Ejercicios Resueltos Ingenieria Sismica

UNIVERSIDAD PRIVADA ANTENOR ORREGO FACULTAD DE INGENIERIA ESCUELA DE INGENIERIA CIVIL

TRABAJO 1  ALUMNO:

Zavaleta Nolasco, Julio

 DOCENTE

Villareal Castro, Genner

 CURSO:

Ingeniería Sísmica

TRUJILLO-2023

1. ANÁLISIS MATRICIAL. Determinar el desplazamiento lateral máximo y fuerzas internas máximas (N, V, V)

a. Esquematizamos los grados de libertad de la estructura 2

5 1

3 B

A

4 6

11

8

C

10

7 9

D

E

F

12

G

H

b. Determinamos las rigideces axiales y en flexión de vigas y columnas 3

0.4 × 0.6 ) 12

EA viga =2173706 ×0.4 × 0.6

EI viga =2173706.(

EA viga =521689 , 44 T

EI viga =15 650,6832T m

2 3

0.4 × 0.4 ) 12

EA columna=2173706 × 0.4 ×0.4

EI columna=2173706.(

EA columna=347 792.96 T

EI columna=4 637,2395 T m

2

c. Ensamblamos la matriz de rigidez de cada elemento 1 2 3 4 5 6

[ [ [

104 337,888 0 0 −104 337,888 0 0 0 1502,465 3 756,164 0 −1 502,465 3 756,164 0 3756,164 12 520,546 0 −3 756,164 6 260,273 K BC = −104 337,888 0 0 104 337,888 0 0 0 −1 502,465 −3 756,164 0 1 502,465 −3 756,164 0 3756,164 6 260,273 0 −3 756,164 12 520,546

4 5 6 7 8 9 104 337,888 0 0 −104 337,888 0 0 0 1 502,465 3756,164 0 −1502,465 3 756,164 0 3 756,164 12520,546 0 −3 756,164 6 260,273 K CE = −104 337,888 0 0 104 337,888 0 0 0 −1 502,465 −3756,164 0 1502,465 −3 756,164 0 3 756,164 6 260,273 0 −3 756,164 12 520,546

7 8 9 10 11 12

104 337,888 0 0 −104 337,888 0 0 0 1 502,465 3756,164 0 −1502,465 3 756,164 0 3 756,164 12520,546 0 −3 756,164 6 260,273 K EG= −104 337,888 0 0 104 337,888 0 0 0 −1 502,465 −3756,164 0 1502,465 −3 756,164 0 3 756,164 6 260,273 0 −3 756,164 12 520,546

] ] ]

1 2 3 4 5 6

4 5 6 7 8 9

7 8 9 10 11 12

0 0 0 1 2 3

[ [ [

2 060,995 0 −3 091,493 −2060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 0 6 182,986 3 091,493 0 3 091,493 K AB= −3 091,493 −2 060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

0 0 0 4 5 6

] ] ]

2 060,995 0 −3 091,493 −2 060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 0 6 182,986 3 091,493 0 3 091,493 K DC = −3 091,493 −2 060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

0 0 0 7 8 9

2 060,995 0 −3 091,493 −2 060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 −3 091,493 0 6 182,986 3 091,493 0 3 091,493 K FE= −2060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

0 0 0 1 2 3

0 0 0 4 5 6

0 0 0 7 8 9

0 0 0 10 11 12

[

2060,995 0 −3 091,493 −2 060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 −3 091,493 0 6 182,986 3 091,493 0 3 091,493 K HG= −2060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

d. Ahora, ensamblamos la matriz de rigidez de la estructura.

1 2 3 4 5 6 7 8 9 10 11 12

]

0 0 0 10 11 12

[

106 398,883 0 3 091,493 −104 337.888 0 117 433,452 3756,164 0 3 091,493 3 756,164 18 703,532 0 −104 337,888 0 0 106 398,883 0 −1 502,465 −3 756,164 0 0 3 756,164 6 260,273 3 091,493 K HG= 0 0 0 −104 337,888 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 −1 502,465 3756,164 0 −3 756,164 6 260,273 0 0 3 091,493 −104 337,888 118 935,917 0 0 0 31 224,078 0 0 0 210 736,771 −1 502,465 −3 756,164 0 3 756,164 6 260,273 3 091,493 0 0 −104 337,888 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 −1 502,465 3756,164 0 −3 756,164 6260,273 0 0 3091,493 −104337,888 3 004,930 0 0 − 115 930,987 31224,078 0 − 0 0 106398,883 −1 502,465 −3756,164 0 1 3 756,164 6260,273 3091,493 −

e. El vector fuerza de los nudos del sistema es:

[]

