Allowable mean pressure [p], in MPa
n1, rpm | Chain pitch, mm | |||||||
12.7 | 15.875 | 19.05 | 25.4 | 31.75 | 38.1 | 44.45 | 50.8 | |
- | - | |||||||
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8.6. Determine the chain pitch
,
where T1 is in N×m.
Round off the pitch to the nearest standard value according to the table 8.1. In our case assume t = 31.75 mm.
8.7. Specify the allowable mean pressure according to table 8.1 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] = 29·1.09=31.61MPa
8.8. Determine the effective mean pressure.
For that we
- find chain speed
;
- find turning (tangential) force
;
- look up the projected hinge area Sh using table 8.2. In our case Sh=262 mm2
Then the effective mean pressure
.
If this inequality is not right it is necessary to increase the pitch t.
;
inequality is right.
8.9. Determine the number of links in the chain
,
where is the chain length in pitches; ; a» (30…50)·t; zS = z1 + z2; . Round off obtained magnitude to even integer numeral.
In our case: at = 45; zS = 26 + 65 = 91; ;
8.10. Specify the centre distance
The slack side of the chain should have a slight sag f» 0.01·a, for which purpose the design centre distance is reduced by 0.2 to 0.4 %.
Assume that a = 1418.76 mm.
8.11. Determine the pitch diameters
- of the driving sprocket ;
- of the driven sprocket .
8.12. Determine the addendum diameters
where d1 is the roller diameter (table 8.2).
8.13. Determine the dedendum diameter
,
.