Calculating the force required to displace a tensioned cable

abhiramv

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.

tiny-tim

hi abhiramv! welcome to pf!

show us what you've tried, and where you're stuck, and then we'll know how to help!

abhiramv

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!

pongo38

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?

abhiramv

Hi Pongo38. Yes, I mean lateral displacement (In my original post I wrote axial displacement of a longitudinally tensioned elastic material).

pongo38

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)

nvn

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.