estructuras 3 dnc repaso
TRANSCRIPT
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REPASO
1
Estructuras 3 – TALLER VERTICAL DNC
RESOLUCIÓN DE PÓRTICOS
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45.20
16
.80
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3
45.20
16
.80
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45.20
16
.80
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45.20
16
.80
6
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L= 7,00
Peso del
hormigón
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L= 7,00
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L= 7,00
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L= 7,00
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L= 7,00
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L= 7,00
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L= 7,00
41,00
16,8
0
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L= 7,00
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SUPERFICIE DE INFLUENCIA
Para C145.20
16
.80
S1= 10,20 x 4,70 = 47,94 m2
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SUPERFICIE DE INFLUENCIA
Para C245.20
16
.80
S1= 10,20 x 3,70 = 37,74 m2
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La altura del dintel la estimamos en d=L/10 d=16,80/10 = 1,60 mA las patas del pórtico le asignamos 1,00 mTodo tiene un espesor de 40 cm
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2210,00 10,00
10,00
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2310,00 10,00
10,00
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Pórtico plano: HIPERESTÁTICO DE TERCER GRADO
TRES ECUACIONES
HA
VA
MA
HB
VB
MB
FX=0
FY=0
Ma=0
SEIS INCOGNITAS
TRES ECUACIONES DE DEFORMACIONES
1
2 3
4
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Pórtico plano: HIPERESTÁTICO DE PRIMER GRADO
HA
VA
HB
VB
CUATRO INCOGNITAS
TRES ECUACIONES
FX=0
FY=0
Ma=0
UNA ECUACIÓN DE DEFORMACIONES
1
2 3
4
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1
2 3
4
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x x
x x
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x x
Jx=b x h3
12=
12
40 x 1603
Jx=b x h3
12=
12
40 x 1003
Dintel
Columna
Jx=b x h3
12=
12
40 x 1003
Columna
xx
xx
1
2 3
4
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x x Jx=b x h3
12=
12
40 x 1603
Dintel
Jx=
Momento de Inercia del DINTEL
40 x 1603
12= 13.653.333 cm4
Jx= 0,1365 m4
30
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x x Jx=b x h3
12=
12
40 x 1603
Dintel
Jx=
Momento de Inercia de la COLUMNA
40 x 1003
12= 3.333.333 cm4
Jx= 0,0333 m4
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RIGIDEZ FLEXIONAL
E x J
L
E:
J:
L:
Módulo de Elasticidad del hormigón
Momento de Inercia
Longitud de la barra
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x x
x x
Rigidez del dintel
Rigidez de la columna
Rigidez del dintel
Rigidez de la columna
+
Rigidez del nudo
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x x
x x
Coeficiente de distribución flexional
Rigidez del dintel
Rigidez de la columnaCoef.col=
Rigidez del nudo
Rigidez del nudo
Coef.dintel=
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x x
Rf dintel=
Rigidez flexional del DINTEL
E x 0,1365
16,00=
Jx= 0,1365 m4L= 16,00 m
E x 0,0085 m3
36
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Rigidez flexional de la COLUMNA
E x 0,0333
7,50=
Jx= 0,0333 m4L= 7,50 m
E x 0,0044 m3
x x
37
Rf col=
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Rigidez flexional del NUDO
38
Rigidez flexional de la COLUMNA
Rigidez flexional del DINTEL+
Rf dintel
E x 0,0085 m3 + E x 0,0044 m3
Rf nudo= Rf col = +
Rf nudo=
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En este caso P3 no existe
Las columnas C2 descargan directamente en
las patas del pórtico (no producen flexión)
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2 3
a b
L
Q Q2
Q3
+
-
2 3
P2
a b
L
Mf
- -
+
47
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Momento total
desequilibrado
en 2
Momento total
desequilibrado
en 3
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Mo= -335,7 tm Mo= -335,7 tm
+335,7 tm x 0,658= 220,9tm
+335,7 tm x 0,658
Tengo en el nudo un momento Mo
Como en realidad no es así, tengo que equilibrar el Mo con otro
momento igual y contrario que se genera en cada barra proporcional a
la rigidez de la misma
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M2= -115 tm Mo= -115 tm
-335,7 tm x 0,342 = -115 tm +335,7 tm x 0,342= +115 tm
Tengo en el nudo un momento Mo
Así se equilibra el nudo
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El M5 y el M6 tienen que ser iguales porque
el pórtico es simétrico
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El Mmáx se corresponde en la mitad del dintel
donde el corte es cero
Determinamos el Mmáx
Ldintel/2 = 16,00/2=8,00 m
Ldintel/2
Mmáx = -114,97 + 109,9 x 8,00 – 33,9 x (8,00-6,90) - 9,50 x (8,00)2 / 2 =
Mmáx = -M2 + Q2 x Ldintel / 2 – C1 x (Ldintel/2-L1) - q x (Ldintel)2 / 2 =
Mmáx = -114,97 + 879,2 – 37,3 – 304,0 = 422,9 tm
Mmáx
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DISTINTAS NOMENCLATURAS PARA NOMBRAR LOS MATERIALES QUE UTILIZAMOS PARA DIMENSIONAR
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Fuerza N menor que cero
COMPRESION
Momento flector nulo
DIMENSIONADO
Área del acero
Tensión del acero
Área del hormigón
Tensión del hormigón
Coeficiente de pandeo
Coeficiente de seguridad
Carga de compresión
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COMPRESION
DIMENSIONADO
N x x = B ( σ´bk x B + σek x A )B
N x x = B ( σ´bk x B + σek x A )B B o = 0,01
B(sección de hormigón) =N x x
( σ´bk + σek x 0,01 )
=H (altura)
b (lado menor)
tabla
Altura de la columna: es dato
Elegimos el ancho de la
columna (por el ancho de la
pared)
Obtenemos de la tabla en
función del valor de
= b x d
Adoptamos por ejemplo el 1%
Es dato
De tablas
Coef. de seguridad: 2,5
4200 kg/cm2Para H30=230 kg/cm2
Elegimos el ancho de la
columna (por el ancho de la
pared)
Obtenemos la otra dimensión
A(sección de acero) = b x d x 0,01
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Fuerza N igual acero
FLEXION SIMPLE
Momento flector (+) (-)
DIMENSIONADO
Brazo de palanca
Tensión del acero
Área del acero
Coeficiente de seguridad
Momento flector
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FLEXION SIMPLE
DIMENSIONADO
Tensión de rotura
del acero
Anec=σek x z
M x =
M
σek x z
Coef. de seguridad a la flexión = 1,75
Tensión admisible
del acero
Anec=M
σadm x z
σek = 4200 kg/cm2
σadm = 2400 kg/cm2
σadm
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N(-)
As1
As2
M(+)
d
d/2 Zs
d1
Momento flector
Esfuerzo normal
Momento
respecto a As1Va con el signo negativo de compresión y la resta
se transforma en suma Ms en mayor que MDistancia entre N y As1
Armadura necesaria por flexión y por compresión
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POR CÁLCULO
POR REGLAMENTO
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As2= colocamos el 50% de As1
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EMPALME DE LAS
BARRAS RECTAS
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… fin
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