Question 6.4: Derive Eq. (6–17). For the specimen of Prob. 6–3, estimate t......

From  S_{f}=a N^{b}

\log S_{f}=\log a+b\log N

Substituting (1,\,S_{u t})

\log S_{u t}=\log a+b\log\left(1\right)

From which     a=S_{u t}

Substituting (10^{3},\,f S_{u t})\,\mathrm{and}a=S_{u t}

\log f S_{u t}=\log S_{u t}+b\log10^{3}

From which

b={\frac{1}{3}}\log f

\therefore S_{f}=S_{u t}N^{(\log f)/3}\qquad1\leq N\leq10^{3}

For 500 cycles as in Prob. 6-3

S_{f}\geq66.2(500)^{(\mathrm{log}\,0.8949)/3}=59.9\ \mathrm{kpsi}