1er tanteo (c= 10 cm)
4200
C= (170)(0.85x10)(15)= 21675 kg 𝜀𝑦 = 2𝑥106 = 0.0021
0.003𝑥20
𝜀𝑠 = 10
= 0.006
εs>εy fs=fy
T= Asfs= (6.97)(4200)= 29274 kg
C≠T Se debe realizar un segundo tanteo
2do tanteo (C=T)
29274
𝐶= = 13.5059 𝑐𝑚
0.85𝑥170𝑥15
C= (0.85)(13.5059)(170)(15)= 29274.0383 kg ≈ T
13.5059𝑥0.85
Mn = TZ = (29274)(30 - ) = 710187.0204 kg . Cm = 7.1019 ton . m
2
Mr = 0.9 Mn = (0.9)(7.1019) = 6.3917 ton . m
Mr > Mu
Marco 4 ( 1era planta)
Datos:
Mmax = 15.41 ton.m
𝑘𝑔
F´c = 250 𝑐𝑚2
𝑘𝑔
Fy = 4200
𝑐𝑚2
𝑘𝑔
F*c = 200 𝑐𝑚2
𝑘𝑔
F”c = 170 𝑐𝑚2
B1 =0.85
Mmax = 15.41 ton . m
Mu = (1.1)(15.41) = 16.951 ton .m
𝑓"𝑐 6000∗𝐵1 170 6000𝑥0.85
Ρb = 𝑓𝑦
∗ = ∗
𝑓𝑦+6000 4200 4200+6000
= 0.0202
Pmax = 0.75 Pb = 0.75 (0.0202) = 0.0152
(𝑃𝑚𝑎𝑥)(𝑓𝑦) 0.0152𝑥4200
q= 𝑓"𝑐
= 170
= 0.3755
Calculo de la base y peralte efectivo, donde d = 2b
Mr = Fr b d2 f”c q (1-0.5 q)
1695100 = (0.9)(b)(2b2)(170)(0.3755)(1-0.5(0.3755))
1695100 = 186.6599 b3
b = 20.8632 cm ≈ 21 cm
d= 42 cm
As = Pbd
D= 42 cm
As = (0.0152)(21)(42)
AS As = 13.4064 cm2
R = 5 cm As = 2vs#6 y 1vs#10 = 13.64 cm2
B= 21 cm
Diseño propuesto
𝑘𝑔
F´c = 250 𝑐𝑚2
𝑘𝑔
Fy = 4200 𝑐𝑚2
D= 42 cm
As = 2vs#6 y 1vs#10
As = 13.64 cm2
AS
R = 5 cm
B= 21 cm
1er tanteo (c = 14 cm)
4200
𝜀𝑦 = 2𝑥106 = 0.0021
0.003𝑥20
𝜀𝑠 = 10
= 0.006
εs>εy
fs=fy
C = (0.85)(14)(170)(21) = 42483 kg
T = (4200)(13.62) = 57204 kg
C≠T Se debe realizar un segundo tanteo
2do tanteo donde C≠T
57204
𝐶 = 0.85𝑥170𝑥21 = 18.8512 𝑐𝑚
C = (0.85)(18.8512)(170)(21) = 57203.9664 ton . m
18.8512𝑥0.85
Mn = TZ = (57204)(42 - ) = 19.4426 ton . m
2
Mr = 0.9 Mn = (0.9)(19.4426) = 17.4983 ton . m
Mr > Mu
Marco 4 (2da planta)
Datos:
Mmax = 13.44 ton.m
𝑘𝑔
F´c = 250 𝑐𝑚2
𝑘𝑔
Fy = 4200 𝑐𝑚2
𝑘𝑔
F*c = 200
𝑐𝑚2
𝑘𝑔
F”c= 170 𝑐𝑚2
B1 =0.85
Mmax = 13.44 ton . m
Mu = (1.1)(13.44) = 14.784 ton .m
𝑓"𝑐 6000∗𝐵1 170 6000𝑥0.85
Ρb = 𝑓𝑦
∗ = ∗
𝑓𝑦+6000 4200 4200+6000
= 0.0202
Pmax = 0.75 Pb = 0.75 (0.0202) = 0.0152
(𝑃𝑚𝑎𝑥)(𝑓𝑦) 0.0152𝑥4200
q= 𝑓"𝑐
= 170
= 0.3755
Calculo de la base y peralte efectivo, donde d = 2b
Mr = Fr b d2 f”c q (1-0.5 q)
1478400 = (0.9)(b)(2b2)(170)(0.3755)(1-0.5(0.3755))
1478400 = 186.6599 b3
b = 19.9334 cm ≈ 20 cm
d= 40 cm
As = Pbd
D= 40 cm
As = (0.0152)(20)(40)
AS As = 12.16 cm2
R = 5 cm As = 2vs#9 = 12.82 cm2
B= 20 cm
1er tanteo (c = 13 cm)
4200
𝜀𝑦 = 2𝑥106 = 0.0021
0.003𝑥27
𝜀𝑠 = 13
= 0.00623
εs>εy
fs=fy
C = (11.05)(170)(20) = 37570 kg
T = (4200)(12.82) = 53844 kg
C≠T Se debe realizar un segundo tanteo
2do tanteo donde C≠T
53844
𝐶 = 0.85𝑥170𝑥20 = 18.63 𝑐𝑚
C = (0.85)(18.63)(170)(20) = 53840.7 ton . m
Mn = TZ = (53844)(32.08225) = 17.27436 ton . m
Mr = 0.9 Mn = (0.9)(17.27436) = 15.55 ton . m
Mr > Mu
Marco 4 (3era planta)
Datos:
Mmax = 5.07 ton.m
𝑘𝑔
F´c = 250 2
𝑐𝑚
𝑘𝑔
Fy = 4200 𝑐𝑚2
𝑘𝑔
F*c = 200 𝑐𝑚2
𝑘𝑔
F”c = 170 𝑐𝑚2
B1 =0.85
Mmax = 5.07 ton . m
Mu = (1.1)(5.07) = 5.577 ton .m
𝑓"𝑐 6000∗𝐵1 170 6000𝑥0.85
Ρb = 𝑓𝑦
∗ = ∗
𝑓𝑦+6000 4200 4200+6000
= 0.0202
Pmax = 0.75 Pb = 0.75 (0.0202) = 0.0152
(𝑃𝑚𝑎𝑥)(𝑓𝑦) 0.0152𝑥4200
q= = = 0.3755
𝑓"𝑐 170
Calculo de la base y peralte efectivo, donde d = 2b
Mr = Fr b d2 f”c q (1-0.5 q)
5577000 = (0.9)(b)(2b2)(170)(0.3755)(1-0.5(0.3755))
b = 14.4029 cm ≈ 15 cm
d= 30 cm
As = Pbd
D= 30 cm
As = (0.0152)(15)(30)
AS As = 6.8 cm2
R = 5 cm As = 2vs#5 y 1vs#6 = 6.81 cm2
B= 15 cm
1er tanteo (c = 10 cm)
4200
𝜀𝑦 = 2𝑥106 = 0.0021
0.003𝑥20
𝜀𝑠 = 13
= 0.006
εs>εy
fs=fy
C = (8.5)(170)(15) = 21675 kg
T = (4200)(6.8) = 28560 kg
C≠T Se debe realizar un segundo tanteo
2do tanteo donde C≠T
28560
𝐶= = 13.1765 𝑐𝑚
0.85𝑥170𝑥15
C = (11.2)(170)(15) = 28560 ton . m
Mn = TZ = (28560)(24.4) = 6.96864 ton . m
Mr = 0.9 Mn = (0.9)(6.96864) = 6.27 ton . m
Mr > Mu
DISEÑO DE
VIGA POR
CORTANTE
Marco 4 nivel 1
As = 2vs#6 y 1vs#10
As = 13.62 cm2
13.62
𝑃 = 21𝑥42 = 0.0154
P>0.015
Se considera la siguiente formula
𝑉𝑐𝑟 = 𝐹𝑟 (0.5)√𝑓 ∗ 𝑐 𝑏 𝑑
𝑉𝑐𝑟 = (0.8)(0.5)√200 (21) (42)
𝑉𝑐𝑟 = 4.9893 𝑡𝑜𝑛
Av = 0.71(2) = 1.42 cm2
Distinguir que formula de Smax se utilizara
1.5 √𝑓 ∗ 𝑐 𝑏 𝑑 = 1.5 √200 𝑥 21 𝑥 42 = 18.71 𝑡𝑜𝑛
Para determinar hasta que distancia de x de la viga el concreto soporta las fuerzas cortantes es
necesario usar triángulos semejantes.
