1 / 15

CHE 333 Class 13

CHE 333 Class 13. Defects in Crystals. Perfect Structure. Perfect Structure for FCC, BCC and HCP crystals – all atom sites filled with an atom. Reality – this does not happen.

mya
Download Presentation

CHE 333 Class 13

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. CHE 333 Class 13 Defects in Crystals.

  2. Perfect Structure Perfect Structure for FCC, BCC and HCP crystals – all atom sites filled with an atom. Reality – this does not happen. Defects occur in the crystal structure. These can range from small to large, that is a missing atom or many missing atoms to form a void in the material. Missing Atom – called a “vacancy”. There is an equilibrium number of vacancies in any Crystal. For aluminum at room temperature it is about one vacancy for every 1010 atoms. At around 500C , the number is one in 104, as the relationship is exponential with Temperature. Processes such as self and substitutional diffusion, in which atoms move around in a Solid, depend on vacancies, so these are a function of temperature, and increase with it. Many high temperature processes are controlled by the movement of vacancies in the Crystal structure.

  3. Vacancy

  4. Line or Two Dimensional Defects 1 2 A dislocation in a crystal structure. An extra half plane of atoms is present in the crystal structure. The end of the extra half plane determines the slip plane, which is also the plane of densest packing, and The dislocation line. The atomic motion is determined by the Burgers vector of the dislocation “b”. The Burgers vector is determined By counting around the dislocation an equal number of steps in each direction. For example,,going clockwise Around the dislocation from atom number 1. One right is followed by 3 down, followed by 3 left, then 3 up. For An equal number of atoms, then only two steps to the right is allowed, to finish at atom 2. The Burgers Vector is then the distance needed to close the loop, as shown in the figure. One of the features of an edge dislocation is that the Burgers vector is 90 degrees to the dislocation line, But parallel to the overall motion of the dislocation.

  5. Screw Dislocation A second type of dislocation is a SCREW dislocation. The properties in this case are that the Burgers vector is parallel to the line, but 90 degrees to the movement of the dislocation line. Note that the Burgers vector is a constant along the dislocation line. However if the line curves Then a mixed dislocation will exist, with both edge and screw characteristics.

  6. Dislocation

  7. Dislocation Movement. Dislocations have a displacement associated with them, the Burgers vector. The Burgers vector is The direction of densest packing in a crystal, so for FCC it is <110>, for BCC it is <111> and for HCP It is <1120>. The combination of a slip plane and a slip direction is called a slip system. So the Movement of a dislocation along a slip plane in a slip direction produces physical displacements. A force is required to move the dislocations, which comes from stressing a material.

  8. Twins – Atomic Arrangement Atomic Structure of Twins

  9. Twins Twins in 70-30 Brass

  10. Twins and Dislocations The same displacements can be achieved by either dislocation motion on Slip planes or by twins.

  11. Grain Boundaries. 1 2 3

  12. Grain SIze ASTM has a grain size number from 1 to 10. It is determined by preparing a Sample for examination at 100 times magnification and measuring the number Of grains per square inch. N=2(n-1) Where N= average number of grains per square inch at x100 mag and N = grain number. Usually charts can be used which are standards against which a sample is compared. The chart most closely resembling the test samples is the grain size.

  13. Pores Al 356 Casting alloy with pores.

  14. Homeworks • Age Hardening. An alloy contains 10% B in A. a a +b solvus temp is 475 C. a  a + Liquid solidus temp is 525 C. Detail the temperature and time needed to age harden this alloy. From this data what would be the maximum service temperature for this alloy.

  15. Homeworks 2. Dislocations. For an edge dislocation, provide a Burgers circuit to indicate how a Burgers vector is determined. From the notes from, Precipitation Hardening, what is the actual distance for the Burgers vector for aluminum and copper? Atomic radius for Fe 0.124nm, Ti 0.145,Al0.143,Cu 0.128,Zn 0.133. Calculate Burgers vectors for these. Also for 0.01mm movement how many Dislocations are needed?

More Related