1. Week 4 Fracture, Toughness, Fatigue, and Creep
Materials Science

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?
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.

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

5. 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

6. 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

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

8. 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

9. 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

10. 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

11. 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

12. 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?

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

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
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)

16. 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

17. 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

18. 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:

19. 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
Steel
Zirconia(ZrO2)
Toughened Zirconia
Critical flaw size (microns)
7000
280
0.45 16
Where Y = Substitute values
SOLUTION

22. 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

23. 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

24. 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 =

25. 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!

26. 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

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