Ejercicio_Dinamica Estructural_Siccha Santos Asly.xlsx

ALUMNA: SICCHA SANTOS, Asly

INGENIERÍA SÍSMICAVII CICLO

ANÁLISIS DINÁMICO - EJERCICIO 2

Datos de la edificación:

Planta Típica

N° Pisos : 3 pisos VP : 6

Altura e/p : 3.8 m VS : 6

F'c : 210 Longitud entre Ejes:

Fy : 4200 A - B : 5.40

Pacab. : 100 B - C : 5.20

S/C : 400 1 - 2 : 7.20

Espesor de losa 20 cm 2 - 3 : 7.40P. Propio Losa: 300

Área Total de paños 135 Longitud de Vigas:

Área Total : 165.00 VP - 1: 7.00

Ɣ Concreto: 2400 VP - 2: 6.80

Columnas: N° = 9 VS - 1: 5.00C-1 : 0.40 x 0.40 m VS - 2: 4.80

Vigas :

kg/cm2

kg/cm2

kg/m2

kg/m2

kgf/m2

m2

m2

kgf/m3

VP (.40 x .60) VP (.40 x .60)

VP (.40 x .60) VP (.40 x .60)

VP (.40 x .60) VP (.40 x .60)

VS

(.4

0 x

.40)

VS

(.4

0 x

.40)

VS

(.4

0 x

.40)

VS

(.4

0 x

.40)

VS

(.4

0 x

.40)

VS

(.4

0 x

.40)

7.2 7.4

5.4

5.2

76.8

4.8

5

A

B

C

321



VP : 0.40 x 0.60 m

VS : 0.40 x 0.40 m



1. Estableciendo el modelo Dinámico:

Según el modelo se tiene: 3 GDL

Entonces la amtriz de rigidez lateral es

2. Metrado de Cargas:

10

0%

DESCRIPCIÓN NIVEL 1 NIVEL 2 NIVEL 3

Peso de la losa 40572.00 Kg 40572.00 Kg 40572.00 Kg

Peso de acabados 13524.00 Kg 13524.00 Kg 13524.00 Kg

Peso VP 23846.40 Kg 23846.40 Kg 23846.40 Kg

Peso VS 11289.60 Kg 11289.60 Kg 11289.60 Kg

Peso C-1 13132.80 Kg 13132.80 Kg 13132.80 Kg

50

% S/C 66000.00 Kg 66000.00 Kg 66000.00 Kg

Peso por Nivel 135364.8 Kg 135364.8 Kg 135364.8 Kg

PESO TOTAL 406094.4 Kg 406.09 Tn

Gráficamente

P3 : 128.80 Tn

P2 : 135.36 Tn

P1 : 135.36 Tn

k3x3

F3

F2

F1

P3 , m3

K3 , u3

P2 , m2

K2 , u2

P1 , m1

K1 , u1

Vb=zucs

R P

P3

P2

P1



P.Total = 399.53 Tn

P3

P2

P1



3. Cálculo de las rigideces por el Método Aproximado "Muto"

Inercias: 213333.33 720000

a. Pórticos Ejes A y C:

Kv = 1000 Kv = 972.973

Kc = 561 561.4 561.4ǩ = 1.73 3.51 1.73

a = 0.46 0.64 0.46

K = 4.71 K = 6.46 K = 4.71

Kc = 561 561.4 561.4

ǩ = 1.73 3.51 1.73

a = 0.46 0.64 0.46

K = 4.71 K = 6.46 K = 4.71

Kc = 561 561.4 561.4

ǩ = 1.78 3.51 1.73

a = 0.60 0.73 0.60

K = 6.12 K = 7.38 K = 6.07

4. Modelo Dinámico:

128.80 Tn

###

47.64 Tn/cm

135.36 Tn###

47.64 Tn/cm

135.36 Tn

###58.70 Tn/cm

IC-1: cm4 IVP: cm4

P3=

m3 =

K3 =

P2=m2 =

K2 =

P1=

m1 =K1 =



1.-Hallando la matriz de masa y rigidez:

Matriz de rigidez Matriz de masas

106.34 -47.64 0 0.14 0 0

-47.64 95.28 -47.64 0 0.14 0

0 -47.64 47.64 0 0 0.13

2.-Calculando las frecuencias:

106.34 -47.64 0 0.14 0 0

= 0-47.64 95.28 -47.64 0 0.14 0

0 -47.64 47.64 0 0 0.13

106.34 -47.64 0 0.14 x 0 0

-47.64 95.28 -47.64 0 0.14 x 0

0 -47.64 47.64 0 0 0.13 x

106.34 - 0.14x -47.64 0

-47.64 95.28 - 0.14x -47.64 = 0

0 -47.64 47.64 - 0.13x

Hallando la determinante:

482647.55 -1330.21 -699.01 1.93 -626.28 1.73 0.91

0.00 0.00 0.00 0.00 -241323.77 313.14 -108108 297.95

### 4.560 x^2 ### ### = 0

= 78.224 = 8.84

= 588.71 = 24.26

= 1157.16 = 34.02

= 0.71 s

Si: w2= x

w21 w1

w22 w2

w23 w3

T1

|𝑲|−𝒘^𝟐 |𝒎|=𝟎|■8(𝒌_𝟏+𝒌_𝟐& −〖 𝒌〗 _𝟐 @█( −〖 𝒌〗 _𝟐@𝟎)&█(𝒌_𝟐+ _𝒌 𝟑@ −〖 𝒌〗 _𝟑 ))■8(𝟎@−〖 𝒌〗 _𝟑@ _𝒌 𝟑 )|−𝒘^𝟐 |■8(𝒎_𝟏&𝟎@█(𝟎@𝟎)&█(𝒎_𝟐@𝟎))■8(𝟎@𝟎@ _𝒎𝟑 )|=𝟎

− 〖𝒘〗^𝟐−



= 0.26 s

= 0.18 s

T2

T3



3.- Modos de vibración:

●MODO 1: = 78.22

106.34 -47.64 0.00 0.14 0.00 0.00

-47.64 95.28 -47.64 0.00 0.14 0.00

0.00 -47.64 47.64 0.00 0.00 0.13

106.34 -47.64 0 10.79 0 0

-47.64 95.28 -47.64 0 10.79 0

0 -47.64 47.64 0 0 10.3

95.55 -47.64 0.00

-47.64 84.48 -47.64

0.00 -47.64 37.37

1

2.01 =

2.56

●MODO 2: = 588.71

106.34 -47.64 0.00 0.14 0.00 0.00

-47.64 95.28 -47.64 0.00 0.14 0.00

0.00 -47.64 47.64 0.00 0.00 0.13

106.34 -47.64 0 81.23 0 0

-47.64 95.28 -47.64 0 81.23 0

0 -47.64 47.64 0 0 77.3

25.11 -47.64 0.00

-47.64 14.04 -47.64

0.00 -47.64 -29.66

1

0.53 =

w21

ø11

ø21

ø31

ø11

ø21

ø31

w22

ø11

ø21

ø31

ø11

ø21

|𝑲|−𝒘^𝟐 |𝒎|{∅_𝑰𝑱 }=𝟎

{∅_𝑰𝑱 }=𝟎− 〖𝒘〗^𝟐− {∅_𝑰𝑱 }=𝟎

=𝟎

{∅_𝑰𝑱 }=𝟎− 〖𝒘〗^𝟐− {∅_𝑰𝑱 }=𝟎

=𝟎



0.85 ø31



●MODO 3: = 1157.16

106.34 -47.64 0.00 0.14 0.00 0.00

-47.64 95.28 -47.64 0.00 0.14 0.00

0.00 -47.64 47.64 0.00 0.00 0.13

106.34 -47.64 0 159.673 0 0

-47.64 95.28 -47.64 0 159.7 0

0 -47.64 47.64 0 0 152

-53.33 -47.64 0.00

-47.64 -64.40 -47.64

0.00 -47.64 -104.29

1

-1.12 =

0.51

Matriz de Eigen Vectores:

ϕ1 1.00 1

= 2.006 0.53 -1.12

2.557 0.85 0.51

4.- Matriz de Frecuencias:

78.22 0 0

= 0 588.71 0

0 0 ###

5.- Matriz de Aceleraciones:

Donde: Z = 0.40 Si : T1 = ### s

U = 1.50 T2 = ### s

Tp = 0.60 T2 = ### s

C = 2.50S = 1.20 Para : C = 2.11

R = 8.00 C = 2.50

w23

ø11

ø21

ø31

ø11

ø21

ø31

Ω2

{∅_𝑰𝑱 }=𝟎− 〖𝒘〗^𝟐− {∅_𝑰𝑱 }=𝟎

=𝟎

𝑆_𝑎=𝑍𝑈𝐶𝑆/𝑅∗𝑔 𝐶=2.5(𝑇𝑝/𝑇)≤2.5



g = 981.00 C = 2.50



0.190

0.225

0.225

A186.42 0 0

= 0 220.725 0

0 0 220.725

6.- Matriz Modal normalizada

MODO 1 MODO 2 MODO 3

1.26 0.52 0.57

ϕ0.792 1.91 1.74

= 1.588 1.00 -1.95

2.024 1.61 0.89

7- Matriz de Participación:

P

0.792 1.59 2.02 0.14 0 0 1

= 1.906 1.00 1.61 0.00 0.14 0 1

1.740 -1.95 0.89 0.00 0 0.13 1

P0.1093 0.2191 0.266 1

= 0.263 0.1386 0.212 1

0.2401 -0.269 0.117 1

P

0.59

= 0.61

0.09

P

0.59 0 0

= 0 0.61 0

Sa1=

Sa2=

Sa3=

𝜙_𝑗𝑛=𝑢_𝑗𝑛/√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗 ^2 〗 )

√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )= √(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=

[𝑃]=[𝜙]^𝑇∙[𝑀]∙{1}



P

0 0 0.09



8.- Matriz de desplazamientos:

U

0.792 ### 1.7403

*0.59 0 0

*186.42 0 0

= 1.588 ### -1.948 0.00 0.61 0 0.00 ### 0

2.024 ### 0.89 0.00 0 0.09 0.00 0 ###

-1

78.2 0.0 0.0

0.0 ### 0.0

0.0 0.0 1157.2

U

0.47 1.17 0.1534

*186.4 0 0

*0.013 ### ###

= 0.944 0.62 -0.172 0.0 ### 0 0.000 ### ###

1.203 0.99 0.0784 0.0 0 ### 0.000 ### ###

U87.7 258 33.856

*0.013 ### ###

= 176 136 -37.9 0.000 ### ###

224 219 17.314 0.000 ### ###

U

1.121 0.44 0.029

= 2.249 0.23 -0.033

2.867 0.37 0.015

Entonces los desplazamientos máximos resultantes son:

U1 = 1.20 cm

U2 = 2.26 cm

U3 = 2.89 cm

9.- Matriz de Fuerzas Laterales:

Fs

106.34 -47.64 0.00

*1.12 0.4 0.029

= -47.64 95.28 ### 2.25 0.2 -0.033

0.00 -47.64 47.64 2.87 0.4 0.015

Fs

12.101 35.62 4.672

= 24.27 -16.56 -5.23

√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=

[𝑈]=[Φ][𝑃][𝐴] [Ω^2 ]^(−1)

[𝐹_𝑠 ]=[𝐾][𝑈]



Fs

29.441 6.6752 2.273



Fuerzas aplicadas:

F1 37.91 Tn

F2 = 29.84 Tn

F3 30.27 Tn

10.- Cortante Basal:

V =

12.1 24.3 29.4

*1

35.6 -16.6 6.7 1

4.7 -5.2 2.3 1

V =

65.81

25.74

1.717

Finalmente:

Vb = 70.69 Tn

11. Modelo Dinámico Final

30.27 Tn

128.80 Tn

0.13 Tn/cm

47.64 Tn/cm

29.84 Tn 135.36 Tn

###

47.64 Tn/cm

135.36 Tn###

37.91 Tn

58.70 Tn/cm

Vb = 70.69 Tn

P3=

m2 =

K3 =

P2=

m2 =

K2 =

P1=m1 =

K1 =

[𝑉]=([𝐹]^𝑇 {1})^𝑇

SICCHA SANTOS, Asly

ANÁLISIS DINÁMICO - EJERCICIO 1

80 Tn

0.0815

7.00 Tn/cm La estructura corresponde

a un colegio de Trujillo

90 Tn

0.0917

11.00 Tn/cm

1.-Hallando la matriz de masa y rigidez:

Matriz de rigidez Matriz de masas

18.00 -7.00 0.09 0

-7 7.00 0 0.08

2.-Calculando las frecuencias:

18.00 -7.00 0.09 0

-7.00 7.00 0 0.08

18.00 -7.00 0.09 0

-7.00 7.00 0 0.08

18.00 - 0.09 x -7.00

-7.00 7 - 0.08x

Hallando la determinante:

0 = [(18.00-0.09x)(7-0.08x)-(49)

Teniendo 77 - -2.11 x + 0.0075 = 0

W2 =

m2 =

K2 =

W1 =

m1 =

K1 =

Si: w2= x

x2

|𝑲|−𝒘^𝟐 |𝒎|=𝟎|■8(𝒌_𝟏+𝒌_𝟐& −〖 𝒌〗 _𝟐@ −〖 𝒌〗 _𝟐&𝒌_𝟐 )|−𝒘^𝟐 |■8(𝒎_𝟏&𝟎@𝟎&𝒎_𝟐 )|=𝟎

− 〖𝒘〗^𝟐=𝟎

− 𝒙 =𝟎=𝟎

(126 −1.47 𝑥−0.64𝑥+〖 0.01𝑥〗^2 )−49

SICCHA SANTOS, Asly

a = 0.007 b = -2.1101 c = 77

= 238.97 = 43.07

Luego:

= 15.46 = 6.56

Período: = 0.41 s = 0.96 s

3.- Modos de vibración:

●MODO 1: = 43.07

18.00 -7.00 - 43.07 0.09 0.00

-7.00 7.00 0.00 0.08

14.05 -7.00 1

-7.00 3.49

= 1.00 cm

= 2.01 cm

●MODO 2: = 238.97

18.00 -7.00 - 239.0 0.09 0.00

-7.00 7.00 0.00 0.08

-3.92 -7.00 1

-7.00 -12.5

= 1.00 cm

= -0.56 cm

Finalmente matriz Eigen Vectores:

ø = 1.00 1.00

2.01 -0.56

w22 w2

1

w2 w1

T2 T1

w21

ø21

ø11

ø21

w22

ø21

ø11

ø21

|𝑲|−𝒘^𝟐 |𝒎|{∅_𝑰𝑱 }=𝟎

{∅_𝑰𝑱 }=𝟎=𝟎

{∅_𝑰𝑱 }=𝟎=𝟎

SICCHA SANTOS, Asly

MODO 1 MODO 2

2.0 cm -0.56 cm

1.0 cm 1.0 cm

T1 = 0.957 s T2 = 0.41 s

4.- Matriz de Frecuencias:

=43.07 0

0 238.97

5.- Matriz de aceleraciones:

Donde: Z = 0.40 Si : T1 = 0.957

U = 1.50 Si : T2 = 0.41

Tp = 0.60

C = 2.50 Para : C = 1.567

S = 1.20 Para : C = 2.50

R = 8.00

g = 9.81 0.141 g

0.225 g

A =138 0

0 221

6.- Matriz Modal Normalizada:

MODO 1 MODO 2

Ω2

Sa1=

Sa2=

𝑆_𝑎=𝑍𝑈𝐶𝑆/𝑅∗𝑔 𝐶=2.5(𝑇𝑝/𝑇)≤2.5

𝜙_𝑗𝑛=𝑢_𝑗𝑛/√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗 ^2 〗 )√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=

SICCHA SANTOS, Asly

0.68 0.33

Finalmente:

ϕ =1.47 3.05

2.96 -1.71

7.- Matriz de Participación:

P =1.47 2.96 . 0.09 0 . 1

3.05 -1.71 0 0.08 1

P =0.135 0.241 . 1

0.28 -0.1396 1

P =0.376

0.141

P =0.376 0

0 0.1406

8.- Matriz de desplazamientos:

U =1.472 3.055

*0.3761 0

*138 0

*0.023 0.000

2.955 -1.712 0 0.14061 0 221 0.000 0.004

U =0.5537 0.43

*138.3 0.00

*### 0.000

1.1113 -0.241 0.00 220.73 ### 0.004

U =76.59 94.8

*0.023 0.000

153.72 -53.14 0.000 0.004

U =1.8 0.4

3.6 -0.2

Entonces los desplazamientos máximos resultantes son:

= 1.8 cm

= 3.6 cm

9.- Matriz de Fuerzas Laterales:

U1

U2

√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=√(∑▒〖𝑚 _𝑗𝑗∙〖𝑢 _𝑗𝑛〗^2 〗 )=

[𝑃]=[𝜙]^𝑇∙[𝑀]∙{1}

[𝑈]=[Φ][𝑃][𝐴] [Ω^2 ]^(−1)

[𝐹_𝑠 ]=[𝐾][𝑈]

SICCHA SANTOS, Asly

Fs =18.00 -7.00

*1.8 0.4

-7.00 7.00 3.6 -0.2

Fs =7.027 8.6974

12.54 -4.3336

= 11.2 Tn

= 13.3 Tn

10.- Cortante Basal:

V =7.027 12.535

*1

8.697 -4.3336 1

V =19.56

4.364

Finalmente:

Vb = 20.04 Tn

11. Modelo Dinámico Final

13.3 Tn P2 = 80.00 Tn

m2 = 0.0815

3.6 cm

11.2 Tn W1 = 90.00 Tn

m2 = 0.09

1.8 cm

20.04 Tn

F1

F2

Tn.s2/cm

u2=

Tn.s2/cm

u1=

[𝑉]=([𝐹]^𝑇 {1})^𝑇

Ejercicio_Dinamica Estructural_Siccha Santos Asly.xlsx

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