George Zucas
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Hello everybody,
I am trying to analyse the lifelines we use at work, though the results I obtain are causing me headaches.
The problem is, our workers often install and use what I would call homemade lifelines. My boss asked me to check whether what they do is reliable. According to what I calculate, most of the professionaly made lifelines are too weak.
Here is what I have done so far:
I checked the safety harnesses we use at work. The rated arrest force is 6 kN (which is also the value mentioned in the standard we are bound to) which means when a user falls, after 6 kN of force is generated in the harness fall smoother ( whatever it is called) gets opened and the fall slows down. So the maximum force the lifeline is subjected to is 6 kN.
There are a few formulas used in analysing the ropes:
H=(w*L^2)/(8*d)
H: Midspan load
w: Unit load on the cable
L: Cable Span
d:sag
T= (H^2+((w*L/2)^2)^0.5
T: Tension in the Rope
What I need is the tension in the rope. When the rope gets loaded, it gets a V shape. So the calculation should be easy. The new rope length is sqrt(L^2-4d^2). The elongation is this new length minusthe initial length. The rope modulus is A*E/Linitial. Then the tension is elongation*modulus.
To translate the values to the harness, I found a formula from the link below which basically does what I do( and I copied some parts from there), it is 4T(s/l). Though I do not know where does it come from. After that I give random values to the sag until 6 kN harness force is reached.
http://sraa.asia/wp-content/uploads/2014/10/HLL-Basic-Calculation_final.pdf
When I do the calculation I get the same results which is 27 something kN for a 8 mm rope. With a safety factor of two the load is 55 kN. 8 mm diameter ropes have a breaking force of around 35 kN. 8 kn is too weak. If I increase the diameter, since the rope modulus is also dependent on the diameter, it is also not enough. So by increasing the diameter I figured only after around 15mm, I can say it is safe, which is a bit crazy to me since most of the official products have a 8 mm rope.
What do I do wrong? I cannot figure it out. The loads can't be that high. The official ones must have an even bigger safety factor since 2 is the minimum.
Also what is the nominal elastic modulus E for wire ropes? I cannot find any data on this. I used the value at the link but it does not specify the type of rope used.
I sumarised the calculation, there are other factors such as the initial sag etc but the results are about the same.
I am trying to analyse the lifelines we use at work, though the results I obtain are causing me headaches.
The problem is, our workers often install and use what I would call homemade lifelines. My boss asked me to check whether what they do is reliable. According to what I calculate, most of the professionaly made lifelines are too weak.
Here is what I have done so far:
I checked the safety harnesses we use at work. The rated arrest force is 6 kN (which is also the value mentioned in the standard we are bound to) which means when a user falls, after 6 kN of force is generated in the harness fall smoother ( whatever it is called) gets opened and the fall slows down. So the maximum force the lifeline is subjected to is 6 kN.
There are a few formulas used in analysing the ropes:
H=(w*L^2)/(8*d)
H: Midspan load
w: Unit load on the cable
L: Cable Span
d:sag
T= (H^2+((w*L/2)^2)^0.5
T: Tension in the Rope
What I need is the tension in the rope. When the rope gets loaded, it gets a V shape. So the calculation should be easy. The new rope length is sqrt(L^2-4d^2). The elongation is this new length minusthe initial length. The rope modulus is A*E/Linitial. Then the tension is elongation*modulus.
To translate the values to the harness, I found a formula from the link below which basically does what I do( and I copied some parts from there), it is 4T(s/l). Though I do not know where does it come from. After that I give random values to the sag until 6 kN harness force is reached.
http://sraa.asia/wp-content/uploads/2014/10/HLL-Basic-Calculation_final.pdf
When I do the calculation I get the same results which is 27 something kN for a 8 mm rope. With a safety factor of two the load is 55 kN. 8 mm diameter ropes have a breaking force of around 35 kN. 8 kn is too weak. If I increase the diameter, since the rope modulus is also dependent on the diameter, it is also not enough. So by increasing the diameter I figured only after around 15mm, I can say it is safe, which is a bit crazy to me since most of the official products have a 8 mm rope.
What do I do wrong? I cannot figure it out. The loads can't be that high. The official ones must have an even bigger safety factor since 2 is the minimum.
Also what is the nominal elastic modulus E for wire ropes? I cannot find any data on this. I used the value at the link but it does not specify the type of rope used.
I sumarised the calculation, there are other factors such as the initial sag etc but the results are about the same.
Last edited: