Three Dimensional Crack Modelling: Techniques … Dimensional Crack Modelling: Techniques and...

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Three Dimensional Crack Modelling:Techniques and considerations from the analysis of pin-loaded tubular joints.

Names:Dr. Richard J. Grant – NEWI, University of Wales, UK.Dr. John Smart – School of Engineering, University of Manchester, UK.

Summary:Experimental and numerical analyses of pin-loaded tubes have been performed and will be reported in this presentation.


Fatigue Damaged TubeTube Specification:• Outside Diameter - 50.8 mm (2”)• Wall thickness - 3.25 mm (10 swg)• Material- Tube Steel 4T45

Pin Specification:• Diameter - 7.94 mm (5/16”)• Material - Core Hardened Steel Dowels.

Loading Conditions:•Mean Load - 18.5 kN•Cyclic - +/- 18 kN


Crack-Tip Elements

2-D Element Topology

• Collapse one side of 8-noded quadrilateral,• Move mid-side nodes to the quarter-point.


Similarly In Three-Dimensions:

• 20-noded bricks - four mid-side nodes to quarter-points,• 27-noded bricks - additionally four mid-faced nodes to the quarter-points:

• 27-noded bricks offered little advantage over 20-noded bricks,• 27-noded brick approx. doubled the numbers of nodes to be shifted c.f. 20-noded.

Note - ABAQUS Version 5.6 was used in this work.


Crack Trajectory

Trajectory…

For a Two-Dimensional Crack:

Crack-tip present position and current direction of growth.

For a Three-Dimensional Crack:

Crack-tip present position, twist and tilt of the crack front and direction of growth.


Tilt & Twist in a 3-D Crack

10 20 30 40 50

Crack Arc Length Ω (degree)-0.1

0

0.1

0.2

0.3

Posi

tion

(inch

)

Crack Trajectory At The:Inside DiameterOutside DiameterPin

βο βο βο βο βο

Crack Profile

It is assumed that for:

• a given crack arc length Ω,• the crack front tilts at an angle β and• twists at an angle γ.


Modelling Process

Crack Focus Block Definition:

• Decide on crack arc length,• Remove appropriate block of elements,• Use peripheral nodes and tip nodes to define focus of elements (tilt and twist).


Almost Coincident NodesProblems Creating The Crack Focus:

• The crack-tip elements must have coincident nodes.

• When equivalencing model individual node definition is reduced to one node.

• Crack surfaces must not be equivalenced.

Solution:

• Define almost coincident nodes.

• Equivalence crack-tip focus with fine tolerance.

• Equivalence model section by section.


LINE definitions

Setting Up The Geometric Zoom of Elements

• Convert the crack-tip almost coincident nodes and the focus block peripheral nodesto GRIDs/POINTs• Connect the almost coincident nodes to the focus block peripheral nodes


Surfaces & Volumes

Focus Definition Geometrically Defined From LINEs

• All geometric definition derived from initial peripheral and crack-tip focus GRIDs/points


The Focus Block

Element Distribution:

• Geometric zoom of elements towards the tube surface to allow for plane stress/plane strain conditions.

• Through-thickness refinement of elements at the crack-tip.

• Optimise refinement compare:

• 1, 2, 4, 8 elements• Work/benefit


Installing the Focus

Careful equivalencing must be made to avoid zipping-up the crack surfaces:

• Define NAMEed sets,

• Equivalence only certain parts of the model at any one time,

• Watch tolerances


Determination of K

)1(42. 2νπ

−=

ErvK I

IStress Intensity Factor Determination

Crack Opening Displacement, COD:

• Calculate KI, KII and KIII and combine to produce Ke

• Ke = Equivalent Stress Concentration Factor (mixed mode) - eg. Proposed by A.P.Parker

others available; but KI dominates.

J-Integral Analysis:

• Total stress intensity factor, K, from J.

)1(42. 2νπ

−=

ErvK II

II

)1(42.ν+

=EE

rvK III

III

4 222IIIIIIe KKKK ++=

)1(.

2ν−=

EJK


Global-local nodal displacements

Method:• Node 1 is a small distance ‘r’ from the crack front • Define vectors a & c.• hence c runs approx. parallel to crack front.• node 3 lies on the crack surface.• follows that normal (to crack surface) vector n = c x a.• also follows that d = n x c.• d is approx. normal to the crack front in the plane of the crack.• Normalise vectors to find direction cosines:

• Find relative global displacements of node 1 with its pair on the opposing crack surface:

• Multiply by the respective direction cosine:

• and hence displacement data:

$ nnn

=

( )... .II IU u u etcδ δ∆ = −

ˆˆ

ˆ

I

II

III

n Ud Vc W

δδδ

∆ = × ∆

∆

...2I

Iv etcδ=


Effect of Twist and Tilt

Change of Ke for a change of Twist (γ) …and for a change of Tilt


Best Tilt and Twist

5 10 15 20 25 30 35 40 45 50Crack Arc Length Ω (degree)

050

100150200250300350400450500550

Stre

ss In

tens

ity F

acto

r (N

/mm

3/2)

Stress Intensity FactorKe (Plane Strain)K from J-integral

Combination of Twist and Tilt:

• For a given crack length will tend towards constant Kealong crack front.

• Select twist and tilt to give flatest curve.

• Comparison of K (from J) and Ke.


Tube Crack Growth Data

170,000 180,000 190,000 200,000 210,000 220,000 230,000Number Of Cycles, N.

0

5

10

15

20

25

30

35

Cra

ck L

engt

h, a

[mm

]

Tube 2:Hole PositionUpper Back LeftUpper Back Right

35

190,000 210,000 230,000 250,000 270,000 290,000Number Of Cycles, N.

0

5

10

15

20

25

30

35

Cra

ck L

engt

h, a

[mm

]


30

35


0

5

10

15

20

25

30

Cra

ck L

engt

h, a

[mm

]



0

5

10

15

20

25

Cra

ck L

engt

h, a

[mm

]



Averaged & Curve Fitted

170,000 180,000 190,000 200,000 210,000 220,000 230,000Number Of Cycles, N.

0

5

10

15

20

25

30

35C

rack

Len

gth,

a [m

m]

Tube 2:Hole PositionUpper Back2nd Order Fit


0

5

10

15

20

25

30

35

Cra

ck L

engt

h, a

[mm

]



0

5

10

15

20

25

30

35

Cra

ck L

engt

h, a

[mm

]



0

5

10

15

20

25

30

35

Cra

ck L

engt

h, a

[mm

]Tube 5:Hole Position

Upper Back2nd Order Fit


Crack Growth Rate

8 10 12 14 16 18 20 22 24 26Crack Length, a [mm]

0

1

2

3

4

5

6

7C

rack

Gro

wth

Rat

e, d

a/dN

[mm

/cyc

le]

(x10

-4)

Mean da/dN with Standard Error bars

Crack Growth Rate, da/dN, for a given crack length, a.


Obeys Paris’ Law?Plot of ln[ K] from J-integral v/s ln[da/dN]

• near surface and mid-wall,

• mid-wall.

6.8 6.9 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7ln[∆K] from J-Int. F.E.M. [N/mm3/2]

-9.4

-9.2

-9

-8.8

-8.6

-8.4

-8.2

-8

-7.8

In[d

a/dN

] (Ex

perim

enta

l) [m

m/c

ycle

]

J-Integral Sampled Thro' Wallat 10%, Max, & 90%.Linear Fit

6.8 6.9 7 7.1 7.2 7.3 7.4 7.5 7.6 7.7ln[∆K] from J-Int. F.E.M. [N/mm3/2]

-9.4

-9.2

-9

-8.8

-8.6

-8.4

-8.2

-8

-7.8

ln[d

a/dN

] (Ex

perim

enta

l) [m

m/c

ycle

]

J-Integral Sampled AtWall Mid-SectionLinear Fit


Published Data

10 100 1,000∆K from J-Int. F.E.M. (MPa.m1/2)

10-7

10-6

10-5

10-4

10-3

da/d

N (m

m/c

ycle

)

data pointsfrom J-integral

Scatterband for highstrength steels withσy = 730-2100 MPa

BS 4360

K v/s da/dN results lie within scatter band for high strength steels


Final Comment…

Why do a pair of cracks take a given path?What governs the crack’s trajectory?• tolerances,• asymmetric loading,• metallurgical defects…


Conclusions• Modelled crack growth in pin-loaded tubes,

• Good correlation with:• crack growth rate,• Published data.

• Unanswered questions regarding crack trajectories.

Three Dimensional Crack Modelling: Techniques … Dimensional Crack Modelling: Techniques and...

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