Diseño Zapata Medianera conectada ACI-318-08

DISEÑO DE ZAPATA DE MEDIANERA Datos: P1= f'c= 210 [Kg/cm²] Pu1= Fy= 4200 [Kg/cm²] σadm= 1.3 [Kg/cm²] L Columna(b)= r= 8 [cm] A Columna(a)= l= 3.8 [m] 1. Predimensionamiento Predimensionamiento 1.1 Dimensionamiento en planta.

44.244 [Tn] 55.328 [Tn] 30 [cm] 20 [cm]

P2= Pu2= L Columna(b)= A Columna(a)=

62.378 [Tn] 78.985 [Tn] 30.0 [cm] 20.0 [cm]

Para la zapata 1, de la medianera: A1= 2B1

 2 × 

=

=

44244 2×1.3

44244

B= 130.4 130.4 [cm [cm]] Adoptamos B1= 135 [cm] .:. A1= 270 [cm]

 =  =

  ∙ 

53328 135 ∙ 270

55328

qu1= 1.52 [Kg/cm²] Para la zapata 2, Aislada:

=

=

 

44244 1.3

B= 219.1 219.1 [cm [cm]] Adoptamos B2= 220 [cm] .:. A2= 220 [cm]

 =

  ∙ 

62378.000 [Tn]

 =

78985 220 ∙ 220

78985.000 [Tn]

qu2= 1.64 [Kg/cm²]

1.2 Dimensionamiento en elevacion. a) Verificacion a corte por flexion Asumimos un ancho de nervio B= 35 [cm] a.1) Para la zapata 1:

 − B 2 270 − 35 = 2 =

 =  ∙   = 1.52 ∙ 135 q1y= 205.2 [Kg/cm]

  =  2 117.5  = 205.2 ∙ 2 M1x= 1416521 [Kg-cm]

m= 117.5 [cm]

 =

  −  ∅

Asumiendo una altura util de d1= 25 [cm]

 =

1.52 ∙ 117.5 − 25 0.75 ∙ 25

vu= 7.50 [Kg/cm²]

 = 0.53 ′  = 0.53 210 vc= 7.68 [Kg/cm²]

vu= 7.50 [Kg/cm²] < vc= 7.68 [Kg/cm²] a.2) Para la zapata 2:

 − B = 2 220 − 35 = 2

 =  ∙   = 1.64 ∙ 220 q2y= 360.8 [Kg/cm]

n= 92.5 [cm]

CUMPLE

  =  2 92.5  = 360.8 ∙ 2 Mx= 1543547.5 [Kg-cm]

 =

  −  ∅

Asumiendo una altura util de d2= 25 [cm]

 =

1.64 ∙ 92.5 − 25 0.75 ∙ 25

vu= 5.90 [Kg/cm²]

 = 0.53 ′  = 0.53 210 vc= 7.68 [Kg/cm²]

vu= 5.90 [Kg/cm²] < vc= 7.68 [Kg/cm²]

CUMPLE

.:. El canto util sera: d1= 25 [cm] h1= 32.5 [cm] Se adopta h1= 30 [cm]

d2= 25 [cm] h2= 32.5 [cm] Se adopta h2= 30 [cm]

3. CALCULO DEL REFUERZO DE ACERO POR FLEXION Para la Zapata 1: Para la armadura As1y, se adoptara el area de acero mayor entre las dos siguientes ecuaciones:

 =

 =

0.85 ∙    ∙  ∙  

1−

1−

2 ∙  0 . 8 5 ∙ ∅ ∙    ∙  ∙ 

1416521.25 [Kg-cm]

0.85 ∙ 210 ∙ 135 ∙ 25 2 ∙ 1416521.25 1− 1− 4200 0.85∙0.9∙210∙135∙25

As1y= 15.87 [cm²] La armadura minima As1 es igual a=

 

 =

 =

0.0018 ∙  ∙ ℎ

0.0018 ∙ 135 ∙ 30

A s1y min= 7.29 [cm²] Entonces el el area de acero necesaria As1y es= 15.87 [cm²] Si usamos varillas de= 16.00 [mm]

° =

15.87  ∙ 1.6 4

N varillas= 8 Entonces se colocaran 8 varillas de diametro 16 mm cada 17 cm. Para la armadura As1x:

La armadura minima As1x es igual a=

 

 =

 =

0.0018 ∙  ∙ ℎ

0.0018 ∙ 270 ∙ 30

A s1x min= 14.58 [cm²] Si usamos varillas de= 12.00 [mm]

° =

14.58  ∙ 1.2 4

N varillas= 13 Entonces se colocaran 13 varillas de diametro 12 mm cada 21 cm.

