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A cylindrical 4340 steel bar is subjected to reversed rotating–bending stress cycling, which yielded the test results presented in animated figure 8. 21. If the maximum applied load is 5,000 n, compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2. 25 and that the distance between loadbearing points is 55. 0 mm

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Based on the maximum applied load, the factor of safety, and the distance between the loadbearing points, the minimum allowable bar diameter is 18.6 mm.

What is the minimum allowable bar diameter?

The diameter is included in the following formula:

Maximum stress / Factor of safety = (16 x Maximum applied load x distance between loadbearing points x 10⁻¹⁰) / (π x diameter³)

Solving gives:

(490 x 10⁶) / 2.25 = (16 x 5,000 x 55.0 x 10⁻¹⁰) / (π x diameter³)

217,777,777.78 = 0.00044 /  (π x diameter³)

diameter = 18.6 mm

Find out more on diameter at https://brainly.com/question/16874040.

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