WO2020119074A1 - 一种齿科用钛合金及其制备方法 - Google Patents

一种齿科用钛合金及其制备方法 Download PDF

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WO2020119074A1
WO2020119074A1 PCT/CN2019/093063 CN2019093063W WO2020119074A1 WO 2020119074 A1 WO2020119074 A1 WO 2020119074A1 CN 2019093063 W CN2019093063 W CN 2019093063W WO 2020119074 A1 WO2020119074 A1 WO 2020119074A1
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alloy
titanium alloy
titanium
dental use
preparation
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French (fr)
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汪涛
费阳
缪润杰
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南京航空航天大学
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    • C—CHEMISTRY; METALLURGY
    • C22—METALLURGY; FERROUS OR NON-FERROUS ALLOYS; TREATMENT OF ALLOYS OR NON-FERROUS METALS
    • C22C—ALLOYS
    • C22C14/00—Alloys based on titanium
    • C—CHEMISTRY; METALLURGY
    • C22—METALLURGY; FERROUS OR NON-FERROUS ALLOYS; TREATMENT OF ALLOYS OR NON-FERROUS METALS
    • C22C—ALLOYS
    • C22C1/00—Making non-ferrous alloys
    • C22C1/02—Making non-ferrous alloys by melting
    • C—CHEMISTRY; METALLURGY
    • C22—METALLURGY; FERROUS OR NON-FERROUS ALLOYS; TREATMENT OF ALLOYS OR NON-FERROUS METALS
    • C22F—CHANGING THE PHYSICAL STRUCTURE OF NON-FERROUS METALS AND NON-FERROUS ALLOYS
    • C22F1/00—Changing the physical structure of non-ferrous metals or alloys by heat treatment or by hot or cold working
    • C22F1/16—Changing the physical structure of non-ferrous metals or alloys by heat treatment or by hot or cold working of other metals or alloys based thereon
    • C22F1/18—High-melting or refractory metals or alloys based thereon
    • C22F1/183—High-melting or refractory metals or alloys based thereon of titanium or alloys based thereon

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  • the invention belongs to the technical field of titanium alloy material design and preparation, in particular to a titanium alloy for dental use and a preparation method thereof.
  • Titanium and titanium alloys have excellent biocompatibility, excellent corrosion resistance and suitable mechanical properties, and have become widely used biomedical materials at home and abroad.
  • the current mainstream medical grades of pure titanium and titanium alloys are TA2 and TC4, and the tensile strength of TA2 is only about 400 MPa, which cannot withstand large loads; although the strength of TC4 alloy is high, the plasticity is poor, and Contains biotoxic elements such as Al and V, and has poor biocompatibility.
  • Zirconium is a neutral element of the same family as titanium, which has high biological safety, and can be infinitely dissolved in both the alpha and beta phases of titanium; due to the different atomic radius of titanium and zirconium, the addition of zirconium in the titanium matrix can The effect of solid solution strengthening significantly improves the strength of the material and improves the corrosion resistance of the substrate.
  • the Ti-Zr alloy with ⁇ single-phase structure is very compatible with the SLA surface treatment technology for dental implants, and has both excellent biocompatibility and mechanical properties. Therefore, in the field of oral restoration, Ti-Zr alloy is a medical material with broad application prospects.
  • Ti-Zr binary alloys Recently, the research on the mechanical properties of Ti-Zr binary alloys is increasing. Bernhard et al. Ti-Zr alloy (Zr: 13wt% ⁇ 17wt%), its tensile strength can reach 953MPa (Bernhard N, Berner S, De Wild M, et al.
  • Steinemann proposed a Ti-Zr alloy (Zr: 5wt% ⁇ 25wt%, O: 0.1wt% ⁇ 0.3wt%) and its preparation in the US patent with the publication date of May 1, 2012, and the publication number of US8168012B2 Process, found that when adding trace oxygen elements and subsequent cold work, the tensile strength of the alloy can be close to 1000MPa, due to solid solution strengthening and work hardening, although the strength of the system Ti-Zr alloy has been greatly improved, but the plasticity sharply decreased. Vicente et al.
  • Ti-Zr alloy does not reflect the strength and plasticity level suitable for dentistry (Vicente FB, Correa DRN, Donato TAG, et al, The influence of small quantities of oxygen in the structure, microstructure, hardness, elasticity modulus and cytocompatibility of Ti-Zr alloys for dental applications, Materials, 2014, 7(1):542–553). Zhou Yunkai et al.
