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

2

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

3

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


4

Solution with Tetrahedrons mesh

ANSYS Workbench

Refinement at the two most significant points

Solution accurate, however tetrahedrons not ok for the CINT command


5

Solution with Hexahedrons mesh

ANSYS Workbench

The model is divided into solids for keep Hexahedrons mesh near the crack front


6

Mesh options

ANSYS Workbench

The Hex dominant option works with this

toroid shape


7

Solution

ANSYS Workbench

Symmetry boundary up to the crack front

Accurate solution


8

CINT,SIFS calculation

ANSYS Workbench

Coordinate systems introduced in Workbench then useful for the CINT calculation in ANSYS Apdl


9


ANSYS Workbench

Coordinate systems 12Coordinate systems 13


10


ANSYS Workbench


11


ANSYS Workbench

- Unreliable values at the boundary of the crack front- Outer contours interact with the boundary of the toroid volume


12

ANSYS Apdl command in the Workbench environment

ANSYS Workbench


13

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


14

Content





material

Table of content


15Pisa, June 15 – July 14, 2015

Sub-clad surface semielliptical crack

The model

Wall

Clad

Crack only on the wall part

16


The model

Hide a body to pick any other surface behind

Create the surface semielliptical crack through the usual steps


17


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


18


SIF results

First result can be obtained without any connection between the two parts


19


SIF results

Then calculation is repeated with connection outside the circle


20


SIF results

Finally the full surface connection can be activated, after imposing plastic properties of the clad

Clad Elastic-Plastic properties


21


SIF results

Finally the full surface connection can be activated, after imposing plastic properties of the clad


22


SIF result comparison

I 510MPa mmK I 426 MPa mmK I 369MPa mmK

Which one is the most realistic result?


23

Content





material

Table of content


24

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


25

Elastic-Plastic steel, Bilinear Isotropic Hardening

CT specimen example


26Pisa, June 15 – July 14, 2015


CT specimen example

27


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


28


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

%


29

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


30

Increasing load – Multiple time step simulation

CT specimen example


31


Non linear analysis – iterative solution

CT specimen example


32


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


33


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


34


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


f.m. with finite element analysis - wall elliptical crack, sub- clad

Documents