Page 57 - 1_usa.qxp

Determine effective installation tension FTV
FTV > FU = 1116,2 N ; choose F TV = 1200 N
Verify allowable tensile load
FTmax = FTV + FU = 2316,2 N
FTzul > FTmax 3800 N > 2316,2 N
Determine necessary elongation
Should the elongation be too high, a larger width must be selec-
ted or ATL10 type of the same width can be used.
Check flexibility
The minimum pulleys diameter are respected.
FU = m • ab + FR = 1116,2 N
A belt width 25 mm is selected as next standard width.
Calculate total pheripheral force FU
Data
Slide mass mc = 50 [kg]
Center distance A = 1600 [mm]
Acceleration ab = 20 [m/s2]
Maximum allowable pulley diameter dw = 110 [mm]
Friction force FR = 100 [N]
Total travel sges = 1100 [mm]
Linear speed v = 5 [m/s]
Shaft diameter d = 28 [mm]
Calculation example
Linear positioning drive
Calculation of total mass in motion
56, 937
86, 101
5 19100
n = ⋅ =
Select belt
From selection graph mass/acceleration of AT belts (preferred
for linear positioning because of the higher stiffness) select a
25 AT10 belt with aluminium pulleys z = 32 (da=100,00 mm and
dw=101,86 mm).
Rpm
Belt mass mR:
Pulley mass mS :
Reduced pulley mass mSred:
kg 50
A = 1600 mm
3200 2 1600 L L L2 1 R = ⋅ = + =
Kg 48,0,0 15,02,3m
R = ⋅ =
( ) ( ) Kg 61,0
104
308,2 28 100
104
B d d
6
2 2
6
2 2
a =
⋅
⋅ ⋅π⋅ −
=
⋅
⋅ρ⋅π⋅ −
Sred R cmmmm+ + + = Kg 81,50 33,0 48,0 50 = + + =
mm
,48 kg
k
k
kg 33,0
100
28
1
2
61,0
d
d
1
2
m
2
2
2
a
2
S =⎟
⎠
⎞
⎜
⎝
⎛ + ⋅ =⎟
⎠
⎞
⎜
⎝
⎛ + ⋅
Nm 85,56
2000
86, 101 2, 1116
M = ⋅
=
mm 18 cm 78,1
12 21,52
2, 1116
b ≈ =
⋅ =
63,0 mm 02,2
952000 2
3200 1200
l ≅ =
⋅
⋅
=Δ ‰
Check belt width
56