# Wire sag formula

by williamlynn
Tags: formula, wire
 P: 4 I am looking for a formula to calculate the deflection(sag) in steel wire that is supported between two points with a known tension. I want to be able to calculate the deflection(sag) at any point along the wire.
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 Quote by williamlynn I am looking for a formula to calculate the deflection(sag) in steel wire that is supported between two points with a known tension. I want to be able to calculate the deflection(sag) at any point along the wire.
Well it depends upon a lot of stuff, like the properties and temperature of the wire, and the span/sag ratio. The wire will take the shape of the catenary (hyperbolic functions, kind of tough to calculate manually), however, if the sag is relatively small in comparison to the span (say the sag is less than about 10 percent of the span), then the catenary curve is very closely approximated by a parbolic curve using the following equation: $$T=wl^2/(8d)$$, where w is the weight of the wire per unit length, l is the horizontal span between supports, d is the sag, and T is the horizontal tension in the cable. So, given T, l, and w, you can easily calculate d (the sag at the low point of the curve). If it is less than 10% or so of l, you plot the parabolic curve, and can then get sags at various points using the parabolic properties of that curve. Temperature and live loadis will affect sag/tension values, but if you're just looking at one tension at a given temperature under the wire dead load weight only, you need not go further.
 P: 4 Thanks for quick response; I really need to be able to find deflection along any point in the wire. I'm lloking at .016" dia wire with spans from 10 feet to 100 feet and wire tension of 30 pounds.
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Wire sag formula

 Quote by williamlynn Thanks for quick response; I really need to be able to find deflection along any point in the wire. I'm lloking at .016" dia wire with spans from 10 feet to 100 feet and wire tension of 30 pounds.
Oh, I was assuming a larger wire with higher tensions and spans. Nonetheless, the same formula applies. But first be sure you're using a very high strength steel, otherewise even your meager 30 pounds will snap it in half, because you are using such small diameter wire.
Second, i'm not sure what your application is, but your sags are going to be very very small for the 10 foot span (the wire will be fiddle tight, use the formula i gave you after calculating w). Even for the 100 foot span. What are you trying to do?
 P: 4 The wire sag is used for turbine machinery alignment; the internal parts of the turbine must be accurately aligned to a centerline. The wire is the centerline reference but the wire sag must be accounted for to get accurate centerline. Alignment of the turbine components has to be within thousants of an inch.
 Sci Advisor P: 5,095 I can't say that I have ever heard of using wire as a centerline reference. That doesn't mean people don't use it though. It seems kind of "hokey." You could machine a holding feature and insert a laser pointer. That's what we use a lot. That way you can calibrate the pointer position and not have to worry about sag.
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 Quote by williamlynn The wire sag is used for turbine machinery alignment; the internal parts of the turbine must be accurately aligned to a centerline. The wire is the centerline reference but the wire sag must be accounted for to get accurate centerline. Alignment of the turbine components has to be within thousants of an inch.
Yow, that's a close tolerance. Let me give you a sample calc based on the parabolic approximation, but don't use it for actual design!

The 0.016 inch diameter steel wire weighs about 0.0007 pounds per foot. So for a 10 foot span between level supports, and 30 pounds tension at a given temperature, the sag at the low point (mid point of the span) is d=wl^2/8T = 0.0003 feet, or about 3/1000 of an inch. For a 100 foot span with that same tension, d = wl^2/8T = 0.03 feet (about 3/8 of an inch). Now to get the wire deflections at other points in the span, let's take the 100 foot span case, using the parabolic approximation y=ax^2 (letting the low point of the curve be at origin (0,0)), then you can solve for 'a' using the condition that y=.03 when x = 50, and get a = .03/2500 = 0.000012; so now, for the 100 foot span case, you have y = .000012x^2, which defines the shape of the curve, and where y is the value measured up from the low point. For example, at x= 0 (low point) y = 0, implying a sag of 0.03 -y = .03'; or at the 1/4 points, where x = 25, y= .0075, and the deflection at that point is .03 -y = .03 - .0075 = .0225 feet.

(For the general case, this specifically is y = wx^2/(2T))

Now please, while the parabolic approximation is extremely good for long spans with appreciable wires sizes and tensions and sags, I don't know how good it is when you're talking such extremely fine tolerances. In which case you might want to use the exact catenary curve equation for level supports y=(T/w)(cosh(xw/T), and compare it to the parabolic approximation of y =wx^2/(2T).

Note also that if the wire is subject to temperature variations, it's tension and sag will change (more sag , less tension, when hot; less sag, more tension, when cold.).

I hope this helps, but again, use it as a guide only.
 Sci Advisor HW Helper P: 8,953 To do accurate surveys before GPS you used metal wires to measure distance. You support the wire along a string of tripods and apply a known tension with a spring balance, there are tables in old surveying books to calculate the sag - it's just a catenary. You also have to do this just before dawn when the temperature is stable and the wire has time to acclimatise. ps. Thank god for RTK-GPS!
 P: 4 Thank You Very Much; Exactly What I Was Looking For

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