Wire Sag Revisited, deflection of a pretensioned extensible cable

[skipm]

In the previous thread (404-60468) EnglishMuffin cited the equation ymax= L*(3*w*L/(64EA))^(1/3) to determine sag of a horizontal extensible cable whose length equals the span from Roark's formulas. He also proposed an equation where the cable is shorter than the span as ymax=w*L^2/(8*A*E*(1-LC/L)).

We make a woven metal wire mesh that is hung on exterior structures that is subjected to wind loading. The mesh is made of interlinked flattened wire spirals that are springy and hinge freely. A link chain fence is a crude approximation of the mesh construction. We have wind tunnel data for loading and load vs. elongation data that a psuedo E value have been derrived from. Using this data, the first equation will estimate the deflection of a strip of mesh hung between anchor points who's length is equal to the span. We would like to be able to estimate the effect of pretension on the mesh's deflection and thus the resulting mesh tension. The problem with the second equation is that as the pretension (mesh stretch) is reduced to zero (equal lengths), the calculated deflection increased to infinity instead of the value of the first equation.

I am looking an equation that will factor in the pretension and estimate the sag. If it would be helpfull, I can supply actual values or clarify details of the application.

Thanks Skip

 

[prex]

The second equation is incorrect: it should be replaced by:
y_max=wL²/8H
where H is the horizontal component of the tension in the wire.
For small deflections H may be taken equal to the total tension in the wire: under that condition the above formula will give (the derivation may be found in many books on the theory of elasticity):
y_max=L(3wL/EA)^1/3/4.
Didn't find the treatment of your situation (initial tension in the wire), but should be not hard to derive from the equations. Will come back if I find something.


prex

http://www.xcalcs.com

Online tools for structural design

 

[EnglishMuffin]

Thread 404-60468 does of course give prex's equation, which always applies whatever the length of LC, but this is not a "replacement" for the equation I gave since it does not incorporate LC. Since it was derived on the assumption that the working tension is the same as the initial tension, I don't claim that the equation I gave for the case where LC<L is 100% correct, as I intimated in the thread. However, it should be fairly accurate if the initial tension is a large percentage of the working tension.

I appear to have derived it as follows :

Make the approximate assumption that the working tension is the same as the initial tension:

Then initially assuming no sag, the cable tension P will be E*A*(L - LC)/L

So from P = W*L^2/(8*ymax) we derive the stated equation.

I'll give it a bit more thought tonight.

 

[skipm]

Thanks for your responses. Unfortunately, the pretensions are not a significant percentage of the mesh strength. The meshs are typically rated above 1000 #/ft tensile and can withstand manys times that before failure. This could exert some tremendous forces on a building. To help keep the forces relatively low (400-500 #/ft @ 90mph wind load) we want the mesh to deflect/sag to releave some of the tension. What we are trying to balance is mesh stiffness, span between supports, clearance to adjacent structure, wind load, and preload. What we are trying to do estimate the reaction to say 100 #/ft preload, will the deflection reduce from 6" to 4" and keep off the building, and then how much stronger do the anchor points need to be. The alternative is to add intermediate stablization supports to reduce the span and thus the deflection and tension but that is not always possible.

Thanks again for the input, Skip

 

[Denial]

For a cable whose horizontal component of tension is H and whose weight per unit length is w, the deflected shape will be part of the catenary
y = c*cosh(x/c)

If the cable runs from x=x₁ to x=x₂ then the length around the catenary is
L_s = c*sinh(x₂/c) - c*sinh(x₁/c)
This is the stretched length of your cable.

The above appears in most texts that cover cables. Unlike what follows, which is the result my own manipulations and so should be used with caution.

Let the cable's effective cross-sectional area be A and its effective Young's modulus be E. Then it can be shown that the amount of stretch in the above cable is
S = cH*{sinh(2x₂/c)-sinh(2x₁/c)}/(4AE) + H(x₂-x₁)/(2AE)

Thus the unstretched length of the cable is given by
L_u = L_s - S

These results are all exact. Use Taylor expansions to come up with polynomial simplifications if you believe your cable is sufficiently taut.

(Trivial closing point. If your cable is so stretchable that the stretch affects the value of w, all bets are off.)

HTH.

 

[EnglishMuffin]

skipm:

I wouldn't get too hung up about "exact" solutions in this case - the "correct" tension formula is also an approximation since in reality the tension varies along the length of the wire/cable very slightly. I don't believe there is a "exact" closed form solution for this problem, although I could be wrong.
Denial's equations are probably "exact" - but they appear to leave a little to the imagination since they are in terms not of ymax but the parameter "c", which is a complicated function of ymax. There is however an approximate solution which is much closer to the truth than my original one (which was never meant to be anything other than a passing "comment").

I continue to use my original symbols, which I think came originally from Roark.

Rothbart (Mechanical Design & Systems Handbook - 1st Ed) gives the following approximate formula for the length of a catenary in terms of ymax :

LS = L*(1+2/3*(2*ymax/L)^2) (assuming ymax << L)

So, in similar vein to my previous derivation , the cable tension P will more correctly be given by:

E*A*(L*(1+2/3(2*ymax/L)^2) - LC)/LC

(Which converges to my previous formula if ymax is small - since whether one uses LC or L for the denominator makes little difference in practice).

So if my math is correct, we obtain the following cubic for ymax :

ymax^3*8/(3*LC*L) + ymax*(L/LC-1) - W*L^2/(8*EA) = 0

This can of course be solved by a variety of methods, including a closed form solution, and should be accurate enough for your needs.

 

[Denial]

Thanks and apologies, EnglishMuffin, I left out the most important part: the fact that
c = H/w

I share your belief that there is no "exact" closed form solution to the problem. Some sort of iterative solution is required. However it can be formulated in such a way that convergence is extremely rapid for any reasonably taut cable, and this fact allows an accurate non-VBA spreadsheet solution that contains (say) five hard-wired iterations. I also believe that assuming a parabolic approximation to the catenary still leaves you without a closed form solution. But I would be delighted to be proved wrong.

I do not have access to Rothbart's book, but I suspect that the formula you quote is applicable only to cables whose two endpoints are at the same height. Again, happy to be corrected.

(Just to avoid any confusion, let me specify the specific problem I believe we are talking about. We have two known fixed points in the XY plane, Y being vertically upwards, and Y₁ not being equal to Y₂. Between these two points we string a cable of known unstretched length, known weight per unit length and known extensibility. The problem is to determine the midspan vertical sag of the cable.)

 

[EnglishMuffin]

Denial:
As far as I am aware, all the formulae and discussion on both threads, except for yours, have referred to the "horizontal" case where the endpoints are at the same height, so your equations are undoubtedly superior, assuming they are correct. I am not clear whether skipm is also interested only in the horizontal case. Now that you have clarified what "c" is - your equation and mine should give approximately the same answer for the horizontal case (assuming the sag is small).

 

[prex]

EnglishMuffin and Denial, the catenary solution is normally used for cables that are not straight in their unloaded condition. For an initially straight wire (may be named horizontal, but the orientation is not relevant if the load is not due to self weight) the approximation normallly used is a parabola (that, as noted by someone above, may be seen as a limiting case of the more general catenary).
However your equations do not explicitly contain the initial tension in the wire, so here is what I propose for the problem originally described by skipm.
Say H_o is the initial tension in the wire, we have y_max=wL²/8(H_o+H).
As the deformed shape is a parabola [&Delta;]L=8y_max²/3L but also [&Delta;]L=HL/EA as this elongation is only due to H(not to H_o).
By combining the above and if I made no errors, the equation sought by skipm should be:
y_max³+3H_oL²/8EA=3wL⁴/64EA, that, as required, reduces to the formula in my previous post for H_o=0.
This is a cubic equation, so, though it is not really in closed form, it can be solved analitically to gain the required closed form.


prex

http://www.xcalcs.com

Online tools for structural design

 

[EnglishMuffin]

prex:

Well, since one can formulate the equations in terms of initial length LC, or initial tension, one could easily modify the equation I gave accordingly, by repacing all the LC terms with a function of the initial tension. Both our equations are cubics in ymax, but my solution has a ymax term which yours lacks. Although both are approximations, it would be interesting to know which is more accurate. It is all a question of what equation one uses for the length of the catenary.

 

[Denial]

I have just derived the cubic equation for y_max as given by EnglishMuffin in his post about five entries above this post. In doing this I had to make the following assumptions:
(1) The deflected shape of the cable is a parabola.
(2) The two ends of the cable are at the same elevation.
(3) The tension throughout the cable is constant and equal to the tension at the cable's midpoint/lowpoint (an assumption entirely consistent with the first two).

