4.18 Mixed methods

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In FE methods we distinguish between displacement-based approaches and

mixed methods. In displacement methods, the structural behavior is solely

governed by the displacements, so approximate displacements suffice. The

classical FE methods are displacement-based methods. But these technologies

are limited in scope, in that incompressible material or simply Kirchhoff plates

and shell structures pose serious problems [26]. Mixed methods can overcome

these difficulties.

In mixed methods separate approximations are chosen for the displacements

and the stresses, so that for example a bar element has four degrees of

freedom: the two end displacements u1, u2 and the stresses σ1, σ2 at the end

cross sections; see Fig. 4.61.

394 4 Plane problems

Fig. 4.59. Spurious displacement field in a water tank and regular field

like support – the volume

does not change

On first glance this approach might look strange, but the mixed approach is

in good agreement with the nature of structural mechanics because structural

mechanics is “mixed”, the displacements ui, the strains εij , and the stresses

σij in a plate are coupled by a system of first-order differential equations:

E(u) − E = E+

C[E] − S = S+ (4.125)

−divS = p

where

E(u) =

1

2

(∇u + ∇uT) =

1

2

_

2 u1,1 u1,2 +u2,1

u2,1 +u1,2 2 u2,2

_

(4.126)

Fig. 4.60. Rubber4.18

Mixed methods 395

4.61. A bar element

and C[ ] is the elasticity tensor

C[E] = 2μE + λ (tr E) I . (4.127)

Here I is the unit matrix, λ = 2μν/(1−2ν), and trE = ε11+ε22 is the trace

of E. The terms on the right-hand side in (4.125) are initial strains ε+

ij and

stresses σ+

ij .

If we call S = {u,E,S} an elastic state, (4.125) can be read as the application

of a differential operator A to the elastic state S.

When the same operator A is applied to the FE solution Sh = {uh,Eh,Sh},

the result is

E(uh) − Eh = E+

h

C[Eh] − Sh = S+

h (4.128)

−divSh = ph .

The operator A satisfies the identity

G(S, ˆ S) = _A(S), ˆ S_ +

_

Γ

Sn•ˆu ds

_ _ _

δWe

−a(S, ˆ S)

_ _ _

δWi

= 0 (4.129)

where the angular brackets denote the scalar product of A(S) and the virtual

elastic state ˆ S

_A(S), ˆ S_ : =

_ l

0

(E(u) − E) •ˆSdΩ +

_

Ω

(C[E] − S) • ˆE dΩ

+

_

Ω

−divS •ˆu dΩ (4.130)

and the strain energy product is

a(S, ˆ S) =

_

Ω

(E(u) − E) •ˆS dΩ +

_

Ω

C[E] • ˆE dΩ

+

_

Ω

S • (E(ˆu) − ˆE ) dΩ . (4.131)

If the initial stresses are zero, σ+

ij = 0, the strains εij are determined by

the stresses

Fig.

for displacements and stresses

with independent functions

396 4 Plane problems

E = C

−1[S] . (4.132)

and the states S contain only two independent quantities—the displacements

ui and the stresses σij :

S = {u,C

−1[S],S} . (4.133)

This is the starting point for the so-called Hellinger–Reissner principle. Mixed

methods are mostly based on this principle.

The FE state

Sh = s1 S1 + s2 S2 + . . . =

_

i

si Si (4.134)

consists of a series of “unit states”, to which weights si are attached. It is best

to split the unit states into u-states Su and σ-states Sσ:

Su = {ϕi, 0, 0} , Sσ = {0,C

−1[Si],Si} . (4.135)

The u-states only represent displacements, while the σ-states only represent

strains + stresses, so the FE solution looks like

Sh =

_n

i=1

ui {ϕi, 0, 0} +

_m

i=1

σi {0,C

−1[Si],Si} , (4.136)

where the ϕi(x) are the unit displacements and the Si(x) the stress states.

A unit state like Su = {ϕi, 0, 0} is stress-free if there exists initial strains

E+

i that annihilate the strains induced by the displacement field ϕi, i.e., if

E = E(ϕi) + E+

i = 0. Hence, the Su can be considered the solutions of

problems A(S) = [−E(ϕi), 0, 0]T .

Similar things can be said about the states Sσ. The strains induced by the

stresses are

E = C

−1[Si] + C

−1[S+] . (4.137)

Normally these strains induce displacements. But if S+ = −Si the strains are

zero and therefore a plate must not undergo any compensating movement,

u = 0, i.e. the Sσ solve problems like A(S) = [0,−S+

i ,−divSi]T .

