Second Order Nonlinear Analysis of Steel Tapered Beams Subjected to

Span Loading

Ali Hadidi1, Bahman Farahmand Azar, Hossein Zonouzi Marand

Department of civil engineering, University of Tabriz, Tabriz, Iran

Abstract

A second-order elastic analysis of tapered steel members with I-shaped sections subjected

to span distributed and concentrated loadings is developed. Fixed-end forces and moments as

well as exact stiffness matrix of tapered Timoshenko-Euler beam are obtained with exact

geometrical properties of sections. The simultaneous action of bending moment, shear and

axial force including P effects is also considered in the analysis. A computer code has

been developed in MATLAB software using a power series method to solve governing

second-order differential equation of equilibrium with variable coefficients for beams with

distributed span loading. A generalized matrix condensation technique is then utilized for

analysis of beams with concentrated span loadings. The accuracy and efficiency of the results

of the proposed method are verified through comparing to those obtained from other

1 Corresponding Author: A. Hadidi

Department of civil engineering, University of Tabriz, Tabriz, Iran

Tel: +98-411-339 25 15, Email: alihadiditabriz@gmail.com

approaches such as finite element methods which indicates the robustness and time saving of

this method even for large scale frames with tapered members.

Keywords: Tapered beams; Span Loadings; Fixed end forces and moments; Stiffness

matrix; Second order analysis

1. Introduction

Tapered members are used in many structures such as sloped frames, bridges and

multi-story buildings as well as mechanical components. Because of the ease of construction,

the use of I-shaped members with a linearly variable depth of web is more practical;

moreover, weight optimization, flexibility in fabrication and design and sometimes satisfying

architectural considerations are some advantages which rolled members cannot provide.

Therefore, an exact analysis of structures containing this kind of members is very important.

On the other hand, according to the AISC-2010 [1], the required strengths of components of

structures should be determined by using a second-order analysis method. This kind of

analysis can consider flexural, shear and axial member deformations and both P and

P effects, and all gravity and other applied loads that may influence the stability of the

structure. Consequently, a direct analysis method for design of structures containing tapered

members which are subject to nodal and span loading is of high significance.

In addition to classic methods, in recent decades, different approaches have been

developed for second-order and buckling analysis of tapered members. Timoshenko and

Young [2] suggested to divide the tapered members to sub-elements for analyzing the

structure. Despite the simplicity of this method, they do not have the required precision for an

exact analysis (as illustrated in example 1).

Numerical methods such as finite element method (e.g. Bathe [3]) and direct integral

method (e.g. Karabalis [4]) have been used for producing stiffness matrix of member. These

methods achieve relatively good precision by choosing fine mesh in modeling; however, it is

time-consuming highly due to its great number of finite elements, especially in the case of

large structures. Therefore, these methods are not conducive to daily use because of high

computational costs.

AlGahtani [5] obtained axial, torsional and flexural stiffness matrix separately, as well

as fixed end forces and moments through solving governing equilibrium equations of member

based on the boundary integral method. This study considers the span loadings on tapered

beams, however, it ignores the second-order effects associated with P and simultaneous

action of shear and bending deformation.

Li et al. [6-8] have solved tapered Timoshenko–Euler beam element using Chebyshev

Polynomial approach considering simultaneous effects of axial force, shear deformation and

P effects with nodal loads. The exact stiffness matrix is derived based on beam-column

theories taking into account second-order effects in governing equilibrium differential

equation. This approach contains less computational effort and is also more time-saving and

practical in advanced analysis of structures; nevertheless, span loadings have not been

considered in this method despite their importance in practical analysis.

This paper, alternatively, takes into account not only nodal loads but also distributed

and concentrated span loads. A practical method for second-order analysis of steel tapered

members is presented considering the simultaneous effect of bending moments, axial forces

and shear deformations (including P ). To fulfill this purpose, a power series method is

used to solve the equilibrium differential equation to obtain exact elastic stiffness matrix,

fixed end forces and moments. In the offered examples, the effect of axial pressure forces and

shear deformations is clearly observable in reduction of element stiffness. Although, only

elements with I-section and linear variation of depth is considered in this study; nevertheless,

the proposed method can be developed for other states. The algorithm presented in this paper

can be used for direct analysis method considering requirements based on AISC code [1].

