LEFM Analysis of a Center Cracked Specimen

26 Pages·2008·0.94 MB·English

LEFM Analysis of a Center Cracked
Specimen
Fergyanto E. Gunawan
Department of Mechanical Engineering
Toyohashi University of Technology
Objectives:
Using (cid:12)nite element analysis for computing the stress intensity factor.
(cid:15)
Using singular element
(cid:15)
Learn ANSYS/APDL programming
(cid:15)
LEFM Analysis of a Center Cracked Specimen
Model Description
Figure1showsacentercrackedtensionspecimen. ForthedatagiveninTable1,computethestressintensity
of the opening mode.
Figure 1: A center cracked tension specimen
Table 1: Data of the center cracked tension specimen.
Parameter Value
crack length (2a) 8.0 in
plate width (2W) 20.0 in
plate length (2L) 20.0 in
thickness (t) 0.1 in
Young’s modulus 1000.0 psi
Poisson’s ratio 0.3
Applied force 1.0 lb
References
[1] D. R. J. Owen and A. J. Fawkes. Engineering Fracture Mechanics: Numerical Methods and Appli-
cations. Pineridge Press Ltd., 1983.
[2] T. L. Anderson. Fracture Mechanics: Fundamentals and Applications. CRC Press, second edition,
1995.
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LEFM Analysis of a Center Cracked Specimen
Introduction of the Fracture Mechanics
In the classical strength material approach to the structural design, the largest stress in the structure is
compared to the material strength via a failure theory for determining the structural safety. However, when
a (cid:13)aw/crack exists, the engineering structure may also fail at a stress lower than the material strength.
Therefore, a new approach that considering the (cid:13)aw is required. The fracture mechanics is a branch in
the solid mechanics that takes into account the in(cid:13)uence of the (cid:13)aw to the material strength. Unlike the
strength material approach that utilizes the material strength, the fracture mechanics approaches utilizing
the fracture toughness in determining the safety of a structure. Figure 2 provides a schematic comparison
of the strength material approach to the fracture mechanics approach.
Applied stress Material Strength
(a) The strength of materials
Applied stress
Flaw size Fracture tougness
(b) The fracture mechanics approach
Figure2: Comparisonofthefracturemechanicsapproachtodesignwiththetraditionalstrengthofmaterial
approach.
Now, we consider an in(cid:12)nite size two-dimensional plate having a crack as depicted in Fig. 3. At the
Figure 3: The stresses ahead of the crack tip [1].
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LEFM Analysis of a Center Cracked Specimen
vicinity of the crack tip, the stresses and displacements can be expressed as
K (cid:18) 3(cid:18) (cid:18)
I
(cid:27) = 1 sin sin cos (1)
x
p2(cid:25)r (cid:18) (cid:0) 2 2 (cid:19) 2
K (cid:18) 3(cid:18) (cid:18)
I
(cid:27) = 1+sin sin cos (2)
y
p2(cid:25)r (cid:18) 2 2 (cid:19) 2
and
K r (cid:18) 3(cid:18)
I
u = (2(cid:20) 1)cos cos (3)
4(cid:22)r2(cid:25) (cid:18) (cid:0) 2 (cid:0) 2 (cid:19)
K r (cid:18) 3(cid:18)
I
v = (2(cid:20)+1)sin sin ; (4)
4(cid:22)r2(cid:25) (cid:18) 2 (cid:0) 2 (cid:19)
where K is the opening mode of fracture, and is the main parameter in the fracture mechanics analysis of
I
linear elastic materials; the parameter (cid:22) is E=(2(1+(cid:23))) and the parameter (cid:20) is obtained by
(3 (cid:23))=(1+(cid:23)) for the plane stress
(cid:20)=8 (cid:0) : (5)
>><
3 4(cid:23) for the plane strain
(cid:0)
>>:
Our focus: To compute the stress intensity factor, K
I
Equations (1){(4) show that K can be expressed in (cid:27) , (cid:27) , u, or v; therefore, once we have the data of
I x y
(cid:27) , (cid:27) , u, or v, we may infer the stress intensity factor. As you have seen in (cid:12)ve previous modules, those
x y
data even for a complex structure can be easily obtained by means of the (cid:12)nite element analysis.
However, Eqs. (1) and (2) reveals that the cracked structure possess a singular stress (cid:12)eld that propor-
tional to 1=pr; in plain English, the stress gradient at the vicinity of the crack tip is extremely
high. Without any special consideration, deliberately use the (cid:12)nite element method for such a problem
would not lead you to accurate data of (cid:27) , (cid:27) , u, or v.
x y
The problem of the high stress gradient can be addressed by two approaches. You have used the (cid:12)rst
approach when dealing with the stress concentration on the plate with hole where the high stress gradient
problem exists on the stress distribution along the ligament. The normal stress, as can be seen in Fig. 4,
graduallyincreasesasthelocationapproachingtheholeedge. Ontheholeedge,thecomputedstressislower
than that given by Peterson. However, when the mesh size is reduced as depicted in Fig. 5, the computed
normal stress steeply increases to a plateau near the exact solution.
Therefore, itisclearthattoaccuratelysimulatethestresssingularity, avery(cid:12)nemeshisrequiredinthe
region near to the crack tip. In addition, one also may use a special element so called the singular element
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LEFM Analysis of a Center Cracked Specimen
2
1.5
n)
xis (i 1
A
−
Y
0.5
FEM
Peterson
0
0 2 4 6 8
SX (psi)
Figure 4: The normal stress along the ligament of the plate with hole.
