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Design of super elevation :- A case study

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THIS ARTICLE IS VERY HELPFUL IN DESIGNING CURVE OF HAUL ROAD IN AN OPENCAST MIN

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Design of super elevation :- A case study

  1. 1. DESIGN OF `HAUL ROAD- A Life Line` of the opencast mine : - By Tikeshwar Mahto Dy. Director of Mines Safety, Bilaspur Region(India) +917898033693 tikeshwarmahto@yahoo.co.in Design of curve to maneuver the speed of the Dumper or other vehicles R Super Elevation CURVE
  2. 2. Super elevation of a haul road curve in Opencast Mines is designed taking into consideration of `Radius of curvature` and the `speed of the dumper at the curve`. It is meant for counter balancing the centrifugal force applied on the dumper or other moving vehicles at the curve of the road. Centrifugal force acts outward on the moving vehicles and tries to topple out of the road. This force is counter balanced by the `Gravity force` or self weight of the vehicles (M*g), when the vehicle is inclined inward by elevating the road at outer side. The amount of elevation of road at the curve depends on the Radius of the curvature(R) and the speed of the vehicle at the curve (V). Smaller the radius of curvature, higher will be the super elevation. Also greater the speed of the vehicles at the curve, higher will be the super elevation. We should see the combined effect of the Radius of the curvature and speed of the vehicles at the curve, while designing the Super elevation. Let us assume that,  Radius of the curvature = R (in mtrs)  Speed of the vehicle at the curve = V (in m/s)  Load Weight of the vehicle(W) = M*g, where, M is the mass of the loaded vehicle and `g` is acceleration due to gravity.  Elevation of the road at outer side = S (in mtrs.)  Width of the road at the curve= B (in mtrs.)  Normal reaction to the ground = N  Co-efficient of skidding resistance between Tyres & ground = µ (approx. =0.05) θ θ θ (M*V2 * Sinθ)/R + M*g Cosθ Figure showing, a dumper taking turn at the curve From the above figure, when the turning vehicle is in equilibrium, Stabilizing forces  Gravity force = M*g, components of which along the plane and normal to the plane are M*g Sinθ & M*g Cosθ , as shown in above Figure.  Frictional force between Tyres & ground= µ*N, towards centre of the curve. Where, µ is co-efficient of frictional resistance between Tyre & ground and N is the normal resistance to super elevated road. B S θ M*g Centrifugal force = (M*V 2 )/R θ W=Mg N µ*N M*g Sinθ (M*V2 * Cosθ)/R
  3. 3. Destabilizing forces Centrifugal force (or pseudo force) = (M*V 2 )/R, outside the road in horizontal direction Components of this force along the super elevated plane and along normal to the plane are (M*V2 * Cosθ)/R & (M*V2 * Sinθ)/R, as shown in above Figure.. In balance condition, when stabilizing & destabilizing forces are equal : M*g Sinθ + µ*N = (M*V2 * Cosθ)/R ;-----------------------------------------------------(1) Here, N= normal reaction to the ground = M*g Cosθ + (M*V2 * Sinθ)/R ----------------(2) Hence, from equation (1); M*g Sinθ + µ[M*g Cosθ + (M*V2 * Sinθ)/R] = (M*V2 * Cosθ)/R ------------(3) Or, g Sinθ + µ[g Cosθ + (V2 * Sinθ)/R] = (V2 * Cosθ)/R ------------------(4) Or, g (Sinθ + µ Cosθ )= (Cosθ – µSinθ) * V2 /R Or, V2 /R*g = (Sinθ + µ Cosθ )/ ( Cosθ –µ Sinθ) [ dividing both side by Cosθ) Or, V2 /R*g = (Tanθ + µ )/(1- µ Tanθ) When, speed of the vehicle is in KM/H, then speed of vehicle `V` in above equation can be written as V*5/18, because 1 KM/H = 5/18 m/s. Hence, (5V/18)2 /R*g = (Tanθ + µ )/(1- µ Tanθ) Or. (25V 2 /324)/R*g = (Tanθ + µ )/(1- µ Tanθ) Assuming (25V 2 /324)/R*g = K , for simplicity Hence, K = (Tanθ + µ )/(1- µ Tanθ) Or, K (1- µ Tanθ) = (Tanθ + µ) Or, K- µ = (K *µ + 1) Tanθ Or, Tanθ = (K- µ)/ (K *µ + 1) Also, from the above figure, Tanθ = S/B …………………………………………… (5) Where, S & B are amount of super elevation and width of haul road. Hence, S/B = (K- µ)/ (K *µ + 1) Or, S = B(K- µ)/ (K *µ + 1) meters. …………………..(6) When super elevation is expressed in % (with respect to Haul road width, B) then % Super elevation = (S/B)*100 = 100(K- µ)/ (K *µ + 1) % -------------(7) Taking B= one way haul road width= 15m, µ = frictional rolling resistance between Tyres & ground= 0.05 (assumed) and g = 9.81m/s 2 =10m/s 2 Then, K= (25V 2 /324)/R*g =25V 2 /3240R Hence, from equation (6) S= B(K- µ)/ (K *µ + 1) meters Or, S= B (25V 2 /3240R - µ)/ [(25V 2 /3240R )*µ + 1] Or, S= B (25V 2 - 3240R* µ)/ [(25V 2 *µ + 3240R] Or, S= 15* (25V 2 - 162R)/ [(1.25V 2 + 3240R] , where µ =0.05 & B= 15m ---------------(8)
  4. 4. Super elevation calculated by the equation (8) taking width of haul road at the curve as, B=15m for different speed and different Radius of curvature are as tabulated below: S.N Radius of Curvature (R) in mtrs. Super Elevation , S= [15* (25V2 - 162R)]/ [(1.25V2 + 3240R] for V= 20 KMPH (in mtrs) Super Elevation , S= [15* (25V2 - 162R)]/ [(1.25V2 + 3240R] for V= 30 KMPH (in mtrs) Super Elevation , S= = [15* (25V2 - 162R)]/ [(1.25V2 + 3240R] for V= 40 KMPH (in mtrs) 1 10 3.82 9.34 16.74 2 20 1.55 4.38 8.25 3 30 0.79 2.69 5.31 4 40 0.41 1.84 3.82 5 50 0.18 1.32 2.92 6 60 0.02 0.98 2.31 7 70 0.00 0.73 1.88 8 80 0.00 0.55 1.55 9 90 0.00 0.41 1.30 10 100 0.00 0.29 1.10 11 110 0.00 0.20 0.93 12 120 0.00 0.12 0.79 13 130 0.00 0.05 0.67 14 140 0.00 0.00 0.57 15 150 0.00 0.00 0.48 16 160 0.00 0.00 0.41 17 170 0.00 0.00 0.34 18 180 0.00 0.00 0.28 19 190 0.00 0.00 0.22 20 200 0.00 0.00 0.18 21 210 0.00 0.00 0.13 22 220 0.00 0.00 0.09 23 230 0.00 0.00 0.06 24 240 0.00 0.00 0.02
  5. 5. Graph showing the variation in super elevation (in mtrs.) on varying the speed of vehicle (V) at the curve and radius of curvature(R): Radius of curvature ---- -> From the above graph and Table, it can be inferred that the curve of haul road need not be super elevated, if radius of curvature is more than 50m, 130m, 240m for corresponding speed of vehicle as 20kmph, 30kmph , 40kmph respectively. Amountofsuperelevation----> 1 2 3 3 2 1 DESIGN OF SUPER ELEVATION IN OPENCAST MINE: ON X-AXIS: Radius of Curvature of Haul Road (in mtrs) ON Y- AXIS : Super elevation at the Curve of Haul Road (in mtrs)
  6. 6. Super elevation (in %) calculated taking the above equation for different speed and different Radius of curvature as tabulated below. Here, S/B = (25V 2 - 162R)/ (1.25V 2 + 3240R) , Or, super elevation in % = (S/B)*100 Or, = (25V 2 - 162R)/ (1.25V 2 + 3240R)*100 Super elevation (in %) S.N. Radius of Curvature Super Elevation (in %) = 100* (25V 2 - 162R)/ (1.25V 2 + 3240R) Super Elevation (in %), = 100* (25V 2 - 162R)/ [(1.25V 2 + 3240R] Super Elevation (in %) = 100* (25V 2 - 162R)/ [(1.25V 2 + 3240R] (R) in mtrs. For V= 20 KMPH (in %) For V= 30 KMPH (in%) For V= 40 KMPH (in %) 1 10 25.47 62.28 111.57 2 20 10.35 29.22 55.03 3 30 5.26 17.94 35.42 4 40 2.71 12.25 25.47 5 50 1.17 8.83 19.45 6 60 0.14 6.54 15.42 7 70 0.00 4.90 12.53 8 80 0.00 3.66 10.35 9 90 0.00 2.71 8.66 10 100 0.00 1.94 7.30 11 110 0.00 1.31 6.19 12 120 0.00 0.78 5.26 13 130 0.00 0.34 4.48 14 140 0.00 0.00 3.80 15 150 0.00 0.00 3.22 16 160 0.00 0.00 2.71 17 170 0.00 0.00 2.25 18 180 0.00 0.00 1.85 19 190 0.00 0.00 1.49 20 200 0.00 0.00 1.17 21 210 0.00 0.00 0.88 22 220 0.00 0.00 0.61 23 230 0.00 0.00 0.37 24 240 0.00 0.00 0.14 25 250 0.00 0.00 0.00
  7. 7. DESIGN OF SUPER ELEVATION IN % : ON X- AXIS : Radius of Curvature of Haul Road (in mtrs) ON Y- AXIS : Super elevation at the Curve of Haul Road (in %) Radius of curvature ---- -> Earlier, an accident occurred in Spain, by derailment of over speeding passenger Train at the Curve and hitting on side wall, in which 88 persons killed. Design of the curve was for maximum speed of 80kmph. But, the estimated speed of the train at the curve was 144-192kmph (more than double the limiting speed of the train at curve). In that case, outward centrifugal force (M*V 2 /R) was about 4 times the designed centrifugal force, which thrown out the train from the track. The snapshot of the accident is attached below in next page. Amountofsuperelevation(in%)----> 1 1 2 32 3
  8. 8. Tikeshwar Mahto Dy. Director of Mines Safety, Bilaspur Region, tikeshwarmahto@yahoo.co.in

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