Diseño Cimentacion Tipo t Avance

DISEÑO DE UNA ZAPATA AISLADA Diseñar la zapata mostrada en la fig: Si la Columna de 70 x 50 lleva 10 fierros de 1" y transmite las cargas PD = 180 tn y PL = 100 tn . La capacidad portante admisible del suelo es qa = 2.5 kg/cm2 ; ademas fy = 4200kg/cm2 , fy = 280 kg/cm2 en la columna y fc = 210 kg/cm2 en la zapata

DATOS:

Zapata Otros Suelof ' c = 210 kg/cm² S/C = 550 kg/m² Df = 1.4 m

Columna PD = 180 Tn 1700 kg/m³f ' c = 280 kg/cm² PL = 100 Tn qa = 2.50 kg/cm²

b = 50 cm db = 2.54 cmt = 70 cm Acero Lv = 150 cm

f y = 4200 kg/cm²

MD,ML

PD, PL

Ld = 58.89 cmht

Tomar Ld = 58.89 cm Df Lv

(Del problema se emplean varillas de Ø1") Øb ( 1") = 2.54 cmr.e. = 7.50 cm (recubrimiento) hchc = 68.93 cm

Tomar hc = 70.00 cm hc = Ld + r.e + Øb Tht = Df - hc

ht = 70.00 cm

Bqm = 2.16 kg/cm²

T

Azap = 129,629.63 cm² Donde:T = 370.00 cm P = Carga de servicioB = 350.00 cm Lv = Volados iguales sin

excentricidad

2.- DETERMINACIÓN DE LA REACCIÓN AMPLIFICADA ( qmu )

Donde:3.26 kg/cm2 Pu = Carga Ultima

3.- VERIFICACION POR CORTE ( Ø = 0.85 )Por Flexión:

Lv = 150.00 cmr.e = 7.50 cm

Øb ( 3/4") = 1.91 cm (Suponiendo varillas Ø3/4")d = 60.59 cm ( d = hc - Øb - r.e. )

Vdu = 101,975.73 kg

g 2 =

1.- DIMENSIONAMIENTO DE LA ZAPATA Cálculo del peralte de la zapata (hc )

Cálculo de la presión neta del suelo ( qm )

Cálculo del área de la zapata ( Az )

btt

Reemplazo los valores que tenemos:

= 1.4 x 180000 + 1.7 x 100000 350 x 370

=

Ld= 0 .08 .db .F y

√f'c

Azap=Pqm

T=√Az+( t1-t2 )2

S=√Az−( t1-t2)2

qm=qa−ght−gchc-s/c

Wnu=PuAzap

Lv=T−t2

Vdu=(WnuxB)(Lv-d )Vc=0.53√f'c bdØVc ³ Vdu

Ø = 0.85 (Coef. De reduccion por corte)

Vc = 162,875.03 kgØVc = 138,443.78 kgØVc > Vdu OK!

Vu = 374,938.21 kg

= 2m +2nbo = 482.36

Vc = 555,426.41 kg Vc= 448,939.43 KgØVc = 472,112.45 kg ØVc= 381,598.52 Kg

ØVc > Vdu OK!1.4

lado menor columna ( b )

m = t + d m = 130.59n = t + b n = 110.59bo = 2*m + 2*nVu = Øvc OK ! Vu = 448,939.43

Øvc = 381,598.52

4.- CALCULO DEL REFUERZO LONGITUDINAL ( Ø = 0.90 )Dirección Mayor:

Lv = 150.00 cm ree = 7.50

Mu = 12,831,081.08 kg-cm 1.91

B = 350.00

d = 60.59a = 3.89 (Valor Asumido)

As = 57.88 2.85As mín = 0.0018 * B * d a = 3.89 # Varilla ( n ) = 20As > As mín OK !! 17.53

As = 57.88 20 Ø 3/4" @ 17.53Aøb

2.85 Aøb # Varilla ( n ) = 15

23.79 n -1 As mín = 44.1 15 Ø 3/4" @ 23.79

As > As mín OK !!

Dirección Menor:

B T = 370 ree = 7.50

As mín = 0.0018 * B * d B = 350 1.91

As > As mín OK !! d = 60.59a = 5.01 (Valor Asumido)

Aøb 2.85# Varilla ( n ) = 21

Aøb 17.65As transv = 61.19 21 Ø 3/4" @ 17.65

n -1 2.85# Varilla ( n ) = 16

23.54

Por Punzonamiento:

bo = 2 x ( t + d ) + 2 x ( b + d ) (perimetro de los

planos de falla)Vc = 0.27 * 2 + 4 * f 'c^.5 * bo * d = 1.06 * f 'c^.5 * bo * d

b c

b c = lado mayor columna ( t )

Vu = 1.1 x f'c x bo x d

Øb ( 3/4") =

Aøb ( 3/4" ) =

Espaciam =

# Varilla ( n ) = As Aøb ( 3/4" ) =

Espaciam = B - 2*r.e - Øb Espaciam =

As tranv = As * T

Øb ( 3/4") =

Aøb ( 3/4" ) =# Varilla ( n ) = As

Espaciam =

Espaciam = B - 2*r.e - ØbAøb ( 3/4" ) =

Espaciam =

d/2

d/2

m = t+d

n =

b+

d

t

b

T

B

bc =

cm

kgkg

cm

cm

cm2

cm2

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cm

Lv=T−t2

Vdu=(WnuxB)(Lv-d )Vc=0.53√f'c bdØVc ³ Vdu

Vu=Pu-Wnu x mnVc=0. 27[2+

4

βc]√ f ' c bo d

β c=DmayorDmenor ,β c≤2→Vc=1.06√ f ' c bo d

Vu≤ØVc; Ø=0 . 85

Mu=(Wnu x B )Lv2

2

As=Mu

ØFy(d-a2)

a=As. Fy0 .85f'cb

Asmin = 46.62 16 Ø 3/4" @ 23.54

As transv > As mín OK !!

Longitud de desarrollo en Traccion ( Ld )

< Lv1

ØbLv1 = Lv - r.e.e

La Zapata es rectangular se debe compartir el Refuerzo adecuadamente de la siguiente manera:

1.00 Øb (3/4") = 1.91 8.500.80 r.e.e = 7.50 ktr = 01.00 fy = 4200

Lado menor Zapata 1.00 f'c = 210 2.5 q = ( C+kt r )/ ØbAøb

Longitud de desarrollo en tracción q= 10.41 Aøb

Lv1 = 142.50 q >= 2.5 ,PONER 2.5 !!

n -1 Ld = 50.04 q < 2.5 ,PONER q !!

Ld < Lv1 OK !!

Espaciamiento del RefuerzoAsc = 61.19

17.65 OK !!

3 x h 210 cm

5.- VERIFICACION DE LA CONEXIÓN COLUMNA - ZAPATA ( Ø = 0.70 )

Para la sección A colum = 70*50 = 3500 cm² ( COLUMNA )

Ø * 0.85 * f 'c * As1 Pu = 422000 # Varilla ( n ) = 6A colum = b *t A1 = 3500Pu < ( Ø * 0.85 * f 'c * A1) Ø * 0.85 * f 'c * A1 = 583100AøbAs mín = 0.005 * A1 Pu < Ø * 0.85 * f 'c * A1 OK !!

Aøb As mín = 17.50As col. > As mín OK !! Aøb ( 3/4" ) = 2.85

USAR As1 = 17.50As col > As min OK !!

Para la sección A zapata = 350*370 = 129500 cm² ( ZAPATA )

Pu < Ø x 0.85 x f 'c x A2/A1 x A1 Pu = 422000A1 = 3500 A2/A1 = ###A2 = 129500

Ø x 0.85 x f 'c x A2/A1 x A1 = 874650 OK !!

ld = Øb * fy * a * b * g * l 3.54 * f 'c^.5 * C + Kr

Asc = 2 * Astrv ( b + 1 ) b = C =

g =b = Lado mayor Zapata l =

a =

# Varilla ( n ) = As

Espaciam = B - 2*r.e.e - Øb

45 cm

# Varilla = As1

cm2 cm

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>

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DISEÑO DE ZAPATA COMBINADA

Diseñar la zapata combinada que soportara las columnas mostradas en la figura. La capacidad portante del suelo es qa = 2 kg/cm2Considere fy = 4200 kg/cm2 y f'c = 210 kg/cm2 para la cimentacion . F'c = 280 kg/cm2 en las columnas.

COL: SECCION REF. PD PLC1 50 50 9 Ø 3/4" 20 15 PL1 PL2C2 40 50 9 Ø 3/4" 40 24

b x t Acero tn tn S/C = 500 kg/m²

h t = 101.00

hc = 54.00

4.55 m 1.40

Df = 1.60 m Otros Columna Zapata1800 kg/m3 S/C = 500 kg/m² f ' c = 210 kg/cm² f ' c = 175 kg/cm²

qa = 2.00 kg/cm2 d(eje-eje)= 4.3 m f y = 4200 kg/cm²

1.- DIMENSIONAMIENTO DE LA ZAPATA ( Az = S*T )

Øb ( 3/4" ) = 1.91 cm raiz(f 'c) 0.08 x Øb x fy / f ´c = 49 cm

.004*øb*fy = 32 cm OK !!ld >= 0.004 * øb * fy

Tomar ld = 49 cm Longitud de desarrollo en compresión

hc = ld + r.e + Øb r.e = 7.50 cmhm = Df - hc h c = 59.00 cm

h t = 101.00 cm

Tomar ht = 101.00 cm

qm = 1.63 kg/cm²

Cálculo del área de la zapata ( Az )

Ps1 = P1D + P1L P1s = 35 Tn P2s = 64 TnPs2 = P2D + P2LRs = P1s + P2s

20 20 30 30

Rs * Xo = P1s* t1/2 + P2s * 455Xo = ??

