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  #1  
Old 12-16-2018, 06:22 PM
guitar344 guitar344 is offline
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Default Maximum tension a string can take

In thoery what is the most a string could take to snap in pounds. Double bass strings have between 60 to 70 pounds in standard tuning.
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Old 12-17-2018, 11:17 AM
Mandobart Mandobart is offline
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This is an interesting question. Warning - mechanical engineer mansplaination ahead:

It depends on the type of string (nylon, steel, silk, etc.) mostly, but also the core shape and configuration.String tension is often expressed in pounds as you state (pounds force more correctly in engineering terminology). In reality, the parameter of concern is tensile stress, which is the force divided by the cross sectional area, pounds force per square inch, or psi. For any material, the end of the elastic region - in the elastic region the material will deform under load, but will return to its previous shape when the load is removed - and the onset of plastic deformation, where further stress or applied tension results in permanent deformation - is called the yield point. The failure point, where the material fails in tension (string snaps) is called the ultimate tensile stress.For most carbon steel the yield point is about 30,000 psi (30 ksi). Ultimate failure is typically around 80 ksi. Applying this to a .010" guitar E string - if it yields at 36 ksi, and its cross sectional area (pi x r squared) is .00007854 square inches, it can take a load of 2.8 lbs before permanently deforming, and a load of 6.2 lbs before snapping. Since D'addario's website lists the normal tension for this string at 16 lbs, its obvious they're not using ASTM A36 for their strings! 16 lbs ÷ 00007854 square inches = 204 ksi!

To further complicate things there is fatigue. This is where cyclic loading eventually results in failure even though loaded below the yield point. An example of this is bending a coat hanger back and forth until it breaks. Most steels have an endurance limit, below which an infinite number of cycles will not cause failure. Interestingly some metals, like aluminum, have no endurance limit. Given enough cycles, it WILL fail.

I know this is a lot to read through to try to get an answer to a simple question. The truth is, its not so simple. BUT - I'm pretty sure that if I attempted to raise the tension on all my strings together I would probably damage the guitar top or neck before they all broke!

Last edited by Mandobart; 12-17-2018 at 11:25 AM.
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Old 12-17-2018, 04:26 PM
HHP HHP is offline
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The "finding out" is likely to be more costly than the possible value of any information obtained.

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Old 12-17-2018, 06:59 PM
Steve DeRosa Steve DeRosa is offline
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Quote:
Originally Posted by Mandobart View Post
This is an interesting question. Warning - mechanical engineer mansplaination ahead:

It depends on the type of string (nylon, steel, silk, etc.) mostly, but also the core shape and configuration.String tension is often expressed in pounds as you state (pounds force more correctly in engineering terminology). In reality, the parameter of concern is tensile stress, which is the force divided by the cross sectional area, pounds force per square inch, or psi. For any material, the end of the elastic region - in the elastic region the material will deform under load, but will return to its previous shape when the load is removed - and the onset of plastic deformation, where further stress or applied tension results in permanent deformation - is called the yield point. The failure point, where the material fails in tension (string snaps) is called the ultimate tensile stress.For most carbon steel the yield point is about 30,000 psi (30 ksi). Ultimate failure is typically around 80 ksi. Applying this to a .010" guitar E string - if it yields at 36 ksi, and its cross sectional area (pi x r squared) is .00007854 square inches, it can take a load of 2.8 lbs before permanently deforming, and a load of 6.2 lbs before snapping. Since D'addario's website lists the normal tension for this string at 16 lbs, its obvious they're not using ASTM A36 for their strings! 16 lbs ÷ 00007854 square inches = 204 ksi!

To further complicate things there is fatigue. This is where cyclic loading eventually results in failure even though loaded below the yield point. An example of this is bending a coat hanger back and forth until it breaks. Most steels have an endurance limit, below which an infinite number of cycles will not cause failure. Interestingly some metals, like aluminum, have no endurance limit. Given enough cycles, it WILL fail.

