure surges or back-pressure on the discharge port.
  • •Rod column buckling is estimated using Euler's ideal column formula as a first-order screening check. Detailed standards-certified structural buckling analysis for production hydraulic cylinders (accounting for stop tubes, guide clearances, cylinder/rod step-changes, and mounting misalignment) requires ISO/TS 13725.

Academic References & Standards

ISO 6020/2standard

Fluid power cylinders - Mounting dimensions for single rod cylinders, 160 bar compact series

International standard for industrial single-rod hydraulic cylinders.

NFPA/T3.6.7 R2standard

Fluid power systems - Cylinder dimensions and column strength

National Fluid Power Association standard.

ISO/TS 13725standard

Hydraulic fluid power — Method for evaluating the buckling load of an oil-hydraulic cylinder

Standard method for evaluating compressive buckling loads in hydraulic cylinders.