Surface Mine Design and Practice

SURFACE MINE DESIGN AND PRACTICE

Engr. Izhar Mithal Jiskani

Department of Mining Engineering, Mehran University of Engineering & Technology


Jamshoro, Sindh


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GEOMETRICAL CONSIDERATIONS

INTRODUCTION TO OPEN PIT MINING The ore deposits being mined by open pit techniques today vary considerably in size, shape, orientation and depth below surface. The initial surface topographies can vary from mountain tops to valley floor. In spite of this there are a number of geometry based design and planning considerations fundamental to them all. The ore body is mined from top to the down in the series of horizontal layers of uniform thickness called benches and after a sufficient floor area has been exposed, mining to the next layer can begin. The process continues until the bottom bench elevation is reached and the final pit outline achieved. To access different bench a road or ramp must be created. The width and steepness of this road or ramp depends upon the type of equipment to be accommodated. Stable slopes must be created and maintained during the creation and operation of the pit. Slope angle is an important geometrical parameter which has a significant economic impact. Open pit mining is very highly mechanized. Each piece of mining machinery has an associated geometry both related to its own physical size, but also with the space it requires to operate efficiently. There is a complementary set of drilling, loading and hauling equipment which requires a certain amount of working space. This space requirement is taken into account when dimensioning the so called working benches. From both operating and economic view points certain volumes must or should at least be removed before others. These volumes have a certain minimum size and an optimum size.


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BASIC BENCH GEOMETRY The basic extraction component is an open pit mine is the “bench”. Bench – nomenclature is shown in Figure 1.

Fig.1: Bench Nomenclature FACE: It is an exposed area from where the overburden or mineral/ore is extracted. CREST: highest point of the face. TOE: lowest point the face. BENCH: the step or floor accommodating the mine machinery. BENCH HEIGHT: each bench has an upper and lower surface separated by a distance “H” equal to the bench height. The bench height is determined by the size of mining equipment and formation of the area. •

The loose/soft rocks allows, bench height up to shovel reach.

•

In hard and very strong rock, bench height is usually 10-40 meters.

BENCH SLOPE: the bench slope or the bench face angle is the inclined plane of the bench made an angle with the horizontal. Or the average angle that a face makes with the horizontal. The exposed sub-vertical surfaces are known as bench faces. The bench faces are described by the toe. The crest and he bench face angle ‘α ∝’. The bench face angle can vary considerably with rock characteristics, face orientation and the blasting practices. In most hard rock pits it varies from about 55 O to 80O. A typical initial design


4


value might be 65O. This should be used with care as the bench face angle can have a major effect on the overall slope angle. BENCH FLOOR: The exposed bench lower surface is called as the bench floor. BENCH WIDTH: The bench width is the distance between the crest and toe measured along the upper surface. BANK WIDTH: It is the horizontal projection of the bench face. There are several types of benches; a working bench is that one which is in process of being mined. The width being extracted from the working bench is called the cut. The width of working bench WB is defined as the distance from the crest of the bench floor to the new toe position after the cut has been extracted as shown in figure 2. After the cut has been removed, a safety bench or catch bench of width SB remains. The purpose of leaving safety benches is to: a) collect the material which slides down from the benches above; and to b) stop the downward progress of the boulders. During primary extraction, a safety bench is generally left on every level. The width of safety bench SB varies with bench height “H”. Generally the width of the safety bench is of the order 2

3 of the bench height. At the end of mine life, the safety benches are

sometimes reduced to a width of about 1

3 of the bench height.

In addition to leaving the safety benches berms (piles) of broken materials are often constructed along the crest. These serve the function of forming a ditch between the berm and the toe of the slope to catch falling rocks. A safety berm is also left along the outer edge of the bench to prevent trucks and other machines from backing over. It serves much the same function as a guard rail on bridges and elevated highways. Normally the pile has a height greater than or equal to the tire radius. The berm slope is taken to be about 35O, i.e. also called the angle of repose. The steps which are followed when considering bench geometry are: i)

Deposit characteristics (total tonnage, grade distribution, value etc) dictate a certain geometrical approach and production strategy.


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ii)

The production strategy yields daily ore-waste production rates, selective mining and blending requirements, numbers of working places.

iii)

The production requirements leads to a certain equipment set (fleet type and size).

iv)

Each equipment set has a certain optimum associated geometry.

v)

Each piece of equipment in the set has an associated operating geometry.

vi)

A range of suitable bench geometries results.

vii)

Consequences regarding stripping ratios, operating v/s capital costs, slope stability aspects etc are evaluated.

viii)

The best of the various alternatives is selected.

In the past, when the rail bound equipments were being extensively used, great attention was paid to bench geometry. Today highly mobile rubber tired/crawler mounted equipments has reduced the detailed evaluation requirements some-what. Fig. 2: Working Bench


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Fig. 2a: Functions of a catch bench


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DEVELOPMENT OF ACCESS ROAD: (ORE ACCESS;) In mining literature, going the initial knowledge about the physical access to the Ore body is of great importance. For this, the question arises that: “How does one actually begin the process of Mining?” Obliviously the approach depends upon the topography of the surrounding ground. To introduce the topic, it is assumed that the ground surface is flat. The overlying vegetation has been removed as has the soil/sand/gravel overburden. In this case it will be assumed that the ore body is 700 feet in diameter, 40feet thick, flat dipping and is exposed by removing the soil overburden. The ore is hard so that drilling and blasting is required. The bench mining situation is shown in figure: 3.

Figure 3: Geometry of the ore body A vertical digging face must be established in the ore body before major production can begin. Further more a “ramp” must be created to allow truck and loader access. A drop cut is used to create the vertical breaking face and the ramp access at the same time. To access the Ore body, the “ramp” shown in figure: 4 will be driven it has an 8% grade and a width of 65 feet. Although not generally the case, the walls will be assumed vertical. To reach the 40ft desired depth, the ramp in horizontal projection will be 500ft in length.


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Figure 4: Ramp access for example ore body The volume of access road or ramp volume is the volume of the waste rock mined in excavating the ramp. :.

Ramp volume = Ramp width (Rw) × Area of ∆abc. 1  Rw ×  × horizontal × H  2  Ramp volume (V)

:.

1  H   = Rw ×  ×  × 100  × H   2  g  1 Rw × × H × 100 × H g 2

Volume of Ramp = V= 50 H2/g × Rw; (ft3)

Example # 01:Determine the volume of Ore body by waste rock to develop access road at the slope of 8%. Depth of cylindrical Ore body is 30 feet and diameter 500 ft, width of ramp is 65 feet. Data: Slope = g = 8% Depth = H = 30 feet Diameter = d= 500 feet Width of ramp = Rw = 65 feet Solution: as

V = Volume of ramp

and V = 50 ×

H2 × Rw g

V = 50 ×

(30) 2 x 65 8

=

2925000 8

V = 365625 ft3 Ans . Example # 02:Repeat example # 01 for Depth (H) = 40 ft and Ramp width (Rw) = 60 ft. Solution: -

V = 50 ×


H2 × Rw g

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V = 50 ×

(40) 2 x 60 8

V = 600000 ft3 Ans


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DEVELOPMENT OF ACCESS ROAD: (ramps) i) In waste rock (Fig. 5a)

. ii) Ramp in ore body (Fig. 5b)

. iii) Ramp starting in waste and ending in ore (Fig. 5c)


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THE PIT EXPANSION PROCESS When the drop cut has reached the desired grade, the cut is expanded laterally. Figure: 6 shows the steps. Initially Fig: 6.A; the operating space is very low/limited; therefore the trucks must turn and stop at the top of the ramp and then back down the ramp towards the loader. When the pit bottom has been expanded sufficiently as shown in fig: 6.B; the truck can turn around on the pit bottom. Latter, as the working area becomes quite larger as shown in fig. 6.C; several loaders can be used at the same time. The optimum face length assigned to a machine varies with the size and type. It is of the range 200-500 feet. Once access has been established the cut is winded until the entire bench/level has been extended to the bench limits. There are three approaches which will be discussed here; they are as follows: 1. Frontal Cuts 2. Parallel Cuts – Drive by 3. Parallel Cuts – Turn & Back The first two apply where there is a great deal of working area available, for example “at the pit bottom”. The mining of the more narrow benches on the sides of the pit is covered under the third approach. Figure: 6 A Step I


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Figure: 6 B Step II

. Figure: 6 C Step III

Figure 6: Detailed steps in the development of a new production level

FRONTAL CUTS: The frontal cut is shown diagrammatically in figure – 7. The shovel faces the bench face and begins digging forward straight ahead and to the side. A niche is cut in the bank wall. For the case shown, double spotting of the trucks is used. The shovel first loads to the left and when the truck is full he proceeds to the other truck on the right. The swing angle varies from 1100 (maximum) to an angle of 100 (minimum). The average swing angle is about 600; hence the loading operation is quite effective. There must be room for trucks to position them around the shovel. The shovel


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penetrates to the point that the face. It then moves parallel to itself and takes another frontal cut as shown in fig: 7.1; With a long face and sufficient bench width, more than one shovel can work the same face, as shown in fig. 7.2.

1.

DRIVE BY CUTS :- (parallel cuts-drive by)

Another possibility when the mine geometry allows is the parallel cut with drive by. This is diagrammatically shown in figure 8. The shovel moves across and parallel to the digging face. For this case bench access for the haul units must be available from both the directions. It is highly efficient for both the trucks and the loader. Although the average swing angle is greater than for the frontal cut, the trucks do not have to back up to the shovel and the spotting is simplified.

2.

TURN AND BACK:- (Parallel cuts-turn and back)

The expansion of the pit at the upper levels is generally accomplished by using parallel cuts. Due to space limitations there is only access to the ramp from one side of the shovel. This means that the trucks approach the shovel from the rear. Then, they stop, turn, and back into the load position. Sometimes there is a room for double spotting of the trucks (fig: 9.1) but sometimes for only single spotting as shown in figure 9.2.


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DETERMINATION OF THE MINIMUM OPERATING ROOM REQUIRED FOR PARALLEL CUTS In determing the width of operating room required for the parallel cut operations; to accommodate the large trucks and shovels involved in loading, the dimension being sought is “the width of working bench WB”. The working bench is that bench mined. The width of working bench “WB” is synonymous with the term “operating room” and is defined as the distance from the crest of the bench providing the floor for the loading operations to the bench toe being created as the parallel cut is being advanced. The minimum amount of operating room varies depending upon whether single or double spotting of trucks is used with the latter obviously requiring some what more. The minimum width of the working bench (wb) is equal to the width of the minimum required safety bench plus the width of the cut. This is expressed as; “WB = SB + WC” The easiest way of demonstrating the principles involved is by way of an example: For this, the following assumptions will be made: •

Beach height = 40 feet

•

A safety beam is required.

•

The minimum clearance b/w the outer truck tire and safety berm = 5 feat.

•

Single spotting is used.

•

Bench face angle = 700

•

Loading is done with a 9yd3 Shovel.

•

Haulage is done by 85 ton capacity Trucks.

•

Truck width = 16 feet

•

Tire Rolling radius = 4 feat.

The general arrangement of working bench (in x-sectional view) is shown in figure 10.


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Figure 10:

. Figure 10.1: Simplified representation of berm.

kl The design shows that:•

Working bench width = 102 feet

•

Cut width = 60 feet

•

Safety bench width = 42 feet.

The basic calculations (justifications) behind these calculations are presented as follows.


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•

Step # 01:-

Highest of the berm should be at least same as the radius of the truck tire, i.e.; = 4 feet.

•

Step # 02:-

The distance b/w the crept and centre line of the truck = Tc = 21: Width of Berm = 8 feet. Clearance distance b/w safety berm and the wheels of truck = 5 feet and; the total width of truck = 16 feet.

•

Step # 03:-

The distance b/w the centre line of the Shovel and centre line of the truck is also called as “Dumping radius” denoted by B; B = 45.5 feet.

