150315579 Surface Mine Design and Practice

Prévia do material em texto

SURFACE MINE 
DESIGN
AND PRACTICE
Engr. Izhar Mithal Jiskani
Department of Mining Engineering,
Mehran University of Engineering & Technology
SURFACE MINE DESIGN AND PRACTICE
Jamshoro, Sindh
Engr. Izhar Mithal Jiskani 2
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.
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SURFACE MINE DESIGN AND PRACTICE
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 55O to 80O. A typical initial design 
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SURFACE MINE DESIGN AND PRACTICE
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 3
2 of the bench height. At the end of mine life, the safety benches are 
sometimes reduced to a width of about 3
1 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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
Fig. 2a: Functions of a catch bench
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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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) = 
H100g
H
2
1Rw
H100
g
H
2
1Rw
Hhorizontal
2
1Rw
××××



×



×××



×××
:. 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 × 
g
H2 × Rw
V = 50 × 
8
65x (30)2 = 
8
2925000
V = 365625 ft3 Ans
.
Example # 02:-
Repeat example # 01 for Depth (H) = 40 ft and Ramp width (Rw) = 60 ft.
Solution: - V = 50 × 
g
H2 × Rw
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SURFACE MINE DESIGN AND PRACTICE
V = 50 × 
8
60x (40) 2 
V = 600000 ft3 Ans
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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
• 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 dimensionsbecome ;
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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.
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SURFACE MINE DESIGN AND PRACTICE
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.
O
O
1
overall 50.4
tan75
5x504x35
(Y)5x50tanθ =








+
→
=
−
• Height of each bench = y = 50 feet.
• Horizontal distance = x = 35 feet
• Distance under slope = x’ =?
• Bench face angle = 750 = θ
:. = x'ytanθ
'' 13.4 
3.732
50 
tan75
50 
tanθ
y x ====
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 11overall
O
overall 0.45 θ =
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SURFACE MINE DESIGN AND PRACTICE
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 1overall
where, Y = 250’
( ) 100' 
tan75
5x50 4x35 X and O +


+=
:. X = 307 feet.
O1
overall 39.18307
250tanθ =


=
−
O
overall 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.
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SURFACE MINE DESIGN AND PRACTICE
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.
( )
O
21
O1
21
OO
1
21
50.4θIRθIR
50.38
103.5
125tanθIRθIR
103.56.726.870X
tan75
25
tan75
5022x35X
124feetY245050Y
x
ytanθIRθIR
==
=


==
=++=



+


+=
=→++=
==
−
−
Figure 16: Inter ramp slope angles after ramp included
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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
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;
( )
O1
overall
O
1
overall
36.98 
332
250tanθ 
feet 332 X 125' 67' 140' X
125''
tan75
505 '4x35 X ,250' Y As;
x
ytanθ
⇒


=∴
=⇒++=
+


+==
=
−
−
Fig. 17: Overall slope angle with working bench included
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SURFACE MINE DESIGN AND PRACTICE
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:-
 O O
2
1
2
1
2
51.6or 51.58θIR
158.6'
200tanθIR
;x
ytanθIR
=



=
=
−
−
As; Y = 250 – 50 = 200 feet and
( ) 


+= Otan75
50 4 35 x 3 X
158.6' 53.6 105 X =+=
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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SURFACE MINE DESIGN AND PRACTICE
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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SURFACE MINE DESIGN AND PRACTICE
• 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,
aφ = friction angle (affected by water),
V= force along potential sliding plane.
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SURFACE MINE DESIGN AND PRACTICE
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 thicknessof 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)
( )
0
000
0
000
110B
70180CA180B
50C
20A5070A
50β 70,
=∠
−=∠+∠−=∠
=∠
=∠=−=∠
==

 i
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 ==∠
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SURFACE MINE DESIGN AND PRACTICE
or
130.54feet
0.766
100
sin50
100b 0 ===
Putting the above value in eqn (a).
( )
2
0
00
2
62.2375
279.0
9396.0
262.0692.17040
2
1
110sin
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
( ) ( )
1.2F
291172.88
349838.4F
291172.88
0.577244323.05130.541600F
0.577Tan30Tanφ
1600lb/ftc
b291172.88lsin50380099.2Wsinβ
b244323.05lcos50380099.2Wcosβ
0
2
0
=
=
×+×
=∴
==
=
=×=
=×=
 
