钢结构基础(第二版)课后习题【完整版】

第一次作业kNN1500解：先计算杆件截面特性56.7159.540097.5483.21120059.5100312583.2110047655312512252.124765526.25252.112508.0100508.0252.12400,12000043423200yyyxxxyyxxyxyxililcmAIicmAIicmIcmIcmAcmlcml对强轴x和弱轴y均为b类截面，查表得742.0miny22321516.20210000742.0101500mmNfmmNANy满足整体稳定要求！9000N=450kN解：先计算杆件截面特性9.13848.69008.12277.9120048.6622.260477.9625.59122.26041225125.591225.1025112206.062506.02512900,12000043423200yyyxxxyyxxyxyxililcmAIicmAIicmIcmIcmAcmlcml对强轴x和弱轴y均为b类截面，查表得349.0miny2232152086200349.010450mmNfmmNANy满足整体稳定要求！第二次作业解：先计算杆件截面特性4.10569.500669.54.158512276.51178.10226819322512212284.1226819327.50284.1121008.04.1581008.0284.12600,12000343423200yyyyyxxyxyxilcmAIicmhIWcmIcmIcmAcmlcml计算梁整体稳定系数8865.03452358.1024.44.14.105176.51178.1024.1584.10543202354.41432022212yyxybfhtWAh进行梁的弹塑性稳定系数折算7519.08865.0282.007.1282.007.1'bb验算梁的整体稳定性22363103.2571076.51177519.041033012'mmNfmmNWMxbx满足整体稳定要求！采用近似计算6993.0235440003454.10507.12354400007.122yybf22363106.2761076.51176993.041033012'mmNfmmNWMxbx验算梁整体稳定性满足稳定性要求！若无跨中侧向支撑9.21069.500121200,1200000yyyyxilcmlcml6.02514.03452358.1024.44.19.210176.51178.1024.1589.21043202354.41432022212yyxybfhtWAh计算稳定系数计算所能承受的荷载kNLWfPfWPLWMxbxbxbx95.132120001076.51172514.03104443第三次作业解:计算构件截面几何特性22.649.1070078.6215.1170015.11808.99408.99401.213402218222280402700,70000422001200yyyxxxxxxxyxililcmAIicmyaAIIcmAcmlcml计算虚轴折算长细比75.6429.4402778.62272120xxxAA781.0minx虚、实轴对应的截面均为b类，则压杆整体稳定验算2232152088000781.0101300mmNfmmNANx满足整体稳定要求！解（1）验算平面内稳定kNNNNNkNlEINilcmAIicmhIWcmIcmAcmlExExEmxxxExxxxxxxxxxx4.84991.13.93491.1'904.03.934990012.013.93491015103.10346510206831.035.551.2715001.278.1403.1034654.31164.663.103465223.10346526.32322.1126418.140641322.12,150062432202034232022363'2151.1904.84999008.01104.311605.141510100904.014080831.0109008.01mmNfmmNNNWMANExxxxmxx面内稳定满足要求！（2）分析面外稳定强于面内稳定可考虑不计算面外稳定性跨中受压翼缘有侧向支撑点时，所需cmblbl512321616161111则可知在梁三分点处设置侧向支撑cmcmll5125003150031解：先计算杆件截面特性48.6324.1065083.3519.1865024.104.142312519.184.1422.471243.1493312404.1223108.402.4712422.4712427.19404.112388.04.142388.0404.12650,65000433423200yyyxxxyyxxyxxyxililcmAIicmAIicmIcmWcmIcmAcmlcmlkNlEINxxExtxmxybyx4.2061565001.1242.471102061.1'0.19784.04400048.6307.14400007.1783.0,913.026202222面内稳定验算22333'2159.1594.2061513008.0110231005.1197107000.114240913.01013008.01mmNfmmNNNWMANExxxxmxx面内稳定满足要求！面外稳定验算223332156.1771023109784.0197107000.114240783.0101300mmNfmmNWMANxbxtxy面外稳定满足要求！解：（A）根据GB50017对不完整框架的压杆计算长度系数修正考虑，两边梁的线刚度应为lIlI5.15.1则梁柱线刚度的比值为（两柱相同）5.215.215.110K根据表5-2的近似公式8125.035.2235.24.13234.100KK解：（B）根据GB50017对不完整框架的压杆计算长度系数修正考虑，两边梁的线刚度应为lIlI0.20.2则梁柱线刚度的比值为（两柱相同）3131210K根据表5-2的近似公式8.0332334.13234.100KK若查表18-1其计算长度系数为0.791解：（A）由于结构对称，只计算一根柱的计算长度系数。梁柱线刚度比为5.115.115.0110K查表5-2，并插值，得未修正的计算长度系数2325.2155.2507.233.207.2239.25.138.01238.0120K由公式计算得计算长度增大系数2247.12111ffllHNHN考虑摇摆柱影响后的框架柱计算长度系数为742.2239.22247.1解：（B）由于结构对称，只计算一根柱的计算长度系数。梁柱线刚度比为0.1110K查表5-2，得未修正的计算长度系数33.2349.20.138.01238.0120K2247.12111ffllHNHN由公式计算得计算长度增大系数考虑摇摆柱影响后的框架柱计算长度系数为853.233.22247.1解：采用侧焊缝，在端部绕角焊。肢背焊缝长度mmmmhfnhNKlfwfwfw2306.2258160287.01060065.07.0311肢尖焊缝长度mmmmhfnhNKlfwfwfw1653.1626160267.01060035.07.03227.10解：由采用的钢材可知，此构件所受静荷载作用。端焊缝所能承受的荷载为：kNflhNwfwff35.2621601602622.17.07.01kNflhNwfwf43.3061604622067.07.02侧焊缝所能承受的荷载为：此焊缝所能承受的最大轴心拉力为：kNNNN78.56843.30635.26221解：⑴先确定焊脚尺寸mmmmhmmf1021286165.1mmxc543008200287.028200287.02计算焊缝群形心计算焊缝有效截面性质46232231031.7415019287.0230012187.05430087.087.05421921921921212mmIIJyx计算危险点焊缝应力22222662661607.1511.12122.14.1111.1211031.7415010604.1111031.741381060mmNfmmNmmNJTrmmNJTrwfyAxA满足强度要求!②按焊缝构造要求，mmmmhmmf1021266165.1462323105.8767.0150200267.0300121267.0100300267.02001212mmIIJyx计算焊缝群截面性质22222662661603.1179.10222.16.689.102105.8715010606.68105.871001060mmNfmmNmmNJTrmmNJTrwfyAxA22222662661603.1179.10222.16.689.102105.8715010606.68105.871001060mmNfmmNmmNJTrmmNJTrwfyAxA满足强度要求!22222662661607.1572.1315.872.131106.6815010605.87106.681001060mmNfmmNmmNJTrmmNJTrwfyAxA③方案1与方案2的焊条用量：32232118000100062122400700821mmVmmV方案2所耗焊条较方案1的少19.6%。7.13如图所示螺栓双盖板连接，构件钢材为Q235钢，承受轴心拉力，螺栓为8.8级高强度摩擦型螺栓连接，接触面喷砂处理螺栓直径20mm，孔径21.5mm。试验算此连接最大承载力。解：单个螺栓承载力kNPnNfRbv25.10112545.029.0kNnNNbvb5.101225.10110kNfANnr7.602205205.212190螺栓群承载力构件承载力此连接承载力kNNNNrb7.602,解：（1）支托受力，螺栓抗拉验算单个螺栓承载力kNfdNbtebt6.41417065.17422