DISEÑO DE ZAPATA DE MEDIANERADatos:

f'c= 210 [Kg/cm²] 44.244 [Tn] 62.378 [Tn]Fy= 4200 [Kg/cm²] 55.328 [Tn] 78.985 [Tn]

1.3 [Kg/cm²] L Columna(b)= 30 [cm] L Columna(b)= 30.0 [cm]r= 8 [cm] A Columna(a)= 20 [cm] A Columna(a)= 20.0 [cm]l= 3.8 [m]

1. Predimensionamiento1.1 Dimensionamiento en planta.

44244

B= 130.4 [cm]

135 [cm]270 [cm]

55328

1.52 [Kg/cm²]

Para la zapata 2, Aislada:

62378.000 [Tn]

B= 219.1 [cm]

220 [cm]220 [cm]

P1= P2=Pu1= Pu2=

σadm=

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

Adoptamos B1=.:. A1=

qu1=

Adoptamos B2=.:. A2=

𝐵=√(𝑃_1/(2×𝜎_𝑎𝑑𝑚 ))𝐵=√(44244/(2×1.3))

𝑞_𝑢1=𝑃_𝑢1/(𝐴_1∙𝐵_1 )𝑞_𝑢1=53328/(135∙270)

𝐵=√(𝑃_2/𝜎_𝑎𝑑𝑚 )𝐵=√(44244/1.3)

𝑞_𝑢2=𝑃_𝑢1/(𝐴_1∙𝐵_1 )

78985.000 [Tn]

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:

205.2 [Kg/cm] 1416521 [Kg-cm]

m= 117.5 [cm]

25 [cm]

vu= 7.50 [Kg/cm²]

vc= 7.68 [Kg/cm²]

vu= 7.50 [Kg/cm²] < vc= 7.68 [Kg/cm²] CUMPLE

a.2) Para la zapata 2:

360.8 [Kg/cm]n= 92.5 [cm] Mx= 1543547.5 [Kg-cm]

qu2=

q1y=M1x=

Asumiendo una altura util de d1=

q2y=

𝑞_1𝑦=𝑞_𝑢1∙𝐵_1𝑞_1𝑦=1.52∙135𝑀_1𝑥=𝑞_1𝑦 𝑚^2/2𝑀_1𝑥=205.2∙〖 117.5〗 ^2/2

𝑣_𝑢1=(𝑞_𝑢1 (𝑚−𝑑_1 ))/(∅𝑑_1 )𝑣_𝑢=(1.52 ∙(117.5−25))/(0.75∙25)

𝑣_𝑐=0.53√( 〖𝑓′〗_𝑐 )𝑣_𝑐=0.53√210

𝑚=(𝐴_1−"B" )/2

𝑞_2𝑦=𝑞_𝑢2∙𝐵_2𝑞_2𝑦=1.64∙220𝑀_2𝑥=𝑞_2𝑦 𝑛^2/2𝑀_2𝑥=360.8∙〖 92.5〗 ^2/2

𝑣_𝑢2=(𝑞_𝑢2 (𝑛−𝑑_2 ))/(∅𝑑_2 )

𝑛=(220−"35 " )/2

𝑞_𝑢2=𝑃_𝑢1/(𝐴_1∙𝐵_1 )𝑞_𝑢2=78985/(220∙220)

𝑚=(270−"35" )/2

𝑛=( _2−"B" )/2𝐴

25 [cm]

vu= 5.90 [Kg/cm²]

vc= 7.68 [Kg/cm²]

vu= 5.90 [Kg/cm²] < vc= 7.68 [Kg/cm²] CUMPLE

.:. El canto util sera: 25 [cm] 25 [cm] 32.5 [cm] 32.5 [cm] 30 [cm] 30 [cm]

3. CALCULO DEL REFUERZO DE ACERO POR FLEXIONPara la Zapata 1:

1416521.25 [Kg-cm]

As1y= 15.87 [cm²]

La armadura minima As1 es igual a=

7.29 [cm²]

Entonces el el area de acero necesaria As1y es= 15.87 [cm²]

Si usamos varillas de= 16.00 [mm]

Asumiendo una altura util de d2=

d1= d2=h1= h2=

Se adopta h1= Se adopta h2=

Para la armadura As1y, se adoptara el area de acero mayor entre las dos siguientes ecuaciones:

A s1y min=

𝑣_𝑢=(1.64 ∙(92.5−25))/(0.75∙25)𝑣_𝑐=0.53√( 〖𝑓′〗_𝑐 )𝑣_𝑐=0.53√210

𝐴_𝑠1𝑦=(0.85∙〖𝑓 ^′〗 _𝑐∙𝐵_1∙𝑑)/𝐹𝑦 [1−√(1−(2∙𝑀_1𝑥)/(0.85∙∅∙〖𝑓 ^′〗 _𝑐∙𝐵_1∙𝑑^2 ))]𝐴_𝑠1𝑦=(0.85∙210∙135∙25)/4200 [1−√(1−(2∙1416521.25)/(0.85∙0.9∙210∙135∙〖 25〗 ^2 ))]

𝐴_(𝑠1𝑦 𝑚𝑖𝑛)=0.0018∙𝐵_1∙ℎ𝐴_(𝑠1𝑦 𝑚𝑖𝑛)=0.0018∙135∙30

𝑁°𝑣𝑎𝑟𝑖𝑙𝑙𝑎𝑠=15.87/(((𝜋∙〖 1.6〗 ^2)/4) )

N varillas= 8

Entonces se colocaran 8 varillas de diametro 16 mm cada 17 cm.

14.58 [cm²]

Si usamos varillas de= 12.00 [mm]

N varillas= 13

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

Para la Zapata 2:

1543547.50 [Kg-cm]

16.95 [cm²]

11.88 [cm²]

Entonces el el area de acero necesaria As1y es= 16.95 [cm²]

Si usamos varillas de= 16.00 [mm]

Para la armadura As1x:

La armadura minima As1x es igual a=

A s1x min=

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

As2y=

La armadura minima As2y es igual a=

A s2y min=

𝑁°𝑣𝑎𝑟𝑖𝑙𝑙𝑎𝑠=15.87/(((𝜋∙〖 1.6〗 ^2)/4) )

𝐴_(𝑠1𝑥 𝑚𝑖𝑛)=0.0018∙𝐴_1∙ℎ_1𝐴_(𝑠1𝑥 𝑚𝑖𝑛)=0.0018∙270∙30

𝑁°𝑣𝑎𝑟𝑖𝑙𝑙𝑎𝑠=14.58/(((𝜋∙〖 1.2〗 ^2)/4) )

𝐴_𝑠2𝑦=(0.85∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑)/𝐹𝑦 [1−√(1−(2∙𝑀_1𝑥)/(0.85∙∅∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑^2 ))]𝐴_𝑠2𝑦=(0.85∙210∙220∙25)/4200 [1−√(1−(2∙1543547.50)/(0.85∙0.9∙210∙220∙〖 25〗 ^2 ))]

𝐴_(𝑠2𝑦 𝑚𝑖𝑛)=0.0018∙𝐵_2∙ℎ_2𝐴_(𝑠2𝑦 𝑚𝑖𝑛)=0.0018∙220∙30

N varillas= 9

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

11.88 [cm²]

Si usamos varillas de= 12.00 [mm]

N varillas= 11

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

4. Verificacion por Adherencia

30 [cm]

4.1 Adherencia Zapata 1:

ld= 31.80 [cm] ld= 42.40 [cm]

m-r= 110.0 [cm]

Para la armadura As2x:

A s1x min=

ld ≥

Donde: Ψt= 1; Ψe=1; Ψs=0.8; λ=1;

𝐹_𝑦/(3.5∙√(〖 ′〗𝑓 _𝑐 )) (𝜓_𝑡∙ _𝜓 𝑒∙ _𝜓 𝑠∙𝜆)/((𝑐_𝑏+𝑘_𝑡𝑟)/𝑑_𝑏 ) 𝑑_𝑏

(𝑐_𝑏+𝑘_𝑡𝑟)/𝑑_𝑏 =2.5

𝑙_𝑑=4200/(3.5∙√210) (1∙1∙0.8∙1)/2.5 𝑑_𝑏𝑙_𝑑=26.5 𝑑_𝑏𝑙_𝑑=26.5 ∙1.2

𝑁°𝑣𝑎𝑟𝑖𝑙𝑙𝑎𝑠=16.95/(((𝜋∙〖 1.6〗 ^2)/4) )

