Calculation examples
Given data
Purely radial load that alternates direction: F_{r} = 12 kN
Half angle of oscillation: β = 15° (fig. 1)
Frequency of oscillation: f = 10 min^{–1}
Maximum operating temperature: +80 °C
Requirements
The bearing must have a basic rating life of 7 000 h.
Calculations and selection
Because a bearing in this application must accommodate an alternating load, a steel/steel radial spherical plain bearing is the appropriate choice. Relubrication is planned after every 40 hours of operation.
If, for the first check, a guideline value of 2 is used for the load ratio C/P (table 1), the required basic dynamic load rating C for the bearing is
C = 2 P = 24 kN
Bearing GE 20 ES, with a C = 30 kN and a sphere diameter d_{k} = 29 mm, is chosen from the product table.
To check the suitability of the bearing using the pv diagram (diagram 1), calculate the specific bearing load using K = 100 from table 2.
p = K (P/C) = 100 (12/30) = 40 N/mm^{2}
and the sliding velocity v using d_{m} = d_{k} = 29 mm, β = 15° and f = 10 min^{–1}
v = 5,82 x 10^{7} d_{m} β f
= 5,82 x 10^{7} x 29 x 15 x 10 = 0,0025 m/s
The values for p and v lie within the permissible operating range I of the pv diagram (diagram 1), for steel/steel radial spherical plain bearings. To calculate the basic rating life for initial lubrication, the values that apply are
b_{1}  =  2 (alternating direction load)

b_{2}  =  1 (operating temperature < 120 °C from table 3) 
b_{3}  =  1,5 (from diagram 2 for d_{k} = 29 mm)

b_{4}
 =  1,1 (from diagram 3 for v = 0,0025 m/s)

b_{5}
 =  3,7 (from diagram 4 for β = 15°)

p
 =  40 N/mm^{2}

v
 =  0,0025 m/s

Therefore
G_{h} = b_{1} b_{2} b_{3} b_{4} b_{5} [330 / (p^{2,5} v)]
= 2 x 1 x 1,5 x 1,1 x 3,7 [330 / (40^{2,5} x 0,0025)]
= 160 operating hours
The basic rating life of the bearing that is relubricated regularly can now be calculated using
f_{β}  =  5,2 (from diagram 5) 
f_{H}
 =  1,8 (from diagram 6 for a relubrication frequency H = G_{H}/N = 160/40 = 4 with the relubrication interval of 40 h) 
G_{hN} = G_{h} f_{β} f_{H} = 160 x 5,2 x 1,8
= 1 500 operating hours
Because this life is shorter than the required rating life of 7 000 h, a larger bearing is chosen and calculations are repeated.
Bearing GE 25 ES, with C = 48 kN and d_{k} = 35,5 mm, is chosen. The values for the specific bearing load lie within the permissible operating range I of the pv diagram (diagram 1)
p = 100 x (12/48) = 25 N/mm^{2}
and the sliding velocity is
v = 5,82 x 10^{7} x 35,5 x 15 x 10 = 0,0031 m/s
As before
b_{1} = 2, b_{2} = 1, b_{5} = 3,7
and now
b_{3}
 =
 1,6 (from diagram 2 for d_{k} = 35,5 mm) 
b_{4}
 =  1,2 (from diagram 3 for v = 0,0031 m/s)

Therefore, the basic rating life for initial lubrication is
G_{h} = 2 x 1 x 1,6 x 1,2 x 3,7 x [330 / (25^{2,5} x 0,0031)]
= 480 operating hours
With
f_{β}  =  5,2 (from diagram 5) 
f_{H}
 =  3 (from diagram 6 for H = G_{H}/N = 480/40 = 12 with the relubrication interval of N = 40 h) 
the basic rating life becomes
GhN = 480 x 5,2 x 3 ≈ 7 490 operating hours
This larger bearing satisfies the rating life requirement.
NOTE:
The SKF Bearing Calculator performs these and many other calculations quickly and accurately. This program can be run any number of times to find the best possible solution. The SKF Bearing Calculator is available online at skf.com/bearingcalculator.
Given data
Purely radial load that alternates direction F_{r} = 16 kN
Half angle of oscillation: β = 5° (fig. 1)
Frequency of oscillation: f = 40 min^{–1}
Maximum operating temperature: +80 °C
Requirements
The bearing must have a basic rating life of 7 000 h and no relubrication should be done.
Calculation and selection
Because the bearing needs to accommodate alternating loads in an application with a small oscillation angle and without relubrication, an SKF Explorer steel/steel plain bearing is chosen.
If, for the first check, a guideline value of 2 is used for the load ratio C/P (Requisite bearing size), the required basic dynamic load rating C for the bearing is
C = 2 P = 32 kN
Bearing GE 20 ESX2LS, with a dynamic load rating C = 44 kN and a sphere diameter d_{k} = 29 mm, is chosen from the product table.
To check the suitability of the bearing using the pv diagram (diagram 7), calculate the specific bearing load using K = 150 N/mm2 (table 2).
p = K (P/C) = 150 x (16/44) = 54,5 N/mm^{2}
and the sliding velocity v (Requisite bearing size) using d_{k} = 29 mm, β = 5° and f = 40 min^{–1}
v = 5,82 x 10^{7} d_{k} β f
= 5,82 x 10^{7} x 29 x 5 x 40
= 0,0034 m/s
The values for p and v lie within the permissible operating range I of the pv diagram (diagram 7), for SKF Explorer steel/steel plain bearings. To calculate the basic rating life, the values that apply are
b_{1}  =  2 (alternating direction load)

b_{2}  =  0,64 (from diagram 8, for T = 80 °C) 
b_{3}  =  1,45 (from diagram 9 for d_{k} = 29 mm)

b_{5}
 =  1,0 (from diagram 10 for β = 5°)

p
 =  54,5 N/mm^{2}

v
 =  0,0034 m/s

G_{h} = b_{1} b_{2} b_{3} b_{5} [5 / (p^{0,6} x v^{1,6})]
= 2 x 0,64 x 1,45 x 1 x [5 / (54,5^{0,6} x 0,0034^{1,6}]
= 7 500 operating hours
Therefore, the selected bearing GE 20 ESX2LS meets the requirements.
NOTE:
The SKF Bearing Calculator incorporates programmes to perform these and many other calculations quickly and accurately. These programmes can be run any number of times to find the best possible solution. The SKF Bearing Calculator is available online at skf.com/bearingcalculator.
Given data
Radial load: F_{r} = 7 kN
Axial load: F_{a} = 0,7 kN
Half angle of oscillation: β = 8° (fig. 1)
Frequency of oscillation: f = 15 min^{–1}
Load frequency: 25 Hz
Maximum operating temperature: +75 °C
Requirements
This bearing must have a basic rating life corresponding to a driven distance of 100 000 km at an average speed of 65 km/h without maintenance.
Calculation and selection
For design reasons, a GE 20 C spherical plain bearing with a steel/PTFE sintered bronze sliding contact surface combination is proposed. From the product table, the basic dynamic load rating C = 31,5 kN and the sphere diameter d_{k} = 29 mm are obtained.
First, the equivalent dynamic bearing load must be determined by
F_{a}/F_{r} = 0,7/7 = 0,1
From diagram 11 factor y = 1,4. The equivalent dynamic bearing load is therefore
P = y F_{r} = 1,4 x 7 = 9,8 kN
To check the suitability of the bearing size using the pv diagram 12 calculate the values for the specific bearing load (using K = 100 from table 2) using
p = K (P/C) = 100 (9,8/31,5) = 31 N/mm^{2}
and the sliding velocity (d_{m} = d_{k} = 29 mm)
v = 5,82 x 10^{7} d_{m} β f
= 5,82 x 10^{7} x 29 x 8 x 15 = 0,002 m/s
The values for p and v lie within the permissible operating range I of the pv diagram 12, for steel/PTFE sintered bronze radial spherical plain bearings, where
b_{1}  =  0,2 (from table 4 for a load frequency over 0,5 Hz and 25 < p < 40 N/mm^{2})

b_{2}  =  1 (from diagram 13 for temperatures < 80 °C) 
The basic rating life for a GE 20 C bearing with the steel/PTFE sintered bronze sliding contact surface combination is
G_{h} = b_{1} b_{2} [1 400 / (p^{1,3} v)]
= 0,2 x 1 x [1400 / (31^{1,3} x 0,002)]
= 1 600 operating hours
This basic rating life corresponds to a distance (at an average speed of 65 km/h) of 1 600 x 65 = 104 000 km. Therefore, the bearing satisfies the rating life requirement.
Given data
Radial load (constant direction)
Operation
case  Load
F_{r}  Time period
t 
I  300 kN  10% 
II  180 kN  40% 
III  120 kN  50% 
The number of press cycles n = 30 per hour, and the movement between the end positions (90°) is made in 10 seconds. The operating temperature is less than +50 °C.
Requirements
A maintenancefree radial spherical plain bearing with a steel/PTFE fabric sliding contact surface combination is required for a rating life of 5 years with 70 h of operation per week.
Calculation and selection
Using a guideline value for the load ratio C/P = 2 (table 1), and with P = F_{rI} the required basic dynamic load rating
C = 2 P = 2 x 300 = 600 kN
From the product table, a GE 60 TXE2LS bearing with a basic dynamic load rating C = 695 kN and a sphere diameter d_{k} = dm = 80 mm is chosen.
First, it is necessary to check that the operation cases I to III fall within the permissible range of the pv diagram 14. The sliding velocity is the same for all three cases. The angle of oscillation is specified as 2β, the time t as the time taken to pass through 2β in seconds. Complete cycle duration is 4β (fig. 1).
v = 8,73 x 10^{6} d_{m} (2β/t)
= 8,73 x 10^{6} x 80 x (90/10) = 0,0063 m/s
The specific bearing load, p = K(P/C), using K = 300 from table 2, is
for case I
p_{I} = K P/C = 300 x (300/695) = 129,5 N/mm^{2}
for case II
p_{II} = K P/C = 300 x (180/695) = 77,7 N/mm^{2}
for case III
p_{III} = K P/C = 300 x (120/695) = 51,8 N/mm^{2}
The values for pI, pII, pIII and v are within the permissible range I of the pv diagram 14.
To make the lifetime estimate for variable loads and/or sliding velocities, the calculation of each load case has to be made separately, with the equation for TX bearings first
G_{h} = b_{1} b_{2} b_{4} (K_{p}/p^{n}v)
The parameters b_{1}, b_{2,} b_{4}, K_{p} and n are as follows
b_{1}  =  1 (from table 5, constant load)

b_{2}  =  1 (from diagram 2, for operating temperature < +50 °C) 
b_{4}  =  1,45 (from diagram 16)


 b_{4 I} = 0,31


 b_{4 II} = 0,48


 b_{4 III} = 1,57

K_{p}
 =  1,0 (from table 6)


 K_{p I} = 40 000 _{ } 

 K_{p II} = 4 000


 K_{p III} = 4 000

n
 =  (from table 6)


 n_{1} = 1,2


 n_{2} = 0,7


 n_{3} = 0,7

for case I
G_{hI} = 1 x 1 x 0,31 x [40 000/(129,5^{1,2}/0,0063)]
= 5 745 operating hours
for case II
G_{hII} = 1 x 1 x 0,48 x [4 000/(77,7^{0,7}/0,0063)]
= 14 477 operating hours
for case III
G_{hIII} = 1 x 1 x 0,57 x [4 000/(51,8^{0,7}/0,0063)]
Using the calculated basic rating lives of the three operation cases, the total basic rating life for continuous operation is
For t_{I}, t_{II} etc., the percentages given in the operating data are inserted (with T = t_{I} + t_{II} + t_{III} = 100%.)
≈ 14 940 operating hours
The required life of five years should be met assuming the machine is operated 70 h/week, 30 cycles/hour and 50 weeks per year, to 525 000 cycles or 2 916 operating hours. (Note that time for a complete cycle is 20 s.)
G_{N, Req} = 5 x 70 x 30 x 50 = 525 000 cycles
G_{h, Req} = (525 000 x 20)/3600 = 2 916 h
Given data
Radial load of alternating direction F_{r} = 5,5 kN
Half angle of oscillation: β = 15° (fig. 1)
Frequency of oscillation: f = 25 min^{–1}
Operating temperature: +70 °C
Requirements
A rod end is needed that provides a basic rating life of 9 000 hours under alternating load conditions.
Calculation and selection
Because the load is alternating, a steel/steel rod end is appropriate. Relubrication is planned every 40 hours of operation. Using the guideline value for the load ratio C/P = 2 from table 1, and as P = Fr, the requisite basic dynamic load rating is
C = 2 P = 2 x 5,5 = 11 kN
The SI 15 ES rod end with a basic dynamic load rating C = 17 kN is selected (product table). The basic static load rating is C_{0} = 37,5 kN and the sphere diameter d_{k} = 22 mm. To check the suitability of rod end size using the pv diagram 1, calculate the values for the specific bearing load (using K = 100 from table 2)
p = K (P/C) = 100 x (300/695) = 32,4 N/mm^{2}
and the mean sliding velocity (d_{m} = d_{k} = 22 mm)
v = 5,82 x 10^{7} d_{k} β f
= 5,82 x 10^{7} x 22 x 15 x 25 = 0,0048 m/s
The values for p and v lie within the permissible operating range I of the pv diagram 1.
Checking the permissible load on the rod end housing
C_{0}  =  37,5 kN

b_{2}  =  1 (from table 3, for temperatures < 120 °C) 
b_{6}
 =  0,35 (from table 7 for rod ends with a lubrication hole)

P_{perm}
 =  C_{0} b_{2} b_{6}

 =  37,5 x 1 x 0,35

 =
 13,125 kN > P

The following values of the factors are used to determine the basic rating life for initial lubrication only
b_{1}  =  2 (alternating load)

b_{2}  =  1 (from table 3, for operating temperatures < 120 °C) 
b_{3}  =  1,3 (from diagram 2 for d_{k} = 22 mm)

b_{4}  =
 1,6 (from diagram 3 for v = 0,0048 m/s)

b_{5}
 =  3,7 (from diagram 4 for β = 15°)

p
 =  32 N/mm^{2}

v
 =  0,0048 m/s

Therefore
G_{h} = b_{1} b_{2} b_{3} b_{4} b_{5} [330 / (p^{2,5} v)]
= 2 x 1 x 1,3 x 1,6 x 3,7 x [330 / (32^{2,5} x 0,0048]
≈ 177 operating hours
The basic rating life for regular relubrication (N = 40 h) with
f_{β}  =  5,2 (from diagram 5) 
f_{H}
 =  1,8 (from diagram 6 for a relubrication frequency H = G_{h}/N = 177/40 = 4,4) 
G_{hN} = G_{h} f_{β} f_{H} = 177 x 5,2 x 2
≈ 1 840 operating hours
The required basic rating life of 9 000 h is not achieved; therefore a larger rod end has to be selected. A SI 20 ES rod end, with C = 30 kN, C0 = 57 kN and d_{k} = 29 mm is selected and the calculation repeated.
The values for the specific bearing load
p = K (P/C) = 100 x (5,5/30) = 18,3 N/mm^{2}
and the mean sliding velocity (d_{m} = d_{k} = 29 mm)
v = 5,82 x 10^{7} d_{k} β f
= 5,82 x 10^{7} x 29 x 15 x 25 = 0,0063 m/s
both lie within the permissible range I. It is not necessary to check the permissible rod end housing load since the basic static load rating of the larger rod end is higher. Also, as before
b_{1} = 2; b_{2} = 1 and b_{5} = 3,7
while
b_{3}  =  1,3 (from diagram 2 for d_{k} = 29 mm) 
b_{4}  =
 1,8 (from diagram 3 for v = 0,0063 m/s)

so that
G_{h} = 2 x 1 x 1,4 x 1,8 x 3,7 x [330 / (18,3^{2,5} x 0,0063]
≈ 681 operating hours
With f_{β} = 5,2 (from diagram 5) and f_{H} = 3,7 (from diagram 6, for H = 681/40 ≈ 17) the basic rating life for regular relubrication (N = 40 h) becomes
G_{hN} = G_{h} f_{β} f_{H} = 681 x 5,2 x 3,7
≈ 13 100 operating hours
Therefore, the larger rod end meets the rating life requirements.