The fracture of a Al bicycle crank arm.

Transcription

1 The frature of a Al biyle rank arm.

2 Failure In hapters 6 and 7 we examined the elasti and plasti behaviour of materials. We learned how the motion of disloations makes it possible for the material to deform. We know from our daily experienes that there is a limit on how muh we an deform a material. Materials eventually fail.

3 Another Mystery When a material splits in two, all the bonds that used to hold the two parts together have been broken. For years people were puzzled beause the fore/stress applied to the material is rarely large enough to break all the bonds. One again, we ask, what is missing? Ship: yli loading from waves. 3 From Fig. 9.0, Callister (original Neil Boenzi, The New York Times.)

4 Stress Conentration Measured frature strengths are less than theoretial values beause of mirosopi flaws whih always exist under normal onditions. An applied stress is amplified at the tip of a rak Magnitude of amplifiation is dependent on: Crak orientation Crak geometry Stress onentration fator (K t ): a measure of the degree to whih an external stress is amplified at the tip of a rak K t max 0 a t

5 Effet of Flaws When the max stress exeeds the yield strength in a dutile material, plasti deformation ours. In brittle materials, leavage/frature ours. Consider an elliptial hole in a plate: Magnitude of stress diminishes with distane from the rak tip 5

6 Demo: Influene of Noth Sharpness J. Hisoks 003 As you an see, both nothes have the same length (a), but different r t. What will happen if we pull the ard? 6

7 Demo: Influene of Noth Length What happens if we have two raks with the same r t but different lengths? J. Hisoks, 003 7

8 Is this Crak Dangerous?

9 Griffith Criterion: The stress at the tip of the rak is large enough to break the bonds in the solid. Whether the rak propagates or not, depends on a simple energy balane: 0 W U el G A work-done by the load if the rak grows Elasti energy released when the rak grows Energy needed to expand the rak (G is the surfae energy per unit area of rak).

10 Griffith Criterion: Example 1: The balloon Let s introdue a rak into the balloon. Inflate the balloon to a diameter of 10m and ask if W U el G A? Let s ignore this term beause d is fixed. If the balloon blows up, we d release all of the energy stored in the walls. Energy needed to expand the rak and ause the balloon to blow up.

11 Griffith Criterion : Example : Real rak with fixed displaement Let s imagine that the rak got bigger by a small amount δa. 0 As before, we ask the question, is W U el G A? a W=0 (displaement of the blok is zero). Elasti energy released when the rak grows: ( 1 1 ) V a at E Energy needed to expand the rak: G da G ta a 0

12 Griffith Criterion : Example : Real rak with fixed displaement The rak will grow if: 0 EG or EG a a a G K K : Toughness[J/m ] a EG : : stress intensity fator ritialstress intensity fator or fraturetoughness[n/m 3/ or Pa [N/m m] 3/ ] a 0 Griffith Criterion: K K

13 Frature Toughness Frature toughness (K ): a property that is a measure of a material s resistane to brittle frature when a rak is present when speimen thikness >>rak dimensions, K beomes independent of thikness. K Y a A unit-less parameter dependent on : Crak geometry Speimen size Manner of load appliation K I (plane strain frature toughness) dereases with inreasing strain rate and dereasing temperature Units: MPa m

14 inreasing Frature Toughness Data 14 Based on data in Table B5, Callister 6e. Composite reinforement geometry is: f = fibers; sf = short fibers; w = whiskers; p = partiles. Addition data as noted (vol. fration of reinforement): 1. (55vol%) ASM Handbook, Vol. 1, ASM Int., Materials Park, OH (001) p (55 vol%) Courtesy J. Cornie, MMC, In., Waltham, MA. 3. (30 vol%) P.F. Beher et al., Frature Mehanis of Ceramis, Vol. 7, Plenum Press (1986). pp Courtesy CoorsTek, Golden, CO. 5. (30 vol%) S.T. Buljan et al., "Development of Cerami Matrix Composites for Appliation in Tehnology for Advaned Engines Program", ORNL/Sub/85-011/, ORNL, (0vol%) F.D. Gae et al., Ceram. Eng. Si. Pro., Vol. 7 (1986) pp

15 How Large a Flaw is Critial? A typial steel struture has a frature toughness of about K I = 75 MPa-m 1/ If the struture is designed to withstand a stress of 750 MPa, how big do flaws in the material have to be to pose a threat? (Assume that Y = 1) Answer: 1 K IC a 0.003m 3mm designy 750 (100) 15 The largest tolerable flaws in many engineering strutures are of this dimension

