Ensuring Fall Protection Compliance: Addressing Queries and Providing Guidance - PDF Download

Dear All,

Please help me in clarifying the concept about the following:

In a lot of write-ups, I have read the below sentence:

"Anchor points for fall arresting systems (including lifelines) must be capable of withstanding a 5000 lb. (2300 kg) load per person attached."

But how do we come to this conclusion? Is it really required to ensure that a lifeline or anchorage point has the strength to withstand 2300 kg?

When a normal person (weight of 60 or 70 kg) falls, how much force is generated? How can we calculate the same?

Hope to get great participation from all of you in the thread...

Regards,
Dipil Kumar V

Dear Dipil,

Sorry to say, I don't have an idea regarding the above-mentioned query. Give me some time; I will check with my seniors, and then I will get back to you.

Thanks & Besafe,
Raghu

Dear Sir,

As we know, the physics formula is F = ma, where F = Force, m = Mass, and a = Acceleration due to gravity (9.8 m/s). The normal length of a safety belt is 2-3 meters. If a person falls from a height, they require 2-3 meters to get support from the anchor point. Let's consider the 3-meter length.

Now, calculate the acceleration for 3 meters. It is 9.8 * 3 = 29.4.

F = ma
F = 70 (normal weight of a person) * 29.4 = 2058.

So, it is recommended to have a capacity greater than this.

Thanks

Dear Vinod Digwal,

Thank you very much for your quick response and easy explanation. Please continue to participate and share your expertise with us.

@ Raghu

Great positive attitude! Keep participating in each thread and strive to address our concerns effectively.

Regards,
Dipil Kumar V

Dear Vinod Digwal,

Thank you for explaining the subject in great detail. It was a very useful piece of information that I lost track almost 25 years ago.

M.V. Kannan

Dear Vinod,

Excellent reply, thanks a lot.

Dear Dipil,

I checked with my seniors, and finally, I got the solution. Please read the below-mentioned points. This information was provided by my seniors, but I didn't receive the calculation:

References in the OSHA and EM-385 are based on ANSI standards for Fall Protection.

All standards allow for engineered fall protection systems (designed by a "qualified" person, usually a Registered Professional Engineer - licensed by a State or Federal Agency) that meets safety factors. All engineered fall protection systems must be tested (drop test of a standard weight). The employer who designs and uses such a system assumes all product and regulatory liability for the use of the system.

However, if the employer chooses not to use an engineered system, then the anchors and fall protection system components must meet the minimum ANSI standard.

Fall protection systems are designed with several goals:

- Arrest the fall before the person reaches the ground.
- Arrest the fall in the shortest fall distance reasonable - maximum 2m (6ft) and preferred 0.75m (2ft).
- Arrest the fall with minimum or no injury to the person, no shock force greater than 900 ft-pounds to the person.
- Keep the system as light as possible without sacrificing durability and integrity.

The different components of the system are designed to different strengths - 1800 lbs (825 Kg) strength webbing and fittings on harness, 3600 lbs (1500 Kg) attachments and lanyard, and 5000 lbs (2300 Kg) on anchors and attachments. These numbers are based on calculated loads of a person weighing up to 310 lbs (150 Kg) free-falling a distance of 6 ft (2m).

These numbers are not flexible. Doing anything less is most often considered negligence on the part of the employer.

Hope this helps.

Thanks & Besafe

Raghu

Dear Friends,

I really appreciate the efforts taken by Mr. Reghuvaran Raghavan in answering this typical question.

I also feel disappointed with the post from Mr. Vinoddigwal. This calculation is just confusing for me. If further clarifications can be provided, it will be of much help for me in understanding the calculations.

It is basic in fall protection that the lanyard length is strictly 1.8 meters to be effective. Now shock-absorbing lanyards are only used. With this, a fall will be arrested at about 1.07 meters, and as such, even the 1.8-meter length cannot be considered before deceleration starts. We cannot use a 3-meter length for the safety belt lanyard at any cost. I would appreciate it if anyone can come up with a calculation that is not even available with OSHA.

Much time has passed since this value was originally defined by OSHA way back in 1970. Unfortunately, there exists no clear explanation as to the origin of the 5,000 lb value.

In general, it is required by OSHA regulations to ensure a lifeline or anchorage point capable of supporting at least 5,000 lbs (2,300 kg) per employee attached. However, it need not be so if it is designed, installed, and used as part of a complete personal fall arrest system that maintains a safety factor of at least two, under the supervision of a qualified person.

