混凝土设计复习题

1．用剪力分配法求图1排架柱顶剪力，并作出弯矩图（20分）。已知：作用在牛腿顶面的弯矩mkNMB165，mkNMC90；各柱惯性矩：4911013.2mmIA，4921052.14mmIA；4911021.5mmIB，4921076.17mmIB；4911013.2mmIC，4921038.15mmIC图1提示：)11(1330nC，03CEIHl，3CHMR，)11(1123323nC，luIIn，HHu解：1．计算剪力分配系数A柱：992.13100.14714.5210n，33.09.106.348.2)1147.01(33.013)11(13330nC10900I1A129003600I1CI2CI2BI1BI2AAEA=∞BCEA=∞lEIuEIRHuH3600MBMCMEECEIHlA96.3548.21052.14109009303B柱：293.01076.171021.599n，33.09.106.376.21293.0133.013)11(13330nCEECEIHlB42.2676.21076.17109009303C柱：138.01038.151013.299n，279.09.126.364.21138.01279.013)11(13330nCEECEIHlC87.5264.21038.15129009303329.087.5242.2696.3596.351111EEEECBAAA，447.087.5242.2696.3542.261111EEEECBABB224.087.5242.2696.3587.521111EEEECBACC2．柱顶剪力计算23.1)1293.01(33.0133.0123)11(112332323nCB218.1)1138.01(279.01279.0123)11(112332323nCCkNCHMRBBB62.1823.19.101653（）kNCHMRCCC50.8218.19.12903（）kNRRRCB12.1050.862.18()kNRVAA33.312.10329.0（）kNRRVBBB10.1462.1812.10447.0（）kNRRVCCC77.1050.812.10224.0（）3．作弯矩图47.某单层单跨厂房，跨度18m，柱距6m，内有两台10t中级载荷状态的吊车，试求该柱承受的吊车竖向荷载Dmin、Dmax和横向水平荷载Tmax。起重机有关数据如下：吊车跨度LK=16.5m，吊车宽B=5.44m，轮距K=4.4m，吊车总质量18.23t，额定起重量10t，最大轮压标准值Pmax，k=104.7KN。解：kNPgQmPgQmmPkkk45.367.104102)1023.18(2)(2)(max,max,21min,48.9338.77114.2450.7611.9936.30AB11.31CM图（kN-m）51.231.6/60.56/64.96/614.4m4.4m5.44m5.44m6m6m查表，9.0，竖向反力影响线如图所示：则有：kNyPrDikQ47.288)696.46.156.01(7.1044.19.0max,maxkNPPDDkk4.1007.10445.3647.288max,min,maxminkNgQmTk69.310)10684.3(9.012.041)(412kNPTDTkk17.107.10469.347.288max,maxmax48.试用剪力分配法求图所示排架，在风荷载作用下各柱的内力。已知基本风压20/45.0mkN，15m高处的14.1z（10m高0.1z），体形系数s示于图中。柱截面惯性距：.105.19,102.7,1038.14,1013.2494493492491mmImmImmImmI上柱高3m，计算单元宽度取B=6m。解：1．求21,qq线性差值：014.1)105.10(10150.114.11zmkNBqmkNBqzskzsk/37.1645.0014.15.0/19.2645.0014.18.00201)(/92.137.14.1)(/07.319.24.12211mkNqrqmkNqrqkQkQ2．求W高度取10.5+2.1=12.6m07.1)106.12(10150.114.11zKNBhhWzk54.7645.007.1]2.11.01.23.1[])6.05.0()5.08.0[(021kNWrWkQ56.1054.74.13.计算参数n,A柱：29.05.10315.01038.141013.29921HHIInuB柱：29.05.10337.0105.19102.79943HHIInu4．在柱顶虚拟加不动铰支座查表计算：A柱：34.0)]115.01(29.01[8)]115.01(29.01[3)]11(1[8)]11(1[3343411nnCB柱：36.0)]137.01(29.01[8)]137.01(29.01[3)]11(1[8)]11(1[3343411nnC风向-0.6-0.5Wq1q2+0.8-0.5I1I2I3I41.2m2.1m10.5m)(3.736.05.1092.1)(1134.05.1007.3112111HCqRHCqRBA故：柱顶剪力)(3.7)(111,1,BBAARVRV5．