1. Topic 6: Case Studies Feb. 3, Feb. 8: Metal casting lab tour (Room 15, IMS building basement) Feb. 10: Quiz 1 Reading assignment: Ch. 1-13 Strengthening strategy: eliminate or stop dislocations –Work hardening –Precipitation hardening –Grain refinement Case study: Al-Li alloys (precipitation hardening) Case study: High strength low alloy (HSLA) steels (grain refining)

2. Overview of engineering methods Engineering: use the learning from science in a practical context Identify and define problems & need Define constraints and success criteria Conduct analysis/investigation Set up specifications or present solutions/alternatives Verify results

3. Case Study I: Al-Li alloys for aerospace applications Before World War I (1914-1918), aircraft frames were made of wood and wings were covered with fabric. The problem with wooden aircraft was short durability. Thus, during WWI, all metal aircraft were developed (made of steels) In 1930s, the availability of strong, corrosion resistant Al alloys made aircraft lighter by replacing steels with Al alloys. Since then the quest for lighter and stronger materials continues Today, there is severe competition between the aluminum industry and composite manufacturers. Carbon fiber / epoxy composites are less dense than Al (composites: 1.8 gm/cc, aluminum: 2.8 gm/cc, steel: 7.8 gm/cc) So Al manufacturers have been hard at work to come up with Al alloys that are lighter and stronger than pure Al

4. The less the weight of the plane, the more cargo weight that can be carried. To be more precise and rigorous, look at the following pie chart showing the weight distribution for a typical passenger aircraft about to take off on a 1000 mile journey The empty plane accounts for 46 % of the total weight and the passengers and freight for only 14.5 %. If we can reduce the weight of the empty plane by 10% and keep the take off weight the same, we could transfer 4.6 % of the total weight (46% x 10%) to passengers & freight. This means the weight carried for profit has gone up by ~ 30% Thus, the key here is to reduce the weight of the empty airplane while maintaining or improving its strength, durability and performance, i.e., find or develop new light weight, high strength materials 46% Manufacturer’s Empty weight 28% Fuel for normal use 4% Fuel reserves 14.5% Passengers plus freight 7.5% Operator’s items Identify & define problem/need

5. Materials developed have to be lighter than existing Al alloys (density ~ 2.8 g/cc) As strong as or stronger than existing Al alloys (UTS = 77,000 psi, yield strength = 62,000 psi) [1 psi = 6890 Pa] As stiff as or stiffer than existing Al alloys (Young’s modulus = 11,000,000 psi) Should be corrosion resistant Should be stable under operating conditions: Be able to withstand high temperatures due to the heat generated via frictional resistance of air. For flight speed at or below Mach 2 (twice the speed of sound, or ~ 1,200 miles per hour at medium altitude), the temperature rise is such that Al alloys perform well. When flight speed is close to Mach 3 or above, materials with greater heat resistance must be used. Stainless steels, Ti alloys and Mo alloys are all qualified in this regard. These alloys are expensive and difficult to manufacture, but their use becomes mandatory for aircraft to fly at Mach 3 or above. Needs to be relatively inexpensive Define constraints and criteria

6. Conduct Analysis/ investigation: Existing materials MaterialsDensity (gm/cc) Young’s modulus (GPa) Tensile strength (MPa) Corrosion resistance Temperature capapility (Celsius) Conventional Al2.876530good< 170 Polymer (e.g., polyester epoxy) 1.1-1.52.0-5.030-150good< 100 Ceramics (Al2O3)3.97380300-1000excellent< 1200 Ceramics (Si3N4)3.18304700-1000excellent< 1400 Glass fibers2.570-853500-4500excellent< 100 Polymer fibers (Kevlar) 1.460-1202700-3600good< 100 Steels (HSLA)7.8207440-650fair< 300-400 Ti Alloys (Ti-6Al-4V) 4.4114990good< 700 Cu Alloy (Cu-30%Zn) 8.53110520excellent< 300 Mg alloy (Mg-9%Al-2%Zn) 1.845230-275fair< 180 Ni Alloy (IN-100) 8.9156680-940excellent< 1000 Graphite fiber1.8-2.0200-3801550-3100Excellent in reducing atmosphere < 2000

7. MaterialSpecific Strength (Tensile strength/density, MPa.cc/gm) Temperature capability (Celsius) Conventional Al189< 170 Mg alloy127-153< 170 Ni alloy76-105< 1000 Ti Alloy225< 700 Steels56-83< 300-400 Cu alloy61< 300 polymer27-136< 100 Graphite fiber775-1550 Glass fiber1400-1800 Specific strength = Strength/density

8. Conclusion: Of the existing materials, only Ti alloys provide advantages over Al alloys in strength, stiffness, specific strength and temperature capability. However, Ti alloys are more expensive and difficult to manufacture than Al alloys Alternatives: (1)Develop polymer matrix composites reinforced with graphite fibers (e.g., epoxy resin/graphite fiber): density = 1.7 gm/cc, Young’s modulus = 150-300 GPa, Tensile strength = 780-1850 MPa), specific strength = 460-1090, all of which better than conventional Al alloys. BUT, not suitable for aircraft with speed of Mach 2 or higher. (2)Carbon/carbon composites have properties better than above and is temperature resistant in non-oxidizing atmospheres (up to 2000 Celsius), but very expensive (3)New Al alloys that are lighter and stronger than existing Al alloys.

9. Development of Al-Li alloys Lithium (Li) is the lightest metal element (density a little more than half that of water!). If we could use Li to replace some Al atoms substitutionally, the new Al alloy will become lighter [note: remember substitutional versus interstitial impurity? do not want interstitial as this will increase the weight] The atomic size difference between host atom (solvent) and solute is an important factor in determining whether the solid solution is substitutional or interstitial For interstitial solid solution to form, the atomic diameter of the interstitial atom must be substantially smaller than that of host atoms. For example, Carbon (1.42 Å in diameter) forms interstitial solid solution in iron (2.48 Å ). The atomic size factor,  =  d i /d = (2.48-1.42)/2.48 = 43 %. For substititional solid solutions to form with appreciable solute concentrations, it requires that  < ± 15 % Li atom diameter ~ 3 Å, Al atom diameter ~ 2.8 Å   = (3-2.8)/2.8 = 7 %. Thus, addition of Li to Al will create a substitutional solid solution, which means we can reduce the density of Al alloys by adding Li element. Second major advantage of adding Li: Al-Li alloys are stronger than conventional Al alloys due to precipitation hardening Lithium in Aluminum is just like salt in water We find that there is solubility limit of Li in Al, which changes with temperature, similar to sugar or salt in water Once Li concentration exceeds solubility limit, precipitates with a composition of Al 3 Li will form. Thus we have opportunities to strengthen Al-Li alloys via precipitation hardening

10. Point Defects (0-dimensional) An avenue for atomic motion within the lattice, in response to an external mechanical or electrical load In stainless steel, carbon, which makes it a steel, is an interstitial impurity in the iron lattice (and chromium, which makes it stainless, is a substitutional impurity) In semiconductors, substitutional impurities are called dopants, and control the amount of charge carriers Intrinsic (vacancies) Extrinsic (interstitial and substitutional impurity atoms) Alter the mechanical properties (by affecting slip and dislocation motion), electronic properties (doping in semiconductors), etc.

11. Al-Li phase diagram Soluble region Precipitation region Solubility limit

12. Al 3 Li precipitates Precipitates have fcc structure with Li atoms at corners of cube, and Al atoms at face centers Two-phase system, with host having lower concentration of Li in substitutional positions in Al fcc lattice, and precipitates also of fcc type but with a higher Li concentration randomly oriented wrt host To maximize precipitation strengthening effects, we want: (1) uniform distribution of precipitates, and (2) large number of precipitates [typically, 20 nm diameter precipitates with average spacing of 40 nm. Thus a dislocation has little chance of moving more than 50 nm before it encounters an obstacle! [Note: 1 nm = 10 Å = 10 -9 m] How do we achieve this? (1) heat (so that Li dissolves in Al), (2) fast cooling (so that precipitate nuclei do not have time to coalesce and grow; i.e., large number of small precipitates rather than small number of large ones) Bottom line: Two-phase system, with homogeneous host having Li in substitutional positions in Al fcc lattice (like salt solution), and precipitates randomly oriented wrt host (like salt precipitates, but not at the bottom of beaker as we have a solid solution)

13. Verify results Impact on properties: Al-Li alloys are 10% lighter, 9% stiffer, 8% stronger for yield strength, and 4% stronger for tensile strength in comparison with conventional Al alloys Al-Li alloys have been produced by Alcoa and used in the vertical stabilizer and tailplanes of Boeing 777 and Airbus A330/340. They result in 650 pounds saving in weight at the additional expense of less than $150,000. If 650 pounds translate into 3 passengers and assuming the average ticket price of $250 and two flights a day, in 100 days $150,000 will be paid off!

14. Case Study II: HSLA steels for car bodies (grain refinement) Steel is the traditional material for car bodies; although there is an incentive to decrease the weight by using aluminum and Al-Li alloys, this is not cost effective Alternative: strengthen steels by small amount amounts of impurities intentionally added (low alloy) HSLA steels: small amounts of niobium added to steel produces niobium carbide precipitates, which result in small grain sizes, and increased strength In Al-Li alloys, Al3Li precipitates “pin” dislocations; in HSLA steels niobium carbide precipitates “pin” grain boundaries, thereby preventing the grains from becoming big

15. Grain refining Grain boundaries are barriers to dislocation movement because: dislocations have to change their directions of motion since the neighboring grain has different orientation the atomic disorder within a grain boundary region will result in a discontinuity of slip planes from one grain to the other A dislocation stops when it reaches a grain boundary; the next dislocation stops behind the first, and so on, creating a “pile-up” or traffic jam of dislocations Smaller the grain size, less distance does the dislocations move (and smaller number of dislocations exerting force on the grain boundary), and so stronger is the material Slip plane Dislocation pileup

16. Crystal structures of iron (or steel) At room temperature steel exists in a body centered cubic (BCC) arrangement When heated above 912 C, BCC  FCC, and reverses when cooled below 912 C BCC FCC

17. Creation of fine grains Heat above 912 C  FCC crystals & grains form “Hot-roll” sample  pancaked grains (Figure in page 122) Grain boundaries held in place by niobium carbide particles Cool below 900 C  BCC crystallites “nucleate” at grain boundaries (just like ice from water, as in page 123) BCC crystallites grow till they bump into each other, finally resulting in fine grained steel

18. Result Ultimate tensile strength (UTS), or just strength, increases by 85 %, yield strength increases by 190 %, no change in Young’s modulus (or stiffness); dislocations can still move somewhat (unlike in Al-Li alloys), but we have stopped the dislocations from moving the grain boundary or the next grain Made possible by 2 important properties: –Phase transformation from BCC to FCC at ~ 900 C –Small NbC precipitates formed at grain boundaries prevent boundaries from moving

19. Summary Strengthening strategy: eliminate or stop dislocations –Work hardening –Precipitation hardening –Grain refinement Overview of engineering methods Case study: Al-Li alloys (precipitation hardening) Case study: High strength low alloy (HSLA) steels (grain refining) Feb. 3, Feb. 8: Metal casting lab tour (Room 15, IMS basement) Feb. 10: Quiz 1 Reading assignment: Ch. 1-13