1. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Pivot and collar bearings are used in machines to support an axial load on a rotating shaft
Provided bearings are not lubricated, or only partially lubricated, the laws of dry friction may be applied to determine the moment M needed to turn the shaft when it supports an axial force P

2. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Frictional Analysis
The collar bearing on the shaft is subjected to an axial force P and has a total contact area π(R22 – R12)
Normal pressure p is considered to be uniformly distributed over this area – a reasonable assumption provided the bearing is new and evenly distributed
Since ∑Fz = 0, then p measured as a force per unit area
p = P/π(R22 – R12)

3. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Frictional Analysis
The moment needed to cause impending rotation of the shaft can be determined from moment equilibrium about the z axis
A small are element dA = (r dθ)(dr) is subjected to both a normal force
dN = p dA
and an associated frictional force,
dF = μs dN = μsp dA = {(μsP)/[π(R22 – R12)]}dA
Normal force does not create a moment about z axis of the shaft, however, frictional force does, dM = r dF

4. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Frictional Analysis
For impending rotational motion
Integrating over entire bearing area Or
This equation gives the magnitude of moment required for impending rotation of the shaft

5. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Frictional Analysis
Frictional moment developed at the end of the shaft, when it is rotating at constant speed, can be found by substituting μk by μs
When R2 = R and R1 = 0, as in the case of a pivot bearing
M = 2/3 μs PR
Apply only to bearings subjected to constant pressure
If the pressure is not uniform, a variation of the pressure as a function of the bearing area must be determined to obtain moment

6. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Example 8.10
The uniform bar has a total mass m. if it is
Assumed that the normal pressure acting at the
contracting surface varies linearly along the length
of the bar, determine the couple moment M
required to rotate the bar. Assume that the bar’s
width a is negligible in
comparison to its length l.
the coefficient of static
friction is equal to μs.

7. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
View Free Body Diagram
Solution
FBD of the bar
Bar has a total weight of W = mg
Intensity wo of the distributed
lead at the center (x = 0) is
determined from vertical force
equilibrium

8. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Solution
Since w = 0 at x = l/2, for distributed load function,
For magnitude of normal force acting on a segment of area having length dx

9. 8.6 Frictional Forces on Collar Bearings, Pivot Bearings and Disks
Solution
For magnitude of the frictional force acting on the same element of area,
For moment created by this force about z axis,
Summation of moments by integration,

10. 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

11. 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

12. 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 Or
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

13. 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

14. 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.

15. 8.7 Frictional Forces on Journal Bearings
View Free Body Diagram
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

16. 8.7 Frictional Forces on Journal Bearings
Solution
Using the simplification
Summing moments about P2,
For more exact analysis,

17. 8.7 Frictional Forces on Journal Bearings
Solution
For radius of friction circle,
Therefore,

18. 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,

19. 8.8 Rolling Resistance
If a rigid cylinder of weight W rolls at constant velocity along a rigid surface, the normal force exerted by the surface on the cylinder acts at the tangent point of contact
Under these conditions, provided the cylinder does not encounter frictional resistance from the air, motion will continue indefinitely
No materials are perfectly rigid and the reaction of the surface on the cylinder consists of a distribution of normal pressure

20. 8.8 Rolling Resistance
Consider the cylinder to be made of a very hard material and the surface on which it rolls to be relatively softer
Due to its weight, the cylinder compresses the surface underneath it
As the cylinder rolls, the surface material in front of the cylinder retards the motion since it is being deformed whereas the material in the rear is restored from the deformed state and tends to push the cylinder forward

21. 8.8 Rolling Resistance
The normal pressures acting on the cylinder in this manner are represented by their resultant forces Nd and Nr
The magnitude of the force of deformation Nd and its horizontal component is always greater than that of restoration Nr and consequently, a horizontal driving force P must be applied to maintain the motion

22. 8.8 Rolling Resistance
Rolling resistance is caused primarily by this effect although the result of surface adhesion and relative micro-sliding between the surfaces of contact
Actual force P needed to overcome these effects is difficult to determine, so we consider the resultant of the entire normal pressure acting on the cylinder
N = Nd + Nr

23. 8.8 Rolling Resistance Force P acts at an angle θ with the vertical
To keep the cylinder in equilibrium, rolling at constant rate, N must be concurrent with the driving force P and the weight W
Summation of moment about A,
Wa = P (r cosθ)
Wa ≈ Pr
P ≈ (Wa)/r

24. 8.8 Rolling Resistance Example 8.12 A 10kg steel wheel has a radius
of 100mm and rest on an
inclined plans made of wood. If θ
is increased so that the wheel
begins to roll down the incline
with constant velocity when θ =
1.2°, determine the coefficient of
rolling resistance.

25. 8.8 Rolling Resistance Solutions FBD of the wheel
View Free Body Diagram
Solutions
FBD of the wheel
Wheel has impending motion
Normal reaction N acts at point A defined by dimension a
Summing moments about point A
Solving

26. Chapter Summary Dry Friction
Frictional forces exist at the rough surfaces of contact
They act on a body so as to oppose the motion or tendency of the motion of the body
A static friction force approaches a maximum value of Fs = μsN
μs is the coefficient of kinetic friction
Solution of the problem requires first drawing of the FBD

27. Chapter Summary Dry Friction
If the unknowns cannot be determined strictly from the equations of equilibrium, and the possibility of slipping can occur, then the friction equation should be applied at the appropriate points of contact in order to complete the solution
It may be possible for slender objects to tip over and this situation should also be investigated

28. Chapter Summary Wedges, Screws, Belts and Bearings
A frictional analysis of these objects can be performed by applying the frictional equation at the points of contact and then using the equations of equilibrium to relate the frictional force to other external force acting on the object
By combining the resulting equations, the forces of friction can be eliminated so that the force needed to overcome the effects of friction can be found

29. Chapter Summary Rolling Resistance
The resistance of a wheel to roll over a surface is caused by deformation between two materials of contact
This effect causes the resultant normal force acting on the rolling body to be inclined so that it provides a component that acts in the opposite direction of the force causing the motion
The effect is characterized using coefficient of rolling resistance

30. Chapter Review

31. Chapter Review

32. Chapter Review

33. Chapter Review

34. Chapter Review

35. Chapter Review

36. Chapter Review

37. Chapter Review