1. Copyright Prentice-Hall Behavior and Manufacturing Properties of Part I Q1.1 (25): What is the difference between a material’s mechanical property and a material’s physical property? Mechanical: properties that involve a reaction to an applied load. Physical: any property that is measurable whose value describes a material state or a measure of the change in material state. Difference: mechanical properties measure reaction to load while physical properties are an inherent measure of state or change in state.

2. Edge Dislocation Movement Q1.2 (25): Why are metals much weaker than their theoretical strength? Shear dislocations help explain why the actual strength of metals is much lower than that predicted by theory, combined with impurity atoms that weaken bonds between atoms. Ductile metals typically fail under shear loads and not normal loads.

3. Solidification of Molten Metal Q1.3 (25): Why is it difficult to get uniform crystalline properties during cooling of molten metals, resulting in random oriented grain structures? Q1.3 (25): Why is it difficult to get uniform crystalline properties during cooling of molten metals, resulting in randomly oriented grain structures? Nucleation of crystals occurs at random sites in the molten metal due to uneven cooling; note that the crystallographic orientation of each site is different. This results in inconsistent directional properties and weakening. Principle: It is difficult to cool materials evenly, under any process and for any shape object; even more difficult for alloys with different melting temperatures!

4. Loading and Unloading of Tensile-test Specimen Figure 2.3 Schematic illustration of the loading and the unloading of a tensile-test specimen. Note that, during unloading, the curve follows a path parallel to the original elastic slope. Q2.1 (20): Give one example of a manu- facturing process that must take this “springback” into account. Answer: Bending of metals, forging, rolling, wherever a load is used to permanently deform metal into a shape.

5. Power Law Constitutive Model where K = strength coefficient n = strain hardening exponent Q2.2 (20): What is the shape/plot of the equation above on a log-log diagram and what does n represent for this shape? Answer: Straight line; n is the slope of the line.

6. Effect of Strain Rate on Tensile Strength of Al Figure 2.8 The effect of strain rate on the ultimate tensile strength for aluminum. Note that, as the temperature increases, the slopes of the curves increase; thus, strength becomes more and more sensitive to strain rate as temperature increases. Source: After J.H. Holloman Q2.3 (20): This important figure depicts a significant limitation to forming aluminum at higher rates and elevated temperatures. What is it? Answer: All the strain rate curves converge at higher rates to similar deformation loads, regardless of temperature. Thus, elevating part temperatures is not helpful for reducing deformation loads at high strain rates; this limits how fast you can deform parts into their final shapes.

7. S-N Curves Figure 2.15 (a) Typical S-N curves for two metals. Note that, unlike steel, aluminum does not have an endurance limit. (b) S-N curves for common polymers Q2.4 (20): If aluminum has no defined endurance limit, how does this non-limit define how aluminum can be used? Answer: Aluminum parts must be designed so that the load stresses fall below that causing failure for the expected number of load cycles; thus, aluminum must be used cautiously in environments where the load is dynamic and cyclic; else, it will fatigue and fail.

8. Effect of Time and Temperature on Yield Stress Figure 4.22 The effect of again time and temperature on the yield stress of 2014-T4 aluminum alloy. Note that, for each temperature, there is an optimal aging time for maximum strength. Q3.2 (20): Aluminum can be aged at temperatures below 200 degrees to attain its greatest strength. T____ F______ Q3.2 (20): Aluminum can be aged at temperatures below 200 degrees to attain its greatest strength. T_X__ F______

9. Hardness of Steel as a Function of Carbon Content Figure 4.24 Hardness of steels in the quenched and normalized conditions as a function of carbon content. Q3.3 (20): What is the difference between quenching and normalizing? Quenching is heating above recrystallization temperature followed by rapid cooling. Normalizing is heating above recrystallization temperature followed by controlled, slower cooling, often in air.

10. Nonferrous Metals and Alloys Q3.4 (20): Why is titanium a good material to be used in the aerospace industry? Titanium has highest strength to weight ratio, good strength at elevated temperatures and good corrosion resistance. These properties are important to aerospace products like jet engines.

11. Wrought Titanium Alloy Properties and Applications Q3.5 (20): What does “wrought” mean in all these slides? Wrought means to be worked, shaped, beaten, etc.

12. Polymer Chains Figure 7.5 Schematic illustration of polymer chains. (a) Linear structure – thermoplastics such as acrylics, nylons, polyethylene, and polyvinyl chloride have linear structures. (b) Branched structure, such as in polyethylene. (c) Cross-linked structure – many rubbers or elastomers have this structure, and the vulcanization of rubber produces this structure. (d) Network structure, which is basically highly cross-linked – examples are thermosetting plastics, such as expoxies and phenolics. Q4.1 (20): What happens to plastics as you move from condition(a) to (d) and why? Polymers become stronger and less ductile with more cross linking. Why? Because there are more atomic bonds in a confined region and in all directions.

13. Chapter 8 Ceramics, Graphite, and Diamond: Structure, General Properties, and Applications Q4.3 (20): This chapter describes hard materials generally confined to applications where the loads are________________. (fill in the blank) How do the components below illustrate that the loads are __(your answer)___ ? Q4.3 (20): This chapter describes hard materials generally confined to applications where the loads are___compressive__. (fill in the blank) How do the components below illustrate that the loads are compressive ? The ball bearings and races are made of hard ceramic material which works because they are subjected only to compressive forces. Carbides, ceramics, etc., are materials are not good in environments where the loads are tensile since they are brittle and fracture easily.

14. Methods of Reinforcing Plastics Figure 9.2 Schematic illustration of methods of reinforcing plastics (matrix) with (a) particles, (b) short or long fibers or flakes, and (c) continuous fibers. The laminate structures shown in (d) can be produced from layers of continuous fibers or sandwich structures using a foam or honeycomb core (see also Fig. 16.50). Q4.4 (20): Why are the reinforcing fibers aligned in different directions in several layers? Because the fibers exhibit strength only in tension or compression along their fiber directions. For the “part” to be strong, the fibers need to be oriented in all possible loading directions which require them to be layered in differing orientations.