Diseño por flexión y corte de una viga peraltada de concreto armado

1. BACH. RONALD J. PURCA

2. Resistencia Concreto f’c=4000 psi
Resistencia de Fluencia del Acero fy=60000 psi
𝑎 𝑣
ℎ
=
56𝑖𝑛
48𝑖𝑛
= 1.17 < 2 (𝑉𝑖𝑔𝑎 𝑎𝑙𝑡𝑎)

3. 2.1 Diseño a Flexión
d = 44.4in
𝑎 =
𝐴 𝑠 𝑓𝑦
0.85𝑓′ 𝑐 𝑏
𝑀 𝑛 = 𝐴 𝑠 𝑓𝑦 𝑑 −
𝑎
2
𝑀 𝑢 = 11984 𝑘𝑙𝑏. 𝑖𝑛 ≤ 0.9𝑀 𝑛
𝐴 𝑠 ≥ 5.4 𝑖𝑛2
𝑐 ≥
𝑎
0.85
≈ 8.1 𝑖𝑛

4. Modelo Reticulado: Bielas (Azul)
Tirantes (Rojo)
1
6
2
71
2
3
4
3 5 4

5. Modelo Reticulado: Bielas (Azul)
Tirantes (Rojo)
307 klb (C)
307 klb (T)
154 klb (C)
154 klb (T)
214klb(T)
214 klb 214 klb

6. 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

7. 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

8. 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

9. 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

10. 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

11. 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)

12. 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)

13. 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

14. 𝑤𝑟𝑒𝑞 =
𝐹𝑢
𝛽235.7
=
263
(0.75)(35.7)
= 9.9𝑖𝑛
BIELA 4
18.8 in > 9.9 in OK

15. TIRANTE 6
𝐴 𝑠 =
𝐹𝑢
0.75𝑓𝑦
𝐴 𝑠 =
154000 𝑙𝑏𝑓
(0.75)(60000𝑝𝑠𝑖)
= 3.4𝑖𝑛2
𝐿 𝑑ℎ =
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

16. TIRANTE 7
𝐴 𝑠 =
𝐹𝑢
0.75𝑓𝑦
𝐴 𝑠 =
307000 𝑙𝑏𝑓
(0.75)(60000𝑝𝑠𝑖)
= 6.8𝑖𝑛2
𝐿 𝑑ℎ =
60000𝑝𝑠𝑖 (1)(1)(1)(1𝑖𝑛)
25 4000𝑝𝑠𝑖
= 38𝑖𝑛
6 Varillas #8 + 2 #8 + 2#6 (7.2in^2) 3 Capas
𝐿 𝑑ℎ =
60000𝑝𝑠𝑖 (1)(1)(1)(3/4𝑖𝑛)
25 4000𝑝𝑠𝑖
= 28.5𝑖𝑛
36-1.5-0.625= 33.9in

17. TIRANTE 5
𝐴 𝑠 =
𝐹𝑢
0.75𝑓𝑦
𝐴 𝑠 =
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 =
𝐴 𝑣
0.0025𝑏
=
(2)0.31
0.0025(14)
= 17.7𝑖𝑛
Usar estribos cerrados #5 @ 6in

18. 𝑠 𝑣 =
𝐴 𝑠 𝑠𝑒𝑛(∝ °)
0.003𝑏𝑠
=
(2)(0.2)𝑠𝑒𝑛(54.3°)
0.003(14)
= 7.6𝑖𝑛
1 Varilla #4 (0.2 in2)
Usar Varillas horizontales #4 @ 7in