ESTRUCTURAS II UNIVERSIDAD AUTONOMA DE YUCATAN
FACULTAD DE ARQUITECTURA
ALBA DANIELA ESPINOSA QUIJANO
ING. LUIS DAVID SANCHEZ RICALDE
PROBLEMAS VIGA DOBLEMENTE ARMADA
Ejercicio 1.
Calcular el momento resistente de una viga de sección rectangular doblemente armada. Determinar si el acero en tensión y compresión se encuentran en estado de sedencia. Datos F’c= 200kg/cm²
Fy= 4200kg/cm² d= 50cm b= 25cm d’= 4cm
As= 30.42cm² 6 barras N. 8 As’= 11.48cm² 4 barras N. 6
Solución
Pmáx ≤ .75 pb
Pmáx= .75 .85B1 F´c fy Pmáx= .75 .85 x .85 x 200 4,200 Pmáx= .75 x .0344x .5882 Pmáx=.0152
6 000 6 000 + fy 6 000 6 000 + 4,200
P= As = 30.42 = 0.0243 Bd 25 x 50 P ˃ Pmáx 0.0243 ˃ 0.0151
P – P´= As – A´s = As- A´s = 30.42-11.48 = 0.0151 Bd Bd Bd 25 x 50
.75 pb= 0.0151 0.0152˃ 0.0151
P-P´ ≥ .85 B1 f´c d´ 6 000 Fy d 6 000 – fy .00151≥ .85 (.85) 200x4
4200x50
6 000 6 000 – 4200
0.7225x 3.80 x 10‾³ x 3.33= 0.00916 0.0151˃ 0.00916
a= (As-A´s) fy = (30.42 – 11.48) 4200 = 18.71 .85 f´c b .85x 200 x 25 Mu= .90[(30.42-11.48)4200(50-.5(18.71))+11.48x4200(50-4)] .90[18.94(170709)+(2217936)] Mu= 4,906,048.014 kg-cm
Ejercicio 2.
Calcular el momento resistente de una viga de sección rectangular doblemente armada. Determinar si el acero en tensión y compresión se encuentran en estado de sedencia. Datos F’c= 200kg/cm²
Fy= 4200kg/cm² d= 50cm b= 25cm d’= 4cm
As= 11.48cm² 4 barras N. 6 As’= 5.74cm² 2 barras N. 6
Solución Pmáx ≤ 0.75 pb
Pmáx = 0.75 85B1 F´c fy
6 000 6 000 + fy
Pmáx= .75 .85 x .85 x 200 4,200
6 000 6 000 + 4,200
Pmáx= .75 x .0344x .5882 Pmáx=.0152 P= As = 11.48 = 0.009184 Bd 25 x 50 P < Pmax 0.0152 ˃ 0.009184
La viga trabaja como simplemente armada.
W=Pfy = 0.009184x4200 = 0.1928 f´c 200 Mu= Ø b d² f´c w (1-0.59w) Mu= 0.90 x 25 x 50² x 200 x .1928 (1- 0.59(.1928)) = 1,922,271.912 kg – cm
Ejercicio 3.
