Dinamica Estructural Ecuacion de Movimiento

Fuerza 150.00

100.00

F(K)

50.00

0.00

-50.00

-100.00

-150.00 0

0.05

0.1

0.15 t(seg)

0.2

0.25

0.3

0.35

15000

10000

5000

Column B Column C Column D

0

-5000

-10000

-15000 0

0.1

0.2

0.3

0.4

0.5

0.6

Desplazamiento 15

10

x(cm)

5

0

-5

-10

-15 -0.1

0

0.1

0.2

0.3 t(seg)

0.4

0.5

0.6

Velocidad 500 400 300

velocidad(cm/s)

200 100 0 -100 -200 -300 -400 -500 -0.1

0

0.1

0.2

0.3 t(seg)

0.4

0.5

0.6

Aceleración 15000

10000

Aceleración(cm/seg2)

5000

0

-5000

-10000

-15000 -0.1

0

0.1

0.2 t(seg)

0.3

0.4

0.5

0.6

Problema 1

Determine el comportamiento dinámico de la torre de la figura, sujeta a la fuerza dinámica sinusoidal F(t) aplicada por 0.30 seg. Sin amortiguación. Determine los máximos: Desplazamiento. Velocidad. Aceleración. Utilice 0.01 como intervalo de tiempo. Datos Fo = w*= W= k= g=

100 k 30 rad/seg 38.6 [K] 100 K/in 386 [in/seg2]

W

k

Idealización

Diagrama de cuerpo libre

F(t)

W

k

Fuerzas

F I =m x¨ t =0.1 x¨ t 

Fs=k∗x t =100xt  F  t = F o sin w  t =100sin 30t  m=

W g

Ecuación de movimiento

m x¨ t K xt =F t 

=

38.6 [K] 386 [in/seg2]

=

0.1

[seg2*K/in]

0 .1 x¨  t 100x t =100sin30t  Ecuación característica

0 . 1S2100=0





k 100 K /in  S 1,2=± − =±ω=± =  1000 rad / seg  m 0 . 1  seg 2∗K /in  Solución complementaria

x  t c =A sin ωt B cos ωt 

ω=  1 0 0 0ra d/ seg Cálculo solución particular

F (t ) = Fo sin( ω t ) = 100 sin( 30t )

x t  p =C 1 sin ω  t C 2 cos ω t

x˙ t  p =C 1 ω  cos ω t −C 2 ω  sin ω t  2 2 x¨  t  p =−C 1 ω  sin ω  t −C 2 ω  cos ω t

x t  p =C 1 sin30t C 2 cos30t  x˙ (t ) p = 30 C1 cos(30t ) − 30C 2 sin( 30t )

I

˙x˙(t ) p = −900 C1 sin( 30 t ) − 900 C 2 cos(30 t ) Reemplazando I en la ecuación de movimiento:

m x¨  t  p Kx  t  p = F  t  0 . 1 x¨  t  p100 x  t  p=100sin 30t  0 . 1 x¨ [−900 C 1 sin  30t −900 C 2 cos 30 t  ]100 [ C 1 sin  30 t C 2 cos  30 t  ] =100 sin  30 t  De esta ecuación:

C 1=1 0 C 2=0 x(t ) p = 10 sin( 30t )

x(t ) p = 10 sin( 30t ) Solución general

x  t = x t c  x  t  p= A sin ωt  B cos ωt 10sin  30 t 

x  t = Asin  1000t B cos  1000t 10sin30t  Grafica de la fuerza aplicada durante 0,3 seg

F (t ) = 100 sin( 30t ) t[seg] 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.3

F[K] 0.00 29.55 56.46 78.33 93.20 99.75 97.38 86.32 67.55 42.74 14.11 -15.77 -44.25 -68.78 -87.16 -97.75 -99.62 -92.58 -77.28 -55.07 -27.94 1.68 31.15 57.84 79.37 93.80 99.85 96.99 85.46 66.30 41.21

Cálculo de A y B, bajo condiciones iniciales de reposo.

x0=0 x˙ 0 =0

x  t =Asin  1000t B cos  1000t 10sin30t 

x˙ t =  1000Acos  1000t −  1000B sin  1000t 300cos30t  x¨ t =−1000A sin  1000t −1000B cos  1000t −9000sin30t  Reemplazando las condiciones iniciales:

x (0) = A sin( 1000 * 0) + B cos( 1000 * 0) + 10 sin( 30 * 0) = 0

x˙ t =  1000Acos  1000∗0−  1000B sin  1000∗0300cos30∗0=0 B=0 300 = − 90 1000

A=−

Comportamiento bajo acción de la fuerza F(t):

x(t ) = − 90 sin( 1000 t ) + 10 sin( 30 t ) x˙ (t ) = −300 cos( 1000 t ) + 300 cos( 30 t ) ˙x˙(t ) = 1000 90 sin( 1000 t ) − 9000 sin( 30 t ) En el sistema métrico:

[

]

x (t ) = − 90 sin( 1000 t ) + 10 sin( 30 t )) * 2.54 = 254 sin( 30 t ) − 2.54 90 sin( 1000 t )

x˙  t =−762cos  1000t 762cos 30t  x¨ t =2540 90sin  1000t −22860sin 30t  Comportamiento libre t > 0.3 seg

x(0.3) = − 90 sin( 1000 * 0.3) + 10 sin( 30 * 0.3) =

x˙ (0.3) = −300 cos( 1000 * 0.3) + 300 cos( 30 * 0.3) =

4.710 (in)

x˙ (0.3) = −300 cos( 1000 * 0.3) + 300 cos( 30 * 0.3) =

26.083 (in/seg)

Ecuación de movimiento t>0,3 seg

0.1˙x˙(t ) + 100 x(t ) = 0 Solución:

x  t = A sin  ωt B cos ωt= A sin  1000t  B cos  1000t  Para las condiciones iniciales de:

x0. 3=4. 7 1 0 x˙ 0 .3=2 6.0 8 3 x (0.3) = A sin( 1000 * 0.30 ) + B cos( 1000 * 0.30 ) = 4.710 x˙ (0.3) = 1000 A cos( 1000 * 0.3) − 1000 B sin( 1000 * 0.3) = 26 .083 -0.06 A -31.56 A

A= B= Comportamiento a vibración libre

-1 B 1.96 B

-1.12 -4.65

x t =−1.115sin  1000t −4.649cos  1000t  x˙ t =−35. 269cos  1000t 147.024sin  1000t  ˙x˙(t ) = 1115 sin( 1000 t ) + 4649 cos( 1000 t ) En el sistema métrico

x t =−2.833sin  1000t −11.809cos  1000t  x˙ t =−89.583cos  1000t 373. 440sin  1000t  x¨  t =2833sin  1000t 11809cos  1000t  Gráfica de: desplazamiento, velocidad, aceleración t[seg] Desplazamiento Velocidad Aceleración [cm] [cm/seg] [cm/seg2] 0 0.00 0.00 0.00 0.01 0.01 3.75 738.04 0.02 0.10 14.29 1336.40 0.03 0.31 29.61 1675.19

= =

4.710 26.083

0.04 0.05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19 0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 0.3 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.4 0.41 0.42 0.43 0.44 0.45 0.46 0.47 0.48 0.49 0.5 0.51 0.52 0.53 0.54

0.70 1.24 1.91 2.64 3.32 3.84 4.08 3.96 3.40 2.40 0.99 -0.74 -2.64 -4.53 -6.20 -7.45 -8.09 -8.01 -7.12 -5.45 -3.10 -0.24 2.88 5.96 8.70 10.78 11.96 12.02 10.89 8.67 5.60 1.97 -1.85 -5.49 -8.59 -10.83 -12.00 -11.98 -10.77 -8.50 -5.38 -1.72 2.10 5.72 8.77 10.94 12.04 11.94 10.66 8.31 5.15

