BELTCONV[1] - CEMA

8/10/2019 BELTCONV[1] - CEMA

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Belt Conveyor Calculations as per CEMA

1.Following six profiles are introduced in the conveyor program me:

2.

Horizontal (i) Inclined (ii) (iii) (iv)

Horizontal + Inclined +

Inclined Horizontal

(v)

Horizontal, Inclined & (vi)

Horizontal Inclined, Horizontal & Inclined

The program calculates all the dimensional details, after receiving the values of certaininput parameters, with respect to the selected profile.

Input parameters to be provided for profile no.: -

(i) c/c Horizontal length (m)

(ii) c/c Horizontal length (m)

Inclination (Deg) / Lift (m)

(iii) & (iv) c/c Horizontal length (m)

Length of horizontal portion (m), Inclination (Deg) / Lift (m)

(v) c/c Horizontal length (m)

Horizontal lengths (m)

Inclination (Deg) / Lift (m)

(vi) c/c Horizontal length (m)

Length of horizontal portion (m)

Base lengths (m)

Lift (m) / Inclination (Deg)

Internal calculations are based on pounds, feet/inches system, but inputparameters are always fed in SI units (ex. metres)

All final dimensional details are determined here itself and the same

values will be carried on in futher calculations.

2. Following material characteristics are entered:-

Volumetric & power densities - These values are taken from CEMA for

respective materials. Sometimes for a material, two values are given, then lowervalue is considered as volumetric density and the higher one as power density.

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Volumetric bulk density is used to check capacity & power density to determine the

final power requirements.

Angles of surcharge & repose are also entered / taken from database as per

CEMA. Five options for angle of surcharge are 0

0

, 10

0

, 15

0

, 20

0

, 30

0

.Values of volume & power densities, angles of surcharge & repose can be

changed if required.

Troughing angle is selected out of 4 alternatives (200,300,350,450).

3. Capacity calculations :-

Basic formula for calculating capacity is -

Capacity (Q) TPH = Area of cross x Belt sppeed x 3600 x vol. bulk densitysection (m2) (m/sec) (T/m3)

x Inclination factor

This comes into picture only for conveyors having inclination & values are provided in

the program as per graph in IS-11592.

Capacity

Belt Speed These three parameters are interdependent. If two

Area are known, third can be calculated.

(Belt width) Area of cross-section can be calculated with known valves

of Belt Width (inches), troughing and surcharge angles & the procedure for calculating

this area is taken from CEMA. Hence, we can conclude that,

If Data given is of :- Program will select

(i) Belt width, capacity - Min. belt speed (m/sec)

(ii) Belt width, Belt speed - Max. capacity (TPH)

(iii) Capacity, Belt speed - Min. belt width (mm)

Belt width range consists of following belt widths in mm :-

400, 500, 650, 750, 800, 900, 1000, 1200, 1400, 1600.

4. Belting Selection :-Belt selection i.e. rating, duty will be as per DUNLOP catalogue

only.Min.belt widths for adequate troughing of empty belts for 350troughing

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angle are kept same as those for 450troughing angle. For determining

top/bottom cover thicknesses, Nirlon catalogue is referred.

Belting gets selected as follows -

Belt width (mm) - Known value/ given by the clientDuty - General/Extra/Heavy

Default option

If we want to change this default option in the beginning itself, we can. If rating

is out of the range of general duty, the program clearly indicates so on the

screen, conveying that the user has to go for higher duty.

Material abrasiveness, temperature - Known parameters

Temperature Grade Factor for calculating belt weight

75 M24 1.1 Std. value

120 HR 1.21 10 % more on std value

180 SHR 1.265 15 %

200 UHR 1.375 25 %

FR 1.32 20 %

This information is provided by Mr.Nair of M/s Ashish

Enterprises on 18.09.98.

Knowing belt width, material density (power density) and troughing angle, it checks

both criteria of adequate load support and adequate troughing of empty belt, & thenselects Primary belting (Belt rating).

If material density exceeds 2.5 T/m3, same values of belt widths for 2.5T/m3areassigned.

It also calculates top cover thickness of the belt -

Loading cycle = 2 x conv. developed length (m)

Belt speed (m/sec)

= _________ sec

If this value lies in between two standard values, select lower value of loading cycle

with respect to material abrasiveness & lump size.Bottom cover thickness with respect

to material abrasiveness is as follows :-

Non-abrasive - 1.5 mm

Abrasive - 2 mm

Very abrasive - 2.5 mmWe can also enter top/bottom cover thicknesses as per requirement.

After selecting min. belt rating, now we can calculate belt weight -

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Wb= constant factor x [top cover thk(mm) +Bottom cover thk(mm)]+Nominal

carcass weight (kg/m2)

X Belt width (mm) X 0.6722

1000

=------lbs/ft

5. Weight of material is calculated as follows -

Wm= Capacity (TPH) x 1000 x 0.6722 = ________ lbs/ft.

3600 x belt speed (m/sec)

6. Idler rating :-

Max. Idler spacing by default is entered as per table in IS-11592. Option

is open to reduce this spacing as per the client requirement.

Belt Width (mm) Carrying Idler spacing (m) for

material density of

Return Idler spacing

(m)

0.4-1.2 T/m3 1.2-2.8 T/m3

400 1.2 1.2 3

500 1.2 1.2 3

600 1.2 1.2 3

650 1.2 1.2 3

750 1.2 1.2 3800 1.2 1.2 3

900 1.2 1.2 3

1000 1.2 1.2 3

1200 1.2 1.2 3

1400 1.0 0.75 3

1600 1.0 0.75 3

Now, Idler load (Il) = (Wb+ Wm)Si

= ___________ lbs

Adjusted load = Il x K1x K2x K3x K4 Belt Speed correction factor

Service factor

Lump Adjustment Environmental & maintenance factor

factor

[K2= 1.06, K3= 1.2] These two values are freezed.

K1= If the values of lump size (inches) and material density fall in between two

standard values, then select the higher values for K1.

K4= Select the roll dia. as it is & higher belt speed to decide the value of K4.

If K1,K2,K3,K4< 1, take it equal to 1.

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If Adjusted load < Idler load, take Al= Il.

Some assumptions are made to decide the value of Ai :-

Pipe OD (mm) Respective OD (inches) adjusted Duty type Ai as per CEMA

76.1 4 A, B , C 2.3

88.9 4

114.3 5 A, B , C ,D 1.8

127 5

139.7 6 C, D 1.5

152.4 6

165.1 7 E 2.4168.3 7

Now with respect to belt width and troughing angle duty type is decided i.e. whether

idler rating is A, B, C, D or E. These values tabulated below are interpolated from

CEMA for belt widths available in India.

