Steel plate and its manufacturing method

Abstract

To provide a steel sheet that can achieve a high yield strength, low yield ratio and low-temperature toughness even when the steel sheet is thick, and a method for producing the steel sheet.SOLUTION: A steel sheet satisfies a predetermined composition and has a Pcm represented by a predetermined formula (1) of 0.30 or less and a Dh represented by a predetermined formula (2) of 21.7 or more. A hard structure composed of bainite and / or island martensite MA has a fraction in the whole steel structure of 23.3-29.4% by area, and an average circle-equivalent diameter of the hard structure is 5.2 μm or less.SELECTED DRAWING: Figure 1

Description

[Technical Field]

[0001] The present invention relates to a steel plate and a manufacturing method thereof, and more particularly to a steel plate exhibiting a lower yield ratio together with high yield strength and good low-temperature toughness, and a manufacturing method thereof. [Background technology]

[0002] Steel plates for construction are primarily required to have the following characteristics to ensure the safety of buildings: (i) high yield strength. With the recent trend toward taller and longer span buildings, it is expected that steel plates will increasingly be designed to bear heavier loads. In order to prevent deformation of buildings under normal conditions, steel plates are required to have high yield strength.

[0003] Furthermore, in emergencies such as earthquakes (when a load exceeding the yield stress is applied), the plastic deformation of the components is thought to absorb the energy of the earthquake and ensure safety. The higher the plastic deformation capacity before a building collapses, in other words, the lower the yield ratio (yield strength / tensile strength), the greater the amount of energy absorbed, so steel plates are required to have a low yield ratio.

[0004] Furthermore, even if a steel plate is plastically designed, if it does not have excellent low-temperature toughness, it may undergo brittle fracture without plastic deformation. Therefore, from the viewpoint of ensuring safety, a certain low-temperature toughness is required to prevent the initiation and propagation of cracks.

[0005] As a steel plate mainly used for building structures, for example, Patent Document 1 discloses a steel plate for thick-walled circular steel pipes with a low yield ratio and a tensile strength of 780 MPa or more, which has a predetermined chemical composition, a hardenability index DI defined by a predetermined formula (1) of 8 inches or more, and satisfies the following requirements (A), (B), and (C): (A) In the microstructure at 1 / 4 of the plate thickness, bainite is 90% or more by area, (B) In the microstructure at 1 / 4 of the sheet thickness, the average circular equivalent diameter d of the region surrounded by high-angle grain boundaries with a misorientation of 15° or more is 4 μm or less; (C) The microstructure at the 1 / 4 sheet thickness region contains 3 to 10 area % of island martensite having an average circle equivalent diameter of 0.5 to 3 μm and a Vickers hardness Hv of 700 or more. [Prior art documents] [Patent documents]

[0006] [Patent Document 1] Japanese Patent Application Laid-Open No. 2013-57105 Summary of the Invention [Problem to be solved by the invention]

[0007] There is a demand for a lower yield ratio along with high yield strength and good low-temperature toughness, but this has been difficult to achieve with conventional technology.

[0008] Furthermore, with conventional techniques, it has been difficult to stably obtain the above-mentioned high yield strength, good low-temperature toughness, and lower yield ratio even in thicker steel plates.

[0009] The present invention has been made in view of the above circumstances, and an object of the present invention is to provide a steel sheet having high yield strength, low yield ratio, and excellent low-temperature toughness, and a method for manufacturing the steel sheet, and further to provide a steel sheet that can achieve the above properties even when the sheet is thick, and a method for manufacturing the steel sheet. [Means for solving the problem]

[0010] Aspect 1 of the present invention is C: 0.040% by mass to 0.060% by mass, Si: 0.20% by mass to 0.30% by mass, Mn: 1.80% by mass to 2.00% by mass, P: more than 0% by mass, 0.010% by mass or less, S: more than 0% by mass, 0.003% by mass or less, Al: 0.040% by mass to 0.080% by mass, Cu:0.40 mass%~0.50 mass%, Ni: 1.40% by mass to 1.50% by mass, Cr:0.90 mass%~1.10 mass%, Mo: 0.17% by mass to 0.23% by mass, V:0 mass%~0.005 mass%, Nb: 0% by mass to 0.005% by mass, Ti: 0.010% by mass to 0.020% by mass, B: 0.0005% by mass to 0.0020% by mass, N: 0.0030% by mass to 0.0065% by mass, and Ca: Contains 0.0005% by mass to 0.0035% by mass, the balance being iron and unavoidable impurities; Pcm represented by the following formula (1) is 0.30 or less, and Dh represented by the following formula (2) is 21.7 or more, The steel plate has a hard structure composed of at least one of bainite and island martensite MA, and the hard structure accounts for 23.3 area % to 29.4 area % of the total steel structure, and has an average equivalent circle diameter of 5.2 μm or less. Pcm=[C]+[Si] / 30+[Mn] / 20+[Cu] / 20+[Ni] / 60+[Cr] / 20+[Mo] / 15+[V] / 10+5×[B]...(1) Dh=DI×([C]+0.91×[Mn]+0.97×[Cr]+0.88×[Mo]) ···(2) In equation (2), DI = 1.16 × ([C] / 10) 0.5 ×(0.7×[Si]+1)×(5.1×([Mn]-1.2)+5)×(0.35×[Cu]+1)×(0.36×[Ni]+1)×(2.16×[Cr]+1)×(3×[Mo]+1)×(1.75×[V]+1)×(200×[B]+1) ···(3) In the formulas (1) to (3), the [element symbol] indicates the content of each element in the steel expressed in mass %, and elements that are not contained are set to zero.

[0011] Aspect 2 of the present invention is The steel plate according to aspect 1, having a plate thickness of 80 mm or more.

