ejemplo holzer b2006

Datos del problema

M = 0.096378151 Ton seg^2/cmK = 55.588 Ton / cmg = 981.000 cm / seg^2M5 0.0964 Ton seg^2/cm

M4 0.3020 Ton seg^2/cm




K5 55.5877 Ton / cm

K4 194.1710 Ton / cm

K3 217.9122 Ton / cm

K2 324.4965 Ton / cm

K1 440.3538 Ton / cm

Normalizando las M's y las K's respecto al ltimo pisodescripcin rigidez 1 masa 1 rigidez 2 masa 2 rigidez 3 masa 3 rigidez 4 masa 4 rigidez 5 masa 5

datos 7.92 3.80 5.84 3.34 3.92 3.18 3.49 3.13 1.00 1.00

MTODO DE HOLZER (clculo de frecuencias y formas modales)SISMO SENTIDO X

descripcin rigidez 1 masa 1 rigidez 2 masa 2 rigidez 3 masa 3 rigidez 3 masa 3 rigidez 4 masa 4 Residuodatos 7.92 3.80 5.84 3.34 3.92 3.18 3.49 3.13 1.00 1.00

frec. j2 0.179 k/M1 1.00 2.2403 3.7448 4.8219 5.8755u1 1.0000 1.2403 1.5044 1.0772 1.0536Fc1 7.9218 7.2404 5.8976 3.7627 1.0536Fi1 -0.6813 -1.3428 -2.1350 -2.7091 -1.054

datos 7.92 3.80 5.84 3.34 3.92 3.18 3.49 3.13 1.00 1.00frec. j2 0.962 k/M

2 1.00 1.7309 1.3997 -0.1976 -5.1813u2 1.0000 0.7309 -0.3313 -1.5972 -4.9838Fc2 7.9218 4.2668 -1.2986 -5.5792 -4.9838Fi2 -3.6550 -5.5654 -4.2806 0.5955 4.984

datos 7.92 3.80 5.84 3.34 3.92 3.18 3.49 3.13 1.00 1.00frec. j2 1.717 k/M

3 1.00 1.2392 -0.2193 -1.5133 2.1096u3 1.0000 0.2392 -1.4585 -1.2940 3.6229Fc3 7.9218 1.3961 -5.7174 -4.5200 3.6229Fi3 -6.5256 -7.1135 1.1974 8.1428 -3.6229


frec. j2 3.252 k/M4 1.00 0.2401 -1.5573 1.0356 -0.4598u4 1.0000 -0.7599 -1.7974 2.5929 -1.4954Fc4 7.9218 -4.4360 -7.0461 9.0571 -1.4954Fi4 -12.3578 -2.6101 16.1032 -10.5525 1.4955

datos 7.92 3.80 5.84 3.34 3.92 3.18 3.49 3.13 1.00 1.00frec. j2 5.195 k/M

5 1.00 -1.0245 0.4991 -0.1510 0.0360u5 1.0000 -2.0245 1.5236 -0.6500 0.1870Fc5 7.9218 -11.8182 5.9726 -2.2707 0.1870Fi5 -19.7400 17.7909 -8.2433 2.4576 -0.1870

0.0000

0.0000

M

O

D

O

1

0.0000

0.0000

0.0000

M

O

D

O

2

M

O

D

O

3

M

O

D

O

4

M

O

D

O

5

M = 0.096378151 Ton seg^2/cm 1 = 10.1694 rad/segK = 55.588 Ton / cm 2 = 23.5535 rad/segg = 981.000 cm / seg^2 3 = 31.4720 rad/segM5 0.0964 Ton seg^2/cm 4 = 43.3094 rad/segM4 0.3020 Ton seg^2/cm 5 = 54.7376 rad/segM3 0.3064 Ton seg^2/cm T1 = 0.6178 segM2 0.3222 Ton seg^2/cm T2 = 0.2668 segM1 0.3662 Ton seg^2/cm T3 = 0.1996 segK5 55.5877 Ton / cm T4 = 0.1451 segK4 194.1710 Ton / cm T5 = 0.1148 segK3 217.9122 Ton / cm ai 1 modo 2 modo 3 modo 4 modo 5 modoK2 324.4965 Ton / cm N5 = 5.8755 -5.1813 2.1096 -0.4598 0.0360K1 440.3538 Ton / cm N4 = 4.8219 -0.1976 -1.5133 1.0356 -0.1510

N3 = 3.7448 1.3997 -0.2193 -1.5573 0.4991N2 = 2.2403 1.7309 1.2392 0.2401 -1.0245N1 = 1.0000 1.0000 1.0000 1.0000 1.0000

localidad Mrida Zona Ssmica 5 Coef. Ao 0.30 Grupo B2 F. Imp 1.00Suelo S2 T* 0.70 2.60 p 1.00 Tipo Est. I

Nivel Diseo ND3 R 4.50 T+ 0.35 C.Corr 0.95 Mat Ct 0.07alt. total hn 15.30 c 1.15Clculo perodo fundamental Ta (aprox) 0.542 seg

Fuerzas Ssmicas para T1 = 0.6178 seg.modo/periodo niveles Mk ik Mk x ik Mk x ik2 Adj Vo1 (Ton) factor Vo1 Fi1 (Ton) chequeo

5 0.0964 5.8755 0.56627 3.32712 0.13 23.422 23.4224 0.3020 4.8219 1.45616 7.02152 0.34 60.231 83.6533 0.3064 3.7448 1.14756 4.29735 0.27 47.466 131.1192 0.3222 2.2403 0.72177 1.61699 0.17 29.854 160.9731 0.3662 1.0000 0.36623 0.36623 0.09 15.148 176.121

= 1.3932 4.25799 16.62921 1.00

METODO DE SUPERPOSICIN MODAL CON UN GRADO DE LIBERTAD POR NIVEL

(Sentido X)

0.165 176.121

Datos del problema:

1 T+ <

T1 =0.6178 < T*

formasmodales

frecuencias

perodos


5 0.0964 -5.1813 -0.49937 2.58740 -0.63 -15.235 -15.2354 0.3020 -0.1976 -0.05967 0.01179 -0.08 -1.820 -17.0553 0.3064 1.3997 0.42892 0.60034 0.54 13.086 -3.9702 0.3222 1.7309 0.55765 0.96525 0.70 17.013 13.0431 0.3662 1.0000 0.36623 0.36623 0.46 11.173 24.217

= 1.3932 0.79377 4.53100 1.00


5 0.0964 2.1096 0.20332 0.42893 0.46 8.532 8.5324 0.3020 -1.5133 -0.45699 0.69155 -1.03 -19.176 -10.6443 0.3064 -0.2193 -0.06720 0.01474 -0.15 -2.820 -13.4642 0.3222 1.2392 0.39922 0.49470 0.90 16.752 3.2881 0.3662 1.0000 0.36623 0.36623 0.82 15.368 18.655

= 1.3932 0.44458 1.99615 1.00


5 0.0964 -0.4598 -0.04432 0.02038 -0.19 -1.445 -1.4454 0.3020 1.0356 0.31273 0.32386 1.33 10.195 8.7503 0.3064 -1.5573 -0.47723 0.74319 -2.03 -15.557 -6.8072 0.3222 0.2401 0.07735 0.01857 0.33 2.522 -4.2861 0.3662 1.0000 0.36623 0.36623 1.56 11.939 7.653

= 1.3932 0.23477 1.47223 1.00


5 0.0964 0.0360 0.00347 0.00012 0.02 0.140 0.1404 0.3020 -0.1510 -0.04560 0.00688 -0.31 -1.837 -1.6973 0.3064 0.4991 0.15294 0.07632 1.04 6.161 4.4642 0.3222 -1.0245 -0.33007 0.33816 -2.25 -13.297 -8.8331 0.3662 1.0000 0.36623 0.36623 2.49 14.753 5.921

= 1.3932 0.14697 0.78772 1.00

0.220 5.921

0.178 24.217

0.192 18.655

0.2084T4 = 0,1451

< T+

5T5 = 0,1148

< T+

7.653

3T3 = 0,1996

< T+

2T2 = 0,2668

< T+

MODO NIVEL Fi5 23.422 NIVEL Fi NIVEL Fi4 60.231 5 29.215 Ton 5 29.251 Ton3 47.466 4 63.236 Ton 4 64.079 Ton2 29.854 3 49.318 Ton 3 52.079 Ton1 15.148 2 38.228 Ton 2 40.552 Ton5 -15.235 1 24.300 Ton 1 30.833 Ton4 -1.8203 13.0862 17.013 31 11.1735 8.5324 -19.176 NOTA:3 -2.820 1.- Analizar la estructura en el sentido ortogonal 2 16.752 2.- Efecto rotacional aplicar el Mtodo Torsin Esttica Equivalente.1 15.3685 -1.4454 10.1953 -15.5572 2.5221 11.9395 0.1404 -1.8373 6.1612 -13.2971 14.753

Las Fuerzas Laterales equivalentes definitivas en cada piso, se determinarn utilizando la combinacin modal: RAIZ CUADRADA DE LA SUMA DE LOS CUADRADOS

1

Usando 3 modos de vibracin Usando 5 modos de vibracinFuerzas Definitivas

3

4

5

Fuerzas Definitivas

2Numero mnimo de Modos a usar (N.V. 1756-01) =

Datos del problema

M = 0.096378151 Ton seg^2/cmK = 37.893 Ton / cmg = 981.000 cm / seg^2M5 0.0964 Ton seg^2/cmM4 0.3020 Ton seg^2/cmM3 0.3064 Ton seg^2/cmM2 0.3222 Ton seg^2/cmM1 0.3662 Ton seg^2/cmK5 37.8926 Ton / cmK4 121.7634 Ton / cmK3 160.0130 Ton / cmK2 274.5387 Ton / cmK1 405.5749 Ton / cm

Normalizando las M's y las K's respecto al ltimo pisodescripcin rigidez 1 masa 1 rigidez 2 masa 2 rigidez 3 masa 3 rigidez 4 masa 4 rigidez 5 masa 5

datos 10.70 3.80 7.25 3.34 4.22 3.18 3.21 3.13 1.00 1.00

MTODO DE HOLZER (clculo de frecuencias y formas modales)SISMO SENTIDO Y


frec. j2 0.206 k/M1 1.00 2.3693 4.3325 6.0296 7.5931u1 1.0000 1.3693 1.9632 1.6972 1.5635Fc1 10.7033 9.9208 8.2901 5.4536 1.5635Fi1 -0.7824 -1.6308 -2.8365 -3.8901 -1.563

datos 10.70 3.80 7.25 3.34 4.22 3.18 3.21 3.13 1.00 1.00frec. j2 1.013 k/M

2 1.00 1.9463 2.0098 0.0798 -6.3753u2 1.0000 0.9463 0.0636 -1.9300 -6.4551Fc2 10.7033 6.8558 0.2685 -6.2020 -6.4551Fi2 -3.8475 -6.5873 -6.4705 -0.2532 6.455

datos 10.70 3.80 7.25 3.34 4.22 3.18 3.21 3.13 1.00 1.00frec. j2 1.835 k/M

3 1.00 1.5151 0.1984 -1.8920 2.2669u3 1.0000 0.5151 -1.3166 -2.0905 4.1589Fc3 10.7033 3.7318 -5.5598 -6.7175 4.1589Fi3 -6.9715 -9.2916 -1.1576 10.8764 -4.1590


frec. j2 3.411 k/M4 1.00 0.6884 -1.7049 0.9042 -0.3750u4 1.0000 -0.3116 -2.3933 2.6091 -1.2792Fc4 10.7033 -2.2578 -10.1065 8.3841 -1.2792Fi4 -12.9611 -7.8487 18.4905 -9.6632 1.2792

datos 10.70 3.80 7.25 3.34 4.22 3.18 3.21 3.13 1.00 1.00frec. j2 6.374 k/M

5 1.00 -0.8655 0.3004 -0.0620 0.0115u5 1.0000 -1.8655 1.1659 -0.3624 0.0735Fc5 10.7033 -13.5156 4.9232 -1.1645 0.0735Fi5 -24.2188 18.4387 -6.0877 1.2380 -0.0735

M

O

D

O

2

M

O

D

O

3

M

O

D

O

4

M

O

D

O

5

0.0000

M

O

D

O

1

0.0000

0.0000

0.0000

0.0000

M = 0.096378151 Ton seg^2/cm 1 = 8.9975 rad/segK = 37.893 Ton / cm 2 = 19.9521 rad/segg = 981.000 cm / seg^2 3 = 26.8574 rad/segM5 0.09638 Ton seg^2/cm 4 = 36.6202 rad/segM4 0.30199 Ton seg^2/cm 5 = 50.0584 rad/segM3 0.30644 Ton seg^2/cm T1 = 0.6983 segM2 0.32217 Ton seg^2/cm T2 = 0.3149 segM1 0.36623 Ton seg^2/cm T3 = 0.2339 segK5 37.89261 Ton / cm T4 = 0.1716 segK4 121.76335 Ton / cm T5 = 0.1255 segK3 160.01298 Ton / cm ai 1 modo 2 modo 3 modo 4 modo 5 modoK2 274.53867 Ton / cm N5 = 7.5931 -6.3753 2.2669 -0.3750 0.0115K1 405.57490 Ton / cm N4 = 6.0296 0.0798 -1.8920 0.9042 -0.0620

N3 = 4.3325 2.0098 0.1984 -1.7049 0.3004N2 = 2.3693 1.9463 1.5151 0.6884 -0.8655N1 = 1.0000 1.0000 1.0000 1.0000 1.0000

localidad Mrida Zona Ssmica 5 Coef. Ao 0.30 Grupo B2 F. Imp 1.00Suelo S2 T* 0.70 2.60 p 1.00 Tipo Est. I

Nivel Diseo ND3 R 4.50 T+ 0.35 C.Corr 0.95 Mat Ct 0.07alt. total hn 15.30 c 1.15

Clculo perodo fundamental Ta (aprox) 0.542 seg

Fuerzas Ssmicas para T1 = 0.6983modo/periodo niveles Mk ik Mk x ik Mk x ik2 Adj Vo1 (Ton) factor Vo1 Fi1 (Ton) chequeo

5 0.0964 7.5931 0.73181 5.55668 0.15 24.210 24.2104 0.3020 6.0296 1.82086 10.97914 0.36 60.239 84.4493 0.3064 4.3325 1.32766 5.75207 0.27 43.923 128.3712 0.3222 2.3693 0.76332 1.80854 0.15 25.253 153.6241 0.3662 1.0000 0.36623 0.36623 0.07 12.116 165.740

= 1.3932 5.00989 24.46265 1.00

METODO DE SUPERPOSICIN MODAL CON UN GRADO DE LIBERTAD POR NIVEL

(Sentido Y)

0.165

formasmodales

frecuencias

perodos

165.740

Datos del problema:

1 T+ <

T1 =0.6983 < T*


5 0.0964 -6.3753 -0.61444 3.91727 -0.60 -15.441 -15.4414 0.3020 0.0798 0.02410 0.00192 0.02 0.606 -14.8353 0.3064 2.0098 0.61591 1.23788 0.60 15.478 0.6422 0.3222 1.9463 0.62703 1.22036 0.62 15.757 16.3991 0.3662 1.0000 0.36623 0.36623 0.36 9.203 25.603

= 1.3932 1.01882 6.74366 1.00


5 0.0964 2.2669 0.21848 0.49528 0.39 8.232 8.2324 0.3020 -1.8920 -0.57136 1.08103 -1.02 -21.527 -13.2963 0.3064 0.1984 0.06081 0.01207 0.11 2.291 -11.0042 0.3222 1.5151 0.48811 0.73952 0.87 18.391 7.3861 0.3662 1.0000 0.36623 0.36623 0.65 13.798 21.185

= 1.3932 0.56227 2.69414 1.00


5 0.0964 -0.3750 -0.03614 0.01355 -0.12 -1.283 -1.2834 0.3020 0.9042 0.27305 0.24688 0.90 9.692 8.4093 0.3064 -1.7049 -0.52247 0.89078 -1.73 -18.546 -10.1372 0.3222 0.6884 0.22177 0.15266 0.73 7.872 -2.2651 0.3662 1.0000 0.36623 0.36623 1.21 13.000 10.735

= 1.3932 0.30243 1.67011 1.00


5 0.0964 0.0115 0.00111 0.00001 0.01 0.060 0.0604 0.3020 -0.0620 -0.01872 0.00116 -0.12 -1.007 -0.9483 0.3064 0.3004 0.09206 0.02765 0.57 4.954 4.0062 0.3222 -0.8655 -0.27883 0.24131 -1.72 -15.005 -10.9991 0.3662 1.0000 0.36623 0.36623 2.26 19.709 8.710

= 1.3932 0.16185 0.63637 1.00

0.216 8.710

0.200 10.735

0.184 21.185

0.170 25.6032T2 = 0,3149< T+

3T3 = 0,2339

< T+

4T4 = 0,1716

< T+

5T5 = 0,1255

< T+

MODO NIVEL Fi5 24.210 NIVEL Fi NIVEL Fi4 60.239 5 29.871 Ton 5 29.899 Ton3 43.923 4 63.973 Ton 4 64.711 Ton2 25.253 3 46.626 Ton 3 50.423 Ton1 12.116 2 34.988 Ton 2 38.876 Ton5 -15.441 1 20.540 Ton 1 31.294 Ton4 0.6063 15.4782 15.757 31 9.2035 8.2324 -21.527 NOTA:3 2.291 1.- Analizar la estructura en el sentido ortogonal 2 18.391 2.- Efecto rotacional aplicar el Mtodo Torsin Esttica Equivalente.1 13.7985 -1.2834 9.6923 -18.5462 7.8721 13.0005 0.0604 -1.0073 4.9542 -15.0051 19.709

