1.44 Partition of unity

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There is a logic built into influence functions: the influence functions for the

support reaction A and B of the beam in Fig. 1.155 add to 1 at every point

x:

ηA(x) + ηB(x) = 1.0 at all points x . (1.597)

212 1 What are finite elements?

Fig. 1.155. The sum of the two influence functions for the support reactions RA

and RB is 1.0 at every point

It cannot be 1.10 or 0.90; it must be exactly 1 otherwise the beam would be

a wonderful machine that could increase or decrease the load at will.

The same logic applies to the nodal unit displacements, because they are

influence functions for the equivalent nodal forces. To perform a unit translation

w = 1 of a slab, the value w = 1 is assigned to the translational degrees

of freedom. The virtual work done by the surface load representing the weight

g is the integral

_

i

_

Ω

g ϕi dΩ . (1.598)

This integral is only equal to the work done by the total weight G

G ·1 =

_

i

_

Ω

g ϕi dΩ , (1.599)

if the nodal unit displacements add to 1 at every point x:

_

i

ϕi(x) = 1 for all x . (1.600)

In other words if the value ui = 1 is assigned to all translational degrees of

freedom of a structure, the structure must undergo a true rigid-body motion

and no point may lag or may rush ahead, i.e., the shape functions must form a

partition of unity in the domain Ω. This is a very important property, because

without it there would be no global equilibrium.

In Euler–Bernoulli beams and Kirchhoff plates it is also required that the

nodal unit displacements represent (pseudo) rotations such as

w(x) = ax (beam) w(x, y) = ax + b y (slab) (1.601)

exactly, because without that property the resulting moments would not be

the same,

_

Mh =

_

M.

In plates the symmetry of the stress tensor σxy = σyx implies M = 0. This

is why in 2-D elasticity only translations such as r = [1, 0]T or r = [0, 1]T

count as rigid-body motions.

A mesh inherits the ability to represent rigid-body motions from the individual

elements. This means that if an individual element has this property,

1.45 Generalized finite element methods 213

then so does the whole mesh if the nodal unit displacements are C0 or C1

respectively (beams and slabs), that is, if the elements are conforming.

The logic behind all this is the following:

1. If rigid-body motions can be represented exactly on the mesh, they also

lie in V +

h , because V +

h contains all deformations that can be “reached” by

the nodal unit displacements—with a proper choice of the coefficients ui

this is possible (V +

h = Vh + rigid-body motions).

2. Because the two load cases p and ph are equivalent with respect to the

ϕi ∈ V +

h they are also equivalent with respect to all rigid-body motions

δWe(p, r) = δWe(ph, r) r = rigid-body motion . (1.602)

3. We have

δWe(p, r) =

_

H ...=

_

V ...=

_

M (1.603)

depending on what kind of rigid-body motion r = a + ω × x is.

4. But (1.602) and (1.603) imply that

→

_

H =

_

Hh ↓

_

V =

_

Vh _

_

M =

_

Mh

(1.604)

which is just the statement that R = Rh.

Next to rigid-body motions, constant stress states are the most important

stress fields which an FE program must be able to represent exactly. Only

then will an FE program have a chance to come close to the exact solution if

the element size h shrinks to zero.

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