Download presentation

Published byRene Cropp Modified over 2 years ago

1
**An introduction to Computationnal Fracture Mechanics**

J.Cugnoni, LMAF-EPFL, 2014

2
**Stress based design vs Fracture Mechanics approach**

Stress concentrator: Critical stress is reached… Stress based criteria (like Von Mises) usually define the onset of “damage” initiation in the material Once critical stress is reached, what happens? In this case, a defect is now present (ie crack) The key question is now: will it propagate? If yes, will it stop by itself or grow in an unstable manner. Stress analysis A crack is formed… Will it extend further? If yes, will it propagate abruptly until catastrophic failure? Fracture mech.

3
**Propagation criteria of a crack**

Crack propagation : stress intensity factor Stress intensity factors KI, KII, KIII measure the intensity of stress singularity at crack tip. “Stress intensity factor”: Constants of the 1/sqrt(r) term in stress field at crack tip: “Critical Stress intensity factor”: Maximum K that a material can sustain, considered as a material property and indentified in standard fracture tests. Units: Pa/sqrt(m), symbol: KIc KIIc KIIIc Crack propagation occurs if K > Kc with K=Kic +KIIc+KIIIc In many materialy, propagation is mode I dominated: KI>KIc Assumed Crack extension dA Stress singularity Fracture test: measure force at failure and calculate KIc From analytical solutions or FE

4
**Propagation criteria of a crack**

Crack propagation : an “energetic” process Extend crack length: energy is used to create a new surface (break chemical bonds). Driving “force”: internal strain energy stored in the system “Energy release rate”: Change in potential energy P (strain energy and work of forces) for an infinitesimal crack extension dA. Units: J/m2, Symbol: G measure the crack “driving force” “Critical Energy release rate”: Energy required to create an additionnal crack surface. Is a material characteristic (but depends on the type of loading). Units: J/m2, symbol: Gc Crack propagation occurs if G > Gc New crack surface dA: Dissipates Ed=Gc*dA Assumed Crack extension dA Potential energy: P0=U0-V0 Potential energy: P1= P0 –Er And Er = G*dA

5
**Computational Fracture Mechanics: Why?**

Using numerical simulation method we can: Apply Finite Element, Finite Volume/Difference, Boundary Element or Meshfree methods To obtain approximations of displacement, force, stress and strain fields in arbitrary configuration. And then? To apply fracture mechanics, we need: to compute fracture mechanics parameters (SIF K, G) in 2D and 3D configurations; to compute J integral in elastic-plastic analyses ; to simulate crack growth (under general mixed- mode conditions); FE simulation of a compact tension fracture test

6
**Overview of the different techniques**

Calculation of K (linear elasticity): Stress or displacement field matching (single / mixed mode) Indirectly from G (Interaction integrals in mixed mode) Calculation of G: Finite difference of potential energy (linear, evtl. non linear) Compliance method (linear) Virtual Crack Closure Technique VCCT (linear, mixed mode) J-integral (non-linear) Simulate crack growth to VCCT criterion, node release, remeshing or XFEM Cohesive elements / interfaces (Damage mechanics)

7
**Calculation of KI: stress matching**

Idea: compare FE stress field at crack tip with theory, fit KI from the numerical stress value In LEFM (for r 0): Step 1: From FE: Extract stress field Syy at =0 (along crack direction) (eq 4.36a)

8
**Calculation of KI: stress matching**

Fitting method 1: Plot (as a function of r, for r 0) Fit region Fit a line over the quasi constant region of the plot, identify KI either as average value or as the intercept at r=0

9
**Calculation of KI: stress matching**

Fitting method 2: Plot as a function of In theory the plot should be linear as: Fit a line over the quasi linear region of the plot, identify KI either as the slope of the line Note : the same method can be applied to get Kii from shear component at theta=0

10
**Calculation of KI: displacement matching**

Idea: compare FE crack displacement opening field with theory, fit KI from the numerical displacement value In LEFM (for r 0): Step 1: From FE: Extract stress field Syy at =0 (along crack direction) (eq 4.40b) x For plane stress: m = shear modulus= For plane strain:

11
**Calculation of KI: displacement matching**

Fitting method 1: Plot (as a function of r=-x, for r 0) Fit region Fit a line over the quasi constant region of the plot, identify either as the average value or as the intercept at r=0

12
**Calculation of KI: displacement matching**

Fitting method 1: Plot uy as a function In theory, the plot should be linear as : Fit a line over the quasi linear region of the plot, the slope is

13
**Improving mesh convergence: singular elements**

Evaluation of K from stress field requires very fine mesh. To better capture the 1/sqrt(r) stress singularity, one can use singular quadratic elements with shifted mid side nodes at ¼ of edge length (Barsoum,1976, IJNME, 10, 25; Henshell and Shaw, 1975, IJNME, 9, ) Using 8 node second order quadrangular elements, if we move the midside nodes at ¼ of edge, we make the Jacobian transformation of the element singular as 1/sqrt(r) along the element edges. By collapsing all nodes of one edge (degenerate a quadrangle to a triangle), the 1/sqrt(r) singularity covers then the whole element. Can be extended into 3D for 20 node hexahedrons

14
**Computing K from G (and vice-versa)**

For linear elastic isotropic materials, the following relations link G to K: For plane stress: For plane strain: m = shear modulus= For Mode I in plane stress and plane strain, this simplifies to: So knowing either K or G, the other can be determined directly. This can be extended to orthotropic material.

15
**Calculation of G: Strain energy method**

compare the strain energy of several FE models with different crack length, calculate the ERR as the derivative of potential energy = U - F For a linear elastic material: the potential energy F = PD = 2U thus = U – F = - U with the strain energy Compute ERR or as the slope of U(a) Different FE models for a0, a1, a2… a Equations are for a unit depth, if not the case, compute use U/b instead of U

16
**Calculation of G: Compliance method**

Idea (similar to experimental test data reduction !) Calculate the compliance C(a)=D(a)/P(a) for different crack lengths a0, a1 (several FE models required) and fit it with an appropriate function For a linear elastic material: the ERR can be obtained from the C(a) Compute ERR where dC/da is obtained from the fitted curve (eq 3.41 & 3.43b) a P, D Different FE models for a0, a1, a2…

17
**Calculation of G: J-integral method**

Using the “J-Integral” approach (see course), it is possible to calculate the ERR G as G = J in linear elasticity. If we know the displacement and stress field around the crack tip, we can compute J as a contour integral: J-integral is path independent for all continuum materials. But G must be perpendicular to the interface if dissimilar materials are used. W=strain energy density u = displacement field s = stress field G = contour: ending and starting at crack surface

18
**Calculation of G: J-integral method**

J-integrals are usually computed from a volume/surface integral in FE, for example in Abaqus with q = virtual crack extension vector: With on G and on C. Using divergence and equilibrium equations we can obtain:

19
**J-integral in 3D: 3D effects**

J-integral can be extend in 3D by computing J on several “slices” of the model. However, there are notable 3D effects affecting crack propagation: in 3D, the external surface is free of normal stress, so in plane stress state For thick specimens, the center of the specimen is closer to plane strain state S22 on external surface Plane stress S22 on symmetry plane Plane strain => higher stresses

20
**J-integral in 3D: 3D effects**

As a consequence, J-integral and G are not constant across the width. This will lead to a slightly curved crack front during propagation Need to be careful about specimen size effects when characterizing G or K In 2D FE simulation: use plane stress for very thin specimens, and plane strain for thick ones. G is max in the center => propagate earlier G is min on the side => propagate last

21
Comparison of methods Results obtained on a relatively coarse 3D mesh to highlight which methods are less mesh sensitive External surface, plane stress Stress fit1 Stress fit2 Displacement fit1 Displacement fit2 From J-integral K from Abaqus K1 3870 4345 4380 5079 4337 4540 (units: Mpa*sqrt(mm)) Inside, plane strain (* please note that the FE data for u2 and s22 were extracted on free surface, a source of error) Stress fit1 Stress fit2 Displacement fit1 Displacement fit2 J-integral K from Abaqus K1 4345 3870 4813 5582 5144 5140 Strain energy, forward finite difference Strain energy, backward finite difference Strain en. centered finite difference Strain energy fit Compliance method J integ. Avg G 322 282 302 302.5 301.5 296 (units: mJ/mm2) Conclusion: K determination is more sensitive to mesh ! G calculation using compliance, J-integral or strain energy fit are the most reliable

22
J-integral in Abaqus Crack tip and extension direction Can be calculated in elasticity / plasticity in 2D plane stress, plane strain, shell and 3D continuum elements. Requires a purely quadrangular mesh in 2D and hexahedral mesh in 3D. J-integral is evaluated on several “rings” of elements: need to check convergence with the # of ring) Requires the definition of a “crack”: location of crack tip and crack extension direction Quadrangle mesh Crack plane Rings 1 & 2

23
**J-Integral in Abaqus: application notes and demonstration**

Create a linear elastic part, define an “independent” instance in Assembly module Create a sharp crack: use partition tool to create a single edge cut, then in “interaction” module, use “Special->Crack->Assign seam” to define the crack plane (crack will be allowed to open) In “Interaction”, use “Special->Crack->Create” to define crack tip and extension direction (can define singular elements here, see later for more info) In “Step”: Define a “static” load step and a new history output for J-Integral. Choose domain = Contour integral, choose number of contours (~5 or more) and type of integral (J-integral). Define loads and displacements as usual Mesh the part using Quadrangle or Hexahedral elements, if possible quadratic. If possible use a refined mesh at crack tip (see demo). If singular elements are used, a radial mesh with sweep mesh generation is required. Extract J-integral for each contour in Visualization, Create XY data -> History output. !! UNITS: J = G = Energy / area. If using mm, N, MPa units => mJ / mm2 !!! By default a 2D plane stress / plane strain model as a thickness of 1. See demo1.cae example file

24
**Singular elements & meshing tips**

To create a 1/sqrt(r) singular mesh: In Interaction, edit crack definition and set “midside node” position to (=1/4 of edge) & “collapsed element side, single node” In Mesh: partition the domain to create a radial mesh pattern as show beside. Use any kind of mesh for the outer regions but use the “quad dominated, sweep” method for the inner most circle. Use quadratic elements to benefit from the singularity. Refine the mesh around crack tip significantly.

25
**J-integral Convergence vs # contours**

26
**Fracture mechanics oriented design**

Stress analysis Perform a stress analysis Locate stress critical regions analysis Crack Assume the presence of a defect in those regions (one at a time) Consider different crack lengths and orientation For each condition, check if the crack would propagate and if yes if it is stable or not Design evaluation Define operation safety conditions: maximum stress / crack length,… before failure occurs Define damage inspection intervals / maintainance plan

27
**Resources & help Abaqus tutorials: Abaqus Help:**

Abaqus Help: See Analysis users manual, section 11.4 for fracture mechanics Presentation and demo files: Computers with Abaqus 6.8: 40 PC in CM1.103 and ~15 in CM1.110

Similar presentations

OK

Illustration of FE algorithm on the example of 1D problem Problem: Stress and displacement analysis of a one-dimensional bar, loaded only by its own weight,

Illustration of FE algorithm on the example of 1D problem Problem: Stress and displacement analysis of a one-dimensional bar, loaded only by its own weight,

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on security features of atm card Ppt online library management system Ppt on blood stain pattern analysis software Mba ppt on body language Ppt on plasma arc machining Ppt on power generation by speed breaker Ppt on save environment photos Cardiovascular system anatomy and physiology ppt on cells Ppt on complex numbers class 11th Ppt on how email works