1.37 Equilibrium

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Statements such as

• Global equilibrium is satisfied

• Equilibrium is usually not satisfied within an element

• Equilibrium is usually not satisfied across interelement boundaries

• Equilibrium of nodal forces and moments is satisfied (?)

do not boost our confidence in the FE method, but they lose much of their

alarmism if we recall that in the FE method the original load case p is replaced

by a work-equivalent load case ph and it is therefore quite natural for the stress

resultants of the load case ph not to maintain equilibrium with the forces of

the load case p.

An FE program commits only one error, and this at the very start: it

replaces the original load case by a work-equivalent load case. Everything else

1.37 Equilibrium 185

1.133. Because the

maintained

that follows is classical structural analysis. An FE program solves the workequivalent

load case exactly. Therefore both the whole structure and every

individual part maintains equilibrium—with the forces of the work-equivalent

load case.

Global equilibrium

This term has a very special meaning in the FE method. After the global

stiffness matrix K is assembled it contains the contributions of the nodal unit

displacement of all the nodes of the structure. Next, those rows and columns

are deleted that belong to fixed degrees of freedom (ui = 0), and the so-called

reduced stiffness matrix is obtained. The entries kij = a(ϕi, ϕj) in the reduced

stiffness matrix are the strain energy products of the nodal unit displacements

in Vh, while the entries in the full matrix K are the strain energy products of

the ϕi in the space V +

h , which is the space Vh plus all nodal unit displacements

of the fixed nodes, the support nodes.

Because all rigid-body motions u0 = a×x+b of the structure do lie in V +

h ,

the resultant force Rh of the substitute load ph coincides in size, direction,

and position with the resultant force R of the original load distribution p.

This follows simply from the fact that p(ϕi) = ph(ϕi), i = 1, 2, . . . n, and that

any rigid-body motion u0 can be written in terms of the ϕi. Hence global

equilibrium in FE terminology means Rh = R.

If A is the resultant support reaction in the load case p, then R+A = 0,

and the same holds for the FE solution Rh + Ah = 0. Because Rh = R,

it follows that Ah + R = 0 must be true as well. In this last equation we

are comparing apples (Ah is from the load case ph) with oranges (R is from

the load case p), but because of global equilibrium, Rh = R, it makes no

difference.

Local equilibrium

Why is it that the Kirchhoff shear vh

n of an FE solution integrated along the

edge of an arbitrary patch Ωp of elements does not balance the original load

acting on that patch? The reason is that the rigid-body motions of the patch

Let us consider a patch Ωp of a slab which is subjected to a uniform surface

load p that vanishes outside the patch. Let Rp the resultant force and let Rp

h

the resultant force of the FE load ph acting on this patch. If the resultant

Fig.

footprint of the nodal unit

displacements extends beyond

the patch, equilibrium is not

extend beyond the patch, see Fig. 1.133.

186 1 What are finite elements?

forces were the same, Rp = Rp

h, the integral of the Kirchhoff shear along the

perimeter of the patch would also be the same, (vn − vh

n, 1)Γ = 0 (we neglect

the corner forces). The equation Rp = Rp

h is valid if and only if

_

Ωp

pu0 dΩ =

_

Ωp

ph u0 dΩ (1.513)

for all rigid-body motions u0 of the patch. The problem is that the rigid-body

motions of the patch do not lie in Vh. For to lift the patch by one unit of

displacement the movement

u0(x) =

_

1 x ∈ Ωp

0 else (1.514)

would have to lie in Vh. But such a discontinuous function is nonconforming.

Imagine that a tablecloth is spread over the slab and that the patch is

lifted while we try to hold down the rest of the tablecloth. This shape is as

close as we can get on Vh to the lift of the patch.

Because both load cases p and ph are equivalent with respect to all ϕi

they are also equivalent with respect to the shape of the “tablecloth”. In a

somewhat symbolic notation this means

_

Ωp+1

p (_ _ ) dΩ =

_

Ωp+1

ph (_ _ ) dΩ , (1.515)

where Ωp+1 denotes that part of the slab where the height of the tablecloth

is not zero. This is Ωp plus one row of elements (probably). According to

our assumptions p is zero outside of Ωp—this simplifies the derivation—and

therefore equilibrium is “almost” established

_

Ωp

p dΩ =

_

Ωp+1

ph (_ _ ) dΩ . (1.516)

Specifically the weight of the load p on the patch Ωp is the same as the weight

of the load “1/2ph + ph + 1/2ph” on the patch Ωp+1, and therefore Rp

h (the

integral of vh

n along the edge of the patch Ωp) cannot be the same as the

weight of p on Ωp, that is

Rp =

_

Ωp

pdΩ     =

_

Γp

vh

nds = Rp

h . (1.517)

The reviewing engineer would like to have

_

Ωp

p dΩ =

_

Ωp

ph dΩ , (1.518)

which would only be true if ph were zero in the neighboring elements, which is

very improbable. We can only hope that the smaller the elements become, the

smaller the ramp becomes, and the closer we come to a true local equilibrium.

1.38 Temperature changes and displacement of supports 187

Nodal forces

The structural equation Ku = f is not an equilibrium condition. To speak

of nodal equilibrium is misleading because the layperson takes this statement

literally. Rather it expresses the equivalence of the FE load case ph with the

original load case p,

ph(ϕi) = fh

i =

_

j

kij uj = fi = p(ϕi) i = 1, 2, . . . n , (1.519)

i.e., virtual work (kN m) is equated, not forces (kN); see Eq. (7.92), p. 515.

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