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Minimum material problem

  1. Dec 15, 2006 #1
    I know calculus can be used to calculate the dimensions of an object and the minimum material which can be used. It is a pressure vessel of a cylindrical shape.
    [tex]v= \frac {4 \pi r^3} {3}[/tex]
    [tex]A=2 \pi r^2 + 2 \pi rh[/tex]

    as the cylinder is hollow the thickness of the walls is found by

    [tex]t= \frac {pd} {4 \sigma }[/tex]

    sigma = [tex]300x10^6 Nm^{-2}[/tex] (steel i think?)
    p = pressure
    d = diameter

    I have found amount of material can be calculated by surface area multiplied by thickness but would like wo see the minimum amount of material used proved by calculus
  2. jcsd
  3. Dec 15, 2006 #2


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    1. The volume of a ball is not the same as the volume of a cylinder..

    2. You have a HOLLOW cylinder; it's surface area is not the same as the surface area of a compact cylinder.

    3. The volume of a cylindrical shell, where the shell has a non-zero thickness is NOT given by the "surface area" multiplied multiplied with the thickness.
    Why should it be that?

    4. In order to find a "minimum amount of material", you need to state what CONDITIONS should hold. Is the height to be constant? Is the net surface area to be constant?
    Or what? You haven't clarified the situation at all.
  4. Dec 15, 2006 #3
    1 of course a silly mistake [tex] V= \pi r^2 h[/tex]
    2 a hollow cylinder with closed ends i might add, with uniform thickness. I dont know if being hollow will have a change on the outcome, although the surface area will be nearly double.
    3 i would say the inner part of wall is less than the outer wall so if i do that multiplication the value will turn out higher than the actual value.

    if i get a formula for the cylinder which im on about, i would say the actual thickness is the integral of that formula with limits of the outer and inner radius.
    4. volume is constant v = [tex]3.3m^3[/tex] although i thought that wouldnt have mattered. r and h variables.
    as a second thought i may stay with my original approximation as this seems too demanding
  5. Dec 16, 2006 #4


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    3.The volume of a hollow cylinder is evidently the difference between the volume of the outer cylinder and the "removed" inner cylinder. Is that really too hard to understand?

    4. Well, if you do not think that condition matters, then obviously, a cylinder of ZERO volume will use the least amount of material.
  6. Dec 16, 2006 #5
    4 i thought the actual value of 3.3m^3 didnt matter.
    do you actually need values to find an equation or can you replace 3.3m^3 with v??
    5 im nearly certain only mentors give constructive replis
  7. Dec 17, 2006 #6

    Gib Z

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    He has just trying to help lazypast, if you don't like it then learn it yourself or ask a mentor directly.
  8. Dec 17, 2006 #7
    ive clarified a the things which have been asked, and still im in need of help. i found his post patronising
  9. Dec 17, 2006 #8


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    I guess the question is something about a thin walled cylindrical pressure vessel. In that case the volume is (to the usual engineering appoximation) the area of the mid-surface plane times the thickness.

    It's impossible to make any sense of the question unless you tell us the constraints that apply. Otherwise, the minimum amount of material is obviously for a vessel of size zero.
  10. Dec 17, 2006 #9
    volume, v=constant
    radius, r = variable
    height, h = variable

    i see why you would use the mid point of the wall thickness, i assume that is just thickness/2 ?

    edit, are there some constraints that you ask of which i am leaving out?
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