Impact Force 7

[brengine]

I know that this isn't an easy problem to solve (because of Impulse, deflection, and stiffness), but does anyone have an approximate formula for this:

1 kg mass, lets say a round steel ball
Dropped from a distance of 0.5m
And it strikes a steel plate.

What is the force seen by the plate?

 

[Haf]

You're right that this is not an easy problem to solve. In fact, I would go one step further and state that it is quite difficult. There was a thread a while back with a similar question. You might want to start there.

Here's the link: thread404-36240

Haf

 

[aberta]

The scatter of results will average around 1.8Kg, but you may want to use a 2 Kg, to be reasonably safe.

 

[brengine]

aberta,
How did to arrive at the 1.8kg? I read that when a "rigid" object is droped from a height just above the contact point, that the deflection of the surface is supposed to be 2X the static deflection. And since F=kx, and k doesn't change, then twice the deflection should result from twice the force. So I guess I was expecting at least 2X the the 1kg mass for any dropped height above zero...but of couse what I was reading made alot of "perfect" assumptions (i.e. "rigid" mass)...so that probably answers my questions right there. Anyway, I would be interested in your formulation method tho.
Thanks,
Ken

 

[GregLocock]

aberta's answer is wrong. The answer will be a force, not a mass, and even if he means 2 kgf it is a ridiculous number.

I don't know of a handy formula, but I'd guess a peak force something of the order of 500 N, but would not be surprised if it was 5 kN, or more.

That's assuming an infinitely thick block of steel of course.

Oh, and the material of the ball is an essential part of the problem.

OK, I've looked at Roark, worth a read, but no simple solution.

A likely approach to my mind would be to start with the momentum equation assuming a perfect bounce

integral F(t) dt= m*2*v

tfinal will depend on F depending on the elastic behaviour of the ball and target.







Cheers

Greg Locock

 

[GregLocock]

Simple approach : work out Hertzian contact stiffness of sphere and plate from Roark.

Then you have a grounded 1 sdof dynamic system with a known initial velocity (v^2=2gs) mass (1.0 kg) and stiffness.

The solution is then straightforward.

This asumes that nothing yields, which is the worst case for maximum force.



Cheers

Greg Locock

 

[aberta]

At Glasgow University in Scotland, experiments were done to determine impact loads caused by placing loads onto structures. The factor of 1.8 came out of these tests.
However, when you are considering impact loads, you cannot ignore stiffness of both materials and Hysteresis damping within the materials. For example, in a hydraulic (water) system operating at 2400 psi, we have valves when at times close very quickly and cause severe water hammer. It makes a frightful noise. We measured these pressures and surprisingly they were around 3900 psi.
If you are letting an object fall 0.5m, you will have to form a mass elastic system with damping properties of both materials. Then integrate step by step to determine the force or deflection. For commonly used materials, the result may surprise you (1.8kg).
As you well know yourself, your questiobn cannot be answered as the question itself is incomplete. Hope this helps you. Regards.

 

[GregLocock]

I'm sorry that does not apply to this case.

First clue: the answer should be a force, not a mass

Second clue: your result seems to ignore the height from which the ball is dropped.

Third clue: get a 2 kg bag of sugar. Balance it on your big toe. Take it off. Get a 1 kgf metal weight or rock and drop it on your toe from 0.5 m height.

Which hurts more? (Oh, please video this)

 

[NARAYANRAN]

here we are talking of energy and force equalance,which is a tough question to answer.

since u have told it is steel ball dropped on a steel ,both are deformable and hence rigid body approximation cannot be applied.
the collisionis moreover in elastic in nature.
I feel if u try to find how crash test or explicit dynamics of structures are done ,This would be a solution to ur problem.

In case u find out let us know
regards

 

[Speedy]

If there is no localised yeilding and everything is perfectly elastic (no damping in both materials) then it should be possible to work out the theoretical max force. I'll let someone else do that.