•

Step # 03b:-

The Maximum duping height (A) is more than sufficient to clear the truck; A = 28 feet.

•

Step # 03c:-

Distance b/w the centre line of Shovel and toe = G ; G= 35.5 feet

•

Step # 04:-

The desired working bench dimensions become ; Total minimum width of working bench = WB = TC +B+G; :. WB = 21 feet + 45.5 feet + 35.5 feet. WB = 102 feet

Note: All the parameters and /or dimensions used above, depends upon the size of the machinery which is used.

WIDTH OF CUT :- (Wc) The corresponding width of cut is:WC =

0.90 × 2 × G = 0.90 × 2 × 35.5 feet

WC =

63.9 feet


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Note: This is applied to the width of the pile of broken material. Therefore, to allow for swell and throw of the material during blasting. The design cut width should be less than this value. Thus a value of 60 feet has been assumed.

WIDTH OF SAFETY BENCH :- (SB) Knowing the width of working bench, and width of cut, the resulting safety bench has a width of; SB = WB – WC ; :. SB = 102-60 = 42 feet. SB = 42 feet. Note: This is of the order of the bench height (40ft) which is a rule of thumb, sometimes employed.


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CUT SEQUENCINGS: Let us consider a pit consisting of four benches as shown in fig: 11

Fig. 11: Initial Geometry of the pit: After the initial geometry of the bench is completed; the mining of first bench is started as shown in figure: 12.

Fig. 12: Mining of bench # 01 The above figure shows that while performing the cut mining operation of bench # 01, the overall slope angle was “θ01”. After the mining of bench # 01, the next bench (i.e.: Bench # 02) is mined, as shown in fig. 13: Fig: 13:- Cut mining from bench # 02


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Where θ = individual slope angle. and θ01 = overall slope angle (while mining Bench # 01) of pit. θ02 =

overall slope angle of pit (while mining bench # 02)

It is observed that; the overall slope angle “θ 0” always keep varying as we will advance the mining from the upper to the lower benches.

i.e.:- θ01 ≠ θ02 It is essential to consider (know) the value of “θ0” for slope stability designs.


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PIT SLOPE GEOMETRY There are a number of “slopes” which enter into “pit design”. Care is taken so that there is no confusion as to how they are calculated and what they mean. One slope has already been introduced. That is the ‘bench face angle” (shown in figure). It is defined as the angle made with the horizontal of the line connecting the Toe to the crest “this def:” will be maintained through this piece of literature. Now consider the slope consisting of “5” such benches (shown in figure). The angle made with the horizontal of the line connecting the lowest most toes to the upper most crests is defined as the overall pit slope. θ overall

  5x50 →(Y) = tan   4x35 + 5x50  tan75O  −1

•

Height of each bench = y = 50 feet.

•

Horizontal distance = x = 35 feet

•

Distance under slope = x’ =?

•

Bench face angle = 750 = θ

:.

   = 50.4 O   

tanθ =  y   x'  x'

=

y tanθ

=

50 50 = tan75 3.732

=13.4 '

Since we have 5 benches, i.e.: -

5 slopes,

X= (4*x) + (5*x’).

X = (4 x 35) + (5 x 13.4) = 140 + 67 Y = 5 x 50 :.



X = 207’

 Y = 250’

Overall pit slope angle = θ overall  y  250  θ overall = tan −1   = tan −1   x  207  θ overall = 50.4O


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INTRODUCING RAMP IN PIT SLOPE GEOMETRY An access ramp with a width of 100 feet; introduced at the half way up bench 3, the overall pit slope becomes; y θ overall = tan −1   x where, Y = 250’  5x50  and X = ( 4x35) +  + 100' O   tan75  :. X = 307 feet.  250  O θ overall = tan −1   = 39.18 307   O θ overall = 39.2 It can be seen that the presence of the ramp n a give section has an enormous impact on the overall slope angle.

Fig. 15: Overall slope angle with ramp included.


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INTER-RAMP SLOPE ANGLES AFTER RAMP INCLUDED The ramp breaks the overall slope angles “ θ overall” into 2 portions as shown in following figure; each of these 2 portions can be described by slope angles. These angles are called as “the Inter ramp (Between the ramps)” angles. y θIR 1 = θIR 2 = tan −1   x    Y = 50 + 50 + 24 → Y = 124feet  50   25  X = ( 2x35) + 2 + O  O   tan75   tan75  X = 70 + 26.8 + 6.7 = 103.5  125  O θIR 1 = θIR 2 = tan −1   = 50.38  103.5  O θIR 1 = θIR 2 = 50.4

Figure 16: Inter ramp slope angles after ramp included


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The inter-ramp wall height is 125 ft for each segment. Generally the inter ramp wall heights and angles for different slope segments would not be the same. From a slope stability view point each inter-ramp segment would be examined separately.


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INTRODUCING WORKING BENCHES IN PIT SLOPE GEOMETRY While active mining is underway, some working benches are included in the overall slope. The fig. ; shows a working bench 125ft in width included as bench 2. The overall slope angle “ θ overall” now becomes; y θ overall = tan −1   x     50  As; Y = 250' , X = ( 4x35) ' + 5 '+125' O   tan75  X =140' + 67' +125' ⇒ X = 332 feet  250  O ∴θ overall = tan −1   ⇒ 36.98 332  

Fig. 17: Overall slope angle with working bench included


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INTER-RAMP SLOPE ANGLES AFTER WORKING BENCH INCLUDED The working bench is treated in the same way as a ramp in terms of interrupting the slope. Therefore from following figure, two inter-ramp angles are shown:y ; θIR 2 = tan −1   x     200  θIR 2 = tan −1    158.6' 

As; Y = 250 – 50 = 200 feet and  50  X = ( 3 x 35) + 4  O   tan75  X =105 + 53.6 = 158.6'

θIR 2 = 51.58O or 51.6 O

As;

for

the

above

inter

ramp

angles,

the

inter

ramp

heights

are,

H1= 50’ and H2 = 2’

Fig. 18: Inter-ramp slope angles after working bench included


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PLANNAR FAILURE

Fig. 19: Perspective view of plannar failure.

