Surface Mine Design and Practice

SURFACE MINE DESIGNAND PRACTICE

Engr. Izhar Mithal Jiskani

Department of Mining Engineering,Mehran University of Engineering & Technology

Jamshoro, Sindh

SURFACE MINE DESIGN AND PRACTICE

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.

Engr. Izhar Mithal Jiskani 2


BASIC BENCH GEOMETRYThe 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 ‘ ’. 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 55O to 80O. A typical initial design



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

sometimes reduced to a width of about 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.



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



Fig. 2a: Functions of a catch bench



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.

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.

Ramp volume (V) =

:. 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 feetDiameter = d= 500 feetWidth of ramp = Rw = 65 feet

Solution:as V = Volume of ramp

and V = 50 × × Rw

V = 50 × =

V = 365625 ft3 Ans

.

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

Solution: - V = 50 × × Rw

V = 50 ×

V = 600000 ft3 Ans



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)





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 AStep I



Figure: 6 BStep II

.

Figure: 6 CStep 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



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.



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.



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.



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



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.



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

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 “θ0” 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.



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.

Height of each bench = y = 50 feet.

Horizontal distance = x = 35 feet

Distance under slope = x’ =?

Bench face angle = 750 = θ

:.

Since we have 5 benches,

i.e.: - 5 slopes, X= (4*x) + (5*x’).

X = (4 x 35) + (5 x 13.4) = 140 + 67 X = 207’

Y = 5 x 50 Y = 250’

:. Overall pit slope angle = θ overall

207

250 tan

x

y tanθ 11

overall

Ooverall 0.45 θ



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;

x

y tanθ 1

overall

where, Y = 250’

100' tan75

5x50 4x35 X and

O

:. X = 307 feet.

O1overall 39.18

307

250tanθ

Ooverall 39.2θ

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.



INTER-RAMP SLOPE ANGLES AFTER RAMP INCLUDED

The ramp breaks the overall slope angles “ 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.

Figure 16: Inter ramp slope angles after ramp included

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.



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 “ overall” now becomes;

Fig. 17: Overall slope angle with working bench included



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:-

As; Y = 250 – 50 = 200 feet and

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



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;



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:

sliding induce to tendingForce

slidingresist toavailable force TotalF

For the case shown in Fig:”20”

βWsin

tanφβ WcoscAF

If there is water present then;

VβWsin

tanφtU-β WcoscAF

where,

U= water pressure along potential failure surface,

= friction angle (affected by water),

V= force along potential sliding plane.



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:-

force Sliding

force frictionalS.F

tanφWcosβcA force Frictional

Fig. 21 (a)

Fig. 21 (b)

B

CAb

sin

sinsin

2

1 2

(a)

A

CBa

sin

sinsin

2

1 2

(b)

C

CAc

sin

sinsin

2

1 2

(c)

as, bb

HB

100sin



or130.54feet

0.766

100

sin50

100b

0

Putting the above value in eqn (a).

2

0

002

62.2375

279.09396.0

262.0692.17040

2

1110sin

50sin20sin54.130

2

1

ft

(V) Volume of (triangle ABC) sliding block = thickness of sliding block

1ft2375.62 31ft2375.62V

(W) Weight of sliding block is = volume × unit weight

s380099.2lbW

160lb/ft2375.62γVW 3

As

Wsinβ

WcosβcosβcAF

A= Area of the failure plane, and

A= length of failure plane × thickness

A= b × 1ft= 130.54ft × 1ft

A= 130.54ft2

1.2F291172.88

349838.4F

291172.88

0.577244323.05130.541600F

0.577Tan30Tanφ

1600lb/ftc

b291172.88lsin50380099.2Wsinβ

b244323.05lcos50380099.2Wcosβ

0

2

0



By using “graphical simplification” of the S.F;

10Y

/1600100160Y

γγH/function height slope Y

40Xor

400function angle SlopeX

30505070φββiX2

22

Now by using slope design chart for plane

failure the safety factor for X= 40 and Y= 10

is;

S.F = 1.6



Example 02:

Determine the limiting pit slope angle (i) using the following data:

i. Inclination of failure plane 0β50

ii. Angle of internal friction 0φ35

iii. Cohesion = 3mkg7800C

iv.3mtons2.5γ

v. Height of slope = 150m

Solution:

48.07Y7.8

375

7800

1502.5

c

γHY

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

:. We know that iX 2

75013i2355050i218

Now, squaring both sides

0

0

22

4.55

4.5560

3324

3324300032460

300060324

750154324

75015218

i

i

i

i

i

i



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



Example 03:

Determine the height of slope using the following data.

i. Slope angle 065i

ii. Inclination of failure plane 0β50

iii. Angle of internal friction 053

iv. Cohesion27800c mkg

v. Unit wt: of rock 3m2.5tγ ;

vi. Safety factor = S.F = 1.4;

Sol:

0

2

22

30X

15225521515

35505065φββiX

By using the graph of “slope design chart for plane failure including various safety

factors”

At 1.4 S.F and30X 0

The value of Y=16

:. As ,

c

HY

49.92mH

49.92m2.5

780016

γ

cYH



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;

dwMw0.;McMw

0ΣM



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’ = 1Rθ

Mc = (Area of Arc) × (cohesion) × (R)

θCRMc

RCRθ2

as, Mw+Mc= 0

wd

θCRS.For

θCRwd

0θCRwd

2

2

2

If the portion is in equilibrium state,

it is stable, and will not slip down.



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.

Figure 23

Soln:

1 vertical Slope

1.5 horizontal Slope

2

3

mKN40C

mKN18.5λ

From fig: 23, consider “ afo Δ ” to find out ‘ 1θ ’

01

11

16.7θ

16.7m

5mTanθ

According to Pythagoras theorem,

17.43mR

303.9R

303.0278.92516.75R

BaseperpHyp222

222

1φ ?; since given slope of cutting plane “ad” is 1 vertical and 1.5 horizontal;

Thus,0

1

11

33,7φ

1.5

1Tanφ

from figure 23; it is clear that:



0

0000

110

39.6φ

39.616.733.790

θφ90φ

Now we consider right angled “oec”

22.6φ

67.490180φ

67.4θ

0.384cosθ

17.43

6.7cosθ

3

003

02

12

2

From obe Δ we get 2 , as follows,

0

00

000

0

00011

0

03

003

00021

03

02

12

56.3Ψ

123.7180Ψ

39.684.1180φθ180Ψ

84.1φ

16.733.790θφ90φ as

123.71φ56.3180φ

22.633.7180φφ180φ

22.6φ17.43

6.7sinφ

Area of sinΨ

sinφsinθR

2

1ΔΔaod 2

1

21

0

002

1

115.85mΔ

56.3sin

39.6sin84.1sin17.43

2

1Δ

In order to find the area of bcd , we must know any one side of bcd . Let’s find “DB”

for sake of ease.

sinΨ

R

sinφ

OD and ODRDB



2

2

1

3222

0

0

4.77mΔ

sinφ

sinφsinφ4.07

2

1Δ

4.07mDB13.3617.43DB

13.36m56.3sin

39.6sin17.43

sinΨ

RsinφOD

Now area of polygon oabc=Ap

2120.62mAp

(as) Area of sector

0

2

1802

1 Raoc

2

0

0

222.86mAs

180

π84.117.43

2

1As

Now, Area of slip mass (A)= As-Ap

2102.24mA

120.62222.86A

Volume of slip mass= 1102.24tAV

3102.24mV

Weight of slip mass= γVW

1.44S.F

6.541891.44180

π80.117.4340S.F

Wd

θCRS.F

1891.44KNW

18.5102.24W

002

2



STRIPPING RATIO

Figure 24 (a)

Vm = Vol. of mining

Vc = Vol. of cone

VT = Vol. of truncated portion

:. Vm = Vc- VT

Fig. 24 (b)



As we know that;

Vol: of cone=hπr

3

1 2

And Vol: of cylinder = hπr 2

:. Volume of circular ore body; Vo

Vo= hπr 2……… (1)

Volume of (small) truncated tip of cone; VT

Δhπr3

1V 2

T …… (2) (:.) Height of small cone is ∆h). and γtanθΔh

as height of big cone, is Hc;

and

or

:. Vol: of bigger cone, Vc

(3) γtanθhπR3

1Vcor

Δhhπr3

1HcπR

3

1Vc

2

22

:. Mined Vol: TC VVVm

(4) Δh πr3

1Hπr

3

1V 2

c2

m

Vol: of waste: 0mw VVV

Where, Vol: of Ore hπrV 2

0

:. Vol: of waste (5) VVV 0mw

:. Stripping Ratio

3

3

m ore. of :Vol

m wasteof :VolSR

(6) V

VSR

0

wor 0

w

W

WSR



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/m3 and that of the waste rock is 2.6 tons/m3.determine the

stripping ratio (SR).

