practica calificada n 04

PRACTICA CALIFICADA N° 04

Graficar el desplazamiento que se presenta en el nudo 4 del pórtico mostrado. (Usar

Sap2000 y resolver de forma manual usando DUHAMEL)

Sección de la columna: 0.25x0.25 m2

Sección de la viga: 0.25x0.40 m2

f c' =280

kg

cm2E=15100√ f c

'

Modelar con las siguientes condiciones.

A. Para una Fuerza Sísmica.



f c' =280

kg

cm2

E=15100√ f c kg/cm 2

I=b∗h3

12

SOLUCIÓN:

Determinación del número de grados de libertad (#gdl).

¿ gdl=3

Determinación de numero de grados de libertad dinámico (#gdlD)

¿ gdlD=1

Determinación de la matriz de Rigidez [K], por el método de Rigidez.

A. Definición del Sistema Global de Coordenadas {Q }−{D}.

B. Definición del sistema local de coordenadas {q }– {d }:

C. Determinación de la Matriz [ A ]:

Se sabe: {d }=[ A ]{D }

{d }=[ A ]1 D1+[ A ]2 D2+[ A ]3 D3 ….(¿)

Primer Estado de Deformación: D1=1 , D2=D3=0, remplazando en (*) tenemos [ A ]1= {d }

[ A ]1=[10100000

] Segundo Estado de Deformación: D2=1 , D1=D3=0 remplazando en (*) tenemos

[ A ]2= {d }.

[ A ]2=[00010001

] tercer Estado de Deformación: D3=1 , D1=D2=0, remplazando en (*) tenemos

[ A ]3= {d }

1

2

3

1

2

3

[ A ]3=[01000100

][ A ]3=[

10100000

00010001

01000100

]D. Determinación de la Matriz de Rigidez de cada elemento en el sistema local {q}-{d} [ k ]:

k 1=k2=[ 12 EI

L3 ( 11+4∅ ) 6 EI

L2 ( 11+4∅ )

6 EIL2 ( 1

1+4 ∅ ) 4 EIL ( 1+∅

1+4∅ )]

1

2

3

∅=3 EIGA ( β

L2 ) GA= E2(1+μ)

A

1

2

3

k 3=[12 EI

L3 ( 11+4 ∅ ) 6 EI

L2 ( 11+4 ∅ )

6 EI

L2 ( 11+4∅ ) 4 EI

L ( 1+∅1+4∅ )

−12 EIL3 ( 1

1+4 ∅ ) 6 EIL2 ( 1

1+4∅ )−6 EI

L2 ( 11+4∅ ) 2 EI

L ( 1−2∅1+4 ∅ )

−12 EIL3 ( 1

1+4∅ ) −6 EIL2 ( 1

1+4 ∅ )6 EIL2 ( 1

1+4∅ ) 2 EIL ( 1−2∅

1+4∅ )

12 EIL3 ( 1

1+4∅ ) −6 EIL2 ( 1

1+4∅ )−6 EI

L2 ( 11+4∅ ) 4 EI

L ( 1+∅1+4∅ ) ]

Cálculo de Las propiedades de sección de los elementos:

- Determinación del Módulo de Elasticidad:

E=15100 x √280 x 10=2526713.280Tn /m2

G= E2(1+μ)

=2526713.2802(1+0.2)

=1052797.200Tn /m2

- Determinación de las Áreas y Momentos de Inercia:

Para barras 1 y 2

I y=0.25 x0.253

12=0.000326 m4

A=0.25 x 0 .25=0.0625 m2

E I y=2526713.280 x0.000326=822.498 Tn .m2

EA=2526713.280 x0.0625=157919.580 Tn

GA=1052797.200 x 0.0625=65799.825Tn . m2

∅=3 E I y

GA ( βL2 )

β=factor de forma por corte para seccionesrectangulares(β=1.2)

L=Longitud del elemento(L1=L2=3m)

∅=3 x822.498 x1.2

65799.825 x 32=0.0050

- Con los datos obtenidos anteriormente se procederá a ensamblar la matriz

de rigidez local.

