Skip to main content

Stacking-fault energy and yield stress asymmetry in molybdenum disilicide

Buy Article:

$60.90 plus tax (Refund Policy)

Abstract:

Stacking-fault energies in MoSi2 due to shear along <331> have been calculated by ab-initio and modified embedded-atom method (MEAM) calculations. The results are used to investigate the configurations of ½:<331> dislocations and their mobility. Shear of ⅙<331> in the {103} plane of MoSi2 produces an antiphase boundary (APB) whose geometry, called APB(1), is different from that produced by ⅙<331> in the opposite direction, APB(2). MEAM calculations show that APB(1) is stable while both types of calculation show that APB(2) is unstable. Both ab-initio and MEAM calculations show that there is a stable fault close to APB(2) with a displacement of about ⅛<331> in the same direction. The calculations also show that there is a stable fault in the {110} plane with a displacement of ¼<111>. The identical fault is produced by a shear of ⅙<331>.There is good agreement between the fault energies calculated by the two methods and also with the experimental value (200-370mJm-2). The agreement between the calculated fault energies in the {013} plane is not so good. One factor is that the relaxation procedures are different; the MEAM method has more flexibility as well as a larger number of atoms, possibly explaining why it gives lower stable fault energies. The {103} planes have an unusual five-layer ABCDE stacking sequence with successive planes offset by 1/5<301>. Shear of 1/10<351> in the correct direction gives a low-energy fault with Mo atoms surrounded by the correct number (ten) of Si nearest neighbours. This vector is close to the 1/8<331> shear that produces a stable fault and may explain its low calculated energy. Various dissociated configurations of ½<331> dislocations are considered on the basis of ⅙<331>, 1/8<331>, ¼<331> and 1/10<351> partials. All can have asymmetrical arrangements which willrespond differently to the direction of the applied stress, explaining why {103}<331> slip is much easier for crystals compressed along \[100] than along \[001].

Document Type: Research Article

DOI: https://doi.org/10.1080/01418610108214430

Publication date: 2001-05-01

More about this publication?
  • Access Key
  • Free ContentFree content
  • Partial Free ContentPartial Free content
  • New ContentNew content
  • Open Access ContentOpen access content
  • Partial Open Access ContentPartial Open access content
  • Subscribed ContentSubscribed content
  • Partial Subscribed ContentPartial Subscribed content
  • Free Trial ContentFree trial content
Cookie Policy
X
Cookie Policy
Ingenta Connect website makes use of cookies so as to keep track of data that you have filled in. I am Happy with this Find out more