Data are given in Table E14.1 for completely reversed strain-controlled fatigue tests on RQC-100 steel. The elastic modulus is E = 200 GPa, and σa and εpa are measured near Nf /2. Determine constants for the strain–life curve.

Question

Data are given in Table E14.1 for completely reversed strain-controlled fatigue tests on RQC-100 steel. The elastic modulus is E = 200 GPa, and [latex]σ_{a}[/latex] and [latex]ε_{pa}[/latex] are measured near [latex]N_{f}[/latex] /2. Determine constants for the strain–life curve.

Accepted Answer

Two least-squares fits are needed to obtain the constants [latex]σ^{\prime }_{f} , b, ε^{\prime }_{f} [/latex] , and c, one for Eq. 14.5 and one for Eq. 14.3(b). Both fits proceed in a manner similar to Ex. 9.1(b). For the first, solve Eq. 14.5 for the dependent variable 2[latex]N_{f}[/latex] , and take logarithms of both sides:

[latex]ε_{ea} = \frac{σ_{a} }{E} =\frac{σ^{\prime }_{f} }{E} (2N_{f} )^{b} , ε_{pa} = ε^{\prime } _{f } (2N_{f} )^{c} [/latex] (a, b) (14.33)

[latex]σ_{a} = σ^{\prime }_{f} (2N_{f} )^{b} [/latex] (14.5)

[latex]2N_{f} =\left(\frac{σ_{a}}{σ^{\prime } _{f} } \right)^{1/b} , log (2N_{f} ) =\frac{1}{b} log σ_{a} − \frac{1}{b} log σ^{\prime } _{f} [/latex]

This has the form y = mx + d, where

[latex]y = log (2N_{f} ), x = log σ_{a} , m =\frac{ 1}{b} , d = −\frac{ 1}{b}log σ^{\prime } _{f} [/latex]

Performing a linear least-squares fit on this basis gives

m = −17.096, d = 50.513

so that the needed constants are

[latex]b =\frac{ 1}{m} = −0.0585, σ^{\prime } _{f} = 10^{−db} = 901MPa[/latex]

Table E14.1

[latex]ε_{a}[/latex]	[latex]σ_{a}[/latex], MPa	[latex]ε_{pa}[/latex]	[latex]N_{f}[/latex] , cycles
0.0202	631	0.01695	227
0.0100	574	0.00705	1 030
0.0045	505	0.00193	6 450
0.0030	472	0.00064	22 250
0.0023	455	(0.00010)	110 000

Source: The author’s data on the ASTM Committee E9 material.

To obtain the remaining two constants, proceed similarly, solving Eq. 14.3(b) for [latex]2N_f[/latex] and taking logarithms of both sides to obtain the form y = mx + d:

[latex]2N_{f} =\left(\frac{\varepsilon _{pa}}{\varepsilon ^{\prime } _{f} } \right)^{1/c} , log (2N_{f} ) =\frac{1}{c} log \varepsilon _{pa} − \frac{1}{c} log \varepsilon ^{\prime } _{f} [/latex]

[latex]y = log (2N_{f} ), x = log \varepsilon _{pa} , m =\frac{ 1}{c} , d = −\frac{ 1}{c}log \varepsilon ^{\prime } _{f} [/latex]

However, the last [latex]ε_{pa}[/latex] value, shown in parentheses in Table E14.1, is judged to be so small that its accuracy of measurement is likely to be poor. Moreover, this point does not lie along the linear trend of the other [latex]ε_{pa}[/latex] vs. [latex]N_{f}[/latex] data on a plot similar to Fig. 14.4. Hence, the fit is done with only the first four [latex]ε_{pa}[/latex] values, giving

m = −1.3456, d = 0.39690

[latex]c = \frac{1}{m} = −0.743, ε^{\prime } _{f} = 10^{−dc} = 1.972[/latex]

Employing E = 200,000MPa as given, the strain–life curve of the form of Eq. 14.4 can finally be written:

[latex]ε_{a} =\frac{σ^{\prime } _{f}}{E} (2N_{f})^{b} + ε^{\prime } _{f} (2N_{f} )^{c} [/latex] (14.4)

[latex]ε_{a} = 0.004505 (2N_{f} )^{−0.0585 } + 1.972 (2N_{f} )^{−0.743} [/latex]

Comments The constants obtained differ somewhat from those in Fig. 14.4 due to the use of an abbreviated set of data. The given [latex]σ_{a}[/latex] and [latex]ε_{pa}[/latex] data may also be employed to fit the cyclic stress–strain curve, with the procedure being the same as in Ex. 12.1, except for the use of amplitude quantities. Again excluding the last point, the results are [latex]H^{\prime } [/latex]=900 MPa, [latex]n^{\prime } [/latex] = 0.0896.

Question 14.1: Data are given in Table E14.1 for completely reversed strain...

Related Answered Questions

Verified Answer:

A notched plate made of the AISI 4340 (aircraft quality) steel of Table 14.1 has an elastic stress concentration factor of kt = 2.80. If the nominal stress is cycled between Smax = 750 and Smin = 50 MPa, how many cycles can be applied before fatigue cracking is expected? ...

Verified Answer:

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? ...

Verified Answer: