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If the compression strength of iron is 50 MPa, or 5 MPa at 1000C, how can it be forged by a falling hammer, weighing say 10 kg, 10x10 cm impact area -> 100N/0.01 m2 = 10 kPa?

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    $\begingroup$ The compression strength is not relevant. You don't break the iron during forging. What happens is a plastic deformation of the forged piece.Each blow deforms the piece by a little bit. Many blows results in a large enough deformation. $\endgroup$
    – nasu
    Jul 24, 2018 at 12:55
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    $\begingroup$ arent these blows fully elastic before the yield point? $\endgroup$
    – Tsayper
    Jul 24, 2018 at 12:57
  • $\begingroup$ Related question by same user : Forge-flattening $\endgroup$ Jul 24, 2018 at 13:43
  • $\begingroup$ I don't know how big a hammer has to be to have a 10x10 cm impact area... Mjölnir? $\endgroup$
    – Floris
    Jul 24, 2018 at 13:52
  • $\begingroup$ A 10 kg hammer is VERY big. A typical blacksmith's hammer is ~ 1 kg. Not many blacksmiths could swing a 10 kg hammer with one hand. Even a 10 kg two-handed sledge hammer would be exhausting to use for more than a few blows! $\endgroup$
    – S. McGrew
    Jul 24, 2018 at 14:01

1 Answer 1

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Having been a quasi-professional blacksmith for the past 13 years, I can provide a partial answer. Your calculation assumes that the pressure exerted by the hammer is just the weight of the hammer divided by the area of the hammer's face. In fact, the pressure should be calculated as the force due to deceleration of the hammer at the moment it strikes the workpiece, divided by the area of impact.

In order to get a sense of the deceleration, you can use the average depth of the impression left by a good hammer blow: around 2 mm. The speed of the hammer at the moment of impact is probably about the speed it would reach after falling freely for about 3 or 4 meters, or about 8 meters per second; and the hammer comes to a dead stop, or even rebounds, after traveling only a couple of millimeters subsequent to initial impact.

Also, a hammer blow is almost never perfectly flat. The hammer face is almost always is slightly tilted relative to the surface it hits, so all of the force due to that deceleration is initially concentrated in a few percent of the area of the hammer's face.

A seat-of-the-pants estimate, against which a properly calculated estimate should be compared, is that the hammer stops in 2 mm compared to a 4000 mm free fall, so the force exerted by the blow is on the order of 2000 times greater than the weight of the hammer. Add that to the slight tilt of the hammer face, and the force is probably more like 20,000 times greater than the weight of the hammer at the first moment of impact.

Another way to estimate the hammer's deceleration is to assume that the workpiece exerts a pressure equal to its yield strength, against the hammer. In the case of the slightly tilted hammer face, that pressure is initially exerted against only a small area so the deceleration is small; and the area increases as the impression size grows, so the deceleration increases. To calculate the deceleration vs time you would need to do an integral.

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  • $\begingroup$ The ratio of distances for acceleration and deceleration is a nice intuitive way to get to approximately the right answer for the force; now you need to divide it by the area to get an estimate of the pressure, which is the thing that is asked for. You conflate "force" and "pressure" in your final paragraph. $\endgroup$
    – Floris
    Jul 24, 2018 at 13:53

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