2. Derivation of element formulation

The general form of members considered in this paper is shown in Fig. 1. which

reveals linearly symmetric tapered web and I-shaped section with constant width and

thickness of flanges. It is assumed that the element is braced laterally while local buckling of

the web and flanges is not taken into account. All elements are initially straight and the cross-

sections of the beam remain plane after deformation. Stiffness matrix and fixed end forces

and moments have been obtained for tapered members in this section. The element

formulation is based on the exact moment of inertia of section with no approximation.

Fig. 1. I-shaped section members with tapered web.

2.1. Geometrical properties of section

The exact formulation of area and inertia moment of the cross-section at location z as

shown in Fig. 1. are obtained as follows:

( ) ( )2 . ( 2 ).f f f wA b t D t t z z (1)

( )3 3 2

( ) ( )

1 1. 2 .( ) 2 .( ).( )

12 12 2

f

w f f w f f w

D tI t D t b t t b t

z

z z (2)

( ) ( ).w wA D tz z (3)

where ( ) 1 .D D s z z , 2 1D Ds

L

and ( )wA z is web area at location z and equal with

overall depth time web thickness (AISC-2010 [1]). In the most of existing researches,

approximate formulation has been used for calculation of inertia moment. This approximation

creates considerable error in structural analysis results including displacements and forces.

Some approximate formulas are shown in Appendix A.

2.2. Members subject to uniform distributed span loading

Tapered beam-column element of this research including uniform distributed span

loadings and applied conventions is shown in Fig. 2.

Fig. 2. Tapered beam of the paper, the used conventions and uniform span loadings

In deriving stiffness matrix and fixed end forces and moments of tapered members, the

simultaneous effect of bending moment and shear and axial force is considered. In so doing,

the equilibrium differential equation of the tapered Timoshenko-Euler beam element is

established, similar to Li at el. [6], in order to compare obtained relations. Eq. (4) is the

governing equation for the equilibrium of tapered Timoshenko–Euler beams without span

loadings (Li et al. [6-8]):

( ) ( ) ( ) 1 1 1. . . . . ( . )y N y N y Q M Q z z z z 0 L z (4)

Where ( ) ( ) ( ). .E I z z z , ( )

( ) ( ) 2

( )

. ..

w

w

AE I

G A

z

z z

z

, ( )

( )

1. w

N

G A z

z

E and G respectively are elastic and shear modulus. In this section, the governing equilibrium

Eq. (4) is re-established taking into account uniform span loadings.

Element deflection consists of two sections, one is induced by the bending deformation, the

other by shear deformation; moreover, the axial deformation due to axial forces is taken into

account separately:

M Qy y y (5)

Second derivation of Eq. (5) Can be expressed as follows:

M Qy y y (6)

Substituting My and Qy terms by relations obtained from Eq. (7) and rearranging terms for

y , y and y , the governing equation for the equilibrium of tapered Timoshenko–Euler

beam is obtained as Eq. (8).

1 1

1 1

1

. . , ,

., ,

. .

. . 1. .( . )

. .

s

M Q

w

s w

s s

w w

dMM M Q M N y Q

d

M Q Ay y

E I G A A

M M Q N y Ay M N y Q M N y

E I G A A

( z)( z) ( z) ( z)

( z) ( z) ( z)

( z) ( z) ( z)

( z) ( z)( z) ( z)

( z) ( z) ( z)

zz

z

(7)

( )

( ) ( ) ( ) 1 ( ) ( ) 1 1 ( )

( )

. . . . . ( ) . ( . )

0

w

s s s

w

Ay N y N y Q M M M Q M

A

L

z

z z z z z z

z

z

z

(8)

Where sM ( z) is the moment created by span loadings at distance z from the left end of the

element and ( ) ( ) ( ). .E I z z z , ( )

( ) ( ) 2

( )

. ..

w

w

AE I

G A

z

z z

z

, ( )

( )

1. w

N

G A z

z

.

For converting Eq. (8) to non-dimensional form Let L

z

:

( )2 2 2

( ) ( ) ( ) 1 ( ) ( ) 1 1 ( )

( )

. . . . . . . ( ) . .( . . )

0 1

w

s s s

w

Ay L N y L N y L Q M M L M Q L M

A

(9)

Eq. (9) can be used for all tapered members with any sections and all forms of span loadings.

Using Eq. (10) for uniform span loadings ( q as Fig. 2.) and I-shaped section, Eq. (9) is

converted to Eq. (11), namely:

2

1

1. , .

2

( . ). , .

s s s

w w w w

M q M q M q

A D s t A s t

( z) ( z) ( z)

( z) ( z)

z z,

z

(10)

2 2 2 2 21( ) ( ) ( ) 1 1 1

1. . . . . . . .( . ) .( . . . . )

2

Dy L N y L N y L Q q L M Q L q L

s (11)

To solve differential equation Eq. (11), power series method is used for y , and :

( )

0

( )

0

( )

0

.

.

.

Mn

n

n

Mn

n

n

Mn

n

n

y y

(12)

Differentiating Eq. (12) term by term with respect to and substituting into Eq. (11), we

arrive at the result and after a further differentiation we have:

2

2 1

0 0 0 0 0

2 2 3 4 211 1 1

0

( 2 )( 1 ) . . . ( 1 ) . . . .

1.( . ). . . . . . .

2

M n M n Mn n n

i n i i n i n

n i n i n

Mn

n

n

n i n i y L N n i y L N y

DL Q q L M L Q q L

s

(13)

Separating out the terms corresponding to 0n , 1n and 2n , and then the equality of

relevant coefficients in two side of Eq. (13), it becomes:

For 0n ,

2 2 2

0 2 0 1 0 0 1 1

12 . . . . . . . ( . ) .

Dy L N y L N y L Q q L M

s (14-1)

For 1n ,

2 2 310 3 1 2 0 2 1 1 1 1 1 16 . 2 . . (2 . . ) . . . ( . ) .

Dy y L N y y L N y L Q q L Q

s (14-2)

Finally for 2n ,

2

0 2 2 1

1 0

2 3 411 1

( 2)( 1) ( 2 )( 1 ) . . ( 1 ) . .

1 2. ( . ) 2 . . .

2

n n

n i n i i n i n

i i

n n

n n y n i n i y L N n i y L N y

DL Q q q L q L

s n

(15)

Where 2

1n

, for 2n and

20

n

, for 2n .

Recurrence relation for 2n using Eq. (15) can be written as:

2 2 3 412 1 1 1

0

2 0

1

1 2. . ( 1 ) . . . ( . ) 2 . . .

2

( 2 )( 1 ) / ( 2)( 1)

n

n i n i n n n

i

n

i n i

i

Dy L N n i y L N y L Q q q L q L

s n

n i n i y n n

(16)

Thus, considering Eq. (16), it can be concluded that any ny (for 4n ) can be expressed in

linear combination of 0y , 1y , 2y , 3y , 1Q and the constant value of 5c as Eq. (17).

0 0 1. 1 2 2 3 3 4 1 5. . . . .y c y c y c y c y c Q c (17)

Hereafter, these parameters must be obtained to determine y for different values of z . It

should also be mentioned that 1M is unknown and appears in ny (for 4n ). From Eq. (7),

and considering: M Qy y y and My ,

1 . .( )

( ) . ( )w

Q q LLy

G A

(18)

Using boundary condition and Eqs. (17,18),

For 0 ,

0(0) 0y y (19)

11 1(0)

(0) . (0)w

QLy y

G A

(20)

For 1 ,

1 1 2 2 3 3 4 1 5 2 1(1)y c y c y c y c Q c (21)

16 1 7 2 8 3 9 1 10 2(1)

