UPTEC-F13026

Examensarbete 30 hpJuni 2013

Smooth Particle Hydrodynamics Applied to Fracture Mechanics

Simon Sticko

Teknisk- naturvetenskaplig fakultet UTH-enheten Besöksadress: Ångströmlaboratoriet Lägerhyddsvägen 1 Hus 4, Plan 0 Postadress: Box 536 751 21 Uppsala Telefon: 018 – 471 30 03 Telefax: 018 – 471 30 00 Hemsida: http://www.teknat.uu.se/student

Abstract

Smooth Particle Hydrodynamics Applied to FractureMechanics

Simon Sticko

A numerical method commonly referred to as smooth particle hydrodynamics (SPH)is implemented in two dimensions for solid mechanics in general and fracturemechanics in particular. The implementation is tested against a few analytical cases: avibrating plate, a bending plate, a modus I crack and a modus II crack. A conclusion ofthese tests is that a better way of treating a shortcoming of SPH called tensileinstability is needed. A study is made on the best choice of a vital parameter called thesmoothing radius, and it is found that a good choice of the smoothing radius isroughly 1.5 times the initial particle spacing.

ISSN: 1401-5757, UPTEC F13 026Examinator: Tomas NybergÄmnesgranskare: Kristofer GamstedtHandledare: Per Isaksson

Contents

1 Introduction 4

2 Theory 52.1 Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Smooth Particle Hydrodynamics . . . . . . . . . . . . . . . . . . . . . . . 52.3 XSPH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.4 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.5 Artificial stress . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.6 Time Stepping . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.7 Cracks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Benchmarking cases 153.1 Vibrating Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153.2 Bending Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.3 Modus I/II Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 Results 254.1 Vibrating Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254.2 Bending Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274.3 Modus I Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294.4 Modus II Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

5 Discussion 365.1 Vibrating Plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.2 Bending plate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.3 Modus I/II Crack . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365.4 General . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385.5 Suggestions for Further Investigations . . . . . . . . . . . . . . . . . . . 38

6 Conclusions 38

References 39

A Used Parameters 40

3

1 Introduction

In many industrial and scientific applications it is necessary to simulate frac-ture and crack propagation. This can be done by the commonly used finiteelement method. If finite elements shall be used it is, however, necessary todefine the path along which the crack will propagate before the problem canbe solved.

Smooth particle hydrodynamics (SPH) is a less commonly used numer-ical method for solving a set of coupled partial differential equations in con-tinuum mechanics. SPH is a meshfree particle method which makes it inter-esting for dynamical problems in fracture mechanics. This is because there isno need to define the path of propagation in advance.

The aim of this master thesis is implementing SPH in two dimensions andto test this implementation against a few analytically known benchmarking-cases. These cases are:

• A vibrating plate

• A bending plate

• A modus I/II crack

This in order to examine if SPH gives reliable results or not.The theory of SPH contains a parameter called the smoothing radius. This

is often mentioned in literature as an important parameter, but it is seldomdiscussed what a good value is. Because of this the above benchmarking casesare in this thesis tested with different values of smoothing radius. This is inorder to find some optimal value.

The SPH formulation used in this thesis is very similar to what was doneby Gray et al. [1], who also treated the case of the vibrating plate. A variationof the smoothing radius was however not made by Gray et al., who also useda different implementation of the boundary conditions.

Attempts at modelling fracture with SPH has also been made by Gray andMonaghan [2]. The way fracture is modelled in this thesis is however quitedifferent from this and is more similar to what was done by Simkins and Li[3] (who worked with a different method called meshfree Galerkin). As far asknown, this way of modelling fracture has not been used previously with SPH.

4

2 Theory

2.1 Problem

We are interested in the following set of coupled partial differential equations:

d xαd t= vα (2.1)

d vαd t=

1

ρ

∂ σαβ

∂ xβ(2.2)

dρ

d t=−ρ

∂ vβ∂ xβ

(2.3)

dσαβd t

=E

1+ν(εαβ +

ν

1−2νεκκδαβ ) (2.4)

where

εαβ =1

2(∂ vα∂ xβ

+∂ vβ∂ xα

), (2.5)

and α,β ,κ ∈ x , y . This system represents how a solid elastic material be-haves in two dimensions.The system represents plane strain but could as wellrepresent plane stress simply by replacing Young’s modulus E and Poisson’sratio ν by E ′ and ν′ according to:

¨

E ′ = E 1+2ν(1+ν)2

ν′ = ν1+ν

.

In addition to solving this set one needs some way of modelling cracks.

2.2 Smooth Particle Hydrodynamics

As can be seen from equations 2.1 to 2.5 we have time derivatives expressed asspatial derivatives. What SPH does is providing a way of approximating spatialderivatives. This is done by starting with expressing the spatial derivative interms of the Dirac function δ(x ):

∂ f

∂ xβ(xi ) =

ˆΩ

∂ f

∂ xβ(x )δ(x − xi )d A,

and then replacing the Dirac function with the so called kernel function, W ,that is:

∂ f

∂ xβ(xi )≈

ˆΩ

∂ f

∂ xβ(x )W (x − xi , h )d A. (2.6)

5

The kernel W has a similar shape to a gaussian as seen in figure 2.1 and is de-pendent on an additional parameter, called the smoothing radius, h . Equa-tion 2.6 is motivated by the fact that the kernel is normalised:

ˆ ∞−∞

ˆ ∞−∞

W (x , h )d x d y = 1

and approaches the Dirac function when this smoothing radius goes to zero:

limh→0

W (x , h ) =δ(x ).

−2 −1 0 1 20

0.1

0.2

0.3

0.4

0.5

0.6

0.7GaussianW

x−xi

h

Figure 2.1: The kernel compared to a normalised gaussian.

There are several kernels used in SPH. Here the (perhaps most commonlyused) cubic spline kernel is utilized. This is defined as

W (x ) =

αk (23− r 2+ 1

2r 3) r ∈ [0, 1]

αk

6(2− r )3 r ∈ (1, 2]

0 r ∈ (2,∞)(2.7)

where

r =

p

xαxαh

and αk has value

αk =15

7πh 2,

in two dimensions. The cubic spline kernel is the one that is plotted (in 1D)in figure 2.1. As can be seen from equation 2.7 W (x − xi ) is zero outside adistance 2h of xi . This is important since one can avoid differentiating the

6

function f in equation 2.6 by moving the differentiation to the kernel by usingGauss’s theorem:

∂ f

∂ xβ(xi )≈

ˆ∂ Ω

f (x )W (x − xi )d l︸ ︷︷ ︸

=0

−ˆΩ

f (x )∂W (x − xi )∂ xβ

d A =−ˆΩ

f (x )∂W (x − xi )∂ xβ

d A

(2.8)where the domain of integration Ω is the circle with radius 2h , centred at xi :

Ω(xi ) = x : ||x − xi ||< 2h,

and the first term is zero because the kernel is zero on the boundary of Ω.In SPH the field variables of the system is represented by particles. A particle

j has properties such as:

• position, x j

• velocity, v j

• stress,σ jαβ

• density, ρ j

• mass per thick-ness, m j

Now, if xi is the position of particle i equation 2.8 is discretized using theseparticles. The integral goes to a sum over all neighbouring particles inside Ωand d A goes to the area, A j , of neighbour j :

∂ f

∂ xβ(x i ) =−

∑

j∈Ωf j

∂W (x j − xi )

∂ xβA j .

That is, one collects all particles within a distance 2h of particle i and weighttheir value of f j , their area and the derivative of the kernel. It might be morehelpful to think of this step in terms of equation 2.6, which is illustrated in fig-ure 2.2. One can think of SPH as centring the kernel at the position of particlei and weight ∂ f

∂ xβof each neighbour with the kernel, even if this is not technic-

ally correct.The area of particle j is (note that m j is mass per thickness):

A j =m j

ρ j

which gives us∂ f

∂ xβ(x i ) =

∑

j∈Ωf j

∂Wi j

∂ xβ

m j

ρ j, (2.9)

where we have used the short hand notation

Wi j =W (xi − x j ).

7

x−xi

hy−yi

h

W (x − xi )

Particle iSurrounding particles

Figure 2.2: Illustration of how equation 2.6 is discretized with particles.

Note that the change in sign in equation 2.9 comes from that W is an evenfunction and thus ∂W

∂ xβis an odd function. Thus

∂Wi j

∂ xβ=−

∂Wj i

∂ xβ. (2.10)

To create an SPH-discretization of equation 2.2, equation 2.9 leads to:

d v iα

d t︸︷︷︸

Of particle i

=1

ρi

∑

j∈Ωσ

jαβ

∂Wi j

∂ xβ

m j

ρ j, (2.11)

where Einstein summation convention is applied over the greek letters. Onecan symmetrize this equation by realizing that the derivative of a constantfunction, 1(x ) = 1, is zero, i.e.

0=∂ 1

∂ xβ=∑

j∈Ω

∂Wi j

∂ x βm j

ρ j(2.12)

according to equation 2.9. Multiplying this zero withσiαβ/ρi gives

0=σiαβ

ρi

∑

j∈Ω

∂Wi j

∂ xβ

m j

ρ j(2.13)

8

and adding this to equation 2.11 gives us

d v iα

d t=∑

j∈Ωm j (

σiαβ +σ

jαβ

ρiρ j)∂Wi j

∂ xβ. (2.14)

Since one can interpret the “force contribution”, F i jα , on particle i from particle

j as

F i jα =mi m j (

σiαβ +σ

jαβ

ρiρ j)∂Wi j

∂ x β, (2.15)

equation 2.14 is now in agreement with Newton’s third law since it is clear fromequation 2.15 that interchanging i and j gives

F i jα =−F j i

α

using equation 2.10. Equation 2.14 is the final SPH-equation representingequation 2.2.

In the same manner, an estimate of∂ v i

α

∂ xleads to

∂ v iα

∂ xβ=∑

j∈Ωv jα

∂Wi j

∂ xβ

m j

ρ j︸ ︷︷ ︸

eq 2.9

−v iα

∑

j∈Ω

∂Wi j

∂ xβ

m j

ρ j︸ ︷︷ ︸

=0 eq 2.12

,

which gives the SPH estimation

∂ v iα

∂ xβ=∑

j∈Ω(v jα − v i

α)∂Wi j

∂ xβ

m j

ρ j. (2.16)

To get an SPH-discretization of equation 2.3 one can reformulate it to

dρ

d t=−

∂

∂ xβ(ρvβ ) + vβ

∂ ρ

∂ xβ

which according to equation 2.9 can be approximated as:

dρi

d t=−

∑

j∈Ωρ j v j

β

m j

ρ j

∂Wi j

∂ xβ+ v i

β

∑

j∈Ωρ j

∂Wi j

∂ xβ

m j

ρ j,

or equivalently:

dρi

d t=∑

j∈Ωm j (v

iβ − v j

β )∂Wi j

∂ xβ. (2.17)

The collected semi-discrete system becomes:

9

'

&

$

%

d x iα

d t= v i

α (2.18)

d v iα

d t=∑

j∈Ωm j (

σiαβ +σ

jαβ

ρiρ j)∂Wi j

∂ xβ. (2.14)

dρi

d t=∑

j∈Ωm j (v

iβ − v j

β )∂Wi j

∂ xβ(2.17)

dσiαβ

d t=

E

1+ν(εiαβ +

ν

1−2νεiκκδαβ )

where εiαβ is determined from

εiαβ =

1

2(∂ v i

α

∂ xβ+∂ v i

β

∂ xα) (2.19)

∂ v iα

∂ xβ=∑

j∈Ω(v jα − v i

α)∂Wi j

∂ xβ

m j

ρ j. (2.16)

It should be emphasised that there are several forms of SPH equations thatcan be developed and not all forms are compatible with each other. For a moredetailed investigation of the subject see for example Bonet and Lok [4].

Since we will frequently use the quantity smoothing radius per mean particledistance we define:

h =h

12(∆x +∆y )

where ∆x and ∆y are the initial particle spacings in the x - and y -direction.This quantity will be called the relative smoothing radius. h will be held con-stant for a given simulation.

2.3 XSPH

According to Gray et al. [1] it is usually better to replace the velocities in theright hand side of our semi-discrete system by the smeared velocity field, v αi ,calculated as

v iα = v i

α+µ∑

j∈Ω(v jα − v i

α)Wi j

m j

12(ρ j +ρi )

, (2.20)

where µ is typically chosen as µ = 0.5. The variable v iα now represents the

velocity of particle i that is weighted by the velocities of the neighbouringparticles. This is sometimes referred to as XSPH.

10

2.4 Viscosity

Typically one needs to add some form of artificial viscosity, probably the mostcommonly used is the so called Monaghan artificial viscosity1:

Πi j =−α(ci + c j )φi j +2βφ2

i j

ρi +ρ j, (2.21)

whereφi j is given by the expression:

φi j = hmin0, (xi − x j ) · (vi − v j )(xi − x j )2+ (h/10)2

.

A good choice for the parameters α and β are2

α= 1 and β=2.

The parameter ci in equation 2.21 is the speed of sound at particle i , that is

ci =

√

√ K

ρi(2.22)

where K is the bulk modulus of the material. This damping is added to equa-tion 2.14 which is now changed to

d v iα

d t=∑

j∈Ωm j (

σiαβ +σ

jαβ

ρiρ j+Πi jδαβ )

∂Wi j

∂ x β−γv i

α. (2.23)

The viscosity in equation 2.21 is good for stabilising the SPH equations, but ifone is interested in the solution at equilibrium as much damping as possiblemight be wanted. To get more damping the term −γv i

α is included. This termprovides damping in the same way as one would introduce damping in an har-monic oscillator. A typical value of γwould be in the range [10−3s−1, 0.1s−1].

2.5 Artificial stress

SPH is subject of a problem called tensile instability (see for example Swegleet al. [7]). This means that during a tensile state the movements of the particlesbecomes unstable. The result of this is that particles tend to clump together.A way to resolve this was proposed by Gray et al. [1]. In this procedure an extraterm is added to equation 2.23, which now becomes

d v iα

d t=∑

j∈Ωm j (

σiαβ +σ

jαβ + (R

iαβ +R j

αβ ) fn

i j

ρiρ j+Πi jδαβ )

∂Wi j

∂ x β−ηv i

α. (2.24)

1See Liu and Liu [5] page 126.2See Monaghan [6] page 551.

11

Where Rαβi is the so called artificial stress3 and fi j is a scaling factor.4 Theartificial stress for a particle i is determined by first rotating our coordinatesystem to determine the principal stresses, σx x

i and σy yi . This can be done by

the transformation

σix x = c 2σi

x x +2s cσix y + s 2σi

y y

σiy y = s 2σi

x x −2s cσix y + c 2σi

y y ,

where

s = sinθ , c = cosθ and θ =1

2tan(

2σix y

σix x −σi

y y

).

The artificial stresses R ix x and R i

y y in this rotated coordinate system are thendetermined by

R ix x =

¨

−ζσix x σi

x x ≥ 0

0 σix x < 0

, (2.25)

and correspondingly for R iy y . The parameter ζ in equation 2.25 typically has

the value ζ = 0.3. The artificial stresses in the original coordinate system arethen obtained by rotating back, i.e. by the transformation:

R ix x = c 2R i

x x + s 2R iy y

R iy y = s 2R i

x x + c 2R iy y

R ix y = s c (R i

x x − R iy y ).

Note that since the artificial stress in equation 2.25 is always negative it willhelp keep particles apart and thus prevent that particles clump together. Theterm fi j is determined by scaling the distance between particle i and j ac-cording to the original particle spacing using the kernel:

fi j =W (ri j )

W (∆x+∆y2h)

where ri j is the distance between particle i and j divided by h . This meansthat if two particles get closer to each other ri j will decrease and fi j will in-crease, which will make the added term in equation 2.24 larger. According toGray et al. [1] a good choice for the parameter n in equation 2.24 is n = 4.

3This is slightly different than what Gray et al. [1] calls artificial stress.4It should be emphasised that n in the exponent of f n

i j now means fi j to the power n .

12

2.6 Time Stepping

The semi-discrete system is time-stepped with Heun’s method, first the fieldvariables are predicted with

x n+1p = x n +∆t ˜v n

v n+1p = v n +∆t

d v n

d t

ρn+1p =ρn +∆t

dρn

d t

¯σn+1p = ¯σn +∆t

d ¯σn

d t

where the subscript p now designates predicted (not particle index), the pre-

dicted derivatives (d v n+1

d t)p , (dρ

n+1

d t)p , (d ¯σn+1

d t)p are then calculated using these

values and the final field variables are calculated as

x n+1 = x n+1p +

∆t

2( ˜v n+1

p − ˜v n )

v n+1 = v n+1p +

∆t

2((

d v n+1

d t)p −

d v n

d t)

ρn+1 =ρn+1p +

∆t

2((

dρn+1

d t)p −

dρn

d t)

¯σn+1 = ¯σn+1p +

∆t

2((

d ¯σn+1

d t)p −

d ¯σn

d t).

Note that position is moved with the smeared velocity from equation 2.20.In addition some extra smearing will be done in the beginning of each

time-step according tov n ← (1−η)v n +η ˜v n . (2.26)

That is, the velocity is assigned a weighted value of itself and the smeared ve-locity from equation 2.20. The parameter η is in the range: η ∈ [0, 1]. If someof the particles in our system are fixed this will introduce additional dampingto the system. The time step is chosen according to:5

∆t =C∆x

c0, (2.27)

where∆x is the initial particle spacing, c0 is the initial speed of sound in thematerial, and C is a dimensionless constant.

5Essentially the same as done by Liu and Liu [5] page 142.

13

2.7 Cracks

A crack, describing a discontinuity, is represented as a curve consisting ofstraight line segments, as can be seen in figure 2.3a. If two particles are neigh-bours or not is determined by if they can “see” each other. A crack can beseen as a wall which breaks that line of sight. For example the “circle”-particlein figure 2.3b is neighbour with the “square”-particle but not with the “star”-particle. This condition is fairly easy to check. The “circle”- and “star”-particleare neighbours if the equation

xcircle+λ1(xstar− xcircle) = x1+λ2(x2− x1)

has solutions for λ1 and λ2 in the range

λ1, λ2 ∈ [0, 1].

This reduces checking the visibility condition to solving a two dimensionallinear equation system.

It is desirable that the crack is a part of the actual material. If all particles ina body would be moving with a constant velocity you would want the crack tobe moving with the body. To get this effect the start and end of each segmentis “floating” half way between a pair of particles, as can be seen in figure 2.3a.

Crack-segment

Particle

(a) Two linked crack segments.

Neighbour of

Not a neighbour of

xcircle

xstar

x1

x2

xsquare

(b) A crack segment and three particles

Figure 2.3: A crack represented as straight line segments, floating betweenpairs of particles.

14

3 Benchmarking cases

For the following benchmarking cases we wish to test the numerical imple-mentation for four different values of the relative smoothing radius:

h = 1.3, h = 1.5, h = 1.7 & h = 1.9,

and at the same time vary the initial number of particles.

3.1 Vibrating Plate

We have a rectangular plate with length L and width D , as seen in figure 3.1.The left end at x = 0 is fixed in a wall and the other end is freely movable. If

L > 10D

the dynamics of this system is well approximated by this one dimensional par-tial differential equation:

E D 2

12ρ

∂ 4u y

∂ x 4=−

∂ 4u y

∂ t 4

u y

x=0= 0

∂ u y

∂ x

x=0

= 0

∂ 2u y

∂ x 2

x=L

= 0

∂ 3u y

∂ x 3

x=L

= 0

where u y is the displacement in the y -direction. If we start with zero displace-ments at time t = 0:

u y

t=0= 0,

and solve our system by the method of separation of variables the solution forthe fundamental mode becomes:

u y (x , t ) = (cosα0+coshα0)(sinλ0 x−sinhλ0 x )−(sinα0+sinhα0)(cosλ0 x−coshλ0 x )(cosα0+coshα0)(sinλ0 L−sinhλ0 L )−(sinα0+sinhα0)(cosλ0 L−coshλ0L )H0 sinω0t

(3.1)where

λ0 =α0

L, ω0 =λ

20D

√

√ E

12ρand α0 = 1.8751

15

Figure 3.1: Geometry of the plate.

To test if SPH captures the dynamics correctly we give the system an initialimpulse which corresponds to the fundamental mode. That is, starting frominitial velocity:

∂ u y

∂ t

t=0= (cosα0+coshα0)(sin(λ0 x−sinhλ0 x )−(sinα0+sinhα0)(cosλ0 x−coshλ0 x )(cosα0+coshα0)(sin(λ0L−sinhλ0L )−(sinα0+sinhα0)(cosλ0L−coshλ0 L )H0ω0,

(3.2)which will make equation 3.1 the analytical solution. The parameter H0 is themaximum displacement in the plate at x = L .

The plate is discretized with particles as seen in figure 3.2. Particles areplaced equally spaced between

x ∈ [0, L −∆x

2] and y ∈ [−(D −

∆y

2), (D −

∆y

2)].

Particles with x ≤ 0 are constrained from moving and the others are free.It might seem strange that particles aren’t placed all the way to x = L andy = ±D

2, but we motivate this by the fact that a particle actually represents

a somewhat larger domain than just a point. In some sense a particle posi-tioned at (x , y ) could be thought of as representing the domain

(x , y ) ∈ [x −∆x

2, x +

∆x

2]× [y −

∆y

2, y +

∆y

2].

The problem is solved only with the damping provided by that in equation2.21. That is, the additional damping parameters γ and η in equations 2.23and 2.26 are both set to zero.

To compare the numerical and analytical solution we solve the problemwith SPH until an end time te nd corresponding to one analytical period, moreprecisely:

te nd =2π

ω0.

During this time the y -displacement of the end particle (marked in figure 3.2)is saved at equally spaced time intervals. This will give us the points seen in

16

figure 3.3. Using these points we can determine the amplitude and frequencyof the oscillations of the numerical solution by making a least square fit to theequation

u y (t ) = Ae −κt sin(ωt ), (3.3)

to determine the three parameters A, κ, and ω. A and ω will then be com-pared to the analytical values of the frequency and amplitude. The parameterκmight seem superfluous, but should be included since the artificial viscos-ity in equation 2.21 introduces damping. This problem is very similar to theproblem studied by Gray et al. [1]with a different implementation of the bound-ary conditions. Gray et al. used h = 1.5.

0 L

−D

D

x/m

y/m

Free Fixed

Tracked

Figure 3.2: Plate discretized with particles

0 2 4 6 8−0.03

−0.02

−0.01

0

0.01

0.02

0.03

time/s

y−di

spla

cem

ent/m

AnalyticalLSQ−fittedData points

Figure 3.3: y -displacement of the end particle as a function of time, togetherwith the analytical curve from equation 3.1 and a least square fit to equation3.3.

17

3.2 Bending Plate

For the same geometry as in figure 3.1 we start from equilibrium:

u y = 0 and∂ u y

∂ t= 0.

But now we prescribe the y -displacements of the particles at the end of theplate to be u p

y . This is illustrated in figure 3.4. The triangular particles arefree to move in the x -direction but are during the time-stepping assigned avelocity in the y -direction according to:

vy (n ) =

¨

u py

N∆tn <N

0 n ≥N. (3.4)

Where n is which time-step and N is the number of time-steps during whichthis loading takes place. The damping parameters γ and η were chosen toγ = 0.7s−1 and η = 0.9 to get as much damping as possible. After loading thesystem and waiting for it to be damped sufficiently the following is checked:

1. The y -displacements of the centreline (particles originally located ony = 0) , which approximately should be6:

u y (x ) =u p

y

2β 3((ξ(x )−α)3−3β 2(ξ(x )−α) +2β 3)

where

ξ(x ) =L − x

L, α=

∆x

2Land β=1-α,

provided that L > 10D .

2. The angle at the end of the plate. Which analytically should be

θe nd =3u p

y

2βL

according to KTH [8]. This angle is calculated as illustrated in figure 3.5.A linear least square fit is made to the x - and y -coordinates of the trian-gular particles in figure 3.4 to determine d y

d x. The angle is then obtained

by

θn um =π

2− tan−1(

d y

d x). (3.5)

6See KTH [8] page 344.

18

0 L

−D

D

x/m

y/m

Free Fixed Prescribed in y

Figure 3.4: Plate with end prescribed.

tan−1(d yd x)

θ

Figure 3.5: End of the plate in figure 3.4 and how the angle θ was determined.

19

3.3 Modus I/II Crack

We have a square body of an elastic material in the domain x , y ∈ [− L2

, L2] as

can be seen in figure 3.6. The body is considered infinitely thin and is there-fore modelled as plain stress. In this body there is a straight crack going from(x , y ) = (− L

2, 0) to (x , y ) = (0, 0).

Free

Prescribed

Crack

L/2−L/2

L/2

x

y

−L/2 0

0

Figure 3.6: Setup for the modus I/II crack.

Two cases will be considered, a modus I and a modus II crack. During thetime-stepping the frame of outermost particles in this body (marked in red)get prescribed velocities in both the x - and y -direction in the same way asin equation 3.4. The final displacements are shown in figure 3.7. As can beseen modus I is an opening crack and modus II is a sliding crack. Expressedin cylindrical coordinates these displacements are

ux (r,θ ) =(1+ν)KI

4πE

p2πr ((2κ−1)cos

θ

2− cos

3θ

2) (3.6)

u y (r,θ ) =(1+ν)KI

4πE

p2πr ((2κ+1)sin

θ

2− sin

3θ

2) (3.7)

in the modus I case and

ux (r,θ ) =(1+ν)KII

4πE

p2πr ((2κ+3)sin

θ

2+ sin

3θ

2) (3.8)

u y (r,θ ) =(1+ν)KII

4πE

p2πr ((2κ−3)cos

θ

2+ cos

3θ

2) (3.9)

in the modus II case7. The parameter κ equals

κ=(3−ν)(1+ν)

,

7See KTH [8] page 238.

20

and KI and KII are constants called the stress intensity factors. This choice ofdisplacements along the frame will make the analytical displacements takethe above values in the whole body. In addition, the stresses in the body willbecome

σx x =KIp2πr

cosθ

2(1− sin

θ

2sin

3θ

2) (3.10)

σy y =KIp2πr

cosθ

2(1+ sin

θ

2sin

3θ

2) (3.11)

σx y =KIp2πr

cosθ

2sinθ

2cos

3θ

2(3.12)

for the modus I case and

σx x =KIIp2πr

sinθ

2(2+ cos

θ

2cos

3θ

2) (3.13)

σy y =KIIp2πr

sinθ

2cos

θ

2cos

3θ

2(3.14)

σx y =KIIp2πr

cosθ

2(1− sin

θ

2sin

3θ

2) (3.15)

for the modus II case, see Williams [9].

−L/2 0 L/2

−L/2

0

L/2

x

y

(a) Modus I

−L/2 0 L/2

−L/2

0

L/2

x

y

(b) Modus II

Figure 3.7: Displacements applied to the frame (exaggerated).

An important parameter when it comes to fracture mechanics is the socalled energy release rate, G , defined as:

G =−∂ U

∂ a, (3.16)

21

where U is the elastic energy per thickness in the body, and a is the length ofthe crack. This says how the potential energy stored in the body is changedwhen the crack grows. Theoretically the energy release rate should have thevalue

G =K 2

I

E+

K 2II

E, (3.17)

that is

G =

¨

K 2I /E in pure modus I

K 2II/E in pure modus II

.

The potential energy per thickness is given by

U =1

2

ˆVσαβεαβd A.

For our discrete system this will be calculated as8

U =1

2

∑

j

σjαβε

jαβ

m j

ρ j(3.18)

where we calculate the strains using:

εx x =1

E(σx x −νσy y ) (3.19)

εy y =1

E(σy y −νσx x ) (3.20)

εx y =1

2Gσx y (3.21)

The issues that will be checked for this problem are:

1. The error-norms for stresses and displacements at equilibrium9, determ-ined as:

e (σαβ ) =1

n

√

√

√

√

n∑

j=1

(σ jαβ −σαβ (x ) j )2, (3.22)

for the stresses and

e (uβ ) =1

n

√

√

√

√

n∑

j=1

(u jβ −uβ (x j ))2, (3.23)

8Note that the m j actually has dimension k g /m in two dimensions and thus is the mass perthickness.

9What is meant by equilibrium is that the norm of the velocities in the system:

|vs y s |=√

√

√

∑

j

v 2j

has decreased enough relative to its value at the time the system had been loaded.

22

for the displacements. In equation 3.22 and 3.23σαβ (x j ) and uβ (x j ) arethe analytical values and the sum goes over all particles in the body. Thefactor 1

nmust be included since the error otherwise will increase when

the number of particles increases.

2. The numerical energy release rate calculated in two ways. Hypothesisfor calculation of crack paths may be divided into those based on propaga-tion and those based on projection. Propagation criteria study the cracktip state of a hypothetical extension of the crack whereas projection basedcriteria study the mechanical state at the tip. Both hypothesis are invest-igated here, namely:

(a) By the procedure illustrated in figure 3.9:

i. At equilibrium determine the energy in the body when the crackhas length a , U (a ), using equation 3.18.

ii. Forcing the crack open one step, waiting for equilibrium to oc-cur and calculating the energy when the crack has length a +∆x , U (a +∆x ).

iii. Opening the crack one step further to determine U (a +2∆x ).

iv. Numerically differentiating these values to determine ∂ U∂ a

usingan O (h 2) estimate of the derivative:

∂ U

∂ a(a )≈

− 32U (a ) +2U (a +∆x )− 1

2U (a +2∆x )

∆x, (3.24)

and thus obtaining G from equation 3.16. This value will bereferred to as G1and is determined from propagation.

(b) At equilibrium taking the marked points in figure 3.8a and:

i. Determine KI by a least square fit of their y -displacement toequation 3.7.

ii. Determine KII by a least square fit of their x -displacement toequation 3.8.

iii. Determine G by using equation 3.17. This value will be referredto as G2 and is determined from the state at the tip (projection).

23

0

0

L/2

−L/2

L/2

x

y

−L/2(a)

0

0

L/2

−L/2

L/2

x

y

−L/2(b)

Figure 3.8: In a) particles that KI and KII are determined from. In b) particlesthat the stresses are plotted for.

1

2

3

x=0

Figure 3.9: How the crack was forced to open to determine the energy releaserate, using equation 3.24.

24

4 Results

4.1 Vibrating Plate

This problem was solved without the artificial stress from section 2.5. Withartificial stress a disturbance can be seen in the bottom of the plate where theplate is attached to the wall, as seen in figure 4.1.

x0/m

y 0/m

0 0.2 0.4 0.6 0.8 1−0.05

00.05

KPa

0 0.2 0.4 0.6 0.8 1

Figure 4.1: Disturbance seen in the von Mises stress when artificial stress isincluded.

Without artificial stress the relative error in frequency is shown in figure4.2a for different values of smoothing radius and particle densities. The para-meter ny describes the number of particles in the y -direction across the heightof the plate. Since all values in figure 4.2a are negative we see that SPH under-estimates the frequency of the plate. That is, the plate oscillates slower thanit should.

The relative error in amplitude is shown in figure 4.2b. Since all values arepositive it is clear that SPH here overestimates the amplitude of the oscilla-tions. In both figures the smallest error is about 3% and the relative error de-creases when ny is increased. The exception is for h = 1.3 where we see thatthe error increases with increasing ny . Overall h = 1.5 seems to be the bestchoice of relative smoothing radius for this problem. Note that figure 4.2aand 4.2b qualitatively looks the same except mirrored since the errors differin sign.

That SPH has problems for h = 1.3 is also visible if one plots the von Misesstress as seen in figure 4.3. The figure is taken at a time close to one quarterof the period when the plate is at its highest position (w > 0). It is possible tosee some spikes on the lower side (y < 0) close to where the plate is attachedto the wall. This looks rather unphysical. Note that the top side (y > 0) closeto the wall in figure 4.3 looks a lot better than the bottom side.

This problem does not seem to exist for a larger smoothing radius. Thecorresponding plot for (ny , h ) = (21, 1.5) is seen in figure 4.4. Here the von

25

Mises stress looks a lot more physical. Neither does the problem seem to existfor (ny , h ) = (9, 1.3) as seen in figure 4.5. It is however present for (ny , h ) =(15, 1.3).

1.3 1.5 1.7 1.9

−0.25

−0.2

−0.15

−0.1

−0.05

h

ωSPH−ω0

ω0

ny=21

ny=15

ny=9

(a) Frequency

1.3 1.5 1.7 1.9

0.1

0.2

0.3

hA

SPH−A

A

ny=21

ny=15

ny=9

(b) Amplitude

Figure 4.2: Relative error in amplitude and frequency for different values ofny and h .

x0/m

y 0/m

0 0.2 0.4 0.6 0.8 1

−0.050

0.05

KPa0 0.5 1 1.5 2 2.5

Figure 4.3: Von Mises stress for ny =21 and h = 1.3 at a time t = 1.5681s .

26

x0/m

y 0/m

0 0.2 0.4 0.6 0.8 1

−0.050

0.05

KPa0 0.5 1 1.5 2

Figure 4.4: Von Mises stress for ny =21 and h = 1.5 at a time t = 1.5681s .

x0/m

y 0/m

0 0.2 0.4 0.6 0.8 1−0.05

00.05

KPa0 0.5 1 1.5 2

Figure 4.5: Von Mises stress for ny =9 and h = 1.3 at a time t = 1.5648s .

4.2 Bending Plate

This problem was also solved without artificial stress, since the same type ofdisturbance as that in figure 4.1 could be seen also here.

The centreline of the plate is seen in figure 4.6 for the case of (ny , h ) =(21, 1.7) after running the simulation for 11 seconds. As can be seen the agree-ment with the analytical solution is very good. The error in end angle at thistime is typically of order:

|θn ume r i c a l −θa na l y t i c a l | ∼ 0.1,

which should be considered very good. It is possible to see some boundaryeffects close to where the plate is attached to the wall (figure 4.7a) and at theend of the plate (figure 4.7b).

At t = 11s the particles are positioned as can be seen in figure 4.8 which isfairly regular. However, if the simulation is continued for an extensive amountof time the particles position themself as seen in figure 4.9. The particlesbetween x = 0 and x = 0.3m have become irregularly spaced. In the bot-tom part of this region (y < 0) particles have also formed lines. It thus seems

27

like when the system has been damped enough the particles start to clump to-gether. You can also see a slight tendency of this in figure 4.8. For x > 0.3m theparticles in figure 4.9 look more regularly spaced. Since this problem occursa study on different values of ny and h is no longer interesting.

0 0.5 1

−0.04

−0.03

−0.02

−0.01

0

x/ meter s

y−displacement/m Numerica l

Analytica l

Figure 4.6: Centreline of plate for (ny , h ) = (21, 1.7)

0 0.05 0.1 0.15−15

−10

−5

0

5x 10

−4

x/ meter s

y−displacement/m

Numerica lAnalytica l

(a) Centreline at beginning of plate.

0.9 0.95 1

−0.04

−0.038

−0.036

−0.034

−0.032

x/ meter s

y−displacement/m

Numerica lAnalytica l

(b) Centreline at end of plate.

Figure 4.7: Boundary effects seen at both ends of the plate.

28

0 0.05 0.1 0.15 0.2 0.25 0.3

−0.05

0

0.05

x/m

y/m

Figure 4.8: Particle positions at t = 11s .

0 0.05 0.1 0.15 0.2 0.25 0.3

−0.05

0

0.05

x/m

y/m

Figure 4.9: Particle positions after a long time, t > 150s .

4.3 Modus I Crack

Let nx be the number of particles across the width of the body. The y - dis-placements for the particles marked in figure 3.8a are shown in figure 4.10for (nx , h ) = (60, 1.7). We see that the numerical solution agrees well with theanalytical solution, except very near the crack tip at x0 = 0. Note that thefour particles closest to the boundary at x0 =−0.1m in figure 4.10 actually areforced to agree with the analytical solution. This is from the way the boundaryconditions are implemented.

The stress components for the particles marked in figure 3.8b are shownin figures 4.11a to 4.11c together with the analytical solutions. It can be seenthat the σy y -component in figure 4.11b agrees very well with the analyticalsolution. For theσx x - andσx y -components the results agree very well exceptvery close to the tip of the crack. Both in figure 4.11a and figure 4.11b we alsosee that there is some problem close to the boundary of the body (close to x0 =0.1m). Even though figures 4.11a to 4.11c are the results for (nx , h ) = (60, 1.7)the qualitative look is the same as for other values of nx and h .

The relative error in energy release rate for the method when the crack isforced to open (propagation) is shown in figure 4.12a. For all values the erroris very large, around 20%-30%, and since all values are negative the energy re-lease rate is always underestimated. As can be seen in the figure 4.12a the errordecreases when nx increases. The exception is the value of (nx , h ) = (40, 1.5)which is much lower than the general trend. From figure 4.12a it is hard to see

29

that any particular value of h would be better.The corresponding error when the energy release rate is calculated by pro-

jection is shown in figure 4.12b. As can be seen these errors are typically afactor 100 smaller compared to the values in figure 4.12a. Note that about halfof the values in figure 4.12b overestimate and about half underestimate theenergy release rate. For each value of nx the smallest error occur for h = 1.5.It should however be emphasized that the relative error in G2 only changesabout 1% when h is varied between 1.3 and 1.9, as can be seen in figure 4.12b.

The error-norms for the x - and y -displacements (see equation 3.23) areshown in figure 4.13a for different values of nx and h . Each ux -error-norm,e (ux , h , nx ), is normalised to the maximum error-norm that occurs for thetested values of nx and h :

e (ux , h , nx )

maxnx ,he (ux , h , nx )

and correspondingly for the u y -error-norm. This is the reason why the largestvalue in figure 4.13a is 1 for both ux and u y . Because of this normalisation,figure 4.13a says nothing about how large the error-norm in equation 3.23 ac-tually is but rather how it changes when varying nx and h . One can not evencompare the error-norm in ux and in u y in figure 4.13a, except by shape ofthe curves. From figure 4.13a it is clear that the error decreases with increas-ing nx and that the displacements favour the lower values of h . It is on theother hand worth noting that even if the error is larger for greater values of hthe error also decreases more rapidly.

The corresponding normalised error-norms (equation 3.22) for the stressesare shown in figure 4.13b. Also here is it clear that the error decreases with in-creasing nx . For σx x and σy y it is clear that h = 1.3 and h = 1.5 is favoured.For theσx y -component the error is however larger for h = 1.3

Overall figure 4.13 and 4.12 together indicates that h = 1.5 seems to be thebest choice of relative smoothing radius also for this problem.

30

−0.1 −0.05 0−4

−2

0

2

4

x0/m

y−

Displacemen

t/mm

numericalanalytical

Figure 4.10: y -displacements for the particles marked in figure 3.8a againsttheir original x -position. In modus I with (nx , h ) = (60, 1.7).

0 0.05 0.10

0.05

0.1

0.15

0.2

x0/m

σxx/MPa

numericalanalytical

(a)σx x

0 0.05 0.10

0.1

0.2

0.3

0.4

0.5

x0/m

σyy/MPa

numericalanalytical

(b)σy y

0 0.05 0.1−0.15

−0.1

−0.05

0

0.05

0.1

0.15

x0/m

σxy/MPa

numericalanalytical

(c)σx y

Figure 4.11: Stress components of the particles marked in figure 3.8b againsttheir original x -position. For nx = 60 and h = 1.7. In modus I.

31

1.3 1.5 1.7 1.9

−0.3

−0.28

−0.26

−0.24

−0.22

−0.2

−0.18

h

nx =40

nx =50

nx =60

(G1−

G)/

G

(a) Relative error in G1

1.3 1.5 1.7 1.9

−4

−2

0

2

4

6

8x 10

−3

h

nx =40

nx =50

nx =60

(G2−

G)/

G(b) Relative error in G2

Figure 4.12: Relative error for G1 and G2 for different values of nx and h . Inmodus I.

1.3 1.5 1.7 1.9

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

h

norm

aliz

ed e

rror

nor

m

ux: nx =40

ux: nx =50

ux: nx =60

uy: nx =40

uy: nx =50

uy: nx =60

(a) Displacements

1.3 1.5 1.7 1.9

0.5

0.6

0.7

0.8

0.9

1

h

norm

aliz

ed e

rror

nor

m

σxx: nx =40σxx: nx =50σxx: nx =60σyy: nx =40

σyy: nx =50

σyy: nx =60

σxy: nx =40

σxy: nx =50

σxy: nx =60

(b) Stresses

Figure 4.13: Normalised error-norms for the displacements and stresses fordifferent values of nx and h .

32

4.4 Modus II Crack

The x -displacements of the particles marked in figure 3.8a are shown in figure4.14. As can be seen the agreement with the analytical solution is very good,although the error clearly gets bigger closer to the crack tip. The stresses forthe particles marked in figure 3.8b are shown in figure 4.15. We essentiallyhave the same case as in modus I. The stresses follow the general trend of theanalytical solution but clearly have some problems close to the edge of thebody at x0 = 0.1 and close to the tip of the crack at x0 = 0. Theσx y -componentin figure 4.15c looks particularly good. It should be noted thatσx x andσy y infigure 4.15a and 4.15b analytically should be antisymmetric around the x -axisbut clearly isn’t close to the crack tip.

The relative error in energy release rate is shown in figure 4.16a for the casewhen the crack is forced to open (propagation). As can be seen the errors arealso here of order 20%-30% and decreases slowly for increasing nx . The excep-tions are h = 1.5 (where the error only changes slightly), and (nx , h ) = (40, 1.3)(which is exceptionally low and also differs in sign from the other values).Note that the lowest errors occur for h = 1.3 for each value of nx .

The same error for when the energy release rate is determined from projec-tion is shown in figure 4.16b. These values are typically a factor 10 lower thanin figure 4.16a and a factor 10 larger than the corresponding modus I case infigure 4.12b. As can be seen the errors in figure 4.16b decrease with increasingnx and the error is smallest for h = 1.9 for all values of nx . The variation with his however very slight. As can be seen the relative error in G2 varies only about0.5% for a given value of nx .

The normalised error-norms (same as in section 4.3) for the displacementsare shown in figure 4.17a and the normalised error-norms for the stresses areshown in figure 4.17b. As can bee seen the errors decreases with increasingnx and are lowest for h = 1.3 for all values of nx . It is also clear that for eachnx the error in general increase with increasing h , even if the dependence ofh becomes less important for larger values of nx . Because of this and sincethe variation in figure 4.16b is so slight h = 1.3 seems the most appropriatechoice for this problem.

33

−0.1 −0.05 0−4

−2

0

2

4

x0/m

x−

Displacemen

t/mm

numericalanalytical

Figure 4.14: x -displacements for the particles marked in figure 3.8a againsttheir original x -position. In modus II. For (nx , h ) = (60, 1.7).

0 0.05 0.1−0.4

−0.2

0

0.2

0.4

x0/m

σxx/MPa

numericalanalytical

(a)σx x

0 0.05 0.1−0.15

−0.1

−0.05

0

0.05

0.1

0.15

x0/m

σyy/MPa

numericalanalytical

(b)σy y

0 0.05 0.10

0.05

0.1

0.15

0.2

0.25

x0/m

σxy/MPa

numericalanalytical

(c)σx y

Figure 4.15: Stress components of the particles marked in figure 3.8b againsttheir original x -position. For (nx , h ) = (60, 1.7). In modus II.

34

1.3 1.5 1.7 1.9

−0.35

−0.3

−0.25

−0.2

−0.15

−0.1

−0.05

0

h

nx =40

nx =50

nx =60

(G1−

G)/

G

(a) Relative error in G1

1.3 1.5 1.7 1.90.03

0.035

0.04

0.045

h

nx =40

nx =50

nx =60

(G2−

G)/

G(b) Relative error in G2.

Figure 4.16: Relative error for G1 and G2 for different values of nx and h . Inmodus II.

1.3 1.5 1.7 1.90.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

h

norm

aliz

ed e

rror

nor

m

ux: nx =40

ux: nx =50

ux: nx =60

uy: nx =40

uy: nx =50

uy: nx =60

(a) Displacements

1.3 1.5 1.7 1.9

0.5

0.6

0.7

0.8

0.9

1

h

norm

aliz

ed e

rror

nor

m

σxx: nx =40σxx: nx =50σxx: nx =60σyy: nx =40

σyy: nx =50

σyy: nx =60

σxy: nx =40

σxy: nx =50

σxy: nx =60

(b) Stresses

Figure 4.17: Normalised error-norms for the displacements and stresses fordifferent values of nx and h .

35

5 Discussion

5.1 Vibrating Plate

The smallest error that SPH provided was about 3%, both for frequency andamplitude, which should be considered a quite good result. That the frequencyis underestimated is not unexpected. Most numerical methods underestim-ate the wave-speed, which should lead to a lower frequency. It would havebeen more worrying if the plate would have vibrated faster than the analyticalsolution. It should also be noted that the analytical solution in equation 3.1is only approximative to start with. Thus one can not expect a perfect agree-ment.

It is interesting that the problems for h = 1.3 occurs on the bottom sideand not on the top side of the plate in figure 4.3. The bottom side should bein a tensile state while the top side should be in a compressed state. Thus oneis lead to suspect that this problem is due to tensile instability. It is slightlytroubling that this problem is seen for (ny , h ) = (21, 1.3) and not for (ny , h ) =(9, 1.3). This leads to the question if the same problem would occur for h = 1.5if ny was increased beyond ny = 21. It would be far more satisfying if onecould find a way of treating tensile instability that doesn’t result in the dis-turbance seen in figure 4.1 on page 25.

5.2 Bending plate

It is somewhat unexpected that particles clump together on the bottom side(y < 0) of the plate. This is the side that should be in a compressed state andif this would be due to tensile instability these problems should start to occuron the top side (y > 0). Nevertheless, that particles clump together is a typicalsign of tensile instability. Thus these problems are suspected to be due to this.

Even though this problem clearly doesn’t result in any convincing resultsfigure 4.6 on page 28 still points to that it would be possible to achieve satis-fying results if one would find a better way to treat tensile instability.

5.3 Modus I/II Crack

I isn’t unexpected that we have problems close to the crack tip. Analyticallythe stresses and strains approach infinity at the tip. Due to the present singu-larity, the numerical problem is difficult. Also, when the number of particlesin the body is increased particles will be initially located close and closer to thetip. So upon increasing the number of particles the peaks of the blue curves infigures 4.11a to 4.11c and figure 4.15a to 4.15c will increase. Thus, it isn’t obvi-ous that increasing the number of particles in the body will make the stressesof the particles closest to the crack tip better, since the problem is self similar.

The estimate of the energy release rate coming from projection clearly gavevery good results compared to the estimate coming from propagation, which

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was close to useless. This can be explained by the fact that the energy calcu-lated using equation 3.18 probably isn’t a very accurate estimate. Since thestrains in equation 3.18 are determined from the stresses through equations3.19 to 3.21 the energy is proportional to the stresses squared:

U ∝σ2.

Because of this, a small error in the stresses will have a big impact on the en-ergy. Since it is clear from the figures in section 4.3 and 4.4 that we have prob-lems both near the crack tip and the boundary it isn’t unexpected that theestimated energy isn’t very good. Since the highest stresses also occur close tothe crack tip these errors will have the biggest effect on the energy estimate.

Another error source is that the differentiation in equation 3.24 is for sim-plicity taken with respect to the initial particle spacing, ∆x . But since theparticles around the crack tip moves slightly in the x -direction during theloading of the structure, the crack might not grown exactly a distance∆x eachtime it is forced open.

It is worth noting that in the current formulation the location of the cracktip is a bit fuzzy. Since each node of the crack is located between a pair ofparticles, the end of the crack-segment closest to origo is initially located at

(x , y ) = (−∆x

2, 0),

even though we assume that the crack tip is located in origo. If we would wantto set up a problem where the crack tip is moved by some amount smaller than∆x/2 this wouldn’t make any difference to the numerical formulation. Thus,it is obvious that the tip is located somewhere in the region

(x , y ) ∈ −∆x

2< x <

∆x

2,−∆y

2< y <

∆y

2,

but we cannot say exactly where.The non-existence of antisymmetry that can be see in figures 4.15a and 4.15b

on page 34 could be explained by tensile instability. If we have an antisymmet-ric case where one side of a symmetry line is in a tensile state and the otherside is in a compressed state this symmetry can be broken through the nu-merics in two ways:

1. The tensile instability should cause the largest problems on the tensileside.

2. The way we tackle tensile instability through the method in section 2.5on page 11 breaks antisymmetry since we add an artificial term only onthe tensile side.

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5.4 General

Looking on all the benchmarking cases studied in this report a value of h = 1.5seems like a good choice.

It isn’t unexpected that we have weird effects close to the boundaries inseveral of the benchmarking problems. Removing the first term in equation2.8 actually relies on the fact that the domain Ω isn’t truncated (see Liu andLiu [5] page 40). Thus some boundary effects are expected.

5.5 Suggestions for Further Investigations

There are several kinds of correction terms that have been suggested to obtainhigher accuracy in SPH, see for example Bonet and Lok [4] or Oger et al. [10] .It would be interesting to see if these corrections would give better results forour benchmarking cases.

There are also several proposed ways to treat tensile instability that havenot been tested in this study. Perhaps the most promising seems to be theuse of additional “stress points” as suggested in Randles and Libersky [11]. Itwould definitely be interesting to see if this would solve the problems experi-enced in sections 4.1 and 4.2.

6 Conclusions

• The results of the benchmarking cases show that SPH is useful for prob-lems in solid and fracture mechanics. It is however necessary to finda better way to treat tensile instability than by the method of artificialstress that has been implemented here.

• A study of the modus I/II crack shows that SPH has problems capturingthe singularities that occur at the tip of the crack, where the stresses andstrains approach infinity.

• It’s recommended to use a smoothing radius equal to 1.5 times the initialparticle spacing.

• The most appropriate way to determine the energy release rate is proba-bly from projection, i.e. from the mechanical state at the tip of the crack.

• In SPH the position of the crack tip is slightly undefined. The tip is lo-cated within some region of width and height proportional to the parti-cle spacing, but it is not possible to say where within this region.

• Even if a problem is analytically antisymmetric on different sides of aline the numerical solution might not be antisymmetric. This could bedue to tensile instability or artificial stress.

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References

[1] J. P. Gray, J. J. Monaghan, and R. P. Swift. SPH elastic dynamics. Computermethods in applied mechanics and engineering, 190:6641–6662, 2001.doi:10.1016/S0045-7825(01)00254-7.

[2] J.P. Gray and J.J. Monaghan. Numerical modelling of stressfields and fracture around magma chambers. Journal ofVolcanology and Geothermal Research, 135(3):259–283, 2004.doi:10.1016/j.jvolgeores.2004.03.005.

[3] D. C. Simkins and S. Li. Meshfree simulations of thermo-mechanicalductile fracture. Computational Mechanics, 38(3):235–249, August 2006.doi:10.1007/s00466-005-0744-8.

[4] J. Bonet and T. S. L. Lok. Variational and momentum preservationaspects of Smooth Particle Hydrodynamic formulations. ComputerMethods in applied mechanics and engineering, 180(1-2):97–115, 1999.doi:10.1016/S0045-7825(99)00051-1.

[5] G.R. Liu and M.B. Liu. Smooth Particle Hydrodynamics : a meshfreeparticle method. World Scientific, 5 Toh Tuck Link, Singapore 596224,2003. ISBN 9789812384560.

[6] J. J. Monaghan. Smoothed Particle Hydrodynamics. An-nual review of Astronomy and Astrophysics, 30:543–574, 1992.doi:10.1146/annurev.aa.30.090192.002551.

[7] J. W. Swegle, D. L. Hicks, and S. W. Attaway. Smoothed Particle Hydro-dynamics Stability Analysis. Journal of Computational Physics, 116:123–134, 1995. doi:10.1006/jcph.1995.1010.

[8] KTH. Handbok och formelsamling i Hållfasthetslära. Institutionen förhållfasthetslära KTH, Stockholm, 6th edition, 2008.

[9] M. L. Williams. On the stress distribution at the base of a stationary crack.Journal of Applied Mechanics, 24:109–114, 1957.

[10] G. Oger, M. Doring, B. Alessandrini, and P. Ferrant. An improved SPHmethod: Towards higher order convergence. Journal of ComputationalPhysics, 225(2):1472–1492, August 2007. doi:10.1016/j.jcp.2007.01.039.

[11] P. W. Randles and L. D. Libersky. Normalized SPH with stresspoints. International Journal for Numerical Methods in Engineering,48:1445–1462, 2000. doi:10.1002/1097-0207(20000810)48:10<1445::AID-NME831>3.0.CO;2-9.

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A Used Parameters

The parameters that were used for the different benchmarking cases are shownin table A.1 to A.3.

Table A.1: Parameters used for the vibrating plate.

E = 1M P a H0 = 0.02L (eq 3.1)ν= 0.3 C = 0.8 (eq 2.27)ρ = 7750k g /m 3 η= 0 (eq 2.26)L = 1m γ= 0 (eq 2.23)D = L/12

Table A.2: Parameters used for the bending plate.

E = 1M P a u py =−0.04L (eq 3.4)

ν= 0.3 C = 0.09 (eq 2.27)ρ = 7750k g /m 3 η= 0.9 (eq 2.26)L = 1m γ= 0.7 (eq 2.23)D = L/12 N = 4840 (eq 3.4)

Table A.3: Parameters used for the modus 1/modus 2 crack.

E = 6M P a KI = KI I = 35.8K P a (eq 3.6 to 3.9)ν= 0.3 N = 8467 (eq 3.4)ρ = 7750k g /m 3 η= 1 (eq 2.26)L = 0.2m γ= 0.05 (eq 2.23)C = 0.09 (eq 2.27)

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