2.053 → 5.80
𝑥 → 4.9893
𝑥 = 1.766 𝑚
Sección 2.393 m
La fuerza del cortante lo soporta el concreto
Smax = 0.5 (d)
Smax = 0.5(42) = 21 cm
Sección 2.68
Vu = 5.8 Vu < 18.71 ton Smax = 0.5 d
Vs = 5.8 -4.9893 = 0.8107
(0.8)(1.42)(4200)(42)
𝑆= = 247.1819
810.7
Smax = 0.5 (42) = 21 cm (RIGE)
Sección 4.42 m
Vu = 12.04 Vu < 18.71 ton Smax = 0.5 d
Vs = 12.04 -4.9893 = 7.0507
(0.8)(1.42)(4200)(42)
𝑆= = 28.4213 𝑐𝑚
7050.7
Smax = 0.5 (42) = 21 cm (RIGE)
“ARMADO FINAL”
Marco 4 nivel 2
As = 2vs#9
As = 12.82 cm2
12.82
𝑃 = 20𝑥40 = 0.016025
P>0.015
Se considera la siguiente formula
𝑉𝑐𝑟 = 𝐹𝑟 (0.5)√𝑓 ∗ 𝑐 𝑏 𝑑
𝑉𝑐𝑟 = (0.8)(0.5)√200 (20)(40)
𝑉𝑐𝑟 = 4.525 𝑡𝑜𝑛
Pendiente
−4.67 − 3.3
𝑚= = −2.9738
2.68
Vu = y = (-2.9738) (x-0) + 3.3
Nunca
Vu > 2.5 (0.8)(20)(40)√200
Vu > 22.627 ton
1.5 √𝑓 ∗ 𝑐 𝑏 𝑑 = (1.5) √200 (20) (40) = 18.576 𝑡𝑜𝑛
Smax = 0.5 (d)
Smax = 0.5(40) = 20 cm
Sección 1.10969 m
Y=0
Vcu = 0
(0.8)(1.42)(4200)(40)
𝑆= =0
0
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 2.1069 m
Y = -2.9738
Vs = Vu = 2.9738 ton = 2973.8 kg
(0.8)(1.42)(4200)(40)
𝑆= = 64.176 𝑐𝑚
2973.8
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 2.68 m
Y = -4.67
Vu = 4.67 ton
Vs = 4.67 -4.525 = 0.145 ton
(0.8)(1.42)(4200)(40)
𝑆= = 1316.193 𝑐𝑚
145
Smax = 0.5 (40) = 20 cm (RIGE)
Pendiente tramo (2.68 – 4.42) m
Puntos (2.68, -6.07) y (4.42, -11.24)
−11.24 − (−6.07)
𝑚= = −2.9713
4.42 − 2.68
Y = (-2.9713)(x-2.68)-6.07 = Vu
Sección 3.68
Y = -9.0413
Vu = 9.0413
Vs = 9.0413 – 4.525 = 4.5163 ton
(0.8)(1.42)(4200)(40)
𝑆= = 42.2576 𝑐𝑚
4516.3
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 4.42
Y = -11.24
Vu = 11.24 ton
Vs = 6.715 ton
(0.8)(1.42)(4200)(40)
𝑆= = 28.42 𝑐𝑚
6715
Smax = 0.5 (40) = 20 cm (RIGE)
“ARMADO FINAL”
Marco 4 nivel 3
As = 2vs#5 y 1vs#6
As = 6.81 cm2
6.81
𝑃 = 30𝑥15 = 0.0152
P>0.015
Se considera la siguiente formula
𝑉𝑐𝑟 = 𝐹𝑟 (0.5)√𝑓 ∗ 𝑐 𝑏 𝑑
𝑉𝑐𝑟 = (0.8)(0.5)√200 (15)(30)
𝑉𝑐𝑟 = 2.5456 𝑡𝑜𝑛
Distinguir que la fórmula de Smax utilizaremos
1.5 √𝑓 ∗ 𝑐 𝑏 𝑑 = (1.5) √200 (15) (30) = 7.6368 𝑡𝑜𝑛
Smax = 0.5 (d)
Smax = 0.5(30) = 15 cm (RIGE)
Calculo de la ecuación de la pendiente
Pendiente
𝑚 = −1.2591
Y + 4.15 = -1.2591 (x – 4.94)
Y = -1.2591x + 2.07
Para saber hasta que distancia x de la trabe el cortante la absorbe el concreto es necesario
sustituirlo Vcr en ecuación de la recta
-2.5456 = -1.2591x + 2.07
X = 3.6658
Se diseñara con Smax = 15 cm
Sección 4.94 m
Vu = 4.15 Vu < 7.6368
Vs = 4.15 – 2.5456
Vs = 1.6044
(0.8)(1.42)(4200)(30)
𝑆= = 89.2147 𝑐𝑚
1604.4
Smax = 0.5 (30) = 15 cm (RIGE)
“ARMADO FINAL”
Marco A nivel 1
𝑉𝑐𝑟 = 4.9893 𝑡𝑜𝑛
Distinguir que formula de Smax utilizamos
1.5 𝐹𝑟 √𝑓 ∗ 𝑐 𝑏 𝑑 = (1.5)(0.8)√200 (21) (42) = 14.9680 𝑡𝑜𝑛
Para determinar hasta que distancia x de la viga la fuerza cortante es absorbida por Vcr.
