The Finite Element Method for the Analysis of Non-Linear and ...

The Finite Element Method for the Analysis of
Non-Linear and Dynamic Systems
Prof. Dr. Eleni Chatzi
Lecture 5 - 22 October, 2010
Institute of Structural Engineering Method of Finite Elements II 1

Setting up the governing equations
From the last Lecture we obtained the Total Lagrangian (TL)
Formulation of the principle of Virtual Displacements:
∫
t+∆t
0 S ij δt+∆t 0 ɛ ij d 0 V = t+∆t R (1)
0 V
where the Green-Lagrange strain component has been defined as:
(
)
t+∆t
0 ɛ ij = 1 3∑
t+∆t
0
2
u i,j + t+∆t
0 u t+∆t
j,i + 0 u k,i t+∆t
0 u k,j
If we denote 0 ɛ ij , 0 u i as increments in the strains and displacements
respectively we can write:
k=1
δ t+∆t
0 ɛ ij = δ t 0ɛ ij + 0 ɛ ij (2)
Institute of Structural Engineering Method of Finite Elements II 2

Setting up the governing equations
The incremental strain in eq (2) can be further decomposed to
0ɛ ij = 0 e ij + 0 η ij (3)
where the linear incremental strains are:
(
)
0e ij = 1 3∑
t
t
0u i,j + 0 u j,i +
2
0u k,i 0 u k,j + 0 u k,i 0u k,j
k=1
where the nonlinear incremental strains are:
0η ij = 1 3∑
0u k,i 0 u k,j
2
k=1
Furthermore, stresses can be written incrementally as:
δ t+∆t
0 S ij = δ t 0S ij + 0 S ij (4)
Institute of Structural Engineering Method of Finite Elements II 3

Setting up the governing equations
Noting that the variation δ t+∆t
0
ɛ ij at configuration time t+∆t is only
dependent on the variation of the increment, hence
δ t+∆t
0
ɛ ij = δ 0 ɛ ij
and plugging eqns (3),(4) into eqn(2) we obtain the Equation of
Motion with Incremental Decompositions
∫
∫
∫
0S ij δ 0 ɛ ij d 0 t
V + 0S ij
δ 0 η ij d 0 V = t+∆t t
R − 0S ij
δ 0 e ij d 0 V
0 V
0 V
0 V
Institute of Structural Engineering Method of Finite Elements II 4

Setting up the governing equations
Remarks
Term ∫0 V t 0S ij
δ 0 e ij d 0 V is known, so it can be moved to the
right hand side of the equation of motion.
Term ∫0 V t 0S ij
δ 0 η ij d 0 V is linear with respect to the
incremental displacements u i .
Term ∫0 V 0 S ij δ 0 ɛ ij d 0 V is highly nonlinear but we can use
Taylor expansion to approximate it
Institute of Structural Engineering Method of Finite Elements II 5

Setting up the governing equations
Since 0 η ij is of higher order we can approximate:
0ɛ ij = 0 e ij + 0 η ij = 0 e ij
Using the above and the Taylor approximation of the highly
nonlinear term ultimately yields
∫
0S ij δ 0 ɛ ij d 0 V =
0 V
∫
0C ijrs e rs δ 0 e ij d 0 V
0 V
∣ ∣∣∣∣t
where 0 C ijrs = ∂t 0S ij
∂ t 0ɛ rs
Institute of Structural Engineering Method of Finite Elements II 6

Setting up the governing equations
Finally, the Linearized Equation of Motion for the (TL) formulation is:
∫
0C ijrs e rs δ 0 e ij d 0 V +
0 V
∫
0 V
∫
t
0S ij
δ 0 η ij d 0 V = t+∆t R −
0 V
t
0S ij
δ 0 e ij d 0 V
Similarly, for the (UL) formulation we get:
∫
tC ijrs e rs δ t e ij d t V +
t V
∫
t V
∫
t τ ij δ t η ij d t V = t+∆t R −
t τ ij δ t e ij d t V
t V
where t 0S ij
, t τ ij are the Piola-Kirchhoff and Cauchy stresses at time t
Institute of Structural Engineering Method of Finite Elements II 7

Element Matrices
The general matrix equations vector depend on the assumed type of
analysis and the approach
A) Material nonlinearity
Static Analysis
t
KU = t+∆t R − t F
Dynamic Analysis, Implicit Integration scheme
M t+∆t Ü + t KU = t+∆t R − t F
Dynamic Analysis, Explicit Integration scheme
M t Ü = t R − t F
Institute of Structural Engineering Method of Finite Elements II 8

