Impact load factor.

[Expengineer]

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.

 

[SlideRuleEra]

I understand that you are not talking about a steel structure, but unless there is a better reason, I like to use a 100% increase in live load (a factor of 2.0) to account for impact loading. This comes from the AISC Manual of Steel Construction, 9th Edition, ASD, Part 5, Paragraph A4-2 (page 5-29) in my copy.

http://www.SlideRuleEra.net

 

[Bagman2524]

I asked a similar question a few weeks ago and someone pointed me to Blodgett's Design of Welded structures (Chp 2.8). I think 2.0 is a good number to use.

 

[aggman]

If you haven't taken a look at Design of Steel Structures by the Lincoln Arc welding foundation you should. I guess I just don't agree that the floor is rigid and their is no measurable deflection. If the material is elastic then it will deflect. The deflection is directly proportional to the load applied. Using energy formulas you should be able to simplify the equation to

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.

 

[woodengineer]

ASCE 7-02 Section 4.7 does address impact loads. The code does not specifically address a solid steel ball dropping from a certain elevation. For example: Elevator loads shall be increased by 100% for impact, and the structural supports shall be designed within the limits of deflection prescribed by Refs 4-1 and 4-2. For your example, it would seem as if it is similar to a crane load. This is found in section 4.10 of the ASCE7-02.

I hope this helps.

Jordan

 

[Expengineer]

aggman,
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.

 

[rb1957]

i'd respond to aggman by asking if a very quick, reasonably conservative answer is good enough, why put in unnecessary work (to count the angels dancing on the pinhead) ?, particularly if you have to make more assumptions about the response of the rebound surface ?

 

[rb1957]

to expengineer, i think it is a reaonable shortcut to ask the effect of the dynamic load is twice the static load ... i remember this relationship from my (long ago) school days.

 

[Denial]

As I understand things, the factor of 2 arises from consideration of dynamic variation in the applied load, being the upper limit that applies in the case of a load that is applied "suddenly". This is not true "impact" in the sense that Expengineer is using the word.

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.

 

[GregLocock]

The factor of two is an unsafe assumption in this case. If you truly wish to regard your target as infinitely stiff, then you need to work out the hertzian contact stiffness of the cylinder (which will be non linear), and then build a time based model of your impact.

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.

 

[aggman]

See page 2.8-3 of design of steel structures for your solution.

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.

 

[rb1957]

i stand corrected ... i agree that 2 is appropiate for a "suddenly" applied load rather than a true impact.

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).