This document provides an example of beam design calculations. It includes predimensioning of the beam, calculation of dead and live loads, determination of required steel reinforcement, and design of the steel layout. Key steps are calculating the bending moments, required steel, steel capacities, development lengths, and spacing of the reinforcement.

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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