a cracked beam finite element for through-cracked tube

COMMUNICATIONS IN NUMERICAL METHODS IN ENGINEERINGCommun. Numer. Meth. Engng 2008; 24:761–775Published online 6 March 2007 in Wiley InterScience (www.interscience.wiley.com). DOI: 10.1002/cnm.988

A cracked beam finite element for through-cracked tube

F. Schoefs∗,†, A. Le Van and M. Rguig

Institute of Civil and Mechanical Research (GeM), UMR-CNRS 6183, Nantes Atlantic University, 2 rue de laHoussiniere, B.P. 92208 44322 Nantes Cedex 03, France

SUMMARY

The reassessment of existing structures leads to analyse the impact of local defects on the structuralbehaviour. This paper proposes a cracked beam finite element with a view to introducing the effectof large through-cracks in the structural analysis for framed structures like jacket offshore platforms.The model parameters are identified for several joint typologies using 3D finite element results. Copyrightq 2007 John Wiley & Sons, Ltd.

Received 15 November 2005; Revised 9 September 2006; Accepted 13 December 2006

KEY WORDS: cracked beam finite element; through-crack; displacement method; offshore structures

1. INTRODUCTION

The study of the impact of through-cracks on the structural integrity of offshore structures suchas jacket platforms is still a challenge. For reassessment purposes, one needs to develop specificcriteria so as to introduce the effect of structural redundancy in presence of a through-crack anddecide whether to repair or not. The purpose of this paper is to build a finite element representingthe behaviour of a through-cracked tube. The paper depicts how to transfer the main mechani-cal effects of a through-crack to the structure behaviour using an equivalent cracked beam finiteelement (CBFE). A number of studies, e.g. [1–5], have been carried out, yet giving rather in-complete solutions mainly for structural components loaded in their plane. Some were devotedto the influence of the joint flexibility in offshore structures subjected to fatigue loadings; see,for instance, [6–8]. The formulations proposed therein showed the significant role of the load-ings as well as the typologies of nodes (T-joint, Y-joint, K-joint or TK-joint). In [7], the flexiblejoints were modelled by means of three springs and their stiffnesses were given in parametric forms.

∗Correspondence to: F. Schoefs, Institute of Civil and Mechanical Research (GeM), UMR-CNRS 6183, NantesAtlantic University, 2 rue de la Houssiniere, B.P. 92208 44322 Nantes Cedex 03, France.

†E-mail: franck.schoefs@univ-nantes.fr

Copyright q 2007 John Wiley & Sons, Ltd.

762 F. SCHOEFS, A. LE VAN AND M. RGUIG

Other authors modelled complex joints without cracks by using thin shell elements, e.g. [9, 10];or to introduce specific limit conditions in buckling analysis [4].

The CBFE proposed in this paper enables one to consider all geometrical typologies of nodeswhich are encountered on jacket structures. First, we describe the cracked beam by introducingfour mechanical parameters (two stiffnesses and two eccentricities). The stiffness matrix is derivedfrom the strain energy as a function of the four parameters. Also, the comparison is made with theforce method. Second, the model parameters are identified using 3D finite element results. Theefficiency of the proposed model is shown through some specific joint typologies.

2. DESCRIPTION OF THE CRACKED BEAM FINITE ELEMENT (CBFE)

This section describes the CBFE with the purpose analysing the behaviour of a through-crackedtube, see Figure 1. The through-crack is modelled as a loss of stiffness which is represented bytwo torsional springs and two eccentricities due to the shift of neutral axis of the tube. Therefore,applying an axial force leads to a momentum at the cracked section. This effect is significant whenanalysing framed structures such as jackets offshore platforms.

2.1. The stiffness matrix K for a cracked tube

The cracked finite element will be next described in the local axis system. The x-axis is directedtowards the tube length (Figure 2) whereas the y and z-axes can be arbitrarily chosen (since thecross-section is circular, any axis in the section is principal). The shift of the neutral axis along they and z-axes is represented by two eccentricities ey and ez and the stiffness loss by two torsionalsprings of stiffness ky and kz between points 1 and 4.

Let us denote by U the nodal generalized displacement vector of the cracked beam element,including rotations �y4 and �z4 at node 4 about y and z axes, respectively (Figure 2). The degreesof freedom �y4 and �z4 are added to the standard displacement vector of displacement in order totake into account the fact that they are variables independent to others

UT =〈u1 v1 w1 �x1 �y1 �z1 u2 v2 w2 �x2 �y2 �z2 �y4 �z4 〉 (1)

where ui , vi , wi are the nodal displacements and � ji the nodal rotations.

chordtubular

bracetubular

through crack

mechanical modelling

Figure 1. Mechanical model for a cracked tube in plane (xy) [2].

Copyright q 2007 John Wiley & Sons, Ltd. Commun. Numer. Meth. Engng 2008; 24:761–775DOI: 10.1002/cnm

CRACKED FINITE ELEMENT FOR THROUGH-CRACKED TUBE 763

Conventionsat node i

(ky, kz)

Figure 2. The cracked beam finite element (CBFE) in 3D space.

The strain energy of the cracked tube is

(du(x)

dx + 1

(d2v(x)

dx + 1

(d2w(x)

(d�x (x)

dx + 1

2ky(�z1 − �z4)

2kz(�y1 − �y4)

where S denotes the cross-sectional area, I the bending second moment of area, J the torsionalsecond moment of area, E the Young modulus and G the shear modulus.

The displacements u(x), v(x), w(x) and the twist rotation �x (x) between points 3 and 2 areinterpolated in a usual way

u(x) = x

(1 − x

v(x) =(1 − 3

l2x2 + 2

(x − 2

lx2 + 1

)�z3

l2x2 − 2

l2x3 − 1

)�z2

w(x) =(1 − 3

l2x2 + 2

)w3 −

(x − 2

lx2 + 1

)�y3

l2x2 − 2

)w2 −

l2x3 − 1

)�y2

�x (x) =(1 − x

)�x3 + x

l�x2

As a matter of fact, points 1 and 4 are at the same position. Hence,

u1 = u4

v1 = v4

w1 = w4

�x1 = �x4

The portion of the beam element between points 3 and 4, which represents the eccentricities, isconsidered as rigid. Therefore,

�x3 = �x4

�y3 = �y4

�z3 = �z4

By assuming small rotations in portions 4–3, the displacement of point 3 is related to that ofpoint 4 by the rigid solid relationship

U3 =U4 + h34 ∧L43 (6)

where U3 = 〈u3, v3, w3〉 is the displacement of node 3, U4 = 〈u4, v4, w4〉 the displacement of node4, h34 =〈�x3, �y3, �z3〉 the rotation vector of the rigid element 3–4 and L43 =〈0, ey, ez〉 the vectorrelating nodes 3 and 4.

Inserting conditions (4) and (5) into Equation (6) leads to

u3 = u1 + ez�y4 − ey�z4

v3 = v1 − ez�x1

w3 = w1 + ey�x1

Relations (3) and (7) give rise to expressions for the terms in the strain energy (2)

dx= NT

1 (x)U

d2v(x)

dx2= NT

2 (x)U

d2w(x)

dx2= NT

3 (x)U

d�x (x)

dx= NT

4 (x)U

�y1 − �y4 = NT5 (x)U

�z1 − �z4 = NT6 (x)U

where vectors Ni (x) are defined as

NT1 (x) =

⟨ −1

l0 0 0 0 0

NT2 (x) =

l2+ 12

l2− 12

)ez 0 0 0

l2− 12

l3x 0 0 0

l2x − 2

NT3 (x) =

⟨0 0

l2+ 12

l3x ey

l2+ 12

)0 0 0 0

l2− 12

l2x + 2

l− 6

NT4 (x) =

⟨0 0 0

l0 0 0 0 0

l0 0 0 0

NT5 (x) = ⟨

0 0 0 0 1 0 0 0 0 0 0 0 −1 0⟩

NT6 (x) = ⟨

0 0 0 0 0 1 0 0 0 0 0 0 0 −1⟩

Thus, the strain energy (2) can be expressed as a function of the nodal displacement vector U

(∫ l

0ESN1(x)N

T1 (x) dx +

0E I N2(x)N

T2 (x) dx +

0E I N3(x)N

T3 (x) dx

+∫ l

0GJ N4(x)N

T4 (x) dx + kyN5(x)N

T5 (x) + kzN6(x)N

T6 (x)

)U (10)

The stiffness matrix K is then derived from its definition W = 12 U

TKU as

K =∫ l

0ESN1(x)N

T1 (x) dx +

0E I N2(x)N

T2 (x) dx +

0E I N3(x)N

T3 (x) dx

+∫ l

0GJ N4(x)N

T4 (x) dx + kyN5(x)N

T5 (x) + kzN6(x)N

T6 (x) (11)

Due to the additional degrees of freedom �y4 and �z4, the stiffness matrix is of dimension 14× 14instead of the usual dimension 12× 12. The stiffness matrix K can be recast as follows:

⎛⎜⎜⎝K11 K12 K1a

K21 K22 K2a

Ka1 Ka2 Kaa

⎞⎟⎟⎠ (12)

where the subscript a recalls the additional degrees of freedom �y4 and �z4 and

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

l0 0 0 0 0

012E I

l30 −ez

0 012E I

0 −ez12E I

l+ (ey2 + ez2)

0 0 0 0 ky 0

0 0 0 0 0 kz

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

K12 = KT21 =

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

l0 0 0 0 0

0−12E I

l30 0 0

0 0−12E I

−6E I

0 ez12E I

l3−ey

l3−GJ

l−ey

l2−ez

0 0 0 0 0 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

l0 0 0 0 0

012E I

l30 0 0

−6E I

0 012E I

0 0 0GJ

0 06E I

0−6E I

l20 0 0

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

KT1a = Ka1 =

⎛⎜⎜⎝

−6E I

l2−ey

l2−ky 0

−eyES

l20 −ez

l20 −kz

⎞⎟⎟⎠ (16)

KT2a = Ka2 =

⎛⎜⎝−ez

−6E I

l20 0 0

⎞⎟⎠ (17)

Kaa =⎛⎜⎝ky + ez2

l+ 4E I

l−eyez

−eyezES

lkz + ey2

l+ 4E I

⎞⎟⎠ (18)

If the crack does not exist, the eccentricities ey and ez tend to 0 whereas the torsional stiffnessestend to infinity. The points 1, 3 and 4 coincide so that one can sum up the lines and columnscorresponding to �y1 and �y4 and those corresponding to �z1 and �z4. As expected, the stiffness matrixof the cracked finite element then reduces to the well-known uncracked beam element.

The equilibrium equation for a CBFE is

K U = F or

⎛⎜⎝K11 K12 K1a

K21 K22 K2a

Ka1 Ka2 Kaa

⎞⎟⎠

⎛⎜⎝U1

⎞⎟⎠ =

⎛⎜⎝F1

⎞⎟⎠ (19)

where UTa =〈�y4 �z4〉 and F1, F2 are loading vectors, respectively, at nodes 1 and 2.

2.2. Comparison with the force method

The force method was described in [1, 2]. This method consists in computing the compliancematrix for a cracked finite element from the stress energy. The stiffness matrix K is then obtainedas the inverse of the compliance matrix. Its expression was given in [1] for the 2D case. Inorder to compare the force method with the CBFE model proposed herein, we have extended theaforementioned expression to the 3D case. The symbolic calculation, conducted using the Matlabsoftware [11], was lengthy and the resulting expression for K was much more complex than(12)–(18). Let us here give only the term K (1, 1) obtained from the force method

K (1, 1) = f1 ES

l f2(20)

f1 = (4E I + lk y)(4E I + lkz) (21)

f2 = 4E I (4E I + lk y + lkz) + 4E I ES(ey2 + ez2) + l ES(ey2ky + ez2kz) + l2kykz (22)

compared to expression for K (1, 1) = ES/ l in (13).However, further tedious computations show that by and large the force method and the CBFE

model do give the same final results, as expected. This has been checked on the example of aT-joint structure composed by a vertical beam clamped at its ends and a horizontal cracked beamwelded onto the vertical beam and loaded at the free end. The parameters ey , ez , ky and kz areidentified following the procedure described in Section 3. It is found that the strain (or stress)energy obtained by the two methods are identical to within less than 0.1%. In the following, onlythe stiffness matrix obtained by the CBFE model is considered.

3. IDENTIFICATION OF MODEL PARAMETERS

3.1. Identification principle

The cracked finite element involves the parameters ey , ez , ky and kz which are to be identified. Theidentification will be carried out on the problem of a cracked tube built in a rigid plate, see Figure 3where the local axis systems relative to the cracked tube and the plate are shown. The y0-axis forthe plate is directed along by the intersection line between the plane of the cracked section andthe symmetry plane (x, y) of the structure. The dimensions of the tube are given standard valuesin offshore engineering: length l = 10.2m, average radius rm = 0.55m and thickness t = 0.02m.On the other hand, the parameter identification is performed with various positions of the tube andthe crack defined by: the angle � between x and y0-axes; the angle � between the crack axis andy0-axis; and the half-opening angle � of the crack.Given a triplet (�, � ,�), the parameters ey , ez , ky and kz are identified by minimizing—in the

least-squares sense—the difference Q between the displacement U2 at the free end of the CBFEmodel and the displacement U2 obtained from 3D finite element computations.

min Q = min ‖U2 − U2‖2 (23)

3.2. 3D finite element computations

For each triplet (�, � ,�), the target displacement U for the minimization problem (23) is obtainedfrom 3D finite element computations using the software CAST3M [12]. A specific routine isadded in CAS3TM in order to generate regular meshes in the neighbourhood of the crack front(Figure 4(a)). The tube mesh is composed of 20-node brick elements, with nel elements along thetube length and nec elements along the circumference of the cross-section.

RemarkOther meshes with 8-node rectangular shell elements are also used, see Figure 4(b). In fact, whereasthe 3D and shell elements give the same global results, only the 3D meshes allow to the stressintensity factors, which are not presented in this work, to be accurately computed. Therefore, onlythe numerical results obtained with 3D elements are presented in this paper.

(a) (b)

Figure 3. Geometry of the cracked tube.

Crack fronts

Surface loading

(a) (b)

Figure 4. Meshes of the cracked tube: (a) 3D FE mesh of the cracked tube (�= 45◦, � = 45◦, � = 40◦)and (b) FE mesh using shell elements (� = 45◦, � = 150◦, � = 30◦).

3.3. Formulas for identifying the model parameters

The displacement U2 in (23) can be expressed in terms of the model parameters by solvingEquation (19) with u1 = 0

U2 = S2F2, S2 =[K22 − K2aK−1aa Ka2]−1 (24)

where S2 denotes the compliance matrix condensed at node 2 of the cracked finite element. Inthe sequel, in order to facilitate the computations, all the equations will be written using ratherthe homogenized displacement vector U∗

2 obtained by multiplying the rotations by the length lof the element, and the homogenized force vector F∗

2 by dividing the moments by l. With thesehomogenized variables, the minimization problem reads

min Q∗ = min ‖S∗2 F

∗2 − U∗

2‖2 (25)

where S∗2 are the homogenized compliance matrix.

For brevity, U∗2, F

∗2 , S

∗2 will be denoted U∗, F∗, S∗ in the sequel. By noting that

S∗T = S∗ ⇒(

�S∗

= �S∗

�ey(26)

it can be shown that minimizing Q with respect to ey , ez , ky and kz leads to the following equationsystem:

F∗T[S∗ �S∗

�ey+ �S∗

�eyS∗

]F∗ − 2U∗T

[�S∗

]F∗ = 0

F∗T[S∗ �S∗

�ez+ �S∗

�ezS∗

]F∗ − 2U∗T

[�S∗

]F∗ = 0

F∗T[S∗ �S∗

�ky+ �S∗

�kyS∗

]F∗ − 2U∗T

[�S∗

]F∗ = 0

F∗T[S∗ �S∗

�kz+ �S∗

�kzS∗

]F∗ − 2U∗T

[�S∗

]F∗ = 0

It has been shown in [2] that applying a shear force or a bending moment at the free end of theelement leads to same results for the minimization problem. Here, we choose to apply a shear forceT y along the y-axis so as to identify ey and kz . The load vector is then F∗T = 〈0 T y 0 0 0 0〉.Solving Equation (27) then yields

ey = l12u

5T yl3/E I − 6(v + l�z)

kz = E I

12T yl3/E I

6(v + l�z) − 5T yl3/E I

In order to identify ez and ky , let us now apply a shear force T z along the z-axis, i.e.F∗T =〈0 0 T z 0 0 0〉. Solving Equation (27) then yields

ez = l12u

5T zl3/E I − 6(w − l�y)

ky = E I

12T zl3/E I

6(w − l�y) − 5T zl3/E I

The study of the convergence of the CBFE model parameters is carried out with respect to themesh refinement, namely the element numbers nel and nec. Figures 5(a) and (b) plot eccentricityey versus nel (nel= 30–50) and nec (nec= 30–200) for � = 90◦, �= 0◦ and �= 16◦. It is foundthat ey varies little with nel—the variation is less than 3%—whereas it varies significantly withnec. The parameter identification is highly sensitive to the robustness of the 3D meshes and tothe assessment of the homogenized displacement at the end of the beam. In case of through-cracks the computation of u is very sensitive to the mesh. This is mainly due to the sensitivityto the longitudinal displacement at the free end of the finite element. Hence, in practical use,the above expressions for ey and ez can be advantageously replaced with the available semi-analytical ones. Without going into details, here we give the semi-analytical formulas for ey and ez

30 35 40 45 500

nec30 70 110 150 190

Figure 5. Eccentricity ey versus element numbers nel and nec.

30 35 40 45 500

1.5E+13

30 35 40 45 50 55 600

1.5E+13

2.5E+13

Figure 6. Stiffness kz versus element numbers nel and nec.

which are retained henceforth

eyth = −(De3 − Di

3(De2 − Di

cos� sin�

� − �

ezth = −(De3 − Di

3(De2 − Di

sin� sin�

� − �

Figure 6 also shows that kz varies little with nel while it almost doubles as nec goes from 30to 200.

100150

102030405060708090

15x 1012 x 1012

ψ[°]

ϕ[°] ϕ[°]

ψ[°]

100150

102030405060708090(a) (b)

Figure 7. Surfaces of (a) ky and (b) kz for � = 90◦.

10 20 30 40 50 60 70 80 90 10 20 30 40 50 60 70 80 900

15x 1012 x 1012

ϕ [°]

ϕ[ ° ](a) (b)

Figure 8. Curves of ky and kz for (a) � = 0◦ and (b) � = 45◦, with �= 90◦.

For specific studies where geometry is fixed, we have selected 50 and 110, respectively, for neland nec. This choice was confirmed by a sensitivity study concerning geometrical parameters �,� and �.

In the objective to give an overview on the main trends which governs parameters variations, weselect in the following nel= 20 and nec= 50. It allows to reduce widely the number of calculations.

3.4. Results: the identified model parameters

A series of computations for identifying ky and kz are undertaken by considering various geometryvalues (�, � and �). Figures 7(a) and (b) depict the surfaces of ky and kz versus � and �, for� fixed to 90◦ (the tube then is perpendicular to the plane). For a fixed � value, ky reaches twolocal maxima at �= 0◦ and 180◦ and one local minimum at �= 90◦; this is most noticeable forsmall � values. On the other hand, kz presents an opposite variation, with two local minima andone local maximum. These variations can be accounted for by simple geometrical considerations.

Indeed, for a given �, the larger � ∈ [0◦, 90◦], the more so is the projected area of the crackon the y-axis and the smaller the projected area on z-axis; hence, the inertia and stiffness of thesection with respect to the y-axis increase whereas those with respect to the z-axis decrease. Theopposite trend occurs for � ∈ [90◦, 180◦].

Figure 8(a) shows the curves of ky and kz for � = 90◦ and �= 0◦. They are obtained by fittingall the numerical results presented in Figure 7 using a polynomial function of degree 5. In thiscase, stiffness ky is higher than kz and the difference between the two values decreases as �increases; this is due to the fact that the projections of the crack surface onto y and z-axes becomecloser as � increases. In the particular case when � = 90◦ and �= 45◦, the projections of the cracksurface are identical and so are stiffnesses ky and kz for the cracked surface, as clearly shown inFigure 8(b). The difference between the unfitted and fitted curves for the last case reaches 6.9%for small crack openings.

4. APPLICATION

A computer program has been developed on the basis of the previous analysis to deal withframed structures. It includes the proposed CBFE together with classical beam elements to modeluncracked tubes and plate elements with rectangular sections.

By means of this program, the CBFE model is assessed through the example of a crackedtube welded to a flexible plate (Figure 9). The tube is of length l = 10.2m, external diameterDe = 1.12m and internal diameter Di = 1.08m. The plate is rectangular, twice as long as thetube; its section is rectangular of width a = 2.2m and thickness b= 0.23m. The tube is weldedperpendicularly to the plate at the middle of the plate. It is assumed that the symmetry plane ofthe crack is the same as that of the structure, i.e. � = 90◦, �= 0◦. The tube is subjected to theforce F = − 106 N applied at its free end and directed along y-axis parallel to the plate length.The results are compared with (i) those from the strength of materials and (ii) those from 3D finiteelement computations using 20-node brick elements as shown in Figure 9.

(a) (b)

Figure 9. Mesh of the cracked tube welded to the plate: (a) mesh of the whole structure and(b) zoom of the cracked region.

20 30 40 50 60 70 80 900

ϕ[°]

CBFE modelStrength of materials3D finite elements

Figure 10. Deflection at the free end of the cracked tube.

Table I. Deflection at the free end by the three methods: CBFE model, strengthof materials and 3D finite elements.

� 20◦ 30◦ 40◦ 50◦ 60◦ 70◦ 80◦ 90◦

CBFE model (mm) 457.5 475.1 502.4 533 583.8 696.2 801.8 873.8Strength of materials (mm) 457.5 475 502.3 532.9 583.8 696.2 801.7 873.73D finite elements (mm) 350.4 446.5 469.7 475.7 568.7 642.6 784.5 949.6Difference CBFE/3D FE (%) 23 6 7 11 3 8 2 8

The deflection can be readily derived from the strength of the materials, to do this one just hasto carefully distinguish the second moment of area for a tube section Itube = �(D4

e − D4i )/64 from

that for the plate section Iplateab3/12. One then obtains

v − Fl3

3E Itube− Fl3

8E Iplate− Fl2

kz(31)

where E = 2.1× 1011 N/m2 is the Young modulus and kz the spring stiffness from the crackedbeam model.

Figure 10 and Table I show the deflection v at the tube free end versus the half-opening � ofthe crack, obtained from the cracked beam model CBFE, the 3D finite elements and the strengthof materials. The cracked beam model and the strength of the materials give almost identicalresults. The difference between 3D finite element results and the two other theories reaches 23%for �= 20◦, yet it does not exceed 11% for other � values. In fact, the 23% difference shouldnot be taken into account, since other computations which are not reported here show that it isdifficult to obtain optimal 3D meshes and good results for small crack opening values, say less orequal to 20◦.

5. CONCLUSIONS

A new finite element has been presented in order to take into account the effect of large through-cracks in the structural analysis of steel-framed structures like the jacket offshore platforms. Thiscracked beam element based on the strain energy involves four parameters (two eccentricities andtwo stiffnesses) which represent the loss of stiffness. These parameters have been identified onthe basis of several joint configurations. Then, the cracked element model has been applied on aT-tubular joint, giving results which are in very good agreement with the strength of materials,and quite satisfactory agreement with 3D finite elements.

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