[0012] Aspect 3 of the present invention is A method for producing a steel sheet according to aspect 1 or 2, comprising: A step of heating a steel slab having the composition according to aspect 1 to 900 to 1250°C; A process of rolling at a finish rolling temperature of 800 ° C to 970 ° C, and then cooling from a cooling start temperature of Ar3 point or higher calculated by the following formulas (4) and (5) to a cooling stop temperature of 400 ° C or lower at an average cooling rate of 1.0 ° C / s or higher; After the cooling, the temperature is within the range of Ac1 point calculated by the following formula (6) to Ac3 point calculated by the following formula (7), and the parameter P calculated by the following calculation method is N A process of performing two-phase region normalizing within a range of 21.04 to 22.12. The method for producing a steel sheet further comprises a step of tempering the steel sheet at a temperature in the range of 300°C to 600°C after the two-phase normalizing. Ar3 points=910-310×[C]+25×[Si]-80×[Mneq]...(4) In equation (4), [Mneq]=[Mn]+[Cr]+[Cu]+[Mo]+[Ni] / 2+10([Nb]-0.02)+1 ···(5) Ac1 point = 723-10.7×[Mn]-16.9×[Ni]+29.1×[Si]+16.9×[Cr]+290×[As]+6.38×[W]...(6) Ac3 points=910-203×[C] 0.5 -15.2×[Ni]+44.7×[Si]+104×[V]+31.5×[Mo]+13.1×[W] ···(7) In the formulas (4) to (7), the [element symbol] indicates the content of each element in the steel expressed in mass %, and elements that are not contained are set to zero. [Parameter P N Calculation method] Parameter P indicates the degree of normalization when the temperature is kept at Tx (℃) for Kx (seconds). xUsing the following formula (8) to calculate the parameter P, which represents the degree of normalization when the temperature of the steel sheet rises over time, according to the procedures shown in [1] to [5] below. N Calculate. P x =(T x +273) × log(K x / 3600+20) ···(8) [1] A temperature rise curve of the steel plate center temperature is created, with the vertical axis representing the steel plate center temperature (°C) and the horizontal axis representing the elapsed time (seconds) from the start of temperature rise. The curve is divided into 1-second intervals and approximated as a rectangle. In the above temperature rise curve, the point when the temperature at the center of the steel sheet is at point Ac1 is set as the starting point, the elapsed time from the start of temperature rise at this starting point is set as t0 (= 0) (seconds), and the temperature at the center of the steel sheet is set as T0 (= Ac1) (°C). [2] Calculate the degree of normalizing P0 after 1 second has elapsed from the starting point. The degree of normalizing P0 is calculated by adding X=0 to equation (8), that is, T x =T0=Ac1(℃), K x Substitute =K0=1 (seconds) to find the answer. [3] The required holding time k1 at the steel sheet center temperature T1 after 1 second has elapsed from the starting point, which is necessary to obtain the normalizing degree P0 after 1 second has elapsed from the starting point, as determined in [2] above, is calculated. The required holding time k1 is calculated by applying X=0 to the following formula (9), that is, P x =P0,T x+1 =T1(℃) to find the answer. k x+1 =(exp(P X / (T X+1 +273))-20)×3600 (9) In equation (9), k x+1 is from the starting point (Ac1) to T x+1 (℃) P x The steel plate center temperature T required to obtain x+1 The required retention time at [4] The conversion holding time at the steel sheet center temperature T1 is set to k1+1=K1 (seconds) from the following formula (10). K x+1 =k x+1 +1 ···(10) In equation (10), Kx+1 is the steel plate center temperature T x+1 indicates the converted retention time at k X+1 is from the starting point (Ac1) to T x+1 (℃) P x The steel plate center temperature T required to obtain x+1 The required retention time at [5] P shown in [2] to [4] above x , k x+1 , K. x+1 The calculation of X = 1, 2, 3, ... is repeated until normalizing in the two-phase region is completed, and P X =P f In search of this P f P N Let's say. [Effects of the Invention]

[0013] According to the present invention, it is possible to provide a steel plate that can achieve high yield strength, a low yield ratio, and excellent low-temperature toughness even when the plate is thick, and a method for manufacturing the steel plate. [Brief explanation of the drawings]

[0014] [Figure 1] Figure 1 is a graph showing the relationship between the hard tissue fraction and vTrs. [Figure 2] FIG. 2 is a graph showing the relationship between the fraction of hard structure and yield strength. [Figure 3] FIG. 3 is a graph showing the relationship between the average circular equivalent diameter of hard tissue and vTrs. [Figure 4] FIG. 4 is a graph showing the relationship between Dh and the average equivalent circle diameter of hard tissue. [Figure 5] FIG. 5 is a diagram for explaining a method for calculating the parameter PN. [Figure 6] FIG. 6 is another diagram illustrating a method for calculating the parameter PN. [Figure 7] FIG. 7 is another diagram illustrating a method for calculating the parameter PN. [Figure 8]FIG. 8 is another diagram illustrating a method for calculating the parameter PN. [Figure 9] FIG. 9 is a graph showing the relationship between PN and hard tissue fraction. DETAILED DESCRIPTION OF THE INVENTION

[0015] The inventors conducted extensive research to develop a steel sheet that satisfies high yield strength, low yield ratio, and good low-temperature toughness. As a result, they discovered that these mechanical properties (strength, yield ratio, and toughness) depend on the fraction and size of a "hard structure" composed of one or more of bainite and island martensite (MA). They also discovered that controlling the fraction and size of this hard structure requires controlling the fraction of reverse-transformed structures during intercritical normalizing in the manufacturing process, as well as chemical composition factors. In particular, for thick-walled steel sheets, the manufacturing process requires time to heat up to the heat treatment temperature, and the thermal history (temperature and furnace time) during heating significantly influences the structure required to satisfy the above properties. In this study, they discovered manufacturing conditions and chemical compositions that can achieve the above properties, and even for thick-walled steel sheets. Below, we first discuss the hard structure that affects the mechanical properties.

[0016] [The percentage of hard structure in the total steel structure is 23.3% to 29.4% by area, and its average equivalent circle diameter is 5.2 μm or less] To achieve both high strength and low-temperature toughness, hard structures are necessary to ensure strength, and the lower limit of their fraction is set to 23.3 area %. On the other hand, hard structures act as fracture initiation points, so if they are present in excess, toughness will decrease. Therefore, the upper limit of the hard film fraction is set to 29.4 area %. Furthermore, toughness is affected not only by the hard structure fraction but also by the hard structure size. If the hard structure size is large, it is more likely to act as a fracture initiation point, and toughness will decrease. Therefore, the upper limit of the average circular equivalent diameter of the hard structure is set to 5.2 μm.

[0017] FIG. 1 is a graph showing the relationship between the hard structure fraction and vTrs, an index for evaluating low-temperature toughness, obtained using data from the Examples described below. It can be seen from FIG. 1 that in order to achieve a vTrs of −13° C. or less, which is considered to be good low-temperature toughness, the hard structure fraction should be 29.4 area % or less. FIG. 2 is a graph showing the relationship between the hard structure fraction and yield strength, obtained using data from the Examples described below. It can be seen from FIG. 2 that in order to achieve a high yield strength of 600 MPa or more, the hard structure fraction should be 23.3 area % or more.

[0018] From the viewpoint of further improving low temperature toughness, the fraction of hard structure is preferably 28.9 area % or less, while from the viewpoint of further improving yield strength, the fraction of hard structure is preferably 23.8 area % or more.

[0019] The hard structure can be generated by performing two-phase region normalizing under predetermined conditions in the manufacturing process, as will be described later.

[0020] Figure 3 is a graph showing the relationship between the average equivalent circle diameter of hard structures and vTrs, an index for evaluating low-temperature toughness, obtained using data from the Examples described below. Figure 3 shows that in order to achieve a vTrs of -13°C or less, which is considered to be good low-temperature toughness, the average equivalent circle diameter of the hard structures should be 5.2 µm or less.

[0021] From the viewpoint of further improving low-temperature toughness, it is preferable that the average equivalent circular diameter of the hard structure is 4.0 μm or less. The smaller the average equivalent circular diameter of the hard structure, the better. However, taking into consideration the manufacturing conditions, etc., the lower limit of the average equivalent circular diameter of the hard structure is about 2.0 μm.

[0022] Furthermore, in order to obtain hard tissue having the above average equivalent circle diameter, the component parameter Dh expressed by the following formula (2) must be 21.7 or more. Dh=DI×([C]+0.91×[Mn]+0.97×[Cr]+0.88×[Mo]) ···(2) In equation (2), DI = 1.16 × ([C] / 10) 0.5 ×(0.7×[Si]+1)×(5.1×([Mn]-1.2)+5)×(0.35×[Cu]+1)×(0.36×[Ni]+1)×(2.16×[Cr]+1)×(3×[Mo]+1)×(1.75×[V]+1)×(200×[B]+1) ···(3) In formulas (2) and (3), the [element symbol] indicates the content of the element in the steel expressed in mass %, and elements that are not contained are set to zero.

[0023] The above Dh is an index that describes the average equivalent circle diameter of the hard structure. The derivation of Dh will be explained below. The hard structure is thought to be formed from cementite that forms between the laths of bainite obtained before intercritical normalizing, i.e., after hot rolling and cooling (hereinafter referred to as "as-rolled and cooled bainite"). Therefore, the size of the hard structure is strongly dependent on the dispersion of cementite in the as-rolled and cooled bainite. Furthermore, the dispersion of cementite is thought to be determined by the number of cementite nucleation sites and the nucleation frequency.

[0024] The number of nucleation sites depends on the lath spacing of the as-rolled and cooled bainite, so it is thought that the higher the hardenability index DI, the smaller the lath spacing, i.e., the more nucleation sites there are. Therefore, the hardenability index DI is used as an index of the number of nucleation sites.

[0025] Furthermore, it is believed that the nucleation frequency depends on the amount of C, an element that forms cementite (FeC), and the amount of X, an element that has a high affinity with cementite and attracts C more easily than Fe. As elements X, Mn, Cr, and Mo were selected after considering their solid solubility in cementite as an index of affinity with cementite, and their carbide-forming ability as an index of ease of attracting C.

[0026] Therefore, it was thought that the size of the hard structure could be estimated using the parameter Dh, which is calculated by multiplying the hardenability index DI, which is an index of the number of nucleation sites, by the nucleation frequency, which is composed of the contents of C, Mn, Cr, and Mo (C + 0.91Mn + 0.97Cr + 0.88Mo).

[0027] Figure 4 is a graph showing the relationship between Dh and the average circular equivalent diameter of hard tissue, obtained using data from the Examples described below. From Figure 4, it can be seen that in order to achieve an average circular equivalent diameter of hard tissue of 5.2 µm or less, Dh should be set to 21.7 or more. Dh is preferably 25.0 or more. There is no particular upper limit for Dh, but considering the range of the specified component composition, it is approximately 34.0.

[0028] Next, the reasons for specifying the ranges of each element and Pcm in the chemical composition of the steel sheet will be explained.

[0029] [C:0.040 mass%~0.060 mass%] C is an essential element for ensuring the strength of the base metal and weld, and must be contained in an amount of 0.040% by mass or more. The C content is preferably 0.045% by mass or more. On the other hand, if the C content is too high, HAZ toughness and weldability will deteriorate. Furthermore, if the C content is excessive, island martensite will be more likely to form. Therefore, the C content is set to 0.060% by mass or less. The C content is preferably 0.055% by mass or less.

[0030] [Si:0.20 mass%~0.30 mass%] Si has a deoxidizing effect and is an element effective in improving the strength of the base metal and weld. To achieve these effects, the Si content is set to 0.20 mass% or more. The Si content is preferably 0.22 mass% or more. However, if the Si content is too high, weldability and toughness deteriorate, so the Si content is set to 0.30 mass% or less. The Si content is preferably 0.28 mass% or less.

[0031] [Mn:1.80 mass%~2.00 mass%] Mn is an element that has the effect of improving hardenability and improving the strength and toughness of steel sheet. To achieve this effect, the present invention contains 1.80 mass% or more of Mn. The Mn content is preferably 1.89 mass% or more. However, if the Mn content is too high, the HAZ toughness and weldability will deteriorate. Therefore, the Mn content is 2.00 mass% or less. The Mn content is preferably 1.99 mass% or less.

[0032] [P: more than 0 mass%, 0.010 mass% or less] P is an element that is inevitably contained in steel, and if the P content exceeds 0.010 mass%, the toughness of the base metal and HAZ will be significantly deteriorated, and resistance to hydrogen-induced cracking will also be deteriorated. Therefore, in the present invention, the P content is limited to 0.010 mass% or less. However, it is industrially difficult to reduce the P content to 0 mass%, so the lower limit is more than 0 mass%.

[0033] [S: More than 0 mass%, 0.003 mass% or less] If there is too much S, it generates a large amount of MnS, which deteriorates weld crack resistance. Therefore, in the present invention, the S content is set to 0.003 mass% or less. However, it is industrially difficult to reduce the S content to 0 mass%, so the lower limit of the S content is more than 0 mass%.

[0034] [Al:0.040 mass%~0.080 mass%] Al is an element that improves the hardenability of B by deoxidizing and fixing free nitrogen. To achieve these effects, the Al content is set to 0.040% by mass or more, preferably 0.045% by mass or more in the present invention. On the other hand, if the Al content is too high, coarse alumina-based inclusions are formed, reducing the toughness of the base material. Therefore, the Al content is set to 0.080% by mass or less, preferably 0.075% by mass or less.

[0035] [Cu:0.40 mass%~0.50 mass%] Cu is an element that is effective in improving hardenability and increasing strength. To achieve this effect, the Cu content must be 0.40% by mass or more. However, if the Cu content exceeds 0.50% by mass, toughness deteriorates, so the Cu content is set to 0.50% by mass or less.

[0036] [Ni: 1.40 mass% to 1.50 mass%] Ni is an element that is effective in improving the strength and toughness of the base metal and weld. To achieve this effect, the Ni content must be 1.40 mass% or more. However, if a large amount of Ni is included, scale defects are likely to occur during rolling, so the Ni content must be 1.50 mass% or less.

