4.14 Equilibrium of resultant forces

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In general the resultant stresses of the FE solution in any cross section do not

balance the exterior load, because FE programs cannot generate the exact

influence functions for the resultant stresses; see Sect. 1.21, p. 86.

The influence function for the horizontal force Nyx (the shear force) in

section A − A in Fig. 4.49 is a dislocation of the upper part of the structure

by one unit of displacement to the right.

To produce this influence function the resultant shear stress σ(i)

xy of the

unit displacement fields ϕi in the section A−A must be applied as equivalent

nodal forces:

fi =

_

A−A

σ(i)

xy dx . (4.102)

The resulting shape (see Fig. 4.49 c) is not the exact influence function, because

the horizontal displacement at the upper left corner is 2.09 units of

displacements instead of the exact 1.0. A simple test confirms this result:

when a point load P is applied at the upper left corner, the stress resultant

in section A − A is exactly

_

A−A

σh

yx dx = 2.09P . (4.103)

This is a large error.

Next we let the section A−A pass right through the center of the elements

and—to our surprise—now the upper part of the wall moves sideways by

exactly one unit of displacement, i.e., the resultant shear force Nxy in the

section A − A is exactly 1.0 P.

2 According to Eurocode

4.14 Equilibrium of resultant forces 379

Fig. 4.49. Shear wall: a) section A − A, b) exact influence function for Nyx in

section A − A, c) FE approximation

Fig. 4.50. Single bilinear element: a) degrees of freedom; equivalent nodal forces

for b) the influence function for Nyx at the center, and c) at the upper quarter

point, all horizontal forces are the same, fi = E a/4 b

380 4 Plane problems

Fig. 4.51. Influence function

Nxy at y = 3.25: a)

bilinear element, b) Wilson’s

element. Lateral deformations

not to scale

To understand this phenomenon it is best to study a single bilinear element

(see Fig. 4.50), and to begin with the shear force Nyx in a horizontal section

which passes through the center of the element. The equivalent nodal forces

that will generate the (approximate) influence function for Nyx are the same

forces that effect a pure shear deformation of the element (see Fig. 4.50 a),

u(x, y) = y + 0.5 b

b

v(x, y) = 0 εxy =

1

2 b

εxx = εyy = 0. (4.104)

Note that at the upper edge u(x, y = 0.5 b) = 1. The proof is given below.

Next it is assumed that the line intersects the element at a distance of 0.25 b

units from the upper edge; see Fig. 4.50 b. The equivalent horizontal nodal

forces do not change, but the vertical forces do, and they effect a rotation

which is responsible for the large error, 2.09 versus 1.0, that was observed

previously at the upper left corner of the shear wall.

To formalize these observations let

fkj

i =

_ l

0

σkj(ϕi)ds (= Nkj(ϕi)) (4.105)

for

4.14 Equilibrium of resultant forces 381

denote the resultant stresses of the unit displacement fields. Then the following

condition can be formulated.

Shape condition An element satisfies the shape condition in a particular

cross section if the shape generated by the equivalent nodal forces fkj

i is in

agreement with the equilibrium conditions.

That is, under the action of the fyx

i (Nyx), the upper edge should slide

sidewards, u = 1, and the fyy

i (Nyy = normal force) should lift the upper edge

by one unit, v = 1, etc. If the shape condition is satisfied, the resultant stress

in the cross section maintains equilibrium with an edge load.

It follows that the shape condition is satisfied if the forces that effect a

unit displacement of the element edge can be recovered in the pertinent cross

section, which is just the same as saying that

_

H = 0,

_

V = 0.

While the bilinear element satisfies the shape condition for Nyx only at the

center, Wilson’s element satisfies the shape condition for Nxy in any horizontal

section; see Fig. 4.51. Regardless of where the cut passes through Wilson’s

element, the fxy

i are always the same, and because the fxy

i at the center yield

the correct shape, any other section has this property as well.

It is evident that the influence functions for the normal forces Nxx and

Nyy in an element satisfy the shape condition. The associated equivalent nodal

forces stretch the element in the horizontal or vertical direction by one unit of

displacement regardless of where the element is intersected. More difficult is

the situation—as seen—for the shear force Nxy while with regard to rotations

(

_

M = 0) the situation is hopeless, because the influence functions (see Fig.

1.60 p. 89) definitely do not lie in Vh.

Proof

We show that the equivalent nodal forces that effect the shear deformation (see

Fig. 4.50 b) are the same equivalent nodal forces fxy

i that provide the influence

function for Nxy at the center. The unit displacement field ϕ7(x) = [u, v]T of

the upper left node has the components

u =

(0.5 a − x)(0.5 b + y)

a b

v(x, y) = 0 (4.106)

εxy =

1

2u,y =

0.5 a − x

2 a b

εxx = . . . εyy = . . . (4.107)

so that with the displacement field u of (4.104), the equivalent nodal force f7

becomes

f7 = a(u,ϕ7) =

_

Ω

2 εxy σxy dΩ = 2E

_

Ω

1

2 b

(0.5 a − x)

2 a b

dΩ = a

4 b

E

(4.108)

382 4 Plane problems

which is the same as the resultant shear stress of the displacement field ϕ7(x):

f7 =

_ +a/2

−a/2

σxy dx = E

_ +a/2

−a/2

(0.5 a − x)

2 a b

dx = a

4 b

E = Nxy . (4.109)

Equivalent nodal forces for influence functions

In principle, it is easy to arrange for an FE program code to display the FE

influence functions on the screen. This can provide a better understanding of

the behavior of the discretized structure. The implementation is simple because

the equivalent nodal forces can be calculated on an element-per-element

basis, as explained in the following.

First let us recall how the influence function G1 for, say, the stress σxx at

the center xc of an element is calculated. The equivalent nodal forces fi are

the stresses σ(i)

xx(xc) of the nodal unit displacement fields ϕi at the center.

This is consequence of

a(G1[xc],ϕi) = (δ1[xc],ϕi) = σ(i)

xx(xc) . (4.110)

If these equivalent nodal forces fi are applied at the four nodes of a bilinear

element, the result is a displacement field Gh1

[xc] that simulates a point

dislocation u(x+c ) − u(x−

c ) = 1 in the horizontal direction at xc.

To obtain the FE influence functions for the resultant stresses

Nx =

_ b

0

σxx dy Nxy =

_ b

0

σxy dy (4.111)

Nyx =

_ a

0

σyx dx Ny =

_ a

0

σyy dx (4.112)

in a vertical or horizontal section of an element, the integrals of these stresses

must be applied as equivalent nodal forces fi. In a horizontal section passing

through the point (x, y) of a bilinear element, we have [96]

_ a

0

σyy dx = E

2 b (−1 + ν2)

·

_

2 b ν (u1 − u3)+

+ a (u2 + u4 − u6 − u8) + 2y ν (−u1 + u3 − u5 + u7)

_

(4.113)

_ a

0

σxy dx = E

4 b (1 + ν)

·

_

a (−u1 − u3 + u5 + u7)+

− 2 (b (u2 − u4) + y (−u2 + u4 − u6 + u8))

_

(4.114)

and in a vertical section

4.15 Adaptive mesh refinement 383

_ b

0

σxx dy = E

2 a (−1 + ν2)

·

_

b (u1 − u3 − u5 + u7)+

+ 2ν (a (u2 − u8) + x (−u2 + u4 − u6 + u8))

_

(4.115)

_ b

0

σxy dy = E

4 a (1 + ν)

·

_

− 2 a (u1 − u7)+

+ b (−u2 + u4 + u6 − u8) + 2x (u1 − u3 + u5 − u7)

_

. (4.116)

Here a denotes the length of the element, b is the width of the element, and

the origin of the coordinate system lies in the lower left corner. To obtain the

nodal force fe

1 for the influence function for Ny in a horizontal section passing

through the point (x, y), for example, let u1 = 1 in (4.113) and let all other

ui = 0.

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