8 0 0 0 0 S 0 F= 0 0 0 0 0 0

1 2 3 4 5 6 7 8 9 10 11 12

f. El vector de fuerzas de empotramiento perfecto de cada elemento será:

[] [] [] [] [] [] []

0 0 E F AB= 0 0 0 0

0 0 0 FE = 1 BC 2 3

0 0 0 0 0 0

1 2 3 FE = 4 DC 5 6

0 0 0 0 0 0

0 0 0 FE = 4 CE 5 6

0 0 0 0 0 0

4 5 6 FE = 7 FE 8 9

0 0 0 0 0 0

0 0 0 FE = 7 EG 8 9

g. Ensamblamos el vector fuerza de empotramiento perfecto del sistema

[]

0 0 0 0 0 E 0 F = 0 0 0 0 0 0

1 2 3 4 5 6 7 8 9 10 11 12

0 0 0 0 0 0

7 8 9 FE = 10 HG 11 12

0 0 0 0 0 0

0 0 0 10 11 12

h. De esta manera, el vector de fuerzas internas del sistema será: S

F=F −F

E

[][][]

8 0 8 0 0 0 0 0 0 0 0 0 0 0 0 S 0 0 0 F= − = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

i.

Resolviendo la ecuación, obtenemos los desplazamientos y pendientes.

[ δ ] =K−1 × [ F ]

[ ] [][ ] δ1 δ2 δ3 δ4 δ5 δ6 −1 =K × δ7 δ8 δ9 δ 10 δ 11 δ 12

j.

8 −0.0000037367 m 0 −0.0000000985 m 0 −0.0000006642 rad 0 −0.0000805041 m 0 −0.0000007317 m 0 0.0000034498 rad = 0 −0.0000782554 m 0 −0.0000040282 m 0 0.0000208549 rad 0 −0.0000769346 m 0 0.0000008302 m 0 0.0000067118 rad

Ahora, calculamos las fuerzas internas de los elementos:

[ F ]ij =[ F ]ijE + [ K ] ij × [ δ ] ij

[] [ ][ ]

0 0 0.00976 T 0 0 0.01142 T 0 0 −0.01361 T .m [ F ] AB= + [ K ] AB × = 0 −0.0000037367 −0.00976 T 0 −0.0000000985 −0.01142 T 0 −0.0000006642 −0.01566 T .m

B

0.00976 T

0.01566T.m T.m 0.01569 0.01142 T 0.00976 T A 0.01361 T.m

[] [ ][ ]

0 0 0.15525 T 0 0 0.84828 T 0 [ F ] DC = 0 + [ K ] DC × = −0.23821 T . m 0 −0.0000805041 −0.15525 T 0 −0.0000007317 −0.84828 T 0 0.0000034498 −0.22755 T . m 0.15525 T C 0.22755 T.m 0.84828 T

0.84828 T

D

0.15525 T 0.23821 T.m

[] [ ][ ]

0 0 0.09681 T 0 0 0.46699 T 0 [ F ] FE= 0 + [ K ] FE × = −0.17745 T . m 0 −0.0000782554 −0.09681 T 0 −0.0000040282 −46699 T 0 0.0000208549 −0.11298 T . m

0.09681 T

E 0.17745 T.m

0.46699 T

0.46699 T 0.09681 T

F

0.17745 T.m

[] [ ][ ]

0 0 0.13781 T 0 0 −0.09625 T 0 0 [ F ] HG= + [ K ] HG × = −0.21709 T . m 0 −0.0000769346 −0.13781 T 0 0.0000008302 0.09625 T 0 0.0000067118 −0.19634 T . m G

0.13781 T

0.09625 T

0.01566 T.m 0.13781 T H 0.21709 T.m

0.09625 T

[] [ ][ ]

0 0 [ F ] BC = 0 + [ K ] BC × 0 0 0

0.01141 T

B

8.00975 T

0.01566 T.m 8.00975 T

−0.0000037367 8.00975 T −0.0000000985 0.01141 T −0.0000006642 = 0.01566 T . m −0.0000805041 −8.00975 T −0.0000007317 −0.01141 T 0.0000034498 0.04141 T . m

C -0.21805 T.m

0.01141 T

[] [ ][ ]

0 −0.0000805041 −0.23462 T 0 −0.0000007317 0.09624 T [ F ]CE = 0 + [ K ] CE × 0.0000034498 = 0.18613 T . m 0 −0.0000782554 0.23462 T 0 −0.0000040282 −0.09624 T 0 0.0000034498 0.29509 T . m 0.23462 T 0.29509 T.m

