analisis - cargas escalera

ANALISIS DE CARGAMATERIALESDatos:

B 420 [Mpa] fyd = 365.22 [Mpa]H 21 [Mpa] fcd = 14.00 [Mpa]

Lmax = 3.7 [m]e = 0.12 [m]

0.03 [m]

Tramo 1 , 3 Tramo D-GLv= 1.17 [m] Lv= 0.7 [m]Lh= 1.5 [m] Lh= 1.6 [m]

Tramo 3 Descanso

Tramo 2Tramo 1

Lv

Lh

hm = 0.26 [m] N peldaños:⁰ hm = 0.20 [m] N peldaños:⁰

37.95 [°] 5.85 23.63 [°] 5a = 0.26 [m] a = 0.32 [m]b = 0.20 [m] Nos damos b = 0.14 [m] Nos damos

b+2a = 65.64 [cm] b+2a = 60.00 [cm]

Descanso = 1.50 [m] Nos damos Descanso = 1.50 [m] Nos damos

1.90 [m] 1.75 [m]

Escalera: Escalera:

Tramo inclinado: Tramo inclinado:

6.41 [KN/m] 5.12 [KN/m]

Altura contrapiso =

α = α =

Longitud Inclinada =


Peso propio de peldaños


a

bhmh1

Alturacontrapiso

hm=e

cosα+b2

e=Lmax30

a=b

tanα

=hm⋅25

Huella 0.66 [KN/m] Huellas 0.66 [KN/m]

Contra huella 0.51 [KN/m] Contra huellas 0.29 [KN/m]

Piso 0.69 [KN/m] Piso 0.63 [KN/m]

Revoque 0.32 [KN/m] Revoque 0.27 [KN/m]

baranda: 0.22[KN/m]

baranda: 0.22[KN/m]

Planta: Perfil:

2.2 0.0

0 0.0

1.5 1.2

0 0 2.0 1.5 2.2

longitud= 7.91 [m] longitud= 1.75 [m]

276.7 61.12

12.01 3.84

1.5 [m] 1.5 [m]

7.609 3.84213

Tramo recto= 4.4

0.35 [km/m] 0.23864 [km/m]

TRAMO INCLINADO

g= 8.81 [KN/m] g= 7.19 [KN/m]q= 4 [KN/m] q= 4 [KN/m]

Pd= 19.62 [KN/m] Pd= 17.18 [KN/m]

Peso de la baranda=

[kg/m²] Peso de la baranda=

[kg/m²]

Area de Influencia=

[m²] Area de Influencia=

[m²]

Altura de baranda=

Altura de baranda=

Tramo inclinado=

[m²] Tramo inclinado=

[m²]

[m²]

Peso de la baranda=

Peso de la baranda=

=AlturaContrapiso⋅22

19.84 [KN/m] 17.42 [KN/m]Escalera:

Tramo Recto:

Losa 3.08 [KN/m]Contrapiso 0.66 [KN/m]Piso 0.60 [KN/m]

Reboque 0.25 [KN/m]

g= 4.59 [KN/m]q= 4 [KN/m]

Pd= 13.29 [KN/m]

13.42 [KN/m]

Pd TOTAL = Pd TOTAL =

Pd TOTAL =

ANALISIS DE CARGAMATERIALES:Datos:

B 420 [Mpa] fyd = 365.22 [Mpa]H 25 [Mpa] fcd = 16.67 [Mpa]

εyd = 1.83

25

Lmax = 4.25 [m] 22

e = 0.14 [m] 18

0.03 [m] 12

0.025 [m]

Tramo:Lv= 1.44 [m]Lh= 2.85 [m]

hm = 0.24 [m] N peldaños:⁰

26.81 [°] 9a = 0.32 [m] contrahuellab = 0.16 [m] huella

b+2a = 0.637 [m]

Descanso = 1.40 [m]

3.19 [m]

Analisis de Cargas:

Tramo inclinado:

5.97 [KN/m]

Contrapiso 0.66 [KN/m]Contra huell 0.33 [KN/m]Huella 0.63 [KN/m]

[ ‰ ]

γHº= [KN/m³]

γmortero= [KN/m³]

γceramica= [KN/m³]

e contrapiso = γyeso(reboque)= [KN/m³]

e reboque =

α =



a

bhmh1

hm=e

cosα+b2

Revoque 0.34 [KN/m]baranda: 0.22 [KN/m]

Planta:

Barandadosecc. tubo pasamanos Ø 50 x 3

área tubo pasamanos = 0.00044 [m^2]secc. tubo travesaños Ø 30 x 3

área tubo pasamanos = 0.00025 [m^2]secc. tubo parante : Ø 50 x 3

área tubo parante = 0.00044 [m^2]altura tubo parante = 0.90 [m]

separación tubo parante = 1.20 [m]

Peso barandado (Tramo inclinado)

vidrio : ( 0.010 x 0.90 ) x ( 1.00 / 0.85 ) x 25.00 = 0.26 [KN/m]

tubo pasamanos : ( 1 x 0.00044 ) x ( 1.00 / 0.89 ) x 78.50 = 0.04 [KN/m]

tubo travesaños : ( 2 x 0.00025 ) x ( 1.00 / 0.89 ) x 78.50 = 0.04 [KN/m]

tubo parante : ( 0.00044 / 1.20 ) x 0.90 x 78.50 = 0.03 [KN/m]

0.26 + 0.04 + 0.04 + 0.03 = 0.37 [KN/m]

= 0.49 [KN/m]

Peso barandado (Descanso)

vidrio : 0.010 x 0.90 x 1.00 x 25.00 = 0.23 [KN/m]

tubo pasamanos : ( 1 x 0.0004 ) x 1.00 x 78.50 = 0.03 [KN/m]

tubo travesaños : ( 2 x 0.0003 ) x 1.00 x 78.50 = 0.04 [KN/m]

g barandado 1 lado =

g barandado 1 lado ADOPTADO

2.85 1.40

1.44

2.88

2.80

2.85

1.40

1.40

tubo parante : ( 0.00044 / 1.20 ) x 0.89 x 78.50 = 0.03 [KN/m]

0.23 + 0.03 + 0.04 + 0.03 = 0.33 [KN/m]

= 0.42 [KN/m]

Tramo Inclinado:g= 8.15 [KN/m]q= 4 [KN/m]

Pd= 18.62 [KN/m]

19.111 [KN/m]Escalera:

Tramo Recto:

Losa 3.54 [KN/m]Contrapiso 0.66 [KN/m]Piso 0.60 [KN/m]

Reboque 0.25 [KN/m]

g= 5.05 [KN/m]q= 4 [KN/m]

Pd= 13.98 [KN/m]

14.401 [KN/m]

g barandado 1 lado =

g barandado 1 lado ADOPTADO

Pd TOTAL =

Pd TOTAL =

SOLICITACIONES

TRAMO: 1 - 3

P1 P2

2.85 1.4

P1 = 19.11 KN/m

P2 = 14.40 KN/m

RA = 39.52 KN

X = 2.07 m

Mmax = 40.87 KN-m

EN APOYOS

M1 M1

1.5 1.5

M1 = 4.30 KN-m

Verificando cuantía mínima

Área a cubrir 3.29

DISENO A FLEXION

H- 25 fcd [Mpa]= 16.67

B- 420 fyd [Mpa]= 365.22

γ c= 1.5 18.00

γ s= 1.15 Rec. mec [cm] = 2.00

b [m] = 1.00

d [cm] = 16.00

Tramo Md [kN m] Armadura

(+) 1 - 3 40.87 7.45 5.39 3 Ø 12

Armadura Transversal

(+) 25% 1.86 1.36 Ø 6 c/ 30

Armadura Longitudinal

(+) 2 #REF! #REF! 5.14 5 Ø 16

Armadura Transversal

(+) 25% #REF! 1.29 Ø 8 c/ 22

Armadura Longitudinal

Asmin = [cm2 /m]

h L [cm]=

As[cm2] As[cm2]

Asmin=0 .04∗f cdf yd

∗b∗h

(-) 1 - 3 #REF! 0.74 0.5 2 Ø 8Armadura Negativa

(-) 2 #REF! #REF! 0.28 4 Ø 12Armadura Negativa

DISENO A FLEXION

TABLA PARA EL DISEÑO DE ESCALERAS A FLEXIÓN

DISEÑO PARA EL TRAMO 1-3Materiales Geometria

fcd = 16.67 [MPa] b = 1.00 [m]fyd = 365.22 [MPa] h = 0.18 [m]zyd = 1.74 recub. = 0.02 [m]

Solicitaciones d1 = 0.16 [m]Nd = 0 [MN] d2 = 0.16 [m]Md = 0.0409 [MN.m]

Diseno con Diagrama Parabolico Rectangularμn = 0.0958 μ3lim= 0.3319αn = 0.1484 α3lim = 0.6680βn = 0.9383 β3lim = 0.7221

x (prof eje)= 0.0237 x3lim = 0.1069Z = 0.1501 Z 3lim= 0.1155

OBSERVACIÓN: Dom.2b (A° Simple)εs1 = 10.0000 %oσs1 = 365.2174 [MPa]

As1 = 7.454

Armadura = ϕ 12 c/15

Asec = 1.864


[ ‰ ]

[cm2]

[cm2]

TABLA PARA EL DISEÑO DE ESCALERAS A FLEXIÓN

MOMENTO NEGATIVO DISEÑO PARA EL TRAMO 1Materiales Geometria

fcd = 14.00 [MPa] b = 1.00 [m]fyd = 365.22 [MPa] h = 0.18 [m]zyd = 1.74 %o hf = 0.00 [m]

Solicitaciones recub. = 0.02 [m]Nd = 0 [MN] d1 = 0.16 [m]Md = 0.0043 [MN.m] d2 = 0.16 [m]

Diseno con Diagrama Parabolico Rectangularμn = 0.0120 μ3lim= 0.3319αn = 0.0176 α3lim = 0.6680βn = 0.9927 β3lim = 0.7221

x (prof eje)= 0.0028 x3lim = 0.1069Z = 0.1588 Z 3lim= 0.1155

OBSERVACIÓN: Dom.2a (A° Simple)εs1 = 10.00 %oσs1 = 365.22 [MPa]

As1 = 0.741


[cm2]

analisis - cargas escalera

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