Materials Engineering (Week 3)

Question 1

When does plastic deformation occur?

Answer 1

Plastic deformation occurs when bonds are
permanently broken during increasing strain.

Plastic deformation = motion of large number of dislocations

Question 2

In crystalline materials, plastic deformation is
related to what?

Answer 2

In crystalline materials, plastic deformation is
related to the presence of defects and
imperfections in the solid

Question 3

What is Plastic deformation by slip?

Answer 3

Plastic deformation by slip where one plane of atoms
slides over adjacent plane by dislocation motion

Question 4

The movement of the dislocation is called…

Answer 4

Slip

Question 5

What is the slip plane
and the slip direction?

Answer 5

The crystallographic plane along which the
dislocation line travels is the slip plane.
The direction it moves in is the slip direction.

Question 6

Slip plane and slip direction form what?

Answer 6

The slip system.

Question 7

The slip system is defined by what?

Answer 7

the crystallographic
structure of the material

Question 8

The slip plane will move: (2)

Answer 8

– Along the path of least resistance
– Along a crystal plane where the density of atoms
is highest

Question 9

Dislocation moves along what? (2)

Answer 9

slip plane in slip direction

Question 10

Dislocation density in metals is typically …
give range

Answer 10

10^3 – 10^10 mm^-2

Question 11

Edge dislocation

Answer 11

Edge dislocation moves in response to
shear stress in stress direction.

Question 12

Screw dislocation

Answer 12

Screw dislocation
moves perpendicular
to stress direction.

Question 13

Slip direction

Answer 13

direction of movement - Highest linear
densities

Question 14

Slip plane

Answer 14

• plane allowing easiest slippage
• Wide interplanar spacings - highest planar densities

Question 15

BC, FFC, HCP
which has the most number of slip systems, and hence is more likely (highest potential) to experience dislocation movement

Answer 15

Body Centered Cubic
Face-Centered Cubic
Hexagonal Close-Packed

Question 16

Stress and Dislocation Motion
Crystals slip due to
.

Answer 16

a resolved shear stress, tau sub(R)
Applied tension can produce such a stress.

Question 17

Stress and Dislocation Motion
Equation for a slanted (diagonal) plane:

Answer 17

normally Applied tensile stress: sigma = F/A

Resolved shearstress: t sub(R) =Fs/As
Fs - shear force acting parallel to the slip direction
As - is the cross sectional area of slippage plane (actually larger)

Fs = f x cos(lambda)
As = A / cos(phi) phi- think water potential symbol

Final equation: Tau = sigma x cos(lambda) x cos (phi)
as F = sigma x A

Question 18

Critical Resolved Shear Stress

When does a slip system occur?

Answer 18

when the resolved shear stress is
larger than a critical value

tau sub(R) > tau sub(CRSS)

CRSS - Critical Resolve Shear Stress

Question 19

What variation can make slippage easier or more difficult?

Answer 19

Crystal orientation can make slip easy or difficult

Tau = sigma x cos(lambda) x cos (phi)
Remember cos(lambda) and cos (phi) are angles

Question 20

Slip Motion in Polycrystals

We have previously been considering
single crystals only
* Many materials are polycrystalline
containing a number of single crystals

what can you say about shear stress in polycrystals?

Answer 20

Slip planes & directions (lambda, phi) change from
one crystal to another.

• tau sub(R) will vary from one crystal to another.
• The crystal with the largest tau sub(R) yields first.
• Other (less favourably oriented) crystals
  yield later.
• Stronger - grain boundaries pin
  deformations
• Deformation of one grain constrained by
  neighbouring grains

Alteration in grain structure can make the material more malleable as majority of the grains are now isotropic

Question 21

Slip in polycrystalline materials

Strength of metals and alloys can be
increased if slip motion is made more difficult

Slip motion in polycrystalline materials (such
as metals and alloys) hindered by (made more difficult):

Answer 21

– Reducing grain size
– Making a Solid solutions (by adding an impurity to it)
– Cold working (form of deformation under ambient or slightly elevated temp)

Question 22

Strategies for Strengthening:
1. Reduce Grain Size
(number of grains)

Answer 22

Each single crystal in a polycrystalline material is known as
a grain. The junction between grains is the grain boundary.
* Grain boundaries are barriers to slip.
* Barrier “strength“ increases with increasing angle of
misorientation.

The greater the angle between the two grains the greater the misalinement and hence the bigger the barrier.

If we consider how the grain size is related to the amount
of grain boundaries
* Smaller grain size: more barriers to slip.

Question 23

Strategies for Strengthening:
1. Reduce Grain Size
(number of grains)

Hall-Petch Equation:

Answer 23

sigma sub(yield) = sigma sub(o) + (k sub(y) x d^1/2)

where k sub(y) and sigma sub(o) are constants
and d is the grain size (length)

Conclude: if the grain size goes down, yield stress goes up

Question 24

Strategies for Strengthening:
2. Solid Solutions

Answer 24

-Impurity atoms distort the lattice & generate stress.
-Stress can produce a barrier to dislocation motion.

If we consider dislocations, the lattice is already
distorted. This distortion is lattice strain

Impurity atoms tend to concentrate at
dislocations in order to lower lattice strain

If the atoms are bunched up above the slip plane,
small impurity atoms will lower the lattice strain

Large impurities concentrate at dislocations on low
density side

It can be a large or small impurity atom (in comparison to the pure metal atoms)

*refer to diagram for these concepts

Question 25

Why impurity atoms reduce slip

Answer 25

Impurity atoms lower the lattice strain For slip to initiate, the bonds between the impurity atoms need to be extended and broken. This increases the lattice strain significantly, indicating that more force is required to both initiate and progress the slip system

Question 26

Strategies for Strengthening: 3. Cold Working (%CW)

Answer 26

Cold working, also known as strain hardening, is when a material is plastically deformed Straining a material of yield strength σ sub(y0) to strain D results results in plastic deformation Releasing the force and then straining again will give an increased yield strength (σ sub(yi)) Room temperature deformation. * Common forming operations change the cross sectional area: Forging, Rolling, Extrusion and Drawing all put the material under strain.

Question 27

Percent cold work: Equation

Answer 27

%CW = A(o) - A(d) / A(o) x 100 A(o) initial cross sectional area A(d) changed cross sectional area

Question 28

Dislocations During Cold Work

Answer 28

-Dislocations entangle with one another during cold work. -Dislocation motion becomes more difficult

Question 29

Result of Cold Work Dislocation density equation Does yield stress increase or decrease as dislocation density increases? and why?

Answer 29

Rho sub(d) = total dislocation length / unit volume Yield stress increases as Rho sub(d) increases. The number of dislocations increases after cold working. * As the dislocation density increases, the chances of dislocations being closer to one another increases. * Dislocations can repulse (if two compressed zones interact/ align) or cancel out one another (if the two dislocations are oppositely sensed i.e tensile above compression below and vise versa for the other dislocation, the tensile and compressive forces cancel each other out and attract each other, Dislocation Annihilation) * On average, dislocation-dislocation strain interactions are repulsive.

Question 30

Therefore to summaries As cold work increases: (3)

Answer 30

Yield strength (sigma sub(y)) increases. * Tensile strength (TS) increases. * Ductility (%EL or %AR) decreases, because of the dislocation, saturation and repulsion.

Question 31

We have shown how cold working increases the yield strength of a material but lowers its ductility. We are now going to consider the opposite where the ductility is increased but, as a consequence, the strength decreases. What is the process the material must undergo?

Answer 31

Annealing (heating without melting) Effects of cold work are reversed!

Question 32

What are the stages of Annealing?

Answer 32

Recovery- Polycrystalline materials possess a range of different lattice strains. * Removal of these lattice strains is achieved by heat treating the material in a process known as annealing. * The increased thermal energy of the atoms in the lattice allow diffusion to occur and reduce the number of dislocations. * The reduction in dislocations is called recovery . Recrystallisation- Annealing at elevated temperature allows atoms to move and form new crystals. * These new crystal grains are small, relatively defect free and consume cold worked grains that possess a high dislocation density Further Recrystallization (as time enfolds) All cold-worked grains are consumed. Recrystallization of cold worked metals can be used to refine the grain structure.

Question 33

Recrystallization temperature,T sub(R)=

Answer 33

temperature at which recrystallization is complete within 1 h

Question 34

Annealing: Grain Growth (why does it happen)

Answer 34

At longer times, larger grains consume smaller ones. Grain boundary area (and therefore energy) is reduced.

Question 35

Dislocations are observed primarily in what materials? (2)

Answer 35

metals and alloys

Question 36

Strength is increased by making ... ... difficult.

Answer 36

dislocation motion

Question 37

Heating (annealing) can reduce ... ... and increase ... ... This decreases the strength

Answer 37

dislocation density grain size

Question 38

Methods to increase strength are: (3)

Answer 38

* decrease grain size * solid solution strengthening * cold work

Question 39

Grain Growth Empirical Relation: Equation

Answer 39

d^n - d sub(o) ^n = Kt k - coefficient dependent on material and T t - time elapsed d - grain diam. at time t. n - exponent typ. ~ 2

Question 40

There is a strong correlation between microstructure and what kinda property

Answer 40

mechanical properties.

Question 41

The development of microstructure of an alloy is related to the characteristics of its phase diagram. Phase diagrams show...

Answer 41

equilibrium states

Question 42

What is the difference between solutions and mixtures

Answer 42

– Solutions – solid solutions, single phase – Mixtures – more than one phase

Question 43

Solubility Limit:

Answer 43

Max concentration for which only a single phase solution occurs. Think of the graph, temp against C(o), composition (wt% e.g sugar)

Question 44

Components:

Answer 44

The elements or compounds which are present in the mixture (e.g., Al and Cu)

Question 45

Phases:

Answer 45

The physically and chemically distinct material regions that result (e.g., alpha and beta).

Question 46

What can you say about phases, and what happens across phase boundaries? Give 2 examples

Answer 46

Phases are homogeneous and have uniform physical and chemical properties. * Across phase boundaries, physical and/or chemical characteristics change abruptly. * Example 1: water and ice are chemically identical but physically dissimilar. * Example 2: polymorphism (metals existing in more than one crystal structure)

Question 47

Microstructure is determined by: (3) (to do with phases) what does microstructure determine?

Answer 47

-number of phases present - proportions of phases - distribution and arrangement of phases Microstructure determines mechanical properties

Question 48

Effect of T & Composition (Co)

Answer 48

Changing T can change number of phases Changing Co can change number of phases

Question 49

Phase equilibria When are equilibrium conditions met?

Answer 49

Equilibrium condition: the Gibbs free energy is at a minimum. Time period required to attain equilibrium can be very long, particularly in solid systems.

Question 50

Time period required to attain equilibrium can be very long, particularly in solid systems. Systems in non-equilibrium where rate of approach to equilibrium is extremely slow are called ...

Answer 50

metastable

Question 51

Phase diagrams indicate phases as function of ... (3)

Answer 51

Phase diagrams indicate phases as function of Temperature, Composition, and Pressure

Question 52

What is the process called when substances goes from solid to gas

Answer 52

Sublimation (reverse called deposition)

Question 53

For this course: -binary systems: just 2 components. -independent variables: T and Co (P = 1 atm is almost always used). Phase Diagrams: What info do you need to work out the number and types of phases, and the compostion (%) of each phase

Answer 53

Number and types of phases: If we know T and composition (Co), then we know: - the number and types of phases present. composition of phases: If we know T and Co, then we know: --the composition of each phase. weight fractions of phases: If we know T and Co, then we know: --the amount of each phase (given in wt%).

Question 54

The Lever Rule (Tie Line)

Answer 54

W sub(L) = S / R + S W sub(alpha) = R / R + S alpha- for the solid Tie line – connects the phases in equilibrium with each other - essentially an isotherm

Question 55

Metal Fabrication in general

Answer 55

How do we fabricate metals? – Blacksmith - hammer (forged) – Molding - cast * Forming Operations – Rough stock formed to final shape Hot working* T high enough for recrystallization * Larger deformations Cold working well below Tm work hardening smaller deformations

Question 56

What are metal alloys?

Answer 56

Metal alloys are mixtures of metal types along with other elements – E.g. Fe with carbon - Weight %, temperature – Define the microstructure via phase diagrams

Question 57

Volume fractions * For multiphase alloys...

Answer 57

* For multiphase alloys, relative phase amounts can be specified in terms of volume fraction rather than mass fraction * Phase volume fractions can be determined from examination of the microstructure * For an alloy consisting of phases α and β: V (alpha) + V (beta) = 1 Therefore V (alpha) = V (alpha) / V (alpha) + V (beta) Now sub in V (x) = W(x) / Rho(x) for where x = alpha and beta respectively

Question 58

What is meant if a system (solution) is binary?

Answer 58

System is: --binary i.e., 2 components: Cu and Ni.

Question 59

What is meant if a system (solution) is isomorphous?

Answer 59

i.e., complete solubility of one component in another; alpha phase field extends from 0 to 100 wt% of a chosen element on the phase diagram.

Question 60

Metal Fabrication Methods (3)

Answer 60

Forming Casting Rolling

Question 61

Metal Fabrication Method Casting

Answer 61

mold is filled with metal – metal melted in furnace, perhaps alloying elements added. Then cast in a mold – most common, cheapest method – gives good production of shapes – weaker products, internal defects – good option for brittle materials

Question 62

Metal Fabrication Method Forming (list all variations): 4

Answer 62

Forging (Hammering; Stamping) Rolling (Hot or Cold Rolling) (I-beams, rails, sheet & plate) Drawing (rods, wire, tubing) die must be well lubricated & clean to reduce friction and prevent impurities Extrusion (rods, tubing) ductile metals, e.g. Cu, Al (hot)

Question 63

Metal Fabrication Method Casting (list all variations): 4

Answer 63

Sand Casting (large parts, e.g.,auto engine blocks) * trying to hold something that is hot * what will withstand >1600ºC? * cheap - easy to mold => sand!!! * pack sand around form (pattern) of desired shape Investment Casting (low volume, complex shapes e.g., jewelry, turbine blades) plaster die formed around wax prototype - pattern is made from paraffin. - mold made by encasing in plaster of paris - melt the wax & the hollow mold is left - pour in metal Die Casting (high volume, low T alloys) Continuous Casting (simple slab shapes) visual filter, molten metal poured in, comes out as a solidified tube like structure.

Question 64

Metal Fabrication Method Joining (list all variations): 2

Answer 64

Powder Metallurgy (materials w/low ductility) point contact at low T (of three seperated atoms) --> densify --> densification by diffusion at higher T. Welding (when one large part is impractical) Heat affected zone: (region in which the microstructure has been changed)