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Total lagrangian formulation of a bar element andpath following methods
Prof. Dr. Eleni ChatziDr. Giuseppe Abbiati, Dr. Konstantinos Agathos
Lecture 5 - 26 October, 2017
Institute of Structural Engineering, ETH Zurich
October 26, 2017
Institute of Structural Engineering Method of Finite Elements II 1
Outline
1 Introduction
2 Weak form
3 Finite element formulation
4 Total Lagrangian formulation of a bar element
5 Path-following Methods
Institute of Structural Engineering Method of Finite Elements II 2
Learnig goals
Understanding how geometrically nonlinear finite elements areformulated
Gaining a basic understanding of path following solutionmethods
Gaining a basic understanding of how the above areimplemented
Gaining a basic understanding of how the above can be used incommercial software (ABAQUS)
Institute of Structural Engineering Method of Finite Elements II 3
Significance of the lecture
Applications:
Structures undergoing large displacements and/orrotations such as cables, arches and shells
Materials such as elastomers and biological/soft tissue
Modeling of plastically deforming materials
Institute of Structural Engineering Method of Finite Elements II 4
Weak form
We recall:
Weak form (TL):∫V
S : δEdV =∫V
F · δudV +∫A
T · δudA
Linearized weak form:
∫V
e·D·δedV +∫V
Si : δηdV =∫V
F·δudV +∫A
T·δudA−∫V
Si : δedV
where:
e = 12
[∇∆u +∇∆uT +∇
(ui)T ∇∆u +∇∆uT∇ui
]η = 1
2
[∇∆uT∇∆u
]
Institute of Structural Engineering Method of Finite Elements II 5
Finite element formulationCoordinates and displacements are interpolated from thecorresponding nodal quantities using FE interpolation:
Xi =N∑
J=1NJX J
i , xi =N∑
J=1NJxJ
i
ui =N∑
J=1NJuJ
i , ∆ui =N∑
J=1NJ∆uJ
i
or in matrix form:
X = NXn, x = Nxn
ui = Nuni , ∆u = N∆un
Index J and superscript n refer to nodal quantities, NJ are the FEshape functions.
Institute of Structural Engineering Method of Finite Elements II 6
Finite element formulation
Typically isoparametric elements are used:
NJ = NJ (ξ1, ξ2, ξ3)
where ξ1, ξ2, ξ3 are the isoparametric coordinates. Then:
∂·∂ξ1
∂·∂ξ2
∂·∂ξ3
=
∂X1∂ξ1
∂X2∂ξ1
∂X3∂ξ1
∂X1∂ξ2
∂X2∂ξ2
∂X3∂ξ2
∂X1∂ξ3
∂X2∂ξ3
∂X3∂ξ3
︸ ︷︷ ︸
J
∂·∂X1∂·∂X2∂·∂X3
⇒
∂·∂X1∂·∂X2∂·∂X3
= J−1
∂·∂ξ1
∂·∂ξ2
∂·∂ξ3
The same transformation can be used for the current configuration.Institute of Structural Engineering Method of Finite Elements II 7
Finite element formulation
Utilizing the above, the quantities ∇u,∇∆u can be computed.
For example:
∂u1∂X1
=N∑
J=1
∂NJ∂X1
uJi
∂NJ∂X1
= J−111∂NJ∂ξ1
+ J−112∂NJ∂ξ2
+ J−113∂NJ∂ξ3
where J−1ij is the element ij of matrix J−1
ij .
Institute of Structural Engineering Method of Finite Elements II 8
Finite element formulation
Utilizing the above, the linear and non-linear part of the strainincrement can be written in terms of the nodal displacementincrements:
e = BL∆un ⇒ δe = BLδ∆un
η = 12∆unT BT
NLBNL∆un ⇒ δη = δ∆unT BTNLBNL∆un
Institute of Structural Engineering Method of Finite Elements II 9
Finite element formulation
By taking into account that:
u = ui + ∆u⇒ δu = δ∆u
Substituting the above in the weak form we obtain:
δ∆unT∫V
BTL DBLdV ∆un + δ∆unT
∫V
BTNLSiBNLdV ∆un =
δ∆unT∫V
NT FdV + δ∆unT∫A
NT TdA− δ∆unT∫V
BTL SidV
Institute of Structural Engineering Method of Finite Elements II 10
Finite element formulation
By taking into account that:
u = ui + ∆u⇒ δu = δ∆u
Substituting the above in the weak form we obtain:
����δ∆unT
∫V
BTL DBLdV ∆un +����
δ∆unT∫V
BTNLSiBNLdV ∆un =
����δ∆unT
∫V
NT FdV +����δ∆unT
∫A
NT TdA−����δ∆unT
∫V
BTL SidV
Institute of Structural Engineering Method of Finite Elements II 10
Finite element formulation
By taking into account that:
u = ui + ∆u⇒ δu = δ∆u
Substituting the above in the weak form we obtain:
∫V
BTL DBLdV ∆un +
∫V
BTNLSiBNLdV ∆un =
∫V
NT FdV +∫A
NT TdA−∫V
BTL SidV
Institute of Structural Engineering Method of Finite Elements II 10
Finite element formulationBy taking into account that:
u = ui + ∆u⇒ δu = δ∆uSubstituting the above in the weak form we obtain:
KT︷ ︸︸ ︷∫V
BTL DBLdV
︸ ︷︷ ︸KL
+∫V
BTNLSiBNLdV
︸ ︷︷ ︸KNL
∆un =
−Ri︷ ︸︸ ︷∫V
NT FdV +∫A
NT TdA
︸ ︷︷ ︸fext
−∫V
BTL SidV
︸ ︷︷ ︸fint
Institute of Structural Engineering Method of Finite Elements II 10
Finite element formulation
The tangent stiffness matrix, residual and force vectors are thenwritten as:
KT = KL + KNL
KL =∫V
BTL DBLdV , KNL =
∫V
BTNLSiBNLdV
−Ri = fext − finti
fext =∫V
NT FdV +∫A
NT TdA, finti =
∫V
BTL SidV
Institute of Structural Engineering Method of Finite Elements II 11
Finite element formulation
Substituting the above, the linearized equilibrium equation can beobtained in terms of the nodal unknowns:
KT∆un = −Ri
KT∆un = fext − finti
Institute of Structural Engineering Method of Finite Elements II 12
Bar element
Next the tangent stiffness matrix and force vector are developed forthe two noded bar element.
