A 1.5-in-diameter bar has been machined from an AISI 1050 cold-drawn bar. This part is to withstand a fluctuating tensile load varying from 0 to 16 kip. Because of the ends, and the fillet radius, a fatigue stress-concentration factor K_{f} is 1.85 for 10^{6} or larger life. Find S_{a} \text { and } S_{m} and the factor of safety guarding against fatigue and first-cycle yielding, using (a) the Gerber fatigue line and (b) the ASME-elliptic fatigue line.
Question : A 1.5-in-diameter bar has been machined from an AISI 1050 co...
We begin with some preliminaries. From Table A–20, S_{u t}=100 kpsi \text { and } S_{y}=84 kpsi \text { . } Note that F_{a}=F_{m}=8 \text { kip. } The Marin factors are, deterministically, k_{a}=2.70(100)^{-0.265}=0.797: Eq. (6–19), Table 6–2, k_{b}=1 \text { (axial loading, see } k_{c} \text { ) }
k_{a}=a S_{u t}^{b} (6–19)
k_{c}=0.85: \text { Eq. }(6-26)
k_{d}=k_{e}=k_{f}=1
S_{e}=0.797(1) 0.850(1)(1)(1) 0.5(100)=33.9 kpsi \text { : } Eqs. (6–8), (6–18), The nominal axial stress components \sigma_{a o} \text { and } \sigma_{m o} \text { are }
S_{e}^{\prime}=\left\{\begin{array}{ll}0.5 S_{u t} & S_{u t} \leq 200 kpsi (1400 MPa ) \\100 kpsi & S_{u t}> kpsi \\700 MPa & S_{u t}>1400 MPa\end{array}\right. (6–8)
S_{e}=k_{a} k_{b} k_{c} k_{d} k_{e} k_{f} S_{e}^{\prime} (6–18)
\sigma_{a o}=\frac{4 F_{a}}{\pi d^{2}}=\frac{4(8)}{\pi 1.5^{2}}=4.53 \text { kpsi } \quad \sigma_{m o}=\frac{4 F_{m}}{\pi d^{2}}=\frac{4(8)}{\pi 1.5^{2}}=4.53 kpsi
Applying K_{f} to both components \sigma_{a o} \text { and } \sigma_{m o} constitutes a prescription of no notch yielding:
\sigma_{a}=K_{f} \sigma_{a o}=1.85(4.53)=8.38 kpsi =\sigma_{m}
(a) Let us calculate the factors of safety first. From the bottom panel from Table 6–7 the factor of safety for fatigue is
n_{f}=\frac{1}{2}\left(\frac{100}{8.38}\right)^{2}\left(\frac{8.38}{33.9}\right)\left\{-1+\sqrt{1+\left[\frac{2(8.38) 33.9}{100(8.38)}\right]^{2}}\right\}=3.66
From Eq. (6–49) the factor of safety guarding against first-cycle yield is
\sigma_{a}+\sigma_{m}=\frac{S_{y}}{n} (6–49)
n_{y}=\frac{S_{y}}{\sigma_{a}+\sigma_{m}}=\frac{84}{8.38+8.38}=5.01
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}=O B / O A From the first panel of Table 6–7, r=\sigma_{a} / \sigma_{m}=1 \text { , }
S_{a}=\frac{(1)^{2} 100^{2}}{2(33.9)}\left\{-1+\sqrt{1+\left[\frac{2(33.9)}{(1)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, n_{f}=O B / O A=S_{a} / \sigma_{a}=S_{m} / \sigma_{m}=30.7 / 8.38=3.66 and we see total agreement.
We could have detected that fatigue failure would occur first without drawing Fig. 6–28 by calculating r_{\text {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(33.9)}\left[1-\sqrt{1+\left(\frac{2(33.9)}{100}\right)^{2}\left(1-\frac{84}{33.9}\right)}\right]=64.0 kpsi
S_{a}=S_{y}-S_{m}=84-64=20 kpsi
The critical slope is thus
r_{\text {crit }}=\frac{S_{a}}{S_{m}}=\frac{20}{64}=0.312
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
n_{f}=\sqrt{\frac{1}{(8.38 / 33.9)^{2}+(8.38 / 84)^{2}}}=3.75
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{(1)^{2} 33.9^{2}(84)^{2}}{33.9^{2}+(1)^{2} 84^{2}}}=31.4 kpsi , \quad 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_{\text {crit }}. From the third row second column panel of Table 6–8,
S_{a}=\frac{2(84) 33.9^{2}}{33.9^{2}+84^{2}}=23.5 kpsi , \quad S_{m}=S_{y}-S_{a}=84-23.5=60.5 kpsi
r_{\text {crit }}=\frac{S_{a}}{S_{m}}=\frac{23.5}{60.5}=0.388
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.
