4.8 Nodal stresses and element stresses

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In a bilinear element of length a and width b as in Fig. 4.10, the stresses are

σxx(x, y) = E

a b (−1 + ν2)

·

_

b (u1 − u3) + aν (u2 − u8)+

+ xν (−u2 + u4 − u6 + u8) + y (−u1 + u3 − u5 + u7)

_

(4.58)

σyy(x, y) = E

a b (−1 + ν2)

·

_

b ν (u1 − u3) + a (u2 − u8)+

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

_

(4.59)

σxy(x, y) =

−E

2 a b (1 + ν)

·

_

b (u2 − u4) + a (u1 − u7)+

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

_

. (4.60)

The stresses in Wilson’s element are

358 4 Plane problems

Fig. 4.30. Analysis of a column with Wilson’s element

σxx(x, y) = − E

a2b2(−1 + ν2)

_

8 b2q1 x + a2ν(b(−4 q4 + u2 − u8) + 8q4 y)

+ a b(b(−4 q1 + u1 − u3) + ν x(−u2 + u4 − u6 + u8)

+ y(−u1 + u3 − u5 + u7)

_

(4.61)

σyy(x, y) = − E

a2b2(−1 + ν2)

_

8 b2νq1 x + a2(b(−4 q4 + u2 − u8) + 8q4 y)

+ a b(bν(−4 q1 + u1 − u3) + x(−u2 + u4 − u6 + u8)

+ νy(−u1 + u3 − u5 + u7)

_

(4.62)

σxy(x, y) = − E

2a2b2(1 + ν)

_

− 8 b2νq3 x + a2(b(−4 q2 − u1 + u7) − 8 q2 y)

+ a b(b(4 q3 − u2 + u4) + x(u1 − u3 + u5 − u7)

+ y(u2 − u4 + u6 − u8)

_

. (4.63)

Remark 4.1. The origin of the coordinate system in these expressions is assumed

to be the lower left corner of the element while in Fig. 4.10 it is the

center of the element.

These equations can be simplified further. Because of (4.47),

4.8 Nodal stresses and element stresses 359

q1 = aν

8 b

[u2 − u4 + u6 − u8] q2 = b

8 a

[u2 − u4 + u6 − u8] (4.64)

q3 = aν

8 b

[u1 − u3 + u5 − u7] q4 = b ν

8 a

[u1 − u3 + u5 − u7] (4.65)

so that

σxx(x, y) = − E

2 a b (−1 + ν2)

_

b(2(u1 − u3) + ν2(−u1 + u3 − u5 + u7))

+ aν(u2 + u4 − u6 − u8) + 2(−1 + ν2)

(u1 − u3 + u5 − u7) y

_

(4.66)

σyy(x, y) = − E

2 a b (−1 + ν2)

_

a(2(u2 − u8) + ν2(−u2 + u4 − u6 + u8))

+ b ν(u1 − u3 − u5 + u7) + 2(−1 + ν2)

(u2 − u4 + u6 − u8) y

_

(4.67)

σxy(x, y) = − E

4 a b(1 + ν)

_

a(u1 + u3 − u5 − u7)

+ b(u2 − u4 − u6 + u8)

_

(4.68)

and here it is seen that the horizontal stresses σxx only depend on y, the

vertical stresses σyy only on x (in a bilinear element this is true only if ν = 0),

and σxy is constant (whereas it varies in a bilinear element). Therefore the

volume forces in Wilson’s element are zero (!):

− σxx,x −σxy,y = 0 (4.69)

−σyx,x −σyy,y = 0. (4.70)

Stress averaging

If the stress distribution is linear, the stresses are discontinuous at the element

edges. This is straightened out by interpolating the stresses at the midpoints

of the elements. Even in the presence of gross stress discontinuities, the results

at the centers are often acceptable; see Sect. 1.22, p. 88. Similar behavior is

shown at the Gauss points; see Sect. 1.25, p. 104.