4.36
4.34
4.32
si) 4.3
p
SX (4.28
4.26
4.24 FE Solution
Exact Solution
4.22
0 50 100 150
Number of elements (NOE)
Figure 5: The e(cid:11)ect of mesh-size to the largest normal stress on the plate with hole.
that designed to capture the singular stress (cid:12)eld.
The aim of the text is to present the most applicable numerical techniques employed by engineers for
the solution of practical fracture problems and their implementation using the ANSYS Parametric Design
Language (APDL).
The Singular Element
Thesingularelementisanelementthatpossessthestrainsingularityof1=pr. Thisbehaviorcanbeachieved
byuseaquadraticelementwherethreeoftheirnodesarejoined|Nodes1,7,and9forthecasethatdepicted
in Fig. 6|and the mid-side nodes are moved to the quarter point adjacent to the crack tip node.
Whenthesingularelementisbeingused,thestressintensityfactorsforthetensionandshearmodescan
be directly obtained by solving Eq. (6) and (7):
K 8(2(cid:20)(cid:0)1)cos2(cid:18) (cid:0)cos32(cid:18)9=4(cid:22) 2(cid:25) 84u2(cid:0)u3(cid:0)3u19; (6)
I><(2(cid:20)+1)sin2(cid:18) sin32(cid:18)>= r L ><4v2 v3 3v1>=
(cid:0) (cid:0) (cid:0)
>: >; >: >;
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LEFM Analysis of a Center Cracked Specimen
Figure 6: The singular element.
and
K 8(cid:0)(2(cid:20)+3)sin2(cid:18) (cid:0)sin32(cid:18)9=4(cid:22) 2(cid:25) 84u2(cid:0)u3(cid:0)3u19; (7)
II><(2(cid:20) 3)cos2(cid:18) cos32(cid:18) >= r L ><4v2 v3 3v1>=
(cid:0) (cid:0) (cid:0) (cid:0)
>: >; >: >;
where u and v is the displacements in x and y directions, respectively, of node i. The index i is 1, 2, or 3.
i i
The best practice for meshing the region surrounding the crack tip in the LEFM is by use a spider-mesh
such as that shown in Fig. 7. For the elasto-plastic analysis, the singular element is not required.
nthet = 6
rrat delr = L npt
Figure 7: Singular elements around the crack-tip.
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LEFM Analysis of a Center Cracked Specimen
ANSYS Implementation of the Singular Element
In ANSYS, the singular element can be generated via by issuing the crack tip location before the mesh is
generated:
ANSYS Main Menu: Preprocessor (cid:66) Meshing (cid:66) Size Cntrls (cid:66) Concentrat KPs (cid:66) Create
In addition, the element is always accessible via a command line:
kscon, npt, delr, kctip, nthet, rrat
where npt is the keypoint number at concentration, delr is L in Eqs. (6) and (7) (see also Fig. 7), kctip =
1,nthetisnumberofelementsincircumferentialdirection,andrratisratioof2ndrowelementsizetodelr.
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LEFM Analysis of a Center Cracked Specimen
Exact Solution of the Center Cracked Tension Specimen
Reference [2] provides the exact solution of the stress intensity factor as:
P a
K = f ; (8)
I
BpW (cid:1) (cid:16)W(cid:17)
where P is the applied load, B is the specimen thickness, and f(a=W) is given by
a (cid:25)a (cid:25)a a 2 a 4
f = sec 1 0:025 +0:06 : (9)
(cid:16)W(cid:17) r4W (cid:1) (cid:16)2W(cid:17)(cid:20) (cid:0) (cid:16)W(cid:17) (cid:16)W(cid:17) (cid:21)
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LEFM Analysis of a Center Cracked Specimen
Finite Element Simulation of the Center Cracked Tension Specimen
Pre-Processing Phase
1. De(cid:12)ne some parameters: ANSYS Pulldown Menu
Parameters (cid:66) Scalar Parameters
Selection: L = 10.0
Accept
Selection: W = 10.0
Accept
Selection: a = 4.0
Accept
Selection: thick = 0.1
Accept
Selection: singularRadius = 0.4*0.1*a
Accept
Selection: young = 1000.0
Accept
Selection: nu = 0.3
Accept
Selection: appliedStress = 1.0
Accept
Close
2. Turn on the keypoint number, the area numbers and the line numbers: ANSYS Pulldown Menu
PlotCtrls (cid:66) Numbering
KEYPOINT Keypoint numbers (cid:2) On
AREA Area Numbers (cid:2) On
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LEFM Analysis of a Center Cracked Specimen
LINE Line numbers (cid:2) On
OK
3. Select an element type: ANSYS Main Menu
Preprocessor (cid:66) Element Type (cid:66) Add/Edit/Delete
Add
Solid 8node 183
OK
Options
Element behavior K3 : Plane strs w/thk
OK
Close
4. De(cid:12)ne the plate thickness: ANSYS Main Menu
Preprocessor (cid:66) Real Constants (cid:66) Add/Edit/Delete
Add
Type 1 PLANE183
OK
Thickness THK thick
OK
Close
5. De(cid:12)ne material properties: ANSYS Main Menu
Preprocessor (cid:66) Material Props (cid:66) Material Models
Structural (cid:66) Linear (cid:66) Elastic (cid:66) Isotropic
EX young
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FG 10