Fly Line Stretch and Viscoelasticity

[Torsten]

Hi Daniel,

how have you measured the bending stiffness and computed the 0.0022 J value?

Greetings,
Torsten

[gordonjudd]

(2) Take several test weights and measure the elongation of the fly line between the markers
l1 = 1m

Torsten,

I have found the stretch on 1 meter of line is hard to measure for smaller initial loads. In my tests I used 9 meters of line and then normalized the values to the equivalent for 1 meter of line by scaling the deflection measured for 9 meters of line by 1/9 to get the equivalent value for 1 meter of line. Even at that, creep makes interpreting the f/d values you get with repeated measurements a real pain.

Does the damping ratio (ζ) you got for your oscillation test ζ=λ/(√λ2+ω2)=1.5/(1.5+26.^2)=.0022 indicate the internal loss due from viscous damping was quite low?

Would that oscillating frequency indicate the equivalent spring constant value of the line was around 26=sqrt(k/.5) , k=26.^2*.5=338 N/m?

Gordy

[gordonjudd]

Does the damping ratio (ζ) you got for your oscillation test ζ=λ/(√λ2+ω2)=1.5/(1.5+26.^2)=.0022 indicate the internal loss due from viscous damping was quite low?

Oops, That should have been
Does the damping ratio (ζ) you got for your oscillation test ζ=λ/(√λ2+ω2)=1.5/sqrt(1.5^2+26.^2)=.058 indicate the internal loss due from viscous damping was quite low?
Gordy

[Merlin]

Hi Torsten

I’m using classic formulas for beams to calculate the energy associated to bending. In the case of the loop, its radius (R) is also the radius of curvature of the loop which allows simplification for equations. The energy per unit length is:
dW/ds = ½ M2/EI, and M=EI/R, so by substitution you have
dW/ds = ½ EI/R2

With the figures I am using, I get 0.00011 J/m

The length of line is the whole line since any piece of it is part of the loop at some time. Since I consider 20 m of line, the final figure is 0.0022 J for the energy associated to bending.

I took the Young modulus from a paper on line flight written by Gatti Bono, 0.5 GPa (or 500 MPa). One can consider other values but for information, the Young modulus of cured resins for composites is in the 4 GPa range. Such material is far from being very flexible. Soft materials have a Young modulus in the 200 MPa range or less.

Merlin

[gordonjudd]

I took the Young modulus from a paper on line flight written by Gatti Bono, 0.5 GPa (or 500 MPa). One can consider other values

Merlin,
You can also get an estimate for the Young's modulus from the measured spring constant of the line, i.e. Ey=k*L/area where L is equal to 1 m when the spring constant dimension is in N/m/. The uncertainty in that calculation is what to use for the cross-sectional area since the line's stiffness comes from the core which has a much smaller diameter than the diameter of line.

To be consistent it probably makes sense to use the outside diameter of the line in that area calculation so it is consistent with getting the linear mass density of the line from its volume mass density. Thus for a seven weight line having a diameter of 0.00138 m and a measured spring constant of 350 N/m the Young's modulus would be Ey=350*1/(pi*.00138^2/4)=.238 Gpa.

For a low stretch (1.5% at 17.6 N) Ridge line with a measured spring constant of 1181 N/m and a line diameter of .0013 m the Young's modulus would be Ey=1181*1/(pi*.0013^2/4)=.890 Gpa. Gatti-Bono's value of .500 Gpa is about in the middle of that range.

Gordy

[Torsten]

Oops, That should have been
Does the damping ratio (ζ) you got for your oscillation test ζ=λ/(√λ2+ω2)=1.5/sqrt(1.5^2+26.^2)=.058 indicate the internal loss due from viscous damping was quite low?
Gordy

Yes, I also get 0.058 for ζ ; k a slightly bit lower 336N/m because you need to compute the natural frequency first (26 s-1 is the frequency of the underdamped oscillator).

I've inserted the found values into the equations of the damped harmonic oscillator and get this result:

"quite low" is not defined, but you don't loose all energy after one oscillation.

Greetings,
Torsten

[Torsten]

Hi Merlin,

we can't assume that the fly line is isotropic (because of the stiffer core vs. coating) - E will be different compared to stretching.

I tried a simple cantilever test with a piece of fly line and I've measured for a test sample (floating line):
length L = 5cm
diameter D = 1.5mm
load P = 0.25 g * 9,81 m/s^2 (I've added a test weight)
deflection delta = 1.3 cm
x = L

Using the equations for an Euler–Bernoulli beam (see https://en.wikipedia.org/wiki/Euler%E2% ... eam_theory)
delta = P*x**2*(3*L - x)/(6*E*I)
I = pi * D**4 / 64
=>
E = 32*P*x**2*(3*L - x)/(3*pi*D**4*delta) = 31.6 MPa

Can you check, if you get similar results?

Thanks,
Torsten

[gordonjudd]

delta = P*x**2*(3*L - x)/(6*E*I)
Can you check, if you get similar results?

Torsten,
I think that equation is for small deflections, so it may not be applicable to a 1.3 cm deflection on a 1.5cm long beam.

With no load (.25g) what did the deflection look like?

I will send you a copy of one of Graig Spolek's papers on measuring the E of different leader materials that may be applicable to fly lines as well.

I will see if I can find an example of one of my measurements on a short length of fly line using his approach if I can find it.

Gordy

[gordonjudd]

I will see if I can find an example of one of my measurements on a short length of fly line using his approach if I can find it

Torsten,
Here is the result I get for the flexural modulus (89Mpa) of a 20.6 cm length of 8wt Bonefish flyline using Spolek's uniformly loaded cantilever approach.

Bonefish_8wt_Ey.jpg (52.52 KiB) Viewed 607 times

I would expect the tensile modulus would be much higher, but I have not measured the spring constant of that line.

The primary difference between flexural modulus and tensile modulus is that one is related to a material’s resistance to bending (flexural modulus) while the other is related to a material’s resistance to tension or compression (tensile modulus).

Gordy

[Merlin]

Hi Torsten

I used a large deflection file to simulate your experiment and I find Ey = 29.0 MPa

The order of magnitude is the same. It means that the figure I used is an order of magnitude too large (500 MPa), and that if I had taken 50 MPa, then the energy associated to bending would have been 0.00022 J. This is to compare to the elastic energy computed before (around 0.00011 J on average).

All in all, such level of energy is really small even if all is lost by internal friction. I am going to run a dynamic simulation for the elastic energy, and we shall see how large the elastic will be in such condition.

Merlin