Solving for impact loading 4

[woodman1967]

I am doing an overhead steel beam that will have a fall arrest system that will run along the bottom flange of the beam. How do you determine the actual force? Up here in Canada we used to have a factor of 1.65 for loading lasting less than 10 min. This is a project for the US.

Basically, if a 200 lbs man fell what would the force be? Would it be as simple as f=ma? Or is there a factor to use?

Thanks

 

[Lion06]

I believe we typically use a load of 1.5 kips in any direction. You can lower it if you meet certain criteria, but that is a good starting point.

 

[JStephen]

http://edocket.access.gpo.gov/cfr_2007/julqtr/pdf/29cfr1910.66.pdf

29 CFR 1910.66, Appendix C, Section I-1

(10) Anchorages to which personal fall arrest
equipment is attached shall be capable
of supporting at least 5,000 pounds (22.2 kN)
per employee attached, or shall be designed,
installed, and used as part of a complete personal
fall arrest system which maintains a
safety factor of at least two, under the supervision
of a qualified person.

Different lanyards are available that limit force to, say, 900 lbs, which would let you use a design load of 1,800 lbs per employee. Start with the lanyards and work back from there. Or, if you don't have much control over how this is actually used, go with the 5,000 lb static load.

 

[asixth]

We had this situation arise in my office the other day. Basically for impact loading, we used Newton's Law of Motion...

1. You can estimate how much the person weighs.
2. You can estimate how long they will freefall and therefore, determine their velocity.
3. You can estimate how long it will take them to go from freefall velocity to zero, assuming the cable has a bit of give in it, I would say 0.2 seconds.
4. With all this known, you can then calculate an acceleration that opposes gravity.

Of course, being engineers that are concerned with safety in design, I would look at whether a safer measure can be put in place such as a catwalk. We don't want to be responsible for anyone breaking their back.

 

[NKT]

"Would it be as simple as f=ma?"

Using f=ma is not necessarily a simple way to do it... the acceleration is difficult to determine and requires simplifying assumptions that make the accuracy questionable.

A simple way is to use conservation of energy; equate the potential energy of the person before the fall (E=mgh) to the potential energy after the fall (E2=kx^2). In your case, k will be the total deflection of the beam and fall-arrest system under a unit load, and "x" will be the deflection of the beam and fall-arrest system once the person comes to a rest at the bottom of the fall. Once you find the deflection, calculate the load required to cause it.

This has been discussed recently, see thread 507-216709.

 

[frv]

NKT-

k would be the stiffness of the beam, not the deflection. x, then, is the deflection. Solve for x, and back out the load required to produce it.

 

[Galambos]

Wow. I'm surprised by the advice on this thread. Please dont calculate fall arrest forces by estimating velocities, or potential energies and please follow JStephen's advice. you must use at least 5 kips unless you are a Qualified Person as determined by OSHA.

to become a Qualified Person, you must have the appropriate fall safety training to first become a Competent Person. Having the PE will then qualify you as a Qualified Person.

if you need more info, there is one book everyone turns to: Introduction to Fall Protection by Dr. Nigel Ellis.

 

[woodman1967]

Thanks everyone for your help,

I went JStephen's advice since his came right from OSHA.

Thanks again
Malcolm

 

[frv]

I agree with you, Galambos. I was just clearing up the energy methodology.