A cylindrical 1045 steel bar (Figure 8.34) is subjected to repeated tension-compression stress cycling along its axis. If the load amplitude is 22,000 N, compute the minimum allowable bar diameter to ensure that fatigue failure will not occur. Assume a factor of safety of 2.0.

Figure 8.34

1045 steel bar

F = 22,000 N

cylindrical cross-section

FS = 2.0

d\(_{min}\) such that failure does not occur

from Figure 8.34: \(\sigma_a\) << fatigue limit = 310 MPa

\( \sigma \) = \(P\over A\)FS

A = \(\pi \over 4 \) D\(^2\)

D = \( \sqrt{F (FS) \over {\pi \over 4}(\sigma)}\)

D = \( \sqrt{22,000N (2.0) \over {\pi \over 4}(310x10^6 {N \over m^2})}\)

D = 0.0134 m = 13.4 mm