1.1 Friction

   

 1.1.1 Introduction

When a body moves or tends to move on another body, a force appears between the surfaces. This force is called force of friction and it acts opposite to the direction of motion. Its line of action is tangential to the contacting surfaces. The magnitude of this force depends on the roughness of surfaces.

 1.1.2 Types Of Friction

There are two types of friction :

(a) Friction in un-lubricated surfaces or dry surfaces - The frictional force which exist when the contacting surfaces are dry is called as dry friction.
(b) Friction in lubricated surfaces- The frictional force which exist when the contacting surface has a film of lubricant on it such that the contacting surface are not in direct contact is called as film friction or friction in lubricated surface.

1.2 Laws Of Dry Friction

A graph between the applied force P and the frictional forces is shown below

The laws of dry friction, are based on experimental evidences, and as such they are empirical in nature :

(a) The friction force is directly proportional to the normal reaction between the surfaces.
(b) The frictional force opposes the motion or its tendency to the motion.
(c) The frictional force depends upon the nature of the surfaces in contact.
(d) The frictional force is independent of the area and the shape of the contacting surfaces.

1.3 Static And Kinetic Friction

Consider a block of weight W subject to a horizontal force P, if P is small the block will not move and the static frictional force F will balance P. Now is we slowly start increasing P, the frictional force F also starts increasing up to a maximum limit, say Fm. If we still increase P the frictional force will no longer be able to keep the block in static position and the block will start moving in the direction of the applied force P. Under this condition the frictional force Fm drops to a new value say Fk, where Fk is the kinetic friction. Unlike Fs (static friction) which is a self-adjusting force till the maximum value if Fm, Fk remains constant.

                                                                       Static and Kinetic Friction

Experimental data shows that Fm is proportional to the reaction force between the two contacting surface (in our case the weight of the block).

A graph between the applied force P and the frictional forces is shown below

  

1.4 Coefficient Of Friction

The ratio between the maximum static frictional force and the normal reaction RN remains constant which is known as coefficient of static friction denoted by Greek letter µ.

${Coefficient\quad of\quad friction =}\frac {\hbox {Maximum static Frictional Force } } {\hbox{Normal reaction}}$  $or, \quad \mu=\frac {F}{R_N} \qquad --- \hbox{ (Eqn.-11.1) }$

The maximum angle Φ which the resultant reaction RN makes with the normal reaction RN is known as angle of friction. It is denoted by Φ

$\tan \phi= \frac {F}{R_N} = \mu$ or,  $\phi = \tan^{-1}\mu \qquad --- \hbox{ (Eqn.-11.2) }$

1.5 Angle of Repose

Angle of repose is that angle of an inclined plane where a block kept on the surface of the inclined plane just starts sliding down. Angle of repose is equal to the angle of friction. To find it value let us consider a mass 'm' resting on an inclined plane. Here if we slowly increase the inclination angle of the plane, a stage will come when the block of mass 'm' will tend to slide down. Let us now try to find the value of this angle by making a free body diagram of the mass 'm'.

Angle Of Repose (Figure f.5.a)

$P=R_N \quad \sin \alpha$ and  $mg=R_N \quad \cos \alpha$

therefore,   $\frac{P}{mg}=\tan \alpha \qquad --- \hbox{ (Eqn.-11.3) }$

$But \qquad \frac{p}{mg}=\frac {F}{R_N}=\mu=\tan\phi$

$than \qquad \tan\alpha=\tan\phi \qquad --- \hbox{ (Eqn.-11.4) }$, thus 

$\hbox{Angle of repose}, \alpha =\hbox{Angle of friction}, \phi$

Least force required to drag a body on a rough horizontal plane

Suppose a block, of mass m, is placed on a horizontal rough surface as shown in Fig.-(11.3) and attractive force P is applied at an angle θ with the horizontal such that the block just tends to move.

Least force required to drag a body on a rough horizontal plane (Fig.- 11.3)

For satisfying the equilibrium conditions the forces are resolved vertically and horizontally.

