En la presente publicación se realizo el análisis dinámico manual de un edificio de forma triangular ubicado en la ciudad de huaraz

1. 2018
INGENIERIA ANTISISMICA
UNIVERSIDAD NACIONAL
“SANTIAGO ANTUNEZ DE MAYOLO”
FACULTAD DE INGENIERIA CIVIL
TRABAJO ESCALONADO
“ANALISIS DINAMICO DE UN EDIFICIO
UBICADO EN LA CIUDAD DE HUARAZ”
ALUMNO:
- ALBA ROSALES ARTHUR
FIC- UNASAM
HUARAZ - 2018
DOCENTE:
- ING. ITA ROBLES LUIS.

2. ANÁLISIS SÍSMICO DE EDIFICIOS USANDO EL MÉTODO ESTÁTICO
1. Coordenadas Globales de la Estructura total:
PRIMER NIVEI
AZOTEA
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3. 2. Calcular la matriz de rigidez Lateral de cada uno de los porticos :
El Portico en modelo estructural
PÒRTICO 1 PÒRTICO 2 PÒRTICO 3
Calculo de la Matriz de Rigidez Lateral del Portico I
Sabemos : [K]=[A]T[K][A]
[K]: Rigidez Lateral del Portico
[K]: Matriz de Rigidez del Portico
[A]: Matriz de Transformacion
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4. Procedimiento a seguir para el calculo de matriz de rigidez:

N°gdl 10
 Grados dinamicos N°gdld 2

Portico 1:
Definicion del Sistema Global de Coordenadas {Q}-{D} :

Definicion del Sistema Local de Coordenadas {q}-{d} :

para 5 y 6
para 1 y 2
para 3y 4
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5. Determnacion de la Matriz de Transformacion [A] :

1° Estado de Deformacion: 2° Estado de Deformacion:
D1=1 , Di=0 D2=1 , Di=0
A11
0
1
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
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
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6. 3° Estado de Deformacion:
4° Estado de Deformacion:
D3=1 , Di=0
D4=1 , Di=0
A13
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
1
0
0
0
0
0

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
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7. 5° Estado de Deformacion: 6° Estado de Deformacion:
D5=1 , Di=0
D6=1 , Di=0
A15
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0

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
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8. 7° Estado de Deformacion: 8° Estado de Deformacion:
D7=1 , Di=0
D8=1 , Di=0
A17
0
0
0
0
0
1
0
0
0
0
0
0
0
0
1
0
0
0
0
0
1

1
0
0
0
0
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
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9. 9° Estado de Deformacion: 10° Estado de Deformacio
D9=1 , Di=0
D10=1 , Di=0
A19
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0
0
0
0
0
0
1

1
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
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10. La Matriz [A] de la estructura:
i 0 25


A1
i 0

A11
i 0

 A1
i 1

A12
i 0


A1
i 5

A16
i 0

 A1
i 6

A17
i 0


A1
i 2

A13
i 0

 A1
i 3

A14
i 0


A1
i 7

A18
i 0

 A1
i 8

A19
i 0


A1
i 4

A15
i 0

 A1
i 9

A110
i 0


barra 1
barra 2
barra 3
A1
0 1 2 3 4 5 6 7 8 9
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 0 -1 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 1 0 0 0 -1 0
0 0 0 0 0 0 0 0 1 0

barra 4
barra 5
barra 6
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11. Determinacion de la Matriz de Rigidez del elemento en el Sistema Local:

Hallando Modulo de Elasticidad, Inercia y Area:
Placa apoyo A : Fc 210

kg
cm2
b 0.2
 m h 0.8
 m
E 15100 Fc
 10
 2.188 10
6



Apa b h
 0.16


E Apa
 3.501 10
5


Iu
b h
3

12
8.533 10
3



 Ik
h b
3

12
5.333 10
4



 Iuk 0

6
0.524

 rad
Ixx
Iu Ik

( )
2
Iu Ik

( ) cos 2
( )

2
 Iuk sin 2
( )

 6.533 10
3




Iyy
Iu Ik

( )
2
Iu Ik

( ) cos 2
( )

2
 Iuk sin 2
( )

 2.533 10
3




E Ixx
 1.43 10
4


E Iyy
 5.543 10
3


Placa Apoyo B: Fc 210

kg
cm2
b 0.2
 m h 2
 m
E 15100 Fc
 10
 2.188 10
6



Apb b h
 0.4


E Apb
 8.753 10
5


Ipb b
h
3
12
 0.133


E Ipb
 2.918 10
5


Viga 1: Fc 210

kg
cm2
b 0.25
 m h 0.40
 m
E 15100 Fc
 10
 2.188 10
6



Av b h
 0.1


E Av
 2.188 10
5


Iv b
h
3
12
 1.333 10
3




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12. E Iv
 2.918 10
3


Para Barra 1 :
0.2
 f 1.2

G
E
2 1 
( )

9.117 10
5



L 3.5
 3 E
 Iyy
 f

G Apa
 L
2

0.011


k1
E Apa

( )
L
0
0
0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

4E Iyy

( ) 1 
( )

L 1 4

( )



















1 10
5

0
0
0
1.485 10
3

2.599 10
3

0
2.599 10
3

6.132 10
3















Para Barra 2 :
L 3.5

3 E
 Ipb
 f

G Apb
 L
2

0.235


k2
E Apb

( )
L
0
0
0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

4E Ipb

( ) 1 
( )

