Press versus hammer, kinetic energy vs pressure

[Benjibear]

Hi there,
I am trying to work out the difference in force exerted by either a press or a power hammer.
I worked out that the kinetic energy of a hammer is calculated as E = ½mv².
Hammer: 0.5 * 30 kg * (15m/sec)sq = 3.375 KJ

How does this relate to the force exerted by a 20 ton press, where the momentum/velocity doesn't really matter that much, or does it, if it is only 0.01 m/s ?
0.5 * 20000 kg * (0.01m/sec)sq = 1 KJ

That doesn't make sense, does it ?
That press would surely move the metal as much as the 30kg hammer blow . . .
There must be another comparison between pressure and kinetic energy.

What about this relation : 1 tonne-force = 9806.65 N = 9807 joule/meter ?
So the 20 ton press would exert a force of 196.13 KJ/meter ?
Still doesn't seem right . . .

Can anyone help me with this ? It's been bothering me for ages.

Thanks, Benjamin

 

[russ_watters]

The force exerted by a press is its weight.

Also, force and energy are not the same thing, so you haven't calculated what you say you want to calculate for the hammer.

 

[xxChrisxx]

Presses can be approximated to static loading. Hammer blows are dynamic, so you'd need to the contact time to work out a force distribustion.

A very crude assumption is that dynamic loads are about twice that of static, so work out the static loading for the hammer (with gravity and an rough acceleration downwards) and double it.

 

[Astronuc]

One has to look at an impulsive load (hammer) and slowly varying load (press), and how the force is applied over the distance.

Also, one can look at impulsive loads applied over different time periods. Changes are that the hammer will cause a nail to heat more than a gradual load applied by a press.

I've worked with hydraulic presses, which I used to press impellers on shafts. I had to be careful when getting close to exceeding the static friction limit, since the shaft would pop as it started moving. Then it was a matter of maintaining pressure as the impeller moved, then backing off quickly as the impeller reached the desired position/location.

Human controlled hammers usually apply more load than necessary, hence the denting of wood or metal.

 

[Dr.D]

The force developed by a hammer blow depends on the energy in the hammer (the kinetic energy) and also the stiffness of the surface that is struck. A very hard surface produces a higher force (such as hardened steel) while a soft surface produces a lesser force (such as wood). In either case, these are impulsive forces.

A 20 ton press can apply a force up to 20 tons, and that will be a steady force. This is the sort of force that is required for metal forming, such as forging or stamping, where time is required for the deformation process to occur.

It is true that the blacksmith did some forging with impulsive blows, but this was largely because he did not have a press available.

 

[Benjibear]

Thank you all for your comments,
it is interesting to see how many perspectives there are on the subject.
So it seems there is no straightforward equation to relate the two types of forces/loads applied in the two forming processes.
I understand that the difference in time, between the impulsive hammer force and the steady pressure of a press, is what makes the comparison difficult.
Maybe it's worth setting up an experiment, where identical pieces of metal are subjected to either process and the resulting deformation is compared.

Basically I create saddle-shaped, negative curvature surfaces in sheet metals for sculpture, and have experimented with both processes. But since I am not using the optimal equipment, i.e. a strong enough power hammer or a fast-acting press, it is hard to say what is best - so now I want to build the right equipment.

Many thanks again, Benjamin

 

[S J Goldthorp]

Benjamin,

You picked a question here that very few metalformers understand.

Firstly your press is it hydraulic or mechanical? Hydraulic is simple as the maximum force is directly proportional to the hydraulic pressure.

If the press is mechanical then the mechanical advantage of the slider crank (or derivative) mechanism and therefore available force is changing throughout the press stroke and the nominal capacity is just the force available at a particular rating point in the press stroke. This rated capacity then used to design the containment (frame) and transmission systems of the press. Mechanical press ratings should always quote X tonnes at Ymm beore BDC.

If the resistance of the workpiece is greater than the mechanical force available from the press on contact the press could stall, break or if close to bottom of stroke deform the press structure and effectively climb over the workpiece.

Mechanical presses (not screw type - another story) usually have large flywheel which store kinetic energy and ensure that the press has enough energy to do the required deformation on the workpiece.

Before we consider hammers its necessary to understand what's happening to the workpiece, so let's use a hydraulic press analogy.

If we squeeze a piece of metal in our hydraulic press load will build up until we reach the point where the pressure on the workpiece is equivalent to the yield stress of the material and at that point the press will start to advance and deform the material, let's call this initial load F1. As we deform the material the force needed to keep it moving will increase, this is due to all kinds of mechanisms but for the moment just accept that it does. Let's say that after a distance x that the force is F2. The work energy used in getting to that point is proportional to the mean force (F1+F2)/2 multiplied by the workpath distance X.