E E 0.23462 T W 0.09624 T

C 0.18613 T.m

[] [ ][ ] 0.09624 T

0 0 [ F ] EG= 0 + [ K ] EG × 0 0 0

0.13781 T

−0.0000782554 −0.13781 T −0.0000040282 0.09624 T 0.0000034498 = 0.28488 T . m −0.0000769346 0.13781 T 0.0000008302 −0.09624 T 0.0000067118 0.19634 T . m

0.09624 T E

G

0.13781 T

0.19634T.m

0.28488 T.m

0.09624 T

k. De esta manera, las reacciones en los apoyos y los diagramas finales de fuerza axial, fuerza cortante y momento flector son los mostrados en la figura: a) C

B

0.01142 T 0.84828 T A

0.00976 T

0.01361 T.m

D

E

G

0.46699 T F

H

0.13781 T

0.15525 T 0.09681 T 0.23821 T.m 0.21709 T.m 0.17745 T.m 0.09625 T

a. Esquematizamos los grados de libertad de la estructura 8

5

D C

4 9

6 2

B

3

A

7

11

E

1 12

F

10

b. Determinamos las rigideces axiales y en flexión de vigas y columnas 3

0.4 × 0.6 ) 12

EA viga =2173706 ×0.4 × 0.6

EI viga =2173706.(

EA viga =521689 , 44 T

EI viga =15 650,6832T m

2 3

0.4 × 0.4 ) 12

EA columna=2173706 × 0.4 ×0.4

EI columna=2173706.(

EA columna=347 792.96 T

EI columna=4 637,2395 T m

2

c. Ensamblamos la matriz de rigidez de cada elemento

1 2 3 10 11 12

[ [ [

104 337,888 0 0 −104 337,888 0 0 0 1 502,465 3 756,164 0 −1 502,465 3 756,164 0 3 756,164 12520,546 0 −3 756,164 6 260,273 K BE= −104 337,888 0 0 104 337,888 0 0 0 −1 502,465 −3 756,164 0 1 502,465 −3 756,164 0 3 756,164 6 260,273 0 −3 756,164 12 520,546

4 5 6 7 8 9

] ] ]

104 337,888 0 0 −104 337,888 0 0 0 1 502,465 3756,164 0 −1502,465 3 756,164 0 3 756,164 12520,546 0 −3 756,164 6 260,273 K CD = −104 337,888 0 0 104 337,888 0 0 0 −1 502,465 −3756,164 0 1502,465 −3 756,164 0 3 756,164 6 260,273 0 −3 756,164 12 520,546

0 0 0 1 2 3

2 060,995 0 −3 091,493 −2060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 0 6 182,986 3 091,493 0 3 091,493 K AB= −3 091,493 −2 060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

1 2 3 10 11 12

4 5 6 7 8 9

0 0 0 1 2 3

1 2 3 4 5 6

[ [ [

2 060,995 0 −3 091,493 −2 060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 0 6 182,986 3 091,493 0 3091,493 K BC = −3 091,493 −2 060,995 0 3 091,493 2 060,995 0 3091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

10 11 12 7 8 9 2060,995 0 −3 091,493 −2 060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 0 6 182,986 3 091,493 0 3 091,493 K ED= −3 091,493 −2060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

0 0 0 10 11 12

2 060,995 0 −3 091,493 −2 060,995 0 −3 091,493 0 115 930,987 0 0 −115 930,987 0 −3 091,493 0 6 182,986 3 091,493 0 3 091,493 K FE= −2060,995 0 3 091,493 2 060,995 0 3 091,493 0 −115 930,987 0 0 115 930,987 0 −3 091,493 0 3 091,493 3 091,493 0 6 182,986

] ] ]

1 2 3 4 5 6

10 11 12 7 8 9

0 0 0 10 11 12

d. Ahora, ensamblamos la matriz de rigidez de la estructura.

1 2 3 4 5 6 7 8 9 10 11 12

[

108 459,878 0 0 −2 060,995 0 −3 091,493 0 0 0 −104 337,888 0 233 364,439 3 756,164 0 −115 930,987 0 0 0 0 0 − 0 3 756,164 24 886,518 3 091,493 0 3 091,493 0 0 0 0 − −2 060,995 0 3 091,493 106 398,883 0 3 091,493 −104 337,888 0 0 0 −3 091,493 −115 930,987 0 0 117 433,452 3 756,164 0 −1502,465 3756,164 0 0 3 756,164 3 091,493 3 091,493 3 756,164 18 703,532 0 −3756,164 6260,273 0 K HG= 0 0 0 −104 337,888 0 0 106 398,883 1502,465 3091,493 −2 060,995 0 0 0 0 −1502,465 −3 756,164 0 112 174,823 −3 756,164 0 − 0 0 0 0 3 756,164 6 260,273 3 091,493 −3756,164 18 703,532 −3 091,493 0 0 0 0 0 0 −2 060,995 0 −3091,493 108 459,878 0 0 0 0 0 0 0 −115 930,987 0 0 2 0 0 0 0 0 0 3 091,493 0 3091,493 0 − 1 2 3 4 5 6 7 8 9 10 11 12

e. El vector fuerza de los nudos del sistema es:

[]

3 0 0 5 0 S 0 F= 0 0 0 0 0 0

1 2 3 4 5 6 7 8 9 10 11 12

f.

