This section is from the book "Machines And Tools Employed In The Working Of Sheet Metals", by R. B. Hodgson. Also available from Amazon: Machines and tools employed in the working of sheet metals.
Although shearing and punching may be considered a true cutting action, it is of a rough character. It has been previously explained that at times a punch and die is made flat, at other times concave or convex.
When a punch is being forced through a metal plate, the first effect of the application of the punch is the crushing of the top surface of the plate within the circle of the hole about to be formed. The tendency of the punch is to draw or drag down the metal surrounding the punch. This dragging down of the metal is prevented by the bottom die, which supports the metal surrounding the punch. A certain amount of compression takes place, which varies according to the temper and thickness of the metal that is being operated upon, stresses are set up, and rupture occurs where the stresses have l>een greatest. When punching thin metals, the action is somewhat different to that of cutting the thicker metals. In the case of thin metal the action may be considered instantaneous, and the top edge of the hole in the scrap metal will appear to be perfectly square, whereas in the case of the thicker metals the shearing action is brought about in a more gradual manner, a series of successive ruptures taking place at various depths in the thickness of the metal plate, until the final shearing occurs. On this account it is difficult to say what actual force or stress is required to cut a given blank, unless experiments are made. But if the ultimate shearing strength of the metal is considered in the calculation, it is possible to arrive at approximate results that will meet all practical requirements. The shearing strength of cold rolled steel may be taken as 60,000 lb. per square inch, mild steel 50,000 lb., wrought iron 45,000 lb., and soft brass 30,0001b. To simplify the calculations the above figures may be reduced to tons, so that to shear off wrought-iron bars 1 square inch area a pressure of 20 tons will be required at the punch. In a previous article some figures were given from Dr. Anderson's"Strength of Materials."Another interesting set of figures taken from the same source will be found in the following table, which is a summary of some experiments relating to the shearing of wrought-iron bars.
Size of bar in inches. | Stress with bar laid flat upon its side, in tons per square inch. | Stress with bar laid on edge, in tons per square inch. | Percentage of less stress required when detruted on flat surface. |
3X1 1/2 | 18.2 | 20.1 | 10 |
4 1/2 x 1 3/8 | 14.3 | 17.9 | 20 |
3X1 | 15.7 | 21.1 | 26 |
5 1/4 X 1 3/4 | 16.7 | 22.6 | 26 |
6 X 1 1/2 | 15 | 18.4 | 18 |
These experiments were made with shear blades at an inclination of 1 in 8. It will be noticed in the table that when the bar of iron was cut on the side or flatways the stress required was considerably less than when the iron bar was cut edgew ays. This is doubtless due to the circumstance of the work to be done being spread over a longer period, since in the case of placing the bar flatways upon the shear blade it only comes into action with a portion of the bar, due to the inclination of the shearing edge.
To find the approximate pressure required to cut any blank, first measure up the area of metal to be sheared, then multiply by the shearing strength of the particular metal; the result will be the pressure or stress required at the punch or at the ram. It will therefore be understood from the foregoing that the ultimate shearing strength of any metal may be used as a constant in any calculations respecting that particular metal.
Taking cold rolled steel, and the blanks to be cut as cylindrical,
Let | D | = | diameter of blank in inches, |
T | = | thickness of metal in inches, | |
π | = | 3 1416, | |
c | = | a constant 26.7 (for steel), | |
p | = | pressure or stress required; | |
then | p | = | D X T X π X C. |
Example.-What stress would be required to punch a blank 6 in. diameter through a steel plate 3/8 in. thick ?
Here P = 6 x .375 x 31416 x 26.7 = 1887 tons.
Next suppose a set of shears has to cut off* wrought-iron bars.
Let | W | = | width of bar in inches, |
T | = | thickness of bar in inches, | |
C | = | a constant 20 (for iron), | |
P | = | pressure or stress required; | |
then | P | = | W x T x C. |
From this it will be seen that in punching a hole (not cylindrical) in a plate the section of metal in square inches to be sheared will be the circumference or length of the curve which forms the outline of the blank, multiplied by the thickness of the plate, the length of curve being obtained by the ordinary rules, according to the shape of the blank. It then remains for this area to be multiplied by the constant C, which has been explained as that stress required to shear one square inch of metal. It must be clearly understood that these calculations refer to punching or shearing metals when the tools are quite flat-that is, having no dip or shear.
Further, it may be mentioned that when providing or designing a machine to perform punching or shearing of a blank or bar, it will be advisable to allow 6 as the factor of safety. In other words, assuming that the pressure required to punch out a blank has been arrived at by calculation, it will be necessary to provide a machine of such strength that it will require six times that pressure or stress to break any shafts or wheels that may happen to be used upon that machine. With reference to the example of cutting the cold rolled steel blank, 6 in. diameter by § in. thick, where a stress of 188.7 tons was required at the ram, it will be advisable to provide a machine having crank shaft and wheels of such strength that it would be necessary to introduce a resistance at the ram or tools equal to 188.7 x 6, or 1,132*2 tons, before any portion of the wheels or shaft would break. The main casting or frame of the machine should necessarily be provided with the same factor of safety.
The punching and shearing machine is an excellent example, demonstrating as it does the great value of the flywheel as a means of storing up energy that may be given out when required to cut a blank. In the case of a machine designed for cutting steel blanks 6 in. diameter by § in. thick, the weight of the flywheel was 15 cwt., its diameter at the edge of the rim being 5 ft. The speed of the counter shaft carrying this flywheel was 116 revolutions per minute. A Bessemer steel crank shaft 6 in. diameter was provided for working the ram, the speed of the crank shaft being brought down to 25 revolutions per minute by means of spur gearing.
According to Professor Goodman, it is usual to store energy in the flywheel equal to the work done in two working strokes of the shear or punch, amounting to about 15 inch tons per square inch of metal sheared or punched through.
 
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