platea ejemplo

DISEÑO DE LA PLATEA DE CIMENTACIÓN

5.00 m

5.00 m

4.00 m 5.00 m

1. Datos :

f´c= 210 kg/cm2 Columnas fy= 4200 kg/cm2 b = 0.40 m

1.80 kg/cm2 h = 0.40 mKo= 5000 tn/m3

C1 C2 C3 40.00 tn 70.00 tn 40.00 tn 25.00 tn 40.00 tn 25.00 tn 3.00 tn 4.00 tn 3.00 tn 2.00 tn 3.00 tn 2.00 tn 2.50 tn 5.00 tn 2.50 tn 2.00 tn 3.00 tn 2.00 tn

C4 C5 C6 65.00 tn/m 90.00 tn/m 65.00 tn 35.00 tn/m 50.00 tn/m 35.00 tn 4.50 tn/m 3.00 tn/m 4.50 tn 2.50 tn/m 2.00 tn/m 2.50 tn 3.00 tn/m 3.50 tn 3.00 tn 2.00 tn/m 2.00 tn 2.00 tn

C7 C8 C9 40.00 tn 70.0 tn 40.00 tn 25.00 tn 40.0 tn 25.00 tn

σt =

PD= PD= PD=PL= PL= PL=

MDX= MDX= MDX=MLX= MLX= MLX=MDY= MDY= MDY=MLY= MLY= MLY=




C1

C9C8C7

C6C5C4

C3C2

3.00 tn 4.0 tn/m 3.00 tn 2.00 tn 3.0 tn/m 2.00 tn 2.50 tn 5.0 tn/m 2.50 tn 2.00 tn 3.0 tn/m 2.00 tn

2. Cálculo del espesor de la platea:

520.00 tn 300.00 tn 820.00 tn 728.00 tn 1238.00 tn 510.00 tn

2.1. Verificación por punzonamiento :

Vc =

Verificación de la columna central C5

n

mm=40+dn=40+d

Pu = 211000.00 Kg

Vu= 211000.00 Kg

2*(40+d)+(40+d)160+4d

Igualando tenemos:

211000.00

d = 40.90 cm

Verificación de la columna lateral C8

n

mn=40+d/2m=40+d

2*(40+d/2)+(40+d)


QDS=QLS= QS=QDU= QU=QLU=

Vu ≤ ØVc , donde :

b0 = b0 =

= 1.1*√210*(160+4d)*d

b0 =

d/2

d/2

d/2d/2

d/2d/2

dbcf o **'*1.1

160+2dPu = 166000 KgVu = 166000 Kg

166000 = 1.1*(210)^(1/2)*(160+2d)*d

gualando : d = 42.50 cm

Verificación de la columna de esquina (C1):

n

mn = 40 + d/2m = 40 + d/2

160 + 2dPu = 98500 KgVu = 98500 Kg

ualando : d (cm) = 28.48 cm

Finalmente el espesor de la platea será :t = 50.77 cm

Asumiendo t (cm) = 60.00 cm d (cm) = 50.14

PRESIONES DEL SUELO EN LOS PUNTOS A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y.

E = 2170000 Tn/m2t = 0.60 m

Ko = 5000 tn/m3

b Ks λ 1.75/λ2.50 1260.02 0.30 5.845.00 1255.00 0.30 5.852.50 1260.02 0.30 5.842.00 1262.53 0.30 5.844.50 1255.56 0.30 5.842.50 1260.02 0.30 5.84

Direccion : X L = 4.00 m < 5.85 m

Direccion Y L = 5.00 m < 5.84 m

b0 =

b0=

Direccion X

Direccion Y

d/2

d/2

x Ko

▲ Se puede utilizar el Metodo Rigido Convencional

Y

ey 5.00

exX

5.00

4.00 5.00

3. Ubicación de la Resultante :Y

A B C D E

2.50

F G H I J

2.50

K L M N O X

2.50

P Q R S T

2.50

U V W X Y2.00 2.00 2.50 2.50

Momentos de inercia :

En ambos casos se cumple que L< 1.75 / λ

C1

C9C8C7

C6C5C4

C3C2

Y´

X ´

750.00 m4607.50 m4

X´= 4.28 m

-0.22 m

Y´= 5.000

0.000

Mx = 0.000

My = -180.000

Con : , encontramos la ecuación :

