混凝土设计复习题.docx

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1、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Mb165kNm，MC90kNm；各柱惯性矩：I1A2.13109mm4，I2a14.52109mm4；11B5.21109mm4，12B17.76109mm4；11C2.13109mm4，12C15.38109mm43提示：C01(-1)nH3REIG'HuHM_312IuC3，C3，nH2,3/1八Il1(—1)ln解：1.计算剪力分配系数A柱：2.13109914.521090.147，3.60.3310.9B柱:C0C柱:C0EI1C0109003E14.5210917.76%0

2、・293，5.212.483.610.935.960.331)1090032.7626.42EIGE17.761092.769込卫90.138,15.381093.612.90.279H31)310.2792.640.138EliC。129003E15.381092.6452.8735.96E0.329,ABC35.9626.4252.8726.42E0.447ABC35.9626.4252.8752.87E0.22435.9626.4252.872.柱顶剪力计算C3B21123(丄1)2n0.3321.233110.33(1)0.2931C3C312213(i1)n1

3、0.27921.218310.2793(1)0.138RbMbcTC3B16510.91.2318.62kNRc归C3CH3C9012.91.2188.50kN（一）rbRc18.628.5010.12kN(J)VaaR0.32910.123.33kN(VbbRRb0.44710.1218.62VccRRc0.22410.128.50作弯矩图10.77kN(~>)3.14.10kN(kAM图(kN-m)47.某单层单跨厂房，跨度试求该柱承受的吊车竖向荷载Dmin、如下：吊车跨度Lk=,吊车宽B=，压标准值Pmax,k=KN。18m，柱距6m,内有两台10t中级载荷状态的

4、吊车，Dmax和横向水平荷载Tmax。起重机有关数据轮距K=，吊车总质量，额定起重量10t，最大轮Pmin,k(mbm2Q)PgPmax,k(m2QJgP(18.231gPmax,k小_2p0^/卒10104.7/□6.45kN查表，0.9，竖向反力影响线如图所示:则有：Dmax0.91.4104.7(10.561.664.96288.47kNDminDmax萱288-47签侦硼1Tk(m2Q)g410.120.9(3.68410)1043.69kNTmax288.4710.17kN104.748.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00.

5、45kN/m2，15m高处的z1.14（10m高z1.0），体形系数s示于图中。柱截面惯性距：94949494rI12.1310mm,丨214.3810mm,I37.210mm,I419.510mm.上柱高3m，计算单元宽度取B=6m。解：1.求q「q2线性差值：z11.141.0(10.510)1.0141510q1kq2k0.81.0140.4562.19kN/m0.51.0140.4561.37kN/mq1rQq1k1.42.193.07kN/m()q2rQq2k1.41.371.92kN/m()风向2.高度取+=15101.0(12.610)1.07Wk[(0.

6、80.5)h1(0.50.6)h2]z0B[1.32.10.11.2]1.070.4567.54KN「QWk1.47.5410.56kN3.计算参数,n11nA柱:I2HuH2.1310914.381090.15B柱:30.2910.54.在柱顶虚拟加不动铰支座查表计算：I3匚HuH竺竺0.3719.5109—0.2910.5A柱：C11B柱：C118[13(丄1)]8[1310.293(-1)]n0.153[11)]3[10.294(11)]n0.373[11)]0.294(丄0.158[13(n1)]8[1曲(盍1)]3[14(丄1)]n0.340.36A柱：Co3

7、13(丄1)n33__110.293(1)0.152.64Raq1HC113.0710.50.3411()Rbq2HC111.9210.50.367.3()故:柱顶剪力Va,1Ra11()Vb,1Rb7.3()5.撤消不动铰支座，反向施加支座反力。B柱：Co313(-1)n11.54110.54()VBVB,1VB,217.327.310.02()2.883110.29(1)0.37于UBH3是10.53UAH310.53EcIlC0Ec14.381092.64EcIlC0Ec19.51092.882.6414.380.40.6所以分配系