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DIBUJO DE ESTRUCTURAS METALICAS.
Para realizar el dibujo de estructuras metálicas se requiere hace el calculo analítico de fuerzas internas en las barras ya sea en tracción compresión, considerando las cargas externas de la estructura cuales se vera un método para su determinación, pero para el calculo d las fuerzas internas existen métodos analíticos pero también existe gráficos para su determinación, para lo cual detallamos a continuació dicho método.
Métodos gráficos para determinación de fuerzas internas en la estructuras. You're Reading a Preview
El método de CREMONA es unos de los procedimientos Unlock full access with a free trial. empleados para la determinación de fuerzas en las barras de estructura sometida a cargas exteriores, suponiendo que las carg Download With Free Trial exteriores actúan en los nudos.
En cada nudo han de estar en equilibrio las fuerzas aplicadas, a como las fuerzas internas correspondiente a las barras que en el mism concurren. El conjunto de fuerzas ha de constituir en cada nudo, u ba polígono cerrado. Ello permite averiguar dos de Read Free Forfuerzas 30this Days Sign up to vote on title desconocidos en un nudo, si se conocen las de las restantes barra Useful Not useful Cancel anytime. concurrentes en el mismo.
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Par
disti
ir si una barra qued
primida
raci
da,
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El método lo explicaremos haciendo uso del siguiente ejemplo, un armadura o cercha la cual esta sometida a cargas. Se procede de l siguiente manera
1. Se dibuja la armadura reducida a sus ejes graficando también la fuerzas aplicadas y las reacciones.
2. Se enumera los campos empleando la notación de BOW. Ca campo esta comprendido entre dos cargas exteriores o dentro d un reticulado de 3 barras 3. Se construye el polígono de fuerzas exteriores (previamente han determinado las reacciones), disponiéndolas en or correlativo (en el mismo orden que se encuentran al recorrer estructura en forma cíclica, a una escala conveniente).
4. Se trazan luego en el polígono de fuerzas (comenzando por u nudo donde solamente existan dos barras), y por cada una d estas, paralelas a las dos barras que concurren en el nu You're Reading a Preview correspondiente. Es muy conveniente recorrer el nudo en sentid horario. Unlock full access with a free trial. Download With Free Trial
ANGULO O PERFIL L (DIN 1028 Y DIN 1029) Este tipo de perfil es uno de los de mayor uso en estruc metálicas para techos, por lo cual se adjunta tablas completas. Read Free Forde 30this Days designa dando el tipo de perfil, las dimensiones los Sign up to vote on title lados y espesor. Useful Not useful
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SELECCIÓN DE UN PERFIL PARA MIEMBROS SOMETIDOS COMPRESIÓN
Los miembros a compresión pueden fallar por pandéala posibilida de pandeo se puede predecir por medio de la formula de Euler par columnas “largas” y la formula de Jonson para columnas “cortas”.
La transición entre una y otra formula se ha tomado en el punto e el que el esfuerzo promedio iguala a la mitad del límite de fluencia.
Para la zona de columnas cortas, el factor de seguridad respecto límite de fluencia varia entre 1.67 para columnas de esbeltez cero hast un valor de 1.92 en el punto en que ya es aplicable la formula de Eule para el rango de aplicación de Euler, el factor de seguridad es constante 1.92. La relación de esbeltez no deberá ser superior a 200.
SELECCIÓN DE UN PERFIL PARA MIEMBROS SOMETIDOS You're Reading a Preview TRACCION. Unlock full access with a free trial.
Los miembros a tracción de no superar el esfuer Downloadademás With Free Trial admisible deberá tener una relación de esbeltez preferiblemente mayor de: en miembros principales Master your semester250 with Scribd Read Free Foron 30this Days 200 en miembros secundarios. Sign up to vote title & The New York Times Useful Not useful Special offer for students: Only $4.99/month.
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El esfuerzo admisible según la especificación del ASIC es 0,6 vece
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Escala 1 cm. ≈ 1 m P4 P3
P3
3 P2
P2 2
7
P1
8 1
6
4
You're Reading a Preview 5 Unlock full access with a free trial.
Download With Free Trial R 1
Luz 17 m
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-1 P1 -2 P2
+4
-3
+5
R A
P3 P4
-6
+8 -7
You're Reading a Preview Unlock full access with a free trial.
*Carga inicial: Download With Free Trial P1 =
L
( Pp + Pv )C t 2 17 P 1 = (28 + 80)5 2 P1 = 4590 Kgf .
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= 2 P 1
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L1 = 10.1 cm. P1' = 46359 Kg − f Donde PP = esfuerzo de la estructura = 28 Kgf Pv = esfuerzo del viento = 1.6X Para h = 6 →X = 50Kgf/m2 Cc = distancia entre cerchas R1 + R2 = P1 + P2 + P3 + P 4 R1 = R2 → R1 = 6 P 1
• R1 = 27540 Kg − f • R2 = 27540 Kg − f
Para la barra (2) L2 = 8.25cm P2 ' = 37867.5Kg − f
Para la barra (3)
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L3 = 8.2cm P3 ' = 37638Kg − f
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Selección del perfil ( ) (3.27m)
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λ=
327 394
=
82.9
≈
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83
De tabla: w = 1.59
Reemplazando: 4(6359)(1.59) 69.4 1062.1 ≤ 1200
≤ 1200
Entonces la selección del perfil: 130x130x14
Calculo de los esfuerzos de la barra (luz) Para la barra (4) L4 = 8.8cm(cremona) P4 ' = 40392 Kg − f
Para la barra (5) L5 = 5.3cm(cremona) P5 ' = 24327 Kg − f
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Lreal = 566 cm.
λ = 200
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40392(26)
≤ 1200 914 1149.0 ≤ 1200
Entonces la selección del perfil es: 150x150x16 Esfuerzo de la barra (6) perfil ( ) L6 = 2cm(cremona) P6 ' = 9180 Kg − f
Lreal = 328 cm. r=
328 200
→ 1.64cm
Tabla r = 225 A = 282 λ =
328 225
= 145 → w = 3.6
Reemplazando 9180(36)
≤ 1200 282 1171.91 ≤ 1200
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Seleccionamos el perfil 75x75x10
Master your semester with Scribd Esfuerzo de la barra (7) perfil (L) & The New York Times L = 2cm(cremona) 7 Special offer for students: Only $4.99/month.
P7 ' = 9180 Kg − f
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Reemplazando: 9180(3.26)
≤ 1200
254 1178 ≤ 1200
Seleccionamos el perfil 120x120x11
CUADRO: 2
BARRA 01
COMPRESIO N TRACCION (Kgf) (Kgf) 46359
-
LONGITU D (m) 3.27
PERFIL
BARR
130x130x14
1’
You're Reading a Preview
02
37867.5
Unlock-full access with a3.27 free trial.
Download With Free Trial
03
37638
-
04
-
40392
Master your semester with Scribd & The New York Times05 24327 Special offer for students: Only $4.99/month.
3.27 3.27
130x130x14
3’
150x150x16
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3.27 Useful Not useful 150x150x16 Cancel anytime. 3.27
2’
130x130x14
4’
5’
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1
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Datos P1’ = 46359 Kgf = 102204.09 lb τadm = 7850 lb/in 2 a = 0.707(14) = 9.898 mm a = 0.389 in. Reemplazando en la formula L1 =
L1 =
p 2aτ adm
+ 4a....................α 102204.09lb
2(0.389 pul g)(7850
+ 1.556
lb pul g 2
)
L1 = 18.29 in * L1 y L2 son las longitudes de la soldadura
Calculo de la longitud (L2)
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Datos Unlock full access with a free trial. P4’ = 40392 Kgf = 89049.11 lb τadm = 7850 lb/in 2 Download With Free Trial a = 0.707(16) = 11.312 mm a = 0.445 in. Reemplazando en α
lbf with Scribd Master your semester L ( . ) ( . in )( ) & The New York Times 89049
2 =
2 0 445
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L2 = 14.52 in.
+
4 0 445
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