- #1

karmatic

- 18

- 0

## Homework Statement

Select a suitable steel rope to suspend a lift from a winch. The weight of the lift is variable as is the weight and diameter of the ropes the lift is suspended from. As an example, I'll use 13mm for the ropes diameter and 0.63kg/m for the rope mass, the total length of the rope at its longest is 37m. The lift mass will be 11,563.1646 kg. I must use at least 3 ropes to suspend the lift. The maximum breaking stress of the ropes is given as 714MPa and I must use a safety factor of 10!

## Homework Equations

P=F/A or P=M/A

Area of a circle=Pie x R

^{2}

## The Attempt at a Solution

First I want to know the total weight of the lift plus ropes to calculate the force acting on the rope.

37 x 3 x 0.63 = 69.93kg

69.93kg + 11,563.1646kg = 11,633.0946kg total weight of lift and ropes.

Next I need to divide the mass by the cross sectional area of the ropes. Now I am unsure of whether to use the area of one rope or to multiply by 3 for the total number of ropes. Either way, the cross sectional area of a single rope is given by

0.0065m x 0.0065m x 3.14 = 1.32665m

^{2}

Now to calculate the total pressure acting over that area I divide the total mass by the area.

P=M/A

P=11,633.0946/1.32665

P=8768.774432Pa

Now I multiply that answer by 10 to calculate the final answer after applying the safety factor of 10 and I get 87,687.774432Pa.

Assuming this is all mostly correct, the problem I am having is that the max stress of 714MPa is way way way higher than the actual stress being applied to the ropes. This is a problem because I am using the rope option with the smallest diameter listed in my problem description, and calculating the total stress using the maximum mass for the lift that I have worked out. So I am assuming I have made a simple mistake in the calculations along the way...

But where is it?!