Week 4

Question 1

Metal forging:

Answer 1

Heating the metal them forming it into the required size and shape

• Shaping of a metal using localized compressive forces (cold, warm or hot forging).

Question 2

Fabrication and types of processing (4)

Answer 2

1. Metal casting
2. Metal Forging
3. Metal rolling
4. Powder metalurgic

Question 3

Metal rolling process:

Answer 3

1. most economical process
2. cold-hot rolling depending onf temprature of operation
3. hot-rolling: sheet metal, rail tracks
4. Cold-rolling: sheets, strips, bars and rods

Question 4

Heat treatment of metals

1. Why is it done? (1-3)
2. What are the results? (2)

Answer 4

1. Metals and alloys may not possess all the desired properties in the finished stock form. Therefore, alloying and heat treatment are methods extensively used to control metal properties.
2. Metallic materials consist of a microstructure of small crystals called grains or crystallites. The size and composition of the grains are the most effective factors that can _determine overall mechanical performance of a metal. _
3. Heat treatment provides an efficient way to manipulate the properties of the metal by controlling the rate of diffusion and the rate of cooling _within the microstructure. _
4. heating and chilling, normally to extreme temperatures ——> desired result such as hardening or softening of the material. Therefore, the _original microstructure is altered. _
5. The resulting phase transformations influences properties like strength, ductility, toughness, hardness, and wear resistance.
6. The resultant material typically exhibits i_mproved manufacturability, and increased strength or hardness. _

Question 5

Cold treatment

Answer 5

also called cold-working is _Strengthening by increase of dislocation density _

_The average distance between dislocations decreases and dislocations start blocking the motion of each other. _

Question 6

Heat treatment of metals:

Answer 6

o Hardening
o Quenching
o Annealing
o Normalizing
o Tempering
o Strengthening
o Surface hardening

Question 7

Name for steel before heating (below crystalization T) (raw steel)

Answer 7

Austenite (it is non-magnetic before cold-working)

Question 8

Name of very hard form of steel crystalline structure,

Answer 8

Martensite** (it becomes **magnetic in cold-working)

Question 9

Hardening:

Answer 9

o Impact strength and hardness by heating up** at certain temperatures (temperature transformation is material dependant) **and cooling it rapidly.

o Steel, for example, is processed by heating and holding the temperature until its carbon is dissolved. Cooling is performed subsequently.

• In this process C atoms will not have sufficient time to escape* and therefore they will *get entrapped and dissipated within the lattice structure. This assists in blocking dislocations movements when stresses are applied.

Question 10

Quenching:

1) process:
2) problem or difficulty
3) Can this problem be solved and how?

Answer 10

1. Cooling metal rapidly by immersing the material in salt water, water, oil, molten salt, air or gas.
2. Quenching results in residual stresses and sometimes cracks.
3. Yes. Residual stresses can be removed by a subsequent process called **annealing. **

Question 11

Annealing:

1. Process
2. Uses (2)
3. More reasons to use it:

Answer 11

1. Annealing consists of heating a metal to a specific temperature and then cooling at a rate that will produce a refined microstructure. The rate of cooling is generally slow (10^oC per hour).
2. Annealing is most often used to (1) soften a metal for cold working, to improve machinability, or (2) to enhance properties like electrical conductivity.
3. Reduces hardness
   
   • Remove residual stresses
   • _ It can improve material toughness_
   • _ Ductility restoration_
   • It is a process used to alter several properties including mechanical, magnetic or electrical through grain refinement.
   • _Process is performed in controlled atmosphere of inert gas to avoid oxidation. _

Question 12

Normalizing:

1. Why is it used for?
2. What is this process similar to and why is it used.
3. Describe process and results

Answer 12

1. Used to provide uniformity in grain size** and **composition throughout an alloy.
2. Similar to annealing but it is mostly carried out to avoid excessive softness.
3. Material is *heated above the austenitic phase* (e.g. 1100^oC) and then is cooled in air. This process results in enhanced hardness and less ductility.
4. Variation in properties of different sections of the part being treated is achieved in normalizing.

Question 13

Tempering:

Answer 13

1. Martensite is very hard and brittle. Then tempering can be applied to hardened steel to r_educe brittleness,_ to _increase ductility and toughness,_ and to _relieve stresses_ in martensite structures (blocking dislocations).
2. Steel is heated to lower critical temperature (350-400^oC) keeping it there for about one hour and then **cooled slowly. **

Question 14

Strengthening is for?

