5. Analysis of the chain drive
Let us carry out the analysis of the chain drive for strength if torque at the driving sprocket T1 = 209.192 N·m; torque at the driven sprocket T2 = 405.215 N·m; rotational speed of the driving sprocket n1 = 272.676 rpm,rotational speed of the driven sprocket n2 = 121.028 rpm; velocity ratio of the chain drive ucd = 2.253; input power of the chain drive P1 = 5.97 kW.
5.1. Determine the number of teeth of the driving sprocket
z1 = 31- 2ucd17
In our case z1 = 31- 22.253 = 26.494
Assume z1 = 26 > 17.
5.2. Determine the number of teeth of the driven sprocket
z2 = z1ucd 120
z2 = 262.253 = 59.578
Assume z2 = 59 < 120.
5.3. Specify the velocity ratio and determine the error
.
The error should be ε 4 %.
In our case ε = 0.71%.
5.4. Determine the service factor Ks
Ks= KKaKlubKKdKten,
where K is used to take into account the load nature and taken as 1 in quiet operation and as 1.2 to 1.5 in the case of shocks and impacts; Ka is the center distance factor assumed as Ka=1 for a = 30t to 50t and Ka=0.8 for a = 60t to 80t; Klub is lubrication factor (Klub=0.8 for immersion lubrication, Klub=1 for drop-feed lubrication and Klub=1.5 for periodic greasing); K accounts for the angle that the shaft centre line makes with the horizontal (K=1 for 60˚ and K=1.25 for > 60˚); Kd is a duty factor (Kd=1 for one-shift operation, Kd=1.25 for two-shift operation and Kd=1.5 for three-shift operation); Kten accounts for the manner of tension control (Kten=1 for drives with chain tighteners, Kten=1.15 for drives with adjustable bases, and Kten=1.25 for fixed-base drives).
In our case we have small shocks and impacts (K=1.2); the center distance is a = 40t(Ka=1); periodic greasing (Klub=1.5); 60˚ and (K=1); one-shift operation (Kd=1); fixed base drive (Kten=1.25).
Ks= 1.2·1·1.5·1·1·1.25 = 2.25
5.5. Approximately determine the allowable mean pressure on the hinges depending on the rotational speed of the smaller sprocket.
In
our case for rotational speed n1=272.676
rpm,
20
MPa.
5.6. Determine the chain pitch
,
where T1 is in Nmm.
Round off the pitch to the nearest standard valu. In our case assume t= 31.75 mm.
5.7. Specify the allowable mean by interpolation [p]. On multiplying it by Kp = 1 + 0.01·(z1-17) = 1 + 0.01·(26-17)=1.09 we get finite magnitude of [p] = 20·1.09=21.8MPa
5.8. Determine the effective mean pressure.
For that we
- find chain speed
;
- find turning (tangential) force
;
- look up the projected hinge area Sh. In our case Sh=179.7 mm2
Then the effective mean pressure
.
If this inequality is not right it is necessary to increase the pitch t.
.
In our case the inequality is right.
5.9. Determine the number of links in the chain
,
where
is the chain length in pitches;
;
a
(30…50)·t ; z
= z1
+ z2
.
Round off obtained magnitude to even integer numeral.
In
our case: at
= 45; z
= 26 + 59 = 85;
;
5.10. Specify the center distance
The slack side of the chain should have a slight sag f 0.01·a, for which purpose the design center distance is reduced by 0.2 to 0.4 %.
Assume that a = 1438.598 mm.
5.11. Determine the pitch diameters
-
of the driving sprocket
;
-
of the driven sprocket
.
5.12. Determine the addendum diameters
where d1 is the roller diameter.
5.13. Determine the dedendum diameter
,
.
5.14. Determine the web thickness of sprocket
C = 0.93∙Bbush= 0.93∙19.05 = 17.72mm
WhereBbush is determined according to table 8.2.
5.15. Determine forces acting to the links
- turning force ;
- centrifugal force Fc = q V2= 3.8 3.7512 = 53.466 N, where q is the mass per meter run of the chain in kg
- load due to chain deflection Ff = 9.81Kfqa
Ff = 9.81 1 3.8 1.423 = 53.047 N.
5.16. Determine the design load on the shaft
Fshaft = Ft + 2∙Ff= 1591.668 + 2 ∙ 53.047 = 1697.762 N.
5.17. Determine the safety factor
,
whereFbr is the breaking load in N; K is the dynamic factor taking into account the load nature; [S] is standard safety factor .
In our case:
Fbr = 88.5 kN, K = 1.2, [S] = 9.4
.
Condition is satisfied.