Programa Examen Final Dinamica Estructural
46
0.50 2.95 0.50 2.95 0.50 3.60 0.50 3.10 0.50 3.10 0.50 E A L IB R E 20% TE V P (0.30x0.30) V P (0.30x0.30) V P (0.30x0.30) V P (0.30x0.30) V P (0.30x0.30) V P (0.30x0.30) V S (0.20x0.30) V S (0.20x0.30) V S (0.20x0.30) V S (0.20x0.30) V S (0.20x0.30) V S (0.20x0.30)
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Transcript of Programa Examen Final Dinamica Estructural
0.50 2.95 0.50 2.95 0.50 3.60
0.50
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VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
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VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
3.25 3.25SOLUCION :
= CORDOVA = 7
= MENDOZA = 7
L3 = Primer nombre (mt) = DERLY = 5
AREA TOTAL = 247 m2
AREA LIBRE 20% = 0.2 54
AREA TECHADA = At – Al= 193
USO: S/C 0.2 Tn/m2
L1 = Apellido materno (mt) L2 = Apellido paterno (mt)
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VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
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1.- PREDIMENSIONAMIENTO DE VIGAS
PERALTE =
L = 4.25 = 0.3
El peralte uniformizamos en ambas direcciones: 0.3
VIGAS DE CIMENTACION
PERALTE =
L = 4.25 0.43El peralte uniformizamos en ambas direcciones: 0.5
fc' = 210
E = 2173706.5119 Tn/m2
2.- CALCULO DE INERCIA DE COLUMNASy 0.50
0.200.50
0.30 1 x 2
0.10 0.30 0.10 0.30
Area (m2) = 0.190 Area (m2) = 0.190Ix (m4) = 0.0040 Ix (m4) = 0.0040Iy (m4) = 0.0028 Iy (m4) = 0.0032
# de columnas= 6.0000 # de columnas= 5.0000
DATOS ADICIONALES:LOSA tabiqueria: 0.15
t W (Tn/m2) ultimo nivel: 0.6*0.15*Atechada0.17 0.280.20 0.30 Acabado: 0.10.25 0.35 ultimo nivel: 0.6*0.1*Atechada0.30 0.42
C°A°: 2.4S/C:Categoria:A,B: 0.5*S/CC: 0.25*S/CAZOTEA: 0.25*S/CDeposito: 0.8*Walmac.
3.- ALTURA PARA CALCULO DE RIGIDEZ
E = 15000√fc’
Df min=1.00 mTalon de zapata=0.60m (min)
VIGA
VIGA.CIM.
14L
12
L
3.00
3.00
con v.c: 3.45sin v.c: 3.95
Df= 1.4
hi= 9.45
4.- METRADOS
PRIMER PISO
VIGA X-X = VIGA CHATA
VIGA X-X = 6.3792 Tn
VIGA Y-Y = VIGA PRINCIPAL
VIGA Y-Y = 14.3640 Tn
ALIGERADO =
Df min=1.00 mTalon de zapata=0.60m (min)
VIGA
VIGA.CIM.
4.030.0122.5159.2387.414.213808.354.34.2 Xxxxxxxx
4.030.0122.5126.5168.4162.435.465.41044.2 Xxxxxxxx
6.34.240.0/21.304.24.0/)87.2894.19(16.3213.0 xxx
ALIGERADO = 48.1800 Tn
COLUMNAS =
COLUMNAS = 26.4942 Tn
TABIQUERIA =
TABIQUERIA = 28.9500 Tn
ACABADOS =
ACABADOS = 19.3000
S / C = Categoria C=
S / C = 9.6500 Tn
W1º = 153.3174
SEGUNDO PISO (Si se repite del primer piso)
VIGA X-X = 6.3792 TnVIGA Y-Y = 14.3640 TnALIGERADO = 48.1800 Tn
COLUMNAS =
COLUMNAS = 25.8480 TnTABIQUERIA = 28.9500 TnACABADOS = 19.3000 TnS / C = 9.6500 Tn
W2º = 152.6712 Tn
TERCER PISO
VIGA X-X = 6.3792 TnVIGA Y-Y = 14.3640 TnALIGERADO = 48.1800 Tn
COLUMNAS =
COLUMNAS = 12.9240 TnTABIQUERIA = 17.3700 Tn
m1 =
m2 =
)60.3189.060.3188.0)60.3(2640.030.0(4.2 xxxx
24.15815.0
24.15810.0
24.15820.0
)20.3189.020.3188.0)20.3(2640.030.0(4.2 xxxx
)60.1189.060.1188.0)60.1(2640.030.0(4.2 xxxx
ACABADOS = 11.5800 TnS / C = 4.8250 Tn
W3º = 115.62 Tn
5.- VERIFICAMOS EN PRIMERA INSTANCIA EL SISTEMA ESTRUCTURAL SIN PLACAS:Ixx= 0.0749 m4Iyy= 0.0440 m4E= 2173706.5119 Tn/m2h1= 3.45 mh2= 3.00 mh3= 3.00 m
27924.3722 Tn/m PRIMER PISO
47582.7914 Tn/m
42469.479529 Tn/m SEGUNDO PISO
72367.477869 Tn/m
42469.479529 Tn/m TERCER PISO
72367.477869 Tn/m
6.- IDEALIZANDO LA ESTRUCTURA TENEMOS: EN LA MAS CRITICA, EN ESTE CASO LA DIRECCION X
11.7982 Tn-seg2/m
K3= 42469.4795 Tn/m
15.5787
K2= 42469.4795 Tn/m
15.6446
K1= 27924.3722 Tn/m
7.- CALCULO DE LA MATRIZ K y m
70393.8517 -42469.4795 0.0000
m3=
K1X =K1Y =
K2X =K2Y =
K3X =K3Y =
m1
m3
m2
K = -42469.4795 84938.9591 -42469.47950.0000 -42469.4795 42469.4795
15.6446 0.0000 0.0000m = 0.0000 15.5787 0.0000
0.0000 0.0000 11.7982
Operando tenemos: (W1)2= 474.1416
T10.2885531621 <
W2 y W3 no tienen solucion real por lo tanto no se puede realizar la verificacion con ellos. Nos daria entender que tenemos que colocar placas.
8.- CONSIDERANDO PLACAS DE:
0.15 NUMERO DE PLACAS =0.90
0.90
NUMERO DE PLACAS =
0.15
9.- METRADO DE PLACAS
PRIMER PISO h = 2.78PESO DE PLACAS = 10.79 Tn
SEGUNDO PISO h = 2.70PESO DE PLACAS = 7.19 Tn
TERCER PISO h = 1.50PESO DE PLACAS = 5.83 Tn
10.- PESO Y MASA TOTAL
WT1º = 164.1066 TnWT2º = 159.8640 TnWT3º = 121.4542 Tn
11.- Sumamos las inercias según el numero del tipo de columnas q existen en el plano
INERCIA TOTAL Ixx = 0.1311 m4INERCIA TOTAL Iyy = 0.1002 m4
E = 2173706.511928 Tn/m2Altura de columnas de entrepiso
H1 = 3.45 m (desde la parte sup.de la zapata hasta el eje de la viga de 1°nivel)H2 = 3.00 m (desde el eje de la viga 1° hasta eje de la viga 2°nivel)H3 = 3.00 m (desde el eje de la viga 2° hasta eje de la viga ultimo nivel)
0.15 X 1.00
63619.8239 Tn/m
83278.2431 Tn/m
96757.799664 Tn/m
126655.798 Tn/m
96757.799664 Tn/m
126655.798 Tn/m
Analizando vemos que la menor rigidez esta en el Eje X: 12.- Datos finales
63619.8239 Tn/m 16.7456
96757.7997 16.3127
96757.7997 12.3933
IDEALIZANDO LA ESTRUCTURA TENEMOS
12.3933 Tn-seg2/m
K3 = 96757.7997 Tn/m
16.3127
K2. = 96757.7997 Tn/m
16.7456
K1 = 63619.8239 Tn/m
13.- CALCULO DE LA MATRIZ K y m
160377.6236 -96757.7997 0.0000K = -96757.7997 193515.5993 -96757.7997
0.0000 -96757.7997 96757.7997
16.7456 0.0000 0.0000m = 0.0000 16.3127 0.0000
K1X =K1Y =
K2X =K2Y =
K3X =K3Y =
K1 = m1 =
K2. = m2 =
K3 = m3 =
m1
m3
m2
0.0000 0.0000 12.3933
14.- HALLAMOS LAS FRECUENCIAS W2:
-3385.42756 99015109.41654 W^4 + -678516685965.64400
(W1)2 = 1026.06101
(W2)2 = 8852.7690363
(W3)2 = 19368.6158557
COMPROBACION:143195.61630 -96757.79966 0
{K}-w2{M}= -96757.79966 176777.77388 -96757.79966 *0 -96757.79966 84041.51774
12776.56989387
15.- VERIFICAMOS POR PERIODOALTURA TOTAL DEL EDIFICIO = 9.45
PERIODO ADMISIBLE = H / 35
PERIODO ADMISIBLE = H / 45
T ADM
0.196 < 0.21
0.067 < 0.21
0.045 < 0.21
16.- CALCULO DE MODOS NORMALESCON (W1)2:
143195.6163 -96757.7997 0{K}-w2{M}= -96757.7997 176777.7739 -96757.7997 *
0 -96757.7997 84041.5177
CON (W2)2:12132.6944 -96757.7997 0
{K}-w2{M}= -96757.7997 49103.0339 -96757.7997 *0 -96757.7997 -12957.2228
CON (W3)2:-163961.4701 -96757.7997 0
{K}-w2{M}= -96757.7997 -122438.8205 -96757.7997 *0 -96757.7997 -143283.2672
W^6 +
W1 =W2 =W3 =
T1 =T2 =T3 =
02 MWK
WT
2
1.0000 1.0000 1.0000
1.4799 0.1254 -1.69461.7039 -0.9364 1.1443
17.- CALCULO DE MODOS NORMALIZADOSCORREGIDO:
0.1063 0.18940.1574 0.02380.1812 -0.1774
0.1063 0.1894 0.11190.1574 0.0238 -0.18970.1812 -0.1774 0.1281
18..- CALCULO DE
19.- ACELERACION APLICADO EN LA BASE: Por condicion del problema usamos este grafico
a g0.3 9.8
0.3
2.94 F(t)
gst = a*g / (W)2
0.0028653267 0.3g/w1
0.0003320995 0.3g/w2
0.000151792 0.3g/w3
Con estos valores nos vamos al abaco0.3059 FAD = 0.94
0.8985 FAD = 1.48
I a I =
l Ф I =
l Ф I =
0.30 g
0.03 0.06
).( 2
1
ki
n
k
k
ijij
am
a
31.321.211.11 mmm
32.322.212.12 mmm
33.323.213.12 mmm
sY
1stg2stg3stg
1Ttd
2Ttd
sY
1.3290 FAD = 1.32
0.002693
0.000492
0.000200
20.- HALLANDO EL DESPLAZAMIENTO CON RESPECTO A LA BASE
0.0017936219 m 0.18
0.0026530404 m 0.27
0.0030544052 m 0.31
1° forma
Según RNE E-030 Rd:
Porticos8
DualH1 = 3.45 m 7H2 = 3.00 mH3 = 3.00 m
0.0027 < 0.007 OK
0.0046 < 0.007 OK
0.0053 < 0.007 OK
Por lo tanto se usaran:
EJE X: 0.15 # PLACAS =0.90
0.90
EJE Y: # PLACAS =
0.15
PLANO DE DISTRIBUCION DE PLACAS:
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3Ttd
stg
gFAD
max
max1g
max2gmax3g
1maxu2maxu3maxu
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007.075.0 dxx R
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VS (0.20x0.30)
VS (0.20x0.30)
21.- HALLANDO CUANDO NOS DA SIN DESPLAZAMIENTO EN LA BASE
22.- FUERZAS APLICADAS AL PORTICO
F1(T)= f1(1-T/TD) f1= 10F2(T)= f2(1-T/TD) f2= 20F3(T)= f3(1-T/TD) f3= 30
23.- APLICACION DE LOS VALORES DESACOPLADOS
Z"1+W1^2*Z1=Ф11f1(t)+Ф21f2(t)+Ф31f3(t)= 8.2098 f(t)Z"2+W2^2*Z2=Ф12f1(t)+Ф22f2(t)+Ф32f3(t)= 3.8597 f(t)Z"3+W3^2*Z3=Ф13f1(t)+Ф23f2(t)+Ф33f3(t)= 5.4350 f(t)
24.- VALORES MAXIMOS
T1=2*3.1416/W1= 0.1961522418 Td/T1= 0.305884854839541T2=2*3.1416/W2= 0.06677906 Td/T2= 0.898485243920644T3=2*3.1416/W3= 0.0451471745 Td/T3= 1.328986822964
25.- DESPLAZAMIENTOS ABSOLUTOSY1MAX=((Ф11Z1MAX)^2+(Ф12Z2MAX)^2+(Ф13Z3MAX)^2)^(1/2) 0.000862011889122866Y2MAX=((Ф21Z1MAX)^2+(Ф22Z2MAX)^2+(Ф23Z3MAX)^2)^(1/2) 0.00126364157845477Y3MAX=((Ф31Z1MAX)^2+(Ф32Z2MAX)^2+(Ф33Z3MAX)^2)^(1/2) 0.00145597066226532
0.0008620119
0.0012636416
0.0014559707
td =
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0.50
11.0
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LIB
RE
20%
FRENTE
FR
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TE
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VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
Yst
0.17
0.24
0.28
25.- VERIFICACIONSegún RNE E-030 Rd:
Porticos8
Dual7
H1= 3.45H2= 3.00H3= 3.00
0.001311757223 < 0.007
0.002211372762 < 0.007
0.002547948659 < 0.007
Yst
0.17
0.24
0.28
007.075.0 dxx R
H
1
2
2
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0.50
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(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
3.3
m LARGO=L1+L2+L3= 19 DEBEMOS HACER QUE SE SEA EN LA DIRECCION VERTICAL
m ANCHO= L1+6 = 13 DEBE SER LA DIRECCION HORIZANTAL
m2
m2
Departamento
0.50 2.95 0.50 2.95 0.50 3.60
0.50
3.10
0.50
3.10
0.50
3.00
0.50
3.05
0.50
2.75
0.50
0.50 2.95 0.50 2.95 0.50 3.00 0.50
0.50
2.75
0.50
2.75
0.50
11.0
0
AR
EA
LIB
RE
20%
FR
EN
TE
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
VIGAS PRINCIPAL: 0.30X0.30m0.30 0.3
de acuerdo a la columna para que encaje
VIGA SECUNDARIA:0.20 0.3
de acuerdo a la columna para que encaje
VIGA DE CIMENTACION:0.30X0.30mEn ambas direcciones
0.3 0.5de acuerdo a la columna para que encaje
Kg. /cm2
0.200.50
3
0.20 0.30
Area (m2) = 0.150Ix (m4) = 0.0031Iy (m4) = 0.0011
# de columnas= 10.0000
Tn/m2
TTn/m2 ESPESOR ESCOGIDO 0.20
Tn/m2
Altura para metrado
Df min=1.00 mTalon de zapata=0.60m (min)
VIGA
VIGA.CIM.
1.50
3.00
1.50
1.50
3.00
1.50
1.58
altura de entrepiso3.15
0.5
0.6
Ingresar 4 decimales
Df min=1.00 mTalon de zapata=0.60m (min)
VIGA
VIGA.CIM.
4.030.0122.5159.2387.414.213808.354.34.2 Xxxxxxxx
4.030.0122.5126.5168.4162.435.465.41044.2 Xxxxxxxx
6.34.240.0/21.304.24.0/)87.2894.19(16.3213.0 xxx
considerar el nudo para la col y no para la viga.
0.25
15.6446
15.5787
Influencia en el ultimo nivel
0.6
11.7982 Tn-seg2/m
5.- VERIFICAMOS EN PRIMERA INSTANCIA EL SISTEMA ESTRUCTURAL SIN PLACAS:
EN LA MAS CRITICA, EN ESTE CASO LA DIRECCION X
70393.8517 -42469.4795K = -42469.4795 84938.9591
0.0000 -42469.4795
15.6446 0.0000m = 0.0000 15.5787
0.0000 0.0000
-2875.48715 38967075.82360W^6 +
02 MWK
(W2)2=
W T21.7747927659484 0.288544651953033 (W3)2=
Tadmisible0.27 NO CUMPLE .DISEÑAR PLACAS
W2 y W3 no tienen solucion real por lo tanto no se puede realizar la verificacion con ellos. Nos daria entender
INERCIA TOTAL DE LAS PLACASIx= 0.00152 m4
6 Iy= 0.05468 m4
INERCIA TOTAL DE LAS PLACASIx= 0.05468 m4
6 Iy= 0.00152 m4
m
m
m
g =9.8 m/s2
16.7456 Tn-s2/m16.3127 Tn-s2/m12.3933 Tn-s2/m
11.- Sumamos las inercias según el numero del tipo de columnas q existen en el plano