Answer 14

o Control of grain size

o Formation of small grain sizes

Question 15

Surface hardening

Answer 15

1. Hardening improves wear resistance of metals but it can lower impact resistance and fatigue life (e.g. surgical instruments versus implant).
2. Two methods are used in surface hardening:
   
   • Heating followed by cooling: done to produce required phase.
   • Thermo-chemical treatment: flame heating, induction heating, laser beam hardening, electron beam hardening, carburizing, nitriding, cyaniding.

Question 16

Strengthening & equation

Answer 16

• Small grains are formed with heat treatment and cooling
• The ability of a metal to deform depends on the ability of dislocations to move —–>_ Restricting dislocation motion makes the material stronger_ —> this is achieved with grain-size reduction. yield strength varies with grain size _d_

σ = σ₀ + (k/d^1/2)

^{d = diameter of grain}

*

Question 17

Oxide layer formation:

Answer 17

Ti + 2H₂O —-> TiO₂ + 4H⁺ + 4e^-

Question 18

Passivation and formation of an insoluble oxide or hydroxyde involving titanium

1. Oxidation:
2. Reduction
3. O₂ depletion and excess H⁺
4. So what happen when the medium is too acidic (H⁺)

Answer 18

1. Oxidation: Dissolution of metal ions – oxide film formed — stop dissolution

Ti + 2H₂O —> TiO₂+ 4H⁺ + 4e^- (0xide layer)

1. Reduction: will consume the oxidation products and keep the environment neutral.

O₂ + 4H⁺ + 4e- —-> 2H₂O (reduction of dissolved O₂)

1. O₂ + 2H₂O + 4e^- —> 4OH (in neutral alcaline sln.)

4H⁺ + 4e^- —> 2H₂ (reduction of hydrogen)

H⁺ + e^- —> H

1. If the environment is too acidic the oxide layer won’t re-passivate.

Question 19

The oxide film thickness depends on: (2)

Answer 19

1. Potential across the interface
2. Solution variables (pH)

Question 20

Oxide film growth will depend on the potential applied to the surface, which will be affected by the environment:

1. Cathodic domain (+)
2. OCP (open circuit potential)
3. Anodic domain (-)

Answer 20

1. Cathodic domain (+): _reduction fo passivity _
2. OCP (open circuit potential): equilibrium potential —> oxide film starts to grow
3. Anodic domain (-): _ oxide film growth_

Question 21

Loosing the oxide layer because no O₂ leads to

Answer 21

leaching of metal ions into the blood

Question 22

Fretting wear is

Answer 22

the repeated cyclical rubbing between two surfaces, which is known as fretting, over a period of time which will remove material from one or both surfaces in contact

Question 23

Reasons for Treatment of surface metals

Answer 23

1. Increase corrosion resistance:
2. Improve blood compatibility:
3. Improve strength of the surface:
4. To enhance biocompatibility:

Question 24

2 ways that we can use to treat metal’s surface ——> Increase corrosion resistance:

Answer 24

1. Passivation method: growth of oxide film by controlling voltage.
2. Anodization: a_pplying high potential_ to the electrolyte which contains the metal specimen.

Question 25

Surface treatment of metals:

• Improve blood compatibility:
• Adsorption of proteins. Effect of voltage on clotting.

1. More (+) voltage:
2. More (-) voltage:

Answer 25

1. More clotting
2. inhibit clotting

Note: Gold is the most electron positive (+Ve) metal = results in clotting effects (awful results with stents)

Question 26

Surface treatment of metals:

• Improve strength of the surface:
  
  • Where do cracks start?
  • what aids in crack growth?
  • What technique improve surface strength?

Answer 26

• Almost all cracks start at the surface.
• Material’s tensile strength aids in crack growth.
• _Ion implantation: PVD technique (Physical Vapor Deposition) _

Question 27

Ion implantation: PVD technique (Physical Vapor Deposition)

Answer 27

N2 → N+ + e-
N+: ion gas (plasma) - **sputtering **

N+ gets accelerated to the plate and is neutralized.

• Penetration of ions
• Scattering of ions
• Sputtering of ions

Question 28

Ion implantation: PVD technique (Physical Vapor Deposition) advantages:

Answer 28

1. Ion implantation creates defects into the structure
2. Creates amorphous regions
3. Deform by creating and moving dislocations
4. it leaves N on the surface and lattice: N is an interstitial atom that will create nitrides
5. It will raise strength
6. Wear resistance becomes higher
7. penetration profile: **100- 500 nm **

Question 29

How tretament of metals surfaces enhance biocompatibility:;

2 treatments that enhance biocompatibility

Answer 29

• Deposition of hydroxyapatite (HA) – mineral of bone as coating.
• (Ca)10(PO4)6(OH)2: crystalline ceramic.
• HA coating is currently done on Ti implants (hip stems, tibial plates). 
• HA enhances bone formation – “bone friendly phase”.
• Other materials used include tricalcium phosphate (TCP).
• Porous plasma sprayed Ti coatings are also commonly used.

1.  Plasma spraying: injection of HA/Ti as coating.
2. Sintering

Question 30

Sintering

Difference Sintering vs Plasma spraying:

Answer 30

Sintering as coating: also employed in a range of applications.

 By definition is a method of making objects from powders by heating

the _powder material below their melting point until they adhere to one anothe_r.

 It can de done with pressure (referred to as hot pressing or hot isostatic pressuring) or without added pressure (nanoparticles)

• Sintering is more granular than spraying

Question 31

Titanium and its alloys General characteristics: (5)

Answer 31

1. Light weight, excellent corrosion resistance and enhanced biocompatibility.
2. Density: 4.5 g/cm3 which is lower in comparison to 316L SS (7.9 g/cm3) and cobalt-chromium (CoCr) alloy (8.3 g/cm3).
3. Although light, Ti provides excellent chemical and mechanical properties comparable to 316L SS and CoCr alloys.
4. Commercially pure titanium (cpTi) and Titanium-6Aluminun- 4Vanadium (Ti6Al4V) are the two Ti-based materials commonly used in the biomedical industry.
5. The American Society for Testing and Materials (ASTM) International has classified cpTi (unalloyed form) into four different grades: Grade- 1, Grade-2, Grade-3, and Grade-4.

Question 32

Relationship between Ti and [O₂] and the grades of Ti

Answer 32

The higher the [O₂] the higher the grade

Grade 1 = 0.18% oxygen

Grade 4 = 0.40% oxygen

Question 33

Ti grade classification is base on:

Answer 33

It is based on the amount of impurities present (such as oxygen, nitrogen and iron). In particular the amount of oxygen present in Ti has a marked effect on the material’s properties.

• For Ti6Al4V the Al and V (vanadium) are responsible for the INCREASE in yield and tensile strength.

Question 34

The cpTi is an allotropic material and exists in two crystallographic forms:

Answer 34

1. Room temperature – 882.5oC: alpha phase (α-Ti): HCP
2. Above 882.5oC: **beta phase **(β-Ti): BCC

Question 35

What happen to

1. the strenght,
2. The ductility
3. the fatigue strength as we increase the amount of impurities in cpTi?

Answer 35

1. an increase in the amount of impurities in cpTi leads to increased strength but
2. reduced ductility.
3. The fatigue strength of cpTi also increases with increase in oxygen concentration.

Question 36

Ti alloys are classified in:

Answer 36

1. α-Ti,
2. β-Ti and
3. α+β-Ti

Question 37

Alpha alloys:

Answer 37

1.  By alloying Ti with elements such as aluminum, gallium or tin, the alloyed Ti retains its alpha structure.
2.  α-Ti are not heat-treatable but are weldable.
3.  Exhibit good creep resistance at elevated temperature and
4. do not undergo a ductile-brittle transformation.
5.  Used for low temperature applications.
6.  Very limited biomedical applications due to low strength in comparison
7. to β-Ti and α+β-Ti.

Question 38

Beta alloys:

Answer 38

1. Formed by alloying Ti with elements such as molybdenum (Mo), tantalum (Ta), vanadium (V), niobium (Nb), tungsten (W), chromium (Cr), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu), Mangnese (Mn).
2.  β-Ti alloys are heat treatable and cold formable.
3.  Capable of undergoing ductile-brittle transformation.
4. Thus not suitable for low temperature applications.
5.  The main advantage is that the beta-phase alloying elements such as** Ta and Mo are more biocompatible than t**he elements used to forma the alpha-phase such (aluminum and tin) and alpha-beta phase (vanadium)
6.  The beta stabilizing elements are classified under 2 categories:
   
   • ** Beta-isomorphous elements: (Mo, Ta, V, Nb, and W).**
   • **Beta-eutectoid elements: (Cr, Fe, Co, Ni, Cu, and Mn). **

Question 39

In an attempt to solve issues associated with vanadium other alloys have been developed and used such as:

Answer 39

**Ti6Al7Nb (Niobium) **as a biomaterial for fabrication of femoral hip stems and fracture fixation devices.

Question 40

Alpha-beta alloys:

Answer 40

1. Alloys containing balanced amounts of both alpha and beta stabilizers.
2. Ti6Al4V is an example of alpha-beta alloy since it contains an alpha- stabilizer (Al) and a beta-stabilizer (V).
3. Exhibit good formability. An important note is that Ti6Al4V is difficult to form even in the annealed condition.
4.  _ α+β-Ti are heat treatable._
5.  High tensile strength
6.  Do not exhibit good creep resistance at elevated temperatures
7. Not weldable if the composition of the **beta-phase is greater than 20%. **

Question 41

The surface chemistry and topography of Ti can be altered by various methods depending on the application. It can be subjected to:

Answer 41

1. heat treatments:
2. o Mechanical treatments: polishing, sand blasting, plasma spraying, physical vapor deposition, sintering.
3. o Chemical treatments: chemical vapor deposition, acid etching.
4. o Electrochemical treatments: anodization (oxide film growth)

Question 42

Stainless steels:

General characteristics and classification:

Answer 42

1. Iron-based alloys which contain at least 10.5% chromium (Cr).
2.  The corrosion resistance of SS is attributed to the formation of Cr oxides (Cr2O3).
3.  The corrosion resistance and properties can be improved by increasing the Cr content.
4. Other alloying elements can also be used to improve mechanical and other physical properties.
5. o Mo addition increases pitting corrosion resistance
6. o Ni increases mechanical strength and pitting corrosion
7. resistance.
8. Based on their microstructure, SS alloys are classified in:
   
   1. Martensitic stainless steel alloys
   2. Ferritic stainless steel alloys
   3. Austenitic stainless steel alloys
   4. Duplex stainless steel alloys

Question 43

Martensitic stainless steels:

Answer 43

1. Body-centered tetragonal structure
2.  Hardened by heat treatments having good mechanical properties and corrosion resistance.
3.  Cr content varies from 10.5% to 18% and
4. the C content can be greater than 1.2% - Cr and C are adjusted to obtain martensitic phase.
5. Other elements can be added to alter toughness (Tu, Nb, Si, Ni).
6.  Applications: **surgical and dental instruments – forceps, pliers, scalpels, curettes, explorers, dental burs, root elevators. **

Question 44

Ferritic stainless steels:

Answer 44

1. Body-centered cubic structure.
2.  Cr content varies from 11-30%. Other elements such as Nb, Si, S, Se, and Mo can be added for specific characteristics.
3.  Cannot be strengthened by heat treatments.
4.  ** **Cold-working is not commonly employed because it reduces ductility.
5.  Limited biomedical applications: handles for instruments and medical guide pins.

Question 45

Austenitic stainless steels:

Answer 45

1. Face-centered cubic structure.
2. Cr, Ni, Mg contents vary from 15-20%, 3-14%, 1-7.5%, respectively.
3.  A variety of other alloying elements (Mo, Nb, Si, Al) can be added for improved corrosion resistance.
4. Cannot be hardened by heat treatment. Cold-working typically performed to harden this alloy.
5.  Possess excellent properties such as high-temperature strength,
6. oxidation resistance and formability.
7. Applications: extensively used for medical implants and devices – **316L stainless steel is the most commonly used stainless steel for implants and devices. **

Question 46

316L stainless steels:

Answer 46

1. Fe (60-65%), Cr (17-20%), Ni (12-14%) and smaller amounts of Mo, Mg, Cu, C, N, H3PO3, Si, S.
2. L: low carbon content (<0.030%) – preferred for corrosion resistance avoiding precipitation of carbides (Cr23C6) at grain boundaries.
3. Typically cold-worked to significantly improve mechanical properties such as yield strength, ultimate tensile strength, and fatigue strength.

 Applications: many uses including _coronary stents, orthopedic implants, fracture fixation devices. _

Question 47

316L Stainless steel

Yield strength and Tensile strength

Annealed vs 30% cold worked

Answer 47

30% cold worked has higher tensile and yield strength than annealed

Question 48

Duplex stainless steels:

Answer 48

1. Two phase alloy that contains equal proportions of ferrite and austenite phases in their microstructure.
2. Carbon contest of less than 0.030%, the amount of Cr and Ni vary from 20-30% and 5-8% respectively. Other minor alloying elements used (Mo, N, Tu, Cu).
3. Applications: **not yet used in medical implants and devices. **

Question 49

Recent developments in stainless steel alloys:

Answer 49

1. Nitrogen strengthened stainless steels have been developed with improved mechanical properties and corrosion resistance when compared to 316L stainless steel.
2.  Nitrogen content can vary from 0.2-1%.
3.  These alloys are ferritic-free and can be cold-worked for improved

tensile strength, fatigue strength, and corrosion in comparison to 316L.

ASTM (American Society for Testing and Materials) F2229 is particularly _attractive because it contains no Ni, which can mitigate patient allergies due to Ni. _

Question 50

Cobalt-chromium alloys:

 Corrosion resistance is a result of:

Answer 50

Corrosion resistance is a result of high bulk chromium content and chromium oxide (Cr2O3).

Question 51

Cobalt-chromium alloys:

General characteristics:

Answer 51

The 4 cobalt-chromium alloys commonly used for biomedical applications are:

o ASTM F75 (Co-28Cr-6Mo casting alloy).
o ASTM F799 (Co-28Cr-6Mo thermodynamically processed

alloy).
o ASTM F90 (Co-20Cr-15W-10Ni wrought alloy).
o ASTM F562 (Co-35Ni-20Cr-10Mo) wrought alloy).

 While ASTM F75 and ASTM F799 alloys possess almost similar compositions, the different processing methods used to make these alloys result in unique mechanical properties.

 The Co content is less in ASTM F90 and ASTM F562 when compared to ASTM F75 and ASTM F799.

 ASTM F562 contains more Ni, whereas ASTM F90 contains more W (tungsten).

Question 52

Nitinol:

Answer 52

NiTi alloy modifies its shape to a preprogrammed structured due to the austenite-to-martensite phase transformations.

Question 53

Nitinol:

General characteristics:

Answer 53

1. Has near equiatomic composition (49%-51% Ni and Ti).  Belongs to the class of shape memory alloys.
2. o They can be plastically deformed at a low temperature but return back to their original pre-deformed shape when exposed to a high temperature.
3.  The shape memory capacity of Ni-Ti is the result of a martensitic- austenitic-transformation.
4.  Broadly used in the design of **self-expanding vascular extents (small diameter at room temperature). **

Question 54

Tantalum:

General characteristics:

Answer 54

1. Used in its commercially pure form (99.9%) or as an alloying element.
2.  Excellent corrosion resistance and biocompatibility because of a stable oxide layer.
3. Also used as a coating on other metallic devices (316L) to improve substrate corrosion resistance and enhance biocompatibility.
4. The elastic modulus of Ta is close to that of 316L SS;
5. however the yield and tensile strengths are lower when compared to the properties of Ti, 316L SS and CoCr alloys.
6.  Although _mechanical properties are lower, _ Ta has been used in applications such as radiographic marker for diagnostic applications, coronary stents, vascular clips, covering for cranial defects, fracture fixation and dental implants.

Question 55

Magnesium:

General characteristics:

Answer 55

1. Known by its light weight and biodegradability.
2.  ** The density, elastic modulus, yield strength, and fracture toughness of Mg are close to that of bone.** In addition, Mg is present naturally in the bone tissue.
3. The mechanical properties appear to be attractive, however the rapid corrosion of pure Mg under physiological conditions limits its use in load bearing applications.
4.  Mg degrades after a few days after in-vivo implantation.
5.  In-vivo the main degradation products of Mg are Mg(OH)2 and hydrogen gas H2.
6.  Applications: _biodegradable coronary stents (growing arteries in children _