FRP bonded connection calculations

Discussion in 'Boat Design' started by boats_designer_fr, Jan 31, 2019.

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  1. gonzo
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    gonzo Senior Member

    He has been posting senseless rants and multiple postings for a while now. He repeatedly hijacks threads with insults and self-aggrandizing claims. Maybe Alzheimer's is setting in.
     
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  2. Ad Hoc
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    Ad Hoc Naval Architect

    Fallguy


    Indeed…those that love to post and pontificate suddenly gone silent...oh well!

    It all starts with:

    1) What materials are you using?

    And

    2)The load scenario.

    So take that Tee joint. Without knowing the properties of the layup, one can only use “nominal” values for composites from ref books or sales literature. Since different layup, clothes resins etc..all have a significant impact on the final mechanical properties. So the only way to know for sure is to make create a simple laminate and using the manufacturers design values, arrive at “acceptable” design values for the layup. But preferably make a test piece, a coupon test. And the obtain the mechanical properties, then you’ll know for sure. So let’s assume you know have “some values”.


    There is a force located on the web, it is unclear whether this is a simple single skin web of a frame or a WTB/BHD…but for now, it doesn’t matter.

    One must identify the load path. Without identifying the load path – loading scenario – it is not possible to analysis. Thus RX has already nicely done this...but for simplicity shown here:

    upload_2019-2-4_11-0-3.png

    So firstly that area noted as “1”, this is taking the pull load or tensile load. So what is the cross sectional sear of that web. The stress then being = F/area

    But it is highly likely that the load will not be 100% perfectly aligned with the fibres for a pure tensile load application. Therefore, one should really check this against the allowable shear stress, not the tensile/yield stress. Again = F/area.

    So we add sufficient “material” to ensure that the stress does not exceed the deign allowable. But, is this a single load application or a dynamic load? If dynamic rather than static, one must consider fatigue strength not simple static strength as the design allowable.

    Then the load has been transferred into the web and is now faced with the joint to the vessel’s structure.

    It is bonded directly to the steel plate (its use is still unknown for now), and also to the laminate that runs over the joint from left to right. So the load path is how much bonded/resin area is there of the laminate to the vertical web? Then how much bonded area is there for the adhesive of the web to the steel plate. Following the same pattern in methodology as above, let’s say you now arrive at the amount of “material” you need to transfer this load, F.

    So now the load F is, as RX’s image a load path for the through thickness of the overlay, and again, is a simple F/Area. But, the steel plate has square corners, this means as the fibres run over the steel plate it will create hard corners what is called a stress concentration. Thus you need to allow for a higher design stress, by providing more “area”, as well as to mitigate this by rounded the corners of the steel FB. And again is this a static or dynamic loading scenario.

    Once the overlay passes your calculations for the amount of material to transfer the load F/2, as shown in RXs image, you follow the load path into the bonding with the main hull structure, in the same fashion. And for the delamination the design values are again different from simple ‘tensile’/yield values.

    Does this help?

    I see usual background noise from those posters who don’t know how to draw simple FF body images of balloons nor know what brittle and ductile means….but hey, it is a free web, they can post as they like.
     
    Last edited: Feb 3, 2019
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  3. rxcomposite
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    rxcomposite Senior Member

    Fallguy- Thanks, I was waiting for the OP to ask question but he left (I think). The diagram I have posted should enable the designer to start calculations. By the way, the bottom orange arrow in my diagram is reversed. It should be pointing inward. Will correct later.

    Taking a cue from AH, I posted the thought process. First is the drawing. This was already posted by the OP. Second is the analysis, which I posted. Now that you have started the calculation process which is in the right path, I will detail the calculation 101 later after I finished what I am doing.
     
  4. TANSL
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    TANSL Senior Member

    "what brittle and ductile means?". Very easy ..., what the Classification Society says.
    As this is a free web, I will also try to help the OP. Procedure :
    1.- calculate the load on the joint.
    2.- check if the connection can support the load.
    3.- If it does not support it, change the union and return to point 2
    Very important note: based on my great knowledge of the Rules of the Lloyd's Register I advise you to do the calculations correctly (make no mistake adding or multiplying). If you do not do them correctly, all my advice will not help you at all.
    Does this help?
     
  5. fallguy
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    fallguy Senior Member

    I think it is easy to delve off into lots of places for all, but I am most curious about this portion of the reply.

    So, let's consider just the shear, static load-a plant hanger! Someone will say it is dynamic in rough seas, don't. A fiberglass and resin overlay of say a 4 oz woven vs a single 1708. Make the plate much smaller 2"x2" or 4" sq. The plant design weight is 100# or 25 psi.

    How do you calculate just the shear strength of the 4 oz woven vs the 1708 to make sure the hook won't break through?

    About to go into surgery, so apologize if not clear.
     
  6. fallguy
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    fallguy Senior Member

    Well, I am bionic friends. I will spare you the gore.

    Let me try again.

    Is the shear calculated by using the some area x in the drawing? It seems like that area is NOT where shear failure would occur. But an actual shear would occur at the perimeter of the plate.
     
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  7. Ad Hoc
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    Ad Hoc Naval Architect

    Welcome back.
    Sorry for delay in replying...been busy.

    Ok..well, a picture explains a lot, so let me go back to the original picture, but now with subtle difference to explain how it works.

    Look at the image below, one of our previous composite designs, it is a WTB joint, but, could just as easily be the joint posted by the OP, and that to which you are referring also (I assume):

    upload_2019-2-5_9-46-5.png

    The vertical load, F, we can now assume is taken by the WTB or Web, as we have calculated that it has sufficient shear area.
    So, where does that load go then..?...it is all about following the load path...like a series of daisy chains.

    So let's assume, for now, that there is no bonding of the WTB/web at its base. So if we take a cut a location No.1, we have just one layer of fibres in the overlay. So what is the shear strength of that overlay of 1 layer of fibre, once bonded in...and is it the resin or the fibre that breaks first? The amount of area being the thickness of the laminate x the length of the WTB/web.
    Then, let's say, again for argument's sake, it fails,....we now investigate this at location no.2, follow the same procedure. Again again for arguments sake, let's say it also fails - insufficient shear area.

    So, now, we come to location no.3 We now have 3 layers, i.e thicker. So doing this same calculation produce as before, does it pass or fail?

    If it passes, it means that the load can transfer from the WTB/web into the laminate without ruption or cracking, so the load is now in the laminate. The first daisy chain as such.
    So how does this load get into the hull structure which is out-of-plane? If we isolate the structure we now effectively have an " L " shaped 'bit' of structure, shown by the red angle:

    upload_2019-2-5_9-56-17.png

    The load must go from the vertical leg of this "angle" into the horizontal part. Depending on how this is arranged, the length of the horizontal part will be subjected to locaised bending, since as the vertical part rises, the extend horizontal part will bend as it attempt to remain in contact with the flat surface. Ok,. lets make this shorter then - solved? Well, not really, as that creates a localised stress concentration at the end of the shorter flange part. So what you need to do is work out the stiffness (interia) of that angled part being subjected to a vertical load at its end. The nagle must be able to take the bending and shear. The number of layers you use will of course dictate how stiff that 'angled' part it and thus when it passes this criterion, as well as its localised thickness at the joint of the L to transfer the shear load, as before.
    So now again for simplicity, you have satisfied yourself that the number of layers to achieve this and the length of the overlay is as shown, the load will then be transferred as shown, through the bond into the hull structure. The daisy chain.

    If we now assume that 3 layers was insufficient that the number of layers required was again for simplicity, a crazy amount, how else can the load be transferred?

    As noted by the contact area of the WTB/web to the hull. So the strength of the adhesive shown, the crestomer, with the applied force F, and the amount of contact area, is the transferred?
    If no, you increase this contact area or change the adhesive to obtain greater mechanical properties, or a combo of both, until this passes. Then you have the same situation as the transfer of the load into the horizontal part of the angle, shown above.

    So why the overlay?

    Well, if you have what's called a hard corner, you are trying to transfer the vertical load over a very short area, and so at the ends it is a location of high stress. So to reduce the amount of stress, i.e to smooth it out gradually, we have the overlay. And think of this a a simple bracket as you would see on any metal type frame structure.

    Now, if the 3 layers originally calculated passed, and the amount of contract area and adhesive passed, you have a fail safe and what is called, structural redundancy.

    Does this make sense?
     
    Last edited: Feb 4, 2019
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  8. Ad Hoc
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    Ad Hoc Naval Architect

    Geess...the paranoia of the background noise is deafening.

    Wasn't even aimed at him, but his paranoia knows no bounds - insecurity i assume?
    But since we are now on this:

    Absolutely NOTHING to do with a classification society. It is basic metallurgy and regarding the release of strain energy.

    And oh:
    This is coming from the same person that espouses:

    As usual flip-flopping to save face and creating an illusion (or is that delusion) of knowledge?
     
