| 摘要 |
A steel for welding includes steel components in which PCTOD is less than or equal to 0.065%, CeqH is less than or equal to 0.225%, FB is greater than or equal to 0.0003%, and Bp is 0.09% to 0.30%. In the steel for welding, in a thickness center portion of a cross-section in a thickness direction, the number of oxide particles having an equivalent circle diameter of 2 μm or greater is less than or equal to 20 particles/mm2 and the number of Ti oxides having an equivalent circle diameter of 0.05 μm to 0.5 μm is 1.0×103 particles/mm2 to 1.0×105 particles/mm2. |
| 主权项 |
1. A steel for welding comprising, as steel components, by mass %,
C: a content [C] of 0.015% to 0.045%, Si: a content [Si] of 0.05% to 0.20%, Mn: a content [Mn] of 1.7% to 2.02%, Ni: a content [Ni] of 0.1% to 0.52%, Ti: a content [Ti] of 0.005% to 0.015%, B: a content [B] of 0.0003% to 0.0015%, N: a content [N] of 0.002% to 0.006%, O: a content [O] of 0.0015% to 0.0035%, P: a content [P] limited to 0.008% or less, S: a content [S] limited to 0.005% or less, Al: a content [Al] limited to 0.004% or less, Nb: a content [Nb] limited to 0.004% or less, Cu: a content [Cu] limited to 0.03% or less, V: a content [V] limited to 0.02% or less, and a balance consisting of Fe and unavoidable impurities, wherein: PCTOD in a following equation 1 is less than or equal to 0.065%, CeqH in a following equation 2 is less than or equal to 0.225%, FB in a following equation 3 is greater than or equal to 0.0003%, and Bp in a following equation 4 is 0.09% to 0.30%, wherein in a thickness center portion of a cross-section in a thickness direction, the number of oxide particles having an equivalent circle diameter of 2 μm or greater is less than or equal to 20 particles/mm2, wherein the number of Ti oxides having an equivalent circle diameter of 0.05 μm to 0.5 μm is 1.0×103 particles/mm2 to 1.0×105 particles/mm2, and wherein:
PCTOD=[C]+[Cu]/22+[Ni]/67+[V]/3 Equation 1CeqH=[C]+[Si]/4.16+[Mn]/14.9+[Cu]/12.9+[Ni]/105+[V]/1.82 Equation 2FB=[B]−0.77×([N]−0.29×([Ti]−2×([O]−0.89×[Al]))) Equation 3; andBp=(884×[C]×(1−0.3×[C]2)+294)×FB Equation 4; wherein: 0 is substituted for ([O]−0.89×[Al]) in the Equation 3 when ([O]−0.89×[Al]) is 0 or less; 0 is substituted for ([Ti]−2×([O]−0.89×[Al])) in the Equation 3 when ([Ti]−2×([O]−0.89×[Al])) is 0 or less; 0 is substituted for ([N]−0.29×([Ti]−2×([O]−0.89×[Al]))) in the Equation 3 when ([N]−0.29×([Ti]−2×([O]−0.89×[Al]))) is 0 or less; and 0 is substituted for FB in the Equation 4 when FB is 0 or less. |
