Fracture Mechanics: Plastic Zone Size and Crack Tip Stress Intensity, Study notes of Mechanics of Materials

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Chapter 3Chapter 3

1

http://www.youtube.com/watch?v=N734bwi9kmw

OO

VERVIEWVERVIEW

Role of Material ThicknessRole of Material Thickness KI

, represents the level of “stress” at the tip of the crack and K

IC

, is the highest

value of K

that a material under veryI

specific (plane-strain) conditions canwithstand without fracture.

^

Ceramics and glasses (brittle materials) have very high yield strengths, giving themno way to relieve this stress by plastic flow (Figure 4(a)).
^

For ductile materials like metals and many polymers the picture is different(Figure 4(b)). The stress still rises as the crack tip is approached, but at the pointthat it exceeds the yield strength the material yields, relieving the stress, and –except for some work hardening – the stress cannot climb higher than

σ

.y

http://www.youtube.com/watch?v=U9GbhHFaTmk

Based on the equation, at r = 0 (at the crack tip)

σ

y^

approaches

infinity. However in practice the stress at the crack tip is limitedto at least the yield strength of the material.

Hence linear elasticity (LEFM) cannot be assumed within acertain distance of the crack.

This nonlinear region is called “crack tip plastic zone”

I

I^

K

K

σ

σ^

=

⇒

=

I^ p

ys

I

y

r K

r K

π

σ

π

σ^

2

2

=

⇒

=

(^

)^

strain

plane

K

r

stress

plane

K

r

I^ ys

p

I ys

p

2

2 2 2 2

1

2 2

ν

πσ πσ

−

Hence= = Where

σ

ys

= yield strength

Example 3.1
^

A thin plate of steel contains a central through-thickness flaw of length 16 mm, which is subjected to astress of 350 MPa applied perpendicularly to the flawplane. The yield stress of the material is 1400 MPa.Calculate the plastic zone size and the effective stress intensity level at the crack tip, making reasonableintensity level at the crack tip, making reasonable assumptions about the state of stress.If, after heat treatment, the yield stress of the steeldropped to 385 MPa, what would the plastic zone sizebe under the applied stress of 350 MPa, and whatconclusions would you draw about the use of LEFM? http://www.tech.plym.ac.uk/sme/interactive_resources/tutorials/fracturemechanics/StressIntensity/So lutions/Solution7.htm

Solution

Need 2 simple assumptions – assumption 1- the plate is largecompared to the size of the crack; this allows us to use the simpleinfinite plate formula for stress intensity factor, i.e;

the length of a central through-thickness crack is defined as 2

a

,

hence we use half the length (8 mm) in the stress intensity equation. Assumption 2

the steel plate is in a state of plane stress

this is

Assumption 2

the steel plate is in a state of plane stress

this is

reasonable, as it is stated to be 'thin'. An accurate assessment ofstress state would require a comparison to be made between platethickness and plastic zone size - if this ratio tends towards 1, planestress prevails, while if it tends towards 15 we are dealing withplane strain.

Plastic zone correction factor to crack length is given by:

PlanePlane

Stress and TransitionalStress and Transitional

StressStress

StatesStates^

For cases where the plastic energy at the crack tip is notnegligible, other fracture mechanics parameters, such as the Jintegral or R-curve, can be used to characterize a material.

The toughness data produced by these other tests will bedependant on the thickness of the product tested and will not be atrue material property.

The requirements for the minimum specimen test piece size for

The requirements for the minimum specimen test piece size for^ LEFM to be valid are very stringent for ductile materials in orderto obtain valid

K

IC

.

Lead to the scale could well exceed the size of the component theresults are to be applied to. Under these circumstances we stillneed a measure of the fracture toughness of these materials inorder to predict and avoid possible failure. Two methods havebeen developed which enable small scale testing to be applied tothe failure of ductile materials. These are the Crack OpeningDisplacement and the J Integral method.

ASTM E561-

Standard Practice for R-Curve

Determination

LEFM - crack extension occurs when the

energy release rate

G equals R, the material's resistance to crack extension.

A plot of R versus crack extension is called a

crack

resistance curve

, or R curve.

The corresponding plot of energy release rate, G versus crack^ extension is called a

driving force curve

.

R and G CurvesR and G Curves^ extension is called a

driving force curve

.

The material resistance to crack extension, R, consists of theenergy to create two new surfaces, 2Gs together with anymechanism which absorbs energy as the crack grows.

The R curve for an ideally brittle material is flat because thesurface energy is a fixed property.

Nonlinear behavior, like ductile fracture, can result in arising R curve as the plastic zone at crack tip increases insize with extension.