fms200904 calc methods conveyor belts 304 en[1]

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Calculation methods ndash conveyor belts

Siegling ndash total belting solutions

conveyor and processing belts

This brochure contains advanced equa-

tions figures and recommendations

based on our longstanding experience

Results calculated can however differ

from our calculation program B_Rex(free to download from the Internet at

wwwforbo-sieglingcom)

Contents

Terminology 2

Unit goods

conveying systems 3

Take-up range for

load-dependent

take-up systems 8

Bulk goods

conveying systems 9

Calculation example

for unit goods conveying 12

These variations are due to the very

different approaches taken while B_Rex

is based on empirical measurements and

requires a detailed description of the

machinery the calculation methodsshown here are based on general simple

physical equations supplemented by cer-

tain factors that include a safety margin

In the majority of cases the safety margin

in calculations in this brochure will be

greater than in the corresponding B_Rex

calculation

Further information on machine design

can be found in our brochure ref no 305

ldquoRecommendations for machine designrdquo



2

Terminology

U n i t

A b b r e v i a t i o n

D e s i g n a t i o n

Key to the abbreviations

Drum and roller width b mmBelt width b0 mm

Calculation factors C ndashDrum and roller diameter d mm

Drive drum diameter dA mmRolling resistance of support rollers f ndash

Tensile force F NMaximum belt pull (on the drive drum) F1 N

Minimum belt pull (on the drive drum) F2

NForce of the tensioning weight FR N

Effective pull FU NTensioning drum weight F TR N

Steady state shaft load on the drive drum FWA NInitial value of the shaft load FWinitial N

Relaxed shaft load on the return drum FWU NAcceleration due to gravity (981ms2) g ms2

Difference in the drum radii (crowning) h mmConveying height h T m

Relaxed belt pull at 1 elongation per unit of width k 1 NmmSupport roller pitch on upper side l0 mm

Transition length lS mmSupport roller pitch on return side lu mm

Geometrical belt length Lg mmLength of conveyor l T m

Mass of the goods conveyed over the entire lengthconveyed (total load) m kg

Mass of the goods conveyed on the top side (total load) m1 kgMass of the goods conveyed on the return side (total load) m2 kgMass of the belt mB kg

Mass of the goods conveyed per m length conveyedon the upper face (line load) m0 kgm

Mass of all rotating drums except for drive drum mR kgMass of the goods conveyed per m length conveyed

on the return side (line load) mu kgmMechanical motor power PM kW

Mechanical power calculated on the drive drum PA kWProduction tolerance Tol

Friction coefficient when running over roller microR ndashFriction coefficient for accumulated conveying microST ndash

Friction coefficient when running over table support micro T ndashBelt velocity v ms

Volume flow for bulk goods conveying V ∙

m3 hTotal take-up range X mm

Belt sag yB mmDrum deflection y Tr mm

Margin for take-up range Z mmMachinersquos angle of inclination α deg

Arc of contact on the drive drum (or snub roller) β degOpening angle on the tensioning drum γ deg

Belt elongation (pre-tensioning with weight) ΔL mmPermitted angle of inclination for unit goods δ deg

Elongation at fitting ε Maximum belt elongation εmax

Drive efficiency η ndashBulk density of goods conveyed ρS kgm3



3

Unit goods conveying systems

mB

mBFU = micro T g ( m + ) + microR g ( + mR ) [N]

2

2

FU = please enquire [N]

Direction conveyed upwards

FU = microR g (m + mB + mR) + g m sin α [N]

Direction conveyed downwards

FU = microR g (m + mB + mR) ndash g m sin α [N]


mB

mB

FU = micro T g ( m + ) + microR g ( + mR ) + microST g m [N]

2

2

FU = micro T g (m1 + m2 + mB) [N]

m = l T Weight of conveyed goods per metre

FU = microR g (m + mB + mR ) [N]

Load examples to

establish the maximumeffective pull Fu [N]

Direction conveyed upwardsFU = micro T g ( m + ) + microR g ( + mR ) + g m sin α [N]


FU = micro T g ( m + ) + microR g ( + mR ) ndash g m sin α [N]

mB

2

mB

2

mB

2

mB

2



4

0 A0 E0 NOVO U1 V1 VH UH V2H U2H E0

T U0 P A0 V5H V10H

micro T (table) 033 033 05 05microR (roller) 0033 0033 0033 0033

microST (accumulated) 033 033 05 05

Siegling Transilon V3 V5 U2 V1 U1 UH U2H 0 U0 NOVO E0

Underside coating A5 E3 V2H V5H A0 T P

Arc of contact β 180deg 210deg 240deg 180deg 210deg 240deg 180deg 210deg 240deg

Smooth steel drumdry 15 14 13 18 16 15 21 19 17

wet 37 32 29 50 40 30 not recommended

Lagged drumdry 14 13 12 16 15 14 15 14 13

wet 18 16 15 37 32 29 21 19 17

F1if the value is larger than C2b0

a stronger belt type (with a higher k 1 value) must be used

Note

If belts have been perforated b0 must be reduced by the total width of the holes at

a typical cross section In the case of extreme temperatures the C2 factors change

Please enquire

Tension member Polyester Aramide

Type Polyester (key letter ldquoErdquo) (key letter ldquoAErdquo)

Examples of E 21 E 31 E 42 E 61 NOVO E 82 E 10M E 122 AE 48H AE 803 AE 1003type classes E 152 E 15M E 183 E 20M E 303 E 443 AE 140H AE 1403

εmax in 20 08

F₁ = FU C1 [N]

PM middot η middot C1 middot 1000

FU = [N]v

If the effective pull FU cannot be calcu-

lated FU can be established from the

motor power installed PM

If effective pull FU can be calculated

Friction coefficients microS for

various coatings (guidelines)

Maximum belt pull F1

Factor C1

(applies to the drive drums)

Factor C2

Checking the Transilon type selected

C2 indicates the max permitted belt pull per unit width for the belt type

C2

= εmax

k 1

You can find details on the maximum elongations in the product data sheets

If these are not available the following can be assumed (but not guaranteed)

F1 le C2 [ ]

b0

Nmm



5

FU middot v

PA = [kW]1000

Siegling Transilon V3 V5 U2 V1 U1 UH 0 U0 NOVOUnderside coating A5 E3 T P

Smooth steel drumdry 25 30 40

wet 50 Not recommended Not recommended

Lagged drum

dry 25 25 30wet 30 40 40

PA

PM = [kW] = the next largest standard motor is selected

η

FU middot C3 middot 180dA = [mm]

b0 β

Minimum diameter of the drive drums dA

Factor C3


Mechanical capacity calculated

on the drive drum PA

Mechanical capacity required PM



Take-up range for screw-

operated take-up systems

The following factors must be taken into

account when establishing the take-up

range

1 The approximate magnitude of elon-

gation at fitting ε of the belt resulting

from the belt load To establish ε see

pages 7 and 8

2 The production tolerances (Tol) of

the belt as regards the length

3 Any external influences that might

necessitate greater elongation

(tensioning) than usual or might

require a safety margin such as for

example the impact of temperature

stop-and-go operation

Guidelines for shaft load

at rest with tensile force F

When you are estimating the shaft loadsplease assess the different levels of belt

pull when the conveyor is at rest and in a

steady state

Guidelines for elongation

at fitting ε for head drives

The minimum elongation at fitting for

head drives is

FU 2 + 2 F2

ε asymp []2 k 1 b0

ndashTol +Tol ε z

At rest

FW1 = FW2 = 2 F F asymp ε k 1 b0 N

Head drive in steady state forces

F2 = F1 ndash FU FWA = F1 + F2

Generally depending on the load elon-

gation at fitting ranging from approx

02 to 1 is suff icient so that normally

a take-up range x of approx 1 of the

belt length is adequate

x



FU 2 + 2 middot F2 + FU

ε = []2 middot k 1 middot b0

Tail drive in steady state forces

F2 = F1 ndash FU

K for head drives = 075

K for return-side drives = 062

K for tail drives = 025

Typical end drum β = 180deg

FW3 = 2 F2 [N]

Typical snub roller β = 60deg

FW6 = 2 F2 sin (β2) [N]

Typical drive drum β = 180deg

FWA = F1 + F2 [N]


at fitting ε for tail drives


return side drives is

Guidelines for elongation at

fitting ε for return-side drives


operating head drives is

Guidelines for steady state

shaft load

FU (C1 ndash K)ε = []

k 1 middot b0Return side drive in steady state

Shaft load when tensioning belts

Tension members made of synthetic

materials display significant relaxation

behaviour As a result the relaxed k 1

value is taken as a basis for calculating

belts in line with ISO 21181 It describes

the probable long-term force-elongation

properties of the belt material that has

been subjected to stress due to deflec-

tion and load change This produces the

calculation force FW

This implies that higher belt forces FWinitial

will occur when tensioning the belt They

will have to be taken into account when

dimensioning the drum and its compo-

nents (bearings) The following value can

be assumed as a reference

FWinitial = FW 15

In critical cases we recommend you

contact application engineers at Forbo

Siegling

Typical drive druml β ne 180deg

FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR

Example for establishing the tension

weight FR [N] at 180deg arc of contract

(F TR = tensioning drum weight [N])

FR = 2 F2 ndash F TR [N]


weight FR [N] at an angle γ according to

the drawing (F TR = tensioning drum

weight [N])

γFR = 2 middot F2 middot cos _ F TR [N] 2

In weight-loaded take-up systems the

tension weight must generate the mini-mum belt pull F2 to achieve perfect grip

of the belt on the drive drum (spring

pneumatic and hydraulic take-up systems

work on a similar principle)

The tension weight must be able to

move freely The take-up system mustbe installed behind the drive section

Reverse operation is not possible The

take-up range depends on the effective

pull the tensile force F2 required elonga-

tion of the belt ΔL the production toler-

ance Tol the safety margin for tensioning

Z and the belt selected

Dimensioning force-dependenttake-up systems

F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL

FU 4 + F TR + FR∆L = middot Lg [mm]

k 1 b0

In force-driven take-up systems the overall elongation of the belt changes according

to the level of the effective pull The change in belt elongation ∆L has to be absorbed

by the take-up system For head drives ∆L is calculated as



9

Longitudinal angle

of inclination δ

Bulk density of some

bulk goods ρS

Volume flow V∙

for belts

lying flat

Goods conveyed Bulk density ρS [103 kgm3]

Ash cold dry 07Soil moist 15 ndash 19

Grain (except oats) 07 ndash 085Wood hard 06 ndash 12

Wood soft 04 ndash 06Wood chips 035

Charcoal 02Pulses 085

Lime lumps 10 ndash 14Artificial fertilizer 09 ndash 12Potatoes 075

Salt fine 12 ndash 13Salt rock 21

Gypsum pulverised 095 ndash 10


Gypsum broken 135Flour 05 ndash 06

Clinker 12 ndash 15Loam dry 15 ndash 16

Loam wet 18 ndash 20Sand dry 13 ndash14

Sand wet 14 ndash 19Soap flakes 015 ndash 035

Slurry 10Peat 04 ndash 06Sugar refined 08 ndash 09

Sugar raw 09 ndash 11Sugarcane 02 ndash 03

Guidelines for the longitudinal angle of

inclination δ permissible in various bulk

goods The machineryrsquos actual angle of

inclination α must be less than δ

These values depend on the particle

shape size and mechanical properties of

the goods conveyed regardless of any

conveyor belt coating

The table shows the hourly volume flow

(m3 h) at a belt velocity of v = 1 ms

Conveyor belt lying flat and horizontal

The belt is equipped with 20 mm high

longitudinal profiles T20 on the belt

edges of the top face

b0 mm 400 500 50 800 1000 1200 1400

Angle of surcharge 0deg 25 32 42 52 66 80 94


Bulk goods conveying systems

Bulk goods δ (approxdeg)

Ash dry 16Ash wet 18

Soil moist 18 ndash 20Grain except oats 14

Lime lumps 15Potatoes 12

Gypsum pulverised 23Gypsum broken 18

Wood chips 22 ndash 24Artificial fertilizer 12 ndash 15

Flour 15 ndash 18


Salt fine 15 ndash 18Salt rock 18 ndash 20

Loam wet 18 ndash 20Sand dry wet 16 ndash 22

Peat 16Sugar refined 20

Sugar raw 15Cement 15 ndash 20



10

b0 mm 400 500 50 800 1000 1200 1400

Troughed angle 20deg




Angle of surcharge 0deg 30 51 95 149 246 360 504Angle of surcharge 10deg 44 74 135 211 345 505 703

Volume flow V∙

for troughed

conveyor belts

in m3 h at a belt velocity of 1 ms

Note

Under real world conditions the theoreti-

cal values for volume flow are hardly ever

reached as they only apply to horizontal

belts with perfectly even loads Uneven

loads and the properties of the goods

conveyed can decrease the amount by

approx 30

In inclined conveying the theoretical

quantity of goods conveyed is slightly

less It is calculated by applying the factor

C6 which depends on the conveying

angle α

Conveying angle α [deg] 2 4 8 10 12

Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076

f = 0025 for roller bearings

f = 0050 for slide bearings

IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C

Establishing the mass of goods conveyed m

Factor C4

Rolling resistance for support rollers f

Additional effective pull for example

from scrapers and cleaning devices is

taken into account by including thefactor C4



11

If maximum sag of 1 is permitted

(ie yB = 001 l0)

Recommendation l0 max le 2b0

lu asymp 2 ndash 3 l0 max

l0 = Support roller pitch on upper side in mm

yB = Maximum conveyor belt sag in mm

F = Belt pull in the place concerned in N

m0 + mB = Weight of goods conveyed and belt in kgm

The support roller pitch depends on the

belt pull and the masses The following

equation is used to calculate it

8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB

Support roller pitches

(ndash) downwards

(+) upwards

FU = g middot C4 f (m + mB + mR ) plusmn g middot m sin α [N]

Calculation as for unit goods

Establishing the effective pull FU



12

Calculation example for unit goods conveying

F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16

m = 1200 kgmicroR = 0033micro T = 033

mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB

mBFU = micro T g (m + ) + microR g ( + mR )

2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6

Snub rollers 3 7 8Drive drum 5

Support rollers 4 9 and various

tension drums 6

Length of conveyor l T = 50 m

Geometrical belt length Lg = 105000 mmBelt width b0 = 600 mm

Total load m = 1200 kg

Arc of contact β = 180deg

v = ca 08 ms g = 981 ms2

Mass rollers mR = 570 kg(all drums except

for 5)

In a goods sorting system conveyor belts

are loaded with goods and sent to thedistribution centre Horizontal conveying

skid plate support return drive systems as

shown on the sketch drive via the top

face of the belt drive drum with lagging

screw-operated tensioning system 14

support rollers Proposed belt type

Siegling Transilon E82 U0V5H MT black

(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm

The belt type has been chosen correctly

F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm

Effective pull FU [N]

Maximum belt pull F1 [N]

Checking the belt type selected



13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm

dA dimensioned at 200 mm

FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW

PM at 55 kW or higher

FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)

FU = 4350 NC1 = 16K = 062k 1 = 8 Nmm for E82 U0V5H black b0 = 600 mm


k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09

Minimum drive drum diameter

Power PA

on the drive drum

Motor power required PM

Minimum elongation at fittingfor return drive



14

Simplified calculation assuming β = 180deg

F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N

F1 = 6960 NF2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N

Governed by minimum belt pull F2 FW3 iscalculated using the equation on page 7

FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N

Shaft load in steady state drum 2

(return drum)

Shaft load in steady state drum drum 1

(return drum)

Shaft load in steady state drum drum 5(return drum)

Shaft load in steady drum 3 (snub roller)



15

To compare rest and steady state modes

please observe the different shaft loads

in drum 1

FW1 at rest = 8640 N

FW1 steady state = 5220 N

Note

When designing machinery both modes

must be taken into account

ndash105 +105 473 200

210

883

Example for a drum with β = 180deg

Arc of contact

(In our example this force is exerted

equally on drums 1 5 and 6 because of

the 180deg arc of contact)

FW = 2 FFW = 2 09 8 600FW asymp 8640 N

At rest tensile forces are defined on the

top and underside by elongation at fittingε The tensile force F is calculated accord-

ing to

F = ε [] k 1 b0 [N]

Tol = plusmn 02 ε = 09 Lg = 105000 mmZ = 200 mm

FW = F12 + F22 ndash 2 F1 F2 cos β

FW = [N]

When β ne 180deg the following applies

when determining FW (F1 = F2 can be

assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+



Forbo Siegling GmbHLilienthalstrasse 68 D-30179 Hannover

Phone +49 511 6704 0 Fax +49 511 6704 305wwwforbo-sieglingcom sieglingforbocom


Because our products are used in so many applications and because of the

individual factors involved our operating instructions details and informati-on on the suitability and use of the products are only general guidelines and

do not absolve the ordering party from carrying out checks and tests them-selves When we provide technical support on the application the ordering

party bears the risk of the machinery functioning properly

R e f n o

3 0 4

- 2

0 4 0 9 middot U D middot R e p

r o d u c t i o n o f t e x t o r p a r t s t h e r e o f o n l y w i t h o u r a p p r o v a l S u b j e c t o f c h a n g e

Forbo Siegling Service ndash anytime anywhere

In the company group Forbo Siegling employs more

than 1800 people worldwide

Our production facilities are located in eight countries

you can find companies and agencies with stock and

workshops in more than 50 countries Forbo Siegling

service centres provide qualified assistance at more

than 300 locations throughout the world

M e t r i k G m b H middot W e r b e a g e n t u r middot H a n n o v e r middot w w w m

e t r i k n e t

T e c h n o l o g i e m a r k e t i n g middot C o r p o r a t e D e s i g n middot T e c h n i c a l C o n t e n t



2

Terminology

U n i t

A b b r e v i a t i o n

D e s i g n a t i o n

Key to the abbreviations

Drum and roller width b mmBelt width b0 mm

Calculation factors C ndashDrum and roller diameter d mm

Drive drum diameter dA mmRolling resistance of support rollers f ndash

Tensile force F NMaximum belt pull (on the drive drum) F1 N

Minimum belt pull (on the drive drum) F2

NForce of the tensioning weight FR N

Effective pull FU NTensioning drum weight F TR N

Steady state shaft load on the drive drum FWA NInitial value of the shaft load FWinitial N

Relaxed shaft load on the return drum FWU NAcceleration due to gravity (981ms2) g ms2

Difference in the drum radii (crowning) h mmConveying height h T m

Relaxed belt pull at 1 elongation per unit of width k 1 NmmSupport roller pitch on upper side l0 mm

Transition length lS mmSupport roller pitch on return side lu mm

Geometrical belt length Lg mmLength of conveyor l T m

Mass of the goods conveyed over the entire lengthconveyed (total load) m kg

Mass of the goods conveyed on the top side (total load) m1 kgMass of the goods conveyed on the return side (total load) m2 kgMass of the belt mB kg

Mass of the goods conveyed per m length conveyedon the upper face (line load) m0 kgm

Mass of all rotating drums except for drive drum mR kgMass of the goods conveyed per m length conveyed

on the return side (line load) mu kgmMechanical motor power PM kW

Mechanical power calculated on the drive drum PA kWProduction tolerance Tol

Friction coefficient when running over roller microR ndashFriction coefficient for accumulated conveying microST ndash

Friction coefficient when running over table support micro T ndashBelt velocity v ms

Volume flow for bulk goods conveying V ∙

m3 hTotal take-up range X mm

Belt sag yB mmDrum deflection y Tr mm

Margin for take-up range Z mmMachinersquos angle of inclination α deg

Arc of contact on the drive drum (or snub roller) β degOpening angle on the tensioning drum γ deg

