diseño escalera

2
0.17 Muro portante de albañileria 0.17 0.70 m metros : 0.25 b = 1.60 2.10 0.40 Ln = 3.70 m DATOS : f'c = 280 Kg/cm² fy = 4200 Kg/cm² P = 30 cm Cp = 18 cm s/c = 600 Kg/m² PREDIMENSIONAMIENTO : Espesor : t = 0.15 m Asumimos : t = 0.17 cm t = 0.19 m Cos Ø = 0.857 Altura eq. : hm = 28.84 cm METRADO DE CARGAS : CM : P.P = hm * b * 2.4 * 1.4 = 1.55 Tn/m Acabado = 0.10 * b * 1.4 = 0.224 Tn/m CV : S/C = s/c * b * 1.7 = 1.632 Tn/m Wu2 = 3.41 Tn/m CM : P.P = t * b * 2.4 * 1.4 0.914 Tn/m Wu1 = 2.77 Tn/m DISEÑO ESCALERA Cos (Ɵ) = 2 + 2 hm = + 2 t = 25 t = 20

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escalera

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Page 1: Diseño Escalera

0.17

Muro portante

de albañileria 0.17

0.70 m

metros : 0.25 b = 1.60 2.10 0.40

Ln = 3.70 m

DATOS :

f'c = 280 Kg/cm²

fy = 4200 Kg/cm²

P = 30 cm

Cp = 18 cm

s/c = 600 Kg/m²

PREDIMENSIONAMIENTO :

Espesor : t = 0.15 m Asumimos :

t = 0.17 cm

t = 0.19 m

Cos Ø = 0.857

Altura eq. : hm = 28.84 cm

METRADO DE CARGAS :

CM : P.P = hm * b * 2.4 * 1.4 = 1.55 Tn/m

Acabado = 0.10 * b * 1.4 = 0.224 Tn/m

CV : S/C = s/c * b * 1.7 = 1.632 Tn/m

Wu2 = 3.41 Tn/m

CM : P.P = t * b * 2.4 * 1.4 0.914 Tn/m

Wu1 = 2.77 Tn/m

DISEÑO ESCALERA

Cos (Ɵ) = 𝑃

𝑃2+𝐶𝑝2

hm = 𝑡

𝐶𝑜𝑠 𝚹 +

𝐶𝑝

2

t = 𝐿𝑛

25

t = 𝐿𝑛

20

Page 2: Diseño Escalera

3.41 Tn/m

2.77 Tn/m

RA RB

1.73 m 2.30 m

Lt = 4.025 m

RB = 6.62 Tn

Xo = 1.943 m

= 6.433 Tn-m

; α = 0.8 = 5.146 Tn-m

ELECCION DEL REFUERZO :

Ø = 1/2 pulg.

Asb= 1.27 cm²

Recub. = 2.00 cm

Peralte efectivo : d= 14.37 cm

Asumimos que : a = 2 cm

As = 10.78 cm a = 1.12 cm

Por lo tanto se usara: N° varillas = 8

Acero = Ø 1/2 @ 25.00

RB =[ 𝑊𝑢1 ∗ 𝐿𝑡2

2 + (Wu2 - Wu1)* L2 * (L1 + 𝐿2

2 )] / Lt

Vx = 𝑅𝐵 − 𝑊𝑢2 ∗ 𝑋𝑜 = 0

(+) Mu max = 𝑅𝐵 ∗ 𝑋𝑜 − 𝑊𝑢2 ∗𝑋𝑜2

2

(-) Mu dis = α ∗ 𝑀𝑢 max

As=𝑀𝑢

Ø 𝑓𝑦 (𝑑−𝑎

2) a =

𝐴𝑠 𝑓𝑦

0.85 𝑓′𝑐 𝑡

d = t − (𝑅𝑒𝑐𝑢𝑏. +Ø𝑣𝑎𝑟

2)

a = 𝑑

5