  • Ti-70wt% Zr alloy had the highest tensile strength of 1216.68MPa, but the elongation was only 7.12% (Zhou YK, Jing R, Ma MZ, et al, Tensile strength of Zr-Ti binary alloy, Chinese Physics Letters, 2013, 30(11): 116201).
  • the hot rolling process has a significant strengthening effect, and its tensile strength can reach up to 1135MPa, but the elongation after break is only 3% (Zhou YK, Liang SX, Jing R, et al, Microstructure and tensile properties of hot-rolled Zr 50 -Ti 50 binary alloy, Materials Science & Engineering A, 2014,621:259-264).
  • the high zirconium content of the above two alloys not only increases the cost of raw materials, but also seriously reduces the plasticity.
  • the zirconium content of the Ti-Zr alloy in the current study is generally greater than 5 wt%, and the raw material cost is high; and the contradiction between high strength and low plasticity cannot be solved by a simple and easy preparation method.
  • the invention provides a titanium alloy for dental use and a preparation method thereof.
  • the titanium alloy has high strength and good plasticity, and has fewer alloying elements added, simple preparation process, and low economic cost .
  • the present invention adopts the following technical solutions:
  • a titanium alloy for dental use including the following weight percent chemical composition:
  • the balance is Ti.
  • the titanium alloy described above has a twin-scale structure at the nanometer scale.
  • a preparation method of titanium alloy for dental use includes the following steps:
  • Step 1 According to the composition ratio, the raw materials are vacuum smelted multiple times to obtain an alloy ingot with a uniform composition
  • Step 2 Place the alloy ingot described in Step 1 in a vacuum furnace, perform homogenization annealing under the protection of argon, and then slowly cool to room temperature;
  • Step 3 The alloy ingot after homogenization and annealing described in Step 2 is thermoformed in different phase regions and cooled to room temperature to obtain a final plate.
  • the heating temperature for homogenization annealing in step 2 is 1000°C to 1100°C
  • the holding time is 6 to 8h
  • the cooling method is furnace cooling.
  • the heating temperature of the thermoplastic forming described in step 3 is 800°C to 1000°C, the amount of deformation is 85% to 90%, and the cooling method is air cooling.
  • the present invention provides a titanium alloy for dental use and a preparation method thereof.
  • the types and contents of alloying elements of the titanium alloy are: Zr (0.5wt% ⁇ 4.5wt%), O (0.05wt% ⁇ 0.4wt%), the tensile strength can be up to 955MPa, reaching the level of ⁇ + ⁇ Ti-6Al-4V alloy; elongation after breaking can be up to 30%, nano-scale twins appear in the alloy, making the alloy in It has high strength and good plasticity at the same time; and the titanium alloy room temperature structure of the present invention is ⁇ single-phase structure, which is compatible with the existing SLA surface treatment process for dental implants; does not contain Al, V and other biological toxic elements , Excellent biocompatibility; less alloying elements added, the preparation process is thermoplastic forming, the preparation process is simple, the economic cost is low, and has high operability; its excellent comprehensive performance fully meets the requirements of clinical application It can be used as a dental implant material in the field of oral restoration and has broad application
  • Ti-1Zr-0.05O alloy of the present invention after SLA treatment
  • FIG. 3 is a surface morphology diagram of the Ti-1Zr-0.30O alloy of the present invention after SLA treatment;
  • FIG. 5 is a nano twin structure appearing in the Ti-3Zr-0.25O alloy of the present invention.
  • FIG. 6 is a surface morphology diagram of the Ti-3Zr-0.25O alloy of the present invention after SLA treatment;
  • Fig. 9 shows the nano twin structure of Ti-4Zr-0.25O alloy of the present invention.
  • Step 1 The weight percentage of each component is: Zr: 1wt%, O: 0.05wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat to 1000°C under the protection of argon, keep the furnace cool to room temperature after 8 hours of heat preservation, and complete the homogenization annealing;
  • Step 4 The homogenized and annealed alloy is subjected to thermoplastic deformation at 950°C with a deformation amount of 90%, and then air-cooled to room temperature to obtain a final plate.
  • Phase analysis X-ray diffraction analysis samples were prepared from the above-mentioned plates by wire cutting, and the alloy phase composition was determined by X-ray diffractometer, with a scanning angle of 20° to 80° and a scanning speed of 10°/min. The X-ray diffraction pattern is shown in Fig. 1, and it is determined that the alloy is composed of ⁇ single phase.
  • Room temperature tensile test cut the tensile sample from the alloy produced by wire cutting, polish the surface oxide scale of the sample, confirm that the sample gauge length is 20mm, and the thickness is 1.5mm, and perform the test on the universal material testing machine
  • the tensile speed is 1mm/min, a group of samples are measured three times in parallel.