Element Matrices
B) TL Formulation
Static Analysis
( t 0K L + t 0
K NL )U = t+∆t R − t 0F
Dynamic Analysis, Implicit Integration scheme
M t+∆t Ü + ( t 0K L + t 0
K NL )U = t+∆t R − t 0F
Dynamic Analysis, Explicit Integration scheme
M t Ü = t R − t 0F
Institute of Structural Engineering Method of Finite Elements II 9

Element Matrices
C) UL Formulation
Static Analysis
( t tK L + t t
K NL )U = t+∆t R − t tF
Dynamic Analysis, Implicit Integration scheme
M t+∆t Ü + ( t tK L + t t
K NL )U = t+∆t R − t tF
Dynamic Analysis, Explicit Integration scheme
M t Ü = t R − t tF
Institute of Structural Engineering Method of Finite Elements II 10

Element Matrices
where the following notation has been used
M = time independent Mass matrix
t
K, t 0K L , t tK L =
linear strain incremental Stiffness matrices
t
0K NL , t tK NL = nonlinear strain incremental Stiffness matrices
t+∆t
R = vector of externally applied nodal point loads at time t + ∆t
t
F, t 0F, t tF = vector of nodal point forces equivalent to the element stresses
U = vector of increments in nodal point displacements
Institute of Structural Engineering Method of Finite Elements II 11

Element Matrices
Institute of Structural Engineering Method of Finite Elements II 12

Element Matrices
where the following notation has been used
H S , H = the shape function matrices (usually denoted by N)
0
f S , 0 f B = vector of surface and body forces at time 0
B L , t 0B L , t tB L =
linear strain displacement transformation matrices
t
0B NL , t tB NL = nonlinear strain displacement transformation matrices
0C, t C = incremental stress strain material property matrices
t
τ , t ˆτ = matrix and vector of Cauchy stresses
t
0S, t Ŝ = matrix and vector of 2nd Piola-Kirchhoff stresses
These matrices depend on the type of Finite Element considered
Institute of Structural Engineering Method of Finite Elements II 13

Structural Elements
Truss and Cable Elements
A Truss Element is a structural element capable of transmitting
stresses only in the direction normal to the cross-sectional area
Consider a truss element
that has an arbitrary
orientation. It is usually
described by two to four
nodes and is subjected to
large displacements and
large strains.
Institute of Structural Engineering Method of Finite Elements II 14

Truss and Cable (Bar) Elements
The nodal point coordinates determine the spatial configuration of the bar
at time 0 and t using:
n∑
n∑
n∑
0 x 1 (r) = N 0 k x k 1
0 x 2 (r) = N 0 k x k 2
0 x 3 (r) = N 0 k x k 3
and t x 1 (r) =
Shape Functions
k=1
n∑
N t k x k 1
k=1
t x 2 (r) =
n = 2 nodes at r 1 = −1, r 2 = 1
k=1
n∑
N t k x k 2
k=1
N1 a = 1 2 (1 − r), N 2 a = 1 (1 + r)
2
n = 3 nodes at r 1 = −1, r 2 = 1, r 3 = 0
t x 3 (r) =
N b 1 = N a 1 − 1 2 (1 − r2 ), N b 2 = N a 2 − 1 2 (1 − r2 ), N b 3 = (1 − r 2 )
k=1
n∑
N t k x k 3
k=1
n = 4 nodes at r 1 = −1, r 2 = 1, r 3 = − 1 3 , r4 = 1 3
N c 1 = N b 1 + 1 16 (−9r3 + r 2 + 9r − 1), N c 2 = N b 2 + 1 16 (9r3 + r 2 − 9r − 1),
N c 3 = N b 3 + 1 16 (27r3 + 7r 2 − 27r − 7), N c 4 = 1 16 (−27r3 − 9r 2 + 27r + 9)
Institute of Structural Engineering Method of Finite Elements II 15

Truss and Cable Elements
Isoparametric Elements ⇒ t u i (r) = ∑ n
k=1 N t k uk i
i=1,2,3
Element Matrices
Since the only stress is the normal stress we consider only the
corresponding longitudinal strain along s: t 0˜ɛ 11.
From the TL Formulation we obtain:
t
0˜ɛ 11 =
3∑
i=1
Incremental Decomposition
0ẽ 11 =
0 ˜η 11 = 1 2
3∑
i=1
d 0 x i
d 0 s
d 0 x i
d 0 s
3∑
i=1
d t u i
d 0 s + 1 d t u i d t u i
2 d 0 s d 0 s
du i
d 0 s + dt u i du i
d 0 s d 0 s
du i du i
d 0 s d 0 s
Institute of Structural Engineering Method of Finite Elements II 16