    Last edited: Feb 4, 2019
  9. Mr Efficiency
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    Mr Efficiency Senior Member

    Lots of squabbles breaking out, meanwhile the OP can't even be bothered answering, I'd suggest leave it alone till he does.
     
  10. Ad Hoc
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    Ad Hoc Naval Architect

    Threads have a life of their own and all deviate..
    Are you now suggesting ignoring a posters request, on a thread, for further information and education merely because the original OP is dormant?
    Nonsense...
     
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  11. Mr Efficiency
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    Mr Efficiency Senior Member

    I don't think so, his post was low on information, and as for embedding a steel plate (in what kind of material ? Not specified !) the biggest problem could be the whole deal will bust out of the "trap"once it starts to rust.
     
  12. Ad Hoc
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    Ad Hoc Naval Architect

    Which is why there is endless background noise and squabbling by those that are offer nothing other than their opines on something has yet to be stated by the OP but already assumed by those squabbling as fact where none exist.
    Absolutely zero engineering rigour - shockingly so. But they prefer others not to point it out to them, hoping the post is sufficient on its own!

    And that's the point, QED.
    At no point has the OP asked for any suggestions nor thoughts on the steel plate. Yet the background noise loafers focus on that because in all their posts they offer nothing to help the OP - except their self-flagellating prowess.

    The OP has only asked about calculating the strength of the joint - NOTHING else.
    The purpose of the steel plate and its effects - if any - has yet to be asked/established.

    Another assumption.
    We have embedded steel tapping plates many times before, none have rusted thus far. It is called treatment and surface preparation.
     
  13. Mr Efficiency
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    Mr Efficiency Senior Member

    This is what he asked...."How can I calculate the bonded composite connection of the type as on the picture? "

    Insufficient information in more ways than one. As you yourself say, the load path isn't specified. Let him come back and answer, or stuff him !
     
  14. rxcomposite
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    rxcomposite Senior Member

    FG-
    1. If the baseplate is secured to the substrate by adhesive, It is cohesion and the tensile strength of the adhesive should not be exceeded. Epoxy adhesive has about 10 MPa and 5 MPa is used as design value. Tensile strength = Vertical Force/Area. In most cases, this would be adequate but when the load is off axis, or there is a tangential force acting on the rod, the bondline at the edge of the plate starts to separate. Adhesive bonding has a very low value of T peel strength of 0.17 MPa. This will start the crack propagation. *this is as AH has noted, if there is no other forces acting.

    2. To add to the strength and increase the moment of inertia, a top cover laminate is added to fix the baseplate to the substrate. The baseplate will be pushing against this laminate and shear will occur in the edges. It is likened to a punch and die and the formula for the punch force to shear clean a piece of laminate with a known thickness is:

    Punching Force (KN) = Perimeter(mm) x Plate thickness/depth (mm) x Shear Strength (kN/mm2)

    Epoxy laminate has about 78 MPa Shear in plane Strength. Multiply that by 2 to get thru thickness (intralaminar shear) strength. You can now re arrange the formula if the thickness is unknown.

    To mitigate shear, the top edge of the baseplate is rounded and the bottom side is filleted and radiused. Fibers do not like sharp corners and will not conform.

    3. From the corner of the baseplate to the bottom connection of substrate, there is a transition. In most cases, this is a 45 degree angle and the resulting vector is 1.4. The 45 degree laminate will be subjected to a pulling force of Vertical load x 1.4. This is the sling theory or as you call it, a suspended plant pot. The tensile strength of the laminate must not be exceeded.

    4. The tapered laminated follows the bracket/spandrel curve. It is to prevent stress riser. The first transition is the rounding of the top of the base plate and the filleting of the bottom part to ease the laminate entry into the 0 degree laminate. Typical rule as posted by AH is 50 mm first layer, followed by 25 mm offset thereafter. The first 50mm rule is to ensure adequate Interlaminar bonding and taper follows the 6:1 rule. There are other rule like 25 + 20+20+20 rule, or the quick and dirty 12:1 rule after the fillet.

    5. Last but not the least is the thermal expansion of dissimilar material like metal and composite. If this is to be used on deck, there will be a large internal stress created by the expansion of the metal. Careful choice of the in plane shear/lap shear property of the adhesive should be considered. Aluminum comes close to FRP. Aluminum and carbon is a no no. Steel and composites creates internal stress but if you keep the steel area low, the stress along with the choice of adhesive is reduced.
     

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    Last edited: Feb 5, 2019
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  15. rxcomposite
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    rxcomposite Senior Member

    I defined the load path as the OP seems not to know where to start. He did specify the material though. Metal and composite. We take it from there.
     
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