Belt elongation (pre-tensioning with weight) ΔL mmPermitted angle of inclination for unit goods δ deg

Elongation at fitting ε Maximum belt elongation εmax

Drive efficiency η ndashBulk density of goods conveyed ρS kgm3



3


mB


2

2







mB

mB


2

2




Load examples to





mB

2

mB

2

mB

2

mB

2



4


T U0 P A0 V5H V10H








Lagged drumdry 14 13 12 16 15 14 15 14 13

wet 18 16 15 37 32 29 21 19 17



Note



Please enquire




εmax in 20 08

F₁ = FU C1 [N]


FU = [N]v








Factor C1


Factor C2



C2

= εmax

k 1



F1 le C2 [ ]

b0

Nmm



5

FU middot v

PA = [kW]1000




Lagged drum

dry 25 25 30wet 30 40 40

PA


η


b0 β


Factor C3











range




pages 7 and 8













steady state




head drives is

FU 2 + 2 F2

ε asymp []2 k 1 b0

ndashTol +Tol ε z

At rest









x






F2 = F1 ndash FU





FW3 = 2 F2 [N]


FW6 = 2 F2 sin (β2) [N]


FWA = F1 + F2 [N]










shaft load




















FWinitial = FW 15



Siegling


FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




3


mB


2

2







mB

mB


2

2




Load examples to





mB

2

mB

2

mB

2

mB

2



4


T U0 P A0 V5H V10H








Lagged drumdry 14 13 12 16 15 14 15 14 13

wet 18 16 15 37 32 29 21 19 17



Note



Please enquire




εmax in 20 08

F₁ = FU C1 [N]


FU = [N]v








Factor C1


Factor C2



C2

= εmax

k 1



F1 le C2 [ ]

b0

Nmm



5

FU middot v

PA = [kW]1000




Lagged drum

dry 25 25 30wet 30 40 40

PA


η


b0 β


Factor C3











range




pages 7 and 8













steady state




head drives is

FU 2 + 2 F2

ε asymp []2 k 1 b0

ndashTol +Tol ε z

At rest









x






F2 = F1 ndash FU





FW3 = 2 F2 [N]


FW6 = 2 F2 sin (β2) [N]


FWA = F1 + F2 [N]










shaft load




















FWinitial = FW 15



Siegling


FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




4


T U0 P A0 V5H V10H








Lagged drumdry 14 13 12 16 15 14 15 14 13

wet 18 16 15 37 32 29 21 19 17



Note



Please enquire




εmax in 20 08

F₁ = FU C1 [N]


FU = [N]v








Factor C1


Factor C2



C2

= εmax

k 1



F1 le C2 [ ]

b0

Nmm



5

FU middot v

PA = [kW]1000




Lagged drum

dry 25 25 30wet 30 40 40

PA


η


b0 β


Factor C3











range




pages 7 and 8













steady state




head drives is

FU 2 + 2 F2

ε asymp []2 k 1 b0

ndashTol +Tol ε z

At rest









x






F2 = F1 ndash FU





FW3 = 2 F2 [N]


FW6 = 2 F2 sin (β2) [N]


FWA = F1 + F2 [N]










shaft load




















FWinitial = FW 15



Siegling


FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




5

FU middot v

PA = [kW]1000




Lagged drum

dry 25 25 30wet 30 40 40

PA


η


b0 β


Factor C3











range




pages 7 and 8













steady state




head drives is

FU 2 + 2 F2

ε asymp []2 k 1 b0

ndashTol +Tol ε z

At rest









x






F2 = F1 ndash FU





FW3 = 2 F2 [N]


FW6 = 2 F2 sin (β2) [N]


FWA = F1 + F2 [N]










shaft load




















FWinitial = FW 15



Siegling


FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t








range




pages 7 and 8













steady state




head drives is

FU 2 + 2 F2

ε asymp []2 k 1 b0

ndashTol +Tol ε z

At rest









x






F2 = F1 ndash FU





FW3 = 2 F2 [N]


