Scantlings methodology for recreational boats

Discussion in 'Class Societies' started by TANSL, Dec 21, 2013.

  1. TANSL
    Joined: Sep 2011
    Posts: 6,309
    Likes: 317, Points: 93, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    Probably I'm not understanding you because I've always thought that any structure, not only the vessels was estimated based on the loads to which they are subjected. High or low temperatures may also generate loads on the beams, but it is a fact that boats are usually not taken into account.
    ISO, like CS, establishes design pressures which, after all, are the loads required to be applied to each panel or reinforcement.
     
  2. Ad Hoc
    Joined: Oct 2008
    Posts: 6,700
    Likes: 681, Points: 113, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    D

    The EU Directives it refers to are those for commercial vessels and others operating in EU waters. The most common being EU 2009/45/EC. Since noted here it is for vessel above 24m. However there is a melting pot of endless rules within the EU directives. The original purpose was to harmonise them all…but clearly this has not worked at all, quite the opposite, and often contrary to others.

    Thus, from the outset what does the vessels SOR say..that shall tell you which rules are applicable.

    Remember, if the vessel is not being Classed, what standards or methods does one apply to design the structure?..that is the crux of whether one uses ISO rules or not. Class is ostensibly a design and build standard, whereas ISO is just a design standard.

    Most non-commercial designers and Joe Blogs down the road, have very little experience of Class and Class rules. Years ago Class rules cost a lot of money and difficult to navigate for those unfamiliar with them. So when Joe Blogs wants to design a boat which…er..hhmm...um rules….shall be applied??..er…um?? In steps ISO. Remember , it is a “design” standard only.

    However today’s Class rules most are now free….so that cost has been removed. Only aspect is familiarly of the rules. One can now elect to buy software that makes it “easier” like the LR SSC rules (there goes the cost element again), or one can plough through by hand. But this now represents another “standard” by which Joe Blogs could use to design the boat he wants, as per the RCD code, as it allows this.

    So ISO came about for those “outside” the world of commercial and Class rules, who needed some “guidance”. Reason being is that Joe Blogs et al, were now over a period of time being forced to comply with some standard to demonstrate that the boat designed is not a piece of crap and will fall apart of hitting the first wave. Some measure or level or standard was needed across the EU for those outside the commercial field.
     
  3. TeddyDiver
    Joined: Dec 2007
    Posts: 2,579
    Likes: 121, Points: 73, Legacy Rep: 1650
    Location: Finland/Norway

    TeddyDiver Gollywobbler

    ISO presumes the design pressures (ie. educated quess) which dictate the structure, however you can allways calculate or test what the real pressures are and build accordingly.
     
  4. TANSL
    Joined: Sep 2011
    Posts: 6,309
    Likes: 317, Points: 93, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    I think you mean the method of assessment, which you can choose, and you used to calculate the mechanical properties of the fibers used. The design pressures are clearly determines by ISO 1225-5, depending on various parameters. Not depend on the designer nor he can fix it, unless he wants / needs to apply a higher load.
     
  5. TeddyDiver
    Joined: Dec 2007
    Posts: 2,579
    Likes: 121, Points: 73, Legacy Rep: 1650
    Location: Finland/Norway

    TeddyDiver Gollywobbler

    Haven't checked it up lately but if I recall it right it was mentioned so in the original RBD (about 20yrs ago.. ). Might also be wrong..
    BR Teddy
     
  6. Baltic Bandit

    Baltic Bandit Previous Member

    Agreed. It would be interesting to see what the thoughts were of what might have gone wrong with the previous iteration of Beau Geste...
     
    1 person likes this.
  7. rxcomposite
    Joined: Jan 2005
    Posts: 2,168
    Likes: 268, Points: 83, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    Now that we have the attention of the most notable and experienced member of the forum, it is time to ask a question about analyzing structural members. It is not discussed in any rulebooks and software but it pops up in some of beam theory design and I would like to know if my interpretation is correct and if I am in the correct practice.

    ISO and LR shows the same basic illustration in stiffener and plate behavior, that is the fixed ends, uniformly loaded beam as in Illustration A. Illustration B is a simply supported beam typical of a long narrow continuous beam/girder in a ship in hogging/sagging condition.

    Illustration A shows the stress direction is nearly vertical near the fixed end or base of stiffener and gradually slants diagonally towards 45 degree as it nears the center. Notice the reversal of compressive and tensile stress as the fixed end reacts in the opposite direction typical of a stiffener bounded by a fixed end. Since the bending moment is calculated to be under the base of the stiffeners, the greatest stress is calculated at this point. For LR software users, a flick of the laminate reversal icon does the trick. For those who tabulate in Excel or ISO, the layup schedule can be reversed or the compressive/tensile strength/modulus can be reversed in the manual input side.

    Illustration B shows the stress direction nearly vertical in the center and gradually slants diagonally to 45 degree as it nears the simply supported end. There is no reaction as the ends simply follow as it is not fixed. This is typical of a girder in global stress analysis.

    Engineers working with steel or alloy would not take this into account as they are working with a material that is homogenous and isotropic but with anisotropic material such as fiber composite, it is sensitive to the direction of stress and its compressive and tensile properties are different even of the same material. Most fiber composite would lose about 73% of its strength/modulus when the stress direction reaches 45 degree.

    Knowledgeable boatbuilders always use 45 degree fiber orientation on the stiffener web to orient a 45 degree truss structure which is always in shear. Good engineering practice as the bridge is always trussed to keep it stable and the ultimate strength of the fibers are utilized. According to illustration A, this diagonal alignment holds true only in the center and loses its strength/modulus as it nears the fixed end so there must be compensation in strength/modulus.

    MY PRACTICE has been to compute the strength/modulus based on the off axis load. This dictates more layers than required and I am sure, at midspan, it is overbuilt.

    For girders running the length of the craft (or crossbeam of a catamaran), illustration B serves as a guide. Again (my practice) has been to include 0/90 degree fabric in the girder web since the midship part always has the greatest design pressure. Not an optimum solution as half of the fibers goes along for the ride. Additional UD fibers in the cap helps in the stiffness department. For the simply supported ends, the diagonal are doing their job in the diagonal positions so most of the time the additional reinforcements are in the mid part. So if it passes the localized panel bearing strength, it will pass the global strength and I leave it at that.
     

    Attached Files:

  8. Ad Hoc
    Joined: Oct 2008
    Posts: 6,700
    Likes: 681, Points: 113, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    Note sure whom you're referring too, but I'll ask away:

    Not sure i could see a question in your text:confused:. It read more like a statement. I'm a bit dumb and simple..if you could ask your question again please :)
     
  9. rxcomposite
    Joined: Jan 2005
    Posts: 2,168
    Likes: 268, Points: 83, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    Not terminated by a question mark but this should be the question.
    You are one of them obviously. First off the diagonal shear slant, i saw it only as text in one of the engineering design book. I have not found other books or illustrations that shows the stress directions. Castigllanos theorem does not seem to fit. The nearest is "Marine Composites" which has some illustrations. The direct question would be "How do I follow the direction of the stress"?

    The following description of the approach is my way of doing things as I interpret the stress analysis. I maybe correct, I may be wrong, or maybe not doing just right.

    Or maybe there is a way of doing global analysis. I see one in LR about longitudinal girders but a simple formula. I see some in ship design and engineering books but it is a generalized 3 moment equation.
     
  10. rxcomposite
    Joined: Jan 2005
    Posts: 2,168
    Likes: 268, Points: 83, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    This one AH. 3M diagram
     

    Attached Files:

  11. TANSL
    Joined: Sep 2011
    Posts: 6,309
    Likes: 317, Points: 93, Legacy Rep: 300
    Location: Spain

    TANSL Senior Member

    I think what you raise is very clear. Whether or not the form of a question, you have an idea about the way to study a structure panels and beams, and want to know if your method is correct or what the experts say on the subject.
    I am not an expert on strength of materials and also my difficulties with English prevent me appreciate the nuances of your disquisition.
    Nevertheless, raise some ideas, as a form of response :
    - I guess the panels are studied according to the theory of plates.
    - The plates are floating perimeters. Although they are embedded, for example , the profiles , which form their contour , also deform so that the contour of the plate is slightly flexible
    - The rule of the three moments, for profiles or continuous beams , must be used .
    - Probably the nodes of a continuous beam , in addition to fixed or supported , should be considered with slight shifts ( springs) .
    - All this is a three-dimensional lattice whose various elements are interacting with each other .
    - Therefore, calculate a beam as an isolated element , although not an accurate model of reality, leads to results that are conservative.
    - The hydrostatic loads exerted on the structure, although this is deformed , are always perpendicular thereto .
    - Shear forces are generated in the plane of the cross section and compressive or tensile stress perpendicular to the cross-sectional plane .
    Do not know if this will help you in any way or I've gone for a field that has nothing to do with what you was talking a bout. In that case, I apologize to everyone for my boldness.
     
  12. Ad Hoc
    Joined: Oct 2008
    Posts: 6,700
    Likes: 681, Points: 113, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    Ok, lets look at this, a simple point load, W.

    Shear in beams-1.jpg

    What occurs….is bends, and bends roughly as shown. If we now ”add” a beam rather than a single line to represents a beam, we can see what is happening to the beam. The central upper surface has compression, and the central lower surface has tension.

    Ok, how do we calculate the stress in that beam owing to the load W?

    The stress is simply = M.I/y

    Where M = the bending moment I = the 2nd Mmt of Inertia of the beam system and y = distance from the neutral axis to the location being investigated.

    So, the second moment of inertia I = ah^2. Where a =area of ‘item’ and h = distance from an assumed (neutral) axis. If the beam is let’s say an I-beam, it has flanges, one upper and one lower. Thus we can idealise the I, into 2 areas separated by a distance, if we assume the I-beam to be symmetrical. Under this basic assumption the I now becomes I = 2ah^2.

    So we can see that if we increase the distance the 2 surfaces are apart the I increases. Similarly if we increase the area of the flange, that too increases the I, or we can do a combination of the two. However, it is clear the further apart the distance the greater the increase as it is proportional to the distance squared.

    So let’s use 2 plates, or flat bars separated by some distance. What occurs?

    Shear in beams-2.jpg

    Well, as can be seen, the upper flat bar (surface) simply bends and clashes with the lower FB. The beam takes no load at all. Huh..??...the implication is the 2 surfaces separated by some distance should be sufficient…yes? Well..what is holding the 2 FBs apart?...er..nothing…hmm!!

    Ok..lets add some vertical supports, to the FBs as shown, does this solve it?...Hmmm…as you can see and by the close up, no, not at all. Why not…the 2 FBs are now supported and being separated, what’s wrong? This now gets to the heart of your question.

    So going back to the beam again as before, we apply a point load, W.

    Shear in beams-3.jpg

    The bending moment is shown by the triangle shape, as such it is easy to calculate the BM of the beam. Then the shear force. The shear force is as shown, two rectangular shapes one above an axis another below, these are equal and opposite. We can calculate the shear easily, since if we assume the force = F, for the beam to remain in equilibrium the force F on the right hand side must be balanced by the same magnitude F on the left hand side, and thus opposite in direction. We can say the left hand side is positive and the right hand side is negative. And since the only apply force to the beam is W, to balance the system, the shear force F = W/2, as can now be seen in the statics diagram at the bottom.

    What this means is now looking at the beam with vertical spacers separating the 2 FBs, is that one must answer the very simple question. How does the reaction load of the simple support (W/2) transfer along the beam? This is the heart of structural analysis..identifying load paths.

    If 2 FBs are just simply pushed apart by small vertical members, how does this “shear force”, which is vertical in direction get from one upper surface FB to the lower surface FB? It can’t, which is why locally the upper FB bent between the 2 small vertical supports. The most important part of structural design is having a shear path. Any beam as it bends has tension, compression AND shear in the system.

    If one ignores the shear one is making a major mistake! The applied load becomes the shear force, in essence…once the structure is “held” somewhere, how and where does the applied load go and how and where does it become a shear force?

    So if we now make the beam an I-Beam with a web that separates the 2 surfaces, what occurs?

    Shear in beams-4.jpg

    We can see that the bean bends and that the upper surface in compression and the lower surface in tension. Between the 2, the web now has lots of “sliding” of the web between the 2 surfaces, otherwise the 2 surfaces would simply slide apart. The shear force is simply the force divide by the area resisting the force.

    I’m condensing a lot here to get the point across!!

    If the web of the beam is solid, like in a classic I-beam, then the web will behave in the same manner as pushing a block of metal or wood on a surface. So looking at this, if we assume a force F applied at the top of a block, and if we assume frictionless contact the block will move.

    Shear in beams-5.jpg

    If we now assume the block is fixed it can’t move, and if we assume no bending at all, the force will simply be transferred from the top location to the base, the “joint” as such. The magnitude is still the same, F. Thus the block is considered “stiff” or “rigid” that no being occurs and so the applied load is simply transferred from the top to the bottom….this is the basics of shear.

    So ideally we want our beams to be the same, in simple language we want them to be “stiff”. The stiffness ensures that the load path is as we expect and we can calculate it and also direct it into locations we want.

    Now going back to our I beam again..the beam bends and has compression and tension and now shear force.

    Shear in beams-6.jpg

    But with a solid web and with a web that is stiff the shear force flows the same principal as with the bock of metal/wood which experiences no bending. The shear force is as shown. (I’ve exaggerated the scale for simplicity).

    If we now look at what is occurring with the shear force in the web, we can see that the web carries all the shear force, and the flanges carry the tensile and compressive loads. Knowing all of this, what other way could we make an I-beam with small supports but actually work? We add diagonals.

    Shear in beams-7.jpg

    Diagonals whilst not being “stiff” enough on their own, when all connected together at an angle can distribute, or transfer the load from the top surface down to the lower surface. Thus the whole system works as one,a nd then "stiff", rather than individual vertical members acting alone and with no load path from one member to another.

    Thus when making a sectional shape in composite with a beam that is very thin, light and very low modulus, how do we transfer the load from the upper surface to the lower…by using 45 degree orientated fibres. Since if the fibres run vertically, they are not effective in transferring the shear across the beam and will also locally buckle, only may be locally if lucky, if they are horizontal (parallel to the flanges), they need to be stiff enough not to buckle or compress, which of course they are not. Thus to get the load from one surface to the other, in simple language - load transfer by identify the load paths that dominate. Remembering that the flanges carry tensile, in-plane loads (or compression). So the load in the web must be able to be transfer from the upper to lower flange. The Sine of 90 (vertical) = 1, or the Sine of 0 = 0 (horizontal). The Sine of 45 = 0.707 (max)

    This is a quick rough n ready step by step...in my haste, there may be some simple mistakes/errors...I'll correct after breakfast if I find them :)
     
  13. Ad Hoc
    Joined: Oct 2008
    Posts: 6,700
    Likes: 681, Points: 113, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    RX

    Does the above answer your question?
     
  14. rxcomposite
    Joined: Jan 2005
    Posts: 2,168
    Likes: 268, Points: 83, Legacy Rep: 1110
    Location: Philippines

    rxcomposite Senior Member

    TANSL- Thank you. What I am trying to get across is the off axis load as I have mentioned in previous post. I am attaching the formula of BV, LR, ISO. It means simply that if it was given by the rules, we should use it to follow the rules to the letter. So if we use a biax or a +45/-45 fabric in a 90 degree load direction (such as a the web in a stiffener), we should adjust the stregth/modulus in the mechanical property input of the fiber/laminate. Knowing then the load direction and the angle of the fiber used is very important as the off axis loss is significant in the equation (see attached).

    AH- Thanks for the lengthy explanation. The first part, I am very much well versed. The last part, I am now grasping or have no full grasp. I will wait for the second part explanation. This is exiting. Attached, the text on Diagonal Stress.
     

    Attached Files:


  15. Ad Hoc
    Joined: Oct 2008
    Posts: 6,700
    Likes: 681, Points: 113, Legacy Rep: 2488
    Location: Japan

    Ad Hoc Naval Architect

    Either I didn't explain myself clearly enough, or I am missing the "2nd" part?:(

    The shear force in a beam is as described. What you have to understand is load paths. So, how does a load that is in-plane and tensile in the flange go from this flange to somewhere else in the member? I load cannot simply go around a corner or change direction at will.

    So If you image you are pulling the upper flange back and forth, this load is then transferred from the flange to the web, as direct shear, back and forth. It is then, if the web is stiff enough, carried down to its base and then transferred as shear into the lower flange which then becomes the back and forth load in tension.

    The web, must be able to transfer the applied loads between the 2 flanged surfaces as well as at the supports (where it is held), where the reactions are.

    Thus, where are the loads coming from and how do you and where do you wish to take that load?
     
Loading...
Forum posts represent the experience, opinion, and view of individual users. Boat Design Net does not necessarily endorse nor share the view of each individual post.
When making potentially dangerous or financial decisions, always employ and consult appropriate professionals. Your circumstances or experience may be different.