(1) . (1)w

QLy c y c y c y c Q c

G A

(22)

Using first relation of Eq. (12), separating out the terms corresponding to 4n , collecting all

the remaining terms under a single summation sign and considering Eq. (21) and (22):

1 2 3 1 1 2 2 3 3 4 1 5

4

1 2 3 6 1 7 2 8 3 9 1 10

4

(1)

(1) 2 3

m

i

i

m

i

i

y y y y y c y c y c y c Q c

y y y y iy c y c y c y c Q c

(23)

Applying below conditions to Eq. (23), coefficients 1c to 10c are derived:

(0) (0)

1 2 3 1 5 10

4 4

(1) (1)

1 2 3 1 1 5 6 10

4 4

(2) (2)

2 1 3 1 2 5 7 10

4 4

(0) : 0 : ,

(1) : 1, 0 :1 , 1

(2) : 1, 0 :1 , 2

m m

i i

i i

m m

i i

i i

m m

i i

i i

Condition y y y Q y c iy c

Condition y y y Q y c c iy c c

Condition y y y Q y c c iy c c

Conditio

(3) (3)

3 1 2 1 3 5 8 10

4 4

(4) (4)

1 1 2 3 4 5 9 10

4 4

(3) : 1, 0 :1 , 3

(4) : 1, 0 : ,

m m

i i

i i

m m

i i

i i

n y y y Q y c c iy c c

Condition Q y y y y c c iy c c

(24)

Now Eqs. (14-1),(14-2),(19),(20),(21) and (22) can be used for obtaining unknown factors

0y , 1y , 2y , 3y , 1Q and 1M in order to find function ( )y z for all values of z . 2Q and 2M can

be found from equilibrium of element as,

2 1 .Q Q q L (25)

2

2 1 1 2 1

1. .( ) .

2M M Q L N q L (26)

On the other hand axial stiffness matrix of element can be derived separately as,

[ ]a a

a

a a

k kK

k k

(27)

where

0

1

( )

a L

Ek

dA

zz

Solving the obtained relations for 1 , 1 , 2 and 2 yields stiffness matrix and fix end forces

and moments in the following form as,

FP K P (28)

where

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

0 0 0 0

0 0

0 0[ ]

0 0 0 0

0 0

0 0

a a

a a

k k

k k k k

k k k kK

k k

k k k k

k k k k

,

1

1

1

2

2

2

{ }

u

u

,

1

1

2

2

0

{ }0

F

F

F

F

F

Q

MP

Q

M

and

1

1

2

2

{ }

N

Q

MP

N

Q

M

(29)

It should be mentioned that by substituting Eqs. (25,26) into Eq. (28), we have,

31 32 33 34 11 12 13 14

41 42 43 44 21 11 22 12 23 13 24 14

2

2 1 2 1 1

. . . .

1. , . .

2

F F F F F

k k k k k k k k

k k k k k k L N k k L k k L N k k L

Q Q q L M M Q L q L

(30)

In Eq. (29) K , , P and FP are elastic stiffness matrix of element, displacement

vector, force vector and fixed end forces and moments vector respectively; 1u and 2u are

axial displacement of element nodes. Component of stiffness matrix K and vector FP are

represented in Appendix B. Stiffness matrix is symmetric and there is no need for calculating

all components of it. This subject is illustrated in example 1.

2.3. Members subject to concentrated span loading

In this part, a static condensation method is used for obtaining the stiffness matrix and

fix end forces and moments of the element. Macguir and Glagher [9] contracted the stiffness

matrix of an element with eliminating certain degrees of freedom. Similar approach is used to

establish formulation of the element with concentrated span loadings. To this purpose, the

element is divided into two parts at the concentrated load location, as Fig. 3., and the stiffness

matrix of each part is calculated from Eq. (29). Then, displacement quantities at division

section are eliminated so it becomes possible to calculate fix end forces and moments

independent of the intermediate displacement vector. Consequently, this approach requires

less number arithmetical operation than those with no condensing. It should be mentioned that

the stiffness matrix obtained from proposed approach is the same that is calculated through

Eq. (29). But in the presence of concentrated span loadings, fix end forces and moments are

obtained, besides, the stiffness matrix of element. The proposed formulation is established as

follows:

Fig. 3. Divided tapered beam with concentrated span loadings

The force-displacement relation of parts (a) and (b) are 1 1 1K P and

2 2 2K P , respectively, where

11 12 1 1

21 22 2 2

k k P

k k P

(31)

and

22 23 2 2

32 33 3 3

k k P

k k P

(32)

1K and 2K are stiffness matrix of part (a) and (b), obtained by Eq. (29).

Assembling Equations (31) and (32) together,

11 12 1 1

21 22 23 2 2

32 33 3 3

0

0

k k P

k k k P

k k P

(33)

which,

2 2 2

22 22 22

P P P

k k k

(34)

After, solving recent relation for 2 and substituting it in the first and third rows of Eq. (33)

the following equation is obtained which is independent of 2 ,

1 1 11 111 12 22 21 12 22 23 12 22 2

1 1 13 332 22 21 33 32 22 23 32 22 2

P k k k k k k k k k P

P k k k k k k k k k P

(35)

Obtained relation is in the form of FP K P where FP is fixed end forces and

moments,

1

12 22 2

1

32 22 2

F k k PP

k k P

(36)

And stiffness matrix K is,

1 1

11 12 22 21 12 22 23

1 1

32 22 21 33 32 22 23

[ ]k k k k k k k

Kk k k k k k k

(37)

Example (2) indicated that matrix K is the same stiffness matrix of whole member with no

division. So it's enough to calculate 12k , 22k and 32k to find fixed end forces and moments. It

should be mentioned that axial stiffness matrix components can be calculated similar to the

previous section.

3. Numerical examples and verification

This section offers some examples to verify the accuracy of obtained relations; in this

measure, the deflection of beam-column elements in the proposed method are compared with

the results of other studies and finite element method. The examples also reveal the symmetry

of the stiffness matrix.

3.1. Beam-column subject to uniformly distributed span loadings

The tapered fixed-hinged beam in Fig. 4 is similar to Li et al. example [6], however, in

this study it is subject to uniform distributed span loadings where maximum deflection and its

location are calculated. Table 1 shows the comparison of the results of the proposed method

obtained in MATLAB [10] software to those of FEM with 50 and 10mm mesh size and also

to the beam subdivided to five segments of equivalent nodal loads. It also indicates the effect

of pressure axial force and shear deflection. To generate the results for verification, the

ABAQUS [11] four node shell element S4R is employed for modeling the beam as Fig. 5.

The results demonstrate an acceptable accuracy of the proposed method. The numbers in

parentheses show the location of maximum deflection along the element and the errors are

calculated compared to FEM with 10mm mesh size. The investigations revealed that choosing

number 14 for polynomial term is acceptable in calculating deflections. In all examples of this

paper, elastic and shear modulus are 206 and 80 KN/mm2 respectively.

Fig. 4. Tapered beam with uniformly distributed span loadings

Fig. 5. Finite element model of tapered beam

Table 1

Maximum Deflection of fixed-hinged beam (mm)

Method Case A Case B Case C Case D Location(mm)*

Proposed method 2.085(1.8%)** 2.065 1.448 1.438 1845

Subdivided Beam 2.210(7.9%) 2.189 1.530 1.507 1800

FEM - mesh 50mm 2.047(0.05%) 2.038 1.396 1.389 1844

FEM - mesh 10mm 2.048( - ) 2.039 1.406 1.400 1844

Case A: Both axial and shear

B: No axial

C: No shear

D: No axial nor shear

* Maximum deflection location from first of beam (mm)

** Errors, calculated compared to FEM with 10mm mesh size.

Stiffness matrix K and fix end forces and moments FP are shown below; it can be seen

that matrix K is symmetric and also reduces computational efforts of calculation.

6

1.1092 0 0 -1.1092 0 0

0 0.0183 34.9736 0 -0.0183 20.7742

0 34.9679 80920.2118 0 -34.9679 23983.432010

-1.1092 0 0 1.1092 0 0

0 -0.0183 -34.9736 0 0.0183 -20.7742

0 20.7799 24000.6566 0 -20.7799 38339.1768

K

6

.8000

-.3505

-241.664810

-.8000

-0.1894

0

P

-3

0

0

0

-0.7212

0

-2.5145 10

6

0

-0.2983

-181.357910

0

-0.2417

96.4047

FP

3.2. Beam-column subject to concentrated span loading

Calculating the deflection at mid-span of steel tapered fixed-hinged beam of Li et al.

[6] is considered in Fig. 6. The results of the proposed method are compared to those of FEM

with 50, 20 and 10mm mesh size and those of Li et al. study. Li et al. divided the member into

two parts and derived stiffness matrix of each part individually, they also utilized FEM by

dividing the element into 10 segments in its length to calculate deflection. Because of the

same axial loads in the two parts of the beam while dealing with elastic analysis, the stiffness

matrix of the whole member in this example is the same with the previous one.

Fig. 6. Tapered beam with concentrated span loadings

Table 2

Deflection at middle of fixed-hinged beam (mm)

Method Case A Case B Case C Case D

Proposed method 0.639(0.2%)* 0.636 0.409 0.407

FEM - mesh 300mm 0.642(0.6%) 0.640 0.367 0.365

FEM - mesh 50mm 0.637(0.2%) 0.635 0.398 0.396

FEM - mesh 10mm 0.638 0.636 0.401 0.399

Li (2 elem.) 0.672(5.3%) 0.669 0.409 0.408

(10 elem.) 0.673(5.5%) 0.671 0.411 0.410

Case A: Both axial and shear

B: No axial

C: No shear

D: No axial nor shear

*Errors: calculated compared to FEM with 10mm mesh.

3.3. Gable frame with tapered members and span loadings

This section takes into account a very applicable single bay gable frame shown in Fig.

7; Similar to previous examples the results were verified by ABAQUS with 50mm mesh size.

To this purpose, maximum deflection and normal stress at the tip and corner of the frame and

also the effects of shear deformations and P on deflections are taken into consideration

(Table 3). These results evidently expose the efficiency of the proposed method in practical

problems. It should also be mentioned that, in the proposed method, the elements are modeled

symmetrically over their longitudinal axes.

Fig. 7. Gable frame with tapered members and span loadings

Table 3

Response of gable frame

Response Method Case A

B C tip corner

Deflection (mm)

Proposed method 11.12 (1.7%)* 2.57(0.4%) 11.09 10.40

FEM - mesh 50mm 10.93 2.58 10.91 10.24

Max. Stress (N/mm2)

Proposed method 53.52 (5.3%)* 58.06(0.5%) - -

FEM - mesh 50mm 50.83 57.75 - -

Case A: Both axial and shear

B: No axial

C: No shear

*Errors: calculated compared to FEM with 50mm mesh

4. Conclusions

This paper investigated second order non-linear analysis of tapered Timoshenko-Euler

beam-column element subjected to span loadings. Thus, a power series method was used to

solve governing differential equations of the member equilibrium. According to examined

examples, the appropriate number of terms for power series became 14 to achieve adequate

precision and speed. The effect of shear displacements and pressure axial forces in the

reduction of beam-column element stiffness are obviously recognizable in the solved

examples. The high speed of the proposed method, its adequate precision and accuracy, and

its simplicity in modeling make it a more effective and practical method compared to others

such as finite element method. The required analysis time of the proposed approach for the

first example is only 0.09 seconds, while it is 0.7 and 17 seconds for the finite element

method with 50 and 10 mm meshes , respectively. The offered method can be applied in 2-D

large scale practical structures and also be generalized to spatial ones.