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SURFACE MINE DESIGN AND PRACTICE
By using “graphical simplification” of the S.F;
( ) ( ) ( ) ( )
( )
10Y
/1600100160Y
 γγH/function height slope Y
40Xor 
400function angle SlopeX
30505070φββiX
2
22
=
×=
==
=
==
−−=−−=
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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SURFACE MINE DESIGN AND PRACTICE
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:
48.07Y
7.8
375
7800
1502.5
c
γHY
=
=
×
==
From the slope design chart, at Y=48 and S.F = 1.0, X=18
:. We know that ( ) ( )φββ −−= iX 2
( ) ( ) 75013i2355050i218 −=−−=
Now, squaring both sides
( ) ( ) ( )
( )
0
0
22
4.55
4.55
60
3324
3324300032460
300060324
750154324
75015218
=
==
=+=
−=
−×=
−×=
i
i
i
i
i
i
Engr. Izhar Mithal Jiskani 33
SURFACE MINE DESIGN AND PRACTICE
Slope design chart for plane failure including 
various safety factors (Hoek, 1970a)
Engr. Izhar Mithal Jiskani 34
SURFACE MINE DESIGN AND PRACTICE
Example 03:
Determine the height of slope using the following data.
i. Slope angle 
065== i
ii. Inclination of failure plane 
0β50=
iii. Angle of internal friction 
053== φ
iv. Cohesion
27800c 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 γ=
49.92mH
49.92m
2.5
780016
γ
cYH
=
=
×
=
×
=∴
Engr. Izhar Mithal Jiskani 35
SURFACE MINE DESIGN AND PRACTICE
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
×==+∴
=
Engr. Izhar Mithal Jiskani 36
SURFACE MINE DESIGN AND PRACTICE
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
RCRθ
2
=
××=
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.
Engr. Izhar Mithal Jiskani 37
SURFACE MINE DESIGN AND PRACTICE
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 ’
0
1
1
1
16.7θ
16.7m
5mTanθ
=
=∴ −
According to Pythagoras theorem,
( ) ( )
17.43mR
303.9R
303.0278.92516.75R
BaseperpHyp
222
222
=
=∴
=+=+=
+=
=1φ ?; since given slope of cutting plane “ad” is 1 vertical and 1.5 horizontal;
Thus,
0
1
1
1
33,7φ
1.5
1Tanφ
=



=
−
from figure 23; it is clear that:
Engr. Izhar Mithal Jiskani 38
SURFACE MINE DESIGN AND PRACTICE
( )
( )
0
0000
11
0
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
00
3
0
2
1
2
2
=
+−=
=
=
=
−
From obe Δ we get 2φ , as follows,
( ) ( )
( )
( ) ( )
( ) ( )
0
00
000
0
000
11
0
0
3
00
3
000
21
0
3
0
2
1
2
56.3Ψ
123.7180Ψ
39.684.1180φθ180Ψ
84.1φ
16.733.790θφ90φ as
123.71φ56.3180φ
22.633.7180φφ180φ
22.6φ
17.43
6.7sinφ
=
−=
+−=−−=
=
+−=+−=
==−=
+−=+−=
==


=
−
Area of sinΨ
sinφsinθR
2
1ΔΔaod 21
×
××==
( ) ( ) ( )( )
2
1
0
00
2
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 =−=∴
Engr. Izhar Mithal Jiskani 39
SURFACE MINE DESIGN AND PRACTICE
( )
( )
( )
2
2
1
322
2
0
0
4.77mΔ
sinφ
sinφsinφ
4.07
2
1Δ
4.07mDB13.3617.43DB
13.36m
56.3sin
39.6sin17.43
sinΨ
RsinφOD
=
×
××=
==−=∴
=
×
==
Now area of polygon oabc=Ap
2120.62mAp =
(as) Area of sector 



