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From tests 24 and 1, the following are tabulated:

Shaft diamater (mm)

F (kg force)

DeltaD.L/ (N.t)

DeltaD . L/(N.t.F)

8.0

9.5

0.000829

0.0000873

8.0

6.8

0.000595

0.0000875

15.8

20.0

0.001460

0.0000730

TABLE II - MISC CALCULATED RESULTS

 

From the above limited data, one can conclude that

 

DeltaD is proportional to F . N.t / L

(wear in millimetres is proportional to force x rpm x time / bearing length)

And that deltaD is independent of D (wear dimension is independent of diameter)

For a steel shaft of diamater Ds metres running at constant angular velocity w radians/second for time t seconds with coefficient of friction f in a wooden bearing of width L metres, subjected to a radial loading of F Newtons, resulting in wear of d metres, then:

Material removed = L . Ds . d cubic metres

Heat Produced = 0.5 f.F.Ds.w.t joules

Now we have found experimentally that:

deltaD is proportional to F.N.t / L

Therefore F .w.t is proportional to L.deltaD

Therefore we can reasonably assume that

0.5 .f.F.Ds.w.t is proportional to L.Ds.deltaD

Thus for any given timber and lubrication treatment operating below certain bearing temperature limits and bearing pressures, then

(Material removed by wear) is proportional to (Heat expended in the Bearing)

 

Since friction coefficient f is also a property of timber type and lubrication treatment, it is possible to adopt the term "wear coefficient" for steel/wood bearings where:

Wear Coefficient Cwear = F.t / (L .w. deltaD) Newtons per square metre

For conditions where Force F and shaft speed w are not constant with time then the intregral can be used in the above calculation...

Values of Cwear for the timbers and lubrication treatments dealt with during the experiements are tabulated below: