1

Product information

2

Recommendations

3

Product data Page ................4 Calculation examples Page ..............57

Example 1

Bearing GE 25 ES having C = 48 kN and d
k = 35,5 mm is chosen. The val-ues for the specific bearing load12p = 100 The torque support of a concretetransporter Given:Purely radial load (alternating direc- tion): F × ––– = 25 N/mm
2 48and the sliding velocityv=5,82

2

r = 12 kNHalf angle of oscillation: β = 15°( âž” fig
3 , page 16 )Frequency of oscillation: f = 10 min
–1 × 10
–7 × 35,5 × 15 × 10=0,0031 m/slie within the permissible operatingrange I of the maximum operating temperature: +80 °CRequirement:A bearing which has a basic rating life of 7 000 h.As the load is alternating, a steel-on-steel spherical plain bearing is the appropriate choice. The intention is to relubricate the bearing after each 40 hours of operation.If, for the first check, a guidelinevalue of 2 is used for the load ratio C/P ( âž” Table , page 21 ), the requiredbasic dynamic load rating C for the bearing is C = 2 P = 24 kN Bearing GE 20 ES having C = 30 kNand a sphere diameter d b
1 =2 (alternating direction load)b pvdiagram
6 , page22
2 =1 (operating temperature <120 °Cfrom . As beforeb Table
5 , page 24 )b
3 =1,5 (from Diagram
11 , page 24 ,for d
1 = 2, b
2 = 1, b
5 = 3,7and nowb
k = 29 mm) b
4 =1,1(from Diagram
12 , page 24 ,for v = 0,0025 m/s)b
3 =1,6 (from Diagram
11 , page 24,
5 =3,7 (from Diagram
13 , page 25 ,for for d
k = 35,5 mm)b
3 β = 15°)p=40 N/mm
4 =1,3 (from Diagram
12 , page 24.
2 for v = 0,0031 m/s)Therefore, the basic rating lifefor initiallubrication330G v=0,0025 m/s Therefore330G
k = 29 mm ischosen from the product table,
h =b
1 b
2 b
3 b
4 b
5 –––––p
2,5 h =2 × 1 × 1,6 × 1,3 × 3,7 × ––––––––––––25
2,5 page62 .To be able to check the suitability ofthe bearing using the v330=2 × 0,0031≈520 hWith f pv diagram
5 , × 1 × 1,1 × 1,1 × 3,7 × –––––––––––––40
2,5 page 22 , it is first necessary to calcu-late the specific bearing load using K = 100 from Table , page 21 × 0,0025≈160 operating hoursThe basic rating lifeof the bearingwhich is to be relubricated regularly can now be calculated usingf
β = 5,2 (from Diagram
14 , page25 P12p =K ––= 100
4 ) and f
H = 3,1 (from Diagram
15 , × ––– = 40 N/mm page 25 for H = 520/40 = 13) thebasic rating lifefor regular relubrica- tion (N = 40 h) becomesG
2 C30and the sliding velocity v using d
hN = 520 × 5,2 × 3,1
m = d
k ß =5,2 (from Diagram
14 , page 25 ) f = 29 mm, β = 15°and f = 10 min
–1 H =1,8 (from Diagram
15 , page 25 ,for a relubrication frequency H = G ≈ 8 300 operating hoursThis larger bearing thus satisfies therating life requirement. v=5,82 × 10
–7 d
m β f=5,82
h /N = 160/40 = 4 with the relu-brication interval of 40 h)G × 10
–7 × 29 × 15 × 10 =0,0025 m/sThese values for p and v lie within thepermissible operating range I of the
hN =G
h f
β f
H = 160 × 5,2 × 1,8 ≈1 500 operating hoursAs this life is shorter than the requiredratinglife of 7 000 h, a larger bearing must be chosen and the calculations repeated. pv diagram , page 22 ,for steel-on-steel spherical plain bearings manufacturer. To cal- culate the basic rating lifefor initial lubrication, the values thatapply are
5 31