Wire Sag Revisited, deflection of a pretensioned extensible cable
Wire Sag Revisited, deflection of a pretensioned extensible cable
(OP)
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
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





RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
ymax=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):
ymax=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
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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.
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
Thanks again for the input, Skip
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
y = c*cosh(x/c)
If the cable runs from x=x1 to x=x2 then the length around the catenary is
Ls = 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
Lu = Ls - 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.
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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.
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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.)
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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).
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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 Ho is the initial tension in the wire, we have ymax=wL2/8(Ho+H).
As the deformed shape is a parabola ΔL=8ymax2/3L but also ΔL=HL/EA as this elongation is only due to H(not to Ho).
By combining the above and if I made no errors, the equation sought by skipm should be:
ymax3+3HoL2/8EA=3wL4/64EA, that, as required, reduces to the formula in my previous post for Ho=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
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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.
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
(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.
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
My equation should have been written as:
ymax3+3HoLymax/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
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
How exactly are you measuring the preload ? Can you confirm that the stated tensions are definitely not working tensions with sag present ?
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
ymax3+3HoL2ymax/8EA=3wL4/64EA
Hope this time is the correct one!
With this formula and your numbers I obtain the following results:
Ho ymax Ho+H
0 3.73 139
100 2.85 181
200 2.12 245
300 1.59 325
400 1.25 415
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 ΔL=8y2/3L=0.0136 and the tension corresponding to this is H=EAΔL/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
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
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
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
Skip
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
Skip
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
RE: Wire Sag Revisited, deflection of a pretensioned extensible cable
Thanks again Prex and EnglishMuffin.
Skip