3

4

5

Fuerzas Definitivas

Numero mnimo de Modos a usar (N.V. 1756-2001) = 2

Las Fuerzas Laterales equivalentes definitivas en cada piso, se determinarn utilizando la combinacin modal: RAIZ CUADRADA DE LA SUMA DE LOS CUADRADOS

1

Fuerzas DefinitivasUsando 3 modos de vibracin Usando 5 modos de vibracin

SENTIDO X

5 0.0964 55.588 1.0000 1.00004 0.3020 194.171 3.1333 3.49313 0.3064 217.912 3.1796 3.92022 0.3222 324.497 3.3428 5.83761 0.3662 440.354 3.7999 7.9218

SENTIDO Y

5 0.0964 37.893 1.0000 1.00004 0.3020 121.763 3.1333 3.21343 0.3064 160.013 3.1796 4.22282 0.3222 274.539 3.3428 7.24521 0.3662 405.575 3.7999 10.7033

Modo SENTIDO X SENTIDO Y SENTIDO X SENTIDO Y1 10.1694 8.9975 0.6178 0.69832 23.5535 19.9521 0.2668 0.31493 31.4720 26.8574 0.1996 0.23394 43.3094 36.6202 0.1451 0.17165 54.7376 50.0584 0.1148 0.1255

FORMAS MODALES (Sentido X)

1 2 3 4 51 1.000 1.000 1.000 1.000 1.0002 2.240 1.731 1.239 0.240 -1.0253 3.745 1.400 -0.219 -1.557 0.4994 4.822 -0.198 -1.513 1.036 -0.1515 5.876 -5.181 2.110 -0.460 0.036

FORMAS MODALES (Sentido Y)

1 2 3 4 51 1.000 1.000 1.000 1.000 1.0002 2.369 1.946 1.515 0.688 -0.8653 4.332 2.010 0.198 -1.705 0.3004 6.030 0.080 -1.892 0.904 -0.0625 7.593 -6.375 2.267 -0.375 0.012

NivelMODOS

CLCULO DE FRECUENCIAS Y FORMAS MODALES MTODO DE HOLZER

k (normalizada)

Nivelmasa

(T seg 2/cm)Rigidez (T/cm) m (normalizada) k (normalizada)

m (normalizada)

NivelMODOS

masa (T seg 2/cm)

Rigidez (T/cm)Nivel

FRECUENCIAS (rad/seg) PERIODOS (seg)

SENTIDO X

SENTIDO Y

MTODO DE SUPERPOSICIN MODAL CON UN GRADO DE LIBERTAD POR NIVEL

MODELOS MATEMTICOS

m5 = 0.0964

m4 = 0.3020

m3 = 0.3064

m2 = 0.3222

m1 = 0.3662

1M

3.1333M

3,1796M

3.3428M

3.7999M

k5 = 55.588

k4 = 194.171

k3 = 217.912

k2 = 324.497

k1 = 440.354

1K

3.4931K

3,9202K

5.8376K

7.9218K

m5 = 0.0964

m4 = 0.3020

m3 = 0.3064

m2 = 0.3222

m1 = 0.3662

1M

3,1333M

3,1796M

3.3428M

3.7999M

k5 = 37.893

k4 = 121.763

k3 =160.013

k2 = 274.539

k1 = 405.575

1K

3.2134K

4.2228K

7.2452K

10.703K

FORMAS MODALES(normalizadas)

Formas Modales (Sentido X)

Modo 1Modo 2Modo 3Modo 4Modo 5

Formas Modales (Sentido Y)

Modo 1Modo 2Modo 3Modo 4Modo 5

COMPARACION DE RESULTADOS

Sentido X Sentido Y5 28.0784 29.2508 29.89914 63.8282 64.0786 64.71063 49.0156 52.0789 50.42302 34.9674 40.5525 38.87561 20.9208 30.8329 31.2939

Nivel

Comparacin de Fuerzas Ssmicas

Mtodo Esttico Equivalente

Superposicin Modal con 1GDL

ejemplo holzer b2006

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