第六章 受压构件的截面承载力
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第六章 受压构件的截面承载力. 问答题. 计算题. 6 问答题. 1. 轴心受压普通箍筋短柱与长柱的破坏形态有何不同 ? 2. 轴心受压长柱的稳定系数 如何确定 ? 3. 轴心受压普通箍筋柱与螺旋箍筋柱的正截面受压承载力计算有何不同 ? 4. 简述偏心受压短柱的破坏形态 ? 偏心受压构件如何分类 ?. 答案. 答案. 答案. 答案. 5. 长柱的正截面受压破坏与短柱的破坏有何异同 ? 6. 为什么要引入附加偏心距 e a ? 7. 为什么采用 ηe i =0.3 h 0 来判别大、小偏心受压构件只是一个近似公式 ? - PowerPoint PPT PresentationTRANSCRIPT
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1.?2.?3.? 4.?? 6
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5.?6.ea7.ei=0.3h0?8.?
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9.?10.? 11.?12.Nu-Mu?
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?? 12
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? M/NM/NMNNM
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eaea12 fc;fc
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ei=0.3h0?b,>b fy fcminmine0b/b0.3,0.3e0b,min=0.3h0 ei e0b,min= 0.3h0ei> e0b,min= 0.3h0
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? 12
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? NMAsAsei>0.3h0 ei0.3h0AsA sAsA sxxxbxxb AsA sAs+A sbh05%
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? =N/ fcbh0b0.3h0NbNb= 1 fc bh0bei>0.3h0NNbei0.3h0ei>0.3h0NNb
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Nu-Mu?1 Mu=0 Nu Nu=0 Mu Mu2 NuMu NuMu3NuNu-Mu
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6-1N3000kN4.2m400mmC35HRB335HPB2352A's6
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,828 As' 4926mm2 = As/A =4926/125600=3.92%
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6-2N1000kNM430kNmbh =400mm500mml05.0mC30HRB335AsA's1e0ei2h
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3
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45
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5
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6-36-2As' 2463mm2AsN=Nu , M= Nu e
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6-4N1200kNbh =400mm600mmaa'45mml04mC40HRB400As' 1520mm2 As1256mm2 h
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l0/h4000/6006.67h1.0ei423mmea20mme0 ei ea42320403mmM=N e0=12000000.403=483.6kNmhM = 483.6kNm
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6-5N5280kNM24.2kNmbh =400mm600mmaa'45mml03mC35HRB400AsA's Nu = N, MuM= Nu e0 e0=M/N=4.58mmea=600/30=20mmei e0 +ea4.58+2024.58mm1e0ei2h
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3
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4AsA's10.8 Asminbh00.002400555444mm =1.212>h/h0=1.081
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Ase'=0.5h-a'-e0-ea h'0=h-a' As 2210mm225mm, As2454mm2, A's2945mm2,
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5
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6-66-2,AsA's14AsA's