Analisis Sismico Yaaaaa

UNASAM FIC ingeniera antissmica 2014-I

HUARAZ- ENERO- 2014

ALUMNA: MENDEZ DE LA CRUZ ALICIA

CODIGO: 072.0709.261

ANALISIS DINAMICO

-ECUACION DIFERENCIAL DEL MOVIMIENTO

-ECUACION DE DUHAMEL

INg. LUIS ItA RObLES


PROBLEMA:

Datos:

- 2=m T-seg2/m

- 503

3 =h

EI T/m

g45.0

I. Determinar la respuesta dinmica a nivel de los desplazamientos cuando 0= , usando la solucin

de la Ecuacin Diferencial del Movimiento.

II. Determinar la respuesta dinmica a nivel de los desplazamientos cuando %5= , usando la

solucin de la Ecuacin Diferencial del Movimiento.

III. Resolver la parte 1, aplicando integral de DUHAMEL

IV. Resolver la parte 2, aplicando integral de DUHAMEL

V. Dibujar Diagrama de Fuerza Cortante y Diagrama de Momento Flector cuando 0= y %5= .


SOLUCION

I. DETERMINACIN DE LA RESPUESTA DINMICA CUANDO 0= , USANDO LA SOLUCIN DE LA ECUACIN DIFERENCIAL DEL MOVIMIENTO.

I.1. Ecuacin Diferencial del Movimiento.

SX =Desplazamiento del suelo rX = Desplazamiento relativo tX = Desplazamiento Total

Fuerza de inercia. )(

+

= rXSXmIF

0=


Fuerza elsticas

Fuerza elsticas

rXrXh

EI == 5033

31V rX=150

rXrXh

EI == 50123

122V rX= 600

= 0 FI+V1 + V2 = 0 Reemplazando valores

0600150)( =++

+

rXrXrXSXm

0750)( =+

+

rXrXSXm

075022 =+

+

rXrXSX

=+

SXrXrX 375 (1)

=+

SXrXrX

2 (2)

De (1) y (2) 2=375 y =19.36 rad/seg


Determinacin de la aceleracin del suelo SX

A. 0 < t < 0.2 seg SX = 0.450.20 SX = 22.5 B. 0.2 < t < 0.4 seg SX = 0.45 SX = 4.5 C. 0.4 < t < 0.6 seg SX = 13.5 22.5 D. t > 0.6 seg SX = 0

I.2. SOLUCIN DE LA ECUACIN DIFERENCIAL DEL MOVIMIENTO.

A. Solucin para : < < . SX = 22.5 A.1. Ecuacin Diferencial del Movimiento (E.D.M)

=+

SXrXrX2

Reemplazando el valor de SX

trXrX 5.222 =+

(1)

Donde =19.36 rad/seg A.2. Solucin E.D.M PXHXtX +=)( (2)

A.3. Solucin Homognea HX

tBSentACosHX += (3)

A.4. Solucin Particular

EDtCtPX ++=2 0=C

DCtPX += 2' en (1) 06.0=D

CPX 2'' = 0=E

tPX 06.0= (4)


Reemplazando (4) y (3) en (2)

ttBSentACostX 06.0)( += (5)

06.0)( +=

tCosBtSenAtX (6)

A.5. Determinacin de A y B en (5) y (6)

0)0(0 == Xt )0(06.000 += A 0=A

0)0(0 ==dt

dXt 06.000 += B 0031.0=B

A.6. Respuesta Dinmica, Reemplazando valores para 0 < t < 0.2 seg ttSentX 06.0)36.19(0031.0)( = 06.036.1906.0)( =

tCostX

Condiciones finales t=0.2

0141.0)2.0( =X

1047.0)2.0( =X

B. Solucin para : . < < . SX = 4.5

B.1. E.D.M.

=+

SXrXrX2


5.42 =+

rXrX (1)

Donde =19.36 rad/seg B.2. Solucin E.D.M PXHXtX +=)( (2)

B.3. Solucin Homognea HX

tBSentACosHX += (3)


B.4. Solucin Particular

CPX = en (1)

0' =PX 5.420 =+ C

3755.4

25.4 ==

C

0'' =PX 012.0=C 012.0=PX (4)


012.0)( += tBSentACostX (5)

tCosBtSenAtX +=

)( (6)

B.5. Determinacin de A y B en (5) y (6)

0141.0)2.0( =X

1047.0)2.0( =X

014.0)2.0(2.0 == Xt 002044.0=A

1047.0)2.0(2.0 =

= Xt 005430.0=B

B.6. Respuesta Dinmica, Reemplazando valores para . < < .

012.0005430.0002044.0)( += tSentCostX

tCostSentX 36.19104.036.19041.0)( +=

Condiciones finales para t=0.4

-0.006866)4.0( =X

0.05217)4.0( =X


C. Solucin para : . < < . SX = 13.5 22.5 C.1. E.D.M.

=+

SXrXrX2


5.135.222 =+

trXrX (1)

Donde =19.36 rad/seg

C.2. Solucin E.D.M PXHXtX +=)( (2)

C.3. Solucin Homognea HX tBSentACosHX += (3)

C.4. Solucin Particular PX

EDtCtPX ++=2 0=C

DCtPX += 2' en (1) 06.0

25.22 ==

D

CPX 2'' = 036.0

25.13 ==

E

036.006.0 = tPX (4)


036.006.0)( ++= ttBSentACostX (5)

06.0)( ++=

tCosBtSenAtX (6)

C.5. Determinacin de A y B en (5) y (6)

-0.006866)4.0( =X

0.05216733)4.0( =X

006866.0)4.0(4.0 == Xt 001035.0=A

0522.0)4.0(4.0 =

= Xt 00505.0=B


C.6. Respuesta Dinmica, Reemplazando valores para . < < .


036.006.0)36.19(00509.0)36.19(001035.0)( ++= ttSentCostX

06.00979.001869.0)( ++=

tCostSentX

D. Solucin para : > . SX = 0 D.1. E.D.M.

=+

SXrXrX2


02 =+

rXrX (1)


D.2. Solucin E.D.M HXtX =)( (2)

D.3. Solucin Homognea HX

tBSentACosHX += (3)

tCosBtSenAtX +=

)(

Determinacin de A y B

-0.00355)6.0( =X

0.13212402)6.0( =X

00355.0)6.0(6.0 == Xt 003489.0=A

13212.0)6.0(6.0 =

= Xt 00686.0=B

D.4. Respuesta Dinmica, Reemplazando valores para t>.

)36.19(006865.0)36.19(003489.0)( tSentCostX +=


II. DETERMINACIN DE LA RESPUESTA DINMICA CUANDO %5= , USANDO LA SOLUCIN DE LA ECUACIN DIFERENCIAL DEL MOVIMIENTO.

II.1. Ecuacin Diferencial del Movimiento.

Debido a que no conocemos la ubicacin, ni la intensidad de la fuerza de amortiguamiento, se obtendr esta por comparacin con la Ecuacin Diferencial del Movimiento del Sistema sin amortiguamiento.

E.D.M de un sistema sin amortiguamiento

=+

SmXrKXrXm

=+

SXrXmK

rXm 2=

mK

=+

SXrXrX2 (1)

Para la estructura

=+

SXrXrX 375 (2)

De (1) y (2) Donde =19.36 rad/seg E.D.M de un sistema con amortiguamiento

=+

+

SmXrKXrXCrXm

=+

+

SXrXm

KrXm

CrX donde

2=mK

y mC 2=

=+

+

SXrXrXrX

22

Donde

=19.36 rad/seg

%5=

%5=


Determinacin de la aceleracin del suelo SX

E. 0 < t < 0.2 seg SX = 0.450.20 SX = 22.5 F. 0.2 < t < 0.4 seg SX = 0.45 SX = 4.5 G. 0.4 < t < 0.6 seg SX = 13.5 22.5

II.2. Solucin De La Ecuacin Diferencial Del Movimiento.

A. Solucin para : < < . SX = 22.5

A.1. Ecuacin Diferencial del Movimiento (E.D.M)

=+

+

SXrXrXrX22


trXrXrX 5.2222 =+

+

(1)


%5=

A.2. Solucin E.D.M PXHXtX +=)( (2)

A.3. Solucin Homognea HX

022 =+

+

rXrXrX

)( tDBSentDACosteHX

+= , 21 =D (3)


A.4. Solucin Particular

EDtCtPX ++=2

DCtPX += 2'

CPX 2'' =

en (1)

tEDtCtDCtC 5.22)2(2)2(22 =+++++

Por comparacin

0=C 06.0=D 0003098.0=E

tPX 06.0= +0.00031 (4)


00031.006.0)()( ++= ttDBSentDACostetX (5)

06.0)()()( +++=

tDCosDBtDSenDAtetDBSentDACos

tetX

(6)

A.5. Determinacin de A y B en (5) y (6)

0)0(0 == Xt )0(06.000 += A +0.0003098 0003098.0=A

0)0(0 ==dt

dXt 06.0)(0 += DBA 003087.0=B

A.6. Respuesta Dinmica, Reemplazando valores para 0 < t < 0.2 seg

=19.36 rad/seg

%5=

34.192)05.0(136.1921 === D 00031.006.0)()( ++= ttDBSentDACos

tetX

[ ] 00031.006.034.19(0031.0)34.19(00031.0968.0)( ++= ttSentCostetX

[ ] 06.034.19(003.0)34.19(06.0968.0)( +=

tSentCostetX



-0.013194)2.0( =X

09874.0)2.0( =X

B. Solucin para : . < < . SX = 4.5 B.1. Ecuacin Diferencial del Movimiento (E.D.M)

=+

+

SXrXrXrX22


5.422 =+

+

rXrXrX (1)


%5= B.2. Solucin E.D.M PXHXtX +=)( (2)

B.3. Solucin Homognea HX

022 =+

+

rXrXrX

)( tDBSentDACosteHX

+= , 21 =D (3)

B.4. Solucin Particular

CPX = en (1)

0' =PX 5.420 =+ C

3755.4

25.4 ==

C

0'' =PX 012.0=C 012.0=PX (4)


012.0)()( += tDBSentDACostetX (5)

)()()( tDCosBtDSenAtetDBSentDACos

tetX +++=

(6)


B.5. Determinacin de A y B en (5) y (6)

-0.013194)2.0( =X

09874.0)2.0( =X

013194.0)2.0(2.0 == Xt 00306.0=A

0987.0)2.0(2.0 =

= Xt 00536.0=B

B.6. Respuesta Dinmica, Reemplazando valores para 0.2 < t < 0.4 seg 012.0)00536.000306.0()( += tDSentDCos

tetX

012.0)05360.01123.0()( +=

tDSentDCostetX


%5=

34.192)05.0(136.1921 === D


-0.00845)4.0( =X

0458.0)4.0( =X

C. Solucin para : . < < . SX = 13.5 22.5 C.1. Ecuacin Diferencial del Movimiento (E.D.M)

=+

+

SXrXrXrX22


5.135.2222 =+

+

trXrXrX (1)


%5= C.2. Solucin E.D.M PXHXtX +=)( (2)


C.3. Solucin Homognea HX

022 =+

+

rXrXrX

)( tDBSentDACosteHX

+= , 21 =D (3)

C.4. Solucin Particular

EDtCtPX ++=2

DCtPX += 2'

CPX 2'' =

en (1)

5.135.22)2(2)2(2 =++++++ tEDtCtDCAC

Por comparacin

0=C 06.02

5.22 ==

D 0363.0225.13 =

+=

DE

0363.006.0 = tPX (4)


0363.006.0)()( ++= ttDBSentDACostetX (5)

06.0)()()( ++++=

tDCosBtDSenAtetDBSentDACos

tetX

(6)

C.5. Determinacin de A y B en (5) y (6)

-0.00845)4.0( =X

0458.0)4.0( =X

0085.0)4.0(4.0 == Xt 00153572.0=A

0458.0)4.0(4.0 =

= Xt 0052611.0=B

C.6. Respuesta Dinmica, Reemplazando valores para 0.4 < t < 0.6 seg

[ ] 0363.006.0)34.19(0052611.0)34.19(00153572.0968.0)( ++= ttSentCostetX


D. Solucin para : > . SX = 0 D.1. Ecuacin Diferencial del Movimiento (E.D.M)

=+

+

SXrXrXrX22


022 =+

+

rXrXrX (1)


%5= D.2. Solucin E.D.M HXtX =)( (2)

D.3. Solucin Homognea HX

022 =+

+

rXrXrX

)( tDBSentDACosteHX

+= , 21 =D (3)

D.4. Determinacin de A y B en (5) y (6)

-0.0022264)6.0( =X

0.10763405)6.0( =X

0022264.0)6.0(6.0 == Xt 005879.0=A

107634.0)6.0(6.0 =

= Xt 0088.0=B

D.5. Respuesta Dinmica, Reemplazando valores para t > 0.6 seg

)34.19(0088.0)34.19(0058795.0(968.0)( tSentCostetX =


III. DETERMINACIN DE LA RESPUESTA DINMICA CUANDO %0= , USANDO LA ECUACIN DE DUHAMEL

dtSent

SXtX )()(0

1)(

=

A. Solucin para : < < .

A.1. Ecuacin de DUHAMEL

dtSent

tX )(0

5.221)( =

A.2. Solucin de la Ecuacin de DUHAMEL

dtSent

tX )(0

5.22)( =

15.22)( ItX

= dtSen

tI )(

01 = (1)

Integrando por partes I1

duvvuI =1 (2)

=u dtSendv )( =

ddu =

dtCosv )( =

Resolviendo = duv

= duv =

ddtCos )( 2

)(

tSen (3)

(2) y (3) en 1

ttSentCostX

02

)()(.5.22)(

+

=

%0=


=

2)(1.5.22)(

tSenttX

2

)(5.223

)(5.22)(

ttSentX =

Reemplazando el valor =19.36 rad/seg

236.19)(5.22

336.19

)(36.195.22)( ttSentX =

A.3. Respuesta Dinmica

ttSentX 06.0)36.19(0031.0)( =

B. Solucin para : . < < .

B.1. Ecuacin de DUHAMEL

dtSent

dtSentX )(2.0

5.41)(2.0

05.221)( =

B.2. Solucin de la Ecuacin de DUHAMEL

ttCostSentCostX

2.0

)(5.42.0

02

)()(.5.22)(

+

=

+

=

)2.0(15.42

)(2

)2.0()2.0(2.05.22)( tCostSentSentCostX

2

)2.0(5.425.4

3)(5.22

3)2.0(5.22

2)2.0(5.4)(

++= tCostSentSentCostX

25.4

3)(5.22

3)2.0(5.22)(

+= tSentSentX

=19.36 rad/seg

B.3. Respuesta Dinmica

012.03

)(5.223

)2.0(5.22)( +=

tSentSentX

=19.36 rad/seg


C. Solucin para : . < < .

C.1. Ecuacin de DUHAMEL

dtSent

dtSendtSentX )(4.0

)5.225.13(1)(4.0

2.05.41)(

2.0

05.221)( =

(1)

C.2. Solucin de la Ecuacin de DUHAMEL, segn el procedimiento de A y B

4.0

2.0

)(5.42.0

02

)()(.5.22)(

+

=

tCostSentCostX

t

tSentCosttCos

4.02

)()(.5.22

4.0

)(5.13

+

+

+

=

)2.0()4.0(5.42

)(2

)2.0()2.0(2.05.22)( tCostCostSentSentCostX

+

2

)4.0()4.0(4.05.22)4.0(15.13

tSentCosttCos

2)2.0(5.4

2)4.0(5.4

3)(5.22

3)2.0(5.22

2)2.0(5.4)(

++= tCostCostSentSentCostX

3)4.0(5.22

2)4.0(9

25.22

2)4.0(5.13

25.13

++ tSentCosttCos

25.13

25.22

3)4.0(5.22

3)(5.22

3)2.0(5.22)(

++= ttSentSentSentX

036.006.03

)4.0(5.223

)(5.223

)2.0(5.22)( ++= ttSentSentSentX

C.3. Respuesta Dinmica

[ ] 036.006.0)4.0()()2.0(35.22)( ++= ttSentSentSentX


D. Solucin para : > .

D.1. Ecuacin de DUHAMEL

dtSendtSendtSentX )(6.0

4.0)5.225.13(1)(

4.0

2.05.41)(

2.0

05.221)( =

(1)

D.2. Solucin de la Ecuacin de DUHAMEL, segn el procedimiento de A y B

4.0

2.0

)(5.42.0

02

)()(.5.22)(

+

=

tCostSentCostX

6.0

4.02

)()(.5.226.0

4.0

)(5.13

+

+

tSentCostCos

+

=

)2.0()4.0(5.42

)(2

)2.0()2.0(2.05.22)( tCostCostSentSentCostX

++ 2

)4.0()4.0(4.02

)6.0()6.0(6.05.22)4.0()6.0(5.13

tSentCostSentCostCostCos

2)2.0(5.4

2)4.0(5.4

3)(5.22

3)2.0(5.22

2)2.0(5.4)(

++= tCostCostSentSentCostX

3)4.0(5.22

2)4.0(9

3)6.0(5.22

2)6.0(5.13

2)4.0(5.13

2)6.0(5.13

+

+

+

tSentCostSentCostCostCos

3

)4.0(5.223

)6.0(5.223

)(5.223

)2.0(5.22)(

++= tSentSentSentSentX

D.3. Respuesta Dinmica

[ ])6.0()4.0()()2.0(35.22)( ++= tSentSentSentSentX


IV. DETERMINACIN DE LA RESPUESTA DINMICA CUANDO %5= , USANDO LA SOLUCIN DE LA ECUACIN DIFERENCIAL DEL MOVIMIENTO.

dtDSent

XteD

tX )()(0

)(1)(

=

21 =D

=19.36 rad/seg

CALCULOS PREVIOS

Se usaran las siguientes abreviaturas

)()( = tDSenteM (1)

)()( = tDCosteN (2)

Integrales a usar:

= DMI )(1 dvtDSenteI = )()(1 (3)

= DNI )(2 dvtDCosteI = )()(2 (4)

Hallando 1I e 2I

dvtDSenteI = )()(1


= duvvuI1 (5)

)( = teu dvtDSendv )( =

%5=


)( = tedu D

dtDCosv

=

)(

Remplazando valores en (5)

=

dtDCos

tD

tDCoste

DI )()()()(11

Usando (2) y (4)

21 IDD

NI

= (6)


= dmnnmI2 (7)

)( = tem dtDCosdv )( =

)( = tedm D

dvtDSenv

=

)(

Remplazando valores en (7)

+=

dtDSen

t

DtDSen

teD

I )()()()(12

Usando (1) y (3)

12 IDD

MI

+= (8)

Sistema de ecuaciones (6) y (8)

21 IDD

NI

= (6)

12 IDD

MI

+= (8)

Resolviendo el sistema para hallar I1 (6) en (8)

+= 11 IDD

M

DD

NI

MDD

N

D

DI2)(

)2)(

2)(2)((1

+=

+


2)(2)(1 D

MNDI

+

+= ()

2)(2)(

)()()()(1

D

tetDSentetDCosDI

+

+=

Resolviendo el sistema para hallar I2 (8) en (6)

+= 22 IDD

N

DD

MI

2)(2)(2 D

MDNI

+

= ()

2)(2)(

)()()()(2

D

tetDSenDtetDCosI

+

=

A. Solucin para : < < . A.1. Ecuacin de DUHAMEL

dtDSent te

DtX )(

05.22)(1)( =

A.2. Solucin de la Ecuacin de DUHAMEL

dtDSent te

DtX )(

0

)(5.22)( =

35.22)( I

DtX

= (1)

dtDSent teI )(0

)(3

= (2)

Se usaran las integrales y constantes del Clculo

)()( = tDSenteM

)()( = tDCosteN

= DMI )(1

= DNI )(2


2)(2)(1 D

MNDI

+

+=

2)(2)(2 D

MDNI

+

=


dpqqpI =3 (3)

=u dtDSentedv )()( =

ddu = =v2)(2)( D

MND

+

+

Reemplazando en (3)

d

D

MND

D

MNDI

+

+

+

+= 2)(2)(2)(2)(3

+

++

+

+= dM

DdN

D

D

D

MNDI )(2)(2)(

)(2)(2)(2)(2)(3

Como = DMI )(1 y = DNI )(2

+

++

+

+= 12)(2)(

22)(2)(2)(2)(3I

DI

D

D

D

MNDI

Reemplazando el valor de I1 Y I2

+

+

++

+

+

+

+=

2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(3 D

MND

DD

MDN

D

D

D

MNDI

++

+

+

+= MNDMDND

DD

MNDI 2)()(2)()(2)2)(2)((

12)(2)(3

+

+

+= MDND

DD

MNDI )2)(2)(()2(2)2)(2)((

12)(2)(3

Reemplazando el valor de M y N

{ +

+

+= 2)2)(2)((

12)(2)(

)()()()(

3DD

tDSentetDCos

teDI


}{ })()()2)(2)(()()()2( 0 tDSenteDtDCosteDt

Evaluando la integral para t= y 0=

+

+

+

+= 2)2)(2)((

)()()2)(2)(()()()2(2)2)(2)((

)2(2)(2)(

)(3

D

tDSenteDtDCos

teD

D

D

D

tDI

Reemplazando valores

+

= 2)9375.374(

)365.19()(968.0)374365.1905.02(

2)9375.374(

)374365.1905.02(9375.374

3743

tCostetI

2)9375.374(

)365.19()()9063.373( tDSente

{ )365.19()(968.00026598.0)365.19()(968.00002664.00002664.005158.03 tSentetCostetI += A.3. Respuesta Dinmica

De (1)

35.22)( I

DtX

=

{+= )365.19()(968.00026598.0)365.19()(968.00002664.00002664.005158.0

3745.22)( tSentetCostettX

+ += )365.19()(968.00031.0)365.19()(968.000031.000031.006.0)( tSentetCostettX

[ ] 00031.006.034.19(0031.0)34.19(00031.0968.0)( ++= ttSentCostetX B. Solucin para : . < < .

B.1. Ecuacin de DUHAMEL

dtDSent te

DdtDSen

teD

tX )(2.0

5.4)(1)(2.0

05.22)(1)( =

dtDSent te

DdtDSen

teD

tX )(2.0

)(5.4)(2.0

0

)(5.22)( =


15.4

35.22)( I

DI

DtX

= (1)


+

+

+= MDND

DD

MNDI )2)(2)(()2(2)2)(2)((

12)(2)(3

(2)

2)(2)(1 D

MNDI

+

+= (3)

)()( = tDSenteM

)()( = tDCosteN


{ }2.0

0

)2)(2)(()2(2)2)(2)((

12)(2)(

5.22)(

+

+

+= MDND

DD

MNDD

tX

t

D

MNDD 2.0

2)(2)(

5.4

+

+

(4)

Evaluando el valor de M y N

0= 2.0= t=

)()()0( tDSenteM = )2.0()2.0()2.0( = tDSen

teM 0

)()()0( tDCosteN = )2.0()2.0()2..0( = tDCos

teN 1

(5)


Reemplazando (5) en (4)

{ }

+

+

+= 2)2)(2)((

)2.0()2)(2)(()2.0()2(2)(2)(

)2.0()2.0(2.05.22)(

D

MDND

D

MNDD

tX

{ }

+

+ 2)2)(2)((

)0()2)(2)(()0()2(5.22

D

MDNDD

+

+

+ 2)(2)(

)2.0()2.0(2)(2)(

5.4

D

MND

D

DD

{ }222

22

22 ))()(()2.0())()(()2.0()2(5.22

)()()2.0()2.0(5.4)(

D

DD

DD

D

D

MNW

MNW

tX

+

+

+

+=

2222 )()()2.0()2.0(5.4

)()(5.4

2)2)(2)((

)0()2)(2)(()0()2(5.22D

D

DD

D MN

DD

MDNDD

++

++

+

2)2)(2)((

)2.0()2)(2)(()2.0()2(5.22)(D

MDND

DWtX

+

=

2)(2)(

5.42)2)(2)((

)0()2)(2)(()0()2(5.22

DD

MDNDD

+

+

2)2)(2)((

)2.0()2)(2)(()2.0()2()2.0(5.22)(D

tDSenDtDCosDte

DWtX

+

=

2)(2)(

5.42)2)(2)((

)(*)2)(2)(()()2()(5.22

DD

tDSenDtDCosDte

DW

+

+



365.19=

3752 = rad/seg

21 =D rad/seg

( 22 )05.0(1(375)1(22)( == D

3742)( =D

%5=

9375.374)3742)365.19*05.0((2)(2)( =+=+ D

9063.373)2)365.19*05.0(374(2)(2)( ==+ D

{ })2.0(0003095.0)2.0(00031.0()2.0(968)( = tDSentDCostetX

{ } 012.00003095.000031.0()(968.0 ++ tDSentDCoste

B.3. Respuesta Dinmica

{ })2.0(34.190031.0)2.0(34.1900031.0()2.0(968)( = tSentCostetX { } 012.034.190031.034.1900031.0()(968.0 ++ tSentCoste

C. Solucin para : . < < .

C.1. Ecuacin de DUHAMEL

dtDSente

DdtDSen

teD

tX )(4.0

2.05.4)(1)(

2.0

05.22)(1)( =

dtDSent te

D)()

4.05.225.13()(1

dtDSente

DdtDSen

teD

tX )(4.0

2.0

)(5.4)(2.0

0

)(5.22)( =


dtDSent te

DdtDSen

t teD

)(4.0

)(5.13)(4.0

)(5.22 +

15.13

35.22

15.4

35.22)( I

DI

DI

DI

DtX

+=

(1)


+

+

+= MDND

DD

MNDI )2)(2)(()2(2)2)(2)((

12)(2)(3

(2)

2)(2)(1 D

MNDI

+

+= (3)

)()( = tDSenteM

)()( = tDCosteN


{ }2.0

0

)2)(2)(()2(2)2)(2)((

12)(2)(

5.22)(

+

+

+= MDND

DD

MNDD

tX

{ } tD

MDND

D

MNDDD

MNDD 4.0

2)2)(2)((

1)2)(2)(()2(2)(2)(

5.224.0

2.02)(2)(

5.4

+

+

++

+

+

t

D

MNDD 4.0

2)(2)(

5.13

+

+

(4)


0= 2.0= 4.0= t=

)()()0( tDSenteM = )2.0(M )4.0(M 0

)()()0( tDCosteN = ))2..0(N )4.0(N 1

(5)



{ }

+

+

+= 2)2)(2)((

)2.0()2)(2)(()2.0()2(2)(2)(

)2.0()2.0(2.05.22)(

D

MDND

D

MNDD

tX

{ }

+

+ 2)2)(2)((

)0()2)(2)(()0()2(5.22

D

MDNDD

+

+

+

+ 2)(2)(

)2.0()2.0(2)(2)(

)4.0()4.0(5.4

D

MND

D

MNDD

{ }

+

+

+

+

++ 2)(2)(

)4.0()4.0(4.05.222)2)(2)((

)2(2)(2)(

5.22

D

MNDDD

D

D

Dt

D

{ }

+

+ 2)2)(2)((

)4.0()2)(2)(()4.0()2(5.22

D

MDNDD

+

+

+ 2)(2)(

)4.0()4.0(2)(2)(

5.13

D

MND

D

DD

Simplificando

{ }222

22

22 ))()(()2.0())()(()2.0()2(5.22

)()()2.0()2.0(5.4)(

D

DD

DD

D

D

MNW

MNW

tX

+

+

+

+=

2)(2)(

)2.0()2.0(5.42)(2)(

)4.0()4.0(5.42)2)(2)((

)0()2)(2)(()0()2(5.22

D

MND

DD

MND

DD

MDND

D

+

++

+

+

+

{ }

+

+

+

+

+2)(2)(

)4.0()4.0(92)2)(2)((

)50(

2)(2)(

5.22

D

MNDDDD

t

{ }2)2)(2)((

)4.0()2)(2)(()4.0()2(5.22

D

MDND

D

+

+

2)(2)(

)4.0()4.0(5.132)(2)(

5.13

D

MNDDD

D

+

++

+


2)2)(2)((

)2.0(*)2)(2)(()2.0()2()2.0(5.22)(D

tDSenDtDCosDte

DWtX

+

=

+

2)2)(2)((

)(*)2)(2)(()()2()(5.22

D

tDSenDtDCosDte

DW

{ }2)2)(2)((

)4.0(*)2)(2)(()4.0()2()4.0(5.22D

tDSenDtDCosDte

DW

+

+

2)(2)(

5.132)2)(2)((

502)(2)(

5.22

DDD

t

+

+

+

+


365.19=

3752 = rad/seg

21 =D rad/seg

( 22 )05.0(1(375)1(22)( == D

3742)( =D

%5=

9375.374)3742)365.19*05.0((2)(2)( =+=+ D

9063.373)2)365.19*05.0(374(2)(2)( ==+ D

{ })2.0(003095.0)2.0(00031.0()2.0(968)( = tDSentDCostetX

{ }tDSentDCoste ++ 003095.000031.0()(968.0

{ } 0363.006.0)4.0(003095.0)4.0(00031.0()4.0(968 + ttDSentDCoste


D. Solucin para : > .

D.1. Ecuacin de DUHAMEL

dtDSente

DdtDSen

teD

tX )(4.0

2.05.4)(1)(

2.0

05.22)(1)( =

dtDSente

D)()

6.0

4.05.225.13()(1

dtDSente

DdtDSen

teD

tX )(4.0

2.0

)(5.4)(2.0

0

)(5.22)( =

dtDSente

DdtDSen

teD

)(6.0

4.0

)(5.13)(6.0

4.0

)(5.22 +

15.13

35.22

15.4

35.22)( I

DI

DI

DI

DtX

+=

(1)

D.2. Solucin de la Ecuacin de DUHAMEL

+

+

+= MDND

DD

MNDI )2)(2)(()2(2)2)(2)((

12)(2)(3

(2)

2)(2)(1 D

MNDI

+

+= (3)

)()( = tDSenteM

)()( = tDCosteN


{ }2.0

0

)2)(2)(()2(2)2)(2)((

12)(2)(

5.22)(

+

+

+= MDND

DD

MNDD

tX


{ } 6.0

4.02)2)(2)((

1)2)(2)(()2(2)(2)(

5.224.0

2.02)(2)(

5.4

+

+

++

+

+

D

MDND

D

MNDDD

MNDD

6.0

4.02)(2)(

5.13

+

+

D

MNDD

(4)


0= 2.0= 4.0= 6.0= t=

)()()0( tDSenteM = )2.0(M )4.0(M )6.0(M 0

)()()0( tDCosteN = ))2..0(N )4.0(N )6.0(N 1

(5)


{ }

+

+

+= 2)2)(2)((

)2.0()2)(2)(()2.0()2(2)(2)(

)2.0()2.0(2.05.22)(

D

MDND

D

MNDD

tX

{ }

+

+ 2)2)(2)((

)0()2)(2)(()0()2(5.22

D

MDNDD

+

+

+

+ 2)(2)(

)2.0()2.0(2)(2)(

)4.0()4.0(5.4

D

MND

D

MNDD

{ }

+

+

++ 2)2)(2)((

)6.0()2)(2)(()6.0()2(5.222)(2)(

)6.0()6.0(6.05.22

D

MDNDDD

MNDD

{ }

+

+

+

+ 2)2)(2)((

)4.0()2)(2)(()4.0()2(5.222)(2)(

)4.0()4.0(4.05.22

D

MDNDDD

MNDD

+

+

+

+ 2)(2)(

)4.0()4.0(2)(2)(

)6.0()6.0(5.13

D

MND

D

MNDD


Simplificando

2)2)(2)((

)2.0(*)2)(2)(()2.0()2()2.0(5.22)(D

tDSenDtDCosDte

DWtX

+

=

+

2)2)(2)((

)(*)2)(2)(()()2()(5.22

D

tDSenDtDCosDte

DW

{ }2)2)(2)((

)4.0(*)2)(2)(()4.0()2()4.0(5.22D

tDSenDtDCosDte

DW

+

+

{ }2)2)(2)((

)6.0(*)2)(2)(()6.0()2()6.0(5.22D

tDSenDtDCosDte

DW

+


365.19=

3752 = rad/seg

21 =D rad/seg

( 22 )05.0(1(375)1(22)( == D

3742)( =D

%5=

9375.374)3742)365.19*05.0((2)(2)( =+=+ D

9063.373)2)365.19*05.0(374(2)(2)( ==+ D

{ })2.0(003095.0)2.0(00031.0()2.0(968)( = tDSentDCostetX

{ }tDSentDCoste ++ 003095.000031.0()(968.0

{ })4.0(003095.0)4.0(00031.0()4.0(968 tDSentDCoste

{ })6.0(003095.0)6.0(00031.0()6.0(968 + tDSentDCoste


V. DIAGRAMA DE FUERZA CORTANTE Y DIAGRAMA DE MOMENTO FLECTOR CUANDO 0= Y %5=

5.1.Diagrama De Fuerza Cortante y Diagrama De Momento Flector Cuando 0= 5.2.1. Funcin de Desplazamiento dinmico

A. 0 < t < 0.2 seg

... DMESolu ... DUHAMELEcuSolu

ttSentX 06.0)36.19(0031.0)( = ttSentX 06.0)36.19(0031.0)( =

014.01max =X , en t=0.2seg 014.01max =X , en t=0.2seg

B. 0.2 < t < 0.4 seg


012.0005430.0002044.0)( += tSentCostX 012.03)(5.22

3)2.0(5.22)( +=

tSentSentX

0.017797392max =X en t= 0.25seg 0.017781612max =X en t= 0.25seg

-0.015

-0.01

-0.005

00 0.05 0.1 0.15 0.2 0.25

X(t)

-0.015

-0.01

-0.005

00 0.05 0.1 0.15 0.2 0.25

X(t)

-0.0200

-0.0150

-0.0100

-0.0050

0.00000 0.1 0.2 0.3 0.4 0.5

X(t)

-0.0200

-0.0150

-0.0100

-0.0050

0.00000 0.1 0.2 0.3 0.4 0.5

DUHAMEL


C. 0.4 < t < 0.6 seg ... DMESolu ... DUHAMELEcuSolu

)36.19(0051.0)36.19(001035.0)( tSentCostX += [ ])4.0()()2.0(35.22)( += tSentSentSentX

036.006.0 + t 036.006.0 + t


D. t > 0.6 seg ... DMESolu ... DUHAMELEcuSolu

)36.19(006865.0)36.19(003489.0)( tSentCostX += [ ])6.0()4.0()()2.0(35.22)( ++= tSentSentSentSentX

0.00784max =X 0.00784max =X

-0.01

-0.008

-0.006

-0.004

-0.002

00 0.2 0.4 0.6 0.8

X(t)

-0.010

-0.008

-0.006

-0.004

-0.002

0.0000 0.2 0.4 0.6 0.8

DUHAMEL

-0.01

-0.005

0

0.005

0.01

0 0.5 1 1.5 2

X(t)

-0.0100

-0.0050

0.0000

0.0050

0.0100

0 0.5 1 1.5

DUHAMEL


5.2.2. Desplazamiento relativo mximo, absoluto

0.0178max =X en t= 0.25seg

5.2.3. Diagrama de fuerza Cortante y Diagrama de Momento Flector

rXrXh

EI == 5033

31V 0178.0150= =2.67 ton

rXrXh

EI == 50123

122V 0178.0600= =10.68 ton

DFC DMF

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

Respuesta Dinamica X(t)


5.2.Diagrama De Fuerza Cortante y Diagrama De Momento Flector Cuando %5= 5.2.1. Funcin de Desplazamiento dinmico

A. 0 < t < 0.2 seg


[ ]tSentCostetX 34.19(0031.0)34.19(00031.0968.0)( += [ ]tSentCostetX 34.19(0031.0)34.19(00031.0968.0)( +=

00031.006.0 + t 00031.006.0 + t

013.01max =X , en t=0.2seg 013.01max =X , en t=0.2seg

B. 0.2 < t < 0.4 seg


012.0)00536.000306.0()( += tDSentDCostetX { })2.0(34.190031.0)2.0(34.1900031.0()2.0(968)( = tSentCostetX

{ } 012.034.190031.034.1900031.0()(968.0 ++ tSentCoste


-0.014

-0.012

-0.01

-0.008

-0.006

-0.004

-0.002

00 0.05 0.1 0.15 0.2 0.25

X(t)

-0.014

-0.012

-0.01

-0.008

-0.006

-0.004

-0.002

00 0.05 0.1 0.15 0.2 0.25

X(t)

-0.02

-0.015

-0.01

-0.005

0

0.2

0.21

50.

230.

245

0.26

0.27

50.

290.

305

0.32

0.33

50.

350.

365

0.38

0.39

5

X(t)

-0.02

-0.015

-0.01

-0.005

0

0.2

0.21

50.

230.

245

0.26

0.27

50.

290.

305

0.32

0.33

50.

350.

365

0.38

0.39

5

DUHAMEL


C. 0.4 < t < 0.6 seg ... DMESolu ... DUHAMELEcuSolu

[ ])34.19(0052611.0)34.19(00153572.0968.0)( tSentCostetX += { })2.0(003095.0)2.0(00031.0()2.0(968)( = tDSentDCostetX

036.006.0 + t { }tDSentDCoste ++ 003095.000031.0()(968.0

{ } 0363.006.0)4.0(003095.0)4.0(00031.0()4.0(968 + ttDSentDCoste

0.0083max =X en t= 0.475seg 0.0083max =X en t= 0.53seg t > 0.6 seg ... DMESolu ... DUHAMELEcuSolu

)34.19(0088.0)34.19(0058795.0(968.0)( tSentCostetX = { })2.0(003095.0)2.0(00031.0()2.0(968)( = tDSentDCostetX

{ }tDSentDCoste ++ 003095.000031.0()(968.0

{ } 0363.006.0)4.0(003095.0)4.0(00031.0()4.0(968 + ttDSentDCoste

{ })6.0(003095.0)6.0(00031.0()6.0(968 + tDSentDCoste

0.00774max =X

-0.01

-0.008

-0.006

-0.004

-0.002

00 0.2 0.4 0.6 0.8

X(t)

-0.01

-0.008

-0.006

-0.004

-0.002

00 0.2 0.4 0.6 0.8

X(t)

-0.006

-0.004

-0.002

0

0.002

0.004

0.006

0.6

0.74

0.88

1.02

1.16 1.

31.

441.

581.

721.

86 22.

142.

282.

422.

56 2.7

2.84

2.98

3.12

3.26 3.

43.

54

x(t)


5.2.2. Desplazamiento relativo mximo, absoluto

0.0167max =X 5.2.3. Diagrama de fuerza Cortante y Diagrama de Momento Flector

rXrXh

EI == 5033

31V r0167.0150= =.2.5 ton

rXrXh

EI == 50123

122V 0167.0600= =10. ton

DFC DMF

-0.02

-0.015

-0.01

-0.005

0

0.005

0.01

0 0.5 1 1.5 2 2.5

Respuesta Dinamica X(t)

Analisis Sismico Yaaaaa

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