I have used the word proportional quite deliberately as the rate of load increase in practice is not linear but for now let’s assume it is and deformation energy is equal to ½ X(F1+F2)

Now let's consider our hammer; its moving parts have a kinetic energy of 1/2mv2 where v is the velocity at impact with the workpiece and m is the mass of the moving hammer. It doesn’t matter whether the hammer is falling under gravity or being driven downwards only its impact velocity counts. If it’s a simple gravity drop hammer then the potential energy of the hammer i.e. mgh where m is the mass, g is gravity and h is the distance dropped through is equal to the kinetic energy at impact (ignoring friction losses)
When the hammer strikes the workpiece its kinetic energy will be converted into work energy to deform the workpiece and when all the energy has been used up it will stop moving; -

So ½ mv2 = ½ X(F1+F2) + losses

The losses are important and we’ll come back to them.

The values of F1 and F2 are entirely dependent on the characteristics of the material being deformed e.g. they will depend on size, material, condition, shape. So if the kinetic energy of our hammer is fixed the higher the workpiece deformation to resistance (F1 and F2) the smaller the deformation X that is achieved by the hammer blow. At first glance it would appear that an infinitely large force could be achieved by our hammer as the workpiece deformation get smaller and approaches zero.

Not so in practice, this is where the losses come in. When the hammer strikes the workpiece it will do plastic work on it but also elastic work on the supporting hammer structure i.e. anvil and hammer (ram). If the hammer does repeated blows on the workpiece the resistance will rise and the workpath will get smaller. As the workpiece resistance (force) increases the elastic energy losses into the hammer structure will increase until we get to the point where all the kinetic energy is dissipated into the structure before we achieve the new F1 load at which workpiece will start to deform. At this point the hammer will bounce off the workpiece without changing its shape. This is the load limit of the hammer.

This load limit is dependent on the type of hammer and work being done but for commercial forging hammers the ratio of hammer energy (kNm) to equivalent press force (tonne) is usually taken as somewhere between 1.5 and 3.5.

Hammer Blow energy (kNm)/Force limit (tonne) = 1.5 to 3.5

Regards

Stephen Goldthorpe.

 

[Benjibear]

Thanks Stephen,
I really appreciate your detailed explanation.
At present I am using a hydraulic press to stretch&compress larger areas of sheet metal/plate, and then use a mechanical hammer for smoothing/planishing and some peripheral stretching.
I am quite impressed with some of the smaller hydraulic forging presses, but building one of those will have to wait. Then there are rolling presses for squeezing the metal along for gentle stretching - that's on my shopping list as well.
Benjamin

 

[S J Goldthorp]

Benjamin,

Sounds interesting, is it work or hobby? I'm a Consulting Engineer and I sort out metal forming related problems for people worlwide though I admit to being more machine than process orientated. Your welcome to any advise that doesn't require research and takes less than 15min to email.

p.s. if its work be careful using a power driven hammer for cold working metal. There are lots of safety and legislative implications.

Stephen

 

[Benjibear]

Thank you very much for the offer, and the warning.
I have already felt the problem of working the metal too cold,
certainly isn't kind to the joints and ligaments.
While it's not a hobby, I don't do it full time, so I think I will be ok
with the odd shock to the system. I am not a blacksmith, so am
only just getting used to heating the metal up efficiently.

Regards, Benjamin

 

[abelli]

Goodmorning,
I know that the mechanical power press constructors give:
1) the diagram of permitted work depending on used number
of slide single stroke per minute;
2) the diagram of eccentric loads on the slide (ellipsis graph).

But I don't know how they can do these ones.
For question 1. I presume that the iter calculation is:
- energy for forming: E0=rated Force x rated distance to BDC
- energy for deformation press: E1=F^2/2xRigidity press
- energy losses: E2= ...? (x% of E0 ?)
Total energy: Et=E0+E1+E2 this is the energy in continuous operation
(permitted work on automatic strokes).

Now, for the press flywheel, assuming a slowdown of 20% during
continous stroking, the energy stored in the flywheel is:
Ev=Et/(1-0.8^2)
So the inertial moment of flywheel is: J=Ev/iω^2
where i: grade of non-uniformity, ω: average speed flywheel in rad/s

Now the questions are:
a) first of all: how can I calculate the drive motor power (kW) ?
b) above all: how can I calculate permitted work on single strokes ??
(I've read somewhere the energy in single strokes is about twice-2- of continuous operation: why? I don't think 2 is a magic number...)

2) In the end how can I drawing diagram of eccentric loads ?


I really thank anyone help me.