ejemplo holzer a2004

Datos del problema

M = 0.082213083 Ton seg^2/cmK = 32.389 Ton / cmg = 981.000 cm / seg^2M5 0.0822 Ton seg^2/cmM4 0.2277 Ton seg^2/cmM3 0.2397 Ton seg^2/cmM2 0.2473 Ton seg^2/cmM1 0.2395 Ton seg^2/cmK5 32.3891 Ton / cmK4 98.2040 Ton / cmK3 134.8944 Ton / cmK2 154.6485 Ton / cmK1 365.6311 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 11.29 2.91 4.77 3.01 4.16 2.92 3.03 2.77 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 11.29 2.91 4.77 3.01 4.16 2.92 3.03 2.77 1.00 1.00

frec. j2 0.197 k/Ma1 1.00 3.2442 5.3560 7.2432 9.0175u1 1.0000 2.2442 2.1118 1.8872 1.7743Fc1 11.2887 10.7156 8.7952 5.7219 1.7743Fi1 -0.5731 -1.9204 -3.0733 -3.9476 -1.774

datos 11.29 2.91 4.77 3.01 4.16 2.92 3.03 2.77 1.00 1.00frec. j2 1.010 k/M

a1 1.00 2.7484 2.7484 0.0795 -8.2349u1 1.0000 1.7484 0.0000 -2.6689 -8.3144Fc1 11.2887 8.3479 0.0000 -8.0920 -8.3144Fi1 -2.9408 -8.3479 -8.0920 -0.2223 8.314

datos 11.29 2.91 4.77 3.01 4.16 2.92 3.03 2.77 1.00 1.00frec. j2 1.981 k/M

a1 1.00 2.1561 0.3971 -2.7756 2.8308u1 1.0000 1.1561 -1.7591 -3.1726 5.6063Fc1 11.2887 5.5202 -7.3262 -9.6194 5.6063Fi1 -5.7685 -12.8464 -2.2932 15.2257 -5.6063


frec. j2 3.813 k/Ma1 1.00 1.0384 -1.7775 0.8730 -0.3104u1 1.0000 0.0384 -2.8158 2.6504 -1.1833Fc1 11.2887 0.1831 -11.7273 8.0361 -1.1833Fi1 -11.1056 -11.9104 19.7635 -9.2195 1.1833

datos 11.29 2.91 4.77 3.01 4.16 2.92 3.03 2.77 1.00 1.00frec. j2 6.410 k/M

a1 1.00 -0.5462 0.2103 -0.0471 0.0087u1 1.0000 -1.5462 0.7565 -0.2574 0.0558Fc1 11.2887 -7.3826 3.1505 -0.7803 0.0558Fi1 -18.6713 10.5331 -3.9309 0.8362 -0.0558

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 = 0.082213083 Ton seg^2/cm 1 = 8.8045 rad/segK = 32.389 Ton / cm 2 = 19.9442 rad/segg = 981.000 cm / seg^2 3 = 27.9329 rad/segM5 0.0822 Ton seg^2/cm 4 = 38.7575 rad/segM4 0.2277 Ton seg^2/cm 5 = 50.2541 rad/segM3 0.2397 Ton seg^2/cm T1 = 0.7136 segM2 0.2473 Ton seg^2/cm T2 = 0.3150 segM1 0.2395 Ton seg^2/cm T3 = 0.2249 segK5 32.3891 Ton / cm T4 = 0.1621 segK4 98.2040 Ton / cm T5 = 0.1250 segK3 134.8944 Ton / cm ai 1 modo 2 modo 3 modo 4 modo 5 modoK2 154.6485 Ton / cm N5 = 9.0175 -8.2349 2.8308 -0.3104 0.0087K1 365.6311 Ton / cm N4 = 7.2432 0.0795 -2.7756 0.8730 -0.0471

N3 = 5.3560 2.7484 0.3971 -1.7775 0.2103N2 = 3.2442 2.7484 2.1561 1.0384 -0.5462N1 = 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 17.43 c 1.15

Clculo perodo fundamental Ta (aprox 0.597 seg

Fuerzas Ssmicas para T1 = 0.7136 seg.modo/periodo niveles Mk ik Mk x ik Mk x ik2 Adj Vo1 (Ton) factor Vo1 Fi1 (Ton) chequeo

5 0.0822 9.0175 0.74136 6.68525 0.16 19.542 19.5424 0.2277 7.2432 1.64940 11.94691 0.35 43.478 63.0203 0.2397 5.3560 1.28408 6.87759 0.27 33.848 96.8682 0.2473 3.2442 0.80239 2.60314 0.17 21.151 118.0191 0.2395 1.0000 0.23946 0.23946 0.05 6.312 124.331