Fig. 20: Dimensions and forces in a rock slope with a potential failure plane. As figure 20 shows the dimensions and forces in a rock slope with potential failure plane. The Mohr-coulomb failure criterion has been used. The following definitions apply: •

i= average slope angle from horizontal in degrees,

•

ß = the angle of discontinuity from the horizontal (degrees),

•

W =block weight;


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•

R = resisting force;

•

C = cohesion,

•

φ =friction angle.

•

Wcosß=

normal force

•

Wsinß=

driving force

•

A= Area of the failure plane.

The factor of safety ‘F’ is defined by the following equation: F=

Total force available to resist sliding Force tending to induce sliding

For the case shown in Fig:”20” F=

cA + Wcos β tanφ Wsin β

If there is water present then; F=

cA + ( Wcos β - U ) tanφt Wsin β + V

where, U= water pressure along potential failure surface, φ a = friction angle (affected by water),

V= force along potential sliding plane.


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Example # 01:The average planned slope angle i = 700, the orientation of the potential failure plane ß = 500 and the friction angle φ = 300. The thickness of he plane is 1ft; the cohesion is 1600 lb/ft3. The unit weight of the rock is 160 lb/ft3, and height of the wall is 100ft. Sol:frictional force Sliding force Frictional force = cA + Wcosβ tanφ

S.F =

Fig. 21 (a)  i = 70,

β = 50

∠A = 70 0 − 50 0 = ∠A = 20 0 ∠C = 50 0

∠B = 180 0 − ( ∠A + ∠C ) = 180 0 − 70 0 ∠B = 110 0

Fig. 21 (b)

as,

∆=

1 2 sin ∠A sin ∠C b × 2 sin ∠B

(a)

∆=

1 2 sin ∠B sin ∠C a × 2 sin ∠A

(b)

∆=

1 2 sin ∠A sin ∠C c × 2 sin ∠C

(c)

sin ∠B =

H 100 = b b


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or

b=

100 100 = = 130.54feet 0 0.766 sin50 Putting the above value in eqn (a).

0 0 1 (130.54) 2 × sin 20 × sin0 50 2 sin 110 1 0.262 ∆ = × 17040.692 × 2 0.9396 = ×0.279

∆=

∆ = 2375.62 ft 2 (V)

Volume of (triangle ABC) sliding block = ∆ × thickness of sliding block = 2375.62 × 1ft

V = 2375.62 ×1ft 3 (W) Weight of sliding block is = volume × unit weight W = V × γ = 2375.62 × 160lb/ft 3 W = 380099.2lbs As F=

cA + Wcosβcosβ Wsinβ

A= Area of the failure plane, and A= length of failure plane × thickness A= b × 1ft= 130.54ft × 1ft A= 130.54ft2 Wcosβ = 380099.2 × cos50 0 = 244323.05lb Wsinβ = 380099.2 × sin50 = 291172.88lb c = 1600lb/ft 2 Tanφ = Tan30 0 = 0.577 (1600 × 130.54) + ( 244323.05 × 0.577 ) ∴F = 291172.88 349838.4 F= 291172.88 F = 1.2


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By using “graphical simplification” of the S.F; X = 2 ( i − β )( β − φ ) = 2 ( 70 − 50 )( 50 − 30 ) X = Slope angle function = 2 400 or X = 40 Y = slope height function = γγH/ Y = (160 × 100 ) /1600 Y = 10 Now by using slope design chart for plane failure the safety factor for X= 40 and Y= 10 is;

S.F = 1.6


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Example 02: Determine the limiting pit slope angle (i) using the following data: 0 Inclination of failure plane = β50

i.

0 Angle of internal friction = φ35

ii.

3 Cohesion = C = 7800 kg m

iii. iv.

γ = 2.5 tons m 3

v.

Height of slope = 150m Solution: γH 2.5 × 150 375 = = c 7800 7.8 Y = 48.07 Y=

From the slope design chart, at Y=48 and S.F = 1.0, X=18 :.

( )( ) We know that X = 2 i − β β − φ 18 = 2 ( i − 50 )( 50 − 35) = 2 13i − 750

Now, squaring both sides

(18) 2 = ( 2) 2 × (15i − 750) 324 = 4 × (15i − 750 ) 324 = 60i − 3000 60i = 324 + 3000 = 3324 3324 i= = 55.4 0 60 i = 55.4 0


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Slope design chart for plane failure including various safety factors (Hoek, 1970a)


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Example 03: Determine the height of slope using the following data. 0 Slope angle = i = 65

i. ii.

0 Inclination of failure plane = β50

iii.

0 Angle of internal friction = φ = 35

iv.

2 Cohesion = c = 7800 kg m 3 Unit wt: of rock = γ = 2.5t m ;

v. vi.

Safety factor = S.F = 1.4; Sol: X = 2 ( i − β )( β − φ ) = 2 ( 65 − 50 )( 50 − 35) 2

15 × 15 = 2 × 255 = 2 × 15

X = 30 0 By using the graph of “slope design chart for plane failure including various safety factors” 0 At X = 30 and S.F = 1.4

The value of Y=16 :.

As

Y=

γH , c

Y × c 16 × 7800 = = 49.92m γ 2.5 H = 49.92m

∴H =


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CIRCULAR FAILURE

Fig. 22 (a): Plane failure in rock with highly ordered structure, such as state

.

Fig. 22 (b): Free body diagram As we know that; ΣM = 0 ∴ Mw + Mc = 0.; Mw = w × d


36


length of Arc ‘AB’ in fig:22(b); = Rθ Area of Arc ‘AB’= length × thickness [For simplicity, here we have considered the value of thickness of Arc as Unit] Area of Arc ‘AB’ = = Rθ × 1 Mc = (Area of Arc) × (cohesion) × (R) = Rθ × C × R Mc = CR 2 θ as, Mw+Mc= 0 wd − CR 2 θ = 0 wd = CR 2 θ or S.F =

CR 2 θ wd If the portion is in equilibrium state, it is stable, and will not slip down.


