Tribo Notes

8/2/2019 Tribo Notes

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4.1. Introduction:

Wear is defined as the gradual loss of material from

the interacting surfaces during the relative motion under the

normal load. Wear is classified as mild wear and severe

wear. Mild wear is therefore generally associated with lowloads where metallic interactions are somewhat inhibited

and the wear debris consists of fine particles and is usually

in the form of oxides. This does not imply that metallic

contacts have never occurred at all, since the resulting

metallic debris would tend to become oxidized at the high

local temperature at the interface. Nevertheless the nature

of the surface asperity interaction is relatively gentle,

resulting in characteristically mild wear and smoothing of

the surfaces.



Fig. 4.1 The transition phenomena in wear

At higher load a much coarser wear process occurs.

The wear debris is of a much larger particle-size, the worn

surfaces are much rougher and the increase in volume

changes the wear rate by several orders of magnitude. This

is the so-called severe wear regime. A starting fact about

these two types of wear behavior is the very rapid transition

from one mode to the other as the load is increased (fig.

4.1). In this figure it should be noted that the wear rate

suddenly changes by more than one hundred times. With

some materials at even higher loads the increasing

temperatures cause metallurgical changes in the materials



such as to increase their inherent hardness. These effects

can then lead to a second transition from the severe wear

back to the mild wear regime.

4.2. Mechanisms of Wear:

We now consider the detailed mechanisms by which

material may be removed from the surface. The most

common mechanisms are:(a) Adhesive wear, (d) Corrosive wear.

(b) Abrasive wear, (e) Erosive wear

(c) Fatigue wear

In some situations more than one of these mechanisms

may be operative at the same time, and this is one of the

reasons for the complexity of wear studies.

4.2.1. Adhesive Wear:

When the two surfaces are having identical hardness

then the contact between surfaces occurs at the tips of the

asperities, which then deform under load. Tile nature of the

adhesion between such asperity interactions is modified by

surface films so that the metallic adhesion characterized in

the simple friction theory is somewhat modified. But as



translation occurs these surface films are to some extent

disrupted and adhesion will occur at a certain proportion of

these contact, as can be appreciated in view of thedistribution of asperity heights discussed in chapter 2.

From these considerations and our knowledge of the

nature of the contact of rough surfaces we can now predict

a ‘wear equation’, If one assumes that the wear particles are

geometrically similar, the wear volume would be expectedto be proportional to the real areas of contact at which

adhesion occurs, and also to the distance of sliding. Since

the real area of contact for the plastic interaction of

asperities is given by

A = W / H

The wear volume V is given by

V∝ A x L ∝ W x L

H

Where L is the distance of s\liding. Thus the adhesive wear

law becomes WV = K W x L

H

Or in words the: ‘laws’ of adhesive wear are:



(a) The volume of Wear is proportional to the distance

of sliding. This relationship has been justified by

experience for a wide range of conditions.(b) The volume of wear is proportional to the applied

load. This has also been shown to be true in many tests for

limited ranges of load, although as wear mechanisms

change with increasing load some abrupt transitions have

been observed; see fig. 4.1.(c) The volume of wear is inversely proportional to the

hardness of the softer material. This has also been shown to

be valid, particularly for pure metals.

Recalling the physical nature of the wear process

arising from adhesion we can give a physical meaning to

the constant of proportionality K. often called the adhesive

wear coefficient or the Archard constant. Surface films

together with the asperity contact forms the adhesive

contacts are only significant on adhesive wear. Since

shearing due to sliding probably formed where the

junctions are stronger than the underlying material, So that

some material from the surfaces is torn away and

eventually released by the continued sliding. So we can see



that K may be probability factor, that is, the factor which

indicates the probability of wear particles being created by

the adhesive effect between the populations of asperities onthe two rubbing surfaces.

Fig. 4·2 Surface contact of an identical

hemispherical asperity

Theory of Adhesive Wear:

Consider two surfaces of an identical material having

the same hardness. Assume that the asperities on both the

surfaces are identical. In nature of such surfaces are in

contact then the deformed area of the asperities is a circular

area of radius 'a' under the application of load W.

Consider anyone asperity in contact when the asperity

is in relative motion the function of asperity will be shear

from the weaker bond of the asperity. Let the load carried



by anyone asperity is W1 = W/n where n is the total

numbers of asperities in contact.

Let the particle separate out from such asperity is of ahemispherical in shape having a radius, ‘a’

The volume of material in dist. is vol. of sphere V/2 =

2/3 π r 3.

V1 = (4/3 π a3) / 2 ( = 2a)X X

V1 = 1 = 1/3 π a2 = 1/3 A1

X 3

Since πa2 is the area of contact, where is proportional to the

ratio W1/H.

Where H = hardness of material.

A1 ∝ W1

But A = A1 + A2 + …………… + An

W = W1 + W2 + …………… + Wn

& also

A1 = A2 = A1 =…………… = An for similar shape of

asperity

W1 = W2 = W1 =…………… = Wn



A = n A1

W = n W1

A1 = A / n ∝ W1 = W / n

A ∝ W

The total volume of wear from the surface, if all the

asperities are effective in removing the material then.

V = n V1

V1 = 1 A1 = 1 W1

X 3 3 H

Therefore volume of wear for displacement is

V1 = 1 W1X3 H

V = n V1

= n . 1 W1X But nW1 = WT = W

3 H

V = W X

3 H

If wear constant K adh can be introduced for considering

the actual numbers of the asperity in contact.



V= K adh . W / 3H.

K adh = 1, When all asperities in contact are effective in

wear = 0.1, When out of 1000 asperities 100 asperities

are effective in wear

= 0.01, When out of 1000 asperities 10 asperities

are effective in wear.

Also,= u . t

= π DN /60 . t

4.2.2. Abrasive Wear:

This type of wear arises from the cuttion acting of

hard surface rubbing on softer materials, as for example,

when hard surface asperities act rather like cutting tools

and remove material from softer materials. Another

example arises when loose debris of any kind is trapped

between sliding surfaces. Such debris may be extraneous,

such as sand particles, or may be the actual wear particles

created by the primary wear process.



One method of reducing the first type of abrasive wear

is to ensure high quality of surface finish of the mating

surfaces particularly the hard surface. With modern production methods this type of wear is no longer as

serious problem. The second type of abrasive wear is more

difficult to eliminate. Suitable sealing and filtration can

reduce correct design of the contact geometry. It is often

desirable to provide grooves or other such recesses on thesurfaces of bearings, which allow the debris to ‘escape’

from the contact geometry.

Fig. 4.3 Wear due to a single conical asperity



Fig. 4.4 Abrasive wear due to Which Olly

Consider two dissimilar surface materials in contactand is in relative motion. Assume that the hard surface

asperities remove the material from relatively soft material

surface.

Let the hard surface asperities are of conical in shape

having radius ‘a’ at the contact and height ‘h’ at the

penetration in soft material. The asperity angle is ‘θ’,

consider anyone such asperity for analysis the volume of

material to be removed from soft material in distance x is

V = The projected area of conical asperity in the

direction of motion x distance x.

= ½ x 2a x h x X but h = a. tan θ

V = a . a tan θ . x



= a2 tan θ . x

For plastic deformation of material

W ∝ A ( area of contact)

= A . H

i.e. W = A . H = A . H/3

The area of contact in the direction of a normal load is A =

π a2 / 2

W = πa2 / 2 . H

a2 = 2W / π H

Substituting value of a2 in equation of volume

V = a2 tan θ . X

Vabr = 2W tan θ . X

πH

Here K abr = X tan θ = 6 tan θ

π π

Vabr = K abr . W.XH

Where Vadh = 1/3 W.X . K adh

H



Vw = Vadh + Vabr = K adh – W.X + K abr + W.X

3 H H

V = K w . W.X

H

Where K w = K abr + K adh /3

In all practical cases the wear occurs due to the

abrasive and adhesive wear. Therefore the general equation

of wear can be given as

Vw = K w . W.XH

4.2.3. Fatigue Wear:

It is well known that if materials are loaded and

unloaded cyclically they exhibit fatigue failure. This type

of failure can occur after large number of loading cycles,

even though the load is less than that which we could

normally expect to produce failure in a single load

application. It is usual to express such behavior bay

logarithmic graph of stress S against the number N of

cycles to failure (the S/N curve, as it is usually called) such

as fig. 4.5. Here we see that the lower the applied cyclical

stress have the longer the life of the material.



Fig. 4.5 A typical S-N fatigue curve

Fig. 4.5 shows a typical S-N fatigue curve. If we

consider the interaction of asperities during the sliding of

one surface over another we can see the possibility of

fatigue mechanisms being broken off the asperities to

produce wear debris. A simple experiment to illustrate this

is to run one's finger a many times along the teeth of a

comb. The teeth (asperities) are continuously being loaded

and unloaded due to repeated traversals by one’s finger and

after many cycles they ultimately break due to fatigue.



4.2.4. Corrosive Wear:

Any clean metal surface reacts with its environment to

form contaminant films, and the rate of formation of suchmm is initially very rapid but decreases as the ‘corrosive’

film thickens. In many instances, such as oxide (rust) films

on steel, these surface films adhere only loosely to the

surface. Rubbing therefore removes the films leaving

exposed ‘clean’ metal, which immediately reacts with itsenvironment to provide new surface films, which are again

removed during ragging. So materials continuously bang

removed from the surface, and wear is taking place. The

chemistry of such reaction is beyond the scope of this book,

but fortunately is not needed for the understanding of the

basic mechanism.

A further effect of corrosive environments is to

enhance the abrasive action of wear debris. Most metal

oxides are harder than the metal itself sot that if metal

debris is created, this becomes oxidized and gives a rate of

abrasion greater than that which would otherwise occur. A

good example of this occurs with relatively soft aluminum,

where the oxidized wear debris is very hard abrasive.



Indeed aluminum oxide is often used as the cutting agent in

girding wheels and the like.

Corrosive effects are not entirely deleterious. In thechapter on friction it has already been shown that the

presence of oxide films in preventing metal-to-metal

contact greatly reduces the coefficient of friction. In other

applications surface films are deliberately produced to

avoid metallic contact. The so-called E.P. (extreme pressure) additives to lubricating oil produce surface films

such as chlorides and sulphides and provide protective

surface layers. In a sense these could be more properly be

called extreme temperature films rather than extreme

pressure films. Their main characteristic is that chemical

stability at the high temperatures of the high pressure

contacts in such situations as hypoid gears as used in motor

car back axles.

4.2.5. Erossive Wear:

Erossive wear is mainly because of the erossion which

is a combined effect of mechanical stresses under the

ambient condition for example the wear on the bearing



surface of a concrete mixture operated in an open

atmosphere.

4.2.6. Fretting:

This is not really a separate mechanism of wear but it

is treated separately because it arises in rather special

circumstances. It shows one particular wear process may be

a complicated combination of several mechanisms of wear,and also demonstrates the deleterious effects of any wear

debris which may become trapped in the contact system.

Fretting effects are associated with the contact of surfaces

in which the sliding motion is an oscillation of relatively

small amplitude, often only a few micrometers. Since

vibrations occur in virtually all machines we find fretting

occurring between surfaces in contact such as bolted

components, splines and components located by friction

rise to small amplitude oscillatory displacements between

the surfaces in contact.

4.3. Tribological Properfies of Plastics:



In recent years there has been a significant growth in

the use of plastics to replace metals in many bearing

applications. In general the friction and wear of plastics can be explained by the adhesion theories already discussed.

The friction coefficients of plastics are not particularly low,

but their main advantage is that they wear at reasonably

low and predictable rates. One notable exception to this

behavior is PTFE (polytetrafluoroethylene) whose frictioncoefficient may be very low, about 0.05. This very low

friction value seems to be associated with the very low

adhesion of this material, because of this it is used in non-

stick kitchenware. It is also reasonably hard due to the

mechanical interlocking of its molecules. So it is

extensively used in bearings, where the loads and speeds

are moddest, sine its is almost self-lubricating and is highly

reliable.



Fig. 4.6. The change in dimension due to wear.

A useful design parameter for all trips of plastics used

in bearings is the ‘pv’ factor (which is the product of thenominal contact pressure and the sliding speed.).

Consider a block of this material sliding on a metal

surface, fig. 4.6. The rate of energy dissipation against

friction is µWv. It is reasonable to assume that the volume

wear rate of the block. V = – dV / dt, is proportional to this

rate of energy dissipation, so

V ∝ µWv.

