Analisis Sismico.docx

ANALISIS DE SISMICO (PORTICO PLANO) EJEMPLO 1

K=[

𝐾𝛿𝛿 𝐾𝜃𝛿

𝐸𝐼 𝐾𝛿𝜃 ]= 3 ℎ 𝐾𝜃𝜃

72 −24 6ℎ = 6ℎ −6ℎ [−6ℎ

−24 6ℎ 6ℎ −6ℎ −6ℎ 24 6ℎ 6ℎ 6ℎ 6ℎ 6ℎ 16ℎ2 2ℎ2 2ℎ2 0 6ℎ 2ℎ2 16 ℎ2 0 2ℎ2 6ℎ 2ℎ2 0 2ℎ2 ℎ2 6ℎ 0 2ℎ2 ℎ2 6ℎ2 ]

 72  24 6 6  6  6  D1  L1   F1  24 24 6 6 6 6   D 2  L 2  F 2   6 6 16 2 2 0   D3   1   0  K     6 6 2 16 0 2 D 4   2     0  6 6 2 0 2 1   D5   3   0        6 0 2 1 6   D6   4   0    6

{𝜃} = −[𝐾𝜃𝜃]−1 [𝐾𝜃𝛿][𝛿𝐿]  0.066567312  0.009190263  0.023348236 .006954794    0.009190263 0.066567312 0.006954794  0.023348236  [ ]^ 1   M .inv  0.023348236 0.006954794 0.179831098  0.032290114    0.006954794  0.023348236  0.032290114 0.179831098   0.4426  0.2459  6 6   6 6  1  0.4426  0.2459   [ ]  [ ]  [L]   6 6 n  0.9836  0.7869      0.9836  0.7869   6 6

𝐾𝐿 = 𝐾𝛿𝛿 − 𝐾𝜃𝛿 𝑇 𝐾𝜃𝜃 −1 𝐾𝜃𝛿  0.066567312   72  24 6 6  6  6  0.00919026 KL    X     24 24  6 6 6 6   0.02334823   0.006954794  54.88  17.58 KL     17.50 11.60  KL 

 0.00919026 0.066567312 0.006954794  0.02334823

EI  54.88  17.58 h^ 2  17.50 11.60 

FRECUNCIAS Y MODOS [(𝐾𝐿) − 𝜔2 (𝑀)]{𝑎} = {0}…….(I) [(𝐾𝐿) − 𝜔2 (𝑀)] = {0}…….(II)  2m 0 0 0 0 0  0 m 0 0 0 0    0 0 0 0 0 0 (M )     0 0 0 0 0 0  0 0 0 0 0 0    0 0 0 0 0 0

 2 m 0 (m)     0 0

 0.02334823 0.006954794 0.179831098  0.03229011

.006954794   6  0.02334823  6 X  0.03229011  6   0.179831098   6

6 6 6  6

𝐸𝐼 𝜔1 = 2.198√ 3 𝑚ℎ 2𝑥106 𝑥(0.30𝑥0.553 /12) 𝜔1 = 2.198√ 3𝑥3.53 𝜔1 = 3817.71 → 𝜔1 = 0.3817

𝑡𝑛 𝑚

𝐸𝐼 𝜔2 = 5.850√ 3 𝑚ℎ 2𝑥106 𝑥(0.30𝑥0.553 /12) 𝜔2 = 5.850√ 1.5𝑥3.53 𝜔2 = 14369.65 → 𝜔2 = 1.4369 𝑡𝑛/𝑚

APLICAR

KL 

𝑚=

 54.88  17.58  10648.83  3396.94  EI  54.88  17.58 , KL  194.02 x      h^3  17.50 11.60   17.50 11.60   3396.94 123596.31 1.5 𝑡/𝑚𝑥7𝑚 9.81 𝑠𝑒𝑐/𝑠𝑒𝑔

= 1.07

2.14 0  [m]     0 1.07 2𝑥106 𝑥(0.30𝑥0.553 /12) 𝜔1 = 2.198√ = 26.60 𝑟𝑎𝑑/𝑠𝑒𝑔 1.07𝑥3.53 2𝑥106 𝑥(0.30𝑥0.553 /12) 𝜔2 = 5.850√ = 78.76 𝑟𝑎𝑑/𝑠𝑒𝑔 1.07𝑥3.53 𝜔1 → 𝑇1 =

