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IMAC-U course 6 th semester Tribology 3. Contact of solid bodies Assoc. Prof. Takeshi YAMAGUCHI [email protected]

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Page 1: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

IMAC-U course 6th semester

Tribology 3. Contact of solid bodies

Assoc. Prof. Takeshi YAMAGUCHI [email protected]

Page 2: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Plane contact Line contact Point contact

Contact types

Elastic contact (Hertzian contact) Plastic contact

Page 3: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Hertzian contact problems

Line contact Point contact

The following assumptions are made in determining the solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming

(implying that the area of contact is much smaller than the characteristic dimensions of the contacting bodies).

3. Each body can be considered an elastic half-space. 4. The surfaces are frictionless.

Page 4: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Contact between two cylinders

Contact area

Contact pressure distribution

p(x,y) =pmax 1−x

a

2−

y

a

2

pmax=3W

2πa2=

3

2pmean

1E′

=12

1−ν12

E1+

1−ν22

E2

(3.1)

(3.2)

(3.3)

(3.4)

(3.5)

Contact pressure distribution

Radius of contact circle

Hertzian maximum contact pressure

Reduced Young’s modulus

Reduced radius of curvature

1R=

1R1

+1R2

E1, n1

E2, n2

Page 5: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Contact between two spheres Contact area

Contact pressure distribution

p=pmax 1−x

b

2

pmax=2

π

W

bL=

4

πpmean

1E′

=12

1−ν12

E1+

1−ν22

E2

(3.6)

(3.7)

(3.8)

(3.9)

(3.10)

Contact pressure distribution

Half width of contact rectangle

Hertzian maximum contact pressure

Reduced Young’s modulus

Reduced radius of curvature

1R=

1R1

+1R2

E1, n1

E2, n2

Page 6: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Plastic deformation due to indentation of a sphere

x

z Starting point of plastic deformation

a -a

W1

0.47a

W3

Plastic region

Elastic region

W2

Plastic region

Elastic region

Elastic region

( ) ( ) ( )

pmean < 1.1 s0 1.1 s0<pmean<3s0 pmean≒3s0

(H) Yield strength

Indentation hardness

Page 7: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Ball-flat contact with normal load and tangential force

tmax contour

T = 0.5mP

J. ホーリング編, トライボロジ,近代科学社(1984)

7

Page 8: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

1. Basic concept of hardness

(1) Touch with a finger Lower contact pressure means softer.

e.g.) Rubber is soft and iron is hard.

(2) Rubbing each other Scratched surface is softer.

e.g.) Glasses can scratch steels.

2. Types of hardness test

(1) ( ) ⇒use hard indenter instead of your finger

(2) Dynamic hardness testing

(3) Scratch hardness testing

Hardness?

Page 9: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Indentation hardness testing methods

Page 10: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

HV = 2 W sin (q/2)

=

L2

Vickers hardness

q

Diamond

pyramid indenter

W

L

Hardness value ( ) means contact pressure necessary for a plastic deformation of surface (plastic

contact).

Page 11: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Vickers hardness of engineering materials

10000

5000 4000 3000

2000

1000

500 400 300

200

100

50 40 30

20

10

5 4 3

2

1

Diamond

Cubic boron nitride (cBN) Boron nitride (BN) Titanium carbide (TiC)

Vic

kers

har

dn

ess

, kgf

/mm

2

Tungsten carbide (WC-Co) Hard chromium plate Martensite

Cementite

Fully hardened tool steel Glass

Titanium Gray iron Annealed mild steel Brass

Nickel Pure iron

Zinc, gold, magnesium Silver Polystyrene Pure aluminum

Vinyl resin Graphite

Tin Lead

Page 12: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Contact between rough surfaces

Normal load

Aa = a×b Apparent contact area

Apparent contact area

Real contact area is ( )

of apparent contact area.

Ar= 𝑨𝒓𝒊𝒊=𝒏𝒊=𝟏

(2.19)

(2.20)

Real contact area

Page 13: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Real contact area

Ar= Ai

n

i=1

= Wi

p0

n

i=1

=W

p0 (3.11)

Plastic flow pressure

Real contact area is proportional to normal load and inverse proportional to plastic flow pressure (≒3s0 or H) .

Page 14: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Plasticity index Y

Ψ =E′2H

σβ

Ψ >

Ψ =

Ψ < Elastic contact

Plastic contact

Plastic and elastic contact

Because of the height distribution of surface asperities, not all asperities contact with plastic deformation. The plasticity index express average contact state of asperities (elastic, elastic and plastic , and plastic)

E’: reduced Young’s modulus s: reduced roughness ([(Rq1)2+(Rq2)2]1/2)

b: radius of surface asperity tip H: hardness

Plastic? Elastic?

(3.12)

Page 15: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Contact stiffness

W

W

W

W

Page 16: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Deflection Sd

No

rma

l lo

ad

P

Contact stiffness

Asperities on the surfaces decrease stiffness when contact with each other.

Stiffness decreases as the number of contact area increases.

Contact stiffness

Asperities on the surfaces

Without asperities on the surfaces

Page 17: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

大西清著、JISにもとづく標準製図法、第13全訂版、理工学社(2010)

Indication of surface roughness and chamfer

Page 18: Tribology - Tohoku University Official English Website...solutions of Hertzian contact problems: 1. ( ). 2. The surfaces are continuous and non-conforming (implying that the area of

Contact between elastic flat surface and rigid body without chamfer

Stress concentration

W Contact geometry Contact pressure distribution

Necessity of chamfer

W

R R

Contact between elastic flat surface and rigid body with chamfer