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# I need to be pointed in the right direction

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## I need to be pointed in the right direction

(OP)
A customer of mine is building a flying shear.  He claims to need 37000 pounds force to cut the metal.  This is achieved by using a 4 inch diameter hydraulic cylinder at 3000 psi.  The shear itself weighs 2000 lbs. The shear has a stoke of 2 inches but it might be better to lift the shear just high enough to clear the sheet metal. The difference between the using the full stroke and barely clearing the sheet metal is that if one barely clears the sheet metal there will be no 'running start' at the sheet metal.  If the shear is lifted 2 inches the shear may be able to build up some kinetic energy which should reduce the need for such high static force. Obviously one can cut metal by putting enough force along a narrow line.  However, it seems to me that the kinetic energy would help too.  If one takes into account kinetic energy wouldn't the need for such a high static force be reduced?

Can anybody point me a source that can be used to find the optimal trade off between static force and kinetic energy.

Thanks

### RE: I need to be pointed in the right direction

Hi PNachtwey

The static force to shear a piece of metal of a given size and shape as well as its properties won't change, what will change is the power required to drive the blade or punch etc.The longer you allow the blade to build up speed the less power you need to achieve it, thereby you can use a smaller motor or whatever is driving the blade.

regards

desertfox

### RE: I need to be pointed in the right direction

(OP)

#### Quote:

The static force to shear a piece of metal of a given size and shape as well as its properties won't change,
what will change is the power required to drive the blade or punch etc.The longer you allow the blade to build up speed the less power you need to achieve it, thereby you can use a smaller motor or whatever is driving the blade.
OK, but there must be formulas for optimizing the stroke and weights and forces applied to accelerate the shear.

Tanks shoot high density projectiles each other at high speed and it is the kinetic energy that does the job.  There is no static force being applied by the tank since they are far apart.  The projectile have kinetic energy. The energy gets converted to work ( force x distance ) over a small area.  When the projectile penetrates the kinetic energy is converted to work to get through the metal, heat and there is still kinetic energy to so the projectile does damage inside.  Shear is force/area.  The projectiles hit with a energy/area.  It seems to me it is the energy per area that is required.  There must be formulas that tie all of this together.

In the case of the shear above it looks like the designer wants be able to cut the sheet metal starting with 0 velocity or kinetic energy. Once the shear starts to cut the kinetic energy will start to increase but the hydraulic pressure/force will go down.

If the shear starts from 2 inches it will gain kinetic energy.  This kinetic energy will be convert to a force x distance in a small area where the shear contacts the sheet metal.

What I am getting at is that there has must be more to than than just a static force.  I am not a mechanical engineer but I know that most dynamic designs should take into account energy/work.  I am just looking for some formulas and I don't find this information on the web.

I find it hard to believe no one has designed a shear or something like it.

So why bother?  I want to reduce the pressure in the hydraulic system from 3000 psi to something less to save energy.I have already got the other designers talked into using two pumps with one motor. One pump for the traverse, and another pump for the shear itself. Each would run at the optimal pressure there jobs. If one pump was used for both the shear and the traverse then the traverse would run at a much higher pressure than what is required and use extra energy.

### RE: I need to be pointed in the right direction

(OP)

#### Quote:

The static force to shear a piece of metal of a given size and shape as well as its properties won't change,
what will change is the power required to drive the blade or punch etc.The longer you allow the blade to build up speed the less power you need to achieve it, thereby you can use a smaller motor or whatever is driving the blade.
OK, but there must be formulas for optimizing the stroke and weights and forces applied to accelerate the shear.

Tanks shoot high density projectiles each other at high speed and it is the kinetic energy that does the job.  There is no static force being applied by the tank since they are far apart.  The projectile have kinetic energy. The energy gets converted to work ( force x distance ) over a small area.  When the projectile penetrates the kinetic energy is converted to work to get through the metal, heat and there is still kinetic energy to so the projectile does damage inside.  Shear is force/area.  The projectiles hit with a energy/area.  It seems to me it is the energy per area that is required.  There must be formulas that tie all of this together.

In the case of the shear above it looks like the designer wants be able to cut the sheet metal starting with 0 velocity or kinetic energy. Once the shear starts to cut the kinetic energy will start to increase but the hydraulic pressure/force will go down.

If the shear starts from 2 inches it will gain kinetic energy.  This kinetic energy will be convert to a force x distance in a small area where the shear contacts the sheet metal.

What I am getting at is that there has must be more to than than just a static force.  I am not a mechanical engineer but I know that most dynamic designs should take into account energy/work.  I am just looking for some formulas and I don't find this information on the web.

