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Old Jan 24th 2017, 09:18 AM   #11
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Originally Posted by Tygra View Post
Actually could I have some further guidance, as I am bit stuck where to start? Could someone start me off. I need some more explantion of how cable thickness increases with length/height with formulas etc.
I suggest starting with an equation for stress in the cable. At the top of the elevator shaft the cable supports weight equal to the weight of the elevator car plus the weight of the cable, and so the stress in the cable is the sum of those two elements divided by the cross-sectional area of the cable.. The weight of the elevator car is a constant, say $\displaystyle W_c$. The weight of the cable is its length times its cross-section area times the density of the cable material $\displaystyle \rho $. So:

Stress = $\displaystyle \sigma = \frac {(W_{car} + W_{cable})}{A} = \frac {(W_{car} + LA \rho g )}{A} $

Rearrange to solve for A, and you'll see that the factor L in the denominator causes A to increase to infinity as L approaches $\displaystyle \frac {\sigma _{max}}{ \rho g} $
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Last edited by ChipB; Jan 24th 2017 at 09:21 AM.
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Old Jan 25th 2017, 12:21 PM   #12
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Moving Mass of a lift increases exponentially

All cables will fail at the same length (more or less) if they do not carry a conveyance. Normally a safety factor is also applied to prevent the rope even coming close to failure. Coming from the mining industry, I have worked on mines where a single lift can be up to 2400m. (This may have been surpassed since breakthroughs in better materials). However, I think that the safety factor used on lifts is a lot higher (maybe 4x?) because the cables have to run through smaller pulleys causing more wear and tear and inspection are less freqient in buildings than on mines.

If you take most cables and see what length they can hold up without any conveyance attached, you will come to about 15000m at a SF of 5 this is about 3000m and at 20 it will be about 750m ~ 500m. The exponential increase happens somewhere around 2400m irrespective of the conveyance weight at SF of 5.

The following formula for moving mass has been derived by considering a cable of length 3000m (Lcrit) and then considering a conveyance of weight
Phi*Mu*Lcrit
where Phi is the weight of the cable per meter and
Mu is a dimensionles number which indicates the proportion of the conveyance weight to the total weight (interestingly enough numbers
bigger than 1 are allowed for this input but you will get more accurate
results if you stick to known values try about 0.24 kg / m and tensile strength of 7.56 kn for starters use Mu of about 0.5-0.9)

Formula for MM is Phi*Mu*(Lcrit)^2/(Lcrit - L)

where MM is Moving Mass, L is the length of the cable (< Lcrit) this formula is obviously designed to give an exponential rise in MM as L approaches Lcrit.

Note that Mu remains constant in the calculation but you can work out the new Mu for a spcecific MM which has been calculated.

Hope this helps.
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Old Jan 26th 2017, 03:42 AM   #13
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Moving Mass of a lift increases exponentially

The formula that I gave can actually still be simplified significantly.

Since Phi*Mu*Lcrit = Mc (Mass of conveyance)

then the formula for moving mass (mm) is

mm = Mc*Lcrit / (Lcrit - L)

where Lcrit is the length at which a specific material will break under it's own weight multiplied by the desired safety factor. (500 in the case of an elevator).

With new advances in material technologies on the mines nowadays there are mine cages going down almost 3000m in a single lift normally the safety factor in this case is about 5
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Old Jan 28th 2017, 08:06 AM   #14
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Yes, thankyou you both have helped. I will need to get back to you both soon with some questions. Thanks again!
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