4. How to Select Chains
When there is no play in the system, the coefficient of shock K = 0.23 (Figure 4.7). The chain tension from the starting torque: Fms = Ts 3 i 3 30 3 1,000 /(d /2) 14 = 0.083 3 60 3 30 3 1,000 / (171.22 /2) = 124.7 kN 14 The chain tension calculated from the braking torque: Fmb = Tb 3 i 3 30 3 1,000 3 1.2 /(d /2) 14 = 0.096 3 60 3 30 3 1,000 3 1.2/(171.22 /2) =173.0 kN 14 Use the larger value (in this case it is Fmb ) to calculate chain tension. F'mb = Fmb 3 Kv 3 Kc 3 Ku 3 K = 173.0 3 1.02 3 1.28 3 0.6 3 0.23 = 31.2 kN Step 4.Calculate the chain tension from motor acceleration and deceleration. Working torque Tm = ( Ts + Tb ) = (0.083 + 0.096) = 0.0895 kN • m 2 2 Load torque T L = M 3 d 3 g (2 3 1,000 3 i ) 1,000
(3,000 3 171.22)
g
=
= 0.02 kN • m
3
30 1,000 14
2 3 1,000 3 60 3
( Im + Il ) 3 n 1
g
Motor acceleration time t s =
3 4
3
375 3 (Tm - Tl) 1,000
(0.015 + 0.00130) 3 1,500
g
=
3 4 = 0.037s
3
375 3 (0.0895 - 0.02)
1,000
( Im + Il ) 3 n 1
g
Motor deceleration time t b =
3 4
3
375 3 (Tm + Tl) 1,000
= (0.015 + 0.00130) 3 1,500 375 3 (0.0895 + 0.02)
g
3 4 = 0.023s
3
1,000
Because t b is smaller than t s , the chain tension due to motor deceleration F b is greater than that of the acceleration. F b = M 3 V + Fw = 3,000 3 6.2 + 29.4 = 42.9 kN t b 3 60 3 1,000 0.023 3 60 3 1,000
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