Dinamica Estructural Homework

ciudad universitaria 07 de diciembre de 2009

UNIVERSIDAD DE EL SALVADOR

FACULTAD DE INGENIERIA Y ARQUITECTURA

ESCUELA DE INGENIERIA CIVIL

DEPARTAMENTO DE ESTRUCTURAS

FUNDAMENTOS DE DINÁMICA ESTRUCTURAL

“TAREA EXAULA”

CATEDRÁTICO:

ING. FREDY ORELLANA

PRESENTADO POR:

GUZMAN MANCIA, LUIS ALEXANDER GM04079

Un rotulo de 10 m de altura, es soportado por una columna con rigidez

igual a 20000 kg y se concentra un peso de 98000 kg. Si se considera

un amortiguamiento del 3 % respecto al crítico. Resolver para las

siguientes condiciones de carga.

A) Para una carga producida por una ráfaga de viento dada por el siguiente

diagrama.

0 0.5 1 1.5 2 2.5 302468

1012 Comportamiento de la Ráfaga de viento

Series2

tiempo (s)

Carga (ton)

Solución… ε = 3 %; w= 98,000 kg; K= 20,000 kg

Frecuencia Natural:ω=√(k¿¿m)=(20000 /98000/981)1/2=14 .15 rad /s¿

Periodo Natural :2 πω

= 2 π14.15

=0. 444 seg

La respuesta del sistema viene dada por: El Método de los

trapecios a través de la integral de Duhamel

Ci=Ci−1∗¿ e−εω ∆t +f O ¿ Si=S i−1∗¿e−εω∆ t+ gO¿

e−εω ∆t=e−0.03∗14.15∗0.1=0.958440 4

∆ t2m∗ωD

= 0.12∗100∗14.1430

=3.53 532 X 10−5

C (t)= 1mWa∫P (τ ) cos(Wt)dτ S(t )= 1

mWa∫

0

t

P ( τ ) sen(Wt )dτ

Ci=Co+ ∇ τ2mWa

.[ Fo+Fi] Si=So+ ∇ τ2mWa

.[Go+Gi ]

U ( t )=Ci . sen (W . t )−Si . cos (W . t)

m = 100

ω = 14.143 e−ξω ∆ τ 0.958459 ξ = 0.03

∆τ = 0.1

t P(t) ωt f i gi C i Si U 1 U 2 U

0 0 0 0 0 0 0 0 0 0

0.1 200 1.4143 31.1716628 197.555884 0.00110202 0.00698423 0.00108855 0.00108855 0

0.2 400 2.8286-

380.566549

123.162908 -0.01134176 0.01774239 -0.00349221 -0.0168804 0.013388186

0.3 600 4.2429-

271.45837

-535.079763 -0.03336285 0.00226189 0.02975298 -0.00102335 0.030776327

0.4 800 5.6572 648.30898 -468.716829 -0.01825537 -0.03253364 0.01069575 -0.02636481 0.03706056

4

0.5 750 7.0715 528.781109 531.874551 0.02316474 -0.02826095 0.01642765 -0.01992514 0.03635278

7

0.6 700 8.4858-

413.429151

564.868425 0.02550392 0.01090528 0.02058051 -0.0064408 0.027021311

0.7 650 9.9001-

577.944506

-297.456127 -0.00999659 0.01907655 0.00457469 -0.01696183 0.021536517

0.8 600 11.3144 188.071001 -569.762493 -0.0225158 -0.011938 0.0213811 -0.00374199 0.02512308

5

0.9 550 12.7287 542.76942 88.8895729 0.00398086 -0.0276057 0.00064338 -0.02724278 0.02788615

4

1 500 14.143-

2.91651288

499.991494 0.02210388 -0.00577064 0.02210351 3.366E-05 0.022069848

1.1 450 15.5573-

444.902284

67.5422639 0.00535812 0.0137989 0.00080422 -0.01364258 0.014446806

1.2 400 16.9716-

120.940958

-381.278487 -0.01421543 0.00203491 0.01355009 -0.00061526 0.014165349

1.3 350 18.3859 313.048216 -156.527361 -0.00665568 -0.01650282 0.00297656 -0.01476051 0.01773706

