SAFE DEAD WEIGHT LOAD FOR LIFE-LINE UNDER DYNAMIC LOADING 10

[peakpilgrim]

Hi

SAFE DEAD WEIGHT LOAD FOR LIFE-LINE UNDER DYNAMIC LOADING

I need to calculate a safe dead weight load for a life-line. The details have been simplfied for clarity.

A man falls two metres on a life-line (Attached). What dead weight load should support him. How is this calculated?

Apologies for the quality of my drawing!

Please advise

Yours sincerely,

peakpilgrim

 

[CANPRO]

It depends on a few things - what kind of fall arrest gear does the user have? This will determine the maximum arrest force (MAF). You also need to determine what kind of friction coefficient you'll get between the roofing material and your counterweight. Check your local codes, there might be a testing requirement to confirm the friction, or this style of fall arrest might not be allowed at all. Once you have the MAF and the friction value, the math is pretty simple. If you need help with the actual calculation, you should find a senior engineer within your company (or hire one) to do/review the work - this is life safety stuff, if you aren't 100% what you're doing then you shouldn't be doing it.

 

[peakpilgrim]

Hi

Thanks for the advice.

As I stated earlier the situation has been simplified for clarity.

The friction will be concrete on concrete

If I knew the answer I would not be asking the question.

I am not in the business of putting peoples lives at risk either.

I am informed that the weight of the block should be of the order of 2.5 tonnes but I want
to verify this independently.

If you are unable to help further then please stay of the thread.

Cheers,

Peakpilgrim.

 

[CANPRO]

peakpilgrim, I did not mean any offence in my first reply, but based on the general nature of your initial post, I get the impression that this type of design is not in your comfort zone.

Once you define your maximum arrest force, you can determine the tension in the cable, and the lateral force applied to the concrete block. Then you can draw a free body diagram of the concrete block. You'll have three forces applied - the tension from the cable, the friction at the base, and the vertical reaction at the base of the block. This basically gets you into a high-school level physics problem - you should be good from there.

 

[Tomfh]

There should be a standard for it based on your area.

Here in Australia you typically design for a static load of ~20kN

If you don’t want the block to move at all it will need to be fairly heavy.

If you’re prepared for it to slide somewhat it can be lighter.

 

[Celt83]

The counterweight will impose a significant load to the roof structure. If it hasn't been already recommend having that checked as well to verify the roof can support it.

As the others have stated the static arrest force is usually defined in a local standard. For the counterweight to rely on friction you should also do a check on overturning to prevent it from rolling over after some friction resistance has been engaged at the base.

Open Source Structural Applications:

https://github.com/buddyd16/Structural-Engineering

 

[Geoff13]

Canadian Standard Z259.16 for design of fall arrest systems has a formula for a conservative ballasted anchor weight. Using some typical values for the parameters gives a weight around the 2.5 tonnes you noted, though this is obviously very sensitive to the coefficient of friction assumed. I could easily see this weight needing to be higher. You should review the fall arrest code for your jurisdiction to see if it provides any requirements.

 

[robyengIT]

in Europe we use systems as per EN 795, were are foreseen anchorage hooks, steel cables etc. Ballasted anchor weights are unknown and not allowed, as far as I know. May be these are more efficient

https://www.safesite.co.uk/products/mobile-man-anchor

 

[TLHS]

It's kind of weird to tell a guy who's just given you useful information to not post. You asked a bunch of internet strangers to give you free advice on something that you're being paid to do.

I'm going to assume that you somehow took offense at Canpro pointing out the serious nature of the work. Nobody knows who you are, what your experience is, or how qualified you are. Pointing out risks is completely fair. Hell, it's a fair reminder for anyone in any situation because sometimes we all overstretch. It seems like you would at least give the people responding to you the benefit of the doubt that they're responding in good faith.

It's an interesting problem, but since you don't want people posting in the thread if they're going to ask you questions you don't like, I probably can't help

 

[Earth314159]

2.5 tonnes is not enough.
Most codes require a load of 5Kips. You then have to factor that by 1.5. The coefficient of friction is probably around about 0.3. You have to multiply that by a resistance factor which i do not believe is noted in any code for such a situation. I would assume about 0.4 to 0.5 for a resistance factor (since you don't experimental data or very solid information to go off of) if I could not find any other source of information. Tf=7.5Kips. Tr=0.4x2.5x2200x0.3=0.66Kips.

In reality, you will get energy dissipation from the sliding of the weight but it probably shouldn't be considered safe.

 

[kissymoose]

For the US, here's a neat article I found about the 5000lb rule: article
He kind of rambles a bit and isn't very clear when describing the numbers, but the OSHA references and overall message are great to learn.
Key points:
•1926.502(d)(15) gives the 5000lb requirement or requires a complete design with SF > 2
•1926.502(d)(16) gives the requirements for the design which includes limiting the arresting force to 900lbs for someone with a body belt, 1800 with a body hardness, fall no more than 6ft, decelerate in less than 3.5ft

•So many fall arrest systems utilize deceleration devices that limit the force to 900lbs, thus the design force would be 1800lbs​

​

•Aka its product specific​

•The 5000lb, prescriptive method only qualifies employees/gear under 310lbs.

•so even the lick it and stick it number can't be used all the time​

If you wanted to use friction (0.75 concrete to concrete, 0.375 with SF=2) with a big block, you'd need 5000/0.375=13500lbf block. That's not gonna happen.
FYI they make non-penetrating fall arrest systems you can just buy.