Version 1.0 - 2020-12-22
116
Translation of original instruction
3440287
│
3442710
│
3442712
GB
K11-
80-10
0-12
5_I
SO-
702-
4_344
0287
_344
2710
_344
2712_
ba-i
nte
grat
ed_
GB.
fm
ATTENTION!
For safety reasons, in accordance with DIN EN 1550, the centrifugal force may be a
maximum of 67% of the initial clamping force.
The formula for the calculation of the total centrifugal force F
c
is:
n is the given speed in r.p.m.. The product
m
B
· r
s
is described as the centrifugal force torque
M
c
.
In case of chucks with split chuck jaws, i.e. with base jaws and top jaws, for which the base
jaws change their radial position only by the stroke amount, the centrifugal torque of base jaws
M
cGB
and the centrifugal torque of top jaws M
cAB
need to be added:
The centrifugal torque of the base jaws M
cGB
can be found in the data of the lathe chuck.
The centrifugal torque of the top jaws M
cAB
is calculated.
The lathe chuck K11-80, 100, 125 has no base jaws and no top jaws.
Example for K11-125:
The centre of gravity radius r
s
of the jaw = 0.04925 m (jaw flush with the outer diameter of
the chuck)
Weight of a jaw = 0.058 kg
Centrifugal moment for one jaw
The lathe chuck has 3 jaws.
Calculation of the total centrifugal force at a rotational speed of 3800 rpm
The total possible clamping force of the chuck at a standstill is
∑
s
17 kN at a tightening torque
of 100 Nm with the lathe chuck key.
An effective clamping forceF
sp
of 12.6 kN remains at the chuck.
see
Clamping force-speed diagram - Lathe chuck K11-80 on page 113
see
Basic safety instructions on page 112
F
c
=
∑
(m
b
·
r
s
)
·
π
·
n
30
(
)
2
=
∑
M
c
·
π
·
n
30
(
)
2
[N]
M
c
= m
B
·
r
s
[ kgm ]
M
c
= M
cGB
+ M
cAB
[ kgm ]
M
cAB
= m
AB
·
r
sAB
[ kgm ]
M
c
= 0.189 kg
·
0.04925 m = 0.0093 kgm
= 0.0093 kgm
·
3 = 0.0279 kgm
F
c
=
∑
(m
b
·
r
s
)
·
π
·
n
30
(
)
2
=
∑
M
c
·
π
·
n
30
(
)
2
[N]
( )
=
0.0279 kgm
·
3.14
·
3800
30
2
= 4413.55 N = 4.4 kN
F
sp
=
∑
s
- F
c
=
17 kN - 4.4 kN =
12.6 kN
