# calculating torsional stiffness hull

Discussion in 'Boat Design' started by Pammie, Mar 17, 2018.

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1. Joined: Oct 2008
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It is referred to as the Bredt-Batho Theory.

Strictly speaking, it is only applicable to circular sections. However it has been developed to thin non-circular sections.
It introduces the term "shear flow" per unit length and is related to the shear strength and thickness of the material. This leads into the shear flow that is bound by the shape. From this we can now calculate, approximately, the Torsion of open cells. and non-circular. Where T = 1/3 {Tau[Sum of (d.t^2)]}.

The issue is not really with the section - box beam - but the whole response to the load. Using the Bredt-Batho theory helps, but doesn't give the whole picture. Especially since tensile and UDL loads on said beams are also present, which begins to complicate matters - in respect of a box beam connection to hulls and what it can or cannot do.

What also complicates matters is that in plate theory deflection are only valid when deflection/thickness < 3/4....and on it goes.

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### rxcompositeSenior Member

Thanks for the lengthy explanation AH. I was wondering why I never heard of the Bredt-Batho formula. My materials book that deals with torsion in thin walls tube shows the same exact illustration and shows the formula as T = 2Aq or q=T/2A. No reference to Bredt-Batho formula.

Since the source of Bredth- Batho appears to be about aircraft structures, I reviewed my composite structure cylinder formula where I can find the optimum helic angle or thickness of wall, it appears to be very much different. I need to review.

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### PammieSenior Member

Sorry for my late reply, I alo have to work .

Diplacement is 1250 kg fully loaded, One hull weighs 220 kg, complete spine 140 kg (an older estimation). Mast load is 29 kN.

Well, maybe calculation is simple if you do it whole day. Maybe I'm making it to difficult. Will post about it tomorrow.

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As noted above, the Bredt-Batho Theory is for non-circular sections.

See here page 147 for ref: http://www.kmp.tul.cz/system/files/hearn_krouc_nekruh_tenkost_profilu.pdf

and page 298 http://freeit.free.fr/Knovel/Structural and Stress Analysis/31961_11.pdf

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Indeed...and so wouldn't expect other to know these calc's!

Do you have a cross sectional (dimensions) through each beam?

6. ### chinaseapiratePrevious Member

If Pammie has that...She has everything, right? All she could do is put all the variables into equation and see that the other design was reasonable or not.

Last edited: Apr 12, 2018
7. ### chinaseapiratePrevious Member

here is another box beam, by my own original calculations - 24" inches wide 16" deep 3/4 inch thick hollow cross section. wrapped with 2 layers 9.8 oz unidirectional 30 degrees plus minus from parallel with beam consisting of 3/4 x 1 1/2 inch strips of "selected" home depot lumber over 1/4 ply frames and local buildup of 24" x 24" at mast step. gives less than 300lbs front beam with deflection ( .6mm). not better, probably worse than original but easy to calc and build. Beam theory Joesph Norwood- high speed sailing/ max deflection = 1/48 x weight x span cubed/MOE/I - formula from Royal Academy of Engineering/ calculations and approval (5 minutes time) - CSP

8. ### chinaseapiratePrevious Member

please observe your own rational when posting about "declarative values". "moonshine" from kurt hughes plan and "tiki 46" from james wharram plans are both homebuilder plans. I am a "professional home builder", yet I didn't rate my building skills any percentage over the average home builder.

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### PammieSenior Member

@ Ad Hoc: well just for pro's and nerds I suppose. Frontbeam: half ellips: 220 x 330 mm, Back: rectangular: 250 x 150 mm.
My problem is mostly not in calculating but in understanding what to calculate. But by trying to I learned a lot and understand your former comments better now.
Calculations: total calculated deflection is 0,5 mm over 9 m. Seems to low. Of course there are a few 'but's besides the ones mentioned before. Let's first check that the method is right, then that the numbers are right.
1. I used torsional moments specified by Germanischer Lloyd. I suppose these are worst case loads. Should have used a combination of torsional loads and longitudinal and transversal bending as specified in global loads section.
2. I calculated with short beam lengths (700 mm back, 900 mm front) because the second moment of inertia of the spine is much bigger than that of the beams. I should have calculated these to.
3. I don't know or I should have calculated deflection from shear force in the foam? Don't know how to do that for hollow sandwich beams.
4. Calculating laminate modulus: the E eff is compensated for fibre orientation.

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Last edited: Apr 12, 2018
10. Joined: Dec 2015
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### PammieSenior Member

@CSP: I'm not that familar with non-SI units so can't comment your laminate. But 300 lbs is around 150 kg? then 0,6 mm seems a lot. At 1250 kg displacement my frontbeam has a max moment (workload) of 23 kNm. At 2,9 m this is 800 kg. I suppose your boat is heavier, wider and has a heavier loaded mast. So your deflection under workload is at least 10 times bigger?

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### gonzoSenior Member

I'm not sure what is the problem about me indicating the page where to find the information???

12. ### chinaseapiratePrevious Member

well a lbs is a lbs in non-Si units. I didn't use your 29kN mast load, I used 2750lbs point load at middle of my beam created when the windward hull is flown. Norwood or Meade Gougeon I don't remember, lost both books. but i used 22 ft beam, 2 x 10^6 MOE for douglass fir and safety factor of 2 for the hell of it. I'm sure it would never break in bending moment due to waves with no rig because that is a smaller load on the beam than mast compression in stayed rigs, in any situation imaginable, plus there is the other beam. What I'm looking for is how to calculate torsion. in a diagram as you have. Ive never done it- I took it on faith of being irrelevant in comparison to the bending moment also created as above by clamping one hull fixed creating a cantilever- that I rember Lock Crowther of his son describing as near foolproof. Anyways I got 1/40 inch deflection= .6mm At 1mm the beam weighs 175 lbs or so., What is your'e beam made out of? aluminum?

Last edited: Apr 12, 2018
13. ### chinaseapiratePrevious Member

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### PammieSenior Member

Well I suppose it all depends on what you do with it: when do you reef, how tight will you pull your sheets? Intende speed/ weight, etc. Did you think about crash loads?
I calculated my beams in foam/glass, but will use mostly carbon for them. I can't remember exactly but I believe weighing 15 - 20 kg each.

15. ### chinaseapiratePrevious Member

I'm going to wait on response to that. The only catamaran I completed was 16 ft long compounded ply wood 1/8 plywood. It bulkhead connections to hull delaminated the 1/8 ply. ripped apart in flooded condition in shorebeak- 6" waves while i was asleep on shore. Except for 1 planing monohull all my other boats had lashed bamboo crossbeams and only the last 70ft one had a decorative 25m^2 lateen sail on it.

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