k 1=k2=[358.387 537.580537.580 1080.536 ]

Para barra N° 03

I x=0.25 x0.403

12=0.00133 m4

A=0.25 x 0 .40=0.10 m2

E I x=2526713.280 x 0.00133=3368.951 Tn .m2

EA=2526713.280 x0.10=252671.328 Tn

GA=1052797.200 x 0.10=105279.720 Tn . m2

∅=3 E I y

GA ( βL2 )

β=factor de forma por corte para seccionesrectangulares(β=1.2)

L=Longitud del elemento(L3=4 m)

∅=3 x3368.951 x1.2

1052797.200 x42=0.0072

- Con los datos obtenidos anteriormente se procederá a ensamblar la matriz

de rigidez local.

k 3=[ 613.995 1227.991227.991 3298.22

−613.995 1227.991−1227.991 1613.743

−613.995 −1227.991227.991 1613.74

613.995 −1227.991−1227.991 3298.219

]E. Ensamblaje de la Matriz de Rigidez [K].

Sabemos que:

[ K ]=∑ [ A ]T [ k ] [ A ]

[ K ]=[ A1 ]T [ k1 ] [ A1 ]+[ A2 ]T [ k2 ] [ A2 ]+[ A3 ]T [k3 ] [ A3 ]

[ K ]=[ [ A1 ]T [ A2 ]T [ A3 ]T ]∗[ [k1 ]00

0[k2 ]0

00

[ k3 ] ]∗[ [ A1 ][ A2 ][ A3 ] ]

[ K ]=[1 0 10 0 00 1 0

0 0 01 0 00 0 1

0 00 10 0]∗[ [358.387 537.580

537.580 1080.536][ 0 ][ 0 ]

[ 0 ]

[358.387 537.580537.580 1080.536]

[ 0 ]

[ 0 ][ 0 ]

[ 613.995 1227.991227.991 3298.22

−613.995 1227.991−1227.991 1613.743

−613.995 −1227.991227.991 1613.74

613.995 −1227.991−1227.991 3298.219

]]∗[10100000

00010001

01000100

][ K ]=[716.774 537.580 537.580

537.580 4378.755 1613.743537.580 1613.743 4378.755]

Determinación de la matriz de rigidez lateral [KL] (por Condensación Estática)

[ KL ]=[ K ]¿−[ K ]LO [ K ]OO−1 [ K ]OL

[ K ]=[ [ K ]¿ [ K ]LO

[ K ]OL [ K ]OO][ K ]=[716.774 537.580 537.580

537.580 4378.755 1613.743537.580 1613.743 4378.755]

[ K ]=[ 620.322 ] Tn /m

Determinación de la matriz masa [M]:

[ M ]=[m11 m 12 m 13m 21 m 22 m 23m 31 m 32 m 33]

Primer estado: D1=1 , D2=D3=0

m 11=3.67

m 21=0

m 31=0

Segundo estado: D2=1 , D1=D3=0

m 22=0

m 12=0

m 32=0

Tercer estado: D3=1 , D1=D2=0

m 33=0

m 13=0

m 23=0

Finalmente:

[ M ]=[3.67 0 00 0 00 0 0]

Condensando:

[ m ]=[ M ]¿−[ M ]LO [ M ]OO−1 [ M ]OL

[ m ]=3.67−[ 0 0 ] [0 00 0]

−1

[00][ m ]=3.67 Tn .

s2

m

Identificación del tiempo de Impulso:

Comparando el tiempo t d con el periodo de vibración.

Sea

T=2 πw

Donde :w=√ KLm

=√ 620.322Tnm

3.67Tn . s2

m

=13.001rad

s

T=2 πw

= 2 π13.001

=0.483 s

T4=0.121<t d=3 s⟹ Impulso de largaduración

Formulación de la Ecuación Diferencial de Movimiento [EDM]:

m X+c X+KX=−m X s

Donde :m=3.67 Tn .s2

m

K=620.322Tnm

C=2 mw ξ=2 x 3.67 x13.001 x0.05=4.771

3.67 X+4.771 X+620.322 X=−3.67 X s

X s=−1.308∗t +3.924

3.67 X+4.771 X+620.322 X=−3.67 (−1.308∗t +3.924 )…… …(¿)

De la ecuación (*), por comparación tendremos:

P(t )=−m∗ ¨Xs(t )−3.67 (−1.308∗t+3.924 )

P(t )=4.800∗t−14.401

Respuesta dinámica aplicando DUHAMEL, para un coeficiente de amortiguamiento:

ξ=5%

.1. FASE (I): 0 ≤ t ≤ td 1=3 s

Sabemos que:

E . D .M ⟹m∗X+C∗X+ K∗X=P(t )

La respuesta dinámica será:

X (t )=1

m∗wD∫

0

t

e−ξ∗wn(t−τ)∗P(t )∗sen (wD (t−τ ) )∗dτ

Determinación de P1(τ ): por comparación de la ecuación (*)

P(t )=−m∗ ¨Xs (τ )=−3.67 (−1.308∗t+3.924 )

Reemplando tendremos :

X (t )=−1wD

∫0

t

e−ξ∗wn(t−τ)∗X (τ)∗sen (wD (t−τ ) )∗dτ

X (τ )=−1.308∗τ+3.924

Gravedad:

g=9.81∗m

s2

Rigidez lateral (de la matriz de rigidez)

[ K ]=[ 620.322 ] Tn /m

Masa (de la matriz de masa)

[ m ]=3.67 Tn .s2

m

Frecuencia natural:

wn2= K

m⟹K=wn

2∗m

wn=√ KLm

=√ 620.322Tnm

3.67Tn . s2

m

=13.001rad

s

wn=13.001 rad /s

Frecuencia amortiguada:

wD=wn∗√1−ξ2

wD=13.001∗√1−0.052

wD=12.984 rad /s

integración de la ecuación de la respuesta dinámica.

X (t )=−1wD

∫0

t


X (t )=1

wD∫0

t

e−ξ∗wn (t−τ )∗(1.308∗τ−3.924)∗τ∗sen (wD (t−τ ) )∗dτ

X (t )=1.308

wD∫

0

t

τ∗e−ξ∗wn(t−τ)∗sen (wD (t−τ ) )∗dτ−3.924wD

∫0

t

e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )∗dτ

∫ A=∫0

t

τ∗e−ξ∗wn (t−τ)∗sen (wD (t−τ ) )∗dτ

∫B=∫0

t

e−ξ∗wn(t−τ )∗sen (wD ( t−τ ) )∗dτ

X (t )=1.308

wD∫ A−3.928

wD∫B ¿

Solucion delbloque A Integración por partes :

∫u∗dv=u∗v−∫ v∗du

u=τ∗e−ξ∗wn (t−τ )

du=(e−ξ∗wn( t−τ )+ξ∗wn∗τ∗e−ξ∗wn (t− τ ) )dτ

dv=sen (wD ( t−τ ) )∗dτ

v=∫ sen (wD (t−τ ))∗dτ

v=cos (wD ( t−τ ) )

wD

∫ A=( τ∗e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )wD

−∫0

t cos (wD (t−τ ) )wD

∗(e−ξ∗wn (t−τ )+ξ∗wn∗τ∗e−ξ∗wn (t−τ ) )∗dτ )∫ A=

τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )wD

−ξ∗wn

wD∫0

t

τ∗e−ξ∗wn (t−τ )∗cos (wD (t−τ ))∗dτ− 1wD

∫0

t

e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )∗dτ

A

B

A2

A1

∫ A=τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )

wD

−ξ∗wn

wD∫ A 1− 1

wD∫ A 2(¿∗¿)

