The breaking force does not tell anything about the young’s modulus or the spring constant. So I selected three different materials with well known ultimate tensile strength (from wikipedia) and then worked out the cross-sectional area required for a breaking strength of 1000 lbf.
Calcs are here and the results for the wire diameters are :
Nylon = 0.0025m (0.099 in)
Steel = 0.0032m (0.126 in)
Graphene = 0.0002m (0.008 in)
Now the “math” part, I tried illustrating with figures and equations here.
Finally, Young’s modulus for the above materials were obtained from wikipedia. And the resulting equations were numerically solved using excel goal seek and the results are posted here for comments / corrections.
The Nylon wire will move 6.63 inches sideways to your 100 lbf, the steel wire would move 4.96 (this is theoretical since it will break) and the graphene would move 6.59 inches.
Hope the above helps. I have the excel sheet setup, so if you know your material properties, I can run the calcs for your material.
A little side trip since there seem to be some valid answers here…
I have been in discussions on musical instrument forums on the topic of tuning metal guitar strings. As guitarists know, when you put a new set of strings on, they tend to go a little flat until they are broken in. Some players have the idea that the string is stretching under tension because it’s new, and it takes a while for it to reach equilibrium. The guys with more of an engineering background say that metal guitar strings do not stretch at all and that the strings go flat because the winds around the tuning post are getting tighter when you first put the string on. It stabilizes once the winds have deformed to hug the post tightly.
So can a nickel alloy wire of 0.017" diameter stretch under tension of under 20 pounds?
That’s pretty good work. My only nit to pick would be that your k value seems to be too low which is resulting from using the post stretched value of the string. I’d use k=EA/l. Since you don’t know l (the unstretched length) you’d need to solve for l to get the spring constant for the wire.
This too low k is then going to result in a larger stretched length.
Iirc the stored strain energy is proportional to displacement squared. So you can’t just start from the 20 foot initially strained state.
It’s been a couple decades since I was in school so maybe my memory is a bit hazy.
I think Octopus (you are replying to his post) was referring to a class of problems in Newtonian Physics/Classical mechanics / Engineering Statics known as Statically Indeterminate Problems. Statically indeterminate - Wikipedia
These problems cannot be solved by simple Newton laws but need some other equations. In simple words, there are more variables than equations.
Examples :
An iron rim is heated and put on a wooden chariot wheel : what is the hoop stress on the rim ?
What is the stress on a railway track on a hot day when a x amount of gap exists per 20 ft of line ?
…,
You’d need to know the cold radius of the wheel and of the rim. The thermal expansion just lets you put it on, but won’t matter after it’s cooled again.
Can it stretch? It does, as must all things under any load at all. To paraphrase both Robert Hooke and Shakespeare, all the world’s a spring. Literally everything deforms under any load, no matter how small. When you walk across a marble floor, the marble deforms under your feet and springs back when you step away. The deformation is infinitesimal, but it’s a predictable deformation.
I think CookingWithGas is not really asking whether the string stretches (strains) at all under a 20-lb load, but rather whether that deformation is permanent. It isn’t; 1000 PSI is nowhere near the yield strength of inconel. This is elastic strain. Permanent deformation is plastic strain.
The engineers are right and the musicians are mistaken on this one; flat steel guitar strings are settling into their mountings, not experiencing plastic deformation. They may deform plastically in wrapping around the posts (if they don’t spring back to straight, they’re plastically deformed). But they don’t “stretch” plastically under a 20-lb tensile load.
Interestingly, nylon strings absolutely stretch. Nylon (like most polymers) undergoes a phenomenon called “stress relaxation” or “creep,” in which a small applied load can cause plastic deformation over a long period of time.
The poster a few entries above this one got the correct strain in a nickel-alloy wire. But this is elastic deformation, not permanent deformation.
An expansion of 0.0032% on a 26" scale is about 0.0008". Guitar tuning is considered acceptable if it’s within 1 cent*, 1/100 of a semitone (1/1200 of an octave, or 0.00083 of an octave). Many tuning devices are accurate within 1 cent, and a precision of 0.1 cent is considered high precision. A lengthening of a string with other factors held constant is proportional to the lowering of pitch (i.e., doubling the length will lower the pitch by one octave). Lengthening by 0.0032% (0.000032) would change the pitch by less than 1/20 of a cent, which would be imperceptible to a musician’s ear and virtually any electronic tuning device.
*The guitar has certain physical characteristics that make it pointless to tune any more precisely than that. For example, the string tension increases simply by depressing a string to play a note, and the increase will vary depending on what point of the string is pressed.
Just to be clear, the elastic deformation of 0.0032% doesn’t change the pitch, no matter how precisely you measure. That deflection is a result of the preload on the headstock. With no pre-tension, the string’s natural frequency is infrasonic and therefore its vibrations are inaudible.
If you were trying to predict the string’s exact natural frequency to a ten-thousandth of a Hertz, these things would matter. But since guitar tuning is a closed-loop operation, all that matters is whether the tuner (or device) hears an acceptable tone.
In other words, you can’t hear an un-stretched string; they’re too floppy. But newly-installed strings go flat because they’re settling into their fittings, not because they’re stretching more. In fact, they go flat because they stretch less. That decreases the applied pre-tension and lowers the natural frequency.
Also: some flattening of the pitch may be due to stress relaxation of the guitar, especially if the guitar itself is brand new.
Pre-tension is the stress state at rest and after tuning.
Headstock preload is the reaction load on the headstock as it resists the pre-tension applies to the string. (The headstock is stressed primarily in bending).
Load and tension would be the dynamic load and tension as the string is plucked and then vibrates freely and the headstock reacts out the dynamic tension in the string.
As discussed at length in response to HoneybadgerDC’s question, plucking the string (displacing a point on the string in a direnormal to the string’s own axis) can increase the tension quite dramatically from its pre-tensioned state.
Does that answer your question about the distinction I’m making?