# Diseño de losas y columnas

Páginas: 11 (2589 palabras) Publicado: 30 de marzo de 2011
Diseño de losa de entrepiso con viga perimetral
wl := 295 kg m f´c := 280
2

kg cm
2

fy := 4200

kg cm
2

Columnas de 40x40 l1 := 7.54m l2 := 6.15m

1. Calculo de hf ln := 7.54m − 0.40m ln 33 = 21.636⋅ cm ln = 7.14 m Por lo tanto hf := 22cm

2. Diseño viga de borde l1 = 40.757⋅ cm 18.5 l1 50 = 15.08⋅ cm

h := 45cm

Por ancho de columna usar b := 40cm

X

40 + 45 − 22 = 6340 + 4⋅ 22 = 128 x := 63cm

3. Calculo de wu ws := hf ⋅ 2400 wf := 0.10⋅ ws wu := 1.2 ws + wf + 1.6⋅ wl 4. Revision a corte A) Como viga Proponiendo v#13 ϕv d := hf − −e 2 ϕv := 1.29cm e := 2cm d = 19.36⋅ cm kg m
3

ws = 528 wf = 52.8

kg m
2

kg m
2

(

)

wu = 1168.96

kg m
2

Vu := wu ⋅ 3.38m⋅ 6.15m Vu = 24299.17 kg ϕ := 0.75 b := 6.15m kg cm ϕVc = 79174.34 kg ϕVc > VuO.K.
2

ϕVc := ϕ⋅ 0.53

f´c⋅ b⋅ d

B) Como losa

c := 40cm + d

c = 59.36⋅ cm
2 2

At := l1⋅ l2 − c

(

)

At = 46.02⋅ m

Vu := wu ⋅ At bo := 4 ⋅ c ϕVc := λ⋅ kg cm
2

Vu = 53794.02 kg λ := 1.0 ⋅ f´c⋅ bo⋅ d

ϕVc = 76893.48 kg ϕVc > Vu O.K.

5. Tabla de resumen (Franja interior) Mo := wu ⋅ l2 ⋅ ln 8
2

Mo = 45812.21 m⋅ kg

Para vano interior M- = 65% M+ = 35%Para vano exterior M-int = 70% M+ = 50% M-ext = 30%

Para franja de columna αf1 := 0 l2 l1 = 0.816 M-int = 75% M+ = 60%

1 3 Is := ⋅ l ⋅ hf 12 2 C1 = (40,45) + (22,23) C2 = (40,23) + (22,63) C1 βt := 2⋅ Is

Is = 545710⋅ cm

4 4 4

C1 := 468466cm C2 := 261198cm βt = 0.429 M-ext = 95.8%

7. Tabla de resumen (Franja exterior) l2 2 wu ⋅ ⋅ ln 2 Mo := 8 Para vano interior M- = 65% M+ = 35%Mo = 22906.1 m⋅ kg

Para vano exterior M-int = 70% M+ = 50% M-ext = 30%

Para franja de columna b := 40cm a hf = 2.045 b ⋅  a
3

a := 45cm b hf = 1.818

hf = 22⋅ cm

ln = 714 ⋅ cm f := 1.21

αf := l2 l1

 ⋅f  ln  hf 
= 0.473

αf = 0.58

αf ⋅ l2 l1

= 0.816

M-int = 77.6% M+ = 69.5%

βt = 0.429 Para viga l2 αf ⋅ = 0.473 l1

M-ext = 96.2%

Viga = 39.1%

Diseñode columnas
Dos niveles cargando la losa solida con viga perimetral anteriormente calculada wu = 1168.96 kg m
2

f´c = 280 ⋅

kg cm
2

L := 2.65m

c := 40cm

Columnas del nivel superior
A) Columna Interior 1. Cargas sobre columna
29779kg-m 29779kg-m 32069kg-m 13744kg-m 7195kg/m 7195kg/m 7.54m 7.54m

R1

R2

7195 R 1 :=

kg

l m 1 kg m 2
2

2 7195⋅ −13744 ⋅ kg⋅ m + 32069⋅ kg⋅ m + ⋅ l1

R 1 = 27125.15 kg

R 2 := wpp := c ⋅ L⋅ 2400
2

l1 kg m
3

R 2 = 29555.52 kg wpp = 1017.6 kg Pu = 57698 kg M u = 2290 m⋅ kg

Pu := R1 + R2 + wpp M u := 32069kg⋅ m − 29779kg⋅ m 2. Momento minimo de diseño emin := 1.5cm + 0.03⋅ c M min := Pu ⋅ emin 3. Propiedades de seccion E := 15100 kg cm 1 3 Is := ⋅ l2⋅ hf 12 1 4 Ic := ⋅c 12 E⋅ Is EILs := ln EILc := E⋅ Ic L
2

emin= 2.7⋅ cm M min = 1558 m⋅ kg M u > M min Diseñar con Mu

⋅ f´c

E = 252671.33 ⋅
4

kg cm
2

Is = 545710⋅ cm

Ic = 213333.33 ⋅ cm

4

EILs = 193.12 × 10 ⋅ kg⋅ cm EILc = 203.41 × 10 ⋅ kg⋅ cm
6

6

4. Calculo de K ψA := ψB := EILc 2 ⋅ EILs 2EILc 2 ⋅ EILs ψA = 0.527 K := 1.25 ψB = 1.053

5. Revision de esbeltez r := 0.3⋅ c K⋅ L r = 27.604 27 > 22 Se considera esbeltez

6.Calculo de δ y Mu EI := 0.25⋅ E⋅ Ic Pc := π ⋅ EI ( K⋅ L) δ := 1− 1 Pu 0.75⋅ Pc M uδ = 2445⋅ kg⋅ m
2 2

EI = 1.348 × 10 ⋅ kg⋅ cm Pc = 1212112 ⋅ kg

10

2

δ = 1.068

M uδ := M u ⋅ δ 7. Diseño de columna ϕ := 0.65 Pn := Pu ϕ M uδ ϕ ton cm
2 2

Pn = 88767 ⋅ kg

Mn := Pn c
2

Mn = 3762⋅ kg⋅ m Mn c
3

= 0.061⋅

= 0.006⋅

ton cm
2

ρ := 0.01

As := c ⋅ ρ Po := 0.8⋅ ϕ⋅ 0.85⋅f´c⋅ c − As + fy⋅ As  

As = 16⋅ cm

2

(2

)

Po = 230980 kg Po > Pn Redusir As
2

Ast :=

As 2

Ast = 8⋅ cm

Usando V#13 Ast NoV := Av

Av := 1.29cm NoV = 6.2

2

Usar 8V#13

S

16⋅ 1.27cm = 20.32⋅ cm 48⋅ 0.95cm = 45.6⋅ cm c = 40⋅ cm

S := 20cm Recubrimiento de 4cm Espacio entre varillas 14cm < 15cm No requiere estribos entre varillas

8. Croquis

8#13...

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