# Calculating the force required to displace a tensioned cable

1. Feb 11, 2013

### 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.

2. Feb 12, 2013

### 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!

3. Feb 21, 2013

### 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!

4. Feb 22, 2013

### 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?

5. Feb 22, 2013

### abhiramv

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

6. Feb 22, 2013

### 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)

7. Feb 23, 2013

### 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.

Last edited: Feb 23, 2013