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**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

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**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)

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**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

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**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

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**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

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**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.

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**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

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**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

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**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,

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**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

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**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

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**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

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**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

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**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.

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**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

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**8.7 Frictional Forces on Journal Bearings**

Solution Using the simplification Summing moments about P2, For more exact analysis,

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**8.7 Frictional Forces on Journal Bearings**

Solution For radius of friction circle, Therefore,

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**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,

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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

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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

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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

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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

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**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

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**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.

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**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

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**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

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**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

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**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

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**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

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Chapter Review

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Chapter Review

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Chapter Review

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Chapter Review

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Chapter Review

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Chapter Review

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Chapter Review

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Chapter Review

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