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LsinhEA

2ee

EA

e

e*

e1

eEA2

eee1

EA)0(Fk

LL

L

L

L2

L

LL

L212

LcothEALsinh

LcoshEA

ee

eeEA

e1

1eEA

1ee1

EA

eeeee1

EAk)L(F

LL

LL

L2

L2

L

L2

LLLL

L222

12

LL

LLL

LL2L

L221

k

)Lsinh(

EA

eeeee

EA

eee1e1

EA

)L(Fk

Then the stiffness Matrix will be:

LcothEA)Lsinh(

EA)Lsinh(

EALcothEA

k ..(A-1)

Eq. (A-1) represent the soil-structure interaction effects and the contribution of the pile and

surrounding soil resistance to Axial load on pile, for a simple bar, the stiffness matrix is :

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L

EA

L

EAL

EA

L

EA

k ..(A-1-1)

Both of equations (A-1), (A-1-1) are symmetric matrix but they differ in presenting of soil-

structure interaction parameter (). If approaches zero, Eq. (A-1) gives Eq. (A-1-1).

A-2: Torsional resistance:

dxP.k)x(dT)x(T)x(T:0Mo =0

0dxP.kdx

)x(dT

We have from strength of material approach:

GJ

)x(Tdd

dx

dGJ)x(T

2

2)(

dx

dGJ

dx

xdT .(4-8)

0P.kdx

dGJ

2

2

..(4-9)

GJ

Pk (4-10) where: P=2R3 ,R: radius of Pile.

xx eBeA)x( ..(4-11)

A-2-1: Applying Boundary Conditions:

Boundary conditions of the problem is that for a finite length of pile L,

at x=0, (0)= 1, at x=L, (L)= 2

1= BA AB 1 ..(A-12-a)

LL2 eBeA

T(x) +dT (x).k

T(x)dxx

Fig. (A-2): Torsional Bar in Soil

Media.

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)x(y.kdx

y(x)dn2

2

..(4-3-4)

AndEI

)x(y.k)

dx

M(x)d(

EI

d

dx

y(x)d n2

2

4

4

0EI

)x(y.k

dx

y(x)dn

4

4

(A-3-5)

This is the D.E. for the flexure problem in the case of one parameter beam on elastic

foundation.

The shear and bending moments become:-

V=3

3

dx

ydEI- (A-3-6)

2

2

dx

yd-EIM (A-3-7)

The general solution of a free field (homogeneous form of a problem)(Eq.(A-3-5)) will be

in the form of exact solution as.

)]xL

sinh().xL

[sin(C)]xL

cosh().xL

[sin(C

)]xL

sinh().xL

[cos(C)]xL

cosh().xL

[cos(C)x(y

43

21

( A-3-8)

where: 4 n

I.E.4

D.kL D: Diameter of Pile (or Width of beam).

)]}xL

sinh().xL

cos()xL

cosh().xL

[sin(C

)]xL

cosh().xL

cos()xL

sinh().xL

[sin(C

)]xL

sinh().xL

sin()xL

cosh().xL

[cos(C

)]xL

cosh().xL

sin()xL

sinh().xL

[cos(C{Ldx

)x(dy

4

3

2

1

. (A-3-9)

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)]}xL

cosh().xL

[cos(C)]xL

sinh().xL

[cos(C

)]xL

cosh().xL

[sin(C)]xL

sinh().xL

[sin(C{L

.2

dx

)x(dy

43

212

2

2

2

( A-3-10)

)]}xL

cosh().xL

sin()xL

sinh().xL

[cos(C

)]xL

sinh().xL

sin()xL

cosh().xL

[cos(C

)]xL

cosh().xL

cos()xL

sinh().xL

[sin(C

)]xL

sinh().xL

cos()xL

cosh().xL

[sin(C{L

.2

dx

)x(dy

4

3

2

13

3

3

3

...(A-3-11)

Now, to find the integration constants in terms of nodal normal displacements and nodal

rotations at each node for a finite length (L) as: -

At x=0 y0=y(0)=C1and )CC(Ldx

)0(dy320

At x=L hs.s.Chc.sChs.cChc.cCy(L)y 4321L And:

)]hs.chc.s(C)hc.chs.s(C)hs.shc.c(C)hc.shs.c(C[Ldx

)L(dy4321L

Or in matrix form: -

L

q(x)M+dM

V(x)M(x)

V+dVKn.y(x)

Fig. (A-3): Beam on Elastic

Foundation.

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4

3

2

1

L

L

0

0

C

C

C

C

]hs.chc.s[

L

]hc.chs.s[

L

]hs.shc.c[

L

]hc.shs.c[

L

hs.shc.shs.chc.c

0LL

0

0001

y

y

..(A-3-12)

i.e. {yi}=[V]{ai}

or {ai}=[V]-1{yi}

where [yi] : Is the nodal displacement vector, [V]: The square matrix in Eq. (A-3-12), {ai}:

Vector of integration constants.

Now, to find the nodal loads depend on exact function (eq. (A-3-8)) as:-

V (x)=3

3 )(dEI-

dx

xy

= )}sh.cch.s(C)ch.csh.s(C)ch.csh.s(C)sh.cch.s(C{)L(EI.2 4321

3

and M(x)=2

2

dx

)x(ydEI-

= )}ch.c(C)sh.c(C)ch.s(C)sh.s(C{L

EI2 43212

2

At: x=0 )C(CL

2EIV(0)V 323

3

0

and 42

2

0 CLEI2)0(MM

At: x = L

VL = V (L)

)}hs.chc.s(C)hc.chs.s(C)hc.chs.s(C)hs.chc.s(C{L

EI2 43213

3

ML = M (L)

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}hc.cChs.cChc.sChs.sC{L

EI2 43212

2

In the matrix form: -

4

3

2

1

2

2

L

L

0

0

C

C

C

C

hc.chs.chc.shs.s

)hs.chc.s(L

)hc.chs.s(L

)hc.chs.s(L

)hs.chc.s(L

1000

0LL

0

LEI2

M

V

M

V

...(A-3-13)

i.e. {Fi}=[G]{ai}

where: {Fi}: vector of nodal loads.

[G]: Is the square matrix in Eq. (A-13-3) then:

{Fi}=[G][V]-1{ai} ..(A-3-14)

The stiffness coefficients in two dimensional beam on elastic foundation in local

coordinate system may be expressed by taking the nodal displacement vector {ai} as unity, then the

stiffness matrix may be expressed as :-

[k]=[G][V]-1

which is yield [6,17]:

1526

5364

2615

6453

TTTT

TTTT

TTTT

TTTT

k (A-3)

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where: T1=))(sin)((sinhL

))cos()sin()sinh()(cosh(EI222

=LEI4 at =0.00

T2=))(sin)((sinhL

))cos()sinh()sin()(cosh(EI222

=

L

EI2 at =0.00

T3=))(sin)((sinhL

))cosh()sinh()sin()(cos(EI4223

3

=

3L

EI12 at =0.00

T4=

))(sin)((sinhL

))cos()sinh()sin()(cosh(EI4223

3

=

3

L

EI12 at =0.00

T5=))(sin)((sinhL

))cos()(sin)((sinhEI2222

222

=

2L

EI6 at =0.00

T6=))(sin)((sinhL

))sin().(sinh(EI4222

2

=

2L

EI6 at =0.00

Finally, to get the three dimensional element stiffness matrix for a pile [Eqs. (A-1), (A-2), (A-3)]

will be superimposed to get the stiffness matrix in 3-D. case as shown in Eq. (4-13).