The strain energy product associated with the Hellinger–Reissner principle

is

a(S, ˆ S) =

_

Ω

[C

−1[S] • ˆS + E(u) • ˆS + S •E(ˆu)] dΩ , (4.138)

hence the stiffness matrix k ij = a(Si, Sj) is

_

0(n×n) A(n×m)

AT(

m×n) B(m×m)

_

=

_

u(n)

σ(m)

_

=

_

p(n)

0(m)

_

, (4.139)

4.18 Mixed methods 397

where

aij =

_

Ω

Sj •E(ϕi) dΩ bij =

_

Ω

C

−1[Si] •Sj dΩ (4.140)

and

p i =

_

Ω

p •ϕi dΩ . (4.141)

The first set of equations corresponds to the equilibrium condition −divS = p.

It is the task of the stress states to balance the external load p:

Aσ = p . (4.142)

The second set of equations

AT u +Bσ = 0 (4.143)

corresponds to the compatibility condition E(u) − E = 0. As long as there

are no initial strains this is an internal affair between the displacements and

stresses, which is why the zero vector appears on the right-hand side.

Of course the compatibility condition is only satisfied in the L2-sense, not

in a strict pointwise sense, because in the second set of equations it is only

required that the error be orthogonal to the test functions Sj , the stress states:

_

Ω

(E(uh) − E) •Sj dΩ = 0 j = 1, 2, . . . m . (4.144)

Hence the difference between E(uh) and the strains Eh—which come from

the stresses Eh = C

−1[S]—is zero only in the weighted L2 sense. The residual

can be interpreted as initial strain E+

h :

E(uh) − E = E+

h . (4.145)

Hence in mixed methods (based on the Hellinger–Reissner principle) the

substitute load case that is work-equivalent to the original load case [0, 0, p ]T

solved by the FE program consists of substitute loads ph, and includes initial

strains in each element, [E+

h , 0, ph].

Now the model has u- and σ-degrees of freedom and the next question

then is how many to choose of each (see Fig. 4.62):

n = number of displacement degrees of freedom

m = number of stress degrees of freedom .

Linear algebra provides the answer. For the system Aσ = p to have a solution,

the right-hand side, the vector p, must be orthogonal to all solutions u of the

adjoint system

398 4 Plane problems

Fig. 4.62. a) Not

stress functions than

solution

ATu = 0 . (4.146)

The vectors u that solve this homogeneous system of equations form the socalled

kernel of the matrix AT , and two vectors are orthogonal if their scalar

product is zero, pTu = 0.

If the system (4.146) is regular, i.e., if the columns of AT are linearly

independent, the problem ATu = 0 has only the trivial solution u = 0, and

the right-hand side p is then orthogonal to the kernel of AT . One need not

worry that there are load cases that the FE program cannot solve.

The columns are linearly independent if and only if

ATu = 0 =⇒ u = 0 (4.147)

i.e., if the fact that the scalar product between all the strains

E =

_n

i=1

ui E(ϕi) (4.148)

and all the stresses Si(x) is zero,

_n

i=1

ui

_

Ω

E(ϕi) •Sj dΩ = 0 j = 1, 2, . . . m , (4.149)

implies that the nodal displacements ui are zero.

Obviously this can only happen if there are more stress states Si than

displacement states ϕi, because otherwise one could easily construct, say, a

12-term displacement field that is so balanced that the scalar product with,

say, all seven stress states is zero, even though the displacement field—or

rather its strains—is not.

enough stress functions,

the Babuˇska-

Brezzi-condition is

violated, b) more

displacement functions,

the condition

is satisfied, c) FE

4.19 Influence functions for mixed formulations 399

Fig. 4.63. Elastic bar a) system and load, b) Green’s function G0 c) strain ε0

Hence the referee must always have more degrees of freedom than that

which he is to check, i.e., there must always be more stress degree of freedoms

than displacement degrees of freedom.

The columns of the matrix AT are linearly independent if and only if there

exists a constant β such that

0 < β|u| < inf

δ

sup

σ

σT ATu

|σ| . (4.150)

This is the Babuˇska–Brezzi condition.

The lower bound β ensures that there is always a large enough distance

from zero. When the elements shrink in size, i.e., if the vectors δ and σ grow

in length, the angle between the two vectors will definitely be greater than

zero.

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