Steel for low-temperature service having excellent surface processing quality
Abstract
Steel for low-temperature service having a high degree of surface processing quality comprises: manganese (Mn): 15 wt % to 35 wt %, carbon (C) satisfying conditions of: 23.6C+Mn≧28 and 33.5C−Mn≦23, copper (Cu): 5 wt % or less (excluding 0 wt %), nitrogen (N): 1 wt % or less (excluding 0 wt %), chromium (Cr) satisfying a condition of: 28.5C+4.4Cr≦57, nickel (Ni): 5 wt % or less, molybdenum (Mo): 5 wt % or less, silicon (Si): 4 wt % or less, aluminum (Al): 5 wt % or less, and a balance of iron (Fe) and inevitable impurities. Stacking fault energy (SFE) of the steel is 24 mJ/m 2 or greater. The SFE is calculated by a formula: SFE (mJ/m 2 )=1.6Ni−1.3Mn+0.06Mn 2 −1.7Cr+0.01Cr 2 +15Mo−5.6Si+1.6Cu+5.5Al−60(C+1.2N) 1/2 +26.3(C+1.2N)(Cr+Mn+Mo) 1/2 +0.6[Ni(Cr+Mn)] 1/2 .
Claims
exact text as granted — not AI-modified1 . Steel for low-temperature service having a high degree of surface processing quality, the steel comprising manganese (Mn): 15 wt % to 35 wt %, carbon (C) satisfying conditions of: 23.6C+Mn≧28 and 33.5C−Mn≦23, copper (Cu): 5 wt % or less (excluding 0 wt %), nitrogen (N): 1 wt % or less (excluding 0 wt %), chromium (Cr) satisfying a condition of: 28.5C+4.4Cr≦57, nickel (Ni): 5 wt % or less, molybdenum (Mo): 5 wt % or less, silicon (Si): 4 wt % or less, aluminum (Al): 5 wt % or less, and a balance of iron (Fe) and inevitable impurities,
wherein stacking fault energy (SFE) of the steel calculated by Formula 1 below is 24 mJ/m 2 or greater,
SFE (mJ/m 2 )=1.6Ni−1.3Mn+0.06Mn 2 −1.7Cr+0.01Cr 2 +15Mo−5.6Si+1.6Cu+5.5Al−60(C+1.2N) 1/2 +26.3(C+1.2N)(Cr+Mn+Mo) 1/2 +0.6[Ni(Cr+Mn)] 1/2 [Formula 1]
where Mn, C, Cr, Si, Al, Ni, Mo, and N refer to contents in wt %.
2 . The steel of claim 1 , wherein the steel comprises austenite in an area fraction of 95% or greater.
3 . The steel of claim 2 , wherein the steel comprises carbides along austenite grain boundaries in an area fraction of 5% or less.
4 . The steel of claim 1 , wherein a value of stress causing twinning in the steel is equal to or a greater than a value of tensile stress corresponding to a tensile strain of 5% of the steel.Cited by (0)
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