  • the specific data of the mechanical properties of the tensile samples are shown in Table 1.
  • the average tensile strength is 510 MPa
  • the average yield strength is 446 MPa
  • the average elongation after break is 30%.
  • SLA (sandblast acid etching treatment) treatment cut the SLA treated sample from the alloy produced by wire cutting, polish the surface of the sample to no obvious scratches and perform ultrasonic cleaning, then perform SLA treatment on the sample surface and scan with Observe the surface morphology by electron microscope.
  • the surface morphology of the sample after treatment is shown in Figure 2.
  • the surface can form a multi-level nested three-dimensional hole structure typical of SLA treatment.
  • Step 1 The weight percentage of each component is: Zr: 1wt%, O: 0.30wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat to 1100°C under the protection of argon, keep the furnace cool to room temperature after holding for 6 hours, and complete the homogenization annealing;
  • Step 4 The homogenized and annealed alloy is subjected to thermoplastic deformation at 1000°C with a deformation amount of 88%, and then air-cooled to room temperature to obtain the final shape.
  • Example 1 The operation steps of the phase analysis are the same as in Example 1.
  • the X-ray diffraction pattern is shown in Figure 1, and it is determined that the alloy is composed of ⁇ single phase.
  • the operation procedure of the room temperature tensile test is the same as that in Example 1.
  • the specific data of the mechanical properties of the tensile samples are shown in Table 1.
  • the average tensile strength is 885 MPa
  • the average yield strength is 730 MPa
  • the average elongation after break is 12.8%.
  • the operation steps of SLA treatment are the same as those in Example 1.
  • the surface morphology of the sample after treatment is shown in FIG. 3.
  • the surface can form a multi-level nested three-dimensional hole structure typical of SLA treatment.
  • Step 1 The weight percentage of each component is: Zr: 2wt%, O: 0.10wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat to 1000°C under the protection of argon, keep the furnace cool to room temperature after 8 hours of heat preservation, and complete the homogenization annealing;
  • Step 4 The above homogenized and annealed alloy is subjected to thermoplastic deformation at 900°C with a deformation amount of 88%, and then air-cooled to room temperature to obtain the final shape.
  • Example 1 The operation steps of the phase analysis are the same as in Example 1.
  • the X-ray diffraction pattern is shown in Figure 1, and it is determined that the alloy is composed of ⁇ single phase.
  • Example 1 The operation steps of the room temperature tensile test are the same as in Example 1.
  • the specific data of the mechanical properties of the tensile samples are shown in Table 1.
  • the average tensile strength is 601 MPa
  • the average yield strength is 508 MPa
  • the average elongation after break is 22.8%.
  • the operation steps of SLA treatment are the same as those in Example 1.
  • the surface morphology of the sample after treatment is shown in FIG. 4.
  • the surface can form a multi-level nested three-dimensional hole structure typical of SLA treatment.
  • Step 1 The weight percentage of each component is: Zr: 3wt%, O: 0.25wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat to 1000°C under the protection of argon, keep the furnace cool to room temperature after 8 hours of heat preservation, and complete the homogenization annealing;
  • Step 4 The above homogenized and annealed alloy is subjected to thermoplastic deformation at 800°C with a deformation amount of 86%, and then air-cooled to room temperature to obtain the final shape.
  • Example 1 The operation steps of the phase analysis are the same as in Example 1.
  • the X-ray diffraction pattern is shown in Figure 1, and it is determined that the alloy is composed of ⁇ single phase.
  • the operation procedure of the room temperature tensile test is the same as that in Example 1.
  • the specific data of the mechanical properties of the tensile samples are shown in Table 1.
  • the average tensile strength is 854 MPa
  • the average yield strength is 799 MPa
  • the average elongation after break is 23.1%.
  • the operation steps of SLA treatment are the same as those in Example 1.
  • the surface morphology of the sample after treatment is shown in FIG. 6, and the surface can form a multi-level nested three-dimensional hole structure typical of SLA treatment.
  • Step 1 The weight percentage of each component is: Zr: 3wt%, O: 0.30wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat to 1100°C under the protection of argon, keep the furnace cool to room temperature after holding for 6 hours, and complete the homogenization annealing;
  • Step 4 The homogenized and annealed alloy is subjected to thermoplastic deformation at 950°C with a deformation amount of 90%, and then air-cooled to room temperature to obtain the final shape.
  • Example 1 The operation steps of the phase analysis are the same as in Example 1.
  • the X-ray diffraction pattern is shown in Figure 1, and it is determined that the alloy is composed of ⁇ single phase.
  • the operation procedure of the room temperature tensile test is the same as that of Example 1.
  • the specific data of the mechanical properties of the tensile specimen are shown in Table 1.
  • the average tensile strength is 955 MPa
  • the average yield strength is 893 MPa
  • the average elongation after break is 10.8%.
  • nitric acid, hydrofluoric acid and distilled water to prepare 50ml of corrosion solution, the specific volume ratio is 25:5:70, drop 3-4 drops of corrosion solution on the sample, and let it dry for about 5 seconds.
  • XJP-300 metallurgical microscope (OM) was used for selective shooting.
  • the metallographic microstructure after thermoplastic deformation is shown in Fig. 7, which is mainly composed of lamellar ⁇ phase and equiaxed ⁇ phase.
  • Step 1 The weight percentage of each component is: Zr: 4wt%, O: 0.25wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat it to 1050°C under the protection of argon, keep the furnace cool to room temperature after 7 hours of heat preservation, and complete the homogenization annealing;
  • Step 4 The homogenized and annealed alloy is subjected to thermoplastic deformation at 1000°C with a deformation amount of 88%, and then air-cooled to room temperature to obtain the final shape.
  • Example 1 The operation steps of the phase analysis are the same as in Example 1.
  • the X-ray diffraction pattern is shown in Figure 1, and it is determined that the alloy is composed of ⁇ single phase.
  • the operation procedure of the room temperature tensile test is the same as that in Example 1.
  • the specific data of the mechanical properties of the tensile samples are shown in Table 1.
  • the average tensile strength is 862 MPa
  • the average yield strength is 723 MPa
  • the average elongation after break is 19.7%.
  • the metallographic observation steps are the same as in Example 5.
  • the metallographic microstructure after thermoplastic deformation is shown in Fig. 8, which is mainly composed of lamellar ⁇ phase and equiaxed ⁇ phase.
  • Example 4 The TEM observation procedure is the same as in Example 4. The TEM morphology of this component alloy is shown in Fig. 9, and a clear nano twin structure can be seen.
  • the titanium alloy described in Example 1 has a tensile strength increased by 27.5%, a yield strength increased by 62.2%, and an elongation after break increased by 20% compared with the TA2 minimum mechanical performance index described in Chinese standard GB/T13810-2007.
  • the titanium alloy Compared with the Ti-14.9Zr alloy described in US8168012B2, the titanium alloy has a tensile strength increased by 35%, a yield strength increased by 30.5%, and the elongation after breaking is equivalent; the titanium alloy described in Example 3 is described in US8168012B2 Compared with Ti-5.1Zr alloy, the tensile strength is increased by 21.2%, the yield strength is increased by 18.4%, and the elongation after fracture is equivalent; the titanium alloy described in Example 4 is compared with the Ti-14.9Zr alloy described in US Patent US8168012B2.
  • the strength is increased by 30.4%, the yield strength is increased by 42.9%, and the elongation after break is increased by 59.3%;
  • the titanium alloy described in Example 5 has an increased tensile strength compared with the minimum mechanical performance index of the TC4 alloy described in Chinese standard GB/T13810-2007 3.2%, yield strength increased by 2.6%, elongation after break increased by 8%; compared with Ti-14.9Zr alloy described in US Patent US8168012B2, the titanium alloy described in Example 6 had a 31.6% increase in tensile strength and a 29.3% increase in yield strength , Elongation after break increased by 35.9%.
  • the strength and plasticity of the titanium alloy of the present invention have been significantly improved. From the test results of TEM, it can be seen that an obvious nano-twist structure appears in the alloy structure, which significantly improves the mechanical properties of the alloy.
  • Step 1 The weight percentage of each component is: Zr: 0.5wt%, O: 0.05wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat to 1000°C under the protection of argon, keep the furnace cool to room temperature after 6 hours of heat preservation, and complete the homogenization annealing;
  • Step 4 The above homogenized and annealed alloy is subjected to thermoplastic deformation at 1000°C with a deformation amount of 85%, and then air-cooled to room temperature to obtain the final shape.
  • Step 1 The weight percentage of each component is: Zr: 4.5wt%, O: 0.4wt%, the balance is Ti;
  • Step 2 According to the composition ratio, the raw material sponge titanium, sponge zirconium and titanium dioxide are vacuum-smelted multiple times to obtain an alloy ingot with a uniform composition;
  • Step 3 Place the above alloy ingot in a vacuum furnace, heat it to 1100°C under the protection of argon, keep the furnace cool to room temperature after 8 hours of heat preservation, and complete the homogenization annealing;
  • Step 4 The above homogenized and annealed alloy is subjected to thermoplastic deformation at 1000°C with a deformation amount of 90%, and then air-cooled to room temperature to obtain the final shape.

