Viga Doblemente Armada
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ESTRUCTURAS II
UNIVERSIDAD AUTONOMA DE YUCATAN
FACULTAD DE ARQUITECTURA
ALBA DANIELA ESPINOSA QUIJANO
ING. LUIS DAVID SANCHEZ RICALDE
PROBLEMAS VIGA DOBLEMENTE ARMADA
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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 6 000 fy 6 000 + fy
Pmáx= .75 .85 x .85 x 200 6 000 4,200 6 000 + 4,200
Pmáx= .75 x .0344x .5882Pmáx=.0152
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
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.75 pb= 0.01510.0152˃ 0.0151
P-P´ ≥ .85 B1 f´c d´ 6 000 Fy d 6 000 – fy
.00151≥ .85 (.85) 200x4 6 000 4200x50 6 000 – 4200
0.7225x 3.80 x 10‾³ x 3.33= 0.009160.0151˃ 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
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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 6 000 fy 6 000 + fy
Pmáx= .75 .85 x .85 x 200 6 000 4,200 6 000 + 4,200
Pmáx= .75 x .0344x .5882Pmáx=.0152
P= As = 11.48 = 0.009184 Bd 25 x 50
P < Pmax
0.0152 ˃ 0.009184 La viga trabaja como simplemente armada.
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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
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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
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
5 m
WL= 20,000 kg/m
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Pmáx ≤ 0.75 pb
Pmáx = 0.75 85B1 F´c 6 000 fy 6 000 + fy
Pmáx= .75 .85 x .85 x 200 6 000 4,200 6 000 + 4,200
Pmáx= .75 x .0344x .5882Pmáx=.0152
As₁ = pbd = 0.0152 x 30 x 50 = 22. 8 cm²
a= As fy = 22.8 x 4200 = 18.78 0.85 f´c b .85 x 200 x 30
M₁ = Ø As fy ( d – 0.5a)
M₁ = 0.90 x 22.8 x 4200 (50-0.5 (18.78))= 3,499,932.24 kg-cm
M₁= 3,499,932.24 kg-cm < 10,791,031 kg-cm
Mu= M₁ + M₂
M₂ = Mu - M₁
M₂ = 10´791,031 kg-cm – 3,499,932.24 kg-cm = 7,291,098.76 kg-cm
M₂ = Ø A’s fy ( d – d’) A’s = Mu Ø fy ( d – d’)
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A’s= 7,291,090.76 = 42.86 cm² 0.90 x 4200( 50-5) As = As₁ +As₂ =22.8 cm² + 42.86 cm² = 65.66 cm² = 7 barras # 11 = 66.96 cm²
A’s = 42.86 cm² = 7 barras # 9 = 44.97 cm²
Ejercicio 4.
50 cm
5 cm
30 cm
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Diseñar la trabe 3-B-C
Datos
Columnas de 40 x 40 cm
Trabe de 25 x 50 cm
Losa de vigueta y bovedilla
Techo de entrepiso para biblioteca
f’c= 300kg/cm2
fy= 4200kg/cm2
d’= 5cm
r=5cm
d=45cm
6m 6m7m 7m
4m
8.5m
4mD
B
C
1 3 4 52
A
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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 8.5Peso propio de la losa 365kg/m2
Acabado inferior 36 kg/m2 Firme de nivel 54 kg/m2 Piso + Adhesivo 35 kg/m2 Sobrecarga 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
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B1= 1.05 – f´c = 1.05 – 300 = .8357 1400 1400
Pmáx = 0.75 .85B1 F´c 6 000 fy 6 000 + fy
Pmáx= .75 .85 x .8357 x 300 6 000 4,200 6 000 + 4,200
Pmáx= .75 x .0507 x .5882Pmáx=.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
M₁ = 0.90 x 25.182 x 4200 (45 -0.5 (16.5904))= 3,888,656.617 kg-cm
M₁= 3,888,656.617 kg-cm < 7, 052, 683.75kg-cm
M₂ = Mu - M₁
M₂ = 7, 052, 683.75kg-cm– 3,888,656.617 kg-cm 3, 164, 027.133 kg-cm
M₂ = Ø A’s fy ( d – d’) A’s = Mu Ø fy ( d – d’)
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A’s= 3, 164, 027.133 = 20.92 cm² 0.90 x 4200( 45-5)
As = As₁ +As₂ = 25.182 + 20.92= 46.102 cm²= 6 barras # 10=47.65 cm²
A’s = 20.92cm² = 8 barras #6= 22.92 cm²
50 cm
5 cm
25 cm