2.6 Linear Constraints

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In addition to the equilibrium equations expressed by Equation (2.27), the solution of a

field problem requires the formulation of boundary conditions. One distinguishes geometric

boundary conditions, which contain the independent variables such as displacements

and temperatures, from natural boundary conditions, which are formulated in terms of

the dependent variables such as stress and heat flux. In this section, the focus is on geometric

boundary conditions. If only one degree of freedom is involved, the constraint is

called a single point constraint, else it is a multiple point constraint. Constraints can be

linear or nonlinear. Examples of truly linear multiple point constraints are the constraint

of a degree of freedom in nonrectangular coordinates, cyclic symmetry conditions and

equations connecting dissimilar meshes. Examples for nonlinear multiple point constraints

are constraints involving finite rotations such as rigid body motions and incompressibility

conditions.

2.6.1 Inclusion in the global system of equations

The multiple point constraint concept is extremely powerful. Therefore, a truly efficient way

must be found to deal with it numerically. One option is to augment the matrix with the

additional equations using Lagrangian multipliers. This, however, leads to larger systems of

92 LINEAR MECHANICAL APPLICATIONS

equations, increased computational times and a nonsymmetric system. This can be avoided

by eliminating one degree of freedom per single point constraint or per multiple point

constraint during the construction of the global stiffness matrix. How this is done will be

explained in the present section.

Assume that our global system of equations contains N degrees of freedom. Equation

l reads

N

_

j=1

aljuj = bl, l= 1, . . . ,N. (2.142)

Assume we have M additional multiple point constraints of the form

N

_

k=1

eikuk = fi, i= 1, . . . ,M. (2.143)

In each of these equations i, we choose one degree of freedom ki , which is suited to be

eliminated and which will be called a dependent degree of freedom. All dependent degrees

of freedom ki, i = 1, . . . ,M must be distinct. Now, Equation (2.143) can be rearranged

by collecting all dependent degrees of freedom on the left-hand side:

M

_

j=1

eikj ukj

= fi −

N

_

k=1

k_∈{k1,... ,kM}

eikuk, i= 1, . . . ,M. (2.144)

This is a set of M equations in M unknowns and can be solved for the dependent degrees

of freedom provided that the equations on the left-hand side are linearly independent. It

results in equations of the form

ukj

=

N

_

k=1

k_∈{k1,... ,kM}

ckj kuk + dkj . (2.145)

Now, assume that there is only one multiple point constraint. Equation (2.145) reduces to

ui =

N

_

k=1

k_=i

cikuk + di (2.146)

which allows us to eliminate ui . Equation (2.146) is of a most general form including

single point constraints (all cik = 0). Substituting Equation (2.146) into Equation (2.142)

yields (no implicit summation in the section)

N

_

j=1

j _=i

(alj + alicij )uj = bl − alidi, l= 1, . . . ,N. (2.147)

The new coefficient a

◦

lj in the global matrix at position (l, j) now reads

a

◦

lj

= alj + alicij, j,l = 1, . . . ,N; j _= i. (2.148)

LINEAR MECHANICAL APPLICATIONS 93

The global stiffness matrix is Hermitian (complex matrices arise because of cyclic symmetry

conditions, cf Section 2.10; if the matrix is real, “Hermitian” can be replaced by

“symmetric”) and consequently,

alj = ajl, j,l = 1, . . . ,N (2.149)

where a stands for the complex conjugate of a. However, for the new coefficient we have

a

◦

jl

= ajl + aji cil _= a

◦

lj, j,l = 1, . . . ,N; j _= i. (2.150)

The Hermitian structure is destroyed! This is a serious drawback since it means that computational

advantages due to the Hermitian structure are lost. The Hermitian structure,

however, can be restored by multiplying row i, which reads

N

_

j=1

j _=i

(aij + aiicij )uj = bi − aiidi (2.151)

by cim and adding it to row m,m = 1, . . . ,N,m _= i. Now, coefficient a

∗

lj at position (l, j)

reads

a

∗

lj

= alj + alicij + aij cil + aiicij cil, j,l = 1, . . . ,N; j, l _= i (2.152)

which coincides with the complex conjugate of

a

∗

jl

= ajl + ajicil + ailcij + aiicilcij, j,l = 1, . . . ,N; j, l _= i (2.153)

which reads

a

∗

jl

= ajl + aji cil + ailcij + aii cilcij, j,l = 1, . . . ,N; j, l _= i. (2.154)

The right-hand side coefficient b

∗

l satisfies

b

∗

l

= bl − alidi + bicil − aiidicil, l= 1, . . . ,N; l _= i. (2.155)

Row i and column i are dropped altogether from the set.

What happens if two multiple point constraints apply? Let us say that there is a second

multiple point constraint of the form

um =

N

_

k=1

k_=m

cmkuk + dm, m_= i. (2.156)

We assume that both constraints were brought in the form of Equation (2.145) such that

cmi = cim = 0. Consequently,

ui =

N

_

k=1

k_=i,k_=m

cikuk + di (2.157)

um =

N

_

k=1

k_=m,k_=i

cmkuk + dm. (2.158)

94 LINEAR MECHANICAL APPLICATIONS

Applying Equation (2.152) twice, first to eliminate multiple point constraint m, and then

to eliminate multiple point constraint i, one obtains

a

∗∗

lj

= a

∗

lj

+ a

∗

lmcmj + a

∗

mj cml + a

∗

mmcmj cml (2.159)

= alj + alicij + aij cil + aiicij cil

+ (alm + alicim + aimcil + aiicimcil)cmj

+ (amj + amicij + aij cim + aiicij cim)cml

+ (amm + amicim + aimcim + aiicimcim)cmj cml (2.160)

j, l = 1, . . . ,N; j, l _= i; j, l _= m;

or taking into account that cim = 0

a

∗∗

lj

= alj + (alicij + almcmj )

+ (aij cil + amj cml)

+ (aimcmj cil + amicij cml)

+ (aiicij cil + ammcmj cml). (2.161)

j, l = 1, . . . ,N; j, l _= i; j, l _= m.

In a similar way one obtains for the right-hand side

b

∗∗

l

= b

∗

l

− a

∗

lmdm + b

∗

mcml − a

∗

mmdmcml (2.162)

= bl − alidi + bicil − aiidicil

− (alm + alicim + aimcil + aiicimcil) dm

+ (bm − amidi + bicim − aiidicim)cml

− (amm + amicim + aimcim + aiicimcim) dmcml (2.163)

l = 1, . . . ,N; l _= i,m;

or, taking into account that cim = 0

b

∗∗

l

= bl + (bicil + bmcml)

− (aiidicil + ammdmcml)

− (alidi + almdm + aimcildm + amicmldi ). (2.164)

l = 1, . . . ,N; l _= i, m.

LINEAR MECHANICAL APPLICATIONS 95

Equations (2.161) and (2.164) cover all possibilities for coefficients a and b. Indeed, in

a finite element code the element matrices are calculated first, Equations (2.21) to (2.24).

Then, these matrices are transferred into the global matrix. This operation is symbolized

by Equations (2.28) to (2.30) and will be looked at in more detail now.

Suppose an entry a in the local matrix corresponds to global degrees of freedom p and

q (row and column). Now, the following possibilities arise:

1. p and q are independent degrees of freedom. Then

apq+ = a (2.165)

where the C-notation + = was used to indicate that the global matrix entry apq is

to be augmented by a.

2. p is a dependent degree of freedom, q is not. Accordingly, row p in the global matrix

is eliminated, column q is not , and a has a similar status as aij in the term aij cil in

Equation (2.152). Since row p is eliminated, the contribution of a is transferred to

whatever position aij in Equation (2.152) has gone to, now substituting p for i and

q for j

alq+ = acpl. (2.166)

This applies to all degrees of freedom l for which cpl _= 0. Because of Equation (2.145)

these are only independent degrees of freedom.

3. q is a dependent degree of freedom, p is not. This contribution is comparable to

alicij in Equation (2.152) and −alidi in Equation (2.155). Now we have

apj+ = acqj ∀j such that cqj _= 0 (2.167)

and

bp+ = −adq . (2.168)

4. p and q are dependent degrees of freedom, p _= q, cf aimcilcmj in Equation (2.161)

and aimcildm in Equation (2.164):

alj+ = acplcqj ∀l, j such that cpl _= 0 and cqj _= 0 (2.169)

and

bl+ = −acpldq ∀l such that cpl _= 0. (2.170)

5. p and q are dependent degrees of freedom, p = q, cf aiicij cil in Equation (2.152)

and −aiidicil in Equation (2.155):

alj+ = acpj cpl ∀l, j such that cpl _= 0, cpj _= 0 (2.171)

bl+ = −adpcpl ∀l such that cpl _= 0. (2.172)

96 LINEAR MECHANICAL APPLICATIONS

For an entry b in the local right-hand side corresponding to global degree of freedom

p, there are only two possibilities:

1. p is an independent degree of freedom

bp+ = b. (2.173)

2. p is a dependent degree of freedom, cf bicil in Equation (2.155):

bl+ = bcpl ∀l such that cpl _= 0. (2.174)

In this way, all entries in the local matrices are transferred to independent degrees

of freedom in the global matrices. Consequently, the elimination of degrees of freedom

involved in multiple point constraints is taken care of implicitly while transferring the local

matrices into the global ones.

Notice that to each entry a in row ip and column iq in the element stiffness matrix

corresponding to global degrees of freedom p and q, respectively, there is a symmetric

entry with the same value a in row iq and column ip. If p is a dependent degree of freedom

and q is an independent degree of freedom, Equation (2.166) applies to entry a in row ip

and column iq :

alq+ = a cpl (2.175)

and Equation (2.167) applies to entry a in row iq and column ip:

aql+ = acpl (2.176)

which keeps the Hermitian structure of the global matrix. In practice, only half of the

Hermitian matrix is calculated and stored.

2.6.2 Forces induced by linear constraints

The introduction of multiple point constraints induces forces. Indeed, imagine a constraint

of the form

ui = uj . (2.177)

Then both degrees of freedom are coupled and behave as if a rigid bar connects both:

degree of freedom i will experience a force F, degree of freedom j will experience the

inverse force. How does this translate to general multiple point constraints of the form in

Equation (2.146)? Recall that the global system contains N degrees of freedom leading

to an N × N stiffness matrix. Adding Equation (2.146) leads to N + 1 equations in N

unknowns. After substitution into the global set, column i was eliminated leading to N

equations in N − 1 unknowns. This is still an overdetermined system and generally has

no solution. This problem was solved by adding multiples of row i to the other rows and

deleting row i afterward. Consequently, row i is not being satisfied. Indeed, the residual

of row i is exactly the multiple point constraint force we are looking for (no implicit

LINEAR MECHANICAL APPLICATIONS 97

summation in this section):

N

_

j=1

j _=i

(aij + aiicij )uj − (bi − aiidi ) = Fi . (2.178)

Accordingly, the addition of multiples of row i to other rows is equivalent to the addition

of multiples of Fi. Indeed, row l reads

N

_

j=1

j _=i

(alj + alicij )uj +



N

_

j=1

j _=i

(aij + aiicij )uj



cil = (bl − alidi ) + (bi − aiidi )cil (2.179)

or

N

_

j=1

j _=i

(alj + alicij )uj − (bl − alidi ) = −Ficil (2.180)

and degree of freedom l experiences the force

Fl = −Ficil . (2.181)

Notice that the force is proportional to the conjugate coefficient in the multiple point

constraint. For instance, if the multiple point constraint reads

ui = 2uj (2.182)

degree of freedom j experiences a force that is twice the force acting on degree of freedom

i.

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