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Abstract

本发明公开了一种齿科用钛合金及其制备方法，属于钛合金材料设计及制备技术领域，该钛合金具有高强度和良好的塑性，而且制备工艺简单。本发明包括：以下重量百分比化学成分：Zr：0.5wt％～4.5wt％，O：0.05wt％～0.4wt％，余量为Ti，合金中出现纳米孪晶结构；采用真空熔炼法获得合金铸锭，经均匀化退火后在不同相区进行热塑性变形，获得最终形状。本发明的钛合金具有较高的强度，抗拉强度最高可达955MPa，而且具有良好的塑性，断后伸长率最高可达30％；该体系合金具有α单相组织，与牙种植体用SLA表面处理工艺兼容，不含Al、V等生物毒性元素，生物相容性优异；所添加合金化元素较少，制备工艺简单，经济成本较低，在口腔修复领域具有广阔的应用前景。

Description

一种齿科用钛合金及其制备方法 技术领域
本发明属于钛合金材料设计及制备技术领域，尤其涉及一种齿科用钛合金及其制备方法。
背景技术
钛及钛合金具有优异的生物相容性、极佳的耐蚀性及适宜的力学性能，已成为国内外应用广泛的生物医用材料。在齿科领域，目前主流的医用纯钛和钛合金牌号为TA2及TC4，而TA2的抗拉强度只有400MPa左右，无法承受较大载荷；TC4合金则虽然强度较高，但塑性较差，且含有Al、V等生物毒性元素，生物相容性也差。锆为与钛同族的中性元素，具有较高的生物安全性，且在钛的α与β相中均能无限固溶；由于钛与锆的原子半径不同，在钛基体中添加锆能起固溶强化的作用，显著提高材料强度，并提高基体耐蚀性。此外，具有α单相组织的Ti-Zr合金与牙种植体用SLA表面处理技术兼容性极佳，同时兼具优异的生物相容性和力学性能。因此，在口腔修复领域，Ti-Zr合金是一种应用前景广阔的医用材料。
近来，关于Ti-Zr二元合金力学性能的研究日益增多。Bernhard等开发了一种名为
Figure PCTCN2019093063-appb-000001
的Ti-Zr合金(Zr：13wt％～17wt％)，其抗拉强度可达953MPa(Bernhard N,Berner S,De Wild M,et al.The binary TiZr alloy-A newly developed Ti alloy for use in dental implants,Forum Implantologicum,2009,5:30-39),但其所述钛合金锆含量较高，加工方式复杂，制备成本较高。Steinemann在公开日为2012年5月1日，公开号为US8168012B2的美国专利中提出了一种Ti-Zr合金(Zr:5wt％～25wt％,O:0.1wt％～0.3wt％)及其制备工艺，发现当添加微量氧元素和进行后续冷加工时，合金的抗拉强度可接近1000MPa,由于固溶强化及加工硬化，该体系Ti-Zr合金虽然强度有较大提升，但塑性急剧降低。Vicente等通过在Ti-Zr合金(Zr:5wt％～15wt％)中添加0.02wt％～0.04wt％的氧元素，发现在该成分范围内合金的晶体结构、显微组织及生物相容性无明显变化，其所述Ti-Zr合金并未体现适用于齿科的强度及塑性水平(Vicente F B,Correa D R N,Donato T A G,et al,The influence of small quantities of oxygen in the structure,microstructure,hardness,elasticity modulus and cytocompatibility of Ti-Zr alloys for  dental applications,Materials,2014,7(1):542–553)。周云凯等研究了不同锆含量(Zr:17wt％～94wt％)的铸态Ti-Zr合金，发现Ti-70wt％Zr合金抗拉强度最高，为1216.68MPa,但伸长率仅为7.12％(Zhou Y K,Jing R,Ma M Z,et al,Tensile strength of Zr-Ti binary alloy,Chinese Physics Letters,2013,30(11):116201)。此外，他们还发现对于Ti-65wt％Zr合金，热轧工艺具有显著的强化效果，其抗拉强度最高可达1135MPa，但断后伸长率仅为3％(Zhou Y K,Liang S X,Jing R,et al,Microstructure and tensile properties of hot-rolled Zr 50-Ti 50binary alloy,Materials Science&Engineering A,2014,621:259-264)。上述两种合金锆含量过高，不仅增加了原料成本，同时严重降低了塑性。Matayshi等研究了不同锆含量的轧制态Ti-Zr合金(Zr:2wt％、10wt％),发现Ti-10wt％Zr合金具有最高抗拉强度，为877MPa，但伸长率仅为6.2％，综合力学性能不够理想(Matayshi Y,Homma T,Effect of Zr addition on recrystallization behavior in rolled Ti-Zr alloys,TMS(The Minerals,Metals&Materials Society)Supplemental Proceedings,2015,979-988)。
