1

THE GENERAL SOLUTION FORA SLAB SUBJECTED TO CENTRE AND EDGE LOADS

AND RESTING ON A KERR FOUNDATION

Frans Van Cauwelaert, Marc Stet1, André Jasienski

(Received October 02, 2000; Revised August 01, 2001)

In this paper the general analytical solution of a concrete slab subjected to interior and edge loads is presented. Inthe mathematical derivation of the closed form analytical solution we consider the pavement slab as a thin plateresting on a Kerr foundation. Kerr’s solutions are a general differential equation expressing the equilibrium of aslab on a subgrade characterised by means of three soil parameters. The effects of transverse shear deformationand the interlocking action of slab and subgrade are considered in calculating the load induced stresses inconcrete slabs. A set of coherent boundary conditions without unnecessary constraints and capable of solving allpossible cases is presented. The presented solution provides the means to determine any stress or deflection atany distance from an interior or edge load.

Keywords: Concrete pavement, thin-plate model, Kerr foundation model, closed form solutions, load-inducedstress

Introduction

The differential equation of the vertical deflection w(x,y) of a thin plate of stiffness Dsubjected to a distributed vertical load q(x,y) and a foundation interface pressure p(x,y) is(Timoshenko, 1951)

),(),(),(22 yxpyxqyxwD −=∇∇

The usual way to solve this equation is based on the assumption of a foundation reactionfunction of the deflection of the plate.

The simplest representation of a continuous elastic foundation has been provided by Winkler(1867) who assumed that the base consisted of closely spaced, independent linear springs.Such a foundation is equivalent to a dense liquid base. The relation between the pressure andthe deflection is then

),(),( yxkwyxp =

where k (N/mm³) is called the foundation or subgrade modulus.

Westergaard’s well-known equations for stresses in concrete slabs (1926, 1948) are based onWinkler’s assumption. It can be seen from Winkler’s equation that for this foundation modelthe displacements are constant for a uniform pressure and zero outside the loaded area. Formost materials this behaviour is not observed in practice. It is therefore generally admittedthat Westergaard’s solution overestimates the values of the deflection and the bending stressin the slab. However, because of their ability to compute edge stresses, the attractiveness ofWestergaard’s solutions have never diminished.

1 Please direct all correspondence to: Marc Stet, Mina Krusemanlaan 90, NL7421LL Deventer, The Netherlands,telephone +31 (0)570 657 562, facsimile +31(0)570 657 783, e-mail marcstet@cs.com

2

In an attempt to find a more accurate representation, Pasternak (1954) assumed shearinteraction between the spring elements and proposed a relation between pressure anddeflection characterised by a vertical modulus k and a shear modulus G (N/mm) in thehorizontal direction

),(),(),( 2 yxwGyxkwyxp ∇−= .

Van Cauwelaert (1999) developed the general solution for a rectangular slab on a Pasternakfoundation subjected to centre and edge loads. However his solution does not perfectly satisfythe edge conditions. Using the two boundary conditions available to take into account thelimiting effect of the edge, he expressed that moment (in the slab) and shear force (in the slaband in the subgrade) vanish at the edge. The correct solution in fact requires three boundaryconditions: vanishing of moment and shear force in the slab, continuity of shear in thesubgrade. But the even character of Pasternak’s solution (differential equation of the fourthorder) makes it impossible to express an odd number of boundary conditions. Note that thisdifficulty does not arise with Westergaard’s solution where the subgrade only reacts in thevertical direction and hence does not require any shear continuity equations.

In the early 60’s, Kerr (1964) presented a subgrade model characterised by 3 parameters, twovertical moduli c and k, and a shear modulus G. His relation between pressure and deflectionwrites

),(),(),(),(1 22 yxwGyxkwyxpcG

yxpck

∇−=∇−

+

This model results in a differential equation of the sixth order. The solution for a limited slabrequires three (one edge) or six (two edges) boundary conditions and hence theaforementioned difficulty could be overcome. However, Kerr (1965) and others (Kneifati,1985; Kerr and El-Sibaie, 1990) proposed only particular analytical solutions of the model.Nevertheless a powerful practical general solution is included in the well known ISLAB2000finite element program (…..reference).

We present in this paper, to our knowledge for the first time, the general analytical solutionfor a slab resting on a Kerr’s foundation. Two solutions are presented. The first solution isgiven in part 1 and satisfies the demands for a slab of infinite extent, e.g. slab interiorconditions. In part 2, this solution is extended to solve a slab of finite extents, e.g. with edgesor joints. Applications of the closed form solutions are presented in part 3. The appendixgives the full mathematical resolution of Kerr’s differential equation.

3

Part 1. Kerr’s foundation model.

1. The general solution of Kerr’s foundation model.

Consider an infinite slab of stiffness D subjected to a load q(x,y) (figure 1).

Load D Slab

c Spring layer

G Shear layer

k Spring layer

Figure 1

The foundation model consists of two spring layers, with constants c and k, and anintermediate shear layer with constant G.

The equilibrium equation of the slab is

(1) qkcG

kq

ck

wwkG

wkD

ck

wkc

GD 2246 11 ∇−

+=+∇−∇

++∇−

Let D/k = l4, G/k = l2g, k/c = f, G/c = l2gf and q be a constant.Then equation (1) simplifies into

(2) qfk

wwglwlfwgfl )1(1

)1( 224466 +=+∇−∇++∇−

Note that for c = ∞ (f = 0) the upper spring layer vanishes and the equilibrium equationtransforms into Pasternak’s equation

(3) qk

wwglwl12244 =+∇−∇

Further note that for G = 0 (g = 0), the shear layer vanishes and Pasternak’s equationtransforms into Westergaard’s equation

4

(4) qk

wwl144 =+∇

The solution of (2) is

(5) ∫ ∫∞ ∞

π=

0 02

)/sin()/cos()/sin()/cos(4dtds

tslsblsyltaltxq

q

double Fourier integral which expresses a uniformly distributed load q over a rectangle 2a.2b,and

(6) ∫ ∫∞ ∞

++++

+++

+π

=0 0 22222322

2 1)(

1)(

)1()(

)/sin()/cos()/sin()/cos(14dtds

fgst

fst

fgf

stts

lsblsyltaltxfg

fk

qw

2. Resolution of Kerr’s equilibrium equation.

For the interested reader, the detailed development of the resolution is given in the appendix.

In order to perform the integration of (6), for example with respect to t, we factorise thedenominator with respect to the same variable t.

=++

+++

+

+++

+++

fgfs

sfg

fst

fs

fgf

stfg

fst

111123

13

12

46224426

(7) ( )( )( )23

222

221

2

1tttttt −−−

=23

21

22

21

21

21

tttttt −+

−+

−=

γβα

The factorisation will enable us to utilise the following integral solutions

(8) ( ) [ ]∫∞

+κ−−κ− −κπ

=κ+0

/)(/)(222 4

)/sin()/cos( laxlax eedttt

ltaltx

when x > a and

(9) ( ) [ ]∫∞

+κ−−κ− −−κπ

=κ+0

/)(/)(222 2

4)/sin()/cos( lxalxa eedt

ttltaltx

when x < a.

5

3. Solutions of the deflection integral.

The solution of the deflection integral depends on the sign of the discriminatordis1 = b2/4 + a3/27where

(10)

+−=

213

31

fgf

fa

(11)

+

+−

+=

fgffgf

fgf

b1

2711

91

2271

3

When dis1 > 0, the solution of the deflection integral is

(12) [ ] dsIBA

BAII

slsblsy

BABAfgf

kq

wo

−+

+−++

+π

= ∫∞

321222 )(3)(

2)/sin()/cos(

6114

where

for x > a

(13)21

/)(/)(

1

11

4 zee

Ilaxzlaxz +−−− −π

=

(14) ( ) [ ]laxvlaxuevu

I lax /)(sin/)(cos2

/)(222 −β−−β

+π

= −α−

( ) [ ]laxvlaxuevu

lax /)(sin/)(cos2

/)(22

+β−+β+

π− +α−

(15) ( ) [ ]laxulaxvevu

I lax /)(sin/)(cos2

/)(223 −β+−β

+π

−= −α−

( ) [ ]laxulaxvevu

lax /)(sin/)(cos2

/)(22

+β++β+

π+ +α−

for x < a

(16)21

/)(/)(

1

1124 z

eeI

lxazlxaz +−−− −−π=

(17) ( )2222

2 vuu

I+

=π

( ) [ ]lxavlxauevu

lxa /)(sin/)(cos2

/)(22

−β−−β+

π− −α−

6

( ) [ ]laxvlaxuevu

lax /)(sin/)(cos2

/)(22

+β−+β+

π− +α−

(18) ( )2232

2 vuv

I+

−=π

( ) [ ]lxaulxavevu

lxa /)(sin/)(cos2

/)(22

−β+−β+

π+ −α−

( ) [ ]laxulaxvevu

lax /)(sin/)(cos2

/)(22

+β++β+

π+ +α−

with

(19) )(3

121 BA

fgf

sz +−+

+=

(20)22

2222 uvuuvu −+=

++= βα

(21) 2sfg3

f12

BAu +

++

+= 3

2BA

v−

=

(22) 332

332

27422742abb

Babb

A +−−=++−=

When dis1 < 0, the solution of the deflection integral is

(23) [ ]dsIIIs

lsblsyfg

fkq

wo

3121112

)/sin()/cos(14γ+β+α

+π

= ∫∞

where

for x > a

(24)21

/)(/)(

1

11

4 zee

Ilaxzlaxz +−−− −π

=

(25)22

/)(/)(

2

22

4 zee

Ilaxzlaxz +−−− −π

=

(26)23

/)(/)(

3

33

4 zee

Ilaxzlaxz +−−− −π

=

for x < a

7

(27)21

/)(/)(

1

1124 z

eeI

laxzlxaz +−−− −−π=

(28)22

/)(/)(

2

2224 z

eeI

laxzlxaz +−−− −−π=

(29)23

/)(/)(

3

3324 z

eeI

laxzlxaz +−−− −−π=

with

(30)delta

zzdelta

zzdelta

zz 22

21

1

21

23

1

23

22

1−

=γ−

=β−

=α

(31) ( ) ( ) ( )22

21

43

21

23

42

23

22

41 zzzzzzzzzdelta −+−+−=

(32)fg3

f1scosuz 2

121

+++−= θ

(33)fg3

f1s

32

cosuz 21

22

+++

+−=π

θ

(34)fg3

f1s

34

cosuz 21

23

+++

+−=π

θ

(35)3a

2u −=

(36) 3θ1= arccos(3b/au)

4. Relations for moments and shear forces.

Bending moments, moments of torsion and shear forces in the slab are given for any x,y bythe classical relations

(37)

+−=

+−= 2

2

2

2

,2

2

2

2

, dywd

dxwd

DMdy

wddx

wdDM slabyslabx µµ

(38)dxdy

wdDMM slabyxslabxy

2

,, )1( µ−=−=

(39)

+−=

+−= 2

2

2

2

,2

2

2

2

, dywd

dxwd

dyd

DTdy

wddx

wddxd

DT slabyslabx

Pasternak’s relations give the shear forces in the subgrade

(40)

+∇= wwcD

dxd

GT subgradex4

,

+∇= wwcD

dyd

GT subgradey4

,

8

Part 2. The finite slab on a Kerr’s foundation.

1.Equilibrium equations.

1.1. Loaded slab.

Consider a slab with two edges, parallel with the y direction, located at distances L1 and -L2

from the axis of the load and resting on an unlimited subgrade. This situation requires 8boundary conditions expressing bending moments, moments of torsion and shear forces atboth edges of the slab and shear forces in the subgrade (shear layer). However, it can beshown (Timoshenko, 1951) that at the edge of the slab the two boundary conditions formoment of torsion and shear force can be reduced to the following unique condition

(41)

−+−= 2

2

2

2

, )2(dy

wddx

wddxd

DT slabx µ

Hence the number of boundary conditions is reduced to six.

Those boundary conditions require 6 particular solutions of the homogeneous equationderived from equilibrium equation (2)

(42) 0)1( 224466 =+∇−∇++∇− wwglwlfwgfl

The 6 particular solutions can be grouped as follows:when the discriminator dis1 is positive

(43) { ++= −∞

∫ lxzp

lxzppart eBeA

slsblsy

Cw //

0

11)/sin()/cos(

( ) ( )}dseFeElxeDeClx lxp

lxp

lxp

lxp

//// /sin/cos α−αα−α +β++β+

where

(44)222 BABA

1fg6

f1k

q4C

+++

=π

and when the discriminator dis1 is negative

(45) { ++= −∞

∫ lxzp

lxzppart eBeA

slsblsy

Cw //

0

11)/sin()/cos(

}dseFeEeDeC lxzp

lxzp

lxzp

lxzp

//// 3322 −− ++++

where

(46)fg

fkq

C+

=14

2π

9

The variables z1, z2, z3, α and β have the same meaning as before.The boundary “constants” Ap … Fp, are determined by the boundary equations.

For more clearness we write the deflection integral of the infinite slab

(47) ∫∞

=0

inf )()/sin()/cos(

dsxws

lsblsyw

and the deflection integral, solution of the homogeneous equation,

(48) ∫∞

=0

)()/sin()/cos(

dsxws

lsblsyw finpart

1.2 Unloaded subgrade beside a loaded slab (free edge).

The equilibrium equation is deduced from Kerr’s equilibrium equation (1), where q = 0 (noload) and D = 0 (no slab)

(49) 02 =∇− wGkw

This equation has two solutions

(50) dseNeMs

)l/sbsin()l/sycos(Cw

0

sg1

l/x

p

sg1

l/x

psubgr,out

22

∫∞ ++−

+=

The deflections must vanish at the infinite. Hence we shall take the solution with the negativeexponent at the positive side of the x-axis and the solution with the positive exponent at thenegative side of the x-axis.

1.3. Unloaded slab besides a loaded slab (joint).

The equilibrium equation of an unloaded slab is evidently the homogeneous equation (42).The solutions are given by (43) or (45). Here also the deflections must vanish at the infinite.Hence the solution with 3 negative exponents will be applied at the positive side of the x-axisand the solution with 3 positive exponents will be applied at the negative side of the x-axis.We note the solution by wout,slab.

2. Boundary equations.

2.1. Boundary equations for a slab with free edges.

Moment and shear force in the slab must be zero at the edge.

For x = L1 and x = -L2

10

(51) ( )∫∞

=+

µ−−=

0inf2

2

2

2

, 0)/sin()/cos(

dswwls

dxd

slsblsy

DM finslabx

(52) ( )∫∞

=+

µ−−−=

0inf2

2

3

3

, 0)2()/sin()/cos(

dswwdxd

ls

dxd

slsblsy

DT finslabx

The shear force in the shear layer at the edges is either zero (subgrade interrupted at the slabedges), either the shear force at the slab side of the edges is equal to the shear force at theother side (continuity of the subgrade).

2.1.1. Shear force zero in the shear layer.

For x = L1 and x = -L2

(53) ( ) 0)/sin()/cos(

10

inf4

, ∫∞

=+

+∇= dswws

lsblsycD

dxd

GT finsubgradex

The 6 boundary constants Ap … Fp are determined by the system of equations (51), (52) and(53).

2.1.2. Shear force inside the edge equal to shear force outside the edge.

We use the solution of equilibrium equation (50) of an unloaded subgrade.

For x = L1 and x = -L2

(54) ( )∫∞

=+

+∇0

inf4 )/sin()/cos(

1 dswws

lsblsycD

dxd

G fin

∫∞

=0

,

)/sin()/cos(dsw

slsblsy

dxd

G subgrout

However, equation (54) introduces two supplementary constants, Mp and Np.

Hence we need two equations more to solve the system of 8 boundary constants. Followingdifferent writers (Eom, Parsons and Hjelmstad, 1998), we consider that the deflection outsidethe edge (deflection of the subgrade) is a fraction δ of the deflection inside the edge(deflection of the slab)

(55) ( )∫∞

=+0

fininf dswws

)l/sbsin()l/sycos(δ ∫

∞

0,

)/sin()/cos(dsw

slsblsy

subgrout

where 0 ≤ δ ≤ 1 .The 8 boundary constants Ap … Fp, Mp, Np are determined by the system of equations (51),(52), (54) and (55).

11

The ratio between the deflections is the consequence of some shear transfer through the shearlayer. On a Winkler foundation a relationship between the deflections would be impossible:the deflection outside the edge is by definition zero.

2.2. Boundary conditions for a slab with joints (unloaded slab besides loaded slab).

The difference between a joint and a free edge is the possibility of load transfer at the joint. Itis generally admitted that notwithstanding the presence of a load transfer medium, dowels forexample, the moment in the slab remains zero at the joint. Hence, boundary equation (51) isnot modified. Further, vertical equilibrium requires that the shear forces are equal in the slabsand in the shear layer at both sides of the joint. As in previous paragraph, we postulate that thedeflection of the unloaded slab is a fraction of the deflection of the loaded slab. As lastboundary equation, we express that the deflection of the shear layer at the unloaded side is afraction of the deflection at the loaded side, the ratio being the same as that between the slabdeflections : slab deflections and shear layer deflections are proportional to each other.

The boundary equations are, for x = L1 and x = L2

Moments are zero at the joints

(56) ( )∫∞

=+

µ−−=

0inf2

2

2

2

, 0)/sin()/cos(

dswwls

dxd

slsblsy

DM finslabx

(57) ( )∫∞

=

µ−−=

0,2

2

2

2

, 0)/sin()/cos(

dswls

dxd

slsblsy

DM slaboutslaboutx

Shear forces in the slabs are equal at the joints

(58) ( )∫∞

=+

µ−−−

0inf2

2

3

3

)2()/sin()/cos(

dswwdxd

ls

dxd

slsblsy

D fin

∫∞

µ−−−=

0

,2

2

3,

3

)2()/sin()/cos(

dsdx

dw

ls

dx

wd

slsblsy

D slaboutslabout

Shear forces in the shear layer are equal at the joints.

(59) ( )∫∞

=+

+∇

0fininf

4 dswws

)l/sbsin()l/sycos(1

cD

dxd

G

( )∫∞

+∇0

,4 )/sin()/cos(

1 dsws

lsblsycD

dxd

G slabout

The slab deflections outside the joints are a fraction of the slab deflections inside

12

(60) ( )∫∞

=+0

fininf dswws

)l/sbsin()l/sycos(δ ∫

∞

0,

)/sin()/cos(dsw

slsblsy

slabout

The contribution of the shear layer in the total deflection outside the joints is a fraction of thecontribution of the shear layer inside the joints

(61) ( )∫∞

=+

+∇0

fininf4 dsww

s)l/sbsin()l/sycos(

1cD

Gδ

( )∫∞

+∇0

slab,out4 dsw

s)l/sbsin()l/sycos(

1cD

G

where 0 ≤ δ ≤ 1 .

The 12 boundary constants are determined by the system of equations (56), (57), (58), (59),(60) and (61).

2.3. Boundary conditions at the joint with full load transfer.

When the load transfer medium assures full shear transfer (e.g. continuously reinforcedconcrete), the surface deflections are equal and the system of equations (56) to (61) can besimplified. Indeed when full shear transfer occurs, the shear forces in both slab and shearlayer must necessarily be equal to the shear forces in an infinite slab (Van Cauwelaert, Stet,1998). Hence, only three boundary equations are required at each joint: the unloaded adjacentneighbour slabs can be ignored.

Moments are zero at the edges of the loaded slab

(62) ( )∫∞

=+

µ−−

0inf2

2

2

2

0)/sin()/cos(

dswwls

dxd

slsblsy

D fin

Shear forces at the slab edges are equal with the shear forces in an infinite slab

(63) ( )∫∞

=+

µ−−−

0inf2

2

3

3

)2()/sin()/cos(

dswwdxd

ls

dxd

slsblsy

D fin

∫∞

µ−−−=

0inf2

2

3

3

)2()/sin()/cos(

dswdxd

ls

dxd

slsblsy

D

Equation (63) can be simplified

(64) ( ) 0)2()/sin()/cos(

02

2

3

3

∫∞

=

µ−−− dsw

dxd

ls

dxd

slsblsy

D fin

Shear forces in the shear layer are equal with the shear forces in the case of an infinite slab

13

(65) ( )∫∞

=+

+∇0

inf4 )/sin()/cos(

1 dswws

lsblsycD

dxd

G fin

∫∞

+∇=0

inf4 )/sin()/cos(

1 dsws

lsblsycD

dxd

G

Equation (65) can be simplified

(66) ( )∫∞

=

+∇0

4 0)/sin()/cos(

1 dsws

lsblsycD

dxd

G fin

The 6 boundary constants are determined by the system of equations (62), (64) and (66).

3. Slab limited by one edge.

If one is principally interested in results at or nearby the edge (or joint), the influence of theother edge, located at a distance great enough, can in general be neglected. Hence, in thiscase, only half of the boundary conditions, namely those at the concerned edge, have to betaken into account.

14

Part 3. Applications

1. Computer programs 1.

Several computer programs have been developed related to the subject of this paper. Aprincipal program handles the 4 cases presented in § 2.1.1., § 2.1.2., § 2.2 and § 2.3 of part 2.In addition, a simple program has been written to perform the numerical integration of thedeflection before any analytical transformation: a program for an infinite slab subjected to auniform circular load. This program allows verifying the validity of the principal program inthe case of a slab of infinite extent. Further a program has been developed handling the 4boundary cases for beams limited by an edge. The limits of the results for a slab tend to theresults for a beam when enlarging one of the sides of the rectangular load. Hence, thoseprograms can be used to verify the slab programs in the case of slabs of finite extent.

1.1. Elementary program for an infinite slab.

The load is considered uniformly distributed over a circular area. Hence the solution can beexpressed as one integral in polar coordinates with

(67)rw

r1

rw

w2

22

∂∂

+∂∂

=∇

- in the hypothesis of Kerr, by transformation of equation (6) in polar coordinates,

(68) ∫∞

+++++=

0246

10

1)1()/()/(

)1( dmgmmffgm

lmaJlmrJf

lkqa

w

- in the hypothesis of Pasternak, by setting f = 0 in equation (68) (f is defined in § 1 of Part 1),

(69) ∫∞

++=

024

10

1)/()/(dm

gmmlmaJlmrJ

lkqa

w

- in the hypothesis of Westergaard by setting g = 0 in equation (69) (g is defined in § 1 ofPart 1),

(70) ∫∞

+=

04

10

1)/()/(dm

mlmaJlmrJ

lkqa

w

Consider a load of 10000 N, a radius of 56,4 mm, a pressure of 1 N/mm² and a slab ofthickness 200 mm, a modulus of 30000 N/mm², a Poisson’s ratio of 0,2.

The deflections computed by the integrals (68), (69) and (70) are presented in table 1.

1 The programs can be obtained by the corresponding author

15

Table 1: Deflections underneath a circular load

k g c w-Westergaard w-Pasternak w-Kerr0.10.10.10.10.1

1.01.91.00.011.9

0.11.0

100010001000

0.027250.027250.027250.027250.02725

0.020960.017620.020960.027160.01762

0.024240.015820.020950.027170.01760

Two series of subgrade characteristics are considered: k = 0.1, g = 1, c = 0.1 with dis1 > 0 andk = 0.1, g = 1.9, c = 1 with dis1 < 0. Note that for c = 1000 (lim f = 0) the results in Kerr’shypothesis tend to the results in Pasternak’s hypothesis; for c = 1000 (lim f = 0) and g = 0.01(lim g = 0) the results tend to those in Westergaard’s hypothesis. This is in agreement with theconclusions of §1 of part 1: the equilibrium equations.

1.2. Verification of the principal program.

1.2.1. Verification of an infinite slab using equation (70).

The deflections computed by equation (12) or (23) for a square load are compared with thosecomputed by (70) for a circular load. Due to the principle of de St Venant, they must nearlybe equal for loads of same weight and pressure. Also the bending stresses are compared inthe axis of the loads. The results are presented in table 2.

Table 2: Comparison between circular and square loads.

k g c distancefrom axis

w(70) σ w(12,23) σ

0.1

0.1

1.0

1.9

0.1

1.0

050100050100

0.024240.024210.024110.015820.015790.01568

0.09482--

0.11114--

0.024260.024220.024130.015830.015800.01569

0.09485--

0.11108--

The results verify both programs.

1.2.2. Verification of the solution for an infinite slab using the program for an infinitebeam.

The load on the beam has a width of 100 mm. The load on the slab has a length of 100 mmand a width of 100000 mm (considered to be infinite). The pressure is 1 N/mm² in both cases.Poisson’s ratio of the slab is set to zero because the solution for the beam is independent ofPoisson’s ratio (state of plane stress). The deflections and the bending stresses are computedin the axis of the loads. The results are presented in table 3.

16

Table 3: Comparison between infinite beam and infinite slab

k g c w-beam w-slab σx-beam σx-slab0.10.1

11.9

0.11.0

0.701030.39402

0.701030.39402

1.771611.48704

1.771611.48704

The results verify both programs.

1.2.3. Verification of the solution for a finite slab using the program for a finite beam.

The same loads as in the previous paragraph are located at the edges of beam and slab.Deflections and stresses are computed at the edges. The results are presented in table 4.

Table 4: Comparison between finite beam and finite slab.

Boundaryconditions

δ k g c w-beam w-slab

§ 2.1.1

§ 2.1.2

§ 2.2

§ 2.3

--

0.00.00.50.51.01.00.00.00.50.51.01.0--

0.10.10.10.10.10.10.10.10.10.10.10.10.10.10.10.1

1.01.91.01.91.01.91.01.91.01.91.01.91.01.91.01.9

0.11.00.11.00.11.00.11.00.11.00.11.00.11.00.11.0

2.049141.039222.049141.039221.524820.722691.214150.553962.049141.039221.366090.692821.024570.519611.02457051961

2.049141.039222.049141.039221.524820.722691.214150.553962.049141.039221.366090.692821.024570519611.024570.51961

The results verify both programs for all cases of boundary conditions..

2. Analysis of the developed model.

2.1. Mechanical behaviour of the model.

Using the principal program for the slab (boundary conditions of § 2.2) we computedeflections and bending stresses in the axis of the load for k = 0.1, c = 1 and values of gvarying from 1.6 to 2.2. In doing so we use successively solution (12), solution (23) and againsolution (12) i.e. the complete area covered by the different analytical solutions. The resultsare presented in table 5.

17

Table 5: Results in a slab for varying values of g.

Infinite slab Jointed slab: edge load (δ = 0.5)g w σx = σy w σy (σx = 0)

1.6 (dis1 > 0)1.7 (dis1 > 0)1.8 (dis1 > 0)1.9 (dis1 < 0)2.0 (dis1 > 0)2.1 (dis1 > 0)2.2 (dis1 > 0)

0.016920.016540.016180.015830.015500.015190.01489

0.122630.118530.114710.110980.107800.104670.10173

0.030070.029250.028480.027750.027060.026410.02580

0.167330.161460.156010.150740.146180.141770.13761

The succession of the results through the different g-values does not show any hiatus ordiscontinuity, although the utilized analytical relations (12) and (23) differ considerably.

2.2. Geometrical behaviour of the model.

Using the different slab programs, we compute the deflections at the edge of a finite slab forloads moving towards the edge. The results are given in table 6 (slab with a free edge andzero shear force in the shear layer), table 7 (slab with a free edge and shear transfer in theshear layer), table 8 (slab with a joint and proportional deflections) and table 9 (slab with ajoint and full shear transfer in slab and in shearlayer). The computations are again performedusing both series of data. The length express’s the distances of the axis of the load to the edge.

Table 6: Deflections for loads moving towards a free edge(no shear transfer in shear layer: boundary conditions 51, 52, 53).

k g c 300 200 100 500.1 1 0.1 0.06164 0.06462 0.06669 0.067270.1 1.9 1 0.03595 0.03882 0.04098 0.04162

Table 7: Deflections for loads moving towards a free edge(shear transfer in shear layer: boundary conditions 51, 52, 54, 55).

δ k g c 300 200 100 500.0 0.1 1 0.1 0.06164 0.06462 0.06669 0.06727

0.1 1.9 1 0.03595 0.03882 0.04098 0.041620.5 0.1 1 0.1 0.04958 0.05202 0.05372 0.05419

0.1 1.9 1 0.02710 0.02933 0.03102 0.031521.0 0.1 1 0.1 0.04181 0.04390 0.04535 0.04576

0.1 1.9 1 0.02195 0.02380 0.02520 0.02563

18

Table 8: Deflections for loads moving towards a jointed edge(proportional deflections: boundary conditions 56, 57, 58, 59, 60, 61).

δ k g c 300 200 100 500.0 0.1 1 0.1 0.06164 0.06462 0.06669 0.06727

0.1 1.9 1 0.03595 0.03882 0.04098 0.041620.5 0.1 1 0.1 0.04109 0.04308 0.04446 0.04484

0.1 1.9 1 0.02397 0.02588 0.02732 0.027751.0 0.1 1 0.1 0.03082 0.03231 0.03334 0.03363

0.1 1.9 1 0.01798 0.01941 0.02049 0.02081

Table 9: Deflections for loads moving towards a jointed edge(full shear transfer in slab and shear layer: boundary conditions 62, 64, 66).

k g c 300 200 100 500.1 1 0.1 0.03082 0.03231 0.03334 0.033630.1 1.9 1 0.01798 0.01941 0.02049 0.02081

As one would normally have expected, the deflections increase regularly when the loadmoves towards the edge.

About the values of the deflections we observe the following:- the values of the deflections of table 6 are the same as the values of table 7 and table 8 forδ = 0: no shear transfer at all in the three cases.- the values of the deflections of table 9 are the same as the values of table 8 for δ = 1: fullshear transfer, in slab and in shear layer, in the two cases.- the values of the deflections of table 7 are higher than the values of table 8 for identicalvalues of δ > 1: shear transfer occurs only in the shear layer in the case of table 7, while itoccurs both in shear layer and in slab in the case of table 8.

This allows us to conclude tha t the different sets of boundary conditions are consistent witheach other.

3. Influence of the different soil parameters.

Using the principal program we briefly analyse the influence of the different soil parameterson the deflection in the axis of the load. We consider a square load of 40000 N, a side of 200mm, a pressure of 1 N/mm² and a slab of thickness 200 mm, a modulus of 30000 N/mm², aPoisson’s ratio of 0,2.In table 10, we analyse the influence of Kerr’s spring constant c for given values of k = 0.1N/mm³ and g = 0.5 in the case of an infinite slab.

19

Table 10. Deflections in an infinite slab for varying c-values

c 0 300 600 900 1200 15000.05 0.142 0.136 0.121 0.102 0.082 0.0640.1 0.115 0.109 0.095 0.077 0.059 0.0440.2 0.101 0.094 0.080 0.063 0.046 0.033

1000 0.093 0.088 0.064 0.046 0.031 0.020

We note that the deflections decrease with increasing values of c. Thus the deflections arehigher than those computed with Westergaard’s model (c = 1000). Further we note that thedeflection basin becomes flatter with decreasing values of c.

In table 11 we analyse for equal centre deflections the relative influence of pairs of k and cvalues (g = 0.5) in an infinite slab.

Table 11. Influence in an infinite slab of pairs of k and c values

k c 0 300 600 900 1200 15000.1 0.1 0.115 0.109 0.095 0.077 0.059 0.0440.2 0.07 0.115 0.109 0.095 0.077 0.059 0.0430.5 0.064 0.115 0.109 0.094 0.075 0.057 0.0401.0 0.066 0.115 0.108 0.091 0.072 0.053 0.037

We note that in the case on an infinite slab, the values of k and c compensate each other in amanner of speaking. This is what one would have expected.

In table 12 we analyse for equal centre deflections the relative influence of pairs of g and cvalues (k = 0.1) in an infinite slab.

Table 12. Influence in an infinite slab of pairs of g and c values

g c 0 300 600 900 1200 15000.1 0.33 0.115 0.105 0.085 0.063 0.044 0.0290.5 0.1 0.115 0.109 0.095 0.077 0.059 0.0441.0 0.057 0.115 0.111 0.100 0.086 0.070 0.055

We note that the influence is more outspoken than in table 11. As noted in table 10, thedeflection basin becomes flatter with decreasing values of c. However the influence of thedecrease of the c values is moderated by the increase of the g values.

In table 13 we analyse for the pairs of k and c of table 11, the deflections for a load located atthe joint of a finite slab with full shear transfer. The deflections are computed along the jointfor increasing distances from the axis of the load.

20

Table 13. Influence in a finite slab with full shear transfer of pairs of k and c values

k c 0 300 600 900 1200 15000.1 0.1 0.126 0.121 0.108 0.091 0.073 0.0560.2 0.07 0.130 0.125 0.112 0.094 0.076 0.0580.5 0.064 0.136 0.130 0.115 0.096 0.076 0.0581.0 0.066 0.139 0.132 0.116 0.095 0.074 0.055

In contrast of the results in table 11, the values of k and c do not compensate each other.Hence the values of the parameters can be discriminated by means of the deflectionsmeasured at the joint.

In table 14 we analyse again for the pairs of k and c of table 11, the deflections for a loadlocated at the joint of a finite slab with full shear transfer. However here the deflections arecomputed orthogonal to the joint for increasing distances from the axis of the load.

Table 14. Influence in a finite slab of pairs of k and c values

k c 0 -300 -600 -900 -1200 -15000.1 0.1 0.126 0.106 0.085 0.066 0.049 0.0350.2 0.07 0.130 0.109 0.087 0.068 0.050 0.0350.5 0.064 0.136 0.112 0.089 0.067 0.048 0.0331.0 0.066 0.139 0.113 0.088 0.065 0.045 0.030.

Here also the values of the parameters can be discriminated by means of the deflectionsmeasured orthogonal to the joint.

4. Determination of the parameters k, G and c.

4.1. Determination by direct soil tests.

The determination of k through a plate-bearing test is known since Westergaard’s time. Thetest consists of a simple application of Winkler’s assumption. Actually the standard requiresthat the test be performed with a plate of 700 mm diameter. The value of k is given by

(71)wq

k∆

=

where q is the applied pressure and ∆w the observed deflection.

The determination of k and G together can be performed by a test described by Pasternak(1954) himself. The test is very similar to Westergaard’s plate bearing test; however thevertical load is applied eccentrically.

Pasternak’s expresses the equilibrium of the vertical forces (load and subgrade reaction) andof the moments. This results in following relations

21

(72)

λλλ

+π+

=)/()/(

212 0

12

aKaK

aa

wwkQ LR

(73)

λλλ

+π−

=)/()/(

41

2 1

23

aKaK

aa

wwkQe LR

whereQ is the applied loade is the eccentricity of the loadk is Westergaard’s subgrade moduluswR is the deflection on the loaded side of the platewL is the deflection on the unloaded side of the platea is the radius of the plateK0, K1, K2 are the modified Bessel functions of the second kind of order zero, one andtwoλ2 = G/kG is Pasternak’s shear modulus.

Divide (72) by (73) in order to obtain an equation in λ/a, then determine k by (72).

We do not know of any test to determine the value of c in situ.

Yao and Huan (1984) give an empirical relation between k and G.

(74) )mm/N(k350000)mm/N(G 3=

4.2. Determination by deflection measurements on beams or plates.

There exist many back-calculation programs for the determination of E (Young’s modulus ofthe slab) and k based on deflection basins measured on beams or slabs on Winkler’sfoundations. Two were very recently published (Khazanovitch , 2000, Fwa, 2000).A recently developed state-of-the-art computer program presents very simple and usefulclosed form algorithms for the computation of stresses and deflections (Stet, 2001). ThePAVERS program utilises an automatic back-calculation program for the determination ofPasternak’s shear modulus. A simple iterative procedure for the determination of k and G (Eis assumed to be known; derived from resonance measurements) based on deflection basinsmeasured on beams. The program utilises closed form formulas of the deflections under anisolated load. For a given value of G, one computes the value of k corresponding to theobserved deflection under the load. Then, with both values, one computes the deflections atthe other distances of the deflection basin and the differences with the measured deflections.If the differences are significant, the process is repeated with a new value of G. An example isgiven in table 15.

22

Table 15. Example of iterative back calculation of k and G.

DistancesG k0 50 100 150 200 250

Σ∆w

Measured deflections 0.21166 0.21110 0.20951 0.20701 0.20374 0.199810 0.134 0.21166 0.21100 0.20912 0.20616 0.20225 0.19753 0.00511

19522 0.119 0.21166 0.21104 0.20928 0.20652 0.20288 0.19850 0.0013145308 0.103 0.21166 0.21109 0.20946 0.20692 0.20360 0.19960 0.0005053212 0.098 0.21166 0.21110 0.20951 0.20704 0.20380 0.19990 0.0001850085 0.100 0.21166 0.21109 0.20949 0.20699 0.20372 0.19979 0.00008

We do not know any back calculation program for the determination of c. The results oftable 11 show us that any pair of values of k and c can be utilised to characterise thedeflection basin of an infinite slab. The value of k corresponding with a value of c equalinfinite is then also a representative value of the subgrade modulus. Hence, as we al readycould have suspected, in the case of an infinite slab we can solve the problem by Pasternak’smodel.

We suggest the following procedure: determine in a first step the values of k and G from adeflection basin observed in the middle of a slab (assumption of an infinite slab on aPasternak’s subgrade). With the obtained value of G, compute, now using Kerr’s model, aseries of pairs of k and c values satisfying the same deflection basin (table 11). Choose out ofthe series, the pair of values that gives the best fit with the deflection basin measured parallelwith the joint (table 13) or orthogonal to the joint (table 14).

Findings

This paper presents the analytical solutions of Kerr’s general differential equation expressingthe equilibrium of a slab on a subgrade characterised by means of three soil parameters. Thisnovel solution, obtained through sound theoretical analysis, is a scientific step forward in theconcerned field and can give a more accurate understanding of the problem taken underconsideration. Without minimizing the tremendous possibilities of Finite Element solutions,analytical solutions are truly the only mean for proper verification. Furthermore, computerprograms based on closed form analytical solutions are often faster and more users friendlythan those based on FEM.

The results of the numerical applications that we have performed at this stage of our research,seem to indicate that Kerr’s modulus is perhaps not an intrinsic characteristic of the subgradebut rather a parameter allowing expressing the slab-soil interaction at an edge or at a joint.The first spring layer of Kerr’s model should rather be linked with the slab than with thesubgrade. Kerr’s model could then be renamed “Slab with variable bottom characteristics on aPasternak subgrade”. Indeed in the case of an infinite slab (interior situation) the value ofKerr’s modulus c can without harm be taken as infinite, the thickness hc of the spring layerthen vanishes and the deflection at the bottom of the spring layer becomes equal with thedeflection on top of the spring layer and on top of the slab: the layer is absorbed in the slabespecially if we dare writing that slabc

0hcEhlim.clim

c

=→∞→

.

The problem is solved assuming a Pasternak subgrade. At a free edge, the solution for theunloaded subgrade besides the slab again is based on Pasternak’s model. It is only at the slab

23

side of a free edge or at a joint that we need Kerr’s solution. We think therefore that Kerr’smodulus is a parameter governing the shear force distribution between slab and subgrade at anedge or at a joint. Hence the difficulty, if not the impossibility, of the direct determination ofthe value of the modulus by a test on the subgrade and the necessity, probably, of doing it bydeflection measurements performed on a real edge or joint.

24

References

Eom, I.S., Parsons, I.D., andHjelmstad, K.D. (1998):” A finite element study of Load Transferbetween Doweled Pavement Slabs. CROW Fourth International Workshop on the Design andEvaluation of Concrete Pavements. Buçaco, Portugal.

Fwa, T.F., Tan, K.H., Lee, S. (2000):”Closed-Form and Semi-Closed-Form Algorithms forBack calculation of Concrete Pavement Parameters.” Non Destructive Testing of Pavementsand Back Calculation of Moduli. Third volume Proceedings ASTM Conference. Tayabji andLukanen Editors. Seatle, 2000.

Kerr, D. (1964): “Elastic and viscoelastic foundation models.” Journal of Applied Mechanics,Vol. 31, Nr. 3.

Kerr, D. (1965): “A study of a new foundation model.” Acta Mechanica, Vol 1 / 2.

Kerr, D. (1976): “On the derivation of well posed boundary value problems in structuralmechanics.” Int. Journal Solids and Structures, Vol 12.

Kerr, D. (1984): “On the formal development of elastic foundation models”. Ingenieur Archiv54 Springer-Verlag 1984.

Kerr, D and El-Sibaie (1990): “Green’s functions for continuously supported plates.”Zeitschrift für Angewandte Mathematik und Physik, Vol. 40.

Khazanovitch L. ( ): ISLAB 2000.

Khazanovitch L , McPeak, T.J., Tayabi, S.D., (2000): “LTPP Rigid Pavement FWDDeflection Analysis and Back calculation procedure.” Non Destructive Testing of Pavementsand Back Calculation of Moduli. Third volume Proceedings ASTM Conference. Tayabji andLukanen Editors. Seatle, 2000.

Kneifati, M.C. (1985): “Analysis of plates on a Kerr foundation model.” Journal ofEngineering Mechanics, ASCE, Vol. 111, No. 11.

Stet, M., Thewessen, B.,, Van Cauwelaert, F., and Verbeek, J.P. (2001): “Standard andrecommended practices for FWD based evaluation and reporting strength (submitted toNATO for enquiry)”. PAVERS computer program; Proceedings first European FWD user’sgroup meeting, Delft University of Technology, 2001.

Timoshenko, S. (1951): “Théorie des Plaques et des Coques.” Librairie Polytechnique Ch.Béranger. Paris, Liège.

Van Cauwelaert, F. (1993): “A Rigorous analytical solution of a concrete slab submitted tointerior and edge loads.” Proc.5th Int. Conf. on Concrete Pavement Design. PurdueUniversity. Indiana 1993.

Van Cauwelaert, F. (1999): “The limited rectangular slab on a Pasternak foundation”. CROW.publicatie 136. Ede,1999. The Netherlands.

25

Van Cauwelaert, F., Stet, M. (1998) ……………… CROW Fourth International Workshopon the Design and Evaluation of Concrete Pavements. Buçaco, Portugal.

Watson, G.N. (1966): “Theory of Bessel Functions”. Cambridge University Press, 1966.

Westergaard, H.M. (1926): “Stresses in Concrete Pavements computed by theoreticalAnalysis.” Proceedings Highway Research Board, Part 1, 1926.

Westergaard, H.M. (1948): “New Formulas for Stresses in Concrete Pavements of Airfields.”Proceedings ASCE, Vol 113, 1948.

Winkler, E. (1867): ”Die Lehre von der Elasticität und Festigkeit.” Prag, Dominicus, 1867.

Yao, Z.K., Huan, L.Y. (1984): ”Experimental analysis on warping stresses in concretepavements.” Journal of East China Highway, n° 3, 1984.

26

APPENDIX

Full mathematical resolution of Kerr’s differential equation

1. Kerr’s model for an infinite slab.

Consider an infinite slab of stiffness D subjected to a load q(x,y) (figure 1).

Load D q Slab p1 p1

cSpring layer

p1 p1

G Shear layer p2

k p2 Spring layer

Figure 1

The foundation model consists of two spring layers, with constants c and k , and an intermediate shear layer withconstant G.

Split the slab deflection into two parts

(1) ),(),(),( 21 yxwyxwyxw +=

where w1 is the contraction of the upper spring layer and w2 the deflection of the rest of the foundation. Thecontact pressures under the slab and the shear layer are given by p1 and p2. Hence, by Winkler’s assumption,

(2) )(),( 211 wwccwyxp −==(3) 22 ),( kwyxp =

The equilibrium equation of the slab is

(4) 14 pqwD −=∇

where D = Eh3/12/(1 - µ2) and ∇4w = d4w/dx4 + 2d4w/dx2dy2 + d4w/dy4.

Vertical equilibrium in the shear layer can be deduced from figure 2.

(5) 021 =++− dydxdy

dTdxdy

dxdT

dxdypdxdyp yx

(6)dy

dT

dxdT

pp yx −−=− 21

27

(Ty+dTy)dx p1 dx Txdy

dyTydx

p2 (Tx + dTx)dy

Figure 2

Tx and Ty are the shear forces per unit length in the shear layer acting on planes orthogonal to the x and ydirections.Pasternak (1954) assumes that the shear forces in the shear layer are proportional with the variation of thedeflection (figure 3)

(7)dy

dwGT

dxdw

GT yx22 ==

dw2 Tx

Tx + dTx

dx

Figure 3

hence

(8)

+=+

22

2

22

2

dywd

dxwd

Gdy

dT

dxdT yx

and

(9) 22

21 wGpp ∇−=−

Both equilibrium equations result in the system

(10) qcwwD =+∇ 14

(11) 22

21 wGkwcw ∇−=−

28

Eliminating w1 and w2 yields the equilibrium equation of the slab on a Kerr’s foundation

(12) qkcG

kq

ck

wwkG

wkD

ck

wkc

GD 2246 11 ∇−

+=+∇−∇

++∇−

Let D/k = l4, G/k = l2g, k/c = f, G/c = l2gf and q be a constant. Equation (12) simplifies into

(13) qfk

wwglwlfwgfl )1(1)1( 224466 +=+∇−∇++∇−

Note that for c = ∞ (f = 0), which requires w1 = 0, the upper spring layer vanishes and the equilibrium equationtransforms into Pasternak’s equation

(14) qk

wwglwl 12244 =+∇−∇

Further note that for G = 0 (g = 0), the shear layer vanishes and Pasternak’s equation transforms intoWestergaard’s equation

(15) qk

wwl 144 =+∇

Kerr’s relation between pressure and deflection (Introduction) can be obtained from equation (4)

(16) wDqpp 41

∇−==From equation (12)

(17) [ ] wwkG

wkc

GDwDq

k1

ck

1 264 +∇−∇−=∇−

+

Hence

(18)

∇−∇−+

= wc

GDwGkw

ck1

1p 62

Equation (18) enables us to compute the vertical stress on the subgrade.

Differentiating (4) yields

(19) wDp 62 ∇−=∇

Replacing ∇6w in (17) results in Kerr’s relation (Kerr, 1984)

(20) wGkwpcG

ck

1p 22 ∇−=∇−

+

2. Resolution of Kerr’s equilibrium equation.

Let

(21) ∫ ∫∞ ∞

π=

0 02

)/sin()/cos()/sin()/cos(4 dtdsts

lsblsyltaltxqq

double Fourier integral which expresses a uniformly distributed load q over a rectangle 2a.2b, and

29

(22) ∫ ∫∞ ∞

π=

0 02

)/sin()/cos()/sin()/cos(),(4 dtdsts

lsblsyltaltxstAqw

Equation (13) is satisfied for

(23) [ ]1)())(1()(1

),(22222322 +++++++

+=

stgstfstfgkf

stA

Hence the solution of the equilibrium equation can be written as

(24) ∫ ∫∞ ∞

++++

+++

+π

=0 0 22222322

2 1)(

1)(

)1()(

)/sin()/cos()/sin()/cos(14 dtds

fgst

fst

fgf

stts

lsblsyltaltxfg

fk

qw

3. Transformation of the sixth order denominator of the integrand.

In order to perform integration (24), for example with respect to t, we reduce by factorisation the order of thedenominator, DEN, of the integrand. The denominator can be dealt with as a cubic equation in t2. The roots canbe real or complex.We first transform the denominator

Let t2 = y2 - s2 - (1+f)/(3fg). Then

(25) bayyDEN ++= 26

where

(26)

+−=2

1331

fgf

fa

(27)

++−

+=fgffg

ffg

fb 12711912271

3

The nature (real or complex) of the roots of (25) depends on the sign of the discriminatordis1 = b2/4 + a3/27.

4. Nature of the roots.

A little bit algebra yields

(28) ( ) ( )

+−+−+=+= 4223

44

32

820141

127

1274

1 fggfffgf

abdis

The sign of (28) depends of the discriminator, dis2, and the roots of the fourth order relation in g between thesquare brackets.

(29) ( ) ( ) [ ]3

4

3224

8121148201

212 fffffdis −=+−−+=

Analysis of the discriminator dis2.

30

If f > 1/8 then (1 – 8f) < 0, dis2 < 0 and dis1 > 0 whatever the value of g.

If f = 1/8 then (1 – 8f) = 0, dis2 = 0 and

(30)

232

43

9

23

32

1

−= g

gdis

Hence if f = 1/8 then dis1 > 0 , unless g = (3/2)3/2. If g = (3/2)3/2 then dis1 = 0.

If f < 1/8 then (1 – 8f) > 0 and the sign of dis1 depends on the value of g2 with respect to the roots g12 and g2

2 of(28).

(31)( ) ( )

ffff

g8

818201 3221

−−−+=

(32)( ) ( )

ffff

g8

818201 3222

−+−+=

One easily shows that 0 < g12 < g2

2. Hence dis1 > 0 when g < g1 or g > g2 and dis1 < 0 wheng1 < g < g2.

The nature of the roots, function of the values of f and g, is presented in table 1.

Table 1: Nature of the roots of the denominator of the integrand.

f g dis1 Nature of the roots

f > 1/8 - dis1 > 0 One real, two complex

f = 1/8 g <> (3/2)3/2 dis1 > 0 One real, two complex

f = 1/8 g = (3/2)3/2 dis1 = 0 Three real, three equal

f < 1/8 g < g1 dis1 > 0 One real, two complex

f < 1/8 g = g1 dis1 = 0 Three real, two equal

f < 1/8 g1 < g < g2 dis1 < 0 Three real

f < 1/8 g = g2 dis1 = 0 Three real, two equal

f < 1/8 g > g2 dis1 > 0 One real, two complex

5. Factorisation of the denominator of the deflection integral .

We factorise the denominator with respect to the variable t.

(33) =++

+++

+

+++

+++

fgfss

fgf

stf

sfg

fst

fgf

st 111123

13

12

46224426

( )( )( )23

222

221

2

1tttttt −−−

=23

21

22

21

21

21

tttttt −

γ+

−

β+

−

α=

31

The system is satisfied for

(34)delta

ttdelta

ttdelta

tt 22

21

1

21

23

1

23

22

1

−=γ

−=β

−=α

(35) ( ) ( ) ( )22

21

43

21

23

42

23

22

41 tttttttttdelta −+−+−=

6. Resolution of the deflection integral when the discriminator is positive .

6.1. Values of the roots.

When the discriminator is positive, the solution of the cubic equation is obtained by the classical algebraicmethod. The roots are

(36)fg

fsBAt

31

)( 221

+−−+=

(37)fg

fs

BABAt

31

322

222

+−−−

−+

+−=

(38)fg

fs

BABAt

31

322

223

+−−−

−−

+−=

where

332

332

27422742abb

Babb

A +−−=++−=

We note that t12 is real and that t2

2 and t32 are complex.

Replacing the ti’ s by their values, one obtains

(39) [ ]221 31

BABA ++=α

(40) [ ]2213)(6

3)()(3

BABABAi

BAiBA

++−

−++−=β

(41) [ ]2213)(6

3)()(3

BABABAi

BAiBA

++−

−−+=γ

6.2.Transformation of the deflection integral.

The deflection integral can now be written as follows

(42) [ ]∫ ∫∞ ∞

++π=

0 0222

)/sin()/cos()/sin()/cos(1614

tslsblsyltaltx

BABAkqw

dtdsttttBA

BAittttttfg

f

−−

−−

++

−−

−−

−+

23

222

223

222

221

2

11

3)(

)(31121

The integral can be solved by (Watson, 1966, par. 13.53)

(43) ( ) [ ]∫∞

+κ−−κ− −κπ=

κ+0

/)(/)(222 4

)/sin()/cos( laxlax eedttt

ltaltx

32

when x > a and by

(44) ( ) [ ]∫∞

+κ−−κ− −−κπ=

κ+0

/)(/)(222

24

)/sin()/cos( lxalxa eedttt

ltaltx

when x < a.

However equations (43) and (44) require that the real part of κ2 is positive.Hence we first have to show that the real parts of t1

2, t22 and t3

2 are negative.We were not able to prove this analytically and thus have to show it numerically for the large range of f and gvalues for which dis1 > 0 . We take s = 0, so that we consider the lowest values of dis1.

The values of t12

are given in table 2 and the values of the real part of t22 and t3

2 are given in table 3.

Table 2: Values of t12

g\f 0.1 0.2 0.5 1 2 5 10 20 50 100

0.1 -110 -60 -30 -20 -15 -12 -11 -10 -10 -10

0.5 -21.5 -11.6 -5.8 -3.9 -2.9 -2.4 -2.2 -2.1 -2.0 -2.0

1 -10.1 -5.2 -2.5 -1.8 -1.4 -1.17 -1.09 -1.05 -1.02 -1.01

5 -0.21 -0.21 -0.21 -0.21 -0.21 -0.21 -0.20 -0.20 -0.20 -0.20

10 -0.10 -0.10 -0.10 -0.10 -0.10 -0.10 -0.10 -0.10 -0.10 -0.10

50 -0.02 -0.02 -0.02 -0.02 -0.02 -0.02 -0.02 -0.02 -0.02 -0.02

100 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01 -0.01

Table 3: Values of ℜ(t22) and ℜ(t3

2)

g\f 0.1 0.2 0.5 1 2 5 10 20 50 100

0.1 -0.041 -0.035 -0.022 -0.012 -0.006 -0.1 0 0 0 0

0.5 -0.21 -0.18 -0.11 -0.06 -0.02 -0.007 -0.002 0 0 0

1 -0.45 -0.39 -0.24 -0.12 -0.05 -0.012 -0.004 -0.001 0 0

5 -1.00 -0.50 -0.20 -0.10 -0.05 -0.016 -0.007 -0.003 0 0

10 -0.50 -0.25 -0.10 -0.05 -0.025 -0.010 -0.005 -0.002 0 0

50 -0.10 -0.05 -0.02 -0.01 -0.005 -0.002 -0.001 0 0 0

100 -0.05 -0.025 -0.01 -0.005 -0.003 -0.001 0 0 0 0

We may conclude that the real parts of the values of the roots are negative. Thus we can transform the deflectionintegral in

(45) [ ]∫ ∫∞ ∞

+++

π=

0 0222

)/sin()/cos()/sin()/cos(16

14ts

lsblsyltaltxBABAfg

fk

qw

dtdsztztBA

BAiztztzt

+−

+−

++

+−

+−

+ 23

222

223

222

221

2

11

3)(

)(3112

where

(46)fg

fsBAz

31

)( 221

++++−=

(47) 323

12

222

BAi

fgf

sBA

z−

−+

+++

=

33

(48) 323

12

223

BAi

fgf

sBA

z−

++

+++

=

6.3. Resolution of the deflection integral.

The integral with respect to t can be split into three parts

(49) ∫∞

+=

021

211)/sin()/cos( dt

zttltaltxI

(50) dtztztt

ltaltxI ∫

∞

++

+=

023

222

2211)/sin()/cos(

(51) dtztztt

ltaltxiI ∫

∞

+−

+=

023

222

2311)/sin()/cos(

6.4. Resolution of integral I1

The solution is obtained by application of relations (43) and (44).For x > a

(52)21

/)(/)(

1

11

4 zeeI

laxzlaxz +−−− −π=

For x < a

(53)21

/)(/)(

1

1124 z

eeIlxazlxaz +−−− −−π=

where

(54) )(3

121 BA

fgfsz +−++=

6.5. Resolution of Integral I2

First we evaluate the values of the complex numbers z2 and z3.We write

(55) 323

12

222

BAi

fgf

sBA

z−

−+

+++

=

(56) 323

12

223

BAi

fgf

sBA

z−

++

+++

=

One deducts the values of z2 and z3

(57) β−α= iz2

(58) β+α= iz3

where

(59)22

2222 uvuuvu −+=β++=α

Hence, for x > a

34

(60)

−+

−π=

+−−−+−−−

23

/)(/)(

22

/)(/)(

2

3322

4 zee

zee

Ilaxzlaxzlaxzlaxz

(61)

−−+

π=

+−+−−−−−

23

/)(

22

/)(

23

/)(

22

/)(

2

3232

4 ze

ze

ze

ze

Ilaxzlaxzlaxzlaxz

(62)

+

+−

π=−β−β

−α−

ivue

ivueeI

laxilaxiLax

/)(/)(/)(

2 4

+

+−

π−+β+β

+α−

ivue

ivuee

laxilaxiLax

/)(/)(/)(

4

Applying eix = cos(x)+ i sin(x)

(63) ( ) [ ]laxvlaxuevu

I lax /)(sin/)(cos2

/)(222 −β−−β

+π

= −α−

( ) [ ]laxvlaxuevu

lax /)(sin/)(cos2

/)(22

+β−+β+

π− +α−

and for x < a

(64) ( )2222

2 vuu

I+

π=

( ) [ ]lxavlxauevu

lxa /)(sin/)(cos2

/)(22

−β−−β+

π− −α−

( ) [ ]laxvlaxuevu

lax /)(sin/)(cos2

/)(22

+β−+β+

π− +α−

6.6. Resolution of the Integral I3

In the same way as for I2 one obtains for x > a

(65) ( ) [ ]laxulaxvevu

I lax /)(sin/)(cos2

/)(223 −β+−β

+π

−= −α−

( ) [ ]laxulaxvevu

lax /)(sin/)(cos2

/)(22

+β++β+π

+ +α−

and for x < a

(66) ( )2232

2 vuv

I+

π−=

( ) [ ]lxaulxavevu

lxa /)(sin/)(cos2

/)(22

−β+−β+π

+ −α−

( ) [ ]laxulaxvevu

lax /)(sin/)(cos2

/)(22

+β++β+π

+ +α−

6.7. General solution of the deflection integral.Using the solutions for I1, I2 and I3, we obtain the complete solution of the deflection integral.

35

(67) [ ] dsIBA

BAII

slsblsy

BABAfgf

kq

wo

−+

+−++

+π

= ∫∞

321222 )(3)(

2)/sin()/cos(

6114

If we had integrated the deflection integral with respect to s, the solution would have written

(68) [ ] dtIBA

BAII

tltaltx

BABAfgf

kq

wo

−+

+−++

+π

= ∫∞

321222 )(3)(

2)/sin()/cos(

6114

where a, x and s in I1, I2 and I3 are respectively replaced by b, y and t.

7. Resolution of the deflection integral when the discriminator is negative.

7.1. Values of the roots.

When the discriminator is negative, we use the so-called trigonometric method.

Consider equation (25). Let y2 = ucosθ. The denominator transforms in

(69) bauuDEN +θ+θ= coscos 33

Let

(70)au

bau 3)3cos(3

2 =θ−=

The solutions of the trigonometric equation are 3θ1= arccos(3b/au), 3θ2 = 3θ1 + 2π,3θ3 = 3θ1 + 4π,The roots of the original equation are then

(71)fg

fsut

31

cos 21

21

+−−θ=

(72)fg

fsut

31

32

cos 21

22

+−−

π

+θ=

(73)fg

fsut

31

34

cos 21

23

+−−

π

+θ=

7.2. Resolution of the deflection integral.

Since the roots are all real, integrals (43) and (44) can be applied, without further analysis, if the roots arenegative.Again we show it numerically covering the required range of f and g values. The limiting values for g,depending on the values of f, are shown in table 4.

36

Table 4: Limiting values for g

f g1 = gmin g2 = gmax

0.01 1.99 5.10

0.05 1.95 2.47

0.10 1.88 1.94

0.125 (3/2)3/2 (3/2)3/2

The values of the roots are computed, in table 5, for s2 = 0.

Table 5: Values of the roots

f G t12 t2

2 t32

0.01 3 -0.38 -30.41 -2.87

0.05 2.1 -0.71 -7.50 -1.79

0.10 1.9 -1.04 -3.14 -1.61

0.125 (3/2)3/2 -1.63 -1.63 -1.63

The roots are negative, hence we may write the deflection integral as follows

(74) ∫ ∫∞ ∞+

π=

0 02

)/sin()/cos()/sin()/cos(14ts

lsblsyltaltxfg

fk

qw

dtdsztztzt

+γ

++β

++

α23

22

22

22

21

22

where

(75)delta

ttdelta

ttdelta

tt 22

21

2

21

23

2

23

22

2

−=γ

−=β

−=α

(76) ( ) ( ) ( )22

21

43

21

23

42

23

22

41 tttttttttdelta −+−+−=

and zi2 = - t i

2 .

The solution of the integrals are given by (43) for x > a and by (44) for x < a.With

(77)21

/)(/)(

1

11

4 zeeI

laxzlaxz +−−− −π=

(78)22

/)(/)(

2

22

4 zeeI

laxzlaxz +−−− −π=

(79)23

/)(/)(

3

33

4 zeeI

laxzlaxz +−−− −π=

when x > a and

37

(80)21

/)(/)(

1

1124 z

eeIlaxzlxaz +−−− −−π=

(81)22

/)(/)(

2

2224 z

eeIlaxzlxaz +−−− −−π=

(82)23

/)(/)(

3

3324 z

eeIlaxzlxaz +−−− −−π=

when x < a.

The solutions of the deflection integrals are

(83) [ ]dsIIIs

lsblsyfg

fk

qwo

3222122

)/sin()/cos(14 γ+β+α+π

= ∫∞

or

(84) [ ]dtIIIt

ltaltxfg

fk

qwo

3222122

)/sin()/cos(14 γ+β+α+π

= ∫∞

where a, x and s in I1, I2 and I3 are respectively replaced by b, y and t.