L 1 4

( )



















2.501 10
5

0
0
0
4.208 10
4

7.365 10
4

0
7.365 10
4

2.122 10
5















Para Barra 3 :
L 3

3 E
 Iyy
 f

G Apa
 L
2

0.015


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13. k3
E Apa

( )
L
0
0
E Apa

( )
L

0
0
0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )


0
12E Iyy

( )
L
3
1 4

( )


6E Iyy

( )
L
2
1 4

( )


0
6E Iyy

( )
L
2
1 4

( )


4E Iyy

( ) 1 
( )

L 1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

2E Iyy

( ) 1 2

( )

L 1 4

( )

E Apa

( )
L

0
0
E Apa

( )
L
0
0
0
12E Iyy

( )
L
3
1 4

( )


6E Iyy

( )
L
2
1 4

( )

0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )

0
6E Iyy

( )
L
2
1 4

( )


2E Iyy

( ) 1 2

( )

L 1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

4E Iyy

( ) 1 
( )

L 1 4

( )






































k3
1.167 10
5

0
0
1.167
 10
5

0
0
0
2.323 10
3

3.484
 10
3

0
2.323
 10
3

3.484
 10
3

0
3.484
 10
3

7.074 10
3

0
3.484 10
3

3.378 10
3

1.167
 10
5

0
0
1.167 10
5

0
0
0
2.323
 10
3

3.484 10
3

0
2.323 10
3

3.484 10
3

0
3.484
 10
3

3.378 10
3

0
3.484 10
3

7.074 10
3
























Para Barra 4 :
L 3

3 E
 Ipb
 f

G Apb
 L
2

0.32


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14. k4
E Apb

( )
L
0
0
E Apb

( )
L

0
0
0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )


0
12E Ipb

( )
L
3
1 4

( )


6E Ipb

( )
L
2
1 4

( )


0
6E Ipb

( )
L
2
1 4

( )


4E Ipb

( ) 1 
( )

L 1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

2E Ipb

( ) 1 2

( )

L 1 4

( )

E Apb

( )
L

0
0
E Apb

( )
L
0
0
0
12E Ipb

( )
L
3
1 4

( )


6E Ipb

( )
L
2
1 4

( )

0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )

0
6E Ipb

( )
L
2
1 4

( )


2E Ipb

( ) 1 2

( )

L 1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

4E Ipb

( ) 1 
( )

L 1 4

( )






































k4
2.918 10
5

0
0
2.918
 10
5

0
0
0
5.687 10
4

8.531
 10
4

0
5.687
 10
4

8.531
 10
4

0
8.531
 10
4

2.252 10
5

0
8.531 10
4

3.071 10
4

2.918
 10
5

0
0
2.918 10
5

0
0
0
5.687
 10
4

8.531 10
4

0
5.687 10
4

8.531 10
4

0
8.531
 10
4

3.071 10
4

0
8.531 10
4

2.252 10
5
























Para Barra 5 :
L 4.29

3 E
 Iv
 f

G Av
 L
2

6.259 10
3




k5
12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )

6E Iv

( )
L
2
1 4

( )

4E Iv

( ) 1 
( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


2E Iv

( ) 1 2

( )

L 1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )


12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )


6E Iv

( )
L
2
1 4

( )

2E Iv

( ) 1 2

( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


4E Iv

( ) 1 
( )

L 1 4

( )





























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15. k5
432.608
927.945
432.608

927.945
927.945
2.671 10
3

927.945

1.31 10
3

432.608

927.945

432.608
927.945

927.945
1.31 10
3

927.945

2.671 10
3
















Para Barra 6 :
L 4.29

3 E
 Iv
 f

G Av
 L
2

6.259 10
3




k6
12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )

6E Iv

( )
L
2
1 4

( )

4E Iv

( ) 1 
( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


2E Iv

( ) 1 2

( )

L 1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )


12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )


6E Iv

( )
L
2
1 4

( )

2E Iv

( ) 1 2

( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


4E Iv

( ) 1 
( )

L 1 4

( )




























k6
432.608
927.945
432.608

927.945
927.945
2.671 10
3

927.945

1.31 10
3

432.608

927.945

432.608
927.945

927.945
1.31 10
3

927.945

2.671 10
3
















Acoplando en un solo Matriz:
i 0 2

 j 0 2


K
i j

k1
i j

 K
i 3
 j 3


k2
i j


i 0 5

 j 0 5


K
i 6
 j 6


k3
i j

 K
i 12
 j 12


k4
i j


i 0 3

 j 0 3


K
i 18
 j 18


k5
i j

 K
i 22
 j 22


k6
i j


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16. APLICAREMOS LA FORMULA GENERAL DE MATRIZ DE RIGIDEZ Y ASI TENDREMOS MATRIZ DE ROGODEZ
KK A1
T
K
 A1


