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)

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

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

• 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’sEmpty weight

28%Fuel for normal use

4% Fuel reserves

14.5%Passengers plus freight

7.5%Operator’s items

Identify & define problem/need

• 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

Conduct Analysis/ investigation: Existing materials

Materials Density

(gm/cc)

Young’s modulus

(GPa)

Tensile strength

(MPa)

Corrosion resistance

Temperature capapility (Celsius)

Conventional Al 2.8 76 530 good < 170

Polymer (e.g., polyester epoxy)

1.1-1.5 2.0-5.0 30-150 good < 100

Ceramics (Al2O3) 3.97 380 300-1000 excellent < 1200

Ceramics (Si3N4) 3.18 304 700-1000 excellent < 1400

Glass fibers 2.5 70-85 3500-4500 excellent < 100

Polymer fibers

(Kevlar)

1.4 60-120 2700-3600 good < 100

Steels (HSLA) 7.8 207 440-650 fair < 300-400

Ti Alloys

(Ti-6Al-4V)

4.4 114 990 good < 700

Cu Alloy

(Cu-30%Zn)

8.53 110 520 excellent < 300

Mg alloy

(Mg-9%Al-2%Zn)

1.8 45 230-275 fair < 180

Ni Alloy

(IN-100)

8.9 156 680-940 excellent < 1000

Graphite fiber 1.8-2.0 200-380 1550-3100 Excellent in reducing atmosphere

< 2000

Material Specific Strength (Tensile

strength/density, MPa.cc/gm)

Temperature capability (Celsius)

Conventional Al 189 < 170

Mg alloy 127-153 < 170

Ni alloy 76-105 < 1000

Ti Alloy 225 < 700

Steels 56-83 < 300-400

Cu alloy 61 < 300

polymer 27-136 < 100

Graphite fiber 775-1550

Glass fiber 1400-1800

Specific strength = Strength/density

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.

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, = di/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 Al3Li will form. Thus we have opportunities to strengthen Al-Li alloys via precipitation hardening

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.

Al-Li phase diagram

Solubleregion

Precipitationregion

Solubilitylimit

Al3Li 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)

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!

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

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

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

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

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

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