A type - Belt Width (mm) Troughing Angle Return Idler

20 30 35 45

400 300 300 300 297 150

500 300 300 300 297 143600 300 300 300 290 127

650 300 295 295 284 119

750 300 282 282 270 102

800 294 275 275 265 94

900 278 259 259 251 78

B type - Belt Width (mm) Troughing Angle Return Idler

20 30 35 45

400 410 410 410 410 220

500 410 410 410 410 212

600 410 410 410 410 192

650 410 410 410 410 184

750 410 410 410 410 167

800 410 410 410 407 163

900 410 410 410 398 156

1000 399 384 384 371 147

1200 382 355 355 344 132

C type - Belt Width (mm) Troughing Angle Return Idler20 30 35 45

400 900 900 900 900 475

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500 900 900 900 900 433

600 900 900 900 900 330

650 900 900 900 900 306

750 900 900 900 900 256

800 900 885 885 878 238900 900 843 843 819 205

1000 872 812 812 785 172

1200 807 750 750 726 129

1400 741 690 690 667

D type - Belt Width (mm) Troughing Angle Return Idler

20 30 35 45

650 1200 1200 1200 1200 600

750 1200 1200 1200 1200 600

800 1200 1200 1200 1200 600

900 1200 1200 1200 1200 6001000 1200 1200 1200 1200 544

1200 1200 1200 1200 1200 435

1400 1191 1108 1108 1056 358

1600 1125 1032 1032 1013 249

1800 1060 986 986 954 167

E type - Belt Width (mm) Troughing Angle Return Idler

20 30 35 45900 1800 1800 1800 1800 1000

1000 1800 1800 1800 1800 1000

1200 1800 1800 1800 1800 1000

1400 1800 1800 1800 1800 912

1600 1800 1800 1800 1800 813

1800 1800 1800 1800 1800 715

2000 1800 1730 1730 1699 616

Now, we get Aivalue, once we fix the combination of [Duty type - Roller dia]

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A, B, C, D, E 4, 5, 6, 7 inches

7. Kx= 0.00068 (Wb+ Wm) + Ai

Si

Kycalculations :- For each portion of the profile, Kyis calculated seperately. Kyselection is as per two tables in CEMA on pg. No. 78, 79.

(i) Conveyor length Lower value if it lies in between two points.

(ii) Wb+ Wm Interpolate

(iii) Percent Slope Lower side

With respect to this data, decide primary value of Ky. But this is for certain

specific spacing. These idler spacings are in ft (3,3.5,4,4.5,5). Converting it to

metres, gives the values, which are nearly equal to generally recommended idler

spacings.

Then corrected Kyvalues are decided from next table in CEMA on pg 80.

Here correct idler spacing is used for a range of (Wb+ Wm).

All the basic parameters required for calculating effective tension are now availableat this stage.

Default roller diameter for carrying, return & impact idlers are also put in the

program with respect to belt width, which are as follows :-

(This information is picked up from one of the enquiries of Grasim cements where the

consultants were Holtec ltd.)

Belt Width (mm) Roller dia for Impact idler

C.I/SACl R.I/SARI Roller dia/Rubber ring OD

400 88.9 76.1 76.1/114.3

500 88.9 76.1 76.1/114.3

600 114.3 89 76.1/114.3

650 114.3 89 76.1/114.3

750 114.3 89 76.1/114.3

800 139.7 114.3 88.9/165

900 139.7 114.3 88.9/165

1000 139.7 114.3 88.9/165

1200 139.7 114.3 88.9/165

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1400 152.4 139.7 114.3/190

1600 152.4 139.7 114.3/190

8. Effective tension calculations :-Idler friction, Tx= Kt[Kxcarrying + Kxreturn] x L

developed = ___ ft

Value calculated 0.015 length (ft)

above always

Resistance of belt dlexure as it moves over the idlers.

Tyb = Tyc + Tyr

for carrying & return sides

Tyc for profile nos -

(i) & (ii) = l x Kyx Wbx Kt= _______ lbs

Tyr for all profiles = l x 0.015 x Wbx Kt

fixed always

Resistance of material flexure as it rides the belt over the idler -Tym for profile nos -

(i) & (ii) = L x Kyx Wm = _______ lbs

(iii) & (iv) = [L1x Ky1+ L2x Ky2+ L3x Ky3] x Wm

(v) & (vi) = [L1x Ky1+ L2x Ky2+ L3x Ky3] x Wm

Force needed to lift the load -

For profiles (ii), (iii), (iv) & (v) Tm= Lift (ft) x Wm(lbs/ft)

= _______ lbs

Resistance of belt to flexure around pulleys & resistance of pulleys to rotate on their

bearings - (Tp)

Following databases are introduced to the program to ease calculations -

Screw VGT HGT

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1800 0.84 0.5 1.2 0.8

2000 0.72 0.42 1 0.7

2100 0.66 0.38 1 0.7

2200 0.62 0.35 0.9 0.6

240

0

0.54 0.54 0.8 0.6

T1(KN) = T1(lbs) x 9.81

2.204 1000

Now following ratio is checked and it should be less than 80 %.

T1 < 80 %

Belt width (m) x TmaxIf it exceeds 80 %, the program displays a message & goes on the higher side to select

new rating. If belting is not available in that duty, then change the duty, appearing on

the screen in belting module. This procedure is continued, till safe belting is selected.

Now with the belt weight of this selected belting, again Teis calculated & checked

whether T1/Tmaxis less than 80 %. Once the belting is finalised, then, depending on

the % value of T1/Tmax, pulley diameters are also decided in conjunction with carcass

thickness of selected belt as follows :-

Carcass thickness (mm)

from - upto & including

60 - 80 % 0 - 60 %

Head(H) Tail/Bend

(T)

Snub(S) H T S

2.8 - 3.5 323 273 219

3.6 - 4.4 400 323 273 323 273 219

4.5 - 5.5 500 400 323 400 323 273

5.6 - 7.0 630 500 400 500 400 323

7.1 - 8.8 800 630 500 630 500 400

8.9 - 11.1 1000 800 630 800 630 500

Pulley diameters are decided only after Belting selection is finalised.

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If wrap angles over head pulley & other pulleys also are not specified, then theprogram will select the following tabulated valves -

Profile No (i) (ii) (iii) (iv) (v) (vi)

S V H S V H S V H S V H S V H SV HHead Pulleya. without

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= ___________ mm/1000 = ________________ m

Drive pulley rpm = 60 x belt speed (m/sec) = ____________ rpm

x pulley dia (m) (value)

Gear box ratio = 1440 = ______________

Drive pulley rpmMotor is selected, based on gear box efficiencies depending on whether KW required at

conveyor pulley shaft is less or greater than 30 KW. Now, if we want to change the

type of gear box, change it here only -

If gear box type is changed, accordingly efficiency will change & absorbed as

well as rated KW of motor will also change.

Once we specify the type of gear box, to arrive at proper size, the program follows the

following path -

Gear Box

Helical

Bevel Helical

Worm Solid Shaft Hollow Shaft

Solid Shaft Hollow Shaft

NU SBN/SCN

FSM It will go Foot Shaft KBN/KCN Foot Shaft

downward while mounted mounted mounted mounted selecting the correct

gear box size SBH/SCH SBA/SCA KBH/KCH KBA/KCA

The program will go from B to C i.e. from double to triple

reduction to select correct gear box size & ratio.

A separate gear box screen is introduced, which on getting the correct data selects

suitable gear box size -

Absorbed Power

Gear box selection based on Rated Power Default Option Type of gear box Worm

Helical

Bevel Helical

Shafting Solid Foot mounting

Hollow

Shaft mounting

Service factor 1.5 (Default value)

option open to changeNow, if gear box type selected is -

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1. Worm

Min. hp required for gear box = Absorbed/Rated motor x Gear box

hp hp service factor

= ___________ hp

Now the gear box ratio calculated, will not be always equal to or tally with thestandard ratios available.

Hence the program always selects the previous /lower gear box ratio.

Now with respect to this gear box ratio, & min. hp requirement, the program

selects the correct gear box size.

Due to reduction in gear box ratio, belt speed of conveyor will rise. Hence the

conveying capacity will increase for same belt width. There will also be a change in

motor power requirement, which increases from its original value.

(From previous page)

Hence gear box screen is continued with -

Required gear box ratio

Actually selected / std gear box ratio Change in speed

Change in hp / New hp

Capacity increased/decreasedHence, the program will always display the lower gear box ratio selected.

But if needed, we can select higher ratio & check whether capacity gets satisfied

or not. As we change ratio, corresponding charges will also be observed on the

screen. Only those ratios available in that particular series will get displayed.

2. HELICAL/BEVEL HELICAL

Min. KW required for gear box = Absorbed/Rated motor KW x Gear box

service factor

Now with this KW and selected gear box ratio, program selects suitable size.

This portion is explained as follows :-

Required gear box ratio = 1440

Drpurpm

Selected ore = ___________ (Let it be X)

Drpurpm changed = 1440

X

Speed changed = x Dr.Dia x Drpurpm changed/60

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KW changed (if any) = Te(N) x speed changed/1000

Capacity increased = Area x increased speed x 3600 x material bulk density x Kf.

11. Coupling selection :-

Input coupling is selected, once motor & gear box are finalised selection

procedure for three types of couplings is introduced in the program. (Fluid, bin

& bush & geared).

If, 1. Fluid Coupling

Check, whether the coupling is with or without delay chamber. See motor KW

selected. Select the coupling with rating equal to or greater than the above

motor KW at 1450 rpm.

2. Pin & bush / Geared Coupling

Input coupling torque = 716 x Rated motor x cplg service factor x 9.81(NM)

1440

Select the greater value among motor shaft dia & gear box input shaft dia with respect

to this greater value, decide the keyway portion in hub, as per the following database -

Shaft dia (mm) [From - upto & including) Keyway portion in hub (mm)

45 - 50 3.8

51 - 58 4.3

59 - 65 4.4

66 - 75 4.9

76 - 85 5.9

86 - 95 5.4

96 - 110 6.4

111 - 130 7.4

131 - 150 8.4

151 - 170 9.4

171 - 200 10.4

201 - 230 11.423 - 30 3.3

31 - 38 3.3

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39 - 44 3.3

Following two conditions are checked -

Shaft dia of motor as well as I/p gear box should bemin.bore of coupling max.bore of coupling

Now, Keyway portion in + 10 mm + Greater value among Hub outside dia

hub the two shaft dias

This value is entered in

the coupling database

The program will select, the correct coupling size satisfying above three conditions of

torque, min & max bore & hub outside diameter criterion.

Output coupling selection :-

O/P Coupling torque = 716 x Rated motor x cplg service factor x 9.81(NM)

(1440/Gear box ratio)

Select the coupling with greater torque next to, calculated above & display the size.

Hold back torque at the = Head Pulley dia (m) Force needed to lift the load (m)

output shaft of gear box 2

- (Idler friction (Tx) + Resistance

of belt flexure over idlers (Tyb)

+ Resistance of material flexure

over belt (Tym)/2)

= ___________ lbs/ft

= __________ x 0.4536 x 0.3048 = ____________ Kg.m

Considering 50 % safety margin

Hold back torque at Gear output shaft = __________ x 1.5 = __________ Kg.m

Equivalent mass of rotating parts of conveyor :-

Belt lengths for various types of take-ups is calculated as follows :-

1. Screw take-up

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Belt length = 2 x developed length (m) + 1 m +/2 [Head pulley dia + Tail

pulley dia]

Should consider

bare pulley dia (m)

2. VGT

Belt length = 2 x developed length (m) + 1 m +

[ Head Pulley dia + Tail Pulley dia + Take up dia ]

2 2

+ 0.025 x developed length (m) x 2 = __________ m

3. HGT

Belt length = 2 x developed length (m) + 1 m +

[ Head Pulley dia + Tail Pulley dia + Take up dia ]

2 2

+ 0.025 x developed length (m) x 2 = __________ m

Equivalent mass of rotating = (i) + (ii) + (iii) + (iv) + (v)

parts of a conveyor = ___________ lbs2.204 = ______________ kgs

(a)

Now,

i. Belt weight = Total belt length (ft) x Wb (lbs/ft) = __________ lbs

ii. Weight of rotating parts = Weight of rotating parts of unit idler (lbs)

of carrying idler x Total developed length (ft)

Carrying idler spacing (ft)

iii. Weight of rotating parts = Weight of rotating parts of unit idler

of return idler x Total developed length (ft)

