The RQC-100 steel of Table 14.1 is subjected to cycling with a strain amplitude of εa = 0.004 and a tensile mean stress of σm = 100MPa. How many cycles can be applied before fatigue cracking is expected?

Question

The RQC-100 steel of Table 14.1 is subjected to cycling with a strain amplitude of [latex]ε_{a}[/latex] = 0.004 and a tensile mean stress of [latex]σ_{m}[/latex] = 100MPa. How many cycles can be applied before fatigue cracking is expected?

Table 14.1 Cyclic Stress–Strain and Strain–Life Constants for Selected Engineering Metals.¹

Material	Source	Tensile Properties				Cyclic σ-ε Curve			Strain–Life Curve
		[latex] σ_{o} [/latex]	[latex] σ_{u} [/latex]	[latex]\overset{\sim }{\sigma }f_{B} [/latex]	% RA	E	[latex]H^{\prime } [/latex]	[latex]n^{\prime } [/latex]	[latex]σ^{\prime }_{f} [/latex]	b	[latex]ε^{\prime }_{f} [/latex]	c
(a) Steels
SAE 1015 (normalized)	(8)	228 (33.0)	415 (60.2)	726 (105)	68	207,000 (30,000)	1349 (196)	0.282	1020 (148)	−0.138	0.439	−0.51
Man-Ten² (hot rolled)	(7)	322 (46.7)	557 (80.8)	990 (144)	67	203,000 (29,500)	1096 (159)	0.187	1089 (158)	−0.115	0.912	−0.606
RQC-100 (roller Q & T)	(2)	683 (99.0)	758 (110)	1186 (172)	64	200,000 (29,000)	903 (131)	0.0905	938 (136)	−0.0648	1.38	−0.704
SAE 1045 (HR & norm.)	(6)	382 (55.4)	621 (90.1)	985 (143)	51	202,000 (29,400)	1258 (182)	0.208	948 (137)	−0.092	0.26	−0.445
SAE 4142 (As Q, 670 HB)	(1)	1619 (235)	2450 (355)	2580 (375)	6	200,000 (29,000)	2810 (407)	0.04	2550 (370)	−0.0778	0.0032	−0.436
SAE 4142 (Q & T, 560 HB)	(1)	1688 (245)	2240 (325)	2650 (385)	27	207,000 (30,000)	4140 (600)	0.126	3410 (494)	−0.121	0.0732	−0.805
SAE 4142 (Q & T, 450 HB)	(1)	1584 (230)	1757 (255)	1998 (290)	42	207,000 (30,000)	2080 (302)	0.093	1937 (281)	−0.0762	0.706	−0.869
SAE 4142 (Q & T, 380 HB)	(1)	1378 (200)	1413 (205)	1826 (265)	48	207,000 (30,000)	2210 (321)	0.133	2140 (311)	−0.0944	0.637	−0.761
AISI 4340² (Aircraft Qual.)	(3)	1103 (160)	1172 (170)	1634 (237)	56	207,000 (30,000)	1655 (240)	0.131	1758 (255)	−0.0977	2.12	−0.774
AISI 4340 (409 HB)	(1)	1371 (199)	1468 (213)	1557 (226)	38	200,000 (29,000)	1910 (277)	0.123	1879 (273)	−0.0859	0.64	−0.636
Ausformed H-11 (660 HB)	(1)	2030 (295)	2580 (375)	3170 (460)	33	207,000 (30,000)	3475 (504)	0.059	3810 (553)	−0.0928	0.0743	−0.7144
(b) Other Metals
2024-T351 Al (Prestrained)	(1)	379	469	558	25	73,100	662	0.07	927	−0.113	0.409	−0.713
2024-T4 Al3 (soln. tr. & age)	(4)	303	476	631	35	73,100	738	0.08	1294	−0.142	0.327	−0.645
7075-T6 Al	(5)	469	578	744	33	71,000	977	0.106	1466	−0.143	0.262	−0.619
Ti-6Al-4V (soln. tr. & age)	(1)	1185	1233	1717	41	117,000	1772	0.106	2030	−0.104	0.841	−0.688
Inconel X (Ni base, annl.)	(1)	703	1213	1309	20	214,000	1855	0.12	2255	−0.117	1.16	-0.749

Notes: ¹The tabulated values either have units of MPa (ksi), or they are dimensionless. ²Test specimens prestrained, except at short lives, also periodically overstrained at long lives. 3For nonprestrained tests, use same constants, except [latex]σ^{\prime }_{f} [/latex] = 900(131) and b = −0.102.
Sources: Data in (1) [Conle 84]; (2) author’s data on the ASTM Committee E9 material; (3) [Dowling 73]; (4) [Dowling 89] and [Topper 70]; (5) [Endo 69] and [Raske 72]; (6) [Leese 85]; (7) [Wetzel 77] pp. 41 and 66; (8) [Keshavan 67] and [Smith 70].