q = 13.756 ± -0.296 X ± 0.000 Y

PUNTO Q/A (Tn/m2) X (m) -0.296 X Y ± 0.000 Y q (Tn/m2) Verificación

A 13.756 -4.50 1.333 5.00 0.000 15.089 27.900B 13.756 -2.50 0.741 5.00 0.000 14.496 27.900C 13.756 -0.50 0.148 5.00 0.000 13.904 27.900D 13.756 2.00 -0.593 5.00 0.000 13.163 27.900E 13.756 4.50 -1.333 5.00 0.000 12.422 27.900F 13.756 -4.50 1.333 2.50 0.000 15.089 27.900G 13.756 -2.50 0.741 2.50 0.000 14.496 27.900H 13.756 -0.50 0.148 2.50 0.000 13.904 27.900I 13.756 2.00 -0.593 2.50 0.000 13.163 27.900J 13.756 4.50 -1.333 2.50 0.000 12.422 27.900K 13.756 -4.50 1.333 0.00 0.000 15.089 27.900L 13.756 -2.50 0.741 0.00 0.000 14.496 27.900M 13.756 -0.50 0.148 0.00 0.000 13.904 27.900N 13.756 2.00 -0.593 0.00 0.000 13.163 27.900O 13.756 4.50 -1.333 0.00 0.000 12.422 27.900P 13.756 -4.50 1.333 -2.50 0.000 15.089 27.900Q 13.756 -2.50 0.741 -2.50 0.000 14.496 27.900R 13.756 -0.50 0.148 -2.50 0.000 13.904 27.900S 13.756 2.00 -0.593 -2.50 0.000 13.163 27.900T 13.756 4.50 -1.333 -2.50 0.000 12.422 27.900U 13.756 -4.50 1.333 -5.00 0.000 15.089 27.900V 13.756 -2.50 0.741 -5.00 0.000 14.496 27.900

IX = IY =

eX (m) =

ey (m) =

W 13.756 -0.50 0.148 -5.00 0.000 13.904 27.900X 13.756 2.00 -0.593 -5.00 0.000 13.163 27.900

Y 13.756 4.50 -1.333 -5.00 0.000 12.422 27.900

27.90 Tn/m2 - Las presiones admisibles del suelo en todos los puntos son menores que el valor qadm(neta) =

OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!OK!

OK!OK!

OK!

4. Acero en la platea :

FRANJAS DE DISEÑO :

FRANJA IV FRANJA V FRANJA VI

FRANJA I2.50

FRANJA II5.00

FRANJA III2.50

2.00 4.50 2.50

FRANJA b b1 PUNTOSq

Q1 (Tn) Q2 (Tn) Q3 (Tn)Carga q´ (tn/m2)

FQ1´ Q2´ Q3´

tn/m2 promed. (tn) modific. (Tn) (Tn) (Tn)I 9.00 2.50 (A+E)/2 13.76 98.5 166 98.5 336.25 14.94 0.93 91.24 153.77 91.24II 9.00 5.00 (K+O)/2 13.46 150.5 211 150.5 558.83 12.42 1.09 164.27 230.30 164.27III 9.00 2.50 (U+Y)/2 13.76 98.5 166 98.5 336.25 14.94 0.93 91.24 153.77 91.24IV 10.00 2.00 (A+U)/2 15.09 98.5 150.5 98.5 324.64 16.23 0.93 92.02 140.60 92.02V 10.00 4.50 (C+W)/2 13.90 166 211 166 584.33 12.99 1.08 178.64 227.06 178.64VI 10.00 2.50 (E+Y)/2 12.42 98.5 150.5 98.5 329.03 13.16 0.95 93.26 142.50 93.26

DIAGRAMA DE MOMENTO FLECTOR

1.- Diseño de la Franja I :

91.24 153.77 91.243.80 4.80

0.20 0.20

14.94 tn/m2

107.6067.44

-+

69.0886.08

ACERO POSITIVO :

d (cm) = 49.19

Mu (tn.m) = 86.08

Iterando adecuadamente :Usar 1 Ø Nº6 @ 0.20 m

As (cm2) = 37.15 a (cm) = 3.49b (m) = 2.50 m

S (cm) = 19.18

ACERO NEGATIVO :

d (cm) = 51.19

Mu (tn.m) = 107.60 Usar 1 Ø Nº6 @ 0.18 m

Iterando adecuadamente :b (m) = 2.50 m

As (cm2) = 39.86 a (cm) = 3.75S (cm) = 17.88

2.- Diseño de la Franja II :

164.27 230.30 164.273.80 4.80

0.20 0.20

12.42 tn/m2

158.95799.62

-+

89.66114.81

ACERO POSITIVO : 143.06

d (cm) = 49.19

Mu (tn.m) = 143.06

Iterando adecuadamente :

As (cm2) = 59.13 a (cm) = 2.78 b (m) = 5.00 m

S (cm ) = 24.10

Usar 1 Ø Nº6 @ 0.25 m

ACERO NEGATIVO :

d (cm) = 51.19

Mu (tn.m) = 158.96 Usar 1 Ø Nº6 @ 0.25 m


As (cm2) = 63.42 a (cm) = 2.99S (m) = 22.47

3.- Diseño de la Franja III :

91.24 153.77 91.243.80 4.80

0.20 0.20

14.94 tn/m2

95.6459.94

-+

53.9569.08

ACERO POSITIVO : 86.08

d (cm) = 49.19Mu (tn.m) = 86.08


As (cm2) = 36.16 a (cm) = 3.40 b (m) = 2.50 m

S (cm ) = 19.70

Usar 1 Ø Nº6 @ 0.20 m

ACERO NEGATIVO :

d (cm) = 51.19

Mu (tn.m) = 95.64 Usar 1 Ø Nº6 @ 0.25 m


As (cm2) = 29.68 a (cm) = 2.55S (m) = 24.01

4.- Diseño de la Franja IV :

92.02 140.60 92.024.80 4.80

0.20 0.20

16.23 tn/m2

83.11 83.11

-+

74.8074.80 74.80

ACERO POSITIVO :

d (cm) = 49.19Mu (tn.m) = 74.80


As (cm2) = 30.43 a (cm) = 3.58 b (m) = 2.00 m

S (cm ) = 18.73

Usar 1 Ø Nº6 @ 0.20 m

ACERO NEGATIVO :

d (cm) = 51.19

Mu (tn.m) = 83.11 Usar 1 Ø Nº6 @ 0.18 m


As (cm2) = 32.69 a (cm) = 3.85S (m) = 17.44

5.- Diseño de la Franja V :

178.64 227.06 178.644.80 4.80

0.20 0.20

12.99 tn/m2

149.59 149.59

-+

134.63 134.63 134.63

ACERO POSITIVO :

d (cm) = 49.19Mu (tn.m) = 134.63


As (cm2) = 52.73 a (cm) = 2.76 b (m) = 4.50 m

S (cm ) = 24.32

Usar 1 Ø Nº6 @ 0.25 m

ACERO NEGATIVO :

d (cm) = 51.19

Mu (tn.m) = 149.59 Usar 1 Ø Nº6 @ 0.25 m


As (cm2) = 56.65 a (cm) = 2.96S (m) = 22.64

6.- Diseño de la Franja VI :

93.26 142.50 93.264.80 4.80

0.20 0.20

13.16 tn/m2

94.76 94.76

-+

75.81 75.81 75.81

ACERO POSITIVO :

d (cm) = 49.19Mu (tn.m) = 75.81


As (cm2) = 30.41 a (cm) = 2.86 b (m) = 2.50 m

S (cm ) = 23.43

Usar 1 Ø Nº6 @ 0.25 m

ACERO NEGATIVO :

d (cm) = 51.19

Mu (tn.m) = 94.76 Usar 1 Ø Nº6 @ 0.20 m


As (cm2) = 36.88 a (cm) = 3.47S (m) = 19.32

Resumen del refuerzo de la platea:

Franja Nº q`modif (tn/m2) b`=b1(m) L (m) Coeficiente M (tn.m) d (cm) As (cm2) As mins (cm) s (cm)

Ø Nº6 @ Ø Nº6 @

I14.94 2.50 3.80 1/8 (-) 107.60 51.19 39.86 23.04 17.88 18.0014.94 2.50 4.80 1/10 (+) 86.08 49.19 37.15 22.14 19.18 20.00

II12.42 5.00 3.80 1/8 (-) 158.957 51.19 63.42 46.07 22.47 23.0012.42 5.00 4.80 1/10 (+) 143.06 49.19 59.13 44.27 24.10 25.00

III14.94 2.50 3.80 1/8 (-) 95.64 51.19 29.68 23.04 24.01 25.0014.94 2.50 4.80 1/10 (+) 86.08 49.19 36.16 22.14 19.70 20.00

IV16.23 2.00 4.80 1/8 (-) 83.11 51.19 32.69 18.43 17.44 18.0016.23 2.00 4.80 1/10 (+) 74.80 49.19 30.43 17.71 18.73 20.00

V12.99 4.50 4.80 1/8 (-) 149.59 51.19 56.65 41.46 22.64 23.0012.99 4.50 4.80 1/10 (+) 134.63 49.19 52.73 39.84 24.32 25.00

VI13.16 2.50 4.80 1/8 (-) 94.76 51.19 36.88 23.04 19.32 20.0013.16 2.50 4.80 1/10 (+) 75.81 49.19 30.41 22.14 23.43 25.00

5.- Verficación de la fuerza cortante en la platea :

Con la fórmula :, que es la fuerza cortante actuante

b (cm) = 100.00d (cm) = 49.19

f´c (kg/cm2) = 210.00Ø = 0.85

También :

ØVc = 32113.03 kg

Es decir : ØVc = 32.11 Tn

Franja Nº q´ (tn/m2) L (m) Vu (Tn)

I14.94 3.80 28.39

14.94 4.80 35.87

II12.42 3.80 23.6012.42 4.80 29.80

III14.94 3.80 28.39

14.94 4.80 35.87

IV16.23 4.80 38.9616.23 4.80 38.96

V12.99 4.80 31.16

12.99 4.80 31.16

VI13.16 4.80 31.59

13.16 4.80 31.59

Luego : Vu max = 38.96 < Ø Vc = 32.11 (OK !!!)

2´**0.1L

qVu

dbcfØØVc **´*53.0*

platea ejemplo

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