综上所述，目前研究中Ti-Zr合金的锆含量普遍大于5wt％,原料成本较高；且无法通过简单易行的制备方法解决高强度和低塑性之间的矛盾。
发明内容
本发明提供了一种齿科用钛合金及其制备方法，所述钛合金在具有高强度的同时，兼具良好的塑性，而且所添加合金化元素较少，制备工艺简单，经济成本较低。
为实现以上目的，本发明采用以下技术方案：
一种齿科用钛合金，包括以下重量百分比化学成分：
Zr：0.5wt％～4.5wt％；
O：0.05wt％～0.4wt％；
余量为Ti。
以上所述钛合金中有纳米尺度的孪晶结构。
一种齿科用钛合金的制备方法，包括以下步骤：
步骤一：根据成分配比，将原料进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤二：将步骤一所述的合金铸锭置于真空炉中，在氩气保护下进行均匀化退火，随后缓冷至室温；
步骤三：将步骤二所述的均匀化退火后的合金铸锭在不同相区进行热塑性成形并冷却至室温，获得最终板材。
以上所述步骤中，步骤二所述的均匀化退火加热温度为1000℃～1100℃，保温时间为6～8h，冷却方式为炉冷。
步骤三所述的热塑性成形加热温度为800℃～1000℃，变形量为85％～90％，冷却方式为空冷。
本发明的有益效果：本发明提供了一种齿科用钛合金及其制备方法，所述钛合金合金化元素种类及含量为：Zr(0.5wt％～4.5wt％)、O(0.05wt％～0.4wt％)，抗拉强度最高可达955MPa，达到α+β型Ti-6Al-4V合金水平；断后伸长率最高可达30％，合金中出现了纳米尺度的孪晶，使合金在具有高强度的同时兼具良好的塑性；而且本发明的钛合金室温组织皆为α单相组织，与现有的牙种植体用SLA表面处理工艺相兼容；不含Al、V等生物毒性元素，生物相容性优异；所添加合金化元素较少，制备工艺为热塑性成形，制备工艺简单，经济成本较低，具有较高的可操作性；其优异的综合性能完全满足了临床应用的要求，可作为牙种植体用材料用于口腔修复领域，具有广阔的应用前景。
附图说明
图1为本发明所述钛合金的X射线衍射图谱；
图2为本发明Ti-1Zr-0.05O合金SLA处理后的表面形貌图；
图3为本发明Ti-1Zr-0.30O合金SLA处理后的表面形貌图；
图4为本发明Ti-2Zr-0.10O合金SLA处理后的表面形貌图；
图5为本发明Ti-3Zr-0.25O合金中出现的纳米孪晶结构；
图6为本发明Ti-3Zr-0.25O合金SLA处理后的表面形貌图；
图7为本发明Ti-3Zr-0.30O合金热塑性变形后金相图；
图8为本发明Ti-4Zr-0.25O合金热塑性变形后金相图；
图9为本发明Ti-4Zr-0.25O合金中出现的纳米孪晶结构。
具体实施方式
下面结合附图和具体实施例对本发明进行具体说明：
实施例1
步骤一：各组分重量百分比为：Zr：1wt％，O：0.05wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1000℃，保温8h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在950℃进行热塑性变形，变形量为90％，随后空冷至室温，获得最终板材。
物相分析：用线切割从上述板材上制取X射线衍射分析试样，用X射线衍射仪确定合金物相组成，扫描角度为20°～80°，扫描速度为10°/min。X射线衍射图谱如图1所示，确定合金由α单相组成。
室温拉伸试验：用线切割从所制合金上切出拉伸试样，打磨净试样表面氧化皮，确定试样标距为20mm，厚度为1.5mm，在万能材料试验机上对试样进行室温拉伸试验，拉伸速度为1mm/min，一组试样平行测定3次。拉伸试样力学性能的具体数据如表1所示，平均抗拉强度为510MPa，平均屈服强度为446MPa，平均断后伸长率为30％。
SLA(喷砂酸蚀处理)处理：用线切割从所制合金上切下SLA处理试样，打磨试样表面至无明显划痕并进行超声清洗，随后对试样表面进行SLA处理，并用扫描电镜观察表面形貌。处理后的试样表面形貌如图2所示，表面可形成SLA处理典型的多级嵌套式三维孔洞结构。
实施例2