The block will therefore wear to a depth d such that

the volume wear V is given by

V= Ad.

Hence its rate

V= Ad.

Where d is the rate of increase of d with time. Thus

d = V ∝ µWv ∝ πν

A AThis shows that the rate of change of dimension of the

block is proportional to the product Pv. This is the rate of

change of bearing clearance with time in any practical



application and as such as in more useful parameter than

the actual volume of material removed.

For any material the allowable value of the pv productmay be defined and fig. 2.7 shows a typical result of PTFE

if the wear rate is to be a dimensional change of 25 mm in

100 hours.

Plastic bearings and particularly PTFE bearings must

be operated within their approved pv ratings. These ratingsare associated with wear and are particularly subject of

thermal effects due to the decomposition of the surface.

Thus at their ambient temperatures the pv factor for such

materials is considerably reduced. Such materials when

used in practical bearing designs are often associated with a

metallic matrix which provides additional strength and

improves the thermal conductivity thus allowing the easier

escape of the heat generated in rubbing. When plastics

slide, problem can arise from the generation of electrostatic

charges. In such 10 situations, designers must incorporate

earth paths to minimize the build-up of charges.



Fig. 4.7 A typical 'pv' curve

4.4. The Measurement of Wear:

In most engineering machinery the rate of wear is

relatively small, typically changes in dimensions of

micrometers per year. Also wear takes place in ‘real time’

and laboratory tests aye to devise conditions where the

wear processes are considerably accelerated so that results

are produced in days rather than years. In so far as such

tests are artificially accelerated their results should viewed

with some caution.

4.4.1. High-Pressure Contact Tests:

These tests produce accelerated results by applying

loads over very small areas of contact. Various geometrical

arrangements are used as shown in fig. 4.8. In each of these



tests specimen A is the material being worn away, the

degree of wear being measured by either a change in

dimensions or a loss of mass from the material., Many suchmachines also measure the frictional force at the contact.

Fig. 4.8 Pin on Ring

Fig. 4.9 Pin on Disc

Such contact geometries can be studied either in the

ordinary atmosphere or in a totally enclosed chamber where

the atmosphere may be controlled, as to such properties as

the gaseous environment, pressure, temperature and



humidity. A very common form of such apparatus is

designed to carry out these tests at various reduced

pressures. In a high vacuum the formation of oxide films isinhibited, so we obtain useful information on the friction

and wear of the materials themselves. UHV apparatus is

also useful for measuring the adhesion between surfaces in

contact, and we have already seen that such information is

very valuable to our understanding of friction.

Fig. 4.10 A simple crossed cylinder wear machine

More complex geometrical arrangements are shown in

fig. 4.11. The Fig. 4.11 (a) shows the four-ball arrangement

in which a rotating ball rubs against three stationary balls to

produce wear scars whose size is an indication of the

volume wear. This arrangement is extremely popular for

industrial testing since the test specimens, the balls, are



readily available at low cost from ball bearing

manufacturers. Fig. 4.11(b) shows the so-called ‘disc

machine’ which is very useful for the study of wear under combinations of rolling and sliding. When the peripheral

speed of both discs is the same one has pure rolling whilst a

difference in speeds implies some additional sliding. This

rather complex roll/ slide process often occurs in

machinery, perhaps the best example being the contact between meshing gear teeth depicted in fig. 4.12. At the

initial contact we have rolling and sliding between the teeth

which becomes pure rolling at the pitch point B followed

by rolling with sliding in the opposite sense during the arc

of disengagement.

Fig. 4.11 The four ball wear machine and disc on disc

wear machine



A. Initial contact - rolling and sliding

B. Pitch point contact-pure rolling

C. Final contact-rolling and sliding

Fig. 4.12 The progress of contact between gear teeth

( Solved Problems )

EX.1. A pin on disc experiment the rate of vol. of wear

observed was 0.1 mm3/min. Pin was placed at 30 mm

radial distance from the centre of disc. The disc was

rotated at 500 rpm. Load on pin is 20 N. Material of pin

is brass, hardness is 120 N/mm2. Material of disc is steel,

hardness 280 N/mm2. Determine the wear constant.

Solution :



Data given:

V = 0.1 mm3/min

r = 30mm N = 500 rpm

W = 20N

H p (pin) = 120 N/mm2

Hd (disc) = 280 N/mm2

V = K w . W.2π rNt

H p

V / t = Wear rate = 0.1 = K w . 20 x 2 x π x 30 x 500

120

K w = 6.3662 x 10-6

Ex.2. Determine the wear constant K w by using a pin on

disc wear measuring test ring. The testing data are as

follows:

(1) Pin diameter at the tip 3.0 mm.

(2) The maximum diameter of the conical pin =

10.0 mm

(3) The cone height of the pin is = 9.0 mm.

(4) Load on the pin = 50 N



(5) The disc diameter = 150 mm.

(6) The pitch radius at pin contact = 60 mm.

(7) r.p.m. of the disc. = 600.(8) The hardness of the pin = 20 N/mm2.

If the experiment is conducted for 130 hours. The

wear reading show the linear in unit of the wear

volume. The final diameter of the pin was 8.0 mm. The

material of the pin is brass and the material of the discis hardened steel. .

If the above tested material is required to use in a

machine for shaft and bearing. Determine the life of the

bearing for a shaft rotating at 140 r.p.m. under a load of

10N. The bearing diameter is 40mm. The maximum

allowable radial wear is 1.0 mm. The length to diameter

ratio of the bearing is 1.0.

Solution:

tan θ = (d2 – d1)/2

r = (10 – 3)/2 = 0.3888

9

θ = 21.25°



K w = 5 (dt3 – d1

3)H

4 tan θ W.R . N.T

t in sec:

Time require = 130 hours

= 130 x 60 x 60 secs.

t = 468000 secs.

K w = 5 (83 – 33)20

4 tan 2125 50 x 60 x 600 x 468000

= 3.7013 x 10-8

Ans = I

Now N = 140 rpm

K w = 3.7013 x 10-8

W = 10 N

D = 40 mm

L = 40 mm

r w = 1.0 mm

Life of the bearings t = (D) . r w . L . H

K w.W.πDN/60

= (40) x 1.0 x 40.20 x 603.7013 x 10-8 x 10 x π x 40 x 140

= 1.47428 x 108 sec. x 2

= 40952.135 hours x 2



= 81904.3 hours.

Ex.3. In an experiment of wear with the plastic bearing

it is found that for the change in dimension of 0.1 mmunder the load of 1.5 N in a slender bearing. The

velocity required is m/sec. For the same operating

conditions, if the load and speed are doubled than what

will be the change of dimensions?

Data given:dt = 0.1 mm

W = 1.5 N

u = 1 m/sec.

dd1 = µ W1 u ,

dt1 A1

&

dd2 = µ W2 u2

dt2 A2

But

A1 = A2

W2 = 2 W1

u2 = 2 W

dt1 = dt2



dd2 = P2 u2

dd1 P1 u1

dd2 = 2.2

0.1

dd2 = 0.4 mm

For the same operating conditions change of

dimension is 0.4 mm if load & pressure both (velocity) are

doubled.

Ex.4. In a crossed cylinder wear measuring experiment

the stationary specimen is brass in contact with a

rotating cylinder made all of hardened cylinder having

radius 50 mm and thickness 25 mm. is rotating at 500

rpm. The rate of Wear was observed to be constant is

found 2mm in 60 hours of a constant operation under a

50 N load.

Determine the wear const. if the hardness of brass

is 120 N/mm2

Solution:

Data given:

H = 120 N/mm2

W = 50 N



R = 50 mm

L = 25 mm

N = 500 rpmt = 60 hours

h = r w = 2 mm

cosθ = R-h = 50.2 = 0.96

R 50

θ = 16.26°

K w = R 2 (θ – sin 2θ / 2) L H

W (π D N / 60) t

t = 60 x 3600 secs.

K w = 502 [16.26 x π/180 – sin {(2 x 16.26 )/ 2}] 25 x

120

50 x [(π x100 x 500)/ 60] x 60 x 3600

= 3.977 x 10-6 Ans.

