Tsubaki Products

Technical Data Large size conveyor chain Selection

9. Calculate chain load

Maximum static load to chain, T_max, during operation can be calculated using the formulae in Table 3.

The formulae are based on mass M (weight W) × coefficient of friction.

Inertial forces are extremely large when suddenly starting or stopping high speed conveyors or when rapidly conveying items using push conveyors or other such systems. Bear these inertial forces in mind when calculating the load and required kW.

Calculations are listed in both SI units and gravimetric units. When calculating tension T in gravimetric units, the mass value (kgf) is the same as the mass value for SI units (kg).

9.1 Description of terms

		SI Unit	Gravity unit
T_MAX	Maximum static load on chain	kN	{kgf}
T'_MAX	Design chain tension	kN	{kgf}
T	Static load on chain	kN	{kgf}
Q	Maximum conveying quantity	t/h	{tf/h}
V	Conveying speed (Chain speed )	m/min	m/min
H	Center distance between sprockets (vertical direction )	m	m
L	Center distance between sprockets (horizontal direction )	m	m
C	Center distance between sprockets	m	m
M	Mass {weight} of moving parts {Weight } (Chain × strands, buckets, aprons, etc. {Weight })	kg/m	{kgf/m}
f1	Coefficient of friction between chain and guide rail (Table 5, Table 6)
f2	Coefficient of friction between material conveyed and casing (Table 7)
f	Material loaded directly on chain 　f=1　Material scraped 　f=f2/f1
g	Gravitational acceleration 　9.80665m/s²
W	Weight of conveyed goods {Weight } Bulk items 　W=16.7^Note×Q/V　{W=16.7×Q/V} Solid Conveyance 　W=Weight of conveyed goods (kg/piece(s))/Loading interval (m) 　{W=Mass Conveyed Item (kgf/each )/Loading interval (m)}	kg/m	{kgf/m}

Note) The coefficient for calculating the mass (weight) per meter of item conveyance is 16.7=1000/60

※If frequent forward and reverse operation is required, take-up is necessary to remove chain slack, so the calculation differs from the below.
When removing slack in a chain by take-up, please use the calculation formula in Q&A6.

9.2Calculate Chain Load (Table 3)

SI Unit

{Gravity unit }

Horizontal conveyance

T₁ = 1.35 ^※1 × M × L₁ × g 1000 ......kN

T₂ = ( L - L₁) × M × f₁ × g 1000 + T₁ ......kN

T₃ = 1.1 ^※2 × T₂ ......kN

T_MAX = (W × f + M ) × L × f₁ × g 1000 + T₃ ......kN

T₁ = 1.35 × M × L₁ ......{kgf}

T₂ = ( L - L₁) × M × f₁ + T₁ ......{kgf}

T₃ = 1.1 × T₂ ......{kgf}

T_MAX = (W × f + M ) × L × f₁ + T₃ ......{kgf}

※1 Refer to Table 4 below
※2 1.1 is for increased load at the driven sprocket.

Horizontal conveyance

T₁ = 1.35 × M × L₁× g 1000 + 0.1^※ × M × L × g 1000 ......kN

T₂ = 1.1 × T₁ ......kN

T_MAX = (W × f + M) × L × f₁ × g 1000 + T₂ ......kN

T₁ = 1.35 × M × L₁ + 0.1^※ × M × L ......{kgf}

T₂ = 1.1 × T₁ ......{kgf}

T_MAX = (W × f + M) × L × f₁ ......{kgf}

※0.1 is the coefficient of roller resistance at return side.

Vertical conveyance

Note) In bucket elevators, 1m is added to center distance (H) to account for shock load when loading.
W_T：Take-up load. Due to the pulley working principle, 1/2 of the take-up load is applied per chain.

T_MAX = (W + M) × H × g 1000 + W_T 2 × g 1000 ......kN

T_MAX = (W + M) × H + W_T 2 ......{kgf}

Inclined conveyance

T₁ = M(Lf₁ - H) × g 1000 ......kN

When T₁<0, T₂=0

T₂ = 1.1 × T₁ ......kN

T_MAX = W(Lf₁ × f + H) × g 1000 + M(Lf₁ + H) × g 1000 + T₂ ......kN

T₁ = M(Lf₁ - H) ......{kgf}

When T₁<0, T₂=0

T₂ = 1.1 × T₁ ......{kgf}

T_MAX = W(Lf₁ × f + H) + M(Lf₁ + H) + T₂ ......{kgf}

Practical Example

T₁= M × L₁ × f₁ × g 1000 ......kN

T₂= T₁× Kc₁ ......kN

T₃= M(L₂f₁ - H) × g 1000 + T₂ ......kN

T₄= T₃× Kc₂ ......kN

When T₃<0, T₄=0

T₅= M × L₃ × f₁ × g 1000 + T₄ ......kN

T₆= 1.1 × T₅ ......kN

T₇= (M + W × f) × L₄ × f₁ × g 1000 + T₆ ......kN

T₈= T₇× Kc₃ ......kN

T₉= W(L₅f₁ × f + H) × g 1000 + M(L₅f₁ + H) × g 1000 + T₈ ......kN

T₁₀ = T₉× Kc₄ ......kN

T_MAX = (M + W × f) × L₆ × f₁ × g 1000 + T₁₀ ......kN

Corner Factor Kc

f1	Angle
f1	30°	60°	90°	120°	180°
0.03	1.02	1.03	1.05	1.06	1.10
0.10	1.05	1.11	1.17	1.23	1.37
0.15	1.08	1.17	1.27	1.37	1.60
0.20	1.11	1.23	1.37	1.52	1.87
0.24	1.13	1.29	1.46	1.65	2.13
0.30	1.17	1.37	1.60	1.87	2.57
0.40	1.23	1.52	1.87	2.31	3.51

T₁ = M × L₁ × f₁ ......{kgf}

T₂ = T₁ × Kc₁ ......{kgf}

T₃ = M(L₂f₁ - H) + T₂ ......{kgf}

T₄ = T3 × Kc₂ ......{kgf}

When T₃<0, T₄=0

T₅ = M × L₃ × f₁ + T₄ ......{kgf}

T₆ = 1.1 × T₅ ......{kgf}

T₇ = (M + W × f) × L₄ × f₁ + T₆ ......{kgf}

T₈ = T₇ × Kc₃ ......{kgf}

T₉ = W(L₅f₁ × f + H) + M(L₅f₁ + H) + T₈ ......{kgf}

T₁₀ = T₉ × Kc₄ ......{kgf}

T_MAX = (M + W × f) × L₆ × f₁ + T₁₀ ......{kgf}

Example using Double Plus Chain

T_MAX = 2.1M(L₁ + L₂) f₁ × g 1000 + (W × L₁ × f₁)
× g 1000 + (W₁ × L₂ × f₃ × g 1000 ) ......kN

T_MAX = 2.1M(L₁ + L₂) f₁ + (W × L₁ × f₁) + (W₁ × L₂ × f₃) ......{kgf}

L₁： Length of conveying portion (m)
L₂： Length of accumulating portion
W₁： Mass of conveyed objects in accumulating portion (kg/m){Mass kgf/m}
f₁： Coefficient of Friction Between Chain and Rail During Conveyance (0.05)
f₃： Coefficient of friction during accumulation (0.2)

Chain load T for calculating required power can be obtained from the following formulae:

Horizontal T = T_MAX - T₁

Vertical T = T_MAX - MH × g 1000

Inclined T = T_MAX - M(H - Lf₁) × g 1000

T = T_MAX - MH

T = T_MAX - M(H - Lf₁)

When H - Lf₁ < 0, T = T_MAX

・Calculating Required Power

1kW = 1kN・m/s

kW = T × V 60

1kW = 102kgf・m/s

kW = T × V 102 × 60

Assuming that the power loss from chain–sprocket engagement and sprocket rotational friction resistance to be 10% (1/0.9 = 1.1)