16 Types of Frature Frature: rak formation and propagation in response to an imposed stress Frature Modes (lassified based on the ability of the material to experiene plasti deformation) Dutile (stable) Exhibit substantial plasti deformation large % RA and % elongation (often > 0%) High energy absorption Brittle (unstable) Little or no plasti deformation low % elongation (often 1%) Low energy absorption Dutility is a funtion of: Temperature Strain rate Stress rate Dutile Material Brittle Material

17 Dutile Frature Dutile frature is indued by plasti deformation. It s aused by damage aumulation. nuleation, growth, oalesene of voids After tensile instability starts, the damage is onentrated in the nek. Note the importane of partiles. Indiates plasti deformation (fibrous) 17

18 Frature Proess a) Initial Neking b) Mirovoids form in the interior of ross setion ) Mirovoids enlarge and oalese to form an elliptial rak (long axis perpendiular to stress diretion) d) Crak propagation e) Rapid propagation of rak around outer perimeter of nek shear frature at 45 relative to tensile diretion

19 Brittle Frature Ours by rapid rak propagation Frature surfaes will have their own distintive patterns; for example: Steel: hevron markings Lines or ridges originating from near the enter of the ross setion Amorphous materials: shiny, smooth surfae Cleavage: rak propagation orresponding to the suessive and repeated breaking of atomi bonds along speifi rystallographi planes Brittle transrystalline steel frature

20 Failure Origins What do you see on the mild steel speimen that suggests why it failed at this loation? Defets are the key to brittle frature 0 Fig. 9.3(b) Callister

21 An Example: Failure of a Pressurized Pipe Failure may be dutile speimen still a single piee extensive deformation brittle many piees little deformation 1 Figures from V.J. Colangelo and F.A. Heiser, Analysis of Metallurgial Failures (nd ed.), Fig. 4.1(a) and (b), p. 66 John Wiley and Sons, In., Used with permission.

22 Impat Frature Testing

23 Impat Testing Tehniques standardized tests used to measure impat energy (noth toughness): Charpy Izod The differene between the tests lies in the way the speimen is supported Izod Charpy

24 Dutile-to-Brittle Transition Ours when the temperature of a material drops Related to the temperature dependene of the measured impat energy absorption A83 Steel Photograph of frature surfaes of A36 steel Charpy V-noth speimens tested at the indiated temperatures

25 Dutile to Brittle Transition Many materials beome brittle at low temperatures. BCC materials (like steel) fail by leavage along the <100> plane Material exhibiting a dutile to brittle transition should only be used above the transition temperature so as to avoid brittle failure Impat energy vs. Temperature Plot 3 general behaviours Influene of arbon ontent on the Charpy energy-temperature behaviour for steel

26 Chapter 8 Review What did you learn? 6

27 Chapter 8 Pratie Problems

28 Pratie Problems 1. A thik glass baking pan is removed from the oven at 00 C and plaed in old water at 0 C. Upon being plaed in old water the pan breaks due to thermal shok. Estimate the minimum rak length present at the surfae of the pan given that the Elasti modulus of the glass is 100 GPa, the oeffiient of thermal expansion is 7 x 10-6 /K, the frature toughness is 5 MPa.sqrt(m) and Y =1. a) a min = 0.1 mm b) a min = 0.5 mm ) a min = 1 mm d) a min = mm

29 Pratie Problems Answer: 0.5 mmor : Now,we an solvefor 16MPa data: using the known First,solvefor design b) design l design a MPa m MPa Y K a a T E

30 Pratie Problems. A speimen with a surfae rak size of.0 mm fratures at a stress of 80 MPa. Estimate the frature toughness, K, assuming that Y=1. a) 6.3 MPa.sqrt(m) b) 1 MPa.sqrt(m) ) 3 MPa.sqrt(m) d) 0.3 MPa.sqrt(m)

31 Pratie Problems Answer: Note: Bearefulof units a K K 10 Y y 3 m a 180MPa 6.3MPa m or a) 10 3 m

32 Pratie Problems 3. A strutural omponent is fabriated from an alloy that has a plane strain frature toughness of 48.6 MPa-m 1/. It has been determined that this omponent fails at a stress of 178 MPa when the maximum length of a surfae rak is 1.08 mm. For a stress of 178 MPa, what is the maximum allowable surfae rak length without frature for this same omponent that is made from another alloy that has a plane strain frature toughness of 47.5 MPa-m 1/? a) 0.57 mm b) 1.03 mm ) 1.08 mm d).09 mm e) 3.1 mm

33 Pratie Problems Answer: 1.03mmor b) : Createtwo equations.solvefor a Using the design stress equation: IC IC IC IC IC a m MPa m m MPa K a K a a K a K a Y K