2015-08-21
264
Determining the pulley width B
| Belt width b, mm | ||||||||||||||
| Pulley width B, mm |
In our case we assume B = 63 mm.
7.17. Determine the pretension in the belt
=1.8∙50∙4.5=405 N,
where s0 =1.8 MPa- tensile prestress.
7.18. Determine tension of the belt
on the tight side F 1= F 0 + 0.5· Ft =405+0.5∙388.06=599.03 N,
on the slack side F 2= F 0 - 0.5· Ft =405-0.5∙388.06=270.97 N
7.19. Calculate the force acting on the shaft and bearings
7.20. Determine tensile stress
.
7.21. Determine bending stress
,
where E is modulus of elasticity of the belt material. For rubberized fabric belts E =100¸200 MPa.
7.22. Determine tensile stress due to action of centrifugal force Fc
,
where r is density of belts. For rubber-impregnated flat belts r = 1100 ¸ 1200 kg /m3.
7.23. Determine the maximum stress
.
For rubberized fabric belts [σ] = 7 MPa.
Condition is satisfied.
7.24. Determine the service life of the belt
,
where s-1 is limit of endurance (for rubberized belts s-1=7 MPa); 
is the number of belt runs per second; 
takes into account the velocity ratio; C l takes into account the nature of the load (for constant load C l =1; for variable load C l=2). We assume that load is variable.
.
Condition is satisfied.
8. ANALYSIS OF THE CHAIN DRIVE
Let us carry out the analysis of the chain drive for strength if torque at the driving sprocket T 1 = 464.5 N·m; torque at the driven sprocket T 2 = 1105 N·m; rotational speed of the driving sprocket n 1 = 122.25 rpm,rotational speed of the driven sprocket n 2 = 48.9 rpm; velocity ratio of the chain drive ucd = 2.5; input power of the chain drive P 1 = 5.947 kW.
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8.1. Determine the number of teeth of the driving sprocket
z1 = 31- 2×ucd ³ 17
and round off obtained magnitude to the nearest integer numeral.
In our case z1 = 31- 2×2.5 = 26
Assume z1 = 26 > 17.
8.2. Determine the number of teeth of the driven sprocket
z2 = z1×ucd £ 120
and round off to the nearest integer numeral
z2 = 26×2.5 = 65
Assume z2 = 65 < 120.
8.3. Specify the velocity ratio and determine the error
.
The error should be ε £ 4 %.
In our case ε = 0 %.
8.4. Determine the service factor Ks
Ks= K×Ka×Klub×Kg×Kd×Kten,
where K takes into account the load nature 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 = 30×t to 50×t and Ka =0.8 for a = 60×t to 80×t; Klub is lubrication factor (Klub =0.8 for immersion lubrication, Klub =1 for drop-feed lubrication and Klub =1.5 for periodic greasing); Kg accounts for the angle that the shaft centre line makes with the horizontal (Kg=1 for g £ 60˚ and Kg=1.25 for g > 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 = 40 t (Ka= 1); periodic greasing (Klub=1.5); g £ 60˚ and (Kg =1); one-shift operation (Kd =1); fixed base drive (Kten =1.25).
Ks= 1.2·1·1.5·1·1·1.25 = 2.25
8.5. Approximately determine the allowable mean pressure on the hinges by means of table 8.1 depending on the rotational speed of the smaller sprocket.
In our case for rotational speed n 1=122.25 rpm, 
29 MPa.
