BACH. RONALD J. PURCA

Resistencia Concreto f’c=4000 psi

Resistencia de Fluencia del Acero fy=60000 psi

𝑎𝑣

ℎ=

56𝑖𝑛

48𝑖𝑛= 1.17 < 2 (𝑉𝑖𝑔𝑎 𝑎𝑙𝑡𝑎)

2.1 Diseño a Flexión

d = 44.4in

𝑎 =𝐴𝑠𝑓𝑦

0.85𝑓′𝑐𝑏

𝑀𝑛 = 𝐴𝑠𝑓𝑦 𝑑 −𝑎

2

𝑀𝑢 = 11984 𝑘𝑙𝑏. 𝑖𝑛 ≤ 0.9𝑀𝑛

𝐴𝑠 ≥ 5.4 𝑖𝑛2

𝑐 ≥𝑎

0.85≈ 8.1 𝑖𝑛

Modelo Reticulado: Bielas (Azul)

Tirantes (Rojo)

1

6

2

7 1

2

3

4

3 5 4

Modelo Reticulado: Bielas (Azul)

Tirantes (Rojo)

307 klb (C)

307 klb (T)

154 klb (C)

154 klb (T)

214

klb

(T)

214 klb 214 klb

Los modelos de bielas y tirantes fallan debido a:

o Aplastamientos de las bielas (extremos)

o Aplastamiento en zonas nodales (caras)

o Fluencia en los tirantes

o Falta de anclaje de tirantes

ZONA NODAL 1 CCT

1

3

263 Klb

154 Klb 214 Klb

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

154

(0.80)(35.7)= 5.4𝑖𝑛

8.0 in > 5.4 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

214

(0.80)(35.7)= 7.5𝑖𝑛

16 in > 7.5 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

263

(0.80)(35.7)= 9.2𝑖𝑛

17.7 in > 9.2 in OK

6

ZONA NODAL 2 CCT

2 2

3

263 Klb

154 Klb

214 Klb

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

154

(0.80)(35.7)= 5.4𝑖𝑛

10 in > 5.4 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

214

(0.80)(35.7)= 7.5𝑖𝑛

14.6 in > 7.5 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

263

(0.80)(35.7)= 9.2𝑖𝑛

17.7 in > 9.2 in OK

ZONA NODAL 4 CCCC

4

2

4

263 Klb 307 Klb

214 Klb

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

307

(1)(35.7)= 8.6𝑖𝑛

10 in > 8.6 in OK

𝑤𝑟𝑒𝑞 =𝑉𝑢

𝛽235.7=

214

(1)(35.7)= 6.0𝑖𝑛

16 in > 6.0 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

263

(1)(35.7)= 7.4𝑖𝑛

18.8 in > 7.4 in OK

1

154 Klb

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

154

(1)(35.7)= 4.3𝑖𝑛

10 in > 4.3 in OK

ZONA NODAL 3 CTTT

3 7

6

263 Klb

307 Klb

214 Klb

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

307

(0.6)(35.7)= 14.3𝑖𝑛

8 in < 14.3 in NO PASA

𝑤𝑟𝑒𝑞 =𝑉𝑢

𝛽235.7=

214

(0.6)(35.7)= 10𝑖𝑛

23.2 in > 10 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

263

(0.6)(35.7)= 12.3𝑖𝑛

18.8 in > 12.3 in OK

4

154 Klb

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

154

(0.6)(35.7)= 7.2𝑖𝑛

8 in > 7.2 in OK

Ancho de biela 3 = 17.7in (Provisto) > 9.9in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

∅ 0.85 𝛽2 𝑓′𝑐 𝑏=

263000 𝑙𝑏𝑓

0.75 0.85 (0.75)(4000𝑝𝑠𝑖)(14𝑖𝑛)= 9.9𝑖𝑛

1

3

BIELA 3 (Nodo 1)

Ancho de biela 3 = 17.7in (Provisto) > 9.9in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

∅ 0.85 𝛽2 𝑓′𝑐 𝑏=

263000 𝑙𝑏𝑓

0.75 0.85 (0.75)(4000𝑝𝑠𝑖)(14𝑖𝑛)= 9.9𝑖𝑛

2

3

BIELA 3 (Nodo 2)

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

154

(1)(35.7)= 4.3𝑖𝑛

BIELA 1

10.0 in > 4.3 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

307

(1)(35.7)= 8.6𝑖𝑛

BIELA 2

10.0 in > 8.6 in OK

𝑤𝑟𝑒𝑞 =𝐹𝑢

𝛽235.7=

263

(0.75)(35.7)= 9.9𝑖𝑛

BIELA 4

18.8 in > 9.9 in OK

TIRANTE 6

𝐴𝑠 =𝐹𝑢

0.75𝑓𝑦

𝐴𝑠 =154000 𝑙𝑏𝑓

(0.75)(60000𝑝𝑠𝑖)= 3.4𝑖𝑛2

𝐿𝑑ℎ =0.02 (1)(1)(60000𝑝𝑠𝑖)

4000𝑝𝑠𝑖= 19𝑖𝑛

6 Varillas #8 (4.74in2) en 2 capas

Si recubrimiento > 2.5in Ldh=0.70(19in)= 13.3in

16+4/tan(54.3°)-1.5-0.625= 16.7in

TIRANTE 7

𝐴𝑠 =𝐹𝑢

0.75𝑓𝑦

𝐴𝑠 =307000 𝑙𝑏𝑓

(0.75)(60000𝑝𝑠𝑖)= 6.8𝑖𝑛2

𝐿𝑑ℎ =60000𝑝𝑠𝑖 (1)(1)(1)(1𝑖𝑛)

25 4000𝑝𝑠𝑖= 38𝑖𝑛

6 Varillas #8 + 2 #8 + 2#6 (7.2in^2) 3 Capas

𝐿𝑑ℎ =60000𝑝𝑠𝑖 (1)(1)(1)(3/4𝑖𝑛)

25 4000𝑝𝑠𝑖= 28.5𝑖𝑛

36-1.5-0.625= 33.9in

TIRANTE 5

𝐴𝑠 =𝐹𝑢

0.75𝑓𝑦

𝐴𝑠 =214000 𝑙𝑏𝑓

(0.75)(60000𝑝𝑠𝑖)= 4.8𝑖𝑛2 1 Varilla #5 (0.31 in2)

Luego: 4.8/ 0.31 /2 = 7.75 Usar 8 estribos cerrados #5

𝑠 =𝐴𝑣

0.0025𝑏=

(2)0.31

0.0025(14)= 17.7𝑖𝑛

Usar estribos cerrados #5 @ 6in

𝑠𝑣 =𝐴𝑠𝑠𝑒𝑛(∝ °)

0.003𝑏𝑠=

(2)(0.2)𝑠𝑒𝑛(54.3°)

0.003(14)= 7.6𝑖𝑛

1 Varilla #4 (0.2 in2)

Usar Varillas horizontales #4 @ 7in