Calculating the force required to displace a tensioned cable

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Discussion Overview

The discussion revolves around calculating the force required to displace a tensioned elastic cable laterally by 5mm. Participants explore the relationship between the required force, tension, and material properties, while addressing the complexities of the problem.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Homework-related

Main Points Raised

  • One participant seeks a method to calculate the force needed for a 5mm lateral displacement of a tensioned elastic material with specified properties.
  • Another participant requests clarification on what the original poster has attempted and where they are facing difficulties.
  • A participant suggests that the question may be poorly worded, indicating a distinction between longitudinal and lateral displacement.
  • Clarification is provided that the original poster is indeed referring to lateral displacement, despite initial wording suggesting otherwise.
  • One participant critiques the definition of stiffness (K) provided, suggesting that it needs to be better defined and proposes using a geometric approach to find an approximate solution.
  • Another participant proposes a formula for calculating the new tension in the cable based on axial elongation, assuming linear behavior, and provides a specific calculation example.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the displacement type and the appropriate method for calculating the required force. There is no consensus on the best approach or the definitions used in the problem.

Contextual Notes

Participants note the need for clearer definitions and assumptions regarding stiffness and displacement types, as well as the potential for different models to apply based on the nature of the displacement.

abhiramv
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Hi,

I'm trying to figure out a method for calculating the force required to displace (axially by 5mm) an elastic material that is tensioned longitudinally at 105 N.

Material properties:
Cross-sectional area: 100mm^2
Stiffness, K = 1652 N/mm
Modulus, E = 0.68 GPa
Length of material, L = 41.9 mm

I need help finding a relationship between required force as a function of displacement, tension and other involved variables.

Thank you! Please let me know if you need any more information.
 
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hi abhiramv! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
Hi tiny-tim, Thanks!

Sorry for the super delayed response.

I'm honestly not sure where to start. I've taken mechanics courses, so I was going to use a beam-bending model, but I wasn't sure how that would apply to an elastic material that is already tensioned.

Any advice would be greatly appreciated!
 
If you mean longitudinal displacement, then the answer is staring you in the face in your question. If you mean lateral displacement, that is a bit more complicated. So...badly worded question?
 
Hi Pongo38. Yes, I mean lateral displacement (In my original post I wrote axial displacement of a longitudinally tensioned elastic material).
 
Your value for k is meaningless unless it is better defined. I assume you mean central lateral force and corresponding 5mm deflection. Consider the triangle of forces at the central node representing equilibrium, and then draw a geometrically similar triangle for the geometry of the situation. Answer is then obvious. (at least it's a good first approximation and you can refine it from there)
 
abhiramv: Is this a schoolwork question? I am currently assuming you want the cable axial force required to elongate your cable axially (longitudinally). If your material remains in a range that is approximately linear, then the tensile force is approximately T2 = k*x, where k = cable axial stiffness, and x = cable axial elongation. E.g., if x = 5 mm, then T2 = k*x = (1652 N/mm)(5 mm) = 8260 N. If the cable was pretensioned to T1 = 105 N, then the new cable tension is T3 = T1 + T2 = 105 + 8260 = 8365 N. Let us know if this is not what you want.
 
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