FW6 = 2 F2 sin (β2) [N]


FWA = F1 + F2 [N]










shaft load




















FWinitial = FW 15



Siegling


FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t







F2 = F1 ndash FU





FW3 = 2 F2 [N]


FW6 = 2 F2 sin (β2) [N]


FWA = F1 + F2 [N]










shaft load




















FWinitial = FW 15



Siegling


FWA

= F1

2 + F2

2 ndash 2 F1 F

2 cos β [N]



8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




8

Establishing FR








weight [N])
















F TR FR

FU F1

F2

F2

γ

F TR FR

FU F1

F2

F2

Establishing belt

elongation ΔL


k 1 b0






9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




9

Longitudinal angle

of inclination δ


bulk goods ρS

Volume flow V∙

for belts

lying flat






























b0 mm 400 500 50 800 1000 1200 1400










Flour 15 ndash 18








10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




10

b0 mm 400 500 50 800 1000 1200 1400






Volume flow V∙

for troughed

conveyor belts


Note







approx 30





angle α


Factor C6 10 099 098 097 095 093


Factor C6 091 089 085 081 076



IT [m] 25 50 5 100 150 200

C4 2 19 18 17 15 13

m = V ∙ δS l T

36 [kg]

v

Factor C


Factor C4







11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




11


(ie yB = 001 l0)










8 F

l0 = [mm]

m0 + mB

yB 800 F

l0 = [mm]

m0 + mB


(ndash) downwards

(+) upwards






12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




12


F1 = FU C1

F1 = 4350 16

F1 asymp 6960 N

FU = 4350 NC1 = 16


mB = 1575 kg (from 25 kgm2 105 m 06 m)

mB


2

2

1575 1575FU = 033 981 (1200 + ) + 0033 981 ( + 570)

2

2

FU asymp 4340 N

End drums 1 2 6



tension drums 6





v = ca 08 ms g = 981 ms2


for 5)









(900026) with k 1 = 8 Nmm

F1 le C2

b0

6960 le 2 8 Nmm 600

116 Nmm le 16 Nmm


F1 = 6960 N

b0 = 600 mm

k 1 = 8 Nmm






13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




13

FU C3 180deg

dA = [mm] b0 β

4340 25 180deg

dA = [mm]

600 180deg

dA = 181 mm


FU v

PA = [kW]

1000

4350 08

PA =

1000

PA asymp 35 kW

PAPM = [kW]

η

35

PM = [kW]

08

PM asymp 44 kW


FU = 4340 NC3 = 25β = 180degb0 = 600 mm

FU = 4350 N

v = 08 ms

PA = 35 kW

η = 08 (assumed)



k 1 b0

4350 (16 ndash 062)

ε = []

8 600

ε asymp 09


Power PA

on the drive drum





14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




14


F1 = 6960 N

F2 = F1 ndash FU

F2 = 6960 ndash 4350F2 = 2610 N


F2 = 6960 ndash 4350F2 = 2610 N


FW2 = 2F1

FW2 = 2 6960 N

FW2 asymp 13920 N

FW1 = 2 F2

FW1 = 2 2610 N

FW1 asymp 5220 N

FW5 = F1 + F2

FW5 = 6960 + 2610

FW5 asymp 9570 N


(return drum)


(return drum)





15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t




15



in drum 1



Note



ndash105 +105 473 200

210

883


Arc of contact




FW = 2 FFW = 2 09 8 600FW asymp 8640 N



ing to

F = ε [] k 1 b0 [N]



FW = [N]



assumed at rest)

2 Tol Lg ε Lg

100 100X = + Z [mm] 2

2 02 105000 09 105000 100 100

X = + 200 [mm]

2

X = 210 + 473 + 200 [mm]

X asymp 883 mm

Shaft load at rest

Take-up range

+

+










R e f n o

3 0 4

- 2












e t r i k n e t











R e f n o

3 0 4

- 2












e t r i k n e t


fms200904 calc methods conveyor belts 304 en[1]

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