455.00 170.00 cm

Rs = 99 TnP1s = 35 Tn

Lz / 2 P2s = 64 Tne = Lz / 2 - Xo Rs = 99 TnLz / 6 Xo = 301.21 cme < Lz/6

Lz = 625.00 cm Lz Lz Lz / 2 = 312.50 cm

e = 11.29 cmLz / 6= 104 cm

g m

g m =

Cálculo del peralte de la zapata (hc )

ld = 0.08 * Øb * Fy


qm = qa - gm*hm -γc*hc - s/c

e Mo = 0

q1,2 = Rs * 1 ± 6 * e

1 2

t1 t2

b1 b2

0Xo

+

X

X

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L = 602 cm

B = 101 cmB =(P1s +P2s) tomamos B = 85 cm

qn x L

Como S/C = 500 kg/cm², Verificamos las presiones del suelo - ADICIONAL ( q 1,2 )

Ps1 = P1D + 50%P1LPs2 = P2D + P2L P1s = 27.5 Tn P2s = 64 TnR1s = P1s + P2s

20 20 30 30

R1s * Xo = P1s* t1/2 + P2s * 455Xo = ??

455.00 170.00 cm

e = Xo1 - Xo P1s = 27.5 Tn R1s = 91.5 TnP2s = 64 Tn

q = Ps + Mc = Ps + Ps x e R1s = 91.5 TnXo1 = 324.26 cm e = 23.05 cm 1454767.9167 cm4

q = 44.56 kg/cm2 AUMENTAMOS EL ANCHO Bq < qm cumple " OK "q > qm no cumple "AUMENTAMOS EL ANCHO B"

Aumentamos el Ancho hc = 150 cmUn nuevo calculo

Ps = 91500 kg q = 1.41 OK !!e = 23.05 cmB = 265 cm

74531250

Ps1 = P1D + P1LPs2 = P2D + 50%P2L P1s = 35 Tn P2s = 52 TnR2s = P1s + P2s

20 20 30 30

R2s * Xo = P1s* t1/2 + P2s * 455Xo = ??

470.00 1.40 mLz / 2e2 = Lz / 2 - Xo P1s = 35 Tn R2s = 87 TnLz / 6 P2s = 52 Tne2 < Lz/6 R2s = 87 Tn

Xo = 288.97 cm B e = 12.24 cm 74531250 cm4

q > qm no cumple "AUMENTAMOS EL ANCHO B" q = 1.62 kg/cm2 OK !!q < qm cumple " OK "

2.- DISEÑO EN SENTIDO LONGITUDINAL P1s = 35 Tn P2s = 64 Tn

qmqn = P1s + P2s 0.2 m 4.50 m 140.00 m

Azqn = 6205.7293 kg/m² ( No amplificada )

Por unidad de Longitudqm = qm1*B Por Unidad de LONGITUD qm = 6205.73 kg/m

L = 2 x Xo

e Mo = 0

I = ( B x L^3 )/12 Az I Az I I =

I =

e Mo = 0

q1,2 = R2s * 1 ± 6 * e2 I = ( B x L^3 )/12I =

Cálculo de la presión neta por unidad de longitud ( qm )

+

0Xo

+

0Xo

kg/cm2

cm4

cm3

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V 55311.978 kg1241.15

+ 1.6297 m-

-8688.022 kg-33758.85

kg - m-27385.1

M

-+

124.12 kg - m

6081.62 kg - m

Encontramos el diagrama de fuerzas cortantes verificamos el corte por Flexion: d = -ree + hc 1.50 x Vu = 67050.27 kg ya que : F = ( 1.4 x D + 1.7 x L )/( D + L )

d = 141.00 cmVu = Vmax - qn x ( b2/2 + d ) = 44700.18 kg F = 1.5

Aporte del Concreto : Vc = 0.53 x f ' c x b x d = 261975.02 kg Vu < 0 x Vc Cumple !!

0.85 x Vc = 222678.77 kg OK !! Aumentamos el Peralte a h = 150 cm. d = h- ree

d = 141 cmVu = Vmax - qn x ( b2/2 + d ) = 44700.18 kg

Aporte del Concreto : Vc = 0.53 x f ' c x b x d = 261975.02 kg Vu < 0 x Vc Cumple !!

0.85 x Vc = 222678.77 kg Vu < O x Vc CUMPLE !!

VERIFICACION POR CORTE ( Ø = 0.85 )

COLUMNA EXTERIOR : COLUMNA INTERIOR :Vu = 1.50 x ( Pu - qnu x m x n ) = 31075.805 kg Vu = 1.50 x ( Pu - qnu x m x n ) = 75697.49 kgVu =31075.80549 kg Vu = 75697.49 kg

El aporte del Concreto : 1 El aporte del Concreto : 1.25

422 cm 493 cm 50

Vc = 1396870.757 kg Vc = 1412870 kgØVc = 1187340.143 kg ØVc = 1200939 kg

Vc = 1.10 x f 'c x bo x d = 948492.49 kg Vc = 1.10 x f 'c x bo x d = 1E+06 kg

Vc = 948492.49 kg Vc = 1106949 kgØVc = 806218.62 kg ØVc = 940906.79 kg

Vu < ØVc CUMPLE !! Vu < ØVc CUMPLE !!

b c = b c =

bo = ( b + ,5*d ) + 2x( t + d ) = bo = ( b + 0,5*d ) + 2x( t + d ) =

Vc = 0.27 x 2 + 4 x f 'c x bo x d Vc = 0.27 x 2 + 4 x f 'c x bo x d b c b c

-

+ +

-

+

kg

kg

-

+

R203

Elaborado por : MARCO ANTONIO PINEDO RUIZ

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4.- CALCULO DEL REFUERZO EN EL TRAMO CENTRAL

X x qn - P1 = 0

X =P1

X = 5.64 Mu =qn x X^2 _ P1 x X

Mu = -91699.11 kg - mqn 2

Calculo de As :Mu = 91699.11 kg - m

As = Mu / ( Ø * fy * ( d - a/2 )) B = 265 r .e .e = 8 db =1.91a = As * fy / ( 0.85 * f 'c * B ) d = 141

a = 2.77 (Valor Asumido)As = 26.06 2.85

As mín = 0.0018 * B * d a = 2.78 # Varilla ( n ) = 9As > As mín OK !! 31Aøb As = 26.06 9 Ø 3/4" @ 31

Aøb 2.85# Varilla ( n ) = 24

n -1 11As mín = 67.26 24 Ø 3/4" @ 11

As > As mín ASUMIR As mín !!

5.- CALCULO DEL REFUERZO POR DEBAJO DE LA COLUMNA INTERIOR

Mu =qn x ( L )^2

Mu = 6081.61 kg - m2

L = 1.91 m 1.4

Calculo de As :Mu = 6081.61 kg - m

As = Mu / ( Ø * fy * ( d - a/2 )) B = 265 r .e .e = 8 db =1.91a = As * fy / ( 0.85 * f 'c * B ) d = 141

a = 0.18 (Valor Asumido)As = 1.71 2.85

As mín = 0.0018 * B * d a = 0.18 # Varilla ( n ) = 1As > As mín OK !! #DIV/0!Aøb As = 1.71 1 Ø 3/4" @ #DIV/0!

Aøb 2.85# Varilla ( n ) = 25

n -1 10As mín = 71.55 25 Ø 3/4" @ 10

As > As mín ASUMIR As mín !!

Diseñar en sentido transversal a cada columna le corresponde una porcion de Zapata

d/2 d/2 d/2

1.205 1.91

Aøb ( 3/4" ) =

Espaciam =

# Varilla ( n ) = AsAøb ( 3/4" ) =

Espaciam = B - 2*r.e - ØbEspaciam =

Aøb ( 3/4" ) =

Espaciam =

# Varilla ( n ) = AsAøb ( 3/4" ) =

Espaciam = B - 2*r.e - ØbEspaciam =

cm

cm2

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14 Ø 5/8"@ 22cm

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ZAPATA EXTERIOR ZAPATA INTERIOR

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qn =Pu

qn = 1.68 kg/cm2 qn =Pu

qn = 1.91 kg/cm2Az Az

Mu =qnu x B x (L^2)

Mu = 1169724 Mu =qnu x B x (L^2)

Mu = 2E+062 2

B = 120.5 r .e .e = 8 db = 1.59 B = 191 r .e .e = 8 db = 1.59d = 141 d = 141a = 0.52 (Valor Asumido) a = 0.64 (Valor Asumido)

As = 2.20 As = 4.34a = 0.52 a = 0.64

As = 2.2 2.85 As = 4.34 2.85# Varilla ( n ) = 11 # Varilla ( n ) = 18

10.00 10.00As mín = 32.54 11 Ø 3/4" @ 10 As mín = 51.57 ### Ø 3/4" @ 10

6.- VERIFICACION DE LA CONEXIÓN ZAPATA - COLUMNA( Ø = 0.70 )

Para la sección A columna = 50*50 = 2500 cm² ( COLUMNA )

Ø * 0.85 * f 'c * As1 Pu = 53500A colum = b *t A1 = 2500Pu < ( Ø * 0.85 * f 'c * A1) Ø * 0.85 * f 'c * A1 = 312375 As mín = 12.50As mín = 0.005 * A1

Pu < Ø * 0.85 * f 'c * A1 CUMPLE !! As col > As min OK !!

AøbAs col. > As mín OK !!

Para la sección B columna = 40*50 = 2000 cm² ( COLUMNA )

Ø * 0.85 * f 'c * As1 Pu = 96800A colum = b *t A1 = 2000Pu < ( Ø * 0.85 * f 'c * A1) Ø * 0.85 * f 'c * A1 = 249900 As mín = 10.00As mín = 0.005 * A1

Pu < Ø * 0.85 * f 'c * A1 CUMPLE !! As col > As min OK !!

AøbAs col. > As mín OK !! Detalle de refuerzos en la Zapata Combinada verificamos primero el espaciamiento maximo

Espaciamiento del Refuerzo

45 cm450 OK !!

3 x h = 450 cm

d/2 d/2 d/2


< Lv1

Øb Ld = 71 Ld1 = 46Lv1 = Lv - r.e.e


Øb (3/4") = 1.911.00 Øb (5/8") = 1.59 6.50 2.5 q = ( C+kt r )/ Øb0.80 r.e.e = 7.50 ktr = 0 q= 8.41

Lado menor Zapata 1.00 fy = 42001.30 f'c = 175 q >= 2.5 ,PONER 2.5 !!

Aøb 1.00 q < 2.5 ,PONER q !!

Longitud de desarrollo en tracción n -1

Lv1 = 132.50 Ld < Lv1 OK !!

