1.21 Why resultant stresses are more accurate

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Near points where stresses become singular it is better to concentrate on

resultant stresses than single values:

Ny =

_ l

0

ny dx =

_ l

0

σyy d dx My =

_ l

0

myy dy . (1.263)

The lower left corner point of the left opening in the plate in Fig. 1.56 is just

such a singular point. An influence function for the singular stress σyy at this

point does not exist, see Sect. 7.6, p. 532.

The stress σyy at the corner point increases steadily when the mesh is

refined while the resultant stress Ny in the cross section A − A is much more

stable [146]. The reason is that the influence function for Ny has a simple

shape. It is the displacement field of the plate if all the points in the cross

section are spread simultaneously; see Fig. 1.56 c. Even a coarse mesh suffices

to approximate this shape, and this is why Ny changes little when the mesh

is refined.

Next consider the plate in Fig. 1.57. The influence function for the stress

σyy at a single point, for example at the lower edge of the plate, certainly

does not lie in Vh, but the influence function for the normal force Ny must lie

in V +

h (= Vh plus the rigid-body motions of the plate), because it represents

a rigid-body motion of the plate, and therefore the FE program finds (as it

must!) the correct result for the stress resultant Ny.

In plate-bending problems the situation is the same, as can be seen in

Equilibrium

The resultant force Rh of the FE solution in a cross section will balance the

external load if the Green’s function for R lies in Vh, see Sect. 1.37, p. 184.

The Green’s function for the sum of the horizontal forces and the vertical

forces are simple movements, ux = 1 and uy = 1 respectively. In the case of

the plate in Fig. 1.59 the influence function for Nx in the cross section A−A

is a rigid-body motion of the part to the right of A−A, i.e., a unit dislocation

of all points on the line A − A.

The equivalent nodal forces fi that try to generate this shape are the

integrals of the stresses of the nodal unit displacements fields ϕi along the

line A − A:

fi =

_

A−A

_

Ω

δxx

1 (y − x) •ϕi(y) dΩy ds =

_

A−A

σ(i)

xx(x) ds . (1.264)

The FE solution (Fig. 1.59 c) is not exact, because displacement fields ϕi in

Vh cannot model step functions.

Fig. 1.58. The influence function for a resultant bending moment is much easier

to approximate than for a single value.

1.21 Why resultant stresses are more accurate 87

Fig. 1.58. Slab: a)

yy; b) influintegral

of myy

The equilibrium condition

_

H = 0 is a simple condition because the influence

function is simple, ux = 1. The condition that the sum of the moments

is zero,

_

M = 0, is more difficult because it involves rotations. If a vertical

force P = 1 kN is applied at the end of the plate in Fig. 1.60, the maximum

bending stresses of the FE solution in the cross section A−A are ±10 kN/m2,

so the bending moment is

Mh =

_ +0.5

−0.5

z · (−_ _20 z_

σxx

) dz = −1.￣6 kNm (1.265)

which is less than the exact value of −2.5 kNm. The reason is that the FE

program cannot model the exact influence function (see Fig. 1.60 b) so it

operates with the shape in Fig. 1.60 c instead, which it obtains when it applies

the moments of the nodal unit displacement ϕi in cross section A − A as

equivalent nodal forces:

fi =

_ +h/2

−h/2

_

Ω

δxx

1 (y − x) •ϕi(y) dΩy · z dz

=

_ +h/2

−h/2

σ(i)

xx(z) · z dz . (1.266)

ence function for the

for m

influence function

X Y

Z

X Y

Z

88 1 What are finite elements?

3117978 kN

6157636 kN

3117978 kN

6157636 kN

6157636 kN 6157636 kN

6157 636 kN 6157636 kN

3117978 kN 31 17978 kN

5.00

4.00

Fig. 1.59. Influfunction

for

x

A-A: a) equivalent

act solution, c) FE

approximation

This pseudorotation lifts node 10 vertically by uy = −1.￣6 m, which is exactly

the bending moment Mh = 1· (−1.￣6) kNm of the FE solution in section A−A

(see Equ. (1.265)).

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