××= 0
2
1802
1 piθRaoc
( ) ( )
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
6.541891.44
180
π80.117.4340
S.F
Wd
θCRS.F
1891.44KNW
18.5102.24W
0
02
2
=
×
×××
=
=
=
×=
Engr. Izhar Mithal Jiskani 40
SURFACE MINE DESIGN AND PRACTICE
STRIPPING RATIO
Figure 24 (a)
Vm = Vol. of mining
Vc = Vol. of cone
VT = Vol. of truncated portion
:. Vm = Vc- VT
Fig. 24 (b)
Engr. Izhar Mithal Jiskani 41
γtanθh Hc
ΔhhHc
+=
+=
SURFACE MINE DESIGN AND PRACTICE
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
Δhπr
3
1V 2T =
…… (2) (:.) Height of small cone is ∆h). and γtanθΔh =
as height of big cone, is Hc;
and
or
:. Vol: of bigger cone, Vc
( )
( ) (3) γtanθhπR
3
1Vcor 
Δhhπr
3
1HcπR
3
1Vc
2
22
+=
+==
:. Mined Vol: ( ) TC VVVm −=
(4) Δh πr
3
1Hπr
3
1V 2c
2
m −=
Vol: of waste: ( ) 0mw VVV−=
Where, Vol: of Ore ( ) hπrV 20
:. Vol: of waste ( ) (5) VVV 0mw −=
:. Stripping Ratio 
( ) ( )( )3
3
m ore. of :Vol
m wasteof :VolSR =
(6) 
V
V
SR
0
w
=
or 0
w
W
W
SR =
Engr. Izhar Mithal Jiskani 42
SURFACE MINE DESIGN AND PRACTICE
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:
3
wast
3
ore
2.6tons/mγ
mtons3.1γ
250mh
50mγ
=
=
=
=
Solution
3
T
2
T
3
C
c
2
C
186924.76mV
Δhπγ
3
1V
4m17044879.2V
HπR
3
1V
=
=
=
=
225.04mR
Tan55
321.4R
Tan55
HR
RTanθH
321.4mH
71.4250ΔhhH
71.4mΔh
50tan55Δh
γtanθΔh
0
0
C
C
C
C
=
=
=∴
=
=
+=+=
=
=
=

( )
3
W
OmW
3
O
22
ore
3
m
TCm
8m14894459.0V
1963495.4816857954.4VVV
1963495.4mV25050πhπγV
8m16857954.4V
186924.76417044879.2VVV
=
−=−=
==×==
=
−=−=
6.36S.R
6086835.74
138725593.6
S.R
3.11963495.4
2.6814894459.0
γV
γV
W
W
S.R
oo
ww
ore
waste
=
=
×
×
=
×
×
==
Engr. Izhar Mithal Jiskani 43
SURFACE MINE DESIGN AND PRACTICE
DETERMINATION OF “SR” BY AREA METHOD
Ao =Area of sections (i), (ii) and (iii)
Area of Sec: (i) =
21250ft5050
2
1
=××
Area of Sec: (ii) =
20ft50050100 =×
Area of Sec: (iii) =
20ft1500501100 =×
( ) ( )
2
O
O
27500ftA
150005000212502A
=
++=
0.36S.R
0.36
275000
10000
Ao
AwS.R
10000ftAw
50005000AwAwAw
AwAw since
5000ft100100
2
1Aw
2
21
21
2
1
=
===∴
=
+=+=∴
=
=××=
Now consider the bench of 25ft thickness.
Engr. Izhar Mithal Jiskani 44
SURFACE MINE DESIGN AND PRACTICE
( )
( )
( )
( ) ( )
2
2
2
22
1
21
2
21
2
2
22
2
2
1
500625ftAw
625500000
312.522500002Aw
312.5ft2525
2
1Aw
150000ft2525.10000100100Aw
2Aw2AwAw
6035.5ftAo
1767.75225002AoAoAo
1767.75ftAo
2570.71255050Ao
2500ft25100Ao
=
+=
×+×=∴
=××=
=××+=
+=
=
+=+=∴
=
×=×+=
=×=
A = ½(b1 + b1) × h = b1h 
Engr. Izhar Mithal Jiskani 45
SURFACE MINE DESIGN AND PRACTICE
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 
Engr. Izhar Mithal Jiskani 46
SURFACE MINE DESIGN AND PRACTICE
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
==
××==
Engr. Izhar Mithal Jiskani 47
SURFACE MINE DESIGN AND PRACTICE
Vol: of waste (Strip 1) 
( )
3
1
1
8.74mVw
1.251.25
2
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:-
3
2
2
6.25mVo 
11.255Vo ore, of Vol.
=
××=
( ) ( )
3
2
2
2
10.3mVw 
1.252
111.256.614Vw waste,of Vol.
=
×+××=
dollars 1.58USN.V
10.311.875costValueN.V
1.65
6.25
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 
Engr. Izhar Mithal Jiskani 48
SURFACE MINE DESIGN AND PRACTICE
( ) ( )( )
( ) ( ) 1.9ins S.R 6.2511.86ins S.R
11.86mVw
1.252
111.258.864Vw
6.25mVo
3
3
2
3
3
3
===
=
+××=
=
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 
2.14
6.25
13.424
Vo
Vw
InsS.R
13.424mVw
0.7812512.6425Vw
1.252
111.2510.114Vw
6.25Vo
4
4
3
4
4
2
4
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;
Engr. Izhar Mithal Jiskani 49
SURFACE MINE DESIGN AND PRACTICE
( ) ( )
( )
( ) ( ) ( )( )
( )
dollars US68.8N.V
49.00-117.8 (overall) value-N
dollars US49.00149 wasteofcost mining overall
dollars US117.81.962ore of valueoverall
0.80.7962
49
Vo
Vwratio strippig overall
62more of : voloverall
1m12.549.5
1552
1159.9 ore of : voloverall &
49m wasteof : voloverall 
1 thickness49m
9.99.9
2
1Δw
3
3
2
=
=
=×=
=×=
====
=
×+=
×××+××=
=∴
×=
×=
 