Calculo del refuerzo minimo : Ld = 71.26As mín = 71.55 Refuerzo montaje en forma de Garrido

2.85 Ø 3/8" para refuerzo principal Ø =< 3/4"

Aøb ( 3/4" ) = Aøb ( 3/4" ) =

Espaciam = Espaciam =

# Varilla = As1

# Varilla = As1


Asc = 2 * Astrv ( b + 1 ) b = C =b = Lado mayor Zapata g =

l =

# Varilla ( n ) = As a =a1 =


Aøb ( 3/4" ) =

cm

cm2 cm2

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11 Ø 3/4"@ 10cm 25 Ø 3/4"@ 10cm

24 Ø3/4"@ 11cm

18 Ø 3/4"@ 10cm

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# Varilla ( n ) = 25 Ø 1/2" para refuerzo principal Ø > 3/4"10.00 con espaciamiento maximo de 45 cm.

25 Ø 3/4" @ 10 0.2Espaciam =

cmcm

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DISEÑO DE ZAPATA CONECTADA

Diseñar la zapata conectada que soportara las columnas mostradas en la figura . La capacidad portante admisible del suelo es qa = 2.0 kg/cm2 . Considere Fy = 4200 kg/cm2 , f'c = 175 kg/cm2 para la cimentacion

DATOS:Zapata Acero

f ' c = 175 kg/cm² f y = 4200 kg/cm² S/C = 500 kg/m²Columna Suelo

f ' c = 210 kg/cm² Df = 1.60 m Lc = 550 cmb1 = 50 cm 1800 kg/m³ db = 1.91 cmt1 = 50 cm 2400 kg/m³b2 = 40 cm qa = 2.00 kg/cm²t2 = 50 cm db = 1.91 cm

COL SECCION REFUERZO PD (tn) PL (tn)1 50 50 9 Ø 3/4" 50 302 40 50 9 Ø 3/4" 40 24

1.- DIMENSIONAMIENTO DE LA ZAPATA Y VIGA DE CIMENTACION

48.51 cm hc = Ld + r.ee + ØbLd = 49 hm = Df - hc

ld >= 0.004 * øb * fy 32.09 cmLd = 32

Cálculo del peralte de la zapata (hc )hc = 59

ht = 101


1.63

57

9249.49

9249.49

Az =Ps

Az =P1 + P' 2

= 54868.437 cm2qn qn

A = 2 x( b )^2 = 54868.44 b = 166 cm

nuevo e =b - t

e = 58P'1 = 9431

2 2 P'2 = 54980A = 2 x( b )^2 = 54980 b = 166

b= 165 tomamos

t=2b t= 330

g m =g c =

ZAPATA EXTERIOR:

qn = qa - gm*hm - gc*hc - s/c

CALCULO DE P'2 : Tanteo e =

Tomamos b = 165 cm -----> A = 165 x 330 e = 57 cm

X

Ld0.08xdbxfy f 'c=

Ld = 0.08x1,91x4200/(175^(1/2)) =

Ld >= 0.004x1.91x4200 =

cm

cm

cm

qn = 2.4-(1800xE^-06)x(101)-(2400xE^06)x(59)-(500xE^04) =

qn =

cmhc = Ld + 10 =

ht = Df - hc =

kg/cm2

( PD + PL ) x e Lc - e

P'2 = =P1xe L =

( 50 + 30 ) x 1000 x 57 550 - 57=

X

==

kg

cm

P'2 =

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49 hc = Ld + r.ee + ØbLd = 49 cm hm = Df - hc

ld >= 0.004 * øb * fy 32 cmLd = 32

Cálculo del peralte de la zapata (hc )hc = 60

ht = 100


2.00

71

9486

9486.00

Az =Ps

Az =P2 + P' 2

= 36743 cm2qn qn

A = ( b )^2 = 36743.00 b = 192 usamos b = 155 cm

nuevo e =b - t

e = 71P'1 = 9486

2 2 P'2 = 44743A = 2 x( b )^2 = 44743 b = 150

b = 150 A2 = ( 150 x 150 )

VIGAS DE CIMENTACION

h =Lc

=550

79 SECCION

7 7 50 80

b = P1

b = 4731 x Lc

DISEÑO DE LA VIGA DE CONEXIÓN

9.6 kg/m Mu = P' 2 x L + Wv x ( L^2 )/2= 7250000Mu = 7250000 kg-cm

1.91 ree = 7.50

As = Mu / ( Ø * fy * ( d - a/2 )) B = 50a = As x fy / ( 0.85 x f 'c x b ) d = 79.00As mín = 0.8 x (f ' c ^0.5) x B * d/ fy a = 5.00 (Valor Asumido)As > As mín OK !! As = 25.07 2.54

a = 14.16 # Varilla ( n ) = 11 Aøb 3

As = 26.67 11 Ø 1" @ 3 n -1

ZAPATA EXTERIOR:

qn = qa - gm*hm - gc*hc - s/c

CALCULO DE P'2 : Tanteo e =

Wv = b x h x g c =

Øb ( 3/4") =

Aøb ( 1" ) =# Varilla ( n ) = As

Espaciam =Espaciam = B - 2*r.e - Øb

Ld0.08xdbxfy f 'c=

Ld = 0.08x1.91x4200/(175^(1/2)) =

Ld >= 0.004x1.91x4200 =

cm

cm

cm

qn = 2.4-(1700xE^-06)x(120)-(2400xE-06)x(60)-(500xE-04) =

qn =

cmhc = Ld + 10 =

ht = Df - hc =

kg/cm2

( PD + PL ) x e Lc - e

P'2 = =P1xe L =

( 40 + 24 ) x 1000 x 50 550 - 50= =

kg

cm

P'2 =

cm

cm

h =x

cm

cm2

cm2

cm

cm2

cm2

cm2

cm2

cm

de 66

As mín = 9.95 11 Ø 1" @ 3

As > Asmin OK !!

cm2 cm

de 66

VERIFICACION POR CORTE APORTE DEL CONCRETO

Vu = P' 2 + Wv x L Vu = 21982 KG Vc = 0.53 x (f ' c ^0.5) x b x d = 23540 L Vu < Ø Vc OK !!

DISEÑO DE LA ZAPATA EXTERIOR

ZAPATA EXTERIOR ZAPATA INTERIOR

qn =Pu

qn = #DIV/0! kg/cm2 qn =Pu

#DIV/0!Az Az

Mu =qnu x B x (L^2)

Mu = #DIV/0! Mu =qnu x B x (L^2)

2 2L = 100 r .e .e =

B = 150 r .e .e = 7.5 B = 250d = 41 db = 1.59 d = 51a = 5.00 (Valor Asumido) a = 5 ( Valor Asumido )

As = #DIV/0! As = #DIV/0!a = #DIV/0! a = #DIV/0!

As = #DIV/0! 1.98 As = #DIV/0!# Varilla ( n ) = #DIV/0! #Varilla n = 14

#DIV/0! Espaciam. = 17.88As mín = 13.5 Asmin = 27

### Ø 5/8" @ #DIV/0! 14 Ø 5/8" @

6.- VERIFICACION DE LA CONEXIÓN ZAPATA - COLUMNA( Ø = 0.70 )

Columna ExteriorPara la sección A columna = 50 x 50 = 2500 cm² ( COLUMNA )

Ø * 0.85 * f 'c * As1 Pu = 121000 # Varilla ( n ) =A colum = b *t A1 = 2500Pu < ( Ø * 0.85 * f 'c * A1) Ø * 0.85 * f 'c * A1 = 312375As mín = 0.005 * A1

Pu < Ø * 0.85 * f 'c * A1 OK !!

AøbAs col. > As mín OK !! As mín = 12.50

Aøb ( 3/4" ) = 2.85USAR : As1 = 12.50

Columna Interior As col > As min OK !!

Para la sección B columna = 50 x 50 = 2500 cm² ( COLUMNA )Pu = 96800

Ø * 0.85 * f 'c * As1 A1 = 2000A colum = b *t Ø * 0.85 * f 'c * A1 = 249900 # Varilla ( n ) =Pu < ( Ø * 0.85 * f 'c * A1) Aøb ( 3/4" ) =As mín = 0.005 * A1 Pu < Ø * 0.85 * f 'c * A1 USAR : As1 =

OK !!

Aøb As mín = 10.00As col. > As mín OK !! As col > As min OK !!


< Lv1

ØbLv1 = Lv - r.e.e

Aøb ( 5/8" ) =

Espaciam =

# Varilla = As1

# Varilla = As1


cm2

cm2

cm

cm2

cm2

cm2

cm

cm2

cm2

cm

cm

cm

cm2

cm2

cm2

cm

cm2

cmcm2

kg

kgcm2

cm2cm2

cm2

kg

kgcm2

cm2

de 66


1.00 Øb (5/8") = 1.59 8.500.80 r.e.e = 7.50 ktr = 01.00 fy = 4200

Lado menor Zapata 1.00 f'c = 175 2.5 q = ( C+kt r )/ ØbAøb

Longitud de desarrollo en tracción q= Aøb

Lv1 = 38.13 q >= 2.5 ,PONER 2.5 !!

n -1 Ld = 46 q < 2.5 ,PONER q !!

Ld < Lv1 AUMENTAR EL T !!

Espaciamiento del Refuerzo

45 cmAs1 = OK !!

3 x h 0 cm

d/2d/2

Asc = 2 * Astrv ( b + 1 ) b = C =

g =b = Lado mayor Zapata l =

a =

# Varilla ( n ) = As


cm

cm

>

12 Ø 5/8"@ 12cm

14 Ø 5/8"@ 22cm

6 Ø 1"@ 7cm

14 Ø 5/8"@ 18cm

de 66

33

cm2

de 66

kg

#DIV/0!

8

( Valor Asumido )

18

4

42.8510.00

OK !!

cm

kg

cm

cm2

cm2

de 66

8.50

q = ( C+kt r )/ Øb

10.09

q >= 2.5 ,PONER 2.5 !!

q < 2.5 ,PONER q !!

DISEÑO DE LOSA DE CIMENTACION

S/C= 450

1700 KS

0.9

Y'

A B C D0.2

1 2 3

E F 4.5

Y9.8

4 5 6X

G H4.5

7 8 9

I L 0.6

O J K X'

0.2 5.4 5.4 0.611.6

DATOS

COLUMNA Pu F'c = 210

1 45 25 70 Fy = 4200

2 60 30 90 2400

3 60 30 90 1.4

4 60 30 90 1

5 100 40 140 S/C= 450

6 70 30 100 17007 60 30 908 70 30 1009 60 30 90

585 275 860

1.- CALCULO DEL ESPESOR DE LA PLACAMETODO DE DISEÑO ASUMIENDO t= 50 cm.