Truss and Cable Elements
Matrix Equivalent Formulation
Strain Displacement Matrices
t
0B L = ( 0 J −1 ) 2 ( 0 x T N Ț rN ,r + t u T N Ț rN ,r )
t
0B NL = ( 0 J −1 )N ,r :independent of orientation
where 0 J −1 = dr
d 0 s and 0 s(r) is the arc length (coordinate) at point
(x 1(r), x 2(r), x 3(r)) given by
n∑
0
s(r) = N k0 s k
k=1
The only non zero stress component is t 0˜S 11
and the tangent Stress
Strain relationship therefore is:
0C 1111 = ∂t 0˜S 11
∂ t 0˜ɛ 11
Institute of Structural Engineering Method of Finite Elements II 17

Truss and Cable Elements
Example
Develop the tangent stiffness matrix and force vector at time t.
Consider large displacement and large strain conditions.
Institute of Structural Engineering Method of Finite Elements II 18

Truss and Cable Elements - Example
Since the element is aligned with the 0 x 1 axis at time 0 we need not introduce
the arc length s in our calculations. From the relevant TL formulation table we
have that the linear and nonlinear components of the Green- Lagrange strain
increments will be given as:
(
)
0e ij = 1 3∑
0u i,j + 0u j,i + ( t t
0u k,i 0u k,j + 0u k,i 0 u k,j )
2
k=1
( 3∑
)
0u k,i 0u k,j
k=1
0η ij = 1 2
Since we are interested in 0e 11, 0η 11, and since the displacements are restricted in
the 0 x 1, 0 x 2 plane the above expressions become
[ (
0e 11 = ∂u1 + ∂t u 1 ∂u 1
+ ∂t u 2 ∂u 2
0η
∂ 0 x 1 ∂ 0 x 1 ∂ 0 x 1 ∂ 0 x 1 ∂ 0 11 = 1 ) 2 ( ) ] 2 ∂u1 ∂u2
+
x 1 2 ∂ 0 x 1 ∂ 0 x 1
(5)
Institute of Structural Engineering Method of Finite Elements II 19

Truss and Cable Elements - Example
From the Geometry the nodal coordinates at time t are:
t u 1 1 = 0,
t u 1 2 = 0,
t u 2 1 = ( 0 L + ∆L)cosθ − 0 L
t u 2 2 = ( 0 L + ∆L)sinθ
The displacement at a point within the element (at a distance ξ from the center)
is given by
t u i =
2∑
k=1
N t k u k i with N 1 = 1 2 (1 − ξ), N2 = 1 (1 + ξ)
2
Also, 0 J = ∂0 x 1
∂ξ
= ∂0 x 2
∂ξ
=
0 L
. Then we obtain
2
∂ t u 1
= ∂N1(ξ)
∂ 0 x 1 ∂ξ
∂ t u 1
∂ 0 x 1
= 0 +
Similarly,
∂ξ t u 1
∂ 0 1 + ∂N2(ξ)
x 1 ∂ξ
[
( 0 L + ∆L)cosθ − 0 L
∂ t u 2
= (0 L + ∆L)sinθ
∂ 0 x 0 1 L
∂ξ t u 2
∂ 0 1 ⇒ (6)
x 1
]
0
J −1 = (0 L + ∆L)cosθ
− 1 (7)
0
L
(8)
Institute of Structural Engineering Method of Finite Elements II 20

Truss and Cable Elements - Example
In addition, the incremental displacements are also written as:
u i =
2∑
k=1
which yields:
N k u k i with N 1 = 1 2 (1 − ξ), N 2 = 1 (1 + ξ)
2
∂u 1
∂ 0 x 1
= 1
0 L (u2 1 − u1 1 ), ∂u 2
∂ 0 x 1
= 1
0 L (u2 2 − u1 2 ) (9)
Substituting expressions (6) - (9) into equation (5) we obtain the matrix equivalent
representation:
[
0e 11 = 1
( [ ] ( 0 )
L + ∆L)cosθ [ −1 0 1 0 + 0 L
− 1 −1
L
0 1
] 0
⎡
( ( 0 )
L + ∆L)sinθ [ ] ] u 1 ⎤
1
u
+
0 −1 0 1 2
0 L
⎢
⎣ u 2 ⎥
1 ⎦ ⇒
u 2 2
⎡
0e 11 = (0 L + ∆L) [ ]
−cosθ −sinθ cosθ sinθ ( 0 L) 2 ⎢
⎣
u 1 1
u 1 2
u 2 1
u 2 2
⎤
⎥
⎦
Institute of Structural Engineering Method of Finite Elements II 21

