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作 者:陶波[1] 伍法权[2] 郭改梅[2] 周瑞光[2]
机构地区:[1]中国石油大学(北京)资源与信息学院,北京102249 [2]中国科学院地质与地球物理研究所,北京100029
出 处:《岩石力学与工程学报》2005年第17期3165-3171,共7页Chinese Journal of Rock Mechanics and Engineering
基 金:国家自然科学基金资助项目(90302011);国家重点基础研究发展规划(973)项目(2002CB412701)
摘 要:通过对伯格斯体与西原体的对比分析,认为西原模型能模拟各种岩石流变特征,而伯格斯体仅适合用于较软岩。由于考虑到荷载与起始流变应力的关系,西原模型的本构关系用分段函数表示。当σ<σs时,西原模型的蠕变不导致无限变形,松驰不导致应力为0;当σ≥σs时,蠕变导致无限变形,松驰也不导致应力为0。伯格斯体忽略荷载与起始流变应力的关系,认为只要有荷载作用于岩体上,就会导致无限变形,这与岩石的实际流变规律相矛盾。此外,利用三轴压缩蠕变试验结果,分别利用伯格斯模型及西原模型,运用最小二乘法对赵各庄煤矿断层破碎带灰黄色糜棱岩的流变曲线进行了拟合,并求取了流变参数。拟合结果表明,西原模型比伯格斯模型更适合于描述岩石的蠕变特性。By comparison of Burgers model with visco-elastoplastic model, this paper considers viscoelastoplastic model is adaptive to all kinds of rock rheological characteristics, while Burgers model is only suitable to rather soft rock. Due to taking the relationship between σ and σs into account, constitutive equations of visco-elastoplastic model are presented by segmented functions. When σ 〈 σs in the visco-elastoplastic model, creep never causes infinite deformation and relaxation wouldn't result in stress disappearing. When σ ≥ σs, deformation caused by creeping will last all along, but stress caused by relaxation never decreases to zero. The Burgers model, regardless of relationship between σ and σs, holds deformation to infinite as long as σis not equal to zero, which is contradictory to the rheological characteristics of rock. Moreover, according to triaxial creep test of rock samples, authors fit the curve of ε -t of the gray-yellow miliolite from fault of Zhaogezhuang Coal Mine respectively by visco-elastoplastic model and Burgers model and obtain their rheological parameters by least squares procedure. The fitting results show that the visco-elastoplastic model is more suitable than the Burgers model to describe rock rheological characteristics.
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