Accepted Answer

Of the various methods given, we will first apply the Morrow equation.
Substituting the constants E, [latex]σ^{\prime } _{f} , b, ε^{\prime } _{f} [/latex], and c from Table 14.1 for RQC-100 steel into Eq. 14.20, we have

[latex]ε_{a } =\frac{σ^{\prime } _{f} }{E} (2N^{∗} )^{b} + ε^{\prime }_{f} (2N^{∗})^{c} [/latex] (14.20)

[latex]ε_{a} = \frac{938}{200,000} (2N^{∗})^{−0.0648} + 1.38(2N^{∗})^{−0.704} [/latex]

Using the given [latex]ε_{a}[/latex] = 0.004 and solving numerically for [latex]N^{∗}[/latex] gives

[latex]N^{∗}[/latex] = 8124 cycles

Since [latex]N^{∗}[/latex] does not include the effect of mean stress, its value must be employed along with [latex]σ_{m}[/latex] = 100MPa to obtain an [latex]N_{f}[/latex] value that does include this effect. Hence, we apply [latex]N^{∗}[/latex] in Eq. 14.23, specifically as [latex]N^{∗}_{mi }[/latex]

[latex]N_{f} = N^{∗}_{mi} \left(1 −\frac{ σ_{m}}{σ^{\prime } _{f} } \right) ^{−1/b} [/latex] (14.23)

[latex]N_{f} = N^{∗}_{mi} \left(1 −\frac{ σ_{m}}{σ^{\prime } _{f} } \right) ^{−1/b} = 8124\left(1 − \frac{100}{938} \right) ^{1/0.0648} = 1426 [/latex] cycles
Second To apply the modified Morrow approach, simply substitute the same material constants into Eq. 14.27:

[latex]ε_{a} =\frac{σ^{\prime } _{f} }{E} \left(1 − \frac{ σ_{m} }{σ^{\prime } _{f}}\right) (2N_{f} )^{b} + ε^{\prime } _{f} (2N_{f} )^{c} [/latex] (14.27)

[latex]ε_{a} =\frac{ 938}{200,000} \left(1 − \frac{σ_{m} }{938}\right) (2N_{f} )^{−0.0648} + 1.38\left(N_{f} \right) ^{−0.704} [/latex]

Substituting [latex]σ_{m}[/latex] = 100MPa and simplifying gives

[latex]ε_{a } = \frac{838}{200,000} (2N_{f})^{−0.0648 } + 1.38(2N_{f})^{−0.704} [/latex]

We then enter this equation with [latex]ε_{a}[/latex] = 0.004 and solve numerically for [latex]N_{f}[/latex] , obtaining

[latex]N_{f}[/latex] = 6597 cycles

Third For the SWT approach, Eq. 14.30, we need the product [latex]σ_{max}ε_{a}[/latex]. Thus, first apply the cyclic stress–strain curve with constants from Table 14.1 to obtain [latex]σ_{a}[/latex]:

[latex]σ_{max} ε_{a} =\frac{\left(σ^{\prime } _{f} \right) ^{2} }{E} (2N_{f} )^{2b} + σ^{\prime } _{f} ε^{\prime } _{f} (2N_{f} )^{b+c} [/latex] (14.30)

[latex]ε_{a} = \frac{σ_{a} }{E} +\left( \frac{σ_{a} }{H^{\prime }}\right) ^{1/n^{\prime }} = \frac{σ_{a}}{200,000} +\left(\frac{ σ_{a} }{903}\right) ^{1/0.0905} [/latex]

Entering this equation with [latex]ε_{a}[/latex] = 0.004 and solving numerically for [latex]σ_{a}[/latex] gives

[latex]σ_{a}[/latex] = 501.2MPa, [latex]σ_{max}[/latex] = [latex]σ_{m}[/latex] + [latex]σ_{a}[/latex] = 100 + 501.2 = 601.2MPa

[latex]σ_{max}ε_{a}[/latex] = 601.2(0.004) = 2.4046MPa

Next, we substitute material constants into Eq. 14.30.

[latex]σ_{max} ε_{a} = \frac{(938)^{2} }{200,000} \left(2N_{f}\right) ^{2(−0.0648) } + (938)(1.38)(2N_{f} )^{−0.0648−0.704} [/latex]

[latex]σ_{max} ε_{a} = 4.399 \left(2N_{f}\right) ^{−0.1296 } + 1294(2N_{f} )^{−0.7688} [/latex]

Entering this equation with [latex]σ_{max} ε_{a}[/latex] = 2.4046MPa and solving numerically yields [latex]N_{f}[/latex] :

[latex]N_{f}[/latex] = 5088 cycles

Fourth For the Walker method, the constant γ can be estimated for this steel from Eq. 9.20. With [latex]σ_{u}[/latex] from Table 14.1, we obtain

γ = −0.000200 [latex]σ_{u}[/latex] + 0.8818 = −0.000200 (758MPa) + 0.8818 = 0.7302

Then apply [latex]N^{∗}[/latex] from the first solution, but now specifically as [latex]N^{∗} _{w}[/latex]. Since [latex]σ_a[/latex] and [latex]σ_{max}[/latex] are available from the third solution, form (a) of Eq. 14.32 is convenient. Solving for [latex]N_{f}[/latex] and substituting the needed values gives

[latex]N_{f} = N^{∗}_{w} \left(\frac{σ_{a}}{σ_{max} } \right) ^{(1−γ )/b} N^{∗}_{w} \left(\frac{1 - R}{2 } \right) ^{(1−γ )/b}[/latex] , (a, b) (14.32)

[latex] N_{f} = N^{∗}_{w} \left(\frac{σ_{a}}{σ_{max} } \right) ^{−(1−γ )/b} = 8124\left(\frac{501.2}{601.2} \right) ^{−(1−0.7302)/(−0.0648)} = 3809 [/latex]cycles

Comment In this example, the Morrow and modified Morrow calculations give values that differ considerably due to the relatively short life involved. The values from the SWT and Walker methods lie between the two Morrow estimates, with the Walker estimate likely being the most accurate of the four.

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