Para la Zapata 2: Para la armadura As2y, se adoptara el area de acero mayor entre las dos siguientes ecuaciones:

 =

 =

0.85 ∙   ∙  ∙  

1−

1−

2 ∙  0 . 8 5 ∙ ∅ ∙    ∙  ∙ 

1543547.50 [Kg-cm]

0.85 ∙ 210 ∙ 220 ∙ 25 2 ∙ 1543547.50 1− 1− 4200 0.85∙0.9∙210∙220∙25

As2y= 16.95 [cm²] La armadura minima As2y es igual a=

 

 =

 =

0.0018 ∙  ∙ ℎ

0.0018 ∙ 220 ∙ 30

A s2y min= 11.88 [cm²] Entonces el el area de acero necesaria As1y es= 16.95 [cm²] Si usamos varillas de= 16.00 [mm]

° =

16.95  ∙ 1.6 4

N varillas= 9 Entonces se colocaran 9varillas de diametro 16 mm cada 25 cm.

Para la armadura As2x:



 =

0.0018 ∙  ∙ ℎ



 =

0.0018 ∙ 220 ∙ 30

A s1x min= 11.88 [cm²] Si usamos varillas de= 12.00 [mm]

° =

11.88  ∙ 1.2 4

N varillas= 11 Entonces se colocaran 11 varillas de diametro 12 mm cada 21 cm.

4. Verificacion por Adherencia 

 ∙  ∙  ∙   3.5 ∙ ′  +  

ld ≥

30 [cm]

Donde: Ψt= 1; Ψe=1; Ψs=0.8; λ=1;

 +  = 2.5 

4.1 Adherencia Zapata 1:  =

4200

1 ∙ 1 ∙ 0.8 ∙ 1  2.5 3.5 ∙ 210  = 26.5 

 = 26.5 

 = 26.5 ∙ 1.2

 = 26.5 ∙ 1.6

ld= 31.80 [cm]

ld= 42.40 [cm]

m-r= 110.0 [cm] m-r= 110.00 [cm] > ld= 31.8 [cm] m-r= 110.00 [cm] > ld= 42.4 [cm]

CUMPLE CUMPLE

4.2 Adherencia Zapata 2:  =

4200

1 ∙ 1 ∙ 0.8 ∙ 1  2.5 3.5 ∙ 210  = 26.5 



= 26.

∙ 1.2

ld= 31.80 [cm] m-r= 85.0 [cm] m-r= 85.00 [cm] > ld= 31.8 [cm]

CUMPLE

5. ANALISIS DE ESFUERZOS EN LA VIGA

Zapata de medianera: Zapata 1 q1x= qu1·A1

 = 1.59 ∙ 270 q1x= 410.40 [Kg/cm]

Zapata de aislada: Zapata 2 q1x= qu2·A2

 = 1.63 ∙ 220 q1x= 360.80 [Kg/cm] Viga: B= 35 [cm] H= 70 [cm] D= 63 [cm]

V1= V2= V3= V4=

M1(-)= 47223.00 [Kg-cm] M1(-)= 2169530.00 [Kg-cm] M3(+)= 21907.00 [Kg-cm]

16900.05 [Kg] - 5766.99 [Kg] - 22621.19 [Kg] 22591.80 [Kg]

6. CALCULO DEL REFUERZO DE ACERO POR FLEXION EN LA VIGA

 =

 =

0.85 ∙    ∙  ∙  

1− 1−

2 ∙  0 . 8 5 ∙ ∅ ∙    ∙  ∙ 

0.85 ∙ 210 ∙ 220 ∙ 25 2 ∙ 47223 1− 1− 4200 0.85∙0.9∙210∙220∙25

As1= 0.20 [cm²]

∙



∙

∙

∙

 =

 =

.





1− 1−





0 . 8 5 ∙ ∅ ∙ 

 ∙

 ∙ 

0.85 ∙ 210 ∙ 220 ∙ 25 2 ∙ 2169530 1− 1− 4200 0.85∙0.9∙210∙220∙25 As2= 9.60 [cm²]

 =

 =

0.85 ∙    ∙  ∙  

1− 1−

2 ∙  0 . 8 5 ∙ ∅ ∙    ∙  ∙ 

0.85 ∙ 210 ∙ 220 ∙ 25 2 ∙ 2169530 1− 1− 4200 0.85∙0.9∙210∙220∙25 As3= 0.09 [cm²]

14 ∙∙  14  = ∙35∙63 4200  =

Asmin= 7.35 [cm²]

6. CALCULO DEL REFUERZO DE ACERO POR CORTE EN LA VIGA  =  =

 ∅ ∙  ∙ 

16900.05 0.75∙35∙63

Vu1= 10.22 [Kg/cm²]

=

≥ 7.68 [Kg/cm²]

 ∙   −  

Asumiendo un diametro= 8 [mm]

 ∙ (0.8)  = = 2  =

1 ∙ 4200 = 10.22 − 7.68 35

1.01 [cm²]

47.51 [cm²]

Entonces se usaran varillas de 8 [mm] cada 35 [cm]

 =  =

 ∅ ∙  ∙ 

5766.99 0.75∙35∙63

Vu2= 3.49 [Kg/cm²]

=

≤ 7.68 [Kg/cm²]

 ∙   −  

Asumiendo un diametro= 8 [mm]

 =

 ∙ (0.8)  = = 2

1.01 [cm²]

1 ∙ 4200 = 10.22 − 7.68 35

0.00 [cm²]

Entonces se usaran varillas de 8 [mm] cada 35 [cm]

Diseño Zapata Medianera conectada ACI-318-08

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