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  • Hyperstatic supported pipe

    Discussion in 'Calculations' started by Virgule, Nov 29, 2011.

    1. Virgule

      Virgule Active Member

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      Hello fellow members,

      My question is regarding to hyperstatic supported beam problems.

      The following links describe the problem.
      http://i1189.photobucket.com/albums/z434/Virgule99/Piping/8pipe.png
      http://i1189.photobucket.com/albums/z434/Virgule99/Piping/hyper_pipe1.png
      http://i1189.photobucket.com/albums/z434/Virgule99/Piping/hyper_pipe2.png


      The pipe is anchored at point A and simply supported at point C. The force FB represents the weight of the pipe and the force FD represents the half weight of a section of pipe (135'') continuing in Z.

      QUESTION 1 : Is aplying the half weight of the 135'' section in Z at point D a good simplification when modeling the effects of this section on the section in X ?
      QUESTION 2 : Could I consider the whole X and Z section as a straight section in X of the same length ?

      Considering the problem as I drew it,
      I have 3 unknown variables : MA, FA & FC
      I have 2 equations coming from equilibrium equations :
      Forces in Y = 0 (equation 1) & Moments in Z around A = 0 (equation 2)​

      So I have a hyperstatic problem of the first degree.

      I thought of solving the problem by using singularity functions (leading to equations 3.1, 4.1 & 4.2 (2nd page)) but that bring up 2 more unknown variables (C1 & C2). So I still have 6 unknowns and 5 equations.

      QUESTION 3 : Is the singularity functions method a proper way of solving this problem ? If yes, how do I get my sixth equation ? If no, how would you proceed ?

      Thanks for your input.
       
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    3. Virgule

      Virgule Active Member

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    4. Virgule

      Virgule Active Member

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      I could use some help.
       
    5. Ramana Rao

      Ramana Rao Well-Known Member

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      What about considering the centre of gravity of the pipe/ pipe sections and then solving for moments. I think this will simplify the problem. I am not sure if this will work, and is only a suggestion.
       
    6. Virgule

      Virgule Active Member

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      If by considering the center of gravity you're refering to calculating the equilibrium ABOUT the center of gravity, this adds unknowns because the weight (applied at the center of gravity) is known and the reactions aren't.
       
    7. Ramana Rao

      Ramana Rao Well-Known Member

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      Not about the centre of gravity. To find out the reactions at the anchor and the support, you could take the CG of different sections of the pipe into consideration, and find the moments individually about the anchor and the support. I feel this will result in two equations with two unknowns. I do not know if this will help you, but it is similar to the bending moment problems of beams we covered in college.
       
    8. maniacal_engineer

      maniacal_engineer Well-Known Member

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      boundary conditions.

      QUESTION 1 : Is aplying the half weight of the 135'' section in Z at point D a good simplification when modeling the effects of this section on the section in X ?

      Applying half the weight is conservative. It would be as if that segment in Z had flexible couplings at each end.

      QUESTION 2 : Could I consider the whole X and Z section as a straight section in X of the same length ?

      No, this would not work, unless you modeled a pin joint at point D.

      Considering the problem as I drew it,
      I have 3 unknown variables : MA, FA & FC
      I have 2 equations coming from equilibrium equations :
      Forces in Y = 0 (equation 1) & Moments in Z around A = 0 (equation 2)​

      So I have a hyperstatic problem of the first degree.

      I thought of solving the problem by using singularity functions (leading to equations 3.1, 4.1 & 4.2 (2nd page)) but that bring up 2 more unknown variables (C1 & C2). So I still have 6 unknowns and 5 equations.

      QUESTION 3 : Is the singularity functions method a proper way of solving this problem ? If yes, how do I get my sixth equation ? If no, how would you proceed ?

      Thanks for your input.

      The missing equation you are looking for is that the slope at A is zero since it is a fixed anchor. Otherwise the moment at A is zero. The slope constraint gives you the extra equation you need to eliminate the extra unknown.
      Also, it would be more accurate to model the weight of the pipe as a distributed load, and do the integrations (integrate load to get shear, integrate shear to get moment, integrate moment to get slope, integrate slope to get displacement. )
      Or you can do castigliano's and set the displacement at A and C to zero. If you want to accurately account for the Z direction segment at D that is the way I would do it.
       
    9. Virgule

      Virgule Active Member

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      @ maniacal

      Question 1 : Thanks for the input, that's what I thought.

      Question 2 : Thanks for the input.

      Question 3 : I already have the slope@A=0 equation, it's equation 3.1. Did I miss something ?
      Thanks for the tip about weight distribution.
       

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