Trabajo de Muto

EJEMPLO 1 Resolver el pórtico mostrado Asumir: 

  



 = 210

 

a = 30 ∗ 60   = 30 ∗ 45    = 760  

   A   

=

 



a  a ada



 = 3 0 ∗



Viga



 = 3 0 ∗



a aa



a=



a aa  aaa



=

 ∗∗

 ∗∗

= 1.5

= 1.18

.+ +

 +  

aa   a   a  

a=

 +

Columnas 1-2 2-3 4-5 5-6 6-7 8-9 9-10 10-11

=

∑  

k 2.26 1.18 1.18 1.77 2.36 0.79 1.18 1.18

a

 = a ∗ 

K

0.65 0.37 0.53 0.47 0.54 0.79 0.37 0.37

0.33 0.37 0.53 0.47 0.54 0.23 0.37 0.37

1.75 7.87 11.27 10.00 11.49 11.01 7.87 7.87

RIGIDES LATERAL ABSOLUTA (K)

 = a ∗   =  ∗ (/)

 = 12   / ℎ ara h = ara h

760 300 cm − Do = 12 ∗ 210 ∗

2

300

= 21.28 /

760 = 600 cm − Do = 12 ∗ 210 ∗

ara h =

6002 760

200 cm − Do = 12 ∗ 210 ∗

2002

= 5.32/

= 47.88 /

CALCULAMO DE Δ. TRABAJAMOS CON LOS CONCEPTOS DE COLUMNAS EN PARALELO Y EN SERIE 3er piso = k = 7.78  11.49  7.78 = 27.23 2do piso = (k = 1.75) … . k = 10.00  7.87 = 17.87 1er piso = (k = 1.75) … . k = 11.27  11.01 = 22.28

APLICAMOS PORTICOS CON MESSANINE Y VIGAS ENTREPISO. COLUMNAS EN SERIE 1

k=

1 1  1 2

remplasmos k=

1 1 1  17.87 22.28

= 9.92

resolvemos 15 ∗ 1.75 1.75  9.92 15 ∗ 9.92 1.75  9.92

Calculamos δ1 =

12.75 22.28

= 0.57 cm

; δ2 =

= 2.25 

= 12.75 

12.75 17.87

= 0.71 cm

; δ3 =

10 27.23

= 0.37cm

PUNTOS DE IMFLEXION:

columns

k

∝1

∝2

∝3

0

1

2

3



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

2.26 1.18 1.18 1.77 2.36 0.79 1.18 1.18

2 -

3/6 --

6/3 2/3 -

0.55 0.45 0.64 0.43 0.45

-

0 -

-

0.55 0.4 0.64 0.43 0.45 0 0.46 0.41

0.46 0.41

0 -

EJEMPLO 2 Aplicando el método de Muto, analizar el pórtico resuelto mediante los métodos del Portal y Voladizo Asumir: VIGAS COLUMNAS Ko E

= 0.3X0.5 m = 0.3X0.4 m = 0.0004 m3 = 2000000 tn/m2

1. MOMENTOS DE INERCIA Iv Ic

= =

bxh3/12 bxh3/12

= =

0.3*0.5^3/12 0.3*0.4^3/12

= =

0.003125 m4 0.0016 m4

2. COEFICIENTE DE RIGIDEZ A FLEXION Kv Kv

= =

Iv/(L*Ko) Iv/(L*Ko)

= =

0.003125/(5*0.0004) 0.003125/(6*0.0004)

= =

1.56 1.30

Kc Kc

= =

Ic/(H*Ko) Ic/(H*Ko)

= =

0.0016/(4*0.0004) 0.0016/(3*0.0004)

= =

1.00 1.33

3. COEFICIENTE "a" 1er Nivel Ḵ

a

=

=

∑ Kv/Kc

0.5 + Ḵ /(2 + Ḵ)

2do Nivel Ḵ

a

=

=

∑ Kv/(2*Kc)

Ḵ /(2 + Ḵ)

1.56/1 (1.56 + 1.3)/1 1.3/1

= = =

1.56 2.86 1.30

(0.5 + 1.56)/(2 + 1.56) = (0.5 + 2.86)/(2 + 2.86) = (0.5 + 1.3)/(2 + 1.3) =

0.58 0.69 0.55

1.56/2*1 (1.56 + 1.3)/2*1 1.3/2*1

= = =

1.17 2.15 0.98

(0.5 + 1.56)/(2 + 1.56) = (0.5 + 2.86)/(2 + 2.86) = (0.5 + 1.3)/(2 + 1.3) =

0.37 0.52 0.33

4. RIGIDEZ LATERAL ESTANDAR "Do" 1er Nivel Do =

12*E*Ko/h2

= 12*2000000*0.0004/(4^2)

= 600 ton/m

Do =

12*E*Ko/h2

= 12*2000000*0.0004/(3^2)

= 1066.67 ton/m

2do Nivel

5. RIGIDEZ LATERAL RELATIVA "D" Y RIGIDEZ LATERAL ABSOLUTA "K" D = a* Kc

COL

K = D*Do

a

Ḵ

Kc

D

Do

K

1 2

1.56

0.58 1.00

0.58

600

348

2 3

1.17

0.37 1.33

0.49 1066.67

522.67

4 5

2.86

0.69 1.00

0.69

600

414

5 6

2.15

0.52 1.33

0.69 1066.67

736

7 8

1.30

0.55 1.00

0.55

600

330

8 9

0.98

0.33 1.33

0.44 1066.67

469.33

6. CORTANTES EN COLUMNA EN PARALELO V = Q*(Ki/∑Ki)

V 12 V 23 V 45 V 56 V 78 V 89

Q1 Q2

= =

15 10

= = = = = =

15*348/(348 + 414 + 330) 10*522.67/(522.67 + 736 + 469.33) 15*414/(348 + 414 + 330) 10*736/(522.67 + 736 + 469.33) 15*330/(348 + 414 + 330) 10*469.33/(522.67 + 736 + 469.33)

= = = = = =

4.78 3.02 5.69 4.26 4.53 2.72

ton ton ton ton ton ton

7. DESPLAZAMENTOS = δ Q/∑Ki 1er Nivel δ1

=

15/(330 + 414 + 348)

=

0.014 m

δ2

=

10/(522.67 + 736 + 469.33)

=

0.006 m

2do Nivel

8. CORRECION POR ALTURA Y UBICACIÓN DEL PUNTO DE INFLEXIÓN COL 1 2  2 3 4 5 5 6 7 8 8 9

Ḵ

α1 — 1.56

1.17 2.86 2.15 1.30 0.98

α2

α3 Yo Y1 Y2 Y3 Y 0.75 — 0.57 0 -0.055 0 0.52 — 1.00 — 1.33 0.45 0 -0.027 0.42 — 0.75 — 0.55 0 0 0 0.55 — 1.00 — 1.33 0.45 0 0 0.45 0.59 0 -0.121 0 0.47 —  0.75 — — 1.00 — 1.33 0.44 0 -0.033 0.41

9. MOMENTO FLECTOR EN COLUMNAS MA = V(1 - Y)*H

V

MB = V(Y*H)

COL 1 2 2 3 4 5 5 6 7 8 8 9

M 1 2 2 1 2 3 3 2 4 5 5 4 5 6 6 5 7 8 8 7 8 9 9 8

V

H 4.78 4.78 3.02 3.02 5.69 5.69 4.26 4.26 4.53 4.53 2.72 2.72

Y 4 4 3 3 4 4 3 3 4 4 3 3

MF 0.52 0.49 0.42 0.58 0.55 0.45 0.45 0.55 0.47 0.53 0.41 0.59

9.85 9.27 3.84 5.24 12.51 10.24 5.75 7.03 8.50 9.63 3.32 4.83

10. FUERZA CORTANTE EN VIGAS POR EQUILIBRIO VIGA 2 5 3 6 5 8 6 9

M 2 5 5 2 3 6 6 3 5 8 8 5 6 9 9 6

MA MB MF L 9.27 3.84 13.11 5.00 10.24 5.75 7.99 5.24 0.00 5.24 5.00 7.03 0.00 3.51 8.50 9.63 9.07 6.00 3.32 9.63 12.95 7.03 0.00 3.51 6.00 4.83 0.00 4.83

Q 4.22 1.75 3.67 1.39