04 Assignment on Curve .pptx layout of simple and transition curve

dfgjffujgfhfnm

1. Assignment (Curve Design Calculations)

2. 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)

3. 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)

4. Possible situations of Ca Carp Carg Ca Cex Cd Carp Carg Ca Cex Cd

5. 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

6. 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 mm Exercise SE = 140 mm Max. Speed = 123.73 Kmph ≈ 120 Kmph

7. 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

8. 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

9. 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

10. 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.

11. Given values – R = 600 m, V max = 110 km/h, V goods = 50 km/h, V eq = 80 km/h. Design a curve ? Cant, Maximum permissible speed and transition length

12. Limiting values for solving the problem:- Maxm Cant (Ca) -165 mm Maxm 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

13. Formulas

14. Solution: Step-1 Calculate actual cant C a for equilibrium speed V eq Check- Ca not to be more than maxm cant permitted Step-2 Calculate maximum permissible speed on the curve Vprm =0.27 Ca and Cd ?  

15. Step-3 If Vprm <Vmax implies PSR of Vprm If Vprm >Vmax implies Vprm is Vmax Check: Cex Cex =Ca- If Cex is more than maxm permitted then Ca to be revised (means ca for Veq is evn not possible) Ca=Cex+  

16. For changed Ca, revise/check Vprm So check Cd Cd= -Ca (normally will be ok) Finalised Vprm and Ca Step 4: Transition length calculation Maxm of L1,L2 and L3 And in no case less than 2/3 L1, 2/3 L2 and ½ of L3  

17. 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/L Finalize Transition length So curve parameters Ca, Vprm and L finalised

18. Values: Step 1 Ca=146.78 mm <165mm so OK So Ca=145 mm Step 2 Vprm =98.09 Kmph say 95 Kmph Step 3 Vprm <Vmax so Vprm =95 kmph Cex=145- =87.67 mm But Cex can not exceed 75 mm  

19. Revise Ca Ca=Cex+ =132.3 mm So Ca= 130 mm Vprm =0.27 Vprm = 94.69 Kmph say 90 Kmph Actual Cd for this Vprm is Cd = -130= 55.76 mm  

20. Transition Length:- Ca=130mm, Cd=55.76mm, Vprm =130Kmph L1=93.6m L2=40.14m L3=93.6m L=100m Check: t=3.6*100/90=4 sec Rate of change of Ca=130/4=32.5 mm/s ok Rate of change of Cd=55.76/4=13.9 mm/s ok Cant Gradient=130/100000=1/770 ok

21. Curve detail: Ca=130 mm Vprm =90 Kmph L=100m

22. 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 Vmax Ca= Ca’-Cd

23. 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)  

24. Step-3 Calculate maximum permissible speed on the curve Vprm =0.27 Ca and Cd ? SR of Vprm to be imposed  

25. For changed Ca, revise/check Vprm So check Cd Cd= -Ca (normally will be ok) Finalised Vprm and Ca Step 4: Transition length calculation Maxm of L1,L2 and L3 And in no case less than 2/3 L1, 2/3 L2 and ½ of L3  

26. 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/L Finalize Transition length So curve parameters Ca, Vprm and L finalised

27. Speed potential of existing curve Known values- R, L, Ca To find V

28. Example Find the speed potential of a curve of R=875 m Ca =100 mm L=100m/ 60m/50m Other limiting values Cd and Cex =75 mm Rca & Rcd =55 mm/s Find maxm speed potential for circular curve and corresponding TL

29. 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

30. 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. Check Rca =Ca/(L/V)=100*105/(3.6*50)= 58.33 mm/s Not Ok

31. 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

32. 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 sec Rcd =42/1.89=22.22 mm/sec so Ok So Speed permitted is 95 Kmph for TL =50m

33. 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

34. 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/Cd Condition-1 Both will be balanced when Ca=Cd (Optimum speed) from transition length consideration

35. 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

36. Case 3 contd Vcircular =0.27*SQRT(R*( Ca+Cd ))= 102.59 Kmph Actual Cd=13.76*(100*100)^2/875-90=67.257 mm For 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.

37. Thank You