Materials Engineering

Front 1

What do the shaded areas represent in stress/strain graph curves? (Think generally, I’m thinking about thedimensions/units).

Back 1

- Energy Stored

- Per unit volume

Front 2

Give one reason why we care about each of resilience and toughness.

Back 2

[Resilience]: Energy absorbed before material permanently deforms

[Toughness]: Total energy absorbed before failure

Front 3

Explain why group 1 and group 2 metals are electropositive.

Back 3

- Filled inner shells shield valence electrons from nuclear charge

-So valence electrons only feel a small effective nuclear charge

-Therefore, they feel a small electrostatic attraction to other nuclei

Front 4

Explain how electrons behave in metallic bonding, and alsogive two properties of metallic bonding that occur because of this behavior.

Back 4

-Electrons are not attached to any particular atom (delocalized)

-They reside between atoms (randomly), providing electrons forces

- Ductility due to non-directional bonds. Electrical Conductivity due to free mobile electrons

Front 5

Explain why group 16 and group 17 non-metals are electronegative.

Back 5

- Large number of valence electrons, not much shielding by inner shells

-So valence electrons see a large effective nuclear charge

- Therefore, they feel a large electrostatic attraction to the nucleus

Front 6

Explain how electrons behave in covalent bonding, and alsogive some properties of covalent bonding that occur because of this behavior.

Back 6

-Electron pairs shared between pairs of atoms

-Electrons locked into a small region of space, mostly between atoms that donated the electrons

-Brittle due to directional bonds and Insulators since there are no free electrons

Front 7

Back 7

- As electron clouds get close, they repel each other

-This deforms the electron cloud, creating the induced dipole

- The two induced dipoles can then attract each other electrostatically

-Those are called Vander Waals Forces

Front 8

Explain why methane is gaseous at room temperature while hexane is not

Back 8

-Methane CH4 is a small molecule

-Each molecule has a small number of Vander Waals bonds thatcan form with neighbouring atoms

-Hexane C6H14 can form more bonds and is therefore held even stronger

-The Hexane requires greater thermal energy to separate so it has a higher boiling point

Front 9

Back 9

-When O-H bonds form, there is an unequal sharing of electrons

-This creates a permanent dipole with Oxygen being partially negativelycharged and Hydrogen being partially positively charged

-These charged species then attract each other

Front 10

Explain why ethane is gaseous at room temperature while ethanol is not

Back 10

-Ethane molecules have only Vander Waals Forces creating intermolecular bonds

-Ethanol has H-bonds between molecules due to the alcohol group

-The ethanol molecules are therefore held more strongly

-They also require more thermal energy to vaporize so they have a higher boiling point

Front 11

What is the electron configuration of aluminium?

Back 11

[Ne] 3(s^2) 3(p^1)

Front 12

Explain the difference between the terms polycrystalline and polymorphic.

Back 12

-Polycrystalline refers to different grains with the same structure

-But with different orientation

-Polymorphic refers to a material with different crystal structures

-Under different conditions such as temperature and pressure

Front 13

Explain one reason why an engineer may want to use a polycrystalline material rather thana single crystal.

Back 13

-At a grain boundary, slip planes do not align and the boundaries interrupt slip planes, making the crystal less brittle

Front 14

Explain one reason why an engineer may not want to use a polymorphic crystal.

Back 14

-If the device is to be used over a large range of conditions it may disintegrate as the crystal structure changes

Front 15

Describe the conditions required for vacancy diffusion to occur.

Hint: think about the conditions required for a single atom to move.

Back 15

-There must be a large number of vacancies for atoms to move into

-The temperature must be high enough that atoms have enough energy to breakexisting bonds and move into a vacancy.

Front 16

Why does interstitial diffusion usually occur more quickly than vacancy diffusion?

Back 16

-Interstitial spaces always exist. It does not rely on a vacancy appearing nearby.

-Interstitial impurities are usual small, light, and therefore move quickly.This speeds up the motion of the impurities through the host.

Front 17

What is diffusion flux, 𝐽? Also, state the units of J

Back 17

-The mass of material passing through a unit area per unit time.

-[J] = kg m^-2s^-1

Front 18

If we wish to diffuse carbon into the surface of steel (surface hardening), give two simplemethods to increase the rate of diffusion and therefore the final carbon content.

Back 18

-Increase partial pressure of carbon in the gas.

-Increase the temperature of the systemAlso, decrease partial pressure of carbon in the gas on side 2.

Front 19

Explain how this slip occurs, referring to slip planes and resolved shear stress.

Back 19

-Tensile force is applied at an angle to the slip plane.

-If the component of the force in the slip plane, and along the slip direction, is largeenough, then slip planes can being to slide over one another

Front 20

What is the main factor in determining which plane is the main slip plane in a crystal?

Back 20

-The densest plane will be the hardest to deform, so will be the slip plane.

Front 21

Describe how dislocations can be moved through a material.

Back 21

-Shear forces can move both types of dislocations.

Front 22

At the yield point, dislocation motion is one of the first causes of plastic deformation.Why are dislocations (relatively) easy to move?

Back 22

-Bonds near the dislocation are already deformed (strained) and therefore weak.

-So, it does not take as much force to break these bonds as unstrained bonds.

Front 23

Reducing grain sizes hardens a material by interrupting dislocation motion and slip planes.Explain how grain boundaries do this.

Hint: Be sure to mention direction of motion and force components.

Back 23

-If slip is able to occur in one grain, the slip planes in the next grain may not beoriented in the same direction.

-The force creating slip in the first grain will have a smaller component lined up withthe slip plane in the second grain.

-This component may not be large enough to cause the second grain to slip

Front 24

Explain how the process of strain hardening (cold working) hardens a material

Back 24

-Strain hardening creates a large number of dislocations.

-The average interaction between dislocations is repulsive,so the dislocations stop each other from being able to move around.

Front 25

What is the third method of hardening a material?Describe how this method works?

Back 25

-Solid solution (alloying)

-Impurity atoms that are bigger or smaller than the host atomsinterrupt slip planes.

Front 26

What is the major negative affect that hardening has on a material?

Back 26

-Reduced ductility.

Front 27

What is recovery, how is it achieved, and why does it weaken a material?

Back 27

-Recovery is the reversal of strain hardening(reduction in number of dislocations).

-It is achieved by heating the material (keeping below 𝑇m of course).

-Removing dislocations allows remaining dislocation to move more easily,reducing the stress required to create plastic deformation.

Front 28

Why might we want to induce recovery in a material?

Back 28

-After cold working, a material may be too brittle to perform more work on.

-Recovery will allow us to perform more forming processes on the material.

Front 29

What is recrystallization, how is it achieved, and why does it weaken a material?

Back 29

-Recrystallization is the growth of many small grains into fewer larger grains.

-It is achieved by heating the material (to about 0.4 𝑇m).

-Removing grain boundaries means there are less barriers to slip and dislocationmotion.

Front 30

Why is ductile fracture “preferred” over brittle fracture?

Back 30

-Ductile materials can absorb some energy and deform before fracturing.

-This gives us some warning, rather than having a very sudden, catastrophic failure!

Front 31

What factors affect how much stress is concentrated?

Back 31

-Depth of the crack, 𝑎.

-Radius of curvature (sharpness) of the tip of the crack 𝜌t.

Front 32

Why is a crack in a brittle material a bigger problem than in a ductile material?

Hint: Think about what a force will do to the crack in a ductile material?

Back 32

-A ductile material is able to deform, absorbing energy and increasing 𝜌t.

-This reduces the stress concentration.

Front 33

For flat objects, what methods can be used to increase the critical stress?

Hint: These may include increasing 𝐾Ic. or decreasing crack sizes.

Back 33

-Polish the surface to reduce crack depths.

-Surface harden by diffusing carbon or nitrogen into the surface.

Front 34

For pieces that must have some shape to them, what other method(s) must be used toincrease the critical stress?

Back 34

-Make sure any features have smooth, well rounded curves (large 𝜌t)

Front 35

What is the main cause of a brittle ductile transition?

Back 35

-Temperature (materials are more brittle at lower temperatures)

Front 36

What test is used to determine where a brittle-ductile transition occurs?Also, state what the test actually measures.

Back 36

-Charpy test.

-Measures energy required to fracture a standard specimen of the material.

Front 37

When is creep a concern to an engineer?

Back 37

-When a material is placed under a high stress (but below the yield point),and at high temperaturesfor a long time.

Front 38

What is the difference between isomorphous and eutectic alloys?

Back 38

-In isoomorphous alloys, the two components are completely soluble in each otherat all compositions (there is only one solid phase).

-In eutectic solids, there are two solid phases.

Front 39

Describe the development of microstructure when an isomorphous alloy cools quickly froma liquid to a solid.

Back 39

-Solid begins to precipitate at the liquidus

-During rapid cooling, coring occurs, in which each new layer of solid forming on theparticles has a different composition

Front 40

What is martensite and how is it formed?

Back 40

-Phase supersaturated with interstitial carbon atoms.

-Formed by rapidly quenching pure austenite.

Front 41

What are the mechanical properties of martensite and one application?

Back 41

-It is incredibly hard.

- Blades.

Front 42

If we do not want martensite present in a steel, describe one process for removing it.

Back 42

-Austenitizing.

-Heat the material above the upper critical temperature

-Hold there long enough for all of the material to transform austenite.

-Cool again, slowly.

Front 43

In order to increase the amount of martensite, we need to increase the severity of aquench. Provide several methods for doing this.

Back 43

-Use water instead of oil.

-Have the water circulating to move heated water away from the object.

-Cool the water to increase the temperature difference between the object and thewater.

Front 44

Provide at least three reasons why Boeing would be investing $30b in Ti for use on theirplanes (as a replacement of steel and aluminum).

Back 44

- Light (compared to steel).

- Very good fatigue characteristics (has a fatigue limit, Al does not).

- Does not corrode when in contact with composites (Al does).

- Does not corrode in air or water (some steels do).

Front 45

Provide at least three reasons why Boeing might be regretting than decision and going backto using Al in some parts of the plane.

Back 45

- Expensive.

- High buy-to-fly ratio (a lot of waste makes it even more expensive!)

- Difficult to machine (slow, need expensive tools).

- Have to buy a lot of material from Russia…always a chance of being shafted!

Front 46

Ceramic properties

Back 46

- Hard

- Brittle

- Good under compression

- Bad under tension

Front 47

Fibers (Ceramic) Properties

Back 47

- Good under tension

- Bond well to ceramics

- Bridge cracks

Front 48

Resin Properties

Back 48

- Ductile

- Able to absorb energy

Front 49

Fibers (Polymers/Resins) Properties

Back 49

- Good under tension

- Hard

- Brittle

- Bonds well to matrix