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How many tons does a sledge produce ?

Dave J

Cast Iron
Joined
Apr 11, 2001
Location
Hammond IN
I had a rush job to straighten out a bent segment on a aluminum frame. The piece is 24" long and 2" X 3" and 1/8" aluminum, most likely 6061.
It had a 2" bow in it (the hard way), I tried a 4 ton porta power, no luck. Jumped up to a 25 ton cylinder still wouldn't budge.

Finally got out the old 20 lb. sledge, worked it back straight not pretty, but it worked..

This brings me to my question.
How much force does a sledge generate ? It was swung horizontal, and it was swung about 3 feet and it wasn't a full power swing.

How would you figure the force generated by the blow ?

Had to be way more than 50,000 lbs of force, because it moved reasonably easy ?

P.S. I am not a sledge kind of a person, this was a last resort. :eek:
Dave J
 
There is a simple formula that escapes me at the moment. It has to do with velocity and the weight of the hammer. Check out Machinery Handbook.

Jim
www.pivotlok.com
Professional Bench Top Positioners
 
Well you know the weight of the sledge hammer, 20lbs. You need to know its speed at the time right before impact, which we don't have. You need the area of the face of the hammer. I think just this would give you the amount of PSI force over the whole face. I think you might also need to know how long the impact last.
I don't actualy remember the formulas. But I remember having to calculate the force exerted on a tennis ball hit at such speed over such time, back in Highschool. It wasn't very complicated, when you had all the data.
 
Force = dP/dt where P is momentum and t is time, so the answer is it depends on:

1) mass of hammer
2) initial speed of hammer
3) final speed of hammer
4) time between final and initial speed

Say you have a 20 lb sledge, moving initially at say 10 ft/s, and bouncing back at .5 ft/s with the impact lasting .01 seconds. All WAG, but probably they're on the right order of magnitude.

Those assumptions give a force of about 1000 lb, so something isn't right. My guess is that the time is too long. If the frame is stiff enough then it can slow the hammer down quite quickly, which means it is subjected to quite a high force, but if it gives then in essence it cushions the hammer blow, lowering the force.

-Justin
 
Any sledge that one man can swing can't move a press fit that a 2 ton press can. I suspect that some other principle made the frame move, like stretching and deforming the metal, or just moving small portions of it at a time. I know that sometimes an impact works better than a push, but don't have the math in my head to explain it. Come on guys, why does that work?
 
I was just beating(heating) on a 2 Inch pin with a 8 pound sledge.I placed a piece of 1.5 inch piece over the top to protect it.I first started by using a 50 ton press and it didn`t move.I then used the sledge and it still didn`t move however the 1.5 inch piece was now over 2 inches,I guess that 8 pound sledge was putting out alot more force than 50 tons?Lets hear from the engineers.
 
I can give you two ferinstances that will illustrate but not illuminate specifics. First, axle bearings that used to be pressed on (the old fords and chevys) that we could not remove with our 12 ton press could be driven off using a bearing splitter and a 12 pound sledge. Second, when driving a large pin into the warped arm of a 966 cat loader years ago, I kept hitting it until it barely moved a few thou each time I smacked it. I then bolted up the hydraulic ram and started bumping the button to the pump until the ram started moving the pin in. The guage read a little over 13 tons. These were full force blows to an object that had almost no spring to it. From that I have always estimated that a big guy with a 12 pound doublejack can equal about 12 tons under the right circumstacnces, but everyting varies...Joe
 
I'm not an engineer but from personal experience you can devolp alot of force with a sledge. Used to knock plastic mold cavities out of their plates. Put some heat to the plate, a piece of aluminium on the cavity, and do your best John Henry impression (don't miss that job
). If that didn't work, they would go on the big press where the gauge would hit between 25-50 tons before the cavivitys would
"pop" and starting sliding. I suspect it might be more of the sudden impact of the sledge blow than the actual force. I'm interested to see what the engineers say.
 
