Diseño y Calculo Estructural de Zapata
Diseño y Calculo Estructural de Zapata
Diseño y Calculo Estructural de Zapata
FACULTAD DE INGENIERIA
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
σn = 20.41 t/m2
𝑃1 + 𝑃2
𝐴𝑍 =
𝜎𝑛
Az = 15.34 m2
R
6.425 m Lv
Hz
C.G.
0.55 m 0.60 m
Xo Xo
Lz
𝑡1 𝑡2
𝑅𝑋0 =𝑃1 +𝑃2 𝐿+
2 2
Xo = 4.72 m
𝐿𝑍 = 2𝑋0
Lz = 9.45 m
2.5.- Cálculo de Lv
𝐿𝑉 = 𝐿𝑍 − (𝑡1 + 𝐿2 + 𝑡2)
Lv = 1.875 m
𝐴𝑍
𝑏=
𝐿𝑍
b= 1.62 m → Usar: b = 2.00 m
𝑃𝑈1 + 𝑃𝑈2
𝑊𝑁𝑈 =
𝐿𝑍
WNU = 49.07 t/m
𝑊𝑁𝑈
𝑊𝑛𝑢 =
𝑏𝑍
Wnu = 24.53 t/m2 → (2.45 kg/cm2)
Z-2 Z-1
𝑉𝑍 = 0 = −𝑃𝑈1 + 𝑊𝑁𝑈𝑋0 Xo
= 3.47 m
𝑋02 𝑡1
𝑀𝑀𝑎𝑥 = 𝑊𝑁𝑈 − 𝑃𝑈1 𝑋0 −
2 2
Mmax = -272.11 t-m = Mu
3.1.3.- Cálculo de la altura de la zapata
𝑀𝑈 = 𝜑𝑓′𝑐𝑏𝑑2𝜔(1 − 0.59𝜔)
ω= 0.067 d
=105.77 cm
𝜑
𝐻𝑧 = 𝑑 + 𝑟 + r= 7.50 cm ; φ= 1 → ( 2.54 cm )
2 ( 5.07 cm2 )
Hz = 115.00 cm d
= 106.23 cm
13.49 t
y1 y3
y2
Vd 3
-156.81 t Vd 1 -106.72 t
𝑡1
𝑦1 =+ 𝑑 y1 = 1.34 cm Vd1 = -91.19 t
2
𝑡2
𝑦2 =+ 𝑑 y2 = 1.36 cm Vd2 = 119.83 t
2
𝑡2
𝑦3 =+ 𝑑 y3 = 1.36 cm Vd3 = -39.88 t
Vdu =119.83 t
𝑉𝑛 = 𝑉𝑑𝑢 φ= 0.75
𝜑
Vn = 159.77 t
𝑉𝐶 = 𝑂. 53 𝑓′𝑐𝑏𝑑
Vc = 162.47 t
𝑉𝑈 = 𝑃𝑈1 − φ= 0.75
𝑊𝑛𝑢𝑚𝑛 Vu =
84.77 t
𝑉𝑑𝑢
𝑉𝑛 =
𝑓′𝑐𝑏0𝑑 c) 𝑉𝐶 = 1.06 𝑓
𝜑
′𝑐𝑏0𝑑
Vn = 113.02 t
𝐷𝑀𝑎𝑦𝑜𝑟 β= 1 Vc = 702.61 t
𝛽=
𝐷𝑀𝑒𝑛𝑜𝑟 Usar: Ecuación c
4
2+
𝛽
CONCRETO ARMADO Bach. TEJADA VILLANUEVA, Richard Eduard
UNIVERSIDAD PERUANA LOS ANDES
FACULTAD DE INGENIERIA
ESCUELA PROFESIONAL DE INGENIERIA CIVIL
𝑉𝑈 = 𝑃𝑈2 −
𝛼𝑠 𝑑
𝑉𝐶 = 0.27 2 + 𝑓′𝑐𝑏0 𝑑
𝑏𝑜
αs = 40 ; αs = 30
φ=
𝑊𝑛𝑢𝑚𝑛 Vu = 0.75
bo : Perimetro de
la sección
157.81 t
critica
Vc = 1085.01 t
Como Vn = 210.41
t
< Vc = 1085.01 t → Ok!
3.1.6.- Diseño por
flexión
𝐴𝑠𝑚𝑖𝑛 = 𝜌𝑡𝑒𝑚𝑝𝑏𝑑
𝑀𝑢 𝑀𝑢 𝐴𝑠𝑓𝑦
𝐴𝑠 = 𝐴𝑠 = 𝑎 𝑎= 0.85𝑓´𝑐𝑏
𝜑𝑓𝑦𝑑 𝜑𝑓𝑦 𝑑 − 2
As = 75.29 → a 8.858 → Iterar
As = cm2 → = cm → Iterar
As = 70.71 → a 8.319 → Iterar
As = cm2 → = cm → Iterar
As = 70.53 → a 8.297 → Converge
cm2 = cm
70.52 a 8.296
cm2 = cm
70.52 a 8.296
cm2 = cm
As = 70.52
< Asmin = 38.24 cm2 →
cm2 As
Usa 1 1 @ 0.14 m (As=70.98
r 4 cm2)
φ
3.1.6.2.- chequeo de cuantia
β1 = 0.85 →
Coeficiente
Cuantia Cuantia de de
Minima diseño reducción
𝐴𝑠 < Cuantia maxima
𝜌𝑚𝑖𝑛 = 𝜌=
𝑓′𝑐
0.0018 𝑏𝑑
6000
𝜌𝑚𝑎𝑥 = 0.750.85𝛽1
𝑓𝑦 6000 + 𝑓𝑦
ρmin = ρ 0.0033 ρmax = 0.0161 → Falla dúctil!
0.0018 = 4
3.1.6.2.-
Refuezo
inferior
𝑊𝑁𝑈𝐿2𝑉
𝑀𝑈 = 2.808 cm
→ 2.850 Iterar
2 → a cm → Iterar
Mu = 2.850 cm →
As =
24.19 cm2 Asmin = 38.24 cm2
3.2.- Diseño en sentido transversal
t1 d/2 d/2 t2 d/2
b1 b2
b1 = 108.12 cm Usar = b1 =
1.10 m b2 = 166.23 cm b2 =
1.70 m
𝑃𝑈1
𝑞𝑁𝑈 = P1U = 170.30 t
𝐵𝑍
Usar 8φ 3/4 @ 0.13 m (As=22.80 cm2)
qNU = 85.15 t/m
𝐿2𝑉
𝑀𝑈 = 𝑞𝑁𝑈
2
Mu = 22.38 t-m
Hz = 1.15 m
𝜑
𝑑 = 𝐻𝑧− 𝑟 −
2d= 106.55 cm
𝑀𝑢
𝐴𝑠 =
𝜑𝑓𝑦𝑑
As = 6.17 cm2
𝐴𝑠𝑚𝑖𝑛 = 0.0018𝑏𝑑
𝑃𝑈2
𝑞𝑁𝑈 = P1U = 293.40 t
𝐵𝑍
qNU = 146.70 t/m
𝐿2𝑉
𝑀𝑈 = 𝑞𝑁𝑈
2
Mu = 35.94 t-m
Hz = 1.15 m
𝜑
𝑑 = 𝐻𝑧− 𝑟 − 𝜑 −
2
d= 104.64 cm
𝑀𝑢
𝐴𝑠 =
𝜑𝑓𝑦𝑑
As = 10.10 cm2
𝐴𝑠𝑚𝑖𝑛 = 0.0018𝑏𝑑