The document provides details about a materials transportation engineering course. It includes: - An overview of the course which covers various material handling methods and equipment used in mining industries. - Course outcomes including being able to describe different material transportation methods, explain principles, and design flow sheets. - A syllabus covering topics like belt conveyors, chain conveyors, fluid transport systems, rope haulage, and hoists over 14 weeks. - Assessment methods including tests, assignments, and a final exam worth 60% of the grade.

1. Lecture 1

2. MATERIAL
TRANSPORTION
ENGINEERING
ASSOC. PROF. DR. HASHIM HUSSINASSOC. PROF. DR. HASHIM HUSSIN

3. Course Synopsis :
•The course covers the material handling /
transportation methods and equipments
that are widely used in mining and mineral
industries.
• Students are introduced to theories
(where related), principles, mechanism
and the performance of the equipments.
•The topics covered are belt conveyors,
chain conveyors and bucket elevators,
screw conveyors and elevators, shaking
and vibratory conveyors, fluid transport,
rope haul systems, monorails and aerial
ropeways, locomotive haulage and hoist
and mine winders.
•Students are also introduced to the basic
of calculation of tonnage, speed, motor
power, and the efficiency of the
equipments.
•Designing of flow sheet using real plant
examples with the awareness of its impact
on environment at cost effective.

4. Course Outcomes (CO) :
At the end of the course, the students should be able to:
 distinguish and describe the materials handling
methods that are widely used in mining and mineral
industries.
 explain the basic concept and principle in materials
transportation
 describe and apply the technology, types and
characteristics of materials transportation equipments
available in the market.
 design a proper flow sheet to transport materials and
ores from one place to another place or from one unit
to another unit.
 evaluate and calculate the equipment performance.
 classify and apply knowledge of materials
transportation to real situations with an awareness of
its impact on environment at cost effective.
 Asses a range of learning resources and to take
responsibility for own learning with appropriate
support.
 explain and work effectively with others as a member
of a group and meet obligations to others.

5. Teaching Plans / Syllabus :
No Topic ContentsTeaching Weeks
1 Introduction to materials handling
materials handling methods, basic concept applied
mechanic
1
2 Belt conveyor
introduction to the equipment, mechanism of the
equipment, calculation of the power, tonnage, speed,
wide, number ply. The efficiency of the belt conveyors.
etc.
2,3
3 Chain conveyor and bucket elevator
introduction to the equipment
mechanism of the equipment
calculation of the power, tonnage, speed, the
efficiency of the chain conveyors. etc.
3,4

6. Teaching Plans / Syllabus :
No Topic ContentsTeaching Weeks
4 Screw conveyors and elevators
 introduction to the equipment
 mechanism of the equipment
 calculation of the power, tonnage, speed, the efficiency of the
chain conveyors. etc.
Shaking conveyor
 introduction to the equipment
 mechanism of the equipment
 calculation of the power, tonnage, speed, the efficiency of the
shaking conveyors. etc.
Vibratory conveyor
 introduction to the equipment
 mechanism of the equipment
 calculation of the power, tonnage, speed, the efficiency of the
vibratory conveyor. etc.
4 & 5

7. Teaching Plans / Syllabus :
No Topic ContentsTeaching Weeks
5 Fluid transport
 introduction to the equipment
 mechanism of the equipment
 calculation of the tonnage, speed, pulp density, slope, efficiency etc.
5
6 Rope haulage system
o introduction to the equipment
o mechanism of the equipment
o calculation of the tonnage, speed, efficiency etc.
Monorails
 introduction to the equipment
 mechanism of the equipment
 calculation of the tonnage, speed, efficiency etc.
Aerial ropeways
 introduction to the equipment
 mechanism of the equipment
 calculation of the tonnage, speed, efficiency etc.
7 & 8

8. Teaching Plans / Syllabus :
No Topic ContentsTeaching Weeks
7 Locomotive haulage
 introduction to the equipment
 mechanism of the equipment
 calculation of the ideal gradient, optimum gradient and work example.
8 & 9
8 Hoist and mine winders
• introduction to the equipment
• mechanism of the equipment
• the mechanics of hoisting
• calculation of the speed, and acceleration, drum torques and duty cycle
diagrams
10,11&
12
9 Designing of flow sheet, industrial examples, both local and
international, with an awareness of health and safety, social, political,
economic and impact on environment.
13 &14
**Week 6 – Mid-semester Break (1 week)

9. Assessment methods %
Test (2 tests)
Quizzes
20
Assignment
2 assignments (individual and group)
20
Final Exam 60
Contribution of assessment
• Exam 60 %
• Course work 40 %

10. References
• Brook, N. 1971. Mechanics of Bulk Materials
Handling. London Butterworths.
• Handbook Society of Mining Engineers. 1979.
New York.
• Hartman H.L., 1987. Introductory Mining
Engineering. New York. John Wiley & Son.
• Ramlu M. A. 1996. Mine Hoisting. A.A
Balkema/Rotterdam/Brookfield.
• Recent journals /publications related to this
subject

11. Objective
• To introduce most of materials transportation
equipments (mechanisms, principles and
calculation) that normally used in mining,
mineral processing plant, and quarrying.
• Give some idea to mineral resources
engineering students to make a preliminary
calculation to narrow the field before selecting
the equipment.