= 1.0365 4.71669 28.35235 1.00

METODO DE SUPERPOSICIN MODAL CON UN GRADO DE LIBERTAD POR NIVEL

(Sentido X)

0.162 124.331

Datos del problema:

1 T1 =0.7136

> T*

formasmodales

frecuencias

perodos


5 0.0822 -8.2349 -0.67701 5.57510 -0.74 -10.900 -10.9004 0.2277 0.0795 0.01810 0.00144 0.02 0.291 -10.6093 0.2397 2.7484 0.65891 1.81092 0.72 10.609 0.0002 0.2473 2.7484 0.67975 1.86819 0.74 10.944 10.9441 0.2395 1.0000 0.23946 0.23946 0.26 3.855 14.800

= 1.0365 0.91920 9.49511 1.00


5 0.0822 2.8308 0.23273 0.65879 0.50 5.182 5.1824 0.2277 -2.7756 -0.63204 1.75426 -1.35 -14.073 -8.8913 0.2397 0.3971 0.09519 0.03780 0.20 2.120 -6.7722 0.2473 2.1561 0.53327 1.14980 1.14 11.874 5.1021 0.2395 1.0000 0.23946 0.23946 0.51 5.332 10.434

= 1.0365 0.46861 3.84011 1.00


5 0.0822 -0.3104 -0.02552 0.00792 -0.10 -0.855 -0.8554 0.2277 0.8730 0.19879 0.17354 0.82 6.659 5.8043 0.2397 -1.7775 -0.42614 0.75745 -1.75 -14.275 -8.4712 0.2473 1.0384 0.25681 0.26666 1.06 8.603 0.1321 0.2395 1.0000 0.23946 0.23946 0.98 8.022 8.154

= 1.0365 0.24341 1.44502 1.00


5 0.0822 0.0087 0.00072 0.00001 0.00 0.068 0.0684 0.2277 -0.0471 -0.01072 0.00050 -0.07 -1.014 -0.9463 0.2397 0.2103 0.05041 0.01060 0.35 4.766 3.8202 0.2473 -0.5462 -0.13509 0.07378 -0.93 -12.770 -8.9501 0.2395 1.0000 0.23946 0.23946 1.65 22.636 13.686

= 1.0365 0.14478 0.32435 1.00

0.216 13.686

0.170 14.800

0.186 10.434

0.2034T4 = 0,1621

< T+

5T5 = 0,1250

< T+

8.154

3T3 = 0,2249

< T+

2T2 = 0,315

< T+

MODO NIVEL Fi5 19.542 NIVEL Fi NIVEL Fi4 43.478 5 22.969 Ton 5 22.985 Ton3 33.848 4 45.700 Ton 4 46.194 Ton2 21.151 3 35.535 Ton 3 38.591 Ton1 6.312 2 26.611 Ton 2 30.744 Ton5 -10.900 1 9.118 Ton 1 25.688 Ton4 0.2913 10.6092 10.944 31 3.8555 5.1824 -14.073 NOTA:3 2.120 1.- Analizar la estructura en el sentido ortogonal 2 11.874 2.- Efecto rotacional aplicar el Mtodo Torsin Esttica Equivalente.1 5.3325 -0.8554 6.6593 -14.2752 8.6031 8.0225 0.0684 -1.0143 4.7662 -12.7701 22.636

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. 1756-98 Rev.

Datos del problema

M = 0.082213083 Ton seg^2/cmK = 31.293 Ton / cmg = 981.000 cm / seg^2M5 0.0822 Ton seg^2/cmM4 0.2277 Ton seg^2/cmM3 0.2397 Ton seg^2/cmM2 0.2473 Ton seg^2/cmM1 0.2395 Ton seg^2/cmK5 31.2930 Ton / cmK4 83.5274 Ton / cmK3 123.9468 Ton / cmK2 141.8793 Ton / cmK1 317.6920 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 10.15 2.91 4.53 3.01 3.96 2.92 2.67 2.77 1.00 1.00

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


frec. j2 0.183 k/Ma1 1.00 3.1214 5.1151 7.0489 8.6314u1 1.0000 2.1214 1.9937 1.9339 1.5824Fc1 10.1522 9.6182 7.8966 5.1619 1.5824Fi1 -0.5340 -1.7216 -2.7347 -3.5795 -1.582

datos 10.15 2.91 4.53 3.01 3.96 2.92 2.67 2.77 1.00 1.00frec. j2 0.966 k/M

a1 1.00 2.6188 2.5508 -0.2414 -7.0486u1 1.0000 1.6188 -0.0679 -2.7922 -6.8072Fc1 10.1522 7.3393 -0.2691 -7.4529 -6.8072Fi1 -2.8129 -7.6084 -7.1838 0.6457 6.807

