losas MOD

ESTRUCTURAS ESTRUCTURA S ESPECIALES

LCAPITULO 6

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EJEMPLO DE DISEÑO

Diseñar una losa reticular de una edificación sin viga de borde o perimetral, mediante el método de pórtico equivalente, conociendo los siguientes datos:

6.1. DATOS f´c =

210 Kg / cm 2

f y =

4200 Kg / cm 2

S/C =

300 Kg /m 2

Columnas =

40 x 40 cm

VISTA EN PLANTA

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VISTA FRONTAL

6.2. PRE – DIMENSIONAMIENTO 6.2.1 LOSA 6.2.1.1. ALTURA MINIMA De acuerdo al código ACI 318S-14 en la sección 8.3.1.1 8.3. 1.1 el Espesor mínimo de losas no preesforzadas en dos direcciones sin vigas interiores tomaremos el siguiente:

ℎ = 33 = 55033 = 16 16.6.677 .. Por tantos se asume h= 0.20 m

6.2.2. CARPETA DE COMPRESION Altura minima minima de carpeta carpeta de compresion compresion para losas losas nervadas nervadas con abacos según según el Código ACI 318S-14 es t= 5cm

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6.2.3. NERVIOS Utilizando casetones de 0.40m x 0.40m x 0.20m bmin ≥ 10cm

 ∗ ℎ = ∗ 20  = 5  

b=

Se asume una base mínima de nervio Código ACI 318S-14).

′ = (Dimensión mínima según

6.2.4. ABACO La dimensión de un ábaco según Código ACI 318S-14 debe ser 1/6 dirección de análisis:

 en cada

1 ∗∗2= 1 ∗ 600 ∗ 2 = 200  = 2 6 6   500 ∗ 2 = 166.67  = 1.7  En la dirección y =  ∗∗2=  ∗ 500 En la dirección x =

Se asume una longitud de ábaco de 2m x 2 m.

6.3. ANÁLISIS DE CARGAS 6.3.1. Cargas muertas D:

Peso cerámica:

2000 Kg / m 3 * 0.015 m = 30 Kg / m

Peso Carpeta:

1800 Kg / m 3 * 0.03 m

= 54 Kg / m

Peso Ala de Viga:

2400 Kg| / m 3 * 0.05 m

= 120 Kg / m

Peso Nervio de Viga:

2400 Kg / m 3 * 0.072 m = 172.8Kg / m

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Peso Cielo Raso:

1250 Kg / m 3 * 0.02 m

Peso Total Losa:

401.8 Kg / m

Peso tabiquería:

100 Kg / m2

Peso total carga muerta:

501.8 Kg / m2

= 25Kg/m

6.3.2. Cargas vivas L: Sobre carga de uso:

300 Kg / m 2

6.5. DETERMINACIÓN DE LA ALTURA EQUIVALENTE: Sección de la losa:

1 2 Σ=

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A

Y

AY

200 250 450

10 22.5

2000 5625 7625

44


̅ = ƩƩ

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̅ =  

̅ =16,94 

  10∗20 50∗5   = 12 +2001017 + 12 + 25022.5  17 =24548.62 Comparación:

 ∗ℎ = 50∗ℎ  =24550 12 12 ℎ =18.06  ℎ ≈   Nota: según el ACI 318S-14/8.3.1.1./ (a). Altura mínima para losas con ábacos es  Ok 

 ≥   ℎ ≤ℎ

6.6. ANÁLISIS DE PÓRTICOS 6.6.1. ANÁLISIS DEL PÓRTICO 1 (A - D) ANALISIS DE PORTICO 1 (A-B)

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MAYORACIÓN DE CARGAS: q = 1.2D + 1.6L q =1.2 (501.8 Kg / m 2 * 2.5 m) + 1.6 (300 Kg / m 2 * 2.5 m) = 2705.4Kg / m q = 2705.4 Kg / m

6.6.1.1. CALCULO DE RIGIDECES EQUIVALENTES: Rigidez de la columna equivalente.

 = ∑∑  1+   = 4   0. 4  = 12  = 0.002133333 ℎ = 3.10  Reemplazando:

 = 4∗∗0.0021333 3.10 ∗ LOSAS RETICULARES

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 =0.0055054 Rigidez torsional Kt:

9   1+     Donde:  = ∑ 1+0.63     =

 ∗0.40 0. 1 8 0. 1 8 =(1+0.63 0.40) 3 =0.0005572 Reemplazando en Kt:

 = 9∗∗0.0005572  0. 4 2.5∗1+ 2.5  =0.0033843 Reemplazando en (Keq):

 = 0.00.055054 1+ 0.00055054 033843  =. Rigidez de la losa

  = 4  LOSAS RETICULARES

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 = 0.001215   = 6.00   = 5.00  Reemplazando:

 = 4∗∗0.001215 6.00  =0.00081  = 4∗∗0.002187 5.00  =0.000972 CALCULO DE FACTORES DE DISTRIBUCIÓN (FD)

=  +  +  0.00081E AB = 0.00081+0.0020959 AB =0.27874 0.00081E BA = 0.00081+0.0020959+0.000972 BA =0.208876 0.000972E BC = 0.000972+0.0020959+0.00081 LOSAS RETICULARES

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BC = 0.250651 CB = 0.250651 CD = 0.208876 DC = 0.27874 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar :

Calculo de momento de empotramiento:

  ∗  = 12  ∗ 6  2705. 4  = 12  = 8116.2 kg-m

 ∗5  2705. 4  = 12  = 5636.25 kg-m  ∗ 6  2705. 4  = 12  = 8116.2 kg-m

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Realizando Cross:

PORTICO 1 (A - D) L(m) = 6.00 F.D. 0.279 M.E. 8116.200 F.T. 0.5 -2262.41226 259.007489 -72.1990238 150.604937 -41.9815253 22.6453806 -6.31245986 5.03040485 -1.40223868 0.96009283 -0.26762842 0.19355255 -0.05395329 0.03823333 -0.01065764 0.00760919 -0.00212108 0.0015102 M (kg*m)

6170.047

0.209 -8116.200 0.5 518.0150 -1131.20613 301.2099 -36.0995119 45.2908 -20.9907626 10.0608 -3.15622993 1.9202 -0.70111934 0.3871 -0.13381421 0.0765 -0.02697664 0.0152 -0.00532882 0.0030 -0.00106054

L(m) = 5.00 =3246.48 kg/m

q 0.251 5636.250 0.5 621.6179725 -310.808986 361.4518487 -180.725924 54.34891351 -27.1744568 12.07297164 -6.03648582 2.304222797 -1.1521114 0.464526113 -0.23226306 0.091759999 -0.04588 0.018262061 -0.00913103 0.00362447 -0.00181223

-8431.543

6162.437

L(m) = 6.00

0.251 -5636.250 0.5 -621.6180 310.8089863 -361.4518 180.7259243 -54.3489 27.17445675 -12.0730 6.03648582 -2.3042 1.152111399 -0.4645 0.232263057 -0.0918 0.045879999 -0.0183 0.00913103 -0.0036 0.001812235 -6162.437

