Cable stiffness lateral direction point load

[lobos22]

Hi all,

I have a vertical cable of length L that has a pretension T, that is applied a horizontal force at its midpoint. I also know the cross-sectional area A, and I can assume small angle approximations.

How can I compute the stiffness in the horizontal direction? I guess I need to have something like F = K*delta, where delta is the horizontal displacement of the midpoint, but I don't know how to approach it.

Thanks.

 

[Compositepro]

The initial stiffness at infinitesimally small displacement is zero.

 

[lobos22]

Hi,

thanks for your quick reply Compositepro. I think that does not have to be like that. If you have something like F = K*delta, delta being 0 does not mean K = 0.

On the other hand, I have found this in the internet:

https://www.thestructuralengineer.i...eparation/calculation-examples/truss-vs-cable

However, I think there are some things I do not see correct there. How is it stated that: DeltaL = delta*sin(theta)? Where does it come from?
And when it goes to the equilibrium for the cable, it rearranges using sin(theta) = delta/L.

I think I'm missing something in the geometry of the problem. Does anyone have any clue about this?

Thank you.

 

[BAretired]

It's easier to see if we rotate the picture 90 degrees.

Initially, F required is very small to get a slight deflection. When F gets large enough, the prestress force T disappears, but the cable stretches as shown below. At large deflection, cable tension becomes the resultant of F/2 and FL/4d. The spring constant is F/d with units of N/mm or #/".

BA

 

[GregLocock]

You've drawn a sagging cable, T<<H. The original question makes me think that T (the pretension) is very much part of the solution.

Cheers

Greg Locock


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[3DDave]

Even so - the idealization is for two springs in line with the displacement applied at their junction, so the restoring force is just from the tension in the springs, just like the restoring force for a pendulum is the tension in the string. That one uses the sin(theta) = theta (radians) for small displacement.

Unlike pendulums, for the cable the restoring force rapidly increases as the spring element is stretched, so it's a non-linear problem at any meaningful deflection.

 

[BAretired]

T is part of the solution, to the extent that it determines the starting length of the cable, which is somewhat less than L. After cable tension exceeds T, it plays no role in the solution, similar to a tension applied to a prestressed bolt after the pretension has been exceeded.

I agree it is a non-linear problem, but F can be found for any deflection 'd', if the A (area) and E (Young's modulus) is known for the cable.

BA

 

[lobos22]

Hi all,

thank you all for your replies. I have worked on this solution. I guess I'm somehow satisfied by this one.

What do you think about it?

Regards

 

[BAretired]

Congratulations on attempting a solution! I have not had time to review your calculations, but your result appears suspect to me, of course I could be mistaken.

BA

 

[Denial]

Lobos22.[&nbsp;] Firstly my apologies for misunderstanding your initial post.
I have done some very untidy algebra and come up with a result that has the centrepoint's
lateral STIFFNESS at a lateral displacement of d being
4K(L-L₀)/L + [8K(3L₀-L)/L³]d² + ...
(plus further terms from a Taylor Series expansion)
where K is the end-to-end axial stiffness of the cable, L is the distance between the two anchorages, and L₀ is the cable's unstretched length.[&nbsp;] My workings were hasty/scrappy, and are not in a form where I am prepared to submit them at this stage.[&nbsp;] If I get some clear air in the near future I will polish them up and submit, but it might be a while.

Compositepro.[&nbsp;] As others have suggested, the lateral STIFFNESS at zero lateral displacement is not zero.[&nbsp;] It is the lateral FORCE that is required for lateral equilibrium that is zero.

 

[BAretired]

If stiffness is F/d, then at F=0 and d=0, stiffness = 0/0 which is undefined.
L₀(1+T/AE) = L, so L₀=L/(1+T/AE)
If cable is stressed to T' (which includes T) then unit strain e = T'/AE and total strain = L₀T'/AE

But total strain = sqrt(4d²+L²)-L = L₀T'/AE
Solve for d
sqrt(4d²+L²) = L + L₀T'/AE
4d²+L² = (L + L₀T'/AE)²


It gets messy here. Have to think about it some more.

BA

 

[Denial]

I am perfectly happy with the first term in my expression above.[&nbsp;] It is quite straight forward to derive, since at VERY SMALL lateral displacements of the midpoint the cable tension DOES NOT change but the direction of the two cable arms that resist the applied lateral force changes linearly.[&nbsp;] [The old sin(x)=x when x is small enough trick.]

 

[Denial]

I found a bit of time to return to what I reported in my 04Sep22@05:39 post above.[&nbsp;] As I suspected, my first term was correct but my second term was wrong.

