Homework #10

I. Glass A will fracture at a lower critical stress than Glass B but have the same fracture toughness (K_1C) as Glass B. - Fracture toughness (K_1C) is a materials constant, so it does not change since the glasses are essentially the same composition and structure (fictive temperature). Critical stress to failure (sigma*) would go down for A because of the longer cracks.

Assume that Glass A ang Glass B are identical in composition and fictive temperature, but Glass A has many cracks up to 2mm maximum length while Glass B only has a few cracks with maximum length of < 1mm. Based on this knowledge, which of these statements is most likely true? A. Glass A will fracture at a lower critical stress than Glass B but have a higher fracture toughness (K_1C) than Glass B. B. Glass A will fracture at a lower critical stress than Glass B and have a lower fracture toughness (K_1C) than Glass B. C. Glass A will fracture at the same critical stress than Glass B and have the same fracture toughness (K_1C) as Glass B. D. Glass A will fracture at a higher critical stress than Glass B and have a lower fracture toughness (K_1C) than Glass B. E. Glass A will fracture at the same critical stress as Glass B but have a higher fracture toughness (K_1C) than Glass B. F. Glass A will fracture at the same critical stress as Glass B but have a lower fracture toughness (K_1C) than Glass B. G. Glass A will fracture at a higher critical stress than Glass B and have a higher fracture toughness (K_1C) than Glass B. H. Glass A will fracture at a higher critical stress than Glass B and have the same fracture toughness (K_1C) as Glass B. I. Glass A will fracture at a lower critical stress than Glass B but have the same fracture toughness (K_1C) as Glass B.

B. Formation of the precipitate decreases the entropy of the system and is exothermic.

Assume that the precipitation of pure copper from an Al-Cu alloy is thermodynamically favorable at a given process temperature (i.e., the transformation is spontaneous). How best would you describe the thermodynamics of this phase transformation at this temperature? A. Formation of the precipitate increases the entropy of the system and is exothermic. B. Formation of the precipitate decreases the entropy of the system and is exothermic. C. Formation of the precipitate decreases the entropy of the system and is endothermic. D. Formation of the precipitate increases the entropy of the system and is endothermic.

B. zero

Assuming it is possible for a liquid to transform into a solid at the equilibrium melting temperature, the change in the Gibbs free energy of the substance as it transforms from a liquid to a solid will be: A. negative B. zero C. positive

B only

Based on these S-N curves which of these alloys would not fail under cyclic loading of 500MPa for 10^5 cycles but would fail under the same loading by 10^6 cycles? A. Ti-5Al-2.5Sn titanium alloy B. 4340 steel C. 1045 steel

No

Based on these S-N curves, would you expect ductile cast iron to fail under cyclic loading of 200 MPa for 10^9 cycles?

D. Heat is absorbed and the entropy of the system increases.

During the phase transformation from liquid to vapor (vaporization): A. Heat is absorbed and the entropy of the system decreases. B. Heat is released and the entropy of the system decreases. C. Heat is released and entropy of the system increases. D. Heat is absorbed and the entropy of the system increases.

A. Heat is absorbed and the entropy of the system increases.

During the phase transformation from solid to liquid (melting): A. Heat is absorbed and the entropy of the system increases. B. Heat is absorbed and the entropy of the system decreases. C. Heat is released and the entropy of the system increases. D. Heat is released and the entropy of the system decreases.

C. crack propagation in a material

Fracture is a direct result of.... A. dislocation glide in a material B. impeded dislocation motion in a material C. crack propagation in a material D. elastic deformations in a material E. crack reduction in a material

A and C

Possible kinetic barriers that prevent phase transformations from occurring quickly include: A. Energy costs associated with interface formation. B. Energy costs associated with the free energy change for the phase transformation. C. Energy costs associated with the long range diffusion of atomic components to the product phase. D. Energy costs associated with dislocation motion in a solid.

B. decreases non-linearly with decreasing temperature.

Solid state diffusion... A. decreases linearly with decreasing temperature. B. decreases non-linearly with decreasing temperature. C. increases linearly with decreasing temperature. D. increases non-linearly with decreasing temperature.

C. Creep

The filament in an incandescent lightbulb is especially susceptible to failure via: A. Fatigue G. Dielectric Breakdown C. Creep D. Brittle Fracture E. Yielding

D. Dislocation motion (climb).

The most likely mechanism for creep in a material under high tensile stresses is: A. Crack propagation. B. Diffusion through the grain boundaries. C. Cyclic loading. D. Dislocation motion (climb). E. Diffusion through the bulk.

C. 4 Phases -- the three different crystalline forms of TiO2 and the amorphous TiO2.

Titanium dioxide (TiO2) is deposited as an amorphous thin film. When annealed, 75% of the film crystallizes. X-ray diffraction analysis reveals that these crystals are a mix of rutile, anatase, and brookite -- three different crystal structures of titanium dioxide. At this point, how many phases are in the TiO2 thin film? A. 1 Phase B. 2 Phases C. 3 Phases D. 4 Phases E. 5 Phases F. Zero Phases

A. Charpy Impact Testing

To measure the fracture toughness of a material, the best technique to use is: A. Charpy Impact Testing B. Dilatometry C. Toughmetry D. Tensile Testing E. Vickers Hardness Testing

True - While precipitates will increase yield strength of the alloy, this usually reduces the plastic zone at the crack tip, leading to reduced fracture toughness.

True/False: Adding precipitates to a metal alloy will likely increase its yield strength but decrease its fracture toughness.

True

True/False: Polycrystalline materials have many mis-oriented crystalline grains in their microstructure because the phase transformations used to create these materials (e.g., solidification) occur via a nucleation and growth process.

False - It is possible to have an amorphous solid with two phases of varying composition. On example, is Pyrex, which has two amorphous phases: an SiO2-rich and a B2O3-rich one.

True/False: Systems containing only amorphous solids CANNOT have more than one phase.

True

True/False: When you plot failure rate data in the form of ln[ln(1/1-P))] versus ln(N) where P is the cumulative probability of failure and N is the number of cycles, a linear fit to that data will give a line having a slope that we can use to estimate the shape parameter for the Weibull distribution of that data set.

A. This material has 3 phases: (1) an Al-Cu-Sn solid solution, (2) a pure Sn phase, and (3) a pure copper phase.

What is the phase assemblage of this Al-Cu-Sn alloy? A. This material has 3 phases: (1) an Al-Cu-Sn solid solution, (2) a pure Sn phase, and (3) a pure copper phase. B. This material has 2 phases: (1) an Al phase and (2) a Cu-Sn phase. C. This material has 4 phases: (1) an Al-Cu-Sn solid solution, (2) a pure Sn phase, (3) a pure copper phase, and (4) another pure copper phase with a different crystal structure. D. This material has 3 phases: (1) an Al phase, (2) a Cu phase, (3) a Sn phase. E. This material has 2 phases: (1) an Al-Cu-Sn solid solution phase and (2) a Cu-Sn phase.

A. The bricks in the external walls of a house.

Which of the components is LEAST likely to suffer from a fatigue failure. A. The bricks in the external walls of a house. B. The anchor cable for an offshore oil rig. C. The axel of a car. D. A diving board. E. The fuselage of an airplane.

Both A and C

Which of the following would increase the time (number of cycles) to fatigue failure in the given material? A. Flame polishing the surface of a glass component B. Cold working a metal alloy C. Shot peening a metal gear D. Reducing the use temperature of a polymer component to below its glass transformation temperature

Both B and C

Which of these components when loaded to below its yield strength is likely to suffer creep failure at room temperature (25 C) operation? (Choose the BEST answer.) A. A steel cable holding up a suspension bridge. B. A nylon cable suspending a yacht from the ceiling in a dry dock. C. A solder joint under tension from the wire bond on a circuit board.

B. 2 phases

White latex paint is an emulsion of acrylate monomers, an initiator, and TiO2 particles that scatter light to give the white color. While drying on the wall, initiators are released and polymerize the monomers into a latex polymer. Excluding the initiator (which effectively become part of the polymer), how many phases are in dried white paint? A. 4 phases B. 2 phases C. 1 phase D. 3 phases

D. have a lower load bearing capacity because of the stress concentration near the holes.

Wooden 4 inch x 4 inch x 12 feet wooden posts hold up the deck that comes off the second floor of your house. On the ground level, you decide to drill 1 inch holes in two of the posts in order to hang a hammock. Compared to the other posts, the posts that now hold the hammock... A. have a higher load bearing capacity because of the stress concentration near the holes. B. have a higher load bearing capacity because the notches will act as barriers to dislocation motion. C. have the same load bearing capacity because both are made of the same material. D. have a lower load bearing capacity because of the stress concentration near the holes. E. have a lower load bearing capacity because the notches will act as barriers to dislocation motion.

D. All of these replies are accurate.

Working as a consultant for an engineering firm, a steam plant manager comes to you for advice. He intends to lower the temperature of the steam running through his steam pipes, but maintain all other operational conditions the same (i.e, steam pressure inside the pipes and the cycling schedules for pressurization and depressurization). He asks you how this change might affect the serviceable lifetime of the steam pipes. All of the following replies are accurate EXCEPT: A. The lower temperature will decrease the creep rate, decreasing the risk for creep failure. B. By not increasing the pressure in the steam pipes, you will not increase the risk for yielding the material. C. By maintaining the same pressure cycling schedule, you will retain about the same fatigue rate as before, and, thus, your risk for fatigue failure will remain the same. D. All of these replies are accurate. E. None of these replies are accurate.

C. the shaper parameter (beta) of the Weibull distribution is much greater than 1. - The shape parameter determines how sharp the transition between parts not failing versus failing. When beta is much larger than 1, this transition is very sharp, and so the time to failure is well define. If the shape is broader, you have parts that fail at all different times, and so it becomes hard to predict the product's lifetime.

You can be highly confident in your prediction of the time to product failure if: A. the shaper parameter (beta) of the Weibull distribution is much less than 1. B. the shaper parameter (beta) of the Weibull distribution is equal to 1. C. the shaper parameter (beta) of the Weibull distribution is much greater than 1. D. the characteristic lifetime (N_0) of the Weibull distribution is very large. E. the characteristic lifetime (N_0) of the Weibull distribution is very small.

C. 1 phase - This is a single phase since it is a solid solution of copper and zinc.

You own a brass tuba composed of 90% copper and 10% zinc. Brass is a substitutional solid solution of copper and zinc. How many phases make up your brass tuba? A. zero phases B. 2 phases C. 1 phase D. 3 phases

C. the shaper parameter (beta) of the Weibull distribution is equal to 1. - The shape parameter determines how sharp the transition is between parts not failing versus failing. When beta is much larger than 1, this transition is very sharp, and so the time to failure is well defined. If the shape is broader, you have parts that fail at all different times, and so it becomes hard to predict the product's lifetime.

You will definitely have low confidence in your prediction of the time to product failure if: A. the characteristic lifetime (N_0) of the Weibull distribution is very small. B. the shaper parameter (beta) of the Weibull distribution is much greater than 1. C. the shaper parameter (beta) of the Weibull distribution is equal to 1. D. the characteristic lifetime (N_0) of the Weibull distribution is equal to 1. E. the characteristic lifetime (N_0) of the Weibull distribution is very large.