1.34 Actio = reactio?

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We expect that the stresses on the two faces of a cut are the same. But this

must not be true in FE analysis. This does not contradict Newton’s principle,

because equilibrium in the FE sense means only that the virtual work done

by the stresses on the two faces and the surface loads or volume forces on the

left- and right-hand side of the cut is the same: δW+

e + δW−

e = 0.

Any load in the neighborhood which senses the movement contributes to

the virtual work and thereby blurs the picture, so that δW+

e + δW−

e = 0 in

general does not imply that the resultant stresses on the two faces are the

same, R+ = R−.

Consider for example the masonry wall and the column in Fig. 1.125. If

the column is modeled with linear elements, concentrated forces will act at

the nodes and line loads at the opposing face if the wall is modeled with CST

elements; see Fig. 1.125. What these different forces though have in common

is that they are work-equivalent with respect to the nodal unit displacements

of the interface nodes (volume forces are absent from this model).

Or imagine that a slab is modeled with Kirchhoff plate elements, and a T

beam (Fig. 1.126) with beam elements. Evidently it is not possible, to simply

178 1 What are finite elements?

1.126.

transfer the nodal moments from the beam to the slab. Reissner–Mindlin

elements would not even tolerate a transfer of the nodal forces.

Add to this that usually the displacements on the two faces are not the

same because the shape functions on the two sides are different. Such inconsistencies

are much more common than users are aware of. But they should only

occur at the interfaces of different structural elements, because a structure

can hardly be modeled with a series of gaps.

Equivalent nodal forces

What is the same though are the equivalent nodal forces at the interface

between two structural components because at any such node N

fL + fR = fN (1.498)

where fN is the equivalent nodal force applied at the node. The components

fi of the two nodal forces (L) and (R) are the strain energy products between

the stress field Sh of the FE solution uh and the nodal unit displacements ϕi

of the node on the left and on the right

fi = a(uh,ϕi) =

_

Ω

Sh •E(ϕi) dΩ =

_

Ω

[σ11 · ε11 + 2σ12 · ε12 + σ22 · ε22] dΩ .

(1.499)

This is also the technique how equivalent nodal forces can be assigned to the

nodes of interelement boundaries if a structure is split into different parts; see

Fig. 1.127.

If no force is applied at the node, fN = 0, then the sum of the two forces,

the two “energy quanta”, is zero. Note that this result is independent of the

shape of the elements on both sides of the interface. Large elements bordering

on small elements possess the same nodal forces as the small elements.

What the small elements miss in size they make good in strains (not the

strains from the FE solution uh but from the fields ϕi) because the smaller an

element gets the larger the strains from the unit displacements of the nodes

will be, this is the 1/h effect (see Fig. 1.68 p. 97) so that

fL

i =

_

small

Sh • E(ϕLi

)

_ _ _

large

dΩ =

_

large

Sh • E(ϕRi

)

_ _ _

small

dΩ = fR

i . (1.500)

Fig. The coupling

between a beam and a slab

is a work-equivalent coupling

but not a mechanical coupling

1.34 Actio = reactio? 179

Fig. 1.127. The equivalent nodal forces on both sides are the same. This is independent

of the shape of the elements on both sides. Even a thin layer of elements

on the left would produce the same forces as the large elements on the right

Note also that the support reactions RA and RB of the plate are uniquely

determined by the stress state in the two shaded elements bordering on the

supports. For example the horizontal component (RA)x of RA is the scalar

product between the stress state Sh in the element ΩA and the strains of

the unit displacement in horizontal direction of the node that attaches to the

support

(RA)x =

_

ΩA

Sh •E(ϕA1

) dΩ . (1.501)

The more the mesh is refined the smaller ΩA gets but the loss in size is easily

balanced by the increase in the stresses Sh—the elements gets closer to the

hot spot.

Example

The plate in Fig. 1.128 a carries a horizontal edge load of magnitude 2P

and is modeled with two bilinear elements. The equivalent nodal forces are

f1 = f2 = P while all other fi (also the vertical components) are zero. The FE

solution is the solution of the load case in Fig. 1.128 g. The stress state Sh of

the FE solution on acting through the strains E(ϕi) of the unit displacement

fields ϕi yields the same equivalent nodal forces fi as the original load

180 1 What are finite elements?

11.49

3.83

3.00

2.00

Fig. 1.128. Plate and shear forces: a) system and loading, b) - e) horizontal nodal

unit displacements, f) equivalent horizontal nodal forces, g) FE solution

Fig. 1.129. The

raw shear stresses

between the elements

must not be the same

1.35 The output 181

fi = a(uh,ϕi) =

_

Ω

Sh •E(ϕi) dΩ =

_

P i= 1, 2

0 i = 3, 4 .

(1.502)

(Of course the plate has more than four degrees of freedom). At the interface

between the two elements the sum of the shear stresses is not zero (d =

thickness)

N(1)

yx + N(2)

yx = d

_ l

0

σ(1)

yx dx + d

_ l

0

σ(2)

yx dx    = 0, (1.503)

because the shear forces must balance the horizontal component of the line

force that acts at the interface. The obvious remedy (see also Fig. 1.129) is to

work with averaged stresses. Let

j = t(1)

h + t(2)

h (= ↓ + ↑) on∂Ω1 ∩ ∂Ω2 (1.504)

the jump in the tractions, the improved (ˆ) averaged tractions are

ˆt

(1) = t(1)

h

− 1

2 j ˆt

(2) = t(2)

h

− 1

2 j . (1.505)

Unlike the resultant stresses N(i)

yx the equivalent nodal forces f(i)

3 on the two

sides of the interface balance

f3 = f(1)

3 + f(2)

3 =

_

Ω1

Sh •E(ϕ3) dΩ +

_

Ω2

Sh •E(ϕ3) dΩ = 0. (1.506)

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