A steel bar undergoes cyclic loading such that \sigma _{max}= 60 \ kpsi \ and \ \sigma _{min}= −20 \ kpsi. \ For \ the \ material \ , \ S_{ut}= 80 \ kpsi, \ S_y= 65 \ kpsi,
a fully corrected endurance limit of S_e=40 \ kpsi, \ and \ f = 0.9.
Estimate the number of cycles to a fatigue failure using:
(a) Modified Goodman criterion.
(b) Gerber criterion.
From the given stresses,
\sigma _{a} =\frac{60-\left(-20\right) }{2} =40 kpsi
\sigma _{m} =\frac{60+\left(-20\right) }{2} =20 kpsi
(a) For the modified Goodman criterion, Eq. (6–46), the fatigue factor of safety based on infinite life is
mod-Goodman \frac{\sigma _a}{S_e} =\frac{\sigma _m}{S_{ut} } =\frac{1}{n} (6–46)
n_f = \frac{1} {\frac{\sigma _a}{S_e} + \frac{\sigma _m}{S_{ut}} } = \frac{1}{\frac{40}{40} + \frac{20}{80} } =0.8
This indicates a finite life is predicted. The S-N diagram is only applicable for completely reversed stresses. To estimate the finite life for a fluctuating stress, we will obtain an equivalent completely reversed stress that is expected to be as damaging as the fluctuating stress. A commonly used approach is to assume that since the modified Goodman line represents all stress situations with a constant life of 10^6 cycles, other constant-life lines can be generated by passing a line through \left(s_{ut},0 \right) and a fluctuating stress point \left(\sigma _m,\sigma _a\right) .
The point where this line intersects the σaaxis represents a completely reversed stress (since at this point \sigma _m= 0 ), which predicts the same life as the fluctuating stress.
This completely reversed stress can be obtained by replacing S_e \ with \sigma _{rev} in Eq. (6–46)for the modified Goodman line resulting in
\sigma _{rev}=\frac{\sigma _m}{1-\frac{\sigma _m}{S_{ut}} } =\frac{40}{1-\frac{20}{80} } =53.3 kpsi
From the material properties, Eqs. (6–14)to (6–16), p. 285, give
a={\left(fS_{ut}\right)^2 } / {S_e} (6–14)
b=- \log \left({fS_{ut}} / {S_e} \right)/3 (6–15)
N=\left({\sigma _{rev}}/{a} \right)^{{1}/{b}}(6–16)
a=\frac{\left(fS_{ut}\right)^2 }{S_e}=\frac{\left[0.9\left(80\right) \right]^2 }{40} =129.6 kpsi
b=-\frac{1}{3} \log \left(\frac{fS_{ut}}{S_e} \right)=-\frac{1}{3} \log \left[\frac{0.9\left(80\right) }{40} \right] =-0.0851
N=\left(\frac{\sigma _{rev}}{a} \right)^{{1}/{b}}=\left(\frac{\sigma _{rev}}{129.6} \right) ^{{-1}/{0.0851}} (1)
Substituting \sigma _{rev} into Eq. (1) yields
N=\left(\frac{53.3}{129.6} \right) ^{{-1}/{0.0851}}=3.4\left(10^4\right) cycles
(b) For Gerber, similar to part (a), from Eq. (6–47),
Gerber \frac{n\sigma _a}{S_e} +\left(\frac{n\sigma _m}{S_{ut}} \right)^2 =1
\sigma _{rev}=\frac{\sigma _a}{1-\left(\frac{\sigma _m}{S_{ut}} \right)^2 } =\frac{40}{1-\left(\frac{20}{80} \right)^2 } =42.7 kpsi
Again, from Eq. (1),
N=\left(\frac{42.7}{129.6} \right)^{{-1}/{0.0851}}=4.6\left(10^5\right) cycles
Comparing the answers, we see a large difference in the results.
Again, the modified Goodman criterion is conservative as compared to Gerber for which the moderate difference in S_{f} is then magnified by a logarithmic S, N relationship.