Force to make 1 mm raised surface ie 1 mm bump

[abdulbadii]

Roughly, how much force does it take to make 1 mm raised surface (1mm bump) of diameter 16 mm circle area out of 1 mm thick, larger area of steel plate ?
is a muscled hand pounding medium M14 (14 mm) bolt onto the steel plate laid over 16mm d. holed base viable ?

[jpanhalt]

If you need any accuracy at all, I suggest using a hydraulic press and press a ball/rounded punch into it.  Commercially, a matching die might be used.  For low volume and with soft steel, you might get by with pressing into very stiff plastic, like polyurethane, or a larger hole in a backing plate.  I did something similar to make a drain in a stainless steel bucket that was slightly thinner material.  I used a 10-ton press and did not have to take it to its limit.

You can also do it by hand with a ball peen hammer.  For backup, use something like a sand bag.  I usually use lead shot in a leather bag.  That forming is by successive blows, not one big blow.

[Infraviolet]

The type of steel is going to be a really crucial point here, is it the softer mild sort or one of the really hard stainless types? The former can be ductile, the latter isn't likely to do much under any force a human can exert and might even prefer to crack than bend.

[abdulbadii]

It'd be ductile, as it's just a steel plate no high carbon

So wonder if e.g. the plate is circle too but far larger area say 2 x the needed bump area, diameter 32 mm, when it's hammered as said and some raised surface has been created, will it get shrink and shorter in diameter ?

[jpanhalt]

I disagree that stainless can't be worked.  Here's a picture of the "dimple" I made in a stainless tank with just a 10-ton press.  It was not near its maximum.  Hand working would be a lot harder.  That picture is of a test I made before actually attacking the tank.

EDIT: I am a bit slow tonight.  The second picture is the dimple in the tank.

[T3sl4co1l]

Ballpark, the force will be on the order of tensile or shear strength (yield) times the area affected.  So, the perimeter of the area times material thickness.  So, 16mm * pi * 1mm * 280MPa = 14kN, or about 1.5 tonne[force].  (Taking a typical 280MPa for mild steel.)

There will be a geometry factor, which I don't know about because I'm not a materials scientist... but this can be looked up.  Probably Machinery's Handbook or the like has standard tables??

(Actually rather curious what that geometry factor is, now.)

Also, this amount of force, I think, won't be a big deal to obtain in pulsed form i.e. hammering.  May need something bigger than a carpentry hammer but I'd be surprised if more than a 1kg hammer was needed?  Doing it by hand of course will require some precision/finesse so it doesn't come out lopsided...

Tim