Pull off testing

Discussion in 'Fiberglass and Composite Boat Building' started by fallguy, May 11, 2018.

  1. fallguy
    Joined: Dec 2016
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    Location: usa

    fallguy Senior Member

    There is no csm in my biax per the tds. Using a matted glass will add weight, but I think I will test some csm fabric as well to see how it behaves.
     
  2. gonzo
    Joined: Aug 2002
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    gonzo Senior Member

    Generically, a crack is a type of flaw in the material.
     
  3. Ad Hoc
    Joined: Oct 2008
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    Ad Hoc Naval Architect

    You can protest as much as you like....you still don't get it.

    Every material is either perfect and has a theoretical value of strength (high) or it is not (low).
    To be 'perfect', all the atoms are arranged equally and no irregularities in the crystal structure exists. This leads to very high strength values for materials.

    Unfortunately real materials are not like this.
    Flaws exist in them owing to atom spacings not being perfectly equal, or the crystal matrix is very subtly different owing to the formation of the material from said irregularities in the atom spacing.

    Thus a flaw is just that, as noted in the post here:
    Pull off testing https://www.boatdesign.net/threads/pull-off-testing.60316/page-5#post-830562

    So, a crack cannot be a flaw. A flaw or void, creates and becomes a crack under and applied load.
     
    rxcomposite likes this.

  4. rxcomposite
    Joined: Jan 2005
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    Location: Philippines

    rxcomposite Senior Member

    Fg-
    There are many ways to find interlaminar shear. More sophisticated methods use the modulus of individual elements and is more accurate. Lacking data, I am using the simplest of approach to demonstrate.

    Your laminate is not optimized. With 12 mm core and a shear strength of 1.2 N/mm2, I forced the laminate to fail by having no safety factor. To fail, it requires 1.8 mm inner skin and 2.4 mm of outer skin. To simplify the analysis, I used 2.4 mm on both sides. Theoretically this is 2 layers skin, which translates to 350 gr/m2 of WR or Biax per laminate.

    The core must fail at 1.2 N/mm2 of shear. Given the I of the core and computing for the required load for the core only, I need 270 Newtons. When laminated, the I increases because of added height. I have to derate the core shear strength by 75%.

    First assumption was to fix the Neutral axis at 1/2 depth as this is a balanced laminate. By locating the midpoint of each element, I can find the second moment of Area, Q. Using the formula outlined in the spreadsheet, I can find the individual shear stress from the core, to the bondline, to the bond between the laminate, There is no shear on top of the laminate.

    The loadings shows that it will shear at failure the core at the center with 1.2 N/mm2 (177 psi), the bondline will shear at 0.5 N/mm2 (68 psi), and the bond between the laminate at 0.3 N/mm2 (37 psi). Note that this is a unit of measure and the analyzed laminate is set at 25mm for simplicity.

    With the stresses derived, you can make a test sample strip by bonding laminate at 25 mm. (1") overlap and cutting to 1" strips. Leave some ear for grips and prepare for lap shear test.

    Will the laminate hold? The WR has a tensile strength of 91 N/mm2 (13,195 psi) and the Biax at 45 degree has a reduced tensile strength of 46 N/mm2 (6,598 psi) I doubt you can break as it needs considerable force to break a 1" strip. By pulling a 25 x 25 mm (1" square) bonded lap joint, you can. Rather than get lost in the translation, you only need to apply about 37 lbs of force to break the interlaminar bond. If you exceed it, your laminate interlaminar strength is in good health.
     

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