1 / 16

ME 3180B - Mechanical Engineering Design - Spring 2005

ME 3180B - Mechanical Engineering Design - Spring 2005. Example #3 (example 7-26 in text). Example #3 : Problem 6-23 Shigley. Given:

jonco
Download Presentation

ME 3180B - Mechanical Engineering Design - Spring 2005

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. ME 3180B - Mechanical Engineering Design - Spring 2005 Example #3 (example 7-26 in text)

  2. Example #3 : Problem 6-23 Shigley • Given: • The figure shows the free-body diagram of a connecting-link portion having stress concentration at three places. The dimensions are: r = 0.25in, d = 0.75in, h = 0.50in, w1=3.75in, and w2 = 2.5in. The forces F fluctuate between a tension of 4 Kip and a compression of 16 Kip. • Neglect column action. Find the least factor of safety if material is cold-drawn AISI 1018 steel.

  3. Example #3: Problem 6-23 Shigley • Solution: • From Table A-20 (Shiggley) • Sut = 64 Kpsi • Syt = 54 Kpsi • From your note: • S’e = 0.504 Sut = 32.3Kpsi • (Note: You could use S’e =0.45Sut in all your calculations. If you do, then set CL = 1 and you will still obtain the same result) • Table 7-4 in your note: • a = 2.7Kpsi, b = -0.265 • CF = Ka = aSbut = 2.7 Kpsi (64Kpsi)-0.265 • CF = Ka = 0.897 • Cs = Kb =1, CL = Kc = 0.923 (Equation 7-22) • Se = CFCsCILCRCTKeS’e (Equation 7-13) = 0.897(1)(0.923)(32.3) Kpsi (The value for Ke is not included here, you can include it if you want to) = 26.7 kpsi

  4. Example #3 : Problem 6-23 Shigley • Solution: See Figure A-15-5 Shigley • Check for yielding at the fillet • If you use 4 Kip, n will be > 4.22. • 4 Kpsi/(2.5*0.5) = 3.2Kpsi • n = 54/3.2 = 16.9 Norton: Fig E-9 (page 998)

  5. Example #3 : Problem 6-23 Shigley • Solution: • Check for fatigue Failure ( Calculate Kt) • Equation 5-26: • Kf = 1 + q(Kt-1) • Figure 9-2: (use r = 0.25in) • q = 0.78 (when r ≥0.16”, use the value at r = 0.16”)

  6. Example #3 : Problem 6-23 Shigley • Solution:

  7. Example #3 : Problem 6-23 Shigley • Solution: • Check for failure due to Soderberg, Goodman and Gerber failure criteria. • Soderberg

  8. Example #3 : Problem 6-23 Shigley • Solution: • Goodman

  9. Example #3 : Problem 6-23 Shigley • Solution: • Gerber

  10. Example #3 : Problem 6-23 Shigley • Solution: See Figure A-15-1 Page 1006 Shigley • Check for yielding at hole

  11. Example #3 : Problem 6-23 Shigley • Solution: • Check for fatigue failure (Calculate Kt ) • q = 0.78 (use r = 0.375’’ in fig. 5-16) Kf = 1+ 0.78(2.5-1) = 2.17 use 0.20 in Fig. A-15-1 or Fig (E-13) to obtain Kt

  12. Example #3 : Problem 6-23 Shigley • Solution: • Let us see what we get with Soderberg Relation • You can do the same for the other two-Gerber and Goodman. • Failure will occur at the discontinuity with the smallest factor of safety

  13. Example #3

  14. Example #3

  15. Example #3

  16. Example #3

More Related