Wire Sag 3

[jerryv]

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

 

[EnglishMuffin]

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 !

 

[duncs]

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.

 

[brad]

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

 

[EnglishMuffin]

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.

 

[EnglishMuffin]

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.

 

[brad]

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&quot; 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

 

[EnglishMuffin]

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.

 

[ZackS]

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

In essence the 7 ends are fused together, then pulled the 960&quot; 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&quot;.

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&quot.

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....

 

[IRstuff]

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&quot; cable with WAY more tension than 5 pounds over 100 ft, and couldn't even get close to 1 ft of sag.

TTFN

 

[EnglishMuffin]

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&quot; dia (see data on Jun 18th, ten posts back - maybe I'm missing something - I've been wrong before.

 

[IFRs]

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

 

[ZackS]

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&quot;

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

 

[EnglishMuffin]

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

 

[ZackS]

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&quot; 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&quot; 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&quot; as usual, clamping the ends as usual, and measuring the gap in the middle. Then we marked the wire past the 960&quot; 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&quot;. 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&quot; equates to a length difference of .280&quot; over 960&quot; of a band.

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

 

[EnglishMuffin]

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