Steel with high mechanical strength and wear resistance

Details Steel with high mechanical strength and wear resistance Steel with high mechanical strength and wear resistance

C - Chemistry, Metallurgy

22

C

C22C 38/44 (2006.01) C21D 8/00 (2006.01) C22C 38/46 (2006.01) C22C 38/50 (2006.01)

Patent

CA 2565162

The invention relates to a steel with high mechanical strength and wear resistance. More specifically, the invention relates to a method of reducing the segregated veins of a steel having high mechanical strength, high wear resistance and the following weight composition: 0.30 % = C = 1.42 %; 0.05 % = Si = 1.5 %; Mn = 1.95 %; Ni = 2.9 %; 1.1 % = Cr = 7.9 %; 0.61 % = Mo = 4.4 %; optionally V = 1.45 %, Nb = 1.45 %, Ta = 1.45 % and V+Nb/2 + Ta/4 = 1.45 %; less than 0.1 % borium, 0.19 % (S + Se/2 + Te/4), 0.01 % calcium, 0.5 % rare earths, 1 % aluminium, 1 % copper; the remainder being iron and impurities resulting from the production thereof. The composition also comprises: 800 = D = 1150, where D = 540(C)?0.25¿ + 245 (Mo + 3 V + 1.5 Nb + 0.75 Ta)?0.30¿ + 125 Cr?0.20 ¿+ 15.8 Mn + 7.4 Ni + 18 Si. According to the invention, all or part of the molydenum is replaced by a double proportion of tungsten, such that W > 0.21 %, and Ti, Zr, C are adjusted so that, after said adjustment, Ti + Zr/2 = 0.2 W, (Ti + Zr/2) X C = 0.07, Ti + Zr/2 = 1.49 % and D is unchanged at 5 %. The invention also relates to the steel obtained and to a method of producing a steel part.

Procédé pour réduire les veines ségrégées d'un acier à haute résistance mécanique et haute résistance à l'usure dont la composition comprend, en poids : 0,30 % = C = 1,42 % ; 0,05 % = Si = 1,5 % ; Mn = 1,95 % ; Ni 2,9%; 1,1 % = Cr = 7,9 % ; 0,61 % = Mo 4,4 %; éventuellement V = 1,45 %, Nb = 1,45 %, Ta = 1,45 % et V+Nb/2 + Ta/4 = 1,45 %; moins de 0,1 % de bore, de 0,19 % de (S + Se/2 + Te/4), de 0,01% de calcium, de 0,5 % de terres rares, de 1 % d'aluminium, de 1 % de cuivre ; le reste étant de fer et des impuretés résultant de l'élaboration. La composition satisfaisait en outre : 800 = D = 1150 avec D = 540(C)0,25 + 245 (Mo + 3 V + 1,5 Nb + 0,75 Ta)0,30 + 125 Cr,020 + 15,8 Mn + 7,4 Ni + 18 Si. Selon le procédé, on substitue tout ou partie du molybdène par une proportion double de tungstène de telle sorte que W > 0,21 %, et on ajuste Ti, Zr, C pour que, après ajustement, Ti + Zr/2 = 0,2 W, (Ti + Zr/2) X C = 0,07, Ti + Zr/2 = 1,49 % et D soit inchangé à 5 % près. Acier obtenu et procédé de fabrication d'une pièce en acier.

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