Dinamica Estructural Ecuacion de Movimiento

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

-150.00

-100.00

-50.00

0.00

50.00

100.00

150.00

Fuerza

t(seg)

F(K

)

0 0.1 0.2 0.3 0.4 0.5 0.6

-15000

-10000

-5000

0

5000

10000

15000

Column B

Column C

Column D

-0.2 -5.55111512312578E-17 0.2 0.4 0.6

-15

-10

-5

0

5

10

15

Desplazamiento

t(seg)

x(cm

)

-0.2 -5.55111512312578E-17 0.2 0.4 0.6

-500

-400

-300

-200

-100

0

100

200

300

400

500

Velocidad

t(seg)

velo

cid

ad(c

m/s

)

-0.2 -5.55111512312578E-17 0.2 0.4 0.6

-15000

-10000

-5000

0

5000

10000

15000

Aceleración

t(seg)

Ace

lera

ció

n(c

m/s

eg2)

Problema 1

DatosFo = 100 kw*= 30 rad/segW = 38.6 [K]k = 100 K/in

g = 386

Idealización Diagrama de cuerpo libre

Fuerzas

m =W

=38.6 [K]

= 0.1g 386

Ecuación de movimiento

[in/seg2]

[seg2*K/in][in/seg2]

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.

W

k

WF(t)

k

m x( t )+Kx ( t )=F ( t )

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

Fs=k∗x ( t )=100x ( t )

F I=m x (t )=0 .1 x ( t )

Ecuación característica

Solución complementaria

Cálculo solución particular

I

De esta ecuación:

Reemplazando I en la ecuación de movimiento:

0 .1 x ( t )+100 x ( t )=100sin(30 t )

0 .1S2+100=0

S1,2=±√− km

=±ω=±√100(K /in)0 .1 (seg2∗K / in)

=√1000(rad / seg )

x ( t )c=A sin(ωt )+B cos(ωt )

ω=√1000(rad / seg )

F ( t )=Fo sin(ω t )=100sin(30 t )

x ( t )p=C1 sin(ω t )+C2 cos(ω t )

x ( t )p=C1 sin(30 t )+C2cos (30 t )

x ( t )p=C1 ωcos(ω t )−C2 ωsin(ω t )

x ( t )p=−C1 ω2sin (ω t )−C2ω

2 cos(ω t )

x ( t )p=−900C1sin(30 t )−900C2cos (30t )

x ( t )p=30C1 cos(30 t )−30C2sin(30 t )

m x( t )p+Kx( t )p=F ( t )

0 .1 x ( t )p+100 x( t ) p=100sin(30 t )

0 .1 x [−900C1sin (30t )−900C2cos(30 t )]+100 [C1sin(30 t )+C2cos(30 t )]=100sin(30t )

C1=10C2=0

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

Solución general

Grafica de la fuerza aplicada durante 0,3 seg

t[seg] F[K]0 0.00

0.01 29.550.02 56.460.03 78.330.04 93.200.05 99.750.06 97.380.07 86.320.08 67.550.09 42.74

0.1 14.110.11 -15.770.12 -44.250.13 -68.780.14 -87.160.15 -97.750.16 -99.620.17 -92.580.18 -77.280.19 -55.07

0.2 -27.940.21 1.680.22 31.150.23 57.840.24 79.370.25 93.800.26 99.850.27 96.990.28 85.460.29 66.30

0.3 41.21

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

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

x ( t )=x ( t )c+x ( t )p=A sin (ωt )+B cos(ωt )+10sin(30 t )

x ( t )=A sin (√1000 t )+B cos(√1000 t )+10sin(30 t )

F ( t )=100sin(30 t )

Reemplazando las condiciones iniciales:

Comportamiento bajo acción de la fuerza F(t):

En el sistema métrico:

