stodola vianello

Datos: g = 980 cm/seg2 Pesos W1 = 135 ton W2 = 131 ton W3 = 127 ton MASAS m1 = 0.137755102 ton-seg2/cm m2 = 0.133673469 ton-seg2/cm m3 = 0.129591837 ton-seg2/cm RIGIDEZ k1 = 359 ton/cm k2 = 296 ton/cm k3 = 127 ton/cm SOLUCIÓN Se puede trabajar con los números exactos o con proporciones m1= 1.063 m2= 1.031 m3= 1 k1= 2.827 k2= 2.331 k3= 1 1ra ITERACION {?} 1 2 3 {Fi} 1.063 2.062 3 m*ω1^2 {V} 6.125 5.062 3 m*ω1^2 {ΔX} 2.167 2.172 3 m*ω1^2 /k {X} 2.167 4.339 7.339 m*ω1^2 /k 2da ITERACION {? cx} 1 2.002 3.387 {Fi} 1.063 2.064 3.387 m*ω1^2 {V} 6.514 5.451 3.387 m*ω1^2 {ΔX} 2.304 2.338 3.387 m*ω1^2 /k {X} 2.304 4.642 8.029 m*ω1^2 /k 3ra ITERACION {? cx} 1 2.015 3.485 {Fi} 1.063 2.077 3.485 m*ω1^2 {V} 6.625 5.562 3.485 m*ω1^2 {ΔX} 2.343 2.386 3.485 m*ω1^2 /k {X} 2.343 4.729 8.214 m*ω1^2 /k 4ta ITERACION {? cx} 1 2.018 3.506 {Fi} 1.063 2.081 3.506 m*ω1^2 {V} 6.65 5.587 3.506 m*ω1^2 {ΔX} 2.352 2.397 3.506 m*ω1^2 /k {X} 2.352 4.749 8.255 m*ω1^2 /k {? cx} 1 2.019 3.51 1 φ1 = 2.019 3.51 (2.352+4.749+8.255) m*ω1^2/k = (1+2.019+3.51) ω1 = 20.4126 rad/seg Periodo Fundamental T = 0.3078 seg Determina la primera frecuencia, el periodo fundamental y su modo de vibración MÉTODO ITERATIVO DE STODOLA VIANELLO Ingresar peso o masa, no ambos

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Datos:

g = 980 cm/seg2

Pesos

W1 = 135 ton

W2 = 131 ton

W3 = 127 ton

MASAS

m1 = 0.137755102 ton-seg2/cm

m2 = 0.133673469 ton-seg2/cm

m3 = 0.129591837 ton-seg2/cm

RIGIDEZ

k1 = 359 ton/cm

k2 = 296 ton/cm

k3 = 127 ton/cm

SOLUCIN

Se puede trabajar con los nmeros exactos o con proporciones

m1= 1.063 m2= 1.031 m3= 1

k1= 2.827 k2= 2.331 k3= 1

1ra ITERACION

{?} 1 2 3

{Fi} 1.063 2.062 3 m*1^2

{V} 6.125 5.062 3 m*1^2

{X} 2.167 2.172 3 m*1^2 /k

{X} 2.167 4.339 7.339 m*1^2 /k

2da ITERACION

{? cx} 1 2.002 3.387

{Fi} 1.063 2.064 3.387 m*1^2

{V} 6.514 5.451 3.387 m*1^2

{X} 2.304 2.338 3.387 m*1^2 /k

{X} 2.304 4.642 8.029 m*1^2 /k

3ra ITERACION

{? cx} 1 2.015 3.485

{Fi} 1.063 2.077 3.485 m*1^2

{V} 6.625 5.562 3.485 m*1^2

{X} 2.343 2.386 3.485 m*1^2 /k

{X} 2.343 4.729 8.214 m*1^2 /k

4ta ITERACION

{? cx} 1 2.018 3.506

{Fi} 1.063 2.081 3.506 m*1^2

{V} 6.65 5.587 3.506 m*1^2

{X} 2.352 2.397 3.506 m*1^2 /k

{X} 2.352 4.749 8.255 m*1^2 /k

{? cx} 1 2.019 3.51

1

1 = 2.019

3.51

(2.352+4.749+8.255) m*1^2/k = (1+2.019+3.51)

1 = 20.4126 rad/seg

Periodo Fundamental

T = 0.3078 seg

Determina la primera frecuencia, el periodo fundamental

y su modo de vibracin

MTODO ITERATIVO DE STODOLA VIANELLO

Ingresar peso o masa, no ambos
Criterio para asumir frecuencias

1 = 2 = 3 = 4 =

1 3 5 7

Condicin del problema

2^2 = 5 2^2 = 2080 2 = 45.607017 rad/seg

1^2 T2 = 0.137767952 seg

3^2 = 15 3^2 = 6240 2 = 78.99367063

1^2

2do MODO DE VIBRACIN

[ X ] 1 1.243 -0.918

[ F1 ] 287.04 346.449 -248.227

[ V ] 359 71.96 -274.489

[ X ] 1 0.243 -2.161

26.262 ton = 0

1

2 = 1.243

-0.918

3er MODO DE VIBRACIN

[ X ] 1 -0.696 -0.067

[ F1 ] 861.12 -581.967 -54.35

[ V ] 359 -502.12 79.847

[ X ] 1 -1.696 0.629

-134.197 ton no es suficiente

Tanteando 2 = 6280

[ X ] 1 -0.715 0.026

[ F1 ] 866.64 -601.687 21.226

[ V ] 359 -507.64 94.047

[ X ] 1 -1.715 0.741

-72.821 ton = 0

3^2 = 6280 3 = 79.24645102 rad/seg

T3 = 0.079286646 seg

1

3 = -0.715

0.026

R = F3 - V3 =

R = F3 - V3 =

^2 =6240rad/seg

^2 =6280rad/seg

R = F3 - V3 =

MTODO ITERATIVO DE HOLTZER

Para calcular las dems frecuencias y modos de

vibracin

^2 =2080rad/seg