# Tribology - Education Essay Example

Q1 - **Tribology** introduction. a) When two metals are in contact, then

Effective modulus (E*) at the interface is given by the following expression

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Where E1 and E2 are elastic modulus and n1 and n2 are Poisson’s ratio of metal 1 and 2 respectively.

In this problem, as both the metals in contact are same (Copper).

Therefore,

E1 = E2 = ECu = 140 GPa and

n1 = n2 = nCu = 0.3

Therefore,

i.e. GPa

Q1. b) Elastic strain at the yield stress or ep is given by

where, H is hardness of the metal and E is Elastic modulus of the metal

Here, the metal is copper and

Hardness H = 700 MPa and

Elastic modulus E = 140 GPa = 140*103 MPa

Therefore, Elastic strain at the yield stress is

Q2. When a conical indenter of a hard material ploughs onto a soft metal’s surface, then normal load (N) is given by the following equation

where ‘r’ is half the width of the scratch and H is hardness of the material.

Here, Normal load N = 10 N

and Hardness H = 700 MPa = 700*106 N/m2

Therefore,

Half the width of the scratch (r) will be

Therefore,

Width of the scratch = 2r = 190 mm

The coefficient of ploughing friction (m) is given by the following equation

0.17

Q3) Life of a ball bearing is given by the following equation

W3L = constant

Where W is Load and L is life in number of cycles

Therefore,

or,

Where L1 and L2 are life of a Ball Bearing at loads W1 and W2 respectively.

Here, W1 = 2 kN, W2 = 8 kN and L1 = 108 cycles

Therefore, Life of the Ball Bearing at 8 kN load will be

Q4.) When two surfaces with average roughness RA1 and RA2 are contact,

then combined average roughness of the two surfaces RA is given as

Here, RA1 = RA, Steel = 0.5 mm and RA2 = RA, Brass = 0.6 mm

Therefore,

Combined Roughness = 0.781 mm

The parameter, M, which characterizes the lubrication regime is given as

where, ft is the thickness of the oil film under operating condition = 4 mm

and RA is combined average roughness = 0.781 mm

Therefore,

As the parameter ‘M’ lies between 5 and 10 therefore, the sliding interface is operating in ‘Hydrodynamic Lubrication Regime’.

Q5) Four important limitations of liquid lubricants are the following.

(i) Not suitable for high temperature applications.

This is because at higher temperatures, these lubricants lose viscosity and therefore lubricating properties due to associated loss in load bearing capabilities. Further, elevated temperature leads to vaporization of the liquid lubricant causing fumes and it may even catch fire if temperature rises beyond the flash point of the lubricant.

(ii) Not suitable for clean applications

Liquid lubricants are not suitable for the applications where cleanliness is required as these lubricants are inherently not clean. Additionally dirt and dust sticks to these lubricants adding further woes for cleanliness considerations.

(iii) Poor action at high load and low speed

At high speed, due to hydrodynamic action, liquid lubricants are able to provide sufficient separation between the mating surfaces. However as the load increases and speed decreases, the lubrication regime is no more hydrodynamic and the lubricant is not able to maintain the desired separation between the mating surfaces.

(iv) The mating surface should be accessible for application of the lubricant during service as liquid lubricant is lost / gets degraded after sometime in service. However, in many applications the mating surfaces are not accessible once the equipment is assembled and put into service. One example is core or near core components in a nuclear reactor. Such components are not accessible once put into service due to high radioactivity. Therefore, liquid lubricants are not suitable for such applications as well.

Q6. Referring to the figure below

‘1’ stands for steel and ‘2’ stands for Aluminium

Therefore,

E1 = 208 GPa = 208*109 Pa v1 = 0.28

E2 = 69 GPa = 69*109 Pa v2 = 0.35

(Note, it appears the values of E for copper and steel are wrongly given in MPa, I have chosen these values in GPa as that is the correct value)

R = 25 mm = 2.5*10-2 m

P = ? (to be determined)

H = 700 MPa = 700*106 MPa

Effective modulus (E*) will be

Therefore, E* = 58.3 GPa = 58.3*106 GPa

Now the Hertz equation gives,

Radius of the indentation (a) on the top surface

……….. (1)

Maximum Pressure (Po) ………. (2)

Maximum Shear Stress (tmax)

………. (3)

Material of the plate first yields when

……… (4)

(Because for metals sy = 3H)

Where ‘H’ is hardness of the material.

