Calculating Stress on a Panel, and Material Selection

Discussion in 'Boat Design' started by Zac Penn, Dec 10, 2015.

  1. SamSam
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    SamSam Senior Member

    If there is no bottom, this thing becomes questionable about holding together. If it
    , at 8.33 lbs per gallon, that is 15,000 lbs. It would be very exciting and memorable to have a tank with 7 1/2 tons of water bursting on a convention floor, if live fish were involved it would be epical.

    You might try googling "making large aquariums".
     
  2. Barry
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    Barry Senior Member

    There are quite a few aluminum concrete form manufactures that might have panels pre made that would suit your application.
    The joining methods for the corners are already engineered and they will be lightweight
    The only design issue that I see is that they use ties between the opposing wall which reduces bending stresses but looking at a few sites "aluminum concrete forms", the size of the cross stringers t seem larger than my calculations required. albeit I do not have the actual measured sizes to compare.

    Another design option would be to have the 4 only 4 x 8 panels, bolt them together,
    under the liner at the bottom, run a cable across to the opposite panel at the midpoint of the lower stringer with a turn buckle to help restrain the deflection and another at the top stringer
    Ie 2 lower cross cables under the liner at the bottom and 2 upper cross cables at the top above the liner. This would enable the stringers to be of less dimension

    I would certainly have an engineer evaluate or design the panels as you do not want a liability problem if the structure breaks
     
  3. gonzo
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    gonzo Senior Member

    Considering that this tank is just holding water, there seems to be no reason to prevent cross bars in it.
     
  4. TANSL
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    TANSL Senior Member

  5. gonzo
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    gonzo Senior Member

  6. TANSL
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    TANSL Senior Member

    No, not exactly that (may be bending moments curve?). There are some other factors to consider. But I wondered how to solve, not how to calculate, the effects of shear stresses in the tank.
     
  7. gonzo
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    gonzo Senior Member

    The bending moment is the second integral
     
  8. TANSL
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    TANSL Senior Member

    No doubt you put interest but you must study a lot more.
    If you have any specific questions, I will gladly clarify.
    Cheers.
     
  9. Barry
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    Barry Senior Member

    [
    http://physics.usask.ca/~chang/course/ep324/lecture/lecture8.pdf

    Attached is a link to a page that shows the use of the integration process to develop a few simple formulas used to calculate some hydrostatic forces applicable to this thread.

    The total, not localized, load on the retaining wall is given by the formula Total Force equals specific weight times height divided by 2 and multiplied by the Area.

    Again, this is the total force which you would divide by two to get the shear at each corner.

    But as the pressure goes from zero at the top to 1.7 psi at the bottom, the actual shear force in the connecting device will be at a maximum at the bottom and diminishing with lessening depth.

    I am not sure on what you are actually looking for here, "how to solve not how to calculate the effects of shear stresses in the tank"

    There are two locations that you would include shear in the calculations as you worked on the design of the tank.
    1) You would take the maximum tensile stress produced by the bending moment which would be at bottom middle of the panel for say a smooth homogeneous cross section, then calculate the shear stress at the bottom as this is will also be the max shear stress due to hydrostatic loading. With these figures you would create a Mohr's Circle to see if the combined bending and shear stresses created are less than the allowable stresses but including a factor of safety

    2) The other area for shear analysis will be at the connecting device. I will assume a bolting schedule just to make the discussion easy ( and the fact that I have almost no experience with composite fasteners )
    The bottom bolt will see the maximum shear. The sizing of this bolt will be easy, cross section area, allowable stress etc
    Depending on the material used where the bolt goes through, a few calculations will be required to ensure that the material can take the tensile and shear stresses where the bolt goes through.

    Not really included but an engineer would also run a quick calculation of the shear flow in the tank wall to confirm that this is within limits. For say an aluminum or steel or plywood wall, shear flow will probably be insignificant but perhaps with composites, with a hard shell and softer spacer material, it might be very relevant

    There becomes a point that the OP has to say what material that the panel will be made of and then the design process can be started. I see an aluminum panel having the advantage of low price, durability, lightweight, easy installation and cheap repair.
     
  10. TANSL
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    TANSL Senior Member

    Barry, my question, therefore Gonzo advise, is simple: remove the cross bars and, if shear stresses occur in the tank walls, how we solve the problem ?. Usually, I would not advise removing structural elements without checking whether they are needed or not.
    I think what I say is rather like what you say
    I appreciate your effort to teach the integrals are very useful. I never doubted it.
    I do not understand what is meant by "integral first of the load" (and still understand less about the negative first integral). Integrating the various existing loads in a range you can get the resultant force but, at this time, I do not know what more can be obtained by the integral of a load. With a single load, on the other hand, few integrations can be done.
    Thank you, Barry, as always, for your kind explanations.
     
  11. Ad Hoc
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    Ad Hoc Naval Architect

    Then perhaps you have not analysed structures and used classical structural theory to investigate the response?

    The bending moment of any structure can be written simply as (without going into too much depth):

    M = EI.(d2y/dx2).

    If one differentiates this, it becomes the slope = dy/dx
    If one differentiates it again, it becomes the deflection = y.

    However, if one integrates the bending moment (the first integral) = -dM/dx = -EI.(d3y/dx3) = shear force.

    Thus integrating the bending moment one arrives at the shear force.

    Simple :)
     
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  12. NavalSArtichoke
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    NavalSArtichoke Senior Member

    You've got differentiation confused with integration in the relations above.

    the integral of the M/EI curve is the slope curve.
    the integral of the slope curve is the deflection curve.

    the derivative of the M curve (dM/dx) is the shear force curve V(x).
    the derivative of the shear force curve (dV/dx) is known as the load curve, which is really used mostly for distributed loadings.
     
  13. gonzo
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    gonzo Senior Member

    Therefore the integral of the load is the shear, etc.
     
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  14. Ad Hoc
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    Ad Hoc Naval Architect

    Doh!!:!:

    Message to self...don't reply to posts when working at the same time as being on the phone too...hahaha!

    Thanks, good catch :)
     

  15. TANSL
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    TANSL Senior Member

    LOL, at the end we will give reason to Gonzo.:p
    Please someone explain him how things are and if it is true what he says. He ignores me (maybe he does well).
    Ad Hoc,:D:D
     
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