calculating torsional stiffness hull

Well I missed one. Built a Filipino stye 32 ft bangka in California, same thing but no bamboo- abs plastic with 2 x 4's jammed inside. its carrer ended when a $2000 55 hsp evinrude jumped of the transom for some reason when i put it in reverse...I'm a sailor who only builds power boats.

 

Deflection is not the most critical factor, unless the beam fails in buckling. Otherwise, you should be calculating shear stresses.

 

1. You can try first by analyzing the loads on the crossbeams. The illustration I have with me is in Russian and the whole article is in Russian. I have a translated copy but not on the captions.
2. You need to know the data or calculated data as posted by AdHoc in post 43.
3. You can derive the mast load by three method a) as formulated by Yades b) by using Skeenes method c) Formula proposed by Dr. Gerritsma.
4. I am not familiar with GL but use LR. If you have access to the software, it is much faster as there is a lot of inputs.
5. The foam shear properties is determined by the load applied and the distance from N.A. The sheathing properties is determined using the Classical Lamination Theory. You will need to know the Tensile, compressive, and shear property of the laminate. The web carries the shear, the upper and bottom cap carries the tension, compression load. Be careful when combining carbon (as spar caps) and glass as web laminate as there is a great difference in strain. Add to that the glass, when used as a web (at 45 degree) loses its tensile and compressive properties.
6. Torsional stress is added to the crossbeam as it is already stressed before any twisting force is applied. I see it in ISO method and LR method.

Calculation is tedious but can be done step by step before you apply the torsional rigidity of hull. Sorry I was answering though it was addressed to AH. have to add some attachments.

 

One point to consider, is that beams are long slender members. They can fail as a structure, even though the material does not fail.

 

The whole discussion is based on the fact that transverse beams are long, hollow, thin-walled beams.
Could you explain a little more the idea that "They can fail as a structure, even though the material does not fail" ?. Thank you.

 

True, but my original question was about the amount of rigidity of the platform. As I understood it rigidity is more difficult to obtain than strength. That is: that mostly a rigid structure has sufficient strength (and ofcourse rigidity and stress hve a more complex relation). I did some calculations on the bending stresses on the part of the hulls before the front beam, and crash stresses in the beams. Both were more than sufficient. What parts do you think are mostly vulnerable for shear stresses?

I don't quite understand what you mean by failing as a structure even though the material does not fail? Not fulfilling its function?

 

@rxcomposite: thanks for your detailed contribution. I will react on them by number.

1. I did identify the forces working on the structure, but should consider more on what the effect would be. The left picture: from top down, that would be middle beam, back beam, front beam. I was wondering however what force P3 represents? Weight of a sailor on a beachcat?
2. I answered Ad Hoc's question in post#48?
3. Well I puzzled it out myself reading an article on a site: small trimaran design. Maybe a start, but I'll check with your sources! The document you added has more variables so I'll check that out also.
4. Software for LR? Hmm. Don't know: more than a year ago I used some ISO12217 spreadsheet (an older version) half knowing what all these numbers meant or why it would be relevant. I learned more (about stability) by studying the (newer) document. Is the underlying concept for LR and GL different? Use (reading) of GL is free.
5. That's your profession I understand? Great! Now I have to pay special attention! Classical Lamination Theory: never heard it as such. Read some colleges dictations. Found CLT on internet and going to read about it. In the spreadsheet I calculate tensile modulus. Did I do that right? Always doubt about serial or parrallel method.
Laminate shear strength? That is interlaminar shear strength? A matrix property?
Glass biax losing tensile and compressive strength: Am I not allowed to calculate 1/2root(2) of a 45 degree layer in the 0 degree direction? Web as in fabric? Does your remark also apply to 2layer stitched biax?
Combing glass with carbon: I used carbon uni with glass bi for the hulls. Is this a problem?
For the beam I'd use the glass for improving impact resistance.
6. OK, I understand and have to check it out. Is this ISO/ GL aspect on beam loading different from identifying the stresses and calculate a representing von Mises stress?
Thanks for your contribution RX!

