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Stress calculations on a disk plate

  1. Dec 15, 2011 #1
    Hello,

    I am trying to determine the calculation to use to determine the force at which a disk plate will fail for the following set up.

    The disk is slid onto a metal bung for a roll, the disk is supported from the inside diameter for around 50mm. A force is then applied on the oppsite side of the support uniformly to the entire face of the disk. Please see image below for clarity. The blue part is the disk and the red is the bung.

    Plate.bmp


    Just wondering what equation I should be using to determine the force to which the disk will fail assuming I know the properties of the disk.

    Any help would be much appreciated, I hope the explanation is in enough detail. I did attempt the problem by treating a section of the disk as a rectangle and analysing that way. Not sure if it was the right way to go about it, I used the bending stress equation to work out the force needed to fail the material.

    For some reason the attachment wont upload, I did create the same thread in another part of the forum but thought I would find more answers here. The link to the other thread is https://www.physicsforums.com/showthread.php?t=560325

    Thanks
     
    Last edited: Dec 15, 2011
  2. jcsd
  3. Dec 25, 2011 #2

    nvn

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    Science Advisor
    Homework Helper

    bellshom: The stress depends on the proportions of the parts, which depend on the dimensions. Therefore, I cannot give you a specific formula, because coefficients are all different for different proportions in plate analysis.

    If we assume your disk is mild steel, with a yield factor of safety of 1.50, and if we assume your diagram is to scale, then we currently have the following.

    Sb = disk allowable bending stress = 167 MPa
    d1 = disk inside diameter = 50 mm
    d2 = disk outside diameter = 186 mm
    t = disk thickness = 10 mm​

    The bending stress on the disk is sigma = Sb = 167 MPa. Therefore, the disk uniformly-distributed allowable pressure is currently p = 0.4490 MPa. Hence, the allowable applied force on the disk (pressure times disk area) is F = p*A = p*(0.25*pi)(d2^2 - d1^2) = 0.4490(0.25*pi)(186^2 - 50^2) = 11 320 N.
     
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