Capítulo V: TRANSICIÓN DEL PERALTE Y SOBREANCHO EN LAS CURVAS HORIZONTALES. 5.1 INTRODUCCIÓN. Cuando un vehículo marcha en recta las fuerzas que actúan sobre él son las de inercia, el peso del vehículo y las reacciones del terreno (normales y debidas al rozamiento por rotación de los neumáticos contra el pavimento). Al entrar el vehículo en la curva se presenta la fuerza centrífuga, que origina dos peligros para la estabilidad del vehículo en marcha: Peligro de deslizamiento transversal. Peligro de vuelco. Se puede demostrar que la condición de deslizamiento transversal es la que primero se produce, y para contrarrestarla, será necesario inclinar la sección transversal de la vía un determinado ángulo. Esta inclinación que se le suministra a la sección transversal de la vía, si se expresa como la diferencia de elevación entre sus bordes (interior y exterior) se le denomina superelevación (S); y si se expresa como la pendiente de la sección transversal de la vía se le llama peralte (e).
5.1.1 RADIO DE CURVATURA MÍNIMO. En un vehículo que no experimenta deslizamiento transversal las fuerzas que actúan sobre él se encuentran en equilibrio, lo cual se observa en la figura 5.1.
FIGURA 5.1. En esta figura se tiene que: - S: superelevación; en metros.
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- : ángulo de inclinación de la sección transversal; en grados sexagesimales. - F: fuerza centrifuga; en m/s. - P: peso del vehículo; en Kg. -R t: reacción debida al rozamiento transversal por rotación, en Kg. Si se realiza una suma de fuerzas sobre el eje x:
F . cos
. P sen
Rt …....(5.1)
En la cual:
F
M .V 2
P .V 2
R
. g R
....(5.2)
donde: R: radio de la curva; en metros. 2
g: aceleración de la gravedad; en m/s . V: velocidad de diseño; en m/s. La formula 5.2 se puede expresar como:
F donde: P: peso del vehículo; en kg. V: velocidad de diseñó en km/h. R: radio de la curva; en metros. f: fuerza centrífuga; en kg.
P .V 2 127,74 R .
....(5.3)
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Rt
. F sen
P . cos
....(5.4)
donde: :coeficiente transversal por rotación. Este coeficiente de rozamiento transversal por rotación ha sido obtenido para diferentes condiciones de estado de los neumáticos y estados de la superficie del pavimento. Depende además de la velocidad a que circulan los vehículos y de las condiciones atmosféricas que imperan más desfavorablemente sobre la vía. Debido al gran número de variables que influyen en su elección, es complicado obtener valores correctos mediante las mediciones de campo; por lo que se han establecido en la práctica valores orientativos que se basan fundamentalmente en el confort que experimenta el conductor al circular por una curva del trazado y en el mantenimiento del equilibrio del vehículo. En la tabla 5.1. se muestran los coeficientes de rozamiento transversal por rotación en función de la velocidad de diseño en las curvas horizontales con pavimento mojado.
Velocidad de diseño (Km/h)
µ
30 40 50 60 80 100 120
0,25 0,21 0,18 0,16 0,14 0,13 0,11
Tabla 5.1 Si se sustituyen las expresiones (5.3) y (5.4) en (5.1), se obtiene:
P .V 2 127,14 R .
Psen
cos
R
En la expresión (5.5) el valor (1 -
tan )
V 2 (1 127,14(
P .V 2 127,14 R tan ) tan )
sen
P cos
....(5.5)
1. Con esta aproximación se esta del lado de la seguridad
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tan
=e
Por lo tanto:
Rmin
V 2 127,14 e
....(5.6)
La expresión (5.6) proporciona el radio mínimo de la curva horizontal para una velocidad de diseño determinada, el coeficiente de rozamiento transversal por rotación correspondiente a esa velocidad y el peralte máximo adoptado. ad optado. No puede utilizarse para calcular valores de peralte para p ara radios mayores que el radio mínimo. Los valores máximos de peralte que se adopten dependen de varias condicionales; entre las cuales se destacan: Condiciones climatológicas. Condiciones topográficas del terreno. Tipo de zona (rural o urbana). Frecuencia de vehículos lentos.
5.1.2 VALORES MÁXIMOS DE PERALTE. El valor máximo de peralte que se adopte debe ser limitado por razones de orden práctico; esto es, si el peralte es muy elevado el vehículo puede deslizar hacia dentro de la curva cu rva por la cual cua l circula cuando la velocidad es muy baja o se ve obligado a detener la marcha. Un peralte demasiado bajo puede resultar inaceptable motivado porque limita la velocidad del vehículo al circular por la curva. Se recomiendan los siguientes valores de peralte máximo (tabla 5.2) en función de diferentes condicionales.
CONDICION Cuando no se forma hielo sobre la vía. Valor más aconsejable en cualquier caso. En regiones de frecuentes nevadas. Para volúmenes de tráfico elevados y áreas urbanas.
emax (%) 12 10 8 6
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En la tabla 5.3 se muestra la relación entre las velocidades de diseño, radio mínimo o grado máximo de curvatura, coeficiente de fricción transversal y peralte máximo adoptado en función de la categoría de la vía en las carreteras de Cuba.
