Maximum impact force that a cable can withstand

[jasonnaylor]

Homework Statement

"
A 1100 kg lift is suspended by a series of 24 steel cables of mean diameter 20mm. The cables are attached to the lift top and are arranged so that torsional vibration is minimised.If the cable is manufactured from steel with an elastic modulii of 185GN/m2 and elastic constant of 200kN/m:
Calculate the maximum impact force that the cables can withstand before catastrophic failure occurs, assuming a safety factor of 3.3.[/B]

Homework Equations

The Attempt at a Solution

[/B]
Rearranging this formula to solve T, the cable tension yields:
(1100).(0.22+9.81)=11033N
Divided by 24 this gives 459N tension per cable.

Safety Factor

“The total stress in a wire rope, in service, is composed of several separate elements. These are reduced to a single tensile load value. When this value exceeds the breaking strength of the wire rope, a failure occurs. The factor to provide a margin of safety between the applied tensile forces and the breaking strength of the rope is defined as the factor of safety.” The max safe working load is obtained by dividing the breaking strength by the safety factor, So conversely the safety factor multiplied by the max safe working load will give the breaking strain (the point at which the cable will fail)
(http://www.usbr.gov/ssle/safety/RSHS/appD.pdf)

Hooke’s law states that stress is proportional to strain up to the elastic limit:

Therefore 200Kn (elastic constant)= stress/strain

rearrange to find the breaking strain = stress/200,000

This is where I've ground to a halt...

 

[CWatters]

Perhaps start with the definitions of...

Stress = Force per unit area
Strain = Extension per unit length

The extension depends on the load and the elastic moduli.

The total tension will be the due to the mass of the lift AND the maximum impact force. Don't substitute values too early.