2015-08-21
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| Fabric plies | ||||
| Б 800 | БКНЛ | ТА-150 | ТК-20 | |
| Nominal strength in N per mm of width | ||||
| Maximum allowable load p0 to a ply in N per mm of width | ||||
| Design thickness δ0 of fabric plies with rubber interlayers, mm | 1.5 | 1.2 | 1.2 | 1.3 |
| Number of plies if belt width b, mm 20-71 80-112 125-560 | 3-5 3-6 3-6 | 3-5 3-6 3-6 | -- -- 3-4 | -- -- 3-4 |
Check the requirement
d=d0× z £ 0.025× d 1;
1.5∙3£0.025×224; 4.5£5.6mm. Condition is satisfied.
7.10. Determine contact angle factor C α
.
7.11. Determine factor Cv, with taks into account the effect of the belt speed
Cv = 1.04 – 0.0004 × V 2=1.04 – 0.0004 ×17.0562=0,9236.
7.12. Determine service factor Cs:
- Cs = 1 for steady operation (belt conveyers, lathes and grinding machines);
- Cs =0.9 in the case of moderate vibration (chain conveyers and milling machines);
- Cs =0.8 in the case of considerable vibration (flight conveyers, planing machines).
The value of Cs is to be reduced by 0.1 in two-shift operation and by 0.2 in three-shift operation.
Let us assume that we have moderate vibration. That is why Cs =0.9.
7.13. Determine factor C Θ, that takes into account the belt position in the space.
- C Θ = 1 for horizontal drives and inclined at up to 60°;
- C Θ = 0.9 for drives inclined at over 60° to 80°;
- C Θ = 0.8 for drives inclined at over 80° to 90°.
In our case we assume C Θ = 1.
7.14. Determine the allowable load to 1mm of ply width, N/mm
[ p ]= p 0 × C a × CV × Cs × C Q=3∙0.966∙0.9236∙0.9∙1=2.409 N/mm.
7.15. Find the belt width
and round off obtained magnitude according to series of standard values: 20; 25; 32; 40; 50; 63; 71; 80; 90; 100; 112; 125; 140; 160; 180; 200 and so on. Assume b =50 mm.
7.16. Determine the pulley width B according to table 7.2.