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Example: A cutting in saturated clay inclined at a slope of 1 vertical and 1.5 horizontal and has a vertical height of 10.0 m. the bulk unit weight of soil is 18.5 KN/m3 and its un-drained cohesion is 40 KN/m2. Determine the safety factor against immediate shear failure along the slip circle as shown in figure. Soln: 1 vertical Slope 1.5 horizontal Slope λ = 18.5 KN m 3 C = 40 KN m 2 From fig: 23, consider “ Δ afo ” to find out ‘ θ 1 ’

Figure 23 ∴ θ 1 = Tan −1

5m 16.7m

θ 1 = 16.7 0 According to Pythagoras theorem, 2

2

Hyp = perp + Base

2

R 2 = ( 5) + (16.7 ) = 25 + 278.9 = 303.0 2

2

∴ R = 303.9 R = 17.43m φ 1 = ?; since given slope of cutting plane “ad” is 1 vertical and 1.5 horizontal;  1  φ 1 = Tan −1    1.5  0 Thus, φ 1 = 33,7 from figure 23; it is clear that:


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φ = 90 0 − ( φ 1 + θ 1 )

(

)

= 90 0 − 33.7 0 + 16.7 0 = 39.6 0 φ = 39.6 0

Now we consider right angled “oec” 6.7 17.43 θ 2 = cos −1 ( 0.384 ) cosθ 2 =

θ 2 = 67.4 0

(

φ 3 = 180 0 − 90 0 + 67.4

)

φ 3 = 22.6 From Δ obe we get φ 2 , as follows,  6.7  0 φ 2 = sin −1   = φ 2 = 22.6  17.43  0 φ 3 = 180 − ( φ 1 + φ 2 ) = 180 0 − 33.7 0 + 22.6 0

(

(

)

φ 3 = 180 − 56.3 = φ 3 = 123.71 0

0

)

0

(

as φ = 90 − ( φ 1 + θ 1 ) = 90 0 − 33.7 0 + 16.7 0 0

φ = 84.1

)

0

(

Ψ = 180 0 − ( θ − φ ) = 180 0 − 84.10 + 39.6 Ψ = 180 − 123.7 0

)

0

Ψ = 56.3 0 Area of Δ1 =

Δaod = Δ 1 =

sinθ × sinφ 1 ×R2 × 2 sinΨ

(

) (

( )

1 sin 84.10 × sin 39.6 0 2 × (17.43) × 2 sin 56.3 0

)

Δ 1 = 115.85m 2 In order to find the area of ∆bcd , we must know any one side of ∆bcd . Let’s find “DB” for sake of ease. ∴ DB = R − OD

and

OD R = sinφ sinΨ


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OD =

(

)

Rsinφ 17.43 × sin 39.6 0 = = 13.36m sinΨ sin 56.3 0

(

)

∴ DB = 17.43 − 13.36 = DB = 4.07m sinφ 2 × sinφ 3 1 2 Δ 2 = × ( 4.07 ) × 2 sinφ 1 Δ 2 = 4.77m 2 Now area of polygon oabc=Ap Ap = 120.62m 2

(as) Area of sector

aoc =

(

1  π  × R 2θ ×  0  2  180 

)

1  π  × (17.43) × 84.10 ×  0  2  180  As = 222.86m 2 As =

Now, Area of slip mass (A)= As-Ap A = 222.86 − 120.62 A = 102.24m 2 Volume of slip mass= V = A × t = 102.24 × 1 V = 102.24m 3 Weight of slip mass= W = V × γ W = 102.24 × 18.5 W = 1891.44KN S.F = S.F =

CR 2 θ Wd

(

)

( 180 )

2 40 × (17.43) × 80.10 × π

S.F = 1.44

1891.44 × 6.54


0

40


STRIPPING RATIO

Figure 24 (a)

Vm

= Vol. of mining

Vc

= Vol. of cone

VT

= Vol. of truncated portion

:. Vm = Vc- VT

Fig. 24 (b)


41


As we know that; 1 2 πr h Vol: of cone= 3 2 And Vol: of cylinder = πr h

:. Volume of circular ore body; Vo 2 Vo= πr h ……… (1)

Volume of (small) truncated tip of cone; VT VT =

1 2 πr Δh 3 …… (2) (:.) Height of small cone is ∆h). and Δh = γtanθ

as height of big cone, is Hc; Hc = h +Δh Hc =h + γtanθ

and or :.

Vol: of bigger cone, Vc

1 1 πR 2 Hc = πr 2 ( h + Δh ) 3 3 1 or Vc = πR 2 ( h + γtanθ ) 3 Vc =

(3)

( ) :. Mined Vol: Vm = VC − VT Vm =

1 2 1 πr H c − πr 2 Δh 3 3

(4)

( ) Vol: of waste: Vw = Vm − V0 ( ) 2 Where, Vol: of Ore V0 πr h ( ) :. Vol: of waste Vw = Vm − V0 :. Stripping Ratio

SR =

Vw V0

(5)

( SR ) = Vol : of waste ( m3

( )

)

Vol : of ore. m

SR =

(6)


3

or

Ww W0

42


Example: A cylindrical ore body with radius of 50m and Depth of 250m is to be excavated by developing a cone. Slopes of the sides of the cone with the horizontal are 550. Unit weight of the ore is 3.1 tons/m 3 and that of the waste rock is 2.6 tons/m3.determine the stripping ratio (SR). Δh = γtanθ Δh = 50tan55 Δh = 71.4m H C = h + Δh = 250 + 71.4

Data: γ = 50m h = 250m γ ore = 3.1 tons m 3

H C = 321.4m

γ wast = 2.6tons/m 3

H C = RTanθ

Solution

∴R =

1 VC = πR 2 H c 3 VC = 17044879.24m 3

HC

Tan55 0

R = 321.4

Tan55 0 R = 225.04m

1 2 πγ Δh 3 VT = 186924.76m 3 VT =

Vm = VC − VT = 17044879.24 − 186924.76 Vm = 16857954.48m 3

(

)

