A s=10.87cm2

A s(min)=14f y

∗b∗d= 144200

∗100∗14.20

A s(min)=4.73 cm2<10.87

espaciamiento :

seusaran fierrosde58

' '

(A s=1.98cm2 )

S5 /8' '

= 1.9810.87

∗100

⟹S5 /8' '=18.2cm

Entonces se usaran∎5 /8' '@18cm .

4.2 . MomentoUltimoNegativo

Mu=1.3 [MD+1.67 (M L+M I ) ]

Mu=[0.226+1.67 (2.169+0.651 ) ]⟹M u=6.416Tn−m

4.2 .1 .Calculo de Acero

MU=0.9 A s∗f y [d− A s∗f y

1.7 f 'c∗b ]6.416∗105=0.9∗4200∗A s[14.20− A s∗4200

1.7∗240∗100 ]6.416∗105=53676 A s−389.12 A s

2

389.12 A s2−53676 A s+6.416∗10

5=0

A s=12.31cm2

A s(min)=4.73 cm2<12.31

Espaciamiento :

Seusaràn fierrosde5/8 ' ' (A s=1.98cm2 )

S5 /8' '

= 1.9812.31

∗100⟹S5/8' '

=16.08cm

Entonces se usaran∎5 /8' '@16 cm

5.DISEÑODETRAMOEN VOLADISO

5.1 .Momento por CargaMuerta

MetradodeCarga

5.2. Momento por Carga viva

M I=I∗M L=0.30∗4.995⟹M I=1.498Tn−m

5.3MomentoUltimo enVoladizo

MU=1.3 [M D+1.67 (M L+M I ) ]

MU=1.3 [3.288+1.67 (4.995+1.498 ) ]⟹MU=18.37Tn−m

5.4 .Cálculo del Acero

MU=ø A s∗f y [d− A s∗f y1.7 f 'c∗b ]18.37∗105

0.9∗4200∗A s[14.20− A s∗42001.7∗240∗100 ]18.37∗105=5367 A s−389.64 A s

2

389.64 A s2−53676 A s+18.37∗10

5=0

A s=48 cm2 espaciamiento :

seusaran fierrosde5 /8' ' (A s=1.98cm2 )

S5 /8' '

= 1.9812.31

∗100⟹S5/8' '

=5cm

entonces seusaran∎5 /8' '@5cm

5.5 . Acero de temperatura

A stemp=0.0018∗100∗20

A stemp=3.6cm2espaciamiento :

seusaran fierrosde3 /8' ' (A s=0.71cm2 )

S3 /8' '

=0.713.6

∗100⟹S5 /8' '

=19.7cm

Entonces s eusaran∎3/8' '@20cm

5.6 . Acero de repartición

A s rep=% A s p

donde :%=121

√S'= 121

√1.825=89.56%