2𝜋 2𝜋 = = 0.21 𝑠𝑒𝑔 𝜔1 26.60

𝜔2 → 𝑇2 =

2𝜋 2𝜋 = = 0.08 𝑠𝑒𝑔 𝜔2 78.76

 1   1  {a}1   , {a}2      0.77 2.58

ESPECTRO DE RESPUESTAS 𝟐

ḿ𝐣 = ∑ 𝐦𝐣𝐢 𝐚𝟐𝐣𝐢 , 𝐢

ḿ1 = (2.14)(1)2 + (1.07)(2.58)2 = 9.26 t − sec/m

ḿ2 = (2.14)(1)2 + (1.07)(−0.77)2 = 2.77 𝑡 − 𝑠𝑒𝑐/𝑚 Ḱ𝐢 = ḿ𝐢 𝛚𝟐𝐢 Ḱ∗𝐣 𝐾1∗ = (9.26)(26.60)2 = 6552.0 𝑡/𝑚 𝐾2∗ = (2.77)(78.76)2 = 17182.69 𝑡/𝑚 𝓛∗𝐈 = {𝐚}𝐓𝐢 [𝐌]{𝐡}𝐱−𝐲 2.14 2.14 0  1  1  1 2.58    ,  1  1 2.58    4.90t  sec 2 / m  1.07   0 1.07 1 2.14  1  1  0.77    2.14  0.77 x1.07  1.32t  sec 2 / m 1.07 

𝑠𝑎𝑖 =

𝑍 𝑈 𝐶𝑖 𝑆 𝑥𝑔 , 𝑅

𝑧 = 0.45 (𝑜𝑠𝑡𝑎),

𝑢 = 1.0 (𝑣𝑖𝑣𝑖𝑒𝑛𝑑𝑎)

𝑠 = 1.05 , 𝑇𝑃 = 0.6 𝑠𝑒𝑔 , 𝑅 = 7 , 𝑔 = 9.81 𝑚/𝑠𝑒𝑔 𝑠𝑎𝑖 =

𝑍 𝑈 𝐶𝑖 𝑆 0.45𝑥1𝑥2.5𝑥1.05 𝑥 𝑔 → 𝑠𝑎1 = 𝑠𝑎2 = 𝑥 9.81 𝑅 7

𝑠𝑎1 = 𝑠𝑎2 = 0.17 𝑔 𝑐𝑖 = 2.5 (

𝑇𝑃 0.6 ) ≤ 2.5 , 𝑐1 = 2.5 ( ) = 7.14 𝑐𝑖 0.21 ∴ 𝑐1 = 2.5 , 𝑐2 = 2.5

{𝐹𝑒𝑠𝑡}𝑚𝑎𝑥

𝑙1∗ 4.90 2.14 0 1.0 = ∗ [𝑀]{𝑎}1 𝑆𝑎𝑖 = [ ][ ] 𝑥0.77𝑥9.81 1.07 2.58 𝑚𝑎 9.26 0

{𝐹𝑒𝑠𝑡}𝑚𝑎𝑥1 =

4.90 2.14 1.0 1.89 𝑡/𝑚 [ ][ ] 𝑥0.77𝑥9.81 = [ ] 2.44 𝑡/𝑚 9.26 1.07 2.58

{𝐹𝑒𝑠𝑡}𝑚𝑎𝑥2 =

1.32 2.14 1.7 𝑡/𝑚 1.0 [ ][ ] 𝑥0.77𝑥9.81 = [ ] −0.65 𝑡/𝑚 2.77 1.07 −0.77

𝑉𝐵𝐴𝑆𝐴𝐿𝑖

𝑙𝑖 ∗ = ∗ 𝑥 𝑆𝑎𝑖 𝑚𝑖

𝑉𝐵𝐴𝑆𝐴𝐿𝑖1

(4.90)2 = 𝑥0.17𝑥9.81 = 4.32 𝑔 9.26

PROBLEMA 2

48

−24 0 0 −6ℎ −6ℎ 24 6ℎ 6ℎ 6ℎ 6ℎ −24 2 2 2 6ℎ 10ℎ 𝐸𝐼 ℎ 2ℎ 0 𝐾𝛿𝛿 𝐾𝛿𝜃 0 𝐾=[ ]= 3= 2 2 𝐾𝜃𝛿 𝐾𝜃𝜃 ℎ 0 6ℎ ℎ 10ℎ 0 2ℎ2 −6ℎ 6ℎ 2 2 2 0 6ℎ ℎ 2ℎ −6ℎ 6ℎ 0 2ℎ2 ℎ2 6ℎ2 ] [  48  24 0 0  6  6  D1  L1   F1  24 24 6 6 6 6   D 2  L 2  F 2   0 6 10 1 2 0   D3   1   0  K     0 6 1 10 0 2 D 4   2     0  6 6 2 0 6 1   D5   3   0        6 0 2 1 6   D6   4   0    6