Proposed formulation can also be derived for each section shape, nonlinear variation

of the section along member, and arbitrary distribution of span loadings. Therefore, member

weight can also be considered in the analysis. Future work should include material

nonlinearity and other conditions of steel design codes in order to achieve more advanced and

direct analysis of frames including tapered members.

Appendix A (Some approximate formula for moment of inertia)

a)

2

z f

zI I

a

where, .

1

f e

f e

I Ia L

I I

and fI and eI are moment of inertia at the ends of

the member (Saffari et al. [12])

b)

2

0 1

n

z

czI I

L

where 0I is moment of inertia at the origin of element and c and

n depend on degree of tapering and shape of the cross section. (Al-Gahtani [5])

c) 21

1

1 ( 1)

n

z

H zI I

H L

where

2 1

2 1

log

log

I In

H H (King at el. [13])

d) 2

1 2 3 4 1 1 2 2 1 3 1 4 2( ), , ( ) , 12 2, 12z f w f f wI z z D t D D l t b t t

(Saka [14])

e) , 1,2,3

n

z f e f

zI I I I n

L

witch determined by the user. (Sap 2000 [15])

Appendix B (Component of stiffness matrix and fix end forces and moments)

0

11 4 1 12 5 10 13 11 14 2 0 3 1 11

2

21 2 1 22 4 10 23 21 24 3 11

31 32 33 34 11 12 13 14

41 42 43 44 21 11 22 12 23 13 24

1

( )

/ , / , , (6 ) /

/( ), /( ), , /( )

. . .

a L

Ek

dA

k k L k k k L c c

k L k L k k k L

k k k k k k k k

k k k k k k L N k k L k k L N k

zz

14.k L

12

2 2

1 3 1 2 0 3 10 7 9 1 13 0 13 6 9 2 10

2

13 1 2 7 3 6 1 13

2 2

1 9 2 3 10 3 13 1 0 1 1 3 8 13 0 4 7

9 3 6 2 7 0 1 7 3 5 9 2 7

( 6 ( ) 6 ( )

( )) /( )

( / ) /( ) ( (2 ( / ) 2

( ) 2 ( ) / (

F

F

Q c q L c q L c c c c c c c c

q DL c c c c

M c c q L L qD Dq c c L c c

c c c c L c c c c N c c

3 6 12 1

2

2 1 2 1 1

) / ) /

1. , . .

2

F F F F F

c c

Q Q q L M M Q L q L

1 1 8 2 0 6 3 1 4 2 7 1

2

5 2 7 3 6 3 1 7 3 5 1 0 2 5 1 6 6 8 13 7 1

2

8 0 9 0 10 1 11 1 12 13

2 2 , 6 , . ( ),

( ) ( ) 6 ( ), / , .( )

. . , . , (0), (1), (0). . (0), (1). . (1)w w

c N L L c

c c c c c c c c c c c c c L L L

L N L A G A G

1 3 1 6 0 2 6 4 7 1 0 4 6 2 7 7 3 6 7 2 7 3 6 3 12

1 7 3 5 1 12 0 1 6 2 5 12

6 6 ( ) /

( ) / 6 ( ) /

c c c c c c c c c c L c c c c

L c c c c L c c c c

2 2

2 7 1 9 0 7 7 0 6 0 6 9 0 5 12 0 7 13 12 7 1 9 12

0 6 9 12

2 12 6 12 / 2 / /

6 /

c c c L c L c Nc

N c

2

3 3 9 1 0 3 7 2 9 0 4 1 12 0 3 13 12

3 9 1 12 0 2 9 12

2 ( 3 ) 12 ( / ) 2 /

/ 6 /

c c c c c L c L

c N c N

2

4 0 1 6 4 5 0 3 6 3 4 7 3 1 7 7 3 5 7 2 7 3 6 3 9

1 7 3 5 1 9 0 9 1 6 2 5 3 1 6 8 0 8 2 6 4 6 4 7 1 8

2 7 3 6 7 8

12 ( ) 2 ( ) ( )

( ) 6 ( ) 6 ( )

( )

c c c c c c c c c c c c c c

c c c c c c c c c c c c c c

c c c c

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