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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.1 CHAPTER 8
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.2 FRAC TURE The Fundamentals Fracture = separation of body into two or more pieces due to application of static stress Tensile, Compressive Shear or torsional Lets talk about the tensile loading of materials Modes of fracture DUCTILEBRITTLE
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.3 Back to the The tensile test
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.4 Ductile fracture in copper nucleating around inclusions
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.5 Cup and Cone fractureBrittle fracture
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.6 Transgranular vs. intergranular fracture (cleavage fracture)
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.7 y x Stress trajectories Professor Inglis (1913) The birth of the term ‘’stress concentration’’ Large structures
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.8 Crack-free silica (glass) fiber ~ 100 microns in diameter Bent to a strain of 7.5% i.e. 5000MPa. Normal strength of glass = 100-200MPa. Griffith (1920) – application of Inglis to cracks and defects HE PROVED HIS POINT!!
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.9 But radius of curvature = t = b 2 /a For circular hole Stress concentration factor b= a = r K t = 1 + 2 (1/1) 1/2 ) = 3 2b Ellipse
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.10 Stress concentration factor vs. Specimen Geometry/configuration So what happens if a crack intersects a hole?
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.11 Griffith and his Energy criterion Crack propagates when favorable,i.e. system reduces its total energy Relaxed material behind crack = Elastic strain energy released Crack having surface energy ( s ) a
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.12 What about ductile materials But for v. ductile materials p >>> s Define the strain energy release rate G c (IRWIN) Hence
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.13 Modes of fracture
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.14 Stress intensity factor AND =
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.15 Plastic zone What about ductile materials consider y (i.e. y means direction not yield)
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.16
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.17
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.18
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.19 To be plane strain Plane strain fracture toughness
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.20
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.21 Design using fracture mechanics Example: Compare the critical flaw sizes in the following metals subjected to tensile stress 1500MPa and K = 1.12 a. K Ic (MPa.m 1/2 ) Al 250 Steel 50 Zirconia(ZrO2) 2 Toughened Zirconia 12 Critical flaw size (microns) 7000 280 0.45 16 Where Y = 1.12. Substitute values SOLUTION
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.22
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.23
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.24 IMPACT TESTING Tensile test vs. real life failures Impact energy measured or notch toughness
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.25 Also HCP Not all materials exhibit DBTT How do we specify a ductile-brittle transition temperature (DBTT)??
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.26
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.27
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.28 FATIGUE Stress amplitude Range of stress Mean stress Failure under repeated cyclic loading Definitions For Reversed cycle fatigue R = -1
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.29 How do you practically make these fatigue measurements ?
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.30 Endurance limit or Fatigue limit vs. fatigue life e.g. Al e.g. steels & Ti Alloys e.g. 35-60% of TS High cycle vs. low cycle fatigue
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.31 STAGE II N f = N i + N p Initiation Propagation How does a fatigue crack form and propagate?
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.32 Striations Beachmarks
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.33
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.34
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.35 One of failure analysis goals = prediction of fatigue life of component knowing service constraint and conducting Lab tests Ignores crack initiation and fracture times
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.36 Can make extrapolations To obtain log A
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.37 QUESTION Eqn. 8.26
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.38 Other effects: a) Mean stress b)stress concentrations c) Surface treatments
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.39 Lead pipes deforming under their own weight CREEP = f (T, t, ) Time dependent and permanent deformation of materials when subjected to load or stress (significant at T = 0.4T m )
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.40
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.41 Effect of temperature and stress
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.42
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ME 240: Introduction to Engineering Materials Chapter 8. Failure 8.43 Data extrapolation methods – e.g. prolonged exposures (years) Perform creep tests in excess of T and shorter time but at same stress level
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