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Fracture

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Fracture

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  1. Fracture Materials Science and Manufacturing IENG 263

  2. Why Study Failure In order to know the reasons behind the occurrence of failure so that we can prevent failure of products by improving design in the light of failure reasons

  3. Mechanical Failure ISSUES TO ADDRESS... • How do flaws in a material initiate failure? • How is fracture resistance quantified; how do different material classes compare? • How do we estimate the stress to fracture? • How do loading rate, loading history, and temperature affect the failure stress? Computer chip-cyclic thermal loading. Hip implant-cyclic loading from walking. Ship-cyclic loading from waves.

  4. What is a Fracture? • Fracture is the separation of a body into two or more pieces in response to an imposed stress • The applied stress may be tensile, compressive, shear, or torsional • Stress can be caused by forces, temperature • Any fracture process involves two steps—crack formation and propagation—in response to an imposed stress.

  5. Fracture Modes • Ductile fracture • Occurs with plastic deformation • Material absorbs energy before fracture • Crack is called stable crack: plastic deformation occurs with crack growth. Also, increasing stress is required for crack propagation. • Brittle fracture • Little or no plastic deformation • Material absorb low energy before fracture • Crack is called unstable crack. • Catastrophic fracture (sudden)

  6. Very Moderately Fracture Brittle Ductile Ductile behavior: (%EL)=100% Large Moderate Small Ductile vs Brittle Failure • Classification: • Ductile fracture is usually desirable! Ductile: warning before fracture, as increasing force is required for crack growth Brittle: No warning

  7. Example: Failure of a Pipe • Brittle failure: --many pieces --small deformation • Ductile failure: --one/two piece(s) --large deformation

  8. Moderately Ductile Failure- Cup & Cone Fracture void growth shearing void necking fracture and linkage at surface nucleation s 50 mm 50 mm • Resulting fracture surfaces (steel) 100 mm particles serve as void nucleation sites. • Evolution to failure: crack occurs perpendicular to tensile force applied

  9. Ductile vs. Brittle Failure cup-and-cone fracture brittle fracture

  10. Transgranular vs Intergranular Fracture Intergranular Fracture Trans-granular Fracture Ductile Fracture Brittle Fracture

  11. Brittle Fracture Surfaces • Transgranular (within grains) • Intergranular (between grains) 304 S. Steel (metal) 316 S. Steel (metal) 160mm 4mm Polypropylene (polymer) Al Oxide (ceramic) 3mm 1mm

  12. Stress Concentration- Stress Raisers σm›σo Suppose an internal flaw (crack) already exits in a material and it is assumed to have a shape like a elliptical hole: The maximum stress (σm) occurs at crack tip: where t = radius of curvature at crack tip so = applied stress sm = stress at crack tip Kt = Stress concentration factor t Theoretical fracture strength is higher than practical one; Why?

  13. Concentration of Stress at Crack Tip

  14. Crack Propagation Cracks propagate due to sharpness of crack tip • A plastic material deforms at the tip, “blunting” the crack. deformed region brittle Effect of stress raiser is more significant in brittle materials than in ductile materials. When σm exceeds σy , plastic deformation of metal in the region of crack occurs thus blunting crack. However, in brittle material, it does not happen. When σm›σy plastic

  15. When Does a Crack Propagate? Crack propagation in a brittle material occurs if Where • σc= Critical stress to propagate crack • E = modulus of elasticity • s= specific surface energy (J/m2) • a = one half length of internal crack For ductile => replace gs by gs + gp where gp is plastic deformation energy sm> sc

  16. Example – Brittle Fracture • Set c = 40Mpa • Solve Griffith Eqn for Edge-Crack Length • Given Glass Sheet with • Tensile Stress,  = 40 Mpa • E = 69 GPa •  = 0.3 J/m • Find Maximum Length of a Surface Flaw • Plan • Solving

  17. Fracture Toughness: Design Against Crack Growth Kc= --Result 2: Design stress dictates max. allowable flaw size. --Result 1: Max. flaw size dictates design stress (max allowable stress). amax fracture fracture no no amax fracture fracture • Crack growth condition: • Largest, most stressed cracks grow first! σc σc

  18. Design Example: Aircraft Wing • Use... --Result: 112 MPa 9 mm constant 4 mm Answer: • Material has KIc = 26 MPa-m0.5 • Two designs to consider... Design B --use same material --largest flaw is 4 mm --failure stress = ? Design A --largest flaw is 9 mm --failure stress = 112 MPa • Key point: Y and KIc are the same for both designs.

  19. SOLUTION Where Y = 1.12. Substitute values Design using fracture mechanics Example: Compare the critical flaw sizes in the following metals subjected to tensile stress 1500MPa and K = 1.12 a. KIc (MPa.m1/2) Al 250 Steel 50 Zirconia(ZrO2) 2 Toughened Zirconia 12 Critical flaw size (microns) 7000 280 0.45 16

  20. KIc= Fracture Toughness • For relatively thin specimens, the value of Kc will depend on specimen thickness. However, when specimen thickness is much greater than the crack dimensions, Kc becomes independent of thickness. • The Kc value for this thick-specimen situation is known as the plane strain fracture toughness KIC Crack Modes Plane strain condition: t/a ratio = large

  21. Fracture Toughness • Brittle materials do not undergo large plastic deformation, so they posses low KIC than ductile ones. • KIC increases with increase in temp and with reduction in grain size if other elements are held constant • KIC reduces with increase in strain rate

  22. Graphite/ Metals/ Composites/ Ceramics/ Polymers Alloys fibers Semicond 100 1 C-C (|| fibers) Steels 7 0 6 0 Ti alloys 5 0 4 0 Al alloys 3 0 ) Mg alloys 0.5 2 0 2 Al/Al oxide(sf) 4 Y O /ZrO (p) 2 3 2 1 C/C ( fibers) 10 3 Al oxid/SiC(w) (MPa · m 5 Si nitr/SiC(w) Diamond 7 4 Al oxid/ZrO (p) 2 6 Si carbide 6 Glass/SiC(w) 5 PET Al oxide Si nitride 4 PP Ic 3 PVC K 2 PC 1 <100> 6 Glass PS Si crystal <111> 0.7 Glass - soda Polyester 0.6 Concrete 0.5 Fracture Toughness KIc=

  23. Design Example: Aircraft Wing • Use... 9 mm --Result: 112 MPa 4 mm Answer: • Material has Kc = 26 MPa-m0.5 • Two designs to consider... Design B --use same material --largest flaw is 4 mm --failure stress = ? Design A --largest flaw is 9 mm --failure stress = 112 MPa • Key point: Y and Kc are the same in both designs. • Reducing flaw size pays off!

  24. TS sy larger e TS smaller sy e e Loading Rate • Increased loading rate... -- increases sy and TS -- decreases %EL • Why? An increased rate gives less time for dislocations to move past obstacles and form into a crack. s

  25. Numerical Problems • Problems 8.1 – 8.10; 8.14 – 8.23; and 8.27