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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/cm
; ademas fy = 4200kg/cm2 , fy = 280 kg/cm2 en la columna y fc = 210 kg/cm2 en la zapata
DATOS:
Zapata Otros Suelo
f ' c = 210 kg/cm² S/C =
550 kg/m² Df = 1.4 m
Columna PD = 180 Tn 1700 kg/m³
f ' c = 280 kg/cm² PL = 100 Tn qa = 2.50 kg/cm²
b = 50 cm db = 2.54 cm
t = 70 cm Acero Lv = 150 cm
f y = 4200 kg/cm²
MD,ML
PD, PL
Ld = 58.89 cm
ht
Tomar Ld = 58.89 cm Df Lv
(Del problema se emplean varillas de Ø1") Øb ( 1") = 2.54 cm
r.e. = 7.50 cm (recubrimiento) hc
hc = 68.93 cm
Tomar hc = 70.00 cm hc = Ld + r.e + Øb T
ht = Df - hc
ht = 70.00 cm
B
qm = 2.16 kg/cm²
T
Azap = 129,629.63 cm² Donde:
T = 370.00 cm P = Carga de servicio
B = 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 cm
r.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
Ø = 0.85 (Coef. De reduccion por corte)
Vc = 162,875.03 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 )
b
t
t
Reemplazo los valores que tenemos:
1.4 x 180000 + 1.7 x 100000
350 x 370
Ld=
- d
b
. F
y
√
f'c
Azap=
P
qm
T= √Az+
(t1-t2 )
2
S = √
Az−
(t1-t2 )
2
qm=qa− g ht − g chc-s/c
Wnu=
Pu
Azap
Lv=
T − t
Vdu= ( WnuxB)(Lv-d )
Vc=0. 53
√
f'c bd
ØVc ³ Vdu
ØVc = 138,443.78 kg
ØVc > Vdu OK!
Vu = 374,938.21 kg
= 2m +2n
bo = 482.
Vc = 555,426.41 kg Vc= 448,939.43 Kg
ØVc = 472,112.45 kg ØVc= 381,598.52 Kg
ØVc > Vdu OK!
lado menor columna ( b )
m = t + d m = 130.
n = t + b n = 110.
bo = 2m + 2n
Vu = Øvc OK! Vu = 448,939.
Øvc = 381,598.
4.- CALCULO DEL REFUERZO LONGITUDINAL ( Ø = 0.90 )
Dirección Mayor:
Lv = 150.00 cm ree = 7.
Mu = 12,831,081.08 kg-cm 1.
B =
350.
d = 60.
a = 3.89 (Valor Asumido)
As = 57.88 2.
As mín = 0.0018 * B * d a = 3.89 # Varilla ( n ) = 20
As > As mín OK !! 17.
As = 57.88 20 Ø 3/4" @ 17.
Aøb
Aøb # Varilla ( n ) = 15
n -1 As mín = 44.1 15 Ø 3/4" @ 23.
As > As mín
OK !!
Dirección Menor:
B T = 370 ree = 7.
As mín = 0.0018 * B * d B = 350 1.
As > As mín OK !! d = 60.
a = 5.01 (Valor Asumido)
Aøb 2.
Varilla ( n ) = 21
Aøb 17.
As transv = 61.19 21 Ø 3/4" @ 17.
n -1 2.
Varilla ( n ) = 16
Asmin = 46.62 16 Ø 3/4" @ 23.
As transv > As mín OK !!
Longitud de desarrollo en Traccion ( Ld )
< Lv
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 - Øb
Aøb ( 3/4" ) =
Espaciam =
ld = Øb * fy * a * b * g * l
d/
d/
m = t+d
n = b+d
t
b
T
B
bc =
cm
kg
kg
cm
cm
cm
2
cm
2
cm
cm
2
cm 2
cm 2
cm²
cm
cm
cm
2
cm
cm
2
cm
2
cm 2
cm 2
cm 2
cm
Vu=Pu-Wnu x mn
Vc=0. 27
[
c
]
√
f ' c b
o
d
c
Dmayor
Dmenor
c
≤ 2 →Vc=1. 06
√
f ' c b
o
d
Vu ≤ØVc; Ø=0. 85
Mu=
( Wnu x B)Lv
2
As=
Mu
ØFy(d-
a
a =
As. Fy
0. 85f'cb
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 PL
C1 50 50 9 Ø 3/4" 20 15 PL1 PL
C2 40 50 9 Ø 3/4" 40 24
b x t Acero tn tn S/C = 500 kg/m²
h t =
hc =
4.55 m 1.
Df = 1.60 m Otros Columna Zapata
1800 kg/m3 S/C = 500 kg/m² f ' c = ### 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
.004øbfy = 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 cm
hm = 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 Tn
Ps2 = P2D + P2L
Rs = P1s + P2s
20 20 30 30
Rs * Xo = P1s* t1/2 + P2s * 455
Xo = ??
455.00 170.00 cm
Rs = 99 Tn
P1s = 35 Tn
Lz / 2 P2s = 64 Tn
e = Lz / 2 - Xo Rs = 99 Tn
Lz / 6 Xo = 301.21 cm
e < Lz/
Lz = 625.00 cm
Lz Lz Lz / 2 = 312.50 cm
e = 11.29 cm
Lz / 6= 104 cm
g m
g m =
Cálculo del peralte de la zapata (hc )
ld = 0.08 * Øb * Fy
Cálculo de la presión neta del suelo ( qm )
qm = qa - gmhm -γchc - s/c
e Mo = 0
q1,2 = Rs * 1 ± 6 * e
1 2
t1 t
b b
0
Xo
X
X
L = 602 cm
B = 101 cm
B =(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%P1L
Ps2 = P2D + P2L P1s = 27.5 Tn P2s = 64 Tn
R1s = P1s + P2s
20 20 30 30
R1s * Xo = P1s* t1/2 + P2s * 455
Xo = ??
455.00 170.00 cm
e = Xo1 - Xo P1s = 27.5 Tn R1s = 91.5 Tn
P2s = 64 Tn
q = Ps + Mc = Ps + Ps x e R1s = 91.5 Tn
Xo1 = 324.26 cm e = 23.05 cm 1454767.9167 cm
q = 44.56 kg/cm
AUMENTAMOS EL ANCHO B
q < qm cumple " OK "
q > qm no cumple "AUMENTAMOS EL ANCHO B"
Aumentamos el Ancho hc = 150 cm
Un nuevo calculo
Ps = 91500 kg q = 1.41 OK !!
e = 23.05 cm
B = 265 cm
74531250
Ps1 = P1D + P1L
Ps2 = P2D + 50%P2L P1s = 35 Tn P2s = 52 Tn
R2s = P1s + P2s
20 20 30 30
R2s * Xo = P1s* t1/2 + P2s * 455
Xo = ??
470.00 1.40 m
Lz / 2
e2 = Lz / 2 - Xo P1s = 35 Tn R2s = 87 Tn
Lz / 6 P2s = 52 Tn
e2 < Lz/6 R2s = 87 Tn
Xo = 288.97 cm
B e = 12.24 cm 74531250 cm
q > qm no cumple "AUMENTAMOS EL ANCHO B" q = 1.62 kg/cm
OK !!
q < qm cumple " OK "
2.- DISEÑO EN SENTIDO LONGITUDINAL P1s = 35 Tn P2s = 64 Tn
qm
qn = P1s + P2s 0.2 m 4.50 m 140.00 m
Az
qn = 6205.7293 kg/m² ( No amplificada )
Por unidad de Longitud
qm = qm1*B Por Unidad de LONGITUD qm = 6205.73 kg/m
L = 2 x Xo
e Mo = 0
I = ( B x L^3 )/
Az I Az I I =
I =
e Mo = 0
q1,2 = R2s * 1 ± 6 * e2 I = ( B x L^3 )/
I =
Cálculo de la presión neta por unidad de longitud ( qm )
0
Xo
0
Xo
kg/cm
cm
cm
4.- CALCULO DEL REFUERZO EN EL TRAMO CENTRAL
X x qn - P1 = 0
X =
P
X = 5.64 Mu =
qn x X^ _
P1 x X
Mu = -91699.11 kg - m
qn 2
Calculo de As :
Mu = 91699.11 kg - m
As = Mu / ( Ø * fy * ( d - a/2 )) B = 265 r .e .e = 8 db =1.
a = As * fy / ( 0.85 * f 'c * B ) d = 141
a = 2.77 (Valor Asumido)
As = 26.06 2.
As mín = 0.0018 * B * d a = 2.78 # Varilla ( n ) = 9
As > As mín OK !! 31
Aøb As = 26.06 9 Ø 3/4" @ 31
Aøb 2.
Varilla ( n ) = 24
n -1 11
As 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 )^
Mu = 6081.61 kg - m
2
L = 1.91 m 1.
Calculo de As :
Mu = 6081.61 kg - m
As = Mu / ( Ø * fy * ( d - a/2 )) B = 265 r .e .e = 8 db =1.
a = As * fy / ( 0.85 * f 'c * B ) d = 141
a = 0.18 (Valor Asumido)
As = 1.71 2.