0.209 8116.200 0.5 -518.014977 1131.206129 -301.209874 36.09951189 -45.2907613 20.99076263 -10.0608097 3.156229929 -1.92018566 0.701119341 -0.38710509 0.133814211 -0.07646667 0.026976643 -0.01521838 0.005328821 -0.00302039 0.001060541 8431.543

0.279 -8116.200 0.5 2262.41226 -259.007489 72.1990238 -150.604937 41.9815253 -22.6453806 6.31245986 -5.03040485 1.40223868 -0.96009283 0.26762842 -0.19355255 0.05395329 -0.03823333 0.01065764 -0.00760919 0.00212108 -0.0015102 -6170.047

Diagramas de Cortantes: (kg)

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Isostatizando: Tramo AB: Para X = 2.86 m  =7739.303192705.4  2705.4∗ =6170.11305+7739.30319 2 X = 0.20m X = 2.86m X = 5.80m

M = -4676.360 kg-m M = 4899.75 kg-m M = -6786.983 kg-m

Tramo BC:

 =6763.52705.4 Para X = 2,5 m  2705.4∗ =6162.44575+6763.5 2 X = 0.20 m X = 2.50m X = 4.80 m

M = - 4863.854 kg-m M = 2291.929 kg-m M = -4863.854 kg-m

Tramo CD: Para X = 3.14 m  =8493.096812705.4  2705.4∗ =8431.49395+8493.09681 2 X = 0.20m X = 3.14m X = 5.80m

M = -6786.983 kg-m M = 4899.75 kg-m M =-4676.360 kg-m

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Diagramando los momentos en la cara de las columnas: (kg-m)

PÓRTICO 2 (A - D)

6.4. MAYORACIÓN DE CARGAS: q = 1.2D + 1.6L q =1.2 (501.8 Kg / m 2 * 5.5 m) + 1.6 (300 Kg / m 2 * 5.5 m) = 5951.88 Kg / m q = 5951.88 Kg / m

CALCULO DE RIGIDECES EQUIVALENTES: Rigidez de la columna equivalente. LOSAS RETICULARES

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 = ∑ ∑ 1+   = 4   0. 4  = 12  = 0.002133333 ℎ = 3.10  Reemplazando:

 = 4∗∗0.0021333 3.10 ∗  =0.0055054 Rigidez torsional Kt:

9   1+     Donde:  = ∑1+0.63     =

 ∗0.40 0. 1 8 0. 1 8 =(1+0.63 0.40) 3 =0.0005572 LOSAS RETICULARES

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Reemplazando en Kt:

 = 9∗∗0.0005572  ∗ 0. 4 2.5∗1+ 2.5  =0.0033843 Reemplazando en (Keq):


  = 4   5.5∗0.18  = 12

 = 0.002673   = 6.00   = 5.00  Reemplazando:


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=  +  +  0.001782E AB = 0.001782+0.0016159 AB =0.52444 0.001782E BA = 0.001782+0.0016159+0.0021384 BA =0.32187 0.0021384E BC = 0.0021384E+0.0016159+0.001782 BC = 0.38625 CB = 0.38625 CD = 0.32187 DC = 0.52444 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar :


  ∗  = 12  ∗6  5951. 8 8  = 12  = 17855.64 kg-m LOSAS RETICULARES

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 ∗5  5951. 8 8  = 12  = 12399.75 kg-m  ∗6  5951. 8 8  = 12  = 17855.64 kg-m

Realizando Cross:

PORTICO 2 (A - D) L(m) = 6.00 F.D. 0.524 M.E. 17855.640 F.T. 0.5 -9364.60607 878.080068 -460.5196977 923.161298 -484.1630933 215.3485547 -112.9421506 80.55127746 -42.24609042 24.64541827 -12.9255873 8.159348087 -4.279268658 2.615947791 -1.371965416 0.849574349 -0.445569529 0.274481506 M (kg*m) 9505.826

LOSAS RETICULARES

L(m) = 5.00 q = 5951.88 kg/m 0.322 0.386 0.386 -17855.640 12399.750 -12399.750 0.5 0.5 0.5 1756.1601 2107.392163 -2107.3922 -4682.303035 -1053.696082 1053.696082 1846.3226 2215.587115 -2215.5871 -230.2598488 -1107.793558 1107.793558 430.6971 516.8365313 -516.8365 -242.0815467 -258.4182656 258.4182656 161.1026 193.3230659 -193.3231 -56.47107532 -96.66153296 96.66153296 49.2908 59.14900385 -59.1490 -21.12304521 -29.57450192 29.57450192 16.3187 19.58243541 -19.5824 -6.462793648 -9.791217704 9.791217704 5.2319 6.278274698 -6.2783 -2.139634329 -3.139137349 3.139137349 1.6991 2.038978437 -2.0390 -0.685982708 -1.019489219 1.019489219 0.5490 0.658755615 -0.6588 -0.222784764 -0.329377808 0.329377808 -18830.018

14960.173

-14960.173

L(m) = 6.00 0.322 17855.640 0.5 -1756.160136 4682.303035 -1846.322596 230.2598488 -430.6971094 242.0815467 -161.1025549 56.47107532 -49.29083654 21.12304521 -16.31869617 6.462793648 -5.231895582 2.139634329 -1.699148698 0.685982708 -0.548963013 0.222784764

0.524 -17855.640 0.5 9364.60607 -878.080068 460.5196977 -923.161298 484.1630933 -215.3485547 112.9421506 -80.55127746 42.24609042 -24.64541827 12.9255873 -8.159348087 4.279268658 -2.615947791 1.371965416 -0.849574349 0.445569529 -0.274481506

18830.018

-9505.826

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Diagramas de Cortantes:

Isostatizando: Tramo AB: Para X = 2.74 m  =16301.6725951.88  5951.88∗ =9506.093+16301.672 2 X = 0.20m X = 2.74m X = 5.80m

M = -6364.7962 kg-m M = 128218.321 kg-m M = -15067.017 kg-m

Tramo BC: Para X = 2.5 m  =14879.75951.88  5951.88∗ =14960.204+14879.7 2 X = 0.20 m X = 2.50m X = 4.80 m

M = - 12103.3016 kg-m M = 3639.421 kg-m M = -12103.3016 kg-m

Tramo CD: Para X = 3.26 m  =19409.6085951.88  5951.88∗ =18829.901+19409.608 2 X = 0.20m X = 3.26m

M = -15067.017 kg-m M = 12818.321 kg-m

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X = 5.80m

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M =-6364.7962 kg-m

Diagramando los momentos :

ANÁLISIS PÓRTICO 3 (A - D)

MAYORACIÓN DE CARGAS: q = 1.2D + 1.6L q =1.2 (501.8 Kg / m 2 * 5.5 m) + 1.6 (300 Kg / m 2 * 5.5 m) = 5951.88 Kg / m q = 5951.88 Kg / m

CALCULO DE RIGIDECES EQUIVALENTES:

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Rigidez de la columna equivalente.

 = ∑∑  1+   = 4   0. 4  = 12  = 0.002133333 ℎ = 3.10  Reemplazando:


9   1+     Donde:  = ∑1+0.63     =

 ∗0.40 0. 1 8 0. 1 8 =(1+0.63 0.40) 3 =0.0005572 Reemplazando en Kt: LOSAS RETICULARES

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 = 9∗∗0.0005572  ∗ 0. 4 2.5∗1+ 2.5  =0.0033843 Reemplazando en (Keq):


  = 4   5.5∗0.18  = 12

 = 0.002673   = 6.00   = 5.00  Reemplazando:


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=  +  +  0.001782E AB = 0.001782+0.0016159 AB =0.52444 0.001782E BA = 0.001782+0.0016159+0.0021384 BA =0.32187 0.0021384E BC = 0.0021384E+0.0016159+0.001782 BC = 0.38625 CB = 0.38625 CD = 0.32187 DC = 0.52444 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar :


  ∗  = 12  ∗6  5951. 8 8  = 12  = 17855.64 kg-m

 ∗5  5951. 8 8  = 12 LOSAS RETICULARES

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 = 12399.75 kg-m  ∗6  5951. 8 8  = 12  = 17855.64 kg-m

Realizando Cross:

PORTICO 3 (A - D) L(m) = 6.00

F.D. 0.524 M.E. 17855.640 F.T. 0.5 -9364.60607 878.080068 -460.519698 923.161298 -484.163093 215.348555 -112.942151 80.5512775 -42.2460904 24.6454183 -12.9255873 8.15934809 -4.27926866 2.61594779 -1.37196542 0.84957435 -0.44556953 0.27448151 M (kg*m) 9505.826

LOSAS RETICULARES

0.322 -17855.640 0.5 1756.1601 -4682.303035 1846.3226 -230.2598488 430.6971 -242.0815467 161.1026 -56.47107532 49.2908 -21.12304521 16.3187 -6.462793648 5.2319 -2.139634329 1.6991 -0.685982708 0.5490 -0.222784764 -18830.018

L(m) = 5.00 q = 5951.88 kg/m 0.386 0.386 12399.750 -12399.750 0.5 0.5 2107.392163 -2107.3922 -1053.696082 1053.696082 2215.587115 -2215.5871 -1107.793558 1107.793558 516.8365313 -516.8365 -258.4182656 258.4182656 193.3230659 -193.3231 -96.66153296 96.66153296 59.14900385 -59.1490 -29.57450192 29.57450192 19.58243541 -19.5824 -9.791217704 9.791217704 6.278274698 -6.2783 -3.139137349 3.139137349 2.038978437 -2.0390 -1.019489219 1.019489219 0.658755615 -0.6588 -0.329377808 0.329377808 14960.173

-14960.173

L(m) = 6.00

0.322 17855.640 0.5 -1756.16014 4682.30304 -1846.3226 230.259849 -430.697109 242.081547 -161.102555 56.4710753 -49.2908365 21.1230452 -16.3186962 6.46279365 -5.23189558 2.13963433 -1.6991487 0.68598271 -0.54896301 0.22278476

0.524 -17855.640 0.5 9364.60607 -878.080068 460.5196977 -923.161298 484.1630933 -215.348555 112.9421506 -80.5512775 42.24609042 -24.6454183 12.9255873 -8.15934809 4.279268658 -2.61594779 1.371965416 -0.84957435 0.445569529 -0.27448151

18830.018

-9505.826

42


CIV- 307

Diagrama de cortantes (kg)

Tramo AB: Para X = 2.74 m  =16301.6725951.88  5951.88∗ =9506.093+16301.672 2 X = 0.20m X = 2.74m X = 5.80m

M = -6364.7962 kg-m M = 128218.321 kg-m M = -15067.017 kg-m

Tramo BC:

 =14879.75951.88 Para X = 2.5 m  5951.88∗ =14960.204+14879.7 2 X = 0.20 m X = 2.50m X = 4.80 m

M = - 12103.3016 kg-m M = 3639.421 kg-m M = -12103.3016 kg-m


M = -15067.017 kg-m M = 12818.321 kg-m M =-6364.7962 kg-m

LOSAS RETICULARES

43


CIV- 307

Diagrama de momentos en la cara de la columna: (kg-m)

PÓRTICO 4 (A - D)

MAYORACIÓN DE CARGAS: q = 1.2D + 1.6L q =1.2 (501.8 Kg / m 2 * 2.5 m) + 1.6 (300 Kg / m 2 * 2.5 m) = 2105.4 Kg / m q = 2705.4 Kg / m

LOSAS RETICULARES

44


CIV- 307

CALCULO DE RIGIDECES EQUIVALENTES: Rigidez de la columna equivalente.

 = ∑ ∑ 1+   = 4   0. 4  = 12  = 0.002133333 ℎ = 3.10 

Reemplazando:


9   1+     Donde:  = ∑ 1+0.63     =

LOSAS RETICULARES

45


CIV- 307


 = 9∗∗0.0005572  0. 4 2.5∗1+ 2.5  =0.0033843 Reemplazando en (Keq):


  = 4   2.5∗0.18  = 12

LOSAS RETICULARES

 = 0.001215   = 6.00   = 5.00  46


CIV- 307

Reemplazando:

 = 4∗∗0.001215 6.00  =0.00081  = 4∗∗0.002187 5.00  =0.000972

CALCULO DE FACTORES DE DISTRIBUCIÓN (FD)

=  +  +  0.00081E AB = 0.00081+0.0020959 AB =0.27874 0.00081E BA = 0.00081+0.0020959+0.000972 BA =0.208876 0.000972E BC = 0.000972+0.0020959+0.00081 BC = 0.250651 CB = 0.250651 CD = 0.208876 DC = 0.27874 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar :


  ∗  = 12 LOSAS RETICULARES

47


CIV- 307

 ∗ 6  2705. 4  = 12  = 8116.2 kg-m  ∗5  2705. 4  = 12  = 5636.25 kg-m  ∗ 6  2705. 4  = 12  = 8116.2 kg-m Realizando Cross:

PORTICO 4 (A - D) L(m) = 6.00 F.D. 0.279 M.E. 8116.200 F.T. 0.5 -2262.41226 259.007489 -72.1990238 150.604937 -41.9815253 22.6453806 -6.31245986 5.03040485 -1.40223868 0.96009283 -0.26762842 0.19355255 -0.05395329 0.03823333 -0.01065764 0.00760919 -0.00212108 0.0015102

LOSAS RETICULARES

L(m) = 5.00 q =2705.40 kg/m 0.209 0.251 0.251 -8116.200 5636.250 -5636.250 0.5 0.5 0.5 518.0150 621.617973 -621.6180 -1131.206129 -310.808986 310.8089863 301.2099 361.451849 -361.4518 -36.09951189 -180.725924 180.7259243 45.2908 54.3489135 -54.3489 -20.99076263 -27.1744568 27.17445675 10.0608 12.0729716 -12.0730 -3.156229929 -6.03648582 6.03648582 1.9202 2.3042228 -2.3042 -0.701119341 -1.1521114 1.152111399 0.3871 0.46452611 -0.4645 -0.133814211 -0.23226306 0.232263057 0.0765 0.09176 -0.0918 -0.026976643 -0.04588 0.045879999 0.0152 0.01826206 -0.0183 -0.005328821 -0.00913103 0.00913103 0.0030 0.00362447 -0.0036 -0.001060541 -0.00181223 0.001812235