5

1.4 300 19.8002 174.34772 244.136996 0.01039206 -0.01249011 0.00845695 -0.00725874 0.01571568

9

1.5 250 21.2145-

178.316632

175.223225 0.00956399 0.00249592 0.00670333 -0.00178026 0.00848359

1.6 200 22.6288-

160.699221

-119.062003 -0.00255672 0.00412038 0.00152204 -0.00331071 0.00483275

1.7 150 24.0431 69.420515 -132.96914 -0.0054415 -0.00478603 0.00482367 -0.00221499 0.00703866

7

1.8 100 25.4574 94.7759629 31.8985402 0.00048747 -0.0079651 0.00015549 -0.007549 0.00770449

4

1.9 50 26.8717-

8.36854989

49.2946992 0.0033828 -0.00481062 0.00333508 0.00080516 0.002529925

2 0 28.286 0 0 0.00295871 -0.00294046 -3.4516E-05 0.00294026 -0.00297477

2.1 0 29.7003 0 0 0.0028358 -0.0028183 -0.00280611 0.00040675 -0.00321286

2.2 0 31.1146 0 0 0.002718 -0.00270123 -0.00080667 -0.00257952 0.001772853

2.3 0 32.5289 0 0 0.00260509 -0.00258901 0.00233681 -0.00114433 0.003481143

2.4 0 33.9432 0 0 0.00249687 -0.00248146 0.0014392 0.00202777 -0.00058857

2.5 0 35.3575 0 0 0.00239315 -0.00237838 -0.00171671 0.00165707 -0.00337377

2.6 0 36.7718 0 0 0.00229373 -0.00227958 -0.00183501 -0.00136772 -0.00046729

2.7 0 38.1861 0 0 0.00219845 -0.00218488 0.0010288 -0.00193088 0.002959679

2.8 0 39.6004 0 0 0.00210712 -0.00209412 0.00199309 0.00067956 0.001313529

2.9 0 41.0147 0 0 0.00201959 -0.00200713 -0.00034963 0.00197682 -0.00232645

3 0 42.429 0 0 0.00193569 -0.00192375 -0.00193539 -3.3662E-05 -0.00190173

3.1 0 43.8433 0 0 0.00185528 -0.00184383 -0.00025705 -0.00182605 0.001568999

3.2 0 45.2576 0 0 0.00177821 -0.00176724 0.00170114 -0.00051464 0.002215777

3.3 0 46.6719 0 0 0.00170434 -0.00169382 0.00074438 0.00152373 -0.00077935

3.4 0 48.0862 0 0 0.00163354 -0.00162346 -0.00134034 0.00092801 -0.00226835

3.5 0 49.5005 0 0 0.00156568 -0.00155602 -0.00108427 -0.0011225 3.82323E-05

3.6 0 50.9148 0 0 0.00150064 -0.00149138 0.00090735 -0.00118788 0.002095227

3.7 0 52.3291 0 0 0.0014383 -0.00142942 0.00126714 0.00067628 0.000590864

3.8 0 53.7434 0 0 0.00137855 -0.00137004 -0.00045495 0.00129329 -0.00174823

3.9 0 55.1577 0 0 0.00132128 -0.00131313 -0.00129998 -0.00023487 -0.00106511

4 0 56.572 0 0 0.00126639 -0.00125858 2.9545E-05 -0.00125824 0.001287783

4.1 0 57.9863 0 0 0.00121379 -0.0012063 0.00120304 -0.00016016 0.001363202

4.2 0 59.4006 0 0 0.00116336 -0.00115619 0.00033229 0.00110802 -0.00077573

4.3 0 60.8149 0 0 0.00111504 -0.00110816 -0.00100589 0.00047817 -0.00148406

4.4 0 62.2292 0 0 0.00106872 -0.00106212 -0.00060578 -0.00087501 0.000269232

4.5 0 63.6435 0 0 0.00102432 -0.001018 0.00074306 -0.00070069 0.001443758

4.6 0 65.0578 0 0 0.00098177 -0.00097571 0.0007785 0.00059448 0.00018402

4.7 0 66.4721 0 0 0.00094098 -0.00093518 -0.00045002 0.0008213 -0.00127132

4.8 0 67.8864 0 0 0.00090189 -0.00089633 -0.00084961 -0.00030074 -0.00054888

4.9 0 69.3007 0 0 0.00086443 -0.00085909 0.00015957 -0.00084433 0.001003901

5 0 70.715 0 0 0.00082852 -0.00082341 0.00082817 2.4011E-05 0.000804155

5.1 0 72.1293 0 0 0.0007941 -0.0007892 0.00010084 0.00078281 -0.00068197

5.2 0 73.5436 0 0 0.00076111 -0.00075642 -0.00073066 0.00021182 -0.00094248

5.3 0 74.9579 0 0 0.00072949 -0.00072499 -0.00031093 -0.00065584 0.000344905

5.4 0 76.3722 0 0 0.00069919 -0.00069488 0.00057832 -0.00039053 0.00096885

5.5 0 77.7865 0 0 0.00067014 -0.00066601 0.00045842 0.0004858 -2.7384E-05

5.6 0 79.2008 0 0 0.00064231 -0.00063834 -0.00039431 0.0005039 -0.00089821

5.7 0 80.6151 0 0 0.00061562 -0.00061183 -0.00053893 -0.00029573 -0.0002432

5.8 0 82.0294 0 0 0.00059005 -0.00058641 0.00020121 -0.00055126 0.000752472

5.9 0 83.4437 0 0 0.00056554 -0.00056205 0.0005552 0.00010697 0.000448228

6 0 84.858 0 0 0.00054204 -0.0005387 -1.8967E-05 0.00053837 -0.00055734

6.1 0 86.2723 0 0 0.00051953 -0.00051632 -0.0005157 6.2578E-05 -0.00057828

6.2 0 87.6866 0 0 0.00049795 -0.00049487 -0.00013665 -0.00047587 0.000339224

6.3 0 89.1009 0 0 0.00047726 -0.00047432 0.00043292 -0.00019966 0.000632578

6.4 0 90.5152 0 0 0.00045743 -0.00045461 0.00025487 0.00037751 -0.00012263

6.5 0 91.9295 0 0 0.00043843 -0.00043573 -0.00032155 0.0002962 -0.00061775

0 1 2 3 4 5 6 7

-0.01

-0.005

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

DIAGRAMA DE RESPUESTA T-U

PERIODO T (Seg)

DESPLAZAMIENTO

Utilizando Matlab 7.0 para generar el diagrama de

desplazamiento (método de los Trapecios)

Definimos los tipos de carga para cada intervalo:

p 1=2000∗t (1 :200 ) ;0 ≤t ≤0.4 p 2=1000−500∗t ;0.4 ≤ t ≤ 2.0

p 3=zeros (1002 :2500) ;2≤t ≤5

SINTAXIS PARA DETERMINAR ACELERACION LINEAL EN MATLAB

Paso a generar: 0.002

t=linspace(0,6,3001)'; p1=2000*t(1:200); p2=1000-500*t(201:1001); p3=zeros*t(1002:2501); p=[p1;p2;p3]; [t,d]=dtrapez(p,100,14.1493,0.03,0.002);

SINTAXIS PARA DETERMINAR ACELERACION LINEAL EN MATLAB

t=linspace(0,6,3001)'; p1=2000*t(1:200); p2=1000-500*t(201:1001); p3=zeros*t(1002:2501); p=[p1;p2;p3]; [t,d,v,a]=dmaclin(p,100,14.1493,0.03,0.002);

B) Aceleración de la base del terreno correspondiente al acelerograma

de “El Centro “de 1940 en los Ángeles, Estados unidos

V= aceleración de la base del terreno (a/g)= sismo Sismo es un vector que contendrá las aceleraciones con espaciamientos

de 0.002s

Sintaxis Para MatLab 7.0

p=-sismo*100*981;[t,d]=dtrapez(p,100,14.1493,0.03,0.002);

Diagrama Desplazamiento:

Diagrama Velocidad

p=-sismo*100*981; [t,d,v,a]=dmaclin (p,100,14.1493,0.03,0.002);