Return idler spacing (ft)

iv. Material weight = Developed length (ft) x Wm= ____________ lbs

v. Equivalent weight of = 2/3 Tail pulley wt + Snub pulley wt (lbs) +

non-driving pulleys + Take-up pulley wt (lbs) + No of bends pulleys

x bend pulley wt (lbs)

G92value of conveyor, reterred = a

to motor shaft x motor rpm (1440) 2

belt speed (m/sec) x 60

= _____________ kg.m2Total G92of conveyor

without motor at motor

shaft

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= (x + y)

Mass of rotating parts of conveyor without drive m1= a

9.81

= _________ kg/m/sec2

= ________ x 2.2043.28

= __________ lbs.sec2

G92value of drive unit = G92of motor

+ Gear Box

+ Input Coupling

+ Output Coupling

Mass of Drive unit reterred = b xx motor rpm (1440)

to belt line 9.81 belt speed (m/sec) x 60

= ___________ Kg/m/sec2= ________ x 2.204

3.28

= ___________ lbs.sec2/ft

Total equivalent mass of conveyor

referred to belt line = A + B

= ________________ lbs.sec2/ft

Component (V) To find out weights of pulley

We, at this stage dont know the shaft diamters of pulleys. Hence assume the

following shaft diameters with respect to pulley diameters -

Pulley Dia Shaft dia to be assumed to

be consider pulley weight

219 50

273 60323 75

400 90

500 100

630 100

800 125

1000 140

Let G- starting force produced, by drive unit considering average starting torque 160 %

of motor rated torque.

G = 1.6 x motor hp x Gear box x 75

Belt Speed (m/sec)

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= _____________ Kg.f

= ________ x 2.204

= _______________ lbs

Acceleration= G - Te= lbs

m lbs.sec2/ft

= _________ ft/sec2

Effective force during starting,

Test= Gx m1

= _________ lbs

T1st= e0,= co-efficient of friction

T2st = 0.25 -- for base head pulley

= 0.85 -- for lagged pulley

= Wrap angle over head pulleyT1st= e

0. T2st

But Test= T1st- T2st = T2st.e

o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)

T1st= Test+ T2st

For calculating tensions of various points in the profile, it is assumed that snub

pulley is always present in the conveyor. Take up travel is considered 3% of

developed length in case of screw take-up & 2.5% in the case of VGT & HGT.

Separate calculations with respect to profile & type of take-up are introduced in

the program.

Profile No. 5 - (v) (Horizontal, inclined & horizontal), HGT (Elecon type)

T3st= T2st+ 100 +x mass of snub pulley x 2.204 + Texternal scraper

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32.2

= T2st+ 100 +x mass of snub pulley x 2.204 + 5 x 2 x Belt width (mm)

32.2 25.4

= __________ lbs

T4st= T3st[0.015 x L4(H) + 0.015 x L4(ft) x Wb(lbs/ft)] + x L4(ft) [Wb(lbs/ft) +

32.2

Wt. of rotating parts of unit return idler (lbs)]

= ___________ lbs

T5st= T4st+ 100 +x mass of bend pulley x 2.204 - H5(ft) x Wb(lbs/ft)

= ________ lbs

T6st= T5st+ 100 +x mass of bend pulley x 2.204 = _________ lbs

32.2

T7st= T7st+ 100 +x mass of take-up pulley x 2.204 = _________ lbs

32.2

T9st= T8st+ 100 +x mass of bend pulley x 2.204 = __________ lbs

32.2

Considering four fall, wire rope arrangement, at take-up tower,

Min. counter weight = (T6st+ T7st) x 2

= _____________ lbs/2.204 = ___________ Kgs.

(Make it a round figure, to ensure no slip)

From this calculated counter weight, fixed normal running full load belt tensions are

worked out as follows :-

T6= Counter weight (lbs)/4 = ________ lbs = T7T5= T6- 100

T4= T5- 100 + H5(ft) x Wb(lbs/ft) = ____________ lbs

T3= T4- [0.015 x L4(g)+ 0.015 x L4x Wb] = ________________ lbs

T2= T3- 100 = ______________ lbsT1= T2+ Te+ 5 x 2 x Belt Width (mm) - H7x Wb+

25.4

(0.015 x L10+ 0.015 x L10x Wb(lbs/ft) = ______________ lbs

T11= T10+ 100 = _______________ lbs

Profile No. 5 (v) VGT Horizontal, Inclined & Horizontal


Test= Gx m1

= _________ lbs

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= Wrap angle over head pulley

T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)

T1st= Test+ T2st


32.2


32.2 25.4

= __________ lbs

T5st= T4st+ 100 +x mass of bend pulley x 2.204 - H5(ft) x Wb(lbs/ft)

32.2

= _______________ lbsT6st= T5st+ 100 +x mass of bend pulley x 2.204 = ___________ lbs

32.2

T7st= T6st+ 150 +x mass of take-up pulley x 2.204 = ___________ lbs

32.2


32.2

Considering four fall, wire rope arrangement, at take-up tower,

Min. counter weight = (T6st+ T7st) x 2 = _____________ lbs/2.204 = __________ kgs.

(Make it a round figure, to ensure no slip)

From this calculated counter-weight, fixed normal running full load belt tensions are

worked out as follows :-

T6= Counter weight (lbs)/4 = _____________ lbs = T7T5= T6- 100

T4= T5- 100 + H5(ft) x Wb(lbs/ft) = ____________ lbs

T3= T4- [0.015 x L4(g)+ 0.015 x L4x Wb] = _____________ lbsT2= T3- 100 = ___________ lbs

T1= T2+ Te+ 5 x 2 x Belt Width (mm) = ___________ lbs

25.4

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T8= T7+ 100 = _____________ lbs

T9= T8+ 100 + H6x Wb= ____________ lbs

T10= T9+ 5 + 2 x Belt Width (mm) - H7x Wb+

25.4

(0.015 x L10+ 0.015 x L10x Wb(lbs.ft)) = _________ lbsT11= T10+ 100 = ___________ lbs

Profile No. 5


Test= Gx m1

= _________ lbs





T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)

T1st= Test+ T2stT3st= T2st+ 100 +x mass of snub pulley x 2.204 + Texternal scraper

32.2

= T2st+ 100 +x mass of snub pulley x 2.204 + 5 x 2 x Belt width (mm) 32.2 25.4

= __________ lbs

T4st= T3st[0.015 x L4(H) + 0.015 x L4(ft) x Wb(lbs/ft)] + x L4(ft) [Wb(lbs/ft) +

32.2

Wt. of rotating parts of unit return idler (lbs)]

