Materials Science and Manufacturing IENG 263 Fracture Materials Science and Manufacturing IENG 263

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

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? Mechanical failure is the change in the structure/material that refrains the specimen to perform its intended operation Flaws include micro-cracks, voids, stress concentrations, etc Ship-cyclic loading from waves. Computer chip-cyclic thermal loading. Hip implant-cyclic loading from walking.

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. (Static stress: constant or slowly changing in magnitude with time). the present discussion will be confined to fractures that result from uniaxial tensile loads

Fracture Modes Ductile fracture Brittle 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) 0. Classification is based on the ability of a material to experience plastic deformation. 1. Ductile materials typically exhibit substantial plastic deformation with high energy absorption before fracture

Ductile vs Brittle Failure • Classification: Very Ductile Moderately Brittle Fracture behavior: Large Moderate (%EL)=100% Small Very ductile: Gold, graphite at room temperature; Polymers, Cu at elevated temp Most of metal alloys, fracture is preceded by necking Ceramics • Ductile fracture is usually desirable! Ductile: warning before fracture, as increasing force is required for crack growth Brittle: No warning

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

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

Ductile vs. Brittle Failure 2. Brittle fracture takes place without any appreciable deformation, and by rapid crack propagation. The direction of crack motion is very nearly perpendicular to the direction of the applied tensile stress and yields a relatively flat fracture surface, as indicated in Figure cup-and-cone fracture brittle fracture

Transgranular vs Intergranular Fracture 1. For most brittle crystalline materials, crack propagation corresponds to the successive and repeated breaking of atomic bonds along specific crystallographic planes (Figure a); such a process is termed cleavage. This type of fracture is said to be transgranular, because the fracture cracks pass through the grains. 2. In some alloys, crack propagation is along grain boundaries this fracture is termed intergranular 3. Scanning electron fractograph showing an intergranular fracture surface. 50 magnification Intergranular Fracture Trans-granular Fracture Ductile Fracture Brittle Fracture

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

Stress Concentration- Stress Raisers 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 σm › σo t The measured fracture strengths for most brittle materials are significantly lower than those predicted by theoretical calculations based on atomic bonding energies. This discrepancy is explained by the presence of very small, microscopic flaws or cracks that always exist under normal conditions at the surface and within the interior of a body of material. These flaws are a detriment to the fracture strength because an applied stress may be amplified or concentrated at the tip, the magnitude of this amplification depending on crack orientation and geometry. This phenomenon is demonstrated in Figure. Due to their ability to amplify an applied stress in their locale, these flaws are sometimes called stress raisers. Theoretical fracture strength is higher than practical one; Why?

Concentration of Stress at Crack Tip

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 Blunt -> make less intense Elastic strain energy is also called as Resilience energy

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 All brittle materials contain a population of small cracks and flaws that have a variety of sizes, geometries, and orientations. When the magnitude of a tensile stress at the tip of one of these flaws exceeds the value of this critical stress, a crack forms and then propagates, which results in fracture σm is max stress at corner of crack (or stress raised by stress raiser)

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

Fracture Toughness: Design Against Crack Growth • Crack growth condition: Kc = • Largest, most stressed cracks grow first! --Result 1: Max. flaw size dictates design stress (max allowable stress). amax no fracture --Result 2: Design stress dictates max. allowable flaw size. amax no fracture Kc is the fracture toughness Y is a dimensionless parameter or function that depends on both crack and specimen sizes and geometries, as well as the manner of load application. Relative to this Y parameter, for planar specimens containing cracks that are much shorter than the specimen width, Y has a value of approximately unity. For example, for a plate of infinite width having a through-thickness crack Y=0; whereas for a plate of semi-infinite width containing an edge crack of length Y=1.1 σc σc

Design Example: Aircraft Wing • Material has KIc = 26 MPa-m0.5 • 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 KIc are the same for both designs. constant 9 mm 112 MPa 4 mm --Result: Answer:

Critical flaw size (microns) 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 Where Y = 1.12. Substitute values SOLUTION

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 KIc = I in above eqn represents mode 1, Similarly, mode II (b) and III (c) Crack Modes Plane strain condition: t/a ratio = large

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

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

Design Example: Aircraft Wing • Material has Kc = 26 MPa-m0.5 • 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 Kc are the same in both designs. 9 mm 112 MPa 4 mm --Result: Pay off -> Yield a profit or result Answer: • Reducing flaw size pays off!

Loading Rate s sy e -- increases sy and TS gives less time for • 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 e sy TS larger smaller That’s why a piece ruptured by impact loading doesn’t show much plastic deformation before getting ruptured

Numerical Problems Problems 8.1 – 8.10; 8.14 – 8.23; and 8.27