混凝土课后答案第6,7章VIP免费

6-1已知一钢筋混泥土矩形截面纯扭构件，截面尺寸b×h=150mm×300mm，作用于构件上的扭矩设计值T=3.60KN.m，采用C30混泥土，采用HPB235级钢筋。试计算其配筋量。『解』查表得：tf=1.43N/2mm,yvf=210N/2mmyf=210N/2mm取s=35mmC=25mm2263150(3)(3300150)2.81251066tbwhbmm42.(15050)(30050)2.510corcorcorAbhmm2()2(100250)700corcorcorubhmm取1.2由公式得：4666640.351.22102.5103.6100.351.432.8125101.21.23.6100.351.432.8125100.31781.21.22.02.510yvstlttcorstlstlfATfWAsAsAs取单肢8即stlA＝50.3即50.31580.3178smm取s＝150mm由纵箍比可知：2..1.221050.3700282.210150ystlyvstlcoryvstlcorstlyfAsfAufAuAmmfs查附表11取410即2314stlAmm结构在剪扭作用下，结构受剪及受扭箍筋最小配筋率为：.min.min0000.1.4350.30.280.190.34210100150svsvcorAbs故满足最小配筋率。配筋图如下：Φ8@1504Φ106-2已知一均布荷载作用下钢筋混泥土矩截面弯、剪、扭构件，截面尺寸为b×h=200mm×400mm。构件所承受的弯矩设计值M=50KN.m，剪力设计值V=52KN,扭矩设计值T=4KN.m。全部采用HPB235级钢筋，采用C20混泥土。试计算其配筋。『解』（1）受弯纵筋计算设钢筋按一排布置，取040035365hmm查表得：tf=1.1N/2mmcf=9.6N/2mmyf=210N/2mm6221050100.19551.09.6200365scMfbh1120.2196s2101.09.60.2196200365733210cSyfbhAmmf（2）受剪及受扭钢筋计算截面限制条件验算：0365,whhmm/365/2001.8254whb36260252104101.459/2003650.86.7100.250.251.09.62.4/tccVTNmmbhWfNmm故截面尺寸满足要求，又22201.459/0.71.1/0.251.09.60.77/tVTNmmNmmNmmbhW故需按计算配置受扭钢筋受剪钢筋计算由公式6-23可得：36601.51.50.9395.52106.71010.510.5410200365ttVWTbh2(20050)(40050)52500corAmm受剪箍筋由公式6-24可得：352100.7(1.50.9395)2003651.10.21391.25210365svlvAs受扭箍筋取1.3，由公式6-27可得：6644100.356.7100.93951.10.10451.21.32105.2510stltAs故得腹板单肢箍筋的需要量为0.21390.10450.2112stlAs取箍筋直径为8（250.3svlAmm）50.32380.2110.211svlAsmm取s=220mm受扭纵筋：21.350.32101000/210220297stlAmm故得腹板纵筋：弯曲受压区纵筋总面积为2297/2149sAmm选用2210(157)SAmm弯曲受拉纵筋总面积为2733297/2882sAmm选用2220218(883)sAmm配筋图如下：7-1已知矩行截面柱b=300mm，h=400mm。计算长度L0为3m，作用轴向力设计值N=300KN，弯矩设计值M=150KN.m，混泥土强度等级为C20，钢筋采用HRB335级钢筋。设计纵向钢筋As及As/的数量。『解』查表得：cf=9.6N/2mmyf=300N/2mm取s=40mm0shha=360mm0/0.5eMNm20aemm0520iaeeemm0/7.55lh要考虑310.5/0.59.6300400/300101.921cfAN取11取21.0220120111(/)1(7.5)11.031400/1400520/360ilheh01.03520535.60.30.3360ieh属于大偏心受压2Φ182Φ102Φ20Φ8@2200/2535.618040675.6iseehamm取0bxh由100(/2)()scysNeafbxhxfAha得22100(10.5)/()560.9scbbysANeafbhfhamm2min0.020.02300400240sAbhmmminssAA选用4142615sAmm由2100(10.5)()cyssNeafbhfAha得2010()/0.385sysscNefAhaafbh11210.480.52s00.52360187.2280shamm由10cysysNafbhfAfA得2109.63003600.5300615/2343300scysyAafbhfAfmm选用52522454sAmm截面配筋图如下：5254142147-2已知条件与题...