結(jié)構(gòu)動力學大作業(yè)參考

1、VV結(jié)構(gòu)動力學作業(yè)題目信息：1、試設(shè)計一個3層框架，給出框架結(jié)構(gòu)的一致質(zhì)量矩陣、一致剛度矩陣，建立框架結(jié)構(gòu)的運動微分方程，求出該框架結(jié)構(gòu)的各階頻率和振型；并采用振型分解法，選定一個正弦動力荷載，求3層框架對于該荷載的位移反應(yīng)；2、試設(shè)計一個3層框架，采用時程分析法，輸入地震波，求出所設(shè)計框架各層的非線性位移時程反應(yīng)，要求畫出所設(shè)計的框架圖、輸入的地震波的波形圖、所求得的位移時程反應(yīng)圖。（每人選擇一條地震波，不要重復(fù)）一、對問題一的解答：自己設(shè)計的三層框架如下圖的cad截圖所示，各界面的尺寸以及跨度也在圖中標識出，梁柱的混凝土都為C30。FQO 。-J-.三X300X60。侯1Xi_rs300X

2、605 Xr*0X6 附300X600E-3LOXX。n O37J=o300X6003聃XM帶女。OXXX- FW /urnLT-1i_n144007200卜L7200圖一：結(jié)構(gòu)布置及尺寸圖我們先假設(shè)的是梁柱不會發(fā)生軸向變形，只考慮角點的轉(zhuǎn)角和平移；該框架有3個平移自由度和9個轉(zhuǎn)動自由度，現(xiàn)在我們對這些梁和柱進行單元編號，對各個自由度分別進行編號處理，編號順序如圖二下圖所示：圖二單元編號及自由度a.計算結(jié)構(gòu)的一致質(zhì)量矩陣表1梁單元的質(zhì)量一致質(zhì)量矩陣，m=450kg/m,L=7.2m梁單元編號水平位移自由度產(chǎn)生的質(zhì)量影響系數(shù)轉(zhuǎn)動自由度產(chǎn)生的質(zhì)量影響系數(shù)Nmiimiimij1m11mL4507.2

3、3240m)55m144里4L21599.6420m45m154處3L21199.74202m11mL4507.23240m55m66mL24L1599.6420m56m65mL23L1199.74203m22mL4507.23240m77ml88mL4L21599.6420m78m87mL3L21199.74204m22mL4507.23240ml99ml88mL.2/八4L1599.6420ml89ml98mLc.2,/ccr3L1199.74205m33mLmi0,i0mii,menmn,104507.23240mL2mL224L21599.63L21199.74204206m33mLm

4、11,11m12,12mugm12,n4507.23240mL2mL24L1599.63L1199.7420420表2.柱單元的一致質(zhì)量矩陣，m=625kg/m,L=4m柱單元各單兀相關(guān)質(zhì)量影響系數(shù)第三層7mLmL_mrm22156928.57mi2m2154321.43420420mLmL4m41-22L523.81m7(13L)309.52420420mLmLm，4m.2L13L309.52mb-(22L)523.81420420mL2mL2m44n74L2380.95m47(3L2)285.714204208llm122156928.5712154321.43420420Lcc,ccLc

5、c1551-22L523.8如8(13L)309.52420420LL2552-13L309.5218(22L)523.81420420L2L25588一4L2380.9558一(3L2)285.714204209LL122156928.57k1-54321.43420420LL165122L523.819(13L)309.52420420%52mL13L309.5229-mL(22L)523.8142042056)9mL4L2380.9569-mL(3-2)285.71420420第二層10h33-mL156928.572332-mL54321.434204202772mL22L523.81

6、mM0-mL(13L)309.52420,42037.3mL13L309.520-mL(22L)523.81420,4207710mL4L2380.95710-mL(3L2)285.71420,42011現(xiàn)2門3吧156928.57m23門2-mL54321.43420420mLmLm,8n2一22L523.81rm211一(13L)309.52，420，420038R83mL13L309.52mMi_mL(22L)523.81420,420m88mirn里4L2380.95m311.mL(3L2)285.71420,42012mLmL_m22m33156928.57m23m3254321.4

7、3420420mLmLm29m)222L523.81,212-(13L)309.52420420mLmLm39mbL13L309.52m12L(22L)523.81420,420mL2mL2m99m12124L380.95m912(3L)285.71420420Q層13l_m33156928.57420310103H22L523.81420m1010W4L2380.9542014m33皿156928.57420L3,1111,3石22L523.811111.4L2380.9542015L33156928.57420312123L22L523.813,12毋4201212L4L2380.9542

8、0由mijm?ij(m)mjm.得:11=9265.6512=964.2913=014=523.8115=523.8116=523.81i7=-309.5218=-309.5219=-309.521,10=0i,ii=01,12=0m22=7444.29m23=964.29m24=309.52m25=309.52m26=309.52m27=0m28=0m29=0m2,10=-309.52m2,11=-309.52m2,12=-309.52m33=9265.71m)34=0m35=0m36=0m37=309.52m38=309.52m39=309.52m3,10=523.81m3,11=523.