Encontrar los valores finales de desplazamientos, derivas de entrepiso, fueras por nivel y cortante basal para la siguiente estructura, suponiendo un 5% de amortiguamiento para un espectro de respuesta basado en EL Sismo el Centro,. Modele y muestre todos los modos de Vibración. f’’c=280kg/cm2

DATOS :

Viga:

b=30 cm h= 50 cm L= 600cm

Columna:

b= 50 cm h= 50 cm L= 400 cm

Inercias de sección:

Inercias de las secciones:

I v=1

12bh3= 1

1230 (50 )3=260416 .67⇒ I v/ L=K=312,500

600=520 .833cm3

Ic=12

bh3=1/12∗50 (50 )3=520,833 . 33⇒ I c/ L=K=520,833 . 33400

=1302. 0833 cm3

∑ Kv=434 . 0278cm3×2=1 041 .6667 cm 3̂

∑ kc1=∑ kc2=∑ kc3=∑ kc4=1302 . 0833×3=3,906 .25cm 3̂

Rigideces de Entrepiso:

Aplicando formulas de wilbur.

Primer entrepiso.

R 1=48 Ec

h 1[( 4 h 1

∑ kc1)+

(h 1+h 2)

∑ Kv1+∑ kc1/12 ]

=48∗15100√280

400 [ 4∗4003,906 . 25

+(400+400)

1041 . 667+3 , 906 . 25 /12 ]1000

=30 . 48 ton/cm

Segundo entrepiso.

R 2=48 Ec

h 2 [(4 h 2

∑ kc2)+

(h 1+h 2)

∑ Kv1+∑ kc1/12

+(h 1+h 2)

∑ K v2

]¿48∗15100√280

400 [4∗4003906. 25

+(400+400 )1041. 667+3906 . .25 /12

+(400+400 )1041. 667 ]1000

=17 . 20 ton/cm

Tercer entrepiso.

R 3=48 Ec

h2 [(4 h3

∑ kc3)+

(h2+h3 )

∑ K v2

+(h2+h3)

∑ K v3

]¿48∗15100√210

400 [4∗4003906. 25

+(400+400 )1041.667

+( 400+400 )1041.667 ]1000

=15. 58 ton/cm

Entrepiso Superior.

R 4=48 Ec

h 3 [(4 h3

∑ kc3)+

(2 h 3+h 2 )

∑ K v2

+(h 1+h 2 )

∑ Kv3

]¿48∗15100√210

306 [4∗4003906 . 25

+(2∗400+400 )1041 .667

+( 400)1041 .667 ]1000

=15 . 58ton/cm

Cuadro resumen:

Eje

A-Ab(cm

) h(cm) Inercia(cm4)Longitud(cm

) Cantidad K(cm^3) ΣK(cm^3) v1 30 50 312500.00 600 2 520.83 1041.67 ΣKv1v2 30 50 312500.00 600 2 520.83 1041.67 ΣKv2v3 30 50 312500.00 600 2 520.83 1041.67 ΣKv3v4 30 50 312500.00 600 2 520.83 1041.67 ΣKv4 c1 50 50 520833.33 400 3 1302.08 3906.25 ΣKc1c2 50 50 520833.33 400 3 1302.08 3906.25 ΣKc2c3 50 50 520833.33 400 3 1302.08 3906.25 ΣKc3c4 50 50 520833.33 400 3 1302.08 3906.25 ΣKc4

R1 30.48 ton/cmR2 17.20 ton/cm

R3 15.58 ton/cmR4 15.58 ton/cm

Peso Propio de Elementos: (masa de cada Elemento)

M v iga=0.3∗0.5∗2.4∗6∗2 /981=2.16981

tonse2

m=4.403669725∗10−3 ton∗s2

cm

M columna=0.5∗0.5∗2.4∗4∗3 /981= 7.2981

tonse2

m=7.339449541 x10−3 ton∗s2

cm

Se considerara la carga distribuida dada en enunciado como el peso por ml de

la losa

Masa de losa nivel 1 :

WL1=2.5∗12/981= 30981

tonse2

m=0.030581039

ton∗s2

cm

Masa de losa nivel 2:

WL2=2.0∗12/981= 24981

tonse2

m=0.024464831

ton∗s2

cm


w L 3=2.0∗12/981= 24981

tonse2

m=0.024464831

ton∗s2

cm


w L 3=1.5∗12/981= 18981

tonse2

m=0.018348623

ton∗s2

cm

Totales: (para masa de losa):

Nivel 1: M 1=Mv+Mc+ML1=0.042321ton∗s2

cm

Nivel 2: M 2=Mv+ Mc+ ML2=0.036208ton∗s2

cm

Nivel 3 : M 3=Mv+Mc+ML3=0.036208ton∗s2

cm

Nivel 4 : M 4=Mv+Mc+ ML3=0.030092ton∗s2

cm

Matriz de rigidez

⟦ K ⟧=[30.48+17.20 −17.10 0 0−17.20 17.20+15.58 −15.58 0

00

−15.580

15.58+15.58−15.58

−15.5815.58

]⟦ K ⟧=[ 47.68 −17.10 0 0

−17.20 32.78 −15.58 000

−15.580

31.16−15.58

−15.5815.58

]Matriz diagonal de masas

⟦ m⟧=[0.042321 0 0 00 0.036208 0 000

00

0.0362080

00.030092

]La ecuacion de Eigen Valores es:

([ K ]−ω2 . [ M ]¿ . [Φ n ]=0

[[ 47.68 −17.20 0 0−17.20 32.78 −15.58 0

00

−15.580

31.16−15.58

−15.5815.58

]−ω2 .[0.042321 0 0 00 0.036208 0 000

00

0.0362080

00.030092

]] . [Φn1Φn2Φn3Φ n4

]=0

El determinante a resolver es:

[47.68−ω2∗0.042321 −17.20 0 0−17.20 32.78−ω2∗0.036208 −15.58 0

00

−15.580

31.16−ω2∗0.036208−15.58

−15.5815.58−ω2 0.030092

]=0

Resolucion de la determinante:

0.000002(ω8−3410.28 ω6+3.66523 x106∗ω4−1.28797∗109∗ω2+7.62188∗1010)=0

Vector polinomial P

>> p=[1,0,-3410.28,0,3.66523*10^6,0,-1.28767*10^9,0,7.62188*10^10];

>> roots(p);

Solucion según matlab.

ω1=8.5769radseg

ω2=23.3943radseg

El periodo es:T={T 1T 2T 3T 4

}={0.73260.26860.18400.1560

}ω3=34.1540

radseg

ω 4=40.2856radseg

A={A 1A 2A 3A 4

}={420 cm /s2

780 cm / s2

680 cm / s2

7 00 cm / s2} Matriz aceleración A=[420 0 0 00 780 0 000

00

6800

0700

]

[47.68−ω2∗0.042321 −17.20 0 0−17.20 32.78−ω2∗0.036208 −15.58 0

00

−15.580

31.16−ω2∗0.036208−15.58

−15.5815.58−ω2 0.030092

]∗[Φn1Φn2Φn3Φn4

]

[44.5667 −17.20 0 0−17.20 30.1164 −15.58 0

00

−15.580

28.49 64−15.58

−15.5813.3663

]∗[Φn1Φn2Φn3Φn 4

]=[0000]

Para la frecuencia. ω1=8.5769 [44.5667 −17.20 0 0−17.20 30.1164 −15.58 0

00

−15.580

28.4964−15.58

−15.5813.3663


]=[0000]

Φ11=1 entonces

44.5667 Φ 11−17.20 Φ21=0 Φ21=2.59110

−17.20 Φ11+30.1164 Φ21−15.58Φ 31=0Φ31=3.90464

0 Φ11−15.58 Φ21−28.4964 Φ 31−15.5841=0Φ 41=4.54814

Para la frecuencia. ω2=23.3943 [24.5180 −17.20 0 0−17.20 12.9636 −15.58 0

00

−15.580

11.3336−15.58

−15.58−0.88915


]=[0000]

Φ12=1 entonces

24.5180 Φ12−17.20 Φ22=0 Φ22=1.42546

−17.20 Φ12+12.9636 Φ22−15.58 Φ32=0 Φ32=0.08210

0 Φ12−15.58 Φ 22−11.3336Φ23−15.5824=0Φ 42=−1.36574

Para la frecuencia ω3=34.1540 [−1.6873 −17.20 0 0−17.20 −9.4565 −15.58 0

00

−15.580

−11.0765−15.58

−15.58−19.5222


]=[0000]

Φ13=1 entonces

−1.6873 Φ13−17.20Φ 23=0 Φ23=−0.09810

−17.20 Φ13−9.4565 Φ23−15.58 Φ33=0 Φ33=−1.04444

0 Φ13−15.58 Φ33−0.88915 Φ34=0Φ 43=0.84130

Para la frecuencia ω 4=40. 285

[−21.0004 −17.20 0 0−17.20 −25.98300 −15.58 0

00

−15.580

−27.61303−15.58

−15.58−33.25720


]=[0000]

Φ13=1 entonces

−1.6873 Φ14−17.20 Φ24=0Φ24=−1.22116

−17.20 Φ14−9.4565 Φ24−15.58 Φ34=0 Φ34=0.93257

0 Φ13+0Φ14−15.58 Φ34−0.88915Φ 44=0 Φ 44=−0.43168

Φ jn=[ 1 1 1 12.59110 1.42546 −0.09810 −1.221163.904644.54814

0.08210−1.36574

−1.044440.84130

0.93257−0.43168

]

Los componentes de la Matriz Modal estan dados por :

Φ jn= Ujn

∑❑

❑

( mjj . U 2 jn )0.5

Modo 1 ∑ ( mjj . U 2 jn )0.5=[0.042321∗(1+2.59112+3.904642+4.548142 )

12 ]=1.3591

Modo 2 ∑ ( mjj . U 2 jn )0.5=[0.036208∗( 1+1.425462+0.082102+1.365742 )

12 ]=0.42138

Modo 3 ∑ ( mjj . U 2 jn )0.5=[0.036208∗( 1+0.098102+1.04442+0.84132)

12]=0.31888

Modo 4 ∑ ( mjj . U 2 jn )0.5=[0.030092∗(1+1.221162+0.932572+0.431682 )

12]=0.32672

Entonces la matriz modal normalizada es:

Φ=[0.73578 2.73155 3.13598 3.060721.90649 3.38284 −0.30763 −3.737672.872983.34646

0.19484−3.241105

−3.275382.63834

2.85437−1.32125

]El vector de coeficiente de participación está definido por:

[ P ]= [ Φ ]T . [ M ] . {1 }

[Φ ]T . [ M ] .{Φ }

Si [ Φ ]T . [ M ] . {Φ }={I } para la matriz modal normalizada, entonces [P] se reduce a:

[ P ]¿ [ Φ ]T . [ M ] . {1 }

[P ]=[0.73578 2.73155 3.13598 3.060721.90649 3.38284 −0.30763 −3.737672.872983.34646

0.19484−3.241105

−3.275382.63834

2.85437−1.32125

]T

.[0.042321 0 0 00 0.036208 0 000

00

0.0362080

00.030092

] o= [0.73578,2.73155,3.13598,3.06072;1.90649,3.38284,-0.30763,-3.73767;2.87298,0.19484,-3.27538,2.85437;3.34646,-3.241105,2.63834,-1.32125];

v1=[0.042321,0.036208,0.036208,0.030092]; m=zeros(4,4) + diag(v1,0); x1=[1,1,1,1]'; p=o'*m*x1

p =(0.30490.14760.08240.0578

) ASUMIENDO QUE LA ESTRUCTURA SE COMPORTA ELÁSTICAMENTE, LA MATRIZ DE DESPLAZAMIENTO ESTA DADA POR:

[ U ]=[ Φ ] . [ P ] . [ D ]=[ Φ ] . [ P ] . [ A ]/ [Ω2]

Matriz frecuencias naturales.

