Question 2.6.12: A steel bar undergoes cyclic loading such that σmax= 60 kpsi...

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)

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

$N=\left(\frac{\sigma _{rev}}{a} \right)^{{1}/{b}}=\left(\frac{\sigma _{rev}}{129.6} \right) ^{{-1}/{0.0851}}$   (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.