[0037] [Cr:0.90 mass%~1.10 mass%] Cr is an element necessary for improving hardenability and improving strength. To achieve this effect, the Cr content is set to 0.90% by mass or more. The Cr content is preferably 0.95% by mass or more. On the other hand, if a large amount of Cr is contained, the weld cracking resistance deteriorates, so the Cr content is set to 1.10% by mass or less. The Cr content is preferably 1.05% by mass or less.

[0038] [Mo:0.17 mass%~0.23 mass%] Mo is an element that improves hardenability and increases strength. To achieve this effect, the Mo content is set to 0.17% by mass or more. However, if the Mo content exceeds 0.23% by mass, weld crack resistance deteriorates. Therefore, the Mo content is set to 0.23% by mass or less.

[0039] [V:0 mass%~0.005 mass%] V is an element that can be unavoidably contained. V has the effect of suppressing recrystallization, which leads to an increase in the acoustic anisotropy of the steel sheet. Therefore, the V content is limited to 0.005 mass% or less. The V content is preferably 0.003 mass% or less.

[0040] [Nb:0 mass%~0.005 mass%] Nb, like V, is an element that can be unavoidably contained. Like V, Nb also has the effect of suppressing recrystallization, which leads to an increase in the acoustic anisotropy of the steel sheet. Therefore, the Nb content is limited to 0.005 mass% or less. The Nb content is preferably 0.003 mass% or less.

[0041] [Ti:0.010 mass%~0.020 mass%] Ti is an element that is effective in preventing coarsening of austenite grains in the HAZ during welding by precipitating as TiN in steel and improving the toughness of the HAZ. To achieve this effect, the Ti content is set to 0.010% by mass or more, preferably 0.012% by mass or more. On the other hand, if the Ti content is excessive, TiN will coarsen and the toughness of the base material will deteriorate, so the Ti content is set to 0.020% by mass or less, preferably 0.018% by mass or less.

[0042] [B:0.0005 mass%~0.0020 mass%] B is an element effective in improving hardenability and increasing base material strength. To achieve this effect, the B content is set to 0.0005% by mass or more. The B content is preferably 0.0008% by mass or more. However, if the B content is excessive, inclusions are formed, deteriorating the base material toughness, so the B content is set to 0.0020% by mass or less. The B content is preferably 0.0018% by mass or less.

[0043] [N:0.0030 mass%~0.0065 mass%] N is an element that precipitates as TiN in the steel structure, suppresses coarsening of austenite grains in the HAZ, and improves the toughness of the HAZ. To achieve this effect, the N content must be 0.0030% by mass or more. However, if the N content is too high, the presence of solute N will actually deteriorate the HAZ toughness, so the N content is set to 0.0065% by mass or less. The N content is preferably 0.0050% by mass or less.

[0044] [Ca:0.0005 mass%~0.0035 mass%] Ca is an element that is effective in neutralizing weld crack resistance by spheroidizing MnS. To achieve this effect, the Ca content is set to 0.0005% by mass or more. However, since too much Ca causes inclusions to coarsen and the base material toughness to deteriorate, the Ca content is set to 0.0035% by mass or less. The Ca content is preferably 0.0030% by mass or less, and more preferably 0.0025% by mass or less.

[0045] The alloy contains the above elements, with the balance consisting of iron and inevitable impurities. Incidental impurities include trace elements that are introduced depending on the conditions of raw materials, materials, manufacturing equipment, etc. For example, elements such as P and S are usually preferable to have as low a content as possible, and are therefore considered inevitable impurities, but their composition ranges are separately specified as above. Therefore, in this specification, the term "unavoidable impurities" that make up the balance refers to a concept that excludes elements whose composition ranges are separately specified.

[0046] [Pcm represented by the following formula (1) is 0.30 or less] Pcm=[C]+[Si] / 30+[Mn] / 20+[Cu] / 20+[Ni] / 60+[Cr] / 20+[Mo] / 15+[V] / 10+5×[B]...(1) In formula (1), the [element symbols] each represent the content of the element in the steel expressed in mass%, and elements that are not contained are represented as zero. Pcm is an index that indicates susceptibility to weld cracking. If Pcm is large, susceptibility to weld cracking increases and crack resistance decreases, so the upper limit is set at 0.30.

[0047] As described above, the steel plate of the present invention exhibits a low yield ratio as well as high yield strength and good low-temperature toughness. Each of these properties will be described below.

[0048] (1) Yield strength (YS) The steel plate has a yield strength (YS) of 600 MPa or more. As a result, as mentioned above, even if the steel plate is designed to bear a large load, sufficient strength can be ensured to prevent deformation of the building under normal conditions. The yield strength is preferably 620 MPa or more, and more preferably 660 MPa or more. The upper limit of the yield strength can be set at 750 MPa, for example, based on the standards for building materials.

[0049] (2) Low-temperature toughness This refers to a ductile-to-brittle fracture transition temperature of -13°C or lower, measured by the method described in the Examples below. The ductile-to-brittle fracture transition temperature is preferably -20°C or lower, and more preferably -25°C or lower.

[0050] (3) Low yield ratio The steel sheet of the present invention exhibits a low yield ratio of 85% or less, preferably 80% or less.

[0051] The steel sheet of the present invention exhibits the above properties even when the sheet thickness is 80 mm or more, preferably more than 80 mm.

[0052] [Manufacturing method] Next, a method for producing a steel plate according to the present invention will be described. The method includes the steps of heating a steel slab having the above-described chemical composition to 900 to 1250°C, A process of rolling at a finish rolling temperature of 800 ° C to 970 ° C, and then cooling from a cooling start temperature of Ar3 point or higher to a cooling stop temperature of 400 ° C or lower at an average cooling rate of 1.0 ° C / s or higher; After the cooling, within the temperature range of Ac1 point to Ac3 point, and the parameter P obtained by the following method N A process of performing two-phase region normalizing within a range of 21.04 to 22.12. After the normalizing, a step of tempering is included in the process at a temperature range of 300°C to 600°C.

[0053] The Ar3 point is calculated by the following formulas (4) and (5): The Ac1 point is calculated by the following formula (6), and the Ac3 point is calculated by the following formula (7). Ar3 points=910-310×[C]+25×[Si]-80×[Mneq]...(4) In equation (4), [Mneq]=[Mn]+[Cr]+[Cu]+[Mo]+[Ni] / 2+10([Nb]-0.02)+1 ···(5) Ac1 point = 723-10.7×[Mn]-16.9×[Ni]+29.1×[Si]+16.9×[Cr]+290×[As]+6.38×[W]...(6) Ac3 points=910-203×[C] 0.5-15.2×[Ni]+44.7×[Si]+104×[V]+31.5×[Mo]+13.1×[W] ···(7) In the formulas (4) to (7), the [element symbol] indicates the content of each element in the steel expressed in mass %, and elements that are not contained are set to zero.

[0054] [Parameter P N Calculation method] Parameter P indicates the degree of normalization when the temperature is kept at Tx (℃) for Kx (seconds). x Using the following formula (8) to calculate the parameter P, which represents the degree of normalization when the temperature of the steel sheet rises over time, according to the procedures shown in [1] to [5] below. N Calculate. P x =(T x +273) × log(K x / 3600+20) ···(8) [1] A temperature rise curve of the steel plate center temperature is created, with the vertical axis representing the steel plate center temperature (°C) and the horizontal axis representing the elapsed time (seconds) from the start of temperature rise. The curve is divided into 1-second intervals and approximated as a rectangle. In the above temperature rise curve, the point when the temperature at the center of the steel sheet is at point Ac1 is set as the starting point, the elapsed time from the start of temperature rise at this starting point is set as t0 (= 0) (seconds), and the temperature at the center of the steel sheet is set as T0 (= Ac1) (°C). [2] Calculate the degree of normalizing P0 after 1 second has elapsed from the starting point. The degree of normalizing P0 is calculated by adding x=0 to equation (8), that is, T x =T0=Ac1(℃), K x Substitute =K0=1 (seconds) to find the answer. [3] The required holding time k1 at the steel sheet center temperature T1 after 1 second has elapsed from the starting point, which is necessary to obtain the normalizing degree P0 after 1 second has elapsed from the starting point, as determined in [2] above, is calculated. The required holding time k1 is calculated by applying x=0 to the following formula (9), that is, P x =P0,T x+1 =T1(℃) to find the answer. k x+1 =(exp(P X / (T X+1 +273))-20)×3600 (9) In equation (9), kx+1 is from the starting point (Ac1) to T x+1 (℃) P x The steel plate center temperature T required to obtain x+1 The required retention time at [4] The conversion holding time at the steel sheet center temperature T1 is set to k1+1=K1 (seconds) from the following formula (10). K x+1 =k x+1 +1 ···(10) In equation (10), K x+1 is the steel plate center temperature T x+1 indicates the converted retention time at k X+1 is from the starting point (Ac1) to T x+1 (℃) P x The steel plate center temperature T required to obtain x+1 The required retention time at [5] P shown in [2] to [4] above x , k x+1 , K. x+1 The calculation of X = 1, 2, 3, ... is repeated until normalizing in the two-phase region is completed, and P X =P f In search of this P f P N Let's say.

[0055] Each step will be described in detail below.

[0056] [Step of heating the steel billet having the above-mentioned composition to 900 to 1250°C] A steel slab having the above-mentioned chemical composition is heated to 900 to 1250°C, at which hot rolling is possible. If the heating temperature is low, the austenite grain size will be small, resulting in poor hardenability and reduced strength. Therefore, the heating temperature is set to 900°C or higher. The heating temperature is preferably 1000°C or higher, and more preferably 1050°C or higher. On the other hand, if the heating temperature is too high, the austenite grain size will become large, resulting in an increase in the bainite block diameter and a deterioration in toughness. Therefore, the heating temperature is set to 1250°C or lower. The heating temperature is preferably 1200°C or lower, and more preferably 1150°C or lower.

[0057] [A process in which rolling is performed at a finish rolling temperature of 800°C to 970°C on the surface, and then cooling is performed from a cooling start temperature of Ar3 point or higher to a cooling stop temperature of 400°C or lower at an average cooling rate of 1.0°C / s or higher] If the finish rolling temperature is below 800°C, the austenite grains will be significantly flattened, increasing acoustic anisotropy. Therefore, the finish rolling temperature is set to 800°C or higher. The finish rolling temperature is preferably set to 810°C or higher, more preferably 820°C or higher. On the other hand, if the finish rolling temperature exceeds 970°C, the effect of refining the austenite grains will be reduced, resulting in a deterioration in toughness. Therefore, the finish rolling temperature is set to 970°C or lower. The finish rolling temperature is preferably set to 960°C or lower, more preferably 950°C or lower.

[0058] After rolling at the above finish rolling temperature, the material is cooled from a cooling start temperature of at least the Ar3 point to a cooling stop temperature of at most 400°C at an average cooling rate of at least 1.0°C / s. The Ar3 point is the temperature at which transformation begins during continuous cooling and is calculated using the above formulas (4) and (5).

[0059] The morphology of bainite and the size of the bainite structure change depending on the cooling conditions after rolling. Therefore, by controlling the cooling conditions after rolling, the morphology of bainite can be controlled and the desired bainite structure size can be obtained. When the cooling start temperature at a predetermined average cooling rate is below the Ar3 point or the cooling stop temperature is above 400°C, bainite with wide lath spacing is formed, and coarse hard structures are formed during the subsequent two-phase normalizing, resulting in a deterioration of toughness. Therefore, the cooling start temperature at a predetermined average cooling rate is set to the Ar3 point or higher. The cooling start temperature is preferably Ar3 point + 20°C or higher, more preferably Ar3 point + 40°C or higher. The upper limit of the cooling start temperature can be, for example, the finish rolling temperature. Furthermore, the cooling stop temperature at a predetermined average cooling rate is 400°C or lower, preferably 390°C or lower, and more preferably 380°C or lower. The lower limit of the cooling stop temperature is not particularly limited and can be, for example, 20°C.

[0060] If the average cooling rate is less than 1.0°C / sec, the structure size of the bainite that is formed becomes coarse, resulting in a deterioration in toughness. The average cooling rate is preferably 1.3°C / sec or more, and more preferably 1.5°C / sec or more. There is no particular upper limit to the average cooling rate.

[0061] After cooling to the cooling stop temperature at the average cooling rate, the material may be cooled to a temperature lower than the cooling stop temperature, for example, room temperature, and then normalized in the two-phase region. Alternatively, the material may be heated from the cooling stop temperature to the heating temperature for normalizing in the two-phase region without being cooled to room temperature.

[0062] [After the cooling, within the temperature range of Ac1 point to Ac3 point, and the parameter P obtained by the following method N The process of performing two-phase region normalizing within the range of 21.04 to 22.12.

[0063] To achieve a low yield ratio, it is effective to perform dual-phase heat treatment to create a two-phase structure consisting of a hard phase and a soft phase. However, with conventional methods, the hardness ratio between the hard and soft phases is not large, making it difficult to consistently achieve a low yield ratio after severe bending, for example, when performing severe bending with a D / t (D: steel pipe diameter, t: steel plate thickness) of 10 to 20. Therefore, the inventors conducted research and found that by performing dual-phase normalizing under the conditions shown below, extremely hard MA is formed in the reverse transformation region, thereby increasing the hardness ratio between the hard and soft structure, a steel plate that satisfies a sufficiently low yield ratio even after severe bending can be obtained. The conditions for dual-phase normalizing are described below.

[0064] In order to secure the aforementioned predetermined fraction of hard structure and achieve a lower yield ratio, two-phase intercritical normalizing is performed. Two-phase intercritical normalizing is performed within the temperature range of Ac1 to Ac3, where the value of the parameter PN, calculated by the following method, is in the range of 21.04 to 22.12. Note that the Ac1 point is the temperature at which reverse transformation begins during continuous heating and is calculated using the above formula (6), and the Ac3 point is the temperature at which reverse transformation ends during continuous heating and is calculated using the above formula (7).

[0065] Above P N is a parameter that correlates with the fraction of reverse-transformed structures, i.e., the fraction of hard structures. When normalizing steel plates with a thickness of less than 80 mm in the two-phase region, the heating time until the steel plate center temperature reaches the soaking temperature is short, so the temperature and time during heating have only a small effect on the degree of two-phase normalizing. In this case, the desired fraction of hard structures can be obtained by controlling the soaking temperature and time. In contrast, for thick plates with a thickness of 80 mm or more, the heating time until the steel plate center temperature reaches the soaking temperature is long, and the temperature effect during heating is significant, so it is difficult to estimate the fraction of reverse-transformed structures based on the soaking temperature and time alone.

[0066] Therefore, in the present invention, assuming that the thick material is to be subjected to two-phase region normalizing, the conditions for two-phase region normalizing are set taking into consideration the influence of the temperature and time during heating on the degree of two-phase region normalizing.

[0067] [Parameter P N Calculation method] P N The calculation method is as follows: The following formula (8) is a parameter P that indicates the degree of normalization when the temperature is kept at Tx (°C) for Kx (seconds). x That is, since the following formula (8) is used when the temperature is maintained at a constant temperature, the degree of normalizing when the temperature of the steel sheet rises over time cannot be calculated by using the following formula (8) as it is. Therefore, the parameter P N Calculate. P x =(T x +273) × log(K x / 3600+20) ···(8)

[0068] [1] As shown in Figure 5, a temperature rise curve of the steel sheet center temperature is created, with the vertical axis representing the steel sheet center temperature (°C) and the horizontal axis representing the elapsed time (seconds) from the start of temperature rise. The curve is divided into 1-second intervals and approximated as a rectangle. In the above temperature rise curve, the point when the temperature at the center of the steel sheet is at point Ac1 is set as the starting point, the elapsed time from the start of temperature rise at this starting point is set as t0 (= 0) (seconds), and the temperature at the center of the steel sheet is set as T0 (= Ac1) (°C). As shown in Figure 5, the time 1 second after the starting point and the temperature at the center of the steel plate are respectively represented as t1 (seconds) and T1 (℃), and the time X seconds after the starting point and the temperature at the center of the steel plate are respectively represented as t X (seconds), T X (℃), and t X The time and the temperature at the center of the steel plate 1 second after t X+1 (seconds), T X+1 (℃). The time at which normalizing in the two-phase region is completed and the temperature at the center of the steel sheet are expressed as t f (seconds), T f Expressed as (℃).

[0069] [2] Calculate the degree of normalizing P0 after 1 second has elapsed from the starting point. The degree of normalizing P0 is calculated by adding X=0 to equation (8), that is, T x =T0=Ac1(℃), K x Substitute =K0=1 (seconds) to find the answer.

[0070] [3] Determine the required holding time k1 at the steel sheet center temperature T1 1 second after the starting point, which is required to obtain the normalizing level P0 determined in [2] above.

[0071] The required holding time k1 is calculated using the following formula (9). The formula (9) is defined as the time when the steel sheet center temperature is increased from the starting point (Ac1) to T x+1 (℃) P x T required to obtain X +1 is the equivalent holding time (seconds) at temperature x+1 This is the formula for calculating k x+1 =(exp(P X / (T X+1 +273))-20)×3600 (9)

[0072] The required holding time k1 is calculated by adding X=0 to the above equation (9), that is, P x =P0,Tx+1 = T1 (°C). Figure 6 shows the case where X = 0, and in Figure 6, the area of ​​the bold framed part showing the degree of normalizing P0 after 1 second has elapsed from the starting point is equal to the area (P value) of the filled part showing the steel sheet core temperature T1 × required holding time k1.

[0073] [4] The conversion holding time at the steel sheet center temperature T1 is set to k1+1=K1 (seconds) from the following formula (10). K x+1 =k x+1 +1 ···(10) In equation (10), K x+1 is the steel plate center temperature T x+1 indicates the converted retention time at k X+1 is from the starting point (Ac1) to T x+1 (℃) P x The steel plate center temperature T required to obtain x+1 The required retention time at

[0074] [5] P shown in [2] to [4] above x , k x+1 , K. x+1 The calculation of X = 1, 2, 3, ... is repeated until normalizing in the two-phase region is completed, and P X =P f In search of this P f P N Let's say.

[0075] A method for determining P1, k2, and K2 in the same manner as above will be explained below with reference to Fig. 7. Fig. 7 shows the case where X=1.

[0076] By substituting T1 (°C) and K1 = k1 + 1 (seconds) into the above formula (8), P1 (the area of ​​the bold frame in Figure 7) when holding at T1 (°C) for K1 (seconds) is calculated. Then, by substituting the above P1 and T2 (°C) into the above formula (9), the required holding time k2 (seconds) at T2 (°C) corresponding to P1 is calculated. In Figure 7, the area of ​​the bold frame indicating the degree of normalizing P1 after 2 seconds has elapsed from the starting point is equal to the area (P value) of the filled-in area indicating the steel sheet core temperature T2 x required holding time k2. From the above formula (10), K2 = k2 + 1 (seconds) is calculated as shown in Figure 3.

[0077] As shown in Figures 6 and 7 above, the calculation is repeated to obtain P x , k x+1 , K. x+1 (X=2,3,4...) are calculated sequentially. Then, as shown in Figure 8, T X is the temperature at the end of normalizing in the two-phase region, Tf (℃) X (=P f ) to P N Let's say.

[0078] The present inventors have N The relationship between the fraction of hard tissue and the P obtained was investigated separately. N 1 is a graph showing the relationship between the P N There is a correlation between the area fraction of hard structure and the area fraction of P. N On the other hand, in order to secure superior low-temperature toughness by suppressing the fraction of hard structures to 29.4 area % or less as mentioned above, P N must be less than or equal to 22.12.

[0079] According to an embodiment of the present invention, the parameter P correlated with the fraction of the hard structure that is a reverse transformation structure N By performing two-phase normalizing so that the value falls within a predetermined range, the fraction of hard structures that affect the properties can be controlled with high precision even in thick steel plates.

[0080] After the normalizing, the material may be cooled to, for example, 100° C. or less, and then subjected to the next tempering step.

[0081] [Step of tempering in a temperature range of 300°C to 600°C after the normalizing] Tempering can eliminate mobile dislocations generated during cooling after two-phase normalizing, thereby improving yield strength. In particular, by setting the tempering temperature in the range of 300°C to 600°C, appropriate strength can be obtained. If the tempering temperature exceeds the Ac1 point, two-phase normalizing occurs, and mobile dislocations are reintroduced during cooling, making it impossible to ensure strength. Furthermore, if the tempering temperature exceeds 600°C, the strength of the hard phase decreases, making it impossible to ensure strength. Therefore, the upper limit of the tempering temperature is 600°C, and a preferable upper limit is 560°C. On the other hand, if the tempering temperature is less than 300°C, mobile dislocations are not sufficiently eliminated, making it impossible to ensure appropriate strength. Therefore, the tempering temperature is 300°C or higher, preferably 400°C or higher. After the tempering, the material can be allowed to cool, for example, to room temperature. [Example]

[0082] The present invention will be described in more detail below with reference to examples. The present invention is not limited to the following examples, and can be practiced with appropriate modifications within the scope of the above-mentioned and below-mentioned aims, and all such modifications are included in the technical scope of the present invention.

[0083] (Steel plate manufacturing) Steels having the chemical compositions shown in Table 1 were produced in a converter or by vacuum melting at 150 kg. In Table 1, a line (-) indicates that the corresponding chemical component was not detected. In addition, Pcm, Ac1, Ac3, Ar3, DI, and Dh in Table 1 were calculated from the above-mentioned formulas (1) to (7).

[0084] After completion of the melting, slabs were obtained by continuous casting or forging. The slabs were heated to the heating temperatures shown in Table 2 and hot rolled. The finishing temperatures for hot rolling are as shown in Table 2. After hot rolling, the slabs were cooled under the conditions shown in Table 2. After cooling to the cooling stop temperatures shown in Table 2, they were allowed to cool to room temperature, and steel plates having the thicknesses shown in Table 2 were obtained.

[0085] Next, two-phase region normalizing and tempering were performed under the heat treatment conditions shown in Table 2. Two-phase region normalizing was performed at the heating temperature shown in Table 2 and with the P N The heating and holding time was set to satisfy the above P N After normalizing in the two-phase region, the specimens were tempered at the heating temperatures shown in Table 2 for about 15 minutes.

[0086] In the production of the above steel sheets, the finish rolling temperatures shown in Table 2 were measured at the surface temperature of the steel sheets. The heating temperature is the atmospheric temperature of the heating furnace, the cooling start temperature and cooling end temperature after rolling are the surface temperatures of the steel sheets, the heating temperature for dual-phase normalizing is the temperature at the center of the thickness of the steel sheets, and the heating temperature for tempering is the temperature at the center of the thickness of the steel sheets. The temperatures at the center of the thickness during dual-phase normalizing and tempering were measured by inserting a thermocouple.

[0087] To evaluate the steel structure of the obtained steel plate, the fraction and size of hard structure were measured, and a tensile test and a Charpy impact test were carried out as described below.

[0088] [Measurement of hard tissue fraction and size] A sample was taken from the steel sheet so that the observation surface was a thickness cross section including both the front and back surfaces of the steel sheet, parallel to the rolling direction and perpendicular to the steel sheet surface. The observation surface was then mirror-polished and then etched with nital solution to reveal the hard structure. The hard structure thus treated was observed as a white structure under a scanning electron microscope (SEM).

[0089] The exposed structure at the t / 4 thickness region was photographed at a magnification of 1000x. The photograph was imported into an image analyzer. The area of ​​the photograph corresponded to 128 μm × 95 μm. The area fraction and average circle equivalent diameter of the hard structure observed as white structure were calculated using the image analyzer.

[0090] [Tensile test] JIS Z2241 No. 4 test pieces were prepared from each steel plate. The No. 4 test pieces were cut from the surface of the steel plate at a depth of t / 4 of the plate thickness, with the longitudinal direction of the test piece perpendicular to the rolling direction. Using these test pieces, tensile tests were performed in accordance with JIS Z2241 to measure the yield strength YS and tensile strength TS. The yield ratio YR was calculated by dividing YS by TS and multiplying the result by 100. A yield strength YS of 600 MPa or higher was evaluated as high strength. A YR of 85% or lower was evaluated as low yield ratio.

[0091] [Charpy impact test] V-notch Charpy impact test specimens according to JIS Z2242 were taken from each steel plate. Twelve specimens were cut from the steel plate at a depth of t / 4 of the plate thickness from the surface, with the longitudinal direction of the specimen aligned with the rolling direction. The twelve specimens were used to determine the absorbed energy and brittle fracture ratio under four test temperature conditions (three specimens per test temperature condition). vTrs was calculated from the relationship between the test temperature and the obtained brittle fracture ratio. A vTrs of -13°C or less was evaluated as having excellent low-temperature toughness.

[0092] The results of these measurements are shown in Table 3.

[0093] [Table 1]

[0094] [Table 2]

[0095] [Table 3]

[0096] Tables 1 to 3 reveal the following: K1, K4 to K11, and K15 to K20 satisfied the specified chemical composition and manufacturing conditions, and the resulting steel sheets had the desired structure, exhibited high strength, a lower yield ratio, and also exhibited excellent low-temperature toughness. In contrast, K2, K3, K12 to K14, K21, and K22 did not satisfy at least one of the chemical composition and manufacturing conditions, as explained below, and therefore did not achieve the desired structure, resulting in inferior yield strength, yield ratio, and / or low-temperature toughness.

[0097] K2 and K3 are normalized in the two-phase region, N Since the test was carried out under conditions below the specified range, an excessive amount of hard structure was formed, resulting in poor low-temperature toughness.

[0098] K12 to K14 are normalized in the two-phase region, N However, because the test was carried out under conditions exceeding the specified range, the hard tissue was insufficient and high strength could not be achieved.

[0099] K21 was not tempered, so it was not possible to achieve high strength.

[0100] In K22, Dh was below the specified range, so the average equivalent circle diameter of the hard structure exceeded the specified range and became large, resulting in poor low-temperature toughness. [Industrial Applicability]

[0101] The steel plate of the present invention can be used, for example, as a steel material for construction, mainly as a thick steel plate with a low yield ratio and high strength suitable for four-sided box column skin plates.

Claims

1. C: 0.040% by mass to 0.060% by mass, Si: 0.20% by mass to 0.30% by mass, Mn: 1.80% by mass to 2.00% by mass, P: more than 0% by mass, 0.010% by mass or less, S: more than 0 mass%, 0.003 mass% or less, Al: 0.040% by mass to 0.080% by mass, Cu: 0.40% by mass to 0.50% by mass, Ni: 1.40% by mass to 1.50% by mass, Cr: 0.90% by mass to 1.10% by mass, Mo: 0.17% by mass to 0.23% by mass, V: 0% by mass to 0.005% by mass, Nb: 0% by mass to 0.005% by mass, Ti: 0.010% by mass to 0.020% by mass, B: 0.0005% by mass to 0.0020% by mass, N: 0.0030% by mass to 0.0065% by mass, and Ca: 0.0005% by mass to 0.0035% by mass, the balance being iron and unavoidable impurities; Pcm represented by the following formula (1) is 0.30 or less, and Dh represented by the following formula (2) is 21.7 or more, A steel plate in which the fraction of a hard structure composed of bainite and island martensite MA in the entire steel structure is 23.3 area % to 29.4 area %, and the average equivalent circle diameter thereof is 5.2 μm or less. Pcm=[C]+[Si] / 30+[Mn] / 20+[Cu] / 20+[Ni] / 60+[Cr] / 20+[Mo] / 15+[V] / 10+5×[B]...(1) Dh=DI×([C]+0.91×[Mn]+0.97×[Cr]+0.88×[Mo])...(2) In formula (2), DI=1.16×([C] / 10) ０．５ ×(0.7×[Si]+1)×(5.1×([M]]-1.2)+5)×(0.35×[[]+1)×(0.36×[Ni]+1)×(2.16×[[]++1)×(3×[[]++1)×(1.75×[V]++1)×(200×[[]+1) ・・・(3) In the formulas (1) to (3), the [element symbol] indicates the content of each element in the steel expressed in mass %, and elements that are not contained are set to zero.

2. The steel plate according to claim 1, having a thickness of 80 mm or more.

3. A method for producing the steel sheet according to claim 1 or 2, A step of heating a steel slab having the component composition according to claim 1 to 900 to 1250°C; A process of rolling at a finish rolling temperature with a surface temperature of 800 ° C. to 970 ° C., and then cooling from a cooling start temperature of Ar3 point or higher calculated by the following formulas (4) and (5) at an average cooling rate of 1.0 ° C. / s or higher to a cooling stop temperature of 400 ° C. or lower; After the cooling, the temperature is within the range of Ac1 point calculated by the following formula (6) to Ac3 point calculated by the following formula (7), and the parameter P calculated by the following calculation method is Ｎ a step of performing two-phase region normalizing within a range of 21.04 to 22.12; The method for producing a steel sheet includes a step of tempering the steel sheet at a temperature in the range of 300°C to 600°C after the two-phase region normalizing. Ar3 points = 910-310×[C]+25×[Si]-80×[Mneq]...(4) In formula (4), [Mneq]=[Mn]+[Cr]+[Cu]+[Mo]+[Ni] / 2+10([Nb]-0.02)+1...(5) Ac1 point = 723-10.7 x [Mn] -16.9 x [Ni] + 29.1 x [Si] + 16.9 x [Cr] + 290 x [As] + 6.38 x [W] ... (6) Ac3 point = 910 - 203 × [C] ０．５ - 15.2 × [Ni] + 44.7 × [Si] + 104 × [V] + 31.5 × [Mo] + 13.1 × [W] ··· (7) In the formulas (4) to (7), the [element symbol] indicates the content of each element in the steel expressed in mass %, and elements that are not contained are set to zero. [Parameter P Ｎ Calculation method] Parameter P indicates the degree of normalization when the temperature is maintained at Tx (°C) for Kx (seconds). ｘ The parameter P, which represents the degree of normalization when the temperature of the steel sheet rises over time, is calculated using the following formula (8) according to the procedures shown in [1] to [5] below. Ｎ Calculate. P ｘ =(T ｘ +273)×log(K ｘ / 3600+20) ・・・(8) [1] A temperature rise curve of the steel sheet center temperature is created, with the vertical axis representing the steel sheet center temperature (°C) and the horizontal axis representing the elapsed time (seconds) from the start of temperature rise. The curve is divided into 1-second intervals and approximated as a rectangle. In the temperature rise curve, the point when the steel sheet center temperature is at point Ac1 is set as the starting point, and the elapsed time from the start of temperature rise at this starting point is t ０ (=0) (seconds), the steel plate center temperature is T ０ (= Ac1) (°C). [2] Degree of normalization P after 1 second has elapsed from the starting point ０ The degree of normalizing P ０ is expressed by adding x=0 to equation (8), that is, T ｘ =T ０ = Ac1 (°C), K ｘ =K ０ = 1 (seconds) to calculate. [3] The degree of normalizing P obtained in [2] above after 1 second has elapsed from the starting point ０ The steel plate center temperature T after 1 second has elapsed from the starting point １ Required retention time k １ The required retention time k １ is expressed by the following formula (9) with x=0, that is, P ｘ =P ０ , T ｘ＋１ =T １ (℃) to calculate. k ｘ＋１ =(exp(P Ｘ / (T Ｘ＋１ +273))-20)×3600 ・・・(9) In equation (9), k ｘ＋１ is from the starting point (Ac1) to T ｘ＋１ The degree of continuous normalizing P that was received until the temperature changed to (℃) ｘ The steel plate center temperature T required to obtain ｘ＋１ The required retention time at [4] Steel plate center temperature T １ The converted retention time at k is calculated from the following equation (10): １ +1=K １ (Seconds). K ｘ＋１ =k ｘ＋１ +1 ・・・(10) In formula (10), K ｘ＋１ is the steel plate center temperature T ｘ＋１ indicates the converted retention time at k Ｘ＋１ is from the starting point (Ac1) to T ｘ＋１ The degree of continuous normalizing P that was received until the temperature changed to (℃) ｘ The steel plate center temperature T required to obtain ｘ＋１ The required retention time at [5] P shown in [2] to [4] above ｘ , k ｘ＋１ , K. ｘ＋１ The calculation of X = 1, 2, 3, ... is repeated until normalizing in the two-phase region is completed, and P Ｘ =P ｆ Looking for this P ｆ P Ｎ Let's say.

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