Aufgaben und Lösungen - Institut für Maschinenwesen

- 10 -
Lösung zur Aufgabe 3
1) Reibbeiwert
15
,
0
7
,
16
tan
2
1
'
42
16
tan
2
1
tan
2
1
2
tan =
°
⋅
=
°
⋅
=
⋅
=
⇒
⋅
= α
µ
µ
α H
H
2) Verteilung des Drehmoments (siehe Kap.4 Blatt 18/23)
1
1
2
tan
−
⋅
⋅
+
⋅
⋅
=
n
H
p
H
A
n
Q
r
F
T
µ
α
µ
Nm
T
T
K
T
Nm
T
T
K
T
T
T
T
Q
K
Q
K
T
T
T
Q
mit
Q
K
Q
K
T
n
Für
K
Q
K
T
n
Für
H
7
,
26
3
1
5
,
0
3
,
53
3
2
1
3
2
)
5
,
0
1
(
1
)
1
(
1
)
1
(
5
,
0
15
,
0
2
7
,
16
tan
7
,
16
tan
2
tan
tan
:
2
1
:
1
2
2
1
1
2
1
1
2
0
1
=
⋅
=
⇒
⋅
=
⇒
=
⋅
=
⇒
⋅
=
⇒
⋅
=
⋅
+
=
⋅
+
=
⇒
+
=
+
=
=
⋅
+
°
°
=
⋅
+
=
⋅
=
⋅
=
=
⋅
=
⋅
=
=
µ
α
α
3) Paßfugendruck
2
2
2
3
2
2
2
2
2
2
3
2
1
1
/
36
,
13
15
,
0
3
,
5
20
2
10
7
,
26
2
/
7
,
26
15
,
0
3
,
5
20
2
10
3
,
53
2
mm
N
mm
mm
Nmm
l
r
T
p
mm
N
mm
mm
Nmm
l
r
T
p
H
p
erf
H
p
erf
=
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
=
=
⋅
⋅
⋅
⋅
=
⋅
⋅
⋅
=
π
µ
π
π
µ
π
4) Axialkraft
N
mm
Nmm
F
r
Q
Q
T
F
n
Q
Q
r
F
T
A
p
H
n
H
A
n
H
p
H
A
10667
20
15
,
0
)
5
,
0
1
(
)
15
,
0
2
7
,
16
(tan
)
5
,
0
1
(
10
.
80
)
1
(
)
2
(tan
)
1
(
)
2
(
1
1
2
tan
2
3
1
1
1
=
⋅
⋅
−
⋅
+
°
⋅
−
⋅
=
⇒
⋅
⋅
−
⋅
+
⋅
−
⋅
=
⇒
=
−
−
⋅
⋅
+
⋅
⋅
=
µ
µ
α
µ
α
µ