On a practical level the % loss of rebound height will give the % of energy absorbed.

Also, max force will be higher with a more elastic system than a equivalent (from same ht and ball wt) damped one, where the force curve will be flatter.

What about shock waves travelling through the material, if these were to resonate.....just a thought.

 

[raviteja]

when a mass is falling throuh a height,first find the time required to reach the ground(plate)from the formula

s=1/2*g*t*t
in this g=9.8 mt/sec2,s=0.5 mt(in this problem)we will gwt the time as 0.3 sec.now we have to the potential energy as m*g*h this is equal to work done per unit time i.e work/time.work=F*.S

m*g*h= F*s/t where m-mass,g-accleration due tograavity,
h-height,Fis the force in kg,t-time in sec.
substtuting all the parameters in the above formule and find the force.that is equal to 2.94 kg.
THE APPROXIMATE DISTANCE REQUIRED FOR IMPACT =2*MASS/WEIGHT IS 0.19 METERS ABOVETHEGROUND

 

[aberta]

Greglocock, why don't you try the experiment yourself and share the results with us. Your comments are not professional.Taking your case of sugar and steel when dropped, you will find that sugar will not bounce back because of its non linear stiffness and damping properties. Steel will bounce when dropped on steel and the response can be represented by relatively simple mass ealstic sytem. This system will become progressively complex as you take into account the shape of the plate and its end constraints. Hence the question is incomplete. The answer I gave earlier can be used as a general guide for a starting point. It represents impact forces normally experienced in shock loading and is only a guide. It was not intended to be an engineered answer.
Rraviteja, you are converging to a lower number.
Speedy correctly pointed out about energy dissipation by shochwaves. This would include sound,damping within steel, thickness of steel plate, energy tranmitted to the strucures supporting the plate or welded to it.
Best regards.

 

[CanEngJohn]

OK after spending an hour or two working on this using Shigley and the Machinist's handbook as a reference and manipulating tons of equations and assuming the steels are identical with a Poisson's ration of 0.3. Note the steel ball would have a diameter of 50 mm for 1 kg.

This gives a surface area of approximatrely A = PI*(13.5543F/E)^(1/3)
And given that Engergy in is totally used
and that the ball lands at the mid point of a 1 meter long steel plate
You can solve mgh = (F^2)/ ((PI* A)E)
or Solving for F (in this case mgh = 1*9.8*0.5 = 4.9 assuming the elastic deformation does not cause a significant increase to the potential energy)

You get F^5 = (4.9 ^3)*(PI^3)*13.5543*E^2
and you get F is approximately 18 kN.

Now can someone else please check? I am solving from Equations 2.88 and 3.29 from Shigley and Mischke (Elastic Strain Energy) and assuming a 1x1 mild steel plate and mild steel ball (E = 207.0 *10^6 and v=0.3).

Noting that in the above experiment we could all easily assume a 1kg steel ball could break a bone in the foot it lands on I don't think the answer is that unreasonable.

 

[GregLocock]

No aberta, you are WRONG. I was being polite, but since you persistently promulgated an incorrect solution I descended to sarcasm. I asked you to /place/ the bag of sugar on your toe, but to /drop/ the rock onto your toe. Please read for comprehension in future.

1) Why do you persist in giving a mass as a solution to a force problem?

2) Why (how) does your solution ignore the height from which the ball is dropped? Don't you think that would have a significant effect?

I fail to see how a general guide that is a factor of 100-1000 low at a rough guess is any help to anyone.

Cheers

Greg Locock

 

[brengine]

CanEngJohn,

I'm not sure were you got: A = PI*(13.5543F/E)^(1/3)
If you just look at the units, you get m^2=m^(2/3)

Same goes for:
mgh = (F^2)/ ((PI* A)E)

You get:
N = N^2/(Nm^2)=N/m^2-->Incorrect

But I don't know the answer either,
Ken

 

[CanEngJohn]

raviteja (Mechanical) writes

m*g*h= F*s/t where m-mass,g-accleration due tograavity,
h-height,Fis the force in kg,t-time in sec.