El vector de fuerzas de empotramiento perfecto de cada elemento será:

[] [] [] [] [] []

0 0 F EAB= 0 0 0 0

0 0 0 FE = 1 BC 2 3

0 0 0 0 0 0

1 2 3 FE = 4 BE 5 6

0 0 0 0 0 0

0 0 0 FE = 4 CD 5 6

0 0 0 0 0 0

4 5 6 FE = 7 ED 8 9

0 0 0 0 0 0

0 0 0 FE = 7 FE 8 9

0 0 0 0 0 0

g. Ensamblamos el vector de fuerza de empotramiento perfecto del sistema:

[]

0 0 0 0 0 E 0 F = 0 0 0 0 0 0

1 2 3 4 5 6 7 8 9 10 11 12

h. De esta manera, el vector de fuerzas internas del sistema será: S

E

[][][]

F=F −F

3 0 3 0 0 0 0 0 0 5 0 5 0 0 0 S 0 0 0 F= − = 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

7 8 9 10 11 12

i.

Resolviendo la ecuación, obtenemos los desplazamientos y pendientes.

[ δ ] =K−1 × [ F ]

[ ] [][ ] δ1 δ2 δ3 δ4 δ5 δ6 −1 =K × δ7 δ8 δ9 δ 10 δ 11 δ 12

j.

3 0.0000859359 m 0 0.0000244003 m 0 −0.0001628829 rad 5 0.0018957418 m 0 0.0000373283 m 0 −0.0002157656 rad = 0 0.0018723506 m 0 −0.0000355964 m 0 −0.0002124172 rad 0 0.0000295245 m 0 −0.0000210537 m 0 −0.0002093806 rad

Ahora calculamos las fuerzas internas de los elementos:

[ F ]ij =[ F ]ijE + [ K ] ij × [ δ ] ij

[] [ ][ ]

0 0 0.32644 T 0 0 −2.82875 T 0 [ F ] AB= 0 + [ K ] AB × = −0.23788 T .m 0 0.0000859359 −0.32644 T 0 0.0000244003 2.82875 T 0 −0.0001628829 −0.74143 T .m 0.32644 T

B 0.74143 T.m 2.82875 T

A

0.32644 T 0.23788 T.m

2.82875 T

[] [ ][ ]

0 0 [ F ] BC = 0 + [ K ] BC × 0 0 0

0.0000859359 −2.55941 T 0.0000244003 −1.49875 T −0.0001628829 = 3.92086 T .m 0.0018957418 2.55941 T 0.0000373283 1.49875 T −0.0002157656 3.75737 T .m

C

B

[] [ ][ ]

0 0 [ F ] ED= 0 + [ K ] ED × 0 0 0 D

E

0.0000295245 −2.49407 T −0.0000210537 1.68595 T −0.0002093806 = 3.74580 T . m 0.0018723506 2.49407 T −0.0000355964 −1.68595 T −0.0002124172 3.75738 T . m

[] [ ][ ]

0 0 0.58645 T 0 0 2.44078 T 0 0 −0.55602 T .m [ F ] FE= + [ K ] FE × = 0 0.0000295245 −0.58645 T 0 −0.0000210537 −2.44078 T 0 −0.0002093806 −1.20332 T . m

E

[] [ ][ ]

0 0 F [ F ] BE= 0 + [ K ] BE × 0 0 0

0.0000859359 5.88585 T 0.0000244003 −1.32999 T −0.0001628829 = −3.17943 T . m 0.0000295245 −5.88585 T −0.0000210537 1.32999 T −0.0002093806 −3.47052 T . m

E

B

[] [ ][ ]

0 0 [ F ]CD = 0 + [ K ] CD × 0 0 0

C

0.0018957418 2.44059 T 0.0000373283 −1.49876 T −0.0002157656 = −3.75738 T . m 0.0018723506 −2.44059 T −0.0000355964 1.49876 T −0.0002124172 −3.73641 T . m

D

EJERCICIO 1 CON SAP2000: a)

b)

c)

d)

Ejercicio 2 con Sap2000: a)

b)

c)

d)