$\Sigma V=0$, and  $mg-P \sin \theta=R_N$

$or \qquad R_N = (mg-p \sin \theta)$

$For \Sigma H=0$

$P \cos \theta=F= \mu R_N= \mu (mg-P \sin \theta)$

$P \cos \theta = \frac{\sin \phi}{\cos \phi}(mg-P \sin \theta)$

$P \cos \theta \cos \phi =mg \sin \phi -Psin \theta \sin\phi$

$P \cos \theta \cos \phi+P \sin \theta sin\phi=mg \sin \phi$

$P(\cos \theta +\cos \phi + \sin \theta \sin \phi)=mg \sin \phi$

$P \cos(\theta - \phi)=mg \sin \phi$

$P=\frac {mg \sin \phi}{\cos(\theta-\phi)}\qquad --- \hbox{ (Eqn.-11.5) }$  

For P to be least, the denominator $\cos (θ-Φ)$ must be maximum and it will be so if

$\cos(\theta - \phi) \qquad or \qquad \theta - \phi =0$

$\theta = \phi \hbox{for least value of P}$

$P_{least}=mg \sin \phi \qquad --- \hbox{ (Eqn.-11.6) }$  

Hence, the force P will be the least if angle of its inclination with the horizontal : θ is equal to the angle of friction Φ.

1.6 Horizontal force required to move the body

Up the Inclined Plane

The force P has been applied to move the body up the plane. Resolving all the forces parallel and perpendicular to the plane and writing equations, we get

$P \cos \alpha = \mu R_N + mg \sin \alpha$, and  $P \sin \alpha = R_N - mg \cos \alpha$,

therefore, $\qquad R_N = P \sin \alpha + mg \cos \alpha$

 

Horizontal Force Required to move the body (Fig.- 11.4)

Substituting the value of RN, We get

$P \cos \alpha = \mu (P \sin \alpha mg \cos \alpha)+mg \sin \alpha$

$Assuming \qquad \mu=\tan \phi$, where Φ is angle of friction.

$Pcos \alpha = \tan \phi (P \sin \alpha +mg \cos \alpha)+mg \sin \alpha$

$or \qquad P(\cos \alpha - \tan \phi \sin \alpha)=mg (\tan \phi \cos \alpha sin \alpha)$

$or \qquad p=mg \frac {(\frac{\sin \phi}{\cos \phi}\cos \alpha + \sin \alpha )}{( \cos \alpha - \frac {\sin \phi}{\cos \phi}\sin \alpha )}$

$=mg \frac{(\sin \phi \cos \alpha + \sin \alpha \cos \phi)}{(\cos \alpha \cos \phi - \sin \phi \sin \alpha)}$

= $mg \frac{\sin ( \alpha + \phi)}{\cos (\alpha + \phi )}=mg \tan (\alpha + \phi) \qquad --- \hbox{ (Eqn.-11.7) }$  

Down the Plane

In this case body is moving down the plane due to the application of force P. Resolving the forces parallel and perpendicular to the plane and writing the equations, we get

$P \cos \alpha = \mu R_N - mg \sin \alpha$ and  $P \sin \alpha = mg \cos \alpha - R_N$ 

therefore, $R_N = mg \cos \alpha - P \sin \alpha$

Down the Plane (Fig.- 11.5)

Substituting for RN, we get

$Pcos \alpha = \mu(mg \cos \alpha-P \sin \alpha)$, $Since, \quad \mu = \tan \phi$ , therefore 

$Pcos \alpha = \tan \phi (mg \cos \alpha-P \sin \alpha)-mg \sin \alpha$ , $or \qquad p(\cos \alpha + \tan \phi \sin \alpha) = mg (\tan \phi \cos \alpha - \sin \alpha)$

$or \qquad P=mg \frac { (\frac {\sin \phi}{\cos \phi}\cos \alpha - \sin \alpha ) } { (\cos \alpha + \frac {\sin \phi}{\cos \phi}\sin \alpha ) }$

$=mg \frac {(\sin\phi \cos \alpha - \cos \phi \sin \alpha)}{(\cos \alpha \cos \phi +\sin \phi \sin \alpha)}$

$or \qquad P = mg \frac {\sin (\phi - \alpha)}{\cos(\phi - \alpha) }=mg \tan (\phi - \alpha) \qquad --- \hbox{ (Eqn.-11.8) }$

