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Zapata Piramidal Datos: Ejercicio 5 Cv ≔ 800 kN Cm ≔ 750 kN qadm ≔ 0.18 MPa

tonnef γc ≔ 2.4 ――― m3 tonnef γs ≔ 1.75 ――― m3

tonnef qadm = 18.355 ――― m2

Cs ≔ 0

Df ≔ 1.9 m b1 ≔ 0.8 m b2 ≔ 0.6 m

1. Determinar La carga neta y el área de zapata hc ≔ 50 cm

Se asumió el hc desde el desplante hasta el cuello de la columna debido a que vamos a dar prioridad al peso de la columna.

hs ≔ Df - hc = 1.4 m tonnef qn ≔ qadm - ((hc ⋅ γc)) - ((hs ⋅ γs)) - Cs = 14.705 ――― m2

tonnef qn = 14.705 ――― m2

Q ≔ Cv + Cm Q = ⎛⎝1.55 ⋅ 10 6 ⎞⎠ N Q = 158.056 tonnef Q A ≔ ― = 10.749 m 2 qn

A = 10.749 m 2

2 ‾‾‾‾‾‾‾‾‾‾‾ b1 - b2 2 ((b1 - b2)) B ≔ ――― + ―――― + A = 3.38 m 2 4

B = 3.38 m

A L ≔ ―= 3.18 m B

L = 3.18 m

2.Mayoración e cargas Pu ≔ 1.2 Cm + 1.6 Cv = 222.298 tonnef Pu tonnef qu ≔ ―― = 20.682 ――― A m2 tonnef qu = 20.682 ――― m2

Pu = 222.298 tonnef

3. Análisis de corte 3.1 Análisis unidireccional Análisis xx hc ≔ 50 cm

rec ≔ 7.5 cm

ϕvarilla ≔ 1.875 cm

ϕvarilla d ≔ hc - rec - ―――= 0.416 m 2 L b2 L1 ≔ ―- ―- d = 0.874 m 2 2

d = 41.563 cm

Axx ≔ L1 ⋅ B = 2.955 m 2

Axx = 2.955 m 2

L1 = 87.439 cm

Vux ≔ qu ⋅ Axx = 61.124 tonnef Análisis yy B b1 B1 ≔ ―- ―- d = 0.874 m 2 2

B1 = 0.874 m

Ayy ≔ B1 ⋅ L = 2.781 m 2 Vuy ≔ qu ⋅ Ayy = 57.507 tonnef Datos: kg f'c ≔ 250 ―― bw ≔ max ((B , L) = 3.38 m cm 2 bw ≔ 338 d ≔ 41.563

f'c ≔ 250

Para fines de calculo no considero las considero adimensionales de los valores de Vc

ϕ ≔ 0.75 Vc ≔ 0.53 ⋅ λ

2

λ≔1

tonnef ‾‾‾ f'c ⋅ bw ⋅ d ⋅ 1 ――― = 117.725 tonnef 1000

Vc = 117.725 tonnef Vu ≔ min ((Vux , Vuy)) = 57.507 tonnef

Vs

Vc ⋅ ϕ = 88.294 tonnef

Demanda Vs Capacidad if Vu ≤ Vc ⋅ ϕ = “Cumple” ‖ “Cumple” ‖ else ‖ “no cumple” ‖

3.2 Análisis bidireccional At ≔ B ⋅ L = 10.749 m 2 d ≔ 41.563 cm A1 ≔ (b1 + d) ⋅ (b2 + d) = 1.235 m 2 A2 ≔ ((At - A1)) = 9.514 m 2 Vbid ≔ qu ⋅ A2 = 196.764 tonnef Vbid = 196.764 tonnef

c ≔ 0.24 m

a ≔ 1.29 m

d ≔ 33.8

f'c ≔ 250

d ≔ 0.338 m

bw ≔ 338

λ≔1

tonnef ‾‾‾ f'c ⋅ bw ⋅ d ⋅ 1 ――― = 198.699 tonnef 1000 0.8 β ≔ ―― = 1.333 0.6

Vc1 ≔ 1.1 ⋅ λ ⋅

2

Vc2 ≔ 239.70 tonnef Vc3 ≔ 387.128 tonnef

Vc ≔ min ((Vc1 , Vc2 , Vc3)) = 198.699 tonnef if Vbid ≤ 0.75 ⋅ Vc = “No Cumple” ‖ “Cumple” ‖ else ‖ “No Cumple” ‖