Engr. Izhar Mithal Jiskani 50
SURFACE MINE DESIGN AND PRACTICE
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 :-
11.00lbs
100
0.55 * 2000lb/st cc ==
b) Copper recovered by mill (RM/st of ore)
8.8lbs
100
8011lbsRM =×=
c) Concentration ratio (r):- the ratio of concentration is defined as
( )
45.45r
45.45
8.8
400
8.8
100202000
r
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.
Engr. Izhar Mithal Jiskani 51
SURFACE MINE DESIGN AND PRACTICE
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 :
( )dii
di
gi
Σ
Σ
n
1i
n
1i
=
=

=ϑ
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
SURFACE MINE DESIGN AND PRACTICE
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
6
22
6
5
22
5
4
22
4
3
22
3
2
22
2
1
22
1
===+=
===+=
===+=
===+=
===+=
===+=
As we know that;
( ) ( ) ( ) ( ) ( ) ( )
1
6
6
2
1
1
1
6
6
2
2
1
1
104.484g
0.0223
0.0100g
316.23
1
223.6
1
282.84
1
223.6
1
316.23
1
282.84
1
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
d
g
d
g
d
g
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
1
di
gi
g
Σ
Σ
=
=
=
Using the data of previous example; calculate the grade of ore by squared weighting 
technique.
Engr. Izhar Mithal Jiskani 53
SURFACE MINE DESIGN AND PRACTICE
 
22
66
22
55
22
44
22
33
22
22
22
11
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%g
108.5003
103.5885g
d
1
d
1
d
1
d
1
d
g
d
g
d
g
d
g
g 
5
5
2
6
2
5
2
2
2
1
2
6
6
2
5
5
2
2
2
2
1
1
=
×
×
=



+


+−−−−+


+





+


+−−−−+



+



=∴
−
−
Engr. Izhar Mithal Jiskani 54
SURFACE MINE DESIGN AND PRACTICE
ORE RESERVE ESTIMATION
by Triangular Method
From above figure:
Area of rectangle abcd= ( ) ( )3213 yyxx −−
Area of 
( ) ( )12121 yyxx2
1ΔA −−=
Area of 
( ) ( )32232 yyxx2
1ΔA −−=
Area of 
( ) ( )31133 yyxx2
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 
Engr. Izhar Mithal Jiskani 55
SURFACE MINE DESIGN AND PRACTICE
mym
mym
mym
800,1100x
1200,1500x
1200,1100x
33
22
11
==
==
==
From given table.
And we know that;
( )( ) ( ) ( ) ( )( ) ( ) ( ) 


−+−−
−+−−
−−−=
311332
233213
3213 yyxxyy
xxyyxx
2
1yyxxA
( ) ( ) ( ) ( ) ( )( )( ) ( )
[ ]
280,000mA
160000
2
1
0160000
2
10
8001200110011008001200
150011001200120011001100
2
1800120011001100A
=
×=
+−−=



