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INELASTIC DYNAMIC ANALYSIS OF SHELLS WITH THE TRIC SHELL ELEMENT

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VII International Conference on Computational Plasticity COMPLAS 2003 E. Oñate and D. R. J. Owen (Eds)
©
CIMNE, Barcelona, 2003
1
INELASTIC DYNAMIC ANALYSIS OF SHELLS WITH THE TRIC SHELL ELEMENT
M. Papadrakakis*, Z. S. Mouroutis*, L. Karapitta* and A.G. Papachristidis*,†
*
Institute of Structural Analysis & Seismic Research National Technical University Athens, Zografou Campus, Athens 15780, Greece e-mail: {mpapadra,zachmour,klucia}@central.ntua.gr
†
4M – VK Civil Engineering Software Inc. Mykinon 9 & Kifisias, GR-15233 Athens, Greece Email: aris@4m.gr, web page: http://www.4m.gr
Key words:
Inelastic, dynamic, shells, TRIC.
Abstract
.
The dynamic analysis of shells has attracted considerable interest in recent years. As analysts are increasingly performing more sophisticated simulations of complex structural models (some problems may comprise hundreds of thousands or even millions degrees of freedom) there is a great need for simple, and at the same time, accurate elements to conduct large-scale computational experiments. Furthermore, most available shell elements lack generality, that is, they are either isotropic or composite. In addition there is a trend in finite element analysis for numerical integration that calls for stiffness and mass matrices containing analytic algebraic expressions. To satisfy these requirements, a lot effort has been devoted to expand and further develop the natural mode finite element method for the analysis of isotropic and laminated composite shell structures. The product of this effort is the TRIC (TRIangular Composite) element, which has been presented in previous papers. The aim of this work is to present the behavior of the TRIC element in geometrical as well as nonlinear dynamic analysis of shells.
M. Papadrakakis, Z. S. Mouroutis, L. Karapitta and A. G. Papachristidis. 2
1 INTRODUCTION
When faced with the challenge of investigating time-dependent nonlinear phenomena of shell structures with the finite element method a major constraint arises which is the high computational cost involved in the simulations. Higher order shell elements have been successfully proposed in the past for linear analysis. However, the extension of this type of shell elements to the nonlinear range and especially to time-dependent problems is not straightforward. Isoparametric finite elements based on higher-order interpolation functions and multiple quadrature loops can prove very expensive and cumbersome when applied to large and complex multilayered shells. Hence, the development of a simple plate and shell finite element including transverse shear deformation, capable of engineering accuracy, competent in the study of intricate nonlinear phenomena and adaptable to many types of material systems including isotropic, sandwich, laminated, composite and hybrid structures remains a challenging task. A shell finite element that has been proved to have all the above-mentioned characteristics in static linear and nonlinear problems is the TRIC shell element initiated by J. Argyris and further developed in a number of subsequent papers
i,ii,iii,iv,v
. The aim of this paper is to formulate a consistent mass matrix that includes both translational and rotational inertia in order to test the efficiency of the TRIC element in linear and materially nonlinear dynamic problems.
2 THE PRINCIPAL OF VIRTUAL WORK IN DYNAMICS
All the implicit time integration schemes developed for linear dynamic analysis can also be applied to nonlinear dynamic response calculations. Using, for example, the Newton Raphson iteration and neglecting the effects of a damping matrix, the governing dynamic equilibrium equation is:
(k)
∆
tt
∆
tt1)(k(k)
∆
tt1)(k
∆
tt
FR
∆
UKUM
++++++
− = +
(1) where k, k+1 represent the iterations within the time step
∆
t. With the Newmark approximations:
()
UU2
β
1U
β∆
t1UU
β∆
t1U
tttt
∆
tt2
∆
tt
+ − ⋅− − =
++
(2) and introducing iterations k, k+1 within the time step
()
UU2
β
1U
β∆
t1
∆
UUU
β∆
t1U
ttt1)(kt(k)
∆
tt21)(k
∆
tt
+ − ⋅− + − =
++++
(3) Eq. (1) becomes:
()
− − − − − − =
+
++++++
U12
β
1U
β∆
t1UU
β∆
t1MFR
∆
U
β∆
tMK
tt
∆
tt(k)
∆
tt2(k)
∆
tt
∆
tt1)(k
2(k)
∆
tt
(4)
M. Papadrakakis, Z. S. Mouroutis, L. Karapitta and A. G. Papachristidis. 3
which can be solved for the incremental displacements at iteration k+1 inside the time steps.
3 THE MASS MATRIX
The computation of the consistent elemental mass matrix necesitates the estimation of matrix
ω
containing the modal functions. More specifically, the displacement vector
u
must be expressed as a function of the natural modes. Then the global elemental mass matrix can be established via:
)1818()183()318()1818(
)1818(
x V x x T x T x e
adV a M
e
=
∫
ω ω ρ
(5) where
α
is the transformation matrix from the local natural coordinate system of each element to the global Cartesian coordinate system and
ρ
is the density of the material. The modal matrix
ω
can be derived by invoking kinematic and geometric arguments. Similarly to static analysis, the rotational inertia forces resulting from antisymmetric deformation are assumed uncoupled from the other forces, and as such they are treated independently. The derivation of the part of the modal matrix that contains the rigid body modes is straightforward
i
and it can be graphically depicted in Figure 1.
Figure 1: Rigid body modes
M. Papadrakakis, Z. S. Mouroutis, L. Karapitta and A. G. Papachristidis. 4
The second component of the modal matrix is related to the axial straining modes (
γ
t
α
,
γ
t
β
and
γ
t
γ
) along the sides of the triangular element. From Figure 2, it can be concluded that the displacements of node
Γ
, for example, are given by:
B
Γ
B’
Γ
’l
α
γ
t
α
/2/sin
Γ
l
α
γ
t
α
/2 l
α
γ
t
α
/2 A l
α
γ
t
α
/2/sin
Β
γ
t
α
>0,
γ
t
β
=0,
γ
t
γ
=0 Figure 2: Axial straining mode
γ
t
α
β β
γ π γ π β
t taa
allu
− Γ + − Γ =
2cossin22cossin2
3
(6)
β β
γ π γ π β
t taa
allv
− Γ + − Γ =
2sinsin22sinsin2
3
(7) or:
β β
γ α γ β
t taa
llu
Γ + Γ =
sin2sinsin2sin
3
(8)
β β
γ α γ β
t taa
llv
Γ − Γ − =
sin2cossin2cos
3
(9) and finally:
β α β β α
γ γ
t ta
yl yl
u
Ω + Ω =
44
223
(10)
β α β β α
γ γ
t ta
x l x l
v
Ω − Ω − =
44
223
(11) Similar expressions can be derived for the other two nodes leading to the following expressions for the inplane displacements of the element due to the axial straining modes along the side of the triangle:
()()()
γα β β α
γβ α γγα
β γβ β γ
α
γ ζ ζ γ ζ ζ γ ζ ζ
t t ta
y yl y yl y ylu
+ Ω + + Ω + + Ω =
444
222
(12)
M. Papadrakakis, Z. S. Mouroutis, L. Karapitta and A. G. Papachristidis. 5
()()()
γα β β α
γβ α γγα
β γβ β γ
α
γ ζ ζ γ ζ ζ γ ζ ζ
t t ta
x x l x x l x x lv
+ Ω − + Ω − + Ω − =
444
222
(13) where
ζ
α
,
ζ
β
,
ζ
γ
, are the area coordinates of the triangular element. By setting any of the axial straining modes equal to 1 and the others equal to zero the corresponding modal functions can be obtained. For the derivation of the expressions of the symmetric and antisymmetric modal functions, the following expression of the vertical (out of the plane) displacement is used
iv
:
()()()
γβ α β α γβ α γγα β
α γβ γβ α γβ α γβ γα β α γβ α
ψ ζ ζ ζ ζ ψ ζ ζ ζ ζ
ψ ζ ζ ζ ζ ψ ζ ζ ψ ζ ζ ψ ζ ζ
Α Α Α
− + − + − + + + =
llllllw
sss
212121212121 (14) where
ψ
s
α
,
ψ
s
β
and
ψ
s
γ
are the natural symmetric bending modes and
ψ
A
α
,
ψ
A
β
and
ψ
A
γ
are the natural antisymmetric bending modes. Thus, for example, for
ψ
s
α
=1 and all the other modes equal to 0 Eq. (14) becomes:
γβ α
ζ ζ
lw
21
=
(15) Since:
∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − = ∂ ∂ − =
x w x w x w z x w zu
γγβ β α α
ζ ζ ζ ζ ζ ζ
(16)
∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ + ∂ ∂ ∂ ∂ − = ∂ ∂ − =
yw yw yw z yw zv
γγβ β α α
ζ ζ ζ ζ ζ ζ
(17) the expressions for u, v become:
()()
β
γγβ α β γγβ
α
ζ ζ ζ ζ
x x zlv y y zlu
+ Ω − = + Ω =
44 (18) In a similar way, the expressions of the other symmetric and antisymmetric modal functions can be deduced. Finally, the expressions for the natural azimuth modes
i
are graphically depicted in Figure 3. Symbolic computation is employed in order to carry out, in a clear way, the tedious but otherwise straightforward matrix multiplications of Equation 1. Consequently, all integrals are evaluated in an exact manner using the formula:

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