diseΓ±o de una viga de gran peralte con tirantes y puntales
TRANSCRIPT
BACH. RONALD J. PURCA
Resistencia Concreto fβc=4000 psi
Resistencia de Fluencia del Acero fy=60000 psi
ππ£
β=
56ππ
48ππ= 1.17 < 2 (ππππ πππ‘π)
2.1 DiseΓ±o a FlexiΓ³n
d = 44.4in
π =π΄π ππ¦
0.85πβ²ππ
ππ = π΄π ππ¦ π βπ
2
ππ’ = 11984 πππ. ππ β€ 0.9ππ
π΄π β₯ 5.4 ππ2
π β₯π
0.85β 8.1 ππ
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
π€πππ =πΉπ’
π½235.7=
154
(0.80)(35.7)= 5.4ππ
8.0 in > 5.4 in OK
π€πππ =πΉπ’
π½235.7=
214
(0.80)(35.7)= 7.5ππ
16 in > 7.5 in OK
π€πππ =πΉπ’
π½235.7=
263
(0.80)(35.7)= 9.2ππ
17.7 in > 9.2 in OK
6
ZONA NODAL 2 CCT
2 2
3
263 Klb
154 Klb
214 Klb
π€πππ =πΉπ’
π½235.7=
154
(0.80)(35.7)= 5.4ππ
10 in > 5.4 in OK
π€πππ =πΉπ’
π½235.7=
214
(0.80)(35.7)= 7.5ππ
14.6 in > 7.5 in OK
π€πππ =πΉπ’
π½235.7=
263
(0.80)(35.7)= 9.2ππ
17.7 in > 9.2 in OK
ZONA NODAL 4 CCCC
4
2
4
263 Klb 307 Klb
214 Klb
π€πππ =πΉπ’
π½235.7=
307
(1)(35.7)= 8.6ππ
10 in > 8.6 in OK
π€πππ =ππ’
π½235.7=
214
(1)(35.7)= 6.0ππ
16 in > 6.0 in OK
π€πππ =πΉπ’
π½235.7=
263
(1)(35.7)= 7.4ππ
18.8 in > 7.4 in OK
1
154 Klb
π€πππ =πΉπ’
π½235.7=
154
(1)(35.7)= 4.3ππ
10 in > 4.3 in OK
ZONA NODAL 3 CTTT
3 7
6
263 Klb
307 Klb
214 Klb
π€πππ =πΉπ’
π½235.7=
307
(0.6)(35.7)= 14.3ππ
8 in < 14.3 in NO PASA
π€πππ =ππ’
π½235.7=
214
(0.6)(35.7)= 10ππ
23.2 in > 10 in OK
π€πππ =πΉπ’
π½235.7=
263
(0.6)(35.7)= 12.3ππ
18.8 in > 12.3 in OK
4
154 Klb
π€πππ =πΉπ’
π½235.7=
154
(0.6)(35.7)= 7.2ππ
8 in > 7.2 in OK
Ancho de biela 3 = 17.7in (Provisto) > 9.9in OK
π€πππ =πΉπ’
β 0.85 π½2 πβ²π π=
263000 πππ
0.75 0.85 (0.75)(4000ππ π)(14ππ)= 9.9ππ
1
3
BIELA 3 (Nodo 1)
Ancho de biela 3 = 17.7in (Provisto) > 9.9in OK
π€πππ =πΉπ’
β 0.85 π½2 πβ²π π=
263000 πππ
0.75 0.85 (0.75)(4000ππ π)(14ππ)= 9.9ππ
2
3
BIELA 3 (Nodo 2)
π€πππ =πΉπ’
π½235.7=
154
(1)(35.7)= 4.3ππ
BIELA 1
10.0 in > 4.3 in OK
π€πππ =πΉπ’
π½235.7=
307
(1)(35.7)= 8.6ππ
BIELA 2
10.0 in > 8.6 in OK
π€πππ =πΉπ’
π½235.7=
263
(0.75)(35.7)= 9.9ππ
BIELA 4
18.8 in > 9.9 in OK
TIRANTE 6
π΄π =πΉπ’
0.75ππ¦
π΄π =154000 πππ
(0.75)(60000ππ π)= 3.4ππ2
πΏπβ =0.02 (1)(1)(60000ππ π)
4000ππ π= 19ππ
6 Varillas #8 (4.74in2) en 2 capas
Si recubrimiento > 2.5in Ldh=0.70(19in)= 13.3in
16+4/tan(54.3Β°)-1.5-0.625= 16.7in
TIRANTE 7
π΄π =πΉπ’
0.75ππ¦
π΄π =307000 πππ
(0.75)(60000ππ π)= 6.8ππ2
πΏπβ =60000ππ π (1)(1)(1)(1ππ)
25 4000ππ π= 38ππ
6 Varillas #8 + 2 #8 + 2#6 (7.2in^2) 3 Capas
πΏπβ =60000ππ π (1)(1)(1)(3/4ππ)
25 4000ππ π= 28.5ππ
36-1.5-0.625= 33.9in
TIRANTE 5
π΄π =πΉπ’
0.75ππ¦
π΄π =214000 πππ
(0.75)(60000ππ π)= 4.8ππ2 1 Varilla #5 (0.31 in2)
Luego: 4.8/ 0.31 /2 = 7.75 Usar 8 estribos cerrados #5
π =π΄π£
0.0025π=
(2)0.31
0.0025(14)= 17.7ππ
Usar estribos cerrados #5 @ 6in
π π£ =π΄π π ππ(β Β°)
0.003ππ =
(2)(0.2)π ππ(54.3Β°)
0.003(14)= 7.6ππ
1 Varilla #4 (0.2 in2)
Usar Varillas horizontales #4 @ 7in