Experimental research on seismic behavior of abnormal joint in reinforced concrete frame 外文翻译.doc
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Experimental research on seismic behavior of abnormal joint
in reinforced concrete frame
Abstract :Based on nine plane abnormal joint s , one space abnormal joint experiment and a p seudo dynamic test of a power
plant model , the work mechanism and the hysteretic characteristic of abnormal joint are put to analysis in this paper. A
conception of minor core determined by the small beam and small column , and a conclusion that the shear capacity of ab2
normal joint depends on minor core are put forward in this paper. This paper also analyzes the effect s of axial compres2
sion , horizontal stirrup s and section variation of beam and column on the shear behavior of abnormal joint . Finally , the
formula of shear capacity for abnormal joint in reinforced concrete f rame is provided.
Key words : abnormal j oint ; minor core ; seismic behavior ; shear ca paci t y
CLC number :TU375. 4 ; TU317. 1 Document code :A Article ID :100627930 (2006) 0220168210
1 Int roduction
For reinforced concrete f rame st ructure , t he joint is a key component . It is subjected to axial
comp ression , bending moment and shear force. The key is whet her the joint has enough shear capaci2
ty. The Chinese Code f or S eismic Desi gn of B ui l di ngs ( GB5001122001) adopt s the following formula
to calculate t he shear capacity of the reinforced concrete f rame joint .
V j = 1. 1ηj f t bj h j + 0. 05ηj N
bj
bc
+ f yv Asvj
hb0 - a′s
s
(1)
Where V j = design value of t he seismic shear capacity of the joint core section ;
ηj = influential coefficient of t he orthogonal beam to the column ;
f t = design value of concrete tensile st rength ;
bj = effective widt h of the joint core section ;
hj = dept h of the joint core section , Which can be adopted as t he depth of the column section in
t he verification direction ;
N = design value of axial compression at t he bot tom of upper column wit h considering the combi2
nation of the eart hquake action , When N 015 f c bc hc , let N = 0. 5 f c
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