5.17 Foundation slabs

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Winkler model

In the Winkler model it is assumed that the soil acts like a system of isolated

springs that move independently and exert a force cw on the underside of the

slab. This leads to the differential equations

EIwIV + cw = p beam (5.95)

KΔΔw + cw = p slab . (5.96)

The strain energy products associated with these two equations are

model and gravity

470 5 Slabs

Fig. 5.52. Concentrated forces acting

on the surface of the half-space

a(w, ˆ w) =

_ l

0

M ˆM

EI

dx + c

_ l

0

w ˆ wdx (5.97)

a(w, ˆ w) =

_

Ω

m•ˆκ dΩ + c

_

Ω

w ˆ wdΩ. (5.98)

Hence to the stiffness matrix K of a beam or plate element must only be

added the mass matrix M,

mij =

_

Ω

ϕi ϕj dΩ , (5.99)

so that

(K + cM)u = f . (5.100)

A consequence of this simple spring model is that under gravity load G, the

slab deflection w = G/c is constant and the bending moments are zero; see

Fig. 5.51. Because the individual springs move independently, the soil outside

the loaded region simply retains its original level; see Fig. 5.51.

Inasmuch as this model does not properly replicate even the most basic

aspects of actual soil behavior, it is no surprise that the modulus of subgrade

reaction c is not a genuine physical quantity, but an artificial (albeit

convenient) tool to simplify the design of foundation slabs. Theoretically the

coefficient not only depends on the soil, but also on the extensions of the

foundation slab. And in principle it is also not a constant, because otherwise

it would not be possible to recover the rapidly increasing soil pressure p near

the edge of a rigid punch from the formula p(x) = c(x)w; see Fig. 1.123 on

p. 175.

Various strategies have been proposed to circumvent these defects. Mostly

these techniques modify the modulus of subgrade reaction c iteratively and

locally in such a way that the shape of the deformed slab and the soil are

approximately the same; see Fig. 5.53.

Half-space model

In a linear elastic isotropic half-space, the displacement field u(x) is the

solution of the system

5.17 Foundation slabs 471

Lij uj := −μui,jj − μ

1 − 2 ν

uj ,ji = pi (5.101)

where the pi are volume forces and μ and ν are elastic constants.

The displacement field due to a concentrated normal force P on the edge of

the elastic half-space was found by Boussinesq [49]. The vertical displacement

is (see Fig. 5.52)

uB := u3(x, y) =

1 + ν

2πE

_

(x3 − y3)2

r3 + 2

1 − ν

r

_

P (5.102)

and the stresses are

σij = − P

2π r

r,i r,j r,2

r

r,i = ∂r

∂yi

= yi − xi

r

. (5.103)

Note that the stresses σij are independent of the modulus of elasticity E. How

this solution can be extended to layered soils was explained in Sect. 2.2, p.

256.

Next let us consider the coupled problem of a foundation slab and the soil.

On the free surface the tractions must be zero, t = 0, while at the interface

between the slab and the soil the deflection must be the same, and the vertical

stresses must have opposite signs,

w = wS t3 − pS = 0 (5.104)

where pS is the soil pressure, which also appears on the right-hand side of the

plate equation with a negative sign because it opposes the load p coming from

the building:

KΔΔw = p − pS . (5.105)

If the soil pressure p(y) is expanded in terms of nodal shape functions ψi(y)

(hat functions)

p(y) =

_

i

ψi(y)p i , (5.106)

the soil deflection at a point x is

wS(x) =

_

i

_

Ω

uB(x, y) ψi(y) dΩy pi =

_

i

ηi(x) pi (5.107)

and the stress in the soil is

σzz(x) =

_

i

_

Ω

3

2π

(x3 − y3)3

r3 ψi(y) dΩy pi =

_

i

θi(x) pi . (5.108)

Hence the coupled problem leads to the system

472 5 Slabs

Fig. 5.53. Deformation patterns

_

K(nn) L(nm)

Iw(

mn)

−J(mm)

__

u(n)

p(m)

_

=

_

f(n)

0(m)

_

FE with soil pressure

wslab

− wsoil = 0 (5.109)

where

k ij = a(ϕi, ϕj) l ij =

_

Ω

ψi ϕj dΩ Jij = ηj(xi) . (5.110)

The matrix Iw results if the n × n unit matrix I is reduced to an m × m

matrix by deleting certain rows and columns. Because only the deflections at

the soil interface are equated, the rotational degrees of freedom in the vector

u are meaningless. So that if the underlined rows

1, 2, 3, 4, 5, 6, 7, 8, 9, . . . (5.111)

in the unit matrix are deleted, the result is the matrix Iw.

With the equivalent nodal forces −Lp of the soil pressure the extended

set of equations becomes Ku = f − Lp, or Ku+ Lp = f.

The Boussinesq solution is based on the linear theory of elasticity, so E

is Young’s modulus, while in soil mechanics a one-dimensional modulus Es is

used instead, which corresponds to consolidation or oedometer testing, and is

therefore also called the constrained modulus. The two moduli are related via

elasticity theory:

E =

(1 + ν)(1 − 2 ν)

(1 − ν) Es , (5.112)

5.17 Foundation slabs 473

Fig. 5.54. Foundation slab: a) model; b) cross section; c) bending moments mxx

in the half-space model; d) Winkler model

where ν is the (drained) Poisson’s ratio. For ν = 0 this becomes E = Es

and for ν = 0.2 Young’s modulus E is 90 percent of Es so there is not much

difference between the two.

Comparisons of the Winkler model and half-space model

Ultimately there is little agreement between results based on the Winkler

model and the half-space model, as in the case of the slab in Fig. 5.54 or the

slab in Fig. 5.55, which is stiffened by a ring of shear walls.

474 5 Slabs

Fig. 5.55. Modified Winkler model: a) plan view and load; b) bending moments

mxx in two sections, constant modulus of subgrade reaction: c) half-space model;

d) in a 1 m strip the modulus of subgrade reaction was raised by a factor of 4

In the half-space model the soil pressure increases towards the edge of the

slab, while in the Winkler model the deflection increases near the edge; see

Fig. 5.55 b. In a first analysis, the modulus of subgrade reaction was assumed

to be constant, c = 10 000 kN/m3 (Fig. 5.55 b), and in a second analysis the

modulus was increased by a factor of 4 in a 1 m strip along the edge. In this

second improved model the agreement of the bending moments mxx with the

half-space model (ES = 50 000 kN/m2) (Fig. 5.55 c and d) is much better.

x

y

6.0

6.0

x

y

6.0

6.0

x

y

6.0

6.0

gravity load g = 9.0 kN/m2

30.0 kN/m

30.0 kN/m

30.0 kN/m

30.0 kN/m

P1 100.0 kN

P2 100.0 kN

x

y

6.0

6.0

5.17 Foundation slabs 475

Fig. 5.56. 3-D model of a multi-story building on an elastic foundation: a) global

view of the deformed model (the deformation are of course not to scale); b) partial

view

476 5 Slabs

3-D models

A 3-D analysis of a multi-story building itself is a complex undertaking. Even

for an experienced structural engineer it is not easy to verify the results of

a 3-D FE analysis with independent models. Too many effects combine—the

modeling of the girders, the floor plates, shear walls, columns, etc. usually has

a much larger influence on the accuracy than questions of mathematical rigor:

’If the mechanical model is false high-precision arithmetic will not save the

analysis’.

Given the large deviations between the Winkler model and the half-space

model it should be no surprise then that the choice of the foundation model has

a great impact on practically all results in a 3-D model of a building, see Fig.

5.56. This shows clearly in numerical studies, [94]. Placing a sophisticated 3-D

model of a building on a simple Winkler foundation cannot be recommended.

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