trabe cantiliver

DISEÑO DE TRABE EN CANTILIVER NTC-DF/2017 TRABE DE ANALISIS :

NERVADURA ENTREPISO TERRAZA

DATOS DE ANALISIS: Estructura del grupo:

B

f'c=

200

kg/cm2

f*c=

160

kg/cm2

f''c= FR=

136

kg/cm2

fy= H= r= d= b=

4200

0.9 30 3 27 15

pmin= 0.002357023

kg/cm2 cm cm cm cm

pb= 0.016190476

pmax= 0.012142857

DISEÑO POR FLEXIÓN Mumax= -48,814 kg-cm Q= -0.0365

(superior en el empotramiento) q= -0.0358 p= -0.00116 pmin pmin= = 0.002 0.00236 36

SOLUCION: ACERO A TENSION (superior): As= 0.95 cm2/m 2 barras no. 2 = 2.06 cm^2 2 barras no. 3 0 barras no. 0 ACERO EN COMPRESION (inferior): Asmin= 0.95 cm2/m 2 barras no. 2 = 1.35 cm^2 1 barras no. 3

Pmax= Pmax= 0.012 0.01214 14

>

A

ok

>

A´s ok

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Diseño por cortante: Preal= Preal= 0.0051 0.0051 cm^2 cm^2

<

Vumax= 714.0 kg

2.5*Fr*b*d*raiz(f*c)=

10245.8 kg

OK se acepta seccion por cortante

VCR=0.5Frbdraiz(f*c) VCR=1.5Frbdraiz(f*c) VCR=Fr*b*d*(0.2+20p)*raiz(f*c) VCR = 6147 Kg VCR = 7131 Kg

si p < 0.015 VCR= 1236.2733 si p > 0.015 VCR= 6147.5 kg

= 2049.2 kg = 6147.5 kg = 1236.3 kg

Si h Suministrar acero diagonal minimo Avmin

Separacion de estribos (Var a dos ramas): ramas): Vu < VCR

1 3 .5

var del #= 2 .0 fy= 5000 kg/cm2 Av= 0.63 cm2 S1= 10.7 cm

Vu < 1.5Frbdraiz(f*c)

Ok

armex 

Smax= 48.3 cm

Solucion: estribos:

ϴ 2.0/8

@ 15.0 15.0 cm en los los apoy apoyos os (L/4) (L/4)

estribos:

ϴ 2.0/8

@ 15.0 15.0 cm en el cent centro ro

1.16

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CROQUIS DE ARMADO:

H= 30 cm

b= 15 cm

L= 0.00 m estribos: ϴ 2/8 @ 15.0 cm

As= 2 barras 2 barras

0 .9 5 5 no. 2 no. 3

cm2/m

Asmin= 2 barras 1 barras

0 .9 5 5 no. 2 no. 3

cm2/m

L= 0.00 m estribos: ϴ 2/8 @ 15.0 cm ARMEX 15X30

L= 0.00 m estribos: ϴ 2/8 @ 15.0 cm