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

Example 4

f
H =2 (from Diagram
15 , page 25 ,forH =G
h /N = 180/40 = 4,5)G The linkages of a conveyorinstallation
hN = G
h f
β f
H = 180 × 5,2 × 2 ≈1 900 operating hoursThe required basic rating life of 9 000his not achieved by the rod end, so that a larger one has to be used. Rod end SI 20 ES, with C = 30 kN, C Given:Radial load of alternating direction: F
r = 5,5 kNHalf angle of oscillation: β = 15°( âž” fig
3 , page 16 )Frequency of oscillation: f = 25 min
= 57 kNand d
–1 k =29 mm is selected and thecalculation repeated.The values for the specific bearingload5,5p =100 Operating temperature: ≈+70 °CRequired:A rod end that will provide a basic rat- ing life of 9 000 hours under condi- tions of alternating load.As the load is alternating, a steel-on-steel rod end is appropriate, and it is to be relubricated after every 40 hours of operation. Using the guideline value for the load ratio C/P = 2 from Table , page 21 , and as P = F × –––– ≈18 N/mm
2 30and the mean sliding velocity (d
m = d
k P
perm =C
b
2 b
6 = 29 mm)v = 5,82
3 –7 r, the requisitebasic dynamic load rating will beC = 2 P = 2 =37,5 × 1 × 0,35 =13,125 kN> PThe following values of the factors areused to determine the basic rating life for initial lubrication:b × 10 × 29 × 15 × 25 =0,0063 m/sboth lie within the permissible range I.It is not necessary to check the permis- sible rod end housing load since the basic static load rating of the larger rod end is higher. Also, as beforeb × 5,5= 11 kN.The rod end SI 15 ES with a basic dy-namic load rating C = 17 kNis se- lected ( page100 ). The basic staticload rating C
= 37,5 kNand thesphere diameter d
1 =2 (alternating load)b
1 = 2, b
2 = 1 and b
5 = 3,7whilstb
k = 22 mm.The first check of size is made usingthe
2 =1 (for operating temperatures < 120 °C, from Table
5 , page 24 )b pv diagram
5 , page 22 , and withK = 100 (from
3 =1,3 (from Diagram
11 , page 24 ,for d
3 = 1,4 (from Diagram
11 , page 24 ,for d Table
4 , page 21 )P5,5p = K––= 100
k = 22 mm)b
k = 29 mm)b
4 =1,6 (from Diagram
12 , page 24 ,for v = 0,0048 m/s)b
4 = 1,8 (from Diagram
12 , page 24 ,for v = 0,0063 m/s)so that330G × –––– = 32 N/mm
2 =3,7 (from Diagram
13 , page 25 ,for C 17and the mean sliding velocity (d
5 β = 15°) p=32 N/mm
2 m = d
k =22 mm)v = 5,82 v=0,0048 m/sTherefore330G
h = 2 × 1 × 1,4 × 1,8 × 3,7 × ––––––––––– 18
2,5 × 0,0063≈710 operating hoursWith f × 10
–7 d
k β f= 5,82 × 10
–7 × 22 × 15 × 25 = 0,0048 m/sp and v both lie within the permissiblerange I of the
h =b
1 b
2 b
3 b
4 b
5 –––––p
2,5 v330=2
β = 5,2 (from Diagram
14 , page25 ) and f
H = 3,7 (from Diagram
15 , × 1 × 1,3 × 1,6 × 3,7 × ––––––––––––32
2,5 page 25 , for H = 710/40 ≈18) thebasic rating life for regular relubrica- tion (N = 40 h) becomesG × 0,0048≈180 operating hoursThe basic rating life for regular relubri-cation (N = 40 h) withf pv diagram
5 , page 22 .Checking the permissible load onthe rod end housingC
hN =710 × 5,2 × 3,7 ≈13 600 operating hoursThus the larger rod end meets therating life requirements.
=37,5 kNb
2 =1 (from Table
5 , page 24 , fortemperatures < 120 °C)b
β =5,2 (from Diagram
14 , page 25 )and
6 =0,35 (from Table
2 , page 20 ,forrod ends with lubrication hole) 34