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displacement stiffness matrix, and its stiffness matrix is

simpler. At the same time, TL method can used to solve

the problems of moderate rotation, but the UL method

can get accurate results even to large rotation problem

under the appropriate load step[1]. Therefore, the UL

method is applied in this paper to distribute structuralnonlinear.

C. Element Stiffness Matrixes for Bar Unit Based on the

Material Nonlinearity and Geometric Nonlinearity

There is derivation of element stiffness matrixe for bar

unit in Ref. [3] with the consideration of geometric

nonlinearity.

t t t

L N K K K (1)

t T t

t V L L T L

K B D B dV (2)

t t T T t L T N N T LV N

t T

t V N T N

K B D B B D B dV

B D B dV

(3)

Where: t

L K is linear stiffness matrixe, and t

N K

is nonlinear stiffness matrixe. L

B is linear strain matrix,

and N

B is nonlinear strain matrix. T

D is linear

elastic matrix.

According to Ref. [6], uilibrium equation described by

geometric nonlinearity holds even for bi-nonlinear

problem if the linear elastic matrix T

D is replaced by

elastic plastic matrix ep

T D in rigidity matrix.

Furthermore, elastic plastic matrix can be regard as an

superposition of elastic matrix and plastic matrix, and the

following expressions are valid.

t t T p ept V L L T L

t T t pt V L L T L

K B D B dV

K B D B dV

(4)

t t p T T ep ep

t L T N N T LV N

t T ept V N T N

t T pt V N L T N

t T T p p

t N T L N T N V

B D B B D B

t

B D B B D B

K dV

B D B dV

K B D B dV

dV

(5)

Where:p

T D is plastic matrix, and its expression as

following,

T

p T T

T T

T

D f f D

D h f D f

. Here

is material constant, and h is plastic modulus.

T

y xy yz x zx z

f f f f f f f

.

Therefore

element stiffness matrixes for bar unit withthe consideration of bi-nonlinearity which is based on UL

method can be expressed as:

t t t p p p

L N

t t t

L N P

K K K

K K K

(6)

t t T T p pt V L T L L T N P

t T T p pt V N T L N T N

K B D B B D B dV

B D B B D B dV

(7)

Here t

P K is plastic correction matrix.

The expression of beam element stiffness matrixes

with the consideration of bi-nonlinearity can be deduced

in a similar way therefore will not be discussed here.

D. Treatment of Nonlinear Space Cable Unit

Cable is a kind of flexible construction whose change

in length includes elastic stretching and sag fix under the

effect of deadweight and tension. The stay guy of guyed

transmission tower has the characteristic of high tension

and small span. According to Ref. [5], the authors use

Ernst formula to modify slack and sag of the cable on the basis of bar unit, that is to modify the elastic modulus of

cable unit.Cable units only endure axial tensile stress. In the

calculation process, if the internal force of cable unit is

negative, making it zero, and removing the contribution

of cable unit stiffness on the entire structure.

E. Condensation of Element Degrees of Freedom and Coordinate Transformation of Characteristic Matrix

There are three types of element existing in the finite

element model of guyed transmission tower, includingordinary beam element, bar element and cable element,

and meanwhile in the model there is beam element with

one rigid joint and one hinged joint. So it should use a

uniform expression form for stiffness matrix of each

element before assembling global stiffness matrix.

The first case: condensation of rod-cable element

degrees of freedom.

Each node of beam element has three linedisplacements and three angular displacements, and each

node of bar element only has three line displacements,

then it just need to expand the rod-cable element stiffness

matrix to 12 * 12 matrix, compared with the beamelement stiffness matrix, adding zero elements in the

rows and columns related with angular displacement to

ensure them maintain the same order of the elementstiffness matrix.

The second case: condensation of degrees of freedom

of beam element with one rigid joint and one hinged joint.

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Assumed that beam element with node i (rigid) and j(hinged), the balance equation of the beam element is:

11 12 13 14

21 22 23 24

31 32 33 34

41 42 43 44

i i

i i

j j

j j

u N k k k k

M k k k k

u N k k k k

k k k k M

(8)

Where:mn

k (m, n = 1-4) is 3 * 3 matrix (partitioning of

element stiffness matrix);i

u ,i

,i

N andi

M is

respectively line displacement vector, angular

displacement vector, node internal force vector and

moment vector of node i; j

u , j

, j

N and j

M is

respectively line displacement vector, angular

displacement vector, node internal force vector and

moment vector of node j.

jM =0then the forth formula of (8) shows

1

44 41 42 43 j i i jk k u k k u

(9)

We can see from (9): hinged end angular displacement

of beam element is not necessarily zero, and it is related

with stiffness matrix, node line displacement and angular displacement of the other side.

Substituting (9) into the first three formulas of (8), it

can get equilibrium equations of beam element with one

rigid joint and one hinged joint after condensation:

1 1 111 14 44 41 12 14 44 42 13 14 44 43

1 1 1

21 24 44 41 22 24 44 42 23 24 44 43

1 1 1

31 34 44 41 32 34 44 42 33 34 44 43

i

i

j

i

i

j

k k k k k k k k k k k k u

k k k k k k k k k k k k

uk k k k k k k k k k k k

N

M

N

(10)

Adding zero elements in the related rows and columns

by condensation in the beam element stiffness matrix,so

as to maintain the original order of the element stiffness

matrix.After condensation , all element stiffness matrixes

become 12 * 12 matrix, and then using the coordinate

transformation matrix 0 , transform the element

characteristic matrix in the local coordinate to the systemglobal coordinate, at last assemble the stiffness matrix,

displacement vector and load vector.

F. Establishing and Solving of Nonlinear Finite Element

Equilibrium Equations

Under initial prestress state, the guyed transmission

tower reach balance state of self-stress by the role of

prestress and gravity. In this process, based on ULmethod, in the overall coordinate system, the load stepfrom time t to , the non-linear incremental

equilibrium equations of the structure is:

t t t t

K d u R F

(11)

Where: t

K is the total tangent stiffness matrix of the

structure at time t , d u is the nodes displacement

increment vector from time t to t+t , t t

R

is the node

load vector at time t+t , t

F is the equivalent nodes

load vector at time t .According to Ref. [3], using Newton-Raphson iterative

method to calculate is better to reflect the merits of UL,

so this method is adopted in this paper. The influence that

deformation has on structural stiffness in the process of

finite element calculate is treated with large deformation

effects and stress stiffening effects.

. CALCULATING EXAMPLE

Taking LM21 (33m) guyed tower in DongChangHa

transmission line as an example, establish a finite element

model. Principals and diaphragm were simulated byBeam189 in the model, auxiliary bars are simulated by

Link8, and stay guys are simulated by Link10. The pre-

stress of stay guys was carried out by using the method of

initial strain simulation, this method can consider thecontribution of cable element stiffness on global stiffness

of the structure. Furthermore, it can solve two problems:

balance of node prestress and the deformation harmony

of cable. Hinge between cross arm and main column was

carried out by mans of freedom degree coupling.

The table below lists the numerical analysis andmeasured results comparison of axial force (see table ).

. CONCLUSION

From Table , we can see most of the numerical

analysis results are larger than the measured values. The

reason for this error is mainly due to the straining of the

bars can not be measured under the effect of deadweight.

TABLE I.COMPARISON OF NUMERICAL ANALYSIS AND TEST RESULTS

Measuring point

number

Value of Numerical

Analysis(MPa)

Measuredvalue

(MPa)Error Remarks

1 37.99 40.11 0.05 Iron274

2 32.37 24.68 -0.31

3 32.38 24.5 -0.32Iron236

4 24.57 22.19 -0.11

5 24.94 22.47 -0.11Iron272

6 33.21 32.74 -0.01 Iron153

7 12.89 12.26 -0.05

8 14.34 13.83 -0.04

9 13.45 13 -0.03

10 13.87 12.94 -0.07

Iron301-303

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Table shows that some member bars have relatively

big errors. The reason is in that all nodes in numerical

analysis are ideal while allowable space has been shown

between node and bolt in practical engineering. For this

reason, element 274 sustains pulling force which is

applied by stay guy in the process of stretch-draw. After this element is out of shape, the member bars connected

with 274 come into effect. This also explains why the

element 274 which is satisfied with the designrequirements produces obvious bending deflection.

Furthermore, detail design of member bars which

connected with stay guy must be taken into consideration

during the design of prestress space steel structure.

R EFERENCES

[1] Dong Chen, and Cimian Zhu, “Feasibility of finite elementmethods for the annlysis of geometrically nonlinear

trusses,” Building Science Research of Sichuan, Chengdu,China, vol. 26, No. 3, September, 2000.

[2] Hongzhou Deng, and Xiaoming Chen, “Experimentalstudy on model of jiangyin long span transmission tower,” Journal of Buiding Structures, Beijing, China, vol. 22, No.6, December, 2001.

[3] Zhihong Zhang, “Theoretical research on large-span tensile

spatial structures composed of cables, bars and beams,”doctoral dissertation, Zhe Jiang University, Hangzhou,China, 2003.

[4] Zuyan Shen, Guoqiang Li, Yiyi Chen, Qilin Zhang,

Yongfeng Luo, “Steel Structure,” China Building Industry

Press, Beijing, China, 2005.[5] Yuanpei Lin, “Cable Stayed Bridge,” China

Communications Press, Beijing, China, 2004.[6] Haojiang Ding, Fubao He, Yiquan Xie and Xing Xu,

“Finite Element Method in Elasticity and Plasticity,”Machinery Industry Press, Beijing, China, 1984.

[7] Xucheng Wang, “Finite Element Method,” TsinghuaUniversity Press, Beijing, China, 2003.

[8] Mingxiang Chen, “Elasticity and Plasticity,” Science Press, Beijing, China, 2007.

[9] Xinmin Wang, “Numerical Analysis of EngineeringStructure Based on ANSYS,” China Communications

Press, Beijing, China, 2007.

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