I find it hard to believe no one has designed a shear or something like it.

So why bother?  I want to reduce the pressure in the hydraulic system from 3000 psi to something less to save energy.I have already got the other designers talked into using two pumps with one motor. One pump for the traverse, and another pump for the shear itself. Each would run at the optimal pressure there jobs. If one pump was used for both the shear and the traverse then the traverse would run at a much higher pressure than what is required and use extra energy.

### RE: I need to be pointed in the right direction

In order to calculate how much kinetic energy is needed you need to know the thickness of the material being cut.

You use work to shear = kinetic energy to shear.

Work to shear is the product of shear force times metal thickness (the distance through which the force moves).

Kinetic energy of the shear is 1/2*m*v^2.  m = mass = W/32.2  v = velocity in ft per sec

F*t = 1/2*m*v^2  You need to know the material thickness t in order to find the shear velocity v.

Once you know the shear velocity you can calculate the hydraulic flow rate gpm because the piston velocity is the same as the shear velocity.  The area of the piston times the piston speed results in the flow rate.

1.5-inch free fall of the 2000-lb shear would cut .081 in. thick material if the force to cut is the quoted 37000 lbs.  The kinetic energy at the end of 1.5 in. free fall is 250 ft-lbs.

Need to know what material is being cut and how thick.

Ted

### RE: I need to be pointed in the right direction

(OP)
Hydtools, what about the energy per area or force per area?
I can calculate the kinetic energy and all of that.

Doesn't the type of shear make a difference?  A straight shear that makes contact with the sheet metal all at once would require more energy than a shear that has an angle to it so it cuts at one edge and finishes at the other edge.  There is less metal in contact with the shear so the force per area would be higher.

However, your calculation is the same that I did that got me thinking about all of this.  You can see that just dropping a 2000 lb shear accelerating at the speed of gravity can do a lot of work.  If the 2000 lbs mass has even 10000 lb of force pushing it down it will have 6 times (1+10000/2000) the acceleration.  The shear would have 6 times the energy of your example so it should be able to cut through anything called sheet metal with much less pressure. So is designing the shear so it can apply 37000 lb while stopped necessary. I think I can get by with half the pressure and still have plenty of energy. I can save energy by reducing either the pressure or how far up I must let raise the shear to get a running start.

I can do the calculations on the shear side.  I need to know what it really takes to cut metal.  There must be a manual of metal properties that says it takes so much energy per area per thickness to cut through metal.
I was hoping that someone would know where to go to get this information.

OK, here is a more basic question.  What cuts metal?  Force?  Shear (Force/area),  energy/area?

If I don't get an answer that makes sense I will go with the current design and turn down the pressure and reduce the stroke if current design is overkill.

### RE: I need to be pointed in the right direction

Hi PNachtwey

When you shear a piece of steel by punching or some other method the tool penetrates the material and generates cracks by which the material subsequently fails.
If you look at a piece of thats been sheared you get a clean edge for about 20% to 40% of the material thickness and the rest is kind of dull and rough.
The clean edge is the tool penetration depth which is typical 20% to 40% of the material thickness depending material being cut, after which the material fails under the cracks generated.
The difficulty is predicting what this penetration might be! which is really what you need to know and I am not aware of any information available.
Certainly if you increase the amount of material penetration
then you can reduce the force required, the difficulty is that nobody can tell you accurately what penetration your liable to get. If you assume 40% pentration then you can calculate the amount of energy required to drive the blade to that point, but what happens if you only achieve 20% pentration in practice then you may fail to shear the material.
I know that punch tools are sometimes ground with an angle to give a progress cut through the steel which will reduce the force required and a smaller capacity machine can be used.
All the books I have looked in give a shearing force based on the material thickness/perimeter and ultimate shear stress for the material to be cut, this being the maximum force required and thus erring on the safe side.

regards

desertfox

### RE: I need to be pointed in the right direction

Here are some illustrations of the mechanics of cutting sheet metal.

http://users.encs.concordia.ca/~mmedraj/mech421/lecture%208%20sheet%20metal%20working.pdf

Yes, having the blade at an angle will reduce the required force to cut.  However, the total energy to make the full cut is the same, or nearly so, to that required for a straight full width cut.

You still need to know the material properties and thickness of what you are cutting.

Ted

### RE: I need to be pointed in the right direction

(OP)
Hydtools, I found that document interesting in that is also explains why pulling occurs and clears up some 'bad think' on my part.

Thanks.

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