混凝土结构设计原理计算题答案

11、分）（）截面配筋图（略），满足要求）验算最小配筋率，受拉钢筋选用配筋满足条件，不需采用双筋截面大弯矩为：单筋截面所能承担的最则计算）计算弯矩设计值解：正截面受弯纵筋计算：25%24.0300/57.145.0/0.452.39%)500200/(2502/4mm2502252224)3mm7.238930044049.02007.160.155.049.0369.0211211369.04404402007.160.1105.238m.kN5.238m.kN8.257N.mm108.257)55.05.01(4404402007.160.1)5.01(mm44006500，mm60)2m.kN5.2388/0.6)254.1152.1(8/)(1min2201620162010220ytssycsbscsbcskqkgffbhAAfhbfAbhfMMbhfMhaAslqgM，满足要求配箍率）验算最小配箍率，即箍筋采用，取，则，），选用双肢箍（）计算箍筋数量分需要按计算配箍筋配箍筋）验算是否需要按计算截面尺寸满足要求，属一般梁验算最小截面尺寸）（剪力设计值）计算支座边缘产生的：解斜截面受剪箍筋计算：%179.021057.124.024.0%252.02002003.50258@200mm200mm2080.484.3052mm3.5082484.044021025.110007.9610006.15225.17.04)(26.1527.96107.9644020057.17.07.03kN6.152kN4.367N104.6734402007.160.125.025.042.2200/440/2kN6.15276.5)254.1152.1(21)211min.max210013030yvtsvsvsvsvyvtsvsvtccwnkqkgffbsAsssAnhfbhfVsnAsAkNVkNNbhfVbhfbhlqgV22、3、2260s02620112160220mm138822022)3(mm1331)35515(36010230)()2(136.0440/352255.0133.0124.0211211124.05152509.110.1102.98kN.m2.988.131230kN.m8.311N.mm108.131)35515(763360)(mm763183mm51553550，mm35)1(ssyssbscsssysssAahfMAA/habhfMMMMahAfMAhaa，受拉钢筋选用配筋计算受压钢筋不能屈服，按单筋矩形截面计算则的钢筋，压区已经配置了由已知条件可知，在受则布置一层钢筋，因已知弯矩较小，假定验算适用条件截面不安全或：截面不安全，不超筋计算满足要求则由已知条件可得：）验算最小配筋率解：m.kN90m.kN7.83N.mm107.83)26.125410(804300)2(m.kN90m.kN7.83N.mm107.83)26.125410(6.1252006.90.1)2(mm5.22541055.0mm6.1252006.90.1804300)2%17.0300/57.145.0/0.45%2.00.89%)450200/(804/mm804,mm41004054，mm4016060101min20MxhAfMMxhbxfMhbfAfxMffbhAAhasyucubcsyuytsss34、5、22min200b10220020131003600mm804)3(mm308600300002.0mm4.562)40560(3609.350300000)(mm9.350402/6009.6102/2mm804022mm30856055.0mm9.593007.160..1300000mm1685603.03.0mm9.610553104.1104.112560/553140011/1400110.11512/0.1101.5103006003007.165.05.0512600/7200/mm55320533mm20mm2030/60030/mm5331030010160mm560406001ssysssisssciicaiasAbhahfeNAAaheeAAahbfNxhehlhehlNAfhleeeehNMeahh，两侧各选用配筋，并验算最小配筋率）计算（，受压钢筋不能屈服，确实为大偏压，可先按大偏压计算，取，取，需计算，则）判断截面类型（)4200@6mm200mm240mm40015smm6mm6mm64/mm804164mm7353503500.6%)314003.1511103.15113503503.1495.09.09.02959.043.1135.0/0.4/φ1minmax22minmin30截面配筋图（略）即箍筋选用，取，，，取，：选配箍筋，选用纵筋：配筋筋只需按构造配置受力钢）计算，查表得，）计算稳定系数解：sbddddAAAkNNkNNAfAhlsscs