撤消不动铰支座，反向施加支座反力。A柱：64.2)115.01(29.013)11(13330nCB柱：88.2)137.01(29.013)11(13330nC于是：64.21038.145.109303clcAECIEHu88.2105.195.109303clcBECIEHu所以分配系数：6.05.1988.238.1464.25.1988.24.05.1988.238.1464.238.1464.2BA故，柱顶剪力：)(32.17)56.103.711(6.0)()(54.11)56.103.711(4.0)(2,2,kNWRRVkNWRRVBABBBAAA6.叠加两个受力状态)(02.103.732.17)(54.01154.112,1,2,1,BBBAAAVVVVVV7.绘制弯矩图17515.44q1=3.07KN/m0.54KNA柱M图（KN·m）211.0538.7q1=1.92KN/m10.02KNB柱M图（KN·m）49.如图所示排架，在A，B牛腿顶面作用有力矩M1和M2。试求A，B柱的内力，已知：M1=153.2KN.M，M2=131.0KN.M，柱截面惯性距同上习题13-2。解：1.计算参数n,A柱：29.05.10315.01038.141013.29921HHIInuB柱：29.05.10337.0105.19102.79943HHIInu2．在柱顶虚拟加不动铰支座A柱：21.1)115.01(29.0129.015.1)11(115.132323nCB柱：32.1)137.01(29.0129.015.1)11(115.132323nC故：)(47.1632.15.100.131)(65.1721.15.102.1533231kNCHMRkNCHMRBA3．撤消不动铰支座，反向施加支座反力)(18.147.1665.17kNRRRBA6.0,4.0BA所以：)(71.06.018.1)(47.04.018.1kNRVkNRVBBAA3．叠加内力M1M2I1I2I3I4)(18.1771.047.16)(18.1747.065.17BAVV4．弯矩图：50.如图所示柱牛腿，已知竖向力设计值kNFv324，水平拉力设计值kNFh78，采用C20（2/54.1mmNftk）混凝土和HRB335级钢筋，试设计牛腿的纵向受力钢筋。（截面宽度b=400mm（另加的条件））解：1．验算截面尺寸mmh760408000Fh=78KNFV=324KN1050500C1=25045040040080017.18KNM=153.2KN.m51.54101.6627.19M图（KN.M）A柱17.18KNM=131KN.m51.5479.4649.39M图（KN.M）B柱kNFkNhabhfFFvstkvshs3243977602505.076040054.1)324785.01(8.05.0)5.01(00截面尺寸满足要求。2．计算纵向受拉钢筋233073030010782.176030085.0250103242.185.0mmfFhfaFAyhyvs选配418，21017mmAs%2.0%32.08004001017bhAs满足最小配筋要求。51.某单层厂房现浇柱下独立锥形扩展基础，已知由柱传来基础顶面的内力：N=920KN，M=276KN.M，V=25KN，柱截面尺寸mmmmhb600400，地基承载力2/200mkNfa，基础埋深1.5m，基础采用C20混凝土，HPB235级钢筋，试设计此基础并绘制基础平面图，剖面和配筋图。解：1.初步确定基础高度和杯口尺寸取插入深度：mmh6001，故，杯口深度为mm65050600；杯口顶面尺寸：宽为mm550275400，长为mm750275600；杯壁厚度取为mmt200；杯底厚度取为mma2001；其他截面尺寸：mmh3002；mma2002；所以，初步确定基础的高度为mm850200506002.确定基础底面积241.55.120200920mdrfFAma增加20%~40%，初步选定尺寸为：mlmB2,5.3；27mA3221.465.326mlbWkNbldrGm2105.125.3202max/2341.485.0252765.32210920mkNWhVMblGFPcc2min/9.881.485.0252765.32210920mkNWhVMblGFPcc（1）22max/2402.1/234mkNfmkNPa（2）2minmax/2004.16129.882342mkNfPPa故，基础底面尺寸满足要求。3．验算基础高度地基净反力的计算。300350200Lb9501150600400b2min,2max,/9.581.485.0252765.32920/9.2031.485.0252765.32920mkNWhVMblFPmkNWhVMblFPccnccn仅需对台阶下基础做冲切验算：mmh510405500冲切面：mmbmmbbt197051