And SURFACE WEAR AND SURFACE FATIGUE Terminology Abrasive wear

One hard and sharp material sliding or rubbing against a material that is softer—material removal

Adhesive wear Two materials in contact that transfer material from one to the other; removed material “adheres” to mating surfaces

Surface Wear – Surface Fatigue Two surfaces contacting one another under pure rolling, or rolling with a small amount of sliding in contact.

Examples of abrasive wear

1. polishing 2. tumbling 3. grinding

Examples of adhesive wear

1. scoring 2. cold welding

Examples of surface wear

1. ball and roller bearings 2. spur and helical gears 3. cams and roller followers

We will concentrate on surface fatigue, which requires that the two contacting machine components are in pure rolling or rolling with a small amount of sliding. The challenge to the engineer in determining how surface fatigue may impact a design, is to find a way to approximate the size and shape of the contact area. For example, a ball rolling along a horizontal plane, theoretically contacts the plane at a single point. However, we know that this cannot be true. For one thing, point loads do not really exist—they are just idealizations the engineer uses in appropriate analysis. However, even if there were point loads, the material from which the ball is made could not withstand the infinite stress created by a point load and an infinitely small area. To find the stress imposed by contact, we will need the customary parameters: force and area. For the general case of contact, the geometry of the contacting area is elliptical and the pressure distribution over this area is considered a semi-ellipsoid:

Contact pressure is a maximum at the center, and zero on the edges of the ellipsoid. The total contact force at the center of the ellipsoid is its volume:

max32 abpFcontact π=

Maximum pressure, therefore, is:

abFp

π23

max =

Average pressure, is the contact load, F, divided by the area.

.23

max avgpp

abFpaverage

=

=π

To determine the contact area, we need to define “geometry constants” that are dependent on radii of curvature of the contacting bodies.

21

2211

2

22

2

11

2211

2cos'11

'112

'11

'11

21

'11

'11

21

−

−+

−+

−=

+++=

θRRRRRRRR

B

RRRRA

θ is the angle between planes containing R1 and R2. Material constants must also be determined for bodies A and B.

Now the patch dimensions, a, and b, are found as follows:

( )

( )3 21

3 21

43

43

AmmFkb

AmmFka

b

a

+=

+=

ka and kb are found depending on the value of: φ = cos-1(B/A) (SEE PAGE 476, NORTON) The pressure distribution can now be determined from:

22

1max

−

−=

by

axpp

At the contact surface the maximum normal stresses occur along the centerline of the ellipsoid and are as follows:

max

max

max

)21(12

)21(12

p

pbaa

pbab

z

y

x

−=

+−+−=

+−+−=

θ

υυθ

υυθ

For two cylinders in contact: The contact patch half width:

1 22 m m FaA lπ+=

+=

21

1121

RRA

Maximum load is determined to be:

max2alpF π=

Average pressure is:

avgpalF =2

Pressure distribution is:

2

max 1xp pa

= −

Maximum applied stresses:

max

max

2 pp

y

zx

υσσσ−=

−==

For two spheres in contact: The contact patch half width:

1 23 .375m ma FA+=

+=

21

1121

RRA

The maximum pressure is:

max 2

32Fpaπ

=

The pressure distribution is:

2 2

max 1 x yp pa a

= − −

Stresses:

( )3

max 32 2 2

3

max2 2 2 2

1

(1 2 ) 2(1 )2

z

x y

zpa z

p z za z a z

σ

σ σ υ υ

= − + +

= = − + + + − + +

IN CLASS ASSIGNMENT: FIND THE PEAK PRESSURE BETWEEN A FLAT CAST IRON SURFACE AND A 100 mm DIAMETER STEEL WHEEL, 5 mm WIDE AND CARRYING A LOAD OF 500 N. THE CAST IRON HAS AN ULTIMATE COMPRESSIVE STRENGTH, SUC, OF 750 M Pa AND IT’S YIELD STRESS Sy, IS 350 M Pa. FIND THE FACTOR OF SAFETY. YOUNG’S MODULUS OF ELASTICITY FOR THE STEEL, ES, IS 207 G Pa AND THE POISSON’S RATIO, ν, IS 0.29 YOUNG’S MODULUS OF ELASTICITY FOR THE CAST IRON, ECI, IS 105 G Pa AND THE POISSON’S RATIO, ν, IS 0.21