As mín = 0.0018 * B * d a = 0.18 # Varilla ( n ) = 1
As > As mín OK !! #DIV/0!
Aøb As = 1.71 1 Ø 3/4" @ #DIV/0!
Aøb 2.
Varilla ( n ) = 25
n -1 10
As 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/
1.205 1.
Aøb ( 3/4" ) =
Espaciam =
Varilla ( n ) = As
Aøb ( 3/4" ) =
Espaciam = B - 2*r.e - Øb
Espaciam =
Aøb ( 3/4" ) =
Espaciam =
Varilla ( n ) = As
Aøb ( 3/4" ) =
Espaciam = B - 2*r.e - Øb
Espaciam =
cm
cm 2
c
m
2
cm
cm
cm
cm 2
cm 2 cm 2
m
m
m
cm 2
cm
cm
cm
cm
cm 2
cm
cm 2
c
m
2
cm
cm
cm 2
cm 2
cm 2 cm 2
cm 2
cm
cm
cm
cm
cm 2
14 Ø 5/8"@ 22cm
ZAPATA EXTERIOR ZAPATA INTERIOR
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/m²
Columna Suelo
f ' c = 210 kg/cm² Df = 1.60 m Lc = 550 cm
b1 = 50 cm 1800 kg/m³ db = 1.91 cm
t1 = 50 cm 2400 kg/m³
b2 = 40 cm qa = 2.00 kg/cm²
t2 = 50 cm db = 1.91 cm
COL SECCION REFUERZO PD (tn) PL (tn)
1 50 50 9 Ø 3/4" 50 30
2 40 50 9 Ø 3/4" 40 24
1.- DIMENSIONAMIENTO DE LA ZAPATA Y VIGA DE CIMENTACION
48.51 cm hc = Ld + r.ee + Øb
Ld = 49 hm = Df - hc
ld >= 0.004 * øb * fy 32.09 cm
Ld = 32
Cálculo del peralte de la zapata (hc )
hc = 59
ht = 101
Cálculo de la presión neta del suelo ( qm )
Az =
Ps
Az =
P1 + P' 2
= 54868.437 cm
qn qn
A = 2 x( b )^2 = 54868.44 b = 166 cm
nuevo e =
b - t
e = 58
P'1 = 9431
2 2 P'2 = 54980
A = 2 x( b )^2 = 54980 b = 166
b= 165 tomamos
t=2b t= 330
g m =
g c =
ZAPATA EXTERIOR:
qn = qa - gmhm - gchc - s/c
CALCULO DE P'2 : Tanteo e =
Tomamos b = 165 cm -----> A = 165 x 330 e = 57 cm
X
Ld
0.08xdbxfy
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 =
cm hc = Ld + 10 =
ht = Df - hc =
kg/cm 2
( PD + PL ) x e
Lc - e
P'2 =
P1xe
L
( 50 + 30 ) x 1000 x 57
X
kg
cm
P'2 =
49 hc = Ld + r.ee + Øb
Ld = 49 cm hm = Df - hc
ld >= 0.004 * øb * fy 32 cm
Ld = 32
Cálculo del peralte de la zapata (hc )
hc = 60
ht = 100
Cálculo de la presión neta del suelo ( qm )
Az =
Ps
Az =
P2 + P' 2
= 36743 cm
qn qn
A = ( b )^2 = 36743.00 b = 192 usamos b = 155 cm
nuevo e =
b - t
e = 71
P'1 = 9486
2 2 P'2 = 44743
A = 2 x( b )^2 = 44743 b = 150
b = 150 A2 = ( 150 x 150 )
VIGAS DE CIMENTACION
h =
Lc
SECCION
b =
P
b = 47
31 x Lc
DISEÑO DE LA VIGA DE CONEXIÓN
9.6 kg/m Mu = P' 2 x L + Wv x ( L^2 )/2= 7250000
Mu = 7250000 kg-cm
ree =
As = Mu / ( Ø * fy * ( d - a/2 )) B = 50
a = As x fy / ( 0.85 x f 'c x b ) d = 79.
As 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.
a = 14.16 # Varilla ( n ) = 11
Aøb 3
As = 26.67 11 Ø 1" @ 3
n -
ZAPATA EXTERIOR:
qn = qa - gmhm - gchc - 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
Ld
0.08xdbxfy
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 =
cm hc = Ld + 10 =
ht = Df - hc =
kg/cm 2
( PD + PL ) x e
Lc - e
P'2 = =
P1xe
L
( 40 + 24 ) x 1000 x 50
kg
cm
P'2 =
cm
cm
h =
x
cm
cm
2
cm
cm
cm 2
cm 2
cm 2
cm 2
cm
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)
L = 100 r .e .e =
B = 150 r .e .e = 7.5 B = 250
d = 41 db = 1.59 d = 51
a = 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.
