Comparative study of wear mechanism of duplex treated M7 HSS drill bits

Abstract

The surface coating of drill bits improves the tool life considerably and enhances the end product quality. The improved coating is desirable for long service operations. One of the methods to achieve this is to develop an interface layer between the coat and the tool base such that the transition of shear properties occurs across the interface layer. The application of duplex treatment (plasma pre‐nitriding before PVD TiN coating) to the tool surface becomes fruitful in this regard. In the present study, the wear properties of the duplex treated and untreated M7 HSS twist drill bits are examined. The wear tests are conducted on a dry base, and the feed rate and rotational speed are kept constant during the drilling tests. SEM and EDS are conducted to analyze the wear properties and metallurgical changes on the tool surface after the tests. It is found that the build up edge forms on the untreated tool surface, which in turn results in adhesion and abrasive wear. In the case of duplex treated drill bit, thinning of TiN coat is observed, but the coat is neither partially nor totally pealed off from the tool surface in the wear land.

A study of cutting forces and surface finish in high-speed machining of AISI H13 tool steel using carbide tools with TiAIN based coatings

“…It has been observed that in HSM ball-nose end milling of H13, the cutting temperature increases with an increase of cutting speed and depth of cut [14]. The increased oxidation stability of TiAlN layers with a high content of aluminium leads to an improvement in abrasion resistance and wear of the tool [15]. The superiority of TiAlN to TiN has been attributed to better oxidation resistance [16].…”

Laser Duplex Treatment of Surfaces for Improved Properties

“…The Laplace inversion of boundary conditions gives Tðx; 0Þ ¼ 0 and vTðx; pÞ vx x¼0 ¼ rVL kp [5] and TðN; pÞ ¼ 0 [6] The analytical solution to eqn [4] gives the result: The Laplace inversion of boundary conditions gives Tðx; 0Þ ¼ 0 and vTðx; pÞ vx x¼0 ¼ rVL kp [5] and TðN; pÞ ¼ 0 [6] The analytical solution to eqn [4] gives the result:…”