Failure of Materials PDF

Failure of Materials How Materials or Components Can Fail 1 Failure of Materials Components must be strong enough to withstand (survive) the stresses acting on them – otherwise they will fail!! Most componen...

Failure of Materials How Materials or Components Can Fail 1 Failure of Materials Components must be strong enough to withstand (survive) the stresses acting on them – otherwise they will fail!! Most components will be OK provided the applied stress is kept below their yield strength However, materials / components often fail in real-life at stresses below the yield stress This is because of 3 main reasons: 2 1  Failure Ductile vs. brittle fracture Principles of fracture mechanics Stress concentration Impact fracture testing Fatigue (cyclic stresses) Cyclic stresses, the S—N curve Crack initiation and propagation Factors that affect fatigue behaviour Creep (time dependent deformation) Stress and temperature effects Alloys for high-temperature use 3 Fracture: separation of a body into pieces due to stress, at temperatures below the melting point. Steps in fracture: crack formation crack propagation 4 2  Brittle vs. Ductile Fracture Ductile materials - extensive plastic deformation and energy absorption (“toughness”) before fracture Brittle materials - little plastic deformation and low energy absorption before fracture 5 This is because of 3 main reasons: Stress Raisers (tiny cracks, flaws, inclusions) Fatigue (cyclic loads) Creep (relatively high temperatures) 6 3  Fracture mechanisms Ductile fracture Accompanied by significant plastic deformation Brittle fracture – Little or no plastic deformation – Catastrophic 7 Ductile vs Brittle Failure Classification: Fracture Very Moderately Brittle behavior: Ductile Ductile Adapted from Fig. 8.1, Callister & Rethwisch 8e. %AR or %EL Large Moderate Small Ductile fracture is Ductile: Brittle: usually more desirable Warning before No than brittle fracture! fracture warning 8 4  Example: Pipe Failures Ductile failure: -- one piece -- large deformation Brittle failure: -- many pieces -- small deformations Figures from V.J. Colangelo and F.A. Heiser, Analysis of Metallurgical Failures (2nd ed.), Fig. 4.1(a) and (b), p. 66 John Wiley and Sons, Inc., 1987. Used with permission. 9 Moderately Ductile Failure Failure Stages: void void growth shearing necking fracture nucleation and coalescence at surface s Resulting 50 50mm mm fracture surfaces (steel) 100 mm particles From V.J. Colangelo and F.A. Heiser, Fracture surface of tire cord wire serve as void Analysis of Metallurgical Failures (2nd loaded in tension. Courtesy of F. ed.), Fig. 11.28, p. 294, John Wiley and Roehrig, CC Technologies, Dublin, nucleation Sons, Inc., 1987. (Orig. source: P. OH. Used with permission. sites. Thornton, J. Mater. Sci., Vol. 6, 1971, pp. 347-56.) 10 5  Moderately Ductile vs. Brittle Failure cup-and-cone fracture brittle fracture Adapted from Fig. 8.3, Callister & Rethwisch 8e. 11 12 6  Ductile Fracture This type of failure is the preferred one for engineers and designers because it warns us that failure is going to occur The material changes shape as it fails – it necks and elongates There is a lot of plastic deformation A ductile fracture surface is uneven, dull and fibrous 13 BrittleArrows Failure indicate point at which failure originated Adapted from Fig. 8.5(a), Callister & Rethwisch 8e. 14 7  Brittle Fracture Surfaces Intergranular Transgranular (between grains) 304 S. Steel (through grains) (metal) 316 S. Steel Reprinted w/permission (metal) from "Metals Handbook", Reprinted w/ permission 9th ed, Fig. 633, p. 650. from "Metals Handbook", Copyright 1985, ASM 9th ed, Fig. 650, p. 357. International, Materials Copyright 1985, ASM Park, OH. (Micrograph by International, Materials J.R. Keiser and A.R. Park, OH. (Micrograph by Olsen, Oak Ridge D.R. Diercks, Argonne National Lab.) 160 mm 4 mm National Lab.) Polypropylene Al Oxide (polymer) (ceramic) Reprinted w/ permission Reprinted w/ permission from R.W. Hertzberg, from "Failure Analysis of "Defor-mation and Brittle Materials", p. 78. Fracture Mechanics of Copyright 1990, The Engineering Materials", American Ceramic (4th ed.) Fig. 7.35(d), p. Society, Westerville, OH. 303, John Wiley and (Micrograph by R.M. Sons, Inc., 1996. Gruver and H. Kirchner.) 3 mm 1 mm (Orig. source: K. Friedrick, Fracture 1977, Vol. 3, ICF4, Waterloo, CA, 1977, p. 1119.) 15 Brittle Fracture This type of failure is NOT the preferred one for engineers and designers because it happens suddenly with no warning The material does not change shape as it fails – there is no necking and no elongation There is little or no plastic deformation A brittle fracture surface is flat, shiny and crystalline 16 8  Low Stress Failures Stress Raisers (tiny cracks, flaws, inclusions) Fatigue (cyclic loads) Creep (relatively high temperatures) 17 Stress Raisers These are present in nearly all components and are due to design features (like sharp corners) or material defects (like scratches) that intensify the stress at these locations Take about 300 mm (~1ft) of ordinary adhesive tape, pull it in two - we find it is quite strong! Now, if we repeat the experiment, but we make a small cut in one of the sides……… We find that the tape breaks very easy now - because the small cut acts as a stress raiser and all stresses concentrate at the (opening) crack tip 18 9  How Much is the Stress Raised? We can easily calculate how much the stress is amplified from the following equation: smax = 2so√(a/rt) where: smax is max. stress caused by flaw; so is the applied stress ‘a’ is ½ the length of an internal flaw or the depth of a surface flaw rt is the flaw tip radius of curvature 19 Ideal vs Real Materials Stress-strain behavior (Room Temp ): s perfect mat’l-no flaws E/10 TSengineering sc sc =    a  where – E = modulus of elasticity – s = specific surface energy – a = one half length of internal crack For ductile materials => replace s with s + p where p is plastic deformation energy 25 25 Fracture Toughness Ranges Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 100 C-C(|| fibers) 1 70 Steels 60 Ti alloys 50 40 Al alloys 30 Mg alloys Based on data in Table B.5, K Ic (MPa · m 0.5 ) Callister & Rethwisch 8e. 20 Composite reinforcement geometry is: f Al/Al oxide(sf) 2 = fibers; sf = short fibers; w = whiskers; Y2 O 3 /ZrO 2 (p) 4 p = particles. Addition data as noted 10 C/C( fibers) 1 (vol. fraction of reinforcement): Al oxid/SiC(w) 3 1. (55vol%) ASM Handbook, Vol. 21, ASM Int., Diamond Si nitr/SiC(w) 5 Materials Park, OH (2001) p. 606. 7 Al oxid/ZrO 2 (p) 4 2. (55 vol%) Courtesy J. Cornie, MMC, Inc., 6 Si carbide Glass/SiC(w) 6 Waltham, MA. 5 Al oxide PET 3. (30 vol%) P.F. Becher et al., Fracture 4 Si nitride Mechanics of Ceramics, Vol. 7, Plenum Press PP (1986). pp. 61-73. 3 PVC 4. Courtesy CoorsTek, Golden, CO. 5. (30 vol%) S.T. Buljan et al., "Development of 2 PC Ceramic Matrix Composites for Application in Technology for Advanced Engines Program", ORNL/Sub/85-22011/2, ORNL, 1992. 6. (20vol%) F.D. Gace et al., Ceram. Eng. Sci. Proc., Vol. 7 (1986) pp. 978-82. 1 Si crystal PS Glass 6 0.7 Glass -soda 0.6 Polyester Concrete 26 0.5 26 13  Design Against Crack Growth Crack growth condition: K ≥ Kc = Ys a Largest, most highly stressed cracks grow first! --Scenario 1: Max. flaw --Scenario 2: Design stress size dictates design stress. dictates max. flaw size. 2 sdesign  Kc 1  Kc  amax    Y amax   Ysdesign   amax s fracture fracture no no fracture amax fracture s 27 27 Design Example: Aircraft Wing Material has KIc = 26 MPa-m0.5 Two designs to consider... Design A Design B --largest flaw is 9 mm --use same material --failure stress = 112 MPa --largest flaw is 4 mm K Ic --failure stress = ? Use... sc = Y amax Key point: Y and KIc are the same for both designs. K Ic = s a = constant Y  --Result: 112 MPa 9 mm 4 mm (s c amax ) = (s A c amax ) B Answer: ( sc )B = 168 MPa 28 28 14  Worked Example An engine piston has a yield strength of 400MPa. Will it experience plastic deformation (i.e. fail) under a stress of 250 MPa if it has an internal crack 40 mm long and tip radius of curvature of 5 mm? smax = 2so√(a/rt) smax = 2 x 250 √(20/5) smax = 2 x 250 x 2 = 1000MPa So because the stress at the crack tip exceeds the yield strength of the material, it will plastically flow (i.e. fail). 