As 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 Exterior
Para 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 = 2500
Pu < ( Ø * 0.85 * f 'c * A1) Ø * 0.85 * f 'c * A1 = 312375
As mín = 0.005 * A
Pu < Ø * 0.85 * f 'c * A1 OK !!
Aøb
As col. > As mín OK !! As mín = 12.
Aøb ( 3/4" ) = 2.
USAR : As1 = 12.
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 = 2000
A 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.
As col. > As mín OK !! As col > As min OK !!
Longitud de desarrollo en Traccion ( Ld )
< Lv
Øb
Lv1 = Lv - r.e.e
Aøb ( 5/8" ) =
Espaciam =
Varilla = As
Varilla = As
ld = Øb * fy * a * b * g * l
3.54 * f 'c^.5 * C + Kr
cm 2
cm
cm
cm
cm 2
cm 2
cm
cm 2
cm 2
cm
cm
cm
cm 2
cm 2
cm
cm
cm 2
cm
cm 2
kg
kg
cm
cm
cm
cm
kg
kg
cm
cm
La Zapata es rectangular se debe compartir el Refuerzo adecuadamente de la siguiente manera:
1.00 Øb (5/8") = 1.59 8.
0.80 r.e.e = 7.50 ktr = 0
1.00 fy = 4200
Lado menor Zapata 1.00 f'c = 175
q = ( C+kt r )/ Øb
Aøb
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 cm
As1 = OK !!
3 x h 0 cm
d/
d/
Asc = 2 * Astrv
( b + 1 ) b = C =
g =
b = Lado mayor Zapata l =
a =
Varilla ( n ) = As
Espaciam = B - 2*r.e.e - Øb
cm
cm
12 Ø 5/8"@ 12cm
14 Ø 5/8"@ 22cm
6 Ø 1"@ 7cm
14 Ø 5/8"@ 18cm
kg
#DIV/0!
( Valor Asumido )
OK !!
cm
kg
cm
cm
cm
q = ( C+kt r )/ Øb
q >= 2.5 ,PONER 2.5 !!
q < 2.5 ,PONER q !!
1.75 5.401 Lc= 5.
Lc= 4.
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.
Iy= 1274.731733 Iy= 1274.
ENTONCES:
q= 1286500 + - 1286500x + - 5146000y
q= 1.132 + - 0.00001x + - 0.00006y
Pu/A = 1.1317 Pu/A = 1.
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.
B 1.132 -290 0.003 490 0.029 1.
C 1.132 250 -0.003 490 0.029 1.
D 1.132 580 -0.006 490 0.029 1.
E 1.132 -580 0.006 245 0.015 1.
F 1.132 580 -0.006 245 0.015 1.
G 1.132 -580 0.006 -205 -0.012 1.
H 1.132 580 -0.006 -205 -0.012 1.
I 1.132 -580 0.006 -490 -0.029 1.
J 1.132 -290 0.003 -490 -0.029 1.
K 1.132 250 -0.003 -490 -0.029 1.
L 1.132 580 -0.006 -490 -0.029 1.
l
1.75/ l <Lc
m
2
qn=qa - gt * ht - gc * hc - S/C
qn= kg/cm
2
X'= SPiXi ex= X' - B/
Pi
Y'= SP
i
Y
i
Pi
ey= Y' - L/
BL
3
/12 = m
4
LB
3
/12 = m
4
11
10
q(kg/cm
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/
m = 40 + d
Vu= 1.4100000 +1.740000 -1.167(40+d)(40+d)
Vu=
Igualamos :
Vc=1.1√210*4(40+d)
COLUMNA 2
Vu= 1.460000 +1.730000 -1.167(40+d)(40+d/2) m = 40 +d
COLUMNA 1
Vc=1.1√210*(120+2d)d
Igualamos n=40 +d/
m = 40 +d/
Vu= 1.445000 +1.725000 -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
E=15000*√F'c
Vu = Pu - (q
nu
* m * n)
Vc=1.1√F'c*b
o
*d
Vu = fVc
fVc=
Vu = Pu - (qnu * m * n)
Vc=1.1√F'c*b
o
*d Vu = Pu - (q
nu
* m * n)
fVc=
Vu=fVc
Vc=1.1√F'cbod
fVc= 1083.95d +13.55d
2
Vu=fVc
13.83d
2
l= 4 3Ks
Et
3
l= m
l= m
n = 40 +d/
