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? 1 How do loading rate, loading history, and temperature affect the failure stress? Ship-cyclic loading from waves. Computer chip-cyclic thermal loading. Hip implant-cyclic loading from walking. CHAPTER 9: MECHANICAL FAILURE

4 Evolution to failure: Resulting fracture surfaces (steel) 50  m particles serve as void nucleation sites. 50  m 100  m MODERATELY DUCTILE FAILURE

5 Intergranular (between grains) Intragranular (within grains) Al Oxide (ceramic) 316 S. Steel (metal) 304 S. Steel (metal) Polypropylene (polymer) 3m3m 4 mm 160  m 1 mm BRITTLE FRACTURE SURFACES

6 Stress-strain behavior (Room T): TS << TS engineering materials perfect materials DaVinci (500 yrs ago!) observed... --the longer the wire, the smaller the load to fail it. Reasons: --flaws cause premature failure. --Larger samples are more flawed! IDEAL VS REAL MATERIALS

7 Elliptical hole in a plate: Stress distrib. in front of a hole: Stress conc. factor: Large K t promotes failure: FLAWS ARE STRESS CONCENTRATORS!

8 Avoid sharp corners! ENGINEERING FRACTURE DESIGN

 t at a crack tip is very small! 9 Result: crack tip stress is very large. Crack propagates when: the tip stress is large enough to make: K ≥ K c WHEN DOES A CRACK PROPAGATE?

10 Condition for crack propagation: Values of K for some standard loads & geometries: K ≥ K c Stress Intensity Factor: --Depends on load & geometry. Fracture Toughness: --Depends on the material, temperature, environment, & rate of loading. GEOMETRY, LOAD, & MATERIAL

12 Crack growth condition: Largest, most stressed cracks grow first! --Result 1: Max flaw size dictates design stress. --Result 2: Design stress dictates max. flaw size. K ≥ K c DESIGN AGAINST CRACK GROWTH

13 Two designs to consider... Design A --largest flaw is 9 mm --failure stress = 112 MPa Design B --use same material --largest flaw is 4 mm --failure stress = ? Use... Key point: Y and K c are the same in both designs. --Result: 9 mm112 MPa 4 mm Answer: Reducing flaw size pays off! Material has K c = 26 MPa-m 0.5 DESIGN EX: AIRCRAFT WING

14 Increased loading rate... --increases  y and TS --decreases %EL Why? An increased rate gives less time for disl. to move past obstacles. Impact loading: --severe testing case --more brittle --smaller toughness LOADING RATE

15 Increasing temperature... --increases %EL and K c Ductile-to-brittle transition temperature (DBTT)... TEMPERATURE

16 Pre-WWII: The Titanic WWII: Liberty ships Problem: Used a type of steel with a DBTT ~ Room temp. DESIGN STRATEGY: STAY ABOVE THE DBTT!

17 Fatigue = failure under cyclic stress. Stress varies with time. --key parameters are S and  m Key points: Fatigue... --can cause part failure, even though  max <  c. --causes ~ 90% of mechanical engineering failures. FATIGUE

18 Fatigue limit, S fat : --no fatigue if S < S fat Sometimes, the fatigue limit is zero! FATIGUE DESIGN PARAMETERS

19 Crack grows incrementally typ. 1 to 6 increase in crack length per loading cycle Failed rotating shaft --crack grew even though K max < K c --crack grows faster if  increases crack gets longer loading freq. increases. crack origin FATIGUE MECHANISM

1. Impose a compressive surface stress (to suppress surface cracks from growing) 20 --Method 1: shot peening 2. Remove stress concentrators. --Method 2: carburizing IMPROVING FATIGUE LIFE

26 Engineering materials don't reach theoretical strength. Flaws produce stress concentrations that cause premature failure. Sharp corners produce large stress concentrations and premature failure. Failure type depends on T and stress: - for noncyclic  and T < 0.4T m, failure stress decreases with: increased maximum flaw size, decreased T, increased rate of loading. -for cyclic  : cycles to fail decreases as  increases. -for higher T (T > 0.4T m ): time to fail decreases as  or T increases. SUMMARY