29 Impact Testing Impact loading: (Charpy) -- severe testing case -- makes material more brittle -- decreases toughness Adapted from Fig. 8.12(b), Callister & Rethwisch 8e. (Fig. 8.12(b) is adapted from H.W. Hayden, W.G. Moffatt, and J. Wulff, The Structure and Properties of Materials, Vol. III, Mechanical Behavior, John Wiley and Sons, Inc. (1965) p. 13.) final height initial height 30 30 15  Influence of Temperature on Impact Energy Ductile-to-Brittle Transition Temperature (DBTT)... FCC metals (e.g., Cu, Ni) Impact Energy BCC metals (e.g., iron at T < 914ºC) polymers Brittle More Ductile High strength materials (s y > E/150) Temperature Adapted from Fig. 8.15, Callister & Rethwisch 8e. Ductile-to-brittle transition temperature 31 31 Design Strategy: Stay Above The DBTT! Pre-WWII: The Titanic WWII: Liberty ships Reprinted w/ permission from R.W. Hertzberg, Reprinted w/ permission from R.W. Hertzberg, "Deformation and Fracture Mechanics of Engineering "Deformation and Fracture Mechanics of Engineering Materials", (4th ed.) Fig. 7.1(a), p. 262, John Wiley and Materials", (4th ed.) Fig. 7.1(b), p. 262, John Wiley and Sons, Inc., 1996. (Orig. source: Dr. Robert D. Ballard, Sons, Inc., 1996. (Orig. source: Earl R. Parker, The Discovery of the Titanic.) "Behavior of Engineering Structures", Nat. Acad. Sci., Nat. Res. Council, John Wiley and Sons, Inc., NY, 1957.) Problem: Steels were used having DBTT’s just below room temperature. 32 32 16  Fracture Toughness (K1c) This is the ability of a material containing flaws to successfully withstand (survive) a stress If we try to bend / break a glass rod –it is strong enough to resist the stress However, with a small cut on the glass surface, it breaks easily – we say it has poor fracture toughness 33 Metals Have Good Fracture Toughness If we repeat the same experiment with a rod of copper, we get a different result The copper bends instead of breaks (even with a flaw present) – it has good fracture toughness 34 17  Stress Concentration Factor (Kt) Stress Concentration Factor is the ratio of the stress at the crack tip to the applied bulk stress: Kt = sm / so It is a measure of how much the stress is amplified by a defect 35 Modes of Fracture 36 18  Stress Intensity Factor (K1) As the name suggests, this is a measure of the intensity of stresses at the tip of a sharp crack It is similar to the stress concentration factor, Kt but the two are not the same because……… 37 Stress Intensity Factor (K1) The Stress Intensity Factor depends on the crack size, the crack shape, and the geometric boundaries of the specimen For a given flawed material, catastrophic failure occurs when the stress intensity factor reaches a critical value, denoted as K1c 38 19  Stress Intensity Factor (K1) This critical value of stress intensity is equal to a property of the material called fracture toughness (K1c) and is found from the equation: K1c = sY √(a) where s is the nominal applied stress, ‘a’ is half the length of an internal crack, and Y is a dimensionless correction factor that accounts for the geometry of the component containing the flaw 39 Stress Intensity Factor and Fracture Toughness The relationship between stress intensity, K1 and fracture toughness, K1c is similar to the relationship between tensile stress and tensile strength Both tensile strength and fracture toughness are properties of a material whereas tensile stress and stress intensity factor are what is being applied to the material or component 40 20  Some Fracture Toughness Values Steels 80-170 Nickel alloys 100-150 Copper alloys 30-120 Titanium alloys 50-100 Aluminium alloys 5-70 GFRP 20-60 CFRP 30-45 Alumina (Al2O3) 3-5 Polyethylene 1-5 Soda glass 0.7 Concrete 0.2-0.4 41 Leak Before Break A pressure cylinder is to be constructed from a steel with a yield strength of 1100 MPa and a fracture toughness of 110 MPa.m1/2. A safety factor of 2 on the yield strength is to be used. If the maximum working pressure is to be 15 MPa and given that the Hoop Stress is: s = [(p) x (r)]/t calculate the maximum wall thickness and minimum vessel diameter required if failure by leak-before-break is to occur. Assume a geometric constant of 1. 42 21  Solution sallowed = 1100 / 2 = 550 MPa K1c = s Y ( a) 2a = 25.4 mm Therefore the maximum wall thickness is 25.4 mm Now s = (p x r) /t so r = (s x t) / p r = (550x106)(25.4x10-3) / (15x106) r = 0.932 m d = 1.864 m For leak-before-break to occur, the maximum wall thickness is 25.4 mm and the min. diameter is 1.864 m 43 44 22