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Question 1

List three types of chemical bonding (IMP):

Answer 1

● Ionic Bonding

● Covalent Bonding

● Metallic Bonding

● Secondary bonding (van der Waals)

Question 2

What is a grain? Why are grains important for hardening and some types of corrosion?

Answer 2

● Crystalline materials consist of atoms arranged in a periodic manner.

● In a single crystal, the repetition extends over the entire extent of the material.

● Polycrystalline materials are comprised of many small crystals or grains.

● Grain boundaries impede the motion of dislocations (strengthening)

● Hence, decreasing grain size leads to a strengthening of the material.

Question 3

Describe the atomic arrangement in polycrystalline materials

Answer 3

● Comprised of many small grains
● The grains have different crystallographic orientation
● There is an atomic mismatch at the regions where the grains meet (grain boundaries)

Question 4

Describe the atomic arrangement in amorphous materials

Answer 4

● There is a lack of ordered or systematic arrangement of atoms

Question 5

What is the difference between solid solution and second phase? (IMP)

Answer 5

● Solid solution:
○ Homogeneous
○ Maintains crystal structure
○ Contains randomly dispersed impurities

● Second phase
○ As solute atoms are added, new compounds/structures are formed, or solute forms local
precipitates

● Whether the addition of impurities results in the formation of solid solution or second phase depends on the nature of the impurities, their concentration, temperature and pressure

Question 6

Describe case hardening:

Answer 6

● Hardening the surface of a metal by exposing it to impurities that diffuse into the surface region and increase the surface hardness.

● Examples: Carburization of steel - diffusion of C atoms increase their concentration and makes steel harder.

Question 7

Define toughness:

Answer 7

● Toughness is the ability to absorb energy up to fracture

● The total area under the stress-strain curve up to fracture

● Unit - energy per unit volume. E.g. J/m3

Question 8

Plastic deformation is:

Answer 8

● Irreversible

● Caused by stress above the yield point

Question 9

Strengthening the metals noticeably affects:

Answer 9

● the yield strength of the material

● the tensile strength of the material

Question 10

What is the reason that a cast alloy has a special factor (increasing the factor of safety compared to a wrought alloy)?

Answer 10

● Airframers hesitate to use castings because of assumed inconsistent mechanical properties and quality. Therefore, the Casting Factor was defined.

● The CF is a number usually ranging from 1.0-2.0 that is attached to a casting based on its criticality.

● In equation form, design property strength = the material allowable/casting factor.

Question 11

Name 3 strengthening mechanism for a metal alloy:

Answer 11

● Grain-size reduction (Hall-Petch Method, grain boundary strengthening)

● Strain hardening (cold working, work hardening)

● Precipitation hardening

● Martensitic transformation

Question 12

Describe precipitation hardening in detail:

Answer 12

● Heat treatment technique used to increase the yield strength of malleable materials (e.g. Al, Ni, Mg, Ti, Stainless Steel)
● Solubility changes with temperature. This can be used for the formation of impurity phases by heating and subsequent quenching
● After quenching, alloy must be kept at elevated temperature for hours to allow precipitation
(formation of intermetallic particles) to take place. This time delay is called aging.
● Solution treatment and aging are sometimes abbreviated “STA”.
● Impurities impede movement of dislocations / other defects
● Plasticity is reduced because of hardening
● In some cases, such as many aluminum alloys, an increase in strength is achieved at the
expense of corrosion resistance (refer to Corrosion-Theory)

Question 13

What influence has annealing on the hardness of a part?

Answer 13

● Hardness decreases

● Increase ductility

● Helps eliminate internal stresses

Question 14

Name an easy way how to check the heat treatment condition of metal without doing a tensile test:

Answer 14

● Check grain size

● Check the designation of the material

Question 15

Describe one hardness test:

Answer 15

● Rockwell hardness:
○ HRA, HRC (diamond cone), HRB (sphere)
○ Application of minor load and subsequent major load
○ Difference in indentation gives hardness value
○ Speed, Reliability, Robustness, good resolution
○ Example: 48 HRC

● Vickers:
○ Size of impression by diamond indenter
○ Example: 610 HV 10

● Brinell:
○ Size of impression by spherical indenter
○ Example: 345 HBW 10/3000

Question 16

The ability of a material to be cold worked as a forming process is limited by the dislocation and grain distortions induced by the process. What thermal treatments are needed to allow further cold work if this limit was reached?

Answer 16

● Recovery

● Recrystallization

Question 17

What is the advantage of ductile fracture modes in terms of safety in aircraft applications?

Answer 17

● For ductile fracture, the crack is stable and resists further extension unless applied stress is increased. There is a lot of plastic deformation and energy absorption (toughness) before fracture

● For brittle fracture, the crack is unstable and propagates rapidly without an increase in applied stress. Low plastic deformation and energy absorption before fracture

Question 18

Why is a ductile material more resistant to crack propagation than a brittle material?

Answer 18

● Cracks with sharp tips propagate easier than cracks having blunt tips

● In ductile materials, plastic deformation at a crack tip blunts the crack

Question 19

Name 2 factors that affect fatigue life:

Answer 19

● Stress

● Quality of the surface

● Thermal cycling

Question 20

Fatigue life is:

Answer 20

● Number of cycles to fail at a specified stress level