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Wire Sag
3

Wire Sag

Wire Sag

(OP)
I need a formula that help me calculate sag in a steel wire due to it’s own weight over a given length.

RE: Wire Sag

Catenary Curve

y = (H/w)*cosh(w*x/H) -(H/w)

where w is the line density (mass/length)
      H is the Horizontal componet of the cable tension (Force/Area)

 

RE: Wire Sag

(OP)
Could you explain why the modulus of elasticity is not a factor?

RE: Wire Sag

Look in Mark's Standard Handbook for the Catenary curve.  The curve represents a "curve in which a flexible chain or cord of uniform density will hang when supported by the two ends".  This means that no stiffness is assumed, hence no Modulus of Elasticity.  

I have used this to also model straps that are restraining uniformly distributed loads.

A short span of steel wire with no tension will act like a beam, while a long span will act like a catenary.  Something in between will be half and half.  The point where one becomes another depends on the thickness of the wire relative to the span.

STF

RE: Wire Sag

2
The modulus of elasticity is a factor ! The equations you need can be found in Roark's Formulas for Stress and Strain.

ymax = l*(3*w*l/(64EA))^(1/3)

P = w*l^2/(8*ymax)

Where ymax is the maximum sag
      l    is the unstretched length
      w    is the weight/unit length
      E    Youngs modulus
      A    Wire area
      P    axial tension

Rich2001 is correct in saying that the curve is a catenary

One interesting point is that the cables on suspension bridges do not hang in catenaries, but parabolas (because of the horizontal road weight).

RE: Wire Sag

Actually, more correctly should have said that P is the horizontally applied stretching force at the ends.

RE: Wire Sag

I should further add that although the curve is in reality a catenary, the two equations I have given appear to be derived on the basis that the curve is a parabola, which is a good approximation if the depth of sag is no greater than 1/8th of the span.
It appears to me that you can use the first equation to determine what the sag would be in a cable that had zero inital tension (this obviously depends on E).
You can then use the second equation to derive the induced tension P. I should note that according to a second reference that I have for the second formula (Blevins formulas for natural frequency), P is not horizontal but is the assumed mean tension in the wire. This tension is actually not constant along the wire, but can be assumed so for shallow sags.
It further appears that one can use the second equation for any tension P.

Blevins also gives this equation:
ymax/L = (5/24)^(1/2)*(1-(1-18/5*(Lc-L)/L)^(1/2))^(1/2)

- where L is the span as before, and Lc is the unstressed cable length. You can use this is conjunction with the formula for P. That should give you everything you need to reduce your sag as required..

RE: Wire Sag

Line sag (ITT Ref. Data with supports at same elev.):

H:=WL^2/8S
S:=WL^2/8H

L: distance between support
S: Sag at mid-span
H: tension (horzontal component of tension)
W: weight of cable per unit length

(parabolic approx.)

The modulus of elasticity is only important where cable stiffness is sufficient to justify its modeling as a beam rather than a cable.
 

RE: Wire Sag

Hacksaw :
Your equation for tension is identical to mine.
I do not agree with your final statement however.

My first reply should probably have read "modulus of elasticity can be a factor, depending on how the problem is formulated".

I provided three equations, which all relate to either extensible or inextensible strings (or cables). None of them have anything to do with beams. Usually, the term "beam" relates to a bar which can resist bending and/or shear.

The first equation applies to the case where an extensible  string (or cable) of initial length LC is fixed between two supports distant L apart, and LC = L.
In this case, ymax = L*(3*w*L/(64EA))^(1/3)
The sag is directly related to E

The second equation applies to the case of an inextensible string, where LC > L
In this case,ymax = L*(5/24)^(1/2)*(1-(1-18/5*(LC-L)/L)^(1/2))^(1/2)
This equation could easily be extended to the case of an extensible string (or cable), in which case you would replace LC with LC*(1+P/(A*E)) where P is given by the next equation - you would then have to iterate to find the solution, but in most cases it would hardly be worth it.

The third equation, which is the same as yours, gives the tension in terms of the sag - and is valid for all cases, as I stated in my second post -including either of the above cases, or the case where the tension has been increased to a higher value.
P = w*l^2/(8*ymax)

So it all depends how the problem is formulated. If the tension is specified, all you need is eqn 3.

If the length of the string (or cable) is specified, then you need eqn 1 or 2 as well.


RE: Wire Sag



I stand corrected, you recognized my error before I could log back in.

It is good to have such able peers at hand to keep me honest!

RE: Wire Sag

Thanks hacksaw - very flattering and thanks for the star!
Did you realise there is another thread on this subject with different answers ?
Thread507-60466

RE: Wire Sag

I have nothing to add to this post's technical content, but I would like to provide an editing tip: you can create subscripts and superscripts by using the code described in the Process TGML link below in Step 2 Options of the reply area.  For example the equation given by EnglishMuffin could look like this:

ymax = L · (5/24)0.5 · (1 - (1 - 18/5 · (LC-L)/L)0.5)0.5



Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.

RE: Wire Sag

Now that you've pointed it out, I do recall using the modulus of elasticity - had I known it was in Roark's, my analysis would have been much simpler!
Thanks for the clarification: I really shot from the hip in my first post.

STF

RE: Wire Sag

Thanks CoryPad :
I'm spending too much time writing and not enough reading !

RE: Wire Sag

(OP)
Thanks EnglishMuffin for all your input.  I tried to put your formula into an Excel spreadsheet and I am having some difficulties with the results.  It appears that no matter what diameter I use the Ymax stays the same. Any thoughts.     

Wire Sag                    

Ymax:     7.223390595 Max. Sag l*(3*w*l/(64*E*A))^(1/3)            
l:        960    Is the Unstretched Length                
Wire dia. 0.015                    
w:      5.01869E-05    Is the weight/unit length (steel)                (PI*((Wire Dia/2)^2)*0.284)        
E:      30000000    Youngs Modulus                
A:      0.000176715    Wire Area PI*((Wire Dia./2)^2)        
P:      0.800390857    Axial Tension    w * I^2 /(8*ymac)            

RE: Wire Sag

Here is the first set of circumstances that JerryV and i are attempting to solve.

A band of 7 clamped cables are pulled 960" from a locked multi-cable "precision" metered source which can be assumed fixed.  The clamped group (to negate cable to cable shifting) is placed under 5.25 lbs. of tension.  It is observed that one cable is the tightest, with a slight sag.  The next cable is 1/2" from the first at the mid point (ymax), the next is 1" from the first (1/2 from the second), fourth is 1.5", fifth is 2.0", sixth is 2.5", and the seventh is 3.0".  This creates a 1/2" gap between each cable.

These gaps translate to approximately .000725" in longer lenght per .5" gap. (this was determined using a basic arc from the 960" end points, this approx is probably debatable as well).

Given .015" diameter cable.  .284 lb/in3. and P = w*l^2/(8*ymax).  Assumption that the tight wire sees all of the 5.25 tension.

It was determined that the top wire sag was .01096".  

There is debate as to why E is not necessary here.  

It seems to me that there would be a variety of regimes for consideration.  Because the wire is not in tension to the point where elongation would occur, then i would not think E is relative in the "Undeformed Regime".  

I would think that the sag of the first cable can be used as the sag of the group.  If additional tension is added to reduce the sag, at what point can it be determined that the "short" cable is under tension and entering a "elastic reginme" where E becomes a factor and elongation can be calculated to the accumulated point that all of the end are tight and of the same length as the long end.

Your thoughts are appreciated, valued, and applied.

My general thought is how the stated equations relate to this specific scenario.

RE: Wire Sag

jerryv - no I think you are quite correct.

You will find this formula in Roark, for a cable with a distributed load on it, and I suppose the only time the wire diameter would actually come into it would be if the weight per unit length were independent of the wire - this would apply in the case of an axially stiff wire with a heavy coating on it, that itself had a very low youngs modulus.

RE: Wire Sag

ZackS

As far as the equations are concerned, if you know the tension, you know ymax, and vica versa. It doesn't matter whether the cable is axially stretched or not, or what the value of E is. The tension depends purely on the loaded geomety and the weight per unit length. Think about the equilibrium of a small element of cable. Bear in mind that all these equations are approximate, based on a parabola rather than a catenary - which is the true shape - but the error is minute with these small sags. They also assume that the tension is constant along the length, which is also not true exactly - but again, the error is minute for small sags. I will study your problem and see whether I understand what you have - in the meantime, someone will probably beat me to it - I'm going to the local Mexican restaurant.

RE: Wire Sag

ZackS :

Let me see if I understand this correctly - I'm a little slow on the uptake, and without a drawing its tough to be sure that I am visualizing it correctly. Can I assume we are effectively talking about this :
The first wire (the shortest) is stretched between two points 960 inches apart. The sag of this wire is .011". Six other wires are connected between these two points, having sags from the horizontal of 1/2", 1", 1.5", 2",2.5", 3".
There is a horizontal force of 5.25 lbf applied between the points.
I am not clear as to whether the .011" is measured, required, or calculated.
I am not clear whether the 5.25 lbf is measured or required.
I am also not clear what it is you want to calculate. The sags, if known, enable  the tension at the end of each cable to be computed. By figuring the slopes at the ends we can compute the necessary total axial force from superposition. It should come out to the measured value - if known.

First - can I get the above clarified ?

And I would not assume at the outset that most of the 5.25 lbf is resisted by tension in the first wire, but it may be so.

RE: Wire Sag

Actually - we'd beter cut to the chase here
You state the following :

Diameter of cable : .015"
Density           :.284 lb/in3
Cable length      : 960 in

This means that the weight per inch of the cable is 5.02*10^-5
Which means that for a sag of 0.01" you would need a tension of :
       
   P = 5.02*10^-5*960^2/(8*.01) = 578 lbf

   Where are you getting the 5.25 lbf from ?  

RE: Wire Sag

And I suppose that I need hardly add that such an enormous tension would produce a stress greatly exceeding the breaking stress of the wire !

RE: Wire Sag

to further englishmuffin you should think of a wire accross a tennis net (a catenery), from experiance u will know that even if u tighten it as much as u can the wire will never be straight and eventually u will rip the post's out or the wire will break. if u look around there are some forumula that only require basic integration to solve for sag and tension etc.

RE: Wire Sag

A few comments:
1) Presuming the loads are due purely to the weight of the cable itself:
A) The loads are directly proportional to the diameter
B) The axial stiffness is directly proportional to the diameter.
Therefore, the diameter is in both the numerator and denominator ("w" and "A" respectively), and hence falls out of the equation (as jerryv observes above).

2) The posted equations appear to be relevant for the situation in which the load is due to weight of the wire itself, and pre-loads are unimportant.  If there is significant pre-tension (such as that utilized in winches), this could effect the results.

3) (Maybe 2(b) . . .) These equations are of course linear approximations; they certainly are not relevant to significant pre-tension (think of guitar strings). It doesn't appear from my understanding of the problem that this is important, but I could be wrong. In such case, however, superposition wouldn't hold.

I'm thinking a bit off-the-cuff here; I hope I've added something of value rather than muddled this . . .

Brad

RE: Wire Sag

bradh:

I posted three equations - one applies when the length of the unstretched cable exactly equals the distance between the supports. (as you say - this equation predicts a deflection independent of the wire diameter, unless it was a wire with an axially weak coating  of non-negligable mass such as some electrical wires). The second equation applies when the length of the cable is specified and is longer than the span. The equation for tension applies in any case, and is probably the only equation you need if the sags are defined. I thought I had made all that fairly clear, but perhaps not! The numbers given by ZackS make no sense, however, unless I am misunderstanding something.

RE: Wire Sag

And at the risk of laboring the point, Brad, if I may call you Brad, the tension equation is applicable to any case, including guitar strings.
One could for completeness add a fourth equation, which would apply in the case that LC < L. I have not found that one anywhere, but as far as I can see it might be :

ymax = w*L^2/(8*A*E*(1-LC/L))

at least to a first order of approximation - but I could be wrong and stand corrected if so.

In addition to electrical wire, another occasion where all the variables in these equations would be required would be the case of an extension spring stretched between two supports, in whch case you could develop new equations with a psuedo E value, - but it would be complicated a bit by the possible pretension and significant bending stiffness in some cases.

RE: Wire Sag

English Muffin,
Let this be a lesson to you young men out there--drinking and posting is not always a good idea (even with only one beer; I'm getting to be a lightweigth . . .).

My apologies English Muffin. I was trying to absorb too many posts, and managed to miss appreciating your most critical posts.

My point 1 still holds, but I'll take back points 2 and 3.

I'm also confused regarding the 5.25 lbf value.
Further, out of pure curiousity--how does one accurately assess the sag 0.011" over a length of 80'?  As the tension is most sensitive to this variable, one must be pretty confident of this sag value (and I can't think how to measure to this level of precision, given other errors).

Brad

RE: Wire Sag

Brad :
Yes - it beats me. I think its a calculated value, although how it was calculated I don't know. The tension required would be so high that the wire would break. I think the two guys that are trying to solve this problem aren't reading this anymore. I hope they come back.

RE: Wire Sag

sorry i took so long, i was running lab tests.

In essence the 7 ends are fused together, then pulled the 960" and placed under a 5.25 lb. load.  The ends are coming of a metering unit that is intended to payoff equal amounts of wire.  Since it can not truly be exact in length, i believe there is a length difference between each wire. All of the ends are fixed together at each end of the 960".

Now, since there is a measureable distance between the centerpoint of each end, i used an arc length approximation to determine what additional length is necessary to allow such gaps in the wire.  And, since they are fixed together at each end, i believe the top wire sees the majority of the 5.25 lbs.  Using the known tension, wire characteristics, and distance i use the non-E formula to calculate sag (.011").  

Now, the issue is the balance between gap due to length difference and gap due to tension/sag.  I have been able to approximate what tension on each end would allow such a gap to exist, however since the ends of each wire are fixed, then i would be inclined to think that the length difference was the cause.

I will continue....

RE: Wire Sag

I've been reading this thread, but I still can't believe that there's only 11 mils of sag over 80 ft.  

I've pulled both 50 lb test fishing line and 1/16" cable with WAY more tension than 5 pounds over 100 ft, and couldn't even get close to 1 ft of sag.

TTFN

RE: Wire Sag

ZackS:

If you are using the tension formula which we all seem to agree on, explain to me why its 5.25 lbf and not 578 lbf for a single steel wire .015" dia (see data on Jun 18th, ten posts back - maybe I'm missing something - I've been wrong before.

RE: Wire Sag

...measurng the sag should be easy, with an inexpensive laser to create a straight baseline and a ruler at the midspan...

RE: Wire Sag

Your right, i had some random error in my calculation, the sag would be more like 1.1 inches. 0.284 lb/in^3 *.015 *.015 *Pi/4 = 5.018*10^-5.   (5.018*10-5 *960^2)/(8/5.25lb) = 1.10"

i will continue later, i am on vacation the rest of the week.

RE: Wire Sag

Phew ! O.K. Now perhaps we can get somewhere. Get back in touch when you can.

RE: Wire Sag

Hello.  It has been a while.  Thanks for your patience and previous help.

A quick recap: In Jerry and my industry we evaluate multi end spool quality by catenary.  Placing the spool on a shaft with tensioning capabilities, and clamping the ends of a band of wire to ensure that they start at even lengths.  Then it is pulled to an equal height roller 960" away and .75 lbs per end is placed on the band of clamped ends, being applied to all of them.  The gap between the tightest and loosest wire is measured, giving a numerical evaluation of the quality of winding that created that spool.  Where quality is related to the length difference between each wire on the spool.

We had wondered what the relationship with wire sag and tension is to this method of evaluating the spool.

The first attempt was to calculate the combination of sag and elongation that would allow the clamped wires to be 3" from the tightest to the loosest wire in a band of 7 ends and estimate what the wire length difference would be between those ends.

I was never able to combine all of those things satisfactorily.

However, today a simple method negated the need to.  One of those Eureka moments where you wonder why you didn't think of it before.

We pulled the wire over the 960" as usual, clamping the ends as usual, and measuring the gap in the middle.  Then we marked the wire past the 960" pivot point and attached 2lb weights to each wire past the clamp.  The clamp was removed and each end was under the same tension and same sag.  Negating these issues.  It could then be measured between the shortest and longest end what the actual difference between the two lengths was.  

In this case .280".  Not too hard, just not done before.  It is important for us to know and develop our machinery, that this type of wire, under these conditions where a previously used measurement of quality that in this case was 4.25" equates to a length difference of .280" over 960" of a band.  

Thank you all for your help.  Particularly EnglishMuffin, for sticking with it.

RE: Wire Sag

Glad it's making (some) sense now. Good luck !

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