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Metals Flashcards

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1
Q

Steel vs aluminium

A

Steel (FE & C) - stronger, cheaper (1/3-1/4), higher energy absorption, high fatigue performance (2x), 210 GPa
Aluminium - low density (1/3) & formability (2/3, springback during forming), 70 GPa

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2
Q

TWIP steels

A

Twinning induced plasticity
30% Mn, 9% Al
Very ductile 88% elongation
Medium-high strength, 650 MPa

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3
Q

Casting process & different types, rate applicability, finish, tolerance, cost & production rate

A

Molten metal, mould cavity (fluid flow), solidification (shrinkage), finishing
Sand - wide application, poor finish & tolerance, low cost & production rate (20)
Investment - wide applicability, good finish & tolerance, medium cost, high production rate (1000)
Permanent mould - Limited applicability, best finish & tolerance, high cost, medium production rate (600)

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4
Q

Forging

A

Produce discrete parts with a set of dies
Finishing required
Elevated temperatures
High equipment cost
Upsetting force = flow stress (1+2friction factorr/3*h)

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5
Q

Rolling & defects

A

Produce flat plates/structure shapes
Fast production rate
High capital investment
Rolling force = friction factorwsqrt(Rh0-hf)sigma
w is width, R is radius of rolls, difference in thickness, sigma is flow stress
Increasing temperature decreases flow stress so decreases rolling force
Defects: not flat, porosity, cracking, crocodile/alligator

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6
Q

Extrusion equation

A

Product long lengths of constant cross section
Elevated T but lead extruded at rtp
Extrusion pressure/flow stress = a + b ln(R)
R = initial/final cross sectional area

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7
Q

Isothermal forging

A

die & workpiece same temperature so low flow stress but expensive

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8
Q

Why additive layer manufacturing?

A

Reduced buy-to-fly ratio
Manufacture parts impossible with conventional techniques
Lower lead times
Freedom of design/creation (change programme instead of tooling)
Customisable
Less scrap

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9
Q

AM technologies

A

Laser (powder bed, blown powder), electron beam (pre-heating for every layer)

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10
Q

Disadvantages of ALM

A

Certain powders expensive to produce
High energy footprint in powder manufacture
Problems with mechanical properties & surface quality
Difficult to use ALM for some metals e.g. aluminium

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11
Q

Metal injection moulding process & advantages

A

Heat & grind metal & binder (35/65 ratio), injection mould for green compact, debind for brown compact, sinter

Lower energy footprint than casting
Great for high precision & small parts (can make large parts)

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12
Q

Hot isostatic pressing process, A&D

A

Steel capsule, 1273K, 1000MPa several hours, strong acids & lengthy mould removal

Makes functionally graded materials
Good mechanical properties, shape complexity, surface quality
Removes internal porosity

Costly, complex, low volume

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13
Q

Spark plasma sintering (field assisted sintering technique)

A

Produces fully dense materials in short periods of time: 10 ms DC pulse in graphite mould, pores close due to high pressure & removes oxide layer

For metals, ceramics, composites

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14
Q

What engineering properties does the bond energy curve determine?

A

melting temperature, stiffness/elastic modulus(but also packing of atoms), thermal expansion coefficient
Covalent>metallic>ionic

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15
Q

Expansion coefficient equations

A

(1/l0)*(dl/dt)

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16
Q

Hexagonal close packed

A

Mg, Zn, Ti, Be, Co
c/a = 1.633, c is distance between hexagonal plates
Coordination no. = 12
6 atoms per unit cell
Brittle at rtp
Casting, rolling, forging, extrusion

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17
Q

Face-centred cubic (examples, a, coordination no., atoms per unit cell, properties, production)

A

Al, Ni, Cu, Fe-gamma, Au, Ag, Pd, Pb, Pt
a = 2sqrt(2)R
coordination no. = 12
4 atoms per unit cell
Ductile, formable
Casting, rolling, extrusion, powder metallurgy

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18
Q

Simple cubic

A

Polonium
Lattice parameter, a = 2R
Coordination no. = 6
1 atom per unit cell

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19
Q

Body centred cubic

A

Cr, W, Fe-alpha
a = 4R/sqrt(3)
coordination no. = 8
2 atoms per unit cell
Brittle at rtp, idfficult to deform at low T

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20
Q

Polymorphism & allotropy

A
  • material has more than 1 crystal structure
  • polymorphic change (reversible with temperature & pressure)
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21
Q

Describe allotropies of iron, tin, Ti

A

Iron:
912C alpha (BCC) >contraction> gamma (FCC)
1394C gamma > delta (BCC)

Tin:
27% decrease in density from BCC to diamond cubic when cooled at 13.2C

884: HCP > BCC

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22
Q

Miller indices

A

reciprocals of the axial intercepts for a plane
[] direction
() plane
Overbar = negative

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23
Q

True or false: Young’s modulus depends on composition & microstructure (heat treatment)

A

False, it depends on atomic bonds & crystal structure

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24
Q

Why is plastic deformation permanent?

A

Atoms have slipped over one another/dislocations move on slip planes (at most close packed planes). Determined by dislocations (defects) in crystals which is sensitive to microstructure & composition

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25
Describe the 4 atomic defects
1) Vacancies - produced by solidification/atomic vibrations, causes creep/progressive deformation (when subjected to stress at elevated T for a long time) 2) Substitutional atoms - alloy to strengthen & for corrosion resistance 3) Interstitial atoms - can change crystal structure (e.g. Fe + C from BCC to BCT) 4) Linear defect - atoms above & below are either squeezed apart or pushed together
26
Define strength & how to increase
Resistance to plastic deformation Increased by obstacles to dislocation e.g. grain boundaries, precipitates, cracks, phase boundaries Through grain boundary refining, work hardening & alloying (precipitation/age hardening, solid solution hardening)
27
Define toughness & how to increase
Resistance to crack propagation(area under stress-strain curve & ductility of metal, related to crystal structure & T) Increased by elevated temperatures & availability of slip systems
28
Slip systems
Von Mises Criterion states that for plastic deformation, at least 5 independent slip systems must be operational FCC – 12 slip systems – 5 independent & operate at all temperatures – ductile at all temperatures, no DBTT BCC – 48 slip systems – highly T dependent HCP – 12 slip systems – T dependent (ductile at high T)
29
Have a look at end of week 6 lecture slides for critical resolved shear stress
tensile stress*cos(plane angle)*cos(plane direction)
30
How to calculate Vickers microhardness
HV = 1.854P/d2, where P is the applied load and d is the diameter in mm
31
Effect of grain size on toughness & why
Reducing grain sizes > grain boundary strengthening > increases strength, toughness & ductility exponentially
32
Temperature for hot forging of metals & considerations
>1000C At wrong temperature, capacity exceeded or defective material produced
33
Define hot working
T/melting T > 0.6 Cold work + annealing combined
34
Define work hardening
Deformation generates lots of dislocations which obstruct each other
35
What happens to dislocation density, flow stress, yield, tensile strength & ductility as strain increases (cold working)?
All increase except ductility which decreases (no dynamic restoration so limited strain due to hardening & loss of ductility)
36
How to overcome work hardening problems?
When yield strength reaches a certain point, apply heat treatment (annealing - stored strain energy revets back to pre-cold worked state)
37
What are the 3 restoration mechanisms and its effect on physical properties?
1) Recovery - dislocations cancel each other out 2) Recrystallisation - driving force is difference in internal energy between strained & unstrained material, grows strain & dislocation free grains 3) Grain growth In this order, decreased strength & increased ductility
38
Define cold working
No dynamic restoration so limited strain due to loss of ductility & hardening Annealing required to restore formability Excellent shape tolerance & surface finish
39
Define warm working
No dynamic recrystallisation but some dynamic recovery Good dimensional tolerances & surface finish Forging, rolling, automotive
40
Define Thermomechanical processing (TMP)
1) Study microstructural changes, deformation simulators, FE models, data analysis/modelling, characterisation for greater understanding & control 2) Reduce process steps & temperature 3) Get desired mechanical properties e.g. small grains & homogeneous composition
41
Advantages of TMP
Get desired mechanical properties. Better reproducibility, reduced tolerances (weight saving), improved properties with simpler alloys (less use of scarce resources & simpler recycling)
42
Through process modelling (TPM)
Work backwards from target properties of microstructure & mechanical
43
What is the Hall-Petch equation?
yield stress = friction stress + kY * (d)^-1/2 kY is locking parameter, d is grain diameter Relative hardening contribution of grain boundaries equation. As grain size decreases, yield stress increases
44
Work hardening vs age hardening
Work hardening - non heat treatable alloys are strengthened by refinement of grain size, solid solution strengthening & cold working Age-hardening - heat treatable alloys are strengthened from precipitation strengthening
45
Which aluminium alloys are heat treatable?
Series 2 (Copper), 7 (zinc) and 6 (magnesium & silicon)
46
Which aluminium alloys are not heat treatable?
Series 1 (99% min Al), 3 (manganese), 5 (magnesium)
47
3xxx series
1 wt.% manganese Annealed: 40 MPa, 30% elongation Cold worked: 185 MPa, 4% elongation Recyclable, drinks cans
48
5xxx series
6 wt% magnesium Annealed: 40-160 MPa, 25% elongation Cold work: 300 MPa, 5% Lightweight, drinks cans Poor weldability, recyclability, repairability, yield point phenomena
49
2xxx, 6xxx, 7xxx applications & properties
2xxx - plane structure & fuselage (resistance to crack propagation) 7xxx - upper wing structures, bridge (compressive stresses, weldable, highest strength) 6xxx - window frames (extrudable, less dense & more weldable than 2xxx)
50
Generic production route
Casting > pre-working treatments > hot working > cold working > post processing > finishing
51
What is homogenisation?
Ingots heated for varying lengths of time (alloy dependent) leading to diffusion & reducing micro segregation Vital pre-hotworking step
52
3 post-working treatments of aluminium alloys
1) Quenching - controlled cooling, can lead to residual stresses/distortion, slow quenching through critical region of 400-290C would reduce achievable strength after ageing, retains elevated temperature microstructure 2) Solution treatment - heat to dissolve alloy additions that contribute to age hardening, T control is critical 3) Ageing - optimise strength & corrosion resistance, 100-190 degrees for 8-16 hours, correct ageing time & T is key
53
Advantages & disadvantages of alloying
- Worse corrosion resistance, surface finish, conductivity - Deformability impaired, density increased But increased strength from 90 MPa to 1 GPa - increased fabrication properties (add Si/Sn for improved castability) - physical metallurgy relatively simple (no allotropic changes with T)
54
T1
Part solution treated & naturally aged
55
T3
Solution treated & cold worked
56
T4
Solution treated & naturally aged
57
T5
Artificially aged
58
T6
Solution treated & artificially aged
59
T7
Solution treated & stabilised
60
T8
Solution treated, cold worked, artificially aged
61
T9
Solution treated, artificially aged, cold worked
62
How does precipitation/age hardening work?
- shear mechanism for coherent precipitates, strength proportional to precipitate size^1/2 - bowing mechanism for incoherent precipitates, strength proportional to 1/distance between particles As T decreases, solubility of a compound decreases so a precipitate is formed > increased strength
63
What is solution treatment?
Alloy is heated for a long time to homogenise solute content. It's then reheated at a lower T to dissolve all useful alloying additions
64
What's artificial ageing?
Heat alloy to a temperature below solubility limit, forms precipitates inside the grains (not on grain boundaries)
65
What's solid solution strengthening?
Add interstitial atoms (stronger) or substitutional atoms (overall strain energy reduced with solute atoms segregating around dislocation, obstructing dislocation motion) Yield strength is proportional to mismatch between atoms^3/2 * solute concentration, C^1/2 As nickel content increases, tensile strength increases
66
3 types of ferrous alloys
1) Iron <0.008% wt. C 2) Steel (iron & carbon) <1.5% wt. 3) Cast iron <2-6% wt.
67
Stainless steels contain
Cr to stabilise BCC, Ni to stabilise FCC Forms austenitic, ferritic, martensitic, duplex (+Ti, Mo)
68
HSLA steels contain (high strength low alloy)
Nb, V, Ti Strengthen with hot rolling & TMCR (thermomechanically controlled) Nb precipitates out for fine austenite & ferrite grains Heat treatment for tough, formable & weldable
69
TRIP steels contain (transformation induced plasticity)
Al, Si
70
Steels, their carbon content & applications
1) Low - ductile, 0.1-0.25%, beams, car bodies, cans 2) Medium - 0.2%, general forgings, shaft, rotors 3) High carbon - strong, 0.5-1.6%, railway lines, car springs, hammers, cutting tools, dies
71
Advantages & disadvantages of steel
Cheap, high strength/stiffness/toughness, easy to join & weld, versatile, recyclable Very dense, poor corrosion resistance (galvanise with zinc)
72
Ferrite - what phase, temperature & arrangement
alpha, <901 & 1400-1534C, BCC, carbon has nearly 0 solid solubility
73
Austenite - what phase, temperature, arrangement
gamma, 910-1400C, FCC, carbon has high solid solubility, white regions
74
Cementite
6.7% wt. carbon
75
Pearlite
Grains of austenite with alternating plates of ferrite & cementite. Has a lamellar structure typical of eutectoid microstructures. Formed from slow cooling
76
What if composition of steel is less than eutectoid composition
Hypoeutectoid - ferrite at grain boundaries of austenite. Rest is perlite
77
What if composition of steel is more than eutectoid composition
Hypereutectoid - cementite at grain boundaries of austenite. Rest is perlite
78
Martensite - formation, grains, strength
Super saturated solid solution of C in Fe Formed from fast cooling Carbon can't diffuse out of Fe to form cementite Needle shape grains > BST structure Very strong due to non cubic structure (fewer slip systems), interstitial carbon, fine grain structure (Hall-Petch equation) Too brittle so temper at 600 degrees (heat treat version of annealing Al or Cu alloys) to allow diffusion > BCC ferrite with Fe3C precipitates (dotty grains)
79
Pearlite vs martensite vs tempered martensite
low hardness, high ductility high hardness, low ductility mid hardness & ductility
80
Strength equation
initial yield stress (in shear) + k*sqrt(p) k is material constant, p is dislocation density Work hardening increases shear strain

Materials (3 decks)

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