Conveyors
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Department of Mining, Dressing and Transport Machines AGH
Belt Conveyors for Bulk Materials
Calculations by CEMA 5th Edition
Piotr Kulinowski, Ph. D. Eng.
Piotr Kasza, Ph. D. Eng.
(
piotr.kulinowski@agh.edu.pl
12617 30 92
B-2 ground-floor room 6
consultations: Mondays 11.00 - 12.00
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Nomenclature of components of a typical belt conveyor
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TYPICAL BELT CONVEYOR TRAVEL PATHS
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Design Considerations
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Characteristics and Conveyability of Bulk Materials
Capacities, Belt Widths, and Speeds
Belt Conveyor Idlers
Belt Tension, Power, and Drive Engineering
Belt Selection
Pulleys and Shafts
Vertical Curves
Steep Angle Conveying
Belt Takeups, Cleaners, and Accessories
Conveyor Loading and Discharge
Conveyor Motor Drives and Controls
Operation, Maintenance, and Safety
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Characteristics and Conveyability of Bulk Materials
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 The angle of repose of a material is the acute angle
which the surface of a normal, freely formed pile makes
to the horizontal.
 The angle of surcharge of a material is the angle to the
horizontal which the surface of the material assumes
while the material is at rest on a moving conveyor
belt.This angle usually is 5 degrees to 15 degrees less
than the angle of repose, though in some materials it may
be as much as 20 degrees less.
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Flowability–angle of surcharge–angle of repose
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Characteristics of BuIk Solid Materials - CEMA
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Material Characteristics
Code
Size
Very fine—100 mesh and under
Fine—1/8 inch and under
Granular—Under 1/2 inch
Lumpy—containing lumps over 1/2‘ inch
Irregular—stringy, interlocking, mats together
A
B
C
D
E
Flowability
Angle of Repose
Very free flowing—angle of repose less than 19°
Free-flowing—angle of repose 20° to 29°
Average flowing—angle of repose 30° to 39°
Sluggish—angle of repose 40° and over
1
2
3
4
Abrasiveness
Nonabrasive
Abrasive
Very abrasive
Very sharp—cuts or gouges belt covers
5
6
7
8
Miscellaneous Characteristics
(Sometimes more than one of
these characteristics may
apply)
Very dusty
Aerates and develops fluid characteristics
Contains explosive dust
Contaminable, affecting use or saleability
Degradable, affecting use or saleability
Gives off harmful fumes or dust
Highly corrosive
Mildly corrosive
Hygroscopic
Interlocks or mats
Oils or chemical present—may affect rubber products
Packs under pressure
Very light and fluffy—may be wind-swept
Elevated temperature
L
M
N
P
0
R
S
T
U
V
W
X
V
Z
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Department of Mining, Dressing and Transport Machines AGH
Example: A very fine material that is freeflowing, abrasive, and contains explosive dust
would be designated:
Class A26N.
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Material characteristics and weight per cubic foot - CEMA
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Average
weight
(lbs/cu ft)
Angle of
repose
(degrees)
Recommended
maximum inclination
(degrees)
Code
Ashes, fly
40-45
42
20-25
A37
Cement, Portland
72-99
30-44
20-23
A36M
60
35
18
B35TY
Coal, lignite
40-45
38
22
D36T
Copper ore
120-150
30-44
20
*D37
Dolomite, lumpy
80-100
30-44
22
D36
Gravel, dry, sharp
90-100
30-44
15-17
D37
Lignite, air-dried
45-55
30-44
100-110
30-44
22
D36
Salt, common dry, fine
70-80
25
11
D26TUW
Sandstone, broken
85-90
30-44
Wood chips
10-30
45
27
E45WY
60
35
18
B35TY
Coal, anthracite, sized
55-60
27
16
C26
Coal, bituminous, mined 50 mesh and under
50-54
45
24
B45T
Coal, bituminous, mined and sized
45-55
35
16
D35T
Coal, bituminous, mined, run of mine
45-55
38
18
D35T
*Coal, bituminous, mined, slack,1/2 inch and under
43-50
40
22
C35T
Coal, bituminous, stripping, not cleaned
50-60
Coal, lignite
40-45
Material
Coal, anthracite, river, or culm, ⅛ inch and under
Rock, soft, excavated with shovel
Coal, anthracite, river, or culm,1/8 inch and under
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*D35
D37
D36T
38
Department of Mining, Dressing and Transport Machines AGH
22
D36T
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Belt Widths
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
The belt widths are as
follows: 18, 24, 30, 36,
42, 48, 54, 60, 72, 84,
and 96 inches.

The width of the
narrower belts may be
governed by the size of
lumps to be handled.
Belts must be wide
enough so that any
combination of prevailing
lumps and finer material
does not load the lumps
too close to the edge of
the conveyor belt.
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Recommended
maximum belt
speeds
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1 m/s = 196,85 ft/min
1 ft = 0.3048 m
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Example: Belt Conveyor Parameters
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Coal, anthracite, sized
Capacity: 1000 tph
Length: 1000 m  3300 feet
Lift: 115 feet
 Belt speed: 600 fpm
 Belt width: 42 inches
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Area of load cross section
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Department of Mining, Dressing and Transport Machines AGH
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1 lb = 0.4536 kg
Belt Conveyor Capacity Table
1 ft3 = 0.0283 m3
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1. Determine the surcharge angle of the material.

2.
3.
4.
5.
6.
Coal, anthracite, sized
27
The surcharge angle, on the average, will be 5 degrees to 15 degrees less than the angle of repose.
(ex. 27° - 12° = 15°)
Determine the density of the material in pounds per cubic foot
(lb/ft3).
Coal, anthracite, sized
55-60
Choose the idler shape.
Select a suitable conveyor belt speed.
Convert the desired tonnage per hour (tph) to be conveyed to the
equivalent in cubic feet per hour (ft3/hr).
(ex. 1000 tph x 2000 / 60 = 33333 ft3/hr)
Convert the desired capacity in cubic feet per hour to the
equivalent capacity at a belt speed of 100 fpm.
(ex. Capacity (eqiuv) = 33333 x (100 / 600 fpm) = 5555 ft3/hr
7. Find the appropriate belt width
8. Selected belt speed may require revision
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Department of Mining, Dressing and Transport Machines AGH
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1 lb = 0.4536 kg
Belt Conveyor Capacity Table
1 ft3 = 0.028 m3
1 ft2 = 0.093 m2
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Department of Mining, Dressing and Transport Machines AGH
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Idler Spacing
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Factors to consider when selecting idler spacing are belt weight, material weight, idler load rating, belt
sag, idler life, belt rating, belt tension, and radius in vertical curves
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Department of Mining, Dressing and Transport Machines AGH
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Example: Belt Conveyor Parameters
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Belt speed: 600 fpm
Belt width: 42 inches
Trough Angle: 35°
5555 [ft3/hr] / 7524 [ft3/hr] = 74%
Idler spacing: 4.5 feet
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The Selection of Idlers
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Idler life is determined by a combination of many factors, such as seals, bearings, shell thickness,
belt speed, lump size/material density, maintenance, environment, temperature, and the proper
CEMA series of idler to handle the maximum calculated idler load.
CEMA
CEMA
CEMA
CEMA
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B load rating based on minimum L10 of 30,000 hours at 500 rpm
C load rating based on minimum L10 of 30,000 hours at 500 rpm
D load rating based on minimum L10 of 60,000 hours at 500 rpm
E load rating based on minimum L10 of 60,000 hours at 500 rpm
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Step No. 1 Troughing Idler Series Selection
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Calculated Idler Load (lbs) = CIL = ((WB + (WM x K1)) x SI) + IML
Where:
 WB = Belt weight (lbs/ft) use actual or estimate from Table 5-5
 WM = Material weight (lbs/ft) = (Q x 2000) / (60 x Vee)
 Q = Quantity of material conveyed (tons per hour)
 Vee = Design belt speed (fpm)
 SI = Spacing of idlers (ft)
 Kl = Lump adjustment factor (see Table 5-6)
 IML = Idler misalignment load (lbs) due to idler height deviation and
belt tension = (D x T) / (6 x SI)
Where:
 D = Misalignment (in.)
 T = Belt tension (lbs)
 SI = Idler spacing (ft)
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Department of Mining, Dressing and Transport Machines AGH
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Step No. 1 Troughing Idler Series Selection
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 WB = Belt weight (lbs/ft)
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Step No. 1 Troughing Idler Series Selection
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 Kl = Lump adjustment factor
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Idler Series Selection
1 lbf = 4.45 N
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Calculated Idler Load (lbs) = CIL = ((11+ (55 x 1.0)) x 4.5) + 0 = 297 lbs
Where:
 WB = 11 lbs/ft
 WM = (1000 x 2000) / (60 x 600 ) = 55 lbs/ft
 Q = 1000 tons per hour
 Vee = 600 fpm
 SI = 4.5 ft
 Kl = 1.0
 IML = 0
Calculated Return Idler Load (lbs) = CILR = (11 x 10) + 0 = 110 lbs
 SI = 10.0 ft
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Department of Mining, Dressing and Transport Machines AGH
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Load ratings for CEMA idlers, lbs
Notes:
1. Troughing idler load ratings (Tables 5.7–5.10) are for three equal length rolls.
2. Load ratings also apply for impact rolls.
3. Troughing idler load ratings are based on a load distribution of 70% on center roll and 15%
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on each end roll for all trough angles.
4. Unequal length rolls or picking idlers are not covered by this standard.
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K2 = Effect of load on predicted bearing L10 life
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L10 (CORRECTED) = L10 (RATING) x K2 x K3A x K3B (IF APPLICABLE)
= 297 / 363 = 0.82  K2 = 2.0
Bearing L10 = (30,000 x 2.0) = 60,000 hours
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K3A = Effect of belt speed on predicted bearing L10 life
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= 600 x 12 / (4 x 3.14) = 573 rpm  K3A = 0.9
K3B = Effect of roll diameter on predicted bearing L10 life
4 inches K3B = 1.0
Bearing L10 = (60,000 x 0.9 x 1.0 ) = 54,000 hours
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K4 = Environmental, maintenance, and other special conditions
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 K4A = Effect of maintenance on
potential idler life.
FAIR K4A = 0.5
 K4B = Effect of environmental
conditions on potential idler life.
Dusty, Wet K4B = 0.6
1 deg F = 0.5556 C
C = 5/9 (F-32)
F = 9/5 * C + 32
 K4C = Effect of operating
temperature on potential idler life.
9/5*26°C+32 = 79 °F K4C = 1.0
Bearing L10 = (54,000 x 0.5 x 0.6 x 1.0 ) = 16,200 hours
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CEMA C Idlers
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= 297 / 791 = 0.4  K2 = 10.0
Bearing L10 = (30,000 x 10.0 x 0.9 x 0.5 x 0.6) = 81,000 hours
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Average weight (lbs) of idler
rotating parts–steel rolls.
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Basic Power Requirements
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 The horsepower, hp, required at the drive of a belt conveyor, is
derived from the pounds of the effective tension, Te, required at the
drive pulley to propel or restrain the loaded conveyor at the design
velocity of the belt V, in fpm:
 Te is the final summarization of the belt tensions produced by forces
such as:
1.The gravitational load to lift or lower the material being
transported.
2.The frictional resistance of the conveyor components, drive, and
all accessories while operating at design capacity.
3.The frictional resistance of the material as it is being conveyed.
4.The force required to accelerate the material continuously as it is
fed onto the conveyor by a chute or a feeder.
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Effective tension,Te
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Te = LKt(Kx + KyWb + 0.015Wb) + Wm(LKy ± H) + T p + Tam + Tac
Where:
L = length of conveyor, ft
Kt = ambient temperature correction factor
Kx = factor used to calculate the frictional resistance of the idlers and the sliding resistance between
the belt and idler rolls, lbs per ft
Ky = carrying run factor used to calculate the combination of the resistance of the belt and the
resistance of the load to flexure as the belt and load move over the idlers. For return run use
constant 0.015.
Wb = weight of belt in pounds per foot of belt length.
Wm = weight of material, lbs per foot of belt length
H = vertical distance that material is lifted or lowered, ft
Tp = tension resulting from resistance of belt to flexure around pulleys and the resistance of pulleys
to rotation on their bearings, total for all pulleys, lbs
Tam = tension resulting from the force to accelerate the material continuously as it is fed onto the
belts, lbs
Tac = total of the tensions from conveyor accessories, lbs
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Kt — Ambient Temperature Correction Factor
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79 °F  Kt = 1.0
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Kx — Idler Friction Factor, lbs/ft
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 The frictional resistance of idler rolls to rotation and sliding resistance
between the belt and the idler rolls.
 Values of Kx can be calculated from the equation:
Wb = 11 lbs/ft
Wm = (1000 x 2000) / (60 x 600 ) = 55 lbs/ft
Si = 4.5 ft
CEMA B4
Kx = 0,00068(11+55)+2.3/4.5 = 0.0449 + 0.51 = 0.555 lbs/ft
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Ky — Factor for Calculating the Force of Belt and Load
Flexure over the Idlers
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 Resistance of the belt to flexure as it moves over idlers
 Resistance of the load to flexure as it rides the belt over the idlers
Table 6-2 gives values of Ky for carrying idlers as they vary with differences in the
weight/ ft of the conveyor belt, Wb ; load, Wm ; idler spacing, Si ; and the percent of
slope or angle that the conveyor makes with the horizontal.
Wb + Wm = 66 lbs/ft
Si = 4.5 ft
Lenght: 3300 feet
Height: 115 feet
Slope: 2 degrees
 Ky = 0.020
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Corrected factor Ky
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 Ky = 0.0214
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Ky vs. Belt Tension
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After estimating the average belt tension and selecting an idler spacing, refer to Table 6-4
to obtain values for A and B for use in the following equation:
Wb + Wm = 66 lbs/ft
Si = 4.5 ft
 Ky = 0.04036
 Ky = 0.01261
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Tp = total of the belt tensions required to rotate each of
the pulleys on the conveyor
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2 x 200 = 400 lbs
3 x 150 = 450 lbs
Tp= 950 lbs
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Pulleys
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Tam — force to accelerate the material continuously as it
is fed onto the belt
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Where:
Q = Capacity of loading point, tph
g = 32.2 ft/sec2
V = design belt speed, fpm
V0 = initial velocity of material as it is fed onto
belt, fpm
Tam = 2.8755 x 10–4 x Q x (V –V0)
= 2.8755 x 10–4 x 1000 x (600 – 0) = 172.53 lbs
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Tac — resistance generated by conveyor accessories
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Conveyor accessories such as trippers, stackers, plows, belt cleaning equipment, and skirtboards
usually add to the effective tension, Te
 Ttr — from trippers and stackers
 Tpl — from frictional resistance of plows
Ttr= 0 lbs
Tpl= 0 lbs
Tbc= 0 lbs
Tbc — from belt-cleaning devices
Tsb — from skirtboard friction
Tsb = Lb (Cshs2 + 6)
Lb = skirtboard length, ft one skirtboard,
=10 ft
hs = depth of the material touching
the skirtboard, in =10 in


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Tsb= 113.8 lbs
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Resistance Calculations
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Te equals the total of the following:
lbs
Tx , idler friction
= L x K x x Kt
= 3300 x 0.555 x 1.0 = 1831.5
+ Tyc , belt flexure, carrying idlers
= L x Ky x W b x K t
= 3300 x 0.0214 x 11 x 1.0 = 776.8
+ Tyr , belt flexure, return idlers
= L x 0.015 x Wb x Kt
= 3300 x 0.015 x 11 x 1.0 = 544.5
Subtotal (A)
LKt(Kx + KyWb + 0.015Wb)
3152.8 lbs
+ Tym , material flexure
= L x Ky x W m
= 3300 x 0.0214 x 55.0 = 3884.1
+ Tm , lift or lower
= H x Wm
= 115 x 55 = 6325
Subtotal (B)
Wm(LKy ± H)
Tp , pulley resistance
10209.1 lbs
950
Tam , accelerated material
172.5
Tac , accessories
(Ttr + Tpl + Tbc + Tsb)
Subtotal (C)
1236,3 lbs
Te = Σ Subtotals (A), (B), and (C)
CEMA Horsepower Formula
113.8
hp = Te x V / 30,000
14598,2 lbs
265,4 hp
If drive efficiency = .94, horsepower at motor shaft = 265.4/.94 = 282.3 hp.
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Wrap Factor, Cw
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Te = T1 - T2 = effective belt tension, lbs
T1 = tight-side tension at pulley, lbs
T2 = slack-side tension at pulley, lbs
e = base of naperian logarithms = 2.718
f = coefficient of friction between pulley surface and belt surface (0.25 rubber surfaced belt driving bare
steel or cast iron pulley; 0.35 rubber surfaced belt driving rubber lagged pulley surface). Values
apply to normal running calculations
θ = wrap of belt around the pulley, radians (one degree = 0.0174 radians)
Cw = wrap factor
Cw = 0.08
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Tension Relationships and Belt Sag Between Idlers
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
 T0 = 1856.2 lbs
where :
 W = weight, (Wb + Wm), lbs/ft of belt and material
 Si = idler spacing, ft
 T = tension in belt, lbs
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Maximum and Minimum Belt Tensions
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Slack Side Tension, T2
T2 = 1167.8 lbs
 T2 = TeCw
or
 T2 = T0 ± Tb ± Tyr
T2 = 1856.2 + 1256 – 544.5 = 2567.7 lbs
Use the larger value of T2
Tension, Tb . The weight of the carrying and/or return run belt for a sloped conveyor
is carried on the pulley at the top of the slope. This must be considered in calculating
the T2 tension, as indicated above.
Tb = 115 x 11 = 1265 lbs
 Tb = H x Wb
where:
Wb = weight of belt, lbs/ft
H = net change in elevation, ft
Return Belt Friction Tension, Tyr . The return belt friction is the belt tension
resulting from the empty belt moving over the return run idlers:
Tyr = 544.5 lbs
 Tyr = 0.015 x L x Wb x Kt
where:
L = length, ft, of conveyor to center of terminal pulleys
Kt = temperature correction factor
Tt = T0
or
Tt = T2 – Tb + Tyr
Tmin = 1856.2 lbs
Tt = Tmin
T1 = T2 + Te = Tmax
Tmax = 17165.9 lbs
© piotr.kasza@agh.edu.pl, piotr.kulinowski@agh.edu.pl
Department of Mining, Dressing and Transport Machines AGH
tel/fax +48126335162
Belt Selection
Item
PIW
EP
ST
Units
LB/In-Width
N/mm or Kn/m
N/mm or Kn/m
Description
Max. Oper. Strength
Breaking Strength
Breaking Strength
System
USA
Metric
Metric
www.kmg.agh.edu.pl
Belt stress = T1 / Belt width = 17166/42 = 409 lbs per inch of width(PIW)
Convert from PIW to EP:
(409 PIW x 10 SF)
5.71
= 716 N/mm
© piotr.kasza@agh.edu.pl, piotr.kulinowski@agh.edu.pl
Department of Mining, Dressing and Transport Machines AGH
tel/fax +48126335162
Bibliography
www.kmg.agh.edu.pl





Continuous handling equipment – Nomenclature, ISO 2148-1974
Belt Conveyors for Bulk Materials, Calculations by CEMA 5th Edition
Kulinowski, Kasza: Wykłady „Conveyors”, www.kmg.agh.edu.pl
www.conveyorbeltguide.com
IDLER CATALOG, Superior Industries, LLC
g - Acceleration due to gravity,
32.2 ft/sec2
© piotr.kasza@agh.edu.pl, piotr.kulinowski@agh.edu.pl
Department of Mining, Dressing and Transport Machines AGH
tel/fax +48126335162
Questions?
www.kmg.agh.edu.pl
© piotr.kasza@agh.edu.pl, piotr.kulinowski@agh.edu.pl
Department of Mining, Dressing and Transport Machines AGH
tel/fax +48126335162