Data:

3wast

3ore

2.6tons/mγ

mtons3.1γ

250mh

50mγ

Solution

3T

2T

3C

c2

C

186924.76mV

Δhπγ3

1V

4m17044879.2V

HπR3

1V

225.04mRTan55

321.4R

Tan55HR

RTan θH

321.4mH

71.4250ΔhhH

71.4mΔh

50tan55Δh

γtanθΔh

0

0C

C

C

C

3W

OmW

3O

22ore

3m

TCm

8m14894459.0V

1963495.4816857954.4VVV

1963495.4mV25050πhπγV

8m16857954.4V

186924.76417044879.2VVV

6.36S.R6086835.74

138725593.6S.R

3.11963495.4

2.6814894459.0

γV

γV

W

WS.R

oo

ww

ore

waste



DETERMINATION OF “SR” BY AREA METHOD

Ao =Area of sections (i), (ii) and (iii)

Area of Sec: (i) =

21250ft50502

1

Area of Sec: (ii) =20ft50050100

Area of Sec: (iii) =20ft1500501100

2

O

O

27500ftA

150005000212502A

0.36S.R

0.36275000

10000

Ao

AwS.R

10000ftAw

50005000AwAwAw

AwAw since

5000ft1001002

1Aw

2

21

21

21

Now consider the bench of 25ft thickness.



2

22

221

21

2

21

22

222

21

500625ftAw

625500000

312.522500002Aw

312.5ft25252

1Aw

150000ft2525.10000100100Aw

2Aw2AwAw

6035.5ftAo

1767.75225002AoAoAo

1767.75ftAo

2570.71255050Ao

2500ft25100Ao

A = ½(b1 + b1) × h = b1h



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

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:-

Vol: of ore (Strip 1) 3

1

1

6.25mVo

11.255Vo

Vol: of waste (Strip 1)

3

1

1

8.74mVw

1.251.252

111.256.364Vw



1.4

Vo

Vw (S.R) Ins S.R

Ins1

1

Value of ore = Vo1* Selling price of 1/m3 of ore

= dollars 11.875US ore of Value

1.96.25

Cost of mining of waste =

1.08.74

waste1m ofcost Mining Vw 31

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:-

32

2

6.25mVo

11.255Vo ore, of Vol.

3

2

22

10.3mVw

1.252111.256.614Vw waste,of Vol.

dollars 1.58USN.V

10.311.875costValueN.V

1.656.25

10.3

Vo

Vw (ins) SR

$ 10.3US110.3 wastemining ofCost

$ 11.875USore of Vol.

1.96.251.9Vo ore of Vol.

2

2

2

For Strip# 03

1.9ins S.R 6.2511.86ins S.R

11.86mVw

1.252111.258.864Vw

6.25mVo

33

23

33



dollars US0.015N.V

11.86-11.875N.V

dollars US11.86111.86 wasteofCost

dollars US11.8751.96.25ore of Value

For Strip # 04:-

dollars -1.55USN.V

13.424-11.875 N.V

$ 11.8751.96.25 ore of ePrice/valu

$ 13.424113.424 wastemining ofCost

2.146.25

13.424

Vo

VwInsS.R

13.424mVw

0.7812512.6425Vw

1.252111.2510.114Vw

6.25Vo

4

4

34

4

24

4

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;

dollars US68.8N.V

49.00-117.8 (overall) value-N

dollars US49.00149 wasteofcost mining overall

dollars US117.81.962ore of valueoverall

0.80.796249

VoVwratio strippig overall

62more of : voloverall

1m12.549.5

15521159.9 ore of : voloverall &

49m wasteof : voloverall

1 thickness49m

9.99.92

1Δw

3

3

2



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 :-

11.00lbs100

0.55 * 2000lb/st cc

b) Copper recovered by mill (RM/st of ore)

8.8lbs100

8011lbsRM

c) Concentration ratio (r):- the ratio of concentration is defined as

45.45r

45.458.8

400

8.8

100202000r

ore ofst recovered/ cu"" of lbs

econcentrat ofcu/st of lbsr

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.



