Calculo Losa de Tablero Puente

CÁLCULO ESTRUCTURAL LOSA DE TABLERO

Datos generales:

Luz del Puente

Luz := 38 m

Carga de Diseño: H2O-S 16-44 N°v := 2

N° de Vias

Espesor capa de rodadura: er := 0.03 m Recubrimiento :

r := 2.5

Ancho losa interior:

s := 2.59 m

Ancho losa exterior

a := 0.91 m

Propiedad de los materiales: materiales:

fc := 250

Kg/cm2

fy := 4200

Kg/cm2

γH°A° := 2400 Kg/m3 γH°S ° := 2300 Kg/m3

Esquema General:

Apoyos sobre vigas de HºPº

a

s

cm

CALCUL O DE LOSA INTERIOR Luz de cálculo

Usaremos la distancia entre ejes L := s L = 2.59 m

Predimensionado del espesor de la losa

t :=

L + 3.05 30

t = 0.188 m

t := 0.18 Factor de impacto

15 L + 38

= 0.3695

⎛ 15 ≥ 0.3, 0.3, 15 ⎞ L + 38 ⎠ ⎝ L + 38

I := if

I = 0.3 Carga muerta

Ancho de calculo: Losa := t⋅

b 100

⋅ γH°A°

Rodadura := er⋅

b 100

⋅ γH°S °

qCM := Losa + Rodadura

b := 100

cm

Losa = 432

Kgr/m

Rodadura = 69

Kgr/m (H° simple)

qCM = 501

Kgr/m

Momento por carga muerta

Aplicando factor de continuidad de 0.8, para tres o mas apoyos. 2

Mcm := 0.8⋅ qCM ⋅

L

8

Mcm = 336.0758 Kgr.m/m

Momento por carga viva

Aplicando factor de continuidad de 0.8, para tres o mas apoyos. P := 7265 Mcv := 0.8⋅

kgr

(Carga de una rueda HS20)

L + 0.61

⋅P

9.75

Mcv = 1907.5282

Kgr.m/m

Momento por impacto

MI := I ⋅ Mcv MI = 572.2585

Kgr.m/m

Momento último

Mu := 1.3[Mcm + 1.67(Mcv + MI)] Mu = 5820.5154

Kgr.m/m

Canto úti l

Diametro del fierro

φ := 1.2

Espesor de la losa

h := t⋅ 100

d := h − r −

φ 2

d = 14.9 cm Cuantía necesaria

Φ := 0.9 ρ :=

fc

⎛

1.18⋅ fy

⎝

1−

1−

2.36⋅ Mu⋅ 100 ⎞

ρ = 0.0075 Cuantía balanceada

β1 := 0.85 ρb := 0.85⋅ β1⋅

2

Φ ⋅ fc⋅ b⋅ d

⎛ 6090 ⎞ fc ⎝ 6090 + fy ⎠ fy

ρb = 0.0255

⎠

Cuantía máxima y mínima

ρmax := 0.75⋅ ρb

ρmax = 0.0191

Verificacion := if (ρ < ρmax , "cumple" , "no cumple" ) Verificacion = "cumple"

ρmin :=

14

ρmin = 0.0033

fy

Cuantía de diseño

ρ := if (ρ ≥ ρmin, ρ , ρmin)

ρ = 0.0075

Ac ero de r efu erzo

As := ρ ⋅ b⋅ d As = 11.1634 cm2

Ao :=

π ⋅φ

2

Ao = 1.131 cm2

4

N°Barras :=

As

N°Barras = 9.8706

Ao

N°Barras := 10 Usar :

N°Barras = 10 de φ = 1.2 cm

(por metro de ancho)

Area de acero proporcionada: Asprop := N°Barras⋅ Ao Asprop = 11.3097 cm2 /m

Ac ero de d is tr ib uc ió n:

Para armadura principal perpendicular al transito:

⎛ 1.22 ≤ 0.67 , 1.22 , 0.67 ⎞ L ⎝ L ⎠

D := if

D = 0.67 AD := D⋅ As AD = 7.4795 cm2

φ := 1.2 Ao :=

π ⋅φ

2

Ao = 1.131 cm2

4

N°Barras :=

AD Ao

N°Barras = 6.6133

N°Barras := 7

Usar :

N°Barras = 7 de φ = 1.2 cm


(En la cara inferior, sobre la armadura positiva) Area de acero proporcionada: ADprop := N°Barras⋅ Ao ADprop = 7.9168 cm2/m

Ac ero po r r etac ci ón y t emp erat ur a

Cuantía mínima por temperatura:

ρtem := 0.0020 As := ρtem⋅ h⋅ b As = 3.6 cm2 /m

φ := 1.0 Ao :=

π ⋅φ

2

4

N°Barras :=

Ao = 0.7854 As Ao

N°Barras = 4.5837

N°Barras := 5

Usar :

N°Barras = 5 de

φ = 1 cm



CALCULO LOSA EXTERIOR

An ch o d e di st ru bu ci on de l a car ga puntual

a 0.3m

X

X := a − 0.3 m E := 0.8⋅ X + 1.14 E = 1.628

m

Momento por Cargas muertas

Qba

Qba Qp2

Qp1 Qa

a

Qp3 QL Qbo

Peso propio losa +capa de rodadura QL := a⋅ (t + er) ⋅ γH°A° MQL :=

a 2

⋅ QL

QL = 458.64

Kgr/m

MQL = 208.6812

Kgr.m/m

Bordillo Alto del Bordillo

hbo := 0.45 m

Ancho del Bordillo

abo := 0.2 m

Qbo := hbo ⋅ abo ⋅ γH°A°

⎛ ⎝

MQbo := a +

abo ⎞ 2

⎠

Qbo = 216

⋅ Qbo

Kgr/m

MQbo = 218.16

Kgr.m/m

Acera Alto

ha := 0.15

m

Ancho

aa := 0.65

m

Qa := ha ⋅ aa ⋅ γH°A°

⎛ ⎝

MQa := a +

aa ⎞ 2 ⎠

Qa = 234 Kgr/m

⋅ Qa

MQa = 288.99 Kgr.m/m

Barandado Alto del Barandado

hb := 0.125 m

Ancho del Barandado

ab := 0.15 m

Qb := 2⋅ hb ⋅ ab⋅ γH°A°

⎛ ⎝

MQb := a + aa −

Qb = 90 Kgr/m

ab ⎞ 2 ⎠

⋅ Qb

MQb = 133.65

Postes Alto del poste:

hp := 0.9

m

Profundidad del poste

bp := 0.2

m

Dimensiones del poste:

ap1 := 0.12 m ap2 := 0.08 m ap3 := 0.10 m N°p := 20

Calculo N° de postes:

Sep :=

Luz − bp N°p − 1

Sep = 1.9895

Verificacion := if(Sep ≤ 2 , "Cumple" , "aumentar postes" ) Verificacion = "Cumple"

Qp1 :=

hp⋅ap1⋅ bp⋅ γH°A° ⋅ N°p

⎛ ⎝

MQp1 := a + aa + ap3 − ap2 − Qp2 :=

ap1 ⎞ 2

⎠

⋅ Qp1 MQp1 = 41.472

0.5⋅ap2⋅ hp ⋅ bp ⋅ γH°A° ⋅ N°p

Qp2 = 9.0947

Luz

⎛ ⎝

2⋅ap2 ⎞

MQp2 := a + aa + ap3 −

Qp3 :=

Qp1 = 27.2842 Kgr/m

Luz

3

⎠

⋅ Qp2

ha⋅ap3⋅ bp⋅ γH°A° ⋅ N°p

⎛ ⎝

MQp3 := a + aa +

ap3 ⎞ 2

⎠

MQp2 = 14.6122

Qp3 = 3.7895

Luz

⋅ Qp3

Kgr/m

Kgr/m

MQp3 = 6.1011

Momento total por carga muerta: Mcm := MQL + MQbo + MQb + MQp1 + MQp2 + MQp3 + MQa Mcm = 911.6665 Kgr.m/m

Momento por Cargas vivas

a

F2v F3h

F3v 0.3m

t

D

F3h := 750.0 kgr/m h3h := hbo −

t

h3h = 0.36 m

2

MF3h := h3h⋅ F3h F2v := 290⋅ aa

⎛ ⎝

MF3h = 270

kgr/m

MF2v := F2v⋅ a +

aa ⎞ 2 ⎠

F2v = 188.5 MF2v = 232.7975

Carga de camión: P = 7265 F3v :=

P

Kgr

F3v = 4462.5307 kgr/m

E

MF3v := F3v⋅ (a − 0.3)

MF3v = 2722.1437

Momento total por carga viva: Mcv := MF3v + MF3h + MF2v Mcv = 3224.9412

kgr.m/m

Impacto:

⎛ 15 ≥ 0.3, 0.3, 15 ⎞ a + 38 ⎠ ⎝ a + 38

I := if

MI := I ⋅ Mcv

I = 0.3

MI = 967.4824

Momento de diseño

Mu := 1.3[Mcm + 1.67(Mcv + MI)] Mu = 10286.918 kgr.m/m Canto úti l

φ := 1.2 d := h − r −

φ 2

d = 14.9 Cuantía necesaria

b = 100 cm

ρ :=

fc 1.18⋅ fy

⎛

1−

⎝

1−

2.36⋅ Mu⋅ 100 ⎞ 2

φ ⋅ fc⋅ b⋅ d

⎠

ρ = 0.0102

Cuantia balanceada

β1 := 0.85 ρb := 0.85⋅ β1⋅

⎛ 6090 ⎞ fc ⎝ 6090 + fy ⎠ fy

ρb = 0.0255

Cuantía máxima y mínima

ρmax := 0.75⋅ ρb

ρmax = 0.0191

Verificacion := if (ρ < ρmax , "cumple" , "no cumple" )

ρmin :=

14

Verificacion = "cumple"

ρmin = 0.0033

fy

Cuantía de diseño

ρ := if (ρ ≥ ρmin, ρ , ρmin)

ρ = 0.0102

Ac ero de r efu erzo

As := ρ ⋅ b⋅ d Ao :=

N°Barras :=

As Ao

π ⋅φ 4

As = 15.2443

cm2

2

Ao = 1.131

N°Barras = 13.4789

N°Barras := 14

Usar :

N°Barras = 14 de φ = 1.2 cm Area de acero proporcionada: Asprop := N°Barras⋅ Ao Asprop = 15.8336 cm2 /m

Ac ero de d is tr ib uc ió n

L := a Para armadura principal perpendicular al transito:

⎛ 1.22 ≤ 0.67 , 1.22 , 0.67 ⎞ L ⎝ L ⎠

D := if

D = 0.67 AD := D⋅ As AD = 10.2137 cm2 Por Tanto:

φ := 1.2cm Ao :=

π ⋅φ

2

Ao = 1.131

4

N°Barras :=

AD Ao

N°Barras = 9.0309

N°Barras := 10

Usar :

N°Barras = 10 de φ = 1.2 cm Area de acero proporcionada: Asprop := N°Barras⋅ Ao Asprop = 11.3097 cm2 /m

(En la carga superior, bajo la armacdura negativa)

Ac ero po r t emp erat ur a

Cuantía mínima por temperatura:

ρtem := 0.0020 As := ρtem⋅ h⋅ b

As = 3.6 cm2 /m

φ := 1 cm

Ao :=

π ⋅φ

2

4

N°Barras :=

As Ao

Ao = 0.7854

N°Barras = 4.5837

N°Barras := 5 Usar :

N°Barras = 5 de φ = 1

cm


Calculo Losa de Tablero Puente

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