Impact load factor.
Impact load factor.
(OP)
I am looking to determine an impact load factor. The closet I could come to describing the problem is as if a solid steel ball was dropped from a certain distance onto a rigid floor of solid steel. Is there a factor or simple equation for this situation? All the reference material I looked at state that this is very complicated and either involves an impact factor or energy calculations. The problem with energy calculations in my case is that the impact is absorbed instantly instead of over a measurable time. The actual impact velocity in my case is from a round steel body, rolled down a 5-degree decline with an elevation of change of about 6 inches. The velocity is not very high but cannot be neglected. The impact is with a rigid structure with no measurable deflection or measurable time over which the body comes to rest.






RE: Impact load factor.
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RE: Impact load factor.
RE: Impact load factor.
Force = Normal Velocity * SQRT(K*weight object / gravity)
K = stiffness of member absorbing the load
This is generated by setting KE = PE
KE = kinetic energy of body moving normal to member
PE = Potential energy of the deflected plate
KE = m*v*v/2
PE = F*F/(2*K)
I don't think this is to much to go through to generate a better solution. It is not that what SRE is telling you is wrong, I just don't see why you should shortcut the solution.
RE: Impact load factor.
I hope this helps.
Jordan
RE: Impact load factor.
I appreciate your reply but it is not as simple as that. I did look at "Design of Welded Structures" section 2.3 to be exact. The problems involve a time for an object to come to rest or distance or deflection. I cannot measure either of these. If you don't mind helping me out, the body is a plain steel roll, impacted squarely on the round side, dropped 12 inches and has a weight of 70,000 lbs. Thanks in advance. I think I calculated once back in school that a 1 pound weight dropped 12 inches exerts a force of approximately 8 pounds. This is much higher than the IF of 2 suggested earlier. I think that even 8 is too low since the shorter the stopping distance and the shorter the stopping time the higher the force. Actually approaching infinity. In practical situations this is not considered.
RE: Impact load factor.
RE: Impact load factor.
RE: Impact load factor.
I would reduce the situation to a time-varying contact problem. Numerous texts will give you formulae from which you will be able to calculate the force versus relative displacement relationship for the (static) contact problem. Use this relationship in a time-stepping solution to the motion of the falling body.
RE: Impact load factor.
thread404-74498
for an earlier discussion, with a reasonably complete description of how to solve it.
Cheers
Greg Locock
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips.
RE: Impact load factor.
F = Wb + SQRT(Wb*Wb + (K * Wb * V*V / g))
Wb = Weight of ball = 70000#
K = Stiffness of floor beams absorbing the load = Force / deflection. (Solve using deflection calculations)
V = ball normal velocity at impact (You can solve for this using free-fall equations, etc.)
g = acceleration of gravity.
This neglects the floor inertia. If you want to include it you can by using the next section in the book. As you can see the impact factor of (2) comes from the above equation when the velocity at impact is zero, but the load is suddenly applied.
The solution I gave in my first post is a variation of this from Solutions to Design of Weldments. I was not trying to upset anyone with my previous post. I guess I have used this approach many times before designing chutes and bins for processing equipment so I felt that wagging an impact factor was an oversimplification. Note that the root of the above equation would have acceleration in them and thus you would need to know the time interval from impact velocity to stop. Since we can re-write the equations using elastic potential energy the time interval cancels.
RE: Impact load factor.
i guess the easiest way to determine K would be to model the structure and apply a unit load. mind you, it's got to be some awfull big structure to support a 70,000 lbs ball.
looking at the equation you could replace v^2*W/g with 2Wh (converting the KE into the original PE).