L(m) = 6.00 0.209 8116.200 0.5 -518.014977 1131.20613 -301.209874 36.0995119 -45.2907613 20.9907626 -10.0608097 3.15622993 -1.92018566 0.70111934 -0.38710509 0.13381421 -0.07646667 0.02697664 -0.01521838 0.00532882 -0.00302039 0.00106054

0.279 -8116.200 0.5 2262.41226 -259.007489 72.1990238 -150.604937 41.9815253 -22.6453806 6.31245986 -5.03040485 1.40223868 -0.96009283 0.26762842 -0.19355255 0.05395329 -0.03823333 0.01065764 -0.00760919 0.00212108 -0.0015102

42


M (Kg*m)

6170.047

CIV- 307

-8431.543

6162.437

-6162.437

8431.543

-6170.047


Tramo AB:

 =7739.303192705.4 Para X = 2.86 m  2705.4∗ =6170.11305+7739.30319 2 X = 0.20m X = 2.86m X = 5.80m

M = -4676.360 kg-m M = 4899.75 kg-m M = -6786.983 kg-m

Tramo BC: Para X = 2,5 m  =6763.52705.4  2705.4∗ =6162.44575+6763.5 2 X = 0.20 m X = 2.50m X = 4.80 m

M = - 4863.854 kg-m M = 2291.929 kg-m M = -4863.854 kg-m


M = -6786.983 kg-m M = 4899.75 kg-m M =-4676.360 kg-m

LOSAS RETICULARES

43


CIV- 307


ANÁLISIS DEL PÓRTICO A (1 - 4)


 = ∑ ∑ 1+  LOSAS RETICULARES

44


CIV- 307

 = 4   0. 4  = 12  = 0.00213333  ℎ = 3.10  Reemplazando:

 = 4∗∗0.00213333 ∗ 3.10  =0.0055054 Rigidez torsional Kt:

9   1+     Donde:  = ∑1+0.63     =


 = 9∗∗0.0005572  0. 4 3∗1+ 3   =0.0025679  LOSAS RETICULARES

45


CIV- 307

Reemplazando en (Keq):

E  = 0.00.055054  1+ 0.00055054 025679  = .  Rigidez de la losa

  = 4   3∗0.18  = 12

 = 0.001458   = 6   =5   = 6 

Reemplazando:

 = 4∗∗0.001458 5  = 0.0011664   = 4∗∗0.001458 6  = 0.000972   = 4∗∗0.001458 5  = 0.0011664  CALCULO DE FACTORES DE DISTRIBUCIÓN (FD)

 − = − =  +  LOSAS RETICULARES

46


CIV- 307

0.0011664  − = − = 0.0011664 +0.0017511  AB = DC =0.39979 − =  +  +   − = 0.0011664  +0.0.00011664 017511  + 0.000972  − =0.29988

− =  +  + 0.00972  − = 0.000972 + 0.0017511 + 0.0011664  − =0.249903 − =0.249903 − =0.2998 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar


 3246.48∗5  ∗ − = 12 = 12 = 6763.5 . − = . .  3246.48∗6  ∗  − = 12 = 12 = 9739.44 . LOSAS RETICULARES

42


CIV- 307

− = . .  3246.48∗5  ∗ − = 12 = 12 = 6763.5 . − = . .

Realizando Cross: PORTICO A ( 1 - 4 ) L (m) =6.00

L (m) = 5.00

L (m) =5.00

q=3246.48 kg/m F.D.

0.400

0.300

0.250

0.250

0.300

0.400

M.E.

6763.500 0.5

-6763.500 0.5

9739.440 0.5

-9739.440 0.5

6763.500 0.5

-6763.500 0.5

-2704.0965

-892.4593

-743.716113

743.7161 892.45934

2704.09646

-446.22967

-1352.04823

371.8580567

-371.8580567 1352.0482

446.229668

178.40586

293.9508

244.9589803

-244.9590

-293.9508

-178.405865

146.97539

89.2029325

-122.47949

122.4794901

-89.20293

-146.975388

-58.761829

9.9794

8.31613276

-8.3161

-9.979359

58.7618286

4.9896797

-29.3809143

-4.15806638

4.15806638 29.380914

-4.98967966

-1.9949102

10.0581

8.381714801

5.0290289

-0.9974551

-4.1908574

-2.0106423

1.5559

1.296609343

0.7779656

-1.00532115

-0.64830467

-0.3110363

0.4959

0.413257045

0.2479542

-0.15551815

-0.20662852

-0.0991339

0.1086

0.090503947

-0.0905

-0.108605

0.0991339

0.0543024

-0.04956695

-0.04525197

0.045251974

0.049567

-0.05430237

-0.0217105

0.0284

0.023696164

-0.0237

-0.028435

0.02171048

0.0142177

-0.01085524

-0.01184808

0.011848082 0.0108552

-0.0142177

-0.0056843

0.0068

0.005673779

F.T.

LOSAS RETICULARES

-8.3817

-10.05806

1.9949102

4.190857401 0.9974551

-5.02902888

-1.2966

-1.555931

2.0106423

0.648304671 1.0053212

-0.77796561

-0.4133

-0.495908

0.3110363

0.206628522 0.1555182

-0.24795423

-0.0057

-0.006809

0.00568434

43


M (kg*m)

CIV- 307

0.0034043

-0.00284217

-0.00283689

3886.467

-8634.223

9499.325

0.00283689 0.0028422 -9499.325

8634.223

-0.00340427 -3886.467

Diagramas de Cortantes:

Isostatizando Tramo 1-2: Para X = 2.02075 m  =7166.64873246.48    = 3886.47+7166.6487   3246.48∗ 2 X = 0.20 m M = -2518.0666 Kg-m X = 4.80 m M = -6886.0026 Kg-m X = 2.2075 m Mmáx = 4023.77045 Kg-m

Tramo 2-3: Para X = 3 m  =9739.443246.48   3246.48∗ =9499.33+9739.44  2 X = 0.20 m X = 5.8 m X=3m

M = - 7616.367 Kg-m M = - 7616.367 Kg -m M máx = 5109.835 Kg -m

Tramo 3-4 :

 =7166.64873246.48  Para X = 2.02075 m   = 3886.47+7166.6487   3246.48∗ 2 LOSAS RETICULARES

44


X = 0.20 m X = 4.80 m X = 2.2075 m

CIV- 307

M = -2518.0666 Kg-m M = -6886.0026 Kg-m Mmáx = 4023.77045 Kg-m

Diagramando los momentos en la cara de las columnas: (kg-m)

ANÁLISIS DEL PÓRTICO B (1 - 4)

CALCULO DE RIGIDECES EQUIVALENTES: LOSAS RETICULARES

45


CIV- 307

Rigidez de la columna equivalente.

 = ∑∑  1+   = 4   0. 4  = 12  = 0.00213333  ℎ = 3.10  Reemplazando:


9   1+     Donde:  = ∑1+0.63     =

 ∗0.40 0. 1 8 0. 1 8 =(1+0.63 0.40) 3 LOSAS RETICULARES

46


CIV- 307

=0.0005572 Reemplazando en Kt:

 = 9∗∗0.0005572  ∗2 0. 4 5.5∗1+ 5.5  =0.0022872  Reemplazando en (Keq):

E  = 0.00.055054  1+ 0.00055054 022872   = .  Rigidez de la losa

  = 4   3∗0.18  = 12

 = 0.001458   = 6   =5   = 6 

Reemplazando:

 = 4∗∗0.002673 5  = 0.0021384   = 4∗∗0.002673 6  = 0.001782   = 4∗∗0.002673 5  = 0.0021384  LOSAS RETICULARES

47


CIV- 307

CALCULO DE FACTORES DE DISTRIBUCIÓN (FD)

=  + 0.0021384  − = − = 0.0021384 +0.0016159  − = − =0.56958687 =  +  +   − = 0.0021384  +0.0.00021384 016159  + 0.001782  − =0.3862507 0.001782  − = 0.001782 + 0.0016159 + 0.0021384  − =0.32187546 − =0.32187546 − =0.3862507 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar :


 5951.88∗5  ∗ − = 12 = 12 =12399.75 . − = . .  5951.88∗6  ∗  − = 12 = 12 =17855.64 . LOSAS RETICULARES

42


CIV- 307

− =. .  5951.88∗5  ∗ − = 12 = 12 = 6763.5 . − = . .

Realizando Cross: L (m) =5.00 F.D. M.E. F.T.

M (ton*m)

0.570 12399.750 0.5 -7062.98485 -1053.69608 600.192702 512.453551 -291.897148 24.5175213 -13.9653527 32.1329403 -18.3031489 6.52009352 -3.71389115 2.81679841 -1.6044682 0.81197217 -0.46250506 0.28561564 -0.16268868 0.09062926

0.386 -12399.750 0.5 -2107.3922 -3531.49242 1024.9071 300.096351 49.0350 -145.948574 64.2659 -6.98267633 13.0402 -9.15157444 5.6336 -1.85694557 1.6239 -0.8022341 0.5712 -0.23125253 0.1813 -0.08134434

5132.782

-16744.335

LOSAS RETICULARES

PORTICO B ( 1 - 4 ) L (m) =6.00 q =5951.88 kg/m 0.322 0.322 17855.640 -17855.640 0.5 0.5 -1756.160136 1756.1601 878.080068 -878.080068 854.0892512 -854.0893 -427.0446256 427.0446256 40.86253553 -40.8625 -20.43126776 20.43126776 53.55490044 -53.5549 -26.77745022 26.77745022 10.86682254 -10.8668 -5.43341127 5.43341127 4.69466402 -4.6947 -2.34733201 2.34733201 1.353286942 -1.3533 -0.676643471 0.676643471 0.476026062 -0.4760 -0.238013031 0.238013031 0.151048769 -0.1510 -0.075524385 0.075524385

17460.584

-17460.584

L (m) =5.00 0.386 12399.750 0.5 2107.392163 3531.492424 -1024.9071 -300.096351 -49.0350426 145.9485741 -64.2658805 6.982676328 -13.040187 9.151574437 -5.63359682 1.856945575 -1.62394433 0.802234098 -0.57123127 0.231252529 -0.18125852 0.081344338

0.570 -12399.750 0.5 7062.98485 1053.69608 -600.1927 -512.45355 291.897148 -24.517521 13.9653527 -32.13294 18.3031489 -6.5200935 3.71389115 -2.8167984 1.6044682 -0.8119722 0.46250506 -0.2856156 0.16268868 -0.0906293

16744.335

-5132.782

43


CIV- 307


Isostatizando : Tramo 1-2:

 =12557.45951.88  Para X = 2.1098 m  5132.78∗  = 5132.78+12557.389  2 X = 0.20 m X = 4.80 m X = 2.2075 m


Tramo 2-3: Para X = 3 m  =17855.65951.88   5132.78∗  = 17855.6+17855.64  2 X = 0.20 m X = 5.8 m X=3m

M = - 14008.494 Kg-m M = - 14008.494 Kg -m M máx = 9322.876Kg -m

Tramo 3-4 :

 =172025951.88  LOSAS RETICULARES

Para X = 2.8902 m

44


CIV- 307

 = 5132.78+12557.389  5132.78∗ 2 X = 0.20 m X = 4.80 m X = 2.2075 m





ANÁLISIS DEL PÓRTICO C (1 - 4)


LOSAS RETICULARES

45


CIV- 307

 = ∑ ∑ 1+   = 4   0. 4  = 12  = 0.00213333  ℎ = 3.10  Reemplazando:


9   1+     Donde:  = ∑ 1+0.63     =


LOSAS RETICULARES

46


CIV- 307

 = 9∗∗0.0005572  ∗2 0. 4 5.5∗1+ 5.5  =0.0022872  Reemplazando en (Keq):

E  = 0.00.055054  1+ 0.00055054 022872   = .  Rigidez de la losa

  = 4   3∗0.18  = 12

 = 0.001458   = 6   =5   =6

Reemplazando:

 = 4∗∗0.002673 5  = 0.0021384   = 4∗∗0.002673 6  = 0.001782   = 4∗∗0.002673 5  = 0.0021384  CALCULO DE FACTORES DE DISTRIBUCIÓN (FD)

LOSAS RETICULARES

47

=  + 0.0021384  − = − = 0.0021384 +0.0016159  − = − =0.56958687 =  +  +   − = 0.0021384  +0.0.00021384 016159  + 0.001782  − =0.3862507 0.001782  − = 0.001782 + 0.0016159 + 0.0021384  − =0.32187546 − =0.32187546 − =0.3862507 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar :


 5951.88∗5  ∗ − = 12 = 12 =12399.75 . − = . .  5951.88∗6  ∗  − = 12 = 12 =17855.64 . LOSAS RETICULARES

42

− =. .  5951.88∗5  ∗ − = 12 = 12 = 6763.5 . − = . .

Realizando Cross: L (m) =5.00 F.D. M.E. F.T.

0.570 12399.750 0.5 -7062.98485 -1053.69608 600.192702 512.453551 -291.897148 24.5175213 -13.9653527 32.1329403 -18.3031489 6.52009352 -3.71389115 2.81679841 -1.6044682 0.81197217 -0.46250506 0.28561564 -0.16268868

LOSAS RETICULARES

0.386 -12399.750 0.5 -2107.3922 -3531.492424 1024.9071 300.0963511 49.0350 -145.9485741 64.2659 -6.982676328 13.0402 -9.151574437 5.6336 -1.856945575 1.6239 -0.802234098 0.5712 -0.231252529 0.1813

PORTICO C ( 1 - 4 ) L (m) =6.00 q =5951.88 kg/m 0.322 0.322 17855.640 -17855.640 0.5 0.5 -1756.160136 1756.1601 878.080068 -878.080068 854.0892512 -854.0893 -427.0446256 427.0446256 40.86253553 -40.8625 -20.43126776 20.43126776 53.55490044 -53.5549 -26.77745022 26.77745022 10.86682254 -10.8668 -5.43341127 5.43341127 4.69466402 -4.6947 -2.34733201 2.34733201 1.353286942 -1.3533 -0.676643471 0.676643471 0.476026062 -0.4760 -0.238013031 0.238013031 0.151048769 -0.1510

L (m) =5.00 0.386 12399.750 0.5 2107.39216 3531.49242 -1024.9071 -300.09635 -49.035043 145.948574 -64.265881 6.98267633 -13.040187 9.15157444 -5.6335968 1.85694557 -1.6239443 0.8022341 -0.5712313 0.23125253 -0.1812585

0.570 -12399.750 0.5 7062.984849 1053.696082 -600.192702 -512.453551 291.8971483 -24.5175213 13.96535266 -32.1329403 18.30314887 -6.52009352 3.713891149 -2.81679841 1.604468196 -0.81197217 0.462505059 -0.28561564 0.162688677

43

0.09062926

-0.081344338

-0.075524385

5132.782

-16744.335

17460.584

M (ton*m)

0.075524385 0.08134434 -17460.584

16744.335

-0.09062926 -5132.782


Isostatizando Tramo 1-2:

 =12557.45951.88  Para X = 2.1098 m  5132.78∗  = 5132.78+12557.389  2 X = 0.20 m M = -2740.341 Kg-m X = 4.80 m M = -13422.97 Kg-m X = 2.2075 m Mmáx = 8114.128 Kg-m

Tramo 2-3: Para X = 3 m  =17855.65951.88   5132.78∗  = 17855.6+17855.64  2 X = 0.20 m X = 5.8 m X=3m

M = - 14008.494 Kg-m M = - 14008.494 Kg -m M máx = 9322.876Kg -m

Tramo 3-4 :

 =17202.0115951.88  LOSAS RETICULARES

Para X = 2.8902 m

44

=16744.335+17202.010  5132.78∗ 2 X = 0.20 m X = 4.80 m X = 2.2075 m




Diagrama de momentos en la cara de la columna : (kg-m)

ANÁLISIS DEL PÓRTICO D (1 - 4)


LOSAS RETICULARES

45

 = ∑ ∑ 1+   = 4   0. 4  = 12  = 0.00213333  ℎ = 3.10  Reemplazando:


9   1+     Donde:  = ∑ 1+0.63     =


LOSAS RETICULARES

46

 = 9∗∗0.0005572  0. 4 3∗1+ 3   =0.0025679  Reemplazando en (Keq):

E  = 0.00.055054  1 + 0.00055054 025679  = . .      Rigidez de la losa

  = 4   3∗0.18  = 12

 = 0.001 0.00145 4588   = 6   =5   = 6  Reemplazando:

 = 4∗∗0.001458 5  = 0.0011 001166 6644   = 4∗∗0.001458 6  = 0.000 0.00097 9722   = 4∗∗0.001458 5  = 0.0011 001166 6644  CALCULO DE FACTORES DE DISTRIBUCIÓN (FD) LOSAS RETICULARES

47

− = − =  + 0.0011664  − = − = 0.0011664 +0.0017511  AB = DC =0.39979 − =  +  +  0.0011664  − = 0.001 0.00116 1664 64  + 0.001 0.00175 7511 11 +  + 0.000 0.00097 9722  − =0.29988 − =  +  +  0.00972  − = 0.000 0.00097 9722  + 0.0017 001751 5111  + 0.001 0.001166 16644  − =0.249903 − =0.249903 − =0.2998 Con los valores encontrados de FD procederemos con el método de Hardy Cross para hallar los máximos momentos y diagramar


 3246.48∗5  ∗  − = 12 = 12 = 67 6763 63.5.5 . .  − =  ..  . .   3246.48∗6  ∗  − = 12 = 12 = 97 9739 39.4.444 . .  − =  ..  . . 

LOSAS RETICULARES

48

 3246.48∗5  ∗  − = 12 = 12 = 67 6763 63.5.5 . .  − =  ..  . . 

Realizando Cross:

F.D. M.E. F.T.

PORTICO D (1 - 4) L (m) = 5.00 L (m) = 6.00 q = 3246.48 Kg/m 0.400 0.300 0.250 0.250 6763.500 -6763.500 9739.440 -9739.440 0.5 0.5 0.5 0.5 -2704.09646 -892.4593 -743.716113 743.7161 -446.229668 -1352.04823 371.858057 -371.8580567 178.405865 293.9508 244.95898 -244.9590 146.975388 89.20293248 -122.47949 122.4794901 -58.7618286 9.9794 8.31613276 -8.3161 4.98967966 -29.3809143 -4.15806638 4.15806638 -1.9949102 10.0581 8.3817148 -8.3817 5.02902888 -0.9974551 -4.1908574 4.190857401 -2.0106423 1.5559 1.29660934 -1.2966 0.77796561 -1.00532115 -0.64830467 0.648304671 -0.3110363 0.4959 0.41325704 -0.4133 0.24795423 -0.15551815 -0.20662852 0.206628522 -0.0991339 0.1086 0.09050395 -0.0905

LOSAS RETICULARES

L (m) = 5.00 0.300 6763.500 0.5 892.45934 1352.0482 -293.95078 -89.202932 -9.9793593 29.380914 -10.058058 0.9974551 -1.5559312 1.0053212 -0.4959085 0.1555182 -0.1086047

0.400 -6763.500 0.5 2704.096464 446.229668 -178.405865 -146.975388 58.76182857 -4.98967966 1.994910197 -5.02902888 2.010642303 -0.77796561 0.311036304 -0.24795423 0.099133902

49

M (kg*m)

0.05430237 -0.02171048 0.0142177 -0.00568434 0.00340427

-0.04956695 0.0284 -0.01085524 0.0068 -0.00284217

-0.04525197 0.02369616 -0.01184808 0.00567378 -0.00283689

3886.467

-8634.223

9499.325

0.045251974 0.049567 -0.0237 -0.0284354 0.011848082 0.0108552 -0.0057 -0.0068085 0.00283689 0.0028422 -9499.325

8634.223

-0.05430237 0.021710482 -0.0142177 0.005684339 -0.00340427 -3886.467


Isostatizando : Tramo 1-2:

 =7166.64873246.48  Para X = 2.2075 m   = 3886.47+7166.6487   3246.48∗ 2 X = 0.20 m X = 4.80 m X = 2.2075 m

M = -6886.0026 Kg-m M = 2518.0666 Kg-m Mmáx = 4023.77045 Kg-m

Tramo 2-3: Para X = 3 m  =9739.443246.48   3246.48∗ =9499.33+9739.44  2 X = 0.20 m X = 5.8 m X=3m

M = - 7616.367 Kg-m M = - 7616.367 Kg -m M máx = 5109.835 Kg -m

LOSAS RETICULARES

50

Tramo 3-4 : Para X = 2.7925 m  =7166.64873246.48   3246.48∗  = 3886.47+7166.6487   2 X = 0.20 m X = 4.80 m X = 2.2075 m



LOSAS RETICULARES

51

DISTRIBUCIÓN DE MOMENTOS EN FRANJA DE COLUMNAS Y FRANJA CENTRAL. Los porcentajes de los momentos negativos y positivos totales a ser resistidos por una franja de columna se obtendrán de las siguientes tablas de la norma ACI 318S-14.