Comportamiento libre t > 0.3 seg

4.710 (in)

x (0 )=0x (0 )=0

x ( t )=A sin (√1000 t )+B cos(√1000 t )+10sin(30 t )x ( t )=√1000 A cos (√1000 t )−√1000 B sin(√1000 t )+300cos(30 t )x ( t )=−1000 A sin( √1000 t )−1000B cos (√1000 t )−9000sin(30 t )

x (0)=A sin(√1000∗0)+B cos (√1000∗0 )+10sin(30∗0 )=0x ( t )=√1000 A cos (√1000∗0 )−√1000B sin(√1000∗0 )+300cos(30∗0 )=0

B=0

A=−300√1000

=−√90

x ( t )=−300cos (√1000t )+300cos (30 t )x ( t )=1000√90sin(√1000 t )−9000sin (30t )

x ( t )=−√90sin (√1000 t )+10sin(30 t )

x (0 .3 )=−√90sin(√1000∗0 .3 )+10sin (30∗0 .3 )=

x (0 .3 )=−300cos(√1000∗0.3 )+300cos(30∗0 .3)=

x ( t )=−762cos (√1000 t )+762cos(30 t )x ( t )=2540√90sin(√1000 t )−22860sin (30 t )

x ( t )=[−√90sin (√1000 t )+10sin(30 t )) ]∗2 .54=254 sin(30 t )−2 .54 √90sin(√1000 t )

26.083 (in/seg)

Ecuación de movimiento t>0,3 seg

Solución:

Para las condiciones iniciales de:

-0.062015200228 A -0.99807521 B = 4.710-31.56190924106 A 1.961092823 B = 26.083

A = -1.1153059B = -4.6492957

Comportamiento a vibración libre

En el sistema métrico

Gráfica de: desplazamiento, velocidad, aceleración

t[seg]

0 0.00 0.00 0.000.01 0.01 3.75 738.040.02 0.10 14.29 1336.400.03 0.31 29.61 1675.19

Desplazamiento [cm]

Velocidad [cm/seg]

Aceleración [cm/seg2]

x (0 .3 )=−300cos(√1000∗0.3 )+300cos(30∗0 .3)=

0 .1 x ( t )+100 x ( t )=0

x ( t )=A sin (ωt )+B cos(ωt )=A sin(√1000 t )+B cos (√1000 t )

x (0 .3 )=4 .710x (0 .3 )=26 .083

x (0 .3 )=A sin(√1000∗0 .30 )+B cos (√1000∗0 .30 )=4 .710x (0 .3 )=√1000 A cos (√1000∗0 .3 )−√1000B sin (√1000∗0 .3)=26 .083

x ( t )=−1 .115sin(√1000 t )−4 .649cos(√1000 t )x ( t )=−35 .269cos(√1000 t )+147 .024sin( √1000 t )x ( t )=1115 sin(√1000 t )+4649cos( √1000 t )

x ( t )=−2 .833sin(√1000 t )−11.809cos( √1000 t )x ( t )=−89.583cos( √1000 t )+373 .440sin(√1000 t )x ( t )=2833sin(√1000 t )+11809cos (√1000 t )

0.04 0.70 46.65 1671.600.05 1.24 61.78 1292.530.06 1.91 71.32 560.850.07 2.64 72.08 -445.560.08 3.32 61.90 -1602.010.09 3.84 40.06 -2751.56

0.1 4.08 7.46 -3724.410.11 3.96 -33.30 -4359.650.12 3.40 -78.17 -4527.060.13 2.40 -122.01 -4145.960.14 0.99 -159.19 -3198.960.15 -0.74 -184.27 -1738.620.16 -2.64 -192.65 114.080.17 -4.53 -181.27 2179.610.18 -6.20 -149.08 4237.280.19 -7.45 -97.35 6048.62

0.2 -8.09 -29.70 7384.030.21 -8.01 48.10 8050.030.22 -7.12 128.62 7913.930.23 -5.45 203.51 6922.920.24 -3.10 264.35 5115.210.25 -0.24 303.52 2621.660.26 2.88 315.20 -342.920.27 5.96 296.03 -3498.170.28 8.70 245.67 -6524.730.29 10.78 167.05 -9096.52

