克拉夫《结构动力学》习题答案汇总

第二章自由振动分析2-1（a）由例22WTgk22()WKTg因此max()()DtkT其中k=0、1、2……TD=0.64sec如果很小，TD=T222200()49.9/0.64sec386/seckipskkipsinin50/kkipsin（b）211lnlnnnvvvv222121()11.2ln0.3330.86210.05292()10.33320.053025.3%(a’)21D2T21DTT249.950/1kkipsin(c)2cmWmg2T4cTg21DTT241WcTg2240.05292000.64sec386/sec10.0529kipscin0.539sec/ckipsinT=TD0.538sec/ckipsin0.54sec/ckipsin2-22km404.472(1/sec)(0)(0)()sin(0)costDDDvvtetvt(0)(0)()sin(0)(0)(0))costDDDvvtetvvvt22(0)(0)()(0)cossinDtDDDvvtevtt21D()(0)cos(0)(0)sintDDDtevtvvt2(0)(0)()(0)cossin1tDDvvtevtt0.055922(2)(4.47)cccm(a)c=00D5.6(1)sin4.470.7cos4.471.384.47vtin(1)5.6cos4.474.47(0.7)sin4.471.69/secvtin(1)1.4vin,(1)1.7/secvin(b)c=2.80.0559(2.8)0.15724.4710.1574.41D(1/sec)(0.157)(4.41)5.60.7(0.157)(4.47)(1)sin4.410.7cos4.414.41te(1)0.764tin(0.157)(4.41)20.157(5.6)4.41(0.7)(1)5.6cos4.41sin4.4110.157te(1)1.10/sectin(1)0.76vin,(1)1.1/secvin第三章谐振荷载反应3-1根据公式有21sinsin1Rtwtwt0.8ww2.778sin0.8sin1.25Rtwtwt将t以80°为增量计算)(tR并绘制曲线如下：080°160°240°320°400°480°560°640°720°800°00.5471.71-0.481-3.2140.3574.33-0.19-4.92404.9241.25wwt)(tR3-2解：由题意得：22mkipssin，20kkipsin，(0)(0)0vv，ww203.162sec2kwradm8wt（a）0c1sincos2Rtwtwtwt将8wt代入上式得：()412.566Rt（b）0.5cksin0.50.03952223.162ccccmw1exp1cosexpsin2Rtwtwtwtwt将8wt代入上式得：()7.967Rt（c）2.0cksin2.00.1582223.162ccccmw1exp1cosexpsin2Rtwtwtwtwt将8wt代入上式得：()3.105Rt3-3解：（a）：依据共振条件可知：1003860.0810.983sec4000kkgwwradmW由2LTVw得：10.9833662.96022wLVfts（b）：122max2221212tgovv1ww0.41.2govin代入公式可得：max1.921tvin（c）：2LTVw45min66Vhfts226611.51336VwradsecL11.5131.04810.983ww0.4代入数据得：122max22212=1.85512tgovvin3-4解：按照实际情况，当设计一个隔振系统时，将使其在高于临界频率比2下运行，在这种情况下，隔振体系可能有小的阻尼。对于小阻尼：211TR又因为：max0.0050.03tgovTRv联立求的：272220125.6secwfrad又因为：wwkkgwmW联立得：222125.68004.6703867wWkkipsing3-5解：按照实际情况，当设计一个隔振系统时，将使其在高于临界频率比2下运行，在这种情况下，隔振体系可能有小的阻尼。对于小阻尼：211TR又因为：max50700tgovTRv联立求的：215221275.36secwfrad又因为：wwkkgwmW联立得：22275.3665006.37638615wWkkipsing3-6(a)由图3-17有，maxsfk，则inlbfinlbffks/260015.0390max(b)由方程3-66有22kEDeq，由图3-17有221kES，因为mc2，所以，又因为，故0713.0)29(4264inlbfinlbfEESDeqmceqeq2222mEDeq)/(78.36)15.0(/102622insradlbfinsradinlbfEcDeq(c)由方程3-78有2，3-7(a)公式同题3.6(b)(b)公式同题3.6(c)(c)通过题3.6与题3.7的比较可知，与无关，故滞变阻尼机理更合理。3-8（原版英文书中为求DE的值）由方程3-66有22kEeqD，当k与不变时，若)(eqeq，则)(DDEE,由题3.7可知%2.141426.00713.0220713.0)29(4264inlbfinlbfEESDeq)/(39.18)15.0(/202622insradlbfinsradinlbfEcDeq%2.141426.00713.022)(DDEE第五章对冲击荷载的反应5-1解：(a)1000386=25.36rad/sec600kkgmW120.15T=0.248sec0.6050.50.248tT*21/t120.94rad/sec0.15t20.94===0.82625.36*2=0.136sec20.94125.36/20.94t(b)**0max2215001(sinsin)sin(20.940.136)0.8489in110001(0.826)pvttk,maxmax10000.8489848.91Sfkvb又10max0.6050.85tpvDinTk5-2解：设无阻尼(a)011()(/)0(1)mvkvptpttttcossincpvAtBtvEtF带入（1）得：0011()0tpkEtFpFEtkt01pptvkt01=cossincpptvvvAtBtkt0011(0)0-ppvBBktkt0111v()(sint)pttktt1tt当2011111121()sinsin()1sincos()2ptvttttttktt（b）1:tt当02001112()1cossin2pptvttktktsin00,1,222ttnnmaxmax10121322sin()/3333vRtpktmax23R5-3解：(a)110242Tttt0cos(1)mvkvptcossincossincpvAtBtvEttFtt带入（1）得：00022ppEFmk0cossinsin2cppvvvAtBtttk0(0)0(0)sin(0)(0)cos(0)020vApvBkB0sin2pvttk0011()sin2224ppttvtkk001()sinsin22222ppvtkk01v()sincos224ptttk(b)max1max00sin/2vtttRtpk1max11sincos=024ttRttttt5-4解：(a)110.15tT,查表得：D=0.50,maxmax00.3954015.8SkpfvkDpDkk(b)10max010101112()()0.150.3532222tpvptdtptptTmkkT,maxmax0.353207.06Sfvkk5-5解：1max0011()()ttvptdtptdtmkm1012340()(424)53ttptdtpppppmax5=0.395404vS,maxmax=0.3954015.8kfvkmax,max15015.82370SMdfkft第六章6-1Solution:2Km(a)简单求和tNP(N)Sin(ωtN)Cos(ωtN)Y(N-1)AN/FY(N-1)BN/F(1)(2)(3)(4)(5)(6)(7)(8)(9)(10)(11)(12)(13)0000.001.000.000.000.000.000.000.000.000.000.0010.1500.590.8140.4540.4529.4029.4023.7823.780.000.000.0020.286.60.950.3126.7667.2182.36111.7663.9234.5329.381.8718.4630.31000.95-0.31-30.9036.3195.10206.8634.53-63.9298.456.2761.8640.486.60.59-0.81-70.06-33.7550.92257.78-19.85-208.54188.7012.01118.5650.5500.00-1.00-50.00-83.750.00257.780.00-257.78257.7816.41161.9660.60-0.59-0.810.00-83.750.00257.7849.25-208.54257.7916.41161.970.10.06366/0.2521Finlbm0.1sec2/klbin(5)(2)(4)(11)(9)(10)(7)(2