Question

The SAE 1015 steel of Table 9.1 is subjected to a stress amplitude of ?a = 160MPa.
Estimate the life for mean stresses ?m of (a) zero, (b) 70 MPa tension, and (c) 70 MPa
compression using:

(a) Morrow equation
(b) SWT equation
(c) Walker equation with ? = 0.735

Chapter 9 Fatigue of Materials: Introduction and Stress-Ra Table 9.1 Constants for Stress-Life Curves for Various Ductile Engineering Metals, From Tests at Zero Mean Stress on Unnotched Axial Specimens Yield Ultimate True Fracture ?? 0K2M)b-ANe Strength Strength Strength Material ?? ?fB o. a) Steels SAE 1015 1020 927 -0.138 (148) (134) 1089 1006 -0.115 (158) (146) 938 897 -0.0648 (136) (131) 1937 1837 -0.0762 (281) (266) 1758 1643 -0.0977 (255) (238) 415 726 228 (33) (60.2) 322 (46.7) (80.8) 683 (99.0) (110) 1584 (230) 1103 (160) (normalized) 990 (144) 1186 (172) 1998 (290) 1634 (237) Man-Ten 557 (hot rolled) RQC-100 758 (roller Q & T) 1757 (255) 1172 (170) SAE 4142 (Q & T, 450 HB) AISI 4340 (aircraft quality) (b) Other Metals 2024-T4 Al 303 (44.0) (69.0) 476 631 900 839 -0.102 (91.5) (131) (122) 2030 1889 -0.104 (295) (274) Ti-6A1-4V 1185 1233 (solution treated (172) (179) and aged) Notes: The tabulated values have units of MPa (ksi), except for dimensionless bB See Table 14.1 for sources and additional properties.

Answer 1

Given SAE 1015 steel amplitude sigma_a = 200 mpa sigma_m = 0 we know that marrow equation sigma_a/se + sigma_m/sigma_f = 1 where sigma_a = amplitude stress se = Endurance limit sigma_m = mean stress sigma_f = True feature stress sigma_f = 726 (from table) substituting all the values we get 200 mpa/se + D/726 = 1 se = 200 Mpa sigma_m = 70 mpa tension 200/se + 70 mpa/726 = 1 se = 224.76 mpa Three empirical equations are only apply for tensile stresses plot for compressive stresses at given is not clear to solve for life of material