A truss rod that is not a truss _rod_ ...

[Downunderwonder]

1 hour ago, tauzero said:

 

Surely it's the bending moment which is significant, and the strings are offset from the centre line of the neck by a greater amount than the truss cable, hence the tension in the truss cable would need to be higher than that in the strings in order to create an identical bending moment.

Yes it is the bending moment. The bridge forces do not induce any moment in the neck though. Carry on...

[tauzero]

8 hours ago, Downunderwonder said:

Yes it is the bending moment. The bridge forces do not induce any moment in the neck though. Carry on...

 

The strings are offset from the centre line of the neck. At the nut end, that's a little over half the thickness of the neck (neck plus nut slots). At the bridge end, there is more of an offset as the strings are further from the centre line. The tension applied at the nut and the bridge causes torque (tension multiplied by offset) that bows the neck.

 

Now, the truss cable is set into the neck, so the offset from the centre line of the neck is smaller. This means that in order to produce an equal and opposite torque, the cable must be under greater tension than the total of the strings.

[3below]

25 minutes ago, tauzero said:

 

The strings are offset from the centre line of the neck. At the nut end, that's a little over half the thickness of the neck (neck plus nut slots). At the bridge end, there is more of an offset as the strings are further from the centre line. The tension applied at the nut and the bridge causes torque (tension multiplied by offset) that bows the neck.

 

Now, the truss cable is set into the neck, so the offset from the centre line of the neck is smaller. This means that in order to produce an equal and opposite torque, the cable must be under greater tension than the total of the strings.

 

I agree with your description and reasoning, as always a diagram helps clarify the situation

 

 

[Downunderwonder]

I thought OP decided on burying the cable so most of the neck thickness could be used as leverage for reversing the bending.

[3below]

28 minutes ago, Downunderwonder said:

I thought OP decided on burying the cable so most of the neck thickness could be used as leverage for reversing the bending.

Due to the complexity of the situation I wonder if an a priori calculation (finite mesh analysis? or first approximations?) is reasonably possible, I suspect it would be quicker to make the neck up and see. 

[tauzero]

22 hours ago, Downunderwonder said:

I thought OP decided on burying the cable so most of the neck thickness could be used as leverage for reversing the bending.

 

Think of the neck as a skyscraper 20m thick. On one side is a girder sticking out 2m, on the other is a channel 2m inside the wall. One cable is attached to the end of the girder, the other to the ground directly below it, and on the other side, a cable is attached to the top of the channel and to the ground directly below. Tension is put into the cable attached to the girder. The integral strength of the structure will resist the bending but it will end a bit. In order to provide an opposite tension of sufficient force to exactly balance that force, the tension in the cable running in the tunnel will need to be 12/8 times (1.5 times) that in the girder cable, as the offset from the centre of the skyscraper is 12m in one case and 8m in the other.

[Downunderwonder]

1 hour ago, tauzero said:

 

Think of the neck as a skyscraper 20m thick. On one side is a girder sticking out 2m, on the other is a channel 2m inside the wall. One cable is attached to the end of the girder, the other to the ground directly below it, and on the other side, a cable is attached to the top of the channel and to the ground directly below. Tension is put into the cable attached to the girder. The integral strength of the structure will resist the bending but it will end a bit. In order to provide an opposite tension of sufficient force to exactly balance that force, the tension in the cable running in the tunnel will need to be 12/8 times (1.5 times) that in the girder cable, as the offset from the centre of the skyscraper is 12m in one case and 8m in the other.

Not so if you run the resisting cable through the building to the opposite side and the girder is minute ( like the nut). To boot the tensioning cable is at an angle to the vertical ( the bridge is higher than the nut ) so it's trivial to resolve to forces from the tensioning to a vertical pure compression and a sideways moment generating loading component.

 

Then the resisting cable acts over a much greater lever than the tensioner. Its force triangle is very much more squat for the same horizontal component = less tension required.

[3below]

On 28/10/2022 at 20:27, 3below said:

Due to the complexity of the situation I wonder if an a priori calculation (finite mesh analysis? or first approximations?) is reasonably possible, I suspect it would be quicker to make the neck up and see. 

 

1 hour ago, Downunderwonder said:

Not so if you run the resisting cable through the building to the opposite side and the girder is minute ( like the nut). To boot the tensioning cable is at an angle to the vertical ( the bridge is higher than the nut ) so it's trivial to resolve to forces from the tensioning to a vertical pure compression and a sideways moment generating loading component.

 

Then the resisting cable acts over a much greater lever than the tensioner. Its force triangle is very much more squat for the same horizontal component = less tension required.

 

As I am now a rusty retired Physicist (and was cra*p at resolving forces, always found it difficult to draw the right diagram in anything complex) I would be interested to see what dimensions and numbers might apply to this problem.  If we can identify where the effective centre line that the forces act about (in a simple model) then the bridge height and nut are easily identifiable.  Body to neck heel treat as a rigid non bending body.  Guestimate the tensioning cable length and distance above (below?) the effective centre line as per my earlier diagram and we might get some order of magnitude answers.  In reality it not simple since the beam (neck) has finite depth and is not isotropic.  Any takers? 

[tauzero]

6 hours ago, Downunderwonder said:

Not so if you run the resisting cable through the building to the opposite side and the girder is minute ( like the nut). To boot the tensioning cable is at an angle to the vertical ( the bridge is higher than the nut ) so it's trivial to resolve to forces from the tensioning to a vertical pure compression and a sideways moment generating loading component.

 

Then the resisting cable acts over a much greater lever than the tensioner. Its force triangle is very much more squat for the same horizontal component = less tension required.

 

So the truss cable runs at a diagonal from the bottom securing point to somewhere on the jutting girder side of the building? I see what you mean, that would exert a greater lateral force.

[Downunderwonder]

Ignoring the cross section shape makes for a much simpler "back of envelope" guestimation. It's then all about the angle of the strings to the neck / body and the available displacement of the cable from vertical.

 

 

X being the difference in 'altitudes' of the strings between nut and bridge ie fixed.

Y is the displacement available to the cable, also fixed.

T is the length of the string.

T' is the vector quantity of the tension in the string.

C' is the compression generated in the neck. 

S' is the sideways force on the neck.The unbalanced parallel forces of the sideways component at the nut and its resistance at the neck heel together require a balancing moment from the neck at the heel. We replace this with the an opposite unbalanced tension of the cable. 

 

Shifting vector S' across to the top of the cable we can imagine dropping a vertical to intersect the cable and the distance back to the cable top scales us the cable tension = T' x  X/Y

 

Note that compression in the neck is not generating moment, only the sideways component of tension must be resisted by the neck and cable. Calculation assumes zero contribution from the neck itself..This is all a bit different to the "girder hanging out of the building" analysis but I feel it's closer because the string is acting in pure tension in the first place, lending itself to vector analysis.

 

Note also that no forces are required of the body at the end so nothing in the analysis changes with the no body stick bass.

 

[tauzero]

5 hours ago, Downunderwonder said:

Ignoring the cross section shape makes for a much simpler "back of envelope" guestimation. It's then all about the angle of the strings to the neck / body and the available displacement of the cable from vertical.

 

 

X being the difference in 'altitudes' of the strings between nut and bridge ie fixed.

Y is the displacement available to the cable, also fixed.

T is the length of the string.

T' is the vector quantity of the tension in the string.

C' is the compression generated in the neck. 

S' is the sideways force on the neck.The unbalanced parallel forces of the sideways component at the nut and its resistance at the neck heel together require a balancing moment from the neck at the heel. We replace this with the an opposite unbalanced tension of the cable. 

 

Shifting vector S' across to the top of the cable we can imagine dropping a vertical to intersect the cable and the distance back to the cable top scales us the cable tension = T' x  X/Y

 

Note that compression in the neck is not generating moment, only the sideways component of tension must be resisted by the neck and cable. Calculation assumes zero contribution from the neck itself..This is all a bit different to the "girder hanging out of the building" analysis but I feel it's closer because the string is acting in pure tension in the first place, lending itself to vector analysis.

 

Note also that no forces are required of the body at the end so nothing in the analysis changes with the no body stick bass.

 

 

Yes, got that now. I was assuming a constant depth shallow channel for the truss string which would place the top attachment for the tension string vertically above it in that diagram rather than offset by Y.