Designing a 9 meter sailing catamaran

I was misreading the rules. I thought it was a "one size fits all formula" since it says "Hand-laminated chopped strand mat (CSM), combined mat/woven roving, woven roving (WR) and crossplied (CP) – 0/90 reinforcement" Since it is Glass content dependent, I changed it to 0.50 Gc, the default value for WR laminated with Polyester. It now gives me value of 197 N/mm2 for tensile.

The ISO tensile value given for biax is 95 N/mm2 without any formula or reference to glass content, The value for Modulus is 0.45 of 14,000 (the modulus at 90 degree), thus 6,300 N/mm2. There is no data for shear modulus G12 and Poissons ratio. Borrowing a shear modulus formula from LR, I get 3.090 Gpa and using the proportional method for the Poisson's ratio resin/fiber, I get 0.315 for Poisson's. I now have 3 material property inputs. Since WR 0/90 has the same properties at 0 and 90 degree, the transverse modulus is 14,000. I now have 4 inputs.

A biax is a 0/90 fiber rotated to 45 degree. There are two types of biax, woven or twill and stitched or cross plied uni. Let us use the formula H3. Using values derived from the ISO formulas for WR and Uni, it was inputted to the H3.

Predicting the result of a +45 degree WR gave me Modulus of 8.28 vs 6.3 for ISO. Shifting to values for UD gave me 11 Mpa versus 11.07 ISO. The H3 formula seems to be accurate for UNI.

To find Tensile strength, I modified the H3 formula to fit the curve. I get 96 N/mm2. This is close to the value of 95 N/mm2 given by ISO. Seems correct or negligible error.

Searching for a simple formula, LR for WR 0/90 is 400*Gc-10. There is no LR formula for Biax so we have to formulate for a biax. The formula I developed is 400*Gc-105. It follows the same slope as LR so it can be substituted for the missing ISO formula. That is if you need to find the strength for different glass content.

If you are using epoxy for the resin and Eglass, The basic polyester formula will not fit. The Volume Fraction, Rule of Mixture, and CLT method needs to be used.

 

I was interested in the angle dependancy of the whole laminate when looking at it as a combination of individual UD plies. At the moment I'm researching a more or less symmetrical quadraxial laminate (symmetrical in direction, not always in fibre weight) to avoid interaction between glass and carbon.
The discussion about biax properties is apart from this.
You mention CLT, but isn't Annex H very much like CLT. Water pressure that generates sigma 1 and 2 (with alpha-beta rule). Both use the same formulas for transforming coordinate systems (xyz) --> (123). Of course CLT is (can be made) more complete (adding temp/moist/other loads)

 

Previous message was reply to #75. So a little outdated.

Great! My average calculated fibre weight is 59 % so far for large area's. From coupontests strength for (epoxy/vacuum) samples is some 10% higher (average - 2 * SD) than the ISO polyester/ no vacuum values. There is some discussion on modulus still. So my line of thought was to use the EL-b method with polyester values. But will have a look at modulus of coupontest again.

 

Yes Pammie. If it is a stack of 0/-45/+45/90 degree UD laminate, it is symmetrical or quasi isotrophic laminate. The -45/+45 is the cross plied biax part.

You can use the H3 formula to predict the property of each ply since that formula is accurate for UD fibers.

Table H2 or H4 which is the long form of CLT can be used. You just have to feed the properties as individual plies to make it easy. To treat the quadriaxial layer as a single ply with a common property is tedious.

The interaction between the glass and carbon is the limiting stress and is covered in H.2.1.10, after table H1.

 

That's good. Expect a slight increase in properties if using epoxy as it has a higher modulus and strength than polyester. Expect a greater increase if the Glass content goes up.

 

Can anyone give me advice on design for daggerboard construction? I have two questions:

1. What is the situation the maximum bending moment occurs for the daggerboard? I know how to calculate bending moment on a daggerboard, but am not sure about the situation to calculate. I could use maximum speed and maximum lift angle but I doubt that is realistic (that is 25 knots; Cl = 1,73 @15 degrees)? Sailing upwind a lets say 10 knots. At maximum lift angle?? Can someone give me a hint??

