筏板基础及侧壁计算书.docx

筏板基础及侧壁计算书一、基本数据：根据 xx 省 xx 护国房地产开发有限公司护国广场岩土工程勘察报告，本工程以层圆 砾层为持力层，地基承载力特征值为 220KPa 。基础形式为筏板基础，混凝土强度等级为 C 40 , fc= 19.1N / mm2 ；受力钢筋均采用 HRB400 级，fy=360 N / mm2 ；根据地质报告，地下水位取 1.700m 。二、地基承载力修正及验算：fa =fak + b (b 3) + d m (d 0.5)= 220 + 0.3 8 (6 3) + 1.5 8 (5.65 0.5)= 289.0kN / m2上部荷载作用下地基净反力（由地下室模型竖向导荷得）f = 61.6kN / m2 f = 289.0kN / m2a地基承载力满足要求。三、地下室侧壁配筋计算：（1）双向板：ly5.175l lx = 8.400m ， ly = 5.175m ，=x8.4= 0.62E土 = rhKa= 8.0 5.175 tan 2 45o = 41.4KN / mE水 = rh = 10.0 3.475 = 34.75KN / mE合 = 1.27E土 + 1.27E水 = 52.6 + 44.1 = 96.7KN / m查静力计算手册，得：2M x max = 0.0072ql2= 0.0072 96.7 5.17522= 18.6KN mM y max = 0.0209ql= 0.0209 96.7 5.1752= 54.1KN m2M x max= 0.0354ql2= 0.0354 96.7 5.1752= 91.7KN mM y = 0.0566ql= 0.0566 96.7 5.175= 146.6KN m配筋计算：取弯矩最大处进行计算。即取 M = 146.6kN m1混凝土强度等级为 C 40 , fc= 19.1N / mm2 ； = 1.0 ；受力钢筋均采用HRB400级，fy=360 N / mm2 ； = 0.523 ； min= 0.2% ；b相对受压区高度：x = h0 (1 21 2M)1 fcbh0= 310 (1 1 2 146600000)1.0 19.11000 3102= 25.8mm xb = b h0 = 0.523 302 = 157.9mm则As =1 fcbx f y= 1.0 19.11000 25.8360= 1368.8mm2实际配筋：内外均16150A = 1340mm2sly5.175l lx = 7.800m ， ly = 5.175m ，=x7.8= 0.66E土 = rhKa= 8.0 5.175 tan 2 45o = 41.4KN / mE水 = rh = 10.0 3.475 = 34.75KN / mE合 = 1.27E土 + 1.27E水 = 52.6 + 44.1 = 96.7KN / m查静力计算手册，得：2M x max = 0.0081ql= 0.0081 96.7 5.17522= 21.0KN m2M y maxMx max= 0 .0194 ql= 0.0351ql 22= 0 .01 96 .7 5 .175= 0.0351 96.7 5.17522= 54 .1KN m= 90.9KN mM y = 0.0542ql= 0.0542 96.7 5.175= 140.4KN m配筋计算：取弯矩最大处进行计算。即取 M = 140.4kN m1混凝土强度等级为 C 40 , fc= 19.1N / mm2 ； = 1.0 ；受力钢筋均采用HRB400级，fy=360 N / mm2 ； = 0.523 ； min= 0.2% ；b相对受压区高度：x = h0 (1 21 2M)1 fcbh0= 310 (1 1 2 140400000)1.0 19.11000 3102= 24.7mm xb = b h0 = 0.523 302 = 157.9mm则As =1 fcbx f y= 1.0 19.11000 24.7s360= 1310mm2实际配筋：内外均16150（2）EF 轴间板：A = 1340mm2lx lx = 4.000m ， ly = 5.175m ，ly= 4.0005.175= 0.77E土 = rhKa= 8.0 5.175 tan 2 45o = 41.4KN / mE水 = rh = 10.0 3.475 = 34.75KN / mE合 = 1.27E土 + 1.27E水 = 52.6 + 44.1 = 96.7KN / m查静力计算手册，得：2M x max = 0.0155ql2= 0.0155 96.7 5.17522= 40.1KN mM y max = 0.0094ql= 0.0094 96.7 5.1752= 24.3KN m2M x max= 0.0386ql2= 0.0386 96.7 5.1752= 100.0KN mM y = 0.0394ql= 0.0394 96.7 5.175= 102.0KN m配筋计算：取弯矩最大处进行计算。即取 M = 102.0kN m混凝土强度等级为 C 40 , fc= 19.1N / mm2 ； = 1.0 ；受力钢筋均采用1HRB400级，fy=360 N / mm2 ； = 0.523 ； min= 0.2% ；b相对受压区高度：x = h0 (1 21 2M)1 fcbh0= 310 (1 1 2 102000000)1.0 19.1 1000 3102= 17.7mm 满足要求。max= 865.3kN / m2抗剪承载力验算：V =5206= 714.2kNs2.7 2.7受剪切承载力截面高度影响系数 hs ：11hs= (800 ) 4 = (800 ) 4 = 0.974h08900.7 hs ft bwh0 = 0.7 0.974 1570 1 0.89 = 952.7kN Vs = 714.2kN满足要求。2、 边柱：取柱轴力最大处，即 10 轴/F 轴 由地下室 PKPM 竖向导荷，知：Fl = 2964 + 146 2 + 75 + 158 2 + 137 + 98 = 3882kN ：抗冲切承载力验算：由上述计算知道um = 4180mm ； c1 = 1245mm ； c2 = 1690mm ； cAB = 874.2mmsI = 7797.3 108 mm4考虑作用在冲切临界截面重心上的不平衡弯矩较小，本计算予以忽略。即： M unb = 0因底板外挑1.2m ，故距柱边 h0 处冲切临界截面的周长2um = 2c1 + 2c2 = 2 1245 + 2 1690 = 5870mm h距柱边 0 处冲切临界截面的最大剪应力 ：2 = Fl+ s M unbcAB =3882max+=kN m2maxum h0I s5.87 0.890743.1/hpt0.7(0.4 + 1.2 ) f s= 0.7 (0.4 + 1.2 ) 0.993 15702= 1091.3kN / m2 满足要求。max= 743.1kN / m2抗剪承载力验算：V =3882= 532.5kNs2.7 2.7受剪切承载力截面高度影响系数 hs ：11hs= (800 ) 4 = (800 ) 4 = 0.974h08900.7 hs ft bwh0 = 0.7 0.974 1570 1 0.89 = 952.7kN Vs = 532.5kN满足要求。3、 角柱：取柱轴力最大处，即 1 轴/J 轴： 由地下室 PKPM 竖向导荷，知：Fl = 2947 + 146 + 137 + 116 + 79 + 167 + 86 + 77 + 25 + 80 = 3860kN抗冲切承载力验算： 由上述计算知道um = 2490mm ； c1 = 1245mm ； c2 = 1245mm ； cAB = 830.0mmsI = 4548.1108 mm4考虑作用在冲切临界截面重心上的不平衡弯矩较小，本计算予以忽略。即： M unb = 0因底板外挑1.2m ，故距柱边 h0 处冲切临界截面的周长2um = 2c1 + 2c2 = 2 1245 + 2 1245 = 4980mm h距柱边 0 处冲切临界截面的最大剪应力 ：2 = Fl+ s M unbcAB =3860max+=kN m2maxum h0I s4.98 0.890870.9/hpt0.7(0.4 + 1.2 ) f s= 0.7 (0.4 + 1.2 ) 0.993 15702= 1091.3kN / m2 满足要求。max= 870.9kN / m2抗剪承载力验算：V =3860= 529.5kNs2.7 2.7受剪切承载力截面高度影响系数 hs ：11hs= (800 ) 4 = (800 ) 4 = 0.974h08900.7 hs ft bwh0 = 0.7 0.974 1570 1 0.89 = 952.7kN Vs = 529.5kN满足要求。筏板承载力计算： 计算模型为倒无梁楼盖，无帽顶板柱帽，采用等代框架法计算。 等代框架法是把整个结构分别沿纵、横柱列划分为具有“等代框架柱”和“等代框架梁”的纵向等代框架和横向等代框架。X 方向 A、等代框架构件尺寸确定： 等代梁：（取 E 轴计算）12000 + 84000取梁截面宽度取板跨中心线间距= 10200mm2梁截面高度取板厚 950mm；本结构采用无帽顶板柱帽。 梁跨度取8400mm ； 梁截面惯性矩为：Ib =1 10.2 0.953 = 0.729m412等代柱：柱截面尺寸800mm 800mm 柱高： 5175mm 柱截面惯性矩：Ib =1 0.8 0.83 = 0.0341m412B、内力计算梁上均布荷载 q = 61.6 10.2 = 628.32kN / m其中 61.6 为地基净反力。 取 q = 630kN / m计算简图如下：分层法和弯矩分配法计算如下： 利用对称性取半跨，柱弯矩传递系数取 0.5，计算简图如下：计算各杆转动刚度及分配系数：iAG= E 0.0341 = 0.00659E5.175iAB= E 0.729 = 0.08679E8.4令iAG = i则 iAB = 13.2i各杆转动刚度：S AG = 4iS AB = 4 13.2i = 52.8iSEF = 13.2i各杆分配系数：uAG= S AG SA=4i56.8i= 0.070uAB= S AG SA= 52.8i = 0.93056.8iuBHu= SBH SB= SCK=4i109.6i=4i= 0.036= 0.036uBAu= uBC= u= SBA SA= SCB= 52.8i109.6i= 52.8i= 0.482= 0.482CK SC109.6iCBCD S109.6iCuDM= SDM SD=4i109.6i= 0.036uDC= uDE= SDE SD= 52.8i109.6i= 0.482uEN= SEN SE=4i70.0i= 0.057uED= SED SE= 52.8i = 0.75470.0iuEF= SEF SE= 13.2i = 0.18970.0i计算各杆件由荷载所产生的固端弯矩：MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mAB12M=F1 ql 2 =BA12121 630.0 8.42 = 3704.4kN m12MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mBC12M=F1 ql 2 =CB12121 630.0 8.42 = 3704.4kN m12MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mCD12M=F1 ql 2 =DC12121 630.0 8.42 = 3704.4kN m12MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mDE1212M=F1 ql 2 =ED121 630.0 8.42 = 3704.4kN m12MF = 1 ql 2 = 1 630.0 8.42 = 14817.6kN mEF33MF = 1 ql 2 = 1 630.0 8.42 = 7408.8kN mFE66WWW.ZHULONG.COM弯矩分配法计算过程见下图：弯矩图如下：筑龙网则弯矩最大值为：M EF = 12485.3kN m ； 由下表：等代框架法板带分配弯矩系数截面柱上板带跨中板带内跨支座负弯矩0.750.25跨中正弯矩0.550.45边跨第一内支座负弯矩跨中正弯矩 边跨支座负弯矩0.750.550.10得：内跨柱上板带分配弯矩： M = 0.75 12485.3 = 9364.0kN m 内跨跨中板带分配弯矩： M = 0.55 12485.3 = 6866.9kN m C、配筋计算：取弯矩最大处进行计算。即取 M = 9364.0kN m1混凝土强度等级为 C 40 , fc= 19.1N / mm2 ； = 1.0 ；受力钢筋均采用 HRB400b级，fy=360 N / mm2 ； 相对受压区高度：= 0.523 ； min= 0.2% ；x = h0 (1 21 2M)1 fcbh0= 890 (1 1 2 9364000000)1.0 19.110200 8902= 55.8mm xb = b h0 = 0.523 890 = 465.5mm则As =1 fcbx f y= 1.0 19.110200 55.8360= 30197.1mm2每米跨度配筋面积为： As1= 30197.1 1000 = 2960.5mm2102002As min = 0.2% 1000 950 = 1900mm实际配筋：HRB400 级钢，直径 25，间距 130mm， 满足要求。Y 方向A、等代框架构件尺寸确定： 等代梁：（取 5 轴计算）梁截面宽度取板跨中心线间距8400mm ； 梁截面高度取板厚 950mm；本结构采用无帽顶板柱帽。梁跨度取分别为 7800mm，8400mm，12000mm。 梁截面惯性矩为：sA = 3776mm2Ib =1 8.4 0.953 = 0.600m412等代柱：柱截面尺寸800mm 800mm 柱高： 5175mm 柱截面惯性矩：Ib =1 0.8 0.83 = 0.0341m412B、内力计算梁上均布荷载 q = 61.6 10.2 = 628.32kN / m计算简图如下：分层法和弯矩分配法计算如下：截取 12 米跨度中点左半跨进行计算，柱弯矩传递系数取 0.5，计算简图如下：计算各杆转动刚度及分配系数：iBH= E 0.0341 = 0.00659E5.175iAB= E 0.60 = 0.10000E6.0iBC= iCD= iDE= iEF= E 0.60 = 0.07143E8.4iFG= E 0.60 = 0.13043E4.6令iBH = i则 iAB = 15.2iiBC = iCD = iDE = iEF = 10.8i iFG = 19.8i各杆转动刚度：S AB = 15.2i = 15.2iSBC = SCD = SDE = SEF = 4 10.8i = 43.2iSFG = 4 19.8i = 79.2iSBH = 4 i = 4i各杆分配系数：uBH= SBH SB=4i62.4i= 0.064uBA= SBA SB= 15.2i = 0.24462.4iuBC= SBC SB= 43.2i = 0.69262.4iuCB= uCD= SCB SC= 43.2i = 0.47890.4iuCK= SCK SC=4i90.4i= 0.044uDC= uDE= SDC SD= 43.2i = 0.47890.4iuDM= SDM SD=4i90.4i= 0.044uED= uEF= SED SE= 43.2i = 0.47890.4iuEN= SEN SE=4i90.4i= 0.044uFE= SFE SF= 43.2i126.4i= 0.342uFG= SFG SF= 79.2i126.4i= 0.627uFP= SFP SF=4i126.4i= 0.031uGF= SGF SG= 79.2i = 0.95283.2iuGQ= SGQ SG=4i83.2i= 0.048计算各杆件由荷载所产生的固端弯矩：MF = 1 ql 2 = 1 630.0 6.02 = 3780.0kN mAB66MF = 1 ql 2 = 1 630.0 6.02 = 7560.0kN mBA33MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mBC12M=F1 ql 2 =CB12121 630.0 8.42 = 3704.4kN m12MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mCD12M=F1 ql 2 =DC12121 630.0 8.42 = 3704.4kN m12MF = 1 ql 2 = 1 630.0 8.42 = 3704.4kN mDE12M=F1 ql 2 =ED12121 630.0 8.42 = 3704.4kN m12M