𝐴_(𝑠1𝑥 𝑚𝑖𝑛)=0.0018∙𝐴_2∙ℎ_2𝐴_(𝑠1𝑥 𝑚𝑖𝑛)=0.0018∙220∙30

𝑁°𝑣𝑎𝑟𝑖𝑙𝑙𝑎𝑠=11.88/(((𝜋∙〖 1.2〗 ^2)/4) )

𝑙_𝑑=26.5 𝑑_𝑏𝑙_𝑑=26.5 ∙1.6

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

4.2 Adherencia Zapata 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 1q1x=

q1x= 410.40 [Kg/cm]

Zapata de aislada: Zapata 2q1x=

q1x= 360.80 [Kg/cm]

Viga:B= 35 [cm]H= 70 [cm]D= 63 [cm]

16900.05 [Kg] 47223.00 [Kg-cm]- 5766.99 [Kg] 2169530.00 [Kg-cm]- 22621.19 [Kg] 21907.00 [Kg-cm] 22591.80 [Kg]

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

qu1·A1

qu2·A2

V1= M1(-)=V2= M1(-)=V3= M3(+)=V4=

𝑙_𝑑=4200/(3.5∙√210) (1∙1∙0.8∙1)/2.5 𝑑_𝑏𝑙_𝑑=26.5 𝑑_𝑏𝑙_𝑑=26.5 ∙1.2

𝑞_1𝑥=1.59∙270

𝑞_1𝑥=1.63∙220

0.20 [cm²]

9.60 [cm²]

0.09 [cm²]

7.35 [cm²]

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

10.22 [Kg/cm²] ≥ 7.68 [Kg/cm²]

As1=

As2=

As3=

Asmin=

Vu1=

𝐴_𝑠1=(0.85∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑)/𝐹𝑦 [1−√(1−(2∙𝑀_1𝑥)/(0.85∙∅∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑^2 ))]𝐴_𝑠1=(0.85∙210∙220∙25)/4200 [1−√(1−(2∙47223)/(0.85∙0.9∙210∙220∙〖 25〗 ^2 ))]

𝐴_𝑠2=(0.85∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑)/𝐹𝑦 [1−√(1−(2∙𝑀_1𝑥)/(0.85∙∅∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑^2 ))]𝐴_𝑠2=(0.85∙210∙220∙25)/4200 [1−√(1−(2∙2169530)/(0.85∙0.9∙210∙220∙〖 25〗 ^2 ))]

𝐴_𝑠3=(0.85∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑)/𝐹𝑦 [1−√(1−(2∙𝑀_1𝑥)/(0.85∙∅∙〖𝑓 ^′〗 _𝑐∙𝐵_2∙𝑑^2 ))]𝐴_𝑠3=(0.85∙210∙220∙25)/4200 [1−√(1−(2∙2169530)/(0.85∙0.9∙210∙220∙〖 25〗 ^2 ))]

𝐴_𝑠𝑚𝑖𝑛=14/𝐹_𝑦 ∙𝐵∙𝐷𝐴_𝑠𝑚𝑖𝑛=14/4200∙35∙63

𝑉_𝑢1=𝑉_1/(∅∙𝑏_𝑜∙𝑑)𝑉_𝑢1=16900.05/(0.75∙35∙63)

Asumiendo un diametro= 8 [mm]

1.01 [cm²]

47.51 [cm²]

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

3.49 [Kg/cm²] ≤ 7.68 [Kg/cm²]

Asumiendo un diametro= 8 [mm]

1.01 [cm²]

0.00 [cm²]

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

Vu2=

𝑠=(𝐴_𝑣∙𝐹_𝑦)/(𝑣_𝑢𝑖−𝑣_𝑐 )𝐵

𝐴_𝑣=(𝜋∙〖 (0.8)〗 ^2)/2=𝑠_1=(1∙4200)/(10.22−7.68)35=

𝑉_𝑢2=𝑉_2/(∅∙𝑏_𝑜∙𝑑)𝑉_𝑢1=5766.99/(0.75∙35∙63)

𝑠=(𝐴_𝑣∙𝐹_𝑦)/(𝑣_𝑢𝑖−𝑣_𝑐 )𝐵

𝐴_𝑣=(𝜋∙〖 (0.8)〗 ^2)/2=𝑠_2=(1∙4200)/(10.22−7.68)35=