Questions to remember

Question 1

What is the function of alloying elements in tool steels?

Answer 1

The alloying elements combine with the carbon to form very hard and wear-resistant carbide compounds.

Question 2

Why must rivets of a 2017 aluminum alloy be refrigerated before they are used?

Answer 2

After being solution heat treated, they precipitation harden at room temperature. Once precipitation hardened, they are too strong and brittle to be driven.

Question 3

What is the chief difference between heat- treatable and non-heat-treatable alloys?

Answer 3

Heat-treatable alloys may be strengthened by a heat treatment wherein a precipitate phase is formed or a martensitic transformation occurs. Nonheat-treatable alloys are not amenable to strengthening by such treatments.

Question 4

Advantages of cold working

Answer 4

1. High quality surface finish
2. Mechanical properties may be varied
3. Close dimensional tolerances

Question 5

Disadvantages of cold working

Answer 5

1. High deformation energy requirements.
2. Large deformations must be accomplished in steps, this may be expensive.
3. Loss of ductility.

Question 6

Advantages of hot working

Answer 6

1. Large deformations possible and repeatable.

2. Deformation energy requirements are relatively low.

Question 7

Disadvantages of hot working

Answer 7

1. Poor surface finish.

2. A variety of mechanical properties is not possible.

Question 8

Advantages of extrusion over rolling.

Answer 8

1. Pieces having more complicated cross-sectional geometry may be formed.
2. Seamless tubing may be produced.

Question 9

Disadvantages of extrusion over rolling.

Answer 9

1. Nonuniform deformation over the cross-section.

2. A variation of properties may result over a cross-section of an extruded piece.

Question 10

List four situations in which casting is the preferred fabrication technique.

Answer 10

1. For large pieces and/or complicated shapes.
2. When mechanical strength is not an important consideration.
3. For alloys having low ductility.
4. When it is the most economical fabrication technique.

Question 11

Describe the full annealing heat treatment procedure and the and the intended final microstructure.

Answer 11

1. Heat to about 50˚C above A3 line (Figure 11.10) (if the concentration of carbon is less than the eutectoid) or above the A1 line (if the concentration of carbon is greater than the eutectoid) until the alloy comes to equilibrium.
2. Furnace cool to room temperature.
   The final microstructure is coarse pearlite.

Question 12

Describe the normalizing heat treatment procedure and the and the intended final microstructure.

Answer 12

1. Heat to at least 55˚C above A3 (if concentration less than eutectoid) or above Acm line (if concentration greater than eutectoid) until alloy completely transforms to austenite.
2. Cool in air.
   Final microstructure is fine pearlite.

Question 13

Describe the quenching heat treatment procedure and the and the intended final microstructure.

Answer 13

1. Heat to a temperature within the austenite region and allow specimen to fully austenitize.
2. Quench to room temperature in oil or water.
   Final microstructure is martensite.

Question 14

Describe the tempering heat treatment procedure and the and the intended final microstructure.

Answer 14

1. Heat a quenched specimen to a temperature between 450 and 650˚C, for the time necessary to achieve the desired hardness.
   Final microstructure is tempered martensite.

Question 15

With regard to the total heat treatment procedure, the steps for the hardening of steel are

Answer 15

1. Austenitize above the upper critical temperature.
2. Quench to a relatively low temperature.
3. Temper at a temperature below the eutectoid.
4. Cool to room temperature.

Question 16

With regard to precipitation hardening, the steps for the hardening of steel are

Answer 16

1. Solution heat treat by heating into the solid solution phase region.
2. Quench to a relatively low temperature.
3. Precipitation harden by heating to a temperature that is within the solid two-phase region.
4. Cool to room temperature.

Question 17

For the hardening of steel, the microstructures that form at the various heat treating stages previously described are

Answer 17

1. Austenite
2. Martensite
3. Tempered martensite
4. Tempered martensite

Question 18

For precipitation hardening, the microstructures that form at the various heat treating stages previously described are

Answer 18

1. Single phase
2. Single phase–supersaturated
   Small plate-like particles of a new phase within a matrix of the original phase.
3. Same as 3.

Question 19

For the hardening of steel, the mechanical characteristics for the various steps in the stages described are

Answer 19

1. Not important
2. The steel becomes hard and brittle upon quenching.
3. During tempering, the alloy softens slightly and becomes more ductile.
4. No significant changes upon cooling to or maintaining at room temperature.

Question 20

For precipitation hardening, the mechanical characteristics for the various steps in the stages described are

Answer 20

1. Not important
2. The alloy is relatively soft.
3. The alloy hardens with increasing time (initially), and becomes more brittle; it may soften with overaging.
4. The alloy may continue to harden or overage at room temperature.

Question 21

What is the principal difference between natural and artificial aging processes?

Answer 21

For precipitation hardening, natural ageing is allowing the precipitation process to occur at the ambient temperature; artificial ageing is carried out at an elevated temperature.

Question 22

Make comparisons of thermoplastic and thermosetting polymers on the basis of mechanical characteristics upon heating.

Answer 22

Thermoplastic polymers soften when heated and harden when cooled, whereas thermosetting polymers harden upon heating, while further heating will not lead to softening.

Question 23

Make comparisons of thermoplastic and thermosetting polymers according to possible molecular structures.

Answer 23

Thermoplastic polymers have linear and branched structures, while for thermosetting polymers, the structures will normally be network or crosslinked.

Question 24

Is it possible to grind up and reuse phenol-formaldehyde?

Answer 24

It is not possible because it is a network thermoset polymer and, therefore, is not amenable to remolding.

Question 25

Is it possible to grind up and reuse polypropylene?

Answer 25

Yes since it is a thermoplastic polymer, will soften when reheated and, thus, may be remolded.

Question 26

Briefly describe the phenomenon of viscoelasticity.

Answer 26

The condition by which at intermediate temperatures the polymer is a rubbery solid that exhibits the combined mechanical characteristics of the two extremes; elastic and viscous behaviour.

Question 27

Briefly explain how molecular weight influences the tensile modulus of a semicrystalline polymer and why.

Answer 27

The tensile modulus is not directly influenced by a polymer’s molecular weight.

Question 28

Briefly explain how degree of crystallinity influences the tensile modulus of a semicrystalline polymer and why.

Answer 28

Tensile modulus increases with increasing degree of crystallinity for semicrystalline polymers. This is due to enhanced secondary interchain bonding which results from adjacent aligned chain segments as percent crystallinity increases. This enhanced interchain bonding inhibits relative interchain motion.

Question 29

Briefly explain how deformation by drawing influences the tensile modulus of a semicrystalline polymer and why.

Answer 29

Deformation by drawing also increases the tensile modulus. The reason for this is that drawing produces a highly oriented molecular structure, and a relatively high degree of interchain secondary bonding.

Question 30

Briefly explain how annealing of an undeformed material influences the tensile modulus of a semicrystalline polymer and why.

Answer 30

When an undeformed semicrystalline polymer is annealed below its melting temperature, the tensile modulus increases.

Question 31

Briefly explain how annealing of a drawn material influences the tensile modulus of a semicrystalline polymer and why.

Answer 31

A drawn semicrystalline polymer that is annealed experiences a decrease in tensile modulus as a result of a reduction in chain-induced crystallinity, and a reduction in interchain bonding forces.

Question 32

Describe 4 strategies for strengthening steel. Name each strategy and provide a brief explanation why this method results in strengthening.

Answer 32

1. Strengthening by reducing the grain size (d in Hall-Petch equation: sigma(y) = sigma(0) + k_y*d^(-1/2)
2. solid-solution strengthening: alloying with hetero atoms that either go interstitial or substitutional, creating strain fields
3. strain hardening –plastic deformation, work hardening or cold work %CW=(Ao-Ad/Ao) x 100
4. precipitation hardening: by introducing harder precipitates, dislocation movement is hindered, increasing strength.