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 4 - 24 October, 2012
Institute of Structural Engineering Method of Finite Elements II 1

The Continuum Mechanics Incremental Equations
The basic Problem
Establish the solution using an incremental formulation. Two main
approaches exist for establishing equilibrium
Lagrangian Formulation: Track the movement of all particles of the
body, in their motion from an initial to a final configuration (pathline)
Eulerian Formulation: The motion of the material through a
stationary control volume is considered (streamlines). Mainly used in
fluid mechanics (but also in large strain plasticity theories - e.g.
generalized plasticity).
Institute of Structural Engineering Method of Finite Elements II 2

Lagrangian vs. Eulerian Formulation - 1D Example
Spatial or Eulerian coordinates (x): These coordinates are used to locate a
point in space with respect to a fixed basis.
Material or Lagrangian coordinates (X): These coordinates are used to label
material points. If we sit on a material point, the label does not change with time.
Example: Assume that the motion is
x = φ(X, t) = X(1 + 2t + t 2 )
The inverse of the map gives us X in terms of x, i.e.,
X = φ −1 (x, t) =
Then, the displacement of the material point X is
x
(1 + 2t + t 2 )
u(X, t) = φ(X, t) − φ(X, 0) = X(2t + t 2 )
The velocity of the material point is (Langrangian Description)
v(X, t) = ∂u = 2X(1 + t)
∂t
Alternatively we can express the velocity in terms of x (Eulerian Description)
ˆv(X, t) = v(φ −1 (x, t), t) =
2x(1 + t)
(1 + 2t + t 2 )
Institute of Structural Engineering Method of Finite Elements II 3

FE Nonlinear Analysis in Solids and Structural Mechanics
How can we evaluate stresses and forces at time t since both the surface and the
volume of the body are unknown ? (In the linear case stiffness and equilibrium
were evaluated based on the initial configuration)
We need to properly map both current strains and stresses to corresponding
measures evaluated at previous configurations
Institute of Structural Engineering Method of Finite Elements II 4

Lagrangian Formulation
In the further we introduce an appropriate notation:
t x i = 0 x i + t u i , i =1, 2, 3
t+∆t x i = 0 x i + t+∆t u i
Increments in displacements from time t to t + ∆t are related as:
u i = t+∆t u i − t u i , i =1, 2, 3
Reference configurations are indexed as e.g.
t+∆t
0 f s i
where the left subscript indicates the reference configuration and the left
superscript indicates at which configuration the quantity occurs.
Note if those quantities are the same the left subscript maybe omitted e.g:
Differentiation is indexed as:
t+∆t
t+∆tτ ij
t+∆t
0 u i
0 u i,j = ∂t+∆t
∂ 0 x j
Institute of Structural Engineering Method of Finite Elements II 5

The deformation gradient, strain and stress tensors
As mentioned we must try to establish a description of the volume we
consider such that we can express the internal virtual work in terms of
an integral over a volume we know!
Further we would like to be able to decompose the stresses and strains
in an efficient manner keeping track of how the volume stretches and
rotates (rigidly).
We consider a body under deformation at times 0 and t:
Institute of Structural Engineering Method of Finite Elements II 6

The deformation gradient
We now consider the change of an infinitesimal gradient vector
Definition
The deformation gradient maps d 0 x onto d t x through the following relation
d t x = t 0Xd 0 x
Institute of Structural Engineering Method of Finite Elements II 7

The deformation gradient
We can write the deformation gradient as the Jacobian of the
current configuration at time t, with respect to the initial
configuration at time 0.
t
0X =
⎡
⎢
⎣
∂ t x 1
∂ 0 x 1
∂ t x 1
∂ 0 x 2
∂ t x 1
∂ 0 x 3
∂ t x 2
∂ 0 x 1
∂ t x 2
∂ 0 x 2
∂ t x 2
∂ 0 x 3
∂ t x 3
∂ 0 x 1
∂ t x 3
∂ 0 x 2
∂ t x 3
∂ 0 x 3
⎤
⎥
⎦
t
0X = ( 0 ∇ t x T ) T , where 0∇ =
⎡
⎢
⎣
It can be shown that t 0X = ( 0 t X) −1
The deformation gradient describes
the stretches and rotations that the
material fibers have undergone from
time zero to time t
∂
∂ 0 x 1
∂
∂ 0 x 2
∂
∂ 0 x 3
⎤
⎥
⎦
and t x T = [ t x 1 t x 2 t x 3
]
Institute of Structural Engineering Method of Finite Elements II 8

The deformation gradient
Then we introduce the Cauchy-Green deformation tensor
The deformation gradient is also used to measure the stretch of a material
fiber and the change in angle between fibers due to the deformation. For
this we use the
t
0C = t 0X T t 0X: Right Cauchy-Green deformation tensor
≠ t 0 B = t 0X t 0 XT : Left Cauchy-Green deformation tensor
The deformation gradient can be decomposed into a unique product of two
matrices
t
0U: Symmetric stretch matrix
t
0R: Orthogonal rotation matrix
t
0X = t 0R t 0 U
This is referred to as a Polar Decomposition
Institute of Structural Engineering Method of Finite Elements II 9

The deformation gradient
Decomposition of the deformation gradient
We continue by rewriting the deformation gradient
X = RU = RUR T R = VR
where U: the right stretch matrix and V: the left stretch matrix
Further it can be shown that:
U = R L ΛR T L
where Λ: the principal stretches and R L : the Direction of principal
stretches
V = R E ΛR T E
where R E : the Base vectors of principal stretches in the stationary
coordinate system
Institute of Structural Engineering Method of Finite Elements II 10

The deformation gradient
Swiss Federal Institute of Technology Page 22
The deformation gradient, strain and stress tensors
Consider a bar under stretch and rotation
We consider a bar under stretch and rotation
2L
t x
1
L
Stretching
⎡
2 0
⎤
0
Method of Finite Elements II
U = ⎢
⎣ 0 1 0 ⎥
⎦
0 0 1
X=
RU
WeDecomposition continue by rewriting (Ex 6.8) the
deformation gradient
It is instructive to consider the
deformation in two steps
X 2 = 0 RU
⎡ 0⎤
⎢
Stretching U = 0 1 0
⎥
⎢
⎥
⎢⎣
0 0 1⎥⎦
0
x 1
⎡0 −1 0
⎤
=
⎢
1 0 0
⎥
Rotation R ⎢ ⎥
⎡ ⎢
⎣0 0 1⎥
⎦ ⎤
R =
⎢
⎣
Decomposition
It is simpler to consider the
deformation in two steps
cosθ −sinθ 0
sinθ cosθ 0
0 0 1
⎡
⎥
⎦ = ⎢
⎣
0 −1 0
1 0 0
0 0 1
⎤
⎥
⎦
Institute of Structural Engineering Method of Finite Elements II 11

The deformation gradient, strain and stress tensors
General Case
Assuming both rotation U and stretch R: X = RU
⎡
⎤
l
L 0 0 ⎡
⎤
cosθ −sinθ 0
U = ⎢ h
⎣ 0
H 0 ⎥
⎦
R = ⎣ sinθ cosθ 0 ⎦
0 0 1
0 0 1
ss Federal Institute of Technology Page 2
deformation gradient, strain and stress tensors
which yields
mple – beam element
⎡
U X =
⎢
⎣
l
L cosθ − h H sinθ 0
l
L sinθ h
H cosθ 0
0 0 1
⎤
⎥
⎦
0
x2, t x2
L
θ
h
H
0
x1, t x1
Institute of Structural Engineering Method of Finite Elements II 12

The deformation gradient
Using the decomposition of the deformation gradient we may rewrite
the right and left Cauchy-Green deformation tensors:
The right Cauchy-Green deformation tensor:
C = X T X = (RU) T RU = U T R T RU = U 2
since R is orthogonal, hence RR T = R T R = I
The left Cauchy-Green deformation tensor:
B = XX T = VR T RV = V 2
Institute of Structural Engineering Method of Finite Elements II 13

The strain tensor
From deformations to strains:
The strain may be understood in terms of the stretch λ = l L .
In a previous lecture we saw that the one dimensional equivalents where:
Using time 0 as reference
Green - Lagrange strain: E = 1 2
Tensor Equivalent:
Using time t as reference
Almansi strain: A = 1 2
Tensor Equivalent:
( l
2
t+∆t
0 ɛ = 1 (C − I)
2
L 2 − 1 )
= 1 2
) (1 − L2
l 2 = 1 ( )
1 − λ
−2
2
t+∆t
t ɛ = 1 2 (I − B−1 )
(
λ 2 − 1 )
Institute of Structural Engineering Method of Finite Elements II 14