Tabla 8.10.5.1-Fracción del momento negativo interior M u en una franja de columna (%)

⁄   ⁄ =   ⁄ ≥  

0.5

1.0

2.0

75

75

75

90

75

45

Para losas sin vigas de borde

 ⁄ = 

Tabla 8.10.5.2- Fracción del momento negativo exterior M u en una franja de columna (%)

 ⁄  > .0 

⁄

 0

≥ ≥

2.5

0

2.5

0

1.0

100 75 100 90

100 75 100 75




2.0 100 75 100 45

α l⁄l = 0  = 0

Tabla 8.10.5.5 -Fracción del momento positivo M u en una franja de columna (%)

⁄   ⁄ =   ⁄ ≥  

LOSAS RETICULARES

0.5

1.0

2.0

60

60

60

90

75

45


 ⁄ = 

52

PORCENTAJES DE DISTRIBUCION DE MOMENTOS Momento negativo Exterior Franja Columna %

Momento negativo Interior Franja Columna %

Momento positivo Interior Franja Columna %

PORTICO

L1

L2

L2/L1

1 (A - D)

6.00

2.50

0.4167

0

0

100

75

60

5.00

2.50

0.5

0

0

100

75

60

6.00

2.50

0.4167

0

0

100

75

60

6.00

5.50

0.9167

0

0

100

75

60

5.00

5.50

1.1

0

0

100

75

60

6.00

5.50

0.9167

0

0

100

75

60

6.00

5.50

0.9167

0

0

100

75

60

5.00

5.50

1.1

0

0

100

75

60

6.00

5.50

0.9167

0

0

100

75

60

6.00

2.50

0.4167

0

0

100

75

60

5.00

2.50

0.5

0

0

100

75

60

6.00

2.50

0.4167

0

0

100

75

60

2 (A - D)

3 (A - D)

4 (A - D)

PORTICO

L1 (m)

L2 (m)

A (1 - 4)

5.00

B (1 - 4)

C (1 - 4)

D (1 - 4)

α

β

Momento negativo Exterior Franja Columna %

Momento negativo Interior Franja Columna %

Momento positivo Interior Franja Columna %

L2/L1

α

3.000

0.6

0

0

100

75

60

6.00

3.000

0.5

0

0

100

75

60

5.00

3.000

0.6

0

0

100

75

60

5.00

5.500

1.1

0

0

100

75

60

6.00

5.500

0.9167

0

0

100

75

60

5.00

5.500

1.1

0

0

100

75

60

5.00

5.500

1.1

0

0

100

75

60

6.00

5.500

0.9167

0

0

100

75

60

5.00

5.500

1.1

0

0

100

75

60

5.00

3.000

0.6

0

0

100

75

60

6.00

3.000

0.5

0

0

100

75

60

5.00

3.000

0.6

0

0

100

75

60

β

La distribución de momentos negativos y positivos totales en la franja de columnas es: Interior=75 %, exterior=100 % y positivo=60 %

LOSAS RETICULARES

53

DISTRIBUCION DE MOMENTOS DEL PORTICOS EN LA DIRECCION X FRANJA DE COL. PORTICO

TRAMO

TRAMO

M(kg-m) %

1 (A - D)

A-B

B-C

C-D

2 (A - D)

A-B

B-C

C-D

3 (A - D)

A-B

B-C

C-D

4 (A - D)

A-B

B-C

C-D

LOSAS RETICULARES

Ap. Externo M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Externo Ap. Externo M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Externo Ap. Externo M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Externo Ap. Externo M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Externo

4676.299 4899.761 6787.027 4863.845 2291.938 4863.845 6787.027 4899.761 4676.299 6364.542 12818.416 15067.121 12103.271 3639.452 12103.271 15067.121 12818.416 6364.542 6364.542 12818.416 15067.121 12103.271 3639.452 12103.271 15067.121 12818.416 6364.542 4676.299 4899.761 6787.027 4863.845 2291.938 4863.845 6787.027 4899.761 4676.299

100 60 75 75 60 75 75 60 100 100 60 75 75 60 75 75 60 100 100 60 75 75 60 75 75 60 100 100 60 75 75 60 75 75 60 100

M

4676.299 2939.857 5090.271 3647.884 1375.163 3647.884 5090.271 2939.857 4676.299 6364.542 7691.050 11300.341 9077.453 2183.671 9077.453 11300.341 7691.050 6364.542 6364.542 7691.050 11300.341 9077.453 2183.671 9077.453 11300.341 7691.050 6364.542 4676.299 2939.857 5090.271 3647.884 1375.163 3647.884 5090.271 2939.857 4676.299

%

0 40 25 25 40 25 25 40 0 0 40 25 25 40 25 25 40 0 0 40 25 25 40 25 25 40 0 0 40 25 25 40 25 25 40 0

FRANJA CENTRAL M

0.000 1959.904 1696.757 1215.961 916.775 1215.961 1696.757 1959.904 0.000 0.000 5127.366 3766.780 3025.818 1455.781 3025.818 3766.780 5127.366 0.000 0.000 5127.366 3766.780 3025.818 1455.781 3025.818 3766.780 5127.366 0.000 0.000 1959.904 1696.757 1215.961 916.775 1215.961 1696.757 1959.904 0.000

54

DISTRIBUCION DE MOMENTOS DEL PORTICOS EN LA DIRECCION Y PORTICO A (1 - 4)

TRAMO 1-2

2-3

3-4

B (1 - 4)

1-2

2-3

3-4

C (1 - 4)

1-2

2-3

3-4

D (1 - 4)

1-2

2-3

3-4

LOSAS RETICULARES

TRAMO

M(Kg-m)

FRANJA DE COL.

FRANJA CENTRAL

%

M

%

M

Ap. Externo

2518.067

100

2518.067

0

0.000

M. Positivo

4023.770

60

2414.262

40

1609.508

Ap. Interno

6886.003

75

5164.502

25

1721.501

Ap. Interno

7616.367

75

5712.275

25

1904.092

M. Positivo

5109.835

60

3065.901

40

2043.934

Ap. Interno

7616.367

75

5712.275

25

1904.092

Ap. Interno

6886.003

75

5164.502

25

1721.501

M. Positivo

4023.770

60

2414.262

40

1609.508

Ap. Externo

2518.067

100

2518.067

0

0.000

Ap. Externo

2740.341

100

2740.341

0

0.000

M. Positivo

8114.128

60

4868.477

40

3245.651

Ap. Interno

13422.970

75

10067.228

25

3355.743

Ap. Interno

14008.494

75

10506.370

25

3502.123

M. Positivo

9322.876

60

5593.725

40

3729.150

Ap. Interno

14008.494

75

10506.370

25

3502.123

Ap. Interno

13422.970

75

10067.228

25

3355.743

M. Positivo

8114.128

60

4868.477

40

3245.651

Ap. Externo

2740.341

100

2740.341

0

0.000

Ap. Externo

2740.341

100

2740.341

0

0.000

M. Positivo

8114.128

60

4868.477

40

3245.651

Ap. Interno

13422.970

75

10067.228

25

3355.743

Ap. Interno

14008.494

75

10506.370

25

3502.123

M. Positivo

9322.876

60

5593.725

40

3729.150

Ap. Interno

14008.494

75

10506.370

25

3502.123

Ap. Interno

13422.970

75

10067.228

25

3355.743

M. Positivo

8114.128

60

4868.477

40

3245.651

Ap. Externo

2740.341

100

2740.341

0

0.000

Ap. Externo

2518.067

100

2518.067

0

0.000

M. Positivo

4023.770

60

2414.262

40

1609.508

Ap. Interno

6886.003

75

5164.502

25

1721.501

Ap. Interno

7616.367

75

5712.275

25

1904.092

M. Positivo

5109.835

60

3065.901

40

2043.934

Ap. Interno

7616.367

75

5712.275

25

1904.092

Ap. Interno

6886.003

75

5164.502

25

1721.501

M. Positivo

4023.770

60

2414.262

40

1609.508

Ap. Externo

2518.067

100

2518.067

0

0.000

55

ANALIZANDO FRANJA MAS ESFORZADA FRANJA 2(A-B)

FRANJA DE COL. PORTICO

TRAMO

TRAMO

M(kg-m) %

2 (A - D)

A-B B-C C-D

LOSAS RETICULARES

Ap. Externo M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Interno Ap. Interno M. Positivo Ap. Externo

6364.796 12818.321 15067.017 12103.302 3639.421 12103.302 15067.017 12818.321 6364.796

100 60 75 75 60 75 75 60 100

M 6364.796 7691.993 11300.263 9077.476 2183.653 9077.476 11300.263 7691.993 6364.796

% 0 40 25 25 40 25 25 40 0

FRANJA CENTRAL M 0.000 5127.328 3766.754 3025.816 1455.768 3025.816 3766.754 5127.328 0.000

56

7. DISEÑO DE ÁBACOS A FLEXION

´ =

Datos 210 kg/cm2

 =

4200 kg/cm2

Distancia en x =

100 m

Distancia en y =

100 m

Recubrimiento =

2.5cm

Peralte =

22.5 cm

M máx. =

11300.263 kg*m

Cuantía Balanceada

6000 )  = ∗∗´ ∗(  6000+ 6000 )  = 0.85∗0.85∗210 ∗ ( 4200 6000+4200 =0.02125 Cuantía máxima

á =0.75∗ á =0.75∗0.02125 á =0.0159375≅0.02 Entonces el área de acero máximo es:

á = á ∗  ∗  á =0.02∗200∗22.5  á = 90  Cuantía mínima

 =0.0018  =  ∗  ∗   = 0.0018∗200  ∗22.5   = 8.10  LOSAS RETICULARES

57

ÁBACO- EJE 2 (Apoyo A) Para

− =6364.796 kgm

 ′ =ѡ0.59ѡ Ø ∗  ∗  ∗  6364.796 ∗100 =ѡ0.59Eѡ 0.9∗200∗22.5 ∗210 ѡ =1.661 ѡ =0.0339 ′  = ѡ ∗  ∗  ∗  210  =0.0339∗200∗22.5∗ 4200  =7.628  

Como

 < 

7.628  < 8.10 2 Usar acero mínimo

= 8.10 

Entonces se utilizará As como área del acero para el cálculo.

Calculo de separación de aceros

  = 8.10 2  =7.162 π ∗Ø π∗1.2cm 4 100  4 = 7.162 = 13.963 

 = Entonces:

: ∅   / 

LOSAS RETICULARES

58

Armadura adicional debido en el ancho efectivo del ábaco, debido a la transmisión de momentos: Verificar la armadura de la losa en la columna exterior según ACI 318S-14/8.4.2.3.2:

Yf = 12  1+ 3 √   =  + 2 =40+ 22.2 5 = 51.25   =  +=40+22.5=62.5 Yf = 2 1 47,75 =0,62356 1 + 3 √ 62.5 =0.62356∗ 6364.796   = 3968.832   

Donde:

Una vez reducido el momento hallamos As con una base de:

 =  + 3ℎ = 40 +325 = 115   ′ =ѡ0.5ѡ Ø ∗  ∗  ∗ 3968.832∗100 =ѡ0.5ѡ 0.9∗115∗22.5 ∗210 ѡ =1.65804 ѡ =0.03687 ′    = ѡ ∗  ∗  ∗  210  =0.03687∗115∗22.5∗ 4200  = 4.770   = 4.770  Que se adiciona y no se compara con el  LOSAS RETICULARES

59


 =

  = 4.770 2 =4.218 π ∗Ø π∗1.2cm 4 100  4 = 4.2176 =23.7 : ∅   / 

ÁBACO- EJE 2 (Apoyo B) Para

− = 11300.263 kgm  ′ =ѡ0.5ѡ Ø ∗  ∗  ∗ 11300.263 ∗100 =ѡ0.5ѡ 0.9∗200∗22.5 ∗210 ѡ =1.634 ѡ =0.06127 ′  = ѡ ∗  ∗  ∗  210  =0.06127∗200∗22.5∗ 4200  = 13.786  

 <  <  8.10  < 13.786 2 < 90  Como

Entonces se utilizará As como área del acero para el cálculo.


 = LOSAS RETICULARES

  = 13.786 2 =12.189 π ∗Ø π∗1.2cm 4 4 60

 =8.203  = 100 12.189 Entonces:

: ∅   / 

Armadura adicional debido en el ancho efectivo del ábaco, debido a la transmisión de momentos: Verificar la armadura de la losa en la columna exterior según ACI 318S-14/8.4.2.3.2:

Donde:

Yf = 12  1+ 3 √   =  + 2 =40+ 22.2 5 = 51.25   =  +=40+22.5=62.5 Yf = 2 1 47,75 =0,62356 1 + 3 √ 62.5  = 0.62356 ∗ 11300.263 =7046.392

Una vez reducido el momento hallamos As con una base de:

 =  + 3ℎ = 40 +325 = 115   ′ =ѡ0.5ѡ Ø ∗  ∗  ∗ 7046.392∗100 =ѡ0.5ѡ 0.9∗115∗22.5 ∗210 LOSAS RETICULARES

61

losas MOD

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