# calculating torsional stiffness hull

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

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1. Joined: Dec 2015
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### PammieSenior Member

I'm trying to calculate the torsional stiffness of a foamcore hull with Bredt's formula. The question is: do I use thickness and shearmodulus of the glass only? (as shearcurrent (right word??) is parallel to the laminate).

In this document www.ppa-europe.eu/db/docs/deliverable_11_d1_2.pdf (section 2.2.3) I read about a variant of Bredt's formula described by Stamm/Witte (in a book). But as I understand it this is specific for flat non hollow foamcore panels with glass on the outside.

Anyone any hints?

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### gonzoSenior Member

Are you trying to calculate G for the complete hull structure or for a panel only?

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### PammieSenior Member

It's the product GI for the the complete hull. G isn't the problem, I is. Maybe I didn't use the right English words: I'm Dutch.

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### gonzoSenior Member

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### PammieSenior Member

That's what I mean, though your link uses K instead of I.
I did split it up in sections. The cockpit section is the most sensitive for torsion.

The hull shown is the middle hull of a catamaran (above the water) for longitudinal rigidity. The crosscut shows just a line; no thickness.

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### Ad HocNaval Architect

The first image you show, indicates your boat has little to no torsional stiffness.
It is all about shear flow. There is no shear flow from the outer 'box' shaped section to the other. The thin line connecting them, shear flows in one direction only = no torsional stiffness.

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### PammieSenior Member

Shear flow: that's what I understood. I realise this section is most problematic, that's why it's only 800 mm long. The question was however: How do I apply Bredt's/ Stamm's formula for sandwich panels. And also: What is the real world value of this calculation?
My calculations is a deflection of 8 mm over 2500 mm width and a section length of 800 mm. Laminate is 0,4 mm carbon on both sides. Deflection of the fully boxed sections is 1,5 mm.
I suppose a middlehull is better for torsion than only 2 beams
The single line is in fact 2 laminate skins so shearflow will be bi-directional, though it doesn't add much to rigidity. I can make a new sketch with thickness, fillets and reinforments. That's good for construction, but doesn't change the configuration.

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### Ad HocNaval Architect

Well, it is more simplistic than that. Are you just wishing to do a calculation as a stand alone calculation, or do you wish to have some useful data? If it is useful data, then for now, you need to ignore the calc' as it is a red herring.

Nope.
The shear flow is in one direction only, as such, there is no torsional stiffness.

Thus, as a simple calculation as a simple stand alone calculation, sure fine you can do it.

BUT....and this is my point, you have no useful data, because the design is inherently flawed, as the ouwer hull will flex badly relative to the middle/centre hull.

So whatever numbers you get in your calculation is not the objective, as the design is flawed. You need to add torsional stiffness into the design before you can calculate what deflections you might get.

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### PammieSenior Member

Thanks Ad Hoc! Of course a good design and usefull data is more inportant. So I'm trying to understand:

How is torsional stiffness of a standard catamaran (only beams) compared to a catamaran with a centre hull and the same beams?
A. a round shaped centre hull
B. a rectangular shaped centre hull
C: two rectangular shaped box constructions (as in my previous picture, but without the connecting structure between the boxes).
All hulls having the same cross-cut area?

Adding torsional stiffness to design: I'm stuck between bridgedeck/ footwell clearance, sitting headroom and total height.
Most monohulls use a cockpit floor separate from hull floor. I can add some 10 cm but I suppose this won't add much?
I could smooth the line connecting the two boxes but then there is still one direction shear flow.
Kurt Hughes Kurt Hughes Multihull Design - Catamarans and Trimarans for Cruising and Charter http://www.multihulldesigns.com/designs_stock/38cat.html uses a flat cockpit floor and bigger size boxconstruction.
Any other suggestions?

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### Ad HocNaval Architect

A - no differenet
B - no different
C - no different.

It all stems from an understanding of torsional stiffness/rigidity. You have a closed cell (outer hull), which is connected to the centre hull (closed section) and then again on the opposite side.

It is the connecting of each closed cell (hull) to the other that is the issue. With a closed cell (hull) being connected to another closed cell (hull) via a single "one cell" (open cell single 'line'), depending upon how you arrange it, it can be shown by simply theory that the angle of twist can increase from a minimum of 5 times that of 15424 times, over what it would be if it were all a 'closed cell'. By that, I mean there is no 'single line (open cell) connection.

Take a normal plastic ruler, hold each end in your hands, and then rotate the end of each hand in the opposite direction (a twist). How easy it is to twist the ruler?!....that is what you have connecting the 2 hulls.

Makes sense?

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### PammieSenior Member

Sense... been thinking whole day about it... Well, your ruler example makes sense.
Your answer to comparing cats with and without centrehulls does not: torsion (and bending) on a structure takes energy. More structure => more energy needed => less movement? Of course angle of twist increases, but I still have the stiffness of the 2 small boxes (instead of one big box)? Is that correct?

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### Ad HocNaval Architect

Hmmm....

Ok...so you understand the ruler. Good.

What is happening here is the the shear flow, is governed by the length of the perimeter times the thickness of that member. So if the ruler's width is say d and has a thickness t its shear flow is dt. (It is a tad more complicated than this, but i'm generalising for simplicity). In the torsion calc it becomes dt^2.

If the ruler was now a channel section of a U shape like this: [_]
The shear flow would be length of the walls and the flange x the thickness. And if all sections equal it would be 3dt^2.

If the ruler is now closed cell, the shear flow is the enclosed area within the closed cell, it is no longer an open cell which uses just the perimeter length. So the shear flow is d^2. (it is actually taken as mean thickness of the member so just a tad less than d).

Thus when the thickness relative to the length is always small, as most closed cells are, you can see as a simple ratio of (d^2)^2/dt^2 = d^3/t^2 is an order of magnitude greater. Therefore the torsional rigidity is the same order of magnitude greater.

So, (again for simplicity) using simple numbers, of the ruler is 50mm wide and 6mm thick and the closed cell is 50x50x6mm, the ratio becomes 47^3/(50x6^2) = 57.7. Thus the closed cell is more than 57 times better at resisting torsion for the same unit length and applied load. Which is also related to the angle of twist or rotation too.

The shear flow is the area bounded by the region that is resisting the torsion. When a closed cell, the 'area' is huge. But, when an open cell, like a flat ruler of single line/plate joining to closed cells, the area is just the perimeter, length x thickness much less!

Make sense now?

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### PammieSenior Member

Yes, it did and does. I was thinking that a partly closed/partly open structure would give enough rigidity. That might even be, but I think your point is that total torsional rigidity is going to be a problem? Is that correct? Also read your comments in another discussion about torsion.
One of the problems is the postion of the front beam (at 5 metre). I didn't give that enough tought. In this other discussion I believe you mentioned 10 mm for a 12 meter cat. Do you think it is feasible with this configuration and a better designed structure (without making it to heavy)?

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### Ad HocNaval Architect

It does, better than a flat panel, but nowhere near as good as a closed cell.
It was one of the biggest issues with container ships. Since they are an open [_] shape, so the sides have []_[] closed cells on each side and the bottom is then also a closed cell, so 3 closed cells. Not perfect, but much better. This containers ships, which are naturally an open cell, have 3 closed cells to connect to each other. Which is same as your case, you need a closed cell, to connect your other closed cells.

Yes!

Without seeing the plan and profile arrangements, hard to say, re: your layout.
As for 10mm for a 12m cat yes, it is roughly linear, the length of boat in metre = acceptable deflection in mm's, under an applied max load case.

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### PammieSenior Member

Thanks Ad Hoc for your explanations. Took me a while to react because of family business..

I was wondering: would it be better to use a 50 cm wide middle hull just for longitudinal rigidity and establish torsional rigidity by 2 beams as big as possible? Is it right that torsional rigidity can be better obtained by 2 beam bending instead of pure torsion? Erik Lerouge used such a configuration:

My configuration (see also picture in the beginning of the discussion): hulls (not shown): height 1,50 max, 1 m wide, 9 m long, 15 mm foam (bottom: foam+ ply), outside lam: 320 g/m2 carbon uni + 600 g/m2biax glass, inside: 820 g/m2 triax glass + all reinforcements (bottom, foils, rudder, beams)
Total width: 5,80 m
original middle hull: 2,5 m wide, 1,45 m height (overall 1 meter + half elliptic footwell 0,45 m), clearance sealevel: 0,35m footwell backside, overall: 0,8 m backside, 0,9m front beam. Mast position/ frontbeam: 5 m (from back). I designed the beams quite small: back: rect. 0,15 *0,25, free length 0,7 m, front: D: 0,4 x 0,3 m, free length 0,9 m. Foam/carbon. Beams can be small because the free length is small. Bending under workload is under 1 mm.
Fully loaded displacement is 1250 kg
I calculated workloads as well as loads specified by GL.

What configuration would be best for torsional rigidity? Do you have examples of sailing multihulls?

Can torsional rigidity be calculated? Estimated by FEM? If I'd make model of paper on a 1:25 scale would that show weakness in design?

Another question: As an alternative for my original middlehull cockpit crosscut (blue line) I have drawn an alternative with a double bottom (white line).

All lines being 15 mm foam with 2 sided laminate. Would you consider this sufficient for calculation (A^2)?

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