This means for force P to be applied $\alpha < \phi$ or else the body will slide down by itself 

1.7 Screw and nut friction

An inclined plane is a perfect analogy to explain screw and nut friction. The various dimension can be seen in the figure below, where α is the helix angle and dm is the screw mean diameter. By geometry

$tan \alpha = \frac {P}{\pi dm}$

Screw And Nut Friction (Fig.- 11.6)

For Upward Motion

$Effort \quad P_0 = mg \tan \alpha \quad \hbox{ without friction } \quad [\Phi=0]$

$Effort \quad P_0 = mg \tan (\alpha + \Phi)\hbox{ with friction }$  

$Efficiency \qquad e_{up}=\frac {\tan \alpha}{\tan (\alpha + \phi ) } \qquad --- \hbox{ (Eqn.-11.10) }$

For Downward Motion

$P_0= mg \tan ( \phi - \alpha) \qquad --- \hbox{ (Eqn.-11.11) }$

 1.7.1 Self locking screws

$P_0 = mg tan (α + Φ) \quad when, P_0 = 0 => tan (α + Φ) = 0 \quad or \quad \alpha = -\phi$

This means that if $\alpha >\phi$ then the mass placed on screw will start moving downwards by its own weight. When α < φ the screw will be self locking.

 1.7.2 Condition for maximum efficiency

Efficiency of screw is given by: $e_{up} = \frac {\tan \alpha} {\tan (\alpha + \phi)} \qquad --- \hbox { (Eqn.-11.13)}$  

Differentiating efficiency equation w.r.t 'α' and equating to zero we get, $\qquad \alpha=\frac{\pi}{4}-\frac {\phi}{2} \qquad --- \hbox{ (Eqn.-11.14) }$

Substituting the value of α in equation of efficiency, we get  $e_{max} = \frac {1-\sin \phi}{1+\sin \phi} \qquad --- \hbox{ (Eqn.-11.12) }$

 

 1.7.3 Screw jack

A screw jack is a simple device used to lift load. A very screw jack is used to lift car tyre in case of deflated tyre. Threads of a screw jack can be square or v-type. A screw jack with a tommy bar reduces the effort required to raise a load considerably. All screw jack are self locking.Let us try to find the effort required to raise a load by screw jack.

Screw Jack

Screw Jack (Fig.- 11.7)

Let,

m = Mass on the jack,

P = Force applied at the screw tangentially in a horizontal plane,

Pe = Horizontal force applied tangentially at the end E of a tommy bar in a horizontal plane, and

L = Horizontal distance between central axis of the screw and the end E of the bar as shown.

r = radius of the screw.

p = pitch of the screw

α = helix angle

In the screw jack nut is stationary and the screw is rotated with the help of tommy bar.

$P_e \times L=P\times r$ or $P_e = \frac {P \times r}{L}$ 

Screw Jack (Fig.- 11.8) 

As shown in the wedge diagram above

$P =mg \tan(\alpha + \phi)=\frac {mg(\tan \alpha + \tan \phi)}{1-\tan \alpha \tan \phi}$, $But \qquad \quad \tan \alpha = \frac {p}{\pi d_m} \qquad and \qquad \tan \phi = \mu$ 

Substituting for tan α and tan Φ

$P=\frac { mg(\frac {p}{\pi d_m}+ \mu ) }{ 1- \frac {p}{\pi d_m} \times \mu }=\frac {mg (p +\mu \pi d_m)}{(\pi d_m - p \mu)}$, $Hence, \qquad \quad P = \frac {mg (p +\mu \pi d_m)}{(\pi d_m - p \mu)}$

$or, P_e = \frac { mg \times r (P +\mu \pi d_m)}{L (\pi d_m - P \mu)}$

Mechanical advantage of the jack with tommy bar

$= \frac {\hbox{Load lifted}}{\hbox{Force applied}} = \frac {mg}{P_e}=\frac {L}{r} \Bigg[ \frac {(\pi d_m - p \mu)} {p + \mu \pi d_m}\Bigg] \qquad --- \hbox{ (Eqn.-11.15) }$

1.8 Pivot And Collar Friction

These are bearing surface which are provided to take the axial load or thrust that comes on a shaft. If the bearing surface is provided at the end of the shaft it is called a pivot and if the bearing surface is provided along the length of the shaft it is called a coller bearing. Pivot bearings can be flat or conical. We will now try to find the frictional moment of pivot bearing. A measure of fictionL moment is a direct measure of the power loss that happens in these kinds of bearings.

 1.8.1 Flat pivot

Let,

W = Axial thrust or load on the bearing,
R = External radius of the pivot,
p = Intensity of pressure, and
µ = Coefficient of friction between the contacting surfaces.
Consider an elementary ring of the bearing surfaces, at a radius r and of thickness dr as shown in Figure

Flat Pivot (Fig.- 11.10)

Axial load on the ring 

$dW = p \times 2 \pi r \times dr$

$\hbox {Total load} \qquad \quad W = \int_0^R p \times 2 \pi r \times dr$

Frictional force on the ring 

$dF = \mu \times dW = \mu \times p \times 2 \pi r \times dr$

Frictional moment about the axis of rotation  

$dM = dF \times r = \mu \times p \times 2 \pi r^2 \times dr$

Total frictional moment  

$M= \int_0^R dM = \int_0^R \mu \times p \times 2 \pi r^2 \times dr \qquad --- \hbox{ (Eqn.-11.19) }$

   Uniform pressure

If the intensity of pressure P is assumed to be uniform and hence constant.

$W = p \times 2 \pi \int_0^R r \times dr = p \times 2 \pi \Bigg[ \frac {r^2}{2} \Bigg]_0^R = p \times 2 \pi \times \frac {R^2}{2}$

$W= p \times \pi R^2$

$M = \mu \times p \times 2 \pi \int_0^R r^2 dr$

$M = \mu p \times 2 \pi \times \Bigg[\frac {r^3}{3} \Bigg]_0^R = \frac {2}{3} \mu \times p \times \pi \times R^3$

But $\qquad \quad p \pi R^2 =W$

$M = \frac {2}{3} \mu WR = \mu W \times \frac {2}{3} R \qquad --- \hbox{ (Eqn.-11.20) }$

The friction force µW can be considered to be acting at a radius of 2/3 R.

   Uniform Rate of Wear

$W = \int_0^R p \times 2 \pi r \times dr$

As the rate of wear is taken as constant and proportional to pr = a constant say c. Substituting for pr = c in the above equation.

$W = \int_0^R 2 \pi \times c \times dr = 2 \pi R \times c$

$c= \frac {W}{2 \pi R} \qquad --- \hbox{ (Eqn.-11.21) }$

Total frictional moment

$M = \int_0^R \mu \times p \times 2 \pi r^2 \times dr = \int_0^R \mu \times 2 \pi \times c \times r \times dr$

$\mu \times 2 \pi \times c \frac {R^2}{2}= \mu \times 2 \pi \times \frac {W}{2 \pi R} \times \frac {R^2}{2}$

$M= \mu W \times \frac{R}{2} \qquad --- \hbox{ (Eqn.-11.22) }$

Thus, the frictional force : µW acts at a distance R/2 from axis.

 1.8.2 Conical pivot

A truncated conical pivot is shown in Figure.
Let 2α = The cone angle,
W = The axial load/thrust,
R1 = The outer radius of the cone,
R2 = The inner radius of the cone, and
p = Intensity of pressure which will act normal to the inclined surface of the cone as shown in Figure

Conical Pivot And Enlarged View Of The Ring (Fig.- 11.11)

Consider an elementary ring of the cone, of radius r thickness dr and of sloping length dl as shown. Enlarged view of the ring is shown in Figure

$M = \int_{R_2}^{R_1} \mu \times p \times 2 \pi r^2 \times \frac {dr}{\sin \alpha}$

Normal load on the ring

$dW_n = p \times 2 \pi r \times dl = p \times 2 \pi r \times \frac {dr}{\sin \alpha}$

Axial load on the ring

$dW = dW_n \times \sin \alpha$

$= p \times 2 \pi r \times \frac {fr}{\sin \alpha} \times \sin \alpha =p \times 2\pi r \times dr$

Total axial load on the bearing

$W = \int_{R_2}^{R_1} dr \times p \times 2 \pi r$

Frictional force on the elementary ring

$dF = \mu dW_n =\mu \times p \times 2 \pi r \times dl = \mu \times p \times 2 \pi r \times dl = \frac {dr}{\sin \alpha}$

Moment of the frictional force about the axis of rotation

$dM = \mu \times p \times 2 \pi r^2 \times \frac {dr}{\sin \alpha}$

R2=0

Full Conical Pivot (Fig.- 11.12)

r=0

$W = p \pi R_1^2$

and $M = \frac {\mu W}{\sin \alpha} \times \frac {2}{3} R_1$ 

In this case of frictional force   $\Bigg(\frac {\mu W }{\sin \alpha} \Bigg)$acts at a radius 2/3 from the axis. 

   Uniform pressure, p is constant

$W = p \times 2 \pi \int_{R_2}^{R_1} r dr = p \times 2\pi\Bigg[ \frac {r^2}{2} \Bigg]_{R_2}^{R_1}$,  $or \qquad \quad W= p \times \frac {2 \pi}{2}( R_1^2 - R_2^2)=p \times \pi ( R_1^2 - R_2^2)$

$And \qquad \quad M=\frac {\mu \times p \times 2 \pi}{\sin \alpha} \int_{R_2}^{R_2} r^2 dr =\frac {\mu \times p \times 2 \pi}{\sin \alpha} \times \frac {1}{3} ( R_1^3 - R_2^3)$

$But \qquad \quad W = p \times \pi ( R_1^2 - R_2^2)$, or  $p= \frac {W}{(\pi R_1^2 - R_2^2)}$

Substituting for p above,

$M= \frac {\mu W}{\sin \alpha} \times \frac {2}{3} \frac {( R_1^3 - R_2^3)} {( R_1^2 - R_2^2)}$

   Uniform wear

$W = \int_r^R p \times 2\pi r \times dr$

Science the rate of wear is uniform, Therefore, pr = constant c

Substituting for pr

$W = \int_{R_2}^{R_1} 2 \pi c \times dr = 2 \pi c \times (R_1 - R_2)$  or  $c= \frac {W}{2 \pi (R_1 - R_2)}$

frictional moment

$M = \int_{R_2}^{R_1} \mu \times p \times 2 \pi r^2 \times \frac{dr}{\sin \alpha}$

$= \int_{R_2}^{R_1} \mu \times 2\pi r \times c \times \frac {dr}{\sin \alpha} \qquad \qquad \Bigg[ since \quad pr = c\Bigg]$

$=\frac {\mu \times 2 \pi c}{ \sin \alpha} \int_{R_2}^{R_1} r dr$

$= \frac {\mu \times 2 \pi c}{ \sin \alpha} \frac {(R_1^2-R_2^2)}{2} = \frac {\mu \times \pi c}{\sin \alpha}(R_1^2-R_2^2)$

$But \qquad \quad c= \frac {W}{2 \pi (R_1 - R_2)}$

$M= \frac {\mu W}{\sin \alpha} \times \frac {(R_1 + R_2)}{2}$

"Thus, the frictional force    $\frac {\mu W}{\sin \alpha}$  acts at a radius= $\frac {R_1+R_2}{2}$ 

For the full conical pivot, $R_2=0$; 

$M = \frac {\mu W}{\sin \alpha} \times \frac {R_1}{2}$

Thus, the frictional force    $\frac {\mu W}{\sin \alpha}$  acts at a radius= $\frac {R_1}{2}$

 1.8.3 Collar bearing

A collar bearing which is provided on to a shaft, is shown in Figure.
Let W = The axial load/thrust,
R1 = External radius of the collar, and
R2 = Internal radius of the collar.
Consider an elementary ring of the collar surface, of radius r and of thickness dr as shown in Figure.

