Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Stress in a hole due to pin force

  1. Feb 18, 2010 #1
    Hi everyone,

    What is the area should be taken to determine the compression stress of pin-hole in tention force (i.e plate with known thickness having a hole for lifting shakle pin) where the hole diameter is greater than the pin diameter (> x3) (theoretically, it is not an area just line of contact between the hole and the pin, which is the thickness of the plate).
    Thank you, Guideon
  2. jcsd
  3. Feb 18, 2010 #2
    Hey man, I would like to help you, but I don't quite understand your description. A diagram would be really helpful here.
  4. Feb 19, 2010 #3
    Thank you,
    Hereby a sketch attached to clarify the question, I hope it would explain it better.


    Attached Files:

  5. Feb 19, 2010 #4
    Oh wow, that's great, thanks for the clarification. When I get home from work today I'll hook you up. Do you have numbers? I could also run a FE solution for you.
  6. Feb 19, 2010 #5


    User Avatar
    Science Advisor

    Your attachment calcs look like you're trying to solve for bearing stress. However, since the pin is much smaller than the socket, the max compressive stress is actually given by:

    [tex] \sigma c_{max} = 0.798 \cdot \sqrt{\frac{p}{K_D \cdot C_E}} [/tex]


    p = load per unit length
    E1 = modulus of elasticity of bottom socket
    E2 = modulus of elasticity of pin
    v1 = poisson's ratio of bottom socket
    v2 = poisson's ratio of pin

    [tex] K_D = \frac{D_2 \cdot D_1}{D_1 - D_2} [/tex]

    [tex] C_E = \frac{1- \nu_1^2}{E_1} + \frac{1- \nu_2^2}{E_2} [/tex]


    Attached Files:

  7. Feb 19, 2010 #6
    I agree with Stewart, though I would also add that the stress in your plate increases due to the hole.

    For an elliptical center hole,

    [tex]\sigma_{A} = \sigma_{c}\left(1+\frac{2a}{b}\right)[/tex]

    Since your hole is circular, a = b = r, thus

    [tex]\sigma_{A} = 3\sigma_{c}[/tex]


    [tex]\sigma_{c}[/tex] is compressive stress
    [tex]\sigma_{A}[/tex] is the actual stress, taking into account the geometry of the hole
    [tex]2a [/tex] is the major axis
    [tex]2b[/tex] is the minor axis
    [tex]r[/tex] is the radius of your circular hole

    Source: Anderson, T.L., Fracture Mechanics Fundamentals and Applications, 3rd Ed., CRC / Taylor and Francis, 2005, p27.
  8. Feb 19, 2010 #7
    The stress concentration formula you're talking about was derived for uniaxial tension. I don't think those boundary conditions apply in this case. Also, since this is a contact stress problem, the stresses will be altered because traction forces are applied to the hole.
  9. Feb 20, 2010 #8
    Thank you all, you are great
    I just want to be sure that I undersdand your answers:
    a) Is the stress equation you sent depends on the units you use in the equation (SI/american)?
    b) The 0.798 factor is dimentionless?
    a) Is the compressive stress you used in the equation for an elliptical center hole is the result of the max compressive stress equation in stewartcs reply? If not, how can I calculate this sress? that was my original question.
    b) According to your important note, actually, the major stress is not where the pin force acts, but at 90 degrees right/left from the hole center (3 times the compressive stress)?

    Thank you all again, Guideon
  10. Feb 22, 2010 #9


    User Avatar
    Science Advisor

    As with all engineering formulas, the equation should be dimensionally consistent.

    The 0.798 is a dimensionless factor.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook