DOI : 10.17577/IJERTV4IS060956
- Open Access
- Total Downloads : 138
- Authors : W. F. Zhang, K. S. Chen, Y. C. Liu, J. Ji, Z. H. Lu
- Paper ID : IJERTV4IS060956
- Volume & Issue : Volume 04, Issue 06 (June 2015)
- Published (First Online): 02-07-2015
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
Nonlinear Vibration Responses Analysis of Suspended-cable Structure based on Three Degrees of Freedom Model
W. F. Zhang, K. S. Chen, Y. C. Liu, J. Ji & Z. H. Lu Northeast Petroleum University
Daqing, Heilongjiang , China
AbstractAccording to its symmetry and mechanical characteristics of suspended-cable structure, the continuous suspended-cable structure is simplified to three degrees of freedom model. Meanwhile, considering that the geometric nonlinear of structure under loading, the expressions of kinetic energy, potential energy and the work done by damping forces and conservative forces of the system are obtained through the geometric relationships of the simplified model and parameter hypothesis. With the selected generalized coordinates, the dynamic and static equilibrium equations can be derived based on the Lagrange formulation. Using the state space formulation method and Runge-Kutta algorithm, the numerical solution of the generalized coordinates under the static and dynamic equilibrium can be calculated by the programming with MATLAB, and then the pictures for the static equilibrium path, the dynamic time-history curve and the phase plan can be plotted. The results show that when the suspended-cable structure vibrates, the vibration amplitude decreases with the increase of time, and the vibration of the suspended-cable structure is always in a stable equilibrium state.
Keywordssuspended-cable structure; three degrees of freedom model ; nonlinear vibration ; Lagrange formulation ; numerical solution
-
INTRODUCTION
Suspension and suspended-cable structure in modern engineering construction are of highly important significance, because of the advantage of reasonable stress characteristic, economic use of materials, convenient construction, simple
-
SESIMPLIFIED MODEL OF THREE DEGREES OF FREEDOM
-
Simplified model
First, single-span suspended-cable structure with continuous distributed mass can be translated to three concentrated mass particles by using the lumped mass method[5], as shown in Fig.1, thus the infinite degrees of freedom problem is simplified to limited degree of freedom. Due to its symmetry, take half-span structure shown in Fig.2 as the research object. Further, three degrees of freedom of springs and masses model is shown in Fig.3. This model contains two linear springs ki with corresponding damping ci (i=1,2) and two concentrated mass m1 and m2, which interconnected by spring. Support A is an immovable hinge and support B is a movable hinge. The system will reach a new equilibrium position under loads P1 and P2. The values of the design parameters are shown in Table 1 of Section 4. Assuming that the parameters are scaled such as l1,0=k1=m1=1, that is these quantities are expressed by using appropriate units rather than a standard MKS system of units.
Fig. 1. Particle model for single-span suspended-cable structure
facilities, strong adaptability and attractive appearance, etc[1], they are widely used in large span buildings, bridges and
other construction of facilities. However, under the service load, the suspended-cable structure will produce large displacement, so the geometric nonlinearity must be
considered in the calculation[2]. At present, the researches on the dynamic response of single span suspension structure are most based on the continuous method and finite element method. In the aspect of simplified model research, nonlinear stability of the simplified model of the double suspension roofs have been studied by Sophianopoulos D.S.[3] based on
Fig. 2. Particle model for half-span suspended-cable structure
-
Geometric relations
In the ABH and AEG in Fig.3., according to trigonometric relations, we will get
the energy method. The continuous suspended-cable structure
is simplified to three degrees of freedom model is this paper, considering that the geometric nonlinear of structure under loading, the dynamic and static equilibrium equations can be derived based on the Lagrange formulation[4]. Static
l x
p sin1,0
x1 cos1,0
v 2 h w 2
(1a)
(1b)
(1c)
characteristics and nonlinear vibration response of the symmetric vibration of suspended-cable structure based on three degrees of freedom model are analyzed.
1 1 1 1 1
In the same way, in BCJ and EFK
p p l2,0 sin2,0
(2a)
And the time derivatives of l1 and l2 are
l v1 x1 v1 p w1 w1
(5)
x l
cos
(2b) 1 l
2
l
2 2,0 2,0 1
l x
v 2 h
-
h w
w 2
(2c)
l x2 v1 v1 p p w2 w1 w2 w1
(6)
2 2 1 2 1 2 1
2
The initial length of the springs AB and BC are respectively
l x2 p
(3)
1,0 1 1
l x2 h h 2
(4)
2,0 2 2 1
x1 x2
v1
A 1,0
p
G
l1,0=1, k1=1, c1
H
B 2,0
I
Initial position
p
J
w
l1 m1=1
1
E
l2,0, k2, c2
m
2 C
K
p+w1- p
w2 w -w +h – h
P1 l2
F
2 1 2 1
Deformed position
P2
Fig. 3. Three degrees of freedom simplified model of suspended-cable structure
-
-
LAGRANGE FORMULATION
-
Kinetic and potential energies, Rayleigh dissipation function and work of conservative forces
2
Setting =P1=P2/P1=m [3], the kinetic energy is
Due to friction in the springs, the linear dissipation has occurred, and the corresponding Rayleigh function is given by
given by
D 1 c 2 1 c 2
(10)
K 1 v2 w 2 1 w 2
(7)
2 1 1 2 2 2
2 1 1 2 2
-
Lagrange formulation
Assuming that the springs are linear and unstressed in initial position. The potential energy of system can be written
d L L D dt z z z
(11)
as
U 1 2 1 k 2
2 1 2 2 2
Where l l l l
(8)
Where L=K-U-W, z is a generalized coordinate of the system, take z=v1z=w1z=w2 respectively. In this problem U and W only change with z, thus the Eq.(11) can be
1 1 1,0
2 2 2,0
simplified to
Assuming conservative forces are vertically applied on the masses m1 and m2, the corresponding work W calculating
from the initial position becomes
d K K U D W 0
dt z z z z z
-
The generalized coordinates z=v1
(12)
W P1w1 P2 w2
(9)
d K K U
D W
w1 w2
dt v
v
-
v
v
v 0
(13a)
1
1 1 1 1
v c v1 x1 k
c x2 v1 0
For any given value of , by setting
zL zL 0
in the
1 1 1 1 l 2 2 2 2 l
Eq.(17), we can obtain
1 2
-
-
The generalized coordinates z=w1
(13b)
FL zL ,0 0
-
-
NUMERICAL RESULTS
(18)
d K K U D W
dt w
w
-
w
w
w 0
(14a)
-
Calculation Parameters
1
1 1 1 1
p w1
Selecting the shape of suspended-cable as a quadratic
w1 1 c11
l1
parabola, namely
x x 2
k
c p p w1 w2 0
(14b)
y 4 f
(19)
l
2 2 2 2
2
l l
-
The generalized coordinates z=w2
d K K U D W
Where f is the span sag, l is the calculation span. According to the common rise-span ratio of suspended-
dt w
w
-
-
-
w
w
w 0
(15a)
cable structure, f
= 1 , 1 , 1
are used respectively,
2
2 2 2 2
l 10 9 8
p p w2 w1
when the numerical results are calculated. Concentrated
w2
k22 c22
0
l2
massed m1, m2 are both located in a quadratic parabola as Eq.(19), and their horizontal distance from left end bearing
(15b)
are l 4
and
l 2 respectively, that is x1=x2 in Fig.3.
The resulting equations of motion can be obtained, namely
1 1 1 1
v c v1 x1
Parameter
Case
1
2
3
rise-span ratio f/l
1/10
1/9
1/8
1,r(º)
16.70
18.43
20.56
2,r(º)
5.71
6.34
7.13
l2,0
0.9626
0.9545
0.9436
1
1
1
0.5
0.5
0.5
0.5
0.5
0.5
k2
1
1
1
c1
0.05
0.05
0.05
c2
0.05
0.05
0.05
l1
Parameters are obtained by simple calculations which are shown in Table 1.
TABLE I. PARAMETER TABLE
k22 c22
x2 v1 0
l2
w
c
p w1
1 1 1 1
l1
p p w1 w2
k22 c22
l2
0
2
w
k
c p p w2 w1 0
2 2 2 2 2 l
(16)
Rewritten in a matrix form as
M L zL FL zL , zL
(17)
1
v1
Where M 1 , z
w
-
Numerical results of static equilibrium
L
L 1
w2
The static equilibrium position can be obtained by
solving Eq.(19), that is
FL zL , zL
v x x v
1 1 1 2 0
v1 x1 x2 v1 0
0 l1
l2
l l
p w1 p p w1 w2
1
0
1 2 1 c11
1
0 k22
(20)
1
p w1 p p w1 w2
0
k22 c22
l1
l2
h h w w
0
l1 l2
0 2 1 2 1 0
h h w w 0
l2
0 2 1 2 1
l2
0
For nonzero values, the result of equation v1, w1, w2
y' v
1 1
can be solved depending on a continuation procedure based
on the algorithm described in reference [6] and using
y' w
2 1
y' w
3 2
MATLAB[7] soft to program. The initial value of is 0, other
y'
v
parameter values are shown in Table 1. With the change of ,
4 1
c
-
v1 x1 k c
-
x2 v1
the curve of equilibrium for three cases can be obtained as
1 1 1 l
2 2 2 2 l
shown in Fig.4.
y'
1
w
2
5 5 5
4 4 4
5 1
c
-
p w1
1 1 1 l
3 3 3
1
2 2 2
k c
-
p p w1 w2
2 2 2 2 l
1 1 1
0 0 0
y'
2
w
0.0 0.1 0.2
v1
0 5 10
w1
0 6 12
w2
6 2
-
h h w w
-
k
-
c
-
-
2 1 2 1
-
Rise-span ratio is 1/10
2 2 2 2 l
5 5 5
4 4 4
2
(22)
3
2
1
0
0.0 0.1 0.2
3
2
1
0
0 5 10
3
2
1
0
0 6 12
According to Runge-Kutta algorithm[8], the solver ode45[9] of ordinary differential equations in MATLAB software is used to program and solve these equations. Specific parameters refer to Table 1. Fig.5 shows the time-
v1 w1 w2
history curve of v1, w1, w2, and phase plane
(v1 , v1)
-
Rise-span ratio is 1/9
5 5 5
4 4 4
3 3 3
2 2 2
(w1 , w1) (w2 , w2 )
0.6
0.4
1
0.2
v
0.0
-0.2
about three cases.
1
0
0.0
0.1 0.2
v1
1
0
0 5 10
w1
1
0
0 6 12
w2
-0.4
0 50 100
4
1
3
t 150 200 250
-
Rise-span ratio is 1/8
Fig. 4. Equilibrium paths of model rise-span ratio with different rise-span ratio
-
Numerical results of dynamic equilibrium
w
2
Dynamic equilibrium equations have been obtained in Section, namely Eq.(16) or (17), this is a second order nonlinear ordinary differential equations. Using the method of state space formulation to select state variables, the first order differential equations can be obtained. If set
0.6
0.4
v'1
0.2
2
w
1
0
0 50 100
6
4
2
0
0 50 100
2
1
t 150 200 250
t 150 200 250
3
2
w'2
1
y1 v1 ; y4 v1
0.0 0
y w ; y w
(21)
-0.2 0
w'1
3 1 5 1
-0.4 -1 -1
4 2
y w ;
Then Eq.(16) can be rewritten as
y6 w 2
-0.6
-0.3 0.0 0.3 0.6
v1
-2
0 1 2 3 4
w1
-2
0 2 4 6
w2
-
Rise-span ratio is 1/10
0.6
0.4
1
0.2
v
0.0
-0.2
-0.4
4
3
w
1
2
1
0 50 100
t 150 200 20
-
-
-
CONCLUSIONS
In this article the continuous suspended-cable structure is simplified to three degrees of freedom nonlinear model, through the analysis of the simplified model, the expression of kinetic energy, potential energy, damping and conservative forces work of the system are obtained, and then the dynamic and static equilibrium equations are presented by using Lagrange formulation. The solver ode45 of ordinary differential equations is used to get the time- history curve and phase plane behavior. It can be concluded that when the suspended-cable structure vibrates, the
0
0 50 100 t 150 200 250
6
2
4
w
2
0
0 50 100 t 150 200 250
vibration amplitude decreases with the increase of time, and the vibration of the suspended-cable structure is always in a stable equilibrium state.
ACKNOWLEDGMENT
The authors acknowledge the valuable support of national natural science foundation of China (51178087), The ministry of education specialized research foundation for
0.6
0.4
0.2
v'1
0.0
-0.2
-0.4
-0.6
-0.3 0.0 0.3 0.6
v1
2
1
w'1
0
-1
-2
0 1 2 3 4
w1
3
2
1
w'2
0
-1
-2
0 2 4 6
w2
the doctoral program of higher education project of China (20122322120004), Heilongjiang natural science foundation of China (E201336), Heilongjiang department of education science and technology and research project (12541068)
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-
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0.6
0.4
1
0.2
v
0.0
-0.2
-0.4
-
Rrise-span ratio is 1/9
-
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-0.3 0.0 0.3 0.6
v1
0 1 2 3 4
w1
0 2 4 6
w2
-
Rrise-span ratio is 1/8
-
Fig. 5. Time-history curve and phase plane behavoir of model with different rise-span ratio