datos 10.15 2.91 4.53 3.01 3.96 2.92 2.67 2.77 1.00 1.00frec. j2 1.861 k/M

a1 1.00 2.0436 0.3494 -2.8750 3.3388u1 1.0000 1.0436 -1.6941 -3.2244 6.2138Fc1 10.1522 4.7314 -6.7102 -8.6067 6.2138Fi1 -5.4207 -11.4417 -1.8964 14.8204 -6.2138


frec. j2 3.511 k/Ma1 1.00 0.9836 -1.6582 0.7823 -0.3115u1 1.0000 -0.0164 -2.6418 2.4405 -1.0938Fc1 10.1522 -0.0744 -10.4637 6.5142 -1.0938Fi1 -10.2265 -10.3893 16.9779 -7.6080 1.0938

datos 10.15 2.91 4.53 3.01 3.96 2.92 2.67 2.77 1.00 1.00frec. j2 5.943 k/M

a1 1.00 -0.5786 0.2261 -0.0479 0.0097u1 1.0000 -1.5786 0.8048 -0.2741 0.0576Fc1 10.1522 -7.1574 3.1876 -0.7316 0.0576Fi1 -17.3095 10.3450 -3.9192 0.7893 -0.0577

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

O

D

O

2

M

O

D

O

3

M = 0.082213083 Ton seg^2/cm 1 = 8.3536 rad/segK = 31.293 Ton / cm 2 = 19.1728 rad/segg = 981.000 cm / seg^2 3 = 26.6157 rad/segM5 0.08221 Ton seg^2/cm 4 = 36.5572 rad/segM4 0.22772 Ton seg^2/cm 5 = 47.5610 rad/segM3 0.23975 Ton seg^2/cm T1 = 0.7521 segM2 0.24733 Ton seg^2/cm T2 = 0.3277 segM1 0.23946 Ton seg^2/cm T3 = 0.2361 segK5 31.29301 Ton / cm T4 = 0.1719 segK4 83.52742 Ton / cm T5 = 0.1321 segK3 123.94680 Ton / cm ai 1 modo 2 modo 3 modo 4 modo 5 modoK2 141.87930 Ton / cm N5 = 8.6314 -7.0486 3.3388 -0.3115 0.0097K1 317.69203 Ton / cm N4 = 7.0489 -0.2414 -2.8750 0.7823 -0.0479

N3 = 5.1151 2.5508 0.3494 -1.6582 0.2261N2 = 3.1214 2.6188 2.0436 0.9836 -0.5786N1 = 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 17.43 c 1.15

Clculo perodo fundamental Ta (aprox) 0.597 seg


5 0.0822 8.6314 0.70961 6.12494 0.16 18.424 18.4244 0.2277 7.0489 1.60516 11.31468 0.35 41.675 60.0983 0.2397 5.1151 1.22631 6.27267 0.27 31.839 91.9372 0.2473 3.1214 0.77201 2.40973 0.17 20.044 111.9811 0.2395 1.0000 0.23946 0.23946 0.05 6.217 118.198

= 1.0365 4.55255 26.36148 1.00

METODO DE SUPERPOSICIN MODAL CON UN GRADO DE LIBERTAD POR NIVEL

(Sentido Y)

0.153

formasmodales

frecuencias

perodos

118.198

Datos del problema:

1T1 = 0,7452

> T*


5 0.0822 -7.0486 -0.57949 4.08459 -0.67 -10.850 -10.8504 0.2277 -0.2414 -0.05497 0.01327 -0.06 -1.029 -11.8793 0.2397 2.5508 0.61154 1.55993 0.71 11.450 -0.4292 0.2473 2.6188 0.64769 1.69614 0.75 12.127 11.6981 0.2395 1.0000 0.23946 0.23946 0.28 4.483 16.182

= 1.0365 0.86424 7.59339 1.00


5 0.0822 3.3388 0.27449 0.91647 0.61 5.407 5.4074 0.2277 -2.8750 -0.65468 1.88222 -1.46 -12.895 -7.4893 0.2397 0.3494 0.08377 0.02927 0.19 1.650 -5.8392 0.2473 2.0436 0.50543 1.03288 1.13 9.955 4.1171 0.2395 1.0000 0.23946 0.23946 0.53 4.717 8.833

= 1.0365 0.44847 4.10030 1.00


5 0.0822 -0.3115 -0.02561 0.00798 -0.11 -0.928 -0.9284 0.2277 0.7823 0.17814 0.13936 0.75 6.456 5.5273 0.2397 -1.6582 -0.39754 0.65920 -1.67 -14.406 -8.8792 0.2473 0.9836 0.24327 0.23928 1.02 8.816 -0.0631 0.2395 1.0000 0.23946 0.23946 1.01 8.677 8.614

= 1.0365 0.23772 1.28528 1.00


5 0.0822 0.0097 0.00080 0.00001 0.01 0.070 0.0704 0.2277 -0.0479 -0.01092 0.00052 -0.08 -0.957 -0.8873 0.2397 0.2261 0.05422 0.01226 0.39 4.752 3.8652 0.2473 -0.5786 -0.14311 0.08281 -1.02 -12.542 -8.6771 0.2395 1.0000 0.23946 0.23946 1.71 20.986 12.308

= 1.0365 0.14044 0.33506 1.00

0.213 12.308

0.200 8.614

0.184 8.833

0.168 16.1822T2 = 0,3277 < T+

3T3 = 0,2361

< T*

4T4 = 0,1719

< T+

5T5 = 0,1321

< T+

MODO NIVEL Fi5 18.424 NIVEL Fi NIVEL Fi4 41.675 5 22.054 Ton 5 22.074 Ton3 31.839 4 43.636 Ton 4 44.122 Ton2 20.044 3 33.875 Ton 3 37.117 Ton1 6.217 2 25.454 Ton 2 29.714 Ton5 -10.850 1 9.000 Ton 1 24.427 Ton4 -1.0293 11.4502 12.127 31 4.4835 5.4074 -12.895 NOTA:3 1.650 1.- Analizar la estructura en el sentido ortogonal 2 9.955 2.- Efecto rotacional aplicar el Mtodo Torsin Esttica Equivalente.1 4.7175 -0.9284 6.4563 -14.4062 8.8161 8.6775 0.0704 -0.9573 4.7522 -12.5421 20.986

3

4

5

Fuerzas Definitivas

Numero mnimo de Modos a usar (N.V. 1756-98 Rev. 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 0.0822 32.389 1.0000 1.00004 0.2277 98.204 2.7698 3.03203 0.2397 134.894 2.9161 4.16482 0.2473 154.649 3.0084 4.77471 0.2395 365.631 2.9127 11.2887

SENTIDO Y

5 0.0822 31.293 1.0000 1.00004 0.2277 83.527 2.7698 2.66923 0.2397 123.947 2.9161 3.96082 0.2473 141.879 3.0084 4.53391 0.2395 317.692 2.9127 10.1522

FRECUENCIAS (wj2) (x K/M) FRECUENCIAS (wj2) (x K/M)

Modo SENTIDO X Modo SENTIDO Y1 0.1968 1 0.18332 1.0097 2 0.96583 1.9805 3 1.86114 3.8129 4 3.51115 6.4104 5 5.9429

FORMAS MODALES (Sentido X)

1 2 3 4 51 1.000 1.000 1.000 1.000 1.0002 3.244 2.748 2.156 1.038 -0.5463 5.356 2.748 0.397 -1.777 0.2104 7.243 0.080 -2.776 0.873 -0.0475 9.018 -8.235 2.831 -0.310 0.009

FORMAS MODALES (Sentido Y)

1 2 3 4 51 1.000 1.000 1.000 1.000 1.0002 3.121 2.619 2.044 0.984 -0.5793 5.115 2.551 0.349 -1.658 0.2264 7.049 -0.241 -2.875 0.782 -0.0485 8.631 -7.049 3.339 -0.312 0.010

NivelMODOS

CLCULO DE FRECUENCIAS Y FORMAS MODALES MTODO DE HOLZER

k (normalizada)

Nivelmasa

(T seg2/cm)Rigidez (T/cm)

m (normalizada)

k (normalizada)

m (normalizada)

NivelMODOS

masa (T seg2/cm)

Rigidez (T/cm)Nivel

SENTIDO X

SENTIDO Y

MTODO DE SUPERPOSICIN MODAL CON UN GRADO DE LIBERTAD POR NIVEL

MODELOS MATEMTICOS

m5 = 0.0822

m4 = 0.2277

m3 = 0.2397

m2 = 0.2473

m1 = 0.2395

k5 = 32,389

k4 = 98,204

k3 = 134,894

k2 = 154,649

k1 = 365,631

1M

2,770M

2,916M

3,008M

2,913M

1K

3,032K

4,165K

4,775K

11,289K

1M

2,770M

2,916M

3,008M

2,913M

1K

2,669K

3,961K

4,534K

10,152K

m5 = 0.0822

m4 = 0.2277

m3 = 0.2397

m2 = 0.2473

m1 = 0.2395

k5 = 31,293

k4 = 83,527

k3 = 123,947

k2 = 141,879

k1 = 317,692

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 27.8217 22.9849 22.07384 45.8726 46.1935 44.12163 37.1506 38.5907 37.11672 26.8278 30.7443 29.71421 14.8424 25.6879 24.4273

Nivel

Comparacin de Fuerzas Ssmicas

Mtodo Esttico Equivalente

Superposicin Modal con 1GDL

ejemplo holzer a2004

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