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Unformatted text preview: CHAPTER 4
Imperfections in solids Types of Crystallographic Defects Vacancy and Interstitial Defects Substitutional and Interstitial Impurities Edge and Screw Dislocations Grain Boundaries WHY STUDY IMPERFECTIONS IN SOLIDS? Dislocations are essentially what make deformation in metals possible. Vacancies play a role in many of the problems and treatments of alloys. Impurities can greatly effect most of the properties of any given material. WHY STUDY IMPERFECTIONS IN SOLIDS? (cont.) For example, an aluminum alloy which has 4 wt% copper (impurity), is significantly stronger than pure aluminum. Also, Iron is significantly strengthened by the addition of less than 1 wt% of carbon (i.e. steel). Learning Objectives Describe vacancy and interstial point defects. Calculate number of vacancies present in a material at a particular temperature. Define substitutional and interstial solid solutions. Learning Objectives Cont. Calculate atomic percent and weight percent of each element present in an alloy. Describe edge, screw and mixed dislocations (linear defects). Define grain boundary and grain size number. POINT DEFECTS There are two types of point defects in metals, vacancy and selfinterstial. An atom missing from a regular atomic site is a vacancy defect. When an atom occupies an empty position in between atoms, it is a selfinterstial defect. Point Defects in Metals Vacancies are several orders of magnitude more common than interstitial defects, since significantly less energy is required for their formation. Hence, generally speaking, much more attention is devoted to vacancies than interstitial defects. However, the numbers of each in a given material increase exponentially with an increase in temperature. Point Defects in Metals cont.
The number of vacancies, Nv, present in any material depends on temperature in the following way: Nv = N exp ( Qv/ k T ) (Eqn. 4.1)
Where: N = Total no. of atomic sites = NA /A (Eqn. 4.2) Qv = Energy required to form vacancy T = Absolute temperature in Kelvin K = Boltzman's constant NA = Avogadro's No. = density A = atomic weight Substitutional and Interstitial Impurities IN SOLIDS Metal Alloys Impurities in Metals Metals are normally used in the form of alloys which are a mixture of two or more metals. Metal alloys are actually solid solutions of two different metals, and not a chemical compound. In an aluminum alloy of 96 wt% Al and 4 wt % Cu, aluminum would be the solvent and copper would be the solute. Impurities in Metals cont. The presence of impurities always hardens and strengthens a material, but can have other potentially less desirable effects as well. Solid solutions Solid solutions of metals are either: 1) Substitutional or 2) Interstitial In the substitutional solid solution, the impurity atom (solute) replaces the solvent atom. Example : Cu Ni solid solution. Solid solutions cont. In the interstial solid solution, the impurity atoms are much smaller in size and instead of replacing the solvent atom they go into the empty space in between atoms. Example: Fe C solid solution . Substitutional Solid Solutions
To make a fully substitutional solid solution, four rules or factors have to be satisfied. These are: 1. Atomic size factor: must have similar atomic radii. ((rsolvent rsolute) / rsolvent) 100% <15% Substitutional Solid Solutions cont.
2. Crystal structure : same crystal structure. 3. Electro negativity : similar electro negativity. 4. Valence: same valences. Group Discussion Determine if Ni would make a complete substitutional solid solution with Cu. Rule #1 (rCu rNi)/rCu 100% = (reference front cover) Rule #2 Crystal structure: (reference front cover) Group Discussion cont. Rule # 3 Electronegativities: (hint: look at Table 2.7 on pg. 19) Rule # 4 Valences: (reference front cover) SPECIFICATION OF COMPOSITION Composition of an alloy can be expressed in weight percent (wt%) or atomic percent (at%). C1= m1/(m1+m2) x 100% (Eqn. 4.3) where C1 is wt%. C1'=nm1/(nm1+nm2) x 100% (Eqn. 4.5) where C1' is at%. Consider 2024 Al alloy which contains only copper and aluminum. For example, if there was 100g of the alloy and it contained 96g of Al and 4g of Cu. Then the respective weight percents would be as follows: Wt% of Al = 96 g / 100 g 100% = 96 wt% Al Wt% of Cu = 4 g / 100 g 100% = 4 wt% Cu SPECIFICATION OF COMPOSITION Specification of Composition cont. To calculate at%, we have to find out the total number of moles (gatom) of each element present. (Which is the total number of grams divided by the molecular weight of the element.) Moles of Al = 96g/26.98 g (atomic wt in g) = 3.56 moles Moles of Cu = 4g/63.55g = .063 moles Specification of Composition cont. Total No. of Moles = moles Al + moles Cu = 3.623 moles at% of Al = (moles Al / Total moles) * 100% = 98.26 at% at% of Cu = (moles Cu / Total moles) * 100% = 1.74 at% Composition Conversions You can convert between at% and wt% by using the following equations: C1'=C1A2/(C1A2 + C2A1) x 100% (4.6a) C2'=C2A1/(C1A2 + C2A1) x 100% (4.6b) C1=C'1A2/(C'1A2 + C'2A1) x 100% (4.7a) C2=C'2A1/(C'1A2 + C'2A1) x 100% (4.7b) Other Conversions You will find additional conversions in your book. For instance, Eqn. 4.10a gives a formula for calculating the average density of an alloy. ave= 100 / (C1/1 + C2/2) DISLOCATIONS Linear Defects There are Three Types of Dislocations Edge Dislocations Screw Dislocations Mixed Dislocations Edge dislocations The edge of an extra plane of atoms within a perfect atomic structure is an edge dislocation line (Ref. Fig. 4.3). The motion of the dislocation is perpendicular (or normal) to the extra plane of atoms, and parallel to the Burgers vector of the dislocation. Edge Dislocations cont. It is the movement of one atomic distance by countless aligned dislocations which manifests itself on a macroscopic scale as a measurable deformation in metals. Screw Dislocations In Fig 4.4(a), the upper front part of the crystal is shifted to the right by one atomic distance with respect to the lower front. The edge of this distortion is a screw dislocation line. As might be expected, screw dislocations are associated with a twisting movement. Screw Dislocations cont. Screw dislocations move in a direction parallel to the extra plane of atoms, but perpendicular to their Burgers vectors, the opposite of how edge dislocations move. Fig 4.4(b) The screw dislocation in (a) as viewed from above. Mixed Dislocations Mixed dislocation line is a mixture of an edge dislocation line and a screw dislocation line (Fig.4.5(a)). Mixed dislocations allow for changes of direction. Fig 4.5 (a) Schematic representation of a dislocation that has edge, screw, and mixed character SCREW DISLOCATION LINE EDGE DISLOCATION LINE Fig 4.5 (b) Top view, where open circles denote atom positions above the slip plane. Fig 4.6 A transmission electron micrograph of a titanium alloy in which dark lines are dislocations. INTERFACIAL DEFECTS Grain Boundaries Grain Boundaries Grain boundaries are the largest crystallographic defect. They result where grains of differing crystallographic orientations meet. Grain boundaries play an important role in the strength and high temperature properties of metals. Grain Boundaries cont. As seen in Figure 3.17, because of the nucleation of many grains of random orientation upon cooling from the liquid to solid state, when the grains grow larger and eventually begin to meet one another, they have varying degrees of mismatch, which produces a discontinuity, called grain boundaries. Fig 3.17 (a) Small crystallite nuclei. (b) Growth of the crystallites; obstruction of some grains that are adjacent to one another is also shown. (c) Upon completion of solidification, grains having irregular shapes have formed. (d) The grain structures as it would appear under the microscope; Dark lines are the grain boundaries. Grain boundary Angles Fig 4.7 shows grain boundaries where there is an atomic mismatch. When the mismatch between adjacent grains is low then the grain boundary is a low angle grain boundary. When the mismatch is large the boundary is a high angle grain boundary. Fig 4.8 shows a small angle tilt boundary made of edge dislocations. Fig 4.7 Schematic diagram showing low-and high-angle grain boundaries and adjacent atom positions. Fig 4.8 Demonstration of how a tilt boundary having an angle of misorientation results from an alignment of edge dislocations. Twinning Another type of grain boundary defect which can occur is called twinning, however it is very different from a normal grain boundary. Twinning results from a sudden movement of many atoms (reference Fig. 4.9). Twinning cont. Almost simultaneously they break their atomic bonds and form them anew with a different near neighbors, such that a new crystallographic structure results. The resultant structure has a high degree of crystalline order, but is not a perfect crystal. Fig 4.9 Schematic diagram showing a twin plane or boundary and the adjacent atom positions (dark circles). Optical Microscopy With the proper sample preparation, grain boundaries are easily visible with the aid of an optical microscope. ASTM E395: Preparation of Metallographic Specimens, gives specific methods for the proper preparation of samples in order to view their grain structure. Fig 4.10 High-purity polycrystalline lead ingot in which the individual grains may be discerned. Sample Preparation First, a sample must be very finely polished to a mirror finish. Next, various acids and etching solutions can be used on the polished surface to etch out details. Sample Preparation cont. For instance, grain boundaries because of their crystallographic imperfection will be more vulnerable to the etching solution and will therefore be effected much quicker than other areas, creating discernable lines (ref. fig. 4.12a). Fig 4.12(a) Section of a grain boundary and its surface groove produced by etching; the light reflection characteristics in the vicinity of the groove are also shown. Sample Preparation cont. Also, the crystallographic orientation of some grains will be more effected by the solution than other orientations, causing some grains to appear darker or lighter than others (reference Fig. 5.15 a & b). Lastly, where dislocations intersect the surface within the grains will be more effected by the etching solution, and some pitting may be observable (reference Fig 7.0). Fig 4.11 (a) Polished and etched grains as they might appear when viewed with an optical microscope. Fig 4.11 (b) Section taken through these planes showing how the etching characteristics and resulting surface texture varying from grain to grain because of difference in crystallographic orientation. Fig 4.11 (c) Photomicrograph of a polycrystalline brass specimen A photomicrograph of etched pits resulting where dislocations intersected the surface of a lithium fluoride single crystal. Using ASTM E11296 Standard Test Methods for Determining the Average Grain Size, one can determine the ASTM standard grain size from a photo micrograph of a properly prepared sample at 100x magnification. Grain Size Number Determination Grain Size Number Determination cont. If N is the number of grains per square inch at a magnification of 100x, then n, the grain size number, is defined as: N = 2 n 1 ln N = (n1) ln 2 or n = (ln N / ln 2) + 1 Fig 4.12(b) Photomicrograp h of the surface of a polished and etched polycrystalline specimen of an iron-chromium alloy in which the grain boundaries appear dark. 1 4 9 3 5 2 6 7 10 6 8 Group Discussion Problem
Calculate the number of vacancies per cubic meter in iron at 850C. The energy per vacancy formation is 1.08 eV/atom. Furthermore, the density and atomic weight for Fe are 7.65 g/cm3 and 55.85 g/mol, respectively. Hint: Nv = N exp ( Qv/ k T ) Hint:
where N = Total no. of atomic sites = NA/A Qv = Energy required to form vacancy T = Absolute temperature in Kelvin K = Boltzman's constant = 8.62 x 105 eV/atomK NA = Avogadro's No. = density A = atomic weight Ans.: 1.18 x 1018 / cm3 = 1.18 x 1024 / m3 Group Discussion Problem
What is the composition, in atom percent, of an alloy that consists of 30 wt% Zn and 70 wt% Cu? Hint: Consider 100 g of the alloy. It contains 30 g Zn and 70 g Cu. To calculate at% we have to find out the total number of moles (gatom) of each element present. Moles of Zn = 30g/AZn (atomic wt in g) = Moles of Cu = 70g/ACu = at% of Zn = Moles of Zn x 100% = 29.4 at% Moles of Zn + Moles of Cu at% of Cu = Moles of Cu x 100% = 70.6 at% Moles of Zn + Moles of Cu Group Discussion Problem Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a germaniumsilicon alloy that contains 15 wt% Ge and 85 wt% Si. The densities of pure germanium and silicon are 5.32 and 2.33 g/cm3, respectively. Hint: Use equation 4.9a on page 71, which gives a result in kg/m3, and the atomic weight of Ge. Ans: N1 = 3.17 x 1021 atoms/cm3 Group Discussion Problem A photomicrograph was taken of some metal at a magnification of 100x and it was determined that the average number of grains per square inch is 10. Compute the ASTM grain size number for this alloy. Hint: If N is the no. of grains per square inch at a magnification of 100 x, then the grain size no., n, is defined as N = 2 n -1 ln N = (n-1) ln 2 or n = (ln N / ln 2) + 1 = 4.3 Homework HW #3 4.2, 4.5, 4.10, 4.14, 4.21 Due 2/14/05 Form Groups ...
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This note was uploaded on 05/03/2010 for the course ME 250-750 taught by Professor Signer during the Summer '10 term at Wichita State.
- Summer '10