结构动力学大作业1

1、结构动力学课程论文结构动力学课程论文一、题目1、试设计一个3层框架，根据实际结构参数，求出该结构的一致质量矩阵、一致刚度矩阵；2、至少采用两种方法求3层框架的频率和振型；3、采用时程分析法，输入地震波，求出所设计的3层框架各层的非线性位移时程反应，要求画出所设计的框架图、输入的地震波的波形图、所求得的各楼层位移时程反应图。二、问题解答1、问题1解答1.1、框架设计框架立面图如下图一所示，梁截面均为400700mm2，柱子的截面均为600600mm2，跨度为7.2m，层高为3.6m，混凝土采用C30。图一 框架立面图设梁、柱均不产生轴向变形，且只考虑在框架的平面内变形，那么有3个平结构动力学课程

2、论文移自由度和12个转角自由度，一共有15个自由度，自由度以及梁柱单元编号如下图二所示：V1V2V3图二 单元编号及自由度方向先计算各个单元的一致质量矩阵和一致刚度矩阵，然后把相关的单元叠加组合计算得到整个结构的一致质量矩阵和一致刚度矩阵。 1.2、结构的一致质量矩阵梁：=0.40.72500=700kg/m, L=7.2m;梁、柱都为均布质量,故：ff ffI1I2I3I4L=4205622L1565415613L22L13L4L-13L-22L-3L-13L-22L-3L4L221vv 23v 4v结构动力学课程论文结构动力学课程论文 柱：=0.60.62500=900kg/m,L=3.6

3、m单元刚度矩阵如下：结构动力学课程论文结构动力学课程论文(m)(n)(p)ijijij由mij=m+m+m+.可计算一致质量矩阵中的各元素：(1)(2)(3)(10)(11)(12)(13)11111111111111m11=m+m+m+m+m+m+m=35040+41203.43=19933.72(10)(11)(12)(13)12121212m12=m+m+m+m=4416.57=1666.28结构动力学课程论文m13=0(10)m14=m15=m16=m17=m14=610.97(10)m18=m19=m1,10=m1,11=m18=-361.03 m1,12=m1,13=m1,14=m

4、1,15=0 (4)(5)(6)(10)(11)(12)(13)(14)(15)(16)(17)2222222222222222222222m22=m+m+m+m+m+m+m+m+m+m+m=35040+81203.43=24747.44(14)(15)(16)(17)23232323m23=m+m+m+m=4416.57=1666.28(10)m24=m25=m26=m27=m24=361.03(14)(10)2828m28=m+m=610.97-610.97=0 同理 m29=m2,10=m2,11=0(14)m2,12=m2,13=m2,14=m2,15=m2.03 ,12=-361(7

5、)(8)(9)(14)(15)(16)(17)(18)(19)(20)(21)3333333333333333333333m33=m+m+m+m+m+m+m+m+m+m+m=35040+81203.43=24747.44(14)m34=m35=m36=m37=0 m38=m39=m3,10=m3,11=m38=361.03 (14)3(18)m3,12=m3,13=m3,14=m3,15=m.97-610.97=0 ,12+m3,12=610(1)(10)(1)444445m44=m+m=2488.32+399.91=2888.23 m45=m=-1866.24m46=m47=0(10)48m

6、48=m=-299.93m49=m4,10=m4,11=m4,12=m4,13=m4,14=m4,15=0(2)(1)(2)(11)56555555=-1866.24m55=m+m+m=2488.32+2488.32+399.91=5376.55m56=mm57=m58=0(11)59m59=m=-299.93 m5,10=m5,11=m5,12=m5,13=m5,14=m5,15=0 (2)(3)(12)666666m66=m+m+m=2488.32+2488.32+399.91=5376.55(3)67m67=m=-1866.24 m68=m69=0(12)6m6,10=m.93 m6,1

7、1=m6,12=m6,13=m6,14=m6,15=0 ,10=-299(3)(13)7777m77=m+m=2488.32+399.91=2888.23m78=m79=m7,10=0(13)7m7,11=m.93 m7,12=m7,13=m7,14=m7,15=0 ,11=-299结构动力学课程论文(4)(10)(14)888888m88=m+m+m=2488.32+399.91+399.91=3288.14(4)89m89=m=-1866.24 m8,10=m8,11=0(14)8m8,12=m.93 m8,13=m8,14=m8,15=0 ,12=-299(4)(5)(11)(15)99

8、999999m99=m+m+m+m=2488.32+2488.32+399.91+399.91=5776.46(5)9m9,10=m.24 ,10=-1866(15)9.93 m9,14=m9,15=0 m9,11=m9,12=0 m9,13=m,13=-299(5)(6)(12)(16)10101010m10,10=m.32+2488.32+399.91+399.91=5776.46 ,10+m,10+m,10+m,10=2488(6)(16)10 m10,11=m=-1866.24m=m.93 m10,15=0 m=m=010,1210,13,1110,1410,14=-299(6)(13

9、)(17)111111m11,11=m.32+399.91+399.91=3288.14,11+m,11+m,11=2488m11,12=m11,13=m11,14=0(17)11m11,15=m.93,15=-299(7)(14)(18)121212m12,12=m.32+399.91+399.91=3288.14 ,12+m,12+m,12=2488(7)12m12,13=m.24 m12,14=m12,15=0 ,13=-1866(7)(8)(15)(19)13131313m13,13=m.32+2488.32+399.91+399.91=5776.46 ,13+m,13+m,13+m,

10、13=2488(8)13m13,14=m.24 m13,15=0 ,14=-1866(8)(9)(16)(20)14m14,14=m+m+m+m.32+2488.32+399.91+399.91=5776.46 ,1414,1414,1414,14=2488(9)14m14,15=m.24 ,15=-1866(9)(17)(21)151515m15,15=m.32+399.91+399.91=3288.14 ,15+m,15+m,15=2488则得：一致质量矩阵（该矩阵为对称矩阵，故下三角省略）单位（kg）结构动力学课程论文019933.721666.2824747.441666.282474

11、7.44M=610.97361.0302888.23610.97361.030-1866.245376.55610.97361.0300-1866.245376.55610.97361.03000-1866.242888.23-361.030361.03-299.930003288.14-361.030361.030-299.9300-1866.245776.46-361.030361.0300-299.9300-1866.245776.46-361.030361.03000-299.9300-1866.243288.140-361.0300000-299.930003288.14-361.0

12、3-361.03-361.03000000000000000000-299.93000-299.93000-299.93-1866.24005776.46-1866.2405776.46-1866.243288.140001.3、结构的一致刚度矩阵各梁、柱均为等截面，故单元刚度矩阵为：-63L3Lv1fs16fv6-3L-3Ls22EI-62= 223f3L-3L2LLLs3v322f3L-3LL2Lv4s4框架梁：C30混凝土E=3107KN/m2，0.400.73EI=310=3.43105kNm2，L=7.2m127结构动力学课程论文7框架柱：0.600.603EI=310=3.2410

13、5KNm2 L=3.6m12结构动力学课程论文结构动力学课程论文结构动力学课程论文(m)+k(n)+k(p)+.可计算一致刚度矩阵中的各元素： 由kij=kijijij(10)+k(11)+k(12)+k(13)=40.833105=3.332105 k11=k11111111(10)+k(11)+k(12)+k(13)=4(-0.833k12=k)105=-3.332105 k13=0 12121212(10)k14=k15=k16=k17=k18=k19=k1,10=k1,11=k14=1.50105k1,12=k1,13=k1,14=k1,15=0(10)+k(11)+k(12)+k(1

14、3)+k(14)+k(15)+k(16)+k(17)=80.833105=6.664105 k22=k2222222222222222(14)+k(15)+k(16)+k(17)=4(-0.833k23=k)105=-3.3321052323232310k24=k25=k26=k27=k24=-0.861105(10)+k(14)=0.861105-0.861105=0 同理 k28=k2828k29=k2,10=k2,11=0结构动力学课程论文(14)=1.50105 k2,12=k2,13=k2,14=k2,15=k2,12(14)+k(15)+k(16)+k(17)+k(18)+k(19

15、)+k(20)+k(21)=80.833105=6.664105 k33=k3333333333333333k34=k35=k36=k37=0(14)k38=k39=k3,10=k3,11=k38=-1.50105(14)+k(18)=1.50105-1.50105=0 k3,12=k3,13=k3,14=k3,15=k3,123,12(1)=0.953105 (1)+k(10)=1.906105+3.60105=5.506105 k=kk44=k44444545k46=k47=0(10)=1.80105k48=k48k49=k4,10=k4,11=k4,12=k4,13=k4,14=k4,1

16、5=0(1)+k(2)+k(11)=1.906105+1.906105+3.60105=7.412105k55=k555555(2)=0.953105 k56=k56k57=k58=0 (11)=1.80105 k59=k59k5,10=k5,11=k5,12=k5,13=k5,14=k5,15=0(2)+k(3)+k(12)=1.906105+1.906105+3.60105=7.412105k66=k666666(3)=0.953105 k67=k67(12)=1.80105 k=k=k=k=k=0 k68=k69=0 k6,10=k6,116,126,136,146,156,10(3)+

17、k(13)=1.906105+3.60105=5.506105k77=k7777k78=k79=k7,10=0(13)=1.80105 k=k=k=k=0 k7,11=k7,127,137,147,157,11(4)+k(10)+k(14)=1.906105+3.60105+3.60105=9.106105k88=k888888(4)=0.953105 k89=k89(14)=1.80105 k=k=k=0 k8,10=k8,11=0 k8,12=k8,138,148,158,12(4)+k(5)+k(11)+k(15)=1.906105+1.906105+3.60105+3.60105k99

18、=k99999999=11.012105 14结构动力学课程论文(5)=0.953105 k9,10=k9,10k9,14=k9,15=0k9,11=k9,12=0(15)=1.80105 k9,13=k9,13(5)+k(6)+k(12)+k(16)=1.906105+1.906105+3.60105+3.60105k10,10=k10,1010,1010,1010,10=11.0121055(6)=0.953105 k(16)k10,11=k10,12=k10,13=0 k10,14=k10,14=1.8010 k10,15=0 10,11(6)+k(13)+k(17)=1.906105+

19、3.60105+3.60105=9.106105 k11,11=k11,1111,1111,11(17)=1.80105 k11,12=k11,13=k11,14=0 k11,15=k11,15(4)+k(7)+k(18)=1.906105+3.60105+3.60105=9.106105 k12,12=k12,1212,1212,12(7)=0.953105 kk12,13=k12,14=k12,15=0 12,13(7)+k(8)+k(15)+k(19)=1.906105+1.906105+3.60105+3.60105k13,13=k13,1313,1313,1313,13=11.012

20、105(8)=0.953105 kk13,14=k13,15=0 13,14(8)+k(9)+k(16)+k(20)=1.906105+1.906105+3.60105+3.60105k14,14=k14,1414,1414,1414,14=11.012105(9)=0.3125105k14,15=k14,15(9)+k(17)+k(21)=1.906105+3.60105+3.60105=9.106105 k15,15=k15,1515,1515,15得到一致刚度矩阵（该矩阵为对称矩阵，故下三角省略）单位（kN/m）3.332K=105-3.3326.6640-3.3326.6641.50-

21、1.5005.5061.50-1.5000.9537.4121.50-1.50000.9537.4121.50-1.500000.9535.5061.500-1.501.800009.1061.500-1.5001.80000.95311.0121.500-1.50001.80000.95311.0121.500-1.500001.80000.9539.10601.50000001.800009.10601.500000001.80000.95311.01201.5000000001.80000.95311.01201.50000000001.80000.9539.106结构动力学课程论文 2

22、 问题2 解答2.1采用振型分解反应谱法，求解框架的频率和振型=0的特征值得到频率和振型： 由K-2Mv在Matlab中导入质量矩阵M和刚度矩阵K,输v,2=eig(K,M);=sqrt(2)可得框架的频率为： T=123.1415=32.861, 109.022, 199.133, 234.897, 299.589, 307.809 , 378.000, 388.414,454.501, 480.646, 583.896 , 637.664, 747.045, 828.365, 1056.507 框架的振型为=123.1415=1 2 3 4 5 6 7 89 10 11 12 13 14

23、15结构动力学课程论文2.2 用Stodola法计算三层框架的频率和振型 此结构的柔度矩阵是f=K-1=D=fm=526123466113564-2919-2009-2009-291910-5-4627-3739-3739-4627-4436-3429-3429-4436453153846316844-648-547-547-648-3237-2622-2622-3237-5313-4034-4034-5313179831712511915-95-55.4-55.4-95-395-408-408-395-2546-1883-1883-25461933.61502.4600.3585.14-40

24、3.97.9103-69.43-330.8-46.98-141.3-151.2-164.3-169.8-202-2022088.51495.3606.51-631828.14-463.5-16.66-39.22-52.13-197.9-226.6-226.6-114.2-168.9-189.520471488.4605.89-7.983-467.4891.72-635.4-193.6-45.69-344.3-30.81-189.2-169.2-112.1-227.41933.61502.4600.3-69.437.9103-403.9585.14-151.2-141.3-46.98-330.8

25、-202-149.4-169.8-164.3-959-507-53.6-141139.935.2281.15567.3-144114.5114.5109.7-71.595.5537.1-466.3-466.3-74.97214.41-174.8143.1345.392-214.1713.23-171.7134.55129.58-99.996.75140.084-885-466-7545.39143.1-175214.4134.6-172713.2-21440.0896.75-99.9129.6-959.1-507.2-53.6281.15235.225139.93-141.1109.67114

26、.46-143.6567.357.99237.09595.548-71.52-959.1-507.2-53.6281.15235.225139.93-141.1109.67114.46-143.6567.357.99237.09595.548-71.52-768.9-687.5-306.5-28.0246.216-14.4511.743129.09-99.9796.6839.588-186.8664.53-159.7122.14-768.9-687.5-306.511.743-14.4546.216-28.0239.58896.68-99.97129.09122.14-159.7664.53-

27、186.8-898.5-828.5-387.3-1.1767.242-15.8945.16858.0437.4995.943-71.48113.72107.89-127.1525结构动力学课程论文V1(1)=DV1(0)迭代过程列表如下 根据D V1(0)526123466113564-2919-2009-2009-291910-5-4627-3739-3739-4627-4436-3429-3429-4436453153846316844-648-547-547-648-3237-2622-2622-3237-5313-4034-4034-5313179831712511915-95-55.

28、4-55.4-95-395-408-408-395-2546-1883-1883-25461933.61502.4600.3585.14-403.97.9103-69.43-330.8-46.98-141.3-151.2-164.3-169.8-202-2022088.51495.3606.51-631828.14-463.5-16.66-39.22-52.13-197.9-226.6-226.6-114.2-168.9-189.520471488.4605.89-7.983-467.4891.72-635.4-193.6-45.69-344.3-30.81-189.2-169.2-112.1

29、-227.41933.61502.4600.3-69.437.9103-403.9585.14-151.2-141.3-46.98-330.8-202-149.4-169.8-164.3-959-507-53.6-141139.935.2281.15567.3-144114.5114.5109.7-71.595.5537.1-466.3-466.3-74.97214.4-174.8143.145.39-214.1713.2-171.7134.5129.5-99.996.7540.08-885-466-7545.39143.1-175214.4134.6-172713.2-21440.0896.

30、75-99.9129.6-959.1-507.2-53.6281.1535.22139.93-141.1109.67114.46-143.6567.357.9937.0195.54-71.52-959.1-507.2-53.6281.1535.22139.93-141.1109.67114.46-143.6567.357.9937.0995.54-71.52-768.9-768.9-687.5-306.5-28.0246.21-14.4511.74129.1-99.9796.6839.58-186.8664.5-159.7122.14-687.5-306.511.743-14.4546.216

31、-28.0239.58896.68-99.97129.09122.14-159.7664.53-186.8-898.5-828.5-387.3-1.1767.242-15.8945.1658.0437.4995.94-71.48113.7107.8-127.1525111111111111111V1(1) V1(1) V1(2) V1(2) V1(3) V1(3) V1(4) V1(4 ) V1(5)11688991257.843091-3558.1-2480.4-2412.9-3571.2-8221.3-6697.9-6710.8-8217.2-12347-9334.2-9331.7-123

32、4810.7810.369-0.03-0.02-0.02-0.03-0.07-0.06-0.06-0.07-0.11-0.08-0.08-0.11949641928321926271926086917330082-3452-2444-2446-3451-7208-5858-5857-7208-9276-7082-7082-9276712580.7504311920.3285-3477-0.037-2465-0.026-2465-0.026-3477-0.037-7335-0.077-5965-0.063-5965-0.063-7335-0.077-9574-0.101-7305-0.077-7

33、305-0.077-9574-0.101693690.7473301840.3252-3454-0.037-2446-0.026-2448-0.026-3454-0.037-7221-0.078-5868-0.063-5868-0.063-7221-0.078-0.1-9304-7103-0.077-7103-0.077-0.1-9304691890.747300910.325-3452-0.04-2444-0.03-2446-0.03-3451-0.04-7209-0.08-5859-0.06-5858-0.06-7209-0.08-9278-0.1-7084-0.08-7084-0.08-

34、9278-0.1则得到第一振型形式为1=（-0.1585 -0.1184 -0.0515 0.005910.00418 0.00419 0.00591 0.01234 0.01003 0.01002 0.01234 0.01588 0.01212 0.01212 0.01588） 再用公式12=(V1)TmV1(0)(V)mV(1)T1(1)1(1)，将数据代入得1=32.75。结构动力学课程论文TT-(1ms)-1(1mr)第二振型，先求第一振型淘汰矩阵的形式S1=（ms是，I质量矩阵的第一列，mr是质量矩阵剩下的各列）再求D2=DS1，然后用求第一振型的格式用D2这个动力矩阵迭代求解第二振

35、型和频率。 求第三阵型时淘汰矩阵S2=S1-122Tm 动力矩阵D3=DS2。 M21nnTm 动力矩阵D3=DS2最后代入数据得Mn更高振型时淘汰矩阵Sn=Sn-1-到T=123.1415=32.75, 106.23, 197.233, 230.1, 299.589, 301.22 , 378.855, 380.414,450.501, 481.64, 581.9 , 631.8, 747.045, 828.289, 1056.665框架的振型为=123.1415= 1 2 3 4 5 6 7 89 10 11 12 13 14 15结构动力学课程论文3、问题三解答3.1、框架资料框架的设计

36、同问题一，梁截面均为400700mm2，柱子的截面均为600600mm2，跨度为7.2m，层高为3.6m，混凝土采用C30，如下图四所示：图四因为框架的质量太小，计算的自振周期太小，导致计算的结果发散，所以结构动力学课程论文 把质量矩阵放大10倍，刚度矩阵保持不变，这样也比较符合在实际中考虑整个楼层质量，这样就得到结构的质量和刚度如下图所示：112253555553图五 结构简图质量矩阵跟刚度矩阵：0002.161.367-1.367kg-1.3672.734-1.367kN/m5M=10502.8080K=10 02.808-1.3673.277003.2、读取地震波我市58号，分配的地震波

37、为USA00001，程序如下所示：fid=fopen(G:动力学EQDB-WAVEUSA00001.ACC);data=fscanf(fid,%g,1,inf);data=data.;fclose(fid);d1=data(1) % total number of datad2=data(2) % time intervald3=data(3) % transfer unitd=data(4:end).*d3; % wave data%以下为结构信息。%自由度数目结构动力学课程论文 plot(TT,d),gridxlabel(时间(s)ylabel(加速度(m/s2)title(地震波加速度曲

38、线)读取地震波如下：并且得到总共有2688个数据时间步长为0.01秒。2.3、分析主程序分析程序采用Matlab,使用wilson-法，结合前面的读取地震波的程序对变量的赋值（没有前面程序赋值，以下程序不可单独使用），主程序如下：cn=3; %自由度数目k1=1.910e5;k2=1.367e5;k3=1.367e5;m1=2.808e2;m2=2.808e2;m3=2.16e2;m0=m3,m2,m1;%以下为地震波的时间参数TT=0.01:0.01:26.88; %0.01为时间间隔，26.88为地震波持续时间%结构动力学课程论文 TT=TT;M=diag(m0);E=0.05;0.10;

39、0.15; % %振型阻尼比dt=0.01;T=1.4*dt;P=d*1,1,1*M; %荷载矩阵S1=0;0;0;V1=0;0;0;A1=0;0;0;%for i=1:2688%k0=k1,k2,k3;K=zeros(cn);for j=1:cn-1K(j,j)=k0(j)+k0(j+1);K(j,j+1)=-k0(j+1);K(j+1,j)=-k0(j+1);endK(cn,cn)=k0(cn);%结构刚度矩阵%v,w=eig(K,M);w=sqrt(w);% w：振动频率矩阵 %Q=1.0/w(1,1)3,1.0/w(1,1),w(1,1);1.0/w(2,2)3,1.0/w(2,2),w(2,2);1.0/w(3,3)3,1.0/w(3,3),w(3,3);a=2*inv(Q)*E;a0=a(2);a1=a(3);C=a0*M+a1*K;%