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Force of Impact

Force of Impact

Force of Impact

(OP)
Can somebody tell me how to find the load you would use to check a timber member if it is hit by a falling object.  I have an idea of how much the object will weigh.  I also know the distance it will fall, so therefore, I know the velocity at the point of impact.  To keep me from digging through my old dusty textbooks, what formula should I use to find the force on the timber member?  I am calculating this using English units.

Thanks for your help in advance.

RE: Force of Impact

2001 National Design Specification (NDS) calls for an impact load duration factor of 2.0 for untreated wood structural members (lumber pressure treated with preservatives or fire retardant chemicals has an unspecified lower value). This 2.0 factor is only for the member, the connections are allowed a factor of 1.6).

www.SlideRuleEra.net idea

RE: Force of Impact

S = 1/2 a t^2 and F = Ma, the "a" being the rate of deceleration.

Mike McCann
McCann Engineering

RE: Force of Impact

I remember calculating the force due to a falling object as:
F=1/2mv(squared) or the equation for Kinetic Energy.  Please verify it for yourself since it has been many years.

RE: Force of Impact

Ismfse-

That's the equation for kinetic energy. Related to the issue, but of no use per se. Slideruleera is right. You need to estimate the deceleration in order to approximate the force of impact.

RE: Force of Impact

Please refer to: Design of Welded Structures, Section 2.8 by Omer Blodgett, Lincoln Arc Welding Institute ($30). It covers exactly the type of problem that you are describing.

RE: Force of Impact

You can solve this with conservation of energy:

1. Assume no losses to damping etc (conservative)
2. Calculate the potential of the object before it falls (E=mgh)
3. Calculate the deflection of the beam to equal the same potential energy under a point load (E=kxx/2)
4. Calculate the force required to produce the deflection.

the spring constant "k" will depend on the beam properties and support conditions

RE: Force of Impact

NKT-

That's actually a pretty good idea.  

You could get the stiffness by applying a unit load on the beam, calculating displacement and taking the reciprocal.

RE: Force of Impact

NKT

I would agree with your method. Depending on the circumstances it may be conservative, as it assumes the object comes to a stop. A large object hitting a single beam will most likely alter direction, not come to a complete stop (or rebound upwards).

Using conservation of energy is definately the first method I would use to check this case. Depending on the results and actual scenario, you might want to look at what portion of the object will strike the beam and what the expected reaction is, in terms of deflection/rotation of the object. That would be a lot more complicated, but it can still be solved by energy methods.

RE: Force of Impact

NKT,

You said exactly what I was going to recommend.  Note that the location of impact will provide different results, because there will be different stiffness at varying locations.  Also note that if you expect rebound of the load, you have to add that energy to the initial energy to compute the deflection & force.  And after all that was done, I would still want a factor of safety of at least 2 unless this is the primary load mechanism, in which case I would want even more.  Good Luck.

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