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Calculating thickness of a cast iron cylinder carrying pressure of 30Nmm^2

  1. Aug 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A cast-iron cylinder of 300mm bore carries a pressure of 30 Newtons per square
    millimetre; what should be its thickness for a unit working stress of 25 newtons per
    square millimetre?


    2. Relevant equations

    ó1 = pd/2t

    t=pd/2tó1


    3. The attempt at a solution

    t=pd/2tó1

    t= 30*300/2*25

    =180mm

    Its got to be wrong as its way too big but im unsure on where im going wrong, any help would be much appreciated.

    Heres another attempt

    ó1/pd=2t

    25/(30*3000) = 2t

    2t = 0.00277778

    t= 0.0014mm (4dp)

    Maybes this one, im not sure.

    Thanks, Joe
     
  2. jcsd
  3. Aug 14, 2012 #2
    I suggest you check the quantities and units in the question. Also I think you have some typos in your relevant equations and solutions.
     
  4. Aug 14, 2012 #3
    Thankyou for your reply.

    This is the best i can get.

    Assuming that the cylinder is a thin cylinder

    o1 = pd/2t

    where o1=Stress on the cylinder

    This stress is the circumferential stress on the cylinder

    p= Pressure

    d=Diameter of he cylinder bore

    t= thickness

    Now we have been given

    o1=25 N/mm^2

    P=30N/mm^2

    d=300mm

    SO

    We know

    o1 = pd/2t

    So the design criteria for the thickness becomes

    SO t=pd/2o1 = (30x300)/(2x25)= 9000/50 = 180

    Its thickness should be 180 mm .
     
    Last edited: Aug 14, 2012
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