Laplace Transform

( ^∞

∫

∞

−

=

. (3.29) Similarly from Eq. (3.29),

β β

π α

α y U e^β d

U_{i i}( , ) ^{i y}

2 ) 1 ,

ˆ (

∫

^∞

∞

−

=

(3.30)

and eventually that,

β α β

π ^U α ^{e α β d d} y

x

u_{i i}( , ) ^{i( x y)} 2

) 1 ,

( ⁺

∞

∞

−

∞

∞

∫ ∫

−

=

(3.31)

giving the inversion formula for the double Fourier transform (Sneddon,1951).

Right now, the generalization to a greater number of variables is obvious. Suppose )

, , (x y z

u_i is a function of the three independent variables x, y, z; then the triple Fourier transforms of the function u_i(x,y,z) are defined to be:

dxdydz e

z y x u

U_{i i}( , , ) ^{i( x y z)} )

2 ( ) 1 , , (

2 3

γ β α

γ π β

α ^{∞ − + +}

∞

−

∞

∞

−

∞

∞

∫ ∫ ∫

−

=

. (3.32) The corresponding inversion formula can then be shown to be:

γ β α γ

β π α

γ β

α d d d

e U

z y x

u_{i i}( , , ) ^{i( x y z)} )

2 ( ) 1 , , (

2 3

+ +

∞

∞

−

∞

∞

−

∞

∞

∫ ∫ ∫

−

=

. (3.33)

3.2.2 Laplace Transform

The function U_i(s) of the variable s is called the Laplace transform of the original function )u_i(z , and will be denoted by L{u_i}. Thus,

dz e z u u

L s

U_{i i i} ^−sz

∞

∫

=

= { } ( )

) (

0 (3.34)

Let )u_i(z be a function that is piecewise continuous on every finite interval in the range z≥0 and satisfies u_i(z) ≤ Me ^Ψz for all z≥0. Then the Laplace transform of u_i(z) exists for all s>Ψ.

Since )u_i(z is piecewise continuous, u_i(z)e^−sz is integrable over any finite interval on the z-axis. From Eq. (3.34), assuming that s>Ψ, we obtain:

Ψ

= −

≤

≤

= ^∞

∫

^{u z e− dz ∞}

∫

^{u z e− dz ∞}

∫

^{MeΨ e− dz} _s^M

u

L _{i i} ^sz _i ^{sz z sz}

0 0

0

) ( )

( }

{

(3.35) where the condition s>Ψ is needed for the existence of the last integral.

Furthermore, the original function u_i(z) in Eq. (3.36) is called the inverse transform or inverse of U_i(s), and will be denoted by L⁻¹(U_i). In another word,

ds e s i U

U L z

u _i ^sz

i c

i c i

i ( )

2 } 1 { )

( ^{1 +}

∫

^∞

∞

−

− =

= π (3.36)

In most physical applications, the variable z can be replaced by the time (t). In this research, the variable z is the position of z-axis.

The constant c in Eq. (3.36) is chosen to be large enough, say c>a, so that the last integral exists. The value of a depends on the behavior of u_i(z)at large z.

Suppose that u_i(x) is continuous for all z≥0, and has a derivative u_i' x( ) that is piecewise continuous on every finite interval. Then the Laplace transforms of the derivative )u_i' x( exist when s>Ψ, and

) 0 ( )) ( ( )}

( '

{u_i z sL u_i z u_i

L = − (3.37)

Similarly,

) 0 ( ' ) 0 ( )) ( ( )}

( ''

{u_i z s²L u_i z su_i u_i

L = − − (3.38)

3.3 Reducing the PDE of (3.15a)-(3.15c) to ODE Using Double Fourier Transformation

The equilibrium equations, as expressed in Eqs. (3.15a)-(3.15c) for an inclined transversely isotropic space subjected to three-dimensional point loads are partial differential equations. Double Fourier transforms can be employed for solving the two independent variables of x and y to reduce the problem of solving partial differential equations to ordinary differential equations. We introduce the double Fourier transforms for displacement functions and derivatives as:

) , , ( )

, , ) (

2 (

1 _{( )}

z u

dxdy e

z y x

u_i ^{i x y} _i α β

π ^{− α+β =}

∞

∞

−

∞

∞

∫ ∫

− ^(3.39a)

) , , ( ) ) (

( ) , , ( )

2 (

1 _{( )}

z u

i dxdy y e

x z y x u

i y

x

i i α β α β

π ∂ ^{α β =}

∂ _{− +}

∞

∞

−

∞

∞

∫ ∫

− ^(3.39b)

) , , ( ) ) (

( ) , , ( )

2 (

1 _{( ) 2 2}

2 2 2

z u

dxdy y e

x z y x u

i y

x

i i α β α β

π ∂ ^{α β =−}

∂ _{− +}

∞

∞

−

∞

∞

∫ ∫

− ^(3.39c)

Hence, the double Fourier transforms for point forces F are given as: _i

dxdy e

F

F_{i i} ^{i( x y)} 2

1 _{α β}

π ^{− +}

∞

∞

−

∞

∞

∫ ∫

−

= (i=x, y, z) (3.40)

Using Eqs. (3.39a)-(3.39c) and Eq. (3.40), the Navier-Cauchy equations (Eqs.

(3.15a)-(3.15c)) can be simplified as the following system of linear ordinary differential equations:

x

))

To obtain the general solution of Eqs. (3.41a)-(3.41c), the homogeneous solution are solved first. Suppose that the solutions of the homogeneous equation can be expressed as the exponential function, namely,

uz

Substituting Eq. (3.44) and it derivatives ⁱ ue^uz dz

homogeneous ordinary differential equations, we obtain

[ ]

The characteristic equation of Eq. (3.45) has six roots. The details and the physical basis of which are given in Appendix B. The characteristic roots can be expressed as:

φ

φ

In order to derive the solution of the homogeneous equation ( Eqs. (3.41a)-(3.41c)), we redefine the three displacement functions (Eqs. (3.49a)-(3.49c)). In the lower half-space, the u_x(H)(α,β,z)^, u_y(H)(α,β,z)^, u_z(H)(α,β,z)^, A , _x^j A_y^j and A of Eqs. _z^j

presented in Appendix C. simplified as:

j j 11 11( , ,u ) D

D α β = , D₂₁(α,β,u_j )=D₂₁^j , D₃₁(α,β,u_j )=D₃₁^j (3.54)

CHAPTER Ⅳ

THREE DIMENSIONAL ELASTIC SOLUTIONS OF A

TRANSVERSELY ISOTROPIC FULL-SPACE SUBJECTED TO POINT LOADS

The analytical solutions for displacements and stresses in an inclined transversely isotropic full-space subjected to a point load in the medium are derived in Chapters 4.

To present the solutions, the ordinary differential equations reduced from partial differential equations by Double Fourier transform respect to variables of x and y are solved first. Three distinct approaches as shown in figure 1.1 were used to derive the solutions in Sec 4.1-4.3. In Sec 4.1(traditional method), we consider the nonhomogeneous part of the ordinary differential equations and solve the homogeneous and particular solution of Eqs. (3.41a)-(3.41c). In Sec.4.2 (imaginary space method), we separate the full-space into three regions of −∞<z<0⁻ (region 2) , 0⁻ <z<0⁺ (imaginary space), and 0⁺ <z<+∞ (region 1) , the point load force is in the region of

+

− < <0

0 z . In regions of −∞<z<0⁻ and 0⁺ <z<+∞, the right-hand side of Eqs.

(3.41a)-(3.41c) does not exist, the equilibrium equations are homogeneous linear equations. Hence, we can solve the boundary-value problem consisting of the three regions. In Sec.4.3(algebraic equation method), the Fourier transform respect to variables of z can reduce the aforementioned ordinary differential equations (Eqs.

(3.41a)-(3.41c)) to algebraic equations. This method, there are three time of Fourier transform respect to variables of x, y and z, is also called Triple Fourier transform

method.

4.1 Traditional Method

In full-space, the coefficients A_x^j₁, A_y^j₁, and A_z^j₁ (j=1-6) of Eqs. (3.50a)-(3.50c) can be determined by assuming the displacements in region 1, u_x1, u_y1, and u_z1 must be finite when z is approaching to ∞. The real part of the {u₄, u₅ , u₆} are positive, therefore, 0A_x⁴₁= A_x⁵₁= A_x⁶₁ = , A⁴_y1= A⁵_y1= A_y⁶₁=0, and A_z⁴₁= A_z⁵₁= A_z⁶₁ =0. Likewise, in region 2, u_x2, u_y2, and u_z2 also must be finite when z is approaching to -∞. The real part of the { u₁ , u₂ , u₃ } are negative, hence, A¹_x2 = A_x²₂ = A_x³₂ =0 ,

3 0

2 2

2 1

2 = _y = _y =

y A A

A , and A¹_z2 = A_z²₂ = A_z³₂ =0.

In order to solve the homogeneous solutions, Eqs. (3.41a)-(3.41c), we define the costants of C_d^j₂ and C_u^j₂ from Eqs. (3.52a)-(3.52c).

j d j j z j

j y

j j

x C

u D

A u

D A u

D A

2 31

1 21

1

11 1

) , , ( )

, , ( )

, ,

( = = =

β α β

α β

α (j=1-3) (4.01a)

j u j j z j

j y

j j

x C

u D

A u

D A u

D A

2 31

2 21

2 11

2

) , , ( )

, , ( )

, ,

( = = =

β α β

α β

α (j=4-6) (4.01b)

Adopting the costants of C_d^j₂ and C_u^j₂, Eqs. (3.50a)-(3.50c) and Eqs. (3.51a)-(3.51c) can be re-written as follows:

for _z>0 (region 1, as shown in Fig.3.1),

z u d z u d z u d H

x z C D e C D e C D e

u_{1( )}(α,β, )= ¹_{2 11}^{1 1} + ²_{2 11}^{2 2} + ³_{2 11}^{3 3} (4.02a)

z u d z u d z u d H

y z C D e C D e C D e

u _{1( )}(α,β, )= ¹₂ ¹₂₁ ¹ + ²_{2 21}^{2 2} + ³_{2 21}^{3 3} (4.02b)

z u d z u d z u d H

z z C D e C D e C D e

u_{1( )}(α,β, )= ¹₂ ¹₃₁ ¹ + ²_{2 31}^{2 2} + ³_{2 31}^{3 3} (4.02c) , and for z<0 (region 2, as also depicted in Fig.3.1),

z u u z u u z u u H

x z C D e C D e C D e

u _{2( )}(α,β, )= ⁴_{2 11}^{4 4} + ⁵_{2 11}^{5 5} + ⁶_{2 11}^{6 6} (4.03a)

z u u z u u z u u H

y z C D e C D e C D e

u _{2( )}(α,β, )= ⁴_{2 21}^{4 4} + ⁵_{2 21}^{5 5} + ⁶_{2 21}^{6 6} (4.03b)

z u u z u u z u u H

z z C D e C D e C D e

u _{2( )}(α,β, )= ⁴_{2 31}^{4 4} + ⁵_{2 31}^{5 5} + ⁶_{2 31}^{6 6} (4.03c) Based on the homogeneous solutions, we can assume the particular solutions with the

forms as:

z j u x j P

x

e j

z B z

u ( , , ) ⁶ ( )

1 )

(

∑

=

β = α

(4.04a)

z j u y j P

y

e j

z B z

u ( , , ) ⁶ ( )

1 )

(

∑

=

β = α

(4.04b)

z j u z j P

z

e j

z B z

u ( , , ) ⁶ ( )

1 )

(

∑

=

β = α

(4.04c) The coefficientsB , _x^j B_y^j, and B_z^j (j=1~6) can be determined by the method of

variation of parameters (Hildebrand, 1976). The point loads (Fx, Fy, Fz) acting at the origin point of the co-ordinate system can be described as Eqs. (3.18a)-(3.18c).

Following the approaches for homogeneous solutions, Eqs. (4.04a)-(4.04c) are expanded as:

for _z>0 (region 1, as shown in Fig. 3.1),

z u x z u x z u x P

x z Be B e B e

u_{1( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.05a)

z u y z u y z u y P

y z B e B e B e

u _{1( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.05b)

z u z z u z z u z P

z z Be B e B e

u_{1( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.05c)

and for z<0 (region 2, as also depicted in Fig. 3.1),

Introducing an imaginary plane along z=0, to separate the full-space into two half-spaces, one is 0⁺ <z<∞ (region 1 in Fig. 3.1), and the other is −∞<z<0⁻ (region 2 in Fig. 3.1). In the lower half-space, the Eqs. (3.42a)-(3.42f) would express as

)

) this case, we may rewrite Eqs. (3.42a)-(3.42f) as

))

When the two half-space are ideally bonded at the interface z=0 such that the material becomes continuous across the interface, we have:

) ( ) ( )

0 , , ( ) 0 , ,

( ₂

1 x y _zx x y P_x x y

zx τ δ δ

τ ⁺ − ⁻ = (4.11a)

) ( ) ( ) 0 , , ( ) 0 , ,

( ₂

1 x y _zy x y P_y x y

zy τ δ δ

τ ⁺ − ⁻ = (4.11b)

) ( ) ( ) 0 , , ( )

0 , ,

( ₂

1 x y _zz x y P_z x y

zz σ δ δ

σ ⁺ − ⁻ = (4.11c)

0 ) 0 , , ( ) 0 , ,

( ₂

1 x y ⁺ −u x y ⁻ =

u_{x x} (4.11d)

0 ) 0 , , ( ) 0 , ,

( ₂

1 x y ⁺ −u x y ⁻ =

u_{y y} (4.11e)

0 ) 0 , , ( ) 0 , ,

( ₂

1 x y ⁺ −u x y ⁻ =

u_{z z} (4.11f)

where −∞<x<∞, −∞< y<∞. The subscripts 1 and 2 mean that z=0 plane is approaching to 0⁺ and 0⁻, respectively.

Following, the double Fourier transforms of Eqs. (4.11a)-(4.11f) can be expressed as:

β π α τ β

α

τ_zx1( , ,0⁺)− _zx2( , ,0⁻)= 2P^x (4.12a)

β π α τ β

α

τ_zy1( , ,0 ) _zy2( , ,0 ) 2P^y

=

− ⁻

+ (4.12b)

β π α σ β

α

σ_zz1( , ,0⁺)− _zz2( , ,0⁻)=2P^z (4.12c)

0 ) 0 , , ( ) 0 , ,

( ₂

1 α β ⁺ − _x α β ⁻ =

x u

u (4.12d)

0 ) 0 , , ( ) 0 , ,

( ₂

1 α β ⁺ − _y α β ⁻ =

y u

u (4.12e)

0 ) 0 , , ( ) 0 , ,

( ₂

1 α β ⁺ − _z α β ⁻ =

z u

u (4.12f)

The general solutions of nonhomogeneous equations are the sum of the forms as: are particular solutions.

Hence, from Eqs. (4.02a)-(4.02c), Eqs. (4.03a)-(4.03c), Eqs. (4.05a)-(4.05c), Eqs.

(4.06a)-(4.06c), Eqs. (4.10a)-(4.10f) and Eqs. (4.13a)-(4.13f), the system of six linear equations (Eqs. (4.12a)-(4.12f)) has six undetermined coefficients C , _d¹₂ C , _d²₂ C , _d³₂

Then, the six undetermined coefficients C , _d¹₂ C , _d²₂ C , _d³₂ C , _u⁴₂ C , _u⁵₂ C can be _u⁶₂ determined. The boundary conditions of the full space consists of two parts, the first term and the second one in the left-hand side of Eq. (4.14) are the components of homogenous solutions and particular solutions, respectively. At the region of

+

−< <0

0 z , the second term in the left–hand side of Eq. (4.14) can be solved as follows:

⎥⎥ where Eq. (4.15) will be demonstrated in Appendix E.

Due to the determinant of fij (i, j =1-6) does not equal to zero, the undetermined

In the method of particular solution for full-space, it is clear that the homogenous solutions do not contribute to the general solution of Eqs. (3.41a)-(3.41c). In the other word, the displacement and stress functions can be obtained alone from the particular solutions in the form:

for _z>0 (region 1, as shown in Fig. 3.1),

z u y z u y z u y P

y

y z u z B e B e B e

u ₁(α,β, )= _{1( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3}

(4.16b)

z u z z u z z u z P

z

z z u z Be B e B e

u₁(α,β, )= _{1( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.16c)

) , , ( )

, ,

( _{1( )}

1 z _{xx P} z

xx α β σ α β

σ = (4.16d)

) , , ( )

, ,

( _{1( )}

1 z _{yy P} z

yy α β σ α β

σ = (4.16e)

) , , ( )

, ,

( _{1( )}

1 z _{zz P} z

zz α β σ α β

σ = (4.16f)

) , , ( )

, ,

( _{1( )}

1 z _{yz P} z

yz α β τ α β

τ = (4.16g)

) , , ( )

, ,

( _{1( )}

1 z _{zx P} z

zx α β τ α β

τ =

(4.16h) )

, , ( )

, ,

( _{1( )}

1 z _{xy P} z

xy α β τ α β

τ =

(4.16i) and for z<0 (region 2, as also depicted in Fig. 3.1),

z u x z u x z u x P

x

x z u z B e B e B e

u ₂(α,β, )= _{2( )}(α,β, )=− ^{4 4} − ^{5 5} − ^{6 6} (4.17a)

z u y z u y z u y P

y

y z u z B e B e B e

u ₂(α,β, )= _{2( )}(α,β, )=− ^{4 4} − ^{5 5} − ^{6 6}

(4.17b)

z u z z u z z u z P

z

z z u z B e B e B e

u ₂(α,β, )= _{2( )}(α,β, )=− ^{4 4} − ^{5 5} − ^{6 6}

(4.17c) )

, , ( )

, ,

( _{2( )}

2 z _{xx P} z

xx α β σ α β

σ = (4.17d)

) , , ( )

, ,

( _{2( )}

2 z _{yy P} z

yy α β σ α β

σ = (4.17e)

) , , ( )

, ,

( _{2( )}

2 z _{zz P} z

zz α β σ α β

σ = (4.17f)

) , , ( )

, ,

( _{2( )}

2 z _{yz P} z

yz α β τ α β

τ = (4.17g)

) , , ( )

, ,

( _{2( )}

2 z _{zx P} z

zx α β τ α β

τ = (4.17h)

) , , ( )

, ,

( _{2( )}

2 z _{xy P} z

xy α β τ α β

τ = (4.17i)

4.2 Imaginary Space Method

In Fig. 4.1, we separate the full-space into three imaginary regions of −∞<z<0⁻,

+

−< <0

0 z , and 0⁺ <z<+∞. The point load force is in the region of 0⁻ <z<0⁺. In region of −∞<z<0⁻ and 0⁺ <z<+∞, the right-hand side of Eqs. (3.41a)-(3.41c) are not existed, the equilibrium equations are homogeneous linear equations. In the method of undetermined coefficients for full-space, it is clear that the particular solutions do not contribute to the general solution of Eqs. (3.41a)-(3.41c). In the other word, the displacement and stress functions can be obtained alone from the homogeneous solutions.

= 0+

z

= 0+

z

= 0−

z

= 0−

z

= 0+

z

= 0−

z

) 0 , ,

1( y + zz x σ

) 0 , ,

2( y − zz x σ

) ( ) (x y P_zδ δ

< −

<

∞

− z 0

+

−< <0 0 z

+∞

<

<

+ z

0

x

y

z

Fig. 4.1 Separate the full-space into three imaginary regions of −∞<z<0⁻,

+

− < <0

0 z and 0⁺ <z<+∞

In order to differentiate varieties of homogeneous solutions of Eqs. (3.50a)-(3.50c) and Eqs. (3.51a)-(3.51c) in Sec 4.1.1 and Sec 4.1.2, we define the costants of C_d^j and

j

Cu from the Eqs. (3.52a)-(3.52b).

j d j j z j

j y

j j

x C

u D

A u

D A u

D

A = = =

) , , ( )

, , ( )

, ,

( ₃₁

1 21

1 11

1 α β α β

β

α (j=1-3) (4.18a)

j u j j z j

j y

j j

x C

u D

A u

D A u

D

A = = =

) , , ( )

, , ( )

, ,

( ₃₁

2 21

2 11

2 α β α β

β

α (j=4-6) (4.18b)

By using the costants of C and _d^j C , Eqs. (3.50a)-(3.50c) and Eqs. (3.51a)-(3.51c) _u^j can obtain as follow:

for z> 0⁺ (region 1, as shown in Fig. 3.2),

z u d z u d z u d

x z C D e C D e C D e

u₁(α,β, )= ¹ ₁₁^{1 1} + ² ₁₁^{2 2} + ³ ₁₁^{3 3} (4.19a)

z u d z u d z u d

y z C D e C D e C D e

u ₁(α,β, )= ¹ ₂₁^{1 1} + ² ₂₁^{2 2} + ³ ₂₁^{3 3} (4.19b)

z u d z u d z u d

z z C D e C D e C D e

u₁(α,β, )= ¹ ₃₁^{1 1} + ² ₃₁^{2 2} + ³ ₃₁^{3 3} (4.19c)

and for z< 0⁻ (region 2, as also depicted in Fig. 3.2),

z u u z u u z u u

x z C D e C D e C D e

u ₂(α,β, )= ⁴ ₁₁^{4 4} + ⁵ ₁₁^{5 5} + ⁶ ₁₁^{6 6} (4.20a)

z u u z u u z u u

y z C D e C D e C D e

u ₂(α,β, )= ⁴ ₂₁^{4 4} + ⁵ ₂₁^{5 5} + ⁶ ₂₁^{6 6} (4.20b)

z u u z u u z u u

z z C D e C D e C D e

u ₂(α,β, )= ⁴ ₃₁^{4 4} + ⁵ ₃₁^{5 5} + ⁶ ₃₁^{6 6} (4.20c)

It is note that the difference between Eqs. (4.02a)-(4.02c), Eqs. (4.03a)-(4.03c) and Eqs. (4.19a)-(4.19c), Eqs. (4.20a)-(4.20c), Eqs. (4.02a)-(4.02c) and Eqs. (4.03a)-(4.03c) are the homogeneous solutions of the nonhomogeneous linear equations of the equilibrium equations, and Eqs. (4.19a)-(4.19c) and Eqs. (4.20a)-(4.20c) are the

homogeneous solutions of the homogeneous linear equations part of the equilibrium equations.

