5.10 Columns

Back

Reliable estimation of the magnitude of the bending moments near columns

and similar point supports is one of the main problems in FE analysis.

Near a point support the bending moments can be split into smooth polynomial

(p) parts and singular parts (s),

mij(x) = mp

ij(x) +P ms

ij(yc, x) yc = center of the column (5.72)

where P is the support reaction. The singular parts (for simplicity it is assumed

that ν = 0 and polar coordinates x → r, ϕ are used)

50.0 kN

x

y

8.0

8.0

X

Y

Z

0.7

140.8

0.4 m

R = 32.6 kN

R

22.3

444 5 Slabs

Fig. 5.25. Support reactions (kN/m) a) produced by rigid supports, EA = ∞, and

b) by masonry walls

57.5

-42.7

-21.3

68.4

60.2

-8.2

78.1 13.9

18.7

11.1

21.7

-10.3

30.1

-28.1

49.5

-21.1

67.1

17.0

55.9

69.9

33.2

396.3

51.5

87.7

46.8

19.8

51.4

76.3

365.3

286.6

487.0

273.3

-115.3

-80.9

104.8

81.9

-34.5

21.2

108.1

26.2

-33.4

43.9

-8.3

-58.2

100.8

-64.2

146.4

127.4

119.2

44.3

901.9

58.5

97.5

162.5

82.8

28.7

93.8

-84.6

51.2

901.9

102.4

497.1

240.6

689.8

460.8

5.10 Columns 445

Fig. 5.26. Slab with intermittent support, slab thickness h = 0.2 m. a) System

and gravity load g; b) bending moments mxx kNm/m; c) principal moments;

d) bending moments in the slab along the interior walls

ms

xx(r, ϕ) = − 1

8 π

[(3 + 2 ln r) cos2 ϕ + (1 + 2 ln r) sin2 ϕ] , (5.73)

ms

yy(r, ϕ) = − 1

8 π

[(3 + 2 ln r) sin2 ϕ + (1 + 2 ln r) cos2 ϕ] , (5.74)

ms

xy(r, ϕ) = − 1

8 π

[(4 + 4 ln r) sinϕ cos ϕ] , (5.75)

are independent of the shape and size of the slab, and of the support conditions

[116]. Only the smooth parts differ.

These singular parts would also be dominant if the single force P were

replaced by the bearing pressure p = P/Ωc at the head of the column,

mij(x) = mp

ij(x) +

_

Ωc

ms

ij(y, x)pdΩy , (5.76)

-4.5

-4.1

3.7

2.8

-2.9

-1.5

x

y

8.0

8.0

-5.2

-5.3

-7.3

-11.1

1m

gravity load g = 6.50 kN/m2

x

y 8.00

8.00

2m

446 5 Slabs

Fig. 5.27. Modeling of a point support

because not too far from the column, this integral would be identical to the

influence of the single force P

_

Ωc

ms

ij(y, x)pdΩy

∼=

P ms

ij(yc, x) . (5.77)

The problem of the FE method is that polynomial shape functions are not

very good at approximating the singular functions ms

ij .

The best strategy would be to add the singular functions to the FE code.

Otherwise the mesh must be refined (see Fig. 5.27), or techniques such as

• spreading the point support onto multiple nodes (multiple nodes model)

• multiple nodes + elastic support

• multiple nodes and plate stiffness K = ∞ near the column

• a single rigid plate element (K = ∞) which can rotate freely placed on an

elastic pinned support

may be employed to ease the burden for an FE program to approximate

singular bending moments with polynomial shape functions; see Fig. 5.28.

Recommendations

To keep a reasonable ratio between effort and accuracy we make the following

recommendations:

• Columns should always be modeled with their natural stiffness.

• If no special coupling elements are used, as in Fig. 5.31 d, the element size

should decrease gradually towards the column center.

5.10 Columns 447

Fig. 5.28. Slab with opening: a) gravity load g = 6.5 kN/m2, single force 50 kN

+ edge load 10 kN/m, b) bending moments (kNm/m) caused by the single force,

c) shear forces (kN/m) caused by the single force in two sections, d) reinforcement

as-x at the bottom (qualitative)

1:3

Fig. 5.29. Widening of the elements above the

column head

x

y

6.0

7.0

4.1

4.2

-60.3

-64.8

x

y

6.0

7.0

4.1

4.2

15.5

15.3

x

y

6.0

7.0

e

g

g g

448 5 Slabs

Fig. 5.30. Square slab with column: a) system; b) influence function for the support

reaction; c) influence function for the bending moment mxx

• It is a good idea to subdivide the region near the column into a patch of

four elements, so that the midpoint of the patch is the column, and to let

the thickness of the elements increase from the edge of the patch to the

center; see Fig. 5.29.

• If possible the center of the elements should coincide with the vertices of

the column cross section.

• It is sufficient if the center node is supported. A multinode model does

not increase accuracy. In particular a multinode model can easily lead to

unilateral rotational constraints, or simulate a rigid joint.

• In no case should a rigid multi-node model be used.

Other strategies and refinement techniques are possible. The important point

is that something should be done to alleviate for an FE program the task to

approximate the influence functions for the bending moments at the columns

(see Fig. 5.30 c), because these are the functions an FE program must approximate

if it is to calculate the bending moments.

Things are different with regard to the support reaction in the column. The

influence function for the support reaction (see Fig. 5.30 b) is much smoother

and more regular. If the focus were only on the column reactions, no mesh

refinement near point supports would be necessary.

x

y

6.0

6.0

X

Y

Z

X

Y

Z

5.10 Columns 449

1.1

x

y

6.0

6.0

1.1

x

y

6.0

6.0

(- 51.0)

(- 38.8)

- 35.2 kNm/m

1 4.2 ( 14.8 ) kNm/m 240 ( 241 ) kN 120 ( 120 ) kN

1 6.5 ( 16.6 ) kNm/m

(- 25.5)

(- 20.8)

- 18.5 kNm/m

(- 17.8)

- 16.8 kNm/m

Fig. 5.31. Hinged square slab with interior column 0.25 m × 0.25 m, length l = 2.75

m. a) Uniform load; b) traffic load on one side; c) FE mesh; d) rigid plate element

Boundary elements

In a BE program the correct singularities (influence functions, Eq. (5.75)),

are built into the code. The column reaction P is so determined that the

deflection of the slab and the compression of the column at the center of the

column are the same, and later—when the bending moments are calculated—

the single force P is replaced by the load bearing pressure p = P/Ωc to avoid

the bending moments becoming singular; see Fig. 5.31.

450 5 Slabs

Fig. 5.32. Slab, h = 0.4 m, columns 0.4 × 0.4 × 2.45 and columns with drop panels

0.2 ×0.2 × 0.2. a) Bending moments mxx, b) bending moments myy, c) bending

moments mxx, d) bending moments myy

Example

As mentioned earlier one possibility for modeling the column–slab interaction

is to use a rigid slab element, K = ∞, which sits on top of the column on a

pinned support so that it can rotate freely.

A hinged square slab 6 m × 6 m, supported at its center by a single column

25 cm × 25 cm with a length of 2.75 m was analyzed with such an element.

The modulus of elasticity was 30 000 MN/m2, the slab thickness was h = 0.2

m, and Poisson’s ratio ν = 0.2. In the first load case the loading consisted of

5.11 Shear forces 451

a constant surface load p = 20 kN/m2, and in the second load case the same

load was applied only to the left portion of the slab. The agreement with a

BE solution is very good (see Fig. 5.31 and the following table).

Uniform load Rigid Elastic

FE BE FE BE

mxx column −37 −43.6 −35.2 −38.8 kNm

mxx span 14.2 14.4 14.8 14.9 kNm

Support reaction 253 257 240 241 kN

One-sided load Elastic

FE BE

mxx column, left −16.8 −17.8 kNm

mxx column, right −18.5 −20.8 kNm

mxx span 16.5 16.6 kNm

Support reaction 120 120 kN

Rigid means stiff supports, EA = ∞, and elastic that the stiffness of the

column was EA = 6.82 · 105 kN/m and the walls had a stiffness of 2.62 · 106

kN/m2, corresponding to a modulus of elasticity of 30 000 MN/m2, a wall

thickness of 24 cm, and a height of 2.75 m.

Drop panels and column capitals

Drop panels and column capitals ensure that the bending moments migrate to

the column capitals, as can be seen in Fig. 5.32. The discontinuity in the slab

thickness ensures that the bending moments in a cross section perpendicular

to the discontinuity jump (see Fig. 5.39).

Навигация