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Surface treatment and tribology

A flat steel block (indentation hardness H = 2.0 GPa) rests on a plate of the same steel, and a normal load of 1.0 kN is applied over an apparent contact area of 10 cm². Assuming every contacting asperity deforms plastically and estimating the real area of contact as Ar = W/H, which is closest to its fraction of the apparent contact area?

  1. AAbout 0.5%
  2. BAbout 50%
  3. CAbout 0.05%
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Answer: C. About 0.05%

Ar = W/H = 1.0×10³/(2.0×10⁹) = 5.0×10⁻⁷ m² = 0.50 mm². The apparent area is 10 cm² = 1000 mm², so the fraction is 0.50/1000 = 5×10⁻⁴, about 0.05%. Even on a surface that looks flat, only a tiny part at the asperity tips actually touches, and those spots carry a high pressure equal to the hardness. About 0.5% is the value obtained by converting 1 cm² as 10 mm² instead of 100 mm², and about 50% is the value obtained by treating the hardness in MPa instead of GPa. The high local pressure at the real contact spots, together with the high temperature generated by sliding, is the starting point for adhesion and wear.

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