Diseño de una viga de gran peralte con tirantes y puntales
19
BACH. RONALD J. PURCA
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Transcript of Diseño de una viga de gran peralte con tirantes y puntales
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