ORE GRADE ESTIMATIONby the constant distance weighting techniques

a) Inverse Distance Techniques:-

The formula to calculate the ore grade by inverse distance technique is :

dii

digi

Σ

Σn

1i

n

1i

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:



First let’s estimate the value of distances d1, d2...d6 by using Pythogona’s theorem.

316.23md100000300100d

223.6md50000100200d

282.84md80000200200d

223.6md50000200100d

316.23md100000100300d

282.84md8000200200d

622

6

522

5

422

4

322

3

222

2

122

1

As we know that;

1

6

6

2

1

1

1

6

6

2

2

1

1

104.484g0.0223

0.0100g

316.231

223.61

282.841

223.61

316.231

282.841

316.23

0.644

223.6

0.023

282.84

1.365

223.6

0.258

316.23

0.165

282.84

0.409

g

ddd

dg

dg

dg

g

:. 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.

2

n

1i

2

n

1i

di

1di

gi

g

Σ

Σ

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

technique.



22

66

2255

2244

2233

2222

2211

9m100001.412d ---------- 316.23md

49996.96md ---------- 223.6md

m79998.4656d ---------- 282.84md

49996.96md ---------- 223.6md

9m100001.412d ---------- 316.23md

m79998.4656d ---------- 282.84md

0.422%g108.5003

103.5885g

d1

d1

d1

d1

dg

dg

d

g

d

g

g

5

5

26

25

22

21

26

62

5

52

2

22

1

1



ORE RESERVE ESTIMATIONby Triangular Method

From above figure:

Area of rectangle abcd= 3213 yyxx

Area of 12121 yyxx

2

1ΔA

Area of 32232 yyxx

2

1ΔA

Area of 31133 yyxx

2

1ΔA

Area of 321 ΔAΔAΔA- abcd of Area ΔA

1

yyxx

yyxx

yyxx

2

1yyxx

3113

3223

1213

3213

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



mym

mym

mym

800,1100x

1200,1500x

1200,1100x

33

22

11

From given table.

And we know that;

311332

2332133213 yyxxyy

xxyyxx

2

1yyxxA

280,000mA

1600002

1

01600002

10

8001200110011008001200

150011001200120011001100

2

1800120011001100A

OR

280,000mΔA

16000021

4004002

1

hb2

1ΔA



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:

2

321

145000mΔA

205000350000

300500213004002

120070021700500

ΔAΔAΔA - abcd of AreaΔA

Average thickness of ore = 3321 CCC

4mt

4m3

12

2

453t

:av

:av

Vol: of ore in Triangular area= :avtΔA

Vol. of ore = 145000×4mVol. of ore = 580000m3.;. Ore reserves = 580000×2.5

Ore reserves = 1450000



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

Grade (%)“g”

Length × grade(t×g)

0.6t1 0.59g1 11gt

1.4t 2 0.48g 2 22gt

1.4t 3 0.6g 3 33gt

4.1t 4 0.56g 4 44gt

1.3t 5 0.32g 5 55gt

:. Average thickness = (tav:)

(1)------- ti

tigi

Σ

Σn

1i

n

1i

0.503m t6.1

3.066 t

1.31.41.41.40.6

0.321.30.561.40.61.40.481.40.590.6 t

:av

:av

: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/m3 by

using the weight age average thickness, when; t1=40m, t2=60m, t3=50m

Solution: the weight age average thickness is calculated as:



0

1

00031

02

000213

00011

02

112

01

111

0

11

332211

w

55.49θ

82.8741.64180φφ180θ

82.8756.3126.56φφθ

41.6426.5668.2φφθ

56.31φ

1.5tan400600

500800tanφ

26.56φ

0.5tan400600

300500tanφ

68.2

2.5tan200400

300800tanφ

:fig above from

(1) 360

θtθtθt

t

Substitute the values of t1, t2, t3, θ 1, θ 2, θ 3 in equation 1

3

152.3 tw

360

9138.5

tw

360

82.87) x (50 55.49) x (60 41.64) x (40

tw

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



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 ( ) =370

High wall angle = ( φ ) = 8740

Pit width =(w) = 15m

Sean thickness = (T) = 1.22 m

Swelling factor = (Ps) = 30 %

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

18.2H2.012

36.62H

2.1338.752.012H

2.132.0121H38.75

1.6193.7481.7251H0.287H

1.222.8251.3H1.3270.287H

1.22tan374

15

100

301Hcot37Hcot7438.75

Ttanθ4

W

100

Ps1HcotθHcotφRe

000


Documents

Surface Mine Design and Practice