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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
Where:
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 Factor Diameter (inches) Pressure Factor Diameter (inches) Pressure Factor Diameter (inches) Pressure Factor 1 500 3 156 5 91 7 64 1¼ 395 3¼ 143 5¼ 86 7¼ 61 1½ 325 3½ 132 5½ 82 7½ 59 1¾ 276 3¾ 123 5¾ 78 7¾ 57 2 240 4 115 6 75 8 55 2¼ 212 4¼ 108 6¼ 72 2½ 189 4½ 101 6½ 69 2¾ 171 4¾ 96 6¾ 66

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
Where:
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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