E=15000*√F'c

gm1=

PD (ton) PL (ton) kg/cm2

kg/cm2

gc= kg/m3

qa = kg/cm2

Ks = kg/cm3

Kg/m2

gm1= kg/m3

l= 4 3Ks

Et3

0.003242 0.003240

1.75 5.401 Lc= 5.4Lc= 4.5

Se debe cumplir la siguiente condición:METODO RIGIDO

AREA DE LOSA

A= B * L = 113.68 B= 11.6 m.L= 9.8 m.

PRESIONES DEL SUELO

Df= 90 cm.

1.167 ht= 40 cm.

2.-PRESION POR DEBAJO DE PUNTOS PERIMETRALES

X'= 578.84 cm. X = 579 cm.

ex = -1 cm.

Y'= 494.30 cm. Y '= 494 cm.

ey = 4 cm.

Entonces :q = ( Pu/a ) + - ( MyX / Iy ) + - ( MxY / Ix )

Donde :Pu= 1286500 kg.

Ix= 909.8189333 Ix= 909.8189

Iy= 1274.731733 Iy= 1274.7317

ENTONCES:q= 1286500 + - 1286500x + - 5146000y

q= 1.132 + - 0.00001x + - 0.00006y

Pu/A = 1.1317 Pu/A = 1.132

3.-PROCESAMOS LA TABLA

PUNTO Pu/A X(cm.) 0.00001X Y(cm.) 0.00006y

A 1.132 -580 0.006 490 0.029 1.167B 1.132 -290 0.003 490 0.029 1.164C 1.132 250 -0.003 490 0.029 1.159D 1.132 580 -0.006 490 0.029 1.156E 1.132 -580 0.006 245 0.015 1.152F 1.132 580 -0.006 245 0.015 1.141G 1.132 -580 0.006 -205 -0.012 1.126H 1.132 580 -0.006 -205 -0.012 1.114I 1.132 -580 0.006 -490 -0.029 1.108J 1.132 -290 0.003 -490 -0.029 1.105

l= m-1 l= m-1

l

1.75/l <Lc

m2

qn=qa - gt * ht - gc * hc - S/C

qn= kg/cm2

X'= SPiXi ex= X' - B/2

Pi

Y'= SPiYi

Pi

ey= Y' - L/2

BL3/12 = m4

LB3/12 = m4

(980)(1160) 1.275*1011 9.098*1010

q(kg/cm2)

K 1.132 250 -0.003 -490 -0.029 1.100L 1.132 580 -0.006 -490 -0.029 1.097

4.-COMPARAMOS LAS PRESIONES CALCULADAS CON LA PRESION NETA DEL SUELO

q<=qn SI CUMPLE

5.-CALCULAMOS EL PERALTE DE LA LOSA A PARTIR DEL CORTE POR PUNZONAMIENTO

COLUMNA 5

n= 40 + d

d/2

m = 40 + d

Vu= 1.4*100000 +1.7*40000 -1.167*(40+d)*(40+d)Vu=

Igualamos :Vc=1.1√210*4(40+d)

COLUMNA 2

Vu= 1.4*60000 +1.7*30000 -1.167*(40+d)*(40+d/2) m = 40 +d

COLUMNA 1

Vc=1.1√210*(120+2d)d

Igualamos n=40 +d/2

m = 40 +d/2

Vu= 1.4*45000 +1.7*25000 -1.167*(40+d/2)*(40+d/2)Vu=

Vc=1.1√210*2(40+d/2)d

Igualamos

Si d = 55

ht = 65

VERIFICAMOS LA CONDICION DE DISEÑO

Vu = Pu - (qnu * m * n)

Vc=1.1√F'c*bo*dVu = fVc

fVc=

Vu = Pu - (qnu * m * n)

Vc=1.1√F'c*bo*d Vu = Pu - (qnu * m * n)

fVc=

Vu=fVc

Vc=1.1√F'c*bo*d

fVc= 1083.95d +13.55d2

Vu=fVc

13.83d2 + 1130.63d - 103633 =0

n = 40 +d/2

E=15000*√F'c

0.002663 0.002660

1.75 6.579 m. Lc= 5.4Lc= 4.5

Se debe cumplir la siguiente condición:

METODO RIGIDO

6.-CALCULAMOS EL FACTOR DE AMPLIFICACION DE CARGA F PARA CADA FRANJA

FRANJA q q prom. b L P CP F

A 1.167 1.1566 245 1160 P1 375500.0 352098 0.938B 1.164 P2 1C 1.159 P3D 1.156E 1.152F 1.141E 1.152 1.1332 450 1160 P4 492000.00 541765.2 1.101F 1.141 P5H 1.114 P6G 1.126G 1.126 1.1084 285 1160 P7 419000.00 392713 0.937H 1.114 P8 1I 1.108 P9J 1.105K 1.100L 1.097A 1.167 1.1372 290 980 P1 375500.00 349351 0.930B 1.164 P4 1J 1.105 P7I 1.108G 1.126E 1.152B 1.164 1.1322 540 980 P2 492000.00 545580 1.109C 1.159 P5J 1.105 P8K 1.100C 1.159 1.1277 330 980 P3 419000.00 391849 0.935D 1.156 P6F 1.141 P9H 1.114K 1.100L 1.097

Efectuamos el diseño de la losa considerando por cuestiones academicas una sola franja central EFGH, en primer lugar determinamos el diagrama de fuerzas cortantes y momentos flectores a partir del modelo estructural.

1.2 P4 1.2 P5 1.2 P6

q1 = b x (qe + qg)2

q1 = 450 x ( 1.153 + 1.126 ) = 5132

l= 4 3Ks

Et3

l= m-1 l= m-1

l

1.75/l <Lc

SP

q2 = b x ( qf + qh )2

q2 = 450 x ( 1.141 + 1.114 ) = 5070.2 5.4 5.4 0.6 2

q = 510149400

123600 1.2 x P4 = 1.2 x(1.4x60000+1.7x30000) = 1620001.2 x P5 = 1.2 x(1.4x100000+1.7x40000) = 249600

10200 1.2 x P6 = 1.2 x(1.4x70000+1.7x30000) = 178800

Verificacion por FlexionVu = 151800 - 510*(20 + 56 ) = 113040

Vc = 0.53 x( 210 ^.5 ) x b x d =126000 29400

151800 ØVc =164515

Vu < ØVc OK

Refuerzo superior :22489412 20614568

Mu = 22489412 kg -cma = 5 cm

( - ) ( - )AS = 111.21 cm2a = 5.81 cm

AS = 112.06 cm2102000 7512000 918000 ( 23 Ø 1" @ 20 cm )

Asmin = 0.0018 x b x h = 52.65 cm2

Refuerzo Inferior :

Mu = 7512000 kg -cma = 1.94 cm

AS = 36.11 cm2a = 5.81 cm

AS = 37.43 cm2

Asmin = 0.0018 x b x h = 52.65 cm2( 19 Ø 3/4" @ 25 cm )

Diseñar la losa de cimentación mostrada considerando un suelo de capacidad portante admisibleqa = 1,2 kg/cm². Todas las columnas son cuadradas de 50 cm x 50 cm. La profundidad de cimentaciónes de 1,10 m., el peso específico del suelo es g = 1650 kg/m³ y la sobrecarga s/c = 500 kg/m² ,fy = 4200 kg/cm² , f'c = 210 kg/cm² y ks = 8 kg/cm³.

1 2 3 4

20.00m

5.05m

5 6 7 8

4.80m

9 10 11 12

4.80m

13 14 15 16

4.80m

17 18 19 200.55m

5.75m 5.50m 5.50m 0.70m

17.45m

CUADRO DE CARGAS

COLUMNAToneladas

PD PL

1 40 152 - 3 60 30

4 50 255 - 9 - 13 70 30

6 - 7 140 7010 - 11 160 7514 - 15 160 85

8 - 12 - 16 90 4017 80 35

18 - 19 85 4020 100 45

SOLUCION

qa = 1.2 kg/cm² 0.00165 kg/cm³

50 cm s/c = 0.05 kg/cm²

50 cm fy = 4200 kg/cm²

Df = 110 cm f'c = 210 kg/cm²

Ks = 8 kg/cm³ 0.25

E = 217371 kg/cm² 0.0024 kg/cm³

CALCULO DEL AREA DE LA LOSA Y EL ESPESOR

A= 20.00*17.45 m2

A = 349.00 m2

Método a usar3 x Ks

E x t³

Lc = { 550 cm480 cm

SiLc <

1.75Þ

l

Lc >1.75

Þl

Tabulando :

t l

10 0.018228569 96 ® METODO FLEXIBLE 020 0.010838772 161 ® METODO FLEXIBLE 030 0.007996715 219 ® METODO FLEXIBLE 040 0.006444772 272 ® METODO FLEXIBLE 050 0.005451614 321 ® METODO FLEXIBLE 060 0.004754875 368 ® METODO FLEXIBLE 070 0.004235738 413 ® METODO FLEXIBLE 0

0CALCULO DE LA PRESION NETA

Asumiendo t = 50.00 cm

qn =

qn = 1,2 - 1650E-6 (110 - 50) - 2400E-6 (50) - 500E-4

qn = 0.931 kg/cm²

Verificación por punzonamiento

Vu = Pu - qn x m x n

Vc =

gs =

bcol =

tcol =

m =

gc =

l = 4√

METODO RIGIDO

METODO FLEXIBLE

1,75/l

qa - gt x ht - gc x hc - s/c

1,1 √ f'c x bo x d

COLUMNA 1

Vu = 1,4 x 40000 + 1,7 x 15000 - 0,931 x (50 + d/2) x (50 + d/2)

Vu = 81500 - 0,931 x ( 2500 + 50 d + 0,25 d²

Vc =

Vc = 31.881 x ( 50 d + 0,50 d² )

ø Vc = 27.099 x ( 50 d + 0,50 d² )

27.099 x ( 50 d + 0,50 d² ) = 81500 - 0,931 x ( 2500 + 50 d + 0,25 d² )

1354,94 d + 13,55 d² = 81500 - 2327,50 - 46,55 d - 0,23 d²

d² + 101,70 d - 5745,46 = 0

® d = 40.43 cm NO OK !