2. I wanted to use the spreadsheet I made for the stringers (12215-5 Annex H). This method is for strips under pressure supported at both ends. The daggerboard is supported at one end. So I have to use a 6 times higher moment? See picture; Text is in Dutch, but I suppose everyone understands the pictures.

Situatie 6 – Gelijkmatig verdeelde belasting bij enkele inklemming

Maximaal moment


Doorbuiging




Situatie 7 – Gelijkmatig verdeelde belasting bij dubbele inklemming


Maximaal moment


Doorbuiging

 

Pammie- The daggerboard is a cantelever design. First you have to find the force exerted on the daggerboard/rudder in your condition @25 knots; Cl = 1,73 @15 degrees. I use the flat plate theory (at an angle of attack) to simplify as plotting the lift and drag on an airfoil is tedious given a NACA airfoil. Second force that will be added is when the boat roll (or in the case of a multihull, when the boat tips over due to a sudden gust of wind). The daggerboard will resist rolling as its plate pushes against the water. Add this load to the 1st calculations. this will be your maximum load.

Now, add a factor of safety so that it won't break.

There are three basic rudder types. The rectangular, tapered, and elliptical. For the rectangular, the load will be uniform. For the tapered, the load diminishes as you calculate the distance from the tip to the base (chord root). For the elliptical, the load is not linear. You have to subdivide the length of the stock into equal parts to analyse the load at various points (x distance to be evaluated).

Moment is force x distance but the Mx formulas available uses the X distance from the tip and calculates the unit force(s) x distance to simplify.

I have this in Excel but it is a wing design for aerospace. Maybe I can simplify it. If you tabulate it, you can also find the vertical shear which is necessary to calculate the rudder stock/thickness of the airfoil at the base. The shear diagram implies that the root is always thicker than the tip so does the airfoil for the tapered and elliptical.

 

Thanks RX,

the second part: is this only rotational? Or also slamming sideways in the water? I suppose i can calculate this the same way as windload on a building?

calculating the first part:
doing this I was questioning my earlier daggerboard area calculation. I read Molland (about rudders, but the same for daggerboards) and Larsson and compared with Firebird catamaran. Originally I planned for 2 degree toed-in asymmetric daggerboards (naca 3410). In my situation I should have an equivalent of 1,5 m2 symmetric board area (for two boards). The 3410 has 1,6 times more lift so one asymmetric board should have 1,5/1,6 = 0,9375 m2 area. Based on the assumption that you need two (symmetric) daggerbards down for maximum lift and can only use one asymmetric at the same time. Correct?

I had problems finding a technical solution for 10 % thickness board so switched to 12 %. So don't know or my math is OK. For a 3412 board with a chord of 0,5 and a length of 1,8 m (out of the case) (Area: 0,9 m2) and a speed of 20 knot's I get a Rn of 4E6. I plot a Cl-alpha graph in xfoil and read a max Cl of 1,87. With a speed of 10,29 m/s, a density of 1025 kg/m3 I get a normal load of 88033 N. The lever is 0,45 (tip losses) * 1,8 = 0,81 m, so max bending moment is 71031 Nm.
References: Firebird Catamaran http://www.firebirdcat.com/technical_information.htm/technical_info_daggerboard.htm mentions side load of 10000 N at 0,5 m = 5000 Nm. Maybe is load in working condition (not maximum)?

With two flanges of 200 mm width (40% of chord) and 20 vs 14 mm thickness I calculate neutral axis and second moment of inertia. From there to max stress in outmost layers (See picture of spreadsheet). However... I don't even come close at safety factor 1... Fibre is carbon: 50 % UTS = 220 MPa in compression, 400 MPa in tension. So I suppose something is wrong in my calculation.. Any tips?
A comparison: I have a description of a sym. daggerboard for a 40 ft catamaran: 1,8 * 0,8 10%. It uses 40 % flanges of 11 layers 300 g/m2 carbon which is lightly less than 5 mm. Speed will be lower? Area is bigger. Lift is less.

