Fall Protection - Query - Pdf Download

dipil

713

Dear All

I tried to pull out the real calculations to my query from different sources... But each reply I am getting leads to a new path or confusing me more...

The latest reply I got as follows:

Regarding the values you mentioned, firstly I want to remind you that the value mentioned by OSHA is a force value 5,000 pounds (22.2 kN) but not 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 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 vlaue of 22200 N as force on the anchrage point:

- The hieght shoudl eqaul 1.8 m

- The mass of the falling object should equal 140kg

- 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 included.

Can anyone come up with actual calcualtions... What the above said calculations is right or not?...

Looking forward to reach into a great coclusion...

Thanks in advance...
22nd November 2010 From India

dipil

713

Dear All
Just sharing two more links related to the topic:
1. Solve problem related to impact force from falling object
2. <link outdated-removed> ( Search On Cite | Search On Google )
Please share your views...
22nd November 2010 From India

dipil

713

Dear Keshav Pillai Sir
Requesting you to kindly come forward with your comments, so that we can colclude this discussion...
Please put light into what's the exact calculation to reach the value 5000 lb force?
I think none of the other forum member having interest in reaching to a great conclusion, hence requesting you to help... I hope you will not disappoint me this time...
I have tried to pull out the exact calculations and not getting the same from anywhere...It is now clear that we don't need always to ensure a 5000 lb capacity anchorage point... Only thing to know that the exact calculations...
Looking forward to hear from you.
Thanks in advance.
3rd December 2010 From India

HSEcsp

"F" is force, "M" is mass, "A" is accleration (here we call it g .i.e 9.8m/s^2), "T" is time required to travel the covered distance, "H" is the distance traveled or hight or simply distance.

F=M x 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 accleration "A" is the rate of change of velocity

so Impluse = M x (V/T) x T = M x V .......... which is the same equation for momentum "p = m x v"

so 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)

this formulas came from conservation of energy and equation of distance traveled after time "T"

.................................................. .................................................. .................................................. ...............
so calculate

person with weight = 100 KG
falling to distance of 1.8 mtr and get arrested without fall arrestor.

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

Impulse = M x V

.i.e V = sqrt ( 2 x g x h) = sqrt ( 2 x 9.8 x 1.8) = 5.9396969 m/s

Impulse = 100 x 5.94 = 594kgm/s

so maximum arresting force or force experienced during arrest might be the sum

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

-----------------------------------------------------

But this figure is also greater than the 22.6KN requirement of standards.
Also I have some doubts of units Please any if anyone could correct then post it.

------------------- ------------------------ ---------------------

also i Heard about one another equation

1/2 x m x g^2

i dont know how this equation is derived. but while computing some what equal result is coming.

1/2 x 100 x 9.8^2 = 4802 ................... suppose the measurment is in pounds and it is converted to kilograms

then 2178.156 kg

i dont know anything about this equation and what are its units etc.

please anyone, if knows, help me.
1st September 2015 From India, Jamnagar

HSEcsp

Hi DIPIL
Hopes you read my post. there are two methods mentioned to calculate the force.
result of first method was 2358Kg
and
result of second method which I am not certain is 2178Kg
so take an average
(2358 + 2178) / 2 = 2268kg
in pounds 5000lbs
in Newton 22.68KN
2nd September 2015 From India, Jamnagar

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