COLUMNA 2 - 3

50 + d/2

50 + d

Vu = 1,4 x 60000 + 1,7 x 30000 - 0,931 x (50 + d) x (50 + d/2)

Vu = 135000 - 0,931 x ( 2500 + 75 d + 0,5 d² )

Vc =

Vc = 15.941 x ( 150 d + 2 d² )

ø Vc = 13.549 x ( 150 d + 2 d² )

13.549 x ( 150 d + 2 d² ) = 135000 - 0,931 x ( 2500 + 75 d + 0,5 d² )

2032,35 d + 27,098 d² = 135000 - 2327,50 - 69,825 d - 0,466 d²

d² + 76,27 d - 4813,25 = 0

® d = 41.03 cm NO OK !

COLUMNA 4

50 + d/2

(a)

1,10 √ 210 x 2 (50 + d/2) d

(b)

Igualando (a) y (b), tenemos:

(a)

1,10 √ 210 x (( 50 + d ) + ( 50 + d/2) x 2 ) x d

(b)


50

50

50 + d/2

50 + d/2

50

50

50

50 + d/2

50 + d

Vu = 1,4 x 50000 + 1,7 x 25000 - 0,931 x (50 + d) x (50 + d/2)

Vu = 112500 - 0,931 x ( 2500 + 75 d + 0,5 d² )

Vc =

Vc = 15.941 x ( 150 d + 2 d² )

ø Vc = 13.549 x ( 150 d + 2 d² )

13.549 x ( 150 d + 2 d² ) = 112500 - 0,931 x ( 2500 + 75 d + 0,5 d² )

2032,35 d + 27,098 d² = 112500 - 2327,50 - 69,825 d - 0,466 d²

d² + 76,27 d - 3996,97 = 0

® d = 35.7 cm OK !

COLUMNA 5 - 9 - 13

50 + d

50 + d/2

(a)

1,10 √ 210 x (( 50 + d ) + ( 50 + d/2) x 2 ) x d

(b)


50

50

50

Vu = 1,4 x 70000 + 1,7 x 30000 - 0,931 x (50 + d) x (50 + d/2)

Vu = 149000 - 0,931 x ( 2500 + 75 d + 0,5 d² )

Vc =

Vc = 15.941 x ( 150 d + 2 d² )

ø Vc = 13.549 x ( 150 d + 2 d² )

13.549 x ( 150 d + 2 d² ) = 149000 - 0,931 x ( 2500 + 75 d + 0,5 d² )

2032,35 d + 27,098 d² = 149000 - 2327,50 - 69,825 d - 0,466 d²

d² + 76,27 d - 5321,16 = 0

® d = 44.18 cm NO OK !

COLUMNA 6 - 7

50 + d

50 + d

Vu = 1,4 x 140000 + 1,7 x 70000 - 0,931 x (50 + d) x (50 + d)

Vu = 315000 - 0,931 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

ø Vc = 13.549 x ( 200 d + 4 d² )

13.549 x ( 200 d + 4 d² ) = 315000 - 0,931 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 315000 - 2327,50 - 93,10 d - 0,931 d²

d² + 50,84 d - 5671,86 = 0

® d = 54.07 cm NO OK !

(a)

1,10 √ 210 x (( 50 + d ) + ( 50 + d/2) x 2 ) x d

(b)


(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

(b)


50

50

COLUMNA 10 - 11

50 + d

50 + d

Vu = 1,4 x 160000 + 1,7 x 75000 - 0,931 x (50 + d) x (50 + d)

Vu = 351500 - 0,931 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

ø Vc = 13.549 x ( 200 d + 4 d² )

13.549 x ( 200 d + 4 d² ) = 351500 - 0,931 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 351500 - 2327,50 - 93,10 d - 0,931 d²

d² + 50,84 d - 6333,97 = 0

® d = 58.13 cm NO OK !

COLUMNA 14 - 15

50 + d

50 + d

Vu = 1,4 x 160000 + 1,7 x 85000 - 0,931 x (50 + d) x (50 + d)

Vu = 368500 - 0,931 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

ø Vc = 13.549 x ( 200 d + 4 d² )

(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

(b)


(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

(b)

50

50

50

50

13.549 x ( 200 d + 4 d² ) = 368500 - 0,931 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 368500 - 2327,50 - 93,10 d - 0,931 d²

d² + 50,84 d - 6642,34 = 0

® d = 59.95 cm NO OK !

COLUMNA 8 - 12 - 16

50 + d

50 + d

Vu = 1,4 x 90000 + 1,7 x 40000 - 0,931 x (50 + d) x (50 + d)

Vu = 194000 - 0,931 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

ø Vc = 13.549 x ( 200 d + 4 d² )

13.549 x ( 200 d + 4 d² ) = 194000 - 0,931 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 194000 - 2327,50 - 93,10 d - 0,931 d²

d² + 50,84 d - 3476,93 = 0

® d = 38.79 cm OK !

COLUMNA 17

50 + d

50 + d/2


(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

(b)


50

50

50

50

Vu = 1,4 x 80000 + 1,7 x 35000 - 0,931 x (50 + d) x (50 + d/2)

Vu = 171500 - 0,931 x ( 2500 + 75 d + 0,5 d² )

Vc =

Vc = 15.941 x ( 150 d + 2 d² )

ø Vc = 13.549 x ( 150 d + 2 d² )

13.549 x ( 150 d + 2 d² ) = 171500 - 0,931 x ( 2500 + 75 d + 0,5 d² )

2032,35 d + 27,098 d² = 171500 - 2327,50 - 69,825 d - 0,466 d²

d² + 76,27 d - 6137,44 = 0

® d = 49.00 cm NO OK !

COLUMNA 18 - 19

50 + d

50 + d

Vu = 1,4 x 85000 + 1,7 x 40000 - 0,931 x (50 + d) x (50 + d)

Vu = 187000 - 0,931 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

ø Vc = 13.549 x ( 200 d + 4 d² )

13.549 x ( 200 d + 4 d² ) = 187000 - 0,931 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 187000 - 2327,50 - 93,10 d - 0,931 d²

d² + 50,84 d - 3349,95 = 0

® d = 37.79 cm OK !

(a)

1,10 √ 210 x (( 50 + d ) + ( 50 + d/2) x 2 ) x d

(b)


(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

(b)


50

50

COLUMNA 20

50 + d

50 + d

Vu = 1,4 x 100000 + 1,7 x 45000 - 0,931 x (50 + d) x (50 + d)

Vu = 216500 - 0,931 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

ø Vc = 13.549 x ( 200 d + 4 d² )

13.549 x ( 200 d + 4 d² ) = 216500 - 0,931 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 216500 - 2327,50 - 93,10 d - 0,931 d²

d² + 50,84 d - 3885,07 = 0

® d = 41.89 cm NO OK !

Þ t = 70 cm

qn = 1,2 - 1650E-6 (110 - 70) - 2400E-6 (70) - 500E-4

qn = 0.916 kg/cm²

COLUMNA 14 - 15

50 + d

50 + d

Vu = 1,4 x 160000 + 1,7 x 85000 - 0,916 x (50 + d) x (50 + d)

Vu = 368500 - 0,916 x ( 2500 + 100 d + d² )

Vc =

Vc = 15.941 x ( 200 d + 4 d² )

(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

(b)


(a)

1,10 √ 210 x 4 x ( 50 + d ) x d

50

50

50

50

ø Vc = 13.549 x ( 200 d + 4 d² )

13.549 x ( 200 d + 4 d² ) = 368500 - 0,916 x ( 2500 + 100 d + d² )

2709,80 d + 54,196 d² = 368500 - 2290 - 91,60 d - 0,916 d²

d² + 50,83 d - 6644,83 = 0

® d = 59.97 cm OK !

® t = 70.00 cm.

3.-PROCESAMOS LA TABLA

PUNTO Pu/A X(cm.) 0.00001 Y(cm.) 0.00006yA 1.132 -580 0.006 490 0.029 1.167B 1.132 -290 0.003 490 0.029 1.164C 1.132 250 -0.003 490 0.029 1.159D 1.132 580 -0.006 490 0.029 1.156E 1.132 -580 0.006 245 0.015 1.152F 1.132 580 -0.006 245 0.015 1.141G 1.132 -580 0.006 -205 -0.012 1.126H 1.132 580 -0.006 -205 -0.012 1.114I 1.132 -580 0.006 -490 -0.029 1.108J 1.132 -290 0.003 -490 -0.029 1.105K 1.132 250 -0.003 -490 -0.029 1.100L 1.132 580 -0.006 -490 -0.029 1.097

6.-CALCULAMOS EL FACTOR DE AMPLIFICACION DE CARGA F PARA CADA FRANJA

FRANJA q q prom. b L P CP FA 1.167 1.1566 245 1160 P1 626450000.0 313389348 0.500B 1.164 P2 1C 1.159 P3D 1.156E 1.152F 1.141E 1.152 1.1332 450 1160 P4 626450000.00 313520765.2 0.500F 1.141 P5 1H 1.114 P6G 1.126G 1.126 1.1084 285 1160 P7 626450000.00 313408213 0.500H 1.114 P8 1I 1.108 P9J 1.105K 1.100L 1.097

(b)


q(kg/cm2)

SP

A 1.167 1.1372 290 980 P1 0.00 161600.8567 0.500B 1.164 P4 1J 1.105 P7I 1.108G 1.126E 1.152B 1.164 1.1322 540 980 P2 626450000.00 313524580 0.500C 1.159 P5J 1.105 P8K 1.100C 1.159 1.1277 330 980 P3 626450000.00 313407349 0.500D 1.156 P6F 1.141 P9H 1.114K 1.100L 1.097

DISEÑO DE PILOTES

Una columna de 30 x 60 , que lleva de refuerzo 8 Ø 3/4" transmite las cargas PD = 80 tn PL = 40 tn . En una edificacion a la estratigrafia mostrada en la figura. Diseñar los elementos a considerar en la cimentacionque permita transmitir la carga de la columna al estrato duro . Considere f ' c =210 kg/cm2 y fy = 4200kg/cm2.

S/C 300 DatosColumna b t 24

cm 30 60 1500

Arena suelta ht= #REF! Cargas PD PL f ' c = 210tn 80 40 Fy = 4200

db 1.91 1800Df = 650

hc= 0.00 m. 2400

Estracto Duro

1.- Cálculo del peralte de la zapata:

Ld = 44 cm hc = Ld + 15 hc =

Se toma hc =

Ld = 0.004 x db x fy Ld = 32 cm

La longitud del Pilote es : L = Df - ( b + t )+ 1 L= 660 cm

Usamos Pilotes de C° A° Prefabricado cuadrados de 40 x 40 . Calculamos su capacidad de carga.

Qu = Qf + Qp Para suelos granulares k = 1 D = 0.4p = 1.6 L = 5.6

12165 kg Z1 = 5.6Ap =0.09

5708.189 kgF.S = 3

Qp = 49572 kgPu =

Qu = 67445.19 kg Qa = Qu/F.S Qa = 22481.7 kg

Calculamos el N° de Pilotes : Dimensiones de la Zapata

kg/m2

f =

g =

gc =

g2 =

Qf1 = k x g x tan x x p x D x ( 15 x L - 112.5 x D )

Qf2 = k x gc x tan x1 x p x ( Z2^2 - Z1^2 )/2

Qp = g x L x( Nq - 1) x Ap

ld = 0 . 08 *db*Fy

√F'c

N° = Pu/ Qa N° = 8.01 Entonces N1° = 9

Az = 3.60 3.60

Verificamos el numero de Pilotes a usar :

N ° = 9.27 Entonces : N° < N1° No Cumple AUMENTAR SECCION !!

Aumentamos (Pilote o la seccion del Pilote a 0.45 x 0.450.45 0.45

Recalculamos la seccion :

Qu = Qf + Qp Para suelos granulares k = 1 D = 0.45p = 1.8 L = 5.6

13176 kg Z1 = 5.6Ap =0.11

6422 kgF.S = 3

Qp = 59982 kgPu =

Qu = 79580 kg Qa = Qu/F.S Qa = 26527 kg

Calculamos el N° de Pilotes : Dimensiones de la Zapata

N° = Pu/ Qa N° = 6.79 Entonces N1° = 9

Az = 4.1 4.1

Verificamos el numero de Pilotes a usar :

P' u = 216726 tn

N ° = 8.17 Entonces : Cumple ,OK !!

Calculamos la capacidad de carca del grupo de Pilotes :

N ° = P'u /Qa =( Pu + Az x ( g t x +g c + S/C) )/Qa

Qf1 = k x g x tan x x p x D x ( 15 x L - 112.5 x D )

Qf2 = k x gc x tan x1 x p x ( Z2^2 - Z1^2 )/2

Qp = g x L x( Nq - 1) x Ap

N ° = P'u /Qa =( Pu + Az x ( g t x +g c + S/C) )/Qa

Q = Ef Qf + Qp

Ef = (1 - tan ^ - 1 x ( D/d ) ) x( ( n^3 -1 ) x m^3 + (m^3 - 1 ) x n^3)/ (90 x m^3 x n^3) =

Qf = 176382 kg entonces Qgu = 0.73 x Qf + QpQp = 539838 kg Qgu = 716221

Qga = Qgu/F.S. Qga = 238740 entonces : Qga > P'u Cumple !!

Una columna de 30 x 60 , que lleva de refuerzo 8 Ø 3/4" transmite las cargas PD = 80 tn PL = 40 tn . En una edificacion a la estratigrafia mostrada en la figura. Diseñar los elementos a considerar en la cimentacionque permita transmitir la carga de la columna al estrato duro . Considere f ' c =210 kg/cm2 y fy = 4200kg/cm2.

Datos

kg/m2

kg/cm2kg/cm2

kg/m2cm

59 cm

60 cm

Usamos Pilotes de C° A° Prefabricado cuadrados de 40 x 40 . Calculamos su capacidad de carga.

0.3249196960.476975533

Z2 = 6.6Nq = 55

180000 kg

kg/m3

Tan x =Tan x1=

0.3249196960.476975533

Z2 = 6.6Nq = 55

180000

Tan x =Tan x1=

0.73

Qga > P'u Cumple !!

DISEÑO DE VIGA DE CIMENTACION EJE 1 - 1

P1 P2 P3 P4

4.400 m. 3.683 m. 2.017 m.

H

11.500 m. B

DATOS

σt = 1.50 f¨c = 210 rec = 0.075 m.

Ko = 5,000 Tn/m3 fy = ### Tn/m3 e piso= 0.15 m.

DF = 1.50 m. Es = ### Kg/cm2 = 2.40 Tn/m3 peso volumetrico del concreto armado

NPT= 0.30 m. = 2.00 Tn/m3 peso volumetrico del concreto simple

S/C = 200 Kg/m2 = 1.70 Tn/m3 peso volumetrico del suelo

Columna 01 0.50 m. 0.30 m. Columna 0 0.60 m. 0.40 m. Columna 01 0.40 m. 0.30 m. Columna### m###m.

Cargas Cargas Cargas Cargas

PD = 40.18 Tn PD = 74.98 Tn PD = 38.61 Tn PD = 10.01 Tn

PL = 5.99 Tn PL = 14.36 Tn PL = 7.29 Tn PL = 1.54 Tn

Momentos longitudinales Momentos longitudinales Momentos longitudinales Momentos longitudinales

MD = 0.47 Tn-m (Horario) MD= -0.42 Tn-m MD = -0.24 Tn-m MD = -0.19 Tn-m

ML = 0.07 Tn-m (Horario) ML = -0.06 Tn-m ML = -0.04 Tn-m ML = -0.03 Tn-m

Msx= 11.18 Tn-m Ms = 12.47 Tn-m Msx = 3.91 Tn-m Msx= 3.41 Tn-m

Psx = 13.52 Tn-m Psx= 1.79 Tn-m Psx = 4.86 Tn-m Psx = 16.59 Tn-m

Momentos Transversales Momentos Transversales Momentos Transversales Momentos Transversales

MD = -0.25 Tn-m MD= -1.02 Tn-m (Horario) MD = -0.25 Tn-m (Horario) MD = 0.00 Tn (Horario)

ML = -0.04 Tn-m ML = -0.28 Tn-m (Horario) ML = -0.07 Tn-m (Horario) ML = 0.00 Tn (Horario)

Msy= 5.25 Tn-m Ms = 16.56 Tn-m Msy = 7.76 Tn-m Msy= 0.00 Tn-m

Psy = 20.22 Tn-m Psy= 30.02 Tn-m Psy = 14.07 Tn-m Psy = 0.00 Tn-m

bw

hf

Kg/cm2 Kg/cm2

ɣconc

ɣcs

ɣs

DIAGRAMA DE CARGAS

ML1= 0.47 + 0.07 ML2= -0.42 + -0.06 ML3 ### +### ML4 -0.19 +###

P1= 40.18 + 5.99 P2= 74.98 + ### P3= ### +### P4=10.01 +###

4.950 m. 4.183 m. 2.367 m.

1. DETERMINACION DE LA PRESION NETA

h= 0.500 m. asumido

σn = ### - 1.20 - 0.20 - 0.30 - 0.19 = 13.11 Tn/m2

σn = ### - 1.20 - 0.20 - 0.30 - 0.19 = 17.61 Tn/m2 para el caso de sismo

2. DIMENSIONAMIENTO EN PLANTA

A. SIN CONSIDERAR EFECTOS DE SISMO

A.1. POSICION DE LA RESULTANTE CONSIDERANDO ALTERNANCIA DE CARGAS

ALTERNANCIA

1 Xg= 4.95 x 82.16 + 9.13 x ### + 11.50 X ### ´+ 0.54 + -0.48 + -0.28 +### ### ) 924.995.16 m

50% ### ### 100%

### 179.14

2 Xg= 4.95 x 82.16 + 9.13 x ### +11.50 X ### ´+ 0.54 + -0.48 + -0.28 + ### ) 949.435.22 m

50% ### ### 50%

182.015 182.02

3 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) 951.685.13 m

50% ### ### 50%

185.550 185.55

4 Xg= 4.95 x 82.16 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) 916.145.05 m

100% ### ### 50%

181.365 181.37

5 Xg= 4.95 x 82.16 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) 958.285.24 m

50% ### ### 100%

182.785 182.78

σn = σt - h*γc - S/C - ep*γCS- er*γS

= =

= =

= =

= =

= =

6 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) 984.975.21 m

50% ### ### 50%

189.195 189.20

ALTERNANCIA

7 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.05 m

100% ### ### 50%

188.545 ###

8 Xg= 4.95 x 82.16 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.08 m

100% ### ### 100%

182.135 ###

9 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.16 m

50% ### ### 100%

186.320 ###

10 Xg= 4.95 x 82.16 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.13 m

100% ### ### 50%

185.010 ###

11 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.15 m

100% ### ### 100%

192.960 ###

12 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.23 m

50% ### ### 100%

189.965 ###

13 Xg= 4.95 x 89.34 + 9.13 x### + ### X### ´+ ### + ### +### + -0.22 ) = ###5.15 m

100% 100% ### 100%

192.960 ###

EL MAYOR: Xg= 5.24 m. EL MENOR: Xg= 5.05 m.

EXCENTRICIDAD e1= 11.50- 5.05 = 0.70 m

e12= 11.50 5.24 = ###m.

2.00 2.00

e1< L/6= 1.983 m OK e12< L/6 = 1.983 m. OK Falta colocar la condicionante.

= =

=

=

=

=

=

=

=

-

A2 DETERMINACION DEL ANCHO "B"

P P= Cargas de servicio aplicadas al 100 %

σn Adoptamos B= 2.20 m.

Az= ### m2 Aumentamos ### por accion de los momentos Az= 16.19 m2 B=### L= 11.90 m.

- considerando solo momentos longitudinales

σ=P´s

±6 * ML P´s= carga mas desfavorable paa Xg mayor y menor

L * B ML= momento longitudinal

1.00*

182.785+

6 *( ### * 0.51 )=

6.98 + 1.79 = 8.77 < σn= ### OK colocar condicionantes

### 2.20 2.20 * ### * 11.90

6.98 - 1.79 = 5.20 < σn= ### OK

σ1= 1.00*

188.545+

6 *( ### * 0.70 )=

7.20 + 2.55 = 9.75 < σn= ### OK colocar condicionantes

### 2.20 2.20 * ### * 11.90

7.20 - 2.55 = 4.65 < σn= ### OK

- considerando solo momentos transversales

σ=P´s

±6 * MT MT= momento transversal

L * B

1.00*

182.785+

6 *( ### -1.30 -0.32 ###)=

6.98 + -0.20 = 6.78 < σn= ### OK colocar condicionantes

### 2.20 ### * ### * 2.20

6.98 - -0.20 = 7.18 < σn= ### OK

σ1= 1.00*

182.785+

6 *( ### -1.30 -0.32 ###)=

6.98 + -0.20 = 6.78 < σn= ### OK colocar condicionantes

### 2.20 ### * ### * 2.20

6.98 - -0.20 = 7.18 < σn= ### OK

Az=

A.2 VERIFICACION DE PRESIONES

B * L2

σ1=

B * L2

σ1=

B. CONSIDERANDO EFECTOS DE SISMO EN DIRECCION LONGITUDINAL X-X

B1 Caso de Momentos Sismicos horarios

P1= ### + 5.99 + ### = 59.69 Tn. P3= ### + 7.29 + 4.86 = ### Tn.

ML1= 0.47 + 0.07 + ### = 11.72 Tn-m ML3=### + ### + 3.91 = 3.63 Tn-m

MT1= ### + -0.04 + 0.00 = -0.29 Tn-m MT3=### + ### + 0.00 = -0.32 Tn-m

P2= ### + 14.36 + 1.79 = 91.13 Tn. P4= ### + 1.54 + ### = ### Tn.

ML2= ### + -0.06 + ### = 11.99 Tn-m ML4=### + ### + 3.41 = 3.19 Tn-m

MT2= ### + -0.28 + 0.00 = -1.30 Tn-m MT4=### + 0.00 + 0.00 = 0.00 Tn-m

P1= 59.69 Tn. P2= ### Tn. P3= ### Tn. P4= ###Tn.

ML1= 11.72 Tn-m. ML2= ### Tn-m. ML3=### Tn-m. ML4=###Tn-m.

MT1= -0.29 Tn-m. MT2= ### Tn-m. MT3=### Tn-m. MT4=###Tn-m.

4.950 m. 4.183 m. 2.367 m.

B.1.1 POSICION DE LA RESULTANTE CONSIDERANDO ALTERNANCIA DE CARGA

ALTERNANCIA

1 Xg= 4.95 x 83.95 + 9.13 x ### + 11.50 X ### ´+ 11.72 + ### + 3.63 + 3.19 ) ###5.62 m

50% ### ### 100%

### 219.55

2 Xg= 4.95 x 91.13 + 9.13 x ### + 11.50 X ### ´+ 11.72 + ### + 3.63 + 3.19 ) ###5.44 m

100% ### ### 50%

### 225.31


e12= 11.50 5.44 = ###m.

2.00 2.00


= =

= =

-

B.1.2 VERIFICACION DE PRESIONES

σ=P

±6 * ML

±6 * MT

L * B

1.00*

219.545+

6 *( ### * 0.13 ) 6.00 * ( -0.29 + -1.30 + -0.32 + ### )=### <σn= ###) OK colocar condicionantes

### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= ###) OK

8.39 + 0.56 - -0.20 =### <σn= ### OK

8.39 - 0.56 + -0.20 =### <σn= ### OK

8.39 - 0.56 - -0.20 =### <σn= ### OK

σ1= 1.00*

225.305+

6 *( ### * 0.31 ) 6.00 * ( -0.29 + -1.30 + -0.32 + ### )=### <σn= ### OK colocar condicionantes

### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= ### OK

σ2= 8.61 + 1.33 - -0.20 =### <σn= ### OK

σ3= 8.61 - 1.33 + -0.20 =### <σn= ### OK

σ4= 8.61 - 1.33 - -0.20 =### <σn= ### OK

B * L2 L * B2

σ1= ´+(

σ2=

σ3=

σ4=

´+(

B.2 Caso de Momentos Sisicos anti-horarios

P1= ### + 5.99 + ### = 32.65 Tn. P3= ### + 7.29 + ### = ### Tn.

ML1= 0.47 + 0.07 + ### = -10.64 Tn-m ML3=### + ### + ### = -4.19 Tn-m

MT1= ### + -0.04 + 0.00 = -0.29 Tn-m MT3=### + ### + 0.00 = -0.32 Tn-m

P2= ### + 14.36 + ### = 87.55 Tn. P4= ### + 1.54 + ### = -5.04 Tn.

ML2= ### + -0.06 + ### = -12.95 Tn-m ML4=### + ### + ### = -3.63 Tn-m

MT2= ### + -0.28 + 0.00 = -1.30 Tn-m MT4=### + 0.00 + 0.00 = 0.00 Tn-m

P1= 32.65 Tn. P2= ### Tn. P3= ### Tn. P4= ###Tn.

ML1= -10.64 Tn-m. ML2= ### Tn-m. ML3=### Tn-m. ML4=###Tn-m.


4.950 m. 4.183 m. 2.367 m.

B.1.1 POSICION DE LA RESULTANTE CONSIDERANDO ALTERNANCIA DE CARGA

ALTERNANCIA

1 Xg= 4.95 x 80.37 + 9.13 x ### + 11.50 X ### ´+ -10.64 + ### + -4.19 + ### ) 683.2804.68 m

50% ### ### 100%

### 146.02

2 Xg= 4.95 x 87.55 + 9.13 x ### + 11.50 X ### ´+ -10.64 + ### + -4.19 + ### ) 676.6764.46 m

100% ### ### 50%

### 151.79


e12= 11.50 4.46 = ###m.

2.00 2.00

= =

= =

-


B.1.2 VERIFICACION DE PRESIONES

σ=P

±6 * ML

±6 * MT

L * B

1.00*

146.025+


### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= ### OK

5.58 + 3.01 - -0.20 =### <σn= ### OK

5.58 - 3.01 + -0.20 =### <σn= ### OK

5.58 - 3.01 - -0.20 =### <σn= ### OK

σ1= 1.00*

151.785+


### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= ### OK

σ2= 5.80 + 3.78 - -0.20 =### <σn= ### OK

σ3= 5.80 - 3.78 + -0.20 =### <σn= ### OK

σ4= 5.80 - 3.78 - -0.20 =### <σn= ### OK

B * L2 L * B2

σ1= ´+(

σ2=

σ3=

σ4=

´+(

C. CONSIDERANDO EFECTOS DE SISMO EN DIRECCION LONGITUDINAL Y-Y

C1 Caso de Momentos Sismicos horarios

P1= ### + 5.99 + ### = 66.39 Tn. P3= ### + 7.29 + ### = ### Tn.

ML1= 0.47 + 0.07 + 0.00 = 0.54 Tn-m ML3=### + ### + 0.00 = -0.28 Tn-m

MT1= ### + -0.04 + 5.25 = 4.96 Tn-m MT3=### + ### + 7.76 = 7.44 Tn-m

P2= ### + 14.36 + ### = ### Tn. P4= ### + 1.54 + 0.00 = ### Tn.

ML2= ### + -0.06 + 0.00 = -0.48 Tn-m ML4=### + ### + 0.00 = -0.22 Tn-m

MT2= ### + -0.28 + ### = 15.26 Tn-m MT4=### + 0.00 + 0.00 = 0.00 Tn-m

P1= 66.39 Tn. P2= ### Tn. P3= ### Tn. P4= ###Tn.


MT1= 4.96 Tn-m. MT2= ### Tn-m. MT3=### Tn-m. MT4=###Tn-m.

4.950 m. 4.183 m. 2.367 m.

C.1.1 POSICION DE LA RESULTANTE CONSIDERANDO ALTERNANCIA DE CARGA

ALTERNANCIA

1 Xg= 4.95 x 112.18 + 9.13 x ### + 11.50 X ### ´+ 0.54 + -0.48 + -0.28 + ### ) ###5.00 m

50% ### ### 100%

### 247.09

2 Xg= 4.95 x 119.36 + 9.13 x ### + 11.50 X ### ´+ 0.54 + -0.48 + -0.28 + ### ) ###4.86 m

100% ### ### 50%

### 252.86

= =

= =


e12= 11.50 4.86 = ###m.

2.00 2.00


C.1.2 VERIFICACION DE PRESIONES

σ=P

±6 * ML

±6 * MT

L * B

1.00*

247.095+

6 *( ### * 0.75 ) 6.00 * ( 4.96 + ### + 7.44 + ### )=### <σn= ###OK colocar condicionantes

### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= OK

9.44 + 3.57 - 2.88 = ### <σn= ###OK

9.44 - 3.57 + 2.88 = 8.75 <σn= ###OK

9.44 - 3.57 - 2.88 = 2.99 <σn= ###OK

σ1= 1.00*

252.855+

6 *( ### * 0.89 ) 6.00 * ( 4.96 + ### + 7.44 + ### )=### <σn= ###OK colocar condicionantes

### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= ###OK

σ2= 9.66 + 4.34 - 2.88 = ### <σn= ###OK

σ3= 11.11 - 4.34 + 2.88 =### <σn= ###OK

σ4= 11.11 - 4.34 - 2.88 =### <σn= ###OK

-

B * L2 L * B2

σ1= ´+(

σ2=

σ3=

σ4=

´+(

C2 Caso de Momentos Sismicos anti-horarios

P1= ### + 5.99 + ### = 25.95 Tn. P3= ### + 7.29 + ### = ### Tn.

ML1= 0.47 + 0.07 + 0.00 = 0.54 Tn-m ML3=### + ### + 0.00 = -0.28 Tn-m

MT1= ### + -0.04 + ### = -5.54 Tn-m MT3=### + ### + ### = -8.08 Tn-m

P2= ### + 14.36 + ### = 59.32 Tn. P4= ### + 1.54 + 0.00 = ### Tn.

ML2= ### + -0.06 + 0.00 = -0.48 Tn-m ML4=### + ### + 0.00 = -0.22 Tn-m

MT2= ### + -0.28 + ### = -17.86 Tn-m MT4=### + 0.00 + 0.00 = 0.00 Tn-m

P1= 25.95 Tn. P2= ### Tn. P3= ### Tn. P4= ###Tn.



4.950 m. 4.183 m. 2.367 m.

