結(jié)構(gòu)動(dòng)力學(xué)大作業(yè)

1、結(jié)構(gòu)動(dòng)力學(xué)課程論文結(jié)構(gòu)動(dòng)力學(xué)課程論文一、題目1、試設(shè)計(jì)一個(gè)3層框架，根據(jù)實(shí)際結(jié)構(gòu)參數(shù)，求出該結(jié)構(gòu)的一致質(zhì)量矩陣、一致剛度矩陣；2、至少采用兩種方法求3層框架的頻率和振型；3、采用時(shí)程分析法，輸入地震波，求出所設(shè)計(jì)的3層框架各層的非線性位移時(shí)程反應(yīng)，要求畫出所設(shè)計(jì)的框架圖、輸入的地震波的波形圖、所求得的各樓層位移時(shí)程反應(yīng)圖。二、問題解答1、問題1解答1.1、框架設(shè)計(jì)框架立面圖如下圖一所示，梁截面均為400700mm2，柱子的截面均為600600mm2，跨度為7.2m，層高為3.6m，混凝土采用C30。圖一 框架立面圖設(shè)梁、柱均不產(chǎn)生軸向變形，且只考慮在框架的平面內(nèi)變形，那么有3個(gè)平結(jié)構(gòu)動(dòng)力學(xué)課程

2、論文移自由度和12個(gè)轉(zhuǎn)角自由度，一共有15個(gè)自由度，自由度以及梁柱單元編號(hào)如下圖二所示：V1V2V3圖二 單元編號(hào)及自由度方向先計(jì)算各個(gè)單元的一致質(zhì)量矩陣和一致剛度矩陣，然后把相關(guān)的單元疊加組合計(jì)算得到整個(gè)結(jié)構(gòu)的一致質(zhì)量矩陣和一致剛度矩陣。 1.2、結(jié)構(gòu)的一致質(zhì)量矩陣梁：=0.40.72500=700kg/m, L=7.2m;梁、柱都為均布質(zhì)量,故：ff ffI1I2I3I4L=4205622L156L22L13L4L-13L-22L-3L-13L-22L-3L4L221vv 23v 4v結(jié)構(gòu)動(dòng)力學(xué)課程論文結(jié)構(gòu)動(dòng)力學(xué)課程論文 柱：=0.60.62500=900kg/m,L=3.6m單元?jiǎng)偠染仃?/p>

3、如下：結(jié)構(gòu)動(dòng)力學(xué)課程論文結(jié)構(gòu)動(dòng)力學(xué)課程論文(m)(n)(p)ijijij由mij=m+m+m+.可計(jì)算一致質(zhì)量矩陣中的各元素：(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結(jié)構(gòu)動(dòng)力學(xué)課程論文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=m1,15=0

4、(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)(8)(9)

5、(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)48m48=m=-2

6、99.93m49=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 m6,11=m6,12

7、=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結(jié)構(gòu)動(dòng)力學(xué)課程論文(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)99999999m

8、99=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)(17)11

9、1111m11,11=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,13=2488

10、(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則得：一致質(zhì)量矩陣（該矩陣為對(duì)稱矩陣，故下三角省略）單位（kg）結(jié)構(gòu)動(dòng)力學(xué)課程論文019933.2824747.2824747.44M=610.97361.23610

11、.97361.030-1866.55610.97361.0300-1866.55610.97361.03000-1866.23-361.03-299.14-361.030-299.9300-1866.46-361.0300-299.9300-1866.46-361.03000-299.9300-1866.140-361.-299.14-361.03-361.03-361.03000000000000000000-299.93000-299.93000-299.93-1866.24005776.46-1866.2405776.46-1866.243288.140001.3、結(jié)構(gòu)的一致剛度矩陣各梁

12、、柱均為等截面，故單元?jiǎng)偠染仃嚍椋?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結(jié)構(gòu)動(dòng)力學(xué)課程論文7框架柱：0.600.603EI=310=3.24105KNm2 L=3.6m12結(jié)構(gòu)動(dòng)力學(xué)課程論文結(jié)構(gòu)動(dòng)力學(xué)課程論文結(jié)構(gòu)動(dòng)力學(xué)課程論文(m)+k(n)+k(p)+.可計(jì)算一致剛度矩陣中的各元素： 由kij=kijijij(10)+k(11)+k(12)+k(13)=40.833105=3.332105

13、k11=k(10)+k(11)+k(12)+k(13)=4(-0.833k12=k)105=-3.332105 k13=0 (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(13)+k(14)+k(15)+k(16)+k(17)=80.833105=6.664105 k22=k22222(14)+k(15)+k(16)+k(17)=4(-0.833k23=k)105=-3.33210k24=k25=k26=k27=k24=-0.861105(10)+k(1

14、4)=0.861105-0.861105=0 同理 k28=k2828k29=k2,10=k2,11=0結(jié)構(gòu)動(dòng)力學(xué)課程論文(14)=1.50105 k2,12=k2,13=k2,14=k2,15=k2,12(14)+k(15)+k(16)+k(17)+k(18)+k(19)+k(20)+k(21)=80.833105=6.664105 k33=k33333k34=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.

15、953105 (1)+k(10)=1.906105+3.60105=5.506105 k=kk44=kk46=k47=0(10)=1.80105k48=k48k49=k4,10=k4,11=k4,12=k4,13=k4,14=k4,15=0(1)+k(2)+k(11)=1.906105+1.906105+3.60105=7.412105k55=k(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+3.6010

16、5=7.412105k66=k(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)+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=k(4)=0.953105 k89=k89(14)=1.80105 k=k

17、=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=k=11.012105 14結(jié)構(gòu)動(dòng)力學(xué)課程論文(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)=0.9531

18、05 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+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,

19、13(7)+k(8)+k(15)+k(19)=1.906105+1.906105+3.60105+3.60105k13,13=k13,1313,1313,1313,13=11.012105(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

20、,15=k15,1515,1515,15得到一致剛度矩陣（該矩陣為對(duì)稱矩陣，故下三角省略）單位（kN/m）3.332K=105-3.3326.6640-3.3326.6641.50-1.5005.5061.50-1.5000.9537.4121.50-1.50000.9537.4121.50-1.9535.5061.500-1.501.800009.1061.500-1.5001.80000.95311.0121.500-1.50001.80000.95311.0121.500-1.80000.9539.10601.10601.80000.95311.01201.80000.95311.012

21、01.80000.9539.106結(jié)構(gòu)動(dòng)力學(xué)課程論文 2 問題2 解答2.1采用振型分解反應(yīng)譜法，求解框架的頻率和振型=0的特征值得到頻率和振型： 由K-2Mv在Matlab中導(dǎo)入質(zhì)量矩陣M和剛度矩陣K,輸v,2=eig(K,M);=sqrt(2)可得框架的頻率為： 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 框架的振型為=123.1415=1 2

22、3 4 5 6 7 89 10 11 12 13 14 15結(jié)構(gòu)動(dòng)力學(xué)課程論文2.2 用Stodola法計(jì)算三層框架的頻率和振型 此結(jié)構(gòu)的柔度矩陣是f=K-1=D=fm=526123466113564-2919-2009-2009-291910-5-4627-3739-3739-4627-4436-3429-3429-44366844-648-547-547-648-3237-2622-2622-3237-5313-4034-4034-53131915-95-55.4-55.4-95-395-408-408-395-2546-1883-1883-25461933.61502.4600.3585

23、.14-403.97.9103-69.43-330.8-46.98-141.3-151.2-164.3-169.8-202-2022088.51495.3606.51-.14-463.5-16.66-39.22-52.13-197.9-226.6-226.6-114.2-168.9-189.5.4605.89-7.983-467.4891.72-635.4-193.6-45.69-344.3-30.81-189.2-169.2-112.1-227.41933.61502.4600.3-69.437.9103-403.9585.14-151.2-141.3-46.98-330.8-202-149

24、.4-169.8-164.3-959-507-53.6-.935.2281.15567.3-.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-.4134.6-.2-21440.0896.75-99.9129.6-959.1-507.2-53.6281.15235.93-141.1109.67114.46-143.6567.357.99237.09595.548-71.52-

25、959.1-507.2-53.6281.15235.93-141.1109.67114.46-143.6567.357.99237.09595.548-71.52-768.9-687.5-306.5-28.0246.216-14.4511.09-99.9796.6839.588-186.8664.53-159.7122.14-768.9-687.5-306.511.743-14.4546.216-28.0239.58896.68-99.97129.09122.14-159.7664.53-186.8-898.5-828.5-387.3-1.1767.242-15.8945.16858.0437

26、.4995.943-71.48113.72107.89-127.1525結(jié)構(gòu)動(dòng)力學(xué)課程論文V1(1)=DV1(0)迭代過程列表如下 根據(jù)D V1(0)526123466113564-2919-2009-2009-291910-5-4627-3739-3739-4627-4436-3429-3429-44366844-648-547-547-648-3237-2622-2622-3237-5313-4034-4034-53131915-95-55.4-55.4-95-395-408-408-395-2546-1883-1883-25461933.61502.4600.3585.14-403.97

27、.9103-69.43-330.8-46.98-141.3-151.2-164.3-169.8-202-2022088.51495.3606.51-.14-463.5-16.66-39.22-52.13-197.9-226.6-226.6-114.2-168.9-189.5.4605.89-7.983-467.4891.72-635.4-193.6-45.69-344.3-30.81-189.2-169.2-112.1-227.41933.61502.4600.3-69.437.9103-403.9585.14-151.2-141.3-46.98-330.8-202-149.4-169.8-1

28、64.3-959-507-53.6-.935.2281.15567.3-.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-.4134.6-.2-21440.0896.75-99.9129.6-959.1-507.2-53.6281.1535.22139.93-141.1109.67114.46-143.6567.357.9937.0195.54-71.52-959.1-507.2-53.6

29、281.1535.22139.93-141.1109.67114.46-143.6567.357.9937.0995.54-71.52-768.9-768.9-687.5-306.5-28.0246.21-14.4511.74129.1-99.9796.6839.58-186.8664.5-159.7122.14-687.5-306.511.743-14.4546.216-28.0239.58896.68-99.97129.09122.14-159.7664.53-186.8-898.5-828.5-387.3-1.1767.242-15.8945.1658.0437.4995.94-71.4

30、8113.7107.8-127.1525111111111111111V1(1) V1(1) V1(2) V1(2) V1(3) V1(3) V1(4) V1(4 ) V1(5)91257.843091-3558.1-2480.4-2412.9-3571.2-8221.3-6697.9-6710.8-8217.2-12347-9334.2-9331.7-1234810.7810.369-0.03-0.02-0.02-0.03-0.07-0.06-0.06-0.07-0.11-0.08-0.08-0.1194964192832-3452-2444-2446-3451-7208-5858-5857

31、-7208-9276-7082-7082-9276.3285-3477-0.037-2465-0.026-2465-0.026-3477-0.037-7335-0.077-5965-0.063-5965-0.063-7335-0.077-9574-0.101-7305-0.077-7305-0.077-9574-0.101.3252-3454-0.037-2446-0.026-2448-0.026-3454-0.037-7221-0.078-5868-0.063-5868-0.063-7221-0.078-0.1-9304-7103-0.077-7103-0.077-0.1-9304.747.

32、325-3452-0.04-2444-0.03-2446-0.03-3451-0.04-7209-0.08-5859-0.06-5858-0.06-7209-0.08-9278-0.1-7084-0.08-7084-0.08-9278-0.1則得到第一振型形式為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)，將數(shù)據(jù)代入得1=32

33、.75。結(jié)構(gòu)動(dòng)力學(xué)課程論文TT-(1ms)-1(1mr)第二振型，先求第一振型淘汰矩陣的形式S1=（ms是，I質(zhì)量矩陣的第一列，mr是質(zhì)量矩陣剩下的各列）再求D2=DS1，然后用求第一振型的格式用D2這個(gè)動(dòng)力矩陣迭代求解第二振型和頻率。 求第三陣型時(shí)淘汰矩陣S2=S1-122Tm 動(dòng)力矩陣D3=DS2。 M21nnTm 動(dòng)力矩陣D3=DS2最后代入數(shù)據(jù)得Mn更高振型時(shí)淘汰矩陣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 ,

34、631.8, 747.045, 828.289, 1056.665框架的振型為=123.1415= 1 2 3 4 5 6 7 89 10 11 12 13 14 15結(jié)構(gòu)動(dòng)力學(xué)課程論文3、問題三解答3.1、框架資料框架的設(shè)計(jì)同問題一，梁截面均為400700mm2，柱子的截面均為600600mm2，跨度為7.2m，層高為3.6m，混凝土采用C30，如下圖四所示：圖四因?yàn)榭蚣艿馁|(zhì)量太小，計(jì)算的自振周期太小，導(dǎo)致計(jì)算的結(jié)果發(fā)散，所以結(jié)構(gòu)動(dòng)力學(xué)課程論文 把質(zhì)量矩陣放大10倍，剛度矩陣保持不變，這樣也比較符合在實(shí)際中考慮整個(gè)樓層質(zhì)量，這樣就得到結(jié)構(gòu)的質(zhì)量和剛度如下圖所示：112253圖五 結(jié)構(gòu)簡圖質(zhì)量

35、矩陣跟剛度矩陣：0002.161.367-1.367kg-1.3672.734-1.367kN/m5M=10502.8080K=10 02.808-1.3673.277003.2、讀取地震波我市58號(hào)，分配的地震波為USA00001，程序如下所示：fid=fopen(G:動(dòng)力學(xué)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=da

36、ta(4:end).*d3; % wave data%以下為結(jié)構(gòu)信息。%自由度數(shù)目結(jié)構(gòu)動(dòng)力學(xué)課程論文 plot(TT,d),gridxlabel(時(shí)間(s)ylabel(加速度(m/s2)title(地震波加速度曲線)讀取地震波如下：并且得到總共有2688個(gè)數(shù)據(jù)時(shí)間步長為0.01秒。2.3、分析主程序分析程序采用Matlab,使用wilson-法，結(jié)合前面的讀取地震波的程序?qū)ψ兞康馁x值（沒有前面程序賦值，以下程序不可單獨(dú)使用），主程序如下：cn=3; %自由度數(shù)目k1=1.910e5;k2=1.367e5;k3=1.367e5;m1=2.808e2;m2=2.808e2;m3=2.16e2;m

37、0=m3,m2,m1;%以下為地震波的時(shí)間參數(shù)TT=0.01:0.01:26.88; %0.01為時(shí)間間隔，26.88為地震波持續(xù)時(shí)間%結(jié)構(gòu)動(dòng)力學(xué)課程論文 TT=TT;M=diag(m0);E=0.05;0.10;0.15; % %振型阻尼比dt=0.01;T=1.4*dt;P=d*1,1,1*M; %荷載矩陣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);%結(jié)構(gòu)剛度矩陣%v,w=eig(K,M);w=sqrt(w);% w：振動(dòng)頻率矩陣 %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;% %結(jié)構(gòu)阻尼矩陣%K1=K+(6/T2)*M+(3/T)*C;% %等效剛度矩陣%if i=1dP