Furthermore, considering the following pertinent continuity and discontinuity conditions at z =0 in the (x,y,z) and (α,β,z)domain are as same as the Eqs.

(4.11a)-(4.11f) and Eqs. (4.12a)-(4.12f), respectively.

Once Eqs. (4.19a)-(4.19c) and Eqs. (4.20a)-(4.20c) are substituted into Eqs.

(4.12a)-(4.12f), the six undetermined coefficients, C¹_d, C_d², C_d³, C_u⁴, C_u⁵, C_u⁶, can

will be demonstrated in Appendix E.

In order to find the undetermined coefficients, C_d¹, C_d², C_d³, C_u⁴, C_u⁵, and C_u⁶. We can solve by the Cramer’s rule, that is, C_d¹ can be obtained from

[ ]

fij by replaceing the first column by the term in the right–hand side of Eq. (4.21). Only the first term (j=1) of Eqs. (4.22a)-(4.22c) make the determinant exist, and the term of j=2-6 must let the determinant equal to zero. So we find:

Eqs. (4.19a)-(4.19c) and Eqs. (4.20a)-(4.20c). They are the same as those from Eqs.

(4.16a)-(4.16i) and Eqs. (4.17a)-(4.17i).

4.3 Algebraic Equation Method (Triple Fourier Transform)

In a full-space problem, u_i(x,y,z) are set as the displacements in a homogeneous and linearly elastic continuum with domain −∞<x ,,y z<∞. Then the triple Fourier transforms of the displacement components are applied to solve the equilibrium equations (Eqs. (3.15a)-(3.15c)). The step by step Fourier transforms for ui (i=x, y, z) are written as:

)

In addition, the Dirac delta function has the following property:

1

dxdydz P e

dxdydz P e

2

dxdydz P e

According to Eqs. (4.24a)-(4.24c) and Eqs. (4.26a)-(4.26c), Eqs. (3.15a)-(3.15c) can be expressed as:

[ ]

(4.27) by matrix operations as:

[ ]

From Eqs. (4.26a)-(4.26c) and Eq. (4.29), the three algebraic equations U_x(α,β,γ),

The characteristic equation (Eq. (4.31)) has six roots, and can be expressed as:

)

) inverse Fourier transforms should be addressed to solve Eqs. (4.30a)-(4.30c) as:

γ γ

To evaluate the integral in Eqs. (4.33a)-(4.33c), the complex γ-plane is first encountered. Since there are six poles in the integrand, we assume the imaginary part of γ , ₁ γ , ₂ γ are positive, and that of ₃ γ , ₄ γ , ₅ γ are negative. ₆

For z>0, a closed contour by adding a large semicirle on the upper γ-plane, as depicted in Fig. 4.2 is employed.

γ

C

R

Imγ

Reγ

1

γ γ

3

γ

4

γ

5

γ

6

Fig. 4.2 A closed contour on the upper γ-plane.

The solutions of Eqs. (4.33a)-(4.33c), as defined by the three displacement functions are:

z i x z i x z i x

x z B e B e B e

u₁(α,β, )= ¹_γ ^γ1 + ²_γ ^γ2 + ³_γ ^γ3 (4.34a)

z i y z i y z i y

y z B e B e B e

u ₁(α,β, )= ¹_γ ^γ1 + ²_γ ^γ2 + ³_γ ^γ3 (4.34b)

z i z z i z z i z

z z B e B e B e

u₁(α,β, )= ¹_γ ^γ1 + ²_γ ^γ2 + ³_γ ^γ3 (4.34c)

, and for _z<0, the other closed contour in the lower γ-plane is shown in Fig. 4.3.

C

R

γ

Imγ

Reγ

1

γ γ

3

γ

4

γ

5

γ

6

Fig. 4.3 A closed contour on the lower γ-plane.

z i x z i x z i x

x z B e B e B e

u ₂(α,β, )=− ⁴_γ ^γ4 − ⁵_γ ^γ5 − ⁶_γ ^γ6 (4.35a)

z i y z i y z i y

y z B e B e B e

u ₂(α,β, )=− ⁴_γ ^γ4 − ⁵_γ ^γ5 − ⁶_γ ^γ6 (4.35b)

z i z z i z z i z

z z B e B e B e

u ₂(α,β, )=− ⁴_γ ^γ4 − ⁵_γ ^γ5 − ⁶_γ ^γ6 (4.35c)

where

j t

j z

j y

j j x

x mR

D P D

P D

iP

B π

γ β α γ

β α γ

β β α

γ α 2

) , , ( )

, , ( )

, , ) (

,

( ¹¹ + ²¹ + ³¹

= (4.36a)

j t

j z

j y

j j x

y mR

D P D

P D

iP

B π

γ β α γ

β α γ

β β α

γ α 2

) , , ( )

, , ( )

, , ) (

,

( ¹² + ²² + ³²

= (4.36b)

j t

j z

j y

j j x

z mR

D P D

P D

iP

B π

γ β α γ

β α γ

β β α

γ α 2

) , , ( )

, , ( )

, , ) (

,

( ¹³ + ²³ + ³³

= (4.36c)

j

Rj γ γ

γ

γ γ γ γ γ γ γ γ γ γ γ β γ

α =

∂

−

−

−

−

−

−

=∂[( )( )( )( )( )( )], )

,

( ^{1 2 3 4 5 6}

(4.36d)

Suppose that u_j =iγ_j , u_x(α,β,z) , u_y(α,β,z) , and u_z(α,β,z) from Eqs.

(4.16a)-(4.16c) and Eqs. (4.17a)-(4.17c) have the same results as Eqs. (4.34a)-(4.34c) and Eqs. (4.35a)-(4.35c).

4.4 Solutions for Displacements and Stresses by Inverse Fourier Transforms

The displacements u_x(x,y,z), )u_y(x,y,z , and u_z(x,y,z) in Eqs. (4.16a)-(4.16c) and Eqs. (4.17a)-(4.17c) also can be solved by the following inverse double Fourier transforms as:

for z>0 (region 1, as shown in Fig. 3.1),

z u x z u x z u x

x z Be B e B e

u₁(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.37a)

z u y z u y z u y

y z B e B e B e

u ₁(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.37b)

z u z z u z z u z

z z Be B e B e

u₁(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (4.37c)

) , , ( )

, ,

( _{1( )}

1 z _{xx P} z

xx α β σ α β

σ =

(4.37d) )

, , ( )

, ,

( _{1( )}

1 z _{yy P} z

yy α β σ α β

σ =

(4.37e) )

, , ( )

, ,

( _{1( )}

1 z _{zz P} z

zz α β σ α β

σ =

(4.37f) )

, , ( )

, ,

( _{1( )}

1 z _{yz P} z

yz α β τ α β

τ =

(4.37g) )

, , ( )

, ,

( _{1( )}

1 z _{zx P} z

zx α β τ α β

τ =

(4.37h)

) , , ( )

, ,

( _{1( )}

1 z _{xy P} z

xy α β τ α β

τ = (4.37i)

and for z<0 (region 2, as also depicted in Fig. 3.1),

z u x z u x z u x

x z B e B e B e

u ₂(α,β, )=− ^{4 4} − ^{5 5} − ^{6 6} (4.38a)

z u y z u y z u y

y z B e B e B e

u ₂(α,β, )=− ^{4 4} − ^{5 5} − ^{6 6} (4.38b)

z u z z u z z u z

z z B e B e B e

u ₂(α,β, )=− ^{4 4} − ^{5 5} − ^{6 6} (4.38c)

) , , ( )

, ,

( _{2( )}

2 z _{xx P} z

xx α β σ α β

σ = (4.38d)

) , , ( )

, ,

( _{2( )}

2 z _{yy P} z

yy α β σ α β

σ = (4.38e)

) , , ( )

, ,

( _{2( )}

2 z _{zz P} z

zz α β σ α β

σ =

(4.38f) )

, , ( )

, ,

( _{2( )}

2 z _{yz P} z

yz α β τ α β

τ =

(4.38g) )

, , ( )

, ,

( _{2( )}

2 z _{zx P} z

zx α β τ α β

τ =

(4.38h) )

, , ( )

, ,

( _{2( )}

2 z _{xy P} z

xy α β τ α β

τ = (4.38i)

The desired displacements of u_x(x,y,z) , )u_y(x,y,z , and u_z(x,y,z) can be obtained by the double inverse Fourier transforms of Eqs. (4.37a)-(4.37c) and Eqs.

(4.38a)-(4.38c) as:

for z>0 (region 1),

β π ^{Be α β B e α β B e α β d}α^d z

y x

u_x { _x ^{i x y uz} _x ^{i x y u z} _x ^{i x y u z}} 2

) 1 , ,

( ^{1 ( ) 1 2 ( ) 2 3 ( ) 3}

1

+ + +

+ +

∞ +

∞

−

∞

∞

−

+ +

=

∫ ∫

^(4.39a)

β π ^{B e α β B e α β B e α β d}α^d z

y x

u_y { _y ^{i x y uz} _y ^{i x y u z} _y ^{i x y u z}} 2

) 1 , ,

( ^{1 ( ) 1 2 ( ) 2 3 ( ) 3}

1

+ + +

+ +

∞ +

∞

−

∞

∞

−

+ +

=

∫ ∫

^(4.39b)

β

In addition, the stress components of Eqs. (3.12a)-(3.12f) are performed by the double Fourier transforms expressed as Eqs. (3.42a)-(3.42f). The desired σ_xx(x,y,z), by the double inverse Fourier transforms as:

for z>0 (region 1),

β

j j x j

z j

y j

x j

xy(α,β)=−i(a₆₆βB +a₆₆αB +a₅₆αB )+ia₅₆B u

τ (4.43f)

In Eqs. (4.43a)-(4.43f), j=1-3 for z>0 (region 1), and j=4-6 for z<0 (region 2).

If we employe a spherical co-ordinate system, which is shown in Fig. 4.4, the variables α, β, and ui can be expressed in terms of (k, θx) as:

cos x

k θ

α = × (4.44a)

sin x

k θ

β= × (4.44b)

α

β

( k , ) θ

x

x

θ

Fig. 4.4 Spherical co-ordinate system (k, θx)

By substiting Eqs. (4.44a)-(4.44b) into Eqs. (3.47a)-(3.47f), we get

φ φ

φ φ

θ θ

φ φ β θ

α _{2 2}

2 2

2 2

sin cos

)) sin (cos

cos (sin

) 1 ( cos sin ) sin

, (

j

j x

x j j x

j A

A A

A k i

u +

+ +

− +

× −

= (j=1-3) (4.45a)

φ φ

φ φ

θ θ

φ φ β θ

α _{2 2}

2 2

2 2

sin cos

)) sin (cos

cos (sin

) 1 ( cos sin ) sin

, (

j

j x

x j j x

j A

A A

i A k i

u +

+ +

+ +

× −

= (j=4-6) (4.45b)

where 0<k<∞, and 0<θx<2π.

According to Eqs. (4.44a)-(4.44b) and Eqs. (4.45a)-(4.45b), D_mn(α,β,γ_j) (m, n=1-3) in Eqs. (3.53a)-(3.53f), B_x^j(α,β,u_j) , B_y^j(α,β,u_j) , B_z^j(α,β,u_j) (j=1-6) in Eqs.

(4.04a)-(4.04c), and U_j(α,β) in Eq. (4.09) can be expressed in terms of k and θx as:

) ( ) , ( ) ,

( _{j j x}

j u k ku

u α β = θ = θ (4.46a)

) ( )

, ( )

, ( )

, ,

( _{j mn mn x} ⁴ _{mn x}

mn u D D k k D

D α β = α β = θ = θ

(4.46b) )

( )

, ( ) ,

( _{j x} ⁵ _{j x}

j U k k U

U α β = θ = θ (4.46c)

) ( )

, ( ) ,

( _x^j _x ¹ _x^j _x

j

x B k k B

B α β = θ = ⁻ θ (4.46d)

) ( )

, ( ) ,

( _y^j _x ¹ _y^j _x

j

y B k k B

B α β = θ = ⁻ θ (4.46e)

) ( )

, ( ) ,

( _x^j _x ¹ _x^j _x

j

x B k k B

B α β = θ = ⁻ θ (4.46f)

Hence, we can obtain:

) ( )

, ( ) ,

( _{xj x} ¹ _{xj x}

xj u k k u

u α β = θ = ⁻ θ (4.47a)

) ( )

, ( ) ,

( _{yj x} ¹ _{yj x}

yj u k k u

u α β = θ = ⁻ θ (4.47b)

) ( )

, ( ) ,

( _{zj x} ¹ _{zj x}

zj u k k u

u α β = θ = ⁻ θ (4.47c)

Furthermore, Eqs. (4.43a)-(4.43f) can be rewritten as:

) ( ) , ( ) ,

( _xx^j _{x xx}^j _x

j

xx α β σ k θ σ θ

σ = = (4.48a)

) ( ) , ( ) ,

( _yy^j _{x yy}^j _x

j

yy α β σ k θ σ θ

σ = = (4.48b)

) ( ) , ( ) ,

( _zz^j _{x zz}^j _x

j

zz α β σ k θ σ θ

σ = = (4.48c)

) ( ) , ( ) ,

( _yz^j _{x yz}^j _x

j

yz α β τ k θ τ θ

τ = = (4.48d)

) ( ) , ( ) ,

( _zx^j _{x zx}^j _x

j

zx α β τ k θ τ θ

τ = = (4.48e)

) ( ) , ( ) ,

( _xy^j _{x xy}^j _x

j

zx α β τ k θ τ θ

τ = = (4.48f)

The exponential terms in Eqs. (4.41a)-(4.41c), Eqs. (4.42a)-(4.42c), Eqs.

(4.43a)-(4.43f) and Eqs. (4.44a)-(4.44f) can be expressed as:

) ( )

( x y u_jz k _{j x}

iα +β + = ×ψ θ (4.49)

Additionally, dαdβ can be replaced by dkdθ_x times the absolute value of Jacobian J as:

dkd x

J d

dα β = θ (4.50)

where:

cos k k sin

sin k cos

k k ) , k (

) , J (

x x

x x

x x x

− =

=

∂

∂

∂

∂ ∂

∂

∂

∂

∂ =

= ∂

θ θ

θ θ

θ β

β θ

α α θ

β

α (4.51)

and therefore, kdkd x

d

dα β= θ (4.52)

Based on Eqs. (4.46a)-(4.46f), Eqs. (4.47a)-(4.47c), Eq.(4.49) and Eq. (4.50), then Eqs. (4.39a)-(4.39c) and Eqs. (4.40a)-(4.40c) can be presented as:

For z>0 (region 1),

x

Also, from Eqs. (4.46a)-(4.46f), Eqs. (4.48a)-(4.48f), Eqs. (4.49) and Eq. (4.52), Eqs.

(4.41a)-(4.41f) and Eqs. (4.42a)-(4.42f) can then be written as:

For z>0 (region 1),

x

x

(4.53a)-(4.53c) and Eqs. (4.54a)-(4.54c) can be expressed as:

For z>0 (region 1),

ω ω

Besides, Eqs. (4.55a)-(4.55f) and Eqs. (4.56a)-(4.56f) can be represented as:

For z>0 (region 1),

ω ω

CHAPTER V

THREE DIMENSIONAL ELASTIC SOLUTIONS OF A

TRANSVERSELY ISOTROPIC HALF-SPACE SUBJECTED TO POINT LOADS

The analytical solutions for displacements and stresses in an transversely isotropic half-space subjected to a point load in the medium are derived in Chapters 5. Three distinct approaches as same as Chapters 4 were used to derive the solutions in Sec 5.1-5.3. In Sec 5.1, we consider the nonhomogeneous part of the ordinary differential equations and solve the homogeneous and particular solution of Eqs. (3.41a)-(3.41c).

In Sec.5.2, we separate the half-space into two regions of 0⁻< z<0⁺ (imaginary space) and 0⁺ <z<+∞ (region 1) , the point load force is in the region of 0⁻ <z<0⁺. In regions of 0⁺ <z<+∞, the right-hand side of Eqs. (3.41a)-(3.41c) does not exist, the equilibrium equations are homogeneous linear equations. Hence, we can solve the boundary-value problem consisting of the two regions. In Sec.5.3, the Laplace transform respect to variables of z can reduce the aforementioned ordinary differential equations (Eqs. (3.41a)-(3.41c)) to algebraic equations..

5.1 Traditional Method

The thesis aims to determine the distribution of stress and displacement in a semi-infinite elastic solid due to the application of an external point load to its surface analytically. In order to derive the homogenous solutions of Eqs. (3.41a)-(3.41c), we

redefine the three displacement functions of Eqs. (3.50a)-(3.50c) and Eqs.

(3.51a)-(3.51c) as follows:

for z>0 (region 1 and 2, as shown in Fig. 5.1),

z u x z u x z u x z u x z u x z u x H

x z A e A e A e A e A e A e

u_{1( )}(α,β, )= ¹₁ ¹ + ²₁ ² + ³₁ ³ + ⁴₁ ⁴ + ⁵₁ ⁵ + ⁶₁ ⁶

(5.01a)

z u y z u y z u y z u y z u y z u y H

y z A e A e A e A e A e A e

u _{1( )}(α,β, )= ¹₁ ¹ + ²₁ ² + ³₁ ³ + ⁴₁ ⁴ + ⁵₁ ⁵ + ⁶₁ ⁶ _(5.01b)

z u z z u z z u z z u z z u z z u z H

z z A e A e A e A e A e A e

u_{1( )}(α,β, )= ¹₁ ¹ + ²₁ ² + ³₁ ³ + ⁴₁ ⁴ + ⁵₁ ⁵ + ⁶₁ ⁶

(5.01c) and for z<0 (region 3, as also depicted in Fig. 5.1),

0 ) , ,

)(

(

2 z =

u_{x H} α β

(5.02a) 0

) , ,

)(

(

2 z =

u_{y H} α β

(5.02b) 0

) , ,

)(

(

2 z =

u_{z H} α β

(5.02c)

In Eqs. (5.01a)-(5.01c), the undetermined coefficients A_x^j₁, A_y^j₁, and A_z^j₁ (j=1-6) can be obtained by assuming the displacements in region 1, u_x1, u_y1, and u_z1 must be finite when z is approaching to ∞. Hence, A_x⁴₁= A_x⁵₁= A_x⁶₁ =0, A⁴_y1= A⁵_y1= A_y⁶₁=0, and 0A_z⁴₁= A_z⁵₁ = A⁶_z1= .

Now, let:

j d j j z j

j y

j j

x C

u D

A u

D A u

D A

2 31

1 21

1 11

1

) , , ( )

, , ( )

, ,

( = = =

β α β

α β

α (j=1-3) (5.03)

where Dij (i, j=1-3) can be written as the second order derterminants in the following, and the complete forms of Dij are presented in Appendix C.

Similar to the full space problem, the general solition of the homogeneous linear equation of Eqs.(3.41a)-(3.41c) can express as follow:

z u d z u d z u d H

x z C D e C D e C D e

u_{1( )}(α,β, )= ¹_{2 11}^{1 1} + ²_{2 11}^{2 2} + ³_{2 11}^{3 3} (5.04a)

z u d z u d z u d H

y z C D e C D e C D e

u _{1( )}(α,β, )= ¹₂ ¹₂₁ ¹ + ²_{2 21}^{2 2} + ³_{2 21}^{3 3} (5.04b)

z u d z u d z u d H

z z C D e C D e C D e

u_{1( )}(α,β, )= ¹₂ ¹₃₁ ¹ + ²_{2 31}^{2 2} + ³_{2 31}^{3 3} (5.04c) The point loads (Fx, Fy, Fz) acts at the point (0, 0, h) of the co-ordinate system can be

described as Eqs. (3.19a)-(3.19c). Hence, the three displacement functions of the particular solutions can be gained from those of full space case (Sec. 4.1.1):

for _z>h (region 1, as shown in Fig. 5.1),

) ( 3 ) ( 2 ) ( 1 )

(

1_P ( , , ) _x ^{u1 z h} _x ^{u2 z h} _x ^{u3 z h}

x z Be B e B e

u α β = ⁻ + ⁻ + ⁻

(5.05a)

) 3 ( ) ( 2 ) ( 1 )

(

1_P ( , , ) _y ^{u1z h} _y ^{u2 z h} _y ^{u3 z h}

y z B e B e B e

u α β = ⁻ + ⁻ + ⁻ _(5.05b)

) ( 3 ) ( 2 ) ( 1 )

(

1_P ( , , ) _z ^{u1 z h} _z ^{u2 z h} _z ^{u3 z h}

z z Be B e B e

u α β = ⁻ + ⁻ + ⁻

(5.05c) and for 0<z<h (region 2, as also depicted in Fig. 5.1),

) 6 ( ) 5 ( ) ( 4 )

(

2 _P ( , , ) _x ^{u4 z h} _x ^{u5 z h} _x ^{u6 z h}

x z B e B e B e

u α β =− ⁻ − ⁻ − ⁻

(5.06a)

) ( 6 ) ( 5 ) ( 4 )

(

2_P ( , , ) _y ^{u4 z h} _y ^{u5 z h} _y ^{u6 z h}

y z B e B e B e

u α β =− ⁻ − ⁻ − ⁻ _(5.06b)

) ( 6 ) ( 5 ) ( 4 )

(

2_P ( , , ) _z ^{u4 z h} _z ^{u5 z h} _z ^{u6 z h}

z z B e B e B e

u α β =− ⁻ − ⁻ − ⁻

(5.06c)

=0 half-space

When the half-space (z>h) and the strip-space (0<z<h) are ideally bonded at the interface z=h such that the material becomes continuous across the interface, we Consider the following boundary conditions:

)

Hence, the double Fourier transforms of Eqs. (5.07a)-(5.07c) can be obtained as:

β π

β π

The Eqs. (5.09a)-(5.09c) can be simplifed as follow:

β π

The system of three linear equations (Eqs. (5.10a)-(5.10c)) has three undetermined coefficients C , ¹_d2 C and _d²₂ C . These coefficients can be associated with _d³₂

[ ]

fij (i,

Eq. (5.11) can be separated into two parts as:

⎥⎥

⎥⎥ express as:

∑

=

By substiting Eqs. (5.14a)-(5.14b) into Eq. (5.12b), we get

)

The general solution of nonhomogeneous equation are the sum of the general solution

of homogeneous equation and particular solution.

Similarly,

z

5.2 Imaginary Space Method

In order to derive the solutions of Eqs. (3.15a)-(3.15c) in half-space, the Eqs.

(4.19a)-(4.19c) and Eqs. (4.20a)-(4.20c) can be rewritten as:

for z> 0⁺ (region 1, as shown in Fig. 5.2),

Introducing an imaginary plane along z=0, to separate the full-space into two half-spaces, one is 0⁺ <z<∞ (region 1 in Fig. 5.2), and the other is 0 z< <0⁺.

= 0+

z

= 0+

z

) 0 , ,

1(xy +

σzz

) ( ) (x y Pzδ δ

+∞

<

<

+ z

0

=0 z

= 0+

z

< +

< 0 0 z

x

y

z

=0 z

Fig. 5.2 Separate the half-space into two imaginary region of 0 z< <0⁺ and +∞

<

+ <z

0 .

Furthermore, considering the following pertinent continuity and discontinuity conditions at z=0 are:

) ( ) ( ) 0 , ,

(x y P_x x y

zx δ δ

τ ⁺ = (5.19a)

) ( ) ( )

0 , ,

(x y P_y x y

zy δ δ

τ ⁺ = (5.19b)

) ( ) ( ) 0 , ,

(x y P_z x y

zz δ δ

σ ⁺ = (5.19c)

where −∞< x<∞, −∞<y<∞. The subscripts 1 and 2 mean that z=0 plane is approaching to 0⁺ and 0⁻, respectively.

d1

C , C_d², C_d³.can be determined from the following system of three linear equations These coefficients can be associated with

[ ]

fij (i, j =1-3) as:

[ ]

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

=

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

=

⎥⎥

⎥

⎦

⎤

⎢⎢

⎢

⎣

⎡

z y x

d d d

d d d

ij

p p p C

C C f f f

f f f

f f f C C C

f 2π

1

3 2 1

33 32 31

23 22 21

13 12 11

3 2 1

(5.20) where fij (i, j =1-3) are presented in Appendix D.

5.3 Algebraic Equation Method (Double Fourier and Laplace Transforms)

Double Fourier transforms with respect to x and y could reduce the desired problem of solving partial differential equations to ordinary differential equations. Sequentiallt, the Laplace transform could reduce the ordinary differential equations to algebraic equations.

In the half-space problem, u_i(x,y,z) are the displacement components in a homogeneous and linearly elastic continuum with the domain of −∞<x,y<∞ and

∞

<

< z

0 . To solve the equilibrium equations (Eqs. (3.41a)-(3.41c)), the technique of double Fourier transforms for the internal force F_i of Eqs. (3.19a)-(3.19c) are utilized.

Hence, the double Fourier transforms of Eqs. (3.19a)-(3.19c) can be obtained as:

) 2 (

) ( ) ( ) 2 (

1 ( ) P z h

dxdy e

h z y x P

F_x = ^∞ _x − ^{−i x+ y} = ^x −

∞

−

∞

∞

∫ ∫

− ^{δ δ δ} _π^δ

π ^{α β (5.21a)}

) 2 (

) ( ) ( ) 2 (

1 _{( )}

h P z

dxdy e

h z y x P

F_y = ^∞ _y − ^{−i x+ y} = ^y −

∞

−

∞

∞

∫ ∫

− ^{δ δ δ} _π^δ

π ^{α β (5.21b)}

) 2 (

) ( ) ( ) 2 (

1 _{( )}

h P z

dxdy e

h z y x P

F_z = ^∞ _z − ^{−i x+ y} = ^z −

∞

−

∞

∞

∫ ∫

− ^{δ δ δ} _π^δ

π ^{α β (5.21c)}

Based on Eqs. (5.21a)-(5.21c), the Navier-Cauchy equations (Eqs. (3.41a)-(3.41c)) can be rewritten as the following system of linear ordinary differential equations:

) 2 (

) , , ( } ) (

) {(

) , , ( } ) (

) {(

) , , ( } 2

{

55 13 56

14

56 14 66

12

2 56 2 55 2 66 2 11

h P z

z dz u

a d a i a

a

z dz u

a d a i a

a

z dz u

ia d dz a d a

a

z x y

x

−

−

= +

+ +

+

+ + +

+

+

− +

πδ β

α α

αβ

β α α

αβ

β α β

β α

(5.22a)

)

The boundary-value problem for the lower half-space z<0 with the normal and shear stresses applied on the ground surface. When z=0 and −∞< x,y<∞, the stresses of σ_zz(x,y,z), τ_yz(x,y,z), and τ_zx(x,y,z) are equal to zero. Utilizing the boundary conditions mentioned abov, we obtain σ_zz(α,β,0)=τ_yz(α,β,0)=τ_zx(α,β,0)=0 after performing the double Fourier transforms. Additionally, Eqs. (3.42c)-(3.42e) can be rewritten as:

55

The Laplace transform for the component of z-axial displacement is adopted. The

step by step Laplace transforms for u_i(α,β,z)(i=x, y, z) are written as: Based on the aforemenioned transforms, the Eqs. (5.22a)-(5.22c) can be expressed as:

[ ]

) conditions in Eqs. (5.26a)-(5.26c).

The characteristic equation (Eq. (5.28)) with six roots can be expressed as:

)

where the real part of the {s₁, s₂ , s₃} are negative and {s₄, s₅ , s₆} are positive.

The A₁, A₂, and A₃ are the same as presented in Eqs. (3.48a)-(3.48c).

Eq. (5.25) can be rewritten as a system of three equations:

)

u_z α β can be obtained by the following inverse Laplace transform of Eqs.

(5.30a)-(5.30c) as:

ds

ds

where c>0 and the path of integration with respect to s is a vertical line parallel to and on the right of imaginary axis in the complex s plane.

Substituting Eqs. (5.26a)-(5.26c) into Eqs. (5.31a)-(5.31c), then u_x(α,β,z) ,

)

(5.32a)-(5.32c) can be integral by the path of the contour in Fig. 5.3 and expressed as:

s

4

Ims

s

5

Res s

6

s

3

s

2

s

1

C

R

Fig. 5.3 A closed contour on right-hand side of s-plane.

) ( 6 ) ( 5 ) ( 4 )

1

( ( , , ) _xs ^{s4 z h} _xs ^{s5 z h} _xs ^{s6 z h}

x z B e B e B e

u α β =− ⁻ − ⁻ − ⁻ (5.34a)

) 6 ( ) 5 ( ) 4 ( )

1

( ( , , ) _ys ^{s4 z h} _ys ^{s5 z h} _ys ^{s6 z h}

y z B e B e B e

u α β =− ⁻ − ⁻ − ⁻ (5.34b)

) ( 6 ) ( 5 ) ( 4 )

1

( ( , , ) _zs ^{s4 z h} _zs ^{s5 z h} _zs ^{s6 z h}

z z B e B e B e

u α β =− ⁻ − ⁻ − ⁻ (5.34c)

When z>h ,the solutions can be integral by the path of the contour in Fig. 5.4 and expressed as:

s

1

s

2

s

3

s

6

s

5

Res

Ims

s

4

C

R

Fig. 5.4. A closed contour on left-hand side of s-plane.

) ( 3 ) ( 2 ) ( 1 )

1

( ( , , ) _xs ^{s1 z h} _xs ^{s2 z h} _xs ^{s3 z h}

x z B e B e B e

u α β = ⁻ + ⁻ + ⁻ (5.35a)

) 3 ( ) 2 ( ) 1 ( )

1

( ( , , ) _ys ^{s1 z h} _ys ^{s2 z h} _ys ^{s3 z h}

y z B e B e B e

u α β = ⁻ + ⁻ + ⁻ (5.35b)

) ( 3 ) ( 2 ) ( 1 )

1

( ( , , ) _zs ^{s1 z h} _zs ^{s2 z h} _zs ^{s3 z h}

z z B e B e B e

u α β = ⁻ + ⁻ + ⁻ (5.35c)

where

j t

j z

j y

j j x

xs mS

s D

P s D

P s D

B P

π

β α β

α β

β α

α 2

) , , ( )

, , ( )

, , ) (

,

( ¹¹ + ¹² + ¹³

= (5.36a)

j t

j z

j y

j j x

ys mS

s D

P s D

P s D

B P

π

β α β

α β

β α

α 2

) , , ( )

, , ( )

, , ) (

,

( ²¹ + ²² + ²³

= (5.36b)

j t

j z

j y

j j x

zs mS

s D

P s D

P s D

B P

π

β α β

α β

β α

α 2

) , , ( )

, , ( )

, , ) (

,

( ³¹ + ³² + ³³

= (5.36c)

Suppose that u=s and u_j = , since s_j B_xs^j(α,β), B_ys^j (α,β), and B_zs^j(α,β) in Eqs. (5.36a)-(5.36c) have the same forms as B_x^j(α,β), B_y^j(α,β), and B_z^j(α,β) in Eqs. (4.07a)-(4.07c); hence, u_x(1)(α,β,z), u_y(1)(α,β,z), and u_z(1)(α,β,z) in Eqs.

(5.34a)-(5.34c) and Eqs. (5.35a)-(5.35c), and u_x(P)(α,β,z) , u_y(P)(α,β,z) , and )

, ,

)(

( z

u_zP α β in Eqs. (5.05a)-(5.05c), and Eqs. (5.06a)-(5.06c) are the particular solutions of the three displacement functions.

Following, we can find the solutions of the second term in the left-hand side of Eqs.

(5.32a)-(5.32c) should be the general solution of Eqs. (5.04a)-(5.04c) as:

When z>0, the solutions can be integral by the path of the contour in Fig. 5.4 and expressed as:

z s x z s x z s x H

x z Ae Ae Ae

u _{( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (5.37a)

z s y z s y z s y H

y z Ae A e A e

u _{( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (5.37b)

z s z z s z z s z H

z z Ae A e Ae

u _{( )}(α,β, )= ^{1 1} + ^{2 2} + ^{3 3} (5.37c)

where

) 0 , , ( )

0 , , ( )

0 , ,

(α β ⁱ_{xy y} α β _xzⁱ _z α β

x i xx i

x A u A u A u

A = + + (5.38a)

) 0 , , ( )

0 , , ( )

0 , ,

(α β ⁱ_{yy y} α β ⁱ_{yz z} α β

x i yx i

y A u A u A u

A = + + (5.38b)

) 0 , , ( )

0 , , ( )

0 , ,

(α β _zyⁱ _y α β _zzⁱ _z α β

x i zx i

z A u A u A u

A = + + (5.38c)

, and in which:

j

When z=0, Eqs. (5.37a)-(5.37c), and Eqs. (5.04a)-(5.04c) can be stated as:

Then Eq. (5.40) can be rewritten as:

2

The general solutions are the sum of the homogeneous solution (Eqs. (5.37a)-(5.37c)) and particular solutions (Eqs. (5.34a)-(5.34c)) as follows:

h Eqs. (5.41a)-(5.41c) and Eqs.(5.42a)-(5.42c) can be expressed as:

)

solutions of u_x(α,β,0), u_y(α,β,0), and u_z(α,β,0) can be expressed as:

3 (5.12a). Hence, Eqs. (5.31a)-(5.31c) can be rewritten as:

z method of Cramer′s rule.

5.4 Solutions for Displacements and Stresses by Inverse Fourier Transforms

The desired u_x(x,y,z), )u_y(x,y,z , and u_z(x,y,z) can be obtained by performing the double inverse Fourier transforms of Eqs. (5.16a)-(5.16c) as follows:

β

β

τxy also can be acquired by the double inverse Fourier transforms as:

For z>0 (region 1),

) ) (

( _{13 11}^j _{23 21}^j _{34 31}^j _{34 21}^j _{33 31}^j _j

j

zz =−iαa D +βa D +βa D −i a D +a D u

σ (5.51c)

) ) (

( _{14 11}^j _{24 21}^j _{44 31}^j _{44 21}^j _{34 31}^j _j

j

yz =−i αa D +βa D +βa D −i a D +a D u

τ _(5.51d)

)) (

) (

( _{56 11}^j ₂₁^j _{55 31}^j ₁₁^j _j

j

zx =−i a βD +αD +a αD −iD u

τ (5.51e)

)) (

) (

( _{66 11}^j ₂₁^j _{56 31}^j ₁₁^j _j

j

xy =−i a βD +αD +a αD −iD u

τ (5.51f)

In Eqs. (5.51a)-(5.51f), j=1-3.

If we take a spherical co-ordinate system, which is shown in Fig. 4.4, the variables α, β, and ui can be expressed in the Eqs. (4.44a)-(4.44b) and Eqs. (4.45a)-(4.45b).

According to Eqs. (3.53a)-(3.53f), D₁₁^j , D₂₁^j , D₃₁^j (j=1-6) and C_d^j (j=1-3) in Eqs.

(5.49a)-(5.49c) and Eqs. (5.50a)-(5.50f) can be presented in terms of k and θx as:

) ( )

,

( ⁴ ₁₁

11 j x

x

j k k D

D θ = θ (5.52a)

) ( )

,

( ⁴ ₂₁

21 j x

x

j k k D

D θ = θ (5.52b)

) ( )

,

( ⁴ ₃₁

31 x

j x

j k k D

D θ = θ (5.52c)

) ( )

,

( _x ⁵ _d^j _x

j

d k k C

C θ = ⁻ θ (5.52d)

Hence, Eqs. (5.51a)-(5.51f) also can be rewritten as:

) ( )

,

( _x ⁵ _xx^j _x

j

xx k θ k σ θ

σ = (5.53a)

) ( )

,

( _x ⁵ _yy^j _x

j

yy k θ k σ θ

σ =

(5.53b) )

( )

,

( _x ⁵ _zz^j _x

j

zz k θ k σ θ

σ = (5.53c)

) ( )

,

( _x ⁵ _yz^j _x

j

yz k θ k τ θ

τ = (5.53d)

) Based on Eqs. (4.44a)-(4.44b), Eqs. (4.45a)-(4.45b), Eq. (4.49), Eq. (4.52), Eqs.

(5.52a)-(5.52d) and Eqs. (5.53a)-(5.53f), then Eqs. (5.49a)-(5.49c) and Eqs.

(5.50a)-(5.50f) can be represented as:

For z>0 (region 1),

x Eqs. (5.54a)-(5.54i) can be expressed as:

For z>0 (region 1),

ω ω

) ( ) ( )

( _{2 5}

8 ω ψ ω ψ ω

ψ = × (5.56b)

) ( ) ( )

( _{3 6}

9 ω ψ ω ψ ω

ψ = × (5.56c)

CHAPTER Ⅵ

ILLUSTRATIVE EXAMPLES

The present analytical solutions demonstrate that there are several factors could affect the displacements and stresses in an inclined transversely isotropic material. These

The present analytical solutions demonstrate that there are several factors could affect the displacements and stresses in an inclined transversely isotropic material. These