A method for solving rectangular beams and discs = Metoda rozwiązywania prostokątnych ortotropowych belek i tarcz

No. 1 2005

A METHOD FOR SOLVING RECTANGULAR BEAMS AND DISCS

Maria OLEJNICZAK Department of Structural Mechanics,

Faculty of Civil and Environmental Engineering ATR, prof. S. Kaliskiego St. 7, 85-796 Bydgoszcz, Poland

An analytic method for solving two-dimensional elasticity problems of orthotropic body has been worked out. A solution for rectangular elements of the beam or disc type has been developed. This solution satisfies the fundamental equations of the elasticity theory of two-dimensional body exactly and the boundary conditions with large accuracy.

Keywords: thick plate, beam, disc, orthotropy, analytic method

1. FORMULATION OF THE PROBLEM

Beams and discs are widely used as structural members. Depending on the kind of reinforcement they may divided into two groups: isotropic or orthotropic. The former comprise all homogenous members like, for instance, steel elements. The reinforced members may be treated as orthotropic with varying degree of orthotropy. Typical examples are beams and discs made of wood, composite fibres and concrete.

Fig.1. A beam as a plate strip 2a3

2a1

2a2

x 3

x1

x2

© University of Zielona Góra Press, Zielona Góra 2005 ISBN 83-89712-71-7

This work focuses on development of a method for solving of orthotropic structural members: beams and discs. Beams are treated as strips cut out of a thick orthotropic plate subject to cylindrical bending (fig. 1). Under axial stretch we face the membrane state.

Equilibrium of such a strip, treated as a beam or a disc depending on the way it is loaded, is governed by a system of two homogeneous differential equa- tions [2,5]:

(

_{31 55}

)

_{1,31 33 3,33} 0

11 , 3

55u + b +b u +b u =

b , (1)

(

13 55

)

3,31 55 1,33 ⁰ 11

, 1

11u + b +b u +b u =

b ,

where: u_j (j=1,3) are the components of the displacement vector and )

5 , 3 , 1 , (k l=

b_kl correspond to the components of the stiffness tensor for an orthotropic material. Specifically, in (1) the following convention has been adopted: b₁₁=b₁₁₁₁, b₁₃=b₁₁₃₃=b₃₁, b₅₅=b₁₃₁₃, b₃₃=b₃₃₃₃.

2. METHOD OF SOLUTION

Following the approach taken in [1,4], we assume the solution to a non- symmetrical problems in the form:

{

∑

^∞

=

+ +

=

1

3 ] 3 [

) ( 1 ] 1 [

) ( 1 3 ] 3 [

) ( 1 ] 1 [

) ( 2

1 ( )cos( ) ( )sin( )

m

m m

m

m x x f x x

f

u δ δ

}

) sin(

) ( ) cos(

)

( _{3 (}^[1]_{) 1 1}^[₍^3]_{) 3 (}^[1]_{) 1}

] 3 [

) (

2 x x f x x

f _m δ_m + _m δ_m

+ ,

{

∑

^∞

=

+ +

=

1

3 ] 3 [

) ( 1 ] 1 [

) ( 3 3 ] 3 [

) ( 1 ] 1 [

) ( 4

3 ( )cos( ) ( )sin( )

m

m m

m

m x x f x x

f

u δ δ (2)

}

) sin(

) ( ) cos(

)

( _{3 (}^[1]_{) 1 3}^[₍^3]_{) 3 (}^[1]_{) 1}

] 3 [

) (

4 x x f x x

f _m δ_m + _m δ_m

+ ,

where the functions: ^{f[ ]}_{( )}

( )

^xj ^{, p=1,4} j

m

p are the unknowns and: ^{[ ]}

( )

j j

m a

m 2

1

2 π

δ = ⁻

the parameters.

Substituting (2) to (1), and separating the variables, we obtain a system of ordinary differential equations for unknown functions f_p^[₍^j_m^]₎(x_j):

( ) ( ) ( )

{

∑

=

−

−

− + + +

3

1

) 3 ]( [

) ( 3 ] [

) ( )

3 ]( [

) ( 2 ] [

) ( )

3 ]( [

) ( 1 ] [

) ( r

j j r m j

m kr j j r m j

m kr j j r m j

m

kr f x B f x C f x

A

( ) }

⁰

) 3 ]( [

) ( 4 ] [

)

( =

+D_kr^j_m f ^j_m ^−r x_j , k=1,4 ,r=1,3. (3)

The coefficients: A_kr^[j₍^]_m),B_kr^[j₍^]_m),C_kr^[j₍^]_m),D^[_kr^j₍^]_m) in (3) can be expressed via the components of the stiffness tensor for the orthotropic material and appropriate parameters:

[ ]( )

[ ]( ) _,

, ₁₃¹ ₅₅ ^[₍^3]₎²

11 1

11m b A m b m

A = =− δ

[ ]( )

[ ]( )

[ ]( ) ,

, ) (

, ₂₂¹ _{13 55 (}^[3]_{) 23}¹ _{55 (}^[3]₎²

11 1

21m b C m b b m B m b m

B = = + δ =− δ

[ ]( )

[ ]( )

[ ]( )

, ,

) (

, ₃₂¹ _{13 55 (}^[3]_{) 33}¹ _{33 (}^[3]₎²

55 1

31m b B m b b m C m b m

C = =− + δ =− δ

[ ]( )

[ ]( )

[ ]( )

, ,

) (

, ₄₂¹ _{13 55 (}^[3]_{) 43}¹ _{33 (}^[3]₎²

55 1

41m b A m b b m D m b m

D = = + δ =− δ (4)

[ ]( )

[ ]( )

[ ]( ) ,

, ) (

, ₁₂³ _{13 55 (}^[1]_{) 13}³ _{11 (}^[1]₎²

55 3

11m b D m b b m A m b m

A = =− + δ =− δ

[ ]( )

[ ]( )

[ ]( )

, ,

) (

, ₂₂³ _{13 55 (}^[1]_{) 23}³ _{11 (}^[1]₎²

55 3

21m b C m b b m B m b m

B = = + δ =− δ

[ ]( )

[ ]( )

[ ]( )

, ,

) (

, ₃₂¹ _{13 55 (}^[1]_{) 33}¹ _{55 (}^[1]₎²

33 1

31m b B m b b m C m b m

C = =− + δ =− δ

[ ]( )

[ ]( )

[ ]( ) .

, ) (

, ₄₂³ _{13 55 (}^[1]_{) 43}³ _{55 (}^[1]₎²

33 3

41m b A m b b m D m b m

D = = + δ =− δ

The remaining parameters have zero values.

We seek the following particular solution to (3):

( ) [

j

]

j m j

m p j j m

p x R x

f^[₍^]₎ = ^[₍^]₎expλ₍^{[ ]}₎ , (5) where R_p^[₍^j]_m) are unknown parameters. The quantities λ^[₍^j_m^]₎ are the roots of the characteristic equation [4]:

, 0 )

( ) )(

(b₁₁λ^[₂¹₍^]_m²₎−b₅₅δ₍^[_m^3]₎² b₅₅λ^[₂¹₍^]_m²₎−b₃₃δ₍^[_m^3]₎² + b₁₃+b₅₅ ²δ₍^[_m^3]₎²λ^[₂¹₍^]_m²₎ = , 0 )

( ) )(

(b₅₅λ^[₂³₍^]_m²₎−b₁₁δ₍^[_m^1]₎² b₃₃λ^[₂³₍^]_m²₎−b₅₅δ₍^[1_m^]₎² + b₁₃+b₅₅ ²δ₍^[1_m^]2₎λ^[₂³₍^]_m²₎= (6)

Each of the above equations is a fourth-order equation (biharmonic equa- tion) and, therefore, each has four roots (two positive and two negative):

[ ]( )

[ ]( ) [ ]( )

[ ]( ).

, _{2 4}

3 1

j m j

m j

m j

m λ λ λ

λ =− =− (7) Numerical computations show that for orthotropic material the roots λ^[₍^j_m^]₎ are all real.

General solution of the system takes the form:

∑

=

ν ν λν

=

4 1

] [

) ( ]

[ ) ( ]

[ )

( ^j exp ^j_{m j}

p m j

m

p R x

f . (8)

Observe that within the set

{

Rp^[₍^j]_m)

}

ν only two groups of coefficients R_1ν^[j₍^]_m) and

] [

) 2 (

j

R_ν m are linearly independent. All the remaining coefficients are linearly re- lated by the equations:

] [

) ( 2 ] [

) ( 2 ] [

) ( 3

j m v j

m v j

m

v R K

R = , R₄^[_v^j]_(m) =R₁^[_v^j₍^]_m)K₁^[_v^j₍^]_m). (9) In the above:

] 1 [ ] 1 1 [ 55 13

55 ] 1 [ 11 ] 3 [

) ( ] 1 [

) ( 55 13

] 3 [

) ( 55 ] 1 [

) ( ] 11

1 [

) (

1 ( ) ( )

2 2 2

v v v m

m v

m m

v m

v K

b b

b b

b b

b

K b =

+

= − +

= −

γ γ δ

λ δ

λ ,

] 1 [ ] 2 1 [ 55 13

55 ] 1 [ 11 ]

3 [

) ( ] 1 [

) ( 55 13

] 3 [

) ( 55 ] 1 [

) ( ] 11

1 [

) (

2 ( ) ( )

2 2 2

v v v m

m v

m m

v m

v K

b b

b b

b b

b

K b =

+

− − + =

− −

= γ

γ δ

λ δ

λ , (10)

] 3 [ ] 1 3 [ 55 13

11 ] 3 [ 55 ] 1 [

) ( ] 3 [

) ( 55 13

] 1 [

) ( 11 ] 3 [

) ( ] 55

3 [

) (

1 ( ) ( )

2 2 2

v v v m

m v

m m

v m

v K

b b

b b

b b

b

K b =

+

= − +

= −

γ γ δ

λ δ

λ ,

] 3 [ ] 2 3 [ 55 13

11 ] 3 [ 55 ]

1 [

) ( ] 3 [

) ( 55 13

] 1 [

) ( 11 ] 3 [

) ( ] 55

3 [

) (

2 ( ) ( )

2 2 2

v v v m

m v

m m

v m

v K

b b

b b

b b

b

K b =

+

− − + =

− −

= γ

γ δ

λ δ

λ ,

where: γ_ν^{[ ]j} =λ_ν^{[ ]j}_{( )m} /δ_m^{[ ]j} .

Making use of (2) and (8) we can determine the components of the displacement vector as:

[ ] [ ]

{

[

^{( )sin( )}

] [

^{( )cos( )}

] }

^,

) cos(

) ( )

sin(

) (

1 ] 1 [ 3 ] 3 [

) ( 2 ] 3 [

) ( 2 1 ] 1 [ 3 ] 3 [

) ( 1 ] 3 [

) ( 1 1

4

1

3 ] 3 [ 1 ] 1 [

) ( 2 ] 1 [

) ( 2 3 ] 3 [ 1 ] 1 [

) ( 1 ] 1 [

) ( 1 1

x x

U R x x

U R

x x

U R x x

U R u

m m

m m

m m m

m m

m m

m m

δ δ

δ δ

ν ν ν

ν ν

ν ν ν

ν

+ +

+ +

=

∑∑

^∞

= = (11)

[ ] [ ]

{

[

^{( )cos( )}

] [

^{( )sin( )}

] }

^.

) sin(

) ( )

cos(

) (

] 1 1 3 [ ] 3 [

) ( 2 ] 3 [

) ( 1 2

] 1 3 [

] 3 [

) ( 1 ] 3 [

) ( 1 1

4

1

] 3 3 1 [ ] 1 [

) ( 2 ] 1 [

) ( 3 2

] 3 1 [

] 1 [

) ( 1 ] 1 [

) ( 3 1

x x

W R x x

W R

x x

W R x x

W R u

m m m m

m m m

m m m m

m m

δ +

δ +

+ δ +

δ

=

ν ν ν

ν

∞

= =ν

ν ν ν

∑∑

ν

We introduced the following notation in (11):

( )







=

= =

− , 3,4

cosh

, 2 , 1 , sinh

)

( [ ]

) ( 2 ] [

) ] (

[ )

( λ ν

ν λ

ν ν ν

j j

m j j

m j

j m

p x

x x

U

( )

. ,

4 , 3 , sinh

1,2 p

, 2 , 1 , cosh

) (

] [

) ( 3 ] [

) ( ]

[ ) ( ]

[ ) (











= ν λ

=

= ν λ

=

− ν ν ν

ν

j j

m j j

m j

m j p

j m p

x x K

x

W . (12)

The components of the strain tensor follow from the kinematic relations:

(

ij ji

)

ij u_, u_, 2

1 +

ε = (13) Substituting (11) to the above, we derive the formulas for the components of the strain tensor:

[ ]

( )

^{cos( )}

[

^{( )sin( )}

]

^,

) cos(

) ( )

sin(

) (

1 ] 1 [ 3 ] 3 [

) ( 2 ] 1 [ ] 3 [

) ( 2 3 ] 3 [ 1 ] 1 [

) ( 2 ] 1 [

) ( 2

1 ] 1 [ 3 ] 3 [

) ( 1 ] 1 [ 1

4

1

] 3 [

) ( 1 3 ] 3 [ 1 ] 1 [

) ( 1 ] 1 [

) ( 1 11



− 

+

 

 ′

+

 +



+

 

 ′

=

∑∑

^∞

= =

x x

U R

x x

U R

x x

U R

x x

U R

m m

m m m

m m

m m

m m

m m

m m

δ δ

δ

δ δ

δ ε

ν ν

ν ν

ν ν

ν ν

ν

[ ]

{

[

^{( )cos( )}

]

^{( )sin( ) ,}

) cos(

) ( )

sin(

) (

1 ] 1 [ 3 ] 3 [

) ( 2 ] 3 [

) ( 2 1 ] 1 [ 1 ] 1 [

) ( 2 ] 3 [ ] 1 [

) ( 2

1 ] 1 [ 3 ] 3 [

) ( 1 1

4

1

] 3 [

) ( 1 3 ] 3 [ 1 ] 1 [

) ( 1 ] 3 [ ] 1 [

) ( 1 33







 ′ +

+

+

 

 ′

+

−

=

∑∑

^∞

= =

x x

W R x x

W R

x x

W R x x

W R

m m

m m

m m m

m m

m

m m

m m m

δ δ

δ

δ δ

δ ε

ν ν ν

ν

ν ν

ν ν

ν

. ) sin(

) ( )

(

) cos(

) ( )

(

) sin(

) ( )

(

) cos(

) ( )

(

1 ] 1 [ 3

] 3 [

) ( 2 ] 1 [ 3 ] 3 [

) ( 2 ] 3 [

) ( 2

3 ] 3 [ 1

] 1 [

) ( 2 1 ] 1 [

) ( 2 ] 3 [ ] 1 [

) ( 2

1 ] 1 [ 3

] 3 [

) ( 1 ] 1 [ 3 ] 3 [

) ( 1 ] 3 [

) ( 1 1

4

1

3 ] 3 [ 1

] 1 [

) ( 1 1 ] 1 [

) ( 1 ] 3 [ ] 1 [

) ( 1 13







 



 



 



 ′ −

+

+



 



 



 



 ′

+ +

+



 



 



 



 ′ −

+



 +



 



 



 



 ′

+

=

∑∑

^∞

= =

x x

W x

U R

x x

W x U R

x x

W x

U R

x x

W x U R

m m

m m

m

m m

m m m

m m

m m

m m

m m

m m m

δ δ

δ δ

δ δ

δ δ

γ

ν ν

ν

ν ν

ν

ν ν

ν ν

ν ν

ν

(14)

Let us introduce the following notation:

) ( )

( _{1 1}^[1]_{( ) 1}

] 1 [

) (

1 x U x

G _{m m}′

= _ν

ν , G₁^[_ν³₍^]_m)(x₃)=δ_m^[1]U₁^[_ν^3]_(m)(x₃), )

( )

( ₁ ^[₂^1]_{( ) 1}

] 1 [

) (

2 x U x

G _{m m}′

= _ν

ν , G₂^[_ν^3]_(m)(x₃)=−δ_m^[1]U₂^[_ν^3]_(m)(x₃), )

( )

( _{3 1}^[3₍^] _{) 3}

] 3 [

) (

1 x W x

H _{m m}′

= _ν

ν , H₁^[_ν^1]_(m)(x₁)=−δ_m^[3]W_1ν^[1₍^]_m)(x₁),

) ( )

( _{3 2}^[3]_{( ) 3}

] 3 [

) (

2 x W x

H _{m m}′

= _ν

ν , H₂^[1_ν^]_(m)(x₁)=δ_m^[3]W₂^[_ν^1]_(m)(x₁), (15)

( )

₁ ^[3] ₁^[1]_{( )}( ₁) ₁^[1₍^] ₎ ( ₁)

] 1 [

) (

1 x U x W x

T _{m m m m}′

+

= _{ν ν}

ν δ ,

( )

_{3 1}^[3₍^] ₎( ₃) ^[1] ₁^[3₍^] ₎( ₁)

] 3 [

) (

1 x U x W x

T_{ν m ν m}′ −δ_{m ν m}

=

( )

₁ ^[3] ₂^[1]_{( )}( ₁) ₂^[1]_{( )}( ₁),

] 1 [

) (

2 x U x W x

T _{m m m m}′

+

= _{ν ν}

ν δ

( )

₃ ^[₂^3]_{( )}( ₃) ^[1] ₂^[3₍^] ₎( ₃)

] 3 [

) (

2 x U x W x

T_{ν m ν m}′ −δ_{m ν m}

= .

Then the functions determining the components of the strain tensor take the form:

[ ]

{ [ ]

[ ( )

^{cos( )}

] [

^{( )sin( )}

] }

^,

) cos(

) ( )

sin(

) (

1 ] 1 [ 3 ] 3 [

) ( 1 ] 3 [

) ( 2 3 ] 3 [ 1 ] 1 [

) ( 2 ] 1 [

) ( 2

1 ] 1 [ 3 ] 3 [

) ( 1 1

4

1

] 3 [

) ( 1 3 ] 3 [ 1 ] 1 [

) ( 1 ] 1 [

) ( 1 11

x x

G R x x

G R

x x

G R x x

G R

m m

m m

m m

m m

m

m m

m m

δ δ

δ δ

ε

ν ν ν

ν

ν ν

ν ν

ν

+ +

+ +

=

∑∑

^∞

= =

(16)

[ ]

{ [ ]

[

^{( )cos( )}

] [

^{( )sin( )}

] }

^,

) cos(

) ( )

sin(

) (

1 ] 1 [ 3 ] 3 [

) ( 2 ] 3 [

) ( 2 1 ] 1 [ 1 ] 1 [

) ( 2 ] 1 [

) ( 2

1 ] 1 [ 3 ] 3 [

) ( 1 1

4

1

] 3 [

) ( 1 3 ] 3 [ 1 ] 1 [

) ( 1 ] 1 [

) ( 1 33

x x

H R x x

H R

x x

H R x x

H R

m m

m m

m m

m m

m

m m

m m

δ δ

δ δ

ε

ν ν ν

ν

ν ν

ν ν

ν

+ +

+ +

=

∑∑

^∞

= =

[ ]

{ [ ]

( )

[

^{( )cos( )}

] [

^{( )sin( )}

] }

^,

) sin(

) ( )

cos(

) (

1 ] 1 [ 3 ] 3 [

) ( 2 ] 3 [

) ( 2 3 ] 3 [ 1 ] 1 [

) ( 2 ] 1 [

) ( 2

1 ] 1 [ 3 ] 3 [

) ( 1 ] 3 [

) ( 1 1

4

1

3 ] 3 [ 1 ] 1 [

) ( 1 ] 1 [

) ( 1 13

x x

T R x x

T R

x x

T R x x

T R

m m

m m

m m

m m

m m

m m

m

δ δ

δ δ

γ

ν ν ν

ν

ν ν ν

ν ν

+ +

+ +

=

∑∑

^∞

= =

The components of the stress tensor follow from the constitutive rela- tions:

kl ijkl ij b ε

σ = (17) Substituting the right-hand sides of (16) to (17), we obtain:

[ ]

{ [ ]

[ ( )

^{cos( )}

] [

^{( )sin( )}

] }

^,

) cos(

) ( )

sin(

) (

1 ] 1 [ 3 ] 3 [

) ( 1 ] 3 [

) ( 2 3 ] 3 [ 1 ] 1 [

) ( 2 ] 1 [

) ( 2

1 ] 1 [ 3 ] 3 [

) ( 1 1

4

1

] 3 [

) ( 1 3 ] 3 [ 1 ] 1 [

) ( 1 ] 1 [

) ( 1 11

x x

Y R x x

X R

x x

X R x x

X R

m m

m m

m m

m m

m

m m

m m

δ δ

δ δ

σ

ν ν ν

ν

ν ν

ν ν

ν

+ +

+ +

=

∑∑

^∞

= =