How Many Steel Strands Are Needed to Safely Lower a Lift into a Mine?

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To safely lower a lift into a mine at a depth of 4000m, the number of high-strength steel strands required must be calculated based on the total load, including the lift's weight and the weight of the miners. Each steel strand has an allowable stress of 600MPa, and the total force exerted must not exceed this limit. The weight of the cable itself, derived from its density, also contributes to the overall load. Understanding the axial stress formula (P/A) is crucial for determining the appropriate number of strands to ensure safety. Proper calculations will ensure that the lift can operate without exceeding the material limits.
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A lift is used to lower miners 4000m into a mine. The lift is supported by many high strength steel strands of 3mm diameter each. The ultimate stress for the steel strands is 1200MPa and the allowable stress is 600MPa. The empty lift weighs 15kN. It has a capacity of 20 persons. The density of steel is 77kN/m3 and each person weighs 800N.
1)Calculate the number of strands required.
2)What is the maximum depth to which the lift can be lowered before the cable snaps?
I am assuming we have to use \epsilon = \sigma/E
I am not sure how to start this question. The biggest problem I have is that I don't know how to use that equation (what each variable stands for), (and for this course we have no textbook, so I can't find any sample problems). Any help is appreciated.
 
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The equation you have noted is for the stress-strain relationship. That is not needed for this problem.. What you need to find out is the force (P) and stress (P/A) in each strand under the given loads to be sure that there are a sufficient number of strands such that the stress does not exceed 600 MPa. You must include the weight of the cable itself, based on the given density and calculated volume... you have enough information given to calculate it's weight, and it's cross section area. You need a basic understanding of P/A axial stress.
 
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