46.65 61.78 71.32 72.08 61.90 40.06 7.46 -33.30 -78.17 -122.01 -159.19 -184.27 -192.65 -181.27 -149.08 -97.35 -29.70 48.10 128.62 203.51 264.35 303.52 315.20 296.03 245.67 167.05 66.25 -54.67 -170.17 -268.80 -340.77 -378.94 -379.54 -342.49 -271.49 -173.56 -58.41 62.52 177.26 274.41 344.36 380.15 378.24 338.83 265.81 166.43 50.55 -70.35 -184.27 -279.91 -347.80

1671.60 1292.53 560.85 -445.56 -1602.01 -2751.56 -3724.41 -4359.65 -4527.06 -4145.96 -3198.96 -1738.62 114.08 2179.61 4237.28 6048.62 7384.03 8050.03 7913.93 6922.92 5115.21 2621.66 -342.92 -3498.17 -6524.73 -9096.52 -10915.38 -12020.56 -10886.88 -8673.55 -5600.08 -1971.25 1853.07 5493.63 8589.38 10833.32 12002.94 11982.23 10773.24 8495.88 5375.99 1722.97 -2100.92 -5716.47 -8765.11 -10944.52 -12038.58 -11938.77 -10655.00 -8314.58 -5149.61

0.55 0.56 0.57 0.58 0.59 0.6 Del gráfico

1.47 -2.35 -5.94 -8.94 -11.05 -12.07

-381.20 -376.79 -335.02 -260.02 -159.24 -42.67

-1473.96 2347.87 5936.86 8937.09 11051.04 12069.06

Desplazamiento Velocidad Aceleración [cm] [cm/seg] [cm/seg2] Máximos

12.039

380.150

12069.064

t  ] =100 sin  30 t 

os30∗0=0

90 sin( 1000 t )

t=

0.3

seg.

w=

31.62

-35.27 -147.02

factor

2.54

-2.83 -11.81

-89.58 -373.44

4000

3000

2000

1000

0

-1000

-2000

-3000 0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

Desplazamiento [cm] 4

3

Desplazamiento (cm)

2

1

0

-1

-2

-3

-4 0

0.1

0.2

0.3

0.4 t(seg)

0.5

0.6

0.7

Velocidad [cm/seg] 100

80

60

velocidad(cm/seg)

40

20

0

-20

-40

-60

-80 0

0.1

0.2

0.3

0.4 t(seg)

0.5

0.6

0.7

Aceleración [cm/seg2] 4000

3000

aceleración(cm/seg2)

2000

1000

0

-1000

-2000

-3000 0

0.1

0.2

0.3

0.4 t[seg]

0.5

0.6

0.7

Problema 2

Determine el comportamiento de la torre mostrada en la figura, sujeta a una carga dinámica impulsiva de duración 0,48 seg. Asuma la amortiguación igual al 10% de la amortiguación crítica. Utilice 0,08 seg. Como intervalo de tiempo. Encuentre el máximo: desplazamiento y velocidad.

Datos

W= k= g=

38.6 100 386

0 0.16 0.32 0.48 0.64

[K] K/in [in/seg2]

0 120 0 -120 0

120 80

Idealización

Diagrama de cuerpo libre

F(t)

F(t) [K]

40

W

k

0 -40 -80 -120 0

0.16

0.48

0.64

t(seg)

Fuerzas m=

0.32

W g

=

38.6 [K] 386 [in/seg2]

F I =m x¨ t =0.1 x¨ t 

Fs=k∗x t =100xt 

ω=  1 00 0ra d/ seg C crit = 2mω C = 0.1C crit = 0.1 * 2mω = 0.2 * 0.1 * 1000 = 0.2 10

F D =C∗ x˙  t =0 . 2  10 x˙  t 

=

0.1 [seg2*K/in]

Ecuación de movimiento

m x¨ t C x˙ t Kx t =F t 

0 .1 x¨  t 0 .2  1 0 x˙ t 1 0 0x t =F t  Ecuación característica

0 . 1S2 0. 2  10 S100=0

(

)

S1, 2 = −ωζ ± ω 1 − ζ 2 i = − 1000 * 0.1 ±

(

)

1000 * 1 − 0.12 i = − 10 ± 990 i

Solución complementaria

x  t c =e−ωζ t  A sin  ω D t  B cos  ω D t   x(t ) c = e −

10t

( A sin(

990 t ) + B cos( 990 t )

Cálculo solución particular

)

0 < t < 0,16 seg

120 t = 750 t[ K ] 0.16 x(t ) p = C1t + C 2

F (t ) =

I

x˙ (t ) p = C1

˙x˙(t ) p = 0 Reemplazando I en la ecuación de movimiento:

m x¨  t  p C x˙  t  p  Kx  t  p = F  t 

De esta ecuación:

C1 =

0 . 1 x¨  t  p 0 . 2  10 x˙  t  p 100 x  t  p =750 t 0 . 1 0 0 . 2  10  C 1 100  C 1 tC 2 =750 t

7.5

C 2=−0 . 015  10=

-0.0474

x  t  p =7 . 5t −0 . 015  10

Solución general

x(t ) = x(t ) c + x(t ) p = e −

10t

( A sin(

)

990 t ) + B cos( 990 t ) + 7.5t − 0.015 10

Cálculo de A y B, bajo condiciones iniciales de reposo.

x0=0 x˙ 0 =0 x(t ) = e −

10t

x˙ (t ) = e −

10t

( A sin( 990 t ) + B cos( [(− 10 A − 990 B ) sin(

)

990 t ) + 7.5t − 0.015 10

(

)

]

990 t ) + − 10 B + 990 A cos( 990 t ) + 7.5

Reemplazando las condiciones iniciales:

x(0) = e − x˙ (0) = e −

10 *0

10 *0

( A sin(

[( −

)

990 * 0) + B cos( 990 * 0) + 7.5 * 0 − 0.015 10 = 0

)

(

)

]

10 A − 990 B sin( 990 * 0) + − 10 B + 990 A cos( 990 * 0) + 7.5 = 0

B=0 . 015  10 7. 5−0 . 015  10∗10 7. 35 A=− =− =−0 . 2336  990  990 Comportamiento bajo acción de la fuerza F(t): Derivando

x(t ) = e −

10t

(− 0.2336 sin(

F (t ) = 750 t[ K ]

)

990 t ) + 0.015 10 cos( 990 t ) + 7.5t − 0.015 10

x˙  t =e− 10 t [ −0. 754  sin  990 t  −7 . 5  cos  990 t  ]7. 5 ˙x˙(t ) = e −

10t

t

[( 238 .366 ) sin(

Desp(pulg) 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16

990 t )

]

Vel(pulg/s) 0 0.010 0.070 0.207 0.417 0.669 0.921 1.136 1.295

Acelerac(pulg/s2) 0 0 1.39 131.7 4.84 199.89 8.84 187.34 11.88 108.21 12.97 -0.84 11.94 -96.63 9.42 -145.92 6.5 -136.33

Cálculo solución particular

F (t ) = 2 *120 −

0,16
120 t = 240 − 750 (t )[ K ] 0.16

x(t ) p = C1t + C 2 x˙ (t ) p = C1

I

˙x˙(t ) p = 0 Reemplazando I en la ecuación de movimiento:

m x¨  t  p C x˙  t  p  Kx  t  p = F  t  0 . 1 x¨  t  p0 . 2  10 x˙  t  p 100 x  t  p=240−750 t 0 . 1 0 0 . 2  10  C 1 100  C 1 tC 2 =240 −750 t 100 C 1 =−750 0 . 2  10 C 1 100 C 2 =240 De esta ecuación: -7.5 C = 1

C 2=

240−0.2∗ 10∗−7.5 = 100

2.4474

x  t  p =−7 . 5t 2 . 4474 Solución general

x(t ) = x(t ) c + x(t ) p = e −

10t

( A sin(

)

990 t ) + B cos( 990 t ) − 7.5t + 2.4474

Cálculo de A y B, bajo condiciones iniciales conocidas

x (0.16 ) = 1.2952 x˙ (0.16 ) = 6.5006 x (t ) = e −

10t

x˙ (t ) = e −

10t

( A sin( 990 t ) + B cos( [(− 10 A − 990 B ) sin(

Reemplazando las condiciones iniciales:

x(0.16) = e −

10 *0.16

( A sin(

)

990 t ) − 7.5t + 2.4474

(

)

]

990 t ) + − 10 B + 990 A cos( 990 t ) − 7.5

)

990 * 0.16) + B cos( 990 * 0.16) − 7.5 * 0.16 + 2.4474 = 1.2952

x(0.16) = e −

x˙ (0.16) = e −

10 *0.16

1 0*0.1 6

[( −

( A sin(

)

990 * 0.16) + B cos( 990 * 0.16) − 7.5 * 0.16 + 2.4474 = 1.2952

)

(

)

− 0.572 A + 0.1907 B = 0.047822 7.8103 A + 17 .393 B = 14 Del sistema

A = 0.1609 B = 0.7333

Comportamiento bajo acción de la fuerza F(t): Derivando

x(t ) = e −

10t

(0.1609 sin(

F (t ) = 240 − 750 (t )[ K ]

)

990 t ) + 0.7333 cos( 990 t ) + 7.5t − 0.015 10

x˙  t =e− 10 t [ −23 . 580  sin   990 t   2. 7426  cos   990 t  ] 7. 5 x¨  t =e− 10 t [ −11. 726  sin   990 t  −750 . 606  cos   990 t  ]

t

Desp(pulg) 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48

]

10A − 990B sin( 990* 0.16) + − 10B + 990A cos( 990* 0.16) − 7.5 = 6.5006

1.295 1.382 1.338 1.139 0.822 0.461 0.139 -0.087 -0.201 -0.228 -0.225 -0.250 -0.345 -0.522 -0.762 -1.024 -1.265

Vel(pulg/s) Acelerac(pulg/s2) 6.510 -136.466 1.51 -341.994 -6.164 -398.828 -13.415 -303.982 -17.66 -109.945 -17.701 100.989 -14.03 250.162 -8.430 290.708 -3.152 220.798 -0.08 78.961 -0.072 -74.716 -2.76 -182.983 -6.855 -211.894 -10.695 -160.371 -12.92 -56.695 -12.908 55.264 -10.93 133.839

Comportamiento a vibración libre t > 0,48 seg

x0.16=1.2952 x˙ 0 .16=6.5006

Ecuación de movimiento t>0,48 seg

0.1˙x˙(t ) + 0.2 10 x˙ (t ) + 100 x(t ) = 0

Solución:

x(t ) c = e −

10t

( A sin(

990 t ) + B cos( 990 t )

)

Para las condiciones iniciales de:

x 0. 48=−1. 265 x˙  0 . 48=−10. 934 x(0.48 ) = e − x˙ (0.48) = e − 0.1247 -6.0659

10 *0.48

10*0.48

[( −

( A sin(

)

990 * 0.48 ) + B cos( 990 * 0.48 ) = −1.265

)

(

)

]

10A − 990B sin( 990 * 0.48) + − 10B + 990A cos( 990 * 0.48) = − 10.934 A A

-0.1803 -3.3529

Del sistema: A= B=

-1.501462 5.977528

B B

= =

-1.265 -10.934

Comportamiento a vibración libre Derivando

x  t =e− 10 t [ −1. 5015  sin   990 t 5 . 97752 cos  990 t  ] x˙ (t ) = e −

10t

x˙ (t ) = e −

10t

t

(− 183 .33 sin( 990 t ) + −66 .145 cos( (2660 .9 sin( 990 t ) + −5559 .19 cos(

Desp(pulg) 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72

) 990 t ) ) 990 t )

Vel(pulg/s) Acelerac(pulg/s2) -1.265 -10.934 1333.839 -1.222 14.516 1130.187 -0.740 31.668 540.045 -0.047 35.274 -176.219 0.581 25.634 -743.344 0.923 7.824 -972.967 0.890 -10.712 -821.750 0.536 -23.154 -389.893 0.030 -25.704 132.331 -0.427 -18.610 544.419 -0.674 -5.597 709.713 -0.647 7.904 597.471 -0.389 16.928 281.456

Gráfica de: desplazamiento, velocidad, aceleración Con un intervalo de 0,02 En el sistema métrico t[seg] Desplazamiento Velocidad Aceleración [cm] [cm/seg] [cm/seg2] 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72 Del gráfico

0 0.06 0.18 0.53 1.06 1.70 2.34 2.89 3.29 3.51 3.40 2.89 2.09 1.17 0.35 -0.22 -0.51 -0.58 -0.57 -0.63 -0.88 -1.33 -1.93 -2.60 -3.21 -3.10 -1.88 -0.12 1.48 2.35 2.26 1.36 0.08 -1.08 -1.71 -1.64 -0.99 Desplazamiento [cm]

Máximos

3.511

0 3.54 12.29 22.46 30.18 32.94 30.33 23.92 16.51 3.85 -15.66 -34.07 -44.84 -44.96 -35.65 -21.41 -8.01 -0.21 -0.18 -7.02 -17.41 -27.17 -32.81 -32.79 -27.77 36.87 80.44 89.60 65.11 19.87 -27.21 -58.81 -65.29 -47.27 -14.22 20.08 43.00 Velocidad [cm/seg] 89.596

0 334.51 507.72 475.85 274.86 -2.13 -245.44 -370.64 -346.29 -868.66 -1013.02 -772.12 -279.26 256.51 635.41 738.40 560.83 200.56 -189.78 -464.78 -538.21 -407.34 -144.01 140.37 339.95 2870.67 1371.71 -447.60 -1888.09 -2471.34 -2087.25 -990.33 336.12 1382.83 1802.67 1517.58 714.90 Aceleración [cm/seg2] 2870.675

31.62

0.63

31.46

750

-0.75 0.0474

-7.5000

-0.2336

238.37

0

750

a

0.16

b

-23.58

2.74

-11.73

-750.61

0.7333

t=

0.48

seg. w=

#DIV/0! #DIV/0!

0.12 -6.07

-0.18 -3.35

#REF! #REF!

-3.81 15.18

-1.265 -10.934

2.22 -4.01

-0.12 -0.08

-183.33

-66.14

2660.95

-5559.19

factor

2.54

t[seg] 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 0.36 0.38 0.4 0.42 0.44 0.46 0.48 0.5 0.52 0.54 0.56 0.58 0.6 0.62 0.64 0.66 0.68 0.7 0.72

Desplazami Velocidad Aceleración ento [cm] [cm/seg] [cm/seg2] 0 0.02 0.07 0.21 0.42 0.67 0.92 1.14 1.3 1.38 1.34 1.14 0.82 0.46 0.14 -0.09 -0.2 -0.23 -0.22 -0.25 -0.34 -0.52 -0.76 -1.02 -1.265 -1.222 -0.74 -0.05 0.58 0.92 0.89 0.54 0.03 -0.43 -0.67 -0.65 -0.39

0 1.39 4.84 8.84 11.88 12.97 11.94 9.42 6.5 1.51 -6.16 -13.42 -17.66 -17.7 -14.03 -8.43 -3.15 -0.08 -0.07 -2.76 -6.85 -10.7 -12.92 -12.91 -10.93 14.52 31.67 35.27 25.63 7.82 -10.71 -23.15 -25.7 -18.61 -5.6 7.9 16.93

0 131.7 199.89 187.34 108.21 -0.84 -96.63 -145.92 -136.33 -341.99 -398.83 -303.98 -109.95 100.99 250.16 290.71 220.8 78.96 -74.72 -182.98 -211.89 -160.37 -56.7 55.26 133.84 1130.19 540.04 -176.22 -743.34 -972.97 -821.75 -389.89 132.33 544.42 709.71 597.47 281.46