F.M. with Finite Element analysis - Wall Elliptical crack, Sub- Clad ...
Transcript of F.M. with Finite Element analysis - Wall Elliptical crack, Sub- Clad ...
Task 6 - Safety Review and LicensingOn the Job Training on Stress Analysis
Pisa (Italy)June 15 – July 14, 2015
F.M. with Finite Element analysis - Wall Elliptical crack, Sub-Clad semielliptical crack and Elastic-Plastic calculation of the
J parameter (ANSYS Workbench/ Apdl)Davide Mazzini – Ciro Santus
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Content
• Calculation example of a wall subsurface Elliptical crack
• The effect of the cladding properties on a semielliptical sub-clad
surface crack, ANSYS Workbench
• J parameter calculation with ANSYS Workbench with Elastic-Plastic
material
Table of content
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Definition of the problem
Half model (symmetry), with an elliptical shape crack
ANSYS Workbench
0
25mm60 mm
30 mm
2 15 mm2 30 mm
100 MPa
tWh
ac
t
W h
2a2b
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Solution with Tetrahedrons mesh
ANSYS Workbench
Refinement at the two most significant points
Solution accurate, however tetrahedrons not ok for the CINT command
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Solution with Hexahedrons mesh
ANSYS Workbench
The model is divided into solids for keep Hexahedrons mesh near the crack front
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Mesh options
ANSYS Workbench
The Hex dominant option works with this
toroid shape
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Solution
ANSYS Workbench
Symmetry boundary up to the crack front
Accurate solution
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CINT,SIFS calculation
ANSYS Workbench
Coordinate systems introduced in Workbench then useful for the CINT calculation in ANSYS Apdl
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CINT,SIFS calculation
ANSYS Workbench
Coordinate systems 12Coordinate systems 13
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CINT,SIFS calculation
ANSYS Workbench
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CINT,SIFS calculation
ANSYS Workbench
- Unreliable values at the boundary of the crack front- Outer contours interact with the boundary of the toroid volume
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ANSYS Apdl command in the Workbench environment
ANSYS Workbench
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Results verification through handbook
ANSYS Workbench
0
25mm30 mm
2 15 mm2 30 mm
100 MPa
th
ac
/ 0/ 2
2 / 0.6
e tc a
a t
AI 0
I
492 MPa mm
(ANSYS Wb, 3D) 475MPa mmOk!possibileeffect of the limited width
mK f a
K
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Content
• Calculation example of a wall subsurface Elliptical crack
• The effect of the cladding properties on a semielliptical sub-clad
surface crack, ANSYS Workbench
• J parameter calculation with ANSYS Workbench with Elastic-Plastic
material
Table of content
Pisa, June 15 – July 14, 2015
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Sub-clad surface semielliptical crack
The model
Wall
Clad
Crack only on the wall part
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Sub-clad surface semielliptical crack
The model
Hide a body to pick any other surface behind
Create the surface semielliptical crack through the usual steps
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Sub-clad surface semielliptical crack
The model
The attachment between the wall and the clad cannot be continuous across the entire surface
A circle can give a dimension of the discontinuity between the two parts
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Sub-clad surface semielliptical crack
SIF results
First result can be obtained without any connection between the two parts
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Sub-clad surface semielliptical crack
SIF results
Then calculation is repeated with connection outside the circle
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Sub-clad surface semielliptical crack
SIF results
Finally the full surface connection can be activated, after imposing plastic properties of the clad
Clad Elastic-Plastic properties
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Sub-clad surface semielliptical crack
SIF results
Finally the full surface connection can be activated, after imposing plastic properties of the clad
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Sub-clad surface semielliptical crack
SIF result comparison
I 510MPa mmK I 426 MPa mmK I 369MPa mmK
Which one is the most realistic result?
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Content
• Calculation example of a wall subsurface Elliptical crack
• The effect of the cladding properties on a semielliptical sub-clad
surface crack, ANSYS Workbench
• J parameter calculation with ANSYS Workbench with Elastic-Plastic
material
Table of content
Pisa, June 15 – July 14, 2015
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J parameter as an option of the Wb automatic Crack tool
Elastic calculation – Relation between KI and J
CT specimen example
2I
22I
(ANSYS) 782MPa mm (ANSYS) 2.786 mJ/mmConversion, validation: 200000MPa, 0.3
2.782mJ/mm OK!
K JE
KJE
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Elastic-Plastic steel, Bilinear Isotropic Hardening
CT specimen example
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Elastic-Plastic steel, Bilinear Isotropic Hardening
CT specimen example
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Elastic-Plastic steel, Bilinear Isotropic Hardening
CT specimen example
2
2
Sameload , Elastic-Plastic material:(ANSYS) 2.748mJ/mm
Previous (just Elastic):(ANSYS) 2.786 mJ/mm
% 1.4%
PJ
J
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Elastic-Plastic steel, Bilinear Isotropic Hardening
CT specimen example
I(II,III)
I
I
Sameload , Elastic-Plastic material: is also calculated,
though theoretically not available(just the output of CINT,SIFS command)
(ANSYS) 800MPa mm
Previous (just Elastic):
(ANSYS) 782MPa mm
% 2.3
PK
K
K
%
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Small Scale Yielding
CT specimen example
I and similar values,with respect to theprevious just elastic calculation,due to the very smallplastic region
K J
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Increasing load – Multiple time step simulation
CT specimen example
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Increasing load – Multiple time step simulation
Non linear analysis – iterative solution
CT specimen example
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Increasing load – Multiple time step simulation
Increasingly discrepancy of KI with respect to J
CT specimen example
Elastic-Plastic, ANSYS WorkbenchS_Y, MPa B, mm
500 12P, kN K_I, MPa mm^0.5 J, mJ/mm^2 K_I, from J Delta % B_min for Pl. Strain
10 800 2.748 777 -2.9 6.0 => Pl. Strain20 1969 11.25 1572 -25.2 24.7 => Pl. Stress30 3712 27.32 2450 -51.5 60.0 => Pl. Stress40 6269 56.85 3535 -77.4 124.9 => Pl. Stress
ReliableUnreliable Equivalence
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Increasing load – Multiple time step simulation
Plastic region size
CT specimen example
2
Imin
Y
2
Ip
Y
10 kN 12mm
2.5 6.0mm
Pla
0.26m
ne st
m
rain
13
P B
KBS
KrS
2.0 mm
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Increasing load – Multiple time step simulation
Plastic region size
CT specimen example
2
Imin
Y
2
Ip
Y
40 kN 12 mm
2.5 124.
Plane stress
1
9 mm
15.9mm
P B
KBS
KrS
2
I
Y
LEFM validity for Pl. Stress satnot isfied:
430 mm 57.9mmKaS
p 5 6mmr
Load <<80%offully plastic valueOk the useof J
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