Question 2.6.10: A 1.5-in-diameter bar has been machined from an AISI 1050 co...

We begin with some preliminaries. From Table A–20,

S_{ut}= 100 \ kpsi \ and \ S_y= 84 kpsi.
Note that F_a= F_m= 8 kip.

The Marin factors are, deterministically,
ka= 2.70(100)^{−0.265}= 0.797 : Eq. (6–19)

, Table 6–2,p. 288

k_b= 1 (axial loading, see k_c)

k_c=0.85 : Eq. (6–26), p. 290

S_e=0.797\left(1\right) 0.850\left(1\right) \left(1\right) \left(1\right) 0.5\left(100\right) =33.9 kpsi: Eqs. (6–8), (6–18), p. 282, p. 287
The nominal axial stress components

(6–8)

(6–18)

\sigma _{a0} \ and \ \sigma _{m0} are \sigma_{a0}=\frac{4F_a}{\pi d^2} =\frac{4\left(8\right) }{\pi 1.5^2}=4.53 kpsi   \sigma _{m0}=\frac{4F_m}{\pi d^2} =\frac{4\left(8\right) }{\pi 1.5^2} =4.53 kpsi.

Applying  K_f  to both components \sigma_{a0} and \sigma _{m0} constitutes a prescription of no notch
yielding:

(a) Let us calculate the factors of safety first. From the bottom panel from Table 6–7

the factor of safety for fatigue is:

From Eq. (6–49) the factor of safety guarding against first-cycle yield is

Thus, we see that fatigue will occur first and the factor of safety is 3.68. This can be seen in Fig. 6–28 where the load line intersects the Gerber fatigue curve first at point B.
If the plots are created to true scale it would be seen that

n_f = {OA}/{OB}.

From the first panel of Table 6–7, r =\sigma _a/\sigma _m= 1,

S_a=\frac{\left(1\right)^2100^2 }{2\left(33.9\right) } \left\{-1+\sqrt{1+\left[\frac{2\left(33.9\right) }{\left(1\right)100 } \right]^2 } \right\} =30.7 kpsi

S_m=\frac{S_a}{r} =\frac{30.7}{1} =30.7 kpsi

As a check on the previous result,

and we see total agreement.
We could have detected that fatigue failure would occur first without drawing Fig. 6–28 by calculating r_{crit}  .

From the third row third column panel of Table 6–7, the intersection point between fatigue and first-cycle yield is

S_m=\frac{100^2}{2\left(33.9\right) } \left[1-\sqrt{1+\left(\frac{2\left(33.9\right) }{100} \right)^2\left(1-\frac{84}{33.9} \right) } \right] =64.0 kspi

S_a=S_y-S_m=84-64=20 kpsi

The critical slope is thus

which is less than the actual load line of r = 1.

This indicates that fatigue occurs before first-cycle-yield.
(b) Repeating the same procedure for the ASME-elliptic line, for fatigue

Again, this is less than n_y= 5.01 and fatigue is  predicted to occur first.

From the first row second column panel of Table 6–8

, with r = 1 , we obtain the coordinates S_a \ and \ S_m of point B in Fig. 6–29 as

S_a=\sqrt{\frac{\left(1\right)^233.9^2\left(84\right)^2 }{33.9^2+\left(1\right)84^2 } } =31.4 kpsi

, S_m=\frac{S_a}{r} =\frac{31.4}{1} =31.4 kpsi

To verify the fatigue factor of safety, n_f={S_a}/{\sigma _a}={31.4}/{8.38}=3.75.

As before, let us calculate r_{crit}.

From the third row second column panel of Table 6–8,

S_a=\frac{2\left(84\right)33.9^2 }{33.9^2+84^2}=23.5  ,  S_m=S_y-S_a=84-23.5=60.5 kpsi

which again is less than r = 1, verifying that fatigue occurs first with n_f = 3.75.
The Gerber and the ASME-elliptic fatigue failure criteria are very close to each other and are used interchangeably. The ANSI/ASME Standard B106.1M–1985 uses ASME-elliptic for shafting.