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Hertzian contact stress SS-2244 / AISI-SAE 4140

  1. Apr 28, 2009 #1
    Hello guys!

    Is there anyone who knows where i can find information about maximum allowed Hertzian contact stress for material SS-2244 / AISI-SAE 4140 or any similar material quality?

    I have been searching for this information in several Swedish books and material tables, without success.
  2. jcsd
  3. Apr 28, 2009 #2


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    Can you give us some more information on the potential failure mode?

    AISI-4140 Low Alloy Steel at room temperature
    F_tu = 150 ksi
    F_ty = 132 ksi
    F_cy = 145 ksi
    F_su = 90 ksi
    F_bru = 219 ksi
    F_bry = 189 ksi

    t = tensile; c = compressive; br = bearing; u = ultimate; y = yield
  4. Apr 28, 2009 #3


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    You need to define what your failure criteria is, especially with contact stresses.
  5. Apr 28, 2009 #4
    A pin made of high grade spring steel is pressed tangential against the inside of a hub.

    Pin diameter (r1) = 2,5 mm
    Hub diameter (r2) = 55 mm (concave, and therefore -55 mm in Hertzian calculations)
    Contact length = 40 mm
    Hub material SS-2244 / AISI-SAE 4140

    The objective is to "backwards" calculate the maximum force F (N) that can be applied without causing plastic deformation on the inside of the hub.

    I´ll draw a simple sketch and add it tomorrow.

    Unfortunately my english is rather bad.

    Attached Files:

  6. Apr 29, 2009 #5
    Thank you minger! Do you have any reference (book/website etc.) for this information?
  7. Apr 29, 2009 #6


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    That seems awfully high for 4140 in the annealed state. Are you sure that isn't in some heat treated condition? The OP didn't mention anything about the final condition so this may be correct.
  8. Apr 29, 2009 #7


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    Oh, the 2.5mm (roughly 0.1") makes a difference. The spec I quoted was for 1" or greater....

    Nope, that is for less than 1". It should be good if it is 4140. It seems that for smaller diameters, AISI-4130 is more commonly used, which is significantly weaker.

    Some comments:

    AISI 4130 is a chromium-molybdenum steel that is in general use due to its well-established heat-treating practices and processes techniques. It is available in all sizes of sheet, plate and tubing. Bar stock of this material is also used for small forgings under one-hal finch in thickness. AISI-4135 is a slightly higher carbon version of AISI 4130, is available in sheet, plate and tubing.

    AISI 4140 is a chromium-molybdenum steel that can be heat treated in thicker sections and to higher strength levels than AISI 4130. This steel is generally used for structural machined and forged parts one-half inch and over in thickness. It can be welded but it is more difficult to weld than the lower carbon grade AISI 4130.

    OK, so agreed Fred, without a little more material information, it's hard to tell. Let me get the other source. OK, 1" round bar has
    Annealed - F_ty = 62ksi
    Normalized - F_ty = 93 ksi

    You're probably looking at a bearing strength, so for bar, forgings
    Annealed - no data
    Normalized - 158 ksi
    Normalized and heat treated (with Ftu = 140) - 225 ksi
    Normalized and heat treated (with Ftu = 160) - 252 ksi
    Normalized and heat treated (with Ftu = 180) - 284 ksi

    edit: Ian, my first source is a MIL spec manual called the MMPDS (Metallic Materials Properties Development and Standardization). It's a slightly more current version of something that was called....(something else can't remember). We just 'have' it, although I do know for fact that it is available from off of a website hosted by a Big10 school that makes a name for itself with engineering.

    I also use the ASMD (Aerospace Materials Database), which we have both in several extremely large old books, and as an online website we need to log into.
  9. Apr 29, 2009 #8
    So a value between 140-284 ksi will work as an approximation for calculations with the 2,5 mm pin?
  10. Apr 29, 2009 #9


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    I would say 158 ksi if its not heat-treated, 225 if it is.

    Since you're doing Hertzian stress, you're probably going to need elastic modulus and Poisson's huh?
  11. Apr 29, 2009 #10
    Yes please! :smile:
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