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作 者:王军[1]
机构地区:[1]中铁第一勘察设计院集团有限公司,兰州730000
出 处:《科学技术与工程》2014年第1期277-279,285,共4页Science Technology and Engineering
摘 要:利用容许应力法计算实体矩形截面大偏心受压构件时需解一个关于y的无二次项的一元三次方程:y3+py+q=0[方程(1)],以往的各类文献对偏心受压构件的介绍,多侧重于构件的受力特点、破坏形态及其与构件的截面尺寸、长度、材料的关系等等,极少有涉及方程(1)的求解论述,实际上方程(1)的解直接决定了截面的检算结果;利用高等数学原理,从数学角度详细分析了方程(1)的解,列举出解存在的八种情况:①p>0,q>0;②p>0,q<0;③p<0,q>0,fmin<0;④p<0,q>0,fmin>0;⑤p<0,q>0,fmin=0;⑥p<0,q<0,fmax<0;⑦p<0,q<0,fmax>0;⑧p<0,q<0,fmax=0。结合工程应用,对大量偏心受压构件进行计算、归纳、总结;提出对八种情况进行取舍的意见。尤其应注意情况③时的解,情况③有两正解,求解不当可能会得到错误结果。When use the allowable stress method to calculate a solid rectangular cross section,which was a compression member with large eccentricity, need to solve a cubic equation about Y which has no quadrat- ic term. y^3+py+q=0 [equation (1) ]. Previous literature on eccentric compression member introduction, focuses on the components of the stress characteristics, failure pattern and the section size, length, materi- al relations and so on, there is little to solve the equation (1) states, in fact the solution of equation (1) directly determines the results of the cross section calculation. Using higher mathematics principle the solu- tion of equation(l) is analyzed detailedly from the mathematical point of view. Enumerative solutions exist eight cases: ①p〉0,q〉0;②p〉0,q〈0;③p〈0,q〉0,fmin〈0;④p〈0,q〉0,fmin〉0;⑤p〈0,q〉0,fmin=0;⑥p〈0,q〈0,fmin〈0;⑦p〈0,q〈0,fmax〉0;⑧p〈0,q〈0,fmax=0, combined with the engineering application calculation, induction, summary of many large eccentric compression member has been done; opinions of eight cases are put forward, especially should pay attention to the solution of the third case, this case has two positive solutions, solving the undeserved may get the wrong result, the designer should pay special attention to it.
分 类 号:U441.1[建筑科学—桥梁与隧道工程]
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