Imperfections and slip / Annealing and cold work

Question 1

Identify the two types of atomic packing and what difference they have

Answer 1

1. Non-dense, Random packing
2. Dense, Ordered packing

Difference: Dense, orderly packed structures tend to have lower energies.

Question 2

What are Crystalline materials

Answer 2

Crystalline materials have atoms packed in a periodic 3D array. This is typically found in metals, many ceramics and some polymers. Materials can also be semi-crystalline with a mixture of crystalline and amorphous phases.

Question 3

What are Amorphous materials

Answer 3

Amorphous or Non-crystalline materials have atoms with no organised periodic packing. This typically occurs within complex structures and rapid cooling processing.

Question 4

Describe the crystal structure within metals and its features.

Answer 4

Atoms tend to be densely packed and nicely ordered:

Typically, only one element is present, similar atomic radius

Metallic bonding is not directional

Sea of electrons shields atomic cores from each other

Question 5

Identify the typical difference in densities between the classification of materials.

Answer 5

In general:
The density of metals > density of ceramics > density of polymers

Metals have orderly close-packing with often large atomic masses
Ceramics have less dense packing resulting in often lighter elements
Polymers have low packing density (often amorphous) but are made up of lighter elements (C,H,O)

And composites will typically have intermediate values depending on what materials make up their fibres.

Question 6

Give examples of engineering applications that require single crystals.

Answer 6

Turbine blades within engines (airplane)

and

Diamond cutting and grinding tools

Question 7

What are polycrystalline materials

Answer 7

A crystalline solid consisting of many crystalline parts that are randomly oriented with respect to each other. The areas where these crystals meet is known as a grain boundary. These grain boundaries will either be anisotropic or isotropic, the main difference being that the properties of isotropic materials are the same in all directions, whereas in anisotropic materials, the properties are direction dependent.

Most engineering materials are polycrystalline

Question 8

What is polymorphism and identify three examples

Answer 8

Polymorphism is when you can have two or more very distinct crystal structures for the same material.

Examples include:

• Titanium: Alpha-Ti and Beta-Ti
• Carbon: Diamond and Graphite
• Iron: (Refer to image)

Question 9

Identify the types of defects that can occur in materials

Answer 9

Point Defects: Vacancy atoms, Interstitial atoms and substitutional atoms.

Line or linear defects: Dislocations

Area defects: Grain boundaries

Question 10

Describe the types of point defects

Answer 10

Question 11

Identify the equation used to calculate the equilibrium concentration for point defects

Answer 11

Question 12

Find the equilibrium number of vacancies in 1 m3 of Cu at 1000C, given:

Density (Rho) = 8.4g/cm³

Q_v = 0.9 eV/atom

A_Cu = 63.5 g/mol

N_A = 6.02 x 10²³ atoms/mol

Answer 12

Question 13

Identify the linear defects in metals

Answer 13

Linear Defects (Dislocations)
Are one-dimensional defects around which atoms are misaligned

•Edge dislocation:
–extra half-plane of atoms inserted in a crystal structure
–b perpendicular to dislocation line

•Screw dislocation:
–spiral planar ramp resulting from shear deformation
–b parallel to dislocation line

• Slip:

− slip between crystal planes result when dislocations move
− produce permanent (plastic) deformation.

Question 14

Describe the Area defect: Grain boundaries

Answer 14

Grain Boundaries are the regions between crystals that are the transition from the lattice of one region to that of the other. Grain boundaries are slightly disordered and the lower density in grain boundaries causes:
–high mobility
–high diffusivity
–high chemical reactivity

Question 15

Identify the two outcomes when adding an impurity to a metal host to form a solid solution

Answer 15

Question 16

Identify the Hume-Rothery rules and conditions for creating substitutional solid solutions

Answer 16

Hume–Rothery rules:

1. Difference in atomic radius of less than 15% difference
2. Proximity in periodic table (similar electronegativities)
3. Same crystal structure for pure metals
4. Valency (higher valency preferred)

If all rules obeyed = full substitutional solid,
If more than 1 obeyed = ‘partial’ substitution

Question 17

State the equations used for calculating weight percentage and atomic percentage.

Answer 17

Question 18

State the Equations for the following values:

Tensile stress

Shear stress

Answer 18

Question 19

State the equations for the following values:

Tensile strain

Lateral strain

Shear strain

Answer 19

Question 20

State the equation used to calculate Youngs’s modulus of elasticity

Answer 20

A piece of copper, 305 mm long, is pulled in tension with a stress of 276 MPa MPa. If . If the deformation is entirely elastic, calculate elongation a) under stress, and b) when the stress is released. E= a) under stress, and b) when the stress is released. E= 110 GPa

Question 21

State Poisson’s ratio

Answer 21

Question 22

Answer 22

Question 23

Define the following terms:

Elastic behaviour:

Plastic behaviour:

Toughness:

Ductility:

Answer 23

Question 24

Using a graph identify yield strengths significance in plastic deformation.

Answer 24

Question 25

What is tensile strengths significance for a stress-strain curve?

Answer 25

Question 26

Graphically present how plastic deformation in metals occurs and explain how.

Answer 26

In metals, we saw that above the yield point plastic deformation occurs. In a perfect single crystal for this to occur every bond connecting two planes would have to break at once, thus requiring a very large amount of energy. However, dislocation allows those bonds along a dislocation line to be broken, rather than needing the entire plane of bonds to be broken at once, therefore requiring much less energy to break. Note: dislocations are defects and are weak spots in the crystal lattice.

Question 27

What dislocation motion do metals have when undergoing plastic deformation

Answer 27

Metal plastic deformation occurs by Slip - under shear stress, an edge dislocation (extra half-plane of atoms) slides over the adjacent plane of atoms. If dislocations can't move, plastic deformation doesn't occur (or require a much larger stress to occur)

Question 28

Define a slip system and how they relate to plastic deformation and dislocation movement

Answer 28

A slip system refers to the combination of a Slip plane and Slip direction Slip plane - The plane of which the easiest slippage will occur (highest planar atomic density(and large interplanar spacings)) Slip direction - Direction of movement (Highest linear densities) A dislocation moves along a slip plane in a slip direction perpendicular to the dislocation line The slip direction is the same as the Burgers vector direction Same resultant plastic deformation with a screw dislocation FCC or BCC high number of slip systems tend to be ductile. HCP have few and tend to be brittle. Temperature can induce different crystal structures and therefore different number of slip systems.

Question 29

Define the dislocation density

Answer 29

Number of dislocations, or dislocation density, is the total dislocation length per unit volume. Units are mm of dislocation per mm3 , or per mm2 * Careful solidification of metals = low densities, 1 x103 mm/mm2 * Poor solidification techniques = high densities, 1 x109 mm/mm2 * In ceramics, densities of 1 x103 mm/mm2 are typical * Single crystal silicon, densities of only 0.1 mm/mm2

Question 30

What are strain fields and how can they be used

Answer 30

Strain fields are formed around dislocations which can affect mechanical strength Strain fields cause atoms to interact – can be used to strengthen

Question 31

How is the resolved shear stress calculated?

Answer 31

Resolved shear stress (T_R ) = the dependence on the applied stress and orientation of stress direction relative to slip plane normal and slip direction. Although a number of slip systems exist in metal crystals, one slip system is generally favoured. This is termed the maximum resolved shear stress, T_R (max) Minimum shear stress required to initiate slip is termed the critical resolved shear stress, T_CRSS The condition for yielding (slipping/plastic deformation), i.e. the yield strength, occurs when T_R \> T_CRSS

Question 32

How to do you strengthen materials to reduce dislocations

Answer 32

Question 33

Define isotropic and anisotropic orientations

Answer 33

Question 34

Why do polycrystals have better resistance to plastic deformation

Answer 34

* Polycrystalline metals are stronger than single crystals – grain boundaries act as barriers preventing dislocation motion. * Slip planes & directions (l, f) change from one grain to another. * T_R will vary from grain to grain. * The grain with the largest T_R yields first. T_R (max). * Other (less favourably oriented) grains yield later. * During plastic deformation mechanical integrity is maintained along grain boundaries (area defects) as the grain boundaries do not come apart. * Each grain is also physically constrained by it’s neighbours which can further increase mechanical stability. * Before deformation, grains are typically randomly oriented and have the same dimensions in both the x and y axis (equiaxed), i.e. isotropic. * Deformation can cause grain elongation along the direction of the stress forming anisotropic grain structure.

