Titanium alloys

Question 1

Give 3 sectors where Ti alloys are used

Answer 1

• civil/military aircraft
• desalination plants (good corrosion resistance)
• power generation

Question 2

Why is Ti so compatible with CFRP

Answer 2

1. Ti and graphite have very similar linear thermal expansion coefficients
   - this reduces residual stresses in the structure
2. Electrochemical compatibility
   - both more cathodic in nature, therefore in saline environment it’s much less likely that galvanic corrosion will occur
3. Undirectional CFRP has similar Youngs Modulus to Titanium (110GPa)
   - therefore similar elastic expansion

Question 3

List 5 advantages of Ti

Answer 3

1. Low density
2. Excellent corrosion resistance
3. High specific strength
4. Good compatibility with CFRP
5. Excellent properties at elevated temperatures

Question 4

List 5 disadvantages of Ti

Answer 4

1. Expensive to machine
2. Low wear resistance
3. Difficult to form
4. Pick up oxygen and hydrogen at temps > 500C
5. Expensive raw product

Question 5

Describe the Kroll process to obtain Titanium sponge

Answer 5

1. Take TiO2 and heat in the presense of Chlorine and Coke (C)
2. This produces TiCl4 (liquid and volatile) and CO2
3. Use fractional distillation to distil it from iron chlorides (FeCl3)
4. Then add this to steel tank with magnesium (solid) where an exothermic reaction occurs (gets very hot)
5. This produces MgCl2 and Ti sponge
   - process is done with excess of Mg
   - takes 5-6 day batch process (10 tonnes per batch)

Question 6

Describe how Ti sponge is processed

Answer 6

1. Ti sponge is compressed into brickets (alloying materials can be added at this stage)
2. 60 brickets are then lined up and electron beam welded together to make an electrode (expensive)
3. Then carry out vacuum arc remelting
4. Grains are very coarse, so hot forging used creating a ‘bloom’

Question 7

Describe the process of vacuum arc remelting

Answer 7

• the electrode is drawn down and consumed by arc in a vacuum
• droplets of titanium melt and drop to bottom forming a melt pool
• ingot produced
• this is done 2 or 3 times for better properties

Question 8

Decribe FFC Cambridge process

Answer 8

1. Take TiO2 and mix with MeOx and press together to sinter it into a cathode
2. Carry out an electrochemical procces:
   - put in molten CaCl2
   - temperature at 800-1100C
   - add 3V
   - have graphite anode which oxygen ions transfer to
   and form CO released into atmosphere
   - takes 12/24hrs
3. This process produces very homogeneous grain characteristics which is very desirable

Question 9

What are the advantages of using FFC Cambridge process

Answer 9

• can produce Ti-10W (Titanium-tungsten) which cannot be produced using kroll process
• this is because tungsten melting temp is too high during vacuum arc remelting at 2000C

Question 10

Give 3 points about the metallurgy of Ti

Answer 10

1. Allotropic metal (exists in 2 different forms)
2. α phase - HCP when temp below 883C
3. β phase - BCC when temp above 883C

Question 11

What are α stabilisers and list some

Answer 11

α stabilisers are elements that promote the formation of α phase i.e. HCP structure:
O, Al, N, C

Question 12

What are β stabilisers and list some

Answer 12

β stabilisers are elements that promote the formation of β phase i.e. BCC structure:
V, Mo, Nb, Fe, Si, Cu, Cr, Mn

Question 13

What happens to properties of Ti when interstitial elements are introduced

Answer 13

• strength rapidly increases

- fracture toughness rapidly decreases

Question 14

If we see martensite and β phase in a Titanium alloy what does this mean

Answer 14

• it does NOT mean that we’re in between Ms and Mf, instead there can be deformation induced martensite below the martensitic deformation temperature

Question 15

Give properties for α-alloys

Answer 15

• not heat treatable
• always α phase (even if take above β line and quench, because material below Ms/Mf line and so martensite will form. this is and α phase structure (only for Ti))
• retains strength at elevated temepatures
• strengthening mechanisms include:
  1. solution strengthening from oxygen
  2. grain refinement
  3. cold work

Question 16

What is the Hall-Petch equation

Answer 16

strain = 231 + 10.54xd^-1/2

d - grain size

Question 17

Uses for α alloys

Answer 17

• Very good at retaining strength at elevated temperatures and therefore good creep resistant
• Also very good corrosion resistance
• used as cladding in chemical plants and will last the lift time of the plant

Question 18

Give properties for near α-alloys

Answer 18

• contains small amount of β stabilisers which widens α+β field range (this is good because want to work in this range rather than β phase field as this induces fast grain growth and so large grains)
• very good creep resistance because:
• silicon pins dislocations (zener pinning)
• large prior beta grain size-inhibits grain boundaries sliding
• α structure - slow diffusion due to HCP
• works up to 600C

Question 19

Give uses for near α-alloys

Answer 19

Compressor blades in a jet engine

Question 20

Give properties for α+β alloys

Answer 20

• most used alloy Ti-6Al-4V
• mixed properties between α and β phases
• can be superplastically formed (can behave as a thermoplastic)
• lots of variation of work treated

Question 21

Uses for α+β alloys

Answer 21

bulkheads in mid-fuselage of F22 fighter

Question 22

Describe the effect of this work treatment of α+β alloys:

forged in β phase, then furnace cooled

Answer 22

get large β grain sizes
forms grain boundary α
then single colony α within the grains

Question 23

Describe the effect of this work treatment of α+β alloys:

forged in β phase, then hot air cooled

Answer 23

get packets of α rather than single colonies

get large β grains

Question 24

Describe the effect of this work treatment of α+β alloys:

forged in β phase, then medium air cooled

Answer 24

get a basket weave structure of lamellar α

Question 25

Describe the effect of this work treatment of α+β alloys:

forged in β phase, then rapidly air cooled

Answer 25

widmanstatten array of α plates

Question 26

Describe the effect of this work treatment of α+β alloys:

forged in β phase, then quenched in water

Answer 26

Fully maternsitic microstructure which is in α phase

Question 27

Describe the effect of this work treatment of α+β alloys:

forged in α+β phase (high β), then air cooled

Answer 27

• have primary α (at forging processing temp)
• predominantly alpha with residual β phase in between
• ‘transformed β’ is just secondary α
• Bimodal structure

Question 28

Describe the effect of this work treatment of α+β alloys:

forged in α+β phase (low β), then air cooled

Answer 28

• more primary α
• the rest is ‘transformed β’ in residual β
• good for superplastic forming as it allows for lots of grains sliding

Question 29

Give properties for near β alloys

Answer 29

• good hardenability (precipition hardening)
• processing in β-single phase cause lots of issues so majority processed in α+β phase
• best example Ti-10V-2Fe-3Al

Question 30

Give properties for β alloys

Answer 30

• cold formable (omega formation surpressed)
• soft and ductile
• any age hardening needs to happen after welding is finished

Question 31

What is the disadvantage of using β alloys

Answer 31

• use of β stabilisers tend to increase density

- have poor high temperature properties due to over ageing