KK
0 1 2 3 4 5 6
0
1
2
3
4
5
6
7
8
5
1.028·10 4
-5.92·10 0 0 0 0 -1.166·10
4
-5.92·10 4
5.92·10 0 0 0 0 8.531·10
0 0 5
5.423·10 -432.608 5
-2.918·10 0 -1.361·10
0 0 -432.608 5
2.172·10 0 5
-1.167·10 1.361·10
0 0 5
-2.918·10 0 5
2.922·10 -432.608
0 0 0 5
-1.167·10 -432.608 5
1.171·10
4
-1.166·10 4
8.531·10 3
-1.361·10 3
1.361·10 0 0 4.424·10
-884.754 3
3.484·10 -927.945 927.945 0 0 2.238·10
4
-8.531·10 4
8.531·10 0 0 3
-1.361·10 3
1.361·10 3.071·10

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17. 9 3
-3.484·10 3
3.484·10 0 0 -927.945 927.945
i 0 1

 j 0 1


KLL
i j

KK
i j


KLL
1.028 10
5

5.92
 10
4

5.92
 10
4

5.92 10
4










i 0 7

 j 0 1


KOL
i j

KK
i 2
 j


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18. KOL
0
0
0
0
1.166
 10
4

884.754

8.531
 10
4

3.484
 10
3

0
0
0
0
8.531 10
4

3.484 10
3

8.531 10
4

3.484 10
3




























i 0 1

 j 0 7


KLO
i j

KK
i j 2



KLO
0
0
0
0
0
0
0
0
1.166
 10
4

8.531 10
4

884.754

3.484 10
3

8.531
 10
4

8.531 10
4

3.484
 10
3

3.484 10
3










i 0 7

 j 0 7


KOO
i j

KK
i 2
 j 2



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19. KOO
5.423 10
5

432.608

2.918
 10
5

0
1.361
 10
3

927.945

0
0
432.608

2.172 10
5

0
1.167
 10
5

1.361 10
3

927.945
0
0
2.918
 10
5

0
2.922 10
5

432.608

0
0
1.361
 10
3

927.945

0
1.167
 10
5

432.608

1.171 10
5

0
0
1.361 10
3

927.945
1.361
 10
3

1.361 10
3

0
0
4.424 10
5

2.238 10
3

3.071 10
4

0
927.945

927.945
0
0
2.238 10
3

1.588 10
4

0
3.378 10
3

0
0
1.361
 10
3

1.361 10
3

3.071 10
4

0
2.302 10
5

2.238 10
3

0
0
927.945

927.945
0
3.378
2.238
9.744
















KL1 KLL KLO KOO
1

 KOL


7.041 10
4

2.678
 10
4

2.678
 10
4

1.402 10
4











Portico 2:
Definicion del Sistema Global de Coordenadas {Q}-{D} :

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20. Definicion del Sistema Local de Coordenadas {q}-{d} :

Determnacion de la Matriz de Transformacion [A] :

Podemos ver que la matriz [A] es la misma del portico 1
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21. A2 A1

A2
0 1 2 3 4 5 6 7 8 9
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 0 -1 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 1 0 0 0 -1 0
0 0 0 0 0 0 0 0 1 0

Determinacion de la Matriz de Rigidez del elemento en el Sistema Local:

Hallando Modulo de Elasticidad, Inercia y Area:
Placa apoyo A : Fc 210

kg
cm2
b 0.2
 m h 0.8
 m
E 15100 Fc
 10
 2.188 10
6



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22. Apa b h
 0.16


E Apa
 3.501 10
5


Iu
b h
3

12
8.533 10
3



 Ik
h b
3

12
5.333 10
4



 Iuk 0

6
0.524

 rad
Ixx
Iu Ik

( )
2
Iu Ik

( ) cos 2
( )

2
 Iuk sin 2
( )

 6.533 10
3




Iyy
Iu Ik

( )
2
Iu Ik

( ) cos 2
( )

2
 Iuk sin 2
( )

 2.533 10
3




E Ixx
 1.43 10
4


E Iyy
 5.543 10
3


Placa Apoyo B: Fc 210

kg
cm2
b 0.2
 m h 2
 m
E 15100 Fc
 10
 2.188 10
6



Apb b h
 0.4


E Apb
 8.753 10
5


Ipb b
h
3
12
 0.133


E Ipb
 2.918 10
5


Viga 1: Fc 210

kg
cm2
b 0.25
 m h 0.40
 m
E 15100 Fc
 10
 2.188 10
6



Av b h
 0.1


E Av
 2.188 10
5


Iv b
h
3
12
 1.333 10
3




E Iv
 2.918 10
3


Para Barra 1 :
0.2
 f 1.2

G
E
2 1 
( )

9.117 10
5



  
You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)

23. L 3.5
 3 E
 Iyy
 f

G Apa
 L
2

0.011


k1
E Apa

( )
L
0
0
0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

4E Iyy

( ) 1 
( )

L 1 4

( )



















1 10
5

0
0
0
1.485 10
3

2.599 10
3

0
2.599 10
3

6.132 10
3















Para Barra 2 :
L 3.5

3 E
 Ipb
 f

G Apb
 L
2

0.235


k2
E Apb

( )
L
0
0
0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

4E Ipb

( ) 1 
( )

L 1 4

( )



















2.501 10
5

0
0
0
4.208 10
4

7.365 10
4

0
7.365 10
4

2.122 10
5















Para Barra 3 :
L 3

3 E
 Iyy
 f

G Apa
 L
2

0.015


k3
E Apa

( )
L
0
0
E Apa

( )
L

0
0
0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )


0
12E Iyy

( )
L
3
1 4

( )


6E Iyy

( )
L
2
1 4

( )


0
6E Iyy

( )
L
2
1 4

( )


4E Iyy

( ) 1 
( )

L 1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

2E Iyy

( ) 1 2

( )

L 1 4

( )

E Apa

( )
L

0
0
E Apa

( )
L
0
0
0
12E Iyy

( )
L
3
1 4

( )


6E Iyy

( )
L
2
1 4

( )

0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )

0
6E Iyy

( )
L
2
1 4

( )


2E Iyy

( ) 1 2

( )

L 1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

4E Iyy

( ) 1 
( )

L 1 4

( )






































You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)

24. k3
1.167 10
5

0
0
1.167
 10
5

0
0
0
2.323 10
3

3.484
 10
3

0
2.323
 10
3

3.484
 10
3

0
3.484
 10
3

7.074 10
3

0
3.484 10
3

3.378 10
3

1.167
 10
5

0
0
1.167 10
5

0
0
0
2.323
 10
3

3.484 10
3

0
2.323 10
3

3.484 10
3

0
3.484
 10
3

3.378 10
3

0
3.484 10
3

7.074 10
3
























Para Barra 4 :
L 3

3 E
 Ipb
 f

G Apb
 L
2

0.32


k4
E Apb

( )
L
0
0
E Apb

( )
L

0
0
0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )


0
12E Ipb

( )
L
3
1 4

( )


6E Ipb

( )
L
2
1 4

( )


0
6E Ipb

( )
L
2
1 4

( )


4E Ipb

( ) 1 
( )

L 1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

2E Ipb

( ) 1 2

( )

L 1 4

( )

E Apb

( )
L

0
0
E Apb

( )
L
0
0
0
12E Ipb

( )
L
3
1 4

( )


6E Ipb

( )
L
2
1 4

( )

0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )

0
6E Ipb

( )
L
2
1 4

( )


2E Ipb

( ) 1 2

( )

L 1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

4E Ipb

( ) 1 
( )

L 1 4

( )






































k4
2.918 10
5

0
0
2.918
 10
5

0
0
0
5.687 10
4

8.531
 10
4

0
5.687
 10
4

8.531
 10
4

0
8.531
 10
4

2.252 10
5

0
8.531 10
4

3.071 10
4

2.918
 10
5

0
0
2.918 10
5

0
0
0
5.687
 10
4

8.531 10
4

0
5.687 10
4

8.531 10
4

0
8.531
 10
4

3.071 10
4

0
8.531 10
4

2.252 10
5
























Para Barra 5 :
You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)

25. L 4.29

3 E
 Iv
 f

G Av
 L
2

6.259 10
3




k5
12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )

6E Iv

( )
L
2
1 4

( )

4E Iv

( ) 1 
( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


2E Iv

( ) 1 2

( )

L 1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )


12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )


6E Iv

( )
L
2
1 4

( )

2E Iv

( ) 1 2

( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


4E Iv

( ) 1 
( )

L 1 4

( )




























k5
432.608
927.945
432.608

927.945
927.945
2.671 10
3

927.945

1.31 10
3

432.608

927.945

432.608
927.945

927.945
1.31 10
3

927.945

2.671 10
3
















Para Barra 6 :
L 4.29

3 E
 Iv
 f

G Av
 L
2

6.259 10
3




k6
12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )

6E Iv

( )
L
2
1 4

( )

4E Iv

( ) 1 
( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


2E Iv

( ) 1 2

( )

L 1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )


12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )


6E Iv

( )
L
2
1 4

( )

2E Iv

( ) 1 2

( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


4E Iv

( ) 1 
( )

L 1 4

( )




























k6
432.608
927.945
432.608

927.945
927.945
2.671 10
3

927.945

1.31 10
3

432.608

927.945

432.608
927.945

927.945
1.31 10
3

927.945

2.671 10
3
















You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)

26. i 0 2

 j 0 2


K
i j

k1
i j


K
i 3
 j 3


k2
i j


i 0 5

 j 0 5


K
i 6
 j 6


k3
i j


K
i 12
 j 12


k4
i j


i 0 3

 j 0 3


K
i 18
 j 18


k5
i j


K
i 22
 j 22


k6
i j


KK A1
T
K
 A1


KK
0 1 2 3 4 5 6
0
1
2
3
4
5
6
7
8
9
5
1.028·10 4
-5.92·10 0 0 0 0 -1.166·10
4
-5.92·10 4
5.92·10 0 0 0 0 8.531·10
0 0 5
5.423·10 -432.608 5
-2.918·10 0 -1.361·10
0 0 -432.608 5
2.172·10 0 5
-1.167·10 1.361·10
0 0 5
-2.918·10 0 5
2.922·10 -432.608
0 0 0 5
-1.167·10 -432.608 5
1.171·10
4
-1.166·10 4
8.531·10 3
-1.361·10 3
1.361·10 0 0 4.424·10
-884.754 3
3.484·10 -927.945 927.945 0 0 2.238·10
4
-8.531·10 4
8.531·10 0 0 3
-1.361·10 3
1.361·10 3.071·10
3
-3.484·10 3
3.484·10 0 0 -927.945 927.945

You created this PDF from an application that is not licensed to print to novaPDF printer (http://www.novapdf.com)

27. i 0 1

 j 0 1


KLL
i j

KK
i j


KLL
1.028 10
5

5.92
 10
4

5.92
 10
4

5.92 10
4










i 0 7

 j 0 1


KOL
i j

KK
i 2
 j


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28. KOL
0
0
0
0
1.166
 10
4

884.754

8.531
 10
4

3.484
 10
3

0
0
0
0
8.531 10
4

3.484 10
3

8.531 10
4

3.484 10
3




























i 0 1

 j 0 7


KLO
i j

KK
i j 2



KLO
0
0
0
0
0
0
0
0
1.166
 10
4

8.531 10
4

884.754

3.484 10
3

8.531
 10
4

8.531 10
4

3.484
 10
3

3.484 10
3










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29. i 0 7

 j 0 7


KOO
i j

KK
i 2
 j 2



KOO
5.423 10
5

432.608

2.918
 10
5

0
1.361
 10
3

927.945

0
0
432.608

2.172 10
5

0
1.167
 10
5

1.361 10
3

927.945
0
0
2.918
 10
5

0
2.922 10
5

432.608

0
0
1.361
 10
3

927.945

0
1.167
 10
5

432.608

1.171 10
5

0
0
1.361 10
3

927.945
1.361
 10
3

1.361 10
3

0
0
4.424 10
5

2.238 10
3

3.071 10
4

0
927.945

927.945
0
0
2.238 10
3

1.588 10
4

0
3.378 10
3

1.361

1.361
3.071
2.302
2.238
















KL2 KLL KLO KOO
1

 KOL


7.041 10
4

2.678
 10
4

2.678
 10
4

1.402 10
4







Portico 3:
Definicion del Sistema Global de Coordenadas {Q}-{D} :

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30. Definicion del Sistema Local de Coordenadas {q}-{d} :

Determnacion de la Matriz de Transformacion [A] :

Podemos ver que la matriz [A] es la misma del portico 1
A3 A1

0 1 2 3 4 5 6 7 8 9
0
1
2
3
0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 0 0 0 0 0
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31. A3
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 1 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 1 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0
0 1 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1 0
0 0 0 1 0 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0
0 0 1 0 0 0 -1 0 0 0
0 0 0 0 0 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 1
0 0 0 0 1 0 0 0 -1 0
0 0 0 0 0 0 0 0 1 0

Determinacion de la Matriz de Rigidez del elemento en el Sistema Local:

Hallando Modulo de Elasticidad, Inercia y Area:
Placa apoyo A : Fc 210

kg
cm2
b 0.2
 m h .8
 m
E 15100 Fc
 10
 2.188 10
6



Apa b h
 0.16


E Apa
 3.501 10
5


Iu
b h
3

12
8.533 10
3



 Ik
h b
3

12
5.333 10
4



 Iuk 0

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32. 6
0.524

 rad
Ixx
Iu Ik

( )
2
Iu Ik

( ) cos 2
( )

2
 Iuk sin 2
( )

 6.533 10
3




Iyy
Iu Ik

( )
2
Iu Ik

( ) cos 2
( )

2
 Iuk sin 2
( )

 2.533 10
3




E Ixx
 1.43 10
4


E Iyy
 5.543 10
3


Placa Apoyo B: Fc 210

kg
cm2
b 0.2
 m h 2
 m
E 15100 Fc
 10
 2.188 10
6



Apb b h
 0.4


E Apb
 8.753 10
5


Ipb b
h
3
12
 0.133


E Ipb
 2.918 10
5


Viga 1: Fc 210

kg
cm2
b 0.25
 m h 0.40
 m
E 15100 Fc
 10
 2.188 10
6



Av b h
 0.1


E Av
 2.188 10
5


Iv b
h
3
12
 1.333 10
3




E Iv
 2.918 10
3


Para Barra 1 :
0.2
 f 1.2

G
E
2 1 
( )

9.117 10
5



L 3.5
 3 E
 Iyy
 f

G Apa
 L
2

0.011


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33. k1
E Apa

( )
L
0
0
0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

4E Iyy

( ) 1 
( )

L 1 4

( )



















1 10
5

0
0
0
1.485 10
3

2.599 10
3

0
2.599 10
3

6.132 10
3















Para Barra 2 :
3.5
3 E
 Ipb
 f

G Apb
 L
2

0.235


k2
E Apb

( )
L
0
0
0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

4E Ipb

( ) 1 
( )

L 1 4

( )



















2.501 10
5

0
0
0
4.208 10
4

7.365 10
4

0
7.365 10
4

2.122 10
5















Para Barra 3 :
L 3

3 E
 Iyy
 f

G Apa
 L
2

0.015


k3
E Apa

( )
L
0
0
E Apa

( )
L

0
0
0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )


0
12E Iyy

( )
L
3
1 4

( )


6E Iyy

( )
L
2
1 4

( )


0
6E Iyy

( )
L
2
1 4

( )


4E Iyy

( ) 1 
( )

L 1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

2E Iyy

( ) 1 2

( )

L 1 4

( )

E Apa

( )
L

0
0
E Apa

( )
L
0
0
0
12E Iyy

( )
L
3
1 4

( )


6E Iyy

( )
L
2
1 4

( )

0
12E Iyy

( )
L
3
1 4

( )

6E Iyy

( )
L
2
1 4

( )

0
6E Iyy

( )
L
2
1 4

( )


2E Iyy

( ) 1 2

( )

L 1 4

( )

0
6E Iyy

( )
L
2
1 4

( )

4E Iyy

( ) 1 
( )

L 1 4

( )






































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34. k3
1.167 10
5

0
0
1.167
 10
5

0
0
0
2.323 10
3

3.484
 10
3

0
2.323
 10
3

3.484
 10
3

0
3.484
 10
3

7.074 10
3

0
3.484 10
3

3.378 10
3

1.167
 10
5

0
0
1.167 10
5

0
0
0
2.323
 10
3

3.484 10
3

0
2.323 10
3

3.484 10
3

0
3.484
 10
3

3.378 10
3

0
3.484 10
3

7.074 10
3
























Para Barra 4 :
L 3

3 E
 Ipb
 f

G Apb
 L
2

0.32


k4
E Apb

( )
L
0
0
E Apb

( )
L

0
0
0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )


0
12E Ipb

( )
L
3
1 4

( )


6E Ipb

( )
L
2
1 4

( )


0
6E Ipb

( )
L
2
1 4

( )


4E Ipb

( ) 1 
( )

L 1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

2E Ipb

( ) 1 2

( )

L 1 4

( )

E Apb

( )
L

0
0
E Apb

( )
L
0
0
0
12E Ipb

( )
L
3
1 4

( )


6E Ipb

( )
L
2
1 4

( )

0
12E Ipb

( )
L
3
1 4

( )

6E Ipb

( )
L
2
1 4

( )

0
6E Ipb

( )
L
2
1 4

( )


2E Ipb

( ) 1 2

( )

L 1 4

( )

0
6E Ipb

( )
L
2
1 4

( )

4E Ipb

( ) 1 
( )

L 1 4

( )






































k4
2.918 10
5

0
0
2.918
 10
5

0
0
0
5.687 10
4

8.531
 10
4

0
5.687
 10
4

8.531
 10
4

0
8.531
 10
4

2.252 10
5

0
8.531 10
4

3.071 10
4

2.918
 10
5

0
0
2.918 10
5

0
0
0
5.687
 10
4

8.531 10
4

0
5.687 10
4

8.531 10
4

0
8.531
 10
4

3.071 10
4

0
8.531 10
4

2.252 10
5
























Para Barra 5 :
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35. L 4.29

3 E
 Iv
 f

G Av
 L
2

6.259 10
3




k5
12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )

6E Iv

( )
L
2
1 4

( )

4E Iv

( ) 1 
( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


2E Iv

( ) 1 2

( )

L 1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )


12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )


6E Iv

( )
L
2
1 4

( )

2E Iv

( ) 1 2

( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


4E Iv

( ) 1 
( )

L 1 4

( )




























k5
432.608
927.945
432.608

927.945
927.945
2.671 10
3

927.945

1.31 10
3

432.608

927.945

432.608
927.945

927.945
1.31 10
3

927.945

2.671 10
3
















Para Barra 6 :
L 4.29

3 E
 Iv
 f

G Av
 L
2

6.259 10
3




k6
12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )

6E Iv

( )
L
2
1 4

( )

4E Iv

( ) 1 
( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


2E Iv

( ) 1 2

( )

L 1 4

( )

12E Iv

( )
L
3
1 4

( )


6E Iv

( )
L
2
1 4

( )


12E Iv

( )
L
3
1 4

( )

6E Iv

( )
L
2
1 4

( )


6E Iv

( )
L
2
1 4

( )

2E Iv

( ) 1 2

( )

L 1 4

( )

6E Iv

( )
L
2
1 4

( )


4E Iv

( ) 1 
( )

L 1 4

( )




























k6
432.608
927.945
432.608

927.945
927.945
2.671 10
3

927.945

1.31 10
3

432.608

927.945

432.608
927.945

927.945
1.31 10
3

927.945

2.671 10
3
















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36. i 0 2

 j 0 2


K
i j

k1
i j


K
i 3
 j 3


k2
i j


i 0 5

 j 0 5


K
i 6
 j 6


k3
i j


K
i 12
 j 12


k4
i j


i 0 3

 j 0 3


K
i 18
 j 18


k5
i j


K
i 22
 j 22


k6
i j


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37. KK A1
T
K
 A1


KK
0 1 2 3 4 5 6
0
1
2
3
4
5
6
7
8
9
5
1.028·10 4
-5.92·10 0 0 0 0 -1.166·10
4
-5.92·10 4
5.92·10 0 0 0 0 8.531·10
0 0 5
5.423·10 -432.608 5
-2.918·10 0 -1.361·10
0 0 -432.608 5
2.172·10 0 5
-1.167·10 1.361·10
0 0 5
-2.918·10 0 5
2.922·10 -432.608
0 0 0 5
-1.167·10 -432.608 5
1.171·10
4
-1.166·10 4
8.531·10 3
-1.361·10 3
1.361·10 0 0 4.424·10
-884.754 3
3.484·10 -927.945 927.945 0 0 2.238·10
4
-8.531·10 4
8.531·10 0 0 3
-1.361·10 3
1.361·10 3.071·10
3
-3.484·10 3
3.484·10 0 0 -927.945 927.945

