Maximum load that a 2mm steel cable can hold.

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SUMMARY

The maximum load that a 2mm steel cable can hold is determined using the ultimate breaking stress of steel and the cross-sectional area of the cable. In this discussion, a participant calculated the maximum force the four cables could withstand to be 10,327.04 N, translating to a maximum mass of approximately 1,053 kg. The calculations utilized the formula F = pA, where p represents the ultimate breaking stress. The participant confirmed that the calculations align with expected values, pending final verification.

PREREQUISITES
  • Understanding of tensile stress and strain concepts
  • Familiarity with the formula for calculating force (F = pA)
  • Knowledge of material properties, specifically the ultimate breaking stress of steel
  • Basic principles of elastic deformation and Young's modulus (E)
NEXT STEPS
  • Research the ultimate breaking stress values for different steel grades
  • Learn about the effects of plastic deformation on materials
  • Study the relationship between diameter and load capacity in cables
  • Explore advanced calculations involving calculus for stress-strain relationships
USEFUL FOR

Mechanical engineers, materials scientists, and students studying structural mechanics will benefit from this discussion, particularly those focused on load-bearing calculations and material properties.

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Homework Statement



A platform (m = 5,0kg) is suspended from 4 steel cables whose length is 3,0m and diameter 2mm.

a) How much further will the platform sink if a mass of 50kg is placed on the platform?

b) What is the maximum load that the wires can hold before breaking?

Homework Equations



\dfrac{\Delta F}{A} = E\dfrac{\Delta L}{L}

The Attempt at a Solution



Part A was easy enough; simply assume that the wires remain in the elastic zone of deformation and that A doesn't change much and plug in the values. However part B baffles me, and I'm not sure where to start.

Surely the cross sectional area A doesn't go to zero as the cable breaks(division by zero)? Am I supposed to use calculus to figure out when E changes(doesn't really make sense since E is a constant for elastic situations)? How would I even do that? Does steel have documented values for when a wire of certain L and A starts deforming plastically? I couldn't find any.

I don't have the right answers for this problem, so I can't just try one of the above methods to find out if it was the correct one, not that any of them make any sense to me. If someone could point me in the right direction, that would be great.
 
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Just as you must be given E to solve part a, you must be given the ultimate breaking stress of the cable to solve part b. The ultimate breaking stress usually assumes no change in area.
 
PhanthomJay said:
Just as you must be given E to solve part a, you must be given the ultimate breaking stress of the cable to solve part b. The ultimate breaking stress usually assumes no change in area.

Alright, so I used Wolfram alpha to find out the ultimate breaking stress of steel and calculated the maximum force the 4 wires could take using F = pA This gave me a value of 10 327.04N and a maximum mass of m = \dfrac{F}{g} = 1.053\cdot 10^3 kg Seems realistic enough, but I can't be sure until Thursday when I have to turn these problems in.
 

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