Twisted Rod Rigging - now what?
- Thread starter jkalucki
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jkalucki
New member
- Thread starter
- #22
Thanks. This is all likely true, but I'm price insensitive when it comes to safety.
My intuition and inclination is to just replace the thing and stop worrying about it. I'm just trying to make the best and most informed decision and also avoid unnecessary boat downtime...in the unlikely case this was some sort of surprise no-big-deal thing and my materials and metallurgical intuition is just wrong.
I'd estimate that the cost to replace a single rod shroud at about $2k +/-, with the second one half that, which is pretty short money given the consequences.
FWIW, they're all only 2 years old and haven't seen any offshore pounding, or even a spinnaker hoist since the install. Worse case maybe they replace both capshrouds, but can't get the material, long delays ensue, and I miss a few races?
Agree 100%, though I doubt you'll find a rigger in the US that would risk the liability of saying it's OK and then risk having the rig come down a month later.
knurd
New member
Sounds like the whole boat is a write off. I'll take it off your hands for free.
If it wants to unwind elastically, then there is a constant twisting load being applied to the swaged fitting and/or the associated turnbuckle. Which is not good, I suspect.
And wouldn't that "urge to untwist" appear when the pin was being inserted into the chainplate?
Yep, if the shear stress is below the yield point/knee you could argue that it's elastic and probably OK. Need accurate stress/strain curve of the annealed rod tho.
The stress is given by r*d(Theta)/dz = G*r*(2pi*r*N)/L = G*2pi*r^2*N/L
My quick calc with generic values for mod of elasticity (30e6 psi) and YS (60 kpsi) shows that the material did yield. But don't take my work for it as I'm multi-tasking now and probaly made some mistake. Replace the rod.
2 years old? Replace the one, then replace it all on your normal schedule. Yeah, 2 years early for that one. Given the weather we've been having in these parts, and the fact that the breeze machine is about to turn on for the summer, you might have to miss a few sailing days. Maybe one of the locals has something in a left over spool...
TJSoCal
Super Anarchist
They say that in theory, theory and practice are the same. But in practice, they're not...
ExOmo
Best Anarchist Ever
YS 60ksi from the data sheet but that would just be annealed mill condition and not the drawn rod as mentioned before...don't have my Navtec data sheets in front of me at the moment...
OK, I did some digging. One assumption that is critical is the actual yield of Nitronic -50 rod as used in rigging. It varies a lot depending on how much cold drawn it is. I used 750 MPa due to a reference in Practical Sailor saying "about 3x 316 s.s.". This is not exactly conclusive!!
But when you do the math and based on the 4 turns guess, it's very very close to yielding due to rotation. Here's my math. The 21 N.m. torque sounds a bit low but it's only -10 rod (1/4") so doesn't take much to torque it.
The calculated shear stress with the strength assumption and 4 turns is 404 MPa and shear strength (when it starts to plastically yield) is 432 MPa i.e. very very close either way. Change any of the assumptions slightly and you are sure to have yielded it.
Since you can't be sure it was just 4 turns I think you must replace it.
But when you do the math and based on the 4 turns guess, it's very very close to yielding due to rotation. Here's my math. The 21 N.m. torque sounds a bit low but it's only -10 rod (1/4") so doesn't take much to torque it.
The calculated shear stress with the strength assumption and 4 turns is 404 MPa and shear strength (when it starts to plastically yield) is 432 MPa i.e. very very close either way. Change any of the assumptions slightly and you are sure to have yielded it.
Since you can't be sure it was just 4 turns I think you must replace it.
| Anglular deflection of a torsion solid shaft: | |||
| α = 32 L T / (G π D4) | |||
| So T = α G pi D^4/(32 L) | |||
| where | |||
| L = length | 13.7 | m | 45 ft |
| T = torsional moment | 20.80 | Nm | |
| G = shear modulus of rigidity | 68947572932 | Pa | Nitronic 50 10e6 psi |
| D = diameter | 0.0064 | m | -10 Rod |
| α = angular shaft deflection | 25.13 | radians | |
| rotation = 4 x 360 | 1440 | deg | |
| The shear stress in a solid circular shaft in a given position can be expressed as: | |||
| τ = T r / J | |||
| where | |||
| τ = shear stress (Pa, lbf/ft2 (psf)) | 404,176,600 | Pa | |
404 | Mpa | ||
| T = twisting moment (Nm, lbf ft) | 20.80 | N.m | |
| r = distance from center to stressed surface in the given position (m, ft) | 0.0032 | m | |
| J = Polar Moment of Inertia of Area (m4, ft4) = π D^4 / 32 | 1.6471E-10 | m^4 | |
| Yield strength of Nitronic 50 for rigging | 750 | Mpa | slight guess, strength varies a lot |
| shear strength | 433 | Mpa |
I'll go out on a limb and say they probably don't cold draw it very much at all.
- Even at 15% cold reduction the difference between yield and tensile failure is very close
- the reduction in area is already 56% with 15% cold reduction; to keep a consistent diameter for rod rigging that much reduction might be too much.
So... just maybe the yield strength is > 750 MPa I guessed. But how much more?
- Even at 15% cold reduction the difference between yield and tensile failure is very close
- the reduction in area is already 56% with 15% cold reduction; to keep a consistent diameter for rod rigging that much reduction might be too much.
So... just maybe the yield strength is > 750 MPa I guessed. But how much more?
If I read it correctly on the internet image of a loos gauge I just seen, the 50 to 55 reading on it translates to approx 300 MPa of stress. Combined with the 435 MPa found by Zonker, this would give you a Von Mises Stress of about 815 MPa (as the lawyers say: this is not professional advice). Given the level of trust in the data acquisition (about 4 turns...) I would absolutely replace, even if the yield stress is significally higher than this, but you probably dont have to worry about the shroud failing in port.
Yes, the tesile stress needs to be combined with the torsion. Another consideration is that peak stress will only happen just near the outer surface of the rod then drop off quickly towards the center. If it did get past yield, it might only be a very small amount of material that did yield. If the typical loading only gets up to maybe 15 percent or so of yield and only a small amount of material is impacted, it could be fine depending on age of the rod.
If it's close to end of life, replace it. If not, some thorough calcs and inspection could save some bucks. Also, just because the material goes past yield (depending by how much) doesn't mean it won't take load anymore.
It's cold drawn steel rod, it only got that strong because they already made it yield a bunch
.
kent_island_sailor
Super Anarchist
I ran into that getting a rigging survey. "If it is old and it is rod, I will fail it right now from my office for free, no need to pay me". I doubt any shop would sign off on this either!
ok probably spending too much time on this, but found this:
my favorite quote is
The primary failure mode seems to be fatigue, which makes sense.
There is NO definitive answer, this is a very low incidence occurrence. Due to liability, pro shops are just going to say 'replace'.
My take is: if rod has no permanent twisting after removing the turns, it's OK. I would examine the rod closely where it goes over the spreader tip. You don't state if the rod bends at that point or continuous straight
My take is: if rod has no permanent twisting after removing the turns, it's OK. I would examine the rod closely where it goes over the spreader tip. You don't state if the rod bends at that point or continuous straight
