Assignment (Curve Design Calculations) - PowerPoint Presentation

Slide1

Assignment

(Curve Design Calculations)

By

Anil

Choudhary

SPTM

Slide2

Slide3

Important Limits

Ca max=165 mm

Cd =100mm if speed more than 100 Kmph

(PCE’s approval)

= 75 mm otherwise

Cex

=75 mm

Rmin

=175 m

Rca

and

Rcd

=35mm/s normally but

upto

55 mm/s

Cant Gradient=1 in 720 but min 1 in 360

(already taken in TL formula given above)

Slide4

Find equilibrium cant for the maximum speed

Find minimum cant required by deducting the cant deficiency from equilibrium cant

Find cant required for booked speed of goods trains.

Add cant excess and find out the maximum cant permissible

The cant to be provided shall be between the two values computed above

Procedure for designing a Curve

(known Vmax, Vg

, Veq, R)

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Possible situations of Ca

Carp

Carg

Ca

Cex

Cd

Carp

Carg

Ca

Cex

Cd

Slide6

Cant to be Provided shall also be less than the Maximum Permissible as per IRPWM

Corresponding to actual cant provided, find maximum speed

Find out the desirable/ minimum transition length

Procedure to find Speed on Curve

Slide7

Find Maximum Permissible Speed for –

BG route having Speed Potential = 130 Kmph

Degree of Curve = 2°

Speed of Goods Train = 65 Kmph

C

d

= 100 mm and C

ex = 75 mmExercise

SE = 140 mm

Max. Speed = 123.73 Kmph ≈ 120 Kmph

Slide8

Ca= GV^2/(127*R)

Goods= 66.53 mm Ca=66.53+75= 141.53

Passenger= 266.14 mm, Ca= 266.14-100=166.14 mm

So Ca 141.53 taking goods train case

Cd for passenger is 266.14-144.53= 121.87 which is more than 100 mm so V to be reduced as per 140 mm Ca.

Vp

= 0.27Sqrf(R(Ca+Cd) =0.27Sqrt (875*(140+100)= 123.729 Kmph

Slide9

Find Desirable and Minimum Transition Length for –

BG route having Speed Potential = 130 Kmph

Degree of Curve = 2°

Speed of Goods Train = 65 Kmph

Exercise

C

d

= 100 mm and

C

ex

= 75 mm

SE = 140 mm

Max. Speed = 120 Kmph

Desirable Length

L

1

= 134.4 m

L

2

= 96/82.99 m (Cd=86.45 mm)

L

3

= 100.8 m

Minimum Length

L

1

= 89.6 m

L

2

= 55.33 m

L

3

= 50.4 m

L=140m and

Lmin

=90m

Slide10

Calculate Shift for –

BG route having Speed Potential = 130 Kmph

Degree of Curve = 2°

Speed of Goods Train = 65 Kmph

Exercise

C

d

= 100 mm; and

C

ex

= 75 mm

SE = 140 mm

Max. Speed = 120 Kmph

Desirable Length of Transition

L

1

= 140 m

Minimum Length of Transition

L

1

= 90 m

0.933 m and 0.385 m

Slide11

Design Case-I

(

Vax,Vmin

and

Veq

Given)

On a Broad Gauge Group 'C' route , a 600 meter radius curve is to be introduced. The maximum sectional speed is 110 km/h and the booked speed of goods train is 50 km/h. Equilibrium speed is fixed as 80 km/h. Design the curve.

Slide12

Given values –

R = 600 m,

V

max

= 110 km/h,

V

goods

= 50 km/h,Veq = 80 km/h.Design a curve ?Cant, Maximum permissible speed and transition length

Slide13

Limiting values for solving the problem:-

Maxm

Cant (Ca) -165 mmMaxm

Cant Excess(

Cex

) - 75 mm

Maxm Cant Deficiency (Cd) - 75 mm ?Cant gradient - 1 in 360 ( 2.8 mm/m)Rate of change of Ca/Cd - 55mm/s

Slide14

Formulas

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Solution:

Step-1

Calculate actual cant C

a

for equilibrium speed

V

eqCheck- Ca not to be more than maxm cant permittedStep-2Calculate maximum permissible speed on the curveVprm=0.27

Ca and Cd ? 

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Step-3

If

Vprm<Vmax implies PSR of

Vprm

If

Vprm

>Vmax implies Vprm is VmaxCheck: CexCex=Ca-

If Cex is more than maxm permitted then Ca to be revised (means ca for Veq is evn not possible)Ca=Cex+ 

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For changed Ca, revise/check

Vprm

So check Cd

Cd=

-Ca

(normally will be ok)

Finalised

Vprm and CaStep 4: Transition length calculationMaxm of L1,L2 and L3And in no case less than 2/3 L1, 2/3 L2 and ½ of L3 

Slide18

Check

rate of change of cant, cant deficiency and cant gradient in transition

-Time of travel(t)(sec)=3.6 L(m)/Vprm(

Kmph

)

-Rate of change of cant= Ca/t

-Rate of change of Cant deficiency=Cd/t-Cant gradient= Ca/LFinalize Transition lengthSo curve parameters Ca, Vprm and L finalised

Slide19

Values:

Step 1

Ca=146.78 mm <165mm so OK

So

Ca=145 mm

Step 2

Vprm=98.09 Kmph say 95 KmphStep 3Vprm<Vmax so Vprm=95

kmphCex=145-=87.67 mm But Cex can not exceed 75 mm 

Slide20

Revise Ca

Ca=Cex+

=132.3 mm So

Ca= 130 mm

Vprm

=0.27

Vprm

=94.69 Kmphsay 90 KmphActual Cd for this Vprm is Cd=-130=55.76 mm 

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Transition Length:-

Ca=130mm, Cd=55.76mm,

Vprm=130KmphL1=93.6mL2=40.14m

L3=93.6m

L=100m

Check:

t=3.6*100/90=4 secRate of change of Ca=130/4=32.5 mm/s okRate of change of Cd=55.76/4=13.9 mm/s okCant Gradient=130/100000=1/770 ok

Slide22

Curve detail:

Ca=130 mm

Vprm=90 KmphL=100m

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Design Case-2

Vmax and

Vmin known

Step-1

(Try to provide cant so that no SR is imposed)

Calculate actual cant

C’a for VmaxCa= Ca’-Cd

Slide24

Step-2

Check:

Cex

Cex

=Ca-

(if Ok Ca is finalized)

If

Cex is more than maxm permitted then Ca to be revised Ca=Cex+ (Revised cant) 

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Step-3

Calculate maximum permissible speed on the curve

Vprm

=0.27

Ca and Cd ?

SR of

Vprm to be imposed  

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For changed Ca, revise/check

Vprm

So check Cd

Cd=

-Ca

(normally will be ok)

Finalised

Vprm and CaStep 4: Transition length calculationMaxm of L1,L2 and L3And in no case less than 2/3 L1, 2/3 L2 and ½ of L3 

Slide27

Check

rate of change of cant, cant deficiency and cant gradient in transition

-Time of travel(t)(sec)=3.6 L(m)/Vprm(

Kmph

)

-Rate of change of cant= Ca/t

-Rate of change of Cant deficiency=Cd/t-Cant gradient= Ca/LFinalize Transition lengthSo curve parameters Ca, Vprm and L finalised

Slide28

Speed potential of existing curve

Known values-

R, L, Ca

To find V

Slide29

Example

Find the

speed potential

of a curve of

R=875 m

Ca =100 mm

L=100m/ 60m/50mOther limiting valuesCd and Cex=75 mmRca & Rcd=55 mm/sFind

maxm speed potential for circular curve and corresponding TL

Slide30

Case-I TL is adequate

Vmax= 0.27SQRT(R*(

Ca+Cd

))

= 105.6542 Kmph say 105 Kmph

Check for adequacy of Transition Length

L1=0.008*Ca*Vmax=84 m (min =56m)L2=0.008*Cd*Vmax=62 m (min=41m) (Cd=73.376 mm)L3=0.72*Ca*V=72 m (min=36m)So for normal limits TL required is 84 m while under limiting condition TL required is 56 m

(So for Vmax of 100 Kmph, TL of 100m is adequate and even 60 m adequate if we go to limiting values)So speed potential of curve is 105 Kmph for TL of 100m and 60m

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Case 2 TL is inadequate

What if

TL is 50 m

TL is more than 36 m so is ok for limiting twist consideration.

TL is also ok from limiting Cd consideration (Min req-41m) for speed of 105 kmph.

Minm

TL required from Rca consideration for 105 Kmph is 56.2 m so given 50 m is inadequate.

CheckRca=Ca/(L/V)=100*105/(3.6*50)= 58.33 mm/sNot Ok

Slide32

Case 2

Contd

Thus

speed restriction

to be imposed from

Rca

Consideration-Rca=Ca/(L/V) V=3.6*Rca*L/Ca

we get V=99 Kmph for Rca of 55mm/sec so Take Vp=95 Kmph

Slide33

Case 2

Contd

Check

Rcd

(actual) for TL=50m

Actual Cd = 13.76*(95*95)/875-100=42 mm

Time for negotiating TL =50/(95*1000/3600)=1.89 secRcd=42/1.89=22.22 mm/sec so OkSo Speed permitted is 95 Kmph for TL =50m

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Case-3 TL is inadequate, to find optimum cant

In the above example

R=875m, Ca=100mm,TL=50 m

Circular curve speed potential-105 kmph

(0.27*SQRT(R*(

Ca+Cd

))TL restricts speed to 95 kmph (V=3.6*Rca*L/Ca)Why to have so high Ca, it can be reworked as

But Ca and Cd are also interrelated i.e. if Ca increased Cd is decreased

Slide35

Case 3

contd

(How to decide Ca and Cd)

Best speed due to

limited TL

can be obtained when

Rca=Ca*V/3.6*L i.e V=3.6*Rca*L/Ca

Rcd=Cd*V/3.6*L i.e V=3.6*Rcd*L/CdCondition-1Both will be balanced when Ca=Cd (Optimum speed) from transition length consideration

Slide36

Case 3

contd

Condition-2

Also overall best result will be when speed permitted in circular is equal to speed permitted on transition.

Combining these:

0.27SQRT(R(

Ca+Ca)=L*

Rca*3.6/Ca Ca=4.4628((L*Rca)2/R)1/3 =91.585 mm - 90 mm say

Slide37

Case 3

contd

Vcircular

=0.27*SQRT(R*(

Ca+Cd

))= 102.59 Kmph

Actual Cd=13.76*(100*100)^2/875-90=67.257 mmFor TL-V1=50*55*3.6/90=110 kmph (For Rca=55) V2=50*55*3.6/67.257=147.19 kmph (Rcd

=55)So speed increased to 102.59 i.e 100 Kmph by reducing Ca from 100mm to 90 mm.

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Thank You