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# 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.

### RE: Cable stiffness lateral direction point load

The initial stiffness at infinitesimally small displacement is zero.

### RE: Cable stiffness lateral direction point load

(OP)
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.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

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

### RE: Cable stiffness lateral direction point load

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.

### RE: Cable stiffness lateral direction point load

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

### RE: Cable stiffness lateral direction point load

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

### RE: Cable stiffness lateral direction point load

Lobos22.  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.  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

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

### RE: Cable stiffness lateral direction point load

I am perfectly happy with the first term in my expression above.  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.  [The old sin(x)=x when x is small enough trick.]

### RE: Cable stiffness lateral direction point load

I found a bit of time to return to what I reported in my 04Sep22@05:39 post above.  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:  "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-L0)/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

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.

### RE: Cable stiffness lateral direction point load

I'm not quite sure where you are heading here, BA.  The idealized mathematics is quite eloquent in this regard.  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
(L-L0)<<6D2/L.

### RE: Cable stiffness lateral direction point load

@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

### RE: Cable stiffness lateral direction point load

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.

### RE: Cable stiffness lateral direction point load

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

### 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 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 + ...

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

I think the lateral stiffness for deflection near 0 is just 4T/L . I can't actually differentiate the tricky bit at the breakfast table, but it looks to me as though it goes to +2EA/L as x->0

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

Greg, I have Taylor'ed your result, and expressed it in terms of K rather than AE for direct comparison with what I derived above.  It Taylors to
4K(L-L0)/L + 12K(L0/L)(D/L)2 - 20K(L0/L)(D/L)4 + ...
where K is (EA/L0).

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

Greg. I succumbed to a quick thought, and looked into what happens to your formula when D heads to infinity: I think it heads to infinity as well. It should asymptotically approach 4EA/L0.

### RE: Cable stiffness lateral direction point load

I haven't checked with Greg's formula, but I plotted a few values assuming a 1/2" 270k tendon with an area of 0.153in^2, an E value of 28.5e6 psi and a minimum yield stress of 37,170psi. I assumed a pretension of zero.

BA

### RE: Cable stiffness lateral direction point load

Time to put this to bed.  I have done some "hand" calculations on a spreadsheet and plotted the resulting stiffness versus displacement values.  On top of this I have plotted the predictions from my 07Sep22@06:35 formula (my third attempt I should admit very 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

Here's my corrected derivation, but it still doesn't agree with yours.

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

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

Greg.

Your eqn (1).  I believe that the 2EA/L term at its start should be 2EA/L0 where L0 is the cable's unstretched length (as opposed to the distance between the support points).  In "normal" configurations (ie where L0 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

#### Quote (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.

It appears lobos22 has abandoned the ship.

By definition, stiffness K = F/d where d is horizontal displacement.
When prestress T = 0, L0 = L. F/d can be calculated for values of d while cable remains below Tyield. Beyond that, it is meaningless.

When prestress T is Tyield, L0 = L/(1+Tyield/AE); if d>0, cable is strained beyond yield.

Applying a prestress of Tyield is pointless. Applying any prestress reduces F, so any prestress seems pointless.

BA

### RE: Cable stiffness lateral direction point load

OK, so you've never designed a flying fox or equivalent. Prestressing cables is a thing. "By definition, stiffness K = F/d " is incorrect in the general case as it is undefined at d=0. It just so happens for linear elastic behaviour when F!=0 K=dF/dx=F/x, but if you think about it for example a cantilever has stiffnesss at x=0.

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

#### Quote (GregLocock)

OK, so you've never designed a flying fox or equivalent. Prestressing cables is a thing. "By definition, stiffness K = F/d " is incorrect in the general case as it is undefined at d=0. It just so happens for linear elastic behavior when F!=0 K=dF/dx=F/x, but if you think about it for example a cantilever has stiffness at x=0.

Cheers

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

Flying fox=children go to hospital. Zip line as they call them these days. Somewhere there is a photo of me, upside down (deliberately), 50 ft above the ground on one of those thing in the rainforest. We also used to get the animal in one of our trees in the old house.

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

Just for fun, I considered a 1/2" dia. 270k tendon 20'-0" (240") long with a prestress of zero and also Ty/2 (half the yield strength of the tendon). The tables and graphs of F versus d are shown below.

BA

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