I regard a cubic equation as a closed form solution. Therefore I must concede that, for a horizontal cable, I was wrong in my earlier assertion that there would not be a closed form solution even if the parabolic approximation was applied.

Thanks for keeping me honest, chaps.

 

[EnglishMuffin]

Denial : Not surprisingly, I agree with you !

 

[prex]

Oops!
My equation should have been written as:
y_max³+3H_oLy_max/8EA=3wL⁴/64EA
So EnglishMuffin it's clear that your equation is the same (with the difference of second order terms): the meaning of your LC term was unclear to me till you explicitly stated it.

prex

http://www.xcalcs.com

Online tools for structural design

 

[skipm]

Thanks again for all the thought you three have put in to this! I have tried the new equations but I am still not getting a usable number from them. I am using the MathCad root function to solve for ymax. It may be me, don't know so here is some actual numbers. Maybe you can prove me wrong. I am only trying to model the horizontal case.

L = 193.5in
w = .1106#/in
E = 11673#/in^2
A = 12in^2

which when entered in the the first equation yield

ymax = 3.73in and a Tension = 139.39#

Front our test fixture I get the following

0# preload => ymax = 3.63in @ 139# (close enough)
100# preload => ymax = 2.26in @ 218#
200# preload => ymax = 1.63in @ 295#
300# preload => ymax = 1.24in @ 378#
400# preload => ymax = 0.99in @ 472#

Once again, it maybe me. These values should give us a common reference. Thanks again for the help.

Skip

 

[EnglishMuffin]

I'll check it out tonight, although I expect somebody will beat me to it. But one question:
How exactly are you measuring the preload ? Can you confirm that the stated tensions are definitely not working tensions with sag present ?

 

[prex]

I'm very sorry, but there's another mistake in my formula, that should read:
y_max³+3H_oL²y_max/8EA=3wL⁴/64EA
Hope this time is the correct one!
With this formula and your numbers I obtain the following results:
[tt]H_o y_max H_o+H
0 3.73 139
100 2.85 181
200 2.12 245
300 1.59 325
400 1.25 415
[/tt]
As you see these are quite different from your measurements, to the point that we cannot consider that formula as giving correct results.
However I think there is something to be explained in your data. Take the last line (preload 400): the measured additional tension is 72. Now with a deflection of 0.99, using your data and assuming a parabola as the deflection line (but that wouldn't change much assuming another shape), the change in length of the wire, excluding the effect of the initial tension is [&Delta;]L=8y²/3L=0.0136 and the tension corresponding to this is H=EA[&Delta;]L/L=10 (very different from the measured 72).
Can you explain how you took your measurements?

prex

http://www.xcalcs.com

Online tools for structural design

 

[skipm]

Hello all, the test fixture is a frame made of 3" channel iron that will hold a 16' long sample, 1 ft wide. The mesh is mounted flush to the bottom of the frame and the frame is laying on the floor. The mesh is anchored to one end of the frame, the other is attached to a horizontally floating carrier that a 1000# tension ring gage is inturn anchored to the frame. With the mesh and frame on the floor and fully supported, the preload is established and then the frame is blocked up to suspend the mesh. The resulting sag is measured and induced tension recorded.

The actual sag/load values I am confortable with, we are currently having internal discussion on the derivations of our E and area values so the equations may not be as close (garbage in => garbage out). I think we are going to test a larger sampling of mesh to derive hopefully a more consistant E value since our present sampling is not depending on where you look. Although the numbers are not exact, they atleast now resemble the tests.

Thanks again

Skip

 

[skipm]

FYI, I have been able to duplicate prex's results. Now I just have to come up with an E that matches the test results.

Skip

 

[skipm]

Another complicating factor is the nature of the mesh. It is a 3D woven, interlinked product that yields a non-linear E profile in the initial loading as the mesh seats. I was attempting to compensate by starting with the mesh under an initial preload to get into the linear E range. Variation/accuracy of this preload can also have a large effect on the sag and thus tension since a small change in length multiplies into relatively large change in sag at when the initial panel length equals the span.

Skip

 

[EnglishMuffin]

Well, I can't improve on Prex's results. I solved the cubic closed form, and got the same numbers (except I think the last one should be 416 rather than 415). I think you are correct, skipm - the problem probably lies with your assumptions about E, and/or A. You have probably noted that at least the tension formula gives reasonable results for the sag. (ie: 3.72 : 2.37 : 1.75 : 1.37 : 1.10). Curiously, they are all off by about .1. Could this have something to do with your method of measurement ?