## Cable stiffness lateral direction point load

## Cable stiffness lateral direction point load

(OP)

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.

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.

## RE: Cable stiffness lateral direction point load

## RE: Cable stiffness lateral direction point load

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.info/education/p...

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.

## RE: Cable stiffness lateral direction point load

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

## RE: Cable stiffness lateral direction point load

Cheers

Greg Locock

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## RE: Cable stiffness lateral direction point load

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.

## RE: Cable stiffness lateral direction point load

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

## RE: Cable stiffness lateral direction point load

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

## RE: Cable stiffness lateral direction point load

BA

## RE: Cable stiffness lateral direction point load

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-L

_{0})/L + [8K(3L_{0}-L)/L^{3}]d^{2}+ ...(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 L

_{0}is the cable's unstretched length. My workings were hasty/scrappy, and are not in a form where I am prepared to submit them at this stage. If I get some clear air in the near future I will polish them up and submit, but it might be a while.Compositepro. As others have suggested, the lateral STIFFNESS at zero lateral displacement is not zero. It is the lateral FORCE that is required for lateral equilibrium that is zero.

## RE: Cable stiffness lateral direction point load

L

_{0}(1+T/AE) = L, so L_{0}=L/(1+T/AE)If cable is stressed to T' (which includes T) then unit strain e = T'/AE and total strain = L

_{0}T'/AEBut total strain = sqrt(4d

^{2}+L^{2})-L = L_{0}T'/AESolve for d

sqrt(4d

^{2}+L^{2}) = L + L_{0}T'/AE4d

^{2}+L^{2}= (L + L_{0}T'/AE)^{2}It gets messy here. Have to think about it some more.

BA

## RE: Cable stiffness lateral direction point load

## RE: Cable stiffness lateral direction point load

This time around I resorted to my algebra software of choice: "Derive - A mathematical assistant" (unfortunately long discontinued). 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-L

_{0})/L + 24K(D/L)^{2}- 40K(D/L)^{4}+ ...where D is the midpoint lateral displacement. See line #30 of the Derive output.

## RE: Cable stiffness lateral direction point load

then L

_{0}= 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.

## RE: Cable stiffness lateral direction point load

_{0}term provided(L-L

_{0})<<6D^{2}/L.## RE: Cable stiffness lateral direction point load

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 L

_{0}, 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

## RE: Cable stiffness lateral direction point load

## RE: Cable stiffness lateral direction point load

BA

## RE: Cable stiffness lateral direction point load

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

## RE: Cable stiffness lateral direction point load

New solution. 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. I had used SIN() where I should have used TAN(), or vice versa. This has been corrected, and the updated workings are attached. The full solution for the lateral stiffness is4K[(4D

^{2}+L^{2})^{3/2}- L^{2}L_{0}] / (4D^{2}+L^{2})^{3/2}and the corresponding Taylor Series is

4K(L-L

_{0})/L + 24K(L_{0}/L)(D/L)^{2}- 120K(L_{0}/L)(D/L)^{4}+ ...BAretired. You ask how I define the lateral stiffness. I take it to be the gradient of the LateralDeflection versus LateralForce diagram: a "tangent stiffness". As does GregLocock above.Compositepro. Greg has expressed more clearly and more succinctly what I was trying to say in my 05Sep22@05:39 post above. I suspect you have been distracted by thinking in terms of bending. There is no bending involved in any aspect of this problem. 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.## RE: Cable stiffness lateral direction point load

dF/dx=4*T/L-2*E*A/L+d/dx(4*x*E*A/L*sqrt(x^2/L^2+1/4))

Luckily Mr Wolfram knows how

dF/dx=4*T/L-2*E*A/L+(2*A*E*(L^2 + 8*x^2))/(L^3*Sqrt[1 + (4*x^2)/L^2])

phew

Cheers

Greg Locock

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

## RE: Cable stiffness lateral direction point load

Taylor'ed your result, and expressed it in terms of K rather than AE for direct comparison with what I derived above. ItTaylors to4K(L-L

_{0})/L + 12K(L_{0}/L)(D/L)^{2}- 20K(L_{0}/L)(D/L)^{4}+ ...where K is (EA/L

_{0}).So we are in agreement on the most important first term, which we could have worked out on the back of a largish postage stamp, but we disagree on the terms that become important at larger lateral displacements. I am going to call stumps at this stage and retire to the pavilion. I have already spent too much time on it, and the OP seems to have taken fright (who could blame him).

## RE: Cable stiffness lateral direction point load

Cheers

New here? Try reading these, they might help FAQ731-376: Eng-Tips.com Forum Policies http://eng-tips.com/market.cfm?

Greg Locock

## RE: Cable stiffness lateral direction point load

_{0}.## RE: Cable stiffness lateral direction point load

Greg Locock

## RE: Cable stiffness lateral direction point load

BA

## RE: Cable stiffness lateral direction point load

sotto voce). And I have further plotted the predictions from Greg's 07Sep@23:42 formula. At low displacements (say D<0.25L) all three are in reasonably close agreement, but at larger displacements Greg's formula takes off like a Saturn rocket.The spreadsheet is attached, hiding in a .zip file to avoid over-zealous antivirus systems.

## RE: Cable stiffness lateral direction point load

extra cable tension b=2EA/L(sqrt(x^2+L^2/4)-L/2) (1)

F/2=(T+b)2x/L (2) similar triangles

So

F/2=(T+2EA(sqrt(x^2/L^2+1/4)-1/2)2x/L

F=4xT/L+8xEA/L(sqrt(x^2/L^2+1/4)-1/2)

dF/dx=4T/L-4EA/L+d/dx(8xEA/L(sqrt(x^2/L^2+1/4))

Wolfram says

d/dx((8 x e A sqrt(x^2/L^2 + 1/4)/L) = (4 e A (L^2 + 8 x^2))/(L^3 sqrt((4 x^2)/L^2 + 1))

so dF/dx=4T/L-4EA/L+(4 E A (L^2 + 8 x^2))/(L^3 sqrt((4 x^2)/L^2 + 1))

which still does not tend towards 4*EA/L for x>>L, that is two half length cables in pure tension in parallel. How odd. (2) is niggling me.

Cheers

Greg Locock

## RE: Cable stiffness lateral direction point load

Your eqn (1). I believe that the 2EA/L term at its start should be 2EA/L

_{0}where L_{0}is the cable's unstretched length (as opposed to the distance between the support points). In "normal" configurations (ie where L_{0}is only slightly less than L) this correction is trivial, and it certainly does not explain your formula's desire to head to infinity.Your eqn (2). I think you have made a similar error to the one I made initially and then reported in my 07Sep22@06:35 post above: using TAN when you should have used SIN. This probably explains the infinity. Your eqn (2) should be

F/2=(T+b)(2x)/sqrt[L^2+(2x)^2]

## RE: Cable stiffness lateral direction point load

Greg Locock

## RE: Cable stiffness lateral direction point load

It appears lobos22 has abandoned the ship.

By definition, stiffness K = F/d where d is horizontal displacement.

When prestress T = 0, L

_{0}= L. F/d can be calculated for values of d while cable remains below T_{yield}. Beyond that, it is meaningless.When prestress T is T

_{yield}, L_{0}= L/(1+Tyield/AE); if d>0, cable is strained beyond yield.Applying a prestress of T

_{yield}is pointless. Applying any prestress reduces F, so any prestress seems pointless.BA

## RE: Cable stiffness lateral direction point load

Greg Locock

## RE: Cable stiffness lateral direction point load

It's true. I have never designed a flying fox.

I was discussing this question with my son earlier today. He pointed out that a guitar string is much stiffer when it is tightened, a concept which is intuitive, so I have to mull it over a bit more.

A cantilever relies on bending whereas a cable is presumed to have zero bending strength. I agree that by my definition, which is similar to spring stiffness, K is undefined at d=0. EDIT: But why should it be defined at d=0? K is not defined at d=0 for a spring either; a spring with no load has no deflection; the spring has not been activated at that point. But as soon as F takes on a value, even a small value, K = F/d applies.

BA

## RE: Cable stiffness lateral direction point load

Greg Locock

## RE: Cable stiffness lateral direction point load

_{y}/2 (half the yield strength of the tendon). The tables and graphs of F versus d are shown below.BA