Post on 25-Feb-2018
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M = Mass of vehicle
V = Speed of vehicle
R = Radius of Road Curve
= Coefficient of friction ( adhesion)between wheels and road
(usually = 0.!)
See "i#. $%.$ for forces actin# on a vehicle&ovin# at a speed of V on a circular curve
of radius R.
Sideslip will occur if M#R
MV
i.e. when R#
V
Mini&u& curvature#VR
=
Fig. 14.1"orces actin# on a vehicle at a curve on level#round
'o "riction
#R
Vtan*
=
+
Fig. 14.2 "orces actin# on a vehicle at a curve on slopin# #round
$
tan* =#R
0.%V
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Fig. 14.3 Chan#e of radial acceleration when enterin# a circular curve fro& a strai#htsection.
r=
Circularcurve
r=R
,
R
r
l
-ran
sition -ransition
"inalse
ction
.nitial
secti
on
Fig. 14.4 "or&ation of transition curves between strai#ht and circular sections
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-he initial re/uire&ent in the desi#n of a transition curve is to find len#th , of the transition curv
&ay be taena) 1s an arbitrary value(say !0 &)
b) Such that the cant is applied at a constant rate (say 0.$ & in $00&)
c) Such that the rate of chan#e of radial acceleration e/uals a chosen value (say 0.2 &3sec2)
4hen the rate of chan#e of radial acceleration is the desi#n criterion
5iven
, = -otal len#th of transition curve
R = Radius of circular curve
V = 6nifor& velocity of vehicle
-he radial acceleration before enterin# the transition curve is 7ero. -he radial acceleration on
circular curveR
V = .
-he ti&e taen to travel alon# the transition curve = V,
-he rate of chan#e of radial acceleration (a ) is
,R
V
V,
)0R
V( 2
=
=a
Ra
VL
2
=
8
S
R
R9S
-$-
6
:
- ;
S
S
,
-ransition
=
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t can be shown that
( )R
LS
%$
= ( ) ( )
tan
SRIB +=
( )
,2 =BT ( ) ( )
,
tan% ++=
SRIT
y
S=
Shift
Circul
arCircular
$
>
8
$
?
@$M:
R$
y
y
' -$
r
R
;
-
Fig. 14.6 Shape function for a transition curve
%
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-> = --$= ,. 1t - = 0 and r = A 1t -$ = ,
and r = R
'ote that the centrifu#al force " at any point on the transition curve is proportional to the
distance of that point fro& the startin# point of the curve.
( )constantrorr$-and-between
rMV" $
a=
1t -$ = R,. 1lso = r R,r
$ =
>? = B rB = B B =r
$B =
R,
B
:y inte#ration =R,
9 C
at = 0 = 0. C = 0 henceR,
=
Cubic Spiral
f we assu&e to be s&all ('oteD in ordinary transition is very s&all)
R,
CB
By
B
By
===
.BR,
By
=
or
Relationship between an !
t can be shown tan )2!
C($
2
C
)$0
EFG($
)$%
$(
yB
+=
==
"or s&all
Since =R,
'ow deflection of points at distances fro& the tan#ent point - can be calculated.
Cubic "arabola
f we assu&e is s&all and also
!
FR,y
2
=
B =
B =FR,
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FR,
y2
=
FR,
= 2
-husFR,
= 2=y . -his is an e/uation for a Cubic Parabola.
'ow it is possible to co&pute offset distances(y) off the tan#ent for distances() alon#the tan#ent.
#ransition Curve Setting$out %ata
R,
= at -$DR
,
R,
,
$ == :M = '-$(&ai&u& offset)
Shift S = :;
= :ME;M
= '-$E(;8EM8)
=FR,
,2E(RERcos$)
= ....)H%I
I
R($JR
FR,
,%
$
$
2
++
#norin# hi#her powers than $
S =
R
FR,
,
$2
= 2
)R,
,(
R
FR,
,
=
KR,
FR,
%R
,S
=
F
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'ow @$-$ ;-$
= R.$= R.
,
R,
,=
@$is the &id point of the transition curve.
Since the deviation of @$fro& the tan#ent is s&all -@$ -: =
,
"ro& theory of si&ple circular curves
: = (R9S) tan
*
- = : 9 -:
'ow you can setEout
Cubic Spiral L with -heodolite(B) Chain() and -ape(y)
Cubic ;arabola L with Chain() and -ape(y)
ither calculateFR,
y2
= orFR,
=y2
= orFR,
B
=
Example 1:Calculate the settin# out data for a N!.0 & transition curve to connect an KO
circular curve Poinin# two strai#hts with an an#le of deflection 0O usin# $! & chords.
a) -o calculate R and S
)
$K0
QK(
$00R=
= N$F.0 &
S =%R
,=
N$F.%
N!
= 0.2N &
b) -o calculate -an#ent len#th
-, = (R9S) tan
09
,
= $F2.K%2 &
c) -o calculate and tabulate the deflection an#les and deflection offsets
S=(rad)
FR,2
)R,
(
2
C
==
= &inR,
!N.FF0&in
Q
$K0.
FR,
=
andFR,
y2
=
R = N$F. & , = N! & = $! 20 %! F0 N! &
N
- = (R9S) tan
*9
,
*
$00
R
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3& 3&R,
!N.F =
FR,y
2
= 3&
$! ! T%U 0.0$0
20 00 T2FU 0.0K%
%! 0! $T2FU
0.K2
F0 2F00 2KT
%U0.FN0
N! !F! F0T0
0U$.20
K
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Fig. 14.&Sa&e and Fig. 14.1'8pposite and
l
R$ W R
(l = , E,$)
CC
C$C$
S2
;2
;
8
8$
@
;
;$
R
S
S$
R$
y
Fig. 14.11
$
$$
%R
,S
,$$ =C
%R
,S
,=C
X = CL C$ and Xy = R$ 9 S$E (R 9 S)
8$@ = 8$89 8; 9 ;@
R$= (X
9 Xy
)$3
9 R9 S2
S2= Y (CL C$)9 JR$ 9 S$E (R 9 S)H
+$3L (RL R$)
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t$
t9,
3
t$9,$3
-
,
-$
R
$
;2
b
a
,$
;%
;
8
8$
@
;
;$
R$
Fig. 14.12
a = rate of chan#e of radial acceleration
aR
V2=L
.a
V2curveofpartsallalo!costa=
R, is a constant for all three transitions.
$0
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(E&
);
@
;
&
a
b
;2
FR,
&b
2
= FR,
&)E(a
2
=
Fig. 14.13
baS +=2FR,
)&2&2( 2
2
+=S
FR,
&)F2( 2 +=
"m
"S
S2is &ini&u& alon# ;@ (on 8$8produced)at ;@ when a 9 b = S2 0FR,
&)F2( =
+
=m
-he transition is bisected by the clearance S2
%KR,
S 22 === ba
%R,S
2
2
=
$$
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#angent lengths
"ro& co£ curve e/uations
t$SinZ = (RL R$)($ECos) 9 R$($ECosZ)
tSinZ = (R$L R)($ECos$) 9 R($ECosZ)Replace R$with (R$ 9 S$) and Rwith (R9S) and
apply in transitions. 4e have
t$SinZ = (R9S) E (R$9S$) CosZ 9 (R$LRLS2) CostSinZ = (R$9S$) E (R 9 S) CosZ E (R$LRLS2) Cos$
-$ = t$9 ,$ - = t9
,
#hroughchainages
1rc;$;=R$($)radE ,$
1rc
;2;% = R()radE ,
-otal curve len#th = -$;$9 ;$;9 ;;29 ;2;%9 ;%-
= ,$9 R$($)radE ,$
+9
9 R()radE ,
+9 ,
= ( ) ( ) ra"ra" RR $$$ ,
, +++
$
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Example 3: 0000N! = o 000020 = o
R$ = 200.00 & R = !0.00 & V = K0 &3h
a = 0.2 &3s2 Chaina#e = KN.N00 &
(-ransition curves with sa&e hand)
0000%!20N!$ === ooo m2$.$%F!00.2
)F0F0
$0K0(
aR
V,
22
2
=
==
mL 22.$$200
2F!N.N$
== m0F!.200%
22.$$
%R
,S
$
$$
=
==
m!FK.2!0%
2$.$%F%R,S
===
= ,E ,$ = $%F.2$ L $$.22 = %.2KF &
(Shift) mS 0$N.02F!N.N%
%.2KF
%R,
%.2KF 22
2 =
==
'ow calculate offsets to the curve fro& tan#ents
usin#FR,
=y
2
= for transition and fro& usual offsets or
deflection an#les for circular sections.
$2
R, = $%F.2$ !0 =2F!N.N &
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Curves of (pposite )an
-$
$
;$
R$8$
=
;
8
R
;%
;2
S2= (a9b)
-
y
b
a
@
S$
S
$
$
Fig. 13.14
;@ = & @;2= E& FR,&
a2
= FR,
&)E(b
2
=
FR,
)&2&2( 2
2
+
=+= baS 0&F2d&
dS
2
=+= #$t
=m
%KR,
S 22 === ba
%R,S
2
2
=
$%
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#hrough chainage
-$-= -$;$9 ;$;9 ;;29 ;2;%9 ;%-
= ,$9 R$$9 9 R9 ,
=
++
+
,
,
$$$$
RRL
= ( )$$$
,
RR
L+++
#angent length
-$$= (R$ 9 S$)tan ,
Z $$+ = $@
-= (R 9 S)tan ,
Z + = @
'ote D
S2= (X9 Xy)$3L (R$9 R)
:ut X = C$9 C C$= ,$ and C=
,
Xy = (R$ 9 S$) 9 (R9 S)
( ) ( )[ ] ( )$$
$$
$2 RRSRSRCCS ++++++=
$!