9、81m3,12=523.81m44=1980.55m)45=-1199.7m46=0m47=-285.71m)48=0m49=0m4,10=0m4,11=0m4,12=0m55=3580.15m56=-1199.7m57=0m)58=-285.71m)59=0m5,10=0m5,11=0m5,12=0m66=1980.55m)67=0m)68=0m69=-285.71m6,10=0m6,11=0m6,12=0m77=2361.5m78=-1199.7m)79=0m7,10=-285.71m7,11=0m7,12=0m88=3961.1m89=-1199.7m8,10=0m8,11=-285.7

10、1m8,12=0m99=2361.5m9,10=0m9,11=0m)9,12=-285.71m10,10=2361.5mi0,11=-1199.7m110,12=0m11,11=3961.1mi1,12=-1199.7m12,12=2361.59265.651 964.290523.81523.81523.817444.29964.29309.52309.52309.529265.710001980.55-1199.70-309.52 -309.52 -309.52000000-309.52-309.52-309.52309.52 309.52 309.52523.81523.81523.81

11、-285.71000000-285.71000000-285.710002361.5 -1199.70-285.71003961.1 -1199.70-285.7102361.500-285.712361.5-1199.703961.1-1199.73580.15 -119971980.552361.5得到一致質(zhì)量矩陣：（對角陣）b.一致剛度矩陣的求解:各個梁柱均為等截面，根據(jù)教科書上所寫公式:fsi6fs22EI6fs3C3Lfs43L TOC o 1-5 h z 63L3Lvi63L3LV23L2L2L2v32一23LL22L2v4框架梁：C30混凝土E3.0107KN/m2270.300

12、.652310=1.6210KNgm,L127.2m樓層號梁單元編號各單元的相關(guān)剛度系數(shù)第1k44k554EIL0.9105k45k542EIL0.45105層2k66k554EIL0.9105k56%52EIL0.45105第3k77k884EIL0.9105k78k872EIL0.45105層4k88k994EIL0.9105k89k982EIL0.45105第5k10,10k14EI1,11l0.9105k10,11k12EI1,10l5-0.45105層6k11,114EI匕2,12L50.910k11,122EIk12,11L50.451030.50 0.552 ,1.5625 10

13、 KN gm , L 4m12框架柱的信息:樓層號柱單元編號各單元相關(guān)系數(shù)第三層7k11k222EI60.2930105,k12k212EI(6)0.2930105L3k14k41k17k71爺3L0.58591052EI5k24k42k27k72書(3L)0.5859105EI31072EI25k44k772L21.5625105,2EIk47k74L20.78125105L38k11k22早60,2930105,k12k21爺(6)0,29301052EI八一5k15k51k18k813L0,5859102EI5k25k52k28k82-p-(3L)0.5859102EIo.k55嚷T2L

14、21.5625105,2EI25k58k85-p-L20.7812510592EI5k11k22-p-60.2930105,k12k21%(6)0,29301052EI5k16k61。k91-p-3L0.585910k26k62k29k92號(3L)0.58591052EI25k66k992L21.5625105,2EIk69k96L3L20.78125105第二層10k22k33號60,2930105,2EI5k23k32L3(6)0.293010k27k72k210k10233L0.58591053k37k73k3,10k10,3L3(3L)0.58591052EI25k77k10,102

15、L21.5625105,2EIk7,10k10,7-p-L20.7812510511k22k33拳60,2930105,k23k32爺(6)0.2930105.2EIc.5k28%2k211k11233L0.5859102EI5k38k83k3,11k11,3-3-(3L)0.5859102EIo.k88k11,112L21.5625105,2EI25k8,11k11,8L20.78125105122EI5k22k33-p-60.2930105,k23k32*(6)0,29301052EI5k29k92k2,12k12,23L0.585910k39k93k3,12k12,3岸(3L)0.585

16、91052EI25k99k12,122L21.5625105,2EIk9,12k12,9L3L20.78125105Q層132EI12EI5k336L30.2930105,2EI24EI5k101032L1.562510,LLk3,10k10,3置3L0,5859105142EI12EI5k33L36L30.293010,kn,112L3I2L24EI1.5625105.k3,11k11,3學3L0.585910515k33省612E_0.2930105,33L3L3k1212理2L2膽1.5625105,L3Lk3,12k12,3-2E3I3L0.5859105由kj=?jm)?jn)?jp

