结构动力学大作业1.docx
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结构动力学大作业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400⨯700mm2,柱子的截面均为
600⨯600mm2,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:
V1
V2
V3
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:
=0.4⨯0.7⨯2500=700kg/m,L=7.2m;梁、柱都为均布质量,故:
⎧f
⎪f⎪⎨⎪f⎪⎩f
I1
I2
I3
I4
⎫
⎪
⎪L⎬=420⎪⎪⎭
5622L⎡156
⎢5415613L⎢⎢22L13L4L⎢
⎣-13L-22L-3L
-13L⎤-22L⎥
⎥-3L⎥
⎥4L⎦
22
1⎫⎧v⎪v⎪⎪2⎪⎨⎬
3⎪⎪v⎪4⎪⎩v⎭
结构动力学课程论文
结构动力学课程论文柱:
=0.6⨯0.6⨯2500=900kg/m,L=3.6m
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)ˆijˆijˆij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)ˆ11ˆ11ˆ11ˆ11ˆ11ˆ11ˆ11m11=m+m+m+m+m+m+m=3⨯5040+4⨯1203.43=19933.72
(10)(11)(12)(13)ˆ12ˆ12ˆ12ˆ12m12=m+m+m+m=4⨯416.57=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=-361.03m1,12=m1,13=m1,14=m1,15=0(4)(5)(6)(10)(11)(12)(13)(14)(15)(16)(17)
ˆ22ˆ22ˆ22ˆ22ˆ22ˆ22ˆ22ˆ22ˆ22ˆ22ˆ22m22=m+m+m+m+m+m+m+m+m+m+m
=3⨯5040+8⨯1203.43=24747.44
(14)(15)(16)(17)ˆ23ˆ23ˆ23ˆ23m23=m+m+m+m=4⨯416.57=1666.28
(10)
m24=m25=m26=m27=m24=361.03
(14)(10)
ˆ28ˆ28m28=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)(14)(15)(16)(17)(18)(19)(20)(21)
ˆ33ˆ33ˆ33ˆ33ˆ33ˆ33ˆ33ˆ33ˆ33ˆ33ˆ33m33=m+m+m+m+m+m+m+m+m+m+m
=3⨯5040+8⨯1203.43=24747.44
(14)
m34=m35=m36=m37=0m38=m39=m3,10=m3,11=m38=361.03(14)
ˆ3ˆ(18)m3,12=m3,13=m3,14=m3,15=m.97-610.97=0,12+m3,12=610
(1)(10)
(1)
ˆ44ˆ44ˆ45m44=m+m=2488.32+399.91=2888.23m45=m=-1866.24
m46=m47=0(10)
ˆ48m48=m=-299.93
m49=m4,10=m4,11=m4,12=m4,13=m4,14=m4,15=0
(2)
(1)
(2)(11)
ˆ56ˆ55ˆ55ˆ55=-1866.24m55=m+m+m=2488.32+2488.32+399.91=5376.55m56=m
m57=m58=0
(11)
ˆ59m59=m=-299.93m5,10=m5,11=m5,12=m5,13=m5,14=m5,15=0
(2)(3)(12)ˆ66ˆ66ˆ66m66=m+m+m=2488.32+2488.32+399.91=5376.55(3)ˆ67m67=m=-1866.24m68=m69=0
(12)ˆ6m6,10=m.93m6,11=m6,12=m6,13=m6,14=m6,15=0,10=-299(3)(13)ˆ77ˆ77m77=m+m=2488.32+399.91=2888.23
m78=m79=m7,10=0
(13)ˆ7m7,11=m.93m7,12=m7,13=m7,14=m7,15=0,11=-299
结构动力学课程论文
(4)(10)(14)ˆ88ˆ88ˆ88m88=m+m+m=2488.32+399.91+399.91=3288.14
(4)ˆ89m89=m=-1866.24m8,10=m8,11=0
(14)ˆ8m8,12=m.93m8,13=m8,14=m8,15=0,12=-299
(4)(5)(11)(15)ˆ99ˆ99ˆ99ˆ99m99=m+m+m+m=2488.32+2488.32+399.91+399.91=5776.46
(5)ˆ9m9,10=m.24,10=-1866
(15)ˆ9.93m9,14=m9,15=0m9,11=m9,12=0m9,13=m,13=-299
(5)(6)(12)(16)ˆ10ˆ10ˆ10ˆ10m10,10=m.32+2488.32+399.91+399.91=5776.46,10+m,10+m,10+m,10=2488
(6)(16)ˆ10ˆm10,11=m=-1866.24m=m.93m10,15=0m=m=010,1210,13,1110,1410,14=-299
(6)(13)(17)ˆ11ˆ11ˆ11m11,11=m.32+399.91+399.91=3288.14,11+m,11+m,11=2488
m11,12=m11,13=m11,14=0(17)ˆ11m11,15=m.93,15=-299
(7)(14)(18)ˆ12ˆ12ˆ12m12,12=m.32+399.91+399.91=3288.14,12+m,12+m,12=2488
(7)ˆ12m12,13=m.24m12,14=m12,15=0,13=-1866
(7)(8)(15)(19)ˆ13ˆ13ˆ13ˆ13m13,13=m.32+2488.32+399.91+399.91=5776.46,13+m,13+m,13+m,13=2488
(8)ˆ13m13,14=m.24m13,15=0,14=-1866
(8)(9)(16)(20)ˆ14ˆˆˆm14,14=m+m+m+m.32+2488.32+399.91+399.91=5776.46,1414,1414,1414,14=2488
(9)ˆ14m14,15=m.24,15=-1866
(9)(17)(21)ˆ15ˆ15ˆ15m15,15=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
0⎡19933.721666.28
⎢24747.441666.28⎢⎢24747.44⎢⎢⎢⎢⎢⎢⎢M=⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
610.97361.0302888.23
610.97361.030-1866.245376.55
610.97361.0300-1866.245376.55
610.97361.03000-1866.242888.23
-361.030361.03-299.930003288.14
-361.030361.030-299.9300-1866.245776.46
-361.030361.0300-299.9300-1866.245776.46
-361.030361.03000-299.9300-1866.243288.14
0-361.0300000-299.930003288.14
⎤
-361.03-361.03-361.03⎥⎥
⎥000⎥
000⎥
⎥000⎥
000⎥
⎥000⎥
000⎥
⎥-299.9300⎥⎥0-299.930⎥
00-
299.93⎥
⎥-1866.2400⎥
5776.46-1866.240⎥
5776.46-1866.24⎥
⎥
3288.14⎦⎥
000
1.3、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:
-63L3L⎤⎧v1⎫⎧fs1⎫⎡6
⎪f⎪⎪v⎪⎢6-3L-3L⎥⎪s2⎪2EI⎢-6⎪2⎪⎥=⎨⎬⎨⎬223
⎢⎥f3L-3L2LLL⎪s3⎪⎪v3⎪⎢22⎥⎪⎪f3L-3LL2L⎣⎦⎪⎩v4⎪⎭⎩s4⎭
框架梁:
C30混凝土E=3⨯107KN/m2,
0.40⨯0.73
EI=3⨯10⨯=3.43⨯105kN·m2,L=7.2m
12
7
结构动力学课程论文
7
框架柱:
0.60⨯0.603
EI=3⨯10⨯=3.24⨯105KN·m2L=3.6m
12
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
ˆ(m)+kˆ(n)+kˆ(p)+....可计算一致刚度矩阵中的各元素:
由kij=kijijijˆ(10)+kˆ(11)+kˆ(12)+kˆ(13)=4⨯0.833⨯105=3.332⨯105k11=k11111111
ˆ(10)+kˆ(11)+kˆ(12)+kˆ(13)=4⨯(-0.833k12=k)⨯105=-3.332⨯105k13=012121212
(10)
k14=k15=k16=k17=k18=k19=k1,10=k1,11=k14=1.50⨯105
k1,12=k1,13=k1,14=k1,15=0
ˆ(10)+kˆ(11)+kˆ(12)+kˆ(13)+kˆ(14)+kˆ(15)+kˆ(16)+kˆ(17)=8⨯0.833⨯105=6.664⨯105k22=k2222222222222222
ˆ(14)+kˆ(15)+kˆ(16)+kˆ(17)=4⨯(-0.833k23=k)⨯105=-3.332⨯10523232323
10k24=k25=k26=k27=k24=-0.861⨯105
ˆ(10)+kˆ(14)=0.861⨯105-0.861⨯105=0同理k28=k2828k29=k2,10=k2,11=0
结构动力学课程论文
ˆ(14)=1.50⨯105k2,12=k2,13=k2,14=k2,15=k2,12
ˆ(14)+kˆ(15)+kˆ(16)+kˆ(17)+kˆ(18)+kˆ(19)+kˆ(20)+kˆ(21)=8⨯0.833⨯105=6.664⨯105k33=k3333333333333333