Truss and Cable Elements - Example
Hence,
t
0 B L = (0 L + ∆L) [ ]
−cosθ −sinθ cosθ sinθ
( 0 L) 2
The same could be obtained by directly applying the formula:
t
0 B L = ( 0 J −1 ) 2 ( 0 x T N Ț ξ N ,ξ + t u T N Ț ξ N ,ξ)
Next, the linear part of the stiffness matrix is then obtained as:
∫
t
0 K t T
L = 0 B L 0 C t 0 B Ld 0 V
0 V
where for the truss element 0 C = ∂t 0 S 11
∂ t 0 ɛ . If we use that the original ratio is equal to the
11
elasticity modulus, we have 0 C = E. Also, 0 V = 0 A 0 L. Then,
⎡
t
0 K L = 0A 0 C (0 L + ∆L) 2
⎢
( 0 L) 3 ⎣
cos 2 θ cosθsinθ −cos 2 ⎤
θ −cosθsinθ
sin 2 θ −sinθcosθ −sin 2 θ
cos 2 ⎥
θ sinθcosθ ⎦
Symm
sin 2 θ
Institute of Structural Engineering Method of Finite Elements II 22

Truss and Cable Elements - Example
In order to derive the nonlinear part of the stiffness matrix we first need to evaluate the
Piola-Kirchhoff stress. We know that the Cauchy stress at time t is equal to t t P
τ =
t A ,
directed along the axis of the element at time t. Using the rotational matrix we can
rotate the stress tensor from the element axis system to the original reference system
( t x 1 , t x 2 ). Denoting the rotated tensor as t ¯τ we have:
t ¯τ = R
[ t τ 0
0 0
]
[
R T cosθ −sinθ
, R =
sinθ cosθ
Then, from Lecture 4 we know that the Piola-Kirchhoff stress is given as:
t
0 [
0 S = ρ
t
0
t t X t ¯τ 0 t X T τ 0
or R
ρ
0 0
]
R T =
]
t ρ t
0 0
ρ
X t 0 S t 0 XT (10)
where the deformation gradient, t 0X, can be obtained as the product of a rotational and
a stretch component as t 0X = RU. For this example the stretch matrix is obviously
(elongation along x):
⎡
⎤
⎡
0 L + ∆L
U = ⎣ 0 0 ⎦
L
⇒ U −1 = ⎣
0 1
0 L
0 L + ∆L 0
0 1
⎤
⎦
Institute of Structural Engineering Method of Finite Elements II 23

Truss and Cable Elements - Example
Then equation (10) can be rewritten as:
[ t
]
τ 0
R
R
0 0
T t ρ
=
0 ρ RU t 0 S RUT ⇒ R
[ t
]
τ 0
t [
ρ
t0
=
0 0 0 ρ U t 0 S S 0
UT ⇒
0 0
[ t τ 0
0 0
]
=
]
R T t ρ
=
0 ρ RU t 0 S UT R T ⇒
0 ρ
t ρ U−1 [ t τ 0
0 0
]
(U −1 ) T
Ultimately, carrying out the calculations yields:
t
0 0 S 11 = ρ 0
t ρ ( L t P
0 L + ∆L )2 t A
Since mass=const
0 ρ 0 L 0 A = t ρ 0 L + ∆L t A ⇒ t 0 0 S 11 = L t P
0 L + ∆L 0 A
Note how the components of t 0S do not depend on rotation, hence the tensor retains only
the S 11 one component. Then, the nonlinear part of the stiffness matrix is derived as:
∫
t
0 K t T
NL = 0 B NL 0 S t 0 B NLd 0 V
0 V
[
where t 0 B NL = ( 0 J −1 −1 0 1 0
) N ,ξ =
0 −1 0 1
]
Institute of Structural Engineering Method of Finite Elements II 24

Truss and Cable Elements - Example
Finally, t 0 K NL =
t P
0 L + ∆L
⎡
⎢
⎣
1 0 −1 0
0 1 0 −1
−1 0 1 0
0 −1 0 1
⎤
⎥
⎦
and, t 0 K N = t 0 K L + t 0 K NL
and the force vector is obtained as (see relevant table for (TL) formulation):
⎡
∫
t
0 F = t Tt
0 B L 0 S 11 d 0 V = (0 L + ∆L)
0 V ( 0 L) 2 ⎢
⎣
⎡ ⎤
−cosθ
t
0 F = t −sinθ
P
⎢
⎣ cosθ
⎥
⎦
sinθ
−cosθ
−sinθ
cosθ
sinθ
⎤
0 L t P
⎥
0 ⎦
0 L + ∆L 0 V ⇒
A
Institute of Structural Engineering Method of Finite Elements II 25