Solucion delbloque A1

∫ A 1=∫0

t

τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )∗dτ




dv=cos (wD ( t−τ ) )∗dτ

v=∫ cos (wD (t−τ ) )∗dτ

v=−sen (wD (t−τ ) )

wD

∫ A 1=−τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )

wD

+ξ∗wn

wD∫

0

t

τ∗e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )∗dτ+ 1wD

∫0

t

e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )∗dτ


wD

+ξ∗wn

wD∫ A+ 1

wD∫B


∫ A 2=∫0

t

e−ξ∗wn (t−τ )∗cos (wD ( t−τ ) )∗dτ


u=e−ξ∗wn (t−τ )

du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ



v=−sen (wD ( t−τ ) )

wD

∫ A 2=−e−ξ∗wn( t−τ )∗sen (wD (t−τ ) )

wD

+ξ∗wn

wD∫

0

t


∫ A 2=−e−ξ∗wn( t−τ )∗sen (wD ( t−τ ) )

wD

+ξ∗wn

wD∫B

Ahora en la ecuación (***)


wD

−ξ∗wn

wD(−τ∗e−ξ∗wn( t−τ )∗sen (wD ( t−τ ) )

wD

+ξ∗wn

wD∫ A+ 1

wD∫B)− 1

wD(−e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )

wD

+ξ∗wn

wD∫B)


wD

+ξ∗wn∗τ∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD2 −

(ξ∗wn )2

wD2 ∫ A−

ξ∗wn

wD2 ∫B+

e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )wD

2 −ξ∗wn

wD2 ∫B

∫ A=wD

2

wD2+(ξ∗wn )2 ( τ∗e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )

wD

+ξ∗wn∗τ∗e−ξ∗wn ( t−τ )∗sen (wD (t−τ ) )

wD2 +

e−ξ∗wn ( t−τ )∗sen (wD (t−τ ) )wD

2 −2∗ξ∗wn

wD2 ∫B)

∫ A=wD

2


wD


wD2 +

e−ξ∗wn ( t−τ )∗sen (wD (t−τ ) )wD

2 )− 2∗ξ∗wn

wD2+(ξ∗wn )2

∫B

Ahora en la ecuación (**)

X (t )=1.308

wD∫ A−3.924

wD∫B ¿

X (t )=1.308

wD [ wD2

wD2+( ξ∗wn)2 ( τ∗e

−ξ∗wn (t−τ )∗cos (wD (t−τ ) )wD

+ξ∗wn∗τ∗e

−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )wD

2 +e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD2 )− 2∗ξ∗wn

wD2+(ξ∗wn )2∫B]−3.924

wD∫B

X (t )=1.308∗wD

2

wD (wD2+ (ξ∗wn )2 ) ( τ∗e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )

wD

+ξ∗wn∗τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )

wD2 +

e−ξ∗wn( t−τ )∗sen (wD ( t−τ ) )wD

2 )− 2∗1.308∗ξ∗wn

wD (wD2+ (ξ∗wn )2 )

∫B−3.924

wD∫B

X (t )=1.308∗wD

2


wD


wD2 +

e−ξ∗wn( t−τ )∗sen (wD (t−τ ) )wD

2 )−( 2∗1.308∗ξ∗wn

wD(wD2+( ξ∗wn )2 )

+3.924

wD )∫B ¿

Solucion delbloque B

∫B=∫0

t

e−ξ∗wn (t− τ )∗sen (wD (t−τ ) )∗dτ






wD


∫B=e−ξ∗wn ( t−τ )∗cos (wD ( t−τ ) )

wD

−ξ∗wn

wD∫

0

t

e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )∗dτ






wD



wD

−ξ∗wn

wD(−e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD

+ξ∗wn

wD∫0

t

e−ξ∗wn (t− τ )∗sen (wD (t−τ ) )∗dτ )∫B=

e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )wD

−ξ∗wn


wD

+ξ∗wn

wD∫B)

∫B=wD

2

wD2+ (ξ∗wn )2 [ e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )

wD

+ξ∗wn∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD2 ]

Finalmente en la ecuación (****)

X (t )=1.308∗wD

2


wD


wD2 +

e−ξ∗wn( t−τ )∗sen (wD (t−τ ) )wD

2 )−( 2∗1.308∗ξ∗wn

wD(wD2+( ξ∗wn )2 )

+3.924

wD )∫B ¿

reemplazando y despejando obtenemos larespuesta dinamica :

X (t )=[ 1.308∗e−ξ∗wn( t−τ )

(wD2+( ξ∗wn )2 ) (τ∗cos (wD (t−τ ) )+

ξ∗wn∗τ∗sen (wD (t−τ ) )wD

+sen (wD (t−τ ) )

wD)−(2∗1.308∗ξ∗wn

(wD2+ (ξ∗wn )2 )2

+ 3.924

(wD2+(ξ∗wn )2 ) )e−ξ∗wn (t− τ ) [cos (wD (t−τ ) )+

ξ∗wn∗sen (wD (t−τ ))wD

]]0

t

X (t )=[ 1.308∗e−ξ∗wn( t−τ )

(wD2+( ξ∗wn )2 ) (τ∗cos (wD ( t−τ ) )+

ξ∗wn∗τ∗sen (wD ( t−τ ) )wD


wD)−(2∗1.308∗ξ∗wn

(wD2+ (ξ∗wn )2 )2

+ 3.924

wD2+( ξ∗wn )2 )e−ξ∗wn (t− τ ) [cos (wD (t−τ ) )+


]]0

t

X (t )=1.308∗t

(wD2+( ξ∗wn )2 )

−(2∗1.308∗ξ∗wn

(wD2+(ξ∗wn )2 )2 + 3.924

wD2+( ξ∗wn )2 )−[ 1.308∗e−ξ∗wn∗t∗sen (wD∗t )

wD∗(wD2+(ξ∗wn)2)

−( 2∗1.308∗ξ∗wn

(wD2+( ξ∗wn )2 )2

+ 3.924

wD2+(ξ∗wn )2 )e−ξ∗wn∗t[cos ( wD∗t )+

ξ∗wn∗sen ( wD∗t )wD

]]wn=13.001 rad /s

ξ=5%

wD=12.984 rad /s

X (t )=1.308∗t

12.9842+(0.05∗13.001 )2−( 2∗1.308∗0.05∗13.001

(12.9842+ (0.05∗13.001 )2 )2+ 3.924

12.9842+(0.05∗13.001 )2 )−[ 1.308∗e−0.05∗13.001∗t∗sen ( wD∗t )12.984∗(12.9842+ (0.05∗13.001 )2)

−( 2∗1.308∗0.05∗13.001

(12.9842+ (0.05∗13.001 )2 )2 +3.924

12.9842+(0.05∗13.001 )2 )e−0.05∗13.001∗t [cos ( wD∗t )+0.05∗13.001∗sen (wD∗t )

12.984 ] ]X (t )=0.007739∗t−0.023277−0.000596∗e−ξ∗wn∗t∗sen (wD∗t )+0.023278∗e−ξ∗wn∗t [cos ( wD∗t )+0.050065∗sen ( wD∗t ) ]

X (t )=0.007739∗t−0.023278−0.001761∗e−0.65∗t∗sen (12.984∗t )+0.023278∗e−0.65∗t∗cos (12.984∗t )

0 0.5 1 1.5 2 2.5 3 3.5

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01 Desplazamiento Vs Tiempo

Tiempo (s)

Desp

laza

mie

nto

(m)

Xmax=−0.04154 m

.2. FASE ( II) : t d 1=3 s≤ t

Sabemos que:

E . D .M ⟹m∗X+C∗X+ K∗X=P(t )


X (t )=1

m∗wD∫

0

t

e−ξ∗wn(t−τ)∗P(t )∗sen (wD ( t−τ ) )∗dτ

Determinación de P1(τ ): por comparación de la ecuación (*)

P(t )=−m∗ ¨Xs (τ )=0

Reemplando tendremos :

X (t )=−1wD

∫0

t


X (τ )=0


X (t )=−1wD

∫0

3

e−ξ∗wn (t−τ )∗X ( τ )∗sen (wD ( t−τ ))∗dτ− 1wD

∫3

t

e−ξ∗wn (t−τ )∗X(τ )∗sen (wD ( t−τ ) )∗dτ

X (t )=[ 1.308∗e−ξ∗wn( t−τ )

(wD2+( ξ∗wn )2 ) (τ∗cos (wD ( t−τ ) )+

ξ∗wn∗τ∗sen (wD ( t−τ ) )wD

+sen (wD ( t−τ ) )

wD)−(2∗1.308∗ξ∗wn

(wD2+ (ξ∗wn )2 )2

+ 3.924

wD2+( ξ∗wn )2 )e−ξ∗wn (t− τ ) [cos (wD (t−τ ) )+


]]0

3

X (t )=e−0.65 (t−3) [0.0023217∗cos (12.984 ( t−3 ) )+0.001162∗sen (12.984 (t−3 ) )+.000596∗sen (12.984 ( t−3 ) )−0.023278∗cos (12.985 (t−3 ))−0.001165∗sen (12.984 ( t−3 ) )−0.000596∗sen (wD∗t )+0.023278∗cos (12.984∗t )+0.001165∗sen (12.984∗t ) ]

FALTAAAAAAAAAAA

X (t )=e−0.65 (t−3) [−0.000061∗cos (12.984 (t−3 ) )+0.000593∗sen (12.984 (t−3 ) ) ]+e−0.65 (t−3) [−0.000582∗sen ( wD∗t )+0.00018∗cos (12.984∗t ) ]

X (t )=e−0.65 (t−3) [−0.020956∗cos (12.984 ( t−3 ) )+0.023278∗cos (12.984∗t )+0.001165∗sen (12.984∗t ) ]

B. Para una Fuerza Puntual.



f c' =280

kg

cm2E=15000√ f c

'

Respuesta dinámica aplicando DUHAMEL, para un coeficiente de amortiguamiento:

ξ=5%

.1. FASE (I): 0 ≤ t ≤ td 1=3 s


X (t )=1

m∗wD∫

0

t

e−ξ∗wn(t−τ)∗P(t )∗sen (wD ( t−τ ) )∗dτ

Determinación de P1(τ ):

P(τ)=P0−P0

3∗τ

Gravedad:

g=9.81∗m

s2

Rigidez lateral (de la matriz de rigidez)

[ K ]=[ 620.322 ] Tn /m

Masa (de la matriz de masa)

[ m ]=3.67 Tn .s2

m

Frecuencia natural:

wn2= K

m⟹K=wn

2∗m

wn=√ KLm

=√ 620.322Tnm

3.67Tn . s2

m

=13.001rad

s

wn=13.001 rad /s

Frecuencia amortiguada:

wD=wn∗√1−ξ2

wD=13.001∗√1−0.052

wD=12.984 rad /s


X (t )=1

m∗wD∫

0

t

e−ξ∗wn(t−τ)∗P(τ )∗sen (wD ( t−τ ) )∗dτ

X (t )=1

m∗wD∫

0

t

e−ξ∗wn(t−τ)∗(−P0

3∗τ+P0)∗τ∗sen (wD ( t−τ ) )∗dτ

X (t )=−P0

3∗m∗wD∫0

t

τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )∗dτ+P0

m∗wD∫

0

t

e−ξ∗wn(t−τ)∗sen (wD ( t−τ ))∗dτ

∫ A=∫0

t

τ∗e−ξ∗wn (t−τ)∗sen (wD ( t−τ ) )∗dτ

∫B=∫0

t

e−ξ∗wn(t−τ )∗sen (wD ( t−τ ) )∗dτ

X (t )=−P0

3∗m∗wD∫ A+

P0

m∗wD∫B ¿

Solucion delbloque A Integración por partes :






A

B


wD

∫ A=( τ∗e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )wD

−∫0

t cos (wD (t−τ ) )wD

∗(e−ξ∗wn (t−τ )+ξ∗wn∗τ∗e−ξ∗wn (t−τ ) )∗dτ )∫ A=

τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )wD

−ξ∗wn

wD∫0

t

τ∗e−ξ∗wn (t−τ )∗cos (wD (t−τ ))∗dτ− 1wD

∫0

t

e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )∗dτ


wD

−ξ∗wn

wD∫ A 1− 1

wD∫ A 2(¿∗¿)