= ___________ lbs

T5st= T4st- (H5(ft) x Wb(lbs/ft) = _______ lbs

T6st= T5st+ 150 +x mass of take up pulley x 2.284 = ________ lbs

32.2

T7st= T6st+ (H5x Wb) + 100 ++ mass of bend pulley x 2.204

32.2

Min. counter weight = (T5st+ T6st) = ________ lbs = ________ Kgs

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2.204

Normal running full load belt tensions :-

T6= Counter weight = ___________ lbs = T5 2

T4= T5-100 + H5x Wb= ____________ lbsT3= T4- [0.015 x l4+ 0.015 x l4x Wb] = _________ lbs

T2= T3- 100 = _________ lbs

T1= T2+ Te+ 5 x 2 x Belt Width (mm)/25.4 = ________ lbs

T7= T6+ (H5x Wb) = ________ lbs + 100

T8= T7+ 5 x 2 x Belt width (mm) - H7x Wb+ (0.015 x l10+ 0.015 x Wb)= _____ lbs

25.4

T9= T8+ 100 = ___________ lbs

Profile No. (vi)

Test= Gx m1

= _________ lbs


T2st = 0.25 -- for base head pulley = 0.85 -- for lagged pulley


T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs

T4st= T3st+[0.015 x l4+ 0.015 x l4x Wb] +x l4x [Wb+ wt. of rotating parts of

32.2 unit return idler wt] = -H5x Wb= _____________ lbs

T5st= T4st+ 100 +x mass of bend pulley x 2.204 - H6x Wb= _________ lbs

32.2

22


23/45

T6st= T5st+ 100 +x mass of bend pulley x 2.204 = ________ lbs

32.2


32.2

T8st= T7st+ 100 +x mass of bend pulley x 2.204 = ________ lbs 32.2

T9st= T8st+ 100 + H7x Wb= ___________ lbs

Assuming four full wire rope arrangement,

min. counter weight = T6st+ T7st= __________ lbs2

Normal running full load belt tension,

T6= Counter weight = T7= ___________ lbs

4

T5= T6- 100

T4= T5- 100 + (H6x Wb) = ___________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] + H5x Wb= __________ lbs

T2= T3- 100

T1= Te+ T2+ 5 x 2 x Belt Width (mm) = _______ lbs

25.4

T8= T7+ 100T9= T8+ 100 + H7x WbT10= T9+ 5 x 2 x Belt Width (mm) - H8x Wb+ [0.015 x l10+ 0.015 x l10x Wb]

25.4 = ___________ lbs

T11= T10+ 100 = __________ lbs

Profile (vi) Inclined, horizontal & inclined VGT

Test= Gx m1

= _________ lbs





T1st= e0. T2st

But Test= T1st- T2st

23


24/45

= T2st.eo- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs

T4st= T3st+[0.015 x l4+ 0.015 x l4x Wb] +x l4x [Wb+ wt. of rotating parts of

32.2 unit return idler wt]

= -H5x Wb= _____________ lbs

T5st= T4st+ 100 +x mass of bend pulley x 2.204 - H6x Wb= _________ lbs

32.2


32.2


32.2

T7= T6+ 100 + H7x Wb =______________ lbs

T8= T7+ 5 x 2 x Belt Width (mm) - H8x Wb+ [0.015 x l10+ 0.015 x l10x Wb]

25.4

T9= T8+ 100 = ___________ lbs

Profile No. (i) Horizontal - Screw take up

Test= Gx m1

= _________ lbs





T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e

0

- 1)T1st= Test+ T2stT3st= T2st+ 100 +x mass of snub pulley x 2.204 + Texternal scraper

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32.2


32.2 25.4

= __________ lbs

Assuming tension at the at tail pulley is at least equal to min. tension T0, for % seq -

Options - 2% T4= 6.25 Si (ft) [Wb+ Wm] = _________ lbs

T5= T4+ 100

T3= T4-[0.015 x l4+0.015 x l4x Wb] = __________ lbs

T2= T3- 100

T1= Te+ T2+ 5 x 2 x Belt Width (mm) = _________ lbs

25.4

Profile (i) Horizontal VGT

Test= Gx m1

= _________ lbs





T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2

= T2st+ 100 +x mass of snub pulley x 2.204 + 5 x 2 x Belt width (mm) 32.2 25.4

= __________ lbs

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26/45

T4st= T3st+ [0.015 x l4+ 0.015 x l4x Wb] + x l4(ft) [Wb+ Wt of rotating parts

32.2 of unit return idler]

T5st= T4st+ 100 ++ mass of the bend pulley x 2.204 - (H5x Wb) = _________ lbs

32.2

T6st= T5st+ 150 ++ mass of take up pulley x 2.204 32.2

T7st= T6st+ 100 ++ mass of bend pulley x 2.204 + (H5x Wb) = ________ lbs

32.2

Min. counterweight = (T5st+ T6st) = ___________ lbs


T6= Counter weight = ___________ lbs = T52

T4= T5- 100 + (H5x Wb) = _________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] = _________ lbs

T2= T3- 100

T1= Te+ T2+ 5 x x Belt width (mm) = _________ lbs

25.4

T7= T6+ 100 + (H5x Wb) = ___________ lbs

T8= T7+ 5 x 2 x Belt Width (mm) + (0.015 x l8+ 2.015 x l8x Wb(lbs/ft)= _____ lbs

25.4

T9= T8+ 100 = _________ lbs

Profile (i)

Considering four full wire rope arrangement, at take up tower,

min. counter weight = (T8st+ T7st) x 2 = __________ lbs

Normal running, full load belt tensions :-

T6= Counter weight = __________ lbs = T7

4T5= T6- 100

T4= T5- 100 + (H5x Wb) = ________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] = __________ lbs

T2= T3- 100 = _________ lbs

T1= T2+ Te+ 5 x 2 x Belt width (mm) + [0.015 x l10+ 0.015 x l10x Wb]

25.4

T11= T10+ 100 = ___________ lbs

Profile (ii) Inclined Screw Take up

26


27/45

Test= Gx m1

= _________ lbs

T1st= e0

,= co-efficient of frictionT2st = 0.25 -- for base head pulley



T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)

T1st= Test+ T2st


32.2 = T2st+ 100 +x mass of snub pulley x 2.204 + 5 x 2 x Belt width (mm)

32.2 25.4

= __________ lbs

T4= min. tension with respect to % seq.

T5= T4+ 100 = ___________ lbs

T3= T4+ H5x Wb- 0.015 x l4x Wb= __________ lbs

T2= T3+ 100

T1= T2+ Te+ 5 x 2 x Belt width (mm) = _________ lbs 25.4

Profile (ii) VGT

Test= Gx m1

= _________ lbs


T2st = 0.25 -- for base head pulley = 0.85 -- for lagged pulley

27


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T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs

T4st= T3st+ [0.015 x l4+ 0.015 x l4x Wb] + x l4x [Wb+ wt of rotating parts


- H5x Wb= ___________ lbs

T5st= T4st+ 100 +x mass of bend pulley x 2.204 - H6x Wb= ___________ lbs

32.2

Min. counterweight = T5st+ T6st

Normal running, full load belt tensions :-T5= T6= Counter weight /2 = __________ lbs

T4= T5- 100 + H6x Wb= _________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] + H5x Wb= ___________ lbs

T2= T3- 100

T1= Te+ T2+ 5 x 2 x Belt Width (mm) - H8x Wb+ [0.015 x l10+ 0.015 x l10x Wb]

25.4 = ______________ lbs

T7= T6+ 100 + H7x Wb

T8= T7+ 5 x 2 x Belt Width (mm) - H8x Wb+ [0.015 x l10+ 0.015 x l10x Wb] =25.4 = ___________ lbs

T9= T8+ 100 = ____________ lbs

Profile (ii) HGT (Elecon type)

Test= Gx m1

= _________ lbs


28


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T1st= e0

. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs

T4st= T3st+ [0.015 x l4+ 0.015 x l4x Wb] + x l4x [Wb+ wt of rotating parts


- H5x Wb= ___________ lbs

T5st= T4st+ 100 +x mass of bend pulley x 2.204 - H6x Wb= ___________ lbs

32.2T6st= T5st+ 100 +x mass of bend pulley x 2.204 = _________ lbs

32.2

T7st= T6st+ 150 +x mass of take up pulley x 2.204 = __________ lbs

32.2

T8st= T7st+ 100 +x mass of bend pulley x 2.204 = ____________ lbs

32.2


32.2 + H7x WbMin. counterweight, assuming four fall wire rope arrangement = T6st+ T7st

2

T6= Counter weight/4 = T7= ____________ lbs

T5= T6- 100 = ____________ lbs

T4= T5- 100 + [H6x Wb] = _________ lbs

T3= T4+ H5x Wb- [0.015 x l4+ 0.015 x l4x Wb] = ___________ lbs

T2= T3- 100,

T1= T2+Te+ 5 x 2 x Belt Width (mm) = __________ lbs 25.4

T8= T7+ 100

T9= T8+ 100 + H7x Wb= ______ lbs

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T10= T9+ 5 x 2 x Belt Width (mm)/25.4 - H8x Wb

+ [ 0.015 x l10+ 0.015 x l10x Wb] = _____________ lbs

T11= T10+ 100 = ______________ lbs

Profile (iii) Screw

Test= Gx m1

= _________ lbs





T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)

T1st= Test+ T2st


32.2 = T2st+ 100 +x mass of snub pulley x 2.204 + 5 x 2 x Belt width (mm)

32.2 25.4

= __________ lbs

T4st= min. tension with respect to % seq.

T5= T4+ 100 = ______________ lbs

T3= T4- [0.015 x l4 + 0.015 xl4x Wb] + H5x Wb= ________________ lbs

T2= T3+ 100

T1= T2+ Te+ 5 x 2 x Belt width (mm) = _________ lbs

25.4

VGT :

30


31/45


Test= Gx m1

= _________ lbs





T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs




32.2

T6st= T5st+ 150 ++ mass of take up pulley x 2.204

32.2


32.2


31


32/45


T6= Counter weight = ___________ lbs = T5 2

T4= T5- 100 + (H5x Wb) = _________ lbsT3= T4- [0.015 x l4+ 0.015 x l4x Wb] = _________ lbs

T2= T3- 100


25.4

T7= T6+ 100 + (H5x Wb) = ___________ lbs


25.4

T9= T8+ 100 = _________ lbs

HGT :


T4= T5- 100 + (H5x Wb) = _________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] = _________ lbs

T2= T3- 100


25.4T7= T6+ 100 + (H5x Wb) = ___________ lbs


25.4

T9= T8+ 100 = _________ lbs

T10= T9+ 5 x 2 x Belt Width (mm) + [0.015 x l10+ 0.015 x l10x Wb] - H7x Wb

= ___________ lbs

T11= T10+ 100 = _____________ lbs

After calculating tensions at various points, with respect to profile, proceed with the

follwing :-

Rated belt tensions = ________________ (KN/m) width

= ________________ KN/belt width (m) (eg. 50 KN/m)

ex. = 50 x 103

9.81

= _________ kgs x 2.204 = ___________ lbs (RBT)

Considering starting tensions is limited to 160 % of the rated belt tension :-

Then, allowable extra belt tension :-

32


33/45

Fa1= 1.6 x (RBT) - T1 = _______________ lbs

Time for acceleration = M1[Belt speed (fpm)]

60 Fa1

t = ________________ secsMax. permissible belt tension = 1.6 x RBT = ____________ lbs

Not to exceed, the max. permisible belt tensions, the time used for acceleration

should not be less than t.

- Known value calculated previously.

Starting time of equipment = Belt speed (fpm)

60 x

t1 = ____________ secs

t1> t Conveyor is safe to start fully loaded with the equipment selected.

Coasting time calculations :-

Mass of Conveyor system M = _______________ lbs.sec2/ft

Kinetic energy of system = MW2

2

Retarding force = Effective tension = Te = _________ lbs

Total work performed during deceleration / retardation = Te x v(fpm) x t

2

= MV2

2

Coasting time, t = MV

Te

= ______________ secs

Radius calculations (during starting conditions)-With material load upto point C :

Profile No. (v)

Resistance of the material, as it rides the belt over the idlers :-

Tym = l1x ky1x Wm= __________ lbs

Effective tension, here at point C,

Te = Tx + Tyb + Tym + Tp + Tam + Tac

= ____________ lbs

Material weight upto point C = l1x Wm= ____________ lbs (A)

As calculated previously, Belt weight = ___________ lbsCarrying roller wt = ___________ lbs Let addition of

Return roller wt = ___________ lbs these be X.

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34/45

X = ___________ lbs

Total mass of rotating parts = 1 ( X + A )

(m1) 32.2

Total equivalent mass of conveyor system = m1+ m2

(M) = _____________ lbs.sec2/ ft

Acceleration,= G - Te

M

= ___________ ft/sec2

T10st= T9st+ (Kyreturn side+ Kyreturn sidex Wb)l1- H7x Wb+ Tbc+l1(Wb+ Wm)

2 9

= T9st+ (0.015 + 0.015 x Wb) x l1 only upto point C - H7x Wb+ 5 x

i.e. the first 2

(ft) horizontal portion

x Belt width (mm) xl1 (Wb+ Wm)

25.4 32.2

= ____________ lbs

T11st= T10st+ 150 +x 2.204 x mass of tail pulley = ______________ lbs

32.2

Test = T11stxl1x [Wb+ Wm+ wt. of rotating parts of unit carrying idler (lbs)]32.2

+ l1[Kx+ Ky1(Wb+ Wm)] + Tam+ Tsb= ____________bs

Radius, Rst= 1.11 x Test

Wb= ft= _________m

3.28

Pulley Shaft diameter calculations :-

Profile No. (ii) Inclined :- with VGT/SCREW/HGT,

SCREW

Test= Gx m1

= _________ lbs

34


35/45





T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs

T4st= min. tension with respect to % seq.

T5= T4+ 100 = ______________ lbs

T3= T4- [0.015 x l4 + 0.015 xl4x Wb] + H5x Wb= ________________ lbs

T2= T3+ 100

T1= T2+ Te+ 5 x 2 x Belt width (mm) = _________ lbs

25.4

VGT :


Test= Gx m1

= _________ lbs



35


36/45



T1st= e0. T2st


o- T2st

= T2st(e0- 1)

T2st= Test/(e0- 1)


32.2


32.2 25.4

= __________ lbs




32.2

T6st= T5st+ 150 ++ mass of take up pulley x 2.204

32.2


32.2




T4= T5- 100 + (H5x Wb) = _________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] = _________ lbs

T2= T3- 100


25.4

T7= T6+ 100 + (H5x Wb) = ___________ lbs


25.4

T9= T8+ 100 = _________ lbs

36


37/45

HGT :


T4= T5- 100 + (H5x Wb) = _________ lbs

T3= T4- [0.015 x l4+ 0.015 x l4x Wb] = _________ lbs

T2= T3- 100


25.4

T7= T6+ 100 + (H5x Wb) = ___________ lbs


25.4

T9= T8+ 100 = _________ lbs

T10= T9+ 5 x 2 x Belt Width (mm) + [0.015 x l10+ 0.015 x l10x Wb] - H7x Wb

= ___________ lbs

T11= T10+ 100 = _____________ lbs

After calculating tensions at various points, with respect to profile, proceed with thefollwing :-

Rated belt tensions = ________________ (KN/m) width

= ________________ KN/belt width (m) (eg. 50 KN/m)

ex. = 50 x 103

9.81

= _________ kgs x 2.204 = ___________ lbs (RBT)

Considering starting tensions is limited to 160 % of the rated belt tension :-

Then, allowable extra belt tension :-

Fa1= 1.6 x (RBT) - T1 = _______________ lbs

Time for acceleration = M1[Belt speed (fpm)]

60 Fa1 t = ________________ secs

Max. permissible belt tension = 1.6 x RBT = ____________ lbs

Not to exceed, the max. permisible belt tensions, the time used for acceleration

should not be less than t.

- Known value calculated previously.

37


38/45

Starting time of equipment = Belt speed (fpm)

60 x

t1 = ____________ secs

t1> t Conveyor is safe to start fully loaded with the equipment selected.Coasting time calculations :-

Mass of Conveyor system M = _______________ lbs.sec2/ft

Kinetic energy of system = MW2

2

Retarding force = Effective tension = Te = _________ lbs

Total work performed during deceleration / retardation = Te x v(fpm) x t

2

= MV2

2

Coasting time, t = MV

Te

= ______________ secs

Radius calculations (during starting conditions)-With material load upto point C :

Profile No. (v)

Resistance of the material, as it rides the belt over the idlers :-

Tym = l1x ky1x Wm= __________ lbsEffective tension, here at point C,

Te = Tx + Tyb + Tym + Tp + Tam + Tac

= ____________ lbs

Material weight upto point C = l1x Wm= ____________ lbs (A)

As calculated previously, Belt weight = ___________ lbs

Carrying roller wt = ___________ lbs Let addition ofReturn roller wt = ___________ lbs these be X.

X = ___________ lbs

Total mass of rotating parts = 1 ( X + A )

(m1) 32.2

Total equivalent mass of conveyor system = m1+ m2

(M) = _____________ lbs.sec2/ ft

Acceleration,= G - Te

M

= ___________ ft/sec2

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39/45

T10st= T9st+ (Kyreturn side+ Kyreturn sidex Wb)l1- H7x Wb+ Tbc+l1(Wb+ Wm)

2 9

= T9st+ (0.015 + 0.015 x Wb) x l1 only upto point C - H7x Wb+ 5 x

i.e. the first 2

(ft) horizontal portion

x Belt width (mm) xl1 (Wb+ Wm)

25.4 32.2

= ____________ lbs

T11st= T10st+ 150 +x 2.204 x mass of tail pulley = ______________ lbs

32.2

Test = T11stxl1x [Wb+ Wm+ wt. of rotating parts of unit carrying idler (lbs)]

+ l1[Kx+ Ky1(Wb+ Wm)] + Tam+ Tsb= ____________bs

Radius, Rst= 1.11 x Test

Wb= ft= _________m

3.28

Drive Pulley Shaft :-

Tensions at drive pulley :- T1= __________ N

(Converting it to Kgs, by dividing N by 9.81)

T2= __________ Kgs

Resultant force acting on the head pulley :-

R1= [T1cos+T2cos(-180-)]2+[T1sin-T2sin(-180-)+Drive pulley weight(kg)]

2

= __________________ Kgs

39


40/45

where,is always equal to conveyor inclination angle


Bending moment, M = R x arm length (a) cm = ___________ Kg.cm

2

T = Torque at pulley shaft = T1- T2x pulley dia (cm) = __________ Kg. cm

2

Equivalent twisting moment, Tw= (Ktx T)2+ (Kbx M)

2= ___________ Kg. cm

Kt= Service factor for torsion = 1.0 & Always

Kb= Service factor for bending = 1.5

Equivalent bending moment, Me= b(M + Tu) = _____________ Kg. cm

Considering shaft material, as C45/EN8 or equivalent :-

Allowable bending stress = fb= 800 Kg/cm2

Allowable shear stress = fs= 650 Kg/cm2

Diameter due to bending moment = dm = 3 32 x Me/TT x bending stress

= _______________ cm

See, which value is greater & then decide the following :-

To consider key effect, increase shaft dia as follows :-

Greater value among the two shaft dia (let it be O) + O = __________ cm x 10

2

= ______________ mm

Adopted shaft dia at brg = (Next available brg in our database) = ________ mcm

at hub = This value + 10 = ________ mcm

Now, checking for deflection due to bending & torsion as well, as per following

procedure :-

Checking for bending deflection :-

Deflection at centre,c = M [ 3lbr2- 4a2]

24EI

40


41/45

I =x (dia at hub)4= ___________ cm4

64

E = 2.1 x 106Kg/cm2

lbr = Brg centres __________ cm, a = arm length = __________ cm

Permissible deflection = lbr = ___________1200

c < permissible deflection (Display message safe/unsafe)

Checking for torsional deflection :-

Angular twist,= 584 x lbr x T G = 0.84 x 106Kg/cm2

G x (dia at hub)4

Permissible twist = 0.08 x 3.28 x lbr (m)(Take care, while putting into program,

lbr is in metres).

< permissible twist, - Dispaly message

Checking for bearing life :-

Brg type :- Display No. (brg + adapter sleeve)

[After displaying type of brg (always double row spherical roller brg)]

Basic dynamic capacity = C = ___________ Kgs

Bearing load = P = 1.2 x Radial load

= 1.2 x R = _________ Kgs

2

Bearing life in millions of revolutions = Lf= (C)10/3

P

= _____________ mill. revolutions

Bearing life in hours = Lfx 106 (Take value calculated during selection

60 x Pulley rpm of gear box ratio)

For Profile Nos. (I), (iv) & (v) - head pulley shaft - calculations

T1 T1= __________ Kgs

T2 T2= __________

W

R = [T1+T2cos(-180)]2+ [-T2sin(-180)+Drive pulley wt (Kgs)]

2= _________ Kgs

Next, each & every step remains the same as explained above for the remaining

profiles.

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Snub pulley shaft dia calculations :-

For profile Nos (ii), (iii) & (vi)

R = [T3cos-T2cos(-180-)]2+[T3sin+T2sin(-180-)+snub pulley wt(kgs)]

2

= ______________ Kgs

M = R x arm length (cm) = ______________ Kg.cm 2

Kb= 1.5

fb= 800 Kg/cm2

dm =332 x M x 1.5/x fb = ______________ cm x 10 = ______________ mm

Adopt -----> next available brg, dia, = _____________ mm (at brg)

+ 10 mm = ______________ shaft dia at hub

Check for bending due to deflection only & calculate bearing life, as explained &

calculated on the previous pages.

For profile nos. (I), (iv) & (v)

R2= (T3- T2cos(-180)]2+[T2sin(-180) + snub pulley wt (kg)]

2

M = R x arm length (cm) = _______________ Kg. cm

2

Kb= 1.5

fb= 800 Kg/cm2`

dm = 332 x M x 1.5/x fb = _________ cm x 10 = ____________ mm

Adopt -------> next available brg dia = ____________ mm (at brg)

+ 10 mm = _____________ shaft dia at hub.

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43/45

Check for bending due to deflection only and calculate bearing life, as explained and

calculated earlier.

For profile nos (I), (iv) & (v)

(-180) R = (T3-T2cos(-180)]2+[T2sin(-180)+snub pulley wt(kg)]

2

T2 W

VGT - There are two bend pulleys. In case of horizontal conveyors with Vgt, wrap

angle over both pulleys is same. But in case of inclined conveyors, wrap angle over

bend pulley nearer to snub pulley is less than, that over bend pulley nearer to tail

side. Hence in such cases, shaft dia is calculated, taking into consideration thegreater wrap angle and same is applicable to other bend pulley. Also in case of profile

nos (iii) and (iv) is assumed that, Vgt will always be placed in the inclined portion only,

near the head pulley.

For profiles (ii), (iii) & (iv)

T7 T6= __________, T7= __________ Kgs

R3= [(T7cos)]2+ [T7sin+ T6+ Bend pulley wt (kg)]

2

W T6 = ________________ Kgs

Remaining calculations same as those for snub pulley as done above.

For profiles (I), (iv) & (v)

R = (T7)3+ [T6+ Bend Pulley wt (kg)]

2 = _____________ Kgs

T6Bend pulley shaft diameter if type of take up is HGT

For profiles (ii), (iii) & (vii)

Please note :- This angle isfor profile nos (ii) & (iii) but it is

2(i.e. inclination 2) for profile (vi) because, it

T9 contains two inclined portions.

W T8

R = (T9cos)2+ [T9sin+ W(bend pulley wt kgs) + T8]

2----> For profiles (ii) & (iii)

For profile (iv) replaceby2(inclination of 2ndportion)

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44/45

Profiles (I), (iv) & (v)

T9 R = (T9)

2+ (T8+ Bend pulley wt)2

W T8Rest is same.

Take up pulley shaft dia calculations :-

For VGT - and all profiles

T6 T5 R = (-T5- T6)2+ [Take up pulley weight (kgs)]2= _________ Kgs

W

For HGT and all profiles

T7R = (-T6- T7)2+ (Take up pulley wt (kgs))2= __________ Kgs

T6

W

Rest of the procedure for calculation of shaft dia is same as explained and calculated

above.

Tail pulley shaft diameter calculations :-

For profiles (ii), (iv) & (vi)

T5Profile (ii) & (iv) screw

T4R = (- T4cos- T5cos)2+ (weight of tail pulley (kgs) -

T4sin- T5sin)2

= ______________ kgs

W

In place of T4and T5, use following tensions in case of

VGT T4= T8& T5T9

HGT T4= T10& T5 T11

For profiles (i), (iii) & (iv)

44


45/45

T5R = (-T4-T5)2+ [Tail pulley wt (kgs)]2 = ____________ kgs

T4

W

Use the following tensions in case of,

VGT T4 T8, T5 T9HGT T4 T10, T5 T11

Documents

BELTCONV[1] - CEMA