Chainplates for a Fiorentino shark drogue

Hi everyone,
Hoping to get some education on designing a set of chainplates for a drogue.
I own a small sail boat, a 1973 C&C 30 MK1 the boat is 30 feet in length, 10 foot beam and displaces 9800lbs loaded.
I recently purchased a Fiorentino Shark drogue size small, this drogue is considered a speed limiting drogue not stopping drogue like the series design. I have been on boats in the past when these drogues were deployed and liked how they handled. A common point for attaching the drogue to the vessel was to utilize a set of aft winches, the reasoning was the drogue exhibited average loads 0f 400 lbs, shock loads of 550 lbs. Fiorentino projects forces could potentially reach 2200 lbs in sever weather (pg 18 of the manual).
My boat unfortunately does not have aft winches, so I would like to come up with a set of chainplates connected to the aft side of the hull that would accommodate the loads with a safety factor. The inside of the boat is very accessible and will include a backing plate. I would like to make the chainplates out of 6mm 316 stainless steel, Aluminum backing plate. They would be straight chainplates with staggered bolt holes.
I have been reading up as much as possible trying to figure out the calculations but I'm struggling where to start and was wondering if I could get some guidance.
I guess my first question would be is there a tensile calculation for 6mm 316 stainless steel? or am I way of base.
I will try to upload a drawing of what I envision to help with any confusion, I included the drogue manual for clarification on the loads.
Thanks for any insight, I really appreciate it.
Cheers

 

I think that you could approach this fairly simply using maximum allowable stress for stainless steel.
Maximum allowable stress could be about 220 MPa typically.
And stress = Force / Area.
If you have a hole in the end of the chainplate to attach the drogue rope to (with a shackle?) then one factor to consider would be the minimum cross section area required of the chainplate that is resisting the load, between the edge of the hole and the end of the chainplate.
And the cross section area of the pin of the shackle, re the allowable stress on it.

I am guessing that you might have two or three bolts securing the chainplate to the hull?
You can use allowable stress again here with these bolts, assuming that they share the load equally.
You mention using 6 mm S/S plate - this is probably on par with (or thicker even?) than the chainplates for your rig?

That is an interesting manual about the Shark drogue, but it looks like you were not able to post a sketch of what you are proposing?

 

Hi Bajansailor
Thanks for the reply, much appreciated. Thanks for the formula that will help me understand things.

I'm planning to use a shackle to attach the drogue rope to the chainplate, I was reading another thread on this site and they expressed having 1 inch of material between the hole and the end of the chainplate. I thought this would be sufficient, they were talking a larger displacement boat. I was thinking of using 1/2 inch or 5/8 inch pin diameter stainless steel shackle. I have not looked up the strengths for the shackles yet, but will do. Thoughts?

I was thinking of securing the chainplate to the hull with six bolts in the same configuration as my shroud chainplates. They are 6mm thick 2 inches wide extending into the deck 12 inches with four 1/2 inch bolts staggered, bottom two 1/2 inch bolts parallel to the end of the chain plate, through bolted to a stainless steel backing plate. Using your calculation above I hope to end up with 12 inch long on the hull and as wide as appropriate. The end of the chainplate that extends past the stern, I was hoping to keep that distance 4 inch or less. I may be wrong but I have a thought that any longer distance past the stern may be weak and inclined to bend. Thought?

I uploaded a very crude drawing of what I'm thinking, I'm definitely a better sailor that artist lol.

Thanks again for your reply, I'm always excited to learn something new and understand forces on my boat.

Cheers,
Tom

 

If you have 1 inch of material between the hole and the end of the chainplate, and the plate is 6 mm thick, then the cross section area resisting the load is approx 150 square mm.
The cross section area of the pin of a 5/8" shackle is approx 200 square mm, and the CSA of the pin of a 1/2" shackle is approx 127 square mm, so if you used a 1/2" shackle pin here, it would be weaker (re CSA) than the chainplate, assuming that they all have the same allowable stress.

Re your sketch above, I am not convinced that you necessarily need so many bolts, or that the plate needs to be so wide - the critical part here is that 150 sq.mm of CSA at the end of the taper where the shackle attaches to the plate.

I am very rusty when it comes to strength calculations, so I will ask @Ad Hoc for his thoughts on this matter.

 

The 25mm x 6mm, as BSailor notes has sufficient strength.
If 316 S/S.

You really need to make sure:
1. The shackle you use is the same or almost the same diameter of the hole you make....no sloppy fit.
2. There is at least 1.5 x diameter of metal 'above' the top of the hole...to prevent 'tear' through load case.
3. The chain plate is attached to structure that can take the loads and transfer them to other structure.
4. From you values above, make sure your shackle has a SWL of 2 tonne min.

In the absence of any other info... that'll keep you going

 

Hi Bajansailor and Ad Hoc for the info and suggestions, I really appreciate the help. I sketched out a chainplate I think is accurate to the suggestions provided. I like the idea of not using a lot of bolts to mount the chainplate to the hull.

Using the Stress formula, I attempted to calculated the area required to withstand 4400 lbs of force on the chainplate but I may be out to lunch with my thinking. Please correct me, I have a brain injury which makes it very difficult to logically use formulas, but I would like to understand the proper way to find the appropriate way to to figure out a suitable length for the chainplate. So using the formula;
Stress= Force/Area
220Mpa= 4400 Lbs/Area
Area= 4400/220
Area= 20 Sq in

I used the recommendation of 1 inch of material between hole and end and 1.5 inch from top of hole to top of chainplate. I used 3 different diameters to round out the end, it looks sort of jankey, wondering if anyone had any suggestions to make it more smooth.