k34=k35=k36=k37=0(14)k38=k39=k3,10=k3,11=k38=-1.50⨯105
ˆ(14)+kˆ(18)=1.50⨯105-1.50⨯105=0k3,12=k3,13=k3,14=k3,15=k3,123,12
ˆ
(1)=0.953⨯105ˆ
(1)+kˆ(10)=1.906⨯105+3.60⨯105=5.506⨯105k=kk44=k44444545
k46=k47=0
ˆ(10)=1.80⨯105k48=k48
k49=k4,10=k4,11=k4,12=k4,13=k4,14=k4,15=0
ˆ
(1)+kˆ
(2)+kˆ(11)=1.906⨯105+1.906⨯105+3.60⨯105=7.412⨯105k55=k555555
ˆ
(2)=0.953⨯105k56=k56
k57=k58=0ˆ(11)=1.80⨯105k59=k59
k5,10=k5,11=k5,12=k5,13=k5,14=k5,15=0
ˆ
(2)+kˆ(3)+kˆ(12)=1.906⨯105+1.906⨯105+3.60⨯105=7.412⨯105k66=k666666
ˆ(3)=0.953⨯105k67=k67
ˆ(12)=1.80⨯105k=k=k=k=k=0k68=k69=0k6,10=k6,116,126,136,146,156,10
ˆ(3)+kˆ(13)=1.906⨯105+3.60⨯105=5.506⨯105k77=k7777
k78=k79=k7,10=0
ˆ(13)=1.80⨯105k=k=k=k=0k7,11=k7,127,137,147,157,11
ˆ(4)+kˆ(10)+kˆ(14)=1.906⨯105+3.60⨯105+3.60⨯105=9.106⨯105k88=k888888
ˆ(4)=0.953⨯105k89=k89
ˆ(14)=1.80⨯105k=k=k=0k8,10=k8,11=0k8,12=k8,138,148,158,12
ˆ(4)+kˆ(5)+kˆ(11)+kˆ(15)=1.906⨯105+1.906⨯105+3.60⨯105+3.60⨯105k99=k99999999
=11.012⨯10
514
结构动力学课程论文
ˆ(5)=0.953⨯105k9,10=k9,10
k9,14=k9,15=0
k9,11=k9,12=0
ˆ(15)=1.80⨯105k9,13=k9,13
ˆ(5)+kˆ(6)+kˆ(12)+kˆ(16)=1.906⨯105+1.906⨯105+3.60⨯105+3.60⨯105
k10,10=k10,1010,1010,1010,10=11.012⨯10
5
5ˆ(6)=0.953⨯105kˆ(16)k10,11=k10,12=k10,13=0k10,14=k10,14=1.80⨯10k10,15=010,11
ˆ(6)+kˆ(13)+kˆ(17)=1.906⨯105+3.60⨯105+3.60⨯105=9.106⨯105k11,11=k11,1111,1111,11ˆ(17)=1.80⨯105k11,12=k11,13=k11,14=0k11,15=k11,15
ˆ(4)+kˆ(7)+kˆ(18)=1.906⨯105+3.60⨯105+3.60⨯105=9.106⨯105k12,12=k12,1212,1212,12ˆ(7)=0.953⨯105kk12,13=k12,14=k12,15=012,13
ˆ(7)+kˆ(8)+kˆ(15)+kˆ(19)=1.906⨯105+1.906⨯105+3.60⨯105+3.60⨯105k13,13=k13,1313,1313,1313,13=11.012⨯10
5
ˆ(8)=0.953⨯105kk13,14=k13,15=013,14
ˆ(8)+kˆ(9)+kˆ(16)+kˆ(20)=1.906⨯105+1.906⨯105+3.60⨯105+3.60⨯105k14,14=k14,1414,1414,1414,14=11.012⨯10
5
ˆ(9)=0.3125⨯105k14,15=k14,15
ˆ(9)+kˆ(17)+kˆ(21)=1.906⨯105+3.60⨯105+3.60⨯105=9.106⨯105k15,15=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)
⎡3.332
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢
K=105⨯⎢
⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎢⎣
-3.3326.664
0-3.3326.664
1.50-1.5005.506
1.50-1.5000.9537.412
1.50-1.50000.9537.412
1.50-1.500000.9535.506
1.500-1.501.80000
9.106
1.500-1.5001.80000.95311.012
1.500-1.50001.80000.95311.012
1.500-1.500001.80000.9539.106
01.50000001.800009.106
01.500000001.80000.95311.012
01.5000000001.80000.95311.012