Se calcula por semejanza de triángulos
3 → 6.39
𝑥 → 4.9893
𝑥 = 2.3424 𝑚
Sección 3.2724 m
Es diseñado con Smax = 0.5 d
Smax = 0.5 (42) = 21 cm
Sección 3.93 m
Vu = 6.93 Vu < 14.968 se usa Smax = 0.5 d y Av = 1.42 cm 2
Vs = 6.93 – 4.9893 = 1.4007 ton
(0.8)(1.42)(4200)(42)
𝑆= = 143.0645 𝑐𝑚
1400.7
Smax = 0.5 (42) = 21 cm (RIGE)
Sección 5.28 m
Vu = 10.65 ton Vu < 14.968 se usa Smax = 0.5 d y Av = 1.42 cm2
Vs = 10.65 – 4.9893 = 5.6607 ton
(0.8)(1.42)(4200)(42)
𝑆= = 35.4003 𝑐𝑚
5660.7
Smax = 0.5 (42) = 21 cm (RIGE)
“ARMADO FINAL”
Marco A nivel 2
As = 2vs#8 y 1vs#5
As = 12.12 cm2
12.12
𝑃 = 20𝑥40 = 0.01515
P>0.015
𝑉𝑐𝑟 = 𝐹𝑟 (0.5)√𝑓 ∗ 𝑐 𝑏 𝑑
𝑉𝑐𝑟 = (0.8)(0.5)√200 (20)(40)
𝑉𝑐𝑟 = 4.525 𝑡𝑜𝑛
−6.38 − 4.62
𝑚= = −2.7990
3.93
Vu = Y = (-2.799)(x-0) + 4.62
NUNCA
𝑉𝑢 < (2.5)(0.8)√200 (20)(40)
𝑉𝑢 < 22627.417 𝑘𝑔
𝑉𝑢 < 22.627 𝑡𝑜𝑛
1.5 𝐹𝑟 √𝑓 ∗ 𝑐 𝑏 𝑑 = 16.97 𝑡𝑜𝑛
Vu > 16.97 ton
Smax = 0.25 d
𝐹𝑟 𝐴𝑣 𝑓𝑦 𝑑
𝑆=
𝐴𝑠
E#3 As = 0.71 cm2 = 1.42 cm2
Sección 1 m
Y = 1.821
Vu = Vs = 1.821 ton
(0.8)(1.42)(4200)(40)
𝑆= = 104.803 𝑐𝑚
1821
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 2 m
Y = -0.978
Vu = 0.978 ton
(0.8)(1.42)(4200)(40)
𝑆= = 195.141 𝑐𝑚
978
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 3 m
Y = -3.777
Vu = 3.777 ton
(0.8)(1.42)(4200)(40)
𝑆= = 50.53 𝑐𝑚
3777
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 3.93 m
Y = -6.38
Vu = 6.38 ton
Vs = 6.38 – 4.525 = 1.855
(0.8)(1.42)(4200)(40)
𝑆= = 102.88 𝑐𝑚
1855
Smax = 0.5 (40) = 20 cm (RIGE)
Tramo (3.93 – 5.28)
Puntos (3.93, -7.77) y (5.28, -11.55)
−11.55 − (−7.77)
𝑚= = −2.8
5.28 − 3.93
Y = Vu = (-2.8)(x-3.93)-7.77
Sección 4 m
Y = -7.966
Vu = 7.966 ton
Vs = 3.441 ton
(0.8)(1.42)(4200)(40)
𝑆= = 55.4629 𝑐𝑚
3441
Smax = 0.5 (40) = 20 cm (RIGE)
Sección 5.28 m
Y = -11.55
Vu = 11.55 ton
Vs = 11.55 – 4.525 = 7.025 ton
(0.8)(1.42)(4200)(40)
𝑆= = 27.1670 𝑐𝑚
7025
Smax = 0.5 (40) = 20 cm (RIGE)
“ARMADO FINAL”
Marco A nivel 3
As = 6.8175 cm2
6.8175
𝑃= 15𝑥30
= 0.01515
P>0.015
𝑉𝑐𝑟 = 𝐹𝑟 (0.5)√𝑓 ∗ 𝑐 𝑏 𝑑
𝑉𝑐𝑟 = (0.8)(0.5)√200 (15)(30)
𝑉𝑐𝑟 = 2.5456 𝑡𝑜𝑛
−3.29 − 0.89
𝑚= = −0.7917
5.28
Vu = Y = (-0.7917)(x-0) + 0.89
NUNCA
𝑉𝑢 < (2.5)(0.8)√200 (20)(40)
𝑉𝑢 < 12727.92 𝑘𝑔
𝑉𝑢 < 12.72792 𝑡𝑜𝑛
1.5 𝐹𝑟 √𝑓 ∗ 𝑐 𝑏 𝑑 = 16.97 𝑡𝑜𝑛
E#3 As = 0.71 cm2 = 1.42 cm2
Sección 1 m
Y = 0.0983
Vu = Vs = 0.0983 ton
(0.8)(1.42)(4200)(30)
𝑆= = 1456.11 𝑐𝑚
98.3
