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Tension, Compression, Shear Problem.

  1. Mar 19, 2010 #1
    1. The problem statement, all variables and given/known data

    A solid steel bar of diameter d1 = 60 mm has a hole of diameter d2 = 32 mm drilled through it. A steel pin of diameter d2 passes through the hole and is attached to supports.

    Determine the maximum permissible tensile load Pallow in the bar.

    -Yield stress for shear in pin is τY = 120 MPa.
    -Yield stress for tension in bar is σY = 250 MPa.
    -Factor of safety n = 2.0.

    2. Relevant equations

    σallow = σY / n

    τallow = τY / n

    In direct tension and compression:

    Pallow = σallowA

    If compression, and a pin is present:

    Pallow = τallowA

    Area of regular cross section in bar is area of circle.

    Area of cross section with pin present requires additional steps. Because I do not have any images, I can only do my best to explain.

    The area of a "circle with a core removed" (side view of a pin through bar), is

    A = 2r2(α - (ab)/r2)

    3. The attempt at a solution

    It would be very hard to show my steps of determining the area, but I know the area is correct. I calculated the area of the cross-section with a pin to be

    A = 0.0010028 m2.

    This is after all the correct conversions and formulas to determine, as outlined in the appendix of my book.

    I recognize that because this bar is in tension, the pin does not factor in. I use the equation:

    Pallow = σallowA

    Pallow = (250MPa / 2.0)(0.0010028)
    Pallow = 0.12535 MPa*m2

    0.12535 MPa*m2 * 106 Pa/MPa = 125350 Pa*m2

    I know a pascal(Pa) is equivalent to 1 N/m2. The m2 cancels out, leaving

    125350 N.

    The solutions in my book says the answer should be 95.6 kN. My answer is 125.35 kN. I was wondering if the book was wrong, or if I'd messed up somewhere?

    Thanks for any help. This is my first of hopefully not many posts. Any feedback regarding both my work and the format of my question would be appreciated! :)

    D
     
    Last edited: Mar 19, 2010
  2. jcsd
  3. Mar 19, 2010 #2

    PhanthomJay

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    You forgot to check the shear allowable in the pin, which may govern. Note also that I get 96.5 kN as the max permissible tensile load in the bar, with a FS = 2.0
     
  4. Mar 19, 2010 #3
    I've gone over the problem with a friend. I saw that I was supposed to account for the shear in the pin. I guess I misunderstood the reading.

    Thanks for the help regardless!

    D
     
  5. Mar 19, 2010 #4

    PhanthomJay

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    Great, that was a nice clearly presented solution methodology.
     
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