Vore = πγ 2 h = π 50 2 × 250 = VO = 1963495.4m 3 VW = Vm − VO = 16857954.48 − 1963495.4 VW = 14894459.08m 3 S.R = S.R =

W waste W ore

=

Vw × γ w Vo × γ o

=

14894459.08 × 2.6 1963495.4 × 3.1

38725593.61

6086835.74 S.R = 6.36


43


DETERMINATION OF “SR” BY AREA METHOD

Ao =Area of sections (i), (ii) and (iii) 1 × 50 × 50 = 1250ft 2 2 Area of Sec: (i) = Area of Sec: (ii) =

100 × 50 = 5000ft 2

Area of Sec: (iii) =

100 × 150 = 15000ft 2

A O = 2(1250) + 2( 5000) + 15000 A O = 27500ft 2 1 × 100 × 100 = 5000ft 2 2 since Aw 1 = Aw 2 Aw 1 =

∴ Aw = Aw 1 + Aw 2 = 5000 + 5000 Aw = 10000ft 2 Aw 10000 ∴ S.R = = = 0.36 Ao 275000 S.R = 0.36 Now consider the bench of 25ft thickness.


44


Ao1 = 100 × 25 = 2500ft 2 Ao 2 =

(

)

50 2 + 50 2 × 25 = 70.71 × 25

Ao 2 = 1767.75ft

2

∴ Ao = Ao1 + 2Ao 2 = 2500 + 2(1767.75) Ao = 6035.5ft 2 Aw = 2Aw 1 + 2Aw 2 Aw 1 =

( 100

2

)

+ 100 2 × 25.10000 × 25 = 150000ft

1 × 25 × 25 = 312.5ft 2 2 ∴ Aw = ( 2 × 250000 ) + ( 2 × 312.5) = 500000 + 625 Aw 2 =

Aw = 500625ft 2 A = ½(b1 + b1) × h = b1h


45


PIT LIMITS The minable material becomes that lying within the pit boundaries. A vertical section taken through such a pit is shown in figure A: The size and shape of the pit depends upon economic factors and design/production constraints. With an increase in price, the pit would expand in size assuming all other factors remained constant. The “pit” existing at the end of mining is called the “final pit” or the ultimate pit. In b/w the birth and death of an open pit mine, there are a series of intermediate pits. This includes a series of procedures based upon: •

Hand Methods,

•

Computer Methods, and

•

Computer assisted hand methods

The above mentioned methods are used for developing the pit limits. Within the pit materials of differing values are found. Economics criteria are applied to assign destination for these materials based on their value (i.e. mill, waste dump, loach dump, stock pile etc). Once the pit limits have been determined and rules established for classifying the in pit materials, then the ore reserves (tonnage and grade) can be calculated. The Net Value Calculation:The term cut-off grade refers to the “grades” for which the destination of pit material changes. It should be noted that “grades” were used rather than “grade” since there may be several possible – destinations. The simplest case is that in which there are two


46


destinations: the mill or the waste dump. One cut of grade is needed for many operations today; there are many possible destinations: the mill or leach dump and the waste dump. Each of the decisions:•

Mill or leach?

•

Leach or waste?

requires a cut-off grade. Cut-off Grade:The grade at which the mineral resources can’t longer be processed at a profit. Break-even cut-off grade:The grade at which the net value is zero; is called Break-even cut-off grade.

Determination of pit limits, on the basis of net value:

Example: Determine the pit limits of an open pit mine as shown in figure: setting price of 1m3 of Ore is US$ 1.9 and mining cost of 1m3 of waste is US$ 1.0. Sol: For Strip # 01:= Vo 1 = 5 ×1.25 ×1 3 Vol: of ore (Strip 1) = Vo 1 = 6.25m


47


1  = Vw 1 = ( 6.364 × 1.25 × 1) +  × 1.25 × 1.25  2  3 Vol: of waste (Strip 1) = Vw 1 = 8.74m

Ins (S.R) =

Vw 1 = Vo 1

Value of ore

S.R = 1.4 Ins

= Vo1* Selling price of 1/m3 of ore

6.25 × 1.9 = Value of ore = 11.875US dollars Cost of mining of waste = = Vw 1 × Mining cost of 1m 3 waste = 8.74 × 1.0 Cost of mining of waste = 8.74 US dollars Net value = value – cost = 11.875 – 8.74 N.V = 3.14 US $.

→ for strip # 01

Calculate for strip, 2, 3 and 4. For Strip #02:Vol. of ore, Vo 2 = 5 × 1.25 × 1 Vo 2 = 6.25m 3

(

Vol. of waste, Vw 2 = ( 6.614 × 1.25 × 1) + 1 × 1.25 2 2 3 Vw 2 = 10.3m

)

Vol. of ore = Vo 2 × 1.9 = 6.25 × 1.9 Vol. of ore = 11.875US $ Cost of mining waste = 10.3 × 1 = 10.3US $ Vw 2 10.3 SR (ins) = = = 1.65 Vo 2 6.25 N.V = Value − cost = 11.875 − 10.3 N.V = 1.58US dollars For Strip# 03


48


Vo 3 = 6.25m 3

(

2 Vw 3 = ( 8.864 × 1.25 × 1) + 1 (1.25) 2 Vw 3 = 11.86m 3

S.R ( ins ) = 11.86

6.25

)

= S.R ( ins ) = 1.9

Value of ore = 6.25 × 1.9 = 11.875 US dollars Cost of waste = 11.86 × 1 = 11.86 US dollars N.V = 11.875 - 11.86 = N.V = 0.015 US dollars For Strip # 04:Vo 4 = 6.25

(

Vw 4 = (10.114 × 1.25 × 1) + 1 × 1.25 2 2 Vw 4 = 12.6425 + 0.78125

)

Vw 4 = 13.424m 3 S.R ( Ins) =

Vw 4 13.424 = = 2.14 Vo 4 6.25

Cost of mining waste = 13.424 × 1 = 13.424 $ Price/value of ore = 6.25 × 1.9 = 11.875 $ N.V = 11.875 - 13.424 N.V = -1.55US dollars As can be seen that the net value changes from (+) to (-) as the pit is expanded, sometimes the N. V become zero, so that this pit position is termed as “Break-even”, which is the location of final pit wall.

Now for overall volume of waste we have;


49


1 ( 9.9) × ( 9.9) 2 = 49m 2 × thickness (1)

Δw =

∴ overall vol : of waste = 49m 3

(

)

& overall vol : of ore = ( 9.9 × 5 × 1) + 1 × ( 5) × ( 5) × 1 2 = ( 49.5 + 12.5) × 1m overall vol : of ore = 62m 3 overall strippig ratio = Vw

= 49 = 0.79 = 0.8 Vo 62 overall value of ore = 62 × 1.9 = 117.8 US dollars overall mining cost of waste = 49 × 1 = 49.00 US dollars N - value (overall) = 117.8 - 49.00 N.V = 68.8 US dollars


50


Example Copper ore is milled to produce a copper concentrate. This mill concentrate is slipped and transported to a smatter and the resulting blister copper is eventually refined. In this example, the following data will be assumed. Mill recovery rate

= 80%

Mill concentrate grade

= 20%

Smelting loss

= 10 lbs/st of concentrate

Refining loss

= 5 lbs/st of blister copper.

The ore is containing 0.55%, All the costs and revenues will be calculated in terms of 1 ton of ore.

Note that ore short ton= 2000 lbs.

Solution:Step # 01:- Complete the amount of saleable copper (lb/s per st of ore) a) Contained copper (cc) is :0.55 cc = 2000lb/st * = 11.00lbs 100 b) Copper recovered by mill (RM/st of ore) RM = 11lbs ×

80 = 8.8lbs 100

c) Concentration ratio (r):- the ratio of concentration is defined as lbs of cu/st of concentrate lbs of " cu" recovered/st of ore 2000 × ( 20 100 ) 400 r= = = 45.45 8.8 8.8 r = 45.45 r=

d) Copper recovered by smelter :- (R.S) As, smelting loss = 10 lbs/st of concentrate.\ Economic Block Models:The block model representation of ore bodies rather than section representation and the storage of the information on high speed computers have offered some new possibilities in open pit mines.


51


ORE GRADE ESTIMATION by the constant distance weighting techniques a) Inverse Distance Techniques:The formula to calculate the ore grade by inverse distance technique is : n

ϑ=

Σ  i =1 n

gi  di 

Σ ( i di ) i =1

Where, gi=

given grade of ore at a point.

g=

estimated grade of ore.

di=

distance b/w known point and point of estimation.

Example: Calculate the estimated grade of ore at point “C”, using inverse distance technique. Known grades of an ore at points C1-C6 ore shown in fig: [in brackets]

Sol: Engr. Izhar Mithal Jiskani

52


First let’s estimate the value of distances d1, d2...d6 by using Pythogona’s theorem. d 1 = 200 2 + 200 2 = 8000 = d 1 = 282.84m d 2 = 300 2 + 100 2 = 100000 = d 2 = 316.23m d 3 = 100 2 + 200 2 = 50000 = d 3 = 223.6m d 4 = 200 2 + 200 2 = 80000 = d 4 = 282.84m d 5 = 200 2 + 100 2 = 50000 = d 5 = 223.6m d 6 = 100 2 + 300 2 = 100000 = d 6 = 316.23m As we know that;  g1  +  g 2  + − − − − − − − +  g 6   d   d   d   1  2  6   g=  1 + 1 +−−−−−−−+ 6   d   d   d   1  2  6    0.409   0.165   0.258   1.365   0.023   0.644   + + + + +  282.84   316.23   223.6   282.84   223.6   316.23   g= 1 + 1 + 1 + 1 + 1 + 1 282.84 316.23 223.6 282.84 223.6 316.23 0.0100 g= = g = 4.484 × 10 −1 0.0223

(

) (

) (

) (

) (

) (

)

:. The estimated grade of ore =g g = 0.45 % b) Inverse Distance Squared weighting Technique:Using this technique, the grade of ore is found, by using following equations. n

g=

gi

Σ ( di ) i =1 n

2

1

Σ ( di ) i =1

2

Using the data of previous example; calculate the grade of ore by squared weighting technique.


53


2

d 1 = 282.84m - - - - - - - - - - d 1 = 79998.4656m 2 2

d 2 = 316.23m - - - - - - - - - - d 2 = 100001.4129m 2 2

d 3 = 223.6m - - - - - - - - - - d 3 = 49996.96m 2 2

d 4 = 282.84m - - - - - - - - - - d 4 = 79998.4656m 2 2

d 5 = 223.6m - - - - - - - - - - d 5 = 49996.96m 2 2

d 6 = 316.23m - - - - - - - - - - d 6 = 100001.4129m 2

 g1   g 2   +  + − − − − + g 5 2  +  g 6 2  d 2  d 2   d5   d6  1   2  ∴ g=       1   1  +  + − − − − + 1 2  +  1 2   2  2     d1   d 2   d5   d6  3.5885 × 10 −5 g= 8.5003 × 10 −5 g = 0.422%


54


ORE RESERVE ESTIMATION by Triangular Method

From above figure:

( )( ) Area of rectangle abcd= x 3 − x 1 y 2 − y 3 Area of Area of Area of

ΔA 1 =

1 ( x 2 − x 1 )( y 2 − y1 ) 2

ΔA 2 =

1 ( x 3 − x 2 )( y 2 − y 3 ) 2

ΔA 3 =

1 ( x 3 − x 1 )( y1 − y 3 ) 2

( ) Area of ΔA = Area of abcd - ΔA 1 + ΔA 2 + ΔA 3 ( x 3 − x 1 )( y 2 − y 1 ) +  1 = ( x 3 − x 1 )( y 2 − y 3 ) − ( x 3 − x 2 )( y 2 − y 3 ) +  − − − − − − (1) 2 ( x 3 − x 1 )( y 1 − y 3 ) +  Example: Calculate the above of ΔA by using ore reserve estimation: Easting (x)m

Northing (y)m

1100

1200

1500

1200

1100

800

Solution:Let


55


x 1 = 1100m, y1 = 1200m x 2 = 1500m, y 2 = 1200m x 3 = 1100m, y 3 = 800m

From given table.

And we know that; A = ( x 3 − x 1 )( y 2 − y 3 ) −

1 ( x 3 − x 1 )( y 2 − y 3 ) + ( x 3 − x 2 )    2 ( y 2 − y 3 )( x 3 − x 1 ) + ( y 1 − y 3 ) 

A = (1100 − 1100 )(1200 − 800 ) − =0−

1 (1100 − 1100 )(1200 − 1200 ) + (1100 − 1500 )  2 (1200 − 800 )(1100 − 1100 ) + (1200 − 800 ) 

1 [ 0 − 16000 + 0] 2

1 × 160000 2 A = 80,000m 2 =

OR

ΔA =

1 ×b×h 2

1 × 400 × 400 2 = 1 × 160000 2 ΔA = 80,000m 2 =


56


Example: Calculate the ore reserves in ∆ area, as shown in following figure. Density of the ore is given as 2.5 tons/m3

Solution:As we know that: ΔA = Area of abcd - [ ΔA 1 + ΔA 2 + ΔA 3 ]

[(

)] (

) (

= ( 500 × 700 ) − 1 × 700 × 200 + 1 × 400 × 300 + 1 × 500 × 300 2 2 2 = 350000 − ( 205000 )

)

∴ ΔA = 145000m 2

C1 + C 2 + C 3 3 Average thickness of ore = 3 + 5 + 4 12 t av: = = = 4m 2 3 t av: = 4m Vol: of ore in Triangular area= ΔA × t av: Vol. of ore = 145000×4m Vol. of ore = 580000m3. ;. Ore reserves = 580000×2.5 Ore reserves = 1450000


57


Calculation Of Thickness Of Ore In A Drill Hole Length of “t” ore body (m)

Grade (%) “g”

Length × grade (t×g)

t 1 = 0.6

g 1 = 0.59

t 1g 1

t 2 = 1.4

g 2 = 0.48

t 2g 2

t 3 = 1.4

g 3 = 0.6

t 3g 3

t 4 = 1.4

g 4 = 0.56

t 4g 4

t 5 = 1.3

g 5 = 0.32

t 5g 5

n

Σ tigi i =1 n

:. Average thickness = (tav:) ∴ t av: =

- - - - - - - (1)

Σ ti i =1

( 0.6 × 0.59) + (1.4 × 0.48) + (1.4 × 0.6) + (1.4 × 0.56) + (1.3 × 0.32) ( 0.6 + 1.4 + 1.4 + 1.4 + 1.3)

3.066 6.1 = 0.503m

t av: = t av:

Calculation Of Reserves Using Weight Age Average Thickness:

Calculate the reserves of ore shown in above fig: having a density of 1.35 tons/m 3 by using the weight age average thickness, when; t1=40m, t2=60m, t3=50m Solution: the weight age average thickness is calculated as:


58


 t 1 + θ1 + t 2 θ 2 + t 3θ 3    60  tw = 

3

− − − − (1)

from above fig :  800 − 300  −1 φ = tan −1   = tan ( 2.5) 400 − 200   0 ϕ = 68.2  500 − 300  −1 φ1 = tan −1   = tan ( 0.5)  600 − 400  φ1 = 26.56 0  800 − 500  −1 φ 2 = tan −1 =   = tan (1.5)  600 − 400  0 φ 2 = 56.31 θ1 = φ − φ1 = 68.2 0 − 26.56 0 = 41.64 0 θ 3 = φ1 − φ 2 = 26.56 0 + 56.310 = 82.87 0

(

θ 2 = 180 0 ( φ1 + φ 3 ) = 180 0 − 41.64 0 + 82.87 0

)

θ1 = 55.49 0 Substitute the values of t1, t2, t3, θ 1, θ 2, θ 3 in equation 1  (40 x 41.64) + (60 x 55.49) + (50 x 82.87)    60 tw =   9138.5    tw =  60  tw =

3

3

152.3 3

tw = 50.77 m ∆ A = Area of abcd – [∆A1, ∆A2, ∆A3] ∆ A = (400 x 500) – [ (½ x 500 x 200) + (½ x 300 x 200) + (½ x 400 x 200) ] ∆ A = 200000 – 120000 ∆ A = 80000 m2 Volume of A = A x tw = 80000 x 50.77 :. Volume of A = 4061600 m3 :. Ore reserves = Volume x density = 4061600 m3 x 1.35 t/m3 = 5483160 tons


59


Example: In a level terrain, determine the max height of high wall that dragline can strip without re-handling, using the following data: Dumping radius

=(Rd)= 47m,

Outside diameter of tub

= (Et) = 11m,

Spoil angle of repose ( θ ) High wall angle = ( φ )

=370

Pit width =(w)

= 15m

Sean thickness = (T)

= 1.22 m

Swelling factor = (Ps)

= 30 %

= 8740

Solution: As we know that Rd = Re + So .:

Re = Rd – So -------> 1 and So = 0.75 Et. Re = 47 – (0.75 Et) Re = 47 – (0.75 ×11) Re = 47 – 8.25 Re = 38.75 m    Ps   W  Re = Hcotφ + cotθ H1 +  +  tanθ  − T     100   4     30  15 0 38.75 = Hcot74 0 + cot37 0 H1 +  +  tan37  − 1.22    100   4  = 0.287H + 1.327(1.3H + 2.825 − 1.22 ) = 0.287H + 1.7251H + 3.748 − 1.619 38.75 = 2.0121H + 2.13 2.012H = 38.75 − 2.13 36.62 H= 2.012 H = 18.2


60

Surface Mine Design and Practice

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