This clearly establishes that the lifeline or anchorage need not be strictly of 5,000 lb (2,300 kg) capacity. It can be substantially reduced by designing a suitable fall arrest system by qualified personnel.

Nowadays, fall arrest equipment is labeled to produce a 900-pound maximum arrest force if the equipment is used in accordance with the manufacturer's recommendations. OSHA also mandates that the maximum arrest force that may be delivered at one's full body harness D-Ring during fall arrest be 1,800 pounds. Under these circumstances, a 2-to-1 safety factor against the anticipated peak dynamic load during a fall arrest event would yield 1,800 pounds (2 times 900 pounds) or 3,600 pounds (2 times 1,800 pounds) respectively. Both these figures are well below the 5,000-pound requirement outlined above. It again establishes that it is not a 5,000 lb capacity that is required of the anchorage all the time.

Regards,

Kesava Pillai

@ Raghu

Thanks for your answer...

Your answer states as follows: "These numbers are based upon calculated loads of a person weighing up to 310 lbs (150 Kg) free falling a distance of 6 ft (2m)."

If we apply the formula brought by Mr. Vinoddigwal into this situation: F=ma a= 2*9.8=19.6 m=150 then F=19.6 * 150 = 2940 kg, which is higher than 2300kg. So, the question arises: can we use this formula for calculating this scenario?

@ Keshav Sir

Thanks a lot for shedding light on the wrong direction the discussion was heading...

Your post stating that "Now shock-absorbing lanyards are only used..." Is this mandatory? We are still using safety harness with a double lanyard only...

Hope to get more participation from all of you...

Regards, Dipil Kumar V

"Now calculate the acceleration for 3 meters... it is 9.8 * 3 = 29.4."

Dear Sir,

I would appreciate a clear understanding of why 9.8 m/s is multiplied by the distance traveled (3m) and not the time taken to cover that 3m?

Thanks in advance.

Dear Friends,

My humble request is not to waste your time over this if you consider it valuable. Even OSHA is not giving an explanation as to how they arrived at the 5000 lb capacity for anchorage, though it is their regulation.

It is true, as Mr. Raghu has mentioned, about the weight of a person in this calculation with OSHA at 310 lb, which is equal to 140.9 kg (140 as rounded off) and not to assume as 100 kg. A fall is arrested at 1.8 meters with the safety harness. In reaching that distance, it will never attain the terminal speed of 9.8 m/s. The calculation part provided by Mr. Vinoddigwal needs authentication. He may be able to advise us as to where he arrived at that calculation. Let us wait and see for clarification.

Regards,
Kesava Pillai

Dear all,

Thank you for the valuable answers. I learned from the various views and am really amazed by the different perspectives on the question and answers.

Apart from this, I have a small detail about fall arrestors. To be frank, I have not gone through the same. I hope you all may provide some useful information on this topic.

Thanks and regards.

Hi Friends,

It was very interesting and informative to go through the discussion on fall protection. But I would like to make a small correction in the calculations shown by Mr. Vinod. It goes like this.

Let's assume that a person weighing 70 Kg falls and is caught by his 1.8m (d) lanyard. During this process, let's assume that the lanyard stretches by 10cm (0.1m, denoted as "s") - (Any lanyard will stretch due to sudden loading to some extent).

The velocity of the falling person when the lanyard starts stretching can be given by:

sqrt(2*g*d) = sqrt(2*9.81*1.8) = 5.94 m/s

His acceleration is calculated as:

a = (v*v)/(2*s) = (5.94*5.94)/(2*0.1) = 176.58 m/s²

The G force is determined as:

a/g = 176.58/9.81 = 18

The force of impact (F) is calculated as:

F = W*G = 70*18 = 1260 Kg.

Therefore, the structure should be capable of withstanding this load. Normally, there is the consideration of a "factor of safety" at 1.5. If we assume the weight of the person as 85 Kg instead of 70 Kg, then the calculation results in 2295 (with a factor of safety of 1.5).

Please find an attachment that specifically addresses the impact load under different conditions.

@HKP

Thanks for your answer and the attachments. It seems to be your first contribution to the forum, and without a doubt, it's a really good one. Once again, thank you for your great participation. Keep sharing your expertise with us. I request other members to give their comments about this post.

Regards, Dipil Kumar V

Dear HKP,

Welcome to our EHS forum. You entered with a good and valuable explanation, mate. We need your contribution in the future as well. Keep sharing.

Thanks & Besafe,
Raghu

Dear HKP,

I appreciate your effort. However, I am sorry to say that this calculation is absolutely invalid.

First, remember that any calculation connected with a fall arrest system is based on a person of a maximum weight of 140 kg. We cannot change the weight and assume it to be 70 kg or 85 kg as we like.

Secondly, the lanyard length remains 1.8 meters only. Elongation is not permitted, and the material should be selected accordingly.

Kindly help by modifying your calculation on this basis, and let us see where we can reach.

Regards,
Kesava Pillai

Dear Sir,

Quite interesting discussions... Thanks for your value additions. I am sure those who are taking these discussions seriously will benefit a lot. Awaiting to reach the conclusion.

Regards,
Dipil Kumar V

Dear Kesava Pillai

I feel there is some misunderstanding with my post. I don’t have access to the OSHAS / EM-385 documents and hence I was not knowing that the weight of person is considered as 310lb or 140 Kg in OSHAS calculation for fall protection. I considered the weight assumed by Mr. Dipil in the first post. Sorry for the inconvenience.

In fact I was only pointing towards the calculation of the total force generated when a fall is arrested. In that matter, when you say that “Elongation is not permitted and the material should be so selected”, I agree with you to some extent only. In fact there is no know material on earth which is 100% rigid and does not deform when a force acts on it. Even concrete and steel have some form of deformation, only thing is that it’ll be of a very minute magnitude. If you refer to the attachment which I have put this will be clear to you. So in this case the lanyard material selected will be of a good strength material with the minimum coefficient of expansion, but there will definitely a small expansion, which has been considered for the calculation. In case we modify the calculation considering the 140 Kg person (He may be a bulky American !!) then the following results emerges

Weight (W) 140 Kg

length of lanyard (d) 1.8 m

Considering 5% Elongation (s) 0.09 m

acceleration due to gravity (g) 9.81 m/sec2

Velocity of the falling person when the lanyard starts loading (v) =sqrt(2*g*d) = 5.942726647



His acceleration (a) at this point =(V*V)/(2*S) = 196.2 m/sec2



G Force =a/g = 20

Force of impact =W*g = 2800 Kg

As you have mentioned in your previous posting --Now shock absorbing lanyards are used—the lanyards absorbs the shock by elongation only, just like a spring. Another thing worth noticing with this calculation is that if you increase the % of elongation ie. Use a more shock absorbing material the force of impact is reduced and vice versa. The force of impact experienced if the material of lanyard elongates by10% is only 1400 Kg. This principle is used for bungee jumping where an elastic chord is used by the person. As you have said if the elongation is not permitted then the value of “s” will be 0 and the force of impact will be infinity. Now regarding the factor of safety if you have access to any of the OSHAS document then the exact value can be used.

@HKP

Thanks for your post and the interest shown in the discussion.

@Keshav Sir, awaiting your comments please.

Dear all, always give your contributions, and let us make this place an assured forum to get great solutions to safety-related matters. Once this assurance is established, participation will increase, and undoubtedly, everyone will benefit greatly. Keep on sharing.

Regards,
Dipil Kumar V

Dear All,

What's the actual difference between deceleration and acceleration? Are we using the same formula a=v^2/2d to calculate both? As per the attachment from BHK, the formula is for deceleration. Do these two terms have any relation in our calculation to reach 2300 Kg? Requesting all to come forward and help in reaching a conclusion.

Regards,
Dipil Kumar V

Dear All,

This discussion is still incomplete.

I have referred to OSHA through the following link: <link outdated-removed> (Search On Cite | Search On Google)

It's stated that lifelines used on rock-scaling operations, or in areas where the lifeline may be subjected to cutting or abrasion, shall be a minimum of 7/8-inch wire core manila rope. For all other lifeline applications, a minimum of 3/4-inch manila or equivalent, with a minimum breaking strength of 5,400 pounds, shall be used.

Here, the value has increased again from 5000 pounds to 5400 pounds.

1. Which formula can we use for fall protection calculations? Is it:

1.1 F=ma

1.2 F = mg(h + x) / x

Where:

m = mass

g = gravity

h = free fall distance (from initial position to just before deployment of energy absorber or other energy-absorbing component)

x = stretch or extension of the fall arrest system during fall arrest

3. What should be the minimum and maximum strength of an anchorage point?

Please help me come to a great conclusion.

This will be of great help.

Thanks in advance and I am really looking forward to reaching a conclusion.

Dear All,

I tried to pull out the real calculations for my query from different sources, but each reply I am receiving leads to a new path or confuses me more.

The latest reply I received is as follows:

Regarding the values you mentioned, firstly, I want to remind you that the value mentioned by OSHA is a force value of 5,000 pounds (22.2 kN) but not a weight value (5000 pounds but not 5000 Ibs). The calculations below will be based on this fact (the force on the anchorage point is 5000 pounds = 22.2 kN = 22200 N).

- The height (h) = 1.8 m
- The gravity acceleration = 9.8
- The mass = 140 kg
- The distance that the falling object will travel after the impact (d) = X
- The force of impact (F) = 22200N

Force of impact = Kinetic Energy / The distance that the falling object will travel after the impact -> F = KE/d -> d = KE/F (note that d is the only parameter that is unknown)

F = 22200N

KE = PE (potential energy) = mxgxh = 140x9.8x1.8 = 2469.6

d = 2469.6 / 22200 = 0.111 meter

So, for you to reach the value of 22200 N as the force on the anchorage point:

- The height should equal 1.8 m
- The mass of the falling object should equal 140 kg
- The distance that the rope of the harness should extend before bringing the falling object to rest should equal 0.111 m

Note that in the above calculations, no safety factor is included.

Can anyone come up with actual calculations? Are the above calculations correct or not?

Looking forward to reaching a great conclusion.

Thanks in advance.

Dear All,

Just sharing two more links related to the topic:

1. [Solve problem related to impact force from falling object](http://www.livephysics.com/tools/classical-mechanics/solve-problem-related-to-impact-force-from-falling-object.html)

2. [Link outdated-removed](https://www.citehr.com/results.php?q=Personal%20Fall%20Arrest%20System%20 Section%20I%20Mandatory%20Sections%20II%20and%20II I%20Non%20Mandatory%201910%2066%20App%20C) | [Search On Google](https://www.google.com/search?q=Personal%20Fall%20Arrest%20System%20Secti on%20I%20Mandatory%20Sections%20II%20and%20III%20N on%20Mandatory%201910%2066%20App%20C)

Please share your views...

Dear Keshav Pillai Sir,

Requesting you to kindly come forward with your comments so that we can conclude this discussion. Please shed light on the exact calculation to reach the value of 5000 lb force. I believe none of the other forum members have an interest in reaching a definitive conclusion, hence I am requesting your assistance. I hope you will not disappoint me this time.

I have tried to derive the exact calculations but have not found them anywhere. It is now clear that we don't always need to ensure a 5000 lb capacity anchorage point. The only thing we need to know is the exact calculations.

Looking forward to hearing from you.

Thanks in advance.

"F" is force, "M" is mass, "A" is acceleration (here we call it 'g', i.e., 9.8m/s^2), "T" is the time required to travel the covered distance, "H" is the distance traveled or height or simply distance.

[ F = M times A ]

Work done = F x Distance traveled = F x H = M x A x H

Impulse = F x T = M x A x T .............. here acceleration "A" is the rate of change of velocity

So, Impulse = M x (V/T) x T = M x V, which is the same equation for momentum "p = m x v"

Therefore, I think the total force experienced during the fall arrest might be the sum of work done and Impulse.

Time "T" = Sqrt ((2 x h) / g)

Velocity "V" = Sqrt (2 x g x h)

These formulas come from the conservation of energy and the equation of distance traveled after time "T".

So calculate:

A person with a weight of 100 KG falling a distance of 1.8 meters and getting arrested without a fall arrestor.

Work done = 100 x 9.8 x 1.8 = 1764 kgm/s^2

Impulse = M x V

That is, V = sqrt ( 2 x g x h) = sqrt ( 2 x 9.8 x 1.8) = 5.9396969 m/s

Impulse = 100 x 5.94 = 594 kgm/s

So, the maximum arresting force or force experienced during arrest might be the sum

1764 kgm/s^2 + 594 kgm/s = 2358 kg or 5198 lb

But this figure is also greater than the 22.6 kN requirement of standards. Also, I have some doubts about units. Please correct them if anyone can.

I also heard about another equation:

[ frac{1}{2} times m times g^2 ]

I don't know how this equation is derived, but while computing, somewhat equal results are coming.

[ frac{1}{2} times 100 times 9.8^2 = 4802 ]

Suppose the measurement is in pounds and it is converted to kilograms, then 2178.156 kg.

I don't know anything about this equation and what its units are, etc.

Please, if anyone knows, help me.

Hi DIPIL,

I hope you read my post. There are two methods mentioned to calculate the force. The result of the first method was 2358 kg, and the result of the second method, which I am not certain about, is 2178 kg. So, to find the average:

(2358 + 2178) / 2 = 2268 kg

In pounds, it is 5000 lbs, and in Newton, it is 22.68 kN.