Plastic Plain Bearings

Calculating the Load The heat that can be dissipated from the This results in the following numerical-value Capacity, Nominal Life and bearing due to the tempera­ ture­­ differential equation:

Clearance is partly dissipated through the ∆ϑ 1 8.33 (p ∙ v) ≤ 37.3 ∙ 10-3 ∙ ∙ + and partly through the shaft. µ s l When calculating the load capacity and the ( ) nominal life of Plastic Plain Bearings, the Q = K ∙ π ∙ l ∙ d ∙ λ ∙ ∆ϑ + [N/cm2 ∙ m/s] (6) following factors, among others, have to be 2 1 s taken into account: µ = coefficient of 2 (as per Table 3) - Sliding velocity + K ∙ 2π ∙ d ∙ λ ∙ ∆ϑ (4) 2 s s = wall thickness of Plastic - Average contact 4 ∙ d - Frictional heat Plain Bearings in mm - Wear (Tables 4 and 5) I = width of Plastic Plain - Bearing clearance. K and K are dimensionless factors indi- 1 2 Bearings in mm. (For Type cating the way in which the bearing heat R13205..00 the value I should be components dissipated­ through the Plastic The average contact pressure is calculated substituted by (I-s )). using the equation Plain Bearing and the shaft are affected by 1 the bearing construction. The p · v values thus obtained for plastic Generally­ the following values can be F plain bearings at an ambient temperature p = [N/cm2] expected: of 20°C or ∆ϑ of 60°C can be read off from l ∙ d (1) K = 1/2 1 Table 6. K = 1/24 where 2 At higher ambient temperatures, the ∆ϑ λ = thermal conductivity of F = bearing load (N) value reduces, as does the p · v value in polyamide l = width of the Plastic Plain accordance with equation (6). = 0.23 W/K · m Bearing (cm) λ = thermal conductivity of steel For Type R13205..00 the value I s = 46 W/K · m should be substituted by (I-s ). Bearing wear 1 s = wall thickness of Plastic d = bearing diameter (cm) If the bearing temperature does Plain Bearings ex­ceed 80°C, hardened and ∆ϑ = temperature differential (K or °C). At low sliding velocities, the average con- ground steel shafts are used, 2 tact pressure is 2,500 N/cm . and the bearing wear is virtually undetecta- This value is used as p in the load capa- Equations (3) and (4) are used to derive the perm. ble. conditions under which the frictional heat city calculations. As the wear "S" increases when can be dissipated from the bearing at a the temperature climbs above 80°C and p = 2,500 N/cm2 (2) temperature differential of ∆ϑ: perm. presumably rises proportionately­ with the ∙· ∙ ∙· length of running time, it can be calculated For bearings with higher sliding velocities, K1 π · λ ∆ϑ (p · v) = + as follows: p has to be limited according to equati- perm. perm. µ ∙ s on (5) or (6). χ ϑ ∙ ϑo S = k ∙ ∙ t [mm] (7) The heat developing in the bearing is calcu- K · ∙ ∙ · · ∙ 2 π λs ∆ϑ (5) ( ϑo ) lated as follows from the work consumed µ ∙ 2 ∙ l by friction: where

Q1 = p ∙· I ∙ d ∙· v ∙· µ (3) v = sliding velocity (m/s) k = 1/6 (constant) χ µ = coefficient of friction This calculation cannot be applied to = 3 (constant) bearings with continuous water or oil ϑ = service temperature The values for the coefficient of friction can . ϑo = 80°C be taken from Table 3. t = time in hours Equation (5) is a quantity equation. The maximum permissible wear rate to be expected with Plastic Plain Bearings is S = 0.2 mm (8)

RE 82 950 Plastic Plain Bearings

2000

1000

500

300 ] 2 200

100 e p [N/cm

essur 50

30 · sec N2 m 20 [ cm ] 150 100 9000

10 6000 4200 · min

Specific bearing pr N2 m 3000 cm 1800 [ ] 5 1200 7 10 15 20900 30 50 70 600 3 420 p·v 2

1 0,1 0,2 0,3 0,5 132 510203050 100 200 500 Circumferential velocity of the shaft v [m/min]

Load capacity of Plastic Plain Bearings as a function of the circumferential velocity of the shaft and the p · v value. Figure 6

Nominal life Bearing clearance The nominal life of Plastic Plain Bearings The bearing clearance must be dimensi- can be calculated from equation (7) as oned such that the bearing will not jam follows: in the event of excessive enlargement of Plastic Plain Bearings as a result of tem- S perature rises or changes in the moisture t = perm. [hours] (9) χ content. Conversely, the clearance must not ϑ − ϑo k · be unnecessarily wide, as this would cause ( ϑ o ) irregular running characteristics. This equation is valid for ϑ > 80°C. Calcu- Experience has shown that, even in the lating the nominal life for worst case, there should still be a minimum ϑ < 80°C would serve no useful pur- clearance of pose as the bearing wear is hardly h = 0.004 ∙· d [mm] (10) detectable and thus not reproducible. 0

RE 82 950 Plastic Plain Bearings

For an increase in temperature of ∆ϑ1 and ϑmax is the maximum temperature arising a relative change in the linear dimensions at the bearing point. This is arrived at by due to a rise of εf in the moisture content, adding ∆ϑ according to equation (6) to the volume of Plastic Plain Bearings will the ambient tem-pera­ ture.­ ϑmax should not increase by exceed 80°C. If the ambient temperature is

20°C, then ∆ϑ1 = ∆ϑ.

∆V = 3 ∙· (εf + α · ∆ϑ1) ∙· V (11) Equation (11) applies when expan­ ­sion is not restricted in any direction. However, where even assuming that the full volume change will occur and that the material will expand εf = linear expansion factor for the moisture content: exclusively to the detriment of the bearing = 0.003 for non water lubricated clear­ance, the calculation­ still provides an bearings additional safety margin. The clear­ance = 0.020 for water lubricated bear- would in this case narrow by ings 3 ∙· s ∙· (ε + α ∙· ∆ϑ ) α = Coefficient of linear expansion ac- f 1 cording to Table 1 1 Taking the minimum permissible clearance = 7 ∙· 10-5 into consideration and remembering that a K great proportion of the change in volume is ∆ϑ = ϑ -20°C taken up by the compensating gap in Plas- 1 max tic Plain Bearings, the minimum permissible clearance can be calculated as follows:

h = 0.004 ∙· d + 3s (εf + α ∙· ∆ϑ1) [mm] (12)

RE 82 950 Plastic Plain Bearings

The axial load that can be taken up by the Permissible p ∙· v value According to Table 3, a value of 0.12 is to bearing collar must never ex-ceed­ 25% of According to Table 6, the perm. p ∙· v value be substituted for µ the radial load capacity. for non-recurring greasing of the Plastic The shaft collar diameter must also Plain Bearing R132052000 is 31 [N/cm2 ∙· · ∆ϑ = 26,2 ∙ 0,12 correspond to dimension "D" in Table 5. 1 8,33 m/s]. -3 · + The values given are valid for a bearing 37,3 ∙ 10 ∙ 0,8 20 2 ( ) temperature of =< 80°C and an ambient For p = 250 N/cm and the permis­ ­sible 2 = 51 K or °C temperature of 20°C. p ∙· v value 31 [N/cm ∙· m/s] a circumferen- tial velocity of 7.5 Calculating the nominal life Calculation example [m/min] can be read off from the chart (Figure 6). In this case, calculating the nominal life A gear drive shaft is to run in Plastic Plain would not serve any useful purpose­ since Bearings. The shaft diameter­ is 20 mm and The circumferential velocity of the drive the arising bearing tem-perature­ does not the drive speed is 100 min-1. Two Plastic shaft is: exceed 80°C. The actual p ∙· v value of 26.2 Plain Bearings R132052000 have been v = d ∙· π ∙· n = 0.02 ∙· 3.14 ∙· · 100 [N/cm2 ∙· m/s] lies below the perm. p ∙· v selected. The load per bearing is 1,000 N. = 6.28 m/min value of 31[N/cm2 ∙· m/s]. The bearings are to be greased once on The calculated value is less than the value Calculation of the required bearing installation. The permissible bearing wear is of 7.5 m/min read off from the chart, i.e. clearance assumed to be 0.2 mm. the proposed lubrication method is ade- In order to achieve long service life the According to equation (12), the required quate. The actual bearing temperature­ should not exceed bearing clearance can be calculated as p ∙· v value is: follows: 80°C (at 20°C ambient temperature). 6.28 The task is to check whether the chosen­ p ∙· v = 250 ∙ · h = 0.004 ∙· d + 3 ∙· s (ε + α ∙· ∆ϑ) bearing type is adequate, what service life 60 f εf is replaced by 0.003 can be expected and what bearing clear- = 26.2 [N/cm2 ∙· m/s] α = 7 ∙· 10-5 ance is required. h = 0.004 ∙· 20 Calculation of the actual bearing + 3 ∙· 0.8 (0.003 + 7 ∙· 10-5 · ∙· 51°) Average contact pressure temperature ∆ϑ = 0.096 mm. According to equation (1), the average After entering the appropriate values in The bearing clearance after mounting contact pressure is: equation (6), the arising bearing should be 0.096 mm. Alternatively, the F 1000 temperature can be calculated as follows: housing bore should be dimensioned as p = = = 250 N/cm2 l · ∙ d 2.0 · ∙· 2.0 follows: p ∙ v ∙ µ ∆ϑ = H7 Housing bore = (d + 2 ∙· s + h) -3 1 + 8,33 H7 H7 37,3 ∙ 10 · ∙ ( s l ) = (20 + 2 ∙· 0.8 + 0.096) = 21.7

RE 82 950