Impact Force
Impact Force
(OP)
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?
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?





RE: Impact Force
Here's the link: Thread404-36240
Haf
RE: Impact Force
RE: Impact Force
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
RE: Impact Force
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
RE: Impact Force
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
RE: Impact Force
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.
RE: Impact Force
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)
RE: Impact Force
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
RE: Impact Force
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.
RE: Impact Force
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
RE: Impact Force
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.
RE: Impact Force
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.
RE: Impact Force
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
RE: Impact Force
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
RE: Impact Force
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
RE: Impact Force
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.
RE: Impact Force
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
RE: Impact Force
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.
RE: Impact Force
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.
RE: Impact Force
It is wrong to assume no damping in the system. If there were no inherent dampings, every elastic member will exhibit infinite amplitudes of vibration. Thus both, the samll ball falling and the large impact plate will self destruct. I hope somebody will agree with this.
Problem is I am not as smart as I was forty years back when I did vibration analysis using computers (you now call them calculators). So I am going to assume no damping in the materials.
I am further going to assume that the ball has no stiffness, just the plate has stiffness.
Consider the case where a weight "W" falls from height "h" on a spring with stiffness "K". Assume the deflection in the plate is "d". The static force produced on the plate is "F".
Hence
W(h+d) = 1.5*F*d
If "ds" is the static deflection due to weight "W", then
F = (d/ds)*W
From these equations we get
F = W(1+sqrt[1+2h/ds])
When h=0, F=2. (P.S. I copied this from a book)
In this case, h=0.5m, assume ds = 10%of (0.05m)=0.005
Hence F=W(1+sqrt[1+2*0.5/0.005])=W*(1+sqrt{200])
=W(1+14.14) = 15.14W
This obviously tells me that I was wrong, provided assumptions are correct.
RE: Impact Force
P=k*y^(3/2)
where k is a constant that can be calculated from the geometry and material properties. Assume that this equation holds for the dynamic case we are now examining, as well as for the static case for which it was derived. (In other words, assume elastic behaviour and no energy losses.)
The equation of motion for the falling sphere after it has made first contact can now be written:
m*y'' = m*g - k*y^(3/2)
where y'' is the second derivative of y with respect to time.
Initial conditions are that at t=0 we have
y=0, and
y' = (2*g*h)^(1/2)
This differential equation can easily be solved numerically for the peak value of y, and then the peak value of the contact force can be calculated directly from that peak y.
HTH
RE: Impact Force
Using Roark's equation and a numerical integration the peak force is 68 kN for a flat plate (Timoshenko 67 kN), and the average during the 0.171 ms of contact is about 37 kN. The plot of deflection vs time is roughly a half sine wave.
Note that in practice a large steel plate may have modal behaviour that affects these results. I suspect this will soften the plate, reducing the peak force.
Cheers
Greg Locock
RE: Impact Force
Would it not be more accurate to calculate the deflection and stresses based upon the strain energy absorbed by the entire structure?
Even better, make sure the crane operator can not drop the load, and not consider this in the calculations?
RE: Impact Force
Most inspection tables (I believe 99%) are made of granite so that you can maintain an even wear surface.
In your scenario it WILL be damaged and you don't have too many other options for materials without having to worry about expensive calibrations on a frequent basis. My advice is to rig a lift assist to prevent the part from slamming into the inspection table.
Finally since it is a cast somponent I would definitely recommend the granite - 6 to 8" thick should do the job.