Question 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,

Table A–20   Deterministic ASTM Minimum Tensile and Yield Strengths for Some Hot-Rolled (HR) and Cold-Drawn (CD) Steels [The strengths listed are estimated ASTM minimum values in the size range 18 to 32 mm ( \frac {3}{4}  to 1\frac {1}{4} in). These strengths are suitable for use with the design factor defined in Sec. 1–10, provided the materials conform to ASTM A6 or A568 requirements or are required in the purchase specifications. Remember that a numbering system is not a specification.] Source: 1986 SAE Handbook, p. 2.15.

k_{a} = 2.70(100)^{−0.265} = 0.797: Eq. (6–19), Table 6–2, p. 279
k_{b} = 1 (axial loading, see k_{c})

Table 6–2   Parameters for Marin Surface Modification Factor, Eq.(6–19)

The nominal axial stress components σ_{ao} and σ_{mo} are

σ_{ao} =\frac {4F_{a}}{πd^{2}} =\frac {4(8)}{π1.5^{2}} = 4.53 kpsi             σ_{mo} =\frac {4F_{m}}{πd^{2 }}=\frac {4(8)}{π1.5^{2}} = 4.53  kpsi

Applying K_{f} to both components σ_{ao} and σ_{mo} constitutes a prescription of no notch yielding:

σ_{a} = K_{f}  σ_{ao} = 1.85(4.53) = 8.38  kpsi = σ_{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

Table 6–7  Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for Gerber and Langer Failure Criteria

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

Langer static yield σ_{a} + σ_{m} =\frac {S_{y}}{n}             (6–49)

n_{y} =\frac {S_{y}}{σ_{a} + σ_{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} = OB/OA.

From the first panel of Table 6–7, r = σ_{a}/σ_{m} = 1,

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}/σ_{a} = S_{m}/σ_{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_{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_{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

Table 6–8   Amplitude and Steady Coordinates of Strength and Important Intersections in First Quadrant for ASMEElliptic and Langer
Failure Criteria

S_{a} =\sqrt {\frac {(1)^{2}33.9^{2}(84)^{2}}{33.9^{2} + (1)^{2}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}/σ_{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(84) 33.9^{2}}{33.9^{2} + 84^{2}} = 23.5 kpsi,  S_{m} = S_{y} − S_{a} = 84 − 23.5 = 60.5   kpsi

r_{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.

17  Use this only for pure torsional fatigue loading. When torsion is combined with other stresses, such as bending, k_{c} = 1 and the combined loading is managed by using the effective von Mises stress as in Sec. 5–5. Note: For pure torsion, the distortion energy predicts that (k_{c})_{torsion} = 0.577.