Ex.5. A 30 mm long brass bearing is used to support a

steel shaft having 50 mm diameter, and a steady radial

load of 60 N. Shaft is rotating at 500 rpm. Shaft surface

is having an average asperity angle 10°. The hardness of

the bearing material is 200 N/mm2. Determine the time

required to exceed the radial wear 2 mm.



Solution:

Data given:

r w = 2 mmL = 30 mm

D = 50 mm

W = 60 N

N = 500 rpm

θ = 10°.

H = 200 N/mm2

(Questions)

1. Write short notes on the following:

(i) Fatigue wear

(ii) Pin on disc wear measurement.

2. Explain cross cylinder wear machine.

Define ‘Wear’. State the condition where ‘wear’ proved

to be boneficial.

3. In pure abrasion wear, show that the wear quantity6 W.X

V = 6 tanθ WX

π H



Where θ = average asperity angle.

W = applied load.

X = sliding distance.H = Hardness of material

4. Explain four ball Wear machine.

5. Classify the wear mechanisms. Explain the methods to

eliminate the wear, of each mechanism.

6. Define fatigue Wear.

7. State the mechanisms of wear. Describe the methods to

eliminate wear.

8. Show that the change in dimension due to wear for the

case of plastic - metal contact is proportional to the

product of pressure and the relative velocity of the

contact surfaces.

9. Define abrasive wear.

10. Define wear. Why do you consider that wear is

beneficial in running in and planned obsolescence?

11. Explain elimination of wear.

12. Show that the volume of wear due to adhesion and

abrasion is

Vw = K w W.x



H

Where K w = wear constant.

13. A steel shaft runs within PTFE bearing at 10 m/sec.After 1000 operating hours the bearings were warn 2 mm

radially. Find out the load on each bearing if the contact

area of each bearing is 0.01 m2. The wear constant may

be taken as 10-21 m2/N.

Suggest the various possibilities to improve the performance of such bearing.

14. In a cross cylinder wear measuring experiment the

stationary specimen is brass in contact with a rotating

cylinder made out of hard steel; having radius 50 mm

and thickness 40 mm is rotating at 960 rpm. The size of

the stationary specimen is 100 mm length, 25 mm width

and 25 mm height. The rate of wear was observed to be

constant, 100 N load. Find out the wear constant K w.

If above tested material is required to use in a machine

for shaft and bearing. Determine the life of the bearing

for a shaft rotating at 250 rpm under load of 25 N. The

maximum allowable radial wear is 0.5 mm and length of



the bearing is 75 mm. Take hardness of the brass = 120

N/mm2.

15. Show that the life of a sleeve bearing due to abrasivewear is

L = b.l.H hrs.

1800 kwWw

Where b is the permissible radial wear of sleeve bearing

l is the length of the sleeve bearing, m

H is the hardness of sleeve bearing, N/m2.

hw is the wear coefficient.

W is the radial load, N

ω is the angular velocity rad/sec.

16. In an experiment on a pin-on-disc test rig the

following observation were made:

(i) Diameter of the pin and its material: 10 mm, Brass

(ii) The pitch diameter of the disc and its material: 120

mm Brass.

(iii) The change in length of the cylindrical pin due towear in 150 hours: 5 mm

(iv) The rotational speed of the disc is: 500 rpm.

(v) The hardness of the pin is: 20 N/mm2



(vi) Load on pin is: 200 N.

If the above tested material is to be used in a machine for

shaft and bearing, determine the life of the brass bearingfor a steel shaft rotating at 150 rpm. under load of 50N.

The permissible radial wear is 1.0 mm. The length/

diameter ratio of the bearing is 1.0. The length of the

bearing 50 mm.

17. In a pin on disc experiment, the test piece pin was madeout of brass and disc was made out of stainless steel. The

pin was located at 50 mm radial distance from the disc

centre. The asperity angle on the disc surface was 3°.

The power required to rotate the disc at constant speed

960 rpm was 500 watts. The testing data are as follows:

Pin diameter at tip = 3.0 mm

The maximum diameter of conical pin = 12.0 mm

The cone height of pin = 12.0 mm.

The hardness of pin material = 20 N/mm2

The wear constant = 3.0 x 10-8.

The wear reading shows the linear increment of the wear

volume. Determine the time required for the experiment

of wear if the final tip diameter of the pin is 10.0 mm.

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