DISEÑO Y CALCULO DE VIGAS

1. DISEÑO DE VIGAS
EJEMPLO

2. h = 45 cms
fy= 2810 kg/m2
Wc= 2 500 kg/m3
Wviva= 500 kg/m3
Recub. = 2.5 cms
t = 0.15 mts
Wacab.= 80kg/m2
PREDIMENSIONAMIENTO
DATOS
b = 1/2h = 25 cms
ÁREAS TRIBUTARIAS
A1 = 1/2 (7)(3.5) = 12.25
A2 = 1.50 (7) = 10.50
At = 22.75 m2f’c= 210 kg/m2

3. CALCULO DE CARGAS
- CARGAS MUERTAS:
Wc = (2500)(0.15) = 375 km/m2
W acabados = 80 km/m2
455 km/m2
- CARGAS VIVAS:
W viva = 500 km/m2
- CARGAS ÚLTIMA:
1.4(455) + 1.7 (500) = 1487
Wd = 22.75(1487) + 1.4 (0.25)(0.45)(2500) = 5 227 kg/m
- CARGAS DISTRIBUIDAS:
7

4. WL2 = (5227)(7)2 =
MOMENTO (-)
MOMENTO (+)
16 16
M(-) = 16,008 Kg/m
WL2 = (5227)(7)2 =
10 10
M(+) = 25,612 Kg/m
ACERO MÁXIMO
As máx = 0.01867 (25)(42.50) = 19.84
As mín= 14.1 (b.d)
ACERO MÍNIMO
fy
= 14.1 (25 x 42.5) = 5.33
2810

5. ACERO REQUERIDO
As (+) = 30.90 cm2
Mu = 0.9 2810As [ 42.5-2810As ]
1.7 (210)(25)
[ ]
As (-) =
a. = 796.25
b. = - 1 074 825
c. = 2 561 200
a. = 796.25
b. = - 1 074 825
c. = 1 600 800
As (-) = 17.05 cm2
As (+) =

6. MOMENTO QUE RESISTE EL ACERO MÁXIMO
Mu (+) = 1819030.235 =
Mu (+) = 0.9 2810 (19.84) [ 42.5 - 2810 (19.84) ]
[ ](1.7)(210)(25)
Mu (+) = 18 190.30
100
Mu (-) = 0.9 2810 (5.33) [ 42.5 - 2810 (5.33) ]
[ (1.7)(210)(25)
]
Mu (-) = 550 261.27 =
100
Mu (-) = 5 502.61

7. MR = 25 612 - 18 190 = 7422
As adicional = Mr - 7 422
O fy d 0.9 (2810)(42.5)
As adicional = 6.91 cm2
As compresión = 1.55
As adicional = 9.19 cm2
As (nuevo) = As más + As ad.
As (nuevo) = 17.05 + 9.19
As (nuevo) = 26.24 cm2

8. ARMADO
* 1/3 As(-) = 8.75 cm2
- CAMA SUPERIOR
* As(Mín) = 5.33 cm2
* As adicional = 9.19 cm2
26.24 - 10.14 = 16.10 BASTÓN
- CAMA INFERIOR
* 1/2 As(mín) = 13.12 cm2
* Corrido = 2 No. 8 + 1 No.7
26.24 - 14.02 = 12.22
* Bastón = 2 No. 8 + 1 No.7
A’
A B
B’

9. CÁLCULO DE LONGITUDES DE DESARROLLO
Ld = Fm 0.06 (Av) (Fy)
* Cama Superior
√ f`c
Ld = 1.4 0.06 (5.07) (2810)
√ 210
[
][
]
Ld = 82.58
* Cama Inferior
Ld = 1 (0.06) (3.88) (2810)
No. 7
[ ]√ 210
Ld = 45.14
Ld = (0.06) (5.07) (2810)
No. 8
[ ]√ 210
Ld = 58.99

10. MOMENTO QUE RESISTE ACERO CORRIDO
[ ]Mu = 0.9 2810 (10.14)
* Cama Superior
* Cama Inferior
[ 42.5 - 2810 (10.14) ]
(1.7) (210) (25)
Mu = 1008 002.92 = Mu (máx) = 10 080.10
100
Mu = 0.9 2810 (14.02)
[ [ 42.5 - (2810) (14.02) ]
](1.7) (210) (25)
Mu = 1 350 394.34 = Mu (mín) = 13 503.94
100

11. 16 008 - 10 080 + 5 227x (x) - 18 295x = 0
∑ Mo y =
a. = 2 613.5
b. = -18 295
c. = 5 928
Y = 0.34
2

12. ∑ Mo x =
13 504 - 25 612 + 5 227x (x) = 0
2
x = 2.15 m
2x = 4.30 m

13. 1. 2Ld + y =
(2) (82.58) + 34 = 199.16 cm Bastón superior
2. 2Ld + 700 =
(2) (82.58) + 700 = 865.16 cm Corrido superior
3. 2Ld + 2x =
(2) (58.99) + 2(4.30) = 126.58 cm Bastón inferior
4. 2Ld + dist. =
(2) (58.99) + 700 = 817.98 cm Corrido superior
RESISTENCIA DEL CONCRETO
Vc = O Vc (b) (d)
Vc = (0.9)(0.53)(√ 210)(25)(42.5)
Vc = 7 344.41(0.55)(√ f’c)

14. (7/2) = d
18 295 = 7 344.41
d = 1.41
V - Vc = V As
18 295 - 7 344.41 = V As
V As = 10 950.59
ESPACIAMIENTO MÁXIMO
S = 2 (5.07) (2810) (1.41)
10 950.59
S = 3.67 42.5 = 21.25
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