−+−−
−+−−
−−−=
OR
280,000mΔA
1600002
1
400400
2
1
hb
2
1ΔA
=
×=
××=
××=
Engr. Izhar Mithal Jiskani 56
SURFACE MINE DESIGN AND PRACTICE
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
3005002
13004002
12007002
1700500
ΔAΔAΔA - abcd of AreaΔA
=∴
−=
××+××+××−×=
++=
Average thickness of ore = 3
321 CCC ++
4mt
4m
3
12
2
453t
:av
:av
=
==
++
=
Vol: of ore in Triangular area= :avtΔA ×
Vol. of ore = 145000×4m
Vol. of ore = 580000m3.
;. Ore reserves = 580000×2.5
Ore reserves = 1450000
Engr. Izhar Mithal Jiskani 57
SURFACE MINE DESIGN AND PRACTICE
Calculation Of Thickness Of Ore In A Drill Hole
Length 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
Σ
n
1i
n
1i
=
=
( ) ( ) ( ) ( ) ( )
( )
0.503m t
6.1
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:
Engr. Izhar Mithal Jiskani 58
SURFACE MINE DESIGN AND PRACTICE
( )
( )
( )
( ) ( )
0
1
000
31
0
2
000
213
000
11
0
2
11
2
0
1
11
1
0
11
332211
w
55.49θ
82.8741.64180φφ180θ
82.8756.3126.56φφθ
41.6426.5668.2φφθ
56.31φ
1.5tan
400600
500800tanφ
26.56φ
0.5tan
400600
300500tanφ
68.2
2.5tan
200400
300800tanφ
:fig above from
(1) 3
60
θtθtθt
t
=
+−=+=
=+=−=
=−=−=
=
=


−
−
==
=
=


−
−
=
=
=


−
−
=
−−−−


 +++
=
−−
−−
−−
ϕ
Substitute the values of t1, t2, t3, θ 1, θ 2, θ 3 in equation 1
3
152.3 tw
3
60
9138.5
 tw
3
60
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
Engr. Izhar Mithal Jiskani 59
SURFACE MINE DESIGN AND PRACTICE
Example:
In a level terrain, determine the max height of high wall that dragline can strip without 
re-handling, using thefollowing data:
Dumping radius =(Rd)= 47m,
Outside diameter of tub = (Et) = 11m,
Spoil angle of repose (θ ) =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
( )
18.2H
2.012
36.62H
2.1338.752.012H
2.132.0121H38.75
1.6193.7481.7251H0.287H
1.222.8251.3H1.3270.287H
1.22tan37
4
15
100
301Hcot37Hcot7438.75
Ttanθ
4
W
100
Ps1HcotθHcotφRe
000
=
=
−=
+=
−++=
−++=




−


+


++=




−


+


++=
Engr. Izhar Mithal Jiskani 60
	INTRODUCTION TO OPEN PIT MINING
	BASIC BENCH GEOMETRY
	DEVELOPMENT OF ACCESS ROAD: (ORE ACCESS;)
	DEVELOPMENT OF ACCESS ROAD: (ramps)
	THE PIT EXPANSION PROCESS
	DETERMINATION OF THE MINIMUM OPERATING ROOM REQUIRED FOR PARALLEL CUTS
	CUT SEQUENCINGS:
	PIT SLOPE GEOMETRY
	INTRODUCING RAMP IN PIT SLOPE GEOMETRY
	INTER-RAMP SLOPE ANGLES AFTER RAMP INCLUDED
	INTRODUCING WORKING BENCHES IN PIT SLOPE GEOMETRY
	INTER-RAMP SLOPE ANGLES AFTER WORKING BENCH INCLUDED
	PLANNAR FAILURE
	Slope design chart for plane failure including various safety factors (Hoek, 1970a)
	CIRCULAR FAILURE
	STRIPPING RATIO
	DETERMINATION OF “SR” BY AREA METHOD
	ORE GRADE ESTIMATION
	ORE RESERVE ESTIMATION
	Calculation Of Thickness Of Ore In A Drill Hole
	Calculation Of Reserves Using Weight Age Average Thickness: