解答7.1有一悬臂柱的受力及支承情况如图7-19所示。采用Q235-B的HN450╳200╳9╳14做成，截面无削弱，验算其截面强度和弯矩作用平面内的稳定性。200N450kN30003000Nx0Mx285y6000图7-20习题7.2图7-19习题7.122解：，N450kN,Mx4500.290kNmfy235N/mm,f215N/mm查型钢截面特性表得：243A97.41cm,Ix33700cm,Wx1500cm,ix18.6cm（1）截面强度验算按边缘屈服准则NM计算：x1AfWxf450103901060.2150.2790.4941.097412151500103215按部分发展塑性准则NMx计算：1.051xAfWxf450103901060.2150.2660.4811.097412151.051500103215NMx按全截面屈服准则1计算：px1.12AfWPxf450103901060.2150.2490.4641.097412151.121500103215（2）弯矩平面内稳定性验算：lox28502x30.6ix186fy30.62350.3290.215按a曲线公式计算：E2.06105122222[(0.9860.152)(0.9860.152)4]21[1.1440.8763]0.9600.2165122.061059741N21129kNEx30.62，其中NMmx1.0mxx1.0NxAfxWx1f(10.8、)NEx450103901060.960974121534501.05150010215(10.8)21129/1.10.2240.2710.4951.07.2图7-20所示I36a热轧普通工字钢截面压弯构件，材料为Q235-B，截面无削弱，承受的荷载轴心压力N350kN，构件平面内两端铰接，两端及跨中点处各设有一侧向支承点。试求该柱所能承担最大弯矩Mx。22解：fy235N/mm,f215N/mm,l0x6000,l0y3000查型钢截面特性表得：243A76.44cm,Ix15796cm,Wx877.6cm,ix12.38cmh360mm,t15.8mm,iy2.69cm（1）强度计算（按边缘屈服准则）NM350103Mxx，计算得：31.0Mx148.5kNmNPMex7644215877.610215（2）平面内稳定性计算（按a曲线）l6000f48.52350x，yx48.5x50.520.215ix123.8E2.061012222x2[(0.9860.152)(0.9860.152)4]21[1.3350.7017]0.9200.540822.061057644N6600kNEx48.52NM，其中mxx1.0NxAfxWx1f(10.8、)NEx350103Mx0.920764421533501.05877.610215(10.8)6600/1.1M0.231x1.0,得M145.2kNm188872684x（3）平面外稳定性计算（按b曲线）l3000f111.52350y，yy111.5y51.20.215iy26.9E2.0610212222y2[(0.9650.3)(0.9650.3)4]21[2.7651.885]0.4832.884320Ahyt2432076.4436111.515.82bb21()1.7521()2.850.6yWx4.4h111.5877.64.4360、0.282用b1.070.97代替bNMtxx，其中ftx1.0yAbWx350103MMx94.8x215,得M102.3kNm0.48376440.97877.6103851272x所以该柱所能