The deformation gradient, strain and stress tensors
In terms of tensor components,
Green - Lagrange strains:
Almansi strains:
(
)
ɛ ij = 1 ∂u i
2 ∂ 0 + ∂u 3∑
j ∂u k ∂u k
x j ∂ 0 +
x i ∂ 0 x i ∂ 0 x j
k=1
(
)
ɛ ij = 1 ∂u i
2 ∂ t + ∂u 3∑
j ∂u k ∂u k
x j ∂ t +
x i ∂ t x i ∂ t x j
k=1
Institute of Structural Engineering Method of Finite Elements II 15

The stress tensors
Finally we need to establish the stresses
We start by introducing the Cauchy stresses:
The Cauchy stress tensor relates forces at the current configuration
to areas at the current configuration
Institute of Structural Engineering Method of Finite Elements II 16

The stress tensors
Since the Cauchy tensor is not known a-priori, forces and areas are
mapped through the deformation gradient to the reference
configuration.
Definition
The Second Piola-Kirchoff stress tensor
t
0S =
0 ρ 0
t
ρ
t X t τ 0 t X T
where t ρ is the mass density of the body at time t and
0 ρ
t
ρ = det(t 0X)
These are so-called work conjugate to the Green - Lagrange strains
The mapping retains the symmetry of the Cauchy tensor
Rigid body motions (translations/rotations) do not induce
strains/stresses
The components of the Piola-Kirchoff stress tensor have little physical
meaning and in practice, Cauchy stresses must be calculated
Institute of Structural Engineering Method of Finite Elements II 17

Total and Updated Lagrangian Formulation
Remember, a configuration C is a snapshot of the set of motions of
all particles
Definition-Initial Configuration
The configuration defined as the origin of Undeformed displacements.
Strain free but not necessarily stress free
Definition-Reference Configuration
Configuration to which stepping computations in an incremental solution
process are referred
Definition-Target Configuration
Equilibrium configuration accepted after completing the an increment step
Institute of Structural Engineering Method of Finite Elements II 18

Total and Updated Lagrangian Formulation
Institute of Structural Engineering Method of Finite Elements II 19

Total and Updated Lagrangian Formulation
Total Lagrangian (TL)
The initial configuration is
the reference configuration.
Stress & Strain measures at
the target configuration
t + ∆t are computed with
respect to the initial
configuration at time 0.
Derivatives and Integrals are
taken with respect to 0 V
(initial conf.)
Updated Lagrangian (UL)
The previous configuration is
the reference configuration.
Stress & Strain measures at
the target configuration
t + ∆t are evaluated with
respect to the configuration
at time t.
Derivatives and Integrals are
taken with respect to t V
Institute of Structural Engineering Method of Finite Elements II 20

Total and Updated Lagrangian Formulation
We originally set out to solve the following equation:
∫
t+∆t V
t+∆t
τ δ t+∆t e ij d t+∆t V = t+∆t R
Two schemes have been formulated for this namely:
The Total Lagrangian (TL) formulation
∫
t+∆t
0 S δt+∆t 0 ɛ ij d 0 V = t+∆t R
0 V
The Updated Lagrangian (UL) formulation
∫
t+∆t
t S δt+∆t t ɛ ij d t V = t+∆t R
t V
Institute of Structural Engineering Method of Finite Elements II 21

Setting up the governing equations
Therefore for the Total Lagrangian (TL) Formulation we obtain :
∫
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 22

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 23

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 24

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 25

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 26

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 27

Total Lagrangian Formulation
Institute of Structural Engineering Method of Finite Elements II 28

Updated Lagrangian Formulation
In practice, it is often sufficient to account for only material non-linearity. In this case
the TL and the UL formulations become identical.
Institute of Structural Engineering Method of Finite Elements II 29

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 30

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 31

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 32

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 33

Element Matrices
Institute of Structural Engineering Method of Finite Elements II 34

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 35

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 36

Truss and Cable (Bar) Elements
The nodal point coordinates determine the spatial configuration of
the bar at time 0 and t using:
0
x 1 (r) =
and t x 1 (r) =
n∑
N k0 x k 1
k=1
n∑
N kt x k 1
k=1
n∑
0
x 2 (r) = N k0 x k 2
k=1
n∑
t
x 2 (r) = N kt x k 2
k=1
n∑
0
x 3 (r) = N k0 x k 3
k=1
n∑
t
x 3 (r) = N kt x k 3
k=1
Institute of Structural Engineering Method of Finite Elements II 37

Truss and Cable (Bar) Elements
Shape Functions
n = 2 nodes at r 1 = −1, r 2 = 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
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 )
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 38

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 39

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 40

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 41

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 42

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 43

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 44

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 45

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 46

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 47

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 48