printing for transition curves lectures

7/25/2019 Printing for Transition Curves Lectures

1/15

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


2/15

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


3/15

-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

=


4/15

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

%


5/15

-> = --$= ,. 1t - = 0 and r = A 1t -$ = ,

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,


6/15

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


7/15

'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


8/15

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


9/15

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

+$3L (RL R$)


10/15

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


11/15

(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

=

$$


12/15

#angent lengths

"ro& co&pound 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 $$$ ,

, +++

$


13/15

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 &


14/15

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

=

$%


15/15

#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 ++++++=

$!

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