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How to calculate potential traction loss on a haul road

  1. Aug 3, 2015 #1
    1. The problem statement, all variables and given/known data
    This is a mining engineering question, but the section I need help with is more general: I am working through a long question regarding a given mine. In previous sections of this question, I have selected equipment based on required machinery outputs etc.
    I have now come to the last question which is "ensure trucks are capable of carrying out this mining safely". Which is a bit of an open-ended one!
    In previous questions I have figured out that I need between 2 and 3 trucks to keep the excavator swinging without stopping. The question didn't ask for it, but to give a decent answer I need more data, so I selected an appropriate truck from the market and used its data (I listed that in assumptions). I feel that I have partially answered the question, but there is data given that I have not used much (ie coefficient of traction) that makes me think the lecturer is looking for calculations showing traction loss. I have been hunting for days and searching through forums and engineering text-books but can't seem to find a clean formula for this purpose. I feel from experience that a coefficient of traction of 0.25 (similar to snow!) would lead to safey problems from traction loss but I can't seem to prove it. I thought stopping distances would do it (see my work below) but unless I messed up the math (quite possible) the stopping distance was not great but still acceptable.
    The data given in the question:

    28. A pit with a straight ramp of 800m, with a gradient of 10%, supplies a ROM pad 1500m away via a level haul road all of loose gravel. Mining of a dipping ore body, with a specific weight of 2.5 t/m3, is required to be carried out at a rate of 10,000 tonnes/8 hour shift to feed the mill. Since truck speed is restricted to 50kmph, assume the average speed of a truck during a round trip is 45kmph, including reversing to load. Waste material of specific weight = 1.85 t/m3 is also to be mined, and deposited on a waste dump 200m from the pit. The waste dump already has a height of 10m, and an area of 100m x 60m from pre-stripped overburden. Excluding the overburden, the strip ratio is 1:4.5.

    Other factors that are working assumptions are:

    · Availability of excavator 83%,

    · Swell factor of 30%

    · Bucket fill factor of 88%

    · Excavator cycle time 45 seconds

    · Rolling Resistance Factor (RRF) = 15 kg/t

    · Coefficient of Friction, m = 0.25

    · Grade Resistance Factor = 20 kg/t

    2. Relevant equations
    As below for equations I have finished. I have not been able to find an equation to calculate if I will lose traction while driving into/out of the pit

    3. The attempt at a solution
    What I have so far for this question:
    Total resistance = grade + rolling resistance% = 10% + (15kg/1000kgx100) = 11.5% uphill (Or 8.5% downhill)
    Truck Performance (I can attach this chart if requested, but I don't think you will need it):
    Using the Hitachi retarder performance data for a EH4500-2 dump truck (using total resistance numbers), a downhill 10% grade for an empty truck would allow a speed of up to 55kh/h with safe retarder use. Using the rim pull chart and the total resistance, a full truck could obtain a maximum speed of up to 13km/h going up the ramps, and 55km/h on the flat (limited to 50km/h as given).

    From this data the following would be within manufacturer recommendations:
    Minimum cycle times:
    Waste:
    13km/h roads = total uphill ramp distance = 800m pit ramp + 100m dump ramp = 900m
    50km/h roads = 2130m-900m = 1130m
    Minimum cycle time for trucks = (0.9km/13km/hx60)+(1.13kmx50km/hx60)+3mins loading = (4.15+1.36)+3 = 5.51mins haul time + 3 mins loading = 8.51mins
    Rom/Ore:
    13km/h roads = total uphill ramp distance = 800m pit ramp
    50km/h roads = 4600m-800m = 3800m
    Safe cycle time for trucks = (0.8km/13km/hx60)+(3.8kmx50km/hx60)+2.25mins loading = (3.69+4.56)+2.25 = 8.25mins haul time+2.25mins loading = 10.5minsTrucks required to do waste circuit safely = 5.51/3 = 1.84
    Trucks required to do rom/ore circuit safely = 8.25/2.25 =

    Rolling resistance = 15kg/t
    Grade resistance factor = 20kg/tCoefficient of Friction, m = 0.25
    Truck gross vehicle mass: downhill empty 200t, uphill full = 470t
    Truck downhill speed: 50km/h = 13.89m/s

    Stopping distance for trucks, using Kaufman and Aultd 1977 SAE formula
    SD= 0.5gt² sinø + Vot + [ (gt sinø + Vo)/(2g{Umin-sinø})]²

    g= 9.81m/s²
    t = 6 seconds (combined reaction time of truck and operator)
    sinø = sin10% (slope)
    Vo= 13.89m/s (vehicle speed of 50km/h)
    Umin = 0.25

    SD=0.5x9.81²sin10%+13.89x6+[( 9.81x6sin10%+13.89)/(2x9.81x{0.25-sin10%)}]²
    SD= 0.5x96.2361x0.1736+13.89+[1.703+13.89)/(19.62x0.0764)]²
    SD= 22.2432+[15.593/1.4990]²
    SD= 130.5m
     
  2. jcsd
  3. Aug 3, 2015 #2
    Usually traction problems are most likely to show up in increased stopping distances when going down hill.

    Try considering a worst case scenario: downhill at the steepest angle, full load, unequal distribution of braking forces, sliding friction, etc.
     
  4. Aug 3, 2015 #3
    Dr Courtney, thankyou for the reply. In my answer I have calculated the worst-case scenario, which is a 10% downward grade with the low traction (SD=130m). I have no idea how I would calculate uneven brake force though with the data I have, are you able to point me in the direction of an equation that I can use to calculate that? The formula I used did not take vehicle weight into consideration, although any vehicle travelling down an incline would be empty.
    Is there an equation that can be use to figure out a what angle of slope a vehicle would lose traction, that takes the coefficient of friction into account?

    To calculate the stopping distance I used a Kaufman and Ault formula based on the SAE stopping distance limitations, its a bit hard to decipher from my answer so I have put it below:

    SD= ½ gt²sinø + Vot + [ gt sinø + Vo
    2g(Umin-sinø)

    SD=Stopping Distance
    g = Gravatational acceleration (9.81m/s²)
    t = elapsed time between perception of need to stop and actual frictional brake contact (given by authors as t= 4.5s for the truck + 1.5s for the operator)
    ø = angle of descent in º
    Umin = coefficient of friction
    Vo= vehicle speed in m/s
     
  5. Aug 3, 2015 #4
    I found an error in my work in the math. Stopping distance actually = 372.9m :wideeyed: With that horrible number I think I can show why its not safe! If anyone has a formula for traction loss, however, it would be very appreciated.
     
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