Coupling_of_IGA_plates_and_3D_FEM_domain

Ins$tute of Mechanics & Advanced Materials

Coupling of IGA plates and 3D FEM domains by a Discontinuous Galerkin Method

V.P. Nguyen1, P. Kerfriden1, S. Claus2, S.P.-‐A. Bordas1

1School of Engineering, Cardiff University, UK 2Department of Mathema$cs, University College London, UK

Ins$tute of Mechanics & Advanced Materials Aim: local/global analysis of thin panels

http://www.supergen-wind.org.uk

Stress analysis with minimium data transfer from CAD model

Hot-‐spot (stress concentra$ons, damage)


• Mixed-‐dimensional analysis: §  use shell/beam descrip$ons, homogenisa$on

§  Full microscale 3D in “hot-‐spots”

•  Isogeometric analysis: minimum CAD to analysis data processing

•  Efficient coupling of heterogeneous models /discre$sa$on

•  (efficient local/global solver)

•  (find “hot-‐spots” with goal-‐oriented model adap$vity)

Building blocks

6.2.2. Cantilever plate: non-conforming coupling

A mesh of 32⇥ 4⇥ 5/ 32⇥ 2 cubic elements is utilized for the mixed dimensional model, cf. Fig. 30. The length ofthe continuum part in the continuum-plate model is 175 mm. The contour plot of the von Mises stress is given Fig. 31where void plate elements were removed in the visualisation.

Figure 30: Cantilever beam subjects to an end shear force: discretisation of the solid and the plate.

Figure 31: Cantilever beam subjects to an end shear force: von Mises stress distribution.

6.2.3. Non-conforming coupling of a square plate

We consider a square plate of dimension L⇥L⇥ t (t denotes the thickness) in which there is an overlapped solid ofdimension L

s

⇥L

s

⇥ t as shown in Fig. 32. In the computations, material properties are taken as E = 103, ⌫ = 0.3 andthe geometry data are L = 400, t = 20 and L

s

= 100. The loading is a gravity force p = 10 and the plate boundaryis fully clamped. The stabilisation parameter was chosen empirically to be 1 ⇥ 106. We use rotation free Kirchho↵NURBS plate elements for the plate and NURBS solid elements for the solid. In order to model zero rotations in arotation free NURBS plate formulation, we simply fix the transverse displacement of control points on the boundaryand those right next to them cf. [47].

In order to find plate elements cut by the boundary surfaces of the solid, we use the level sets defined for thesquare which is the intersection plane of the solid and the plate. The use of level sets to define the interaction of finite

35

[Nguyen et al. 2013]

6.2.2. Cantilever plate: non-conforming coupling

A mesh of 32⇥ 4⇥ 5/ 32⇥ 2 cubic elements is utilized for the mixed dimensional model, cf. Fig. 30. The length ofthe continuum part in the continuum-plate model is 175 mm. The contour plot of the von Mises stress is given Fig. 31where void plate elements were removed in the visualisation.

Figure 30: Cantilever beam subjects to an end shear force: discretisation of the solid and the plate.

Figure 31: Cantilever beam subjects to an end shear force: von Mises stress distribution.

6.2.3. Non-conforming coupling of a square plate

We consider a square plate of dimension L⇥L⇥ t (t denotes the thickness) in which there is an overlapped solid ofdimension L

s

⇥L

s

⇥ t as shown in Fig. 32. In the computations, material properties are taken as E = 103, ⌫ = 0.3 andthe geometry data are L = 400, t = 20 and L

s

= 100. The loading is a gravity force p = 10 and the plate boundaryis fully clamped. The stabilisation parameter was chosen empirically to be 1 ⇥ 106. We use rotation free Kirchho↵NURBS plate elements for the plate and NURBS solid elements for the solid. In order to model zero rotations in arotation free NURBS plate formulation, we simply fix the transverse displacement of control points on the boundaryand those right next to them cf. [47].

In order to find plate elements cut by the boundary surfaces of the solid, we use the level sets defined for thesquare which is the intersection plane of the solid and the plate. The use of level sets to define the interaction of finite

35

CAD model

Analysis mesh

IGA / plate Solid FE

???


MIXED-DIMENSIONAL COUPLING IN FINITE ELEMENT MODELS 737

Figure 8. Stress contours in 3D–2D mixed-dimensional cantilever model loaded by a terminalshear force Fz . (2D contours illustrated relate to top surface of model). Abaqus C3D20R brick

elements and S8R shell elements.

