Goodman Diagram Design Example

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  • เผยแพร่เมื่อ 6 ก.ค. 2024
  • This walks through a basic design example of a fluctuating axial stress on a steel pin at elevated, non-zero mean stress, using the modified Goodman diagram to describe fatigue failure and factor of safety.

ความคิดเห็น • 29

  • @Soffex
    @Soffex 9 ปีที่แล้ว +10

    I can't thank you enough for spending the time to put together such a thorough explanation of a Goodman diagram. It's been a while since my machine design course in college and this was a great refresher. Keep up the good work

    • @tc285202
      @tc285202  9 ปีที่แล้ว +1

      Speeding944T Cheers! Glad it was helpful!

  • @hrishikeshmurgudkar8266
    @hrishikeshmurgudkar8266 9 ปีที่แล้ว

    Thank you very much Dr. Cyders for sharing this knowledge by simplifying it to very basic level.

  • @satyanchandra6348
    @satyanchandra6348 8 ปีที่แล้ว

    Dr. Cyders, thanks for the demo - I had forgotten some of this, and was a great review.

    • @tc285202
      @tc285202  8 ปีที่แล้ว

      +Satyan Chandra Cheers! Glad it was helpful

  • @rezaafshar3832
    @rezaafshar3832 9 ปีที่แล้ว

    Thank you very much for sharing the knowledge!..it is a very clear explanation of the Goodman diagram.

  • @riding.through
    @riding.through 6 ปีที่แล้ว

    Thank you Dr. Cyders!

  • @isagumus1
    @isagumus1 5 ปีที่แล้ว

    You are an amazing one sir ! Thank you so much for your well explanation !

  • @garethmoore9417
    @garethmoore9417 7 ปีที่แล้ว

    Excellent lecture on fatigue, really helped me understand the chapter in the Shigley textbook. First rate!!

  • @sister1ist
    @sister1ist 7 ปีที่แล้ว

    Great job Sir .i wish i could put more likes for your videos ! they are helping me a lot.

  • @emadmazhari1538
    @emadmazhari1538 2 ปีที่แล้ว

    The video is super useful. Thank You!

  • @RobleroMari
    @RobleroMari 7 ปีที่แล้ว

    Thanks Dr. Cyders

  • @eng9hussein
    @eng9hussein 7 ปีที่แล้ว

    thank you doctor

  • @miguelaphan58
    @miguelaphan58 5 ปีที่แล้ว

    ..exelent job, indeed thanks so much

  • @zfyl
    @zfyl 7 ปีที่แล้ว +1

    Hi, im from Europe and my question is: how can be the safety factor be as high as 2,5,8? Or maybe just the difference of EU/US standards and i just need to use in EU the reciprocal of these safety factors?

  • @micaelav5973
    @micaelav5973 7 ปีที่แล้ว +1

    Please can you explain me the origin of the equation σ=256963N^-0.146

  • @danpt2000
    @danpt2000 6 ปีที่แล้ว +1

    I m confused as to how the Yield Line is determined. The value of Sigma, sub m I understand, but the sigma a of the Yield Line, I'm not sure how that value was determined.

  • @Ayazmpengr
    @Ayazmpengr 8 ปีที่แล้ว +1

    Nice video Sir.. I don't understand one point.
    How you have calculated the value of FOS=6.4 @ D=0.533in??

    • @Ayazmpengr
      @Ayazmpengr 8 ปีที่แล้ว +1

      +Ayaz Mehmood: N=S_f(N)/Sig_a =35900/5650=6.4 ..? Is it right?

  • @xtkj301x
    @xtkj301x 9 ปีที่แล้ว

    That it means the @ symbol?. I've seen in several books.

    • @Oxupexu
      @Oxupexu 5 ปีที่แล้ว

      You mean lowercase Sigma? It is the symbol for normal stress. Lowercase Tau is the symbol for shear stress.

  • @josephk2414
    @josephk2414 5 ปีที่แล้ว

    why the volume is low?

  • @sudhansugrahacharya7094
    @sudhansugrahacharya7094 3 ปีที่แล้ว

    hii sir, how to find the correction factors for size, loading, Sfinsh, temp and reliability

    • @brandynbailey7828
      @brandynbailey7828 3 ปีที่แล้ว

      If you look in Norton or Shigley there are formulas that you can follow and make the right decision

  • @johncullen6157
    @johncullen6157 8 ปีที่แล้ว

    Can anyone share how C(reliability) of 95% comes out to .868 not .950?
    Thanks

    • @tc285202
      @tc285202  8 ปีที่แล้ว +2

      +John Cullen The reliability factor comes from the normal standard distribution - some data has shown that strength varies stochastically with a standard deviation of about 8% of the mean value. If you use that standard deviation, you can calculate how much you need to reduce the stress level to maintain a 95% confidence that you're below the actual strength of the material. A single-tailed 95% confidence interval with a standard deviation of 0.08 puts you at a correction factor of 0.868. If you google around, you'll find more complete explanations with graphics.

    • @AnnLaustsen87
      @AnnLaustsen87 8 ปีที่แล้ว

      +Dr. Cyders Thank you!!!

    • @johncullen6157
      @johncullen6157 8 ปีที่แล้ว

      +Dr. Cyders Thanks! that helps a lot

  • @angeljohanngarcia
    @angeljohanngarcia 7 ปีที่แล้ว

    Thanks Dr. Cyders