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.
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