17、)得:見下表：Xl0e5k11=0,879k12=-0.879k13=0k14=0.5859k15=0.5859k16=0.5859k17=0.5859k18=0.5859k19=0.5859k1,10=0k1,11=0k1,12=0k22=1.758k23=-0.2930k24=-0.5859k25=-0.5859k26=-0.5859k27=0k28=0k29=0k2,10=0.5859k2,11=0.5859k2,12=0.5859k33=1.758k34=0k35=0k36=0k37=-0.5859k38=-0.5859k39=-0.5859k3,10=0k3,11=0k3,12=0k

18、44=2.4625k45=0.45k46=0k47=0.7813k48=0k49=0k4,10=0k4,11=0k4,12=0k55=3.3625k56=0.45k57=0k58=0.7813k59=0k5,10=0k5,11=0k5,12=0k66=2.4625k67=0k68=0k69=0.7813k6,10=0k6,11=0k6,12=0k77=4.025k78=0.45k79=0k7,10=0.781k7,n=0k7,12=0k88=4.925k89=0.45k8,10=0k8,11=0.7813k8,12=0k99=4.025k9,10=0k9,11=0k9,12=0.7813k10

19、,10=4.025k10,11=0.45k10,12=0k11=4.925k11,12=0.45k12,12=4.025得到一致剛度矩陣：（對稱陣，單位：10e8N/m）0.879-0.87900.58590.58590.58590.58590.58590.58590001.758-0.2930-0.5859-0.5859-0.58590000.58590.58590.58591.758000-0.5859-0.5859-0.58590002.46250.4500.7813000003.36250.4500.781300002.4625000.78130004.0250.4500.781004

20、.9250.4500.781304.025000.78134.0250.4504.9250.454.025由特征方程kw2 m ?0可以求出振形(f)和頻率 w我們使用matlab求解，結(jié)果如下：T113.4353.6488.6669784.1957.41041.21216.51496.31571.92186.82846.4w1，2，3，4，5，6，7，8，9，10，11，120.00910.0029-0.00460-0.000800.00010.0001-0.00250.0009-0.00130.00020.0046-0.00560.00820.00060.00170-0.00050-0.0

21、001-0.0003-0.0050.0007-0.0002-0.008-0.0063-0.0006-0.00030.00020-0.00010.0006-0.00010.0024-0.004-0.0007-0.00180.0038-0.0031-0.0075-0.0096-0.0039-0.01030.01050.00270.01540.0108-0.0004-0.00090.00120.00880.00570.00140.0013-0.00080.0093-0.00190.01520.0102-0.0006-0.00140.0036-0.0079-0.00160.00630.00770.01

22、120.0063-0.00740.01570.0109-0.0012-0.0020.00020.002-0.00340.00110.0101-0.0023-0.0060.0143-0.00020.0146-0.0007-0.0007-0.001-0.00470.0060.0048-0.0059-0.0028-0.00790.0040.00160.0144-0.0011-0.0016-0.00060.0045-0.0033-0.006-0.00480.0057-0.0128-0.00760.00170.0152-0.00040.0011-0.0017-0.00160.0036-0.0066-0.

23、00290.00970.01020.0061-0.01170.0118-0.00040.0005-0.00080.0024-0.00630.0072-0.0036-0.00110.0071-0.0029-0.0110.0112-0.00040.0011-0.0015-0.00240.0036-0.00260.0082-0.00870.0025-0.0119-0.01110.0119作用一個250kn為幅值的正弦荷載，F(xiàn)=250sin10t,求該框架的位移反應(yīng)將結(jié)構(gòu)質(zhì)量集中到各層，此結(jié)構(gòu)用層剪切模型簡化為框架等效多質(zhì)點體系mi0.50.5251030.30.67.225103213.98103k

24、gm20.50.5251030.30.67.225103213.98103kgm30.50.5251030.30.67.2251033210.2310kgki12EI121.56251058.789104kN/mk212EII3-_5121.56251058.789104kN/m12EIT3-121.5625105438.789104kN/m由于結(jié)構(gòu)的質(zhì)量集中到各層，因此結(jié)構(gòu)的質(zhì)量矩陣為對角矩陣。13.9813.98103kg10.237.031253.51563.51567.031253.51561083.51563.5156用matlab計算體系的頻率方程:76.2382209.35242

25、92.0874-0.00310.0065-0.0045振形-0.00550.00170.0062-0.0066-0.0061-0.0042250sin10t令荷載向量為1.0028616-0.00055020.0018786廣義質(zhì)量Mn-0.00055021.01171550.00052680.00187860.00052681.00094345829.0641-17.7255-102.05315皿、t-17.725544342.4885179.20205廣義剛度Knk-102.05315179.2020585394.21745-0.775sin10t廣義荷載Pn1.625sin10t-1.125sin10t假設(shè)初始速度為零，初始位移為零振形一、0.135sin10t0.131sin76.24t單位mm振形二：y20.038(sin10t0.048sin209.35t)單位mm振形三：y30.013(sin10t0.034sin292.1t)單位mm問題二的解答框架仍