{𝜃} = −[𝐾𝜃𝜃]−1 [𝐾𝜃𝛿][𝛿𝐿]  0.108920815  0.013030404  0.038088874 0.010691614   0.013030404 0.108920815 0.010691614  0.038088874  [ ]^ 1   M .inv  0.038088874 0.010691614 0.185098563  0.034413632    0.010691614  0.038088874  0.034413632 0.185098563   0 6  0.16438  0.41096  0 6   1  0.16438  0.41096   [ ]  [ ]  [L]   6 6 n  0.90411  0.73973       6 6  0.90411  0.73973

𝐾𝐿 = 𝐾𝛿𝛿 − 𝐾𝜃𝛿 𝑇 𝐾𝜃𝜃 −1 𝐾𝜃𝛿  0.108920815  0.013030404  0.038088874 0.010691614   0     48  24 0 0  6  6  0.013030404 0.108920815 0.010691614  0.038088874  06 KL    X     24 24  6 6 6 6   0.038088874 0.010691614 0.185098563  0.034413632  6     0.010691614  0.038088874  0.034413632 0.185098563   6  37.15  15.12 KL     15.12 10.19  KL 

EI  37.15  15.12 h^ 2  15.12 10.19 

FRECUNCIAS Y MODOS [(𝐾𝐿) − 𝜔2 (𝑀)]{𝑎} = {0}……. (I) [(𝐾𝐿) − 𝜔2 (𝑀)] = {0}……. (II)  2m 0 0 0 0 0  0 m 0 0 0 0    0 0 0 0 0 0 (M )     0 0 0 0 0 0  0 0 0 0 0 0    0 0 0 0 0 0

 2 m 0 (m)     0 0

𝐸𝐼 𝜔1 = 2.198√ 3 𝑚ℎ 2𝑥106 𝑥(0.30𝑥0.453 /12) 𝜔1 = 2.198√ 5𝑥3.53 𝜔1 = 10.13 → 𝜔1 = 0.10

𝑡𝑛 𝑚

𝐸𝐼 𝜔2 = 5.850√ 3 𝑚ℎ 2𝑥106 𝑥(0.30𝑥0.453 /12) 𝜔2 = 5.850√ 2.5𝑥3.53 𝜔2 = 38.14 → 𝜔2 = 0.38 𝑡𝑛/𝑚

APLICAR

KL 

 37.15  15.12  7207.97589  2934.22027 EI  37.15  15.12 , KL  106.26     h^3  15.12 10.19   15.12 10.19   2934.22027 1977.40932 

𝑚=

2.5𝑡/𝑚𝑥7𝑚 = 1.78 9.81 𝑠𝑒𝑐/𝑠𝑒𝑔

6 6 6  6

3.56 0  [m]     0 1.78 2𝑥106 𝑥(0.30𝑥0.453 /12) 𝜔1 = 2.198√ = 5.09 𝑟𝑎𝑑/𝑠𝑒𝑔 1.78𝑥3.53 2𝑥106 𝑥(0.30𝑥0.453 /12) 𝜔2 = 5.850√ = 13.56 𝑟𝑎𝑑/𝑠𝑒𝑔 1.78𝑥3.53 𝜔1 → 𝑇1 =

2𝜋 2𝜋 = = 1.23 𝑠𝑒𝑔 𝜔1 5.09

𝜔2 → 𝑇2 =

2𝜋 2𝜋 = = 0.46 𝑠𝑒𝑔 𝜔2 13.56

MODO 1

[(𝐾𝐿) − 𝜔2 (𝑀)]{𝑎} = {∅}  37.15  15.12 3.56 0   a11  ( 5 . 09 )^ 2  15.12 10.19   0 1.78 a22          𝑎11 = 1 (37.15) − (5.09)2 𝑥(3.56)(1) − 15.12𝑎21 = 0 (37.15) − (5.09)2 𝑥(3.56)(1) = 15.12𝑎21 −55.09 = 15.12𝑎21 → 𝑎21 =

−55.09 = −3.64 15.12

𝑎21 = −3.64

 1  {a}1     3.64 MODO 2

[(𝐾𝐿) − 𝜔2 (𝑀)]{𝑎} = {∅}  37.15  15.12 3.56 0   a11  15.12 10.19   (13.56)^ 2 0 1.78 a 22         𝑎11 = 1 (37.15) − (13.56)2 𝑥(3.56)(1) − 15.12𝑎22 = 0 (37.15) − (13.56)2 𝑥(3.56)(1) = 15.12𝑎22

−617.44 = 15.12𝑎21 → 𝑎21 =

−617.44 = −40.83 15.12

𝑎22 = −40.83

 1  {a}2     40.83

ESPECTRO DE RESPUESTAS 𝟐

ḿ𝐣 = ∑ 𝐦𝐣𝐢 𝐚𝟐𝐣𝐢 , 𝐢

ḿ1 = (3.56)(1)2 + (1.78)(−3.64)2 = −20.02 t − sec/m ḿ2 = (3.56)(1)2 + (1.78)(−40.83)2 = −2963.85 𝑡 − 𝑠𝑒𝑐/𝑚 Ḱ𝐢 = ḿ𝐢 𝛚𝟐𝐢 Ḱ∗𝐣 𝐾1∗ = (−20.02)(5.09)2 = −5.18.68 𝑡/𝑚 𝐾2∗ = (−2963.85)(13.56)2 = −544973.76 𝑡/𝑚 𝓛∗𝐈 = {𝐚}𝐓𝐢 [𝐌]{𝐡}𝐱−𝐲 3.56 0  1  1  1  3.64     ,  1  1  3.64  0 1.78 1

3.56 1.78   2.91t  sec 2 / m  

3.56  2  1  40.83    3.56  40.83x1.78  69.11t  sec 2 / m 1.78 

𝑠𝑎𝑖 =

𝑍 𝑈 𝐶𝑖 𝑆 𝑥𝑔 , 𝑅

𝑧 = 0.45 (𝑜𝑠𝑡𝑎),

𝑢 = 1.0 (𝑣𝑖𝑣𝑖𝑒𝑛𝑑𝑎)

𝑠 = 1.05 , 𝑇𝑃 = 0.6 𝑠𝑒𝑔 , 𝑅 = 7 , 𝑔 = 9.81 𝑚/𝑠𝑒𝑔 𝑠𝑎𝑖 =

𝑍 𝑈 𝐶𝑖 𝑆 0.45𝑥1𝑥1.22𝑥1.05 𝑥 𝑔 → 𝑠𝑎1 = 𝑥𝑔 𝑅 7

𝑠𝑎1 = 0.08𝑔 𝑠𝑎2 =

0.45𝑥1𝑥2.5𝑥1.05 𝑥 𝑔 = 0.17𝑔 7

𝑐𝑖 = 2.5 (

𝑇𝑃 0.6 0.6 ) ≤ 2.5 , 𝑐1 = 2.5 ( ) = 1.22 , 𝑐2 = 2.5 ( ) = 3.26 𝑐𝑖 1.23 0.46 ∴ 𝑐1 = 1.22 , 𝑐2 = 2.5

{𝐹𝑒𝑠𝑡}𝑚𝑎𝑥

𝑙1∗ −2.91 3.56 1.0 0 = ∗ [𝑀]{𝑎}1 𝑆𝑎𝑖 = [ ][ ] 𝑥0.08𝑥9.81 1.78 −3.64 𝑚𝑎 −20.02 0

{𝐹𝑒𝑠𝑡}𝑚𝑎𝑥1 =

−2.91 3.56 0.41 𝑡/𝑚 1.0 [ ][ ] 𝑥0.08𝑥9.81 = [ ] −0.74 𝑡/𝑚 −20.02 1.78 −3.64

{𝐹𝑒𝑠𝑡}𝑚𝑎𝑥2 =

−69.11 3.56 0.14 𝑡/𝑚 1.0 [ ][ ] 𝑥0.17𝑥9.81 = [ ] −2.82 𝑡/𝑚 −2963.85 1.78 −40.83

PROBLEMA 3 Hallar los máximos momentos flectores y cortantes de la estructura mostrada, debido al movimiento sísmico. Datos:

U = 1.0, Z = 0.45, Ti = 0.6, S = 1.05, R = 8

Ic = 900 𝑝𝑢𝑙𝑔4 , Iv = 850 𝑝𝑢𝑙𝑔4 , 𝐴𝑎𝑟𝑚𝑎𝑑 = 5 𝑝𝑢𝑙𝑔2 , 𝐸 = 29000 ksi