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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.
Sección de la columna: 0.25x0.25 m2
Sección de la viga: 0.25x0.40 m2
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
e−ξ∗wn(t−τ)∗X (τ)∗sen (wD (t−τ ) )∗dτ
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
∫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=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
∫ A 1=−τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )
wD
+ξ∗wn
wD∫ A+ 1
wD∫B
Solucion delbloque A2
∫ A 2=∫0
t
e−ξ∗wn (t−τ )∗cos (wD ( t−τ ) )∗dτ
∫u∗dv=u∗v−∫ v∗du
u=e−ξ∗wn (t−τ )
du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ
dv=cos (wD ( t−τ ) )∗dτ
v=∫ cos (wD (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
e−ξ∗wn (t−τ )∗sen (wD (t−τ ) )∗dτ
∫ A 2=−e−ξ∗wn( t−τ )∗sen (wD ( t−τ ) )
wD
+ξ∗wn
wD∫B
Ahora en la ecuación (***)
∫ A=τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )
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)
∫ A=τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )
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
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+(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+ (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 )
+3.924
wD )∫B ¿
Solucion delbloque B
∫B=∫0
t
e−ξ∗wn (t− τ )∗sen (wD (t−τ ) )∗dτ
u=e−ξ∗wn (t−τ )
du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ
dv=sen (wD ( t−τ ) )∗dτ
v=∫ sen (wD (t−τ ))∗dτ
v=cos (wD ( t−τ ) )
wD
∫u∗dv=u∗v−∫ v∗du
∫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
u=e−ξ∗wn (t−τ )
du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ
dv=cos (wD ( t−τ ) )∗dτ
v=∫ cos (wD (t−τ ) )∗dτ
v=−sen (wD (t−τ ) )
wD
∫u∗dv=u∗v−∫ v∗du
∫B=e−ξ∗wn ( t−τ )∗cos (wD ( t−τ ) )
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(−e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )
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+ (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 )
+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
+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∗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 )
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 )=0
Reemplando tendremos :
X (t )=−1wD
∫0
t
e−ξ∗wn(t−τ)∗X (τ)∗sen (wD (t−τ ) )∗dτ
X (τ )=0
integración de la ecuación de la respuesta dinámica.
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 ) )+
ξ∗wn∗sen (wD (t−τ ))wD
]]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.
Sección de la columna: 0.25x0.25 m2
Sección de la viga: 0.25x0.40 m2
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
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 ):
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
integración de la ecuación de la respuesta dinámica.
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 :
∫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
A
B
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=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
∫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=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
∫ A 1=−τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )
wD
+ξ∗wn
wD∫ A+ 1
wD∫B
Solucion delbloque A2
∫ A 2=∫0
t
τ∗e−ξ∗wn (t−τ )∗cos (wD ( t−τ ) )∗dτ
∫u∗dv=u∗v−∫ v∗du
u=e−ξ∗wn (t−τ )
du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ
dv=cos (wD ( t−τ ) )∗dτ
v=∫ cos (wD (t−τ ) )∗dτ
A2
A1
v=−sen (wD ( t−τ ) )
wD
∫ A 2=−τ∗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
∫ A 2=−τ∗e−ξ∗wn ( t−τ )∗sen (wD ( t−τ ) )
wD
+ξ∗wn
wD∫B
Ahora en la ecuación (***)
∫ A=τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )
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)
∫ A=τ∗e−ξ∗wn (t− τ )∗cos (wD (t−τ ) )
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
+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∗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 ) )
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
u=e−ξ∗wn (t−τ )
du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ
dv=sen (wD ( t−τ ) )∗dτ
v=∫ sen (wD (t−τ ))∗dτ
v=cos (wD (t−τ ) )
wD
∫u∗dv=u∗v−∫ v∗du
∫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
u=e−ξ∗wn (t−τ )
du=ξ∗wn∗e−ξ∗wn (t−τ ) dτ
dv=cos (wD ( t−τ ) )∗dτ
v=∫ cos (wD (t−τ ) )∗dτ
v=−sen (wD ( t−τ ) )
wD
∫u∗dv=u∗v−∫ v∗du
∫B=e−ξ∗wn ( t−τ )∗cos (wD ( t−τ ) )
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(−e−ξ∗wn (t−τ )∗sen (wD ( t−τ ) )
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 ) )+
ξ∗wn∗τ∗sen (wD (t−τ ) )wD
+sen (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 ) )+
ξ∗wn∗τ∗sen (wD (t−τ ) )wD
+sen (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