How to calculate potential traction loss on a haul road

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1. Aug 3, 2015

markk

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
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. Aug 3, 2015

Dr. Courtney

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.

3. Aug 3, 2015

markk

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

4. Aug 3, 2015

markk

I found an error in my work in the math. Stopping distance actually = 372.9m 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.