∫ A 1=∫0

t

τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )∗dτ







wD


wD

+ξ∗wn

wD∫

0

t

τ∗e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )∗dτ+ 1wD

∫0

t



wD

+ξ∗wn

wD∫ A+ 1

wD∫B


∫ A 2=∫0

t

τ∗e−ξ∗wn (t−τ )∗cos (wD ( t−τ ) )∗dτ






A2

A1


wD

∫ A 2=−τ∗e−ξ∗wn ( t−τ )∗sen (wD (t−τ ) )

wD

+ξ∗wn

wD∫

0

t



wD

+ξ∗wn

wD∫B

Ahora en la ecuación (***)


wD

−ξ∗wn

wD(−τ∗e−ξ∗wn( t−τ )∗sen (wD ( t−τ ) )

wD

+ξ∗wn

wD∫ A+ 1

wD∫B)− 1

wD(−τ∗e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )

wD

+ξ∗wn

wD∫B)


wD

+ξ∗wn∗τ∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD

−(ξ∗wn )2

wD2 ∫ A−

ξ∗wn

wD2 ∫B+

τ∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )wD

2 −ξ∗wn

wD2 ∫B

∫ A=wD

2

wD2+(ξ∗wn )2 ( τ∗e−ξ∗wn (t−τ )∗cos (wD ( t−τ ) )

wD


wD

+τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )

wD2 −

2∗ξ∗wn

wD2 ∫B)

∫ A=wD

2


wD


wD

+τ∗e−ξ∗wn ( t−τ )∗sen (wD (t−τ ) )

wD2 )− 2∗ξ∗wn

wD2+(ξ∗wn )2

∫B

Ahora en la ecuación (**)

X (t )=−P0

3∗m∗wD∫ A+

P0

m∗wD∫B ¿

X (t )=−P0

3∗m∗wD [ wD2

wD2+ (ξ∗wn )2 ( τ∗e

−ξ∗wn( t−τ )∗cos (wD (t−τ ) )wD

+ξ∗wn∗τ∗e

−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )wD

+τ∗e

−ξ∗wn (t−τ )∗sen (wD (t−τ ) )wD

2 )− 2∗ξ∗wn

wD2+( ξ∗wn )2

∫B]− P0

m∗wD∫B

X (t )=−P0∗wD

2

m∗wD (wD2+(ξ∗wn )2) ( τ∗e−ξ∗wn ( t−τ )∗cos (wD ( t−τ ) )

wD

+ξ∗wn∗τ∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ))

wD

+τ∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD2 )+ 2∗P0∗ξ∗wn

3∗m∗wD (wD2+(ξ∗wn )2 )

∫B−P0

m∗wD∫B

X (t )=−P0∗wD

2

3∗m∗wD (wD2+(ξ∗wn )2 ) ( τ∗e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )

wD

+ξ∗wn∗τ∗e−ξ∗wn( t−τ )∗sen (wD (t−τ ) )

wD

+τ∗e−ξ∗wn (t− τ )∗sen (wD ( t−τ ) )

wD2 )+( 2∗P0∗ξ∗wn

m∗wD(wD2+( ξ∗wn )2 )

−P0

m∗wD )∫B ¿

Solucion delbloque B

∫B=∫0

t

e−ξ∗wn (t− τ )∗sen (wD ( t−τ ) )∗dτ





v=cos (wD (t−τ ) )

wD


∫B=e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )

wD

−ξ∗wn

wD∫

0

t

e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )∗dτ






wD



wD

−ξ∗wn

wD(−e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD

+ξ∗wn

wD∫0

t

e−ξ∗wn (t− τ )∗sen (wD (t−τ ) )∗dτ )∫B=

e−ξ∗wn ( t−τ )∗cos (wD (t−τ ) )wD

−ξ∗wn


wD

+ξ∗wn

wD∫B)

∫B=wD

2

wD2+ (ξ∗wn )2 [ e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )

wD

+ξ∗wn∗e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )

wD2 ]

Finalmente en la ecuación (****)