By equating equations (3) and (4),

……………. (5)

or, N

The corresponding contact width is 2a i.e.

mm

Maximum Pressure (Po) will be,

GPa

Average pressure will be MPa

Maximum Shear Stress will be

GPa

Maximum Shear Stress is located at a depth‘d’, which is given as

d = 0.48*a = 0.48*657.8 mm = 315.7 mm

below the free surface.

Q7. a) Assumptions made for deriving Archard’s adhesive wear equation are the following.

(i) When two surfaces in relative motion force against each other a contact area pa2 supports the load pa2H.

(ii) The two materials get welded in this region and thus the relative motion causes a fragment of material from this volume.

(iii) As the two surfaces travel against each-other, more and more of such wear fragments are produced.

(b) Derivation of equation for wear volume per unit slid distance

The Archard’s model is schematically shown in the following figure.

Contact area pa2 supports the load pa2H

When distance traveled is ‘2a’

Wear volume produced will be (2pa3)/3

Thus wear volume per unit length dQ = {(2pa3)/3}/(2a) = (pa2)/3

Thus total wear, Q, in unit slid distance is Q = n*(pa2)/3

Total load is N = n*pa2H

Therefore,

Q/N = 1/3H

i.e., Total wear per unit distance Q = N/3H

(c) Based on this equation total wear is directly proportional to sliding distance and applied load and inversely proportional to the hardness of the material getting worn out.

(d) This equation gives much higher value for the volume of wear than that observed in the experiments. The difference is very high and is 4 to 7 order of magnitude higher than that observed in the experiments. This is because only a very small fraction of the asperity contact results in wear of the material and therefore, in stead of the equality sign there should be a proportionality sign. When this is done a constant is introduced in the equation whose value is determined experimentally. Thus the revised equation is

Q = k(N/3H)

Here, ‘k’ is a constant, known as wear coefficient. This is a constant between mating surfaces as long as the underlying wear mechanism remains unchanged.

Q8. Increasing surface hardness of a steel component

(a) Two methods that do not change composition of the surface

(i) Transformation Hardening:

When surface is heated, the material on the surface transforms into austenitic phase and when the heat source is removed, the underlying material acts as heat sink causing the quenching of the material on the surface. This heat treatment is known as hardening. The surface gets hardened to very high value due to formation of martensite, which is a body centered tetragonal structure, rich in carbon. This structure is asymmetric and rich in dislocations and therefore, is much harder than the case.

There are many techniques such as Laser Surface Hardening, Induction Hardening, Flame Hardening etc., which are used for surface hardening of a steel component. In case of Laser Surface Hardening, a line shaped beam is scanned over the area to be hardened and a thin case of approximately 0.5 mm gets hardened. In case of Induction hardening or Flame hardening one gets a deeper case but the gross heat input is much more, therefore, there is a chance of thermal distortion of the component.

(ii) Shot Blasting

In this technique, the surface to be hardened is blasted by metal shots and the pressure imparted by the metal shots causes hardening of the surface by mechanical action. Another very advanced version of such a technique is laser pinning, in which high intensity laser beam is used to ablate a coating placed onto the steel surface and as the coating ablates, it imparts tremendous shock wave into the material and thus causes what is known as laser shock hardening of material

(b) Two methods that change composition of the surface

(i) CarburizingTribology

In this technique, the component is heated in an atmosphere of carbonaceous material. The carbonaceous material may be either solid or gas. In case of solid material, the component is covered by charcoal or graphite coating and then heated in a furnace at elevated temperature ~ 750 oC. Such a treatment produces a carbon rich case or high carbon case over a mild steel substrate due to diffusion of carbon, the high carbon case is much harder as compared to mild steel. Besides, it can be easily heat treated to give much higher hardness.

This treatment can be given in carbon rich gaseous atmosphere as well.

(ii) Nitriding

In this treatment a nitrogen rich layer is produced by heating the component in nitrogen rich atmosphere (normally the atmosphere is maintained so that nascent nitrogen is produced during heat treatment and goes into the steel component). This nitrogen produces fine needles of nitrides, which imparts hardness to the surface of the material. This treatment is carried out at much lower temperature that the temperature of carburizing.