Calculating the force required to displace a tensioned cable

In summary: Hi Pongo38! Thanks for the response. I think I got it now. :)In summary, an elastic material is tensioned longitudinally and the cable axial force required to elongate the cable axially is 8365 N.
  • #1
abhiramv
3
0
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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  • #2
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:
 
  • #3
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!
 
  • #4
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?
 
  • #5
Hi Pongo38. Yes, I mean lateral displacement (In my original post I wrote axial displacement of a longitudinally tensioned elastic material).
 
  • #6
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)
 
  • #7
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.
 
Last edited:

1. What is the formula for calculating the force required to displace a tensioned cable?

The formula for calculating the force required to displace a tensioned cable is F = T x sin(theta), where F is the force, T is the tension in the cable, and theta is the angle between the cable and the direction of force.

2. How do you determine the tension in a cable?

The tension in a cable can be determined by using the formula T = (F x L) / (2 x sin(theta)), where T is the tension, F is the force applied, L is the length of the cable, and theta is the angle between the cable and the direction of force.

3. Can the force required to displace a tensioned cable be calculated using only the angle and length of the cable?

No, the force required to displace a tensioned cable also depends on the tension in the cable. Therefore, the tension must be known in order to accurately calculate the force using the formula F = T x sin(theta).

4. How does the angle of displacement affect the force required to displace a tensioned cable?

The force required to displace a tensioned cable increases as the angle of displacement increases. This is because a larger angle results in a larger component of the force acting in the direction of the cable, increasing the tension and therefore the force required to displace it.

5. Is there a limit to the amount of force that can be applied to a tensioned cable before it breaks?

Yes, every cable has a maximum tension or breaking point that should not be exceeded. If the force required to displace the cable exceeds this limit, the cable may break or become permanently damaged.

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