Calculation of Artificial Friction Conveying Coefficient f, And a Comparison Between ISO and CEMA

7/24/2019 Calculation of Artificial Friction Conveying Coefficient f, And a Comparison Between ISO and CEMA

1/9

Calculation of Artificial Friction Conveying Coefficient

f

and a Comparison between

I S 0

and

CEMA

lshwar6 Mulanl

In conveyor design the Frs t step is capacity and the second step is belt wi dth an d speed. The third step is

decision on 'ff, where distinction arises between DIN/ISO and CEMA or calculating resistances. Then

again both the methods are virtuall y similar. This topic helps in understanding D m design style and its

comparison to CEMA.

his topic isfro m the book Engineering Science and Application Design for Belt Conveyorwherein the

issue has been analysed in detaiL It presents a workable calculation model to calculate the value off in

design office based on conveyo r param eters.

CEMA uses IOr Ky and 0.01 5 instead o ff . Thus f is a sum effect of Kx Ky and 0.015 Hence knowledge

on f is also applicable t o

Kx

and Ky. This helps to understand their relationship.

PREFACE

This topic explains the design procedure for calculation

of artificial friction coefficient for conveying f abbrevi-

ated as conveying friction coefficient f for ease of

writing. The value of f is very important for calculating

main resistance FH, while designing belt conveyors as per

European standards. The belt conveyors all over the

world are designed by two methods:

1)

DIN/ISO style,

and 2) CEMA style. The choice is often mentioned by

the buyer. Therefore, it is essential that the engineers

concerned with conveyor design are equally competent

in both of the methods to promote business needs.

The CEMA publication provides specific values for

factors corresponding to f, so design as per CEMA has

easy input. However, DIN/ISO publications do not

mention the direct value of f, which

is

left to the

designer. Hence, the designer often finds himself partly

guessing on this crucial value, without which he cannot

proceed in design. Also, this issue becomes somewhat

disputable because the value marginally differs from

person to person. This topic provides a unique calcula-

tion model, the first of its kind, suitable for use in a

design office that is setup to calculate the value of f.

Thus, this topic

will

be of great help to engineers who

need to design belt conveyors as per DIN/ISO style.

The CEMA method uses the f in different forms: Kx, Ky

and 0.015. The

f

is the sum effect of

Kx,

Ky and 0.015 of

CEMA. Therefore, knowledge of f is equally relevant to

CEMA factors. The conveyor parameters which increase/

decrease

f

will also increase/decrease

Kx,

Ky and 0.015.

Thus information in

this

topicwill also help to consider a

better value for CEMA parameters, their understanding,

and clarity about comparative aspects. This topic indi-

rectly promotes the interchangeability of both of the

concepts to the advantage of both methods. This topic is

taken from the international book Engineering Science

and Application Design for Belt Conveyors by Ishwar G .

Mulani from Pune, India. The author has contributed to

this topic.

For such a calculation model to be of ~ractical alue. it

needs to be

in

conjunction with commercially acceptable

standards and norms. Therefore, this calculation proce-

dure considers the guidelines of DIN/ISO as a basis. The

calculation model further analyzes the issue and provides

a calculation model so that the designer can calculate the

right value of f, as a design input in the DIN/ISO method.

Thus, the proposed calculation method helps for better

design in conformity with DIN/ISO.

The readers who are used to USA practice will also find

this topic very useful to get new insight into the subject

and to comply with the respective standards

in

a more

exact and easy manner. This is due to the fact that

although calculation style could be different, the under-

lying considerations and principles are the same.

Although this article is pertaining

to

design issue,

its

information is equally interesting and useful

to

plant

engineers, because it explains the intrinsic phenomenon

Pune India


2/9

6 CONVEYOR BELT OPERATING TENSIONS FOR DESIGN

in conveying and effect of conveyor features on power

consumption, particularly for longer conveyors.

INTRODUCTION

This article deals with the calculation of conveying fric-

tion coefficient

f, which is required for calculating the

main resistance while designing belt conveyor, as per

European practices. This unique calculation model is the

first of its kind to calculate value o ff in design office set-

up. The reader may want to become familiar with DIN

22101/ISO 5048 for easier understanding of the subject.

MAIN RESISTANCE

The belt conveyor resistance is made up of many resis-

tances such as main resistance, slope resistance, skirt

board resistance, scraper resistance, pulley rotational

resistance, etc. s the name implies, main resistance is

one of the most significant resistances in a belt conveyor.

The main resistance is defined as the resistance to belt

travel due to the motion alone on idlers (accounting for

misalignment in installing the idlers but not accounting

for fonvard/backward tilt of idlers). The carrying run

belt travels along with material whereas the return run

belt travels without material. The belt travel encounters

three types of friction resistances, namely: (1) idler rota-

tional resistance, (2) belt flexure resistance, and

(3) material flexure resistance.

The aforesaid resistances are frictional in nature and

their total value is main resistance. The clear under-

standing of each of these frictional resistances is neces-

sary for correct calculation of an artificial friction

coefficient for conveying.

Idler Ratatlonal

Resistance

The belt travel compels the idlers to rotate, and hence, the

belt is subjected to rotational resistance of idlers. The rota-

tional resistance is a sum of the following three resistances.

Bearing frictional resistance. One of the rota-

tional resistances is due to bearing frictional resistance

which is pF, where p is the friction coefficient of the

bearing and F is radial force. The typical values for the

coefficient of friction are 0.0015 for ball bearings 6200/

6300, and 0.0018 for taper roller bearings 30200, as

referred to bearing bore radius/diameter. The rotation

effort F1 at periphery is given by the following fonnula:

bearing bore

roller diameter

bearing bore

roller diameter

The bracketed portions signify the friction coefficient,

referred to as roller periphery, and their typical values for

rollers of 139.7 outer diameter fitted with 6205 or 30205

bearings are 0.0015 x

25 139.7

=

0.0027 or 0.0018

x

25 139.7

=

0.00034. In general, this can be safely taken

s 0.0004 considering variations in ratio of bearing bore

and idler diameter. This amounts to only 2% of the basic

value of the conveying coefficient f

=

0.02.

Idler misalignment or sliding resistance. The other

main cause for rotational resistance is due to the inherent

visible/nonvisible misaligned position of idlers. This

results in very passive (slow) sliding of belt on idlers, and

D

Value

of

0

is

exaggerated

FIGURE1

thereby this resistance. It is also proportional to weight of

the belt and material resting on idlers.

The idlers need to be installed such that their axis of

rotation is exactly perpendicular to the belt travel path.

The misalignment (mostly invisible) would be present

within the permissible limits in accordance with the toler-

ances for structural fabrication, erection, etc. Figure

shows the exaggerated misaligned position of the idler

with respect to belt path by an angle

8

Belt velocity v can

be replaced by two vector components OA AB as shown

in the figure. Component

v

cos 8 is perpendicular to the

idler axis and causes the idler rotation. Component v sin 8

is parallel to the idler

axis

and cannot cause any rotation.

The belt is sliding in the direction of OA on the idler, there-

fore, the idler will apply sliding friction force F on belt in

the direction of AO, i.e., OD.

Force F on belt, direction OD

=

po i ; g where p, is

the sliding friction coefficient between belt and idler.

is the mass of

belt plus m teri l

if any is carried by it.

Resolving F along and across the belt,

Force

OC =

os 0

=

p,+

MI

g- cos 0

(This force results into side thrust on the belt.)

F o r c e C D = F . s i n O = h . M s . g l s i n O

This force is acting against the belt travel and causes

hidden sliding resistance as a part of rotational resistance.

Therefore, the resistance to belt motion

=

(yo sin 8)

M:.gN.

The value (pot sine) signifies equivalent sliding friction

coefficient for the belt on the idlers. Now, sin

8 =

offset

gradient of idlers.

Therefore,

Equivalent sliding friction coefficient

=

b)(offset

gradient)

The value of p, is nearly up to 0.4 for the rubber belt

on the idler shell (steel). The offset/misalignment is

considered at the rate of 3 mm for 1,000 mrn length of

idler base. From actual belt path, as the tentative

average aggregate of all idlers, the typical value of the

equivalent sliding friction coefficient is

Equivalent sliding friction coefficient

=

(po-sin 0) =

(po) (offset gradient)

=

0.40 x 3+1,000= 0.0012

The idler sliding coefficient is applicable to the moving

mass of the belt plus material (but not for idlers rotating

mass). For further details, refer to the book

ngineering


3/9

CALCULATION OF ARTIFICIAL FRICTION CONVEYING COEFFICIENT

7

Y

elt thickness exaggerated)

L Denting exaggerated)

FIGURE

Science and Application Design for Belt Conveyors,

authored by Ishwar

G

Mulani.

Sealresistance

The idler rotation causes shearing

cum churning of grease. This shearing/churning force

increases with shearing speed rotational speed). The

smaller diameter idler needs to rotate faster for the same

belt speed and will encounter more resistance to rota-

tion. The value will be less for bigger diameter. This

resistance is independent of load on the idler.

Some of the seals are contact type. This contact area

creates frictional resistance, which is independent of load

and speed. The grease churning/shearing resistance plus

seal contact resistance are together named as sealing

resistance.

The seal resistance is not related to radial load idler

load), and hence, its value is constant whether the

conveyor is running at full load or empty.

The idler rotational total resistance is the sum effect of

aforesaid frictions sections, which can be expressed as

under.

Idler rotational resistance

=

combined coefficient for

bearing and sliding resistance) x load)+ sealing

resistance of constant magnitude)

The effect of seal resistances at roller periphery is

inversely proportional to roller diameter. The typical

values with reference to roller periphery can range from

N

to 4

N

per roller 3

N

to 12

N

total for three rollers of

an idler set) depending upon design.

Belt

Flexure Resistance

The belt is supported on spaced idlers and is hanging

freely between the idlers.

As

the belt travels on idlers, it

is subjected to two types of flexing actions repetitively:

flexing of its thickness and flexing of its shape. The

flexing of its thickness is denting or indenting) flexure

and flexing of shape is bending flexure shape can only

change by bendindunbending of the belt).

Belt denting is shown in Figure 2. Any point on the belt

gets dented when on the idler and it gets undented the

moment it leaves the idler. Suppose a belt is moving at

2.5 mps on idlers spaced at 1.25-m pitch, then every

point on the belt will be subjected to one denting-

undenting cycle every half second.

The bending flexure of the belt occurs due to sag of the

belt and slight opening of the belt between the idlers as it

travels. Figure 3 shows concave and convex bending of

the belt on a longitudinal axis as it travels. In addition,

the belt is also subjected to bending flexure along its

width as shown in Figure 4. Both of the above

phenomena cause bending flexure of the belt. Its

frequency is the same as that for denting flexure.

ll the energy input during denting and bending is not

recovered during undenting and unbending reverse

bending), but some energy is lost in these repetitive

Vertical sag exaggerated)

FIGURE

3

op view of belt exaggerated)

FIGURE

4

Schematic depiction

of

material flexure

exaggerated)

FIGURE

5

flexing cycles. This is synonymous to internal friction of

belt texture and opposes the cause, which is belt travel.

The belt flexure resistance is proportional to magni-

tude of belt flexure and thereby it is influenced by the

following parameters.

The belt denting flexure is effected by hardness and

thickness of belt texture structure). The hard and

thin texture w ll have less depth of dent and would

cause less drag.

The higher tension in the belt keeps it straight with

less sag and less variation in shape. Therefore, a

belt with higher tension has less bending flexure

and moves easily on the idlers.

In a given conveyor of specific belt construction

and belt tension, the magnitude of belt flexures and

thereby its resistance is proportional to material

weight + belt weight.

Materla l Flexure Resistance

The material has a concave shape between the idlers and

a convex shape directly on the idler, as shown in

Figure 5. This causes subtle relative oscillating motion of

material particles among themselves and also on the belt,

in longitudinal direction. This is known as longitudinal

flexure of material.

Likewise, when seen from the top, the belt shape along

with material slightly opens between the idlers and

slightly closes near the idlers. This causes internal

rubbing within material and also on belt in transverse

direction. This phenomenon is known as transverse

flexure of material.


4/9

58

CONVEYOR BELT OPERATING TENSIONS FOR DESIGN

Thus belt travel causes longitudinal as well as trans-

verse flexure of material, which is frictional in nature. The

friction opposes the cause, which is belt travel. Therefore,

the belt travel encounters the material flexure resistance,

which is influenced by the following parameters.

The higher tension in the belt keeps it straight with

less sag and less variation in shape. Therefore,

material flexure resistance is remarkably less when

it is conveyed by belt under higher tension.

Contrarily, more effort is needed to move a slack

belt loaded with material.

The material of rough surface and/or interlocking

shapes causes more intense friction and hence

more flexure resistance. The smooth and rounded

lumps/particles have less flexure resistance.

For a given belt conveyor with specific material and

belt tension, the material flexure resistance is

proportional to material weight per meter on the

belt. (The quantity as well as intensity of friction,

which is occurring in the body of bulk material, is

proportional to material weight.

BASIS FOR ARTIFICIAL FRICTION COEFFICIENT

OF CONVEYING

f

As described in the foregoing sections, the main resis-

tance is comprised of frictional resistances: (1) idler rota-

tional resistance, (2) belt flexure resistance, and (3)

material flexure resistance.

The above frictional resistances are related to the

following moving masses: (1) idler rotating mass per

meter length of conveyor, (2) belt mass per meter length

of conveyor, and (3) material mass per meter length of

conveyor.

These moving masses' relation to earlier said frictional

resistances is not one to one, but it is cumulative

in

effect as below.

The idler bearing supports the weight of rotating

parts belt material resting on the belt. Hence, in

a broad sense the idler rotating resistance is

proportional to the sum total weight of

ll

three

masses.

The magnitude of belt flexure is effected by the

belt's own weight material weight resting on a

belt. Therefore, belt flexure resistance is

proportional to the sum total weight of belt mass

material mass.

The material flexure resistance depends upon

material self-weight only, as it is at the top.

Figure

6

shows the equivalent model for the depiction of

the above.

Thus there could be three different friction coefficients

applicable to three different values of mass. However, to

avoid complexities, the practice is to use single value of

artificial/hypothetical friction coefficient f applied to the

total moving mass force to calculate main resistance FH.

The complete mass is assumed to

be

moving on the

bearing inner race and is multiplied by the artificial fric-

tion coefficient f to give same frictional resistance as

occurring actually. The above depiction of mass forces

shows direct addition. In fact, these are to be added as

otion

Material mass (Mm)

Tractive pull Belt mass (2Mb)

Idler rotating mass (Mc

M r

FIGURE6

vector quantities to account for conveyor inclination,

which makes slight difference.

The main resistance,

where

L

is conveyor length.

DIN I SO STIPULATIONS I N BRIEF FOR VALUE OF

f CONVEYOR WITH POSITIVE POWER

Referring to the DIN/ISO standard, the basic value of

conveying friction coefficient is

f =

0.02. This value can

be

used for conveyors of positive power in the following

situations.

The material being conveyed has average flexure

resistance.

2 Belt sag between the carrying idlers is limited to a

maximum of 1.0% at any point along conveying

route.

3 The standard aligned stationary conveyors with

good maintenance conditions.

4 All idlers are equipped with antifriction bearings.

The carrying idlers are of 3 roller type,with trough-

ing angle up to 35 .

5 The idler diameter is not less than 108mm

6. The maximum pitch of carrying idler and return

idler is 1.5 m and 3.0 m, respectively.

7

Calculation of main resistance when the capacity is

in the range of 70% to 100%.

8

Belt speed is nearly 5.0 mps.

9 For ambient temperature of 20C and above (if we

ignore

4% increase in

f

for 0C to 20C, then basic

value 0.02 can also be considered for temperature

0C and above).

The value increases/decreases as per actual condi-

tions. Refer to the respective standard for specific infor-

mation. The above stipulations are unquantified. The

following exercise provides quantified effect.

DIVISION OF f INTO CONSTITUENT

COEFFICIENTS

Relatlve Proportion of Mwl ng Masses

It is necessary to establish relative proportion of moving

masses applicable to average conveyors, for use in the

next section. Following are the average reference values

for moving masses.

Mm Mb =

77%

of Mm 2Mb Mc Mr)

Mm

1.5

Mb

=

81%of (Mm 2Mb Mc Mr)


5/9

C LCUL TION OF RTIFICI L FRICTION CONVEYING COEFFICIENT 9

The basis for the above has not been included here due to

space limitations. One can also use the exact proportion

applicable to his/her specific calculation, if need be.

DerivsHon

of

Average Proportlom

Value

of

Consmuents

of f

This topic derives the proportion of constituents of f for

usual application needs. Referring to the amcle by

C Spaans in a Trans Tech publicationBelt Conveyor Tech-

nology the formula for FH using a constituent of f is as

below:

FH = Roller resistance (fdc

fbc fm

(Mb Mm) - g

(f& fb,) .Mb .

where fdc, fbc and fm are friction coefficients for belt

denting flexure, belt bending flexure, and material

flexure, respectively, for carrying run. The aforesaid

value of main resistance is per meter of conveyor.

fdr and fbr are friction coefficients for belt denting

flexure and belt bending flexure, respectively, for

return run.

Using the above formula and its pertaining informa-

tion, the following calculation model has been created by

this exercise to enable the calculation of f in design office

set-up.

~ i r i t l ~ ,his exercise decides roller resistance, belt

flexure resistance and material resistance, as a propor-

tion (percentage) of the total value, as a general basis.

Subsequently, the quantified effect on

f

has been calcu-

lated under the Quantified Effect on f section which

follows, due to various parameters of the conveyor. Rear-

ranging the formula,

FH = Roller resistance (fdc .Mb+

fdc

,,Mm

fdr ,.

Mb

fbc..Mb+fh..Mm+fbr..Mb+fm.-Mb+fm-.Mm) g

The next step is to see the approximate relationship

between fdcand f&, and fbcand fbr. The general formula

for belt denting flexure coefficient is

where cl is constant, D is roller diameter, and lr is the

belt and idler contact length.

All

parameters are identical

for carrying and return run except load per meter and

idler pitch.

In general, the ratio of p, and p, can be considered 2.5.

The ratio of load per meter on carrying and return could

be in the range of 8:l to 20:l.

For ratio 8:1,

=

1.5 i.e.,

f d r =

0.66 fdc

f d r

For ratio 20: 1,

As can be seen, the nature of the formula ensures a

nearly stable relation between fdc and fh. This is consid-

ered here as fdr = 0.6 fdc This marginal variation in rela-

tionship does not have an effect on the final result

because

fdr

is multiplied to only Mb whereas the fdc is

multiplied to (Mm

Mb

and is governing.

The relationship between fbcand fbr s in direct propor-

tion to average sag on the carrying side and average sag

on the return side. In general, this average sag can be

considered in the ratio 2:1

Therefore fbr= 0.5 fbc. There could be some variation

in this relation but its effect on the total result is not

significant because fb, is multiplied to only Mb whereas

fk is multiplied to both Mm Mb) and is governing the

result. Now, substituting the value of fd, in t e r n of fdc,

and fbr in terms of fk,

. FH

= Roller mistance [fdc (Mm 1.6 Mb) + fk

.

(Mm 1.5 Mb)

fm

(Mm Mb)]

-

g

Now,

Roller resistance = 0.0004 (Mm

+

2Mb

+

MC

+

Mr)

g

+

seal resistance yl sin (Mm + 2Mb) g

For this analysis, considering seal resistance equal to

1.5 times the bearing resistance and sin

=

3

+

1000:

Roller resistance = 0.001 (Mm 2Mb MC ~ r )

0.0012 x (Mm 2Mb) g

FH = 0.001 (Mm

Mb

Mc Mr) g

+

0.0012 x

Mm

2Mb)

.

g [fdc

.

Mm 1.6 Mb) fbc

.

Mm + 1.5 Mb)

fm

(Mm Mb)] .

Now,

Equating this with the above FH value, the equation for f

is as below:

: f =

0.001 x (Mm 2Mb Mc Mr) x g

(Mm+2Mb+Mc+Mr)xg

0.0012 x (Mm 2Mb) x g

(Mm+2Mb+Mc+Mr)xg

fdcx (Mm 1.6Mb) x g

(Mm+2Mb+Mc+Mr)xg

fbcx (Mm 1.5Mb) x g

(Mm+2Mb+Mc+Mr)xg

Using various proportion values from the Relative

Proportion of Moving Masses section:

f

=

0.001 0.0012 x 0.85 0.83 fdc 0.81 fbc

+ 0.77

Typical values for

fdc

=

0.012, fk

=

0.0035 and fm

=

0.008

for sag less than 1 at any point on carrying

run.

:. f = 0.00202+ 0.83 x 0.012 0.81 x 0.0035

0.77 x 0.008


6/9

60 CONVEYOR BELT OPERA TING TENSIONS FOR

DESIGN

TABLE

1

elt Width

Material Depth (mm)/% Increase

(mm) h-35 -40

-

450

Average considered

8 16

:. f = 0.00202 + 0.00996 0.00283

+

0.00616 = 0.02097

Roller resistance coefficient = 0.00202 + 0.02097 =

0.096 .5% of total value off

Belt denting flexure resistance coefficient= 0.00996 +

0.02097 = 0.475 r 47% of total value off

Belt bending flexure resistance coefficient= 0.00283

+

0.02097 = 0.135 13.5% of total value off

Material flexure resistance coefficient= 0.00616

+

0.02097= 0.2937 0% of total value off

We can use the previous proportions to assess the

implication on f due to variation in application condition.

If need be, the designer can follow the procedure to find

out actual proportion and consequent implications.

QuantWed Effect

on

f

(Calculatlan

for f

Various standards mention only * effect (unquantified)

on the value of f for deviation from base conditions.

Therefore, it leaves the value of f open to judgement/

manipulations. This calculation model provides quanti-

fied effect, and hence values of f, in accordance with

specific features of the belt conveyor. Analysis depth has

been limited up to engineering needs for the solution.

Increment

in

f for material difficult to flexure. It has

been considered that very difficult material has a 66%

increase in material flexure resistance. Accordingly, the

revised percentage figures are 9.5 + 47 + 13.5 + (30

x

1.66) = 119.8. Therefore increase has been mentioned as

20% in basic value 0.02.

Increment in f for deeper troughing angle. The 40

and 45 trough idlers are normally of deep trough type.

The comparative depth of material for 3S0, 40 and 45'

idlers are as per Table for 15 surcharge angle. Other

parameters being constant, the material flexure resis-

tance increases in proportion to the square of material

depth on the belt.

The of constituents, 40 troughing

=

9.5 + 47

+

13.5 + (30 x 1.08 x 1.08) = 105%; i.e., 5% increase in f.

The of f consti tuents, 45 troughing = 9.5 + 47 +

13.5 + (30 x 1.16

x

1.16)

=

110%; i.e., 10% increase in f.

In addition, the deeper troughing angle also affects the

force on side roller, and thereby the denting flexure. This

exercise is not presented due to space limitation. Its

result is as below:

For 40 trough idlers: 9.5 + (47 x 1.009) + 13.5 + 30 =

100.423; i.e., 0.42% increase. (This is ignored).

For 45 trough idlers: 9.5 + (47 x 1.020) + 13.5 + 30

=

100.94; i.e., 0.94%, say

1%

increase. Accordingly, total

implications in f

=

0.02 are 5% for 40 trough idlers and

10 +

=

11% for 45 trough idlers.

Increment in f for smaller diameter idlers. The other

parameters being unchanged, the idler diameter will

primarily affect the belt denting coefficient. The formula

for denting flexure coefficient is as below (where c2 is

constant and

D

s idler diameter):

5

.

f, for roller 88.0

mm

108.0 =

.

-

, for roller 108.0

mm

- 88.9

Therefore, roller diameter of 88.0 mm instead of 108mm

would change the f constituent as below,

9.5+(47x1.14)+13.5+30=106.6;i.e.,6.6 .Smd

diameter also increases the roller rotational coefficient

slightly. Therefore total implication has been considered

as 7.5%.

Increment if f d ue to poorly aligned stationar

conveyor. For p r l y aligned stationary conveyor,

a d i

tional misalignment offset gradient is considered as 5/

1,000; i.e., 5

mm

offset for 1,000 mm length of idler

base. This results in additional rotational resistance due

to more sliding. Looking from the top, the belt

will

be in

shallow wavy line, which tends to reduce idler misalign-

ment with respect to the actual path of the belt. Consid-

ering deviation from the actual belt path as 80% of

dimensional deviation,

(A) Additional resistance due to sliding

=

(Equivalent

sliding friction coefficient) (mass) - g

= {pl (offset gradient)) . Mm 2.Mb) g

=

(0.4. (0.80

x

5

+

1000)) (0.85 (Mm 2.Mb Mc

Mr)) .g

(B) Basic resistance force = f (Mm 2.Mb + Mc + Mr g

=0 .02-(Mm+2.Mb+Mc+Mr) .g

Ratio of increase for resistance = (additional resistance,

A) (basic resistance force,B =

0.068

This means, say, 6.5% increase in basic value of =

0.02, for poorly aligned conveyor.

From the above generalized formula for percentage

increase in f = (0.4 x 0.85 .02) x (offset gradient

increase)

x

100= 1,700 x (offset gradient increase).

The aforesaid increase in offset gradient is with respect

to the actual path of the belt.

Increment in f du e to poor ali nment in shiftable

conveyor. The shiftable conveyor fength is made up of

modules, placed side by side, as shown in Figure 7. It will

have more misalignment affecting f

The

AB

is the conveying line. The true path of the

belt will be lying somewhere in between the straight line

AB and modules orientation, in a very shallow wavy

form. The idler misalignment affecting the tractive pull is

with reference to the actual path of the belt. The approxi-

mate implication on f due to misalignments is as

below. The misalignment gradient 3

+

1,000 is already

counted in the basic value off .

Arrangement-X. This arrangement is statistically

the least possible and hence has not been analysed.

Arrangement-Y. The misalignment with reference

to belt line is considered as 80% of dimensional misalign-

ment. Accordingly, additional gradient

= (0.8 x 100

+


7/9

CALCULATION OF ARTIFICIAL FRICTION CONVEYING COEFFICIENT 6

Frame

module

6,250

l

B B B

Arrangement X Arrangement Y Arrangement Z

Arrangement-X: Statistically least probable

~rrangement-Y: tatistically

Arrangement-Z: Statistically probable

Exaggerated)

FIGURE 7

6,250) - (3 1,000) = 0.0098.

Therefore, increase in

f

= 1,700 x 0.0098 = 16.6667, say 17 .

Arrangement-Z. The

50

of modules have proper

placement and 50 of modules have misaligned place-

ment. As the belt will be in wavy line, even properly

placed modules will contribute to the misalignment

gradient, which is considered as 20 of the dimensional

misalignment. For balance 50 of modules, it is consid-

ered 80 o the dimensional gradient. Accordingly, for

50

of modules, the increase in offset gradient =

0.0098

(as per

Y .

Percentage increase in f for these modules = 1,700 x

0.0098

=

16.6667

For balance 50 of modules, increase in offset gradient

= (0.20 X 100 6,250) - 0.003 = 0.0002.

Percentage increase in

f

for this 50 of modules

=

1,700 x 0.0002 = 0.34

Hence,

the average mean effect = 0.5 x (16.66 0.34) = 8.5

This explains the effect on f in shiftable conveyors. In

general, one can consider a 12.5 increase in value of

f

for average shiftable conveyors, and approximately

17.5 for a poorly used system.

Increment in f due to belt sag. The belt sag between

two carrying idlers (and also between two return idlers)

is to be limited within 1 of respective idler spacing at

any point along conveyor route. It

is

essential to design

all high-performance conveyors according to this sag

value. If the designer opts for higher sag value, its likely

effect is as below.

The belt sag average value along conveyor length

affects the belt bending flexure and material flexure coef-

ficients. Now, average sag

wiU

be proportionate to allow-

able design sag. Hence, it can be said that belt bending

flexure and material flexure relation to design sag

wiU

be

the same as for average sag.

While assessing the response of a particular conveyor

to sag, all other conditions are identical and only sag

values are manipulated.

For all other conditions being constant except sag, the

relation between

fb

and sag is as below.

fb

=

c3 (sag

),

where c3 is constant. Therefore,

increase in

fb

is

50

for

1.5

sag instead of

1 .

Simi-

larly, increase in fb is 100 for 2 sag instead of 1 .

Now sag also affects the material flexure coefficient.

ll

other conditions being identical, the relationship

between f

and sag is as below.

[

value also increases with sag, which is approximately

accounted by 50 addition into the first part result.

fun tion of sag I-

: = c, .

h i p i

.

for 1.5 sag

A = 1.225

-

f,forl.O%sag O

i.e., 22.5 11= 33.5 increase in f

.

, for2.0 sag

O

- -

f,,, for 1.0 sag O

i.e., 41 20.5 = 61.5 increase in

f

The implication in f for 1.5 sag instead of 1.0 sag =

9.5 +47.0+ (13.5 x 1.5) + (30x 1.335)

=

116.8; i.e.,

increase, say

15

to

19 .

The implication in f for 2.0 sag instead of 1.0 sag

=

9.5

+

47.0

+

(13.5 x 2.0)

+

(30 x 1.615)

=

131.9; i.e.,

increase, say 27 to 34 increase.

The aforesaid increase is 50 for 3 sag. The decrease

is 10 for sag value 0.65 .

Increase in f due to thicklsoft rubber cover on belt.

The belt denting flexure is occurring on the carrying as

well as the return run. The belt denting flexure on the

carrying run plays a dominant part because the same is

getting multiplied to a much larger quantity (Mm Mb),

and it has been considered for analysis.

The belt denting flexure on the carrying ru is affected

the by the bottom cover. The denting flexure will

increase due to the combined effect of hardness as well

as thickness of the bottom cover. If the belt cover is soft

(less hard) but it is thin, the depth of dent will be less and

it may not affect the denting flexure. Also, if rubber has

usual hardness but much thicker cover, it results in a

deeper dent and affects flexure resistance. This exercise

has not been presented here due to space limitation;


8/9

6 CONVEYOR BELT OPERATING TENSIONS FOR DESIGN

however, the end result is as below (increase in basic

value f

=

0.02).

Nil Bottom cover thickness less than 2 rnm

5.5% Bottom cover thickness 2 mm to 3 mm and

rubber hardness 65 shore

11% Bottom cover thickness more than 3 mm and


Nil Bottom cover thickness up to 4 mm and rubber

hardness 75 shore

5.5% Bottom cover thickness 5 mm to 6 mm and


11%

Bottom cover thickness 8 mm above and rubber

hardness 75 shore

Incremenvdecrement for belt

speed.

The conveying

friction coefficient value f = 0.02 can be decreased for

conveyor speed lesser than 5.0 mps. These reduction

factors with respect to speed are based on a ContiTech

Germany publication (slightly changed here for safety in

design):

10%reduction on 0.02 for 2.75 rnps


9/9

C LCUL TION OF RTIFICI L FRICTION CONVEYING COEFFICIENT

6

Belt cover thickness and hardness do not have any

implication.

Condition can be considered clean and hence,

implication nil.

The evaluated value of f = 0.02 basic value

-

0.002 ten

percent for belt speed 0.001 five percent for temp.

The total of t is is 0.019. The design can be done with

f = 0.019 or 0.02.