 

My guess is gonzo is referring to situations where a structure can fail without the loads causing stresses which exceed the allowable stress levels until after failure. Perhaps the simplest example is classic buckling of a column/strut in compression when the local stresses are much less than allowable. Another example is buckling of the surface of a thin walled section in torsion.

 

To fully clarify the issue it is best to let Gonzo explain what he is referring to.

I also do not understand what you mean. Would a more obvious explanation be possible?. What kind of section?

 

Hi Pammie

Sorry snowed under with paid work at the moment

Once i get a few mins at the weekend to look at this in detail, i'll reply more fully.

BTW
Since you're vessel is composite/GRP, your main design driver will be deflections. Having a low modulus your pass/fail criterion will be excessive deflection not the stress per se.

 

Not sure what Gonzo means but in can be interpreted in many ways. In design, ultimate strengths are usually given in material properties but designer use the yield point. When the yield strength is approached (which is around 60% of the ultimate), the element/structure is considered to have failed. If it is a cored structure, if buckling occurs, it is considered failed. In long slender beam, the Euler's conditions must be satisfied.

 

Buckling is a failure due to geometry and not a material failure. I think that beams in a multihull are slender beams.

 

Gotta be careful what you're referring to here Gonzo.

Buckling is also influenced by the material properties, not just geometry.

Since it is proportional to "E", the Young's modulus of the material in question.

 

Yes, you are right. A rubber beam would definitely buckle. I was not rigorous enough.

 

Pammie- We will have to breakdown the calculation step by step. First we have to define your boat. It is a catamaran with a center pod or cuddy cabin supported by the crossbeam. The pod does not touch the water, it is hanging.

CROSSBEAMS

1. Mastbeam- the main beam that supports the mast and the hull. It is a simply supported beam. The shear and moment diagram must be drawn for each load diagram.

1a. With just the mast acting as a point load, the mastbeam must be able to handle the load. The mast is stayed so there is a force acting on the mast pulling it downward. It is not a weight per se but a force. Like a bow with the strings drawn and the arrow wedge in between the string and the center of the bow stock. You will find that on a 10 meter boat with about 2,000 kg displacement, there is already 6,000 kg. acting on it.

1b. The crossbeam support a pod whose sides do not extend to the hull, nor to the hull to beam connection. This is added distributed load and must be added to the mast load. The point of shear also moves. The shear and moment diagram will change. The crossbeam will now be stressed with the upper cap in compression, the lower cap in tension. Since the material is composite, the tensile and compressive properties is not the same. One will be thicker than the other.

1c. Even if you do not intend to fly the hull, it will happen from a gust of wind or a mistake in seamanship. Check if the mastbean can handle the load.

2a. An outside force (wind) acting on a hull at a certain distance from the mast base serves as a lever pushes the outboard hull out of the water. Half of the beam now acts a cantelever supporting the weight of the hull out of the water (like a fishing rod with a fish on the outer end). The other hull is being pressed upon and digs deeper into the water.

2b. The crossbeam will deflect into an S curve. In the outer hull, the upper cap of the crossbeam will be in tension and the opposite will occur in the second half. So what worked in the static loading may not work anymore as the stresses are reversed.

2c. The height of lever from the mast is assumed to be the center of effort and the pivot point is assumed the neutral axis of the hull (at the centerline of mast) , approximately 1/3 from the base or slightly above waterline.

3. Aft beam-The aft beam located aft, serves to stabilize the whole structure and prevents the whole structure from twisting. It does not bear the load of the mastbeam, hence it is sized smaller.

3a. If the hull is lifted/lowered at the center line of the crossbeam and the centerline of the mastbeam is assumed the pivot point, a twisting force will be imposed on the mastbeam. For the sake of demonstration, it is assumed the hull is rigid, the mastbeam to hull connection is fixed and the aftbeam to hull connection pivots.

3b. Since it is composite, the overall elastic modulus is low and every part of the structure bends and twists. The main beam twist (combined load as it is already experiencing bending load), the twisting of the main beam is dependent on how the hull twist (the main topic in your post). The aftbeam also twist and bend as it is firmly fixed to the hull.

I will deal with the principle of crossbeam composite design later. In the meantime, the image I attached is not a good quality but the diagram is self explanatory. The second image I believe was originally posted by Angelique.