 

Tanks for a very interesting thread. I probably can't help much. You are on a more advance level. But just a thought. About 80kN seem slightly to much. And does reflect maximum theoretical lift. So that calculation is likely good. Only it does not seem likely that the sailplan Will generate more than about 5-10kN max sideway force. Likewise even if the boat is at around full displacement of about what, 1,5 to 2 tonnes and slaming 1 G sideways, it will not generate more than 20kN distributed between boards, rudders and hull. I think that you van probably "reef" the board with åt least 40% because they generate exess lift at those speeds. My guess you need max area at light winds and low speeds. With exess lift, the boat will have less drift, and there for less angle of attack and less lift. Reducing lift to balance rigg load. Byt I'm no expert. Just my guess.

 

Thanks Niclas, but doubt that I'm more advanced. Missing the overview sometimes. Boat is 1350 kg max and has almost 60 sqm upwind sail area.

The "big" size of the board is for light winds/ low speeds indeed. How do you know the sideway force of the sailplan? Never calculated that because I don't know the coefficients.
Reefing the boards would be necessary when this is used as safety factor, so how foolproof do I need to make it. My concern was in having the boards at the height that is good for the static situation, while breaking it because a sudden increase in angle. (Eg hard steering while surfing from a wave).

Addition (later): I see that Larsson describes coefficients of sails!

 

Pammie, you got a load of 88033 N. Because this lifting force is about evenly distributed over the whole area of the dagger board. Why didn't you use the formula for "Maximal Moment" you cited as "Situatie 6" in your post #81? (I don't know, if that would be appropriate.)

 

Well, maybe my English terminology is not correct... In fact it is the same formula. Molland uses force on a distance L/2. Q in the equation above is force per length. So QL^2/2 goes to F/L * L^2/2 to FL/2.

 

My mistake, sorry. It was because we usually use Q = q * l . (A line load is signalled by a lower case letter and a load in general (or a point load) by an upper case letter.)

 

Pammie- You have to isolate the calculations. It is an iterative process.

First you have to determine the load under the conditions it will operate. There are three basic types of rudder/centerboard/daggerboard.

1. Rectangular

2. Tapered

3. Elliptical.

Each have a different shear and moment diagram.

The major load is the angle of attack when moving forward. Like I said, I use only a inclined plate theory for this which is Force = Ap x P x Cd. Ap=projected area, P is pressure of air/water at a given velocity and density of fluid, Cd =drag coefficient of the shape of the body. Streamlined symmetrical body is 0.04. You may use the CL of a symmetrical airfoil if you wish.

The most important load is the plate moving forward at a certain angle of attack. This will not be the only load because there are other conditions which add to the load. When moving forward at maximum speed, there is a max load on the plate. If at this point, a sudden gust of wind rotates the hull, an added load will be imposed on the plate. There is also heave and pitch. The greatest load is pitch on the rudder. Being away from the LCoG, it can sometimes reach up to 5g depending on the condition. See LR - DNV/GL rule.

At this point, I like to correct myself. Call this the Load Factor, not Safety Factor. reserve the safety factor for the calculations of the strength of the materials.

Given the shape and calculated the load, you now can draw the shear and moment diagram. Because it is an airfoil, the thickest part is about 15% of the chord (varying width). With the thickness derived, you can calculate the shear and I section.

I lost my original spreadsheet while doing this because I saved the edited spreadsheet forgetting to revise the name to a different version. I have only the tapered profile to show. I did not erase the calc for the thicknesses of the caps bottom and top plus the web thickness just to show. I don't use this as it is too simplistic in approach. This spreadsheet is for preliminary sizing of a composite spar. Note, this is for the spar sizing, not the skin of the airfoil.

I know you have trouble viewing my Excel. I included the formulas just in case.

The spreadsheet shows the comparison with a standard cantelever diminishing load formula. I can't find the trapezoidal load formula so I used this one. This may account for the difference in curves for the moment diagram.

 

It seems you like to do the calculations yourself. I guess I do too as it is better for learning and for making sure there are no misstakes. But I have used a very useful online calculator to get fast good approximations. Hope it can be useful for you as well. Sure it is for keelboats. And it will overestimate heel, thus giving slightly wrong estimates. It can to some extent be coutered by entering a very hing righting moment, keeping heeling angle lower.

SailPowerCalc http://cdn2.hubspot.net/hub/209338/news/SailPowerCalc/SailPowerCalc.htm