Ω=[8.5769 0 0 00 23.3943 0 000

00

34.15400

040.2856

]Matriz fuerza

P=[0.3049000

00.34276

00

00

0.08240

000

0.578]

[ U ]=[0.73578 2.73155 3.13598 3.060721.90649 3.38284 −0.30763 −3.737672.872983.34646

0.19484−3.241105

−3.275382.63834

2.85437−1.32125

]∗[0.3049000

00.14

00

00

0.08240

000

0.578]∗[420 0 0 0

0 780 0 000

00

6800

0700

] /[8.57692 0 0 00 23.39432 0 000

00

34.15402

00

40.28562]o=[0.73578,2.73155,3.13598,3.06072;1.90649,3.38284,-0.30763,-3.73767;2.87298,0.19484,-3.27538,2.85437;3.34646,3.241105,2.63834,-1.32125];o=[0.73578,2.73155,3.13598,3.06072;1.90649,3.38284,-0.30763,-3.73767;2.87298,0.19484,-3.27538,2.85437;3.34646,-3.241105,2.63834,-1.32125];

v1=[0.042321,0.036208,0.036208,0.030092]; m=zeros(4,4) + diag(v1,0); x1=[1,1,1,1]'; p=o'*m*x1

p =(0.30490.14760.08240.0578

) P=zeros(4,4) +diag(p,0); a=[420,780,680,700];% A=zeros(4,4)+diag(a,0) w1=[8.5769,23.3946,34.1540,40.2856] W2=zeros(4,4)+diag(w1,0) W=W2^2 U=o*P*A/W

U= 1.2808 0.5746 0.1506 0.0763 3.3187 0.7116 -0.0148 -0.0932 5.0012 0.0410 -0.1573 0.0711 5.8254 -0.6818 0.1267 -0.0329

LOS DESPLAZAMIENTOS MÁXIMOS.

U1=U.^2; U2=[1,1,1,1]'; U3=U1*U2; Umax=U3.^0.5

Umax=

1.4139 3.3955 5.0043 5.8666

LAS DERIVAS DE ENTREPISO.

U4=U';

fil1=U4(5:8)-U4(1:4);

fil2=U4(9:12)-U4(5:8);

fil3=U4(13:16)-U4(9:12);

fil4=U4(13:16);

der=[fil1;fil2;fil3;fil4]

der =

2.0379 0.1370 -0.1654 -0.1695

1.6824 -0.6707 -0.1425 0.1643

0.8242 -0.7228 0.2840 -0.1041

5.8254 -0.6818 0.1267 -0.0329

MATRIZ DE FUERZAS LATERALES EN CADA NUDO.

[ F ]=[ K ] ⌊U ⌋

[ F ]=[[ 47.68 −17.20 0 0−17.20 32.78 −15.58 0

00

−15.580

31.16−15.58

−15.5815.58

] ]⌊ 1.3723 0.6225 0.1993 0.08173.5558 0.7710−0.0196−0.09985.3584 0.0444−0.20820.0762

6.2415−0.7386 0.1677−0.0353

⌋

k1=[47.68,32.78,31.16,15.58];

k2=[-17.20,-15.58,-15.58];

K=zeros(4,4)+diag(k1,0)+diag(k2,-1)+diag(k2,1);

F=U*K

F =

51.1857 -5.5398 -5.4490 -1.1576

145.9976 -33.5246 -10.0962 -1.2214

237.7512 -82.2262 -6.6482 3.5590

289.4824 -124.5211 15.0839 -2.4870

UTILIZANDO L ARAIZ DE LA SUMA DE LOS CUADRADOS LA FUERZA LATERAL EN CADA NUDO ES:

F1=F.^2; F2=[1,1,1,1]'; F3=F1*F2; Fc=F3.^0.5

Fc =

51.7851 150.1420

251.6817 315.4984

VECTOR DE CORTANTE BASAL

[ V ]=[ [ F ]T {1 } ]T

R=[1,1,1,1]'

V=[F'*R]'

V = [724.4170 -245.8116 -7.1095 -1.3070]

CORTANTE BASAL actuante en la estructura es:

V1=V.^2;

V2=[1,1,1,1]’;

Vb=(V2*V1’).^0.5;

CORTANTE BASAL = 765.02 ton

Dinamica Estructural Homework

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