Lecture 14 Flashcards

1
Q

Give examples of applications of friction

A
  • Bearing surfaces
  • Brake pad & disc
  • Nuts & bolts
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2
Q

What is waviness?

A

All irregularities which are longer than the roughness sampling length

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3
Q

What is roughness?

A

Fine irregularities on the surface texture, superimposed on the waviness

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4
Q

Why does roughness usually occur?

A

Inherent action of production method

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5
Q

What are asperities?

A

Undulations (peaks/troughs) at a microscopic level

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6
Q

What is Ra?

A

Arithmetic average roughness

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7
Q

How is Ra calculated?

A

L
Ra = (1/L) $ lzl dx
0
$ is an integral

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8
Q

How does friction act in relation to motion?

A

Friction opposes motion

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9
Q

What are mu(s) and mu(k)?

A
mu(s) = coefficient of static friction 
mu(k) = coefficient of kinetic friction
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10
Q

How do mu(s) and mu(k) compare?

A

mu(k) < mu(s)

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11
Q

What is the 1st law of friction?

A

Friction force is proportional to the normal force

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12
Q

What is the 2nd law of friction?

A

Friction force is independent of apparent area of contact

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13
Q

What is the 3rd law of friction?

A

Friction force is independent of sliding velocity

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14
Q

When does the 3rd law of friction break down?

A

At very high sliding speeds - mu(k) falls with increasing velocity

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15
Q

What happens to asperities at low loads? (In the adhesion model)

A

They deform elastically

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16
Q

What happens to asperities at high loads? (In the adhesion model)

A

The tips are deformed plastically and form weld junctions

17
Q

What are welds called in the adhesion friction model?

A

Adhesion junctions

18
Q

What is the friction force in the adhesion friction model?

A

The force needed to fracture the adhesion junctions

19
Q

What is the actual contact area?

A

a

The total junction area of asperities

20
Q

How is the approximate value of mu = 0.5 obtained?

A

P = asigmay
Welds will fail when F (shear force) overcomes max shear stress in material
F = a
tau(max) = a*sigmay/2
So F = P/2 = 0.5P

21
Q

Why is mu large for hard metal contacts?

A

Small but strong welds form, so lots of energy is required to break them

22
Q

Why is mu large for soft metal contacts?

A

Weak welds but large area means lots of energy is required to break the number of welds present

23
Q

How is low friction achieved with metals?

A

One hard metal surface and one soft metal surface gives a weak weld of small area

24
Q

How are bearings often made?

A

A thin film of soft metal between two hard metals

25
Q

Why do polymers make good low friction bearings?

A

The polymer chains orientate parallel to the sliding surface

They shear easily, so mu is low

26
Q

What are the drawbacks to using polymers in bearings?

A

Molecules peel off easily, so there is heavy wear

Creep allows junction growth in the static state, so mu(s) becomes very large in comparison to mu(k)

27
Q

How might excessive contact stress or deformation cause component failure?

A
  • Overload
  • Wear
  • Rolling Contact Fatigue
  • Seizure
  • Loss of Tolerance
28
Q

What is seizure?

A

Component surfaces locally weld under high contact stress

29
Q

What assumptions are used in Hertz Theory?

A

-The size of the contact area is small compared with the size of the curved bodies
-Both contacting surfaces are smooth and frictionless
-The deformation is elastic
-The gap between the undeformed surfaces can be approximated by
h = Ax^2 + By^2

30
Q

What is the highest point on the pressure profile?

A

The centre

31
Q

Where is the maximum shear stress beneath a circular point contact?

A

In the subsurface

32
Q

What is the effect of Q < mu*P?

A

No sliding

Regions of stick and microslip

33
Q

What is the effect of Q = mu*P?

A

Sliding will occur

34
Q

What do points 1 and -1 mean for the x/a axis on the contact stress distribution?

A

They are the contact boundary

35
Q

What is a?

A

Half the width of the contact length

36
Q

What is the shape of the contact region for spheres?

A

A circle

37
Q

What is the shape of the contact region for cylinders?

A

An oblong