12. Introduction
• Material transportation/handling methods are
normally classified into 3 main groups ( from the
standpoint of mechanics involved);
 Continuous methods
 Semi-continuous, or small batches methods and
 Batch methods

13. • Some example of material transportation/handling
equipments and techniques:
– Belt conveyor (penghantar tali sawat)
– Chain conveyor (penghantar rantai)
– Screw conveyor and feeder (penghantar skru dan penyuap)
– Suction pipe (paip sedutan)
– pressure pipe (paip tekanan)
– suction pneumatic
– pressure pneumatic
– Monorails
– bucket elevator
– aerial ropeway
– Chute
– shaker conveyor
– Locomotive
– Hoists and lift

14. Belt Conveyor (penghantar tali sawat)

16. Chain conveyor (penghantar rantai)

17. Bucket elevator (pengangkat timba)

19. Shaker Conveyor (penghantar bergetar)

20. Suction Pipe (paip sedutan)

21. Screw Conveyor (penghantar skru)

22. Locomotive Haulage

23. Hoist

24. Mine Winder (pengangkat lombong)

25. Chute

26. Lecture 2

27. Belt Conveyor (Talisawat Penghantar)
Belt conveyor is basically an endless strap
stretched between two drums.
Suitable for very short distances and low
outputs.
It is necessary to support the top strand of the
conveyor at regular intervals to prevent undue
sagging and to reduce the spillage of material
which may occur if the belt does not run truly,
while maintaining a high carrying capacity.
 these requirements usually met by trough
idlers.

28.  Idler (Pemelahu)
Consist of three separate rollers to support the
belt and also bend it into a trough shape.
The two outer rollers are tilted upwards at an
angle of 25o
to 30o
.
For very wide belts a design using five
separate rollers is sometimes used.
For narrow belts two angled idlers only may be
used.
• impact idlers = Pemelahu hentaman
• trough idlers = Pemelahu paluh
• return idlers = Pemelahu kembali

29. Idlers for small conveyor belt
Top strand
(Lembar
atas )
Bottom strand
(Lembar bawah)

30. Material Belt Conveyor
(Talisawat Penghantar)
Normal Trough Idler (three idlers)
Pemelahu paluh yang biasa (tiga pemelahu)

31. Belt Conveyor
Trough idler
Idler for wide belt conveyor (five idlers) Pemelahu untuk
talisawat yang lebar (lima pemelahu)

32. Return Idlers
(Pemelahu Kembali )
Trough Idler
(Pemelahu Paluh)
Feed Chute
(Pelongsor suapan)
Impact Idler
(Pemelahu
Hentaman )
Drive Drum
(Gelendung Pemacu)
Idler Drum (Gelendung
Pemelahu )
Bottom Strand
(Lembar bawah)
Top Strand
(Lembar atas )

33. Exercise 1

34. Lecture 3

35. • The driving drum relies on the friction between drum and belt
to provide the drive to the belt.
• If the two tensions in the belt at the driving drum are P1 and P2
with P1 the bigger tension in the top strand, the limiting ratio of
tensions when about slip is about to occur is given by:
θ
P1
P2
P1 / P2 = eµθ
Log10 P1 / P2 = 0.434µθ
µ is coefficient of friction or coefficient of grip between the belt & drum
θ Is the angle of wrap

36. • For a long conveyor with a large hauling duty P1 requires
to be large.
• Value of P1 possible isobtained by three method:
– Increasing µ by lagging the driving drum with a
suitable rubber-like material.
– Increasing the value of θ
– Increasing the value of P2
No slip P1/ P2 ≤
eµθ

37. The value of θ can be increased by:
1. Using a snub
pulley
θ
θ = 250o
2. Using more than one
driving drum.
θ = θ1 + θ2 ;
θ=400~450o
θ2
θ1
θ1
θ2
θ3
θ = θ1 + θ2
+ θ3;
θ= ~600o

38. The value of P2 can be increased by
3. pre-
tensioning belt
•By a screw tightening device at the tail end of
the belt conveyor

39. 2. Take-up loop
Tension
Rail mounted
carriage
Drivi
ng
drums
• as part of a belt storage or take-up loop

40. Driving
drums
Weighted roller
3. Gravity tensioning
device
•For higher powered conveyors a gravity operated tension
devise may be used, either by pulled via steel cables on
the loop take-up carriage or by a heavy roller mounted in
a frame

41. Pemelahu Kembali
(Return Idlers)
Pemelahu Paluh
(trough idler)
Gelendung Pemacu
(Drive drum)
Gelendung Pemelahu
(Idler drum)
Lembar bawah
(Bottom strand)
Lembar atas
(top strand)
Pemelahu
Hentaman
(impact
idler)
Pelongsor suapan
(feed chute)
Berat
Pengambang
(Counter weight)
Kapi ambil
(take-up
Pulley)

42. Factors affecting the use of belt conveyors
1. A straight line plan is usually required (some small deviations of
a few degrees are possible). If the line of the conveyor system
must be angled it is often necessary to use separate conveyors
but some systems have a complex belt lacing which permits a fix
angle at some point in the conveyor, the material being
discharged from one section of the belt to the other as if the two
sections were separate conveyor.
2. The angle of inclination of the conveyor is limited by the friction of
the material on the belt, to about 25o
. The maximum gradient used
for a conveyor must allow for restarting with the belt loaded on the
incline and sufficient frictional grip must be provided to overcome
the component of the weight tending to pull the material down the
belt and also to accelerate the material.

43. 3. The maximum lump size is limited to about half
the belt width.
4. The carrying capacity of the belt depends on how
the material can be piled up on the belt width. As
the belt is continuously passing over sets of
supporting idlers, the materials is slightly
disturbed all the time and tends to spread out on
the belt. The carrying capacity of the conveyor is
given by the equation;
T= a (m2
) b (t/m3
) v (m/s) = abv (t/s)
where T is the carrying capacity, a is the average
cross-sectional area of material, b is the bulk
density and v is the speed of the conveyor belt.

44. For a belt of width W the value of the area a varies approximately
between w2
/10 (high loading) and w2
/12 depending on the nature of the
material. A blocky type material such as coal, broken rock, or ore can be
piled onto the belt as shown in Fig.(a) whereas a smooth material such
as particle tends to run out over the belt as shown in Fig.(b).
The value of b the bulk density in t/m3
is numerically equal to the relative
density (g/cm3
) but it relates to the density of broken material including
air spaces, and not to the solid relative density
(a
)
(b)

45. Material Density (t/m3
)
Coal
Solid coal
Gravel
Dry ashes
Wet ashes
Broken sandstone
Solid sandstone
Broken limestone
Slag
Dry sand
0.8
1.35
1.4 – 1.7
0.55 – 0.65
0.7 – 0.8
1.35
2.4
1.45
1.35
1.6
5. The belt strength affects the maximum force which can be
taken by the belt, and the value of the maximum force
depends on the power required and the drive head frictional
grip. The power required by a belt conveyor can be divided
into three components.
i. power for the empty belt, We
ii. Power to convey the material, Wm
iii. Power to raise the material, Wr
The total power required by a belt is then W = We + Wm ±

46. • The value of Wr is written as plus or minus as if the material is being
lowered and help to run the belt thus the power requirement for this is
negative.
• The power as calculated above is the power required at the driving drum
of the conveyor, and so the motor power required will be greater
because of power losses in the gearing at the drive head.
• Assuming an efficiency of 90% for this gearing the otor power is then
given by
W = WT / 0.9
• The power required to drive the empty belt depends on the total force
required to move the empty belt, and on the belt speed.
• The force required : Ne = total weight on idlers x friction coefficient
Ne = Mi g µe
= mi ( l + lx ) g µe
And then the power required We = Ne v
We = mi ( l + lx ) g µev
Note: l the length of the conveyor is increased by lx = 45 m to allow for end
pulley friction

47. • The power required to convey the material; Wm =mml g µmv
• The value of mm the mass of material per unit length is obtained from
mm = T tan/s
v m/s
Therefore,
Wm = T l g µm v
v
Or Wm = T l g µm
the value of µm the friction coefficient is again 0.03 for well maintained
conveyors but sometimes raised to 0.04 if the conditions are unfavourble.

48. • The power required to raised the material at the rate T
through height h is obtained directly as
Wr = T g h kW
Effective belt tension Pe = WT kN
v
The maximum tension P1 is obtained from the formula
Pe = P1 – P2
For no slip to occur P1 < eµθ
= n
P2
P2 = P1/n
P1–P2 = P1 – P1/n = P1 (n-1)/n
P1= (n/n-1) Pe

49. The calculated value of P1 is then used to find the belt stree f (kN/m per ply).
Fabric Density (kg/m2
) Stress (kN/m ply
U.S. cotton
Rayon and cotton
Rayon, cotton, nylon
Nylon and cotton
Steel reinforced
(Steel cords in rubber and
fabric belt)
0.814
0.930
1.043
1.220
1.395
1.744
1.19
1.63
0.930
1.02
3.0
16.4 – 42.3
4.25
5.25
5.75
7.0
8.75
10.5
12.25
15.8
7.0
9.65
35.0
80-450

50. • Exersice 2
A conveyor is 600 m long and conveys coal of bulk density
0.8 t/m3
up a gradient of 1 in 60 at the rate of 220t/h.
Determine suitable speed and strength for the installation.
(Note: assume the area of the material is w2
/11, width of the
belt conveyor for transporting coal is 0.75 m, total mass
acting on the idlers 60 w = 45kg/m, µm =0.04, Ø= 440o
)