Calculating Impact Force

The heigh you are releasing the mass accounts for the energy input you are giving to it. The force that will be developed to stop the mass depends on where it is resting during the impact. If you release it in a pillow, the resulting force will be much smaller then if you release it in a steel table because the pillow deforms much more to absorb the same energy. The more deformation or displacement of the mass until it comes to a stop, the less the force required to stop it because it has more room to dissipate the same energy. Be certain that force and energy are quite different physical entities, do not mix it. No surface is perfectly rigid, but may be perceived as infinitely rigid when compared to your mass. Supposing you are hitting a target made of concrete or stone, for example, it is most likely that you will have to calculate the force based on the elastic deformation of the released mass itself. The area (diameter) of the mass will result in a tension in the contact area (Force/area) and is usefull to check if there is permanent deformation or not, but will not be included in the force calculation. Basic formula: Ec=mv^2 / 2 kinectic energy Ep=mgh potential energy W=Fd(in direction of force displacement) work performed by a force F and of course F=ma Ideally, at the moment of hitting target: Ec = Ep initial = W required to stop mass You may measure deceleration travel and calculate force. Not really hard to do. Well, if its an elastic colision (if it rebounces), than some other equations regarding conservation of movement quantity apply, but as you described the problem I see this is not the case.

hello bhrescobar, thank you ,thats clear as mud. there is an elastic collisions with a small bounce. the test is for polypropylene encasing a .500 rebar the plastic is .250-.325 thick. i know that absorbing some energy changes some numbers but i really just hope for a ballpark #

oh, and the test dart is 5lb 1.5 diam. 10 inch long mild steel

Your question is not clear. I know you have it straight in your mind, but help us understand it.

You have a 5 lb piece of 0.5 in dia rebar. It is encased in polypropylene that is 0.25-0.325 thick. You're dropping it from 10', 11', and 12'.

What is it landing on?

You ask for the force calculation.

Is that the force felt by the rebar or by whatever it's landing on?

Are you looking for a peak force, as in a break test, or an average force as if it were doing some kind of work?

ok, sorry if its clear as mud,, I'm dropping a dart 1.5 inch in diameter 10inch long weighing 5lbs the drop height is 10,11,12,feet . it is landing on a manhole step that is .500 rebar overmolded with polypropylene. the test is to simulate pounding it into a wall with a sledgehammer.

sorry I'm slow to respond gotta sleep sometime.

Well, unfortunately, force doesn't relate very well to driving it with a hammer. Sorry to be so difficult, but you've got to provide more explanation.

why not? when you swing a hammer and strike the shoulder of the step to drive it in to a hole in the concrete. ok,ok,the diff. is the hole is missing and of course the variation in swing speed . but what were doing is just trying to verify there are no voids in the plastic... this is a repeatable impact test to verify the plastic wont shatter at installation of product....

OK, I get that you think it's simple and that someone should just spit out a number. Perhaps it is and perhaps they will.

Good luck.

tvp45, I'm so sorry if i sounded flip,that was not the intension, i know it is not simple by far, it least for me. its been 25 years since i studied math,even though i do plenty. the algebraic equations are beyond me .I'm no theoretical physicist or even a book taught engineer..just a shade-tree genius trapped in a dropout body

and i do appreciate all the great minds that are collecting here

Not at all. I'm flip all the time. But, I (and maybe others) need some more info. Tell me in some detail what you want to test for and how you want to do the test. I gather these are manhole "steps" that get driven into the wall? Do you drive them with a 5 lb hammer. Is the end where the hammer hits also covered with polypropylene? How many times do you typically have to hit it? How good a test do you want (5% false "good", 25%, ?)?

excellent question the bar is encapsulated, min. .1875 to max. .3125 . 3-5 strikes to drive it into the wall. with a 3-5 lb sledge . test is to determine if there are are air bubbles in the finished part,and the plastic itself ,impact Resistance or deformation,or if it just shatters...... if we cant break it here it should be fine for the field.. test.....

but of course you can break anything with a big enough hammer,,or low enough intelligence

OK. To sort of get us on the same page, try this experiment. Take a 8d cut nail and start it into a 2x4 turned so the nail will get all the way through. Get it started, then set the 2x4 on a concrete block. Now pound the nail. You'll notice the huge difference in felt force when the nail first contacts the concrete. That's your question. You're driving the bar into a hole and part of the effort goes into the bar going deeper; that lessens the impact force. How far into the hole does it go each time?

that lessens the impact force

AAARRRGGHHH NO.
Surely the impact force is the same , it's the effect of it which is changing due to the changing conditions after each blow?
EG. the force is now being dissipated as friction between the sides of the nail and the hole as well as at the tip.
If you lift a weight a fixed distance you get a fixed potential energy, when you drop it, that's what you'll get back, or a whicker less...never more else CR4 goes mad and shouts at you as an over-unity nut.
The big question is 'how is that force dissipated'
Oooh sorry...scampers back to secret cat nest to peer out and watch in silence.
Del

Sorry Del but this time intuition doesn't play out. You are confusing kinetic energy with force. The kinetic energy is the same. The time over which it is released is the key factor in this equation.

Although I used a different formula as dkwarner because I thought it would illustrate the relationship more clearly, as he indicates, the formula for impulse is F=mv/t. and here you can see clearly what happens as t gets smaller. This also illustrates why there are no inelastic collisions.

You don't. Like I mentioned in my post an accelerometer will also work, that's what I use. I have this all set up because I study breaking things in my Tae Kwon Do practice. I suppose if it were important enough you could rent the equipment or hey, maybe send it to me and I could characterize it for him.

Oooooh no I agree with TVP45 here. There's a huge difference between force and energy. If the head of the nail moves say ½ an inch when you hit it the force between the head and the hammer is on average only 1/10 the average force if the nail head only moves 1/20 th of an inch (ignoring things like the change in re-bound).

This come to the heart of the problem. Suppose the OP drops his 5 lb "dart" from a height of 100" and from the point of impact the C of G (centre of gravity) of the dart moves say 1/10 of an inch before the dart comes to a stop then the average force of the impact is 5000 lb.

(If you drop your camera from a height of 1 m onto a solid concrete floor and the shell deforms by 1 cm (very conservative!) when it lands then the insides of the camera experience a 100g deceleration.)

If his dart has an area of contact of 2 in² then the average pressure is 2500 lb/in². But if the dart lands slightly on its edge and the average area of contact is only 1/10 th of a square inch then the average pressure is 50,000 lb/in².

polypropylene encasing a .500 rebar

weight 1.5 inch in dia. dropping from 10ft,11ft,and 12ft.

oh, and the test dart is 5lb 1.5 diam. 10 inch long mild steel

Make up your mind!

There seems a huge amount of overthink here.

You want ball park.

5lb lifted 12feet...hmmmm
How about 60 ft pounds?

Can't get much simpler.
Or am I missing something?
As long as the test method is consistent it doesn't matter a damn what you call it.
Del

Yes Del, I think you are! He wants to know the force produced on impact, not the energy dissipated.

Cheers...........Codey

If you assume there is no energy lost due to air resistance, deformation of the projectile, rebound, the planet being moved by the impact, temperature rise etc, Then all the energy is going into the test 'piece+projectile'.

He wanted a ball park figure...what exactly is required then?
The only other factor I can see is the shape/area of the impact, but that's a function of the test piece anyway, so just measure it. and divide by it if you feel so inclined.
I still can't see the problem....
Del

sorry dell, theres a little more math here

Well if you want to apply Netons Laws of Motion, it's still easy to work out the velocity at impact, then you have your ½mv² .
I'm going to hide in my secret cat nest.
Del

I'm thinking your ballpark is reasonably close after some heated discussion and basic theory verifying tests... call it blunt force impact test i guess.

thats one cool cat !!

Since you haven't answered my questions, here's the ballpark number: 7,108 pounds.

Another case of the blind leading the blind...

KE=MA=FORCE * TIME

So, the first part is easy enough. The hard part is to asses TIME, specifically, how long between initial contact and the instant of reversal of the mutual elastic deformation. That will be your peak force. We are also assuming that the rate of deceleration is constant, which is not a given but should probably get you fairly close.

How to get the dt part of the dv/dt of deceleration? Maybe use high speed photography or a strain gauge hooked to a logging oscilloscope. If you happen to have a high end sound level meter that has an accelerometer probe as is typically used for vibration analysis (HK makes these, I have one), then you can run the output of the meter to a logging scope and get a beautiful profile of the deceleration curve. You will then be able to determine the time points I mention earlier and plug them into the formula. You would also be able to determine if the deceleration curve were linear and possibly break it down to determine the different forces at different points if it isn't.

So you see, if your deceleration time were 1 second, then F=ma=1/2MV would be the answer. It will of course be much less than that which will multiply F by 1/T.

Neither of those equations is right. KE (kinetic energy) = 0.5*M*V². (V = velocity)

Assuming MA = mass*acceleration, MA = force.

For the problem posted, no need to calculate velocity explicitly, as energy on impact (ignoring losses) = M*h where h = height.

As was said way back in #1, average force on impact = KE/d where d = distance to stop. So if d can be measured or estimated can get force. E.g. if d = 0.1 ft, M = 5 lb, h = 12 ft, force = M*h/d = 600 lbforce.

This works OK s long as force and mass are in lb, but if you want to calculate V, need to be careful with units, and bring g into it. Also as somebody pointed out, force = momentum/time (that's time to come to a stop) but even if you could estimate the time (harder than distance, I think) you still need to watch the units, maybe use slug-foot-second system.

Cheers.........Codey

Sorry, but MA=0.5MV^2, do the calculus, they are equivalent representations.

I know it looks funny KE=FORCE * TIME but if you make time=1 it might help to see the relationship. When time becomes less than 1 (second) then force must go up to conserve KE and vs-vs.

Not quite with you rcapper.

If A is acceleration, MA=0.5MV^2 means A has units ft²/sec², not ft/sec² as it should.

If A is distance (height) MA is energy of sorts (ft.lbf) but need to put in g to get the units (and the figures) right. For free fall, M*g*h = 0.5*M*V², giving the well-known V = √(2*g*h). If M is in slugs, M*g*h and 0.5*M*V² are in ft.lbf.

Units of KE are ft.lbf, units of FORCE * TIME are lbf*sec so I don't see they can be the same. If you make t = 1 second it still has units of time.

Cheers........Codey

Finally ... SOMEONE who understands the difference (F = ma) and work (Work = Integral (F . s), that is the dot product of force and displacement).

There's no way to calculate what the 'force of impact' will be because it'll vary over the time during which the target will be decelerating the ball.

Case in point: driving a car into a wall vs. driving it into a giant cotton ball with nothing behind it. The force that the wall will exercise on the car will be much greater than that exercised by the cotton ball, and that despite the properties of the car, including its speed.

That being said, the problem here is likely that the question's improperly formulated. For our information, what's the purpose of the experiment/ calculation? How well the target can resist impact? Or how well the moving objective can?

Yeah I pretty much gave up. I mean really if you need to know exactly what is going on you must plot the deceleration curve that results on impact and analyze it moment by moment. At that point 1/2MV^2 is not a practical input you're going to have to use MA. You try to explain something and the first thing you get is some rocket scientist who is eager to misinterpret what you say because he's more interested in trying to make you wrong than to answer the question. I try to answer a question, as asked, with a broad enough answer that they can get a clue and if the don't understand maybe ask a better question. If you can't see where I'm pointing with my brief answer, I don't have time to educate you and I'm done. Oh well, I suppose we all do our best.

Hello DreadZontar,

Check out post number 42 for information on why this test was used.

Take care...........

thanks bb i know you all are not that bad

Hi mrhem,

Your not so bad either my friend!

Take care.

Hi Doogleass,

Bloomin' good idea my friend!

GA to you Sir.

=

I thought that as you say the OP wanted to try the solidity of the edge of a step. Can't understand why, as it may be used maybeten times in its life?

I just do not see why all this effort if being put into a very simple job?

Take care..

wow, thankyou all that felt like responding, the examples spelled out shur help.. me understand the math (it had been a whyl) all the info is in post 11,14,20 for the people who still have ?`s.. i think you've all been very gracious with your knowledge and have answered my question sufficiently ...

i don't talk to strangers.. stranger than me !!!!!

Hello mrhem,

I for one appreciate your thanks!

Maybe you can see that it is not all as simple as you first thought? Plus you never said as far as I am aware, why this test had to be done?

Take care.

According to ASTM D4508 Impact is recorded as in-lb. The answer to 5lb at 10ft, 11ft, and 12ft are respectively 600in-lb, 660in-lb, 720in-lb.

Hello Guest,

If you look at this realistically, and dare I say 'honestly', the figures you list in inch pounds [in-lb].....................

Is IDENTICAL TO THE DETAILS I HAVE ALREADY POSTED!

You have changed the NOMENCLATURE to [in-lb] thus multiplying my figures and, some other members figures, by 12 !

What you write is a waste of time.

The address I found your info' on was:

http://www.astm.org/Standards/D4508.htm

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I have copied and pasted part of the 'STANDARD' below.

Firstly, this 'Standard' you mention is for the testing of small plastic pieces only! So therefore IS NOT applicable to this thread.

The next piece I paste, clearly says the 'Standard' refers to the Joules figure, and is refereed to in square brackets as 'in-lb', because the joules amount is very small and 'ft-lb' was not appropriate. It also says the figure in '[]' are for reference only!

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ASTM D4508

1. Scope

1.1 The purpose of this test method is to provide an impact test that can be performed on small specimens of plastics of different thicknesses. This test method is especially suited for observing the effects of microcracks caused by weathering, or by exposure to solvents or other hostile environments, on the surface of plastic materials. It is not meant to be used as a replacement for any existing impact test, but can be used to measure impact on coupons machined from finished parts that cannot be tested by the drop-weight, Izod, or Charpy method because of shape or thickness limitations

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1.3 Round-robin testing has indicated that materials that break at total energy values of less than 0.17 joules [1.5 in.-lbf] have within-laboratory coefficients of variation of approximately 30 %. Therefore, such values are considered out of the normal testing range for this test.

1.4 The values stated in SI units are to be regarded as standard. The values given in brackets are for information only.

1.5 There is no ISO equivalent to this test method

No offence.

The OP requested the FORCE of impact, not the ENERGY of impact.

FORCE is measured in pounds or Newtons (and a few other less common units).

ENERGY is measured in many units, including in-lb, ft-lb, N-m (=Joules=Watt-sec), kWh, etc.

As I indicated previously, the force can not be calculated unless you know either the time or the distance moved during impact.

Hello dkwarner,

How are you doing?

I wish the 'Guest', whom posted the piece about a standard ASTM D4508 had actually read the first paragraph, of that standard..............

Hey, take care my friend.

bb

well bb, we had some bad polypropylene, and had a costumer shattering the plastic off the rebar trying to install the manhole steps .. also if there are large air bubbles aka. voids it can crack. we made changes to our process to fix the void prob. but needed a test to check the impact resistance throughout the run.....

O, i new it was far from simple .. even the brains are batting around big words like a ball of string.. thank you everyone. i hope i can contribute to someones enlightenment someday

knowledge is power

Hello mrhem,

I thank you for the reply post and part answer.

I look forward to hearing from you as a 'poster' to help other.

Take care

http://books.google.com/books?id=GGAOAAAAYAAJ&pg=PA430&lpg=PA430&dq=%22force+of+a+blow%22&source=bl&ots=SIxvmYZ50f&sig=vcKeFaVDGM_OW4CwRzBoK3YMdLM&hl=en&ei=wKR9SvaXCtmLtgfAyd3cAQ&sa=X&oi=book_result&ct=result&resnum=2#v=onepage&q=%22force%20of%20a%20blow%22&f=false

look on page 430, 431.

As I understand it, you have a piece of 1/2" rebar bent into a "U" shape, the bottom of the "U" covered with a plastic, .250 to .325 thick. The two legs of the step are free of plastic coating. You want to know if the plastic coating will remain intact after driving the step into a concrete wall (Pre drilled) with a sledge hammer, right?

yes, but the entire u bent rebar in encapslated in plastic, right

Hello mrhemmightdo,

You have now moved the 'goal posts'.

Hitting this thing with a 5Lb hammer is not the same as dropping a 5Lb weight onto it!

When hitting it with a 5Lb hammer you are swinging, you will be putting, adding more force than there may be in a 5Lb steel block dropping onto it.

Have you thought of using a block of wood against the item being knocked into the wall or floor, then hitting it?

Take care.

bb,unfortunately we cant control how its installed in the field. and the ballpark thing was because of the difference in swinging a hammer and dropping a weight.

and to answer everyones ? in the test, the step doesn't move when impacted like it is the field being installed. its set on a steel plate.... were testing for cracking.

Hello mrhemightdo,

Is what you make these steps you mentioned? And am I right in thinking you have a team which fits them?

I suggest if you cannot get your 'team' to do as instructed by way of fitting these, that perhaps you should think of finding other individuals to replace the people who do not listen to you? That is what I would do with my team. You talk as if you have no control at all. Well, if it is your company you do have control.

You asked about an 'impact test' when in fact it seems you are worried about the possible damage to the steps and or the steel or plastic covered steel? If you are just worried of the possible damage it is a well tried practice to cushion the item you are hitting with a block of wood or something softer than the step and steel.

I wish you luck. But will say one thing about your 'attitude' as you write your replies and answers. Do not be so cocky. Remember me, or we at this site have no idea what you are wanting to do or have have in mind or have planned. So, do not assume we can see what you are doing OK? Meant in the kindest way.

When corresponding, we are all equal. If anything often the 'askers' are less 'know it all', simply because they ARE the questioners who are trying to solve something. I say that not in any kind of pompous way at all, and anyone on here will probably tell you I am far from pompous. But I feel this has to be said. Do not be so 'short' in any future requests, it does not come over very well.

Hi Randall,

How are you?

I thank you for your reply, and accept your 'ticking off' with grace, though I will stay with my first thoughts.

I think a "Fitting kit" would be a very good idea.

Problem is, I and others it seemed, found the OP's explanation a little 'wanting' and the thread was quite long before I realised what was needed. I had the idea that the re-bar ran along the step 'angle' edge, to stop the step from wearing, and that the test was to hit the re-bar and concrete together.

I think perhaps the re-bar is the problem, and it sounds as if the bar was rusty before the plastic was applied?

Take care.

"the step doesn't move when impacted like it is the field being installed."

Then all the calculations were useless! Your dropping hammer will be useful for comparing one batch of plastic to another, but not much for predicting what will happen in the field. In this case, Doogleass's suggestion is the best. Get your biggest, clumsiest, collegue to do the hammering.

Here. If you don't want the engineering, buy a selection of these. Put them on the real thing and see what the shock is. Then duplicate it with the dart. Use the left over ones to see what the package express services are doing when you're not around!

I know I am late, but just saw that one.

You got many answers, some are correct of course. It simply is called the "force of a blow", as in driving a pile into the ground.

The KE is W*S, here 5 lbs * 12 feet= 60ftlbs e.g.

The energy is equal to the energy absorbed by a pile being driven a distance into the ground, which is calculated: mean Force * distance.

The impact force is not accurately known at any moment however the average can be calculated by the distance, see this formula:

F av = W*S /d where d is the distance the pile moves down on impact.

There is no need to use time or velocity or such.

It was correctly stated in one of the answers before, that the force depends on the movement it generates.

If that movement is small the force is high and vice versa. If there is no movement at all, which is not really possible, than the force approaches infinity. If your object that is dropped bounces back then likely there was no movement or only a small indentation on the surface being impacted.

You can compare that with dropping a tennis ball. The ball deforms a distance 'd' and bounces back, where the force is not that high. If you drop a steel ball (or steel rod) on a hardened steel surface of a solid and heavy body then the ball or rod bounces back just like the tennis ball where both parts likely deform to some extend. The deformation can, but must not be permanent.

In conclusion, no answer to the impact force is possible without knowing the distance the rod travels during its deceleration to zero velocity.