Figure 9. Transverse shear stresses !13 (!xz) obtained by method of Reference [5].

Copyright ? 2000 John Wiley & Sons, Ltd. Int. J. Numer. Meth. Engng 2000; 49:725–750

• Reference coupling §  displacement-‐recovery, Stress-‐recovery

§  Equality of work provides coupling on dual quan$ty

• Discrete treatment § Mul$-‐point constraints [Monaghan et al 1998,

McCune et al. 2000, Shim et al. 2002, Song et al. 2012]

§  Transi$on elements [Surana 1979, Cofer 1991, Gmur et al, 1993, Dohrmann et al. 1999, Wagner et al. 2000, Garusi et al. 2002, Chavan et al. 2004]

§ Mortar methods -  Penalty formula$ons [Blanco et al. 2007]

-  Lagrange mul$plier-‐based mortar [Rateau et al. 2003, Combescure et al. 2005]

-  Hybrid itera$ve method [Guguin et al. 2013]

Some coupling methods

Thus, the following constraint equation holds for each point of the shell cross section

f = xI ! x0 ! ! ("2a2 + "3a3) = 0 (7)

Based on the above defined assumptions warping cannot be decribed. This may be crucial

in some situations.

3 Finite element formulation

Based on the above described kinematical assumptions the element is developed. The beam

node in the transition cross–section is called ’reference node’. Furthermore, the base vectors

A2 and A3 define the orientation of the cross section. It is assumed that the shell nodes to

couple (’coupling nodes’) lie in this plane. The vectors A2 and A3 are used to specify the

section coordinates, see eq. (3). In the current configuration the base vectors a2 und a3 of

the beam element together with the convective coordinates (0, "2, "3) and the parameter !

define the coupling nodes.

The mechanical model of the cross section can be considered as a sum of rigid beams which

allow only for axial deflections. The boundary conditions are clamped at the reference node

and jointed at the coupling node, see Fig. 3.

clamped bounded

rigid beam, axial free

hinged bounded

Transition elements

Fig. 3: Transition elements in a beam cross–section

The implementation of the constraint equation (7) in a transition element is done via the

Penalty and the Augmented Lagrange Method. Furthermore a consistent linearization is

derived for the element with respect to finite rotations. The transition is formulated between

5

Adapted from [Wagner et al. 2000]

Chapitre 5. Raccord 3D/coque

Le probleme consiste a determiner, dans le cadre de l’elasticite linearisee isotrope, l’equilibrede cette structure simultanement, sous l’e↵et d’un chargement volumique f de L2(⌦), et sousune condition d’encastrement sur une partie �u, de mesure non nulle, de son bord mince.

e

Γu

ω

f

Fig. 5.7 – Probleme modele

5.2.2.2 Modelisation Arlequin

Ce probleme est modelise par la superposition d’un modele coque et d’un modele vo-lumique, qui occupent respectivement l’adherence des ouverts connexes ⌦coq et ⌦3d. Parcommodite, nous designons par !coq la surface moyenne du premier et nous notons !0

coq lesous-domaine correspondant de !0. En outre, comme au §5.1.2.1, nous supposons que levoisinage de la condition d’encastrement est represente par le modele 3D.

ωcoq

ω3d

sc

Fig. 5.8 – Modelisation Arlequin

Les relations de comportement sont celles des paragraphes 5.1.1.3 et 5.1.2.1, et l’ensemble deschamps de deplacement cinematiquement admissibles du modele tridimensionnel est definipar (5.17), tandis que celui du modele coque est donne par l’expression suivante :

W coq =n

vcoq = v0 + ⇠3(v1⌧ 1 + v2⌧ 2) ; v0 2 H1(!0

coq), v1, v2 2 H1(!0coq), |⇠3| <

e

2

o

(5.59)

106

[Rateau et al. 2003]

[McCune et al. 2000]


•  Introduc$on •  Automa$c coupling §  Problem statement §  IGA §  Discrete coupling strategy •  Numerical examples

•  Conclusion


§  Kinema$cs:

§  Equilibrium:

§  Cons$tu$ve rela$on:

§  Primal vibra$onal formula$on:

Problem statement: uncoupled solid

Figure 2: Coupling of a two dimensional solid and a beam.

Figure 3: Coupling of a three dimensional solid and a plate.

6

as(us,us?) :=

Z

⌦s

✏(u) : Cs : ✏(us?) d⌦ = ls(us?)

KA0 Z

⌦s

�s : ✏s(us?) d⌦ =

Z

⌦s

b · us? d⌦+

Z

�t

t · us? d�

�s = Cs : ✏s in ⌦s

✏s =1

2(rus +rTus)

us=

¯u on �

su


§  Kinema$cs: §  Equilibrium:

§  Cons$tu$ve rela$on:

§  Primal VF:

Problem statement: uncoupled beam



6

Z

⌦b

✓NM

◆·✓

v?,x

w?

,xx

◆dl =

Z

⌦b

0

@p,x

p,y

m

1

A ·

0

@v?

w?

w?

,x

1

A dl +X

P2Pntm

0

@NTM

1

A

|P

·

0

@v?

w?

w?

,x

1

A

|P✓NM

◆=

✓ES 00 EI

◆✓v,x

w,xx

◆in ⌦b

v = v on �

b

v

w = w on �

b

w

w,x

=

¯✓ on �

b

✓

ab(⇥,⇥b

?

) :=

Z

⌦b

✓v?,x

w?

,xx

◆T

✓ES 00 EI

◆·✓

v,x

w,xx

◆dl = lb(⇥b

?

)

KA0


•  Solid:

•  EB-‐Beam, use as test and trial in 2D VF

Primal coupling (strong/weak)

as(us,us?) = ls(us?) +

Z

�?

us? · (�s(us) · ns) d�

ab(⇥b,⇥b?) = lb(⇥b?) +

Z

�?

ub(⇥b?) · (�b(⇥b) · nb) d�

ub(⇥b) =

✓v � w

,x

yw

◆

R


•  Solid:

•  EB-‐Beam, use as test and trial in 2D VF

• Primal coupling §  Kinema$cs: - For us:

§  V. Work equality for any KA field:

Primal coupling (strong/weak)

as(us,us?) = ls(us?) +

Z

�?

us? · (�s(us) · ns) d�

ab(⇥b,⇥b?) = lb(⇥b?) +

Z

�?

ub(⇥b?) · (�b(⇥b) · nb) d�

Choice for space?

For us

Z

�?

ub? · (�s(us) · ns � �b(⇥b) · ns) d� = 0

ub(⇥b) =

✓v � w

,x

yw

◆

R



6

Z

�?

�us � ub(⇥)

�· �? d�

us = ub(⇥)


•  Introduc$on •  Automa$c coupling §  Problem statement §  IGA §  Discrete coupling strategy •  Numerical examples

•  Conclusion

Ins$tute of Mechanics & Advanced Materials B-‐splines

Ins$tute of Mechanics & Advanced Materials Descrip$on of geometry by B-‐splines

00.1

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10

0.1

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

00.1

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0.80.9

10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

N3,3(!)

M3,3(")

N3,33,3 (!, ")

Figure 4: A bivariate cubic B-spline basis function with knots vectors ! = H = {0, 0, 0, 0, 0.25, 0.5, 0.75, 1, 1, 1, 1}.

!

"

x

yz

(!, ")

0, 0, 0

0,0,0 1, 1, 1

1,1,1

0.5

0.5

Figure 5: A bi-quadratic B-spline surface (left) and the corresponding parameter space (right). Knot vectors are! = H = {0, 0, 0, 0.5, 1, 1, 1}. The 4! 4 control points are denoted by red filled circles.

12

⌅ = {⇠1, ⇠2, . . . , ⇠n+p+1}

x(⇠) =nX

i

Ni,p(⇠)Bi

x(⇠, ⌘) =nX

i

mX

j

Ni,p(⇠)Mj,p(⌘)Bij

Ins$tute of Mechanics & Advanced Materials IsoGeometric Analysis (IGA)

−0.5 0 0.5 1 1.5−0.5

0

0.5

1

1.5

−0.5

00.5

11.5

−0.50

0.51

1.5

00.1

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

00.1

0.20.3

0.40.5

0.60.7

0.80.9

10

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

N3,3(!)

M3,3(")

N3,33,3 (!, ")

Figure 4: A bivariate cubic B-spline basis function with knots vectors ! = H = {0, 0, 0, 0, 0.25, 0.5, 0.75, 1, 1, 1, 1}.

