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Go Back   SailNet Community > On Board > Gear & Maintenance
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  #21  
Old 04-26-2008
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Alex,
Thanks kiddo. Got it this time.
Cheers

A
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  #22  
Old 04-27-2008
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Quote:
Originally Posted by xuraax View Post
From the web, Young's Modulus for 1x19 wire rope is 107.5 kN/mm^2.

Using the table of breaking loads for different diameters of AISI-316 wire rope as given in the book "Principles of Yacht Design" I got the following table for wire stretch for a 2000mm wire loaded to 5% of breaking load:

diameter(mm)____breaking strength(kN)___delta L(mm)

3_______________7.7__________________1.01
4______________13.8__________________1.02
5______________21.6__________________1.02
6______________30.o__________________0.99
7______________40.9__________________0.99
8______________53.5__________________0.99
10_____________69.1__________________0.82
11_____________83.5__________________0.82
12____________120.2__________________0.99
14____________160.1__________________0.97

This practically confirms what Alex is saying.

regards
Right; I said the same thing. If the shroud length is constant; the amount of stretch required is the same. But you can't apply this universally to all shrouds because different boats have different lengths of rigging wire. If you plug in 4000 mm for your base length; the delta L will double to get the proper tension. Since that's true you can't use the 1mm/5% rule to get exact tension. If I tighten an intermediate 1/4" shroud using that rule the tension will likely be ~40-50% of breaking load because the shroud lenth is much shorter than the upper shroud; which goes from the masthead to the deck.

Giu-

I'll get back to you with the data; I am aboard my boat tonight using a different computer. I should still have the spreadsheat I was doing the calc's on; but if not I will make up a new one. I was just using the modulus for 316 stainless and an approximate breaking strength for each size. Please don't use those numbers I posted as "actual"; I was only trying to make the point that stretch is also dependent on wire length (and this is independent of the max strength of each wire diameter).
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  #23  
Old 04-27-2008
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Giulietta is just really nice Giulietta is just really nice Giulietta is just really nice Giulietta is just really nice Giulietta is just really nice
It's ok, Keel..its best to use a Loos gauge.

Go here and download the stretch calculator, instead.

And instead of stretch use the force. I use a Loos gauge...I bought 3 from the old sailnet...


Loos Gauge

PRICES

Last edited by Giulietta; 04-27-2008 at 09:12 AM.
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  #24  
Old 04-27-2008
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Quote:
Originally Posted by KeelHaulin View Post
Right; I said the same thing. If the shroud length is constant; the amount of stretch required is the same. But you can't apply this universally to all shrouds because different boats have different lengths of rigging wire. If you plug in 4000 mm for your base length; the delta L will double to get the proper tension. Since that's true you can't use the 1mm/5% rule to get exact tension. If I tighten an intermediate 1/4" shroud using that rule the tension will likely be ~40-50% of breaking load because the shroud lenth is much shorter than the upper shroud; which goes from the masthead to the deck.

Keel,

As I understand it, if you mark off 2m of a shroud 6m long and then tension it to 10% of its breaking strength you should find that the marks that you made are now 2.002m apart.

If you mark off another 2m of a shroud that is 10m long and you tension this shroud to the same amount your 2 marks will again measure in at 2.002m apart.

this sound logical to me but hey...I am no expert, I could be wrong.

regards
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Old 04-27-2008
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Hello Xuraax, Since u r in Malta and sail u r doing the Marzamemi race next weekend! Good luck with your rig tuning and d race if you are doing it.

Regards

Mike
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Old 04-28-2008
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Quote:
Originally Posted by Mikelivingstone View Post
Hello Xuraax, Since u r in Malta and sail u r doing the Marzamemi race next weekend! Good luck with your rig tuning and d race if you are doing it.

Regards

Mike
Hi Mike,

Pleasant surprise. Yes I am doing the Marzamemi race. Are you?

regards
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Old 04-28-2008
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Don't know yet, I crew for Willy on Bordeaux 3, But I have to be in Brussels on Sunday so I will probably have to miss it. Which boat r u on?

Mike
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Old 04-28-2008
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i crewed once on Bordeaux 3 about 3 years ago. I am on Sailaway an Elan 37.
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Old 04-28-2008
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Saw u practising on Sat afternoon, give my regards to rayair.
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Old 04-28-2008
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OK; here is the way to solve for stretch at 15% breaking strength. It's a simple derivation; it really only uses two well known engineering equations to solve it.

Strain is defined using the greek symbol epsillon; I'm using E

Strain is defined as:

E = dL / Lo

Change in length is dL; Lo is the Original Length.

For elastic conditions; we can use the definition of Young's Modulus to determine how much extension (strain) exists in a length of wire for a given amount of force applied.

Young's Modulus - E = Applied Stress / Strain = S/E = S/(dL/Lo)

Young's Modulus is a material specific constant measured by a testing applied force vs extension.

For type 316 Stainless; E = 28,000 kPSI

We can solve this for the change in length; dL

dL = Lo * (S/E)

The only real "variable" in this equation is Lo. For all wire diameters we want the same amount of stress (15% of breaking) so we can say that S is a constant.

Let's calculate S for some different wire diameters to prove it's relatively constant:

S = (F/Ao) - "F" is the tension force in the wire; Ao is the original cross sectional area of the wire.

(Breaking strengths taken from loosco.com for 1x19 type 316 wire)

For 7/16 wire - S = 15% * (20,000#)/(Pi * (7/32")^2) = 15% * 133,040 PSI = 19,956 PSI

For 1/4" wire - S = 15% * (6900#)/(Pi * (1/8")^2) = 15% * 140,560 PSI = 21,084 PSI (within 5%)

For 5/16 wire - S = 15% * (10,600#)/(Pi * (5/32")^2) = 15% * 138,200 PSI = 20,730 PSI (within 4%)

Using Young's Modulus for type 316; the 15% breaking load equation becomes:

dL = Lo * (20,600 PSI) / (28,000 kPSI) = Lo * .000736.

Use inches or mm for the shroud length; multiply by .000736 and get the length you need to stretch the cable.

For a shroud 55' long: dL = 660" * .000736 = 0.485"

For a shroud 20' long: dL = 240" * .000736 = 0.177"

It's -fairly- independent of wire diameter; but clearly dependent on length! You could use this for type 316; (but of course the standard disclaimer applies); and it does not take into account deflection of the rig or hull when you tighten the shrouds. Again; you should use an appropriate tension gauge to determine the actual rig tension.

Last edited by KeelHaulin; 04-28-2008 at 11:30 PM.
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