On the drawing the chainplate measures 3 inches wide x 9 inches long with three 1/2 inch bolts mounted to the hull. I was wondering instead of using hex bolts, would it be possible to use Allen head countersunk bolt, making the assembly flush?

Please add any critiques, it helps me to learn and understand the forces acting on my boat.

Thanks for any guidance, I really appreciate it.

Cheers,
Tom

 

Hi Tom,

The first load path is through the chain plate itself, and assuming you have, as my no.1 - the shackle 'about' same diameter as the hole, in this case 0.51inch or 12mm or so - then the following can be said.
In effect you have double shear.



So given the 25mm distance from hole to outer edge, and 6mm thick = 25 x 6 = 150mm^2.
And noting no.1, then this acts in double shear, which means you effectively have 150 x 2 = 300mm^2 of area.

The shear stress = force/area.

Since you noted 316 S/S with 220MPa, i'd say this is slightly incorrect and more of a sales value.
316 is generally accepted around 170MPa with LR and 195MPa with DNV.

If we assume the worst case, take 170MPa. This gives the allowable shear stress to be 98MPa.

Therefore the amount of force required to shear through the chain plate is simply = 98 x 300 = 29.4kN or roughly 3 tonne.
So, taking your 200lb snatch load = 900kg or 8.9kN, we must also add a FoS to this value.
If you want a once or twice load/year then FoS=3 should suffice.
If you think this may occur more frequently then a FoS = 5 is more appropriate.
So 8.9 x 3 = 26.7kN
So 8.9 x 5 = 44.5kN.

Therefore given the allowable load of 29.4kN, only one satisfies this, the FoS of 3.

One may argue that this is to just yeild, not UTS. But one never designs to UTS, that is a fools game.
Always design to yeild.

The second issue is that of bearing stress.

Given the thickness of 6mm and the diameter of 25mm the bearing stress is:

FoS = 3 => 178MPa
FoS = 5 => 297MPa.

In the FoS 3 case, it is marginal....meaning over time this may start to crush/ripple the hole joint.
In the FoS 5 case, it will tear through.

So, you can kill 2 birds with one stone...increase the amount of area for the bearing, which then increases the double shear area too.

You can elect to add a doubler, but this doesn't seem possible with just 25mm distance for welding....thus increasing the thickness.

At bearing stress at:
8mm thick => 133MPa
10mm => 107MPa.

Now back to the double shear.
8mm thick => 39.2kN
10mm => 49.0kN

Thus, for the small amount of cost, if it were me, i'd elect for the 10mm thick plate.
It 'covers' you for all scenarios.

 

Amen! CAN I GET AN AMEN FOR THAT!

Lug Analysis | Engineering Library https://engineeringlibrary.org/reference/lug-analysis-air-force-stress-manual

 

Hi Everyone,
Firstly I want to say thanks for all the help with suggestions in designing a successful chainplate. I really appreciate the help.
I must apologize for not being on the site and continuing to build this chainplate. On the 7th of Aug. I had set out on a solo circumnavigation of PEI Canada, for a Veterans charity (I'm a medically released Canadian Military Vet). On the 3rd day out I encountered some severe weather, I was on deck rigging boom preventers when the boat carried out an accidental gybe. The boom contacted my head sending me into the cockpit where I laid unconscious for about 2 hours. After coming to, it took another 36 hours to return to port after which I was transported to hospital. Luckily i was tethered, but foolish for not taking care to avoid the boom.
Everything is getting back to normal, I ended up with a severe concussion and a slight brain bleed. Lol, I thought my head was harder than that.
My boat is out for the season, I'm really jonesing to get back to sailing, although it may be with a helmet. In the meantime I'm going to go over all the previous post to make sure my notes are accurate and I understand everyone's suggestions.
I hope my absence will not deter members from imputing thier thoughts, I do sincerely appreciate everyone's help and it motivates me knowing the part will be strong and safe.
Cheers,
Tom

 

Hi Ad Hoc,
Thank you for the detailed explanation of the forces, I can understand everything except for one detail, which I'm hoping you can clarify. I cannot see how you came to the conclusion of 98 MPa, I was wondering if you could share the equation with me. The rest I totally understand.
I will definitely use your suggestion of 10mm. My forestay chainplate is quite thin, maybe 3/16". At the anchor point for the forestay wire they added a 4mm doughnut to thicken the connection point. This is somethin I have to replace as the weld is severely corroded and slightly cracked. So I really like the recommendation of 10mm.
Thanks again for you input, I really appreciate it.
Cheers,
Tom

 

Tom

I highly recommend not having a fixed attachment for the drogue.
Imo have an aft fairlead going to a forward winch. Making it possible to adjust the warp length to keep the drogue and vessel in the same wave phase. Ie crest and crest. Shock loads are created when the boat accelerates down a wave and the drogue is slowly climbing.

Plus-you will never haul it in by hand.

 

Tom,
Tensile/sqrt3.

So, 170/sqrt(3) = 170/1.73205 = 98MPa

It is from the Von Mises shear stress.