!

"

x

yz

(!, ")

0, 0, 0

0,0,0 1, 1, 1

1,1,1

0.5

0.5

Figure 5: A bi-quadratic B-spline surface (left) and the corresponding parameter space (right). Knot vectors are! = H = {0, 0, 0, 0.5, 1, 1, 1}. The 4! 4 control points are denoted by red filled circles.

12

⌅1 = {0, 0, 0, 0.5, 1, 1, 1}

⌅2=

{0,0,0,0.5,1,1,1}

N2,3(⇠)

u(x(⇠, ⌘)) =X

i

X

j

Ni,p(⇠)Mj,p(⌫)Uij

�1

1

�1

1 ⇠

⌦1 ⌦2

⌦3 ⌦4

⌘

⌘

⇠

0 0.5 1

1

0.5

Parametric domain

Physical domain

Parent domain (integra$on)

x(⇠, ⌘) =nX

i

mX

j

Ni,p(⇠)Mj,p(⌘)BijM

2,2(⌘)

(⇠, ⌘)|⌦i = �((⇠, ⌘))

References: [Kagan et al. 1998, Cirak et al. 2000, Hughes et al. 2005, Cofrell et al. 2009]


•  Introduc$on •  Automa$c coupling §  Problem statement and reference §  IGA §  Discrete coupling strategy •  Numerical examples

•  Conclusion


• Penalty formula$on

§ 

Mortaring non-‐conforming discrete spaces

JXK = Xs �Xb

us = ub(⇥)

In discrete space, poten$ally discon$nuous

a ⌘ as + ab u ⌘ (us,⇥b)ap(uh,u?) =: a(uh,u?) + ↵

Z

�?

JuhK · Ju?Kd� = l(u?)



§ 

§  Lack of consistency:


JXK = Xs �Xb

hXi� = �Xs + (1� �)Xb

us = ub(⇥)


J(� · ns) · u?K = J� · nsK · hu?i� + Ju?K h� · nsi�

a ⌘ as + ab u ⌘ (us,⇥b)

� =1

2

ap(uex,u?)� l(u?) = �Z

�

?

J�(uex) · ns · u?K 6= 0

ap(uh,u?) =: a(uh,u?) + ↵

Z

�?




§ 


§  Nitsche method: add to penalty formula$on and symmetrise


JXK = Xs �Xb

hXi� = �Xs + (1� �)Xb

us = ub(⇥)


an(uh,u?) = a(uh,u?)�Z

�?

Ju?K⌦�(uh) · ns

↵�d�

�Z

�?

JuhK h�(u?) · nsi� d�+ ↵

Z

�?


J(� · ns) · u?K = J� · nsK · hu?i� + Ju?K h� · nsi�```

a ⌘ as + ab u ⌘ (us,⇥b)

� =1

2


�

?

J�(uex) · ns · u?K 6= 0

ap(uh,u?) =: a(uh,u?) + ↵

Z

�?




§ 


§  Nitsche method: add to penalty formula$on and symmetrise


JXK = Xs �Xb

hXi� = �Xs + (1� �)Xb

us = ub(⇥)



�?


↵�d�

�Z

�?


Z

�?


J(� · ns) · u?K = J� · nsK · hu?i� + Ju?K h� · nsi�J(� · ns) · u?K = J� · nsK · (I�⇧b) hu?i� + Ju?K h� · nsi�

```

a ⌘ as + ab u ⌘ (us,⇥b)

� =1

2


�

?

J�(uex) · ns · u?K 6= 0

ap(uh,u?) =: a(uh,u?) + ↵

Z

�?



• Coercivity:

Stability


�?


↵�d�

�Z

�?


Z

�?


t(u) :=��(us) + �(ub)

�· ns

in discrete space, poten$ally discon$nuous

Related work: [Griebel et al. 2002, Dolbow et al. 2009]

an(u,u) = a(u,u) + ↵

Z

�?

JuK · JuKd��Z

�?

JuK · t(u) d�

an(u,u) � Cc kuk2X


• Coercivity:

§  Parallelogram ineq. :

§  ``Trace inequality” (assump$on)

→ 

Stability


�?


↵�d�

�Z

�?


Z

�?


t(u) :=��(us) + �(ub)

�· ns

kt(u)k2�? C2 a(u,u)

an(u,u) �✓1� C2 ✏

2

◆a(u,u) +

✓↵� 1

2✏

◆kJuKk2�?

in discrete space, poten$ally discon$nuous

✏ =1

C2) ↵ >

C2

2


an(u,u) � a(u,u) + ↵kJuKk2�? �✓

1

2✏kJuKk2�? +

✏

2kt(u)k2�?

◆

an(u,u) = a(u,u) + ↵

Z

�?

JuK · JuKd��Z

�?

JuK · t(u) d�

an(u,u) � Cc kuk2X


•  Solve numerically for regularisa$on parameter s. t.

→ 

→ 

Eigenvalue problem for regularisa$on parameter

↵ >�1

2

�Kuncoupled

��1H

a(u,u) = [u]T Kuncoupled [u]

kt(u)k2�? =

Z

�?

�(�(us) + �(ub)) · ns

�·�(�(us) + �(ub)) · ns

�d� = [u]T H [u]

�1 largest eigenvalue of


kt(u)k2�? < 2↵ a(u,u)

References on embedded interfaces and implicit boundaries using Nitsche [Hansbo et al. 2002, Dolbow et al. 2009, Sanders et al. 2011, Burman et al. 2012, Chouly et al. 2013]

1

2

[u]T H [u]

[u]T Kuncoupled [u]< ↵


•  Introduc$on •  Automa$c coupling §  Problem statement and reference §  IGA §  Discrete coupling strategy •  Numerical examples

•  Conclusion

Ins$tute of Mechanics & Advanced Materials Examples

where I = D

3/12 is the moment of inertia. The exact displacement field of this problem is [63]

u

x

(x, y) =Py

6EI

(6L� 3x)x+ (2 + ⌫)

✓y

2 � D

2

4

◆�

u

y

(x, y) = � P

6EI

3⌫y2(L� x) + (4 + 5⌫)

D

2x

4+ (3L� x)x2

� (88)

and the exact stresses are

�

xx

(x, y) =P (L� x)y

I

; �

yy

(x, y) = 0, �

xy

(x, y) = � P

2I

✓D

2

4� y

2

◆(89)

In the computations, material properties are taken as E = 3.0⇥ 107, ⌫ = 0.3 and the beam dimensions are D = 6 andL = 48. The shear force is P = 1000. Units are deliberately left out here, given that they can be consistently chosenin any system. In order to model the clamping condition, the displacement defined by Equation (88) is prescribed asessential boundary conditions at x = 0,�D/2 y D/2. This problem is solved with bilinear Lagrange elements (Q4elements) and high order B-splines elements. The former helps to verify the implementation in addition to the ease ofenforcement of Dirichlet boundary conditions (BCs). For the latter, care must be taken in enforcing the Dirichlet BCsgiven in Equation (88) since the B-spline basis functions are not interpolatory.

Figure 13: Timoshenko beam: problem description.

The mixed continuum-beam model is given in Fig. 14. The end shear force applied to the right end point is F = P .

Figure 14: Timoshenko beam: mixed continuum-beam model.

Lagrange elements In the first calculation we take l

c

= L/2 and a mesh of 40⇥ 10 Q4 elements (40 elements in thelength direction) was used for the continuum part and 29 two-noded elements for the beam part. The stabilisation

26

where I = D

3/12 is the moment of inertia. The exact displacement field of this problem is [63]

u

x

(x, y) =Py

6EI

(6L� 3x)x+ (2 + ⌫)

✓y

2 � D

2

4

◆�

u

y

(x, y) = � P

6EI

3⌫y2(L� x) + (4 + 5⌫)

D

2x

4+ (3L� x)x2

� (88)

and the exact stresses are

�

xx

(x, y) =P (L� x)y

I

; �

yy

(x, y) = 0, �

xy

(x, y) = � P

2I

✓D

2

4� y

2

◆(89)

In the computations, material properties are taken as E = 3.0⇥ 107, ⌫ = 0.3 and the beam dimensions are D = 6 andL = 48. The shear force is P = 1000. Units are deliberately left out here, given that they can be consistently chosenin any system. In order to model the clamping condition, the displacement defined by Equation (88) is prescribed asessential boundary conditions at x = 0,�D/2 y D/2. This problem is solved with bilinear Lagrange elements (Q4elements) and high order B-splines elements. The former helps to verify the implementation in addition to the ease ofenforcement of Dirichlet boundary conditions (BCs). For the latter, care must be taken in enforcing the Dirichlet BCsgiven in Equation (88) since the B-spline basis functions are not interpolatory.

Figure 13: Timoshenko beam: problem description.

The mixed continuum-beam model is given in Fig. 14. The end shear force applied to the right end point is F = P .

Figure 14: Timoshenko beam: mixed continuum-beam model.

Lagrange elements In the first calculation we take l

c

= L/2 and a mesh of 40⇥ 10 Q4 elements (40 elements in thelength direction) was used for the continuum part and 29 two-noded elements for the beam part. The stabilisation

26

Analy$cal solu$on available

�?


parameter ↵ according to Equation (55) was 4.7128⇥ 107. Fig. 15a plots the transverse displacement (taken as nodalvalues) along the beam length at y = 0 together with the exact solution given in Equation (88). An excellent agreementwith the exact solution can be observed and this verified the implementation. The comparison of the numerical stressfield and the exact stress field is given in Fig. 15b with less satisfaction. While the bending stress �

xx

is well estimated,the shear stress �

xy

is not well predicted in proximity to the coupling interface. This phenomenon was also observedin the framework of Arlequin method [64] and in the context of MPC method [38]. Explanation of this phenomenonwill be given subsequently.

0 10 20 30 40 50−0.07

−0.06

−0.05

−0.04

−0.03

−0.02

−0.01

0

x

w

exactcoupling

(a) transverse displacement

0 5 10 15 20 25−400

−200

0

200

400

600

800

x

stre

sses

alo

ng y

=0.3

sigmaxx−exactsigmaxx−couplingsigmaxy−exactsigmaxy−coupling

(b) stresses

Figure 15: Mixed dimensional analysis of the Timoshenko beam: comparison of numerical solution and exact solution.

B-splines elements are used to discretise the continuum part (with bi-variate B-splines elements) and the beam part(with uni-variate B-splines elements). Such a mesh is given in Fig. 16. Dirichlet BCs are enforced using the leastsquare projection method see e.g., [65]. Note that Nitche’s method can also be used to weakly enforce the DirichletBCs. However, we use Nitsche’s method only to couple the di↵erent models. In what follows, we used the followingdiscretisation– 16⇥4 bi-cubic continuum elements and 4 cubic beam elements. The stabilisation parameter ↵ accordingto Equation (55) was 5.5 ⇥ 109. Comparison between numerical and exact solutions are given in Fig. 17. Again, anexcellent estimation of the displacement was obtained whereas the shear stress is not well captured. An explanation forthis behavior is given in Fig. 18. The error of the continuum-beam model consists of two parts: (1) model error whenone replaces a continuum model by a continuum-beam model and (2) discretisation and coupling errors. When theformer is dominant, the coupling method is irrelevant as the same phenomenon was observed in the Arlequin method,in the MPC method [64, 38] and mesh refinement does not cure the problem. In other words, the error originatingfrom the Nitsche coupling is dominated by the model error and therefore we cannot draw any conclusions about the”coupling error”. In order to alleviate the model error, we computed the shear stress with ⌫ = 0.0 and the results aregiven in Figs. 19 and 20. With ⌫ = 0.0, the numerical solution is very much better than the one with ⌫ = 0.3.

27

Q4 elements 2-‐noded cubic elements

Deflec$on of neutral axis Stress profile


32x4 bi-‐cubic B-‐spline elements 8 cubic B-‐spline elements (patch extends throughout the 2D domain)

6.1.2. Timoshenko beam: non-conforming coupling

In this section, a non-conforming coupling is considered. The B-spline mesh is given in Fig. 21. Refined meshes areobtained from this one via the knot span subdivision technique. We use the mesh consisting of 32⇥ 4 cubic continuumelements and 8 cubic beam elements. Fig. 22 gives the mesh and the displacement field in which l

c

= 29.97 so thatthe coupling interface is very close to the beam element boundary. A good solution was obtained using the simpletechnique described in Section 5.

Figure 21: Mixed dimensional analysis of the Timoshenko beam with non-conforming coupling. The continuum partis meshed by 8⇥ 2 bi-cubic B-splines and the beam part is with 8 cubic elements.

(a) mesh

0 10 20 30 40 50−0.07

−0.06

−0.05

−0.04

−0.03

−0.02

−0.01

0

0.01

x

w

exactcontinuumbeam

(b) displacement field

Figure 22: Mixed dimensional analysis of the Timoshenko beam with non-conforming coupling: (a) 32⇥ 4 Q4 elementsand 8 quartic (p = 4) beam elements and (b) displacement field.

6.1.3. Frame analysis

In order to demonstrate the correctness of the solid-beam coupling in which the beam local coordinate system is notidentical to the global one, we perform a plane frame analysis as shown in Fig. 23. Due to symmetry, only half model isanalysed with appropriate symmetric boundary conditions. We solve this model with (1) continuum model (discretisedwith 7105 four-noded quadrilateral elements, 7380 nodes, 14760 dofs, Gmsh [66] was used) and (2) solid-beam model(cf. Fig. 24). The beam part are discretised using two-noded frame elements with three degrees of freedom (dofs) pernode (axial displacement, transverse displacement and rotation). Note that continuum element nodes have only two

30

6.1.2. Timoshenko beam: non-conforming coupling

In this section, a non-conforming coupling is considered. The B-spline mesh is given in Fig. 21. Refined meshes areobtained from this one via the knot span subdivision technique. We use the mesh consisting of 32⇥ 4 cubic continuumelements and 8 cubic beam elements. Fig. 22 gives the mesh and the displacement field in which l

c

= 29.97 so thatthe coupling interface is very close to the beam element boundary. A good solution was obtained using the simpletechnique described in Section 5.

Figure 21: Mixed dimensional analysis of the Timoshenko beam with non-conforming coupling. The continuum partis meshed by 8⇥ 2 bi-cubic B-splines and the beam part is with 8 cubic elements.

(a) mesh

0 10 20 30 40 50−0.07

−0.06

−0.05

−0.04

−0.03

−0.02

−0.01

0

0.01

x

w

exactcontinuumbeam

(b) displacement field

Figure 22: Mixed dimensional analysis of the Timoshenko beam with non-conforming coupling: (a) 32⇥ 4 Q4 elementsand 8 quartic (p = 4) beam elements and (b) displacement field.

6.1.3. Frame analysis

In order to demonstrate the correctness of the solid-beam coupling in which the beam local coordinate system is notidentical to the global one, we perform a plane frame analysis as shown in Fig. 23. Due to symmetry, only half model isanalysed with appropriate symmetric boundary conditions. We solve this model with (1) continuum model (discretisedwith 7105 four-noded quadrilateral elements, 7380 nodes, 14760 dofs, Gmsh [66] was used) and (2) solid-beam model(cf. Fig. 24). The beam part are discretised using two-noded frame elements with three degrees of freedom (dofs) pernode (axial displacement, transverse displacement and rotation). Note that continuum element nodes have only two

30


dofs. The total number of dofs of the continuum-beam model is only 5400. The stabilisation parameter is taken to be↵ = 107 and used for both coupling interfaces. A comparison of �

xy

contour plot obtained with (1) and (2) is given inFig. 25. A good agreement was obtained.

Figure 23: A plane frame analysis: problem description.

Remark 6.1. Although the processing time of the solid-beam model is much less than the one of the solid model, onecannot simply conclude that the solid-beam model is more e�cient. The pre-processing of the solid-beam model, if notautomatic, can be time consuming such that the gain in the processing step is lost. For non-linear analyses, where theprocessing time is dominant, we believe that mixed dimensional analysis is very economics.

6.2. Continuum-plate coupling

6.2.1. Cantilever plate: conforming coupling

For verification of the continuum-plate coupling, we consider the 3D cantilever beam given in Fig. 26. The materialproperties are E = 1000 N/mm2, ⌫ = 0.3. The end shear traction is t = 10 N/mm in case of continuum-plate modeland is t = 10/20 N/mm2 in case of continuum model which is referred to as the reference model. We use B-splineselements to solve both the MDA and the reference model. The length of the continuum part in the continuum-platemodel is L/2 = 160 mm. A mesh of 64 ⇥ 4 ⇥ 5 tri-cubic elements is utilized for the reference model and a mesh of32 ⇥ 4 ⇥ 5/ 16 ⇥ 2 cubic elements is utilized for the mixed dimensional model, cf. Fig. 27. The plate part of themixed dimensional model is discretised using the Reissner-Mindlin plate theory with three unknowns per node and theKirchho↵ plate theory with only one unknown per node. The stabilisation parameter was chosen empirically to be5⇥103. Note that the eigenvalue method described in Section 4.3 can be used to rigorously determine ↵. However sinceit would be expensive for large problems, we are in favor of simpler but less rigorous rules to compute this parameter.Fig. 28 shows a comparison of deformed shapes of the continuum model and the continuum-plate model and in Fig. 29,the contour plot of the von Mises stress corresponding to various models is given.

31

Figure 24: A plane frame analysis: solid-beam model.

Figure 25: A plane frame analysis: comparison of �xy

contour plot obtained with solid model (left) and solid-beammodel (right).

Figure 26: Cantilever beam subjects to an end shear force: problem setup.

32





32

�xy

(normalise

d)

�xy

(normalise

d)

Q4 elements

Q4 elements

Cubic B-‐spline elements


Figure 27: Cantilever beam subjects to an end shear force: typical B-spline discretisation.

Figure 28: Cantilever beam subjects to an end shear force: comparison of deformed shapes of the continuum modeland the continuum-plate model.

33

•  3D/plate coupling (Kirchhoff)

§





32

Figure 27: Cantilever beam subjects to an end shear force: typical B-spline discretisation.

Figure 28: Cantilever beam subjects to an end shear force: comparison of deformed shapes of the continuum modeland the continuum-plate model.

33

10.0 20.0 30.0 40.0 50.0

von Mises stress

0.429 51

(a) reference model

10.0 20.0 30.0 40.0

von Mises stress

5.17 48.6

(b) mixed dimensional model, Mindlin plate

10.0 20.0 30.0 40.0

von Mises stress

5.12 48.6

(c) mixed dimensional model, Kirchho↵ plate

Figure 29: Cantilever beam subjects to an end shear force

34

10.0 20.0 30.0 40.0 50.0

von Mises stress

0.429 51

(a) reference model

10.0 20.0 30.0 40.0

von Mises stress

5.17 48.6

(b) mixed dimensional model, Mindlin plate

10.0 20.0 30.0 40.0

von Mises stress

5.12 48.6

(c) mixed dimensional model, Kirchho↵ plate

Figure 29: Cantilever beam subjects to an end shear force

34

32x4x5 tri-‐cubic B-‐spline elements

16x2 bi-‐cubic B-‐spline elements

Full 3D

MDA

Ins$tute of Mechanics & Advanced Materials Figure 32: Square plate enriched by a solid. The highlighted elements are those plate elements cut by the solid

boundaries. The plate is fully clamped ans subjected to a gravity force.

elements with some geometry entities is popular in XFEM, see e.g., [58]. Fig. 33 plots the deformed configurationof the solid-plate model and the one obtained with a plate model. A good agreement can be observed. In order toshow the flexibility of the non-conforming coupling, the solid part was moved slightly to the right and the deformedconfiguration is given in Fig. 34. The same discretisation for the plate is used. This should serve as a prototype formodel adaptivity analyses to be presented in a forthcoming contribution.

Figure 33: Square plate enriched by a solid: transverse displacement plot on deformed configurations of plate model(left) and solid-plate model (right).

7. Conclusions

We presented a Nitsche’s method to couple (1) two dimensional continua and beams and (2) three dimensionalcontinua and plates. A detailed implementation of those coupling methods was given. Numerical examples using loworder Lagrange finite elements and high order B-spline/NURBS isogeometric finite elements provided demonstrate thegood performance of the method and its versatility. Both classical beam/plate theories and first order shear beam/platemodels were presented. Conforming coupling where the continuum mesh and the beam/plate mesh is not overlapped

36

Figure 32: Square plate enriched by a solid. The highlighted elements are those plate elements cut by the solidboundaries. The plate is fully clamped ans subjected to a gravity force.



7. Conclusions


36

Figure 32: Square plate enriched by a solid. The highlighted elements are those plate elements cut by the solidboundaries. The plate is fully clamped ans subjected to a gravity force.



7. Conclusions


36

Load: weight

Fully clamped


• Versa$le coupling for mixed-‐dimensional analysis with non-‐conforming discre$sa$ons (IGA/FEM)

•  Future work § Weighted averages in the Nitsche Plate/3D coupling §  Cheap way to evaluate the lower bound on the regularisa$on parameter §  Efficient and weakly intrusive local/global solver §  Damage in solid region

Conclusion

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