material science homework

ENGR 1210: FA 2015; HW 6

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Homework 6 on chapter 6; (Total points: 100)

Reading assignments: Chapter 6 from textbook, notes and slides

Student Name: ____________________ MTSU ID: ______________

Honor pledge: I _______________________________ acknowledge that I have neither received nor given any unauthorized help/aid during this assignment.

Part 1 Key Engineering Terms: fill in the blanks – 30 points

1. Hot working: _____permanent____ deformation of metals and alloys ____above____ the

recrystallization temperature.

2. Cold working: ___permanent____ deformation of metals and alloys __below___ the

recrystallization temperature.

3. Hardness and ductility exhibit ___inverse____ relationship.

4. Ductility: The measure of degree of ____plastic/permanent_____ deformation that has

been sustained at fracture.

5. Elastic deformation: if a metal is deformed by a force ____returns_____ to its original

dimensions after force is removed.

6. Engineering stress: average _____uniaxial______ force divided by original cross-

sectional area.

7. Engineering strain: change in length of the sample divided by unit ___original___ length.

8. Shear stress: ratio of ___shear____ force divided by area on which it acts.

9. Modulus of elasticity: stress divided by strain in the ____elastic_____ region of an

engineering stress-strain diagram.

10. Yield strength: the stress _____at____ which material just begins to plastically deform.

11. Offset yield strength: value of stress at ___0.2% or 0.002____ strain.

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12. Resilience: it is the capacity of material to absorb energy when deformed

___elastically____.

13. Toughness: Property that is indication of material’s resistance to ___brittle fracture___

when a crack is present.

14. Ultimate tensile strength: the ___maximum_____ stress into engineering stress strain

curve before material fractures.

15. True stress: load divided by ___instantaneous____ cross-section area.

16. Hardness: Measure of material’s resistance to localized ___plastic______ deformation.

17. Dislocation: a ____line_____ defect where an array of atoms are missing resulting in

lattice distortion.

18. Slip: the process by which plastic deformation is produced by ___dislocation motion___.

19. Critical resolved shear stress: the minimum amount of shear stress required to initiate

___slip____.

20. Twin boundary: a special grain boundary across which the grains show

______mirror_______ lattice symmetry.

21. Recovery: the first stage in the annealing process that results in removal of

____residual_____stresses and formation of ____low_____ energy dislocation

configurations.

22. Grain boundary acts as a ____barrier_____ to dislocation movement.

23. Recrystallization: the second stage of annealing process in which new grains start to grow

and dislocation density ____decreases_____ significantly.

24. Solid solution strengthening: alloying materials with _____solute_____ atoms that go

into Substitutional or interstitial solid solution.

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25. Strain hardening: it is a phenomenon by which ductile material becomes harder and

stronger as it is _____plastically/permanently_______ deformed.

26. Superplasticity: the ability of some metals to deform plastically by ___1000% to

2000%____ at ________high_______ temperatures and low loading rates.

Part 2 Learning concepts/reflections/problems – 50 points

1. (a) Explain the three stages of annealing treatment in 2-3 sentences each.

(b) Explain any five properties that we can measure using a stress-strain curve under

uniaxial tensile loading. 10 points

(a) Recovery: reduction in the dislocation density by annihilation of dislocations; rearrangement

of atoms to form perfect crystal lattice; atoms diffuse to regions of tension.

Recrystallization: new strain free and equiaxed grains form that have low dislocation densities

and small size; eventually all cold work is consumed; parent cold-work grains are replaced

Grain growth: small grains shrink and ultimately disappear; large grains continue to grow; at

longer times, average grain size increases.

(b) Modulus of elasticity: stress divided by strain in the elastic region of an engineering stress-

strain diagram

Poisson’s ratio: the ratio of lateral strain and axial/linear strain

Yield strength: the stress above which permanent deformation just begins

Ultimate tensile strength: the maximum stress in the engineering stress-strain curve before

material fractures

Ductility: it is a measure of degree of plastic deformation that has been sustained at fracture

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Modulus of resilience: it is the strain energy per unit volume required to stress a material from an

unloaded state to point of yielding

Toughness: a material property indicative of its resistance to fracture when a crack is present

2. What are the 4 strengthening mechanisms of metals? Explain each mechanism in 4-5

sentences. 10 points

Reduction in grain size: Grain boundaries are barriers to slip. Barrier "strength" increases with

increasing angle of misorientation. Smaller the grain size more the barriers to slip.

Solid solution strengthening: Impurity atoms distort the lattice & generate lattice strains. These

strains can act as barriers to dislocation motion. Small impurities tend to concentrate at

dislocations (regions of compressive strains) - partial cancellation of dislocation compressive

strains and impurity atom tensile strains. Reduce mobility of dislocations and increase strength.

Large impurities tend to concentrate at dislocations (regions of tensile strains).

Precipitation hardening: Hard precipitates are difficult to shear. Large shear stress is required to

move a dislocation toward a precipitate and shear it. Dislocation moves around and advances but

the precipitate acts as pinning sites.

Work hardening: Deformation at room temperature (for most metals). Common forming

operations reduce the cross-sectional area. Processes such as forging, drawing, rolling, extrusion

involved in inducing cold work. Dislocation density increases and dislocations accumulate along

the defects such as grain boundaries increasing the plastic deformation before fracture thus

strengthening the material.

3. Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of

10.0 mm. A tensile force of 1500 N produces an elastic reduction in diameter of 6.7 × 10–

4 mm. Compute the elastic modulus of this alloy, given that Poisson’s ratio is 0.35.

10 points

This problem asks that we calculate the modulus of elasticity of a metal that is stressed in tension. Combining

Equations for elastic modulus and poisson’s ratio results in

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From the definition of Poisson's ratio, (Equation 6.8) and realizing that for the transverse strain,

leads to

Therefore, substitution of this expression for εz into the above equation for E yields

Incorporation of values for F, ν, d0, and ∆d given in the problem statement (and realizing that ∆d is negative)

allows us to calculate the modulus of elasticity as follows:

4. A specimen of a 4340 steel alloy with a plane strain fracture toughness of 54.8 MPa

(50 ksi ) is exposed to a stress of 1030 MPa (150,000 psi). Will this specimen

experience fracture if the largest surface crack is 0.5 mm (0.02 in.) long? Why or why

not? Assume that the parameter Y has a value of 1.0. 10 points

We calculate design stress.

𝜎𝜎𝑐𝑐 = 𝐾𝐾𝐼𝐼𝐼𝐼 𝑌𝑌√𝜋𝜋𝑎𝑎

Substituting we get,

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𝜎𝜎𝑐𝑐 = 54.8

1�𝜋𝜋(0.5 × 10−3)

𝜎𝜎𝑐𝑐 = 1382.68 𝑀𝑀𝑃𝑃𝑎𝑎

Since applied stress is less than design stress, the material will NOT fail.

5. (a) A 10-mm-diameter Brinell hardness indenter produced an indentation 2.50 mm in

diameter in a steel alloy when a load of 1000 kg was used. Compute the HB of this

material.

(b) What will be the diameter of an indentation to yield a hardness of 300 HB when a

500-kg load is used? 10 points

(a) We are asked to compute the Brinell hardness for the given indentation. It is necessary to use the equation in

Table 6.5 for HB, where P = 1000 kg, d = 2.50 mm, and D = 10 mm. Thus, the Brinell hardness is computed as

(b) This part of the problem calls for us to determine the indentation diameter d that will yield a 300 HB

when P = 500 kg. Solving for d from the equation in Table 6.5 gives

Part 3 Materials design – 20 points

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1. Why are nanocrystalline materials stronger? Explain answer based on dislocation

activity. 5 points

As grain size decreases, grain boundary density increases which can prevents slip and the ability

of the metal to resist movement of dislocations increases. Any produced dislocations will quickly

pile up at the boundaries and create dislocation entanglement resulting in increased strength of

the metal.

2. What are the five important factors that affect the recrystallization process in metals?

5 points

1. the extent of deformation of the metal prior to recrystallization; 2. the temperature used for the recrystallization process; 3. the length of time of the recrystallization process; 4. the initial grain size of the metal; 5. the composition of the metal, in terms of metal purity.

3. Consider casting a cube and a sphere on the same volume from same metal. Which one

would solidify faster? Why? 5 points

For a cube and a sphere of the same volume (4π/3R3=a3)

R = 0.62 a

Radius of the sphere side of the cube

Surface area of the sphere = 4πR2 = 4.82a2

Surface area of the cube = 6a2

Since the cube (of the same volume as the sphere) has a larger surface area, it will lose heat more rapidly and will cool faster.

4. Why is it difficult to improve both strength and ductility simultaneously? 5 points

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One can improve strength by i) applying higher %CW, ii) introducing impurities, iii) reducing

grain size, iv) applying heat treatment procedures. However, all of the above approaches will

result in a reduction of dislocation motion either due to interaction/interference with existing

dislocations, grain boundaries, impurity elements in the solvent matrix, and or harder second

phase. The ensued reduction in dislocation motion will strengthen the material but at the same

time reduces the amount of plastic/permanent deformation in the materials, resulting in brittle

fracture behavior upon continued loading.

Bonus questions: -10 points

1. Why are cast metal sheet ingots hot rolled first instead of being cold rolled?

Hot rolling is applied first because it is more efficient in reducing ingot sheet thickness than cold

working.

2. Why does slip in metals usually take place on closed packed planes and in closed packed

directions?

Slip usually takes place on the most densely packed planes because the atoms on these planes are

in close proximity and hence require less shear energy for displacement.

Slip typically occurs along the closest-packed directions because minimal energy is required to

force the atoms to change positions.