This time around I resorted to my algebra software of choice:[&nbsp;] "Derive - A mathematical assistant" (unfortunately long discontinued).[&nbsp;] This achieved a full algebraic solution, which you can locate at line #23 in the Derive output that I will attach to this post.

The first three terms of the Taylor's Series expansion of this full expression are
4K(L-L₀)/L + 24K(D/L)² - 40K(D/L)⁴ +[&nbsp;]...
where D is the midpoint lateral displacement.[&nbsp;] See line #30 of the Derive output.

 

[BAretired]

If T is just enough to ensure the cable is straight, say 1,000# and the tendon is 1/2" 270k with E of 28.5e6psi:
then L₀ = 240/(1=1000/0.153*28.5e6) = 239.94", so close to 240" that the effect of T could be neglected.
Are we agreed? Can T be neglected, or should we continue to include the effect of T?

For 1/2" - 270k tendon, yield is 37,170#, so a 240" long tendon would grow to 240(1+37,170/0.153*28.5e6) = 242.04", an increase of only 2". 'S' would be 121.02" and 'd' would be sqrt(121.02^2-120^2) = 15.68". Theta = cos^-1(240/242.04) = 0.129924787 radians.

Cable tension at yield Ty = 37,170#; H = Ty cos(theta) = 36,857#; V = Ty.sin(theta)= 4,815#; F = 9,631#; F/d = 614.25 at yield.
A spring constant 'K' could be calculated for stress values between 0 and Fy to obtain a 'K' curve. It will not be linear.

 

[Denial]

I'm not quite sure where you are heading here, BA.[&nbsp;] The idealized mathematics is quite eloquent in this regard.[&nbsp;] The initial tension can never be ignored if you want the lateral stiffness at zero deflection (unless the initial tension is zero, in which case the lateral stiffness is also zero). If you want the lateral stiffness when the midpoint's lateral displacement is a finite value D, then you can neglect the L-L₀ term provided
[&nbsp;][&nbsp;][&nbsp;](L-L₀)<<6D²/L.

 

[BAretired]

@Denial,
How do you define lateral stiffness? I define it as F/d, but at zero deflection, F is zero and d is zero, and 0/0 is undefined.

I have an expression which includes the term L₀, and I can work it out that way if the OP insists, but I am of the impression that T is a nominal tension to ensure the cable is straight. Could be 500# or 1,000#, whatever is needed to keep it straight. Whether T is included or not, the lateral stiffness at zero deflection is zero.

BA

 

[Compositepro]

An ideal cable, by definition, has zero bending stiffness. So the bending stiffness of a cable is only due to tension. But, as I said before, at zero deflection, the initial bending stiffness is zero.

 

[BAretired]

I agree the stiffness is zero at zero deflection, but that changes quickly as soon as there is a small deflection. The initial tension T would have an immediate effect at that stage of loading, but if T is a nominal force to ensure straightness of cable, the lateral stiffness at that stage would be of little interest to anyone.

BA

 

[GregLocock]

There may be a little confusion here, stiffness = dF/dx, so it is not necessarily 0 at x=0

Cheers

Greg Locock


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[Denial]

New solution.[&nbsp;] When checking my earlier solution to see what it threw up as D tended to infinity (it should have thrown up 4K) I discovered an error.[&nbsp;] I had used SIN() where I should have used TAN(), or vice versa.[&nbsp;] This has been corrected, and the updated workings are attached.[&nbsp;] The full solution for the lateral stiffness is
4K[(4D²+L²)^3/2 - L²L₀] / (4D²+L²)^3/2
and the corresponding Taylor Series is
4K(L-L₀)/L + 24K(L₀/L)(D/L)² - 120K(L₀/L)(D/L)⁴ +[&nbsp;]...

BAretired.[&nbsp;] You ask how I define the lateral stiffness.[&nbsp;] I take it to be the gradient of the LateralDeflection versus LateralForce diagram:[&nbsp;] a "tangent stiffness".[&nbsp;] As does GregLocock above.

Compositepro.[&nbsp;] Greg has expressed more clearly and more succinctly what I was trying to say in my 05Sep22@05:39 post above.[&nbsp;] I suspect you have been distracted by thinking in terms of bending.[&nbsp;] There is no bending involved in any aspect of this problem.[&nbsp;] The answer would be the same if the two halves of the cable were replaced by extensible bars with hinges at each end of each bar and an appropriate pretension applied to the bars before the lateral load appears on the scene.
[&nbsp;]