When the power transmission ratio of the drive section is η,

kW = T × V 60 × 1.1 × 1 η

kW = T × V 102 × 60 × 1.1 × 1 η

Table 4. Catenary Load Graph

Catenary Load

T₁ = 1.35 × M × L₁ × g 1000 ......kN

1.35 factor in the above formula is worked out as follows:

When catenary sag is 10%, δ = 0.10L₁

Then from the graph
δ L₁ = 0.10 → 2T₁ ML₁ = 2.7
T₁ = 1.35 × M × L₁ × g 1000

Chain Length at Catenary

δ L₁ = 0.10 → S L₁ = 1.027

S = 1.027L₁

Table 5.Rolling Friction Factor f1 Between Chain and Rail f₁

Roller Diameter (mm)	Lubricated		Dry
Roller Diameter (mm)	R, F-roller	S,M,N-roller	R, F-roller	S,M,N-roller
D＜65	0.08	0.16	0.15	0.24
65≦D＜100	0.08	0.15	0.14	0.23
100≦D	0.08	0.14	0.13	0.22
RF-214(exception)	0.12	0.15	0.18	0.22

Note)

1. Lubricant ISO VG100–150 (SAE30–40)
2. Conditions: Clean and room temperature
3. The friction factor f₁ between top roller and material conveyed is the same as that of R roller in the above.

Series	f₁
Plastic Roller Series	0.08(No lubrication required )
Bearing Roller Series	0.03(Lubricated )
Bearing Bush Series	0.14(Lubricated ), 0.21(Dry )
EPC78	0.1(Lubricated ), 0.2(water-lubricated ), 0.25(Dry )

Table 6.Sliding Coefficient of Friction f1 Between Chain and Rail f₁

Temperature of Conveyed Material ℃	Lubricated	Dry
Ambient Temperature ～400	0.20	0.30
400～600	0.30	0.35
600～800	0.35	0.40
800～1000	-	0.45

Table 7.Coefficient of Friction f2 Between Material Conveyed and Casing f₂

Conveyed object	f2	Apparent Specific Gravity
Scale	0.67	1.54
Red iron ore	0.47	2.99
Pyrites	0.58	1.54
Slag	0.48	0.90
Scrap	0.73	0.54
Lead ore powder	0.77	3.26
Zinc ore powder	0.79	1.93
Nickel ore powder	0.45	0.92
Chrome powder	0.51	1.14
Alumina	0.55	0.83
Magnesia	0.84	1.48
Gypsum	0.64	0.77
Quartz powder	0.55	1.24
Feldspar	0.55	1.36
Dolomite	0.55	1.62
Clay	0.63	0.77
Foundry Sand	0.41	1.59
Phosphate rock	0.42	1.51
Quicklime	0.46	1.53
Slaked lime	0.63	0.69
Asbestos	0.58	0.19
Limestone	0.47	0.35～0.55
Cement	0.54	0.60～0.75
Cement Clinker	0.46	1.30
Charcoal	0.41	0.44
Carbon	0.53	0.30
Pitch	0.41	0.70
Soda Ash	0.45	0.52
Alum	0.63	1.01
Polyethylene	0.52	0.34
Rubber powder	0.53	0.39
Soap material	0.27	0.65
Urea	0.63	0.64
Ammonium chloride	0.79	0.67
Calcium chloride	0.43	0.68
Sulphurated calcium	0.64	1.01
Calcium carbonate	0.49	0.88
Wood chip	0.74	0.36
Rice	0.4	0.77
Barley	0.71	0.39
Wheat	0.43	0.73
Soybean	0.41	0.68
Corn	0.4	0.71
Starch	0.57	0.71
Sugar	0.47	0.68
Rock Salt	0.57	1.09
Mixed Feed	0.5	0.55
Coal	-	0.30～0.70
Coke	-	0.30～0.70