4.7 Supports

Back

Fixed supports are perfectly rigid by nature. Such supports can be point

supports or line supports. While point supports are ambiguous in nature, line

supports are legitimate constraints, see Fig. 4.20 and Fig. 4.21.

At a roller support, the displacements normal to the support are zero,

uT n = 0. If the roller support is inclined, kinematic constraints couple the

horizontal and vertical displacements. A slight misalignment of the supports

can lead to dramatically different results. Plates are very sensitive to geometrical

constraints or other such incompatibilities.

Care should also be taken not to constrain a structure unintentionally in

the horizontal direction, because it could easily suggest a load bearing capacity

that later fails to materialize because of nonexistent abutments.

Therefore a correct assessment of the support characteristics is very important.

Only if the supports of the continuous beam in Fig. 4.22 are perfectly

rigid will the support reactions agree with the ratios 0.36:1:1:0.36 that we expect

in a continuous beam. If instead the four rigid supports are replaced by

four 0.24m × 0.24m × 2.88 m columns with vertical stiffness

348 4 Plane problems

Fig. 4.20. Wall: a) loading, b) displacements, c) principal stresses, d) qualitativ

representation of the horizontal reinforcement as-x cm2/m

4.7 Supports 349

-1834.0

1086.4

1725.9

1094.6

555.2

-878.4

-1797.2

x

y

14.0

21.0

-2546.8

-113.4

-31.6

50.3

90.7

352.9

499.2

x

y

14.0

21.0

-4855.6

-2348.5 x

y

14.0

21.0

Fig. 4.21. Stresses and reinforcement: a) σxx b) τxy c) σyy d) part of the reinforcement

as-y cm2/m in the vertical direction

350 4 Plane problems

Fig. 4.22. Influence of the support stiffness: a) rigid support, b) stresses σxx,

xx kN/m2 c) soft support (brickwork columns), d) stresses σ

4.7 Supports 351

Fig. 4.23. A shear wall: a) on soft supports the wall tends to behave like a long

flat arch held taut by a tie rod; b) hard supports allow the plate to carry the load

directly to the intermediate supports

k = EA

h

=

30 000MN/m2 · 0.24 · 0.24m2

2.88m

= 6.0 · 105kN/m , (4.54)

the support reactions are much more evenly distributed, as indicated by the

ratios 0.72:1:1:0.72.

The softer the supports, the closer the structural behavior of the plate to

a long flat arch held taut by a tie rod; see Fig. 4.23.

Whenever possible, the support reactions should be compared with the

support reactions of an equivalent beam model (see Fig. 4.24) and the reinforcement

should be checked by working with approximate lever arms z [103].

In the following formulas d denotes the width of the wall, l the length of

the span, and zF and zS the lever arms in the span and at the intermediate

352 4 Plane problems

Fig. 4.24. Shear wall and equivalent beam

supports:

One span

zF = 0.3 · d · (3 − d/l) 0.5 < d/l < 1.0

zF = 0.6 ·l d/l ≥ 1.0

Two-span beams and last span of a continuous beam

zF = zS = 0.5 · d · (1.9 − d/l) 0.4 < d/l < 1.0

zF = zS = 0.45 ·l d/l ≥ 1.0

Interior spans of continuous beams

zF = zS = 0.5 · d · (1.8 − d/l) 0.3 < d/l < 1.0

zF = zS = 0.4 ·l d/l ≥ 1.0

Cantilever beam with length lk

zS = 0.65 · lk + 0.10 · d 1.0 < d/lk < 2.0

zF = 0.85 · lk d/l ≥ 2.0

If M is the bending moment of an equivalent beam, the tensile force T = M/z

and the reinforcement are

As = T

βs/γ

= T kN

28.6 kN/cm2 . (4.55)

4.7 Supports 353

Fig. 4.25. Plate analysis of a beam with bilinear elements (Q4): a) system and

load, where the point supports were modeled by keeping two nodes fixed, and the

single forces were input as nodal forces; b) principal stresses, for which the support

reactions agree with the beam theory

Here 28.6 kN/cm2 is the allowable steel stress.

A check of the plate in Fig. 4.22 with these formulas shows a good agreement

Span 1 Support Span 2

Moment (kN/m2) 64 −80 20

Lever arm z (m) 1.8 1.8 1.6

Tensile force T (kN) 35.6 44.4 12.5

As (cm2) 1.24 1.55 0.43

As FE (cm2) 1.35 1.60 0.60

Point supports

True point supports are not compatible with the theory of elasticity, because

the exact support reaction would be zero. But if a node is kept fixed, it

is not a point support (see Sect. 1.16, p. 55, and 1.24, p. 99). Instead the

resulting support reaction agrees with the beam solution (see Fig. 4.25). Only

the stresses signal that the solution is singular.

354 4 Plane problems

Fig. 4.26. Symmetric structures and symmetric loads

Symmetries

If the system and load are symmetric the FE analysis can be restricted to parts

of a structure. Loads that happen to lie on the symmetry axis, are halved as

is the stiffness of supports that lie on the axis (see Fig. 4.26).

Displaced point supports

The situation is the same as for point supports. According to the theory of

elasticity, no force is necessary to displace a single point of a plate while an

FE program produces the beam solution.

Tangential supports

A shear wall that is connected to floor plates experiences a deformation impediment

in the tangential direction (see Fig. 4.27). Without this constraint

(see Fig. 4.27 b and c; principal stresses and reinforcement), the structural

behavior of the wall is similar to an arch held taut by a tie rod, while if the

upper and lower floor plate constrain the wall, the distribution of the stresses

in the wall is much more homogeneous and the reinforcement required is only

about half as much as before.

4.7 Supports 355

Fig. 4.27. Shear wall: a) system and load, b) principal stresses, c) reinforcement—

only qualitatively—without tangential support, d) and e) with such a support at

the upper and lower edge

356 4 Plane problems

Fig. 4.28. The transfer of the shear force in the chords requires the compressive and

tensile forces to be inclined. The shear forces lead to asymmetric bending moments

in the chords

-

+

Fig. 4.29. Internal forces and bending moment

distribution in the chords

Beam-like elements

Door or window lintels carry shear forces (see Fig. 4.28). The normal forces

(compression and tension) in these elements are

C = T = M

z

z = vertical distance of the lintels, (4.56)

and it can be assumed—if the concrete is not cracked—that the shear force

V is evenly carried by both elements Vu = Vl = 0.5 V , so that the bending

moment in one element becomes (see Fig. 4.29)

4.8 Nodal stresses and element stresses 357

Melement = 0.5 V

l

2 l = length of the element. (4.57)

If the concrete in the lower element cracks, the stiffness will shrink, and it can

then be assumed that all the shear force is carried by the upper beam which

makes it necessary to provide hanger reinforcement to carry the shear force

to the upper element.

To investigate how Wilson’s element can represent such bending states,

the column in Fig. 4.30 was analyzed. The height of the column is 3 m and

its width is 50 cm. It is fixed at the base and it can slide horizontally at the

top where a horizontal force P = 20 kN is applied. The structural system

corresponds to the system in Fig. 4.30 c. As can be seen from the following

table

Elements

Width (m) × Height (m) M (kNm) u (mm)

0.500 × 0.600 25 0.74

0.250 × 0.250 27.6 0.76

0.125 × 0.150 28 0.76

exact 30 0.72

with elements of size 25 cm × 25 cm, two elements across the width of the

column, good results are achieved. The error in the bending moments at the

top is about 8%.

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