It is convenient to write:
u21 = u1
1 + l cos θ − L
u22 = u1
2 + l sin θ
Institute of Structural Engineering Method of Finite Elements II 13
Bar element
FE shape function for a two noded bar:
N =[
1− ξ2
1 + ξ
2
]Jacobian: J = L
2
Shape function derivative:
N,1 = N∂X1
= J−1∂N∂ξ
= 1L[−1 1
]
Institute of Structural Engineering Method of Finite Elements II 14
Bar element
By considering displacements along both directions:
u1
u2
=
1− ξ2 0 1 + ξ
2 0
0 1− ξ2 0 1 + ξ
2
︸ ︷︷ ︸
N
·
u11
u12
u21
u22
and
∂u1∂X1∂u2∂X1
= 1L
−1 0 1 0
0 −1 0 1
︸ ︷︷ ︸
N,1
·
u11
u12
u21
u22
Institute of Structural Engineering Method of Finite Elements II 15
Bar element
Displacement derivative:
∂ui∂X1
= N,1uni = 1
L[−1 1
]·
u1i
u2i
= u2i − u1
iL
For i = 1:
∂u1∂X1
= u21 − u1
1L = u1
1 + l cos θ − L− u11
L = l cos θ − LL = l cos θ
L − 1
For i = 2:
∂u2∂X1
= u22 − u1
2L = u1
2 + l sin θ − u12
L = l sin θL
Institute of Structural Engineering Method of Finite Elements II 16
Bar element
By taking into account that ∂ (·)∂X2
= 0 the Green-Lagrange strain atthe previous step (only component 11 is of interest) assumes theform:
E11 = ∂u1∂X1
+ 12
(∂u1∂X1
)2+ 1
2
(∂u2∂X1
)2
=( l cos θ
L − 1)
+ 12
( l cos θL − 1
)2+ 1
2
( l sin θL
)2
= l2 − L2
2L2
Institute of Structural Engineering Method of Finite Elements II 17
Bar element
By taking into account that ∂ (·)∂X2
= 0 the linear part of the strainincrement (only component 11 is of interest) assumes the form:
e11 = ∂∆u1∂X1
+ ∂u1∂X1
∂∆u1∂X1
+ ∂u2∂X1
∂∆u2∂X1
Institute of Structural Engineering Method of Finite Elements II 18
Bar element
e11 =[1
L
[−1 0 1 0
]︸ ︷︷ ︸
∂∆u1∂X1
+
+( l cos θ
L − 1)
︸ ︷︷ ︸∂u1∂X1
1L
[−1 0 1 0
]︸ ︷︷ ︸
∂∆u1∂X1
+
+( l sin θ
L
)︸ ︷︷ ︸
∂u2∂X1
1L
[0 −1 0 1
]]︸ ︷︷ ︸
∂∆u2∂X1
∆u11
∆u12
∆u21
∆u22
Institute of Structural Engineering Method of Finite Elements II 19
Bar element
e11 = lL2
[− cos θ − sin θ cos θ sin θ
]︸ ︷︷ ︸
BL
∆u11
∆u12
∆u21
∆u22
︸ ︷︷ ︸
∆un
Institute of Structural Engineering Method of Finite Elements II 20
Bar element
The nonlinear part of the strain increment is:
η = 12
[(∂∆u1∂X1
)2+(∂∆u2∂X1
)2]=[∂∆u1∂X1
∂∆u2∂X1
]·
∂∆u1∂X1∂∆u2∂X1
∂∆u1∂X1∂∆u2∂X1
= 1L
−1 0 1 0
0 −1 0 1
︸ ︷︷ ︸
BNL=N,1
·
∆u11
∆u12
∆u21
∆u22
︸ ︷︷ ︸
∆un
Institute of Structural Engineering Method of Finite Elements II 21
Bar element
If a linear stress strain relation is employed then:
S11 = E E11 = E l2 − L2
2L2
where E is Young’s modulus
Then:
∂S11∂E11
= E
Institute of Structural Engineering Method of Finite Elements II 22
Bar element
Linear part of the stiffness matrix:
KL =∫V
BTL EBLdV
Employing that V = AL, where A the cross section of the bar in thereference configuration, we obtain:
KL = EA l2
L3
cos2θ cosθsinθ −cos2θ −cosθsinθ
sin2θ −sinθcosθ −sin2θcos2θ sinθcosθ
Symm sin2θ
Institute of Structural Engineering Method of Finite Elements II 23
Bar element
Similarly the nonlinear part of the tangent stiffness matrix is:
KNL =∫V
BTNLS11BNