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PRÁCTICA CALIFICADA 3TRABAJO
ALUMNO : CESAR ZAPATEL
CURSO :
ANÁLISIS ESTRUCTURAL I
DOCENTE :
CICLO :2014-
I
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2014- I
CHICLAYO, 28 DE MAYO DEL 2014
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PRÁCTICA CALIFICADA 3TRABAJO
ANÁLISIS ESTRUCTURAL I
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CHICLAYO, 28 DE MAYO DEL 2014
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PRACTICA CALIFICADA Nº3 2014 –1CURSO ANALISIS ESTRUCTURAL 1 SECCIÓN A
Ing Cesar Zapatel Soraluz DURACIÓN 120MINCICLO VII
ESCUELA Ingenieria Civil FECHA 5/24/2014FECHA DE ENTREGA 21-May-14 FECHA EXPOSICION 24-May-14Se permite cualquier elemento de consulta, Software previa autorizacion
TRABAJO ESCALONADO PREGUNTA No 1.-(10 ptos)La fig es una planta tipica de un techo aligerado DESTINADO A OFICINAS(S/C=250kG/M2)considerar tabiqueria flotante 100 kg/cm2
Col 0,40x0,60mAZOTEA existe solo parapeto en su perimetro, S/C=100kg/m2
Parapeto h=1,20m2,1m V-B(0,25X50) V-B(0,25X50)
Aducto
SH muro, h=2L(a ejes)= 3m
L1 1pta 1m
V-1(30X60) V-1(30X60)V-1(30X60)
V-B(0,25X50) V-B(0,25X50) V-B(0,25X50)B
L(a ejes)=
L2 2
V-1(30X60) V-1(30X60) V-1(30X60)
CL(a ejes)= L(a ejes)=
4.5 51 V-C(30X50) . 2 V-C(30X50) 3
EVALUACIÓN
SEM. ACADE.
PROFESOR (ES)
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EDIFICIO DE 2 NIVELES N+3,6, N+6,60Fig 1.-PLANTA TECHO ALIGERADO H=20CM N+3,6
Se solicita1,-Determinar las cargas por gravedad del Port Eje 22,-Determinar el metrado de Cargas de la Columna B-23,-Determinar el metrado de Cargas del Port Eje 3 4,-Determinar el metrado de Cargas de la Columna B-3Realizar el DCL de cada uno indicando los Coeficiente de rigidez,
PREGUNTA No 2.-(10 ptos)La fig representa el Portico del Eje 2, de la pregunta anterior. Se solicitaa) Realizar el analisis Estructural para las Fzas horizontalesb) Resolver el Portico del Eje 2 según los combos de la norma E-060
15t
3.0 m7t
3.6
L2 L1
Fig.-Elevacion Portico Eje 2 Edificio 2 niveles
Alumno L1(m) L2(m) Alumno L1(m) L2(m)1 6 5.5 13 7.1 6.452 6.1 5.65 14 7.25 6.63 6.2 5.8 15 7.35 6.754 6.3 5.95 16 7.25 6.95 6.4 6.1 17 7.15 7.056 6.5 6.25 18 7.05 7.27 6.6 6.4 19 6.95 7.358 6.7 6.55 20 6.85 7.259 6.8 6.7 21 6.75 7.15
10 6.9 6.85 22 6.65 7.0511 7 7 23 6.55 6.9512 7.1 7.15 24 6.45 6.85
Se puede cambiar las dimensiones de los elementos estructurales si no cumple con el desplazamiento max Smax=h*(7/1000)Los desplazamientos por muto multiplicar por Factor=0,75R=0,75*8=6
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4060
V-B(0,25X50)h=1,20m 1m
DescansoVC(0,15*0,20)muroh=1,2m
V-B(0,25X50)
h=1,20m1.15
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N+6,6
N+3,6
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TRABAJO ESCALONADO PREGUNTA No 1.-(10 ptos)La fig es una planta tipica de un techo aligerado DESTINADO A OFICINAS(S/C=250kG/M2)considerar tabiqueria flotante 100 kg/cm2 40
Col 0,40x0,60m 60AZOTEA existe solo parapeto en su perimetro, S/C=100kg/m2
Parapeto h=1,20m2.1 V-B(0,25X50) V-B(0,25X50)
A h=1,20m 1mducto
SH muro, h=2L(a ejes)= 3
L1 16.7 pta 1m w1 Descanso
V-1(30X60) V-1(30X60) VC(0,15*0,20)V-1(30X60) muro
h=1,2m
B2 B3V-B(0,25X50) V-B(0,25X50) V-B(0,25X50)
B
L(a ejes)=L2
6.55 2
V-1(30X60) V-1(30X60) V-1(30X60)
CL(a ejes)= L(a ejes)= h= 1.2 m
4.5 5 1.151 V-C(30X50) 2 V-C(30X50) 3
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1) DETERMINACIÓN LAS CARGAS POR GRAVEDAD DEL PORTICO EJE 2
EJE 2-V2(030X0,60) atat1 = 4.75at2 = 4.45
0.36.55 6.7
C B AViga Longitudinal 2.25 2.5
Secc Viga
1 NIVEL :
Carga permanente Rectangular longitud total de la viga
Kg/m2 at(m) ParcialP.P. T.A 300 4.45 1335 kg/mP. Enlucido 100 4.75 475 kg/mP.P. de la viga 2400 0.3 0.6 432 kg/mTabiqueria mobil 100 4.45 445 kg/m
total 2687 kg/m2.687 Tn/m
Carga permanente Rectangular longitud PARCIAL de la vigamuros paralelos viga
W Kg/m2 Altura(m) Kg/m %Weje1 %Weje2 W=SumWw1=14*15= 210 2 420 224 196 420
%Weje1=(P/L)*b%W1 w1 %W1 L(m) %Weje2=(P/L)*a
WD =
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3
a b2.1 2.4
L(m) 4.5
Ejes 1 2
420
2.1 2.44.5
224 196 t/meje 1 eje 2
Carga permanente Rectangular longitud PARCIAL de la vigamuros perpendiculares -Viga Chata
kg/m3,kg/m2 b hpeso de viga chata 2400 0.25 0.2 120 kg/m
PESO MURO 210 2 420 kg/m
CARGA SOBRE EJE 2 de la VCH0.462
1.55 2.95
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0.42 tn/m0.12 tn/m1.1
L(m) 1 2.44.5
0.27 0.270.30 0.16
0.57 0.43 Teje 1 eje 2
Sobrecargakg/m2 at total
OFICINA 250 4.75 1187.5 kg/mWL= 1.1875 tn/m
2 NIVEL : Carga permanente Rectangular longitd de la viga
Kg/m2 at(m) ParcialP.P. T.A 300 4.45 1335 kg/mP. Enlucido 100 4.75 475 kg/mP.P. de la viga 2400 0.3 0.6 432 kg/m
total 2242 kg/m2.242 Tn/m
SOBRECARGA OFICINA Carga permanente Rectangular longitd de la viga
kg/m2 at total100 4.75 475 kg/m
WD =
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WL= 0.475 tn/m
Fig.- PORTICO EJE 22.24 tn/m
0.43 t3m
0.196 tn/mD 2.69 tn/m
3m 3.6m
6.55 m 6.7 C B A
0.48 tn/m3m
L 1.19 tn/m3.6m
6.55 m 6.7 mC B A
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2) Metrado de cargas columna B2 :
ANCHO V2TRIBUTARIO 3.05
31.47 m2 2.056.625
VB 2.32.975
4.75
Elem L1 L2 A(m2) VB 0.3 0.6Col 4.75 6.63 31.47 V2 0.25 0.5V2 6.03 1.81 COL 0.4 0.6VB 4.35 1.09Techo 28.57
2° NIVEL:Aligerado 28.57 300.00 8572.13 kgAcabado 31.47 100.00 3146.88 kgV2 2400.00 0.30 0.60 6.03 2602.80 kgVB 2400.00 0.30 0.60 4.35 1879.20 kgParapetos 0.00 0.00tanqueElev 0.00 0.00Columna 2400.00 0.40 0.60 3.00 1728.00 kgSobrecarga 100.00 31.47 3146.88 kg
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PD = 17929.0 kgPL = 3146.9 kg
1° NIVEL:Aligerado 28.57 300 8572.13 kgAcabado 31.47 100 3146.88 kgTab Movil 28.57 100 2857.38 kgV2 2400.00 0.30 0.60 6.03 2602.80 kgVB 2400.00 0.30 0.60 4.35 1879.20 kgVCH 192.15 kgW1 131.64 kgColumna 2400 0.4 0.6 3.6 2073.60 kg
Sobrecarga 250 31.46875 7867.1875 kg
Total pesosPD = 21455.77 kg PD 17929.00 3146.875 PLPL = 7867.19 kg 21455.766 7867.19
0.59 t 39384.77 11014.065.2 1.5 Total: 50398.83 kg
0.43 t AT =A1+A2 62.94 m2
0.20 Peso/m2 0.80 tn/m2
3.7 3
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VCH 0.19 0.24W 0.13 0.46
0.32 0.69
6.7B A
DCL EJE 2
3) METRADO CARGAS DEL PORTICO EJE 3
EJE 3at1 = 3.65at2 = 3.25
6.55 6.7 0.4
1 NIVEL : 2.5 1.15
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a) Carga permanente
Kg/m2 at(m) ParcialP.P. T.A 300 3.25 975 kg/mP. Enlucido 100 3.65 365 kg/mP.P. de la viga 2400 0.3 0.6 432 kg/mTabiqueria mobil 100 3.25 325 kg/m
total 2097 kg/m2.097 Tn/m
b) Sobrecargakg/m2 at total
oficina 250 3.65 912.5 kg/mWL= 0.9125 tn/m
2 NIVEL: CARGA PERMANENTE
Kg/m2 at(m) ParcialP.P. T.A 300 3.25 975 kg/mP. Enlucido 100 3.65 365 kg/mP.P. de la viga 2400 0.3 0.6 432 kg/m
total 1772 kg/m1.772 Tn/m
SOBRECARGA
kg/m2 at total peso de parapeto + Vc(15x20)100 3.65 365 kg/m 0.354 tn/m
WD =
WD =
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WL= 0.365 tn/m 2mC B A
w= 0.772 6.55 M 6.7 mw = 0.460 w = 0.460
METRADO DE CARGAS DE ESCALERA
PESO Wpp= [(1.2t+0.5c/p)x1m] ϒ t (m) = 0.15ESCALERA Wpp= 672 c/p (m) = 0.19
Enlucido 100 ϒ = 24000.312 tn/m 772 kg/m 0.772 tn/m
Actua sobre eje 0.46 tn/m4.15m PESO wpp= 2400kg/m3(0.15x1)= 360kg +100(enlucido)
1.15m 1m DESCANSO Wpp= 460kg/m0.460 tn/m 1.26tn 1.28tn Wpp= 0.46 tn/m
0.354 tn/m1.772 tn/m
0.96tn/m 3mWD 2.451 tn/m
3.6m
6.55 6.7C B A
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0.350tn/m
3mWL 0.9125 tn/m
3.6m
6.55 m 6.7 mC B A
4) Metrado de cargas columna B3 :
ANCHOTRIBUTARIO 3.05
24.18125 m2 2.36.625
0.952.975
3.65
Aligerado : 0.2375 0.575 0.915 0.8925Suma = 2.62 m2
ALIGERADO 21.56125 m2
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1° NIVEL:Aligerado 21.56125 300 6468.375 kgAcabado 24.18125 100 2418.125 kgViga 1(0.3x0.6 2400 0.3 0.6 3.05 1317.6 kgViga 2(0.3x0.6 2400 0.3 0.6 2.975 1285.2 kgVigaB(0.25x0. 2400 0.25 0.5 0.95 285 kgVigaB(0.25x0. 2400 0.25 0.5 2.3 690 kgVC(0.15x0.20) 2400 0.15 0.2 6.375 459 kgCrecta 2400 0.4 0.6 3.6 2073.6 kgTab. Móvil 100 21.56125 2156.125 kgSobrecarga 250 24.18125 6045.3125 kg
PD = 17153.025 kgPLD = 6045.3125 kg
NIVEL 2Aligerado 21.56125 300 6468.375 kgAcabado 24.18125 100 2418.125 kgViga 1(0.3x0.6 2400 0.3 0.6 3.05 1317.6 kgViga 2(0.3x0.6 2400 0.3 0.6 2.975 1285.2 kgVigaB(0.25x0. 2400 0.25 0.5 0.95 285 kgVigaB(0.25x0. 2400 0.25 0.5 2.3 690 kgColumna 2400 0.4 0.6 3 1728 kgParapeto 354 1.2 424.8 kgSobrecarga 100 24.18125 2418.125 kg %Total pesos
PD = 14617.1 kgPLD = 2418.125 kg
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%Total pesos 14617.1 +2418.125
17153.0256045.3125
Total: 40233.5625 kg
AT = 24.18125 m2
ENTONCES 1.66 tn/m2
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c/p (m) = 0.19
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PREGUNTA N°2.- (10 ptos)La figura representa el Portico del Eje 2, de la pregunta anterior. Se solicita:a) Realizar el análisis Estructural para las fuerzas horizontalesb) Resolver el Portico del Eje 2 según los combos de la norma E-060
103 m
53.6 m
L2 = 6.55 L1 = 6.7
Elementos b(m) h(m) I(m4) K=I/L(m3) K/Ko= Fza hor QVigas 0.3 0.6 0.0054 10 10Lv= 6.55 0.00082 1.02 5 15Lv= 6.7 0.00081 1.00Columna 0.4 0.6 0.0072hc= 3 0.00240 2.98hc= 3.6 0.00200 2.48148148
Ko=Kmin 0.000806Kv1 Kv2
Ko(m3)= 0.000806E(tn/m2)= 2000000 D=a*k KcDo(tn/m)= 12EKo/h^2 K=D*Do EmpotPara h(m)= 3 V=K*SiDo(tn/m)= 2149.25 Kv1 Kv2Para h(m)= 3.6 S2Do(tn/m)= 1492.53731 Kc S1Yo=Y1+Y2+Y3=Yo(Ejemplo) Kv3 Kv4Cuando no existe un cambio significativo entre las alturas de los pisos consecutivos(menos del 30%) y cuando el valor de K>1 , puede observarse en la tabla 3 del anexo 2
k=SumKv/2Kc k=SumKv/Kca=k/(2+k) a=(0,5+k)/(2+k)
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que Y2=Y3=0
MUTO
10tn Kv 1.02 Kv 1.00Kc 2.98 30*60viga 2.98 2.98
k 0.34 0.68 Q2= 0.343m a 0.15 0.25 10 0.14
D 0.44 0.76 SumK= 0.43K 938 1623 3481 920V 2.70 4.66 S2=Q/SumK 2.64
5tn Yo 0.4 0.4 0.00287 0.4Kc 2.48148148 1.02 2.48 1.00 2.48148148
k 0.41 0.82 Q2= 0.403.6m a 0.38 0.47 15 0.38
D 0.94 1.16 SumK= 0.93K 1401 1730.3 4523 1392V 4.65 5.74 S1=Q/SumK 4.62
0.7 0.64 0.0033 0.7
6.55 6.7
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4.24 4.764.85 8.39 4.76
4.85 4.15 h = 31.39 1.33 h = 3.6
2.70 4.66 2.64 VIGASCOLUMNAS
6.59 8.165.02 7.44 4.99
3.23 5.59 3.178.25 2.27 6.44 2.18
4.65 DMF(t*m) 5.74 4.62DFC(t)
11.71 L= 6.55 13.22 L= 6.7 11.63
37.14NIVEL Si(m) Si(mm) 0.75R 0.75R*Si(mm) 17.24
2 0.00287 2.87279632 6 17.24 19.90
1 0.0033 3.31663033 6 19.90
La deformacion Horizontal maxima db la estructura Sadm= 7*h/1000
h (mm) = 3600 Sadm= 25.2 1 nivelh (mm) = 3000 Sadm= 21 2 nivel
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El portico tiene un desplazamiento horizontal Sh(mm 37.14
4.85 4.24 4.15 4.76
L = 6.55 L = 6.7
1.39 1.39 1.33 1.33
8.25 6.59 6.44 8.16
L = 6.55 L = 6.7
2.27 2.27 2.18 2.18
1.39 1.39 1.33 1.331.39 0.06 1.331.39 0.06 1.332.27 0.09 2.183.65 0.15 3.51
3.65 0.15 3.51
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6.55 m 6.7 m
REACCIONES
b) Resolver el Portico del Eje 2 según los combos de la norma E-060
Elementos b(m) h(m) I(m4) K=I/L(m3) K/Ko=Vigas 0.3 0.6 0.0054Lv= 6.55 0.00082 1.02Lv= 6.7 0.00081 1.00Columna 0.4 0.6 0.0072hc= 3 0.00240 2.98hc= 3.6 0.00200 2.48148148
Ko=Kmin 0.000806
Σ k = 4.00 Σ k = 5.00 Σ k = 3.98
k = 2.98 k = 2.98 k = 2.983 m
k = 1.02 k = 1.00
k = 2.48 k = 2.48k = 2.48 3.6 m
Σ k = 6.48 Σ k = 7.48 Σ k = 6.46
6.55 6.7
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1° NIVEL 0.16 0.14 0.13 0.15
2° NIVEL 0.26 0.2046 0.20 0.25
9.61 9.61 10.05 10.052.687 2.687
6.55 m 6.70 m
+
0.17 0.440.196
36.70 m
+0.485
0.36 0.44
3.7 36.70 m
WD 1° NIVEL
WD 2° NIVEL
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8.02 8.02 8.39 8.392.242 2.242
6.55 m 6.70 m
0.16 0.14 0.13 0.15
-9.61 9.61 -10.59 10.941.52
0.760.03 0.03
0.02 0.015-1.70
-0.850.12 0.11
0.06 0.06-0.009
-8.02 10.51 -11.29 9.31
0.26 0.20 0.20 0.25
-8.02 8.02 -8.39 8.392.05
1.02-0.13 -0.13
-0.07 -0.065
1° NIVEL
2° NIVEL
WD
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-2.09-1.05
0.21 0.210.11 0.10
-0.026-5.93 9.12 -9.35 6.31
4.25 4.25 4.44 4.441.1875 1.1875
6.55 m 6.70 m
+
0.17 0.440.196
36.70 m
+0.485
0.36 0.44
3.7 36.70 m
WL 1° NIVEL
WL 2° NIVEL
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1.70 1.70 1.78 1.780.475 0.475
6.55 m 6.70 m
0.16 0.14 0.13 0.15
-4.25 4.25 -4.98 5.330.67
0.330.05 0.05
0.03 0.026-0.83
-0.410.06 0.06
0.03 0.03-0.004
-3.52 4.69 -5.28 4.55
0.26 0.20 0.20 0.25
-1.70 1.70 -1.78 1.780.43
0.22
1° NIVEL
2° NIVEL
WL
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-0.03 -0.03-0.01 -0.014
-0.44-0.22
0.05 0.040.02 0.02
-0.006-1.26 1.93 -1.98 1.34
GRÁFICOS WD & WL
GRAFICA WD GRAFICA WL
9.12 -9.35 1.93 -1.98-5.93 6.31 -1.26 1.34
B C D B C D
10.51 -11.29 4.69 -5.28-8.02 9.31 -3.52 4.55
B C D B C D
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6.55 6.7 6.55 6.7
COMBOS
2 NIVEL L = 6.55
DATOS DEL ANALISIS ESTRUCTURAL TRAMO IZQSiempre +
Datos wD wL MD ML SNudo B 2.242 0.475 -5.93 -1.26 4.85Nudo C 2.242 0.475 9.12 1.93 4.24
El sismo tiene que considerar +/-SD=Valor muerto o permanente L=Valor permanente o sobrecargaS=Valor de sismo= +/- Sismo
COMBINACIONES (NORMA PERU )COMB1=1,4D+1,7L COMB4=0,9D+SCOMB2=1,25(D+L)+S COMB5=0,9D-SCOMB3=1,25(D+L)-S
Tramo NUDO B Tramo NUDO CCOMBO wU1 Mu1 COMBO wU1 Mu1
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1 3.9463 -10.4307886 1 3.9463 16.0538941COMB wU2 Mu2 COMB wU2 Mu2
2 3.39625 -4.13 2 3.39625 18.0592319COMB wU3 Mu3 COMB wU3 Mu3
3 3.39625 -13.83 3 3.39625 9.57325313COMB Wu4 Mu4 COMB Wu4 Mu4
4 2.0178 -0.48 4 2.0178 12.4515766COMB wU5 Mu5 COMB wU5 Mu5
5 2.0178 -10.18 5 2.0178 3.96559778
2 NIVEL L = 6.7
DATOS DEL ANALISIS ESTRUCTURAL TRAMO IZQSiempre +
Datos wD wL MD ML SNudo B 2.242 0.475 -9.35 -1.98 4.15Nudo C 2.242 0.475 6.31 1.34 4.76
El sismo tiene que considerar +/-SD=Valor muerto o permanente L=Valor permanente o sobrecargaS=Valor de sismo= +/- Sismo
COMBINACIONES (NORMA PERU )COMB1=1,4D+1,7L COMB4=0,9D+SCOMB2=1,25(D+L)+S COMB5=0,9D-SCOMB3=1,25(D+L)-S
Tramo NUDO B Tramo NUDO CCOMBO wU1 Mu1 COMBO wU1 Mu1
1 3.9463 -16.4654144 1 3.9463 11.1029474
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COMB wU2 Mu2 COMB wU2 Mu22 3.39625 -10.02 2 3.39625 14.3133888
COMB wU3 Mu3 COMB wU3 Mu33 3.39625 -18.32 3 3.39625 4.79736568
COMB Wu4 Mu4 COMB Wu4 Mu44 2.0178 -4.27 4 2.0178 10.4351084
COMB wU5 Mu5 COMB wU5 Mu55 2.0178 -12.57 5 2.0178 0.91908531
1° NIVEL L = 6.55
DATOS DEL ANALISIS ESTRUCTURAL TRAMO IZQSiempre +
Datos wD wL MD ML SNudo B 2.687 1.1875 -8.02 -3.52 8.25Nudo C 2.687 1.1875 10.51 4.69 6.59
El sismo tiene que considerar +/-SD=Valor muerto o permanente L=Valor permanente o sobrecargaS=Valor de sismo= +/- Sismo
COMBINACIONES (NORMA PERU )COMB1=1,4D+1,7L COMB4=0,9D+SCOMB2=1,25(D+L)+S COMB5=0,9D-SCOMB3=1,25(D+L)-S
Tramo NUDO B Tramo NUDO C
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COMBO wU1 Mu1 COMBO wU1 Mu11 5.78055 -17.2088175 1 5.78055 22.6904995
COMB wU2 Mu2 COMB wU2 Mu22 4.843125 -6.17 2 4.843125 25.592201
COMB wU3 Mu3 COMB wU3 Mu33 4.843125 -22.67 3 4.843125 12.4132934
COMB Wu4 Mu4 COMB Wu4 Mu44 2.4183 1.04 4 2.4183 16.0491637
COMB wU5 Mu5 COMB wU5 Mu55 2.4183 -15.47 5 2.4183 2.87025616
1° NIVEL L = 6.7
DATOS DEL ANALISIS ESTRUCTURAL TRAMO IZQSiempre +
Datos wD wL MD ML SNudo B 2.687 1.1875 -11.29 -5.28 6.44Nudo C 2.687 1.1875 9.31 4.55 8.16
El sismo tiene que considerar +/-SD=Valor muerto o permanente L=Valor permanente o sobrecargaS=Valor de sismo= +/- Sismo
COMBINACIONES (NORMA PERU )COMB1=1,4D+1,7L COMB4=0,9D+SCOMB2=1,25(D+L)+S COMB5=0,9D-SCOMB3=1,25(D+L)-S
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Tramo NUDO B Tramo NUDO CCOMBO wU1 Mu1 COMBO wU1 Mu1
1 5.78055 -24.7890402 1 5.78055 20.7642675COMB wU2 Mu2 COMB wU2 Mu2
2 4.843125 -14.28 2 4.843125 25.4779002COMB wU3 Mu3 COMB wU3 Mu3
3 4.843125 -27.16 3 4.843125 9.16339363COMB Wu4 Mu4 COMB Wu4 Mu4
4 2.4183 -3.72 4 2.4183 16.5326903COMB wU5 Mu5 COMB wU5 Mu5
5 2.4183 -16.60 5 2.4183 0.21818376