dear all,
thanks for the valuable answers. I learnt from the various views. really amazed on the views of the question & answers in various angles.
apart form this I have a small detail about the fall arrestors. to be frank, i have not gone thro' the same. hope you all may get some useful info on the same.
thanks & regards.
From India, Delhi
68thanks for the valuable answers. I learnt from the various views. really amazed on the views of the question & answers in various angles.
apart form this I have a small detail about the fall arrestors. to be frank, i have not gone thro' the same. hope you all may get some useful info on the same.
thanks & regards.
From India, Delhi
Hi Friends
It was very interesting and informative to go through the discussion on fall protection. But I would like to put 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 and during this process let's assume that the yard stretchs by 10cm (=0.1m say "s") (Any lanyard will stretch due to sudden loading to some extent).
Velocity of the falling person when the lanyard start stretching can be given by
[sqrt(2*g*d)]= [sqrt(2*9.81*1.8)]=5.94 m/s
His acceleration is a=(v*v)/(2*s) =(5.94*5.94)/(2*0.1) = 176.58 m/s2
G force = a/g = 176.58/9.81 = 18
Force of impact F = W*G = 70*18 = 1260 Kg.
So the structure should be capable of withstanding this load. then normally there comes the "factor of safety" @1.5. If we assume the weight of the person as 85 instead of 70 then the calculation comes to 2295 (with a factor of safety of 1.5).
Also pl find an attachment which exactly deals with the impact load on different conditions.
From India, Mumbai
It was very interesting and informative to go through the discussion on fall protection. But I would like to put 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 and during this process let's assume that the yard stretchs by 10cm (=0.1m say "s") (Any lanyard will stretch due to sudden loading to some extent).
Velocity of the falling person when the lanyard start stretching can be given by
[sqrt(2*g*d)]= [sqrt(2*9.81*1.8)]=5.94 m/s
His acceleration is a=(v*v)/(2*s) =(5.94*5.94)/(2*0.1) = 176.58 m/s2
G force = a/g = 176.58/9.81 = 18
Force of impact F = W*G = 70*18 = 1260 Kg.
So the structure should be capable of withstanding this load. then normally there comes the "factor of safety" @1.5. If we assume the weight of the person as 85 instead of 70 then the calculation comes to 2295 (with a factor of safety of 1.5).
Also pl find an attachment which exactly deals with the impact load on different conditions.
From India, Mumbai
@HKP
Thanks for your answer and the attachements... It's seems your first contribution to the forum and no doubt, its really good one...
Once again thanks for your great participation... Keep on share your expertise with us...
Requesting to other members to give your comments about this post...
Regards,
Dipil Kumar V
From India
Thanks for your answer and the attachements... It's seems your first contribution to the forum and no doubt, its really good one...
Once again thanks for your great participation... Keep on share your expertise with us...
Requesting to other members to give your comments about this post...
Regards,
Dipil Kumar V
From India
Dear HKP, Welcome to our EHS forum u enter/Start with Good/Valuable explanation mate. . .We need ur contributioni in future also. . .keep sharing. . . Thanks & Besafe Raghu
From United States, Fpo
From United States, Fpo
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 fall arrest system is based on a person of maximum weight of 140 kg. We can not change the weight and assume it to be 70 kg or 85 kg as we like.
Second the lanyard length remains 1.8 meters only. Elongation is not permitted and the material should be so selected.
Kindly help by modifying your calculation on this basis and let us see where we can reach.
Regards,
Kesava Pillai
From India, Kollam
I appreciate your effort. However I am sorry to say that this calculation is absolutely invalid.
First remember that any calculation connected with fall arrest system is based on a person of maximum weight of 140 kg. We can not change the weight and assume it to be 70 kg or 85 kg as we like.
Second the lanyard length remains 1.8 meters only. Elongation is not permitted and the material should be so selected.
Kindly help by modifying your calculation on this basis and let us see where we can reach.
Regards,
Kesava Pillai
From India, Kollam
@ Keshav Pillai
Dear Sir
Quite interesting discussions... Thanks for your value additions...
I am sure those who taking these discussions seriously will benefict a lot...
Awaiting to reach to the conclusion...
Regards,
Dipil Kumar V
From India
Dear Sir
Quite interesting discussions... Thanks for your value additions...
I am sure those who taking these discussions seriously will benefict a lot...
Awaiting to reach to the conclusion...
Regards,
Dipil Kumar V
From India
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.
From India, Mumbai
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.
From India, Mumbai
@HKP
Thanks for your posting and the interest shown into the discussion...
@ Keshav Sir
Awaiting for your comments please...
Dear all give your contributions always and let us make this place an assured forum to get great solutions to safety related matters... One this assurance will come, participation also will get more and no doubt all will benefict a lot...
Keep on sharing...
Regards,
Dipil Kumar V
From India
Thanks for your posting and the interest shown into the discussion...
@ Keshav Sir
Awaiting for your comments please...
Dear all give your contributions always and let us make this place an assured forum to get great solutions to safety related matters... One this assurance will come, participation also will get more and no doubt all will benefict a lot...
Keep on sharing...
Regards,
Dipil Kumar V
From India
Dear All
Whats the actual differance between deceleration and acceleration? Are we using the same formula a=v2/2d to calculate the both... As per the attachement from BHK the formula is for deceleration...
Did these two terms have any relation in our calculation to reach 2300 Kg?
Requesting to all come forward and help in reaching to a conclusion...
Regards,
Dipil Kumar V
From India
Whats the actual differance between deceleration and acceleration? Are we using the same formula a=v2/2d to calculate the both... As per the attachement from BHK the formula is for deceleration...
Did these two terms have any relation in our calculation to reach 2300 Kg?
Requesting to all come forward and help in reaching to a conclusion...
Regards,
Dipil Kumar V
From India
Dear All
This discussion still in-complete...
I have refer OSHA through following linK:
<link outdated-removed> ( Search On Cite | Search On Google )
It's stating 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 again increase from 5000 pound to 5400 pounds...
1. Which formula we can use for fall protection calculations. is that
1.1 F=ma
1.2 F = mg(h + x) / x
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 in coming into a great conclusion...
This will be of great help...
Thanks in advance and really looking forward to reach into a conclusion...
From India
This discussion still in-complete...
I have refer OSHA through following linK:
<link outdated-removed> ( Search On Cite | Search On Google )
It's stating 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 again increase from 5000 pound to 5400 pounds...
1. Which formula we can use for fall protection calculations. is that
1.1 F=ma
1.2 F = mg(h + x) / x
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 in coming into a great conclusion...
This will be of great help...
Thanks in advance and really looking forward to reach into a conclusion...
From India
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PF query.doc
WORKERS PROTECTION.pdf
Fall protection.ppt