Collar Bearing (Fig.- 11.13)

Axial load on the ring

$dW = p \times 2 \pi r \times dr$

Total axial load

$W = \int_{R_2}^{R_1} p \times 2 \pi r \times dr$

Frictional force on the ring

$dF = \mu \times p \times 2 \pi r \times dr$

Frictional moment on the ring

$dM = \mu \times p \times 2 \pi r^2 \times dr$

Total frictional moment

$M =\int_{R_2}^{R_1} \mu \times p \times 2 \pi r^2 \times dr$

Substituting for p from above

$M= \mu W \times \frac {2}{3} \frac {( R_1^3 - R_2^3)} {( R_1^2 - R_2^2)}$

   Uniform Wear

Since the rate of wear is uniform

pr = constant c

Substituting for pr above,

$W= \int_r^R c \times 2 \pi dr = c \times 2 \pi \int_{R_2}^{R_1} dr = c \times 2 \pi \times (R_1-R_2)$

$c = \frac {W} {2 \pi (R_1-R_2)}$

$M= \int_{R_2}^{R_1} \mu \times p \times 2 \pi r \times dr = \mu \times 2 \pi \times c \int_{R_2}^{R_1} r \times dr$ $=\mu \times2 \pi \times c\frac {(R_1-R_2)}{2}$

substituting for c

$M = \mu W \times \frac {(R_1+R_2)}{2}$

   Uniform Pressure

uniform pressure, p is constant

$W = p \times 2 \pi \int_{R_2}^{R_1} r \times dr = p \times 2 \pi \frac {( R_1^2 - R_2^2 )}{3}$        $W = p \times \pi ( R_1^2 - R_2^2 )$

$M = \mu \times p \times 2 \pi \int_{R_2}^{R_1} r^2 \times dr = \mu \times p \times 2 \pi \frac {( R_1^3 - R_2^3 )}{3}$

There is a limit to the bearing pressure on a single collar and it is about 40 N/cm2. Where the axial load is more and pressure on each collar is not to be allowed to exceed beyond the designed limit, then more collars are provided as shown in Figure.

Number of collars: $n= \frac {\hbox{Total axial load}} {\hbox{Permissible axial load on each collar}}$

It may be pointed out there is no change in the magnitude of frictional moments with more number of collars. The number of collars, as given above, only limit the maximum intensity of pressure in each collar.

1.9 Clutch

   A clutch is aa mechanical device used to engage and disengage the driving shaft from the driven shaft at the will of the user. A very common use of clutch is in automobiles, however clutches are used in multiple applications.

 1.9.1 Types Of Clutches

Clutches can be classified into two types as follows:
(a) conical clutch, and
(b) the place or disc clutches can be of single plate or of multiple plates.

 1.9.2 Conical Clutch

Cone clutch: The general design and working of a cone clutch is described in the below interactive module.

 

Cone Clutch

Conical Clutch (Fig.- 11.14)

Torque transmitted by a cone clutch - A cone clutch is very similar to the conical bearing

$T = \frac {2 \pi \mu}{3} \times W \times cosec \alpha \Bigg[ \frac { (r_1)^3 - (r_2)^3 } { (r_1)^2 - (r_2)^2 } \Bigg]$

 1.9.3 Advantages and disadvantages of cone clutch

Advantages

(a) Small axial force is required to keep the clutch engaged.
(b) Simple design.
(c) For a given dimension, the torque transmitted by cone clutch is higher than that of a single plate clutch.

Disadvantages

(a) One pair of friction surface only.
(b) The tendency to grab.
(c) The small cone angle causes some reluctance in disengagement.

For calculating frictional moments or torque transmitted on account of friction in clutches, unless otherwise specifically stated uniform rate of wear is assumed. For torque transmitted formulae of conical pivot can be used.

 1.9.4 Single Plate Clutch

The general design and working of a single plate clutch is explained in below interactive module

Single Plate Clutch

Single Plate Clutch (Fig.- 11.18)

If the axial thrust is W, T is the torque transmitted by the clutch, p is the Intensity of axial pressure with which the contact surfaces are held together, r1 and r2 is the External and internal radii of friction faces, and μ is the coefficient of friction, then 
Total frictional torque on the friction surface
(i) Considering uniform pressure,

$T = \frac {2}{3} \times \mu.W \Bigg[ \frac {(r_1)^3 - (r_2)^3}{(r_1)^2 - (r_2)^2} \Bigg]$

(ii) Considering uniform wear,

$T = \frac {1}{2} \times \mu.W (r_1+r_2)$

 1.9.5 Multi Plate Clutch

To increase the capacity of clutch to transmit torque more than one friction surface is used. When the number of surface friction is more than one we it a multi-plate clutch.

1.10 10. Journal Bearing

The portion of a shaft, which revolves in the bearing and subjected to load at right angle to the axis of the shaft, is known as journal as clearly indicated in Figure. The whole unit consisting of the journal and its supporting part (or bearing) is known as journal bearing.

Journal Bearing

 

1.11 Rolling Friction

The frictional resistance arises only when there is relative motion between the two connecting surfaces. When there is no relative motion between the connecting surfaces or stated plainly when one surface does not slide over the other question of occurrence of frictional resistance or frictional force does not arise. When a wheel rolls over a flat surface, there is a line contact between the two surfaces, Friction parallel to the central axis of the cylinder. On the other hand when a spherical body rolls over a flat surface, there is a point contact between the two. In both the above mentioned cases there is no relative motion of slip between the line or point of contact on the flat surface because of the rolling motion. If while rolling of a wheel or that of a spherical body on the flat surface there is no deformation of depression of either of the two under the load, it is said to be a case pure rolling. In practice it is not possible to have pure rolling and it can only be approached. How-so-ever hard the material be, either the rolling body will be deformed as happens in case of a car or cycle tyre, indicated in Figure or the flat surface gets depressed or deformed. When the road roller passes over unsettled road or kacha road as shown in Figure.

Rolling Friction (Fig.- 11.22)

At times when a load or a heavy machine or its part is to be shifted from one place to another place, for a short distance, and no suitable mechanical lifting device is available, the same is placed on a few rollers in the form of short pieces of circular bars or pipes as shown in Figure and comparatively with a less force the load is moved. The rollers, roll and the one which becomes free at the rear side is again placed in the front of the load and so on. Because of the reduced friction, it requires less force.

Rolling Friction (Fig.- 11.23)

1.12 Ball and roller bearings

When a shaft revolves in a bush bearing, there is sliding motion between the journal and the bearing surface, resulting in loss of power due to friction. If between the journal and the bearing surface, balls of rollers are provided, instead of sliding motion, rolling motion will take place. To reduce the coefficient of rolling friction the balls or rollers are made of chromium steel or chrome-nickel steel and they are further heat treated with a view to make them more hard. They are finally ground and polished with high precision.

Such an arrangement as mentioned above is provided in Ball and Roller bearings.

 1.12.1 Ball Bearing

A ball bearing is shown in Figure. As may be seen, it consists of mainly the following parts :

Ball Bearing

Ball Bearing (Fig.- 11.24)

The inner ring is tight press-fitted to the shaft and the outer ring is press-fitted into a fixed housing which supports the whole bearing. The cage is meant for keeping balls at fixed and equal distance from one another. Balls and races are made of best quality chromium steel containing high percentage of carbon with a view to reduce the value of coefficient of rolling friction, to the barest minimum, to make the bearing really an anti-friction.

 1.12.2 Roller Bearing

A roller bearing is shown in Figure. There is no difference between ball bearing and the roller bearing except that in roller bearing rollers, instead of balls are used. In roller bearings, there is a line contact instead of point contact as in ball bearings. The roller bearings are widely used for more load carrying capacity than that of ball bearings.

Roller Bearing

Roller Bearing (Fig.- 11.25)

The rollers may be cylindrical, straight or tapered. When tapered rollers are used, the bearings are called tapered roller bearings and those in which needles are used are called needle bearings.

1.13 Tapered Roller Bearing

arts of a tapered roller bearing as well as its assembly are shown in Figure. The rollers are in the form of frustum of a cone. The contact angle is between 12degrees to 16degrees for thrusts of moderate magnitude and 28degrees to 30degrees for heavy thrust.

Tapered Bearing

 

1.14 Needle Bearing

The rollers of small diameter are known as needless. It is : 2 to 4 mm. Ratio between length and diameter of needle is 3-10 : 1. Needle bearings are not provided with cage.

Needle Bearing

Needle Bearing (Fig.- 11.27)