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1、拉弯和压弯构件(精)第六章拉弯和压弯构件1. 一压弯构件的受力支承及截面如下图(平面内为两头铰支支承)。设资料为Q235(fy235N/mm2),计算其截面强度和弯矩作用平面内的稳固性。y-300x12MA=80kN.mMB=120kN.m-376x10NxxN=800kN60006000y-300x12图压弯构件受力表示图解:截面面积A109.6cm2,Ix31536.34cm4,Iy5403.13cm4;Wx1576.81cm3,Wy360.2cm3;展转半径:ixIx16.96cm,iyIy7.02cm。AA(1)强度验算(右端截面最不利):NMB8000001201060.6351.0
2、AfyWxfy109602351576810235强度验算合格(2)平面内稳固验算:长细比:xlox70.75,依据b类截面查表得0.747。ix0.65MA0.883mx0.35MB2EA22.0610510960NEx270.7524447.2kN因此有:NmxMx0.4160.3180.7341.0xAfyxWxfy10.8NNEx平面内整体稳固验算合格2. 某压弯缀条式格构构件,截面如下图,构件平面内外计算长度lox29.3m,loy18.2m。已知轴压力(含自重)N=2500kN,问能够蒙受的最大偏爱弯矩Mx为多少。设钢材牌号为Q235,N与Mx均为设计值,钢材强度设计值取fd205
3、N/mm2。/1xL125x10151yy4I63a5x418001-1图缀条构件横截面解:I63a型钢截面几何特征:A154.59cm2,Ix94004cm4,Iy1702.4cm4,Wx2984.3cm3,Wy193.5cm3;ixIx24.66cm,AiyIy。3.32cmA24.37cm2;Ix361.67cm4,Ix0573.9cm4,角钢L125x10截面特征:AIy0149.5cm4;ix3.85cm,ix04.85cm,iy02.48cm。组合截面几何特征:A2154.59309.18cm2,Ix2507763cm4,W27864.03cm3,ixIx90.06cm,iy24.
4、66cm。xA(1)平面内整体稳固计算:虚轴方向长细比:l oxx32.53,i x换算长细比:oxx227A32.53227309.1835.06A1x224.37依据b类截面查表得:x0.918,mx1。NEx2EA22.061053091851076.2kNox235.062因此由:NmxMx88.10.0376Mxfd205xANWx1xNEx可得:2Mmax3109kNm(2)平面外整体稳固计算:实轴方向长细比:loy0.727,yiy依据b类截面查表得:y0.727,tx1.0,1.0。因此由:NtxMx111.20.00359Mxfd205yAWx可得:Mmax26128kNm综
5、合(1)、(2)可知,能够蒙受的最大偏爱弯矩为Mmax310.9kNm。(3)分肢稳固验算:柱子偏爱距:M3109e01.24mN2500分肢所受最大轴力:N1N(90124)2972.2kN1801800分肢在弯矩作用平面内长细比1x55.9,分肢在弯矩作用平面32.2外长细比1y18007.30。246.6依据b类截面查表得:1x0.828N2972200232.2N/mm2fd205N/mm21xA10.82815459分肢稳固验算不合格。这说明,分肢将在整体损坏以前失稳。应该减小M值。N1A1fd0.828154592052624.01kNe0N1cy22624.0118098.93c
6、mN902500MmaxNe025000.98932473.25kNm亦即能蒙受的最大弯矩为Mmax2473.25kNm。(4)缀条稳固验算:由于斜缀条擅长横缀条,且前者的计算内力大于后者,故只须验算斜缀条。柱段计算剪力:VmaxAfdfy74.57kN852353单个缀条受力:VN12cos45o52.73kN斜缀条长细比:180024.5cos45o102.7依据b类截面查表得:0.538折减系数:00.60.0015102.70.754斜缀条正应力:N0.5385273053.3N/mm2fd205N/mm2A024370.754缀条知足要求。综上所述,该压弯缀条式构件所能肩负的最大弯矩Mmax2473.25kNm。4