【作者】 梁莎莎；

【导师】 杨昇；

【作者基本信息】 郑州大学 ， 凝聚态物理， 2013， 硕士

【摘要】 研究表明在铝合金中加入Ti、Zr、B等元素,并合理分配它们的比例份额可以提高铝合金的耐热性,同时保持较好的导电性及强度。但对结果的理论解释研究较少。通过扫描电镜测试及透射电镜测试研究了在纯铝中添加Ti、Zr、B等元素后的生成物及其对铝合金微观组织的影响,实验表明在原子数目比为(Ti+Zr)：B>1：2的3种不同样品中不仅有六方结构的TiB2、ZrB2存在,还有立方结构的Al3Ti和Al3Zr存在；而在原子数目比为(Ti+Zr)：B=1：2的3种不同样品中只发现了六方结构的TiB2、ZrB2晶粒。另外金相结果显示原子数目比为(Ti+Zr)：B>1：2的3种样品的晶粒度普遍小于原子数目比为(Ti+Zr)：B=1：2的3种样品。即当原子数目比为(Ti+Zr)：B=1：2时,细化程度不明显,对电子的散射较少,导电性能受影响较少。通过EET理论中的键距差法(Bond Length Difference,简称BLD方法)分别计算了TiB2、ZrB2、Al3Ti和Al3Zr的键能、共价电子数。结果表明TiB2的最强键键能、最强键的共价电子数远大于Al3Ti的最强键键能、最强键的共价电子数,所以Ti元素优先和B元素结合成TiB2,当B元素消耗完,Ti元素再与Al结合成Al3Ti。ZrB2的最强键键能、最强键的共价电子数远大于Al3Zr的最强键键能、最强键的共价电子数,所以Zr元素优先和B元素结合成ZrB2,当B元素消耗完,Zr再与Al结合成Al3Zr。即六方结构的TiB2、ZrB2优先于立方结构的Al3Ti和Al3Zr生成。这就解释了扫描电镜的实验结果：当原子数目比为(Ti+Zr)：B=1：2时,只发现了六方结构的晶粒(TiB2、ZrB2)；当原子数目比为(Ti+Zr)：B>1：2时,既有六方结构的晶粒(TiB2、ZrB2)存在,又有立方晶格结构的晶粒(Al3Ti、Al3Zr)存在。分别计算α-Al与TiB2、ZrB2、Al3Ti和Al3Zr的最相似面的电子面密度差。Al3Ti的(111)面和α-Al的最相似面(111)面的电子面密度差1.11%远小于TiB2的(0001)面与α-Al的最相似面(110)面的电子面密度差40.88%；Al3Zr的(111)面和α-Al的最相似面(111)面的电子面密度差20.05%远小于ZrB2的(0001)面与α-Al的最相似面(110)面的电子面密度差87.09%。因为电子面密度差越小,两种物质电子的连续性越好,越有助于α-Al的异质形核,细化效果越显著。所以原子数目比为(Ti+Zr)：B>1：2的样品的晶粒度普遍小于原子数目比为(Ti+Zr)：B=1：2的样品的晶粒度,与金相实验的结果相符合。而晶粒越小,晶界对电子的散射就越严重,进而降低合金的导电性。所以,控制Ti、Zr、B的添加量,使原子数目比为(Ti+Zr)：B=1：2,此时只有TiB2、ZrB2生成,铝合金的导电性较好。分别计算a-Al、TiB2、ZrB2、Al3Ti和Al3Zr的派-纳力,计算结果为：0.05002、14.03272、7.20934、0.43760、0.21664。TiB2、ZrB2、Al3Ti和Al3Zr派-纳力较大,位错无法从TiB2、ZrB2基体切过,只能采取绕过机制。当它们大量细小弥散地分布在铝基体中时,会对位错起到很好的钉扎效果,提高再结晶温度,使铝合金保持良好耐热性。综合分析添加不同含量Ti、Zr、B后,对铝合金的导电性、耐热性的影响,可知应使Ti、Zr、B的原子数目比为(Ti+Zr)：B=1：2,此时铝合金既保持了良好的耐热性,又未降低导电性。更多还原

【Abstract】 Research shows that adding Ti, Zr and B in the aluminum alloy, and rational allocation of their proportional share can improve the heat resistance of aluminum alloy, while maintaining good electrical conductivity and strength. But the theory research is less.The product and the effect on the microstructure of aluminium alloy of adding Ti, Zr and B in pure aluminium are researched by scanning electron microscopy and Transmission electron microscope. Experiments show that there are hexagon TiB2, ZrB2and cubic Al3Ti, Al3Zr in the presence of3different samples of the atomic number ratio (Ti+Zr):B>1:2; while only hexagon TiB2, ZrB2is found in3different samples of atomic number ratio (Ti+Zr):B=1:2. In addition the metallographic results show that grain size of3samples of atomic number ratio (Ti+Zr):B>1:2is generally less than that of3samples of the atomic number ratio (Ti+Zr):B=1:2. When the atomic number ratio (Ti+Zr):B=1:2, refining effect is not obvious, electron scattering is less, the conductive property is less influenced.By the bond length difference method (referred to as BLD) of EET, the bond energy and covalent electron number of TiB2, ZrB2, Al3Ti and Al3Zr were calculated. The results show that the energy of the strongest bond of TiB2, the covalent electron number of the strongest bond is far greater than the energy of the strongest bond of Al3Ti, the covalent electron number of the strongest bond, so TiB2has priority, when B is consumed, Ti and Al bind to Al3Ti The energy of the strongest bond of ZrB2, the covalent electron number of the strongest bond is far greater than the energy of the strongest bond of Al3Zr, the covalent electron number of the strongest bond, so ZrB2has priority, when B is consumed, Zr and Al are combined into Al3Zr. Hexagon lattice structure of TiB2, ZrB2prior to the cubic lattice structure of Al3Ti and Al3Zr. This explains the results of scanning electron microscope:when the atomic number ratio (Ti+Zr):B=1:2, only hexagon lattice existence (TiB2, ZrB2) is found; when the atomic number ratio (Ti+Zr):B>1:2, there are six square lattice (TiB2, ZrB2) and cubic lattice (Al3Ti, Al3Zr).Surface electron density differences of α-Al with TiB2, ZrB2, Al3Ti an Al3Zr were calculated. Surface electron density difference (1.11%) of Al3Ti (111) and α-Al (111) is far less than that(40.88%) of TiB2(0001) and α-Al (110); Surface electron density difference (20.05%) of Al3Zr (111) and α-Al (111) is far less than that (87.09%) of ZrB2(0001) and α-Al (110). The lower the surface electron density is, the better the coherent relationship between the two substances, the more heterogeneous nucleation in α-Al, the better the effect of refinement. So the grain size of the atomic number ratio (Ti+Zr):B>1:2is generally less than that of the atomic number ratio (Ti+Zr):B=1:2. This is consistent with the result of metallographic experiment. And the smaller the grain, the electron scattering from grain boundary is more serious, and the conductivity of the alloy is reduced. Therefore the electrical conductivity of aluminum alloy is better when the atomic number ratio is (Ti+Zr):B=1:2and only TiB2and ZrB2generate.The Peierls-Nabarro forces of α-Al, TiB2, ZrB2, Al3Ti and Al3Zr were calculated. The results are as follows:0.05002,14.03272,7.20934,0.43760,0.61224. The Peierls-Nabarro forces of TiB2, ZrB2, Al3Ti and Al3Zr are so large that dislocations can not cut them and can only take the bypass mechanism. When a large of fine TiB2, ZrB2, Al3Ti and Al3Zr disperse in the aluminum matrix, dislocation will be pinned well and recrystallization temperature will be improved and aluminum alloy will have good heat resistance.After analyzing the influence of conductivity and heat-resistance of aluminum alloy of adding different content Ti, Zr and B, we should make the atom ratio of Ti, Zr and B (Ti+Zr):B=1:2. Then the aluminum alloy can keep good heat-resistance and does not reduce the conductivity.更多还原