0.3 11.96 66.25 -10915.380.31 12.02 -54.67 -12020.560.32 10.89 -170.17 -10886.880.33 8.67 -268.80 -8673.550.34 5.60 -340.77 -5600.080.35 1.97 -378.94 -1971.250.36 -1.85 -379.54 1853.070.37 -5.49 -342.49 5493.630.38 -8.59 -271.49 8589.380.39 -10.83 -173.56 10833.32

0.4 -12.00 -58.41 12002.940.41 -11.98 62.52 11982.230.42 -10.77 177.26 10773.240.43 -8.50 274.41 8495.880.44 -5.38 344.36 5375.990.45 -1.72 380.15 1722.970.46 2.10 378.24 -2100.920.47 5.72 338.83 -5716.470.48 8.77 265.81 -8765.110.49 10.94 166.43 -10944.52

0.5 12.04 50.55 -12038.580.51 11.94 -70.35 -11938.770.52 10.66 -184.27 -10655.000.53 8.31 -279.91 -8314.580.54 5.15 -347.80 -5149.61

0.55 1.47 -381.20 -1473.960.56 -2.35 -376.79 2347.870.57 -5.94 -335.02 5936.860.58 -8.94 -260.02 8937.090.59 -11.05 -159.24 11051.04

0.6 -12.07 -42.67 12069.06Del gráfico

Máximos 12.039 380.150 12069.064

Desplazamiento [cm]

Velocidad [cm/seg]


0 .1 x [−900C1sin (30t )−900C2cos(30 t )]+100 [C1sin(30 t )+C2cos(30 t )]=100sin(30t )

t = 0.3 seg. w = 31.6227766

x ( t )=√1000 A cos (√1000∗0 )−√1000B sin(√1000∗0 )+300cos(30∗0 )=0

x ( t )=[−√90sin (√1000 t )+10sin(30 t )) ]∗2 .54=254 sin(30 t )−2 .54 √90sin(√1000 t )

-35.2690691 -2.83287697 -89.5834356-147.02364 -11.8092111 -373.440046

factor 2.54

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8

-3000

-2000

-1000

0

1000

2000

3000

4000

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

-4

-3

-2

-1

0

1

2

3

4

Desplazamiento [cm]

t(seg)

Des

pla

zam

ien

to (

cm)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

-80

-60

-40

-20

0

20

40

60

80

100

Velocidad [cm/seg]

t(seg)

velo

cid

ad(c

m/s

eg)

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

-3000

-2000

-1000

0

1000

2000

3000

4000


t[seg]

acel

erac

ión

(cm

/seg

2)

Problema 2

Datos0 0

0.16 120W = 38.6 [K] 0.32 0k = 100 K/in 0.48 -120

g = 386 0.64 0

Idealización Diagrama de cuerpo libre

Fuerzas

m =W

=38.6 [K]

= 0.1g 386

[in/seg2]

[seg2*K/in][in/seg2]

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.

WF(t)

k

Fs=k∗x ( t )=100x ( t )

F I=m x (t )=0 .1 x ( t )

ω=√1000(rad / seg )

0 0.16 0.32 0.48 0.64

-120

-80

-40

0

40

80

120

t(seg)

F(t

) [K

]

FD=C∗x ( t )=0 .2√10 x ( t )

Ccrit=2mω

C=0 .1Ccrit=0 .1∗2mω=0 .2∗0.1∗√1000=0 .2√10

Ecuación de movimiento

Ecuación característica

Solución complementaria

Cálculo solución particular 0 < t < 0,16 seg

I

De esta ecuación:

7.5

-0.0474


m x( t )+C x ( t )+Kx( t )=F ( t )

0 .1 x ( t )+0 .2√10 x ( t )+100 x( t )=F( t )

0 .1S2+0 .2√10 S+100=0

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=750 t

x ( t )p=7 .5 t−0 .015√10

x ( t )c=e−ωζ t (A sin(ωD t )+B cos(ωD t ))

S1,2=−ωζ±(ω√1−ζ2 ) i=−√1000∗0 .1±(√1000∗√1−0 .12) i=−√10±√990 i

x ( t )c=e−√10 t ( A sin(√990 t )+B cos(√990 t ))

F ( t )=1200 .16

t=750 t [K ]

x ( t )p=C1 t+C2x ( t )p=C1x ( t )p=0

0 .1(0)+0.2√10(C1 )+100(C1 t+C2 )=750 t

C1=

C2=−0 .015√10=

Solución general

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


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

t Desp(pulg) Vel(pulg/s) Acelerac(pulg/s2)0 0 0 0

0.02 0.010 1.391783771 131.695964192490.04 0.070 4.837802864 199.888381123710.06 0.207 8.841762311 187.342847943590.08 0.417 11.88240951 108.21128477677

0.1 0.669 12.96929258 -0.83985216686660.12 0.921 11.93960221 -96.628564060580.14 1.136 9.418994448 -145.922952920050.16 1.295 6.500555873 -136.33478124196

x (0 )=0x (0 )=0

B=0.015√10A=−7 .5−0.015√10∗10

√990=−7 .35

√990=−0 .2336

x ( t )=x ( t )c+x ( t )p=e−√10 t (A sin (√990t )+B cos (√990 t ))+7 .5 t−0.015√10

x ( t )=e−√10 t (A sin( √990t )+B cos (√990t ))+7 .5 t−0.015√10x ( t )=e−√10 t [ (−√10 A−√990B )sin (√990t )+(−√10 B+√990 A )cos (√990 t )]+7 .5

x (0)=e−√10∗0 (A sin(√990∗0 )+B cos (√990∗0))+7 .5∗0−0.015√10=0x (0)=e−√10∗0 [ (−√10 A−√990 B ) sin(√990∗0 )+(−√10B+√990 A )cos(√990∗0 )]+7 .5=0

x ( t )=e−√10 t (−0 .2336sin(√990 t )+0 .015√10cos(√990 t ))+7 .5 t−0 .015√10

x ( t )=e−√10 t [ (−0 .754 ) sin(√990 t )+(−7 .5 )cos (√990 t )]+7 .5

x ( t )=e−√10 t [ (238.366 )sin( √990 t ) ]

F ( t )=750 t [K ]

Cálculo solución particular 0,16 <t <= 0,48 seg

I

De esta ecuación:-7.5

2.4474

Solución general

Cálculo de A y B, bajo condiciones iniciales conocidas



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

x ( t )p=−7 .5 t+2 .4474

x ( t )p=C1 t+C2x ( t )p=C1x ( t )p=0

0 .1(0)+0.2√10(C1 )+100(C1 t+C2 )=240−750 t

C1=

C2=240−0 .2∗√10∗(−7 .5)100

=

F ( t )=2∗120−1200 .16

t=240−750( t )[ K ]

100C1=−7500 .2√10C1+100C2=240

x ( t )=x ( t )c+x ( t )p=e−√10 t (A sin (√990t )+B cos (√990 t ))−7 .5 t+2 .4474

x ( t )=e−√10 t (A sin( √990t )+B cos (√990t ))−7 .5 t+2 .4474x ( t )=e−√10 t [ (−√10 A−√990B )sin (√990t )+(−√10 B+√990 A )cos (√990 t )]−7 .5

x (0 .16 )=1.2952x (0 .16 )=6 .5006

x (0 .16 )=e−√10∗0 .16 (A sin (√990∗0 .16 )+B cos (√990∗0 .16 ))−7 .5∗0 .16+2 .4474=1 .2952

Del sistema

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

t Desp(pulg) Vel(pulg/s) Acelerac(pulg/s2)0.16 1.295 6.510 -136.4660.18 1.382 1.51 -341.994

0.2 1.338 -6.164 -398.8280.22 1.139 -13.415 -303.9820.24 0.822 -17.66 -109.9450.26 0.461 -17.701 100.9890.28 0.139 -14.03 250.162

0.3 -0.087 -8.430 290.7080.32 -0.201 -3.152 220.7980.34 -0.228 -0.08 78.9610.36 -0.225 -0.072 -74.7160.38 -0.250 -2.76 -182.983

0.4 -0.345 -6.855 -211.8940.42 -0.522 -10.695 -160.3710.44 -0.762 -12.92 -56.6950.46 -1.024 -12.908 55.2640.48 -1.265 -10.93 133.839

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 (√990t )+(−750 .606 ) cos( √990t ) ]

x (0 .16 )=e−√10∗0 .16 (A sin (√990∗0 .16 )+B cos (√990∗0 .16 ))−7 .5∗0 .16+2 .4474=1 .2952x (0 .16 )=e−√10∗0 .16 [ (−√10 A−√990B )sin (√990∗0 .16 )+ (−√10B+√990 A )cos(√990∗0 .16) ]−7 .5=6 .5006

−0 .572 A+0 .1907B=0 .0478227 .8103 A+17 .393B=14

A=0 .1609B=0.7333

x ( t )=e−√10 t (0 .1609sin(√990 t )+0 .7333cos (√990 t ))+7 .5 t−0 .015√10

F ( t )=240−750( t )[ K ]

Comportamiento a vibración libre t > 0,48 seg

Ecuación de movimiento t>0,48 seg

Solución:

Para las condiciones iniciales de:

0.1247 A -0.1803 B = -1.265-6.0659 A -3.3529 B = -10.934

Del sistema:

A = -1.501462B = 5.977528

x (0 .48 )=−1 .265x (0 .48 )=−10 .934

x (0 .16 )=1.2952x (0 .16 )=6 .5006

0 .1 x ( t )+0 .2√10 x ( t )+100 x( t )=0

x ( t )c=e−√10 t (A sin(√990 t )+B cos(√990 t ))

x (0 .48 )=e−√10∗0 .48 (A sin(√990∗0 .48 )+B cos (√990∗0. 48))=−1 .265

x (0 .48 )=e−√10∗0 .48 [ (−√10 A−√990 B ) sin(√990∗0 .48 )+ (−√10B+√990 A )cos(√990∗0 .48 )]=−10 .934

Comportamiento a vibración libreDerivando

t Desp(pulg) Vel(pulg/s) Acelerac(pulg/s2)0.48 -1.265 -10.934 1333.839

0.5 -1.222 14.516 1130.1870.52 -0.740 31.668 540.0450.54 -0.047 35.274 -176.2190.56 0.581 25.634 -743.3440.58 0.923 7.824 -972.967

0.6 0.890 -10.712 -821.7500.62 0.536 -23.154 -389.8930.64 0.030 -25.704 132.3310.66 -0.427 -18.610 544.4190.68 -0.674 -5.597 709.713

0.7 -0.647 7.904 597.4710.72 -0.389 16.928 281.456

x ( t )=e−√10 t [ (−1.5015 ) sin(√990 t )+5 .97752cos (√990 t )]x ( t )=e−√10 t (−183 .33sin(√990 t )+−66 .145cos(√990 t ))x ( t )=e−√10 t (2660 .9 sin(√990 t )+−5559 .19cos(√990 t ))

Gráfica de: desplazamiento, velocidad, aceleraciónCon un intervalo de 0,02En el sistema métrico

t[seg]

0 0 0 00.02 0.06 3.54 334.510.04 0.18 12.29 507.720.06 0.53 22.46 475.850.08 1.06 30.18 274.86

0.1 1.70 32.94 -2.130.12 2.34 30.33 -245.440.14 2.89 23.92 -370.640.16 3.29 16.51 -346.290.18 3.51 3.85 -868.66

0.2 3.40 -15.66 -1013.020.22 2.89 -34.07 -772.120.24 2.09 -44.84 -279.260.26 1.17 -44.96 256.510.28 0.35 -35.65 635.41

0.3 -0.22 -21.41 738.400.32 -0.51 -8.01 560.830.34 -0.58 -0.21 200.560.36 -0.57 -0.18 -189.780.38 -0.63 -7.02 -464.78

0.4 -0.88 -17.41 -538.210.42 -1.33 -27.17 -407.340.44 -1.93 -32.81 -144.010.46 -2.60 -32.79 140.370.48 -3.21 -27.77 339.95

0.5 -3.10 36.87 2870.670.52 -1.88 80.44 1371.710.54 -0.12 89.60 -447.600.56 1.48 65.11 -1888.090.58 2.35 19.87 -2471.34

0.6 2.26 -27.21 -2087.250.62 1.36 -58.81 -990.330.64 0.08 -65.29 336.120.66 -1.08 -47.27 1382.830.68 -1.71 -14.22 1802.67

0.7 -1.64 20.08 1517.580.72 -0.99 43.00 714.90

Del gráfico

Máximos 3.511 89.596 2870.675

Desplazamiento

[cm]

Velocidad

[cm/seg]

Aceleración

[cm/seg2]

Desplazamiento [cm]

Velocidad [cm/seg]


31.6227766

0.63245553

31.4642654

750

-0.75377836 -7.50000.0474

-0.2336

238.365647 0

a 0.16086 b 0.7333

-23.5801713 2.74257002

-11.7259024 -750.605536

t = 0.48 seg.w =

#DIV/0!

#DIV/0!

0.12467825 -0.18025592 -1.265 2.21833487 -0.11926068-6.06588745 -3.35289038 -10.934 -4.01330438 -0.08248946

#REF! -3.81371312#REF! 15.1829223

-183.330503 -66.1449991

2660.94576 -5559.19076

factor 2.54

t[seg]

0 0 0 00.02 0.02413412 1.39178377 131.6959640.04 0.06951467 4.83780286 199.8883810.06 0.20673694 8.84176231 187.3428480.08 0.41663776 11.8824095 108.211285

0.1 0.66881484 12.9692926 -0.839852170.12 0.92111589 11.9396022 -96.62856410.14 1.136352 9.41899445 -145.9229530.16 1.29522166 6.50055587 -136.3347810.18 1.38241948 1.5138435 -341.99387

0.2 1.33781267 -6.16409431 -398.827510.22 1.13882746 -13.4151964 -303.9823040.24 0.82160704 -17.6552721 -109.9452930.26 0.46096446 -17.7013266 100.9885590.28 0.13860238 -14.0349497 250.162436

0.3 -0.08739466 -8.42953372 290.7077150.32 -0.20086076 -3.15236803 220.7980820.34 -0.2284423 -0.08206768 78.96134050.36 -0.22482836 -0.0721034 -74.71562170.38 -0.24954577 -2.76247297 -182.982821

0.4 -0.34475345 -6.85470128 -211.8936090.42 -0.52198634 -10.6951879 -160.3713510.44 -0.76160247 -12.9182877 -56.69510520.46 -1.02362486 -12.9083453 55.26440860.48 -1.265 -10.9342991 133.839128

0.5 -1.222 14.5162889 1130.186890.52 -0.74033052 31.6679579 540.0447710.54 -0.04687422 35.2741935 -176.2193670.56 0.58122014 25.6340029 -743.3438090.58 0.92348174 7.82435374 -972.967294

0.6 0.88950159 -10.7124029 -821.7504070.62 0.53633171 -23.1539847 -389.8930540.64 0.0302325 -25.7035849 132.3312480.66 -0.4267189 -18.6100751 544.4193520.68 -0.67431508 -5.59687082 709.7128

0.7 -0.64745968 7.90392191 597.4708910.72 -0.38852122 16.9284492 281.456303

Desplazami

ento [cm]

Velocidad

[cm/seg]

Aceleración

[cm/seg2]

Dinamica Estructural Ecuacion de Movimiento

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Transcript of Dinamica Estructural Ecuacion de Movimiento