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Formulas and Data

How to compute tonnage requirements:

  1. General - When pressure per square inch is known:
    psi x area of work/2000 = 2 tons of ram force required
    Example: Where it is known that 100 psi is needed to do a job on a 5" x 8" wide piece.
    100 x 5" x 8"/2000 = 2 tons
  2. Press Fit - To determine the force required to press fit two round pieces together such as a shaft pressed into a bushing, use the following formula:
    F = D x ϖ x L x I x P/2
    F = force required in tons
    D = diameter of the part to be pressed in inches
    L = length of part to be pressed in inches (Note: the length of the interference fit only.)
    I = interference in inches (usually .002" to .006")
    P = pressure factor (See table below).
    Diameter (inches)Pressure FactorDiameter (inches)Pressure FactorDiameter (inches)Pressure FactorDiameter (inches)Pressure Factor

    Example: A steel shaft 2" in diameter pressed into a hole 3" long. The interference fit between the two diameters is .006".
    2" x 3.14 x 3" x .006" x (240/2) = 13.56 tons

  3. Punching - A quick guide to determine tonnage requirements for punching steel is:
    Diameter x thickness x 80 = tons (where 80 is constant for steel. Use 65 for brass.)
    Example: A 3" hole in .250" stock: 3" x .250" x 80 = 60 tons

    For noncircular holes, instead of the diameter, use 1/3 of the total length of cut.
    Example: A rectangular hole 4" x 6" in .250" stock: (4" + 6" + 4" + 6"/3) x .250" x 80 = 133.3 tons

  4. Deep Drawing - Deep-drawing calculations can be complex. The press, dies, material, radius, and part shape all have bearing. For drawing round shells, the following formula is a simple guide:
    C x T x Ts = tons
    C = circumference of the finished part; T = material thickness in inches; and
    Ts = tensile strength of the material.
    Example: To draw a 5" diameter cup of .040" stock with a tensile strength of 46,000 psi would require the following tonnage:
    (5 x 3.1416) x .040 x (46000/2000) = 14.44 tons
    A 20-ton press would be recommended
  5. Straightening - The pressure required to straighten a piece of metal depends on its shape. Below is an approximate formula with a further definition for different shapes.

    Where F is the ram force in tons; 6 is a constant; U is ultimate strength of the material in psi; Z is the section modulus (see below); and L is the distance between the straightening blocks in inches.

    Example: A 2" diameter shaft, 18" between the blocks, 100,000 psi ultimate strength.

    How to determine strokes per minute for a hydraulic press
    The number of strokes per minute for a hydraulic press is determined by calculating a separate time for each phase of the ram stroke. The rapid advance time is calculated, then the pressing time, (the work stroke); then, if there is no dwell time, the rapid return.

    The basic formula for determining the length of time in seconds for each phase of the stroke:

    Example: a hydraulic press with a 600 IPM rapid advance, 60 IPM pressing speed, and 600 IPM rapid return. The work requires a 3" advance, 1" work stroke, and 4" rapid return.

    60 ÷ 2.199 = 27 cycles per minute.

* Electrical actuation and valve shift time varies depending on the type of hydraulic circuit. One half second is a reasonable average figure.
1. These formulae are intended as guidelines only. Please consult a qualified manufacturing engineer for recommendations concerning your specific requirements.
2. Based on steel shaft and cast iron bushing (with OD/ID > 2).

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