Cold formable spring steel wire excellent in cold cutting capability and fatigue properties and manufacturing process thereof

摘要

Disclosed is a cold formable spring steel wire excellent in cold cutting capability and fatigue properties, in which the steel wire satisfies given composition, has an average globular carbide particle size [√(ab)]:1.0 μm or less with aspect ratio (a/b, a: major axis of carbide, b: minor axis of carbide) being 2 or less, a ratio (area %) of the globular carbide in the steel: (0.1 to 3)×amount (mass %) of C in the steel, an amount (mass %) of Cr in the globular carbide: [0.4×amount (mass %) of Cr in the steel] or less, hardenability factor (Dic): between 110 mm and 450 mm, and tensile stress of 2000 MPa or more.

主权项

1. A spring steel wire, comprising the following composition:
C: 0.45-0.70% (“%” herein means “mass %”), Si: 1.9-2.5%, Mn: 0.15-1.0%, and Cr: 1.0-2.0%, with the proviso that Cr+Si: 3.5% or more, and Cr/Si: 0.95 or less; wherein: P: 0.015% or less (exclusive of 0%); S: 0.015% or less (exclusive of 0%); wherein the metallic structure of the steel obtained from said composition satisfies: an average globular carbide particle size with aspect ratio (a/b, a: major axis of carbide, b: minor axis of carbide) being 2 or less [√(ab)]: 1.0 μm or less; a ratio (area %) of the globular carbide in the steel: (0.1 to 3)×amount (mass %) of C in the steel; an amount (mass %) of Cr in the form of Cr carbide in the composition: [0.006×amount (mass %) of Cr in the steel] to [0.4×amount (mass %) of Cr in the steel]; tensile strength: 2100 MPa or more; and hardenability factor (Dic) represented by the following formulas (1)-(3): 110 mm≦Dic≦450 mm, in the case where the C content is not less than 0.45% nor greater than 0.55%,Dic=25.4×(0.171+0.001⁡[C]+0.265⁡[C]2)×(3.3333⁡[Mn]+1.0)×(1.0+0.7⁡[Si])×(1.0+0.363⁡[Ni])×(1.0+2.16⁡[Cr])×(1.0+0.365⁡[Cu])×(1.0+1.73⁡[V])(1) in the case where the C content is greater than 0.55% but not greater than 0.65%,Dic=25.4×(0.115+0.268⁡[C]-0.038⁡[C]2)×(3.3333⁡[Mn]+1.0)×(1.0+0.7⁡[Si])×(1.0+0.363⁡[Ni])×(1.0+2.16⁡[Cr])×(1.0+0.365⁡[Cu])×(1.0+1.73⁡[V]),and(2) in the case where the C content is greater than 0.65% but not greater than 0.70%,Dic=25.4×(0.143+0.2⁡[C]×(3.3333⁡[Mn]+1.0)×(1.0+0.7⁡[Si])×(1.0+0.363⁡[Ni])×(1.0+2.16⁡[Cr])×(1.0+0.365⁡[Cu])×(1.0+1.73⁡[V])(3) (in which, [C], [Mn], [Si], [Ni] [Cr], [Cu], and [V] represent an amount (mass %) of each element in the steel).