How to simulate impact force? 1

[padan]

I need simulate a 12 ton impact force on a piece of metal for (shear strength). Can a I use a vertical slide hammer type devise to do this? If so, how heavy should the slide hammer be and from what height should it be dropped from? Thanks.

 

[fwz]

You forget three formulary F=m*a, H=v2/2*g, v=a*t, v--- speed g¡ªgravity a--acceleration if you let hammer free drop, acceleration of gravity is about 10m/s2, you can use a hammer 1200kg, from nearly 5 meters free drop, it be produce 12 ton impact, but too high so you must change ¡°a¡±&¡±m¡± to get desired result. i think the best way is hydraulic pressure, its safe and more efficient.

 

[brmech]

It is difficult to simulate the shear force from an impact force of a desired magnitude. Materials have different material properties for the very short time duration of an impulse force. I know for aluminum you can look up what the dynamic modulus is. If you review your principles of impulse and momentum you will quickly see that you do not know what the delta t is, thus will not know what the impulse force is to balance your conservation of momentum equation. The big unknown is the delta t, which you can determine experimentally by using accelerometers. There are many variables that effect the delta t such as geometry of the colliding parts and their respective materials, so you will have to come up with an expected statistical range of the delta t for the actual impact scenario you are simulating.

 

[3DDon]

Padan,

I've got 6 years of automotive testing and I can say without a doubt that a droping weight is the most reliable, consistant, and cheepest way to simulate an inpact. Using the formulas that FWZ has provided, cut off a piece of steel, weigh it, and then back calculate the drop height. Try and make the weight round in shape so that you can drop it through a loose fitting piece of plastic pipe. You can punch holes through the sides of the pipe at different heights for a pin that the weight can set on. Each hole position is a different Joules impact force.

Or you can do it in reverse by punching a hole through the center of the weight and inserting linear bearings. Then you need a piece of shafting for the linear bearing to ride on. Slide the weight up the shaft to a pre determined height and let it go. The only drawback to this method is you can not have different drop heights because holes through the shafting would destroy the linear bearings in the drop weight.

In either case, you will obviously need some form of a stand for holding either the plastic tube or the shafting. No problem. Depending on the orientation of the device under test, a swinging weight could also be developed, but the math isn't as simple.

My #1 choice is a droping weight through a plastic tube. It's the cheepest and simplest the design and build.

Don

 

[brad]

I agree with brmech on this point.
With all due respect, Don and fwz are missing the crucial point--Padan explicitly stated that he needs to simulate a 12 ton impact FORCE. Don himself refers to a "joules impact force", which is an oxymoron; joules are an energy unit.

All these formulas and Don's description of the test merely are describing how to get a specific kinetic energy associated with said test fixture.

One does not know, based on these details, what the dynamic force on the structure is.

As brmech points out, this force is a function both of the kinetic energy of the falling mass and the energy absorption characteristics of the falling mass and the piece of metal being tested.

If there is any doubt about this, just consider the following: use the drop tests described, and drop the given mass first onto a piece of steel, then onto a piece of rubber. One knows intuitively that the dynamic forces on the rubber will be less than those on the steel. This is because the rubber's compliance allows the dropped mass to decelerate over a longer distance, resulting in a smaller overall dynamic force.

Knowing the specifics of your structure, you'll have to do some calculations to get a "ballpark" force for this test. brmech didn't give you any equations (and I won't either), because this is very much a problem-specific answer--there are no general-purpose equations to arrive at your answer.
Pull out the dynamics book and some structural mechanics book(s), and do your best to figure this out.

Good luck,
Brad

 

[Suilven]

A couple of points to lookout for, we were involved in testing some passive fire protection materials for offshore use where we were requested to conduct tests aiming to simulate a 5kJoule impact. This was to simulate the sort of impacts from flying objects (e.g scafold tubes and fire extinguishers) in an explosion.

One material we were asked to test claimed that they could meet this figure and on examination they had gone back to F=ma, E=½mv², etc and dropped a large flat block weighing 5000 kg from 10cm with little damage. When we tested using a 32kg mass of he same diameter as a fire extinguisher, drop from 7 metres it punched a hole straight through despite being less than half the impact energy!

In another test on a flexible fire protection jacket wrap system - the system survived intact but the steel backing panel on the test bed (that represented the actuator we were trying to protect) was badly damaged.

The moral of the story is that the test needs to be representitive of the speed of impact and size/shape of impact object expected.

 

[brad]

Suilven--
Which is why I'd rather have a 5 lb bean bag dropped on my head from 10 feet than a 1 lb lawn dart!

Total agreement--once one figures out the force, there is still stress to deal with.
Brad

 

[Hagan]

Didn't the original question suggest that what we are looking for is shear strength of the material?

I am no expert in these matters but I would have thought that what you need to do is to apply a steady and increasing force to a known cross section of the material in a shearing manner whilst looking for its yield point (similar to a tesile test with a tensometer)

Impact type tests will not give shear strength for all the reasons pointed out above.

Rich

 

[brad]

I think we're all waiting for Padan to speak up with more details. By stating "impact force", I presumed that we were talking a dynamic event. Lacking any better information, I am presuming that he is wanting to see a dynamic shear penetration on a sheet of metal.
Please login and speak up Padan!
Brad

 

[3DDon]

Bradh,

Let's recap shall we?

padan stated: "I need simulate a 12 ton impact force on a piece of metal for (shear strength). Then you stated: "All these formulas and Don's description of the test merely are describing how to get a specific kinetic energy associated with said test fixture."

The fact that Padan indicated "sheer strength" would lead one to assume an arm or lever like that of a turn signal switch, or even a flag pole. No where did he say "lawn dart" or "penetration". And he especially didn't say "sheet metal".

Padan asked a "how do I..." type of question. And I think between FWZ and myself, we fairly well covered it. So Joules isn't the unit of measurment that you would have picked. Oh well. Go tell it to Lear Corporation in Dearborn, MI. I'm sure that if you call them, they'll change the units of all of their impact test fixtures to something more to your liking. You want the number?

Good greif...





Don Shoebridge
Dayton, Ohio
The birthplace of aviation
Check out my webpage

http://www.geocities.com/donshoebridge

 

[brad]

It's still back to the same point Don--

Padan asked how to get a specific force value (12 tons).
Quoting Don "So Joules isn't the unit of measurement that you (bradh) would've picked." This is an inaccurate statement. In fact, Joules isn't the unit of measurement that PADAN picked. I'm not arbitrarily choosing force units, I'm following through with Padan's question which explicitly requested force units.

Neither you nor fwz answered that question. You answered how to get a specific kinetic energy. That is fundamentally different from a dynamic impact LOAD. This is an important distinction. Now I will grant you that maybe the question was not worded correctly. I will also acknowledge that I don't know what Padan is actually trying to do anymore than anybody else here (hence my encouragement for Padan to clarify).

A drop test is used when one wants a specific impact energy. If one wants a specific impact FORCE, a drop test may still be appropriate. However, as brmech, Sulliven, and I have tried to point out, a variety of factors other than just kinetic energy must be taken into account.

To be completely clear, it very well may be that the impact test is the best approach for Padan. However, the formulas given by fwz will do nothing within themselves to predict the force on the structure.

Impact energy does NOT a priori dictate impact force--this may seem a minor point to some, but this is vitally important if one truly is concerned with the peak FORCE.

You imply that my assertion runs counter to standard test procedures. It does not. Most engineers at Lear, and other engineering companies, would agree with me when I say Joules is a unit of energy, not force.

By your tone, it appears that you've taken my comments personally. I certainly did not intend in my comments to insult anybody; merely to clarify what I feel is an important point. I'm happy to continue discussing the issue. I'm even happy if people are passionate about their views; let's just keep it professional.