According to the FE algorithm, the weighted average of the error in the

stresses is zero on each patch Ωi,

_

Ωi

(σ − σh) • εi dΩ = 0 i = 1, 2, . . . . (4.71)

where εi = [ε(i)

xx, ε(i)

yy, γ(i)

xy ]T are the strains that belong to the unit displacement

fields ϕi, and the patch Ωi is the support of the field (where the strains

360 4 Plane problems

Fig. 4.31. Gauss points: a) the stresses at the Gauss points are extrapolated to

the boundary, b) contour lines of the stresses σxx, the jumps are a measure of the

quality of an FE mesh (Bathe)

are nonzero). Hence we may expect that the error in the stresses σ − σh =

[σxx − σh

xx, σyy − σh

yy, τxy − τh

xy]T changes its sign on Ωi repeatedly. Good

candidates for these zeros are obviously the centers of the elements and the

Gauss points; see Sect. 1.25, p. 104.

For the same reason the stresses at the edge of an element are the least reliable,

and they are often replaced by values extrapolated from the Gauss points

out to the edge; see Fig. 4.31. In the next step the solution is “improved” by

averaging the stresses between the elements and at the nodes, because this is

what a user wants to see. But one has to be careful.

In between two elements with different thickness (see Fig. 4.32 a), the

normal force N(1)

n = σ(1)

nn · t1 = σ(2)

nn · t2 = N(2)

n (orthogonal to the common

edge) is the same, but the strains are discontinuous, ε(1)

nn        = ε(2)

nn. If the thickness

is the same but the modulus of elasticity changes, the stresses σtt parallel to

the common edge are discontinuous. Because the strains εtt parallel to the

edge are the same, if we let ν = 0, then

σ(1)

tt = E1 εtt        = E2 εtt = σ(2)

tt . (4.72)

This means that eventually the reinforcement parallel to the edge is different

(see Fig. 4.33). Initial stresses σ0

ij would complicate things even more. A good

FE program will average the stresses only if the elements have the same elastic

properties and the elements are load free, and no supports interfere with the

stress distribution.

4.8 Nodal stresses and element stresses 361

Fig. 4.32. Stress distribution in a vertical cross section of a shear wall with varying

thickness under gravity load

not the same

Fig. 4.33. a) The thickness

of the plate changes; b) the

modulus of elasticity changes,

and therefore the stresses

parallel to the interface are

362 4 Plane problems

pulls at the support:

system, b) noraveraging

(each time

with two elements)

A small example will illustrate the implications a smoothing process can

have when it ignores the structural context. If two elements—one placed upon

the other—are fixed along their common boundary, the gravity load will produce

compression in the upper element and tension in the lower element; see

Fig. 4.34.

If the nodal stresses are averaged, the stresses at the fixed nodes will be

zero. Hence if the system consists of just two elements, the stresses at all

the nodes will be zero, and therefore—if the results are extrapolated into the

interior—the stresses will be zero everywhere. If the two elements are split

into two elements each, the stresses will still be halved. Only if very many

elements are used will the averaging process leave no trace.

The misalignment of the contour lines of the raw stresses at the interelement

boundaries offers a good visual control of the quality of a mesh (see

Fig. 4.31). But very often these discontinuities are so strong that the user is

irritated, and therefore program authors tend to display only the averaged

stresses. Nevertheless stress discontinuities in the range 5 to 15% are in no

way unusual, and not a warning sign. Even discrepancies of up to 40% can

be tolerated if the design is based on the stresses at the midpoints of the

elements.

Averaging the stresses at the edges and the nodes is the simplest way

to improve the results. More sophisticated methods use an L2 projection to

improve the stresses; see Sect. 1.31, p. 147.

What is seen on the screen is often not the raw output; see Fig. 4.35. Hence

to judge an FE program one must know which filters the program uses, how

it displays the result, and what smoothing algorithms are employed.

Fig. 4.34. The upper

element presses

the lower element

a)

mal stresses in the

element without averaging,

c) with

on the support and

4.9 Truss and frame models 363

Fig. 4.35. Only the tensile stresses are displayed. The shear wall is fixed on the

left- and right-hand side

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