CHAPTER 3: CALCULATE AND DESIGN SCREENING SYSTEM
3.4 Calculate the transmission system
3.4.1. Calculate and design the chain system
Due to low velocity, the working condition, the parameters of the chain such as power, the revolution, working environment, tensile force, and the expectation of life when the chain is working, so that we can choose various kinds of chain as rolling chain.
The input parameters of the chain transmission are:
✓ P𝑚 = 11 (kW)
✓ n𝑚 = 970 (rpm)
✓ u𝑐ℎ𝑎𝑖𝑛 𝑑𝑟𝑖𝑣𝑒 = 2
3.4.1.1 Choose the number of teeth for driving and driven sprockets:
{z1 = 29 − 2u = 25 teeth z2 = z1∙ u = 50 teeth
Choosing Z2 = 50 teeth as satisfying the condition Zmax ≤ 120 teeth
3.4.1.2 Determine the conditional coefficient for using chain drive according to the formular in page 83, document [1] :
K = K0∙ Ka ∙ Kdc ∙ Kb∙ Kr∙ Klv = 1,2 ∙ 1,25 ∙ 1 ∙ 1 ∙ 1 ∙ 1 = 1,5 (3.23) By which:
All of coefficients are looked up in the table 4.11, page 84, document [1] :
Ko = 1 - The coefficient takes into account the influence of the transmitter position: when the line connecting the two centers of the sprocket merges with the horizontal line at a smaller angle 60°
Ka = 1,25 - The coefficient takes into account the effect of axial distance or chain length, the longer the chain, the fewer times each link is matched in a unit of time, so the less wear the chain will wear.
Kdc = 1 - The coefficient takes into account the effect of the ability to adjust the chain tension: the shaft is adjustable.
Kb = 1 - Coefficient considering lubrication conditions: drip lubrication.
Kr = 1,2 - Dynamic load factor: load with impact.
Kbt = 1 - The coefficient takes into account the working mode: working one shift.
3.4.1.3 Calculating power
Pt =
K ∙𝑧01 𝑧1 ∙𝑛01
𝑛1 ∙ 𝑃
𝐾𝑥 ≤ [P] (3.24)
By which:
K𝑧 =Z01
Z1 =25
11 = 2.27: sprocket tooth coefficient Kn =𝑛01
𝑛1 =600 500
= 1.2: rotation coefficient, the value collected in table 4.12, document [1] , page 85
46 Kx = 1: single roller chain
P = 10.36 (kw): computing power
Looking up the table 4.12, in page 85 document [1] , with 𝑛01 = 600(rpm), I get:
⇒ Pt ≤ [P] with 18.65 < 25.7
⇒ { pc = 12.7 [P] = 3.98 (kw)
3.4.1.4 Determine the average velocity v of chain and tangential force 𝐹𝑡: v = n1pcz1
60000 =500 ∙ 12.7 ∙ 25
60000 = 2.65 (m/s) Ft =1000 ∙ P
v =1000 ∙ 0.01036
2.65 = 3.92 (N) 3.4.1.5 Calculating and check the chain pitch
pc ≥ 600√ P1∙ K z1∙ n1∙ [P0] ∙ Kx
3 (3.25)
By which:
P1 = 10.36 (kW) = 0.01036 (W): computing power
𝑧1 = 25 (teeth): The number of teeth of driving sprocket
[p0] = 26 (MPa): The allowable pressure is collected via the table 4.14 in page 86 document [1] , when n1 = 600(rpm) and pc = 12.7
⇒ pc > 2.05 so the chosen chain pitch satisfy the condition
3.4.1.6 Choose the preliminary axle distance and calculate the number of chain links With pc = 12.7 then a = (30 ÷ 50) ∙ pc = 400 (mm)
✓ The number of chain link is:
X =2 ∙ a
Pc +Z1+ Z2
2 + (Z1− Z2 2π )
2
∙Pc
a (3.26)
By which:
{z1 = 25 (teeth) z2 = 50 (teeth)
⇒ 𝑋 ≈ 100.99 Choose X = 100 (links)
Then the length of chain drive is L = X ∙ pc = 1270 (mm)
✓ Recalculate the axle distance a = 0,25 ∙ Pc[X −Z1+ Z2
2 + √(X −Z1+ Z2
2 )
2
− 8 ∙ (Z1+ Z2 2π )
2
] (3.27)
a = 0,25 ∙ 12.7 [76 −11 + 22
2 + √(76 −11 + 22
2 )
2
− 8 ∙ (22 − 11 2π )
2
]
47
⇒ a = 364 (mm) when the value is decreased by (0.002 ÷ 0.004)a
3.4.2. Checking the allowable safety factor and checking the number of chain bumps in a second
3.4.2.1 Checking the allowable safety factor
s = Q
Ft + Fv+ Fo ≥ [s] (3.28)
By which:
Checking the table 4.16 in page 87 document [1] , I get Q = 22,7 (kN): Destructive load
Fv = qm∙ v2 = 0.9 ∙ 2.652 = 6.32025 (N) qm is the mass of 1m chain (kg)
FO = Kf∙ a ∙ qm∙ g = 3 ∙ 0.364 ∙ 0.9 ∙ 9.81 = 9.6413(N) Via the table 4.15 in page 87 document [1] , I have:
[s] = 9.3 when { pc = 12.7 (mm) n = 749.62 (rpm)
⇒ s = 22700
11.93 + 1.21104 + 9.959 = 1142.02 ≥ 9.3
⇒ s ≥ [s] so that the allowable safety factor is satisfied.
