Topic/content. By the end of this course : Why study? MATERIALS ENGINEERING SME 3623

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1 MATERIALS ENGINEERING SME 3623 WWII Liberty Ships fracture into two halves Dr. Norhayati Ahmad Department of Materials Engineering Faculty of Mechanical Engineering, Universiti Teknologi Malaysia. Hp: Off: Room : C Topic/content 1. Introduction 2. Metal fracture 3. Metal creep 4. Metal fatigue 5. Metal wear 6. Corrosion 7. Polymer 8. Ceramic 9. Composite 10. Materials selection and case studies Why study? An engineer ~ will be exposed to a design problem involving materials properties required deterioration during service cost Knowledge needed materials characteristics Structure property relationship Processing techniques By the end of this course : Able to : 1) explain, analyse and differentiate the failure mechanisms (fracture, creep, fatigue, corrosion) of materials 2) Apply the theory of fracture mechanics in failure analysis 3) Relate structure, properties and processing of non metallic materials (polymer, ceramic, composite)

2 References: Callister W.D., Materials Science and Engineering An introduction, 7 th edition, Wiley, Smith W.F., Foundation of Materials Science and Engineering, 4 th edition, McGraw Hill, Fontana M.G., Corrosion Engineering, 3 rd edition, McGraw Hill, Assessment 2x Test = 40% Assignment = 20% Final Exam = 40% Attendance : 80% Dieter G.E., Mechanical Metallurgy, 3 rd edition, REVIEW OF MATERIALS PROPERTIES Properties are the way the material responds to the environment and external forces Mechanical Properties : response to mechanical forces, such as Strength Toughness Hardness Ductility Elasticity Fatigue, Creep etc Physical Properties Density, melting point,, etc Electrical & MagneticProperties e.g conductivity Thermal Properties e.g thermal expansion Optical Properties absorption, transmission and scattering of light Chemical Properties corrosion resistance such as oxidation, corrosion, materials composition -Strength : Ability to support load tension, compression,shear -Hardness : Resistance to penetration/ scratches -Toughness : Ability to resist impact force. -Ductility: Ability to change shape. Opposed to brittleness

3 Type of loading Common States of Stress Simple tension: cable F F Ao = cross sectional area (when unloaded) σ = F Ao σ σ Ski lift (photo courtesy Torsion (a form of shear): drive shaft P.M. Anderson) Ac M M Fs Ao τ τ = F s Ao 2R Note: τ = M/AcR here. 10 OTHER COMMON STRESS STATES (1) Stress-Strain Testing Simple compression: Typical tensile test machine Typical tensile specimen Ao extensometer specimen Adapted from Fig. 6.2, Canyon Bridge, Los Alamos, NM (photo courtesy P.M. Anderson) Balanced Rock, Arches National Park (photo courtesy P.M. Anderson) σ = F Ao Note: compressive structure member (σ < 0 here). gauge length 11 Adapted from Fig. 6.3, (Fig. 6.3 is taken from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, p. 2, John Wiley and Sons, New York, 1965.) 12

4 Linear Elastic Properties Young s Moduli: Comparison Modulus of Elasticity, E: (also known as Young's modulus) Hooke's Law: σ = E ε σ Linearelastic E ε F F simple tension test E(GPa) 10 9 Pa Metals Alloys Tungsten Molybdenum Steel, Ni Tantalum Platinum Cu alloys Zinc, Ti Silver, Gold Aluminum Magnesium, Tin Graphite Ceramics Polymers Composites /fibers Semicond Diamond Si carbide Al oxide Si nitride <111> Si crystal <100> Glass -soda Concrete Graphite Polyester PET PS PC PP HDPE PTFE Carbon fibers only CFRE( fibers)* Aramidfibers only AFRE( fibers)* Glass fibers only GFRE( fibers)* GFRE* CFRE* GFRE( fibers)* CFRE( fibers) * AFRE( fibers) * Epoxy only Wood( grain) Based on data in Table B2, Composite data based on reinforced epoxy with 60 vol% of aligned carbon (CFRE), aramid (AFRE), or glass (GFRE) fibers LDPE 14 TS σ y engineering stress Tensile Strength, TS Maximum stress on engineering stress-strain curve. Typical response of a metal strain engineering strain Metals: occurs when noticeable necking starts. Polymers: occurs when polymer backbone chains are aligned and about to break. Adapted from Fig. 6.11, F = fracture or ultimate strength Neck acts as stress concentrator 15 Tensile strength, TS (MPa) Tensile Strength : Comparison Metals/ Alloys Steel (4140) qt W (pure) Ti (5Al-2.5Sn) a Steel (4140) a Cu (71500) cw Cu (71500) hr Steel (1020) Al (6061) ag Ti (pure) a Ta (pure) Al (6061) a Graphite/ Ceramics/ Semicond Diamond Si nitride Al oxide Si crystal <100> Glass-soda Concrete Graphite Polymers Nylon 6,6 PC PET PVC PP LDPE HDPE Composites/ fibers C fibers Aramid fib E-glass fib AFRE( fiber) GFRE( fiber) CFRE( fiber) wood( fiber) GFRE( fiber) CFRE( fiber) AFRE( fiber) wood ( fiber) Room Temp. values Based on data in Table B4, a = annealed hr = hot rolled ag = aged cd = cold drawn cw = cold worked qt = quenched & tempered AFRE, GFRE, & CFRE = aramid, glass, & carbon fiber-reinforced epoxy composites, with 60 vol% fibers. 16

5 Plastic tensile strain at failure: Engineering tensile stress, σ Adapted from Fig. 6.13, Ductility smaller %EL larger %EL % EL L L = f o x 100 L o L o A o A f L f Energy to break a unit volume of material Approximate by the area under the stress-strain curve. Engineering tensile stress, σ Adapted from Fig. 6.13, Toughness small toughness (ceramics) large toughness (metals) very small toughness (unreinforced polymers) Another ductility measure: Engineering tensile strain, ε A A %RA = A o - o f x 100 Engineering tensile strain, Brittle fracture: elastic energy Ductile fracture: elastic + plastic energy ε Hardness Resistance to permanently indenting the surface. Large hardness means: --resistance to plastic deformation or cracking in compression. --better wear properties. Table 6.5 Hardness: Measurement e.g., 10 mm sphere apply known force measure size of indent after removing load D d Smaller indents mean larger hardness. most plastics brasses Al alloys easy to machine steels file hard cutting tools nitrided steels diamond increasing hardness 19 20

6 True Stress & Strain Note: S.A. (cross-sectional area) changes when sample stretched True stress σt = F A i σ = σ 1+ ε True Strain εt = ln λi ( λ ) o ε T T = ln 1 ( ) ( + ε) Adapted from Fig. 6.16, 21 Design or Safety Factors Design uncertainties mean we do not push the limit. Factor of safety, N Often N is σy between σworking = 1.2 and 4 N Example: Calculate a diameter, d, to ensure that yield does not occur in the 1045 carbon steel rod below. Use a factor of safety of 5. d σ 220, 000N π ( d 2 / 4) working σy = N 5 d = m = 6.7 cm 1045 plain carbon steel: σ y = 310 MPa TS = 565 MPa F = 220,000N L o 22 Dislocation Motion Strength is linked to dislocation mobility If dislocation mobility is easy, low forces will lead to easy movement Strengthening Mechanism of Metals Dislocation motion is analogous to the locomotion of a caterpillar. Caterpillar moves by repeated lifting and shifting of leg pairs. 24

7 Dislocations & plastic deformation Cubic & hexagonal metals - plastic deformation by plastic shear or slip where one plane of atoms slides over adjacent plane by defect motion (dislocations). Dislocation moves along slip plane in slip direction perpendicular to dislocation line Slip direction same direction as Burgers vector Edge dislocation Adapted from Fig. 7.2, Screw dislocation If dislocations don't move, deformation doesn't occur! Adapted from Fig. 7.1, Deformation Mechanisms Slip System Slip plane - plane allowing easiest slippage Wide interplanar spacings - highest planar densities Slip direction - direction of movement - Highest linear densities Adapted from Fig. 7.6, Slip System Deformation Mechanisms FCC Slip occurs on {111} planes (close-packed) in <110> directions (close-packed) => total of 12 slip systems in FCC in BCC & HCP other slip systems occur 27 28

8 Obstacles to dislocation motion Plastic deformation is due to the motion of a large number of dislocations. The motion is called slip. Thus, the strength (resistance to deformation) can be improved by putting obstacles to slip. The number of dislocations per unit volume is the dislocation density, in a plane they are measured per unit area. Solid solution Substitutional interstitial Grain boundaries Precipitation strengthening Cold work/ strain hardening Strategies for Strengthening 1. Solid Solutions 1. Solid Solutions 2. Reduce Grain Size 3. Precipitation Strengthening/ precipitation hardening 4. Cold Work (%CW) / strain hardening Impurity atoms distort the lattice & generate stress. Stress can produce a barrier to dislocation motion. Smaller substitutional Larger substitutional impurity impurity A C B D Impurity generates local stress at A and B that opposes dislocation motion to the right. Impurity generates local stress at C and D that opposes dislocation motion to the right. 32

9 Strengthening in Copper Tensile strength & yield strength increase with wt% Ni. Tensile strength (MPa) wt.% Ni, (Concentration C) Empirical relation: Alloying increases σy and TS. σ ~ C y Yield strength (MPa) wt.%ni, (Concentration C) 1/ 2 Adapted from Fig (a) and (b), 2. Reduce Grain Size Materials with finer grain size are stronger than materials with coarse grains Grain boundaries are barriers to slip. Barrier "strength" increases with Increasing angle of misorientation. Smaller grain size: more barriers to slip. Hall-Petch Equation: σ Adapted from Fig. 7.14, (Fig is from A Textbook of Materials Technology, by Van Vlack, Pearson Education, Inc., Upper Saddle River, NJ.) yield = σ + k d o 1/ 2 y Strengthening by Alloying small impurities tend to concentrate at dislocations reduce mobility of dislocation increase strength 3. Precipitation Strengthening/ Precipitation Hardening Hard precipitates are difficult to shear. Ex: Ceramics in metals (SiC in Iron or Aluminum). Side View Top View precipitate Unslipped part of slip plane S Slipped part of slip plane Large shear stress needed to move dislocation toward precipitate and shear it. Dislocation advances but precipitates act as pinning sites with spacing S. Adapted from Fig. 7.17, Result: σ y ~ 1 S 35 36

10 The particles can be precipitates, which are natural. They can also be things like dispersed oxide or carbide particles which are not natural. Particle hardening is generally a more better way to strengthen a materials than solid solution hardening. Precipitates and dispersed phases are usually more effective barriers to dislocation penetration than single solutes. Age hardening/precipitation hardening 1. solution treatment Reheat the alloy up to a temperature where only one solid phase exists (above the solvus). Don t exceed the eutectic temperature. 2. Quench Rapidly cool to room temperature or below. This result in a supersaturated nonequilibrium structure. The second phase does not form, because diffusion is so slow. 3. Aging Reheat o a temperature Diffusion a short distance Result in a fine precipitate There is an optimum aging time Al-Cu 4. Cold Work (%CW) / strain hardening Strengthening by increase of dislcation density. Room temperature deformation. Ductile metals become stronger when they are deformed plastically at temperatures well below the melting point. Common forming operations change the cross sectional area: -Forging force -Rolling roll die Ad Ao Ao blank Ad Adapted from Fig. roll 11.8, force -Extrusion -Drawing Ao container die holder Ao die die Ad tensile force force A % o A CW = d x 100 Ao ram billet container extrusion die 39 Ad Ti alloy after cold working: 0.9 µm Dislocations During Cold Work Dislocations entangle with one another during cold work. Dislocation motion becomes more difficult. Adapted from Fig. 4.6, (Fig. 4.6 is courtesy of M.R. Plichta, Michigan Technological University.) 40

11 Impact of Cold Work As cold work is increased Yield strength (σ y ) increases. Tensile strength (TS) increases. Ductility (%EL or %AR) decreases. The purposes of strain hardening To enhance strength Reduce ductility Shape products Adapted from Fig. 7.20, The effect of strain hardening can be removed by annealing heat treatment - during annealing three stages take place: Revovery Recrystallization Grain growth 41 Recovery Annihilation reduces dislocation density. Scenario 1 Results from diffusion Scenario 2 extra half-plane of atoms 3. Climbed disl. can now move on new slip plane 2. grey atoms leave by vacancy diffusion allowing disl. to climb 1. dislocation blocked; can t move to the right atoms diffuse to regions of tension extra half-plane of atoms Dislocations annihilate and form a perfect atomic plane. 4. opposite dislocations meet and annihilate Obstacle dislocation τ R New grains are formed that: -- have a small dislocation density -- are small -- consume cold-worked grains. 33% cold worked brass Recrystallization 0.6 mm 0.6 mm New crystals nucleate after 3 sec. at 580 C. Adapted from Fig (a),(b), (Fig (a),(b) are courtesy of J.E. Burke, General Electric Company.) 43 44

12 Further Recrystallization All cold-worked grains are consumed. 0.6 mm 0.6 mm At longer times, larger grains consume smaller ones. Grain boundary area (and therefore energy) is reduced. 0.6 mm 0.6 mm Grain Growth After 4 seconds After 8 seconds Adapted from Fig (c),(d), (Fig (c),(d) are courtesy of J.E. Burke, General Electric Company.) 45 After 8 s, 580ºC Empirical Relation: exponent typ. ~ 2 grain diam. at time t. d After 15 min, 580ºC n d n o = Kt Adapted from Fig (d),(e), (Fig (d),(e) are courtesy of J.E. Burke, General Electric Company.) coefficient dependent on material and T. elapsed time Ostwald Ripening 46 T R º T R = recrystallization temperature Adapted from Fig. 7.22, Recrystallization Temperature, T R T R = recrystallization temperature = point of highest rate of property change 1. T m => T R T m (K) 2. Due to diffusion annealing time T R = f(t) shorter annealing time => higher T R 3. Higher %CW => lower T R strain hardening 4. Pure metals lower T R due to dislocation movements Easier to move in pure metals => lower T R º 47 48

13 Summary Dislocations are observed primarily in metals and alloys. Strength is increased by making dislocation motion difficult. Particular ways to increase strength are to: -- solid solution strengthening -- decrease grain size -- precipitate strengthening -- cold work Heating (annealing) can reduce dislocation density and increase grain size. This decreases the strength. Failure Criteria Review 49 Metal Fracture A: Very ductile. Soft metals (e.g Pb, Au) at room temp. Other metals,polymers, glasses at high temp. B: Mederately ductile fracture, typical for ductile metals C: Brittle fracture, cold metals, ceramics. %AR or %EL Large Moderate Small Ductile: warning before fracture Brittle: No warning Ductile fracture (a) Necking (b) Cavity Formation (c) Cavity coalescence to form a crack (d) Crack Propagation (e) Fracture

14 Creep Creep is a time-dependent and permanent deformation of materials when subjected to a constant load at a high temperature (>0.4Tm), Example : turbine blades, steam generators. t r = time to rupture or rupture lifetime If a material is kept under a constant load over a long period of time (for example, carry a load permanently), it undergoes permanent deformation. Creep rate increase with increase in temp. Creep rate increase with temp and stress Fatigue Components (e.g tools, dies, gears, cam shaft, springs.etc) failure because of rapidly fluctuating (cyclic or periodic) loads in addition to static loads Fatigue testing apparatus Cyclic stress may be caused by fluctuating mechanical loads (such as in gear teeth or thermal stress (such as on tool, die..) Parts fails at stress level below that at which failure would occur under static loading S-N curve for Ferrous Metal and S-N curve for non Ferrous Metal

15 Fatigue :Failure under cyclic stress. Beachmarks may represent an 8hr daily shift : For a shaft operating at 3000 rpm, total number of cycles per day is. Cracks that cause fatigue failure almost always initiate/nucleate at component surface at some stress Concentration (scratches, dents, fillets, keyways, threads, weld beads/spatter..) On very smooth surfaces, SLIP steps can act as stress raisers. Beach marks DO NOT indicate the crack growth per stress cycle Corrosion Destructive of a material due to electrochemical attack from the environment.