1. 8.7 Frictional Forces on
Journal Bearings
When a shaft or axle is subjected to lateral
loads, a journal bearing is used for support
Well-lubricated journal bearings are subjected to
the laws of fluid mechanisms, in which the
viscosity of the lubricant, speed of rotation and
the amount of clearance between the shaft and
the bearing is used to determine the frictional
resistance of the bearing
When the bearing is not lubricated or is only
partially lubricated, analysis of the frictional
resistance can be based on the laws of dry
friction

2. 8.7 Frictional Forces on
Journal Bearings
Frictional Analysis
Consider a typical journal bearing support
As the shaft rotates in the direction shown, it rolls
up against the wall of the bearing to some point A,
where slipping occurs
If the lateral load acting at
the end of the shaft is P,
the bearing reactive force
R acting at A is equal and
opposite to P

3. 8.7 Frictional Forces on
Journal Bearings
Frictional Analysis
Moment needed to maintain constant
rotation of the shaft can be found by the
summation of moments about the z axis
of the shaft
∑ M z = 0;
M − ( R sin φk )r = 0
Or M = Rr sin φk
The dashed circle with radius rf is called
the friction circle and as the shaft
rotates, the reaction R will always be
tangent to it

4. 8.7 Frictional Forces on
Journal Bearings
Frictional Analysis
If the bearing is partially lubricated,
µk is small and therefore
µk = tanΦk ≈ sinΦk ≈ Φk
Frictional resistance
M ≈ Rrµk

5. 8.7 Frictional Forces on
Journal Bearings
Example 8.11
The 100mm diameter pulley fits loosely on a
10mm diameter shaft for which the coefficient of
static friction is µs = 0.4. Determine the minimum
tension T in the belt needed
to (a) raise the 100kg block
and (b) lower the block. Assume
that no slipping occurs between
the belt and the pulley and
neglect the weight of the pulley.

6. 8.7 Frictional Forces on
Journal Bearings
Solution
Part (a)
FBD of the pulley
When the pulley is subjected to
belt tensions of 981N each, it
makes contact with the shaft at
point P1
As tension T is increased, the
pulley will roll around the shaft to
point before motion P2 impends
Friction circle’s radius, rf = r sinΦs

7. 8.7 Frictional Forces on
Journal Bearings
Solution
Using the simplification
sin φs ≈ (tan φs ≈ φs )
rf ≈ rµ s = (5mm)(0.4) = 2mm
Summing moments about P2,
∑ M P2 = 0;981N (52mm) − T (48mm) = 0
T = 1063N = 1.06kN
For more exact analysis,
φs = tan −1 0.4 = 20.8o

8. 8.7 Frictional Forces on
Journal Bearings
Solution
For radius of friction circle,
rf = r sin φs = 5 sin 21.8o = 1.86mm
Therefore,
∑ M P2 = 0;
981N (50mm + 1.86mm) − T (50mm − 1.86mm) = 0
T = 1057 N = 1.06kN

9. 8.7 Frictional Forces on
Journal Bearings
Solution
Part (b)
When the block is lowered, the
resultant force R acting on the
shaft passes through the point
P3
Summing moments about this
point,
∑ M P3 = 0;
981N (48mm) − T (52mm) = 0
T = 906 N