X (t )=−P0∗wD

2

3∗m∗wD (wD2+(ξ∗wn )2 ) ( τ∗e−ξ∗wn (t−τ )∗cos (wD (t−τ ) )

wD

+ξ∗wn∗τ∗e−ξ∗wn( t−τ )∗sen (wD (t−τ ) )

wD2 +

τ∗e−ξ∗wn (t− τ )∗sen (wD ( t−τ ) )wD

2 )+( 2∗P0∗ξ∗wn

m∗wD(wD2+( ξ∗wn )2 )

−P0

m∗wD )∫B ¿

reemplazando y despejando obtenemos lar espuestadinamica :

X (t )=[ −P0∗e−ξ∗wn ( t−τ )

3∗m∗(wD2+(ξ∗wn )2 ) (τ∗cos (wD (t−τ ) )+



wD)+( 2∗P0∗ξ∗wn

3∗m∗(wD2+ (ξ∗wn )2 )2

−P0

m∗(wD2+( ξ∗wn )2) )e−ξ∗wn (t−τ ) [cos (wD (t−τ ) )+

ξ∗wn∗sen (wD ( t−τ ) )wD

]]0t

X (t )=[ −P0∗e−ξ∗wn ( t−τ )

3∗m∗(wD2+(ξ∗wn )2 ) (τ∗cos (wD (t−τ ) )+



wD)+( 2∗P0∗ξ∗wn

3∗m∗(wD2+ (ξ∗wn )2 )2

−P0

m∗(wD2+( ξ∗wn )2) )e−ξ∗wn (t−τ ) [cos (wD (t−τ ) )+

ξ∗wn∗sen (wD ( t−τ ) )wD

]]0t

X (t )=−P0∗t

3∗m∗(wD2+ (ξ∗wn )2 )

+( 2∗P0∗ξ∗wn

3∗m∗(wD2+( ξ∗wn )2)2 −

P0

m∗(wD2+(ξ∗wn )2 ) )−[−P0∗e−ξ∗wn∗t∗sen ( wD∗t )

3∗m∗wD∗(wD2+( ξ∗wn )2)

+( 2∗P0∗ξ∗wn

3∗m∗(wD2+(ξ∗wn )2 )2−

P0

m∗(wD2+( ξ∗wn )2 ))e−ξ∗wn∗t [cos ( wD∗t )+

ξ∗wn∗sen ( wD∗t )wD

]][ m ]=3.67 Tn .

s2

m

wn=13.001 rad /s

ξ=5%

wD=12.984 rad /s

X (t )=−4∗t

3∗3.67∗(12.9842+(0.05∗13.001 )2)+( 2∗4∗0.05∗13.001

3∗3.67∗(12.9842+(0.05∗13.001 )2 )2− 4

3.67∗(12.9842+(0.05∗13.001 )2 ) )−[ −4∗e−ξ∗wn∗t∗sen (wD∗t )3∗3.67∗12.984∗(12.9842+ (0.05∗13.001 )2 )

+( 2∗4∗ξ∗wn

3.67∗(12.9842+(0.05∗13.001 )2 )2− 4

3.67∗(12.9842+(0.05∗13.001 )2 ) )e−ξ∗wn∗t [cos (wD∗t )+0.05∗13.001∗sen ( wD∗t )

12.984 ]]X (t )=0.007739∗t−0.023277−0.007739∗e−ξ∗wn∗t∗sen ( wD∗t )+0.023277∗e−ξ∗wn∗t [cos ( wD∗t )+0.050065∗sen ( wD∗t ) ]

X (t )=0.007739∗t−0.023277−0.006573∗e−ξ∗wn∗t∗sen ( wD∗t )+0.023277∗e−ξ∗wn∗t∗cos ( wD∗t )

0 0.5 1 1.5 2 2.5 3 3.5

-0.05

-0.04

-0.03

-0.02

-0.01

0

0.01 Desplazamiento Vs Tiempo

Tiempo (s)

Desp

laza

mie

nto

(m)

Xmax=−0.04254 m

practica calificada n 04

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