Example

Select value of f for a 1,000-mm belt

x

3,500-m long

inclined (up) conveyor for conveying (-175-mm lime-

stone, bulk density 1,350 kg/m3, from mine to plant. The

conveyor design capaaty is 1,325 mtph and speed

3.5 mps. It has 152.4 mm 0.d. x 3 roll x 35 trough

carrying idlers, at 1.5 m pitch. The return idler pitch is

3.0 m. The ambient temp. is 2C to 45C. The conveyor

has steel cord belt of cover thickness 8

and 5 mm.

Solution. The designer analyzes the requirement

and makes following decisions:

Belt speed 2.75 < v

=

3.5 mps < 3.75 mps.

Therefore, the implication is -10% on f.

The belt width is more than 800 mm. Therefore, no

implication for belt width.

Limestone is somewhat difficult for flexure.

Therefore, the implication is +lo% on f.

Idler diameter and troughing angle do not have any

implication. Also no implication due to

temperature.

The maintenance quality is likely to be average,

hence consider increase of half of +6.5%; i.e.,

+3.25%.

Conveyor

will

include a reliable take-up device for

the correct tension all the time. For such a long

conveyor, the design sag is likely to result in less

than 0.65%. Accordingly, 10% reduction in basic

value off.

For belt bottom cover thickness 5

mm

and rubber

hardness around 75 shore-A, the increase is +5.5%.

The evaluated value of f = (0.02 basic value)

-

(0.002

decrease 10% for speed) (0.002 increase 10% for mate-

rial flexure) - (0.002 decrease 10% for sag) (0.00065

increase 3.25% for maintenance)

+ (0.0011 increase

5.5% for thick rubber cover). The total of this is 0.01975.

The design can be done with f = 0.02.

T H I S C L C U L T I O N M O D E L N D C E M

CEMA calculates the main resistance of DIN/ISO in the

following form (notations Kx, Ky, 0.015, Wm Wb are

as per CEMA).

Carrying idlers' rotation resistance is accounted for by

factor Kx. However, the difference with respect to f is

that Kx also includes multiplication by Wb Wm).

Carrying run belt flexure material flexure resistance

is accounted for by factor Ky.

Return run idlers' rotational resistance

+

belt flexure

resistance is accounted for by factor 0.015 (CEMA stipu-

lations, reference 2nd edition, are at variance whether

return idlers' rotational resistance is part of 0.015 or Kx.

If i t is part of Kx then the calculation of return belt

tension at various points

will

not be feasible. Hence,

return idler rotation resistance is mentioned as part of

0.015).

The aforesaid information provides basic comparison.

Further elaboration is not included due to space limita-

tion. The article helps to understand various phenomena

and their implications. It can

be

useful while choosing/

adjusting the values of

Kx,

Ky and 0.015 for calculation

by CEMA method. For example, if idlers' alignment is

poor in a specific case, the designer can think of taking

somewhat more of a value of Kx. Likewise, while

choosing a specific value of Ky, one can take into consid-

eration issues such as troughing angle, belt cover thick-

ness, cover hardness, sag, etc., which will have parallel

effects (although magnitude/percentage increase in f are

not to

be

used directly for Ky, as its multiplier is

W b

Wm instead of total mass).

Following are the commonly used engineering symbols

in

this

article:

General notation for force, N

General notation for coefficient of friction

General notation of mass, kg

Earth surface gravity acceleration, 9.81

m 2

Inclination for conveyor or conveyor portion, as

applicable, degree

Artificial friction coefficient for conveying, or

conveying friction coefficient

Belt speed, m/s

Belt mass (either of carry or return), kg/m

Carry side idler set rotating mass, per meter of

conveyor length, kg/m

Material mass (which is on belt), kg/m

Return idler sets rotating mass per meter length of

conveyor, kg/m

Carry side 3 roller idler set troughing angle

Notations other than the above are clarified wherever

they are mentioned in the article.

R E F E R E N C E S

Conveyor Equ ipmen t Man ufacturers Association CEMA). Belt

Conveyors for Bulk M aterials 2nd ed. Boston, Mass. CBI

Publishing Company, Inc.

Mulani, Ishwar G. 2001. Engineering Science and Application

Design for Belt Conve yors Pune, India: Ishwar G. Mulani.

Spaans, C. 1991. Calculation of the Main Resistance of Belt

Conveyors. In Belt Conveyor Technology Vol. 1/94. Edited

by

Reinhard H. WoN bier. TransTech Publications.

Standards:

DIN

2 2 1 0 1 a n d

IS0

5048.

Documents

Calculation of Artificial Friction Conveying Coefficient f, And a Comparison Between ISO and CEMA