331 Exam 1 harder

A metal has an elastic modulus of 200 GPa. If it is subjected to an elastic stress of 300 MPa, what is its strain?

Use Hooke's law: ε = σ/E = 300 MPa / 200,000 MPa = 0.00150, or 0.150%.

A tensile specimen has an original cross-sectional area of 80 mm² and is loaded with 16,000 N. What is the engineering stress?

σ = F/A0 = 16,000 N / 80 mm² = 200 N/mm² = 200 MPa.

A specimen is originally 50 mm long and stretches to 53.5 mm. Calculate engineering strain and percent elongation.

ε = (53.5 - 50)/50 = 0.070 = 7.0%. Therefore, percent elongation = 7.0%.

A specimen changes from 50 mm to 53.5 mm. Calculate its true strain.

εT = ln(L/L0) = ln(53.5/50) = ln(1.07) ≈ 0.06766.

A tensile specimen carries 20,000 N. Its original area is 100 mm², but its instantaneous area has decreased to 80 mm². Calculate engineering stress and true stress.

Engineering stress = 20,000/100 = 200 MPa. True stress = 20,000/80 = 250 MPa.

Why is true stress greater than engineering stress when a tensile specimen has decreased in cross-sectional area?

Engineering stress uses the original area A0, while true stress uses the smaller instantaneous area Ai. Since the same force is divided by a smaller area, true stress is larger.

A material has a yield strength of 350 MPa and an elastic modulus of 175 GPa. What elastic strain corresponds to the yield strength?

ε = σ/E = 350/175,000 = 0.00200, or 0.200%.

A material has an engineering stress of 500 MPa at an engineering strain of 0.25. Assuming the deformation is uniform, calculate true strain.

εT = ln(1 + εe) = ln(1.25) = 0.2231.

A specimen has an original area of 120 mm² and is cold worked until its area is 90 mm². What is the percent cold work?

%CW = [(A0 - Ad)/A0] × 100 = [(120 - 90)/120] × 100 = 25%.

A metal is cold worked from 200 mm² to 140 mm². If its original yield strength was 250 MPa and a provided material chart gives 410 MPa at 30% cold work, what is the new yield strength?

%CW = [(200 - 140)/200] × 100 = 30%. From the provided chart, the new yield strength is 410 MPa.

A material is cold worked by reducing its cross-sectional area from 100 mm² to 60 mm². Predict the changes in yield strength, tensile strength, hardness, ductility, and dislocation density.

40% cold work. Yield strength increases, tensile strength increases, hardness increases, ductility decreases, and dislocation density increases.

Two steel samples have grain sizes of 0.100 mm and 0.025 mm. Which should have the higher yield strength and why?

The 0.025 mm sample should have the higher yield strength. Smaller grains provide more grain boundaries, which act as barriers to dislocation motion.

Using Hall-Petch, calculate the yield strength if σ0 = 150 MPa, ky = 0.60 MPa√m, and d = 0.0004 m.

σy = σ0 + ky d^(-1/2). √0.0004 = 0.02 m^1/2, so d^(-1/2) = 50 m^(-1/2). σy = 150 + (0.60)(50) = 180 MPa.

Using Hall-Petch, calculate the yield strength if σ0 = 100 MPa, ky = 0.80 MPa√m, and d = 0.0001 m.

√0.0001 = 0.01. Therefore d^(-1/2) = 100. σy = 100 + (0.80)(100) = 180 MPa.

Two materials have identical composition, but Material A has half the grain diameter of Material B. Which one should have the higher yield strength according to Hall-Petch?

Material A. Since σy = σ0 + ky d^(-1/2), decreasing d increases d^(-1/2), increasing yield strength.

A material's grain size decreases from 100 μm to 25 μm. By what factor does d^(-1/2) change?

d decreases by a factor of 4. Since d^(-1/2) is proportional to 1/√d, it increases by √4 = 2. Therefore the Hall-Petch strengthening contribution doubles.

Why does adding a substitutional alloying element generally strengthen a metal?

The solute atoms have different sizes and/or bonding characteristics than the host atoms, producing lattice distortions that interfere with dislocation motion.

Why can an interstitial atom be especially effective at strengthening a metal?

Interstitial atoms fit into spaces between host atoms and produce significant local lattice distortion, creating strong obstacles to dislocation movement.

A metal's dislocation density increases dramatically during cold working. Explain why this makes the metal stronger.

Dislocations interact with and obstruct one another. As dislocation density increases, additional dislocations have more difficulty moving, so a higher applied stress is required for further plastic deformation.

A cold-worked metal is heated below its recrystallization temperature. Which annealing stage is occurring and what is the primary effect?

Recovery. Internal stresses are relieved and dislocations rearrange or partially annihilate, with relatively little change in grain structure.

A cold-worked metal is heated to a temperature where new strain-free grains begin forming. What stage is this?

Recrystallization.

During recrystallization, what happens to strength and ductility?

Strength and hardness decrease significantly while ductility increases significantly.

A material is heated well beyond the recrystallization stage. Large grains begin replacing smaller grains. What stage is occurring?

Grain growth.

Why does grain growth generally decrease strength?

Larger grains mean fewer grain boundaries per unit volume, so there are fewer barriers to dislocation motion.

A severely cold-worked metal is annealed through all three stages. Rank its strength from highest to lowest during the process: immediately after cold work, after recovery, after recrystallization, after grain growth.

Highest immediately after cold work; slightly lower after recovery; much lower after recrystallization; generally lowest after significant grain growth.

A material has strong atomic bonding and a very deep interatomic potential-energy well. Predict its elastic modulus, melting temperature, and coefficient of thermal expansion relative to a weakly bonded material.

It should have a higher elastic modulus, higher melting temperature, and lower coefficient of thermal expansion.

Why does stronger atomic bonding generally produce a higher elastic modulus?

Stronger bonds resist changes in interatomic spacing more strongly, so greater stress is required to produce a given elastic strain.

Two materials have similar crystal structures, but Material A has much stronger atomic bonding than Material B. Which should generally have the higher melting temperature and why?

Material A. More energy is required to separate strongly bonded atoms, so its melting temperature is generally higher.

A material has a high coefficient of thermal expansion. What does this suggest about the shape of its interatomic potential-energy curve?

It generally indicates a more asymmetric and relatively shallow potential-energy well, allowing average interatomic spacing to increase more as temperature rises.

A material has no long-range periodic atomic arrangement but still has short-range atomic order. Is it crystalline or amorphous?

Amorphous. Amorphous materials have short-range order but lack long-range periodic order.

A material consists of many grains, each having a different crystallographic orientation. Is it a single crystal or polycrystalline?

Polycrystalline.

Why can a single crystal have strongly direction-dependent mechanical properties?

The atomic arrangement and available slip systems depend on crystallographic direction, so properties can vary with loading direction.

A student writes that [110] represents a crystal plane. Is this correct? Explain.

No. [110] represents a crystallographic direction. A plane is represented using parentheses, such as (110).

A student writes that (111) represents a crystallographic direction. Is this correct? Explain.

No. (111) represents a crystallographic plane. Directions use square brackets, such as [111].

A material has a vacancy concentration of Nv/N = exp(-Qv/kT). What happens to vacancy concentration if temperature doubles, assuming Qv and k remain constant?

Vacancy concentration increases because the magnitude of the negative exponent decreases as T increases. The increase is exponential rather than linear.

Why do vacancies become more numerous at high temperatures even though creating a vacancy requires energy?

Higher temperatures provide greater thermal energy, making it more likely that atoms have enough energy to leave their normal lattice sites. The equilibrium vacancy concentration therefore increases.

An alloy has atoms whose radii differ by 8%, the same crystal structure, similar electronegativities, and compatible valences. Based on Hume-Rothery rules, would extensive substitutional solid solubility be expected?

Yes. The alloy satisfies the major Hume-Rothery requirements, including an atomic-size difference below about 15%, compatible crystal structure, electronegativity, and valence.

An alloy has atoms whose radii differ by 25%, and their crystal structures are different. Would extensive substitutional solid solubility be expected?

No. The large atomic-size difference and different crystal structures violate important Hume-Rothery criteria.

An edge dislocation has its Burgers vector perpendicular to the dislocation line. Is this consistent with an edge dislocation?

Yes. For a pure edge dislocation, the Burgers vector is perpendicular to the dislocation line.

A screw dislocation has its Burgers vector parallel to the dislocation line. Is this consistent with a screw dislocation?

Yes. For a pure screw dislocation, the Burgers vector is parallel to the dislocation line.

Why does dislocation motion require much less stress than moving an entire plane of atoms simultaneously?

A dislocation allows plastic deformation to occur incrementally as bonds break and reform locally, rather than requiring all atoms in an entire plane to move at once.

A material has many available slip systems and relatively nondirectional bonding. Would you expect it to be relatively ductile or brittle?

Relatively ductile, because dislocations can move more easily when many favorable slip systems are available and bonding is nondirectional.

Why are ceramics generally less ductile than metals?

Ceramics often have directional covalent bonds and/or ionic bonding constraints, plus fewer favorable slip systems, making dislocation motion much more difficult.

A tensile specimen reaches a maximum engineering stress and then the engineering stress decreases while the specimen continues deforming. What phenomenon has likely begun?

Necking. Localized reduction in cross-sectional area has begun.

Why can engineering stress decrease during necking even though the material is still being plastically deformed?

Engineering stress is calculated using the original area. During necking, the actual load-carrying area becomes much smaller, so the measured load may decrease even though the true stress on the remaining material can continue increasing.

A material has a stress-strain curve with a large area under the entire curve before fracture. What property does this indicate?

High toughness. Toughness is related to the total energy absorbed before fracture and corresponds to the area under the entire stress-strain curve.

A material has a steep linear elastic stress-strain curve but fractures after very little plastic deformation. What can you conclude about its stiffness and ductility?

It has relatively high stiffness because the elastic slope is steep, but low ductility because it undergoes little plastic deformation before fracture.

Material A has E = 200 GPa and Material B has E = 70 GPa. Both experience a stress of 140 MPa while still elastic. Which has the greater strain and by how much?

Material A: ε = 140/200,000 = 0.00070. Material B: ε = 140/70,000 = 0.00200. Material B has 2.86 times the strain and is therefore less stiff.

A specimen has F = 25,000 N, A0 = 100 mm², L0 = 50 mm, Ai = 75 mm², and L = 60 mm. Calculate engineering stress, true stress, engineering strain, and true strain.

Engineering stress = 25,000/100 = 250 MPa. True stress = 25,000/75 = 333.3 MPa. Engineering strain = (60-50)/50 = 0.20. True strain = ln(60/50) = 0.1823.

A metal is strengthened by reducing grain size, adding alloying atoms, and cold working. What common physical mechanism explains why all three methods increase strength?

All three methods hinder dislocation motion. Grain boundaries block dislocations, alloying atoms create lattice distortions that obstruct them, and cold work increases dislocation density so dislocations interfere with one another.