I know this is a lot to read through to try to get an answer to a simple question. The truth is, its not so simple. BUT - I'm pretty sure that if I attempted to raise the tension on all my strings together I would probably damage the guitar top or neck before they all broke!
Best, most thorough, and most mathematically/scientifically detailed answer to the OP question I've seen in its long and recurring history ; looks like there's plenty of possibilities for several years' worth of empirical experimentation - I'm looking forward to reading the response...
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Old 12-17-2018, 11:56 PM
guitar344 guitar344 is offline
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Quote:
Originally Posted by Mandobart View Post
This is an interesting question. Warning - mechanical engineer mansplaination ahead:

It depends on the type of string (nylon, steel, silk, etc.) mostly, but also the core shape and configuration.String tension is often expressed in pounds as you state (pounds force more correctly in engineering terminology). In reality, the parameter of concern is tensile stress, which is the force divided by the cross sectional area, pounds force per square inch, or psi. For any material, the end of the elastic region - in the elastic region the material will deform under load, but will return to its previous shape when the load is removed - and the onset of plastic deformation, where further stress or applied tension results in permanent deformation - is called the yield point. The failure point, where the material fails in tension (string snaps) is called the ultimate tensile stress.For most carbon steel the yield point is about 30,000 psi (30 ksi). Ultimate failure is typically around 80 ksi. Applying this to a .010" guitar E string - if it yields at 36 ksi, and its cross sectional area (pi x r squared) is .00007854 square inches, it can take a load of 2.8 lbs before permanently deforming, and a load of 6.2 lbs before snapping. Since D'addario's website lists the normal tension for this string at 16 lbs, its obvious they're not using ASTM A36 for their strings! 16 lbs ÷ 00007854 square inches = 204 ksi!

To further complicate things there is fatigue. This is where cyclic loading eventually results in failure even though loaded below the yield point. An example of this is bending a coat hanger back and forth until it breaks. Most steels have an endurance limit, below which an infinite number of cycles will not cause failure. Interestingly some metals, like aluminum, have no endurance limit. Given enough cycles, it WILL fail.

I know this is a lot to read through to try to get an answer to a simple question. The truth is, its not so simple. BUT - I'm pretty sure that if I attempted to raise the tension on all my strings together I would probably damage the guitar top or neck before they all broke!
Assuming the double bass low E takes about three times the tension to break that be around 200 pounds of tension. Double bass Low E for instance on average is about 35 to 40 percent.
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Old 12-18-2018, 01:17 PM
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Rodger Knox Rodger Knox is offline
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Quote:
Originally Posted by Mandobart View Post
For most carbon steel the yield point is about 30,000 psi (30 ksi). Ultimate failure is typically around 80 ksi. Applying this to a .010" guitar E string - if it yields at 36 ksi, and its cross sectional area (pi x r squared) is .00007854 square inches, it can take a load of 2.8 lbs before permanently deforming, and a load of 6.2 lbs before snapping.
Excellent explanation, with a slight addition. When the string is tensioned to the yield point, the string stretches in accordance with the modulus of elasticity and the diameter decreases in accordance with Poisson's ratio. This reduction in diameter will reduce the load that the string can carry, and the reduction in diameter happens quite quickly beyond the yield point.
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Old 12-18-2018, 01:25 PM
rockabilly69 rockabilly69 is offline
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I know one thing, that with an Elixer Phospher Bronze light gauge set (012) the G-string will snap after a few times of being tuned up to A Learned that the hard way as two of my songs are in the tuning DADAAD.
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Old 12-18-2018, 09:34 PM
guitar344 guitar344 is offline
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Originally Posted by rockabilly69 View Post
I know one thing, that with an Elixer Phospher Bronze light gauge set (012) the G-string will snap after a few times of being tuned up to A Learned that the hard way as two of my songs are in the tuning DADAAD.
Try a Dean Markley. The wound G 22 can reach C.
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Old 12-29-2018, 02:58 AM
rockabilly69 rockabilly69 is offline
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Try a Dean Markley. The wound G 22 can reach C.

Are they a coated string like elixer?
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Old 12-29-2018, 12:54 PM
MC5C MC5C is offline
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Typical music/piano wire has a UTS in the 380 - 425 Ksi range (Ksi is thousands of pounds per square inch, so tensile strength is directly related to cross-sectional area), ten times higher than the steel in the (excellent) example. So punching in the numbers a .010" string might go as high as around 30 lbs of tension. Remember that the tensile strength is directly related to the cross sectional area of the core wire, so making assumptions about hex cored wound G strings is going to be hard, I would expect them to be a pretty small core wire. Area is Pi R squared, so a very small increase in diameter has a large impact to area, and so tensile strength.

Failures are going to happen first where the wire is already stressed, so the kink at the tuning post, the break over the nut, where you play on the third fret the time, where the string breaks over the saddle, and where the tight winding around the end bead are all going to break before the unmarked straight middle of the string.
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Old 01-04-2019, 02:19 PM
Denny B Denny B is offline
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Quote:
Originally Posted by Steve DeRosa View Post
Best, most thorough, and most mathematically/scientifically detailed answer to the OP question I've seen in its long and recurring history ; looks like there's plenty of possibilities for several years' worth of empirical experimentation - I'm looking forward to reading the response...

It's déjà vu all over again.
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