01.50000000001.80000.9539.106
⎤⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎥⎦
结构动力学课程论文2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
ˆ}={0}的特征值得到频率ω和振型φ:
由[K]-ω2[M]{v
在Matlab中导入质量矩阵[M]和刚度矩阵[K],输[v,ω2]=eig(K,M);ω=sqrt(ω2)
可得框架的频率为:
[]
ωT={ω1ω2ω3........ω14ω15}=
{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[φ]=[{φ1}{φ2}{φ3}......{φ14}{φ15}]=
φ1φ2φ3φ4φ5φ6φ7φ
8
φ9φ10φ11φ12φ13φ14φ15
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=
⎡52612⎢34661⎢
⎢13564⎢
⎢-2919⎢-2009⎢
⎢-2009⎢-2919⎢
10-5⨯⎢-4627
⎢-3739⎢
⎢-3739⎢
⎢-4627⎢-4436⎢
⎢-3429⎢-3429⎢⎢⎣-4436
453153846316844-648-547-547-648
-3237-2622-2622-3237-5313-4034-4034-5313
179831712511915-95-55.4-55.4-95-395-408-408-395-2546-1883-1883-2546
1933.61502.4600.3585.14-403.97.9103-69.43-330.8-46.98-141.3-151.2-164.3-169.8-202-202
2088.51495.3606.51-631828.14-463.5-16.66-39.22-52.13-197.9-226.6-226.6-114.2-168.9-189.5
20471488.4605.89-7.983-467.4891.72-635.4-193.6-45.69-344.3-30.81-189.2-169.2-112.1-227.4
1933.61502.4600.3-69.437.9103-403.9585.14-151.2-141.3-46.98-330.8-202-149.4-169.8-164.3
-959-507-53.6-141139.935.2281.15567.3-144114.5114.5109.7-71.595.5537.1
-466.3-466.3-74.97214.41-174.8143.1345.392-214.1713.23-171.7134.55129.58-99.996.75140.084
-885-466-7545.39143.1-175214.4134.6-172713.2-21440.0896.75-99.9129.6
-959.1-507.2-53.6281.15235.225139.93-141.1109.67114.46-143.6567.357.99237.09595.548-71.52
-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
-768.9-687.5-306.511.743-14.4546.216-28.0239.58896.68-99.97129.09122.14-159.7664.53-186.8
-898.5⎤-828.5⎥⎥-387.3⎥
⎥-1.176⎥7.242⎥
⎥-15.89⎥45.168⎥
⎥58.04⎥37.49⎥
⎥95.943⎥
⎥-71.48⎥113.72⎥
⎥107.89⎥-127.1⎥
⎥525⎥⎦
结构动力学课程论文
V1
(1)=DV1(0)迭代过程列表如下根据
DV1(0)
⎡52612⎢34661⎢
⎢13564⎢
⎢-2919⎢-2009⎢
⎢-2009⎢-2919⎢
10-5⨯⎢-4627
⎢-3739⎢
⎢-3739⎢
⎢-4627⎢-4436⎢
⎢-3429⎢-3429⎢
⎢-4436⎣
453153846316844-648-547-547-648
-3237-2622-2622-3237-5313-4034-4034-5313
179831712511915-95-55.4-55.4-95-395-408-408-395-2546-1883-1883-2546
1933.61502.4600.3585.14-403.97.9103-69.43-330.8-46.98-141.3-151.2-164.3-169.8-202-202
2088.51495.3606.51-631828.14-463.5-16.66-39.22-52.13-197.9-226.6-226.6-114.2-168.9-189.5
20471488.4605.89-7.983-467.4891.72-635.4-193.6-45.69-344.3-30.81-189.2-169.2-112.1-227.4
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