# Calculation examples

The following calculation examples illustrate the methods used to determine the requisite bearing size or the basic rating life for spherical plain bearings and rod ends.
1. Torque support of a concrete transporter - Radial spherical plain bearing, steel/steel

Given data

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 dk = 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/mm2

and the sliding velocity v using dm = dk = 29 mm, β = 15° and f = 10 min–1

v = 5,82 x 10-7 dm 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

 b1 = 2 (alternating direction load) b2 = 1 (operating temperature < 120 °C from table 3) b3 = 1,5 (from diagram 2 for dk = 29 mm) b4 = 1,1 (from diagram 3 for v = 0,0025 m/s) b5 = 3,7 (from diagram 4 for β = 15°) p = 40 N/mm2 v = 0,0025 m/s

Therefore

Gh = b1 b2 b3 b4 b5 [330 / (p2,5 v)]

= 2 x 1 x 1,5 x 1,1 x 3,7 [330 / (402,5 x 0,0025)]

= 160 operating hours

The basic rating life of the bearing that is relubricated regularly can now be calculated using

 fβ = 5,2 (from diagram 5) fH = 1,8 (from diagram 6 for a relubrication frequency H = GH/N = 160/40 = 4 with the relubrication interval of 40 h)

GhN = Gh fβ fH = 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 dk = 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/mm2

and the sliding velocity is

v = 5,82 x 10-7 x 35,5 x 15 x 10 = 0,0031 m/s

As before

b1 = 2, b2 = 1, b5 = 3,7

and now

 b3 = 1,6 (from diagram 2 for dk = 35,5 mm) b4 = 1,2 (from diagram 3 for v = 0,0031 m/s)

Therefore, the basic rating life for initial lubrication is

Gh = 2 x 1 x 1,6 x 1,2 x 3,7 x [330 / (252,5 x 0,0031)]

= 480 operating hours

With

 fβ = 5,2 (from diagram 5) fH = 3 (from diagram 6 for H = GH/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.

2. Linkages of a flap opening system - SKF Explorer steel/steel plain bearing

Given data

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 ESX-2LS, with a dynamic load rating C = 44 kN and a sphere diameter dk = 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/mm2

and the sliding velocity v (Requisite bearing size) using dk = 29 mm, β = 5° and f = 40 min–1

v = 5,82 x 10-7 dk 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

 b1 = 2 (alternating direction load) b2 = 0,64 (from diagram 8, for T = 80 °C) b3 = 1,45 (from diagram 9 for dk = 29 mm) b5 = 1,0 (from diagram 10 for β = 5°) p = 54,5 N/mm2 v = 0,0034 m/s

Gh = b1 b2 b3 b5 [5 / (p0,6 x v1,6)]

= 2 x 0,64 x 1,45 x 1 x [5 / (54,50,6 x 0,00341,6]

= 7 500 operating hours

Therefore, the selected bearing GE 20 ESX-2LS 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.

3. Attachment of a shock absorber of an off-highway vehicle - Spherical plain bearing, steel/PTFE sintered bronze

Given data

Axial load: Fa = 0,7 kN
Half angle of oscillation: β = 8° (fig. 1)
Frequency of oscillation: f = 15 min–1
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 dk = 29 mm are obtained.

First, the equivalent dynamic bearing load must be determined by

Fa/Fr = 0,7/7 = 0,1

From diagram 11 factor y = 1,4. The equivalent dynamic bearing load is therefore

P = y Fr = 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/mm2

and the sliding velocity (dm = dk = 29 mm)

v = 5,82 x 10-7 dm 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

 b1 = 0,2 (from table 4 for a load frequency over 0,5 Hz and 25 < p < 40 N/mm2) b2 = 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

Gh = b1 b2 [1 400 / (p1,3 v)]

= 0,2 x 1 x [1400 / (311,3 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.
4. A 320-bar hydraulic cylinder on a fully automatic press for building industry waste - Radial spherical plain bearing, steel/PTFE fabric

Given data

 Operation case Load Fr 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 maintenance-free 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 = FrI the required basic dynamic load rating

C = 2 P = 2 x 300 = 600 kN

From the product table, a GE 60 TXE-2LS bearing with a basic dynamic load rating C = 695 kN and a sphere diameter dk = 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 dm (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

pI = K P/C = 300 x (300/695) = 129,5 N/mm2

for case II

pII = K P/C = 300 x (180/695) = 77,7 N/mm2

for case III

pIII = K P/C = 300 x (120/695) = 51,8 N/mm2

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

Gh = b1 b2 b4 (Kp/pnv)

The parameters b1, b2, b4, Kp and n are as follows

 b1 = 1 (from table 5, constant load) b2 = 1 (from diagram 2, for operating temperature < +50 °C) b4 = 1,45 (from diagram 16) b4 I = 0,31 b4 II = 0,48 b4 III = 1,57 Kp = 1,0 (from table 6) Kp I = 40 000 Kp II = 4 000 Kp III = 4 000 n = (from table 6) n1 = 1,2 n2 = 0,7 n3 = 0,7

for case I

GhI = 1 x 1 x 0,31 x [40 000/(129,51,2/0,0063)]

= 5 745 operating hours

for case II

GhII = 1 x 1 x 0,48 x [4 000/(77,70,7/0,0063)]

= 14 477 operating hours

for case III

GhIII = 1 x 1 x 0,57 x [4 000/(51,80,7/0,0063)]

Using the calculated basic rating lives of the three operation cases, the total basic rating life for continuous operation is For tI, tII etc., the percentages given in the operating data are inserted (with T = tI + tII + tIII = 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.)

GN, Req = 5 x 70 x 30 x 50 = 525 000 cycles
Gh, Req = (525 000 x 20)/3600 = 2 916 h
5. Linkages of a conveyor installation - Rod end, steel/steel

Given data

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 C0 = 37,5 kN and the sphere diameter dk = 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/mm2

and the mean sliding velocity (dm = dk = 22 mm)

v = 5,82 x 10-7 dk 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

 C0 = 37,5 kN b2 = 1 (from table 3, for temperatures < 120 °C) b6 = 0,35 (from table 7 for rod ends with a lubrication hole) Pperm = C0 b2 b6 = 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

 b1 = 2 (alternating load) b2 = 1 (from table 3, for operating temperatures < 120 °C) b3 = 1,3 (from diagram 2 for dk = 22 mm) b4 = 1,6 (from diagram 3 for v = 0,0048 m/s) b5 = 3,7 (from diagram 4 for β = 15°) p = 32 N/mm2 v = 0,0048 m/s

Therefore

Gh = b1 b2 b3 b4 b5 [330 / (p2,5 v)]

= 2 x 1 x 1,3 x 1,6 x 3,7 x [330 / (322,5 x 0,0048]

≈ 177 operating hours

The basic rating life for regular relubrication (N = 40 h) with

 fβ = 5,2 (from diagram 5) fH = 1,8 (from diagram 6 for a relubrication frequency H = Gh/N = 177/40 = 4,4)

GhN = Gh fβ fH = 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 dk = 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/mm2

and the mean sliding velocity (dm = dk = 29 mm)

v = 5,82 x 10-7 dk 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

b1 = 2; b2 = 1 and b5 = 3,7

while

 b3 = 1,3 (from diagram 2 for dk = 29 mm) b4 = 1,8 (from diagram 3 for v = 0,0063 m/s)

so that

Gh = 2 x 1 x 1,4 x 1,8 x 3,7 x [330 / (18,32,5 x 0,0063]

≈ 681 operating hours

With fβ = 5,2 (from diagram 5) and fH = 3,7 (from diagram 6, for H = 681/40 ≈ 17) the basic rating life for regular relubrication (N = 40 h) becomes

GhN = Gh fβ fH = 681 x 5,2 x 3,7
≈ 13 100 operating hours

Therefore, the larger rod end meets the rating life requirements. 