3.4.2.2 Checking the number of chain bumps in a second i =4 ∙ V
L =4 ∙ n1∙ Z1∙ Pc
Pc∙ X ∙ 60 =n1∙ Z1
15 ∙ X =500 ∙ 25
15 ∙ 100 = 8.3 ≤ [i] = 40 (3.29) By which:
⌈i⌉ = 40 (per second) is collected via the table 4.17 in page 88 document [1]
3.4.3. Calculate the applying force on shaft
Fr = Km. Ft = 1,15 ∙ 3.92 = 4.508(N) (3.30) By which:
Km = 1,15: chain weight coefficient when the chain is horizontal or when the angle of inclination between the centerline of the axis and the horizontal is less than 40o.
3.4.4. Determine the parameters for the chain transmission - Pitch dimeter
{
Driving sprokect: d1 = pc sin180
𝑧1
≈ 101.33 (mm)
Driven sprocket: d2 = pc sin180
𝑧2
≈ 202.26 (mm) - Addendum diameter
48 {
Driving sprokect: da1 = pc∙ (0.5 + cotg(π
z1)) ≈ 106.88 Driven sprocket: da2 = pc∙ (0.5 + cotg(π
z2)) ≈ 208.21 Calculate and Design
Parameters Values Parameters Values
Type of chain Roller chain Pitch diameter Driving sprocket, mm Driven sprocket, mm
101.33 202.26 Pitch chain, mm 12.7
Axle distance, mm 364 Addendum diameter Driving sprocket, mm Driven sprocket, mm
106.88 208.21 Length of chain, mm 1270
Number of links 100 The number of teeth:
Driving sprocket 𝐳𝟏 Driven sprocket 𝐳𝟐
25 50
Dedendum diameter Driving sprocket, mm Driven sprocket, mm
88.63 182.26 Force on shaft 𝐅𝐫, N 4.508
Calculate and Check Parameters Allowable
values
Computed values Comments The number of driving
revolution 𝐧𝟏, 𝐫𝐩𝐦 600 500 Satisfy
The number of bumps 8.3 40 Satisfy
Safety factor s 1142.02 9.3 Satisfy
Table 3. 6: Parameters for chain transmission 3.4.5. Calculate and check the bearings
Since there is no axial force applied to the bearing, so I choose the ball bearing that supports a row for all shafts.
3.4.5.1 Calculate and check the bearings for driving shaft
According to the appendix I.3.1 in page 647 document [1] , with d = 20mm I choose the bearing with low load.
With Fr = 1800 (N).
Applying the formula 9.10 in page 230 document [1] :
Q = (XVFr + YFa)KtKσ = 2340(N) (3.31) TCVN 189-85
Signal d D T B r C 𝑪𝟎
204 20 47 14 14 1.5 10 6.3
49 By which:
Following the table 9.4 in page 231 document [1] , i have: {X = 1 Y = 0 V = 1 as the innner ring rotates
K𝜎 = 1.3: characteristic of load, checking from the table 9.3, in page 230 document [3]
Kt = 1 when t ≤ 1000C: coefficient affected by temperature, in page 231 document [1]
3.4.5.2 The total of working hours
The working life of bearings when the load change constantly:
LhE = KHE∙ L = 0.25 ∙ 20000 = 5000(hours) (3.32) The working life of bearings calculated by the number of revolutions:
Lℎ =60Lℎn
106 = 75(millionrevolutions) (3.33) By which:
KHE = 0.25: collected via table 6.14 document [3]
3.4.5.3 Checking the ability of dynamic load:
Ctt = Q0m√L = 2340. √753 = 9868N < 10000N (3.34) And: Q < 6.3kN
⇒ Bearings satisfy dynamic load and static load condition.
50