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Homework Help: Torus pressure vessel stress

  1. Nov 12, 2005 #1
    I am a novice at stress analysis. For my competition, I need to calculate the stress on a torus in vacuum filled with air at pressure P.

    Here's is what I have managed to do. I have attached an image below.

    The torus section is an angular section of a curved cylinder. Its angular dimensions are [tex]d\psi[/tex] in reference to the torus cross section at an angle [tex]\psi[/tex] to the horizontal and [tex]d\theta[/tex] in reference to the center of the torus. The major radius of the torus is R and the minor Radius is r.

    Summing the forces in the radial direction
    \displaystyle\frac{1}{2}(r+dr)d\psi[R-rcos(\psi+d\psi)+R-rcos\psi]d\theta*\sigma_r - \displaystyle\frac{1}{2}dr[R-rcos(\psi+d\psi)+R+drcos\psi-rcos(\psi+d\psi)]d\theta*\sigma_{t1}+\displaystyle\frac{1}{2}\pi[(r+dr)^2-r^2]d\theta*\sigma_{t2}=0

    Here [tex]\sigma_{t1}[/tex] is the tangential stress with reference to the cross section of the torus and [tex]\sigma_{t2}[/tex] is the tangential stress with reference to the center of the torus and [tex]\sigma_r[/tex] is the radial stress.
    \displaystyle\frac{1}{2}(r+dr)d\psi[R-rcos(\psi+d\psi)+R-rcos\psi]d\theta[/tex] is the area between [tex]R_{rt}[/tex] and [tex]R_{rb}[/tex](See figure)
    \displaystyle\frac{1}{2}dr[R-rcos(\psi+d\psi)+R+drcos\psi-rcos(\psi+d\psi)]d\theta[/tex] is the area between [tex]R_{rt}[/tex] and [tex]R_{lt}[/tex]

    Am I correct upto this point? If I am, how do I process further?

    Attached Files:

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    Last edited: Nov 12, 2005
  2. jcsd
  3. Nov 12, 2005 #2
    Can anyone help please?
  4. Nov 12, 2005 #3


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    Even for someone with experience, you have picked a problem that is not so trivial. :biggrin: Out of curiosity, have you reviewed the solution for a straight pipe or spherical vessel. That should give you an idea the basic theory.

    The radial stress is easy. Since the system is in equilibrium, the inside radial stress = gas pressure, and the outside in a vacuum must = zero. That's pretty straight forward.

    Finding the tangential stresses in the hoop and longitudinal (wrt azimuthal axis) is not so easy.

    Give me some time, but if anyone else wants to jump in, they are more than welcome. :biggrin:

    This type of problem would be encountered in the internal stresses of an elbow bend of a pipe.
  5. Nov 12, 2005 #4
    Thank you for the reply.

    Yes, I have done that.
    Last edited: Nov 12, 2005
  6. Nov 12, 2005 #5


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    I'll work on this, but it's not often I work with toroidal coordinates.

    I don't think you want terms like cos (x + dx).
  7. Nov 13, 2005 #6
    Thank you

    Since [tex]\lim_{x\rightarrow 0} cos x = 1 - \lim_{x\rightarrow 0} sinx[/tex],
    cos (x+dx) = cosxcosdx - sinx sindx = cos x (1-sindx) - sinx sin dx = cos x (1 - dx) - sin x dx[/tex]

    because for extremely small angles dx, sin dx = dx.

    Provided that the equation is correct, can this substitution work?
  8. Nov 13, 2005 #7


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    With respect to cos (x + dx), I was thinking along the lines of your latex.

    Bascially, cos (dx) ~1 and sin (dx) ~ dx, but we don't usually see such use of the differential.

    In spherical coordinates for examples, on has r sin[itex]\,\theta[/itex] d[itex]\,\phi[/itex], r d[itex]\,\theta[/itex], and dr.

    I believe one needs to work in terms of the major and minor radii, R and r respectively, and the azimuthal angle which sweeps around the plane containing the major radius, and the azimuthal ange which sweeps around the plane containing the minor radius/diameter. One also needs the wall thickness, t, so the wall extends from r to r+t when measured from the toroidal axis.

    I think I have an idea on this, but I have to work throught the gory details. :biggrin:
    Last edited: Nov 13, 2005
  9. Nov 8, 2011 #8
    Rationale and solution there:

    High pressure vessels
    by Donald M. Fryer, John F. Harvey
    http://books.google.fr/books?id=SZD...page&q="Pressure vessel" Torus stress&f=false

    and there in chapter 13 page 804 (Pdf 11/98)
    Mechanical Design of Process Equipment.pdf
    by Coulson & Richardson

    It's done by cutting out a part of the torus with cylindrical symmetry and deducing one stress from the vertical equilibrium. The other stress comes from the equilibrium of a surface element, the stresses acting on one curvature each.
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