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Calculating the force required to displace a tensioned cable 
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#1
Feb1113, 07:02 PM

P: 3

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: Crosssectional 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. 


#2
Feb1213, 02:32 PM

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P: 26,157

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
Feb2113, 10:27 AM

P: 3

Hi tinytim, 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 beambending 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
Feb2213, 02:40 PM

P: 696

Calculating the force required to displace a tensioned cable
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
Feb2213, 02:53 PM

P: 3

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



#6
Feb2213, 04:00 PM

P: 696

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
Feb2313, 11:27 AM

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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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