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.

 

[GregLocock]

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


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[Denial]

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).

 

[GregLocock]

Hmm, a factor of 2 on the constants after the first. How odd. Definitely a 3 pipe problem!

Cheers

Greg Locock


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[Denial]

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.

 

[GregLocock]

OK I better have another look, with a bit more enthusiasm since the maths is doable.

Cheers

Greg Locock


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[BAretired]

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

 

[Denial]

Time to put this to bed.[&nbsp;] I have done some "hand" calculations on a spreadsheet and plotted the resulting stiffness versus displacement values.[&nbsp;] On top of this I have plotted the predictions from my 07Sep22@06:35 formula (my third attempt I should admit very sotto[&nbsp;]voce).[&nbsp;] And I have further plotted the predictions from Greg's 07Sep@23:42 formula.[&nbsp;] 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.

 

[GregLocock]

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


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[Denial]

Greg.

Your eqn (1).[&nbsp;] 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).[&nbsp;] 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).[&nbsp;] I think you have made a similar error to the one I made initially and then reported in my 07Sep22@06:35 post above:[&nbsp;] using TAN when you should have used SIN.[&nbsp;] This probably explains the infinity.[&nbsp;] Your eqn[&nbsp;](2) should be
F/2=(T+b)(2x)/sqrt[L^2+(2x)^2]

 

[GregLocock]

Agree on the second one, I used fairly similar triangles, not similar triangles! Not so sure about the first one, I'll try it both ways.

Cheers

Greg Locock


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[BAretired]

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 T_yield. Beyond that, it is meaningless.

When prestress T is T_yield, L₀ = 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

 

[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 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


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[BAretired]

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

 

[GregLocock]

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


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[BAretired]

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

BA