Velocidad de diseño (km/h)
R min min (m)
Gcº
µ
30 40 50 60 80 100
22,92 46,77 81,85 109,14 208,35 343,78
50º00’ 24º30’ 14º00’ 10º30’ 5º30’ 3º20’
0,25 0,21 0,18 0,16 0,14 0,13
emax (m/m ) 0,06 0,06 0,06 0,10 0,10 0,10
TABLA 5.3 Es conveniente aclarar que en la parte recta del trazado la pendiente de la sección transversal recibe el nombre de factor de bombeo (b); y a la diferencia de elevación entre su eje y cada uno de sus bordes bombeo (B). Si se quiere determinar el bombeo en recta la expresión a utilizar será:
B
a 2
.b ....(5.7)
donde: B: bombeo; en metros. a: ancho de la vía en recta; en metros. b: factor de bombeo; en m/m. De acuerdo al tipo de pavimento se recomiendan los siguientes valores de factor de bombeo (tabla 5.4).
Tipo de pavimento Alta calidad Mediana calidad
Factor de bombeo (%) 1–2 1,5 – 3
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5.1.3 RADIOS MÍNIMOS PARA CURVAS SIN PERALTE. En la figura 5.2 se representa la sección transversal de una vía que conserva en la curva la misma sección transversal que en el tramo recto; o sea, su sección parabólica normal con peralte igual al factor de bombeo. Se quiere determinar el radio mínimo de curva necesario, para que el vehículo circulando a la velocidad de diseño, conserve en la curva la sección parabólica correspondiente a la parte recta del trazado y no ocasione peligro a la circulación vehicular. En la figura se observa que en la mitad derecha el peralte esta orientado hacia el interior de la curva y se considera negativo; mientras que el dirigido hacia el exterior se considera positivo. Si se realiza una sumatoria de fuerzas sobre el eje x en la parte correspondiente al peralte negativo, se obtiene que:
Rt F . cos
. P sen
....(5.8)
donde:
Rt
( P . cos
F .sen ) ....(5.9)
Si se sustituye (5.3) y (5.9) en (5.8); se obtiene:
P . cos
P .V 2
sen 127,14 R
R
R
P .V 2 127,14 R
V 2 (1
. tan )
127,14
tan
V 2 (1
.e)
127,14(
e)
cos
. P sen
....(5.10)
De acuerdo a la experiencia internacional, la sección en curva puede mantener la sección parabólica normal correspondiente al tramo recto, cuando (e + ) no sea mayor que el valor 0,015. Cuando se satisface esta condición para: e b
1%
0,01m / m
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y para e = - 0,025
FIGURA 5.2 0,015 0,25
0,040
Si se sustituyen estos valores de coeficiente de fricción en la expresión (5.10) se obtiene: para
= 0,025: 1 e.
1 0,01.0,025
1,00025
1
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para e = 0,025;
= 0,040 ........ R min
V2 1,907
por lo que: R min
V2 1,907
....(5.12)
En la tabla 5.5 se muestran para las distintas velocidades de diseño el radio de curvatura a partir del cual la vía en curva conserva la misma sección transversal que los tramos rectos del trazado.
Velocidad de diseño (km /h)
R min min sin peralte (m)
30 40 50 60 80 100
472 839 1311 1888 3356 5244
TABLA 5.5 Estos valores, aproximados por exceso, se encuentran en las tablas del ANEXO I; y en la parte correspondiente al peralte (e) aparece la simbología CN, que significa que la sección transversal del pavimento en la curva se mantiene con la corona normal del tramo recto del trazado.
5.1.4. CONDICIÓN DE VUELCO EN LA CURVA. Es la segunda condición de peligro que puede producirse en los vehículos cuando circulan por una curva; y se produce cuando la resultante de las fuerzas en el centro de gravedad del móvil, sale fuera
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Figura 5.3 ab = aj + jb P . ab = F . cb…. (5.13) donde:
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a1
ab
2
. ....(5.16) H sen
. cos
si se realiza un análisis similar para el segmento cb; se obtiene:
cd H . cos
....(5.17)
y:
bd jm
a1
. am sen
2
. ....(5.18) sen
Por lo tanto:
a1
cb H . cos
2
. sen
....(5.19)
Si se sustituyen estos valores de las expresiones (5.16) y (5.19) en la expresión (5.13); se obtiene:
P (
a1 2
. ) H sen
. cos
P .V 2 127,14 R
( H . cos
a1 2
. ) sen
De la cual se despejan los valores límites para que no ocurra el vuelco del vehículo.
V 11,28
R
R a1
2 H .e
2 H a1 .e
V 2 2 H a1 .e 127 14( a1
2 H e)
.... (5.20)
....(5.21)
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mínimo para cada velocidad; el otro extremo es cero,(radio de curvatura infinito), ya que no es necesario peraltar los tramos rectos del trazado. Para los radios comprendidos entre estos valores extremos de peralte (e = emax y e = 0) ; para una velocidad de diseño determinada, el peralte se distribuirá de forma tal que exista una relación entre el factor de fricción transversal y la razón de peralte aplicada. Son varios los métodos que pueden utilizarse para obtener el peralte, pero en todos ellos se sigue el criterio de mantener una relación adecuada entre los valores del factor de fricción transversal y el valor del peralte; de forma tal que contrarresten a la fuerza centrífuga. Para una velocidad de diseño determinada, existen cuatro métodos para la distribución del peralte; estos son: METODO I : La razón de peralte peralte es directamente proporcional al grado de curvatura. METODO II: La razón de peralte es tal que un vehículo que viaje a la velocidad de diseño, tiene la totalidad de la fuerza centrífuga equilibrada por el peralte, hasta curvas donde se requiere el valor máximo de peralte; con el peralte máximo en todas las curvas fuertes del trazado. METODO III: Es idéntico al método II, excepto que se basa en la velocidad promedio de marcha. METODO IV: La distribución de peralte esta en razón curvilínea con el grado de curvatura; adoptando valores entre los métodos I y III. En la figura 5.4 se ha construido la gráfica de peralte contra grado de curvatura y fricción transversal contra grado de curvatura, para un vehículo que viaja a la velocidad de diseño de 100Km/h. En la figura 5.5 la gráfica de fricción lateral contra grado de curvatura para un vehículo que viaja a la velocidad promedio de marcha. Para la construcción de estas gráficas se han calculados: las fuerzas centrífugas (F), los peraltes (e) y el coeficiente coeficiente de fricción fricción transversal transversal ( ); para cada uno de estos estos cuatro métodos, métodos, tanto para para la velocidad
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velocidad uniforme; ya sea en la recta o en curva con cualquier curvatura hasta la máxima correspondiente. Aunque la velocidad uniforme es el propósito de la gran mayoría de los conductores, ya que ella es fácil de obtener en las carreteras bien diseñadas, cuando los volúmenes no son pesados; existe una tendencia por parte de algunos conductores de viajar a mayor velocidad en las rectas y en curvas suaves, que en las curvas fuertes del trazado; sobre todo después de haber sido demorados por la presencia de vehículos que se mueven más lentamente.
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Producto de la marcada diferencia en la fricción lateral requerida para las diferentes curvas, resulta un movimiento errático de los vehículos; ya sea a la velocidad de diseño o a la velocidad promedio de marcha.
Gc
R (m)
METODO1 F e
METODO II F e
0º20’ 0º30’ 0º40’ 0º50’ 1º00’ 1º30’ 2º00’ 2º30’ 3º00’
3437,76 2291,84 1718,88 1375,10 1145,92 763,95 572,96 458,36 381,97
0,017 0,025 0,034 0,042 0,051 0,076 0,102 0,127 0,152
0,010 0,014 0,019 0,024 0,029 0,043 0,057 0,071 0,086
0,017 0,011 0,015 0,018 0,022 0,033 0,045 0,056 0,065
0,017 0,025 0,034 0,042 0,051 0,076 0,102 0,127 0,152
3º20’
343,78
0,178 0,100
0,078
0,178
0,023 0,034 0,046 0,057 0,069 0,100 0,100 0,100
-0,006 -0,009 -0,012 -0,015 -0,018 -0,024 0,002 0,027 0,052
0,100 0,100 0,078
METODO III F e
METODO IV F e
0,017 0,025 0,034 0,042 0,051 0,076 0,102 0,127 0,152
0,017 0,025 0,034 0,042 0,051 0,076 0,100 0,100 0,100
0,000 0,000 0,000 0,000 0,000 0,000 0,002 0,027 0,052
0,017 0,025 0,034 0,042 0,051 0,076 0,102 0,127 0,152
0,017 0,025 0,034 0,042 0,046 0,069 0,069 0,083 0,093 0,099
0,000 0,000 0,000 0,000 0,002 0,007 0,019 0,034 0,053
0,178
0,100
0,078
0,178
0,100
0,078
TABLA 5.7 5.1.5.3 METODO III. Se muestra que al utilizar la velocidad promedio de marcha, a volúmenes bajos, se alcanza el peralte máximo casi en el punto medio de la variación del grado de curvatura. En la figura 5.5 se observa que a la velocidad promedio de marcha no se requiere fricción lateral hasta esa curvatura y a partir de aquí, la
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5.1.6 DETERMINACIÓN DEL PERALTE PARA VELOCIDAD DE DISEÑO DE 100 KM/H. KM/H. Se desea determinar la distribución parabólica del peralte contra el grado de curvatura, para una velocidad de diseño de 100km/h y su correspondiente de marcha de 86km/h (figura 5.6). Sobre un sistema de coordenadas se toman como abscisas los grados de curvatura y como ordenadas los peraltes. Como se conoce que la pendiente, que es la forma usual de expresar los peraltes, es una unidad sin dimensión; ya que es el cociente de dividir dos unidades de la misma clase, se hace necesario representar los grados de curvatura en la misma unidad que los peraltes. Por eso, los grados de curvatura entre Gc = 0º00' y Gc = 3º20', se han representado; además, en radianes. Con estas condiciones es posible dibujar el gráfico de la figura 5.6 y determinar las ecuaciones que permitan relacionar el peralte con el grado de curvatura. Para ello: Se sitúa el vértice de la parábola de acuerdo a sus coordenadas(Gmax; e max); o sea, (3º20'; 0,10) y en radianes (0,0581776; 0,10). Se calcula el grado de curvatura para la velocidad promedio de marcha (86 km/h); en el punto donde el peralte alcanza el valor de 0,10 m/m, por la fórmula (5,6):
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pendOQ
m
0,10 0,034378
2,908837m / m
Si imaginariamente se concibe la otra rama de la parábola, esta será tangente en un punto a la recta PR, cuya pendiente es igual a la de la recta OQ, pero de signo contrario. Se determinan los puntos de comienzo de la parábola y el ev (ordenada máxima) de la misma. Como se conoce el vértice de la parábola(Vp), por las condiciones de grado y peralte máximo, se pueden determinar los puntos de comienzo de la parábola (A y B). Como la parábola es tangente a las rectas OQ y PR, cuyas pendientes son conocidas, se puede calcular el ev:
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y en radianes: Gc = 0,0105785 Por lo tanto, desde el origen (0,00 ; 0,00), hasta el punto A(0,0105785; 0,030771), la variación del peralte será lineal y calculada de acuerdo al método III, para la velocidad de marcha de 86 km/h. Se calcula el peralte desde el origen hasta el comienzo de la parábola; por la expresión: e
V2 127,14R
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2
(0,0105785 - 0,0581776) = - 4p(0,030771 - 0,10) 0,00226567
4 p
0,069229
0,032727
p = 0,00818179 Sustituyendo el valor de p en la expresión (5.26); se obtiene: 2
(x - 0,0581776) = - 0,032727 (y - 0,10) ....(5.27)
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ya que se estarían dotando de valores de peralte a tramos rectos de la vía, donde técnicamente no se necesitan. Lo que en definitiva se ha propuesto para evitar la dualidad de sección transversal que se produce en el PC y PT de las curvas circulares simples, es desarrollar parte del peralte en el tramo recto anterior al PC y posterior al PT; y parte dentro de la curva circular. Se plantea que entre el 60 y 80 % del peralte debe ser desarrollado en el tramo recto y; consecuentemente, entre el 20 y 40 % del peralte dentro de la curva circular simple. Por lo tanto se tendrá un tramo de la curva circular simple donde aparentemente, van existir condiciones de inseguridad debido a que los peraltes en ese tramo son menores que el máximo requerido para esa velocidad de diseño y grado de curvatura. Sin embargo, si se analiza el movimiento
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pmin . a/2 . b
N
pmax . a/2 . b ….(5.33)
Cuando el peralte es mayor que el 2 % y menor que el 3 %; el PC y el PT de la curva circular simple se encuentran entre las secciones transversales B y C. Cuando el peralte es igual al 2 %(figura 5.8), el
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Giro de la sección transversal por el eje de la vía. Giro de la sección transversal por el borde interior de la vía.
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En la sección transversal D (en curva), la sección transversal ha alcanzado el peralte total; que se mantendrá en el resto de la curva hasta comenzar el proceso inverso a la salida de la misma.
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En la figura 5.7 se muestra un esquema de este giro: Determinación del peralte máximo:
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La cual es menor que la máxima permisible de 1/150 = 0,00667 m/m, como era de esperarse.
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Los valores de emax; W; B; S; N; LR; tramo CD; dist PC C = dist PT C ; estaciones básicas para el desarrollo de la superelevación y los factores, no varían con relación al ejemplo anterior. Aunque no lo realizaremos, la pendiente relativa del borde exterior con relación al eje elevado en el tramo CD, se mantiene por debajo de las especificaciones máximas (P = 1/150). En la tabla 5.22,se encuentra el prerregistro de superelevación.
ESTACION B 365 + 4,63 366 + 0,00 PC 366 + 2,51 C 366+3,63
ESTACION D 367+3,63
ESTACION
DIST. 0,00 5,37 7,88 9,00
DIST.
390 + 0,00 D 389 + 3,39
BI
BE
-
-
0,061 0,064 0,067
0,036 0,053 0,06
PENDIENTE TRANSVERSAL
Z entre BI Y Eje
Z entre BI Y BE
0,025
.3,70 = 0,0925
.6,70 = 0,1675
PENDIENTE TRANSVERSAL POR EL SEMIANCHO DE LA VIA. -
BI
BE
0,02.3,06 0,02.3,20 0,02.3,35
10,00
DIST.
B 391 + 2,39 PT 390 + 4,51 C 390 + 3,39
ESTACION
PENDIENTE TRANSVERSAL POR EL SEMIANCHO DE LA VIA. -
0,02.3,20 0,02.3,35
DIST.
0,01193.3,00 0,01751.3,00 0,02.3,00
0,01751.3,00 0,02.3,00
PENDIENTE TRANSVERSAL
3,39 10,00
0,0217 0,025
-
-
0,064 0,067
0,053 0,06
Z entre BI Y Eje
Z entre BI Y BE
.3,47 = 0,075 .3,70 = 0,0925
.6,47=0,14 .6,70=0,1675
TABLA 5.22 Desde la estación A hasta la estación C, no experimenta cambio alguno con relación a la tabla 5.20 del ejemplo anterior. Para calcular la diferencia de elevación entre el borde interior y el eje de la vía; y entre el borde exterior y el borde interior en la EST 390 + 0,00, se procede de la forma siguiente: Z entre el borde interior y el eje: Z
factor ( 2).dist C 390
Z (0,0005.3,39
0,02)(3,00
Z 0,075m Z entre el borde interior y el borde exterior:
b.a/2
w(390)
0,47)
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Z (0,0005.3,39 Z
0,02). 6,00
0,47
0,14m
Estos valores aparecen en la tabla 5.22. El registro de superelevación aparece en la tabla 5.23.
CARRETERA CARRETERA VD = 80 k/h Curva No = 4 (Clotoide) b = 2% a = 6,00 m = 24º12’ B = 0,06 m CD = 10,00 m/m Gc = 2º00’ AB= BC= 9,00 m pend(C-D) = 0,002375 m/m Rc = 572,96 m e = 0,025 m/m Dc = 242,00 m S = 0,1675 m CON ENSACHE= 0,70 m GIRO = BI PRIMER TANGENTE CIRCULAR SEGUNDA TANGENTE EST. BI EJE BE EST. BI EJE BE EST. BI EJE BE A 364+5,63 R-0,06 B 365+4,63 R-0,06 366+0,00 R0,061 PC 366+2,51 R0,064
R R R
R
PC R-0,06 366+2,51 C R 366+3,63 D R+0,036 367+3,63
R-0,053
R-0,064
R
R-0,053
R-0,067
R
R-0,06
R-0,0775 R+0,015
R-0,09
392+0,00 B 391+2,39 PT 390+4,51
368+0,00
R-0,0775 R+0,015
R+0,09
370+0,00 272+0,00 . . . .
R-0,0775 R+0,015 R-0,0775 R+0,015 . . . . . . . .
R+0,09 R+0,09 . . . .
388+0,00 D 389+3,39
R-0,0775 R+0,015
R+0,09
R-0,0775 R+0,015
R+0,09
390+0,00
R-0,070
R+0,005 R+0,070
C 390+3,39 PT 390+4,51
R-0,067
R
R-0,064
R
R+0,06
R+0,053
TABLA 5.23
A 392+1,39
R-0,06
R
R-0,06
R-0,06
R
R-0,05 R-0,05
R-0,06
R
R
R-0,064
R
R+0,053
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Así, por ejemplo, para calcular lo que se eleva el borde interior; el borde exterior y el eje elevado (que son las columnas correspondientes que aparecen en la tabla 5.23)en la EST 367 + 3,63(D), se procede de la siguiente forma: - Para el borde interior: S ( BI ) B
W ( D).e( D) S( BI)
0,06
(0,70.0,025)
0,0775m
- Para el eje elevado: S( EJE)
(0,0925 S( BI) S(EJE)
0,0925 0,0775
0,015m
El valor de 0,0925 es la diferencia de elevación entre el borde interior y el eje elevado en la EST 367 + 3,63 (D); el cual aparece en la tabla 5.22; y el valor de S(BI) es el anteriormente anteriormente calculado. -
Para el borde exterior:
S ( BE )
0,1675
S ( BI )
S(BE)
0,1675 0,0775
0,09m
El valor de 0,1675 es la diferencia de elevación entre el borde interior y el borde exterior en la EST 367 + 3.63 (D); el cual aparece en la tabla 5.22;y el valor de S(BI) S(BI) es el anteriormente anteriormente calculado para el borde interior. Por ejemplo, para calcular lo que se eleva el borde interior; el borde exterior y el eje elevado en la EST 390 + 0,00, se procede de la siguiente forma: Para el borde interior: S ( BI ) B
W (390).e(390) S( BI)
Para el eje elevado:
0,06
0,070m
(0,47.0,0217)
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Para el borde exterior: S(BE)
0,140 S(BI)
0,140 0,070
S(BE)=0,07m. Estos valores aparecen en la tabla 5.23 sin aproximar al centímetro, para la mejor comprensión de los cálculos.
5.5.3 GIRO POR EL BORDE EXTERIOR. Calcular la curva circular simple del ejemplo anterior, si el giro se realiza por el borde exterior de la vía. Los valores de emax; W; B; S; N; LR; tramo CD ; dist PC C = dist PT C ; estaciones básicas para el desarrollo de la superelevación y los factores, no varían con relación al ejemplo anterior. En la tabla 5.24,se encuentra el prerregistro de superelevación.
ESTACION
DIST.
B 365 + 4,63 366 + 0,00 PC 366 + 2,51 C 366 + 3,63
0,00 5,37 7,88 9,00
D 367 + 3,63 B 391 + 2,39 PT 390 + 4,51 C 390 + 3,39 390 + 0,00 D 389 + 3,39
10,00 0,00 7,88 9,00 3,39 10,00
PENDIENTE TRANSVERSAL 0,01193 . 3,00 0,01751 . 3,00 0,02. 3,00 0,025 . 3,00
0,01751 . 3,00 0,02 . 3,00 0,02169 . 3,00 0,025 . 3,00
0,02 0,02 . 3,06 0,02 . 3,20 0,02 . 3,35 0,02 . 6,70 0,02 0,02 . 3,20 0,02 . 3,35 0,02169. 6,47 0,025 . 6,70
Z entre BE Y Eje 0,0358 + 0,0525 + 0,06 + 0,075
0,525 + 0,06 + 0,065 0,075
Z entre BI Y BE 0,06 0,061 = 0,097 0,064 = 0,116 0,067 = 0,127 0,1675 0,06 0,064 = 0,116 0,067 = 0,127 0,140 0,1675
TABLA 5.24 - Para calcular la diferencia de elevación entre el borde exterior y el eje de la vía; y entre el borde exterior y el borde interior en la EST 366 + 0,00, se procede de la forma siguiente: Z1 entre el borde exterior y el eje:
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Z1
0,0358m
Z2 entre el borde interior y el borde exterior: Z2 Z2
b a / 2 W (366) 0,02 6,00 / 2 0,06 Z2
Z1 0,0358
0,097m
- Para calcular la diferencia de elevación entre el borde exterior y el eje de la vía; y entre el borde exterior y el borde interior en la EST 390 + 0,00, se procede de la forma siguiente: entre el borde exterior y el eje Z 1
Factor ( 2).dist (C 390) Z 1
(0,0005.3,39 Z1
b .a / 2
0,02).3,00
0,065m
Z2 entre el borde interior y el borde exterior: Z 2
Factor ( 2).dist (C 390) Z 2
(0,0005.3,39 Z2
b . a W (390)
0,02). 6,00
0,47
0,14m
Estos valores aparecen en la tabla 5.24. El registro de superelevación se encuentra en la tabla 5.25; y se calculó de la forma siguiente: Por ejemplo, para calcular lo que se deprimen el borde interior y el eje de la vía, con relación al eje calculado en la estación EST 366 + 0,00,se procede de la siguiente forma: - Para el borde interior: S(BI) = 0,097 + (a/2) . 0,02
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CARRETERA CARRETERA VD = 80 k/h Curva No = 4 (Clotoide) b = 2% a = 6,00 m = 24º12’ B = 0,06 m CD = 10,00 m Gc = 2º00’ AB= BC= 9,00 m pend(C-D) = 0,002375 m/m Rc = 572,96 m e = 0,025 m/m Dc = 242,00 m S = 0,1675 m CON ENSACHE= 0,70 m GIRO = BE PRIMER TANGENTE CIRCULAR SEGUNDA TANGENTE EST. BI EJE BE EST. BI EJE BE EST. BI EJE BE A 364+5,63 R-0,06
R
R-0,06
B 365+4,63 R-0,12
R-0,06
R-0,06
366+0,00 R-0,15
R-0,096
R-0,06
PC 366+2,51 R0,176
R-0,112
R-0,06
PC 366 +2,51 C 366 +3,63 D 367 +3,63
R-0,176
R-0,112
R-0,06
A 392+1,39
R-0,06
R-0,187
R-0,12
R-0,06
392+0,00
R-0,07 R-0,01
R-0,06
R-0,2275 R-0,135
R-0,06
B 391+2,39
R-0,12 R-0,06
R-0,06
368 +0,00
R-0,2275 R-0,135
R-0,06
PT 390+4,51
R0,176
R+0,06
370 +0,00 372 +0,00 . . . .
R-0,2275 R-0,135
R-0,06
R-0,2275 . . . .
R-0,06 . . . .
388 +0,00 D 389 +3,39
R-0,2275 R-0,135
R-0,06
R-0,2275 R-0,135
R-0,06
390
R-0,20
R-0,135 . . . .
R-0,125
R-0,06
R
R0,112
R-0,06
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S ( EJE )
0,0358 (a / 2)b S ( EJE )
0,0358
(6,00 / 2)0,02
0,096m.
Para calcular lo que se deprimen el borde interior y el eje de la vía, con relación al eje calculado en la EST 390 + 0,00,se procede de la siguiente forma: - Para el borde interior:
S ( BI )
0,140
(a / 2)b S(BI)
0,140
(6,00 / 2)0,02
0,200m
- Para el eje deprimido:
S ( EJE )
0,065
( a / 2)b S(EJE)
0,065
(6,00 / 2)0,02
0,125m.
Estos valores aparecen en la tabla 5.25 sin aproximar al centímetro, para la mejor comprensión de los cálculos.
5.6 EJEMPLOS COMBINADOS DE SUPERELEVACIÓN Y SOBREANCHO EN LAS CURVAS DE TRANSICIÓN. 5.6.1 GIRO POR EL EJE.
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Por la expresión (5.36) se determina la superelevación máxima:
S a.e
0,45m
(7,00 0,50).0,06
La longitud de los tramos AB = BC = N; se determino por la expresión (5.30): N = 14,00 m Por la expresión (5.41) se determina la longitud del tramo CD: CD
ls
BC
75,00 14,00
61,00m
Por la expresión (5.38) se comprueba la pendiente del tramo CD: pend(CD)
S/ 2 B
0,45 / 2 0,07
CD
61,00
0,002541m / m
menor que la máxima permisible de 1/200 = 0,005 m/m. - Cálculo de las estaciones básicas para el desarrollo de la superelevación: No varían con relación a las determinadas en el epígrafe 5.3.1.1. Por lo tanto: A LA ENTRADA DE LA CURVA: EST A = 189 + 4,71 EST B = 190 + 8,71 (TS) EST C = 192 + 2,71
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- Cálculo de los factores: No varían con relación a los calculados en el epígrafe 5.3.1.1: Factor (1) = b/BC = 0,02/61,00 = 0,00142857 m/m/m Factor (2)
e b
0,06 0,02
CD
61,00
0,0006557m / m / m
- Cálculo del prerregistro de superelevación: En la tabla 5.26 se observa el prerregistro de superelevación. Así; por ejemplo, para calcular lo que se eleva y deprime el borde exterior e interior, respectivamente, en la estación EST 192 + 0,00; se procede de la forma siguiente: - Para el borde exterior: S 192 S 192
Factor (1). DIST ( B 192).a / 2 0,00142857.11,29(7,00 / 2) S 192
ESTACION TS(B) 190 + 8,71 192 + 0,00 C 192 + 2,71 194 + 0,00 196 + 0,00 198 + 0,00
DIST. 0,00 11,29 14,00 17,29 37,29 57,29
0,056m
PENDIENTE TRANSVERSAL 0,01613 . 3,50 0,2 . 3,50 0,03134 . 3,50 0,4445 . 3,50 0,05756 . 3,50
0,02 . 3,51 0,02 . 3,52 0,03134 . 3,55 0,4445 . 3,82 0,05756 . 3,99
BI 0,0702 0,0704 0,111 0,170 0,230
BE 0,056 0,07 0,110 0,156 0,201
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S 192
Factor (1). DIST ( B 192). a / 2 W (192)
S 192
(0,00142857.11,29).((7,00 / 2) S 192
0,01)
0,0702m
Estos valores calculados aparecen en la tabla 5.26 para dichas estaciones. Para calcular lo que se eleva y deprime el borde exterior e interior, respectivamente, en la EST 212 + 0,00 ;se procede de la forma siguiente: - Para el borde exterior: S 212 S 212
( Factor ( 2). DIST (C 212)
b).a / 2
(0,0006557.49,71 0,02).7,00 / 2 S 212
0,184m
- Para el borde interior: S 212 S 212
Factor (2). DIST (C 212)
b . a / 2 W (212)
0,0006557.49,71 0,02 . 7,00 / 2 S 212
0,49
0,21m
Estos valores calculados aparecen en la tabla 5.26 para dichas estaciones. El registro de superelevación aparece en la tabla 5.27; el tramo AB se calcula por el concepto de
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CARRETERA CARRETERA VD = 80 k/h b = 2% a = 7,00 m B = 0,07 m pend(A-C) = 0,005 m/m AB= BC= 14,00 m pend(C-D) = 0,002541 m/m e = 0,06 m/m S = 0,45 m CON ENSANCHE = 0,50 PRIMERA RECTA EST. BI EJE BE A 189+4,71 190+0,00 TS(B) 190+8,71
EST.
R-0,07
R
R-0,07
R-0,07
R
R-0,04
R-0,07
R
R
PRIMERA CLOTOIDE BI EJE BE
TS(B) 190+8,71 R-0,07 192+0,00 R0,0702 C 192+2,71 R0,0704
Curva No = 4 (Clotoide) = 20º00’ Gc = 2º00’ Rc = 572,96 m GIRO = EJE EST.
SEGUNDA RECTA BI EJE BE
A 219+7,71 219+0,00 ST(B) 218+3,71
ES E ST.
CIRCULAR BI EJE
BE
R
R
R
R+0,056 R+0,056
200+0,00
R-0,24
R
R+0,21
218+0,00
R
R+0,07
202+0,00
R-0,24
R
R+0,21
C 216+9,71
R
R
R-0,07
R-0,07
R
R-0,03
R-0,07
R
R
SEGUNDA CLOTOIDE EST. BI EJE BE
SC(D) 198+3,71
R-0,24
R-0,07
R+0,21
ST(B) 218+3,71
R-0,07
R
R-0,18
R
R-0,018
R0,0704
R
R-0,07
R
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TS(B)190+8,71 192 + 0,00 C 192 + 2,71
0,00 11,29 14,00
ESTACION
DIST.
194 + 0,00 196 + 0,00 198 + 0,00 SC(D)198+3,91
17,29 37,29 57,29 61,00
ESTACION
DIST.
ST(B)218+3,71 218 + 0,00 C 316 + 9,71
0,00 3,71 14,00
ESTACION
DIST.
216 + 0,00 214 + 0,00 212 + 0,00 CS(D)210+8,71
9,71 29,71 49,71 61,00
0,01613.3,50 0,02 . 3,20
0,02 . 3,51 0,02 . 3,52
PENDIENTE TRANSVERSAL 0,3134.3,55 0,4445.3,82 0,05756.3,99 0,06 . 4,00
0,03134.7,05 0,4445.7,32 0,05756.7,49 0,06 . 7,50
PENDIENTE TRANSVERSAL POR EL SEMIANCHO DE LA VIA. 0,00530.3,50 0,02 . 3,52
0,00530.3,50 0,02 . 3,50
PENDIENTE TRANSVERSAL 0,02637.3,56 0,03948.3,85 0,05259.3,99 0,06 . 4,00
0,02637.7,06 0,03948.7,35 0,05259.7,49 0,06 . 7,50
TABLA 5.28
Z entre el borde interior y el eje:
-
-
0,0702 0,0704
0,056 0,07
Z entre BI Y Eje
Z entre BI Y BE
0,1113 0,170 0,230 0,24
0,221 0,326 0,431 0,45
BI
BE
-
-
0,018 0,0704
0,018 0,07
Z entre BI Y Eje
Z entre BI Y BE
0,094 0,152 0,210 0,24
0,186 0,290 0,394 0,45
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CARRETERA CARRETERA b = 2% B = 0,07 m AB= BC= 14,00 m e = 0,06 m/m S = 0,45 m EST. A 189+4,71 190+0,00 TS(B) 190+8,71
EST. TS(B) 190+8,71
VD = 80 k/h a = 7,00 m pend(A-C) = 0,005 m/m pend(C-D) = 0,002541 m/m
Curva No = 4 (Clotoide) = 20º00’ Gc = 2º00’ Rc = 572,96 m
CON ENSANCHE = 0,50
PRIMERA RECTA BI EJE
BE
EST. A 219+7,71
R-0,07
R
R-0,07
R-0,07
R
R-0,04
R-0,07
R
R
PRIMERA CLOTOIDE BI EJE
GIRO = BI
219+0,00 ST(B) 218+3,71
CIRCULAR BI EJE
BE
EST.
R
SC(D) 198+3,71
R-0,10
BE
EST.
R+0,14
R+0,35
ST(B) 218+3,71
R-0,07
R
192+0,00 C 192+2,71
R-0,0702
R
R+0,056
200+0,00
R-0,10
R+0,14
R+0,35
R-0,0704
R
R+0,07
202+0,00
R-0,10
R+0,14
194+0,00
R-0,0715
R+0,04
R+0,15
204+0,00
R-0,10
196+0,00
R-0,084
R+0,086
R+0,242
206+0,00
198+0,00 SC(D) 198+3,71
R-0,098
R+0,132
R+0,333
R-0,010
R+0,14
R+0,35
SEGUNDA RECTA BI EJE
BE
R-0,07
R
R-0,07
R-0,07
R
R-0,03
R-0,07
R
R
SEGUNDA CLOTOIDE BI EJE
BE
R-0,07
R
R
R-0,18
R
R+0,037
R+0,35
218+0,00 C 216+9,71
R-0,0704
R
R+0,07
R+0,14
R+0,35
216+0,00
R-0,072
R+0,022
R+0,114
R-0,10
R+0,14
R+0,35
214+0,00
R-0,84
R+0,068
R+0,206
208+0,00
R-0,10
R+0,14
R+0,35
R-0,096
R+0,114
R+0,30
210+0,00 CS(D) 210+8,71
R-0,10
R+0,14
R+0,35
212+0,00 CS(D) 210+8,71
R-0,10
R+0,14
R+0,35
R-0,10
R+0,14
R+0,35
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El valor de 0,170 es la diferencia de elevación entre el borde interior y el eje elevado en la EST 196 + 0,00; el cual aparece en la tabla 5.28; y el valor de S(BI) es el anteriormente calculado.
- Para el borde exterior: S(BE)
0,326 S( BI) S( BE)
0,326 0,084
0,24m
El valor de 0,326 es la diferencia de elevación entre el borde interior y el borde exterior en la EST 196 + 0,00;el cual aparece en la tabla 5.28 ; y el valor de S(BI) es el anteriormente calculado para el borde interior. Estos valores aparecen en la tabla 5.29 sin aproximar al centímetro, para la mejor comprensión de los cálculos.
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Z1
0,156m
Z2 entre el borde interior y el borde b orde exterior:
Z 2 Z 2
Factor (2).dist (C 196) (0,0006557.37,239 Z2
b . a W (196)
0,02). 7,00
0,32
0,326m
Estos valores aparecen en la tabla 5.30.
ESTACION TS(B)190+8,71 192 + 0,00 C 192 + 2,71 194 + 0,00 196 + 0,00 198 + 0,00 SC(D)198+3,71
DIST. 0,00 11,29 14,00 17,29 37,29 57,29 61,00
PENDIENTE TRANSVERSAL 0,01613 0,02 0,03134 0,4445 0,05756 0,06
Z entre BE Y Eje . 3,50 = 0,056 . 3,50 = 0,07 . 3,50 = 0,110 . 3,50 = 0,156 . 3,50 = 0,201 . 3,50 = 0,21
Z entre BI Y Eje + 0,02.3,51 = 0,126 + 0,02.3,52 = 0,14 . 7,05 = 0,221 . 7,32 = 0,326 . 7,49 = 0,431 . 7,50 = 0,45
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CARRETERA CARRETERA b = 2% B = 0,07 m AB= BC= 14,00 m e = 0,06 m/m S = 0,42 m EST. A 189+4,71 190+0,00 TS(B) 190+8,71
VD = 80 k/h a = 7,00 m pend(A-C) = 0,005 m/m pend(C-D) = 0,002541 m/m
Curva No = 4 (Clotoide) = 20º00’ Gc = 2º00’ Rc = 572,96 m
CON ENSANCHE = 0,50
PRIMERA RECTA BI EJE
BE
EST. A 219+7,71
R-0,07
R
R-0,07
R-0,110
R-0,03
R-0,07
R-0,14
R-0,07
R-0,07
PRIMERA CLOTOIDE
GIRO = BE
219+0,00 ST(B) 218+3,71
CIRCULAR
SEGUNDA RECTA BI EJE
BE
R-0,07
R
R-0,07
R-0,09
R-0,02
R-0,07
R-0,14
R-0,07
R-0,07
SEGUNDA CLOTOIDE
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El paseo correspondiente al lado deprimido (borde interior), podrá mantener su pendiente de proyecto (sección transversal típica), si se cumple que: Pendiente del paseo > e p donde: e p: peralte en una sección cualquiera de la curva. Cuando la condición anterior no se cumple, debe suministrarse al paseo una pendiente igual al peralte de la sección transversal, con lo cual se consigue un plano inclinado con una pendiente igual a ep. Para el paseo correspondiente al borde exterior de la curva, se recomienda que la diferencia algebraica entre la pendiente del paseo y la pendiente de la sección transversal (e p), cumpla con la condición:
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10- Woods, K. Highway Engineering Handbook. Tomo II. Instituto del Libro. La Habana, 1970. 11- Benítez, R. Proyecto de carreteras. Empresa de producciones del MES. La Habana, 1988. 12- Seelye, E. Desing. Jonh Wiley and Son. USA, 1956. 13- Jeuffroy, G. Proyecto y construcción de carreteras. Editorial Científico Técnica. La Habana, 1977. 14- Oglesby, C. H y Hewes, L. Ingeniería de carreteras. Compañía Continental S. A. México, 1960. 15- Ruhle, F. G. Norma de diseño geométrico de caminos rurales. Dirección Nacional de vialidad. Buenos Aires, 1967. 16- Manual de Proyecto geométrico de carreteras. México, 1977.