The problem with knocking out a large pin is the weight of the pin itself absorbing alot of the force and with a constant hydraulic force it doesn`t matter.The pin I was trying to dislodge was cut flush with the bucket so it was only 3 inches in length and therefore not heavy.The 50 ton hyd.press didn`t flatten out my spacer but the sledge did and it covered the entire spacer so it wasn`t hitting a smaller diameter. :confused:
 
I had a stuck bolt on an old mower. I had a socket on the bolt with a 3 foot long pipe over the handle of the socket wrench. Could not budge it. Put the impact gun on it and it came loose as if it were finger tight. It has to be something to do with all the force delivered at one momment.
 
Even better than

Dust buster

Work truck

Paint guns

Butcher block care

or --- moving a machine Eskimo style???

Surprising that an engineer hasn't posted. Lots of them here, surely one who will calculate this out and explain why a twenty pound sledge will do what a 50 ton press cannot do.
 
The peak impact force is probably almost unlimited with a sledge (limit is due to elasticity of the sledge face and part). Energy is of course a different deal, it is constant for a given swing.

But if you assume a steady acceleration during the swing, and same (negative acceleration, i.e. deceleration) during the impact, peak force is proportional to acceleration distance/deceleration distance.

If the thing hit does NOT move, the impact force is high, because the "stop distance" is so small.

If it does move, the peak force is lower, since the stop distance is less.

The sledge (or impact tool of any sort) kind of "automatically adapts" its peak force to the behaviour of the part struck. If part does not move, impact force is highr, but shorter. if it does move, longer and lower. Just what you want, actually.

The press can only do what it does.

Additionally, the effect of "stick-slip" is in tehre, and the sledge impact can "adapt" to exceed the peak force necessary to break the stick-slip and move the part.
 
This is a brain teaser, I like that. The basic formula is F=MA. We know that the mass (sledge) is 20#. What we don't know is how much area was the 50 tons applied to and how much of the face of the hammer struck the piece in question. In either case if you work with the formula you would note that the acceleration of the sledge to equal 50 tons is almost an impossibility, unless that the area that the sledge struck the object is extremely small and the area of the press is extremely large.

Some have mentioned the bounce back. That is a method of measuring how much force was absorbed by the object. An example would be if you struck a concrete sidewalk with a sledge hammer, you would find that the bounce back to be relatively small, therefore the concrete would be absorbing the majority of the force. We know that the concrete may not break on the first strike but will break after several blows. Why is that?

Something that hasn't been mentioned is, where is that energy going or what is it being transformed too? When two bodies collide the energy is transformed into heat. Plasticity of the object is at work here due to the almost instantaneous transfer of force and heat. The same thing could have been done with a smaller sledge with greater acceleration and more rapid strikes. Remember that a lot of things where and are still being bent with drop hammers when a giant press isn't practical or available.
 
We used to replace wheels used to support the barrel washer in a sand and gravel plant.Shafts were 4"-6" diameter.Sometimes the sixty ton press wouldn`t move them but 14lb sledge hammer would.Don`t ask me why.
regards,Mark
 
As has been said, the force when the hammer comes to rest = mass x acceleration, or deceleration in this case. Therefore the force is critically dependent on the deceleration rate. If you know that, you know the force. Of course we don’t know the deceleration rate.

Even if we knew the contact force, I suspect there’s a lot more to it than that. The internal shock waves are probably very significant (think of the destructive effect of water hammer). Perhaps the shock wave forces instantaneously distort the struck object so that it’s momentarily released from the tight grip.

I'm reminded of the ancient joke:
Gent watching a road-mender swinging a sledge:
'I say, my man, do you get paid money simply for bringing a hammer down onto some rocks?'
'No, guv. I gets paid for lifting it up. It comes down on its own'
 
People think of a 50 ton press being really large,it isn`t.If I put a 3/8 bolt laying flat inside the effects of 50 tons is minimal.Now let me take a sledgehammer to it and I will flatten it out.The contact point will be near the same since it is round.
 
How about using the acceleration of gravity from about 8-9'? typically my heavy sledge swing is a pendulum swing from bottom to top and letting gravity do most of the work on the way down. I can reach almost 7' with both hands together, flat footed, given grip....

OK, I just went out and measured the peak of my 16# hammer against the roof line and it's right at 9'.

Will that work as a starting point?

Kevin D
 








 
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