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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
snub
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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
Effective force during starting,
Test= Gx m1
= _________ lbs
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T1st= e0,= co-efficient of friction
T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
o- T2st
= T2st(e0- 1)
T2st= Test/(e0- 1)
T1st= Test+ T2st
T3st= 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
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
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] = _____________ 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
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 pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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
T1st= e0,= co-efficient of friction
T2st = 0.25 -- for base head pulley = 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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+ 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
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T6st= T5st+ 100 +x mass of bend pulley x 2.204 = ________ lbs
32.2
T7st= T6st+ 160 +x mass of bend pulley x 2.204 = ________ lbs
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,= co-efficient of friction
T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st
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= T2st.eo- 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+ 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
T6st= T5st+ 100 +x mass of bend pulley x 2.204 = ________ lbs
32.2
T7st= T6st+ 160 +x mass of bend pulley x 2.204 = ________ lbs
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,= co-efficient of friction
T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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
= T2st+ 100 +x mass of snub pulley x 2.204 + 5 x 2 x Belt width (mm)
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,= co-efficient of friction
T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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
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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
Normal running full load belt tensions :-
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
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Test= Gx m1
= _________ lbs
T1st= e0
,= co-efficient of frictionT2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
o- T2st
= T2st(e0- 1)
T2st= Test/(e0- 1)
T1st= Test+ T2st
T3st= 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
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
T1st= e0,= co-efficient of friction
T2st = 0.25 -- for base head pulley = 0.85 -- for lagged pulley
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= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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+ 0.015 x l4x Wb] + x l4x [Wb+ wt of rotating parts
32.2 of unit return idler]
- 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
T1st= e0,= co-efficient of friction
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T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0
. T2st
But Test= T1st- T2st = T2st.e
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+ 0.015 x l4x Wb] + x l4x [Wb+ wt of rotating parts
32.2 of unit return idler]
- 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
T9st= T8st+ 100 +x mass of bend pulley x 2.204 = __________ lbs
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,= co-efficient of friction
T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
o- T2st
= T2st(e0- 1)
T2st= Test/(e0- 1)
T1st= Test+ T2st
T3st= 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= 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 :
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Profile (i) Horizontal VGT
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 pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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+ 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
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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
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
HGT :
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
25.4T7= 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
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 :-
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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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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
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T1st= e0,= co-efficient of friction
T2st = 0.25 -- for base head pulley
= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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= 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 :
Profile (i) Horizontal VGT
Test= Gx m1
= _________ lbs
T1st= e0,= co-efficient of friction
T2st = 0.25 -- for base head pulley
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= 0.85 -- for lagged pulley
= Wrap angle over head pulley
T1st= e0. T2st
But Test= T1st- T2st = T2st.e
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+ 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
Normal running full load belt tensions :-
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
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HGT :
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
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.
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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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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
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where,is always equal to conveyor inclination angle
= Wrap angle over head pulley
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
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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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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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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)
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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
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