Diseñar por flexión una viga rectangular doblemente armada. Datos
WL=20 ton/m = 20,000 kg/m Claro= 5m F’c= 200kg/cm²
Fy= 4200kg/cm² d= 50cm b= 30cm d’= 5cm
WL= 20,000 kg/m
5m
Solución
Wu= 1.4 D + 1.7 L Wu= 1.4 x 0.3 x 0.55 x 2300 + 1.7 x 20 000 kg/ m² = 34, 531. 3 kg / m Mu= Wl²= 34 531.3 x 5² = 107,910.31 kg-m = 10,791,031.5 kg-cm 8 8
Pmáx ≤ 0.75 pb
Pmáx = 0.75 85B1 F´c fy
6 000 6 000 + fy
Pmáx= .75 .85 x .85 x 200 4,200
6 000 6 000 + 4,200
Pmáx= .75 x .0344x .5882 Pmáx=.0152 As = pbd = 0.0152 x 30 x 50 = 22. 8 cm² ₁
a= As fy = 22.8 x 4200 = 18.78 0.85 f´c b .85 x 200 x 30 M = Ø As fy ( d – 0.5a) ₁
M = 0.90 x 22.8 x 4200 (50-0.5 (18.78))= 3,499,932.24 kg-cm ₁
M = 3,499,932.24 kg-cm < 10,791,031 kg-cm ₁
Mu= M + M ₁
₂
M = Mu - M ₂
₁
M = 10´791,031 kg-cm – 3,499,932.24 kg-cm = 7,291,098.76 kg-cm ₂
M = Ø A’s fy ( d – d’) ₂
A’s =
Mu
Ø fy ( d – d’)
A’s= 7,291,090.76
= 42.86 cm²
0.90 x 4200( 50-5) As = As +As =22.8 cm² + 42.86 cm² = 65.66 cm² = 7 barras # 11 = 66.96 cm² ₁
₂
A’s = 42.86 cm² = 7 barras # 9
= 44.97 cm²
50 cm
5 cm
30 cm
Ejercicio 4.
Diseñar la trabe 3-B-C
6m
7m
7m
6m
A 4m B
8.5m C 4m D 1
2
3
Datos
Columnas de 40 x 40 cm Trabe de 25 x 50 cm Losa de vigueta y bovedilla Techo de entrepiso para biblioteca 2
f’c= 300kg/cm
fy= 4200kg/cm2 d’= 5cm
r=5cm
4
5
d=45cm b=25cm L= 8.5m Área de la Losa= (7)(8.5)= 59.5m2 WL= 365kg/m2 WL= 350kg/m2 = (350)(59.5)= 20825kg/m2= 20825 = 2,450kg/m 8.5 Vigueta de 20-6 y bovedilla de WD= 365kg/m2= 365x59.5= 21717.5 = 2,555 8.5 Peso propio de la losa 365kg/m2 Acabado inferior Firme de nivel Piso + Adhesivo Sobrecarga
36 kg/m2 54 kg/m2 35 kg/m2 40 kg/m2 530 kg/m2
WD= (530) (59.5) = 31535 kg/m2 = 31535 = 3,710 kg/m 8.5 Solución Wu = 1.4 D + 1.4 D + 1.7 L Wu= 1.4 (0.25 x 0.50 x 2300) + 1.4(3710) + 1.7(2450) = 9, 761.5kg/m Mu= Wl² 10 Mu= 9,761.5x8.5² = 70, 526.83kg-m = 7, 052, 683.75kg-cm 10
Pmáx ≤ 0.75 pb
B1= 1.05 – f´c = 1.05 – 300 = .8357 1400 1400 Pmáx = 0.75 .85B1 F´c fy
6 000 6 000 + fy
Pmáx= .75 .85 x .8357 x 300 4,200
6 000 6 000 + 4,200
Pmáx= .75 x .0507 x .5882 Pmáx=.022384 As1 = pbd= (0.022384) (25) (45) = 25.182cm2
a= As fy = 25.182 x 4200 = 16.5904 0.85 f´c b .85 x 300 x 25 M = 0.90 x 25.182 x 4200 (45 -0.5 (16.5904))= 3,888,656.617 kg-cm ₁
M = 3,888,656.617 kg-cm < 7, 052, 683.75kg-cm ₁
M = Mu - M ₂
₁
M = 7, 052, 683.75kg-cm – 3,888,656.617 kg-cm ₂
M = Ø A’s fy ( d – d’) ₂
A’s =
Mu
Ø fy ( d – d’)
3, 164, 027.133 kg-cm
A’s=
3, 164, 027.133
= 20.92 cm²
0.90 x 4200( 45-5) As = As +As = 25.182 + 20.92= 46.102 cm²= 6 barras # 10= 47.65 cm² ₁
₂
A’s = 20.92cm² = 8 barras #6=
22.92 cm²
50 cm
5 cm
25 cm
