STARBOARD!!
Join Date: Mar 2006
Posts: 1,662
Thanks: 0
Thanked 2 Times in 2 Posts
Rep Power:
11
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; 04-29-2008 at 12:30 AM.