ME 3180B - Mechanical Engineering Design - Spring 2005

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

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