three dimensional crack modelling: techniques … dimensional crack modelling: techniques and...
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Industry Sector RTD Thematic Area Date Deliverable Nr
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.
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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
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Crack-Tip Elements
2-D Element Topology
• Collapse one side of 8-noded quadrilateral,• Move mid-side nodes to the quarter-point.
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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.
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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.
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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.
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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).
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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.
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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
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Surfaces & Volumes
Focus Definition Geometrically Defined From LINEs
• All geometric definition derived from initial peripheral and crack-tip focus GRIDs/points
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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
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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
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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
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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=
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Effect of Twist and Tilt
Change of Ke for a change of Twist (γ) …and for a change of Tilt
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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.
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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
]
Tube 3:Hole PositionUpper Back LeftUpper Back Right
30
35
150,000 170,000 190,000 210,000 230,000 250,000Number Of Cycles, N.
0
5
10
15
20
25
30
Cra
ck L
engt
h, a
[mm
]
Tube 4:Hole PositionUpper Back LeftUpper Back Right
110,000 130,000 150,000 170,000 190,000 210,000Number Of Cycles, N.
0
5
10
15
20
25
Cra
ck L
engt
h, a
[mm
]
Tube 5:Hole PositionUpper Back LeftUpper Back Right
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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
150,000 170,000 190,000 210,000 230,000 250,000Number Of Cycles, N.
0
5
10
15
20
25
30
35
Cra
ck L
engt
h, a
[mm
]
Tube 4:Hole PositionUpper Back2nd Order Fit
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
]
Tube 3:Hole PositionUpper Back2nd Order Fit
110,000 130,000 150,000 170,000 190,000 210,000Number Of Cycles, N.
0
5
10
15
20
25
30
35
Cra
ck L
engt
h, a
[mm
]Tube 5:Hole Position
Upper Back2nd Order Fit
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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.
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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
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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
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Final Comment…
Why do a pair of cracks take a given path?What governs the crack’s trajectory?• tolerances,• asymmetric loading,• metallurgical defects…
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Conclusions• Modelled crack growth in pin-loaded tubes,
• Good correlation with:• crack growth rate,• Published data.
• Unanswered questions regarding crack trajectories.
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