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83
ELA-flex SD
For the initial drive selection, use the selection graphs illustrated
in the relative ELA-flex SD
cataloge section. For initial pulley
choice, it is recommended to use the driver pulley with maximum
diameter allowable in the application.
Safety factors
Belt selection is made according to a constant working load. In
case of peak loads and vibrations a safety factor c
1
must be con-
sidered.
Transmission with steady load
c
1
= 1,0
Transmission with peak or fluctuating loads:
Light
c
1
= 1,4
Medium
c
1
= 1,7
Heavy
c
1
= 2,0
For speed up driver factor c
2
must be considered:
i = from 0,66 to 1
c
2
= 1,1
i = from 0,40 to 0,66 c
2
= 1,2
i < 0,40
c
2
= 1,3
The resulting total safety factor is:
c
0
= c
1
·
c
2
The necessary data for drive calculation are:
• Power to be transmitted
P [kW]
• Driver rpm
n
1
[min
-1
]
• Motor starting torque
M
ab
[Nm]
• Required center distance
A [mm]
• Maximum driver pulley diameter d
w1
[mm]
Drive calculation
Select type of belt
Calculate drive ratio
Determine belt width
with
β [°] =
wrap angle
Calculate teeth in mesh
A drive is correctly tensioned when the belt slack side is ten-
sioned in all working conditions. It is also important to use
the minimum necessary tension to minimize shaft loads. Belt
tension is dependent also on belt length L
R
and its number of
teeth Z
R
. According to belt number of teeth, following tension
is suggested:
2 shafts drive
Z
R
< 75
F
TV
= 1/3 F
U
75
<
Z
R
< 150
F
TV
= 1/2 F
U
Z
R
> 150
F
TV
= 2/3 F
U
More than 2 shafts drive
F
TV
>
F
U
Determine installation tension
Verifiy allowable tensile load
Calculate shaft load
Belt lenght for drive with ratio i
#
1
Calculate belt lenght
Belt lenght for drive with ratio i = 1
and more precisely:
In order to ensure the correct drive installation tension, it is
recommended to use the special belt tension meter available
from ELATECH
®
.
driven
driver
n
n
i
=
tzA2 d A2 L
w
R
⋅ + ⋅ = ⋅π+ ⋅ =
( )
( )
2
k
g
k
g
R
t z z
A4
1
A2 z z
2
t
L
π
+ + + ⋅ ≈
( )
− ⋅⎟⎠
⎜⎝
β− + + ⋅ +β⋅
=
k
g
k
g
R
z z
180
1 z z
2
t
2
sin A2 L
k
e
z
360
z
⋅ β=
spez
e k
0
Mz z
cM 100
b
⋅ ⋅
=
(
)
1 i
F2 F
cos F2 F
Tv
Wsta
Tv
Wsta
=
⋅ =
β⋅
⋅ =
( )
⋅π⋅
− ⋅
⋅ =β
A 2
z z t
arccos
2
k
g
spez
e k
0
Pzz
c 1000 P
b
⋅ =
The allowable tensile load of the belt must be highter than the
total corrected peripheral force.
F
Tzul
> c
0
·
F
U
with
d
M 2000
F
W
U
=
for