Sistemas de múltiples grados de libertad

Universidad Nacional Autónoma de NicaraguaUNAN- Managua.

Recinto Universitario “Rubén Darío”Facultad de Ciencias e Ingenierías

Departamento de construcciónIngeniería Sismorresistente

Tema: Sistemas de múltiples grados de libertad

Título del Trabajo:

Solución a sistemas con uno a más grados de libertad.

Profesor: Dr. Ing. Edwin Antonio Obando.

Fechas de Entrega: 22 de octubre del 2015.

Estudiantes:

Joel Enrique santana Peña. Greybin Josué Borge Castro.

“No basta tener un buen ingenio, lo principal es aplicarlo bien”René Descartes

Ingeniería Sismorresistente | Borge & SantanaEjercicio 1

BORGE & SANTANA 2

Cálculo de kt= 3ℎ = (3)(29 10 . )(28.1 )(10 12) = 1414.76 /La masa en función de la gravedad

Masa del torrecilindro= 18.97 = 189.7= 189.7 + 300032.2 12 = 8.255 . / Periodo de vibración de la estructura= 2 = 2 . . /. / = 0.479951

interpolando,

0.471369 0.162835

0.479951 x

0.501593 0.1372490.471369 − 0.4799510.471369 − 0.501593 = 0.162835 −0.162835 − 0.137249= = 0.155570


BORGE & SANTANA 3

Cálculo de Dmax= ( ) ( ) ; = 0.155570 ∗ = 0.061248= 0.061248 ( 0.4799512 ) = 0.357 10 Cálculo de la fuerza lateral= ∗= 1414.76 ∗ 0.357 10 = 0.5051 Cálculo del momento basal por estática Vb=Fo= ∗ ℎ= 0.5051 ∗ 10 12 = 60.612 . Cálculo del esfuerzo flexionante máximo

= = 60.612 . (6.625 2 )28.1 = 7.145 /


BORGE & SANTANA 4

= 100= 326.32 /= 2= 2 326.32= =− == 0 00 00 0 ; = + − 0− + −0 −2 − 0− 2 −0 − − 0 00 00 0 12 = 000⎩⎨⎧ 2 − 0− 2 −0 − − ⎣⎢⎢

⎡ 0 00 00 0 12 ⎦⎥⎥⎤⎭⎬⎫ = 000

⎩⎨⎧2 − − 0− 2 − −0 − − 12 ⎭⎬

⎫ = 000 Encontrar el Determinante por Sarrus.

⎣⎢⎢⎡2 − − 0− 2 − −0 − − 12

2 − −− 2 −0 − ⎦⎥⎥⎤


BORGE & SANTANA 5

(2 − 2)(2 − 2) − 12 2 + [(− )(− )(0)] + [(0)(− )(− )]− (− )(− ) − 12 2 − [(− )(− )(2 − 2)] − [(0)(0)(2 − 2)] = 0(4 − 4 2 + 4) − 12 2 − − 12 2 − [2 − 2] = 0

4 − 2 2 − 4 2 + 2 4 + 4 − 12 6 − + 12 2 − 2 + 2= 0− 92 2 + 3 4 − 12 6 = 0 ∆ =− 92 ∆ + 3 ∆2 − 12 ∆3= 0 Raíces de ∆ desde MatLab∆ = 9411.5 =∆ = 5043.6 =∆ = 675.7 == √9411.5 = 97.01289 /= √5043.6 = 71.0183 /= √675.7 = 25.9942 /Calculo de Periodos naturales.= 2

= 2 = 297.01289 = 0.065= 2 = 271.0183 = 0.0885= 2 = 225.9942 = 0.242 Modos de Vibración.( − 2)∅ = 0


BORGE & SANTANA 6

2 − 0− 2 −0 − − 0 00 00 0 12 2 ∅∅∅ = 0

⎩⎨⎧ 2 − 0− 2 −0 − − ⎣⎢⎢

⎡ 2 0 00 2 00 0 12 2⎦⎥⎥⎤⎭⎬⎫ ∅∅∅ = 0

Si =1305.25 −652.64 0−652.64 1305.25 −652.640 −652.64 652.64 − (0.2588)(675.7) 0 00 (0.2588)(675.7) 00 0 (0.1294)(675.7) ∅∅∅ = 01,130.407 −652.64 0−652.64 1,130.407 −652.640 −652.64 565.203 ∅∅∅ = 01,130.407∅11 − 652.64∅21 = 0 ∅ = 11,130.407 − 652.64∅21 = 0∅ = 1,130.407652.64∅ = 1.732−652.64∅11 + 1,130.407∅21 − 652.64∅31 = 0−652.64(1) + 1,130.407(1.732)− 652.64∅31 = 0∅ = 1,305.225652.64∅ = 2.0∅ = ∅∅∅ = 1.01.7322.0

Si =1305.25 −652.64 0−652.64 1305.25 −652.640 −652.64 652.64 − (0.2588)(5,043.6) 0 00 (0.2588)(5,043.6) 00 0 (0.1294)(5,043.6) ∅∅∅ = 0−0.646 −652.64 0−652.64 −0.646 −652.640 −652.64 0.323 ∅∅∅ = 000


BORGE & SANTANA 7

−0.646∅12 − 652.64∅22 = 0 ∅ = 1.0−0.646(1.0)− 652.64∅22 = 0∅ = − 0.646652.64∅ = −0.001−652.64∅12 − 0.646∅22 − 652.64∅32 = 0−652.64(1.0) + 1,130.407(−0.001)− 652.64∅32 = 0∅ = −1.0∅ = ∅∅∅ = 1.0−0.001−1.0

Si =1305.25 −652.64 0−652.64 1305.25 −652.640 −652.64 652.64 − (0.2588)(9,403.5) 0 00 (0.2588)(9,403.5) 00 0 (0.1294)(9,403.5) ∅∅∅ = 0−1,128.338 −652.64 0−652.64 −1,128.338 −652.640 −652.64 −564.169 ∅∅∅ = 000−1,128.338∅13 − 652.64∅23 = 0 ∅ = 1.0−1,128.338(1.0)− 652.64∅23 = 0∅ = 652.641,128.338∅ = 1.729−652.64∅13 − 0.646∅23 − 652.64∅33 = 0−652.64(1.0) − 1,128.338(−1.729)− 652.64∅33 = 0∅ = 1.989∅ = ∅∅∅ = 1.0−1.779−1.01.989


BORGE & SANTANA 8

Resultados de los Modos de Vibración…

Expansión Modal de las fuerzas sísmicas.Cálculo de desplazamiento= ∅ ( )Modo 1 Modo 2 Modo 3

∅ = ∅∅∅ = 2.01.7321.0 ; ∅ = ∅∅∅ = −1.0−0.0011.0 ; ∅ = ∅∅∅ = 1.989−1.7291.0= ∅ + ∅ + ∅∅ + ∅ + ∅= (0.2588)(1.0) + (0.2588)(1.732) + (0.1294)(2.0)(0.2588)(1.0) + (0.2588)(1.732) + (0.1294)(2.0)= 0.622= ∅ + ∅ + ∅∅ + ∅ + ∅= (0.2588)(1.0) + (0.2588)(−0.001) + (0.1294)(−1.0)(0.2588)(1.0) + (0.2588)(−0.001) + (0.1294)(−1.0)

2 -1 1.989

1.732 0.001 -1.729

1 1 1

T1=0.242 seg T2=0.088 seg T3=0.065 seg


BORGE & SANTANA 9

= 0.333= ∅ + ∅ + ∅∅ + ∅ + ∅= (0.2588)(1.0) + (0.2588)(−1.729) + (0.1294)(1.989)(0.2588)(1.0) + (0.2588)(−1.729) + (0.1294)(1.989)= 0.044

Pseudo-aceleraciones

Para = .T (periodo) PSA

0.237944 0.3506180.242 X

0.2532 0.2908820.237944 − 0.2420.237944 − 0.2532 = 0.350618 −0.350618 − 0.290882= 0.334736 /Para = .

T (periodo) PSA0.082727 1.299758

0.088 X0.088032 1.4772620.082727 − 0.0880.082727 − 0.088032 = 1.299758 −1.299758 − 1.477262= 1.476191 /

Para = .T (periodo) PSA

0.064519 1.1996780.065 X

0.068656 1.405843


BORGE & SANTANA 10

0.064519 − 0.0650.064519 − 0.068656 = 1.199678 −1.199578 − 1.477262= 1.223648 /

= ( )= (0.131786 / ) 0.2422= 1.955 10= (0.581177 / ) 0.0882= 1.140 10= (0.481751 / ) 0.0652= 5.156 10

= 0.622 ∗ 2.01.7321.0 ∗ 1.955 10 = 2.432 102.106 101.216 10= 0.333 ∗ −1−0.0011 ∗ 1.140 10 = −3.796 10− 3.796 103.796 10= 0.044 ∗ 1.989−1.7291 ∗ 5.156 10 = 4.512 10− 3.922 102.269 10


BORGE & SANTANA 11

Calculo de SRSS= + += (1.216 10 ) + (3.796 10 ) + (2.269 10 )= 1.274 10= + += (2.106 10 ) + (−3.796 10 ) + (−3.922 10 )= 2.106 10= + += (2.432 10 ) + (−3.796 10 ) + (4.512 10 )= 2.462 10


BORGE & SANTANA 12

Demuestre que los periodos y modos de Vibracion del sistema mostrado son :

Solución:

Matrices M y K del sistema.

= 0 00 00 0 ; = + − 0− + −0 − Masas en función de la gravedad= ; = /= / = 0.407747 ∗ /= 20081 / = 0.203874 ∗ / Matriz K del sistema

= 200 + 200 −200 0−200 200 + 80 −800 −80 80 = 400 −200 0−200 280 −800 −80 80 Determinante[ − ] = 0200 + 200 −200 0−200 200 + 80 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 = 0

Figura 1

Figura 2


BORGE & SANTANA 13

400 −200 0−200 280 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 = 000400 −200 0−200 280 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 = 000400 − 0.407747 −200 0−200 280 − 0.407747 −800 −80 80 − 0.203874 = 0 Resolviendo determinante:[400 − 0.407747 ] ∗ [(280 − 0.407747 )(80 − 0.203874 )] − [(−80)(−80)] −[−200] [(−200)(80 − 0.203874 )] − [(−80)(0)] − [0] [(−200)(−80)] − [(280 −0.407747 )(0)] = 0[(400 − 0.407747 ) ∗ (22400 − 57.08472 − 32.61976 + 0.0831290119( ) ) − 6400) −(−200) ∗ (−16000 + 40.7748 ) + (0)] = 0[8,960,000 − 22,833.888 − 13,047.904 + 33.25160476( ) − 2,560,000 −9,133.5328 + 23.27612333( ) + 13.30060928( ) − 0.0338956052( ) +2609.5808 − 3,200,000 + 8,154.96 ] = 0−0.0338956052( ) + 69.82833737( ) − 34,250.784 + 3,200,000 = 0

Multiplicando por − . queda:−( ) + 2,060.100032( ) − 1,010,478.609( ) + 94,407,519.24 = 0Raíces:= 121.956 = 562.882 = 1375.262= √121.956 = √562.882 = √1375.262= 11.0434 = 23.7251 = 37.0845 Periodos de vibración

Si = , entonces:= = == . = . = .= 0.5689 = 0.2648 = 0.1694


BORGE & SANTANA 14

Modos de vibración de la estructura[ − ] ∗ ∅ =400 −200 0−200 280 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0Para ∅400 −200 0−200 280 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0

400 −200 0−200 280 −800 −80 80 − 121.956 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0400 −200 0−200 280 −800 −80 80 − 49.727193 0 00 49.727193 00 0 24.863658 ∗ ∅∅∅ = 0123→→→ 350.272807∅ −200∅ 0∅−200∅ 230.272807∅ −80∅0∅ −80∅ 55.136342∅ = 0

Si ∅ = 1 en Ec1 entonces:350.272807∅ − 200∅ = 0350.272807 ∗ (1) − 200 ∗ ∅ = 0∅ = 350.272807200 = 1.7514Sustituyendo valores de ∅ y ∅ en Ec2−200∅ + 230.272807∅ − 80∅ = 0−200 ∗ (1) + 230.272807 ∗ (1.7514) − 80∅ = 0∅ = 203.299880 = 2.5412

Para ∅400 −200 0−200 280 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0


BORGE & SANTANA 15

400 −200 0−200 280 −800 −80 80 − 562.882 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0400 −200 0−200 280 −800 −80 80 − 229.513447 0 00 229.513447 00 0 114.757005 ∗ ∅∅∅ = 0123→→→ 170.486553∅ −200∅ 0∅−200∅ 50.486553∅ −80∅0∅ −80∅ −34.757005∅ = 0

Si ∅ = 1 en Ec1 entonces:170.486553∅ − 200∅ = 0170.486553 ∗ (1) − 200 ∗ ∅ = 0∅ = 170.486553200 = 0.8524Sustituyendo valores de ∅ y ∅ en Ec2−200∅ + 50.486553∅ − 80∅ = 0−200 ∗ (1) + 50.486553 ∗ (0.8524) − 80∅ = 0∅ = −158.32839980 = −1.962

Para ∅400 −200 0−200 280 −800 −80 80 − 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0400 −200 0−200 280 −800 −80 80 − 1375.262 0.407747 0 00 0.407747 00 0 0.203874 ∗ ∅∅∅ = 0400 −200 0−200 280 −800 −80 80 − 560.758955 0 00 560.758955 00 0 280.380165 ∗ ∅∅∅ = 0123→→→ −160.758955∅ −200∅ 0∅−200∅ −280.758955∅ −80∅0∅ −80∅ −200.380165∅ = 0

Si ∅ = 1 en Ec1 entonces:


BORGE & SANTANA 16

−160.758955∅ − 200∅ = 0−160.758955 ∗ (1) − 200 ∗ ∅ = 0∅ = −160.758955200 = −0.8038Sustituyendo valores de ∅ y ∅ en Ec2−200∅ − 280.758955∅ − 80∅ = 0−200 ∗ (1) − 280.758955 ∗ (−0.8038) − 80∅ = 0∅ = 25.67404880 = 0.3209 Desplazamientos reales de la estructura.= ∅ ( )

Modo 1 Modo 2 Modo 3

∅ = ∅∅∅ = 2.54121.75141 ∅ = ∅∅∅ = −1.9620.85241 ∅ = ∅∅∅ = 0.3209−0.80381= ∅ + ∅ + ∅(∅ ) + (∅ ) + (∅ )= (0.407747)(1) + (0.407747)(1.7514) + (0.203874)(2.5412)(0.407747)(1) + (0.407747)(1.7514) + (0.203874)(2.5412)= 0.551242= ∅ + ∅ + ∅(∅ ) + (∅ ) + (∅ )= (0.407747)(1) + (0.407747)(0.8524) + (0.203874)(−1.962)(0.407747)(1) + (0.407747)(0.8524) + (0.203874)(−1.962)= 0.238653= ∅ + ∅ + ∅(∅ ) + (∅ ) + (∅ )


BORGE & SANTANA 17

= (0.407747)(1) + (0.407747)(−0.8038) + (0.203874)(0.3209)(0.407747)(1) + (0.407747)(−0.8038) + (0.203874)(0.3209)= 0.2101 Pseudoaceleraciones (PSA)

Interpolación para = .T(período) (PSA)0.567977 0.0746843970.5689 X0.604394 0.057055368

(0.567977 − 0.5689)(0.567977 − 0.604394) = (0.074684397 − )(0.074684397 − 0.057055368)= = 0.074237 ∗ 1 2.54 = 0.029227 /Interpolación para = .T(período) (PSA)0.2532 0.2908828140.2648 X0.269435 0.304566669

(0.2532 − 0.2648)(0.2532 − 0.269435) = (0.290882814 − )(0.290882814 − 0.304566669)= = 0.300665 ∗ 1 2.54 = 0.118372 /Interpolación para = .T(período) (PSA)0.163884 0.6143757290.1694 X0.174392 0.475749077

(0.163884 − 0.1694)(0.163884 − 0.174392) = (0.614375729 − )(0.614375729 − 0.475749077)= = 0.541606 ∗ 1 2.54 = 0.213231 /


BORGE & SANTANA 18

Cálculos para

= ( )= ( ) 2= (0.029227 / ) 0.56892= 2.396 10= ( ) 2= (0.118372 / ) 0.26482= 2.102 10= ( ) 2= (0.213231 / ) 0.16942= 1.550 10

Cálculos para= ∅ ( )= ∅ ( )

= 0.551242 ∗ 2.54121.75141 ∗ 2.396 10 = 3.356 102.313 101.321 10


BORGE & SANTANA 19

= ∅ ( )= 0.238653 ∗ −1.9620.85241 ∗ 2.102 10 = −9.842 104.276 105.016 10= ∅ ( )= 0.2101 ∗ 0.3209−0.80381 ∗ 1.550 10 = 1.045 10−2.618 103.257 10

Desplazamientos totales por nivel de piso.= ( ) + ( ) + ( )= (1.321 10 ) + (5.016 10 ) + (3.257 10 )= 1.450= (2.313 10 ) + (4.276 10 ) + (−2.618 10 )= 2.367= (3.356 10 ) + (−9.842 10 ) + (1.045 10 )= 3.499

Engineering

Sistemas de múltiples grados de libertad