Question 35

Discuss the following method of preventing dislocation movement and therefore strengthen metals: Reduce grain size

Answer 35

The size of grains, average grain diameter, influences mechanical properties. Adjacent grains normally have different atomic alignment and packing densities are not as high at the grain boundary interface. * Grain boundaries are barriers to slip. The slip plane has to change direction across the grain boundary * Barrier "strength" increases with increasing angle of misorientation. * If dislocations accumulate at grain boundaries stress can build up and induce new dislocations in adjacent grains. Fine-grained material (smaller grains) is harder and stronger than coarsely grained counterparts as the higher total grain boundary area impedes dislocation Grain size can be regulated by solidification rate from the liquid phase and through heat treatment – annealing. Toughness as well as strength improved by grain size reduction.

Question 36

Express the Hall-Petch equation used to calculate yield strength and use it to resolve the following issue

Answer 36

Question 37

Discuss the following method of preventing dislocation movement and therefore strengthen metals: Formation of solid solutions

Answer 37

Lattice strain exists around dislocations: The addition of large or small atoms into vacancies within the lattice structure can alter the strain fields around dislocations. * Impurity atoms distort the lattice & generate lattice strains. * These strains can act as barriers to dislocation motion. * Small impurities tend to concentrate at dislocations (regions of compressive strains). * Impurity partially cancels dislocation compressive strains by imposing tensile strains on the crystal lattice. * This results in reduced dislocation mobility and increases strength. * Large impurities tend to concentrate at regions of tension caused by dislocations. * Large impurities impose compressive strains on the crystal lattice. * In addition to reduced dislocation mobility, due to increased lattice stability, higher stress is needed to first initiate plastic deformation.

Question 38

Discuss the following method of preventing dislocation movement and therefore strengthen metals: Precipitation strengthening

Answer 38

Question 39

Discuss the following method of preventing dislocation movement and therefore strengthen metals: Cold working (strain hardening)

Answer 39

* Ductile metals become harder and stronger as they are plastically deformed. * Typically at room temperature (for most metals), but certainly well below the melting temperature (i.e. CW has a very low energy input). * Dislocation structure in Ti after cold working. * Dislocations entangle with one another during cold work. * Areas of compression and tension around the dislocation (within the lattice) repel one another. * Dislocation motion becomes more difficult resulting in a harder and stronger metal As CW procedures are typically carried out at room temp, energy from the mechanical processes can be stored as stress within the crystal lattice. • Electrical conductivity and corrosion properties can also be adversely modified by cold work.

Question 40

Express the equation used to calculate the degree of plastic deformation is expressed as a percentage of the cold work

Answer 40

Question 41

Answer 41

Question 42

Express the equation used to calculate the dislocation density (pd)

Answer 42

Question 43

Identify the three methods of annealing

Answer 43

There are three annealing stages: 1. Recovery 2. Recrystallization 3. Grain Growth

Question 44

Identify and discuss the annealing method: Recovery

Answer 44

Question 45

Identify and discuss the annealing method: Recovery

Answer 45

Question 46

Identify and discuss the annealing method: Recrystallization

Answer 46

• New grains are formed that: - have low dislocation densities - are small in size - consume and replace parent cold-worked grains

Question 47

Identify and discuss the annealing method: Grain Growth

Answer 47

• At longer times, average grain size increases. - Small grains shrink (and ultimately disappear) - Large grains continue to grow An empirical relationship exists between grain diameter (d), temperature (“K”) and time (t)

Question 48

Define Recrystallization Temperature

Answer 48

* Hot working → deformation above T_R * Cold working → deformation below T_R (typically room temperature)

Question 49

A cylindrical rod of brass originally 10 mm (0.39 in) in diameter is to be cold worked and the circular cross section maintained during deformation (i.e. drawn). A cold-worked tensile strength in excess of 380 MPa (55,000 psi) and a ductility of at least 15 %EL are required. Furthermore, the final diameter must be 7.5 mm (0.30 in). Describe the process required to meet the above specification.

Answer 49