C.1.1 POSICION DE LA RESULTANTE CONSIDERANDO ALTERNANCIA DE CARGA

ALTERNANCIA

1 Xg= 4.95 x 52.14 + 9.13 x ### + 11.50 X ### ´+ 0.54 + -0.48 + -0.28 + ### ) 681.1815.75 m

50% ### ### 100%

### 118.48

2 Xg= 4.95 x 59.32 + 9.13 x ### + 11.50 X ### ´+ 0.54 + -0.48 + -0.28 + ### ) 674.5785.43 m

100% ### ### 50%

= =

= =

### 124.245.43 m


e12= 11.50 5.43 = ###m.

2.00 2.00


C.1.2 VERIFICACION DE PRESIONES

σ=P

±6 * ML

±6 * MT

L * B

1.00*

118.475+

6 *( ### * 0.00 ) 6.00 * ( -5.54 + ### + -8.08 + ### )=1.25 <σn= 17.61 O colocar condicionantes

### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= 17.61 OK

4.53 + 0.00 - -3.28 = 7.81 <σn= 17.61 OK

4.53 - 0.00 + -3.28 = 1.25 <σn= 17.61 OK

4.53 - 0.00 - -3.28 = 7.80 <σn= 17.61 OK

σ1= 1.00*

124.235+

6 *( ### * 0.32 ) 6.00 * ( -5.54 + ### + -8.08 + ### )=2.23 <σn= 17.61 O colocar condicionantes

### 2.20 2.20 * ### * 11.90 11.90 * ###* ###

<σn= 17.61 OK

σ2= 4.75 + 0.77 - -3.28 = 8.79 <σn= 17.61 OK

σ3= 4.75 - 0.77 + -3.28 = 0.70 <σn= 17.61 OK

σ4= 4.75 - 0.77 - -3.28 = 7.26 <σn= 17.61 OK

= =

-

B * L2 L * B2

σ1= ´+(

σ2=

σ3=

σ4=

´+(

CARGAS DE SERVICO

RESUMEN DE PRESIONES

- CARGAS DE GRAVEDAD - CARGAS DE GRAVEDAD + SISMO LONGITUDINAL HORARIO

σ1= 8.77 Tn/m2 ### Tn/m σ1= 9.15 Tn/m2 20.12 Tn/m

σ´1= 9.75 Tn/m2 ### Tn/m σ´1=10.13 Tn/m2 22.29 Tn/m

- CARGAS DE GRAVEDAD + SISMO LONGITUDINAL ANTIHORARIO - CARGAS DE GRAVEDAD + SISMO TRANSVERSAL HORARIO

σ1= 8.79 Tn/m2 ### Tn/m σ1= 15.89 Tn/m2 34.96 Tn/m

σ´1= 9.77 Tn/m2 ### Tn/m σ´1=16.88 Tn/m2 37.13 Tn/m

- CARGAS DE GRAVEDAD + SISMO TRANSVERSAL ANTIHORARIO

σ1= 7.81 Tn/m2 ### Tn/m

σ´1= 8.79 Tn/m2 ### Tn/m

PRESIONES Y CARGAS ADOPTADAS

σ1= ### Tn/m2 ### Tn/m

σ2= ### Tn/m2 ### Tn/m

W1 = W1 =

W´1 = W´1 =

W1 = W1 =

W´1 = W´1 =

W1 =

W´1 =

W1 =

W´1 =

ENVOLVENTE DE PRESIONES

### Tn ### Tn ### Tn 11.55

6.950 4.183 2.367

34.96 Tn/m

### Tn/m

13.90 m.

AGREGAR MOMENTOS CALCULADOS POR EL CYPECAD

DISEÑO DE VIGA DE CIMENTACIONa) Mayoracion de las cargas

FACTOR DE MAYORACION

FPM= 1.40 * ( 40.18 + ### + 38.61 + ### ) + 1.70 * ( 5.99 + ### + 7.29 * ### ) 278.90 = 1.445

### + ### + 38.61 + ### + 5.99 + ### + 7.29 + 1.54 192.96

Presion ultima: ### Tn/m * 1.445 = ### Tn/m2

Carga ultima: ### Tn/m * 2.20 = ### Tn/m2 = ### m

b) Determinacion de la seccion transversal por rigidez

- Ancho de columnas 0.40 m. ### adoptado H=1.20

entonces 0.45 m. Ko= 5,000 Tn/m3 ###

B= 2.20 m. f´c=### kg/cm2

0.45 m. (asumido) Ec= ### Tn/m2

4.95 m. H=### m. asumido. B= 2.20 m.

determinacion de la inercia de la seccion y verificacion de la rigidez

yt= 0.377542 m. posicion del ejec centroidal (teorema de steiner) Eje Y

VER

IFIC

AR

bw

λL=

bw= hf

hf=

Lm=

Tabla de calculo de centro de gravedad en sentido "Y"

Ig= 0.123137 m. 0.45

VER

IFIC

AR

Teorema de Steiner. X1 0.438 m.### 2

Figura Area Ӯ Ӯ Area

Ïi,x = Iix + Ai(YiG)2 X2 1.100 m. 0.875 0.875 1 ### 0.225 ###

Ïi,y = Iiy + Ai(XiG)2 X3 1.538 m.### 1 3

2 ### 0.825 ### ### 0.56

Ix,total = Σ Ïi,x Y1 0.225 m. 3 ### 0.225 ###

Iy,total = Σ Ïi,y Y2 0.825 m. 1.325 X ### ###

Y3 0.225 m. Yg= 0.378 m. =### m.simetrico

0.007 Cailculo del momento de inercia con respecto al centro de gravedad

0.025 4 Ko* B 0.3184 Figura Area (Yi - Yg)(Yi-Yg)^2

0.016 4 * E * I 1.5759 ### ### 1 ### 0.025 ### -0.15 0.02 0.02 0.20

0.006 0.00 ### 2 ### 0.006 ### 0.45 0.20 0.08 0.01

0.010 0.44 ### 3 ### 0.087 ### -0.15 0.02 0.02 0.20

0.087 0.12 0.40

c) VERIFICACION POR CORTANTE

Se toma el mayor valor del cortante actuante ds= 3/8 " diametro de estribo

Vu= ### Tn Valor calculado con el software CYPECAD dbl= 5/8 " diametro longitudinal asumido

B= 2.20 m. recubrimiento es 7 cm + db/2, verificar

Col.= 0.60 Donde se produce el cortante Vu d = h- rec Rec= 9.25 cm.

d= ### cm 0.85 x Vc 60,413.97 d = ### cm.

Wu= ### Tn/m 0.85 x Vc 51.35 Tn

cortante ultimo Vu 73.38 Tn. - tn = Tn.

### Tn. ≥ Vud si entonce OK colocar condicionante O.K. VALOR DE hf CORRECTO

d) VERIFICACION POR PUNZONAMIENTO

Se analiza la zona donde se aplica la mayor carga

dbt= 5/8 " diametro transversal asumido

d/2 0.60 d/2

d/2 d/2 0.50 dl= ### cm. D5/= 1.59 cm.

C-2 ### ### C-1 dt= ### cm.

Σ=

Xg

I1X= m4

I1Y= m4 λ= λ= (Xi-XG)(Xi-XG)^2 Ii,x Ii,y ï1X ï1Y

I2X= m4 λL= valor de H correcto u

optimizar HI2Y= m4

I3X= m4

I3Y= m4

Vc = 0.53 x f ' c^(1/2) x b x d =

cortante tomado por el concreto ØVc= ØVc

d/2 d/2

columnas 2 columnas 2

m= 95.754 cm. m= 67.877 cm. Lado mayor/lado menor 1.50 (C-2)

n= 74.17 cm. n= 64.167 cm. Vu = Pu − wnum × n 1.67 (C-1)

bo = 339.84 cm. bo = 199.92 cm. ØVcp2=

Ao = 7,101.80 Ao = 4,355.47

Pu= 129.38 Tn. Pu= ### Tn.

Vup= 112.06 Tn. Vup= 55.81 Tn.

189.08 Tn. 104.78 Tn.

164.64 Tn. 96.85 Tn.

Vup < OK Vup < OK condicionante colocar

e) DETERMINACION DEL REFUERZO

colocar diagramas de momento flector y cortantes del cypecad.

bw= 0.45 m

As positivo DETERMINACION DEL ACERO MINIMO POSITIVO

As.min1= 1.2 * Mag Mag=

0.9*fy*(0.95*d) Yt

d= ### cm.

1.20 m. yt= 0.80 m. Posicion del eje centroidal (teorema de steVERIFICAR PARECE QUE HAY ERROR

Ig= 0.11525 m4 Calculo con figura anterior o software

Mag= ### kg-m

Asmin1= 12.60 cm2 entonces el acero minimo serAsmin= 12.60 cm2

Asmin2= 12.04 cm2

hf= ### m. 4 Ø 3/4 " + 1 Ø 5/8 " = 13.38 cm2 falta calcular

bo = ( b + 0,5*d ) + 2x( t + d ) =

βc = βc =

βc =

Φ * 1.1 * √fć * b * d

cm2. cm2.

ØVcp1= ØVcp1=

ØVcp2= ØVcp2=

ØVcp ØVcp

As.min2=0.7*(fć^(1/2)*bw*d/fy Ig * 2 *fć^(1/2)

Distribuir en una capa

b < 4bw= 1.80 m

As negativo

DETERMINACION DEL ACERO MINIMO NEGATIVO

As.min1= 1.2 * Mag Mag=

0.9*fy*(0.95*d) Yt

yb= 0.40 m. Posicion del eje centroidal (teorema de steiner)

Ig= 0.1153 m4

Mag= ### kg.m d= ### cm.

Asmin1= 25.20 cm2 entonces el acero minimo serAsmin= 25.20 cm2

Asmin2= 24.07 cm2 db= 3/4 " Area### = 2.85 cm2

S= 0.20 m. 4 Ø 3/4 " + 1 Ø 5/8 " = 13.38 cm2 falta calcular

9.00 var. Area de acero colocad 25.65 cm2 adoptamos### Ø 3/4 " ´@ 0.20 m.

As.min2=0.7*(f´c^(1/2)*2*bw*d/fy Ig * 2 *f´c^(1/2)

colocar condicionantes

colocar condicionantes

ENVOLVENTE DE PRESIONES

recubrimiento es 7 cm + db/2, verificar

falta calcular

Diseño Cimentacion Tipo t Avance

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