chainplate design

[66jwa]

How can I calculate the strength of my chain plate? It measures 3/8" by 1 1/2" and has a 1/2" clevis. I want to drill it out to 5/8" for a larger clevis. What is the calculation? It is 316 SS, attached properly for this discussion.

TIA

Tom

[68Morgan30]

The most basic calculation is tensile area in the minimum cross section.
minimum tensile area = .38x(1.5-.688)= .31 sq inches
Not knowing the exact condition or pedegree of the steel you should be conservative. 300 series stainless steel plate has a yield strength of 30 KSI and ultimate strength of 75 ksi. so the plate would start to yield at 30000 x .31 = 9200 lbs. now what is the load that you would expect? typically you might tension a shroud to 20-30% of its breaking strength. depending on the boat, shroud wire diameter this could be 1000-3000 lbs pre-tension. strains on the mast will cause tensions higher than the pre-tension. a conservative approach would be to make sure the chain plate wont fail before the shroud parts. so determine your shroud wire diameter, material and new breaking strength and compare this number to the 9000 lbs tensile capability of your chain plate. big safety factors are nice.

there are other techniques for analysing the shear tear out of the pin. shear stress allowables are typically 40% reductions on the minimum yield strength.

I hope this helps....

[Chris Krumm]

I'm no engineer...

...but 68MORGAN30's post looks right on. I've seen formulas in strength of materials books that assume elastic failure in shear when shear stress is 50% of yield stress. Many bolt and hardware manufacturers charts quote shear stress allowables at 50-60% of their tensile stress allowables.

68MORGAN30's suggestion you look at making sure shroud parts before chainplate fails makes me ask why you need to replace with a bigger clevis in the first place. The whole system should be oversized with safety factors, but even then systems are often engineered with a "sacrificial link."

rilling your chainplate out and adding a 5/8" dia.clevis gives you 1.5-.625/2 = .328 sq. in. x-section in chainplate in tension. 5/8" dia pin in double shear gives you (.625/2)^2*3.14*2=.614 sq. in., so your clevis shear stress allowable would be a right in the area of a 40-50% reduction on minimum yield strength - but it's much less likely to fail before the chainplate than it would have been if you'd left it at .5" dia.

Chris

[Chris Krumm]

Formula error in last post: drilling your chainplate out and adding a 5/8" dia.clevis gives you (1.5-.625)*.375 = .328 sq. in. x-section in chainplate in tension.

Chris

[66jwa]

Thank you for your responses.

The fixed cable is 3/8" 1X19 with swage jaws with 1/2" pins. I want to add a turnbuckle at one end, but the turnbuckles for this wire size only come with 5/8" clevis pins, to match the thread diameter.

I wonder if this arrangement was sized right in the first place. 3/8 breaking strength is 17,500, which exceeds the yield strength of the current chainplate arragement using your calculations [.375X(1.5-.5)]X30,000=11,250lbs Am I doing this right? What is the yield strength of 3/8" 1X19?

The application is "water stays" for a 34' trimaran, running from the main hull near the waterline to the outboard end of the cross arm.

One suggestion has been to make a custom adapter which keeps the 1/2" pin, but I want to keep the solution simple, and I want to confirm the engineering on what exists before I do anything.

How do you determine the strength of the clevis pin, or is that not pertinent for this analysis?

[68Morgan30]

http://www.hallspars.com/details/rodnwire.html
3/8 1-19 316 stainless wire-14,476 lbs breaking load according to this web site.

http://www.sanlo.com/product/gsstrand.htm
this web site says 14,800 for 1-19 316 ss, and 17,500 for type 304 stainless

The pin is in shear so your maximum allowable stress before the metal starts to yield is 57% of the min rated yield strength. Your actual stress level should be smaller than 1/2 or 1/4 of this level to be safe.

[68Morgan30]

http://boatbuilding.com/content/engineer/page2.html

aint the web great?

[Chris Krumm]

Tom -

68MORGAN30's last web link brings up the point you need to be sure drilling the chainplate out larger doesn't compromise the single shear loading at the end of the chainplate (essentially your clevis ripping out the end of the chainplate. Most chainplate holes are drilled farther back from the end than they are from each side for this reason.

5/8" dia pin in double shear gives you (.625/2)^2*3.14*2=.614 sq. in total shear area at the clevis. Assuming yield strength for 316 SS at 30 KSI, shear would be 30 Ksi*.60= 18 Ksi. 18,000*.618=11,124 lbs allowable shear for pin WITHOUT any additional safety factors. The area left in tension across the plate, both sides of the hole, would have a yield strength of .328 sq. in*30 Ksi =9840 lbs, again, without any safety factors. To match that in single shear at the chainplate end, you'd be looking for 1.67*.328 sq. in = .55 sq. in.

Most calculations for boat rigging I've seen adopt 2x -3x safety factors, but less for racing machines (but theater riggers I work with typically go with 10x when they're dealing with human lives!). Wire rope typically only has breaking strength given, nothing for yield strength.