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38. i 0 1

 j 0 1


KLL
i j

KK
i j


KLL
1.028 10
5

5.92
 10
4

5.92
 10
4

5.92 10
4










i 0 7

 j 0 1


KOL
i j

KK
i 2
 j


KOL
0
0
0
0
1.166
 10
4

884.754

8.531
 10
4

3.484
 10
3

0
0
0
0
8.531 10
4

3.484 10
3

8.531 10
4

3.484 10
3




























i 0 1

 j 0 7


KLO
i j

KK
i j 2



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39. KLO
0
0
0
0
0
0
0
0
1.166
 10
4

8.531 10
4

884.754

3.484 10
3

8.531
 10
4

8.531 10
4

3.484
 10
3

3.484 10
3










i 0 7

 j 0 7


KOO
i j

KK
i 2
 j 2



KOO
5.423 10
5

432.608

2.918
 10
5

0
1.361
 10
3

927.945

0
0
432.608

2.172 10
5

0
1.167
 10
5

1.361 10
3

927.945
0
0
2.918
 10
5

0
2.922 10
5

432.608

0
0
1.361
 10
3

927.945

0
1.167
 10
5

432.608

1.171 10
5

0
0
1.361 10
3

927.945
1.361
 10

1.361 10

0
0
4.424 10

2.238 10

3.071 10

0
















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40. KL3 KLL KLO KOO
1

 KOL


7.041 10
4

2.678
 10
4

2.678
 10
4

1.402 10
4











CONDENSACIÓN DINÁMICA
KLE
1
N°porticos
ip
C
T
ip
  KLip
 Cip


 



 KLip ........... (1)
i 0 1

 j 0 1


KLL
i j

KK
i j


i 0 7

 j 0 1


I). Calculo del centro de masa:
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41. X
4.5 4

4.5
2
 0.5 4
 1
 4.5
1
3















4.5 4

1
2
4
 1




Y
4.5 4
 2
 0.5 4
 1

2
3
 4








4.5 4

1
2
4
 1




4. Determinar La Matriz de Transformacion de Desplazamiento Para Cada Portico:
Portico 1:

0
 r Y 2.067


C1
cos( )
0
0
cos( )
sin( )
0
0
sin( )
r
0
0
r






1
0
0
1
0
0
0
0
2.067
0
0
2.067








Portico 2:

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42. 2
 r X
 2.508



C2
cos( )
0
0
cos( )
sin( )
0
0
sin( )
r
0
0
r






0
0
0
0
1
0
0
1
2.508

0
0
2.508









Portico 3:

 r 4 Y

( ) 1.933


C3
cos( )
0
0
cos( )
sin( )
0
0
sin( )
r
0
0
r






1

0
0
1

0
0
0
0
1.933
0
0
1.933








Placa 4:

75.963756532
180
1.326

 r 2.4334

C4
cos( )
0
0
cos( )
sin( )
0
0
sin( )
r
0
0
r






0.243
0
0
0.243
0.97
0
0
0.97
2.433
0
0
2.433








5. Calcular La Matriz De Rigidez Lateral Del Edificio:
KLE C1
T
KL1
 C1
 C2
T
KL2
 C2

 C3
T
KL3
 C3



KLE
1.408 10
5

5.356
 10
4

4.311
 10
12


1.64 10
12


9.388 10
3

3.571
 10
3

5.356
 10
4

2.803 10
4

1.64 10
12


8.582
 10
13


3.571
 10
3

1.869 10
3

4.311
 10
12


1.64 10
12


7.041 10
4

2.678
 10
4

1.766
 10
5

6.717 10
4

1.64 10
12


8.582
 10
13


2.678
 10
4

1.402 10
4

6.717 10
4

3.516
 10
4

9.388 10
3

3.571
 10
3

1.766
 10
5

6.717 10
4

1.007 10
6

3.83
 10
5

3.571
 
1.869 10

6.717 10

3.516
 
3.83
 10

2.004 10













6. Definir El Vector Carga Nodal En El Edificio:
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43. Factor de Zona:(3) Parametros del Suelo: (Intermedio)
Z 0.4
 S 1.2
 T 1 T

Factor de Amplificacion Sism ico: Categoria de Edificacion:(Colegio)
C 2.5
Tp
T
 

Tp
C 2.5
 U 1.5

Entonces : C 2.5

Factor de Reducción de Carga:
R 6
 ESTRUCTURA IRREGULAR:
T =0.1*#Pisos
T 0.1 2
 0.2


Metrado de cargas:
Primer Piso:
Cargas muertas:
Aligerado de h=25cm
entonces su peso es
350kg/m2
Peso de la Losa: Pl 4 4.5

4 1

2







350
 7 10
3



Peso del Piso Terminado: Ppt 4 4.5

4 1

2







100
 2 10
3



Peso de Tabiquería: Ppta 13.1
( ) 0.15
 3.25
 1800
 1.15 10
4



Peso de las Vigas: Pv 0.3 0.45
 14
 2400
 4.536 10
3



Peso de las Columnas: Pc 2 0.35
 0.35
 3.25
 2400
 1.911 10
3



Peso de la Placa: Ppl 0.2 17
 3.25
 2400
 6.432 10
3



md1 Pl Ppt
 Pv
 Pc
 Ppl
 Ppta
 3.337 10
4



Carga VIva: sobrecarga para aulas
de un colegio
300kg/m2
Sobrecarga: ml1 4 4.5

4 1

2







250
 5 10
3



P1
md1 0.5 ml1


( )
1000
35.874

 Tn
Segundo Piso:
Cargas muertas:

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44. Peso de la Losa: Pl 4 4.5

4 1

2







350
 7 10
3



Peso del Piso Terminado: Ppt 4 4.5

4 1

2







100
 2 10
3



Peso de Tabiquería: Ppta 13.1
( ) 0.15
 1.5
 1800
 5.306 10
3



Peso de las Vigas: Pv 0.3 0.45
 14
 2400
 4.536 10
3



Peso de las Columnas: Pc 2 0.35
 0.35
 1.5
 2400
 882


Peso de la Placa: Ppl 0.2 17
 1.5
 2400
 2.969 10
3



md2 Pl Ppt
 Pv
 Pc
 Ppl
 Ppta
 2.269 10
4



Carga VIva: Se considera como
azotea, por tanto la
sobrecarga es
100kg/m2
Sobrecarga: ml2 4 4.5

4 1

2







100
 2 10
3



P2
md2 0.25 ml2


( )
1000
23.192

 Tn
P P1 P2
 59.066

 Tn
V
Z U
 C
 S
 P

( )
R
17.72

 Tn
Po 0.7 T
 V
 2.481

 Po 0.15V

Suma :
h1 3.5
 h2 6.5

Wh h1 P1
 h2 P2

 276.309


F1
P1 h1

( ) V Po

( )

Wh
6.925

 F2
P2 h2

( ) V Po

( )

Wh
8.314


SIsmo en el Eje "X": SIsmo en el Eje "Y":
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45. i 0 5

 i 0 5


Qx
i 0

0
 Qx
0 0

F1
 Qx
1 0

F2
 Qy
i 0

0
 Qy
2.0
F1
 Qy
3 0

F2

Qy
0
0
6.925
8.314
0
0

















Qx
6.925
8.314
0
0
0
0

















7. Determinacion de los Desplazamientos del Edificio:
Desplazamiento debido a "Qx" Desplazamiento debido a "Qy"
Dx KLE
1

Qx

5.935 10
4


1.431 10
3


2.478
 10
5


5.975
 10
5


9.88
 10
6


2.382
 10
5

























 Dy KLE
1

Qy

2.478
 10
5


5.975
 10
5


2.118 10
3


5.107 10
3


3.718 10
4


8.963 10
4


























8. Determinar los Desplazamientos Laterales en los Porticos que forman Parte del
Edificio:
Portico 1:
Desplazamiento en "X" Desplazamiento en "Y"
d1px C1 Dx

5.731 10
4


1.382 10
3











 d1py C1 Dy

7.435 10
4


1.793 10
3












Portico 2:
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46. Desplazamiento en "X" Desplazamiento en "Y"
d2px C2 Dx

0
0







 d2py C2 Dy

1.186 10
3


2.859 10
3












Portico 3:
Desplazamiento en "X" Desplazamiento en "Y"
d3px C3 Dx

6.126
 10
4


1.477
 10
3











 d3py C3 Dy

7.435 10
4


1.793 10
3












Placa:
Desplazamiento en "X" Desplazamiento en "Y"
d4px C4 Dx

9.586 10
5


2.311 10
4











 d4py C4 Dy

2.954 10
3


7.121 10
3












9. Calcular las Fuerzas Laterales Que Actuan En Cada Portico:
Portico 1:
Fuerza en "X" Fuerza en "Y"
q1px KL1 d1px

3.347
4.019







 q1py KL1 d1py

4.343
5.214








Portico 2:
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47. Fuerza en "X" Fuerza en "Y"
q2py KL2 d2py

6.925
8.314








q2px KL2 d2px

0
1.168 10
15










Portico 3:
Fuerza en "X" Fuerza en "Y"
q3px KL3 d3px

3.578

4.296








 q3py KL3 d3py

4.343
5.214








Placa:
Fuerza en "X" Fuerza en "Y"
q4px KL4 d4px
 
 KL4 q4py KL4 d4py
 
 KL4
10.- Hallando Centro de Rigidez:
Aplicando un Torsor Unitario al Ultimo Nivel:
F
0
0
0
0
0
1

















[F] = [KLE][D]
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48. Dx1
Dx2
Dy1
D KLE
1

F

8.276
 10
7


2.176
 10
6


3.114 10
5


8.186 10
5


1.241 10
5


3.264 10
5


























Dy2
Dz1
Dz2
Determinar las Excentricidades:
Piso 1:
ex1
D
2 0


D
4 0

2.508



ey1
D
0 0

D
4 0

0.067



Piso 2:
Para segundo piso Sumanos el angulo obtenido del primer piso
ex2
D
3 0


D
5 0

D
4 0


1.817



ey2
D
1 0

D
5 0

D
5 0

D
4 0


 

0.028



Ubicacion del Centro de Rigidez:
Piso 1:
XCr ex1 X
 2.665 10
15




YCr ey1 Y
 2


Piso 2:
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49. CR2 CM2
d xr
CR2
CM2



CR2
xcm1 D
4 0

( )
ey1
( )
 8.276
 10
7




ycm1 D
4 0

( )
ex1
( )
 3.114
 10
5




cm2
xcm1
ycm1






8.276
 10
7


3.114
 10
5












CR2 XCR2 YCR2

( )
CM2
X
Y






2.508
2.067








Para segundo piso Sumanos los angulos obtenidos:
XCr ex2 X
 0.691


YCr ey2 Y
 2.039


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