(desde la parte sup.de la zapata hasta el eje de la viga de 1°nivel)(desde el eje de la viga 1° hasta eje de la viga 2°nivel)(desde el eje de la viga 2° hasta eje de la viga ultimo nivel)
m1 =m2 =m3 =
3
12HEI
K
Tn-seg2/m
W^2 + 595613358846280.00000 = 0
32.0322
94.0892
139.1712
a11 0.0000a21 = 0.0000a31 0.0000
m
0.27 SIN PLACAS
0.21 CON PLACAS
OK
OK
OK
a11 0.0000a21 = 0.0000a31 0.0000
a12 0.0000a22 = 0.0000a32 0.0000
a13 0.0000a23 = 0.0000a33 0.0000
9.40498650335
0.1119 5.27904227746-0.18970.1281 8.93400067094
-6.2578
-0.6718
-1.7277
Por condicion del problema usamos este grafico
0.06 es el ultimo T del grafico de pulsotd =
=
=
=
0.30 g
0.03 0.06
gY s 30.0
cm
cm
cm
2° formasegún NORMA E-030 para estructuras de C°A°lim. Desplaz. Lateral (cm) de entre piso:
0.18 < 2.410.27 < 2.100.31 < 2.10
0.31
0.27
6
0.18
6
0.50 2.95 0.50 2.95 0.50 3.60
0.50
3.10
0.50
3.10
0.50
3.00
0.50
3.05
0.50
2.75
0.50
0.50 2.95 0.50 2.95 0.50 3.00 0.50
0.50
2.75
0.50
2.75
0.50
11.0
0
AR
EA
LIB
RE
20%
FRENTE
FR
EN
TE
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
Yst
0.17
0.24
0.28
0.50 2.95 0.50 2.95 0.50 3.60
0.50
3.10
0.50
3.10
0.50
3.00
0.50
3.05
0.50
2.75
0.50
0.50 2.95 0.50 2.95 0.50 3.00 0.50
0.50
2.75
0.50
2.75
0.50
11.0
0
AR
EA
LIB
RE
20%
FRENTE
FR
EN
TE
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
Yst
0.17
0.24
0.28
0.06 es el ultimo T del grafico de pulsoDEL ABACO fdmax1= 1 z1max=fdmax1*f01/(w1^2)=DEL ABACO fdmax2= 1.5 z2max=fdmax2*f02/(w2^2)=DEL ABACO fdmax3= 2 z3max=fdmax3*f03/(w3^2)=
0.50 2.95 0.50 2.95 0.50 3.60
0.50
3.10
0.50
3.10
0.50
3.00
0.50
3.05
0.50
2.75
0.50
0.50 2.95 0.50 2.95 0.50 3.00 0.50
0.50
2.75
0.50
2.75
0.50
11.0
0
AR
EA
LIB
RE
20%
FRENTE
FR
EN
TE
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VP
(0.
30x0
.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30) VS (0.20x0.30) VS (0.20x0.30)
VS (0.20x0.30)
VS (0.20x0.30)
OK
OK
OK
DEBEMOS HACER QUE SE SEA EN LA DIRECCION VERTICAL
DEBE SER LA DIRECCION HORIZANTAL
1 y" x"0.28157895 0
2 y" x"0.28157895 0.202631578947368
3 Y" X"0 0
W0.30
3.00
3.08
ALTURA LIBRE DE ENTREPISO = 3
SE CAMBIA DEPENDIENDO , ES IGUAL CASI SIEMPRE
VIGA CHATA
DATOS fy= 4200 CANTIDAD
f`c= 210 4
C.A 2.4 4
3
2
VIGA PRINCIPAL 1
CANTIDAD LONGITUD 2
6 4
10 4.25
0 3
0 3.1
0.0000-42469.479542469.4795
0.00000.0000
11.7982
W^4 + -124055081859.70900 W^2 + 50365980725437.90000 =
W T
02 MWK
4127.782988 64.2478248 0.097793194126241
T2 Tadmisible0.097796078 < 0.27 OK
W T8949.544703 94.6020333 0.0664150629884064
T3 Tadmisible0.066417022 < 0.27 OK
11.47991.7039
10.1254
-0.9364
1-1.69461.1443
0.007
okokok
cm
cm
cm
z1max=fdmax1*f01/(w1^2)= 0.00800126869653213z2max=fdmax2*f02/(w2^2)= 0.000653985633877897z3max=fdmax3*f03/(w3^2)= 0.000561216361797465
0.2
0.25
0.30.35
0.40.5
VIGA CHATA
LONGITUD
2.8
2.45
3.1
3
2.9
2.55
0
