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Bearing Forces and Bending Moments

By deflecting the moment in the joint, the additional moment MZ acts on the fork. For the joint shown in Fig. 21, M2 and MZ are calculated by dividing the moments vectors:

Fig. 21:

Bild 21a

Position 0° and 180°
M2 = M1 · cos ß
M
Z = M1 · sin ß

Bild 21b

Position 90° and 270°
M2 = M1 · 1/cos ß
M
Z = M1 · tan ß

 

7.1 Bearing Forces in Z-Arrangement

The additional moment MZ exerts forces on the bearing which apply bending stresses to the shaft. Fig. 22 shows the additional moments and bearing forces in the 0° and 90° position.

Fig. 22:

side view

Bild 22a Bild 22b

top view

Bild 22c Bild 22d
Position at 1= 0° = 180°
Propeller shaft centre under bending stress.
Position at 1= 90° = 270°
Input and output shaft under bending stress
A = B = C = D = 0 Formel 10

The bearing forces swing between zero and maximum twice per rotation.

 

7.2 Bearing Forces in W-Arrangement

According to Fig. 23, in this arrangement the following additional moments and bearing forces apply:

Fig. 23:

side view

Bild 23a Bild 23b

top view

Bild 23c Bild 23d
Position 1= 0° = 180°
Propeller shaft centre and input and output shafts under bending stress.
Position 1= 90° = 270°
Input and output shafts under bending stress.
Formel 11 Formel 10

The bearing forces swing between minimum and maximum twice per rotation.

 

7.3 Displacement Force on Propeller Shafts with Length Extension

To displace the sliding piece under the effect of torque, a displacement force L is required which must be supported by the bearing A, B, C, D.

The maximum displacement force is:

Formel 12

where

µ = Coefficient of friction. For hardened, nitrated and/or phosphatized parts, µ = 0,1 can be assumed; for rilsan-coated parts, µ = 0,06
M1 = drive torque
dt = reference diameter of sliding profile (see table)
= angle between tooth flank and centre point beam (see table)
C = profile overlap (tooth engagement length, see table)

Table

Profile to DIN 5480 dt·cos
[m]
Cmin
[m]
38 x 2
52 x 2,5
55 x 2,5
62 x 2
65 x 2,5
75 x 2,5
90 x 2,5
95 x 2
0,0310
0,0427
0,0452
0,0503
0,0539
0,0626
0,0758
0,0789
0,072
0,100
0,105
0,075
0,125
0,145
0,175
0,085

This gives the bearing forces:

Formel 13

Usually only axial forces are significant.

 
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