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?

 

[aberta]

One way to warm up in cold Canadian climate is to read eng-tips. It is good for the soul as well.
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.

 

[Denial]

Roark (Table 33, case 1, page 516 in his Fifth Edition) gives an equation relating the contact force (P) to the distance that the sphere has moved closer to the plate than its first contact. The equation has y proportional to P raised to the power 2/3. Switching this around, we arrive at a non-linear spring equation for the contact:
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

 

[GregLocock]

Roark and Timoshenko pretty much agree (thank goodness) for the two extreme cases, ie r1=r2 and r1=infinity

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

 

[franklineng]

I am working on an inspection table for a machined cast component weighing 100 lbs. I am assuming the casting could be dropped accidently at a height of 24 inches. It would strike plate on the top of the table. I've seen the formulae in Roark's: ratio of impact deflection/static deflection. Because the table is for inspection purposes I am assuming, the impact deflection is very small (0.005 in). Using the formulae the ratio is 220. Wouldn't the structure have to be quite massive?

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?

 

[CanEngJohn]

Franklineng -

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.