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- - 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
- - 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
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Where
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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)
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Pressure
Factor
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Diameter
(inches)
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Pressure
Factor
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Diameter
(inches)
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Pressure
Factor
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Diameter
(inches)
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Pressure
Factor
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1
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500
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3
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156
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5
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91
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7
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64
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1¼
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395
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3¼
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143
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5¼
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86
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7¼
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61
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1½
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325
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3½
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132
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5½
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82
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7½
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59
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1¾
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276
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3¾
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123
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5¾
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78
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7¾
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57
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2
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240
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4
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115
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6
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75
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8
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55
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2¼
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212
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4¼
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108
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6¼
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72
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2½
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189
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4½
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101
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6½
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69
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2¾
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171
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4¾
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96
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6¾
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66
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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
- - 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
- - 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
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Where
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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
- - 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.
: A 2" diameter shaft, 18" between the blocks,
100,000 psi ultimate strength.
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:
: 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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