步骤一：各组分重量百分比为：Zr：1wt％，O：0.30wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1100℃，保温6h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在1000℃进行热塑性变形，变形量为88％，随后空冷至室温，获得最终形状。
物相分析操作步骤同实施例1，X射线衍射图谱如图1所示，确定合金由α单相组成。
室温拉伸试验操作步骤同实施例1，拉伸试样力学性能的具体数据如表1所示，平均抗拉强度为885MPa，平均屈服强度为730MPa，平均断后伸长率为12.8％。
SLA处理操作步骤同实施例1，处理后试样表面形貌如图3所示，表面可以形成SLA处理典型的多级嵌套式三维孔洞结构。
实施例3
步骤一：各组分重量百分比为：Zr：2wt％，O：0.10wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1000℃，保温8h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在900℃进行热塑性变形，变形量为88％，随后空冷至室温，获得最终形状。
物相分析操作步骤同实施例1，X射线衍射图谱如图1所示，确定合金由α单相组成。
室温拉伸试验操作步骤同实施例1，拉伸试样力学性能的具体数据如表1所示，平均抗拉强度为601MPa，平均屈服强度为508MPa，平均断后伸长率为22.8％。
SLA处理操作步骤同实施例1，处理后试样表面形貌如图4所示，表面可以形成SLA处理典型的多级嵌套式三维孔洞结构。
实施例4
步骤一：各组分重量百分比为：Zr：3wt％，O：0.25wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1000℃，保温8h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在800℃进行热塑性变形，变形量为86％，随后空冷至室温，获得最终形状。
物相分析操作步骤同实施例1，X射线衍射图谱如图1所示，确定合金由α单相组成。
室温拉伸试验操作步骤同实施例1，拉伸试样力学性能的具体数据如表1所示，平均抗拉强度为854MPa，平均屈服强度为799MPa，平均断后伸长率为23.1％。
TEM观察：从轧制后的板材上用电火花线切割切下一块1.5mm厚的小圆片，将其用502胶水粘在样品座表面，然后在水砂纸上进行减薄，减薄至100μm左右后，用丙酮把胶水溶化后取下薄片试样。接着使用模具在薄片试样上裁剪下直径30mm的圆片，使用双喷离子减薄法获得最终的测试样品，电解溶液为5％高氯酸+35％正丁醇+60％甲醇混合液。样品准备完毕后，使用型号为JEM-1200EX的透射电镜(TEM)进行观察，加速电压控制在120kV。合金的TEM形貌如图5所示，出现了明显的纳米孪晶结构。
SLA处理操作步骤同实施例1，处理后试样表面形貌如图6所示，表面可以形成SLA处理典型的多级嵌套式三维孔洞结构。
实施例5
步骤一：各组分重量百分比为：Zr：3wt％，O：0.30wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1100℃，保温6h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在950℃进行热塑性变形，变形量为90％，随后空冷至室温，获得最终形状。
物相分析操作步骤同实施例1，X射线衍射图谱如图1所示，确定合金由α单相组成。
室温拉伸试验操作步骤同实施例1，拉伸试样力学性能的具体数据如表1所示，平均抗拉强度为955MPa，平均屈服强度为893MPa，平均断后伸长率为10.8％。
金相观察：用电火花线切割机从热轧后的板材上切下尺寸为10mm×10 mm×1.5mm小试样，然后使用XQ-2B型镶嵌机，往里面添加酚醛树脂镶嵌料进行热镶嵌。依次使用耐水砂纸、0#到7#细砂纸分别对镶嵌好的试样进行粗磨和细磨，磨至划痕统一。然后在金相抛光机上进行抛光，全程滴加三氧化二铬悬浮溶液，抛光到试样表面没有明显划痕为止，抛光结束后依次使用清水和酒精进行清洗并吹干。使用硝酸、氢氟酸和蒸馏水来配制50ml腐蚀溶液，具体的体积比为25:5:70，在试样上滴3-4滴腐蚀溶液，静置5秒钟左右吹干。用XJP-300型金相显微镜(OM)进行选域拍摄。热塑性变形后的金相显微组织如图7所示，主要由片状α相和等轴状α相组成。
实施例6
步骤一：各组分重量百分比为：Zr：4wt％，O：0.25wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1050℃，保温7h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在1000℃进行热塑性变形，变形量为88％，随后空冷至室温，获得最终形状。
物相分析操作步骤同实施例1，X射线衍射图谱如图1所示，确定合金由α单相组成。
室温拉伸试验操作步骤同实施例1，拉伸试样力学性能的具体数据如表1所示，平均抗拉强度为862MPa，平均屈服强度为723MPa，平均断后伸长率为19.7％。
金相观察步骤同实施例5，热塑性变形后的金相显微组织如图8所示，主要由片状α相和等轴状α相组成。
TEM观察步骤同实施例4，该成分合金的TEM形貌如图9所示，可以看到明显的纳米孪晶结构。
表1实施例1～6所述钛合金与其他文献所述钛合金力学性能对比
Figure PCTCN2019093063-appb-000002
表1中6个实施例所述钛合金与其他文献所述钛合金的力学性能对比仅选择断后伸长率符合GB/T 13810-2007最低水平(10％)的合金。
实施例1所述钛合金与中国标准GB/T 13810-2007所述TA2最低力学性能指标相比，抗拉强度提高27.5％，屈服强度提高62.2％，断后伸长率提高20％；实施例2所述钛合金与美国专利US8168012B2所述Ti-14.9Zr合金相比，抗拉强度提高35％，屈服强度提高30.5％，断后伸长率相当；实施例3所述钛合金与美国专利US8168012B2所述Ti-5.1Zr合金相比，抗拉强度提高21.2％，屈服强度提高18.4％，断后伸长率相当；实施例4所述钛合金与美国专利US8168012B2所述Ti-14.9Zr合金相比，抗拉强度提高30.4％，屈服强度提高42.9％，断后伸长率提高59.3％；实施例5所述钛合金与中国标准GB/T 13810-2007所述TC4合金最低力学性能指标相比，抗拉强度提高3.2％，屈服强度提高2.6％，断后伸长率提高8％；实施例6所述钛合金与美国专利US8168012B2所述Ti-14.9Zr合金相比，抗拉强度提高31.6％，屈服强度提高29.3％，断后伸长率提高35.9％。本发明钛合金的强度和塑性都有了明显地提高，从TEM的测试结果看合金结构中出现了明显的纳米孪晶结构，使合金的力学性能有了明显的提升。
实施例7
步骤一：各组分重量百分比为：Zr：0.5wt％，O：0.05wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1000℃，保温6h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在1000℃进行热塑性变形，变形量为85％，随后空冷至室温，获得最终形状。
实施例8
步骤一：各组分重量百分比为：Zr：4.5wt％，O：0.4wt％，余量为Ti；
步骤二：根据成分配比，将原料海绵钛、海绵锆以及二氧化钛进行多次真空熔炼，获得成分均匀的合金铸锭；
步骤三：将上述合金铸锭置于真空炉中，在氩气保护下加热至1100℃，保温8h后炉冷至室温，完成均匀化退火；
步骤四：将上述均匀化退火后的合金在1000℃进行热塑性变形，变形量为90％，随后空冷至室温，获得最终形状。
以上所述仅是本发明的优选实施方式，应当指出：对于本技术领域的普通技术人员来说，在不脱离本发明原理的前提下，还可以做出若干改进和润饰，这些改进和润饰也应视为本发明的保护范围。

Claims (8)

  1. 一种齿科用钛合金，其特征在于，包括以下重量百分比化学成分：
    Zr：0.5wt％～4.5wt％；
    O：0.05wt％～0.4wt％；
    余量为Ti。
  2. 根据权利要求1所述的齿科用钛合金，其特征在于，所述钛合金结构中有纳米尺度的孪晶结构。
  3. 一种齿科用钛合金的制备方法，其特征在于，包括以下步骤：
    步骤一：根据成分配比，将原料进行多次真空熔炼，获得成分均匀的合金铸锭；
    步骤二：将步骤一所述的合金铸锭置于真空炉中，在氩气保护下进行均匀化退火，随后缓冷至室温；
    步骤三：将步骤二所述的均匀化退火后的合金铸锭在不同相区进行热塑性成形并冷却至室温，获得最终板材。
  4. 根据权利要求3所述的齿科用钛合金的制备方法，其特征在于，所述原料为海绵钛、海绵锆以及二氧化钛。
  5. 根据权利要求3所述的齿科用钛合金的制备方法，其特征在于，步骤二所述的均匀化退火加热温度为1000℃～1100℃，保温时间为6～8h。
  6. 根据权利要求3或5所述的齿科用钛合金的制备方法，其特征在于，步骤二所述冷却方式为炉冷。
  7. 根据权利要求3所述的齿科用钛合金的制备方法，其特征在于，步骤三所述的热塑性成形加热温度为800℃～1000℃，变形量为85％～90％。
  8. 根据权利要求3或7所述的齿科用钛合金的制备方法，其特征在于，步骤三所述冷却方式为空冷。
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