PSI - Issue 2_B

Udaya B Sathuvalli et al. / Procedia Structural Integrity 2 (2016) 1771–1780

1778

8

Sathuvalli, Rahman, Wooten and Suryanarayana/ Structural Integrity Procedia 00 (2016) 000–000 where � � �⁄� , � �� � � �� ⁄� , and � �� � � + � �� and the function F 5/3 is obtained from Eq. (4). It is convenient to rewrite the term � �� by invoking Eq. (11), so that   2 2 9 16 fp      (22)

where  is the dimensionless plasticity index introduced by Greenwood and Williamson (1966),   * E H s R  

(23)

and H = 3  yp is the hardness of the wearing surface. The work required to shear a single contact equals the force required to shear the contact times the sliding distance (2 a c ),     , 2 c shear yp co c W A a    (24)

where  yp (=  yp /2) is the yield strength in shear and A co is the average area of the fully plasticized contacts. Substituting Eqs. (24) and (21) in Eq. (19) and rearranging we obtain

5 2 4 5 3 * 3 3 3 C E R s F d    2

  fp

2 N N a

5

yp

 (25) From the definition of wear efficiency in Eq. (1)), � � � � �� �� � ⁄ � ⁄ � � � ���⁄���� �� � ⁄ �� ⁄ . By combining Eqs. (25) and (3), we obtain the following expression for the wear efficiency,     5 2 4 5 * 2 3 3 3 3 5 3 0 . yp fp yp co c C E R s F d K F d A a      . (26)  3 w c yp co c A a  

It remains to estimate the magnitude of A co and a c .

5.1. Average radius and area of the fully plasticized contacts The average radius of the fully plasticized contacts is given by     1 fp c fp s s d a N aN z dz    

(27)

where N fp is the number of fully plasticized contacts and a in the integrand is given by Eq. (5). By setting  = z s – d (see Fig. 2), the above integral evaluates to     1 2 c fp fp a N Rs N F d  . (28) Since Eq. (3) implies that � �� � �� � �� �� ����� � , the average contact radius of the plasticized contacts becomes     1 2 c fp o fp a Rs F d F d        . (29)

The total area of the fully plasticized contacts is given by

 

 



   . R N z dz   

A dA 



fp

fp

s

s

d

d

fp

fp

As before, by making the substitution  = z s – d the expression for the fully plastic contact area becomes