I would like to point out that you do not know t. This is the time of the impact and not the time of descent in an impact scenario. Also you need to know the distance over which the work is done. Which can be assumed to be the elastic deformation of the plate. This is why I headed to elastic strain energy theory. But on further contemplation when I had time....

If you equate energy then F max might be solved

m*g*h = F * d where d is F/E (the deformation of the plate is equal to the force divided by the elastic modulus.)

then m*g*h*E = F^2

and substituting we have F = 31,836 N. Which doesn't quite jive with what I put up earlier but I am pretty sure I made a mistake with the earlier equations. But to say it is 2 kgf would not be accurate I believe.

 

[CanEngJohn]

KenBolen (Mechanical)

Actually the 13.5543 also has units which I did not post. That is why I gave a reference for where I got the equations. I double checked the units and it seemed right but I wasn't sure the numbers were accurate. I only took an hour at lunch to approximate it and already had Shigley open.


Also E is in MPa which is N/m^2 so A*E is m^2 * N/m^2 which is N and hence N^2 / N = N. Also the true formula is

F^2 * l / (2*A*E) but I assumed a 1 m plate meaning l is 0.5 cancelling out the 2 but leaving a unit of m in the upper formula resulting in N.m = N.m.

Which I just realised this is the error in my first post. I cancelled out a 2 when it should have been multiplied by 2. This changes my result to 42 kN which I should double check again.

I like my new answer from the energy balance. Much simpler. I know I make mistakes when I do math on paper quickly so I am still expecting somone to tell me what else I did wrong with the strain energy equations.

 

[GregLocock]

Not very surprisingly Timoshenko covers this problem in "Theory of Elasticity".

Unfortunately I didn't copy all the right equations down, so I'll just solve the simpler case of a steel ball bouncing off its twin. If you think about that case it is likely that this is also the same as the steel ball bouncing off an infinitely stiff flat plane.

The time of the contact is

t=2.94(1.25*sqrt(2)*pi*rho*(1-nu^2)/E)^0.4*R*v^-0.2 eqn 244

R=0.031 m
v=3.132 m/s
rho=7843 kg m-3
E=210*10^9 N m-2
nu=0.3

t=0.1493 ms

The average force, F during the contact is 2*m*v/t (ie the change of momentum divided by the time)

F=42 kN, ie a little over 4 tons force.

In practice the time of contact with a flexible target will be longer so the average force will be lower. On the other hand a flat plate is likely to be stiffer than a twin of the sphere, so the time would be shorter. So maybe it is the right answer after all.

Timoshenko actually gives a direct solution for a ball on a flat plate, and the peak force rather than the average force, but it is spread over two pages. This solution assumes that the contact time is long compared with the period of the lowest modes of vibration.

Cheers

Greg Locock

 

[CanEngJohn]

In my energy balance equation I forgot to include the first moment of Inertia of the plate.

My above equation should have been d = F/EI. What I need to know is what is the first moment of inertia of a steel plate 1x1x0.01.

I vaguely remember an integral based on mass and gemoetry but I haven't been doing this in over a decade. I believe it works out to Ml^2/3 and in this case l = 0.5 so (78.33/2)*.25/3 = 2.55

But this makes it larger which I find harder to believe.

I haven't done or looked at this stuff in 10 years. Still waiting for someone to tell me where I messed up. I looked into my Mark's Handbook (6th edition) this morning but it was just as useful as what I am looking at now.

I know this is going to now keep me up until I think I have it right. Thank KenBolen.

 

[Speedy]

This discussion brought to mind 'shotpeening'. (Back to the real world of damping and plastic deformation.)
I did a quick search and found this;

http://www.shotpeener.com/learning/curve.pdf

There is an equation on page 19 which calculates the dimple diameter from shot diameter and density, shot velocity and yield strength of material treated. I assume it is empirical.