steigemann fulland on the computation of the pure neumann problem

7/27/2019 Steigemann Fulland on the Computation of the Pure Neumann Problem

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On the Computation of the Pure Neumann Problem in2-dimensional Elasticity1

Martin Steigemann1, Markus Fulland2

1Institute of Analysis and Applied Mathematics, University of Kassel2Westfalisches Umwelt Zentrum, Universitat Paderborn

Abstract. The accurate computation of stress intensity factors (SIFs) plays a decisive rolein the determination of crack paths. The calculation of SIFs with the help of singular weightfunctions leads to pure Neumann problem for anisotropic elasticity in a plane domain with a

crack. Here a method is presented to overcome the specific numerical difficulties which ariseswhile calculating these solutions with Finite Element methods. The accuracy and advantage of this method are shown by a numerical example, the calculation of SIFs of a compact tensionspecimen.

Key words. Two-dimensional elasticity, finite elements, Neumann problem, stress intensityfactors, anisotropic solids, weight functions

AMS Subject Classification. 65N22, 65N30, 74G15, 75R10, 74S05

1. Introduction. For many problems in elasticity theory it is important to compute solutionsof the pure Neumann problem

−∇ · σ(u) = f in Ω, σ(n)(u) = g on ∂ Ω

where Ω ⊂ R2 is some open domain. In particular, if the domain has a crack, in fracture

mechanics one is often interested in computing the stress intensity factors K j . Stress intensityfactors (SIFs) play an essential role while computing the path of the crack in the solid underan external load. The accuracy of the computed crack path depends highly on the SIFs. Forisotropic solids, there are various methods to compute these factors, but up to now, it is notclear, if this all works for anisotropic ones. Nevertheless, for every homogeneous anisotropicsolid, K I and K II can be computed very accurately by evaluating an integral over so-calledweight functions ζ j and the applied loads, namely

K j =

Ω

ζ j · f dx +

∂ Ω

ζ j · g ds, j = 1, 2.

These weight functions, introduced by Mazya and Plamenevsky [14], are composed of singulareigenfunctions of the elasticity operator in the half-plane with a semi-infinite crack and somespecial solutions of the Neumann problem in Ω. Eigenfunctions of the elasticity operator areknown analytically for isotropic homogeneous solids and can be computed with arbitrary pre-cision for anisotropic ones. The solution of the Neumann problem depends on the domain and

1In: International Journal of Fracture, 146, 265-277 (2007)

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has to be computed numerically.

A solution of the pure Neumann problem is uniquely determined up to a rigid motion only andthe Green’s formula implies, that the data f and g must be orthogonal to such motions. Thedirect Galerkin discretization leads to a linear system of equations with a stiffness matrix A

with three-dimensional kernel and a right-hand side orthogonal to this kernel. Computing afinite element solution from this system brings several numerical difficulties.

First of all, the linear system cannot be solved by a standard solver like the Conjugate Gradi-ent method. One has to modify the method or has to use an other approach like the MinimalResidual method.Moreover, a discrete solution only exists, if the discrete right-hand side is orthogonal to thekernel of A. While computing the discrete right-hand side using quadrature formulas and with

finite precision, the orthogonality condition is not fulfilled precisely. The accuracy of the nu-merical solution and the stress intensity factors will highly depend on this precision, while usingthe Minimal Residual approach.

To avoid these difficulties, a widely used basic approach to fix the solution is to prescribe thevalue on at least two nodes. This eliminates the kernel of the stiffness matrix and one can applya conventional solver. The linear system is solvable for all right-hand sides. From a mathemat-ical point of view, the computed finite element solution is only a weak solution in the Sobolevspace H 1(Ω) and it is not obvious that such a solution is continuous or exists at a single point.Besides this, the numerical solution will have a peak at these nodes and this can destroy thecomputation of SIFs.

We want to follow another idea to solve the Neumann problem, presented in [3] for the Laplaceequation. There exists a unique solution of the Neumann problem, if we look for a solution ina subspace of H 1(Ω), complementary to the space of rigid motions. Such a subspace H R wasintroduced in [6] and with a projection

P : H 1(Ω) −→ H R

an arbitrary solution of the Neumann problem can be fixed uniquely in this subspace. This alsoworks in the discrete case. Using a discrete analogon to P , the unique numerical solution inH R can be found from any numerical solution of the Neumann problem. The discrete projec-tion can also be used to improve the accuracy of the orthogonality conditions of the discrete data.

In this paper we will consider a bounded domain with a crack and our main interest is to computethe displacement field and the stress intensity factors K I and K II under an external load. First,we will construct a subspace H R, complementary to rigid motions and a projection P onto it.On H R we recall the existence and uniqueness of a weak solution.We will show, that the Galerkin discretization leads to a linear system with a three-dimensionalnull-space, while using Lagrangian bilinear finite elements. It turns out, that the continuousorthogonality conditions are equivalent to the discrete ones. We give a discrete analogon of the projection P to compute the finite element approximation of the unique solution in H Rand we will show the advantages of this projection method. For a Compact Tension specimen,well-known in fracture mechanics, we will present numerical results, containing the displacement

field and stress intensity factors.

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2. The Neumann problem. Let G be a domain in the plane R2 with compact closure G

and Lipschitz boundary Γ, containing the origin. We consider the pure Neumann problem of

2-dimensional elasticity theory in the domain Ω := G \ Ξ, where Ξ := x ∈ G : x1 ≤ 0, x2 = 0is a rectilinear edge cut:

L (∇x)u(x) = −∇ · σ(u) = f (x), x ∈ Ω

N (∇x)u(x) = σ(n)(u) = 0, x ∈ Ξ+ ∪ Ξ−

N (∇x)u(x) = σ(n)(u) = p(x), x ∈ Γ

(1)

n = (n1, n2)⊤ is the outward normal, u = (u1, u2)⊤ the displacement vector, f = (f 1, f 2)⊤

the vector of body forces and p = ( p1, p2)⊤ denotes the vector of surface load. With Ξ+

and Ξ− we denote the upper and lower sides of the crack, considered to be tension-free. To

omit technical difficulties, we also assume, that Ω can be divided in two Lipschitz domainsΩ− = x ∈ Ω : x2 < 0 and Ω+ = x ∈ Ω : x2 > 0. The strain tensor in cartesian coordinatesis given by

εij(u; x) =1

2

∂ui

∂x j+

∂u j

∂xi

, i, j = 1, 2

and by Hooke’s Law, for the stress tensor the relation holds

σij(u; x) =2

k,l=1

aklijεkl(u; x), i, j = 1, 2.

The real constants aklij , called elastic moduli, satisfy the following symmetry and positivityconditions

aklij = alk ji = a jkli ,

2i,j,k,l=1

ζ ijaklijζ kl ≥ c0

2i,j=1

|ζ ij|2, c0 > 0

for any symmetric real 2 × 2 matrix ζ . In an anisotropic material there are 6 different elasticmoduli, for an isotropic material holds

a1111 = a2222 = λ + 2µ a1122 = a2211 = λ a1212 = a2121 = µ a1211 = a1222 = 0

with the Lame constants λ and µ.

3. The variational problem and weak solutions. We recall the definition of a weaksolution. With H m(Ω) we denote the usual Sobolev space of order m with norm

u; H m(Ω) =

mk=0

∇ku; L2(Ω)2

1/2

,

here we use the notation

∇ku; L2(Ω)2 := |α|=k ∂ αx u; L2(Ω)2.

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Ω

Ξ

Γ

x1

x2

Figure 1: Homogeneous elastic body Ω with a rectilinear edge cut Ξ

The expression (u, v)Ω := Ω

u · v dx indicates the inner-product of the Lebesgue space L2(Ω) and

H 1/2(∂ Ω) denotes the space of traces equipped with the norm

u; H 1/2(∂ Ω) := inf

v; H 1(Ω) : v ∈ H 1(Ω) and u = v on ∂ Ω

.

In our notation, we do not distinguish between scalar and vector-valued functions. The vectorspace of rigid motions is

R = u ∈ C ∞(R2) : u = a + b · xR, a ∈ R2, b ∈ Rwith vR(x) := (v2(x), −v1(x))⊤ for every field v. On the Lipschitz domain G for fields u ∈ H 2(G),v ∈ H 1(G) we have Green’s formula

(L u, v)G + (N u, v)∂G = a(u, v) with a(u, v) =

G

2i,j=1

2k,l=1

aklijεkl(u)εij(v) dx.

Note that 12a(u, v) is the elastic energy. As considered, Ω can be divided in two Lipschitz domains

Ω+ and Ω− and applying Green’s formula to each, we find

a(u, v)Ω = a(u, v)Ω+ + a(u, v)Ω− = (L u, v)Ω + (N u, v)∂ Ω + (N u, v)γ + + (N u, v)γ −

with γ := x ∈ G : x1 > 0, x2 = 0. If u ∈ H 2(Ω), v ∈ H 1(Ω), the traces on γ + and γ − coincide.Taking into account the direction of the outward normal on γ + and γ −, we find the equation

(N u, v)γ + = −(N u, v)γ −

and Green’s formula still holds on Ω.For right-hand sides f ∈ L2(Ω), g ∈ H 1/2(∂ Ω) with g = p on Γ and g = 0 on Ξ, we callu ∈ H 1(Ω) a weak solution of the Neumann problem, if the relation

a(u, v) = (f, v)Ω + (g, v)∂ Ω =: F (v) (2)

holds for all v ∈ H 1(Ω).

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Lemma 1 For u ∈ H 1R(Ω) there exists a constant c > 0 with

ε(u); L2(Ω) ≥ cu; H 1(Ω)

For a proof see [9], [12] or [6]. The existence and uniqueness of a weak solution follows now ina standard way from the Lax-Milgram Theorem.

Although a weak solution of the Neumann problem exists only for self-balanced data, the pre-vious arguments show, that the variational problem

a(u, v) = (f, v)Ω + (g, v)∂ Ω for all v ∈ H 1R(Ω) (4)

is well-posed in H 1R(Ω) for any f ∈ L2(Ω) and g ∈ H 1/2(∂ Ω) with g = 0 on Ξ. The next resultshows that if f and g do not satisfy the orthogonality conditions (3), a field u ∈ H 1(Ω) solves aNeumann problem with modified right-hand sides, if u satisfies (4).

Lemma 2 Let u ∈ H 1(Ω) be a solution of problem (4) with f ∈ L2(Ω), g ∈ H 1/2(∂ Ω)with g = 0 on Ξ. Then u solves the Neumann problem (1) with right-hand sides P ∗Ωf

and P ∗∂ Ωg, where

P ∗Ωf := f −1

|Ω|

Ω

f dx +

∂ Ω

g ds

,

P ∗∂ Ωg := g −1

2|Ω|

(f, xR)Ω + (g, xR)∂ Ω −

1

|Ω|

Ω

xR dx

Ω

f dx +

∂ Ω

g ds

nR

Proof: By the Gauss Theorem, for u ∈ H 1(Ω) holds Ω

rot(u) dx =

Ω

∇ · uR ds =

∂ Ω

n · uR ds = −

∂ Ω

nR · u ds

and a simple calculation proves

(f, P Rv)Ω + (g, P Rv)∂ Ω = (P ∗Ωf, v)Ω + (P ∗∂ Ωg, v)∂ Ω for all v ∈ H 1(Ω).

From H 1(Ω) = H 1R(Ω) ⊕ R we have a(u, v) = a(u, P Rv) for all v ∈ H 1(Ω) and because u is asolution of problem (4) there holds

a(u, v) = a(u, P Rv) = (f, P Rv)Ω + (g, P Rv)∂ Ω = (P ∗Ωf, v)Ω + (P ∗∂ Ωg, v)∂ Ω (5)

for all v ∈ H 1(Ω). In particular we have for v ∈ C ∞0 (Ω)

(u,L v)Ω = a(u, v) = (P ∗Ωf, v)Ω,

which implies L u = P ∗Ωf in the distributional sense. Moreover, we have L u ∈ L2(Ω) and theGauss theorem in the weak formulation implies

a(u, v) = (L u, v)Ω + (N u, v)∂ Ω

for all v ∈ H 1(Ω). Together with (5) this leads to

(L u − P ∗Ωf, v)Ω + (N u − P ∗∂ Ωg, v)∂ Ω = (N u − P ∗∂ Ωg, v)∂ Ω = 0 for all v ∈ H 1(Ω).

Since v ∈ H 1(Ω) is arbitrary, this implies N u = P ∗∂ Ωg.

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4. Finite Element Solutions. Consider a triangulation

T h = T

(r)

: r = 1, . . . , Rhof Ω into isoparametric rectangles. In the following, we always use bilinear Lagrangian finiteelements and for simplicity, we assume that Ω has no reentrant corners. We also neglect curvedparts of the boundary and assume

Rhr=1

T (r)

= Ω.

With xhk , k = 1, . . . , N h we denote the vertices of the triangulation and

φhi

N

i=1

with φhi (xh

k) = δ i,k for i, k ∈ 1, . . . , N

is a nodal basis of the space of shape functions P . As we allow isoparametric bilinear elements

T (r) we assume that there exist constants h(r)2 > h

(r)1 > 0 and x

(r)0 ∈ T (r) with

Bh(r)1

(x(r)0 ) ⊂ T (r) ⊂ B

h(r)2

(x(r)0 ),

h(r)2

h(r)1

< C for all 1 ≤ r ≤ Rh (6)

where C does not depend on r and Bh(x) is the circle with radius h around x. Moreover, weassume that every element T (r) is convex and we define the discretization parameter as

h := max2h(r)2 : 1 ≤ r ≤ Rh .

This conditions ensure, that the elements T (r) are not too degenerated and are needed forstandard error estimates (see e.g. [8]). To approximate vector fields, we need a vector-valuedfinite element space and we define a nodal basis ϕh

i i=1,...,2N by

ϕh2i−1(x) = φh

i (x)e1, ϕh2i(x) = φh

i (x)e2, i = 1, . . . , N

where e j denotes the j-th unit vector. We define the finite element space

V h :=

vh : vh(x) =

2N i=1

viϕhi (x)

⊂ H 1(Ω) ∩ C 0(Ω ∪ ∂ Ω)

and the interpolation operator

Πu : H 1(Ω) ∩ C 0(Ω ∪ ∂ Ω) −→ V h with Πu :=2N i=1

u(xhi )ϕh

i (x).

Because the domain Ω has a crack Ξ, there is a difference between Ω and Ω ∪ ∂ Ω. We cannotrequire, that functions are continuous over the crack and in Ω.

In the following we assume, that the data f and g are self-balanced. The Galerkin approximationproblem is to find uh ∈ V h with

a(uh, v) = (f, v)Ω + (g, v)∂ Ω for all v ∈ V h.

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Denoting the coefficient vector of uh ∈ V h by u := (u1, . . . , u2N )⊤, the problem is reduced to

solve the system of linear equations

Au = f + g

with the stiffness matrix

Ai,j = a(ϕhi , ϕh

j ), i, j = 1, . . . , 2N

and the discrete right-hand side

(f + g)i =

Ω

f · ϕhi dx +

∂ Ω

g · ϕhi ds, i = 1, . . . , 2N

or f + g = (f 1, φh1)Ω + (g1, φh1)∂ Ω, (f 2, φh1)Ω + (g2, φh1)∂ Ω, . . . , (f 2, φhN )Ω + (g2, φhN )∂ Ω⊤.

Because of R ⊂ V h, there is no reason why the discrete problem should have a unique solutionor is always solvable. To prove similar solvability conditions as in the continuous case, we needsome more definitions. With x = (x1, x2, . . . , x2N )

⊤ := (xh1,1, xh

1,2, . . . , xhN,1, xhN,2)⊤ we denote

the vector of vertices and xR = (x2, −x1, . . . , x2N , −x2N −1)⊤ is the rotated one, similar to thecontinuous case. The mean value of a function uh ∈ V h can be written as

Ω

uh dx =N i=1

u2i−1

u2i

(φh

i , 1)Ω = b⊤1 u + b⊤2 u

and the integral over the rotation of uh ∈ V h is Ω

rot(uh) dx =

Ω

∂ 1uh,2 − ∂ 2uh,1 dx =N i=1

u2i(∂ 1φhi , 1)Ω − u2i−1(∂ 2φh

i , 1)Ω = b⊤Ru

with the vectors

b1 := ((φh1 , 1)Ω, 0, (φh

2 , 1)Ω, 0, . . . , (φhN , 1)Ω, 0)⊤

b2 := (0, (φh1 , 1)Ω, 0, (φh

2 , 1)Ω, . . . , 0, (φhN , 1)Ω)⊤

bR := (−(∂ 2φh1 , 1)Ω, (∂ 1φh1 , 1)Ω, . . . , −(∂ 2φhN , 1)Ω, (∂ 1φhN , 1)Ω)⊤.

With the definitions c1 := (1, 0, 1, 0, . . . , 1, 0)⊤ and c2 := (0, 1, 0, 1, . . . , 0, 1)⊤, we can formulatethe solvability conditions of the system of linear equations.

Lemma 3 The linear system of equations Au = f + g with Ai,j = a(ϕi, ϕ j) and f i + gi = (f, ϕh

i )Ω + (g, ϕhi )∂ Ω is solvable if and only if

(f + g)⊤v = 0 for all v ∈ ker(A)

The kernel is of the form

ker(A) = spanc1, c2, xR

8



while using bilinear first order Lagrangian elements. The continuous compatibility conditions

(f j , 1)Ω + (g j , 1)∂ Ω = 0, (f, xR)Ω + (g, xR)∂ Ω = 0, j = 1, 2,

are equivalent to the discrete compatibility conditions

c⊤1 (f + g) = 0, c⊤2 (f + g) = 0, x⊤R(f + g) = 0.

Proof: For the first conclusion, we only have to prove the representation of the kernel. Becausewe use bilinear first order Lagrangian elements, the representation

1 =N

i=1φhi (x), xR =

N

i=1 xh2ie1 − xh

2i−1e2

φhi (x)

is valid. For k ∈ 1, . . . , 2N we find

0 = a(e1, ϕhk) =

N i=1

a(φhi e1, ϕh

k) = (Ac1)k and 0 = a(e2, ϕhk) =

N i=1

a(φhi e2, ϕh

k) = (Ac2)k.

The same argumentation leads to 0 = a(xR, ϕhk) = (AxR)k. Again, by construction of the nodal

basis we have

(f j, 1)Ω + (g j , 1)∂ Ω =N

i=1 (f j, φh

i )Ω + (g j , φhi )∂ Ω

= c⊤ j (f + g), j = 1, 2

and

(f, xR)Ω + (g, xR)∂ Ω =N i=1

x2i((f 1, φh

i )Ω + (g1, φhi )∂ Ω) − x2i−1((f 2, φh

i )Ω + (g2, φhi )∂ Ω)

= x⊤R(f + g)

The lemma is proved.

The system of linear equations has only a solution, if the discrete right-hand side is orthogonalto the kernel of system matrix A. We have proven, that this condition is fulfilled, if the dataare self-balanced.Solving the linear system numerically, there is another problem. The system is not uniquelysolvable and the system matrix is only semi-definite. We can not apply a standard method tosolve like Conjugate Gradient. Nevertheless, such problems can be solved by the Minimal Resid-ual method (MinRes) developed in [19]. But again, the computed solution is not necessarily thenormalized unique one. For the continuous problem, we have solved this problems with the helpof projections. It is obvious to translate this ideas to the discrete case.

The discrete projector P is given by

P = I −

c1b⊤1 + c2b⊤2

(1, 1)Ω +

1

2(1, 1)ΩxR − c1b⊤1 + c2b⊤2 (1, 1)Ω xRb

⊤

R

9



and we prove b⊤1 Pu = b⊤2 Pu = 0 which is equivalent to (P Ruh,j, 1)Ω = 0, j = 1, 2. A shortcalculation shows

b⊤1c1b⊤1

(1, 1)Ω=

1

(1, 1)Ω

N i=0

(φhi , 1)Ω

b⊤1 = b⊤1 , b⊤2

c2b⊤2(1, 1)Ω

= b⊤2 , b⊤1 c2b⊤2 = b⊤2 c1b⊤1 = 0

and we find b⊤1 Pu = b⊤2 Pu = 0. For our choice of finite elements it is clear that

1 =N i=0

φhi (x), |Ω| = (1, 1)Ω =

N i=0

(φhi , 1)Ω

and therefore, for the derivatives of the shape functions hold

0 = ∂ k1 =N

i=0 ∂ kφhi (x)T (r) , 0 =

N

i=0(∂ kφhi , 1)Ω, k = 1, 2.

With

−2 =(rot(xR), 1)Ω

(1, 1)Ω= −

1

(1, 1)Ω

N i=1

x2i−1(∂ 1φh

i , 1)Ω + x2i(∂ 2φhi , 1)Ω

=

1

(1, 1)Ωb⊤RxR

we get b⊤RPu = 0 what means (rot(P Ruh), 1)Ω = 0. Finally, the discrete analogon of theprojector P ∗ is

P⊤ = I −b1c⊤1 + b2c⊤2

(1, 1)Ω+

1

2(1, 1)ΩbR

x⊤R − x⊤R

b1c⊤1 + b2c⊤2

(1, 1)Ω

With the same arguments one can easily proof c⊤P⊤(f + g) = 0 for all c ∈ ker(A).

We will take great advantage of the discrete projectors in practical computations as we will seein the last paragraph. If one can compute some discrete solution u, the approximation of theunique solution u ∈ H 1R(Ω) is obtained just by multiplication with P. Moreover, the projectorP⊤ can be used to improve the precision of the discrete orthogonality conditions.

5. Numerical examples. For computations we use the C++ library deal.II developed atthe Numerical Methods group at University of Heidelberg. deal.II is open source and for moreinformations see www.dealii.org. We use structured meshes (see Figure 3) and only global re-finement. We don’t want to deal with local refinement or hanging nodes.

In Fracture Mechanics, one of the main interests is the computation of the stress intensity factorsK I and K II . If we consider the elastic solid Ω under an external surface load g ∈ H 1/2(∂ Ω)without body forces and tension-free crack Ξ, the stress intensity factors can computed by

K I =

∂ Ω

ζ 1 · g ds, K II =

∂ Ω

ζ 2 · g ds

see e.g. [4], [14]. The so-called weight functions ζ j are solutions of the homogeneous problem

L (∇x)ζ j(x) = 0, x ∈ Ω

N (∇x)ζ j(x) = 0, x ∈ ∂ Ω

10



with a singularity at the crack tip

ζ j(x) = r−1/2Ψ j(ϑ) + ζ j(x)

Ψ j are smooth functions, which are known analytically for isotropic solids, see [18], and canbe computed numerically for anisotropic ones [20]. Because r−1/2Ψ j are solutions of the homo-geneous elasticity problem in the whole plane with a semi-infinite crack, the remainder ζ j is asolution of the problem

L (∇x)ζ j(x) = 0, x ∈ Ω

N (∇x)ζ j(x) = 0, x ∈ Ξ+ ∪ Ξ−

N (∇x)

ζ j(x) =

p(x), x ∈ Γ

(7)

with p := −N (∇x

) r−1/2Ψ j. Moreover, the right-hand side g, defined by g = p on Γ andg = 0 on Ξ, is self-balanced and in H 1/2(∂ Ω). Therefore, a unique solution ζ j ∈ H 1R(Ω) exists.For more details about weight-functions we refer to [10] and [17, p. 278].

We take into account, that we only have to compute the solutions ζ j with finite energy by theFinite Element method. There is no need to compute solutions with singularities.

Solving the linear system. As mentioned above, we use MinRes [19] to solve the systemAu = b := f + g with u0 = 0. MinRes is a Krylov subspace method and computes in everyiteration step

b − Auk2 = min, uk ∈ Kk(A; b), k ≤ 2N (8)

where Kk(A; b) = spanb, Ab, . . . , Ak−1b is the Krylov subspace of dimension k. For moredetails about Krylov subspace methods see [15] and for a detailed representation of MinRes [21]also.We know, that a solution u exists, if the data are orthogonal on the kernel of A. But whilecomputing b by using quadrature formulas and in floating point arithmetic, the right-hand sideis not exactly orthogonal, which is of great importance for numerical computations. MinRescomputes a numerical solution, even the orthogonality conditions do not hold, but whether thisis a solution of the linear system, depends on the accuracy of the orthogonality conditions. Wewill show this in an example.

Compact Tension specimen. We consider the classical Compact Tension (CT-) specimensubjected to two symmetric forces F = 4500N in x2-direction applied in the two holes accordingto Figure 2. All other faces are traction free.We select the length units to w = 72mm and a = 17mm. The thickness of the specimen is10mm. To compare our method with well-known results, we choose the material properties

λ = 56023N

mm2µ = 26364

N

mm2or E = 70656

N

mm2ν = 0.34

corresponding to aluminium alloy 7075-T651. First, we give results for the computed displace-ment. In [16, p.18], the crack opening at the front face at point x0 := (−a − 0.25w, 0) is givenby

δ 0 = 0.1121 mm ± 0.5%

11



Computing the mean value of the last solution and the integral over its rotation we have

b⊤1 u + b

⊤2 u = −0.000393 b

⊤Ru = 0.000011

These integrals do not really vanish and the computed solution is not the normalized and uniqueone. In contrast, the integrals of the projected solution are

b⊤1 Pu + b⊤2 Pu = 4.14E − 18 b⊤Ru = 1.95E − 19.

Applying the symmetric force F , the discrete orthogonality conditions are fulfilled,

c⊤1 g = 1.54E − 18, c⊤2 g = 9.33E − 19, x⊤Rg = 3.98E − 17

and there is no need to project the data to improve this results.

Figure 3: Used mesh with 686 DOFs

Next, we compute the stress intensity factors, given in [22] by

K I = 251.4N

mm3/2, K II = 0.

N

mm3/2

with an accuracy of ±0.5% for 0.2 ≤ aw ≤ 1.0. Table 6 and Table 8 show the computed results

without projecting the data or the solutions. As one can see, the orthogonality conditions arenot fulfilled in the first iteration steps and MinRes computes no solution. To fulfill the internalstopping criterion with eps = 1. × 10−10, numerous iteration steps are needed, but the residualnorm Auk − b2 does not become small. The computed stress intensity factors are far fromthe exact values, not until using 37 758 DOFs. Computing K I with a finite element solutionwith 2 366 462 DOFs, we have an error

|K I − K I,h | = 1.4484, which means, the relative error is ≤ 0.54%.

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In contrast, Table 7 and Table 9 show the results while projecting the data and the solutions.The residual norm decrease linear, up to the effect of rounding errors, using more then 149 246

DOFs. Computing K I in the last iteration step, the error is

|K I − K I,h | = 0.2216, which means, the relative error is ≤ 0.088%.

This shows one advantage using discrete projections.In the final iteration step, the relative error of the computed values of K I differ only by 0.45%,but the number of degrees of freedom in the finite element solution is very large and computingsuch a solution needs several time. Projecting the data and the solution, the relative error inK I is less than 0.87% with only 37 758 DOFs and such a finite element solution can be com-puted in a fractional amount of time. Without projecting, this error is about 2%. This is a realadvantage of using projections and is of great interest while computing crack paths for example.Approximating the path of a crack in a solid by a polygon, one has to compute SIFs in numerousapproximation steps and using projections will save a lot of computing time. We will show thisin upcoming papers.

An example for an anisotropic solid. After we have shown some of the advantages andthe accuracy of our projection method, we discuss an anisotropic example. We consider theCT-specimen, subjected to the same symmetric forces and again, we want to compute the stressintensity factors K I and K II . The only difference to the example before is the change of theelastic behavior of the solid. In [2, p. 2349] an orthotropic material with two planes of elasticsymmetry is given. We denote this material axes by y1 and y2 and the angle between the materialaxes and the crack by α, see Figure 4. For α = 0 the material axes and the coordinate systemcoincide. If we change the angle α, the crack lies not in a plane of elastic symmetry and we haveto rotate the elastic moduli also. Of course, this is not the case if we consider an isotropic solid.Enforcing plane stress conditions, the resulting elastic moduli for different angles α are given inTable 2.

x1

x2

y1

y2

Ξ+

Ξ−α

Figure 4: Material and coordinate axes

We use the same meshes and numbers of DOFs as in the isotropic example and compute thestress intensity factors K I and K II using our projection method. The results are given in Table3. We do not present the orthogonality conditions and the numerical results without usingprojections for all angles and give them only for α = 0 in Table 10 and Table 11. For all otherangles the difference in results look nearly the same as for α = 0.An important fact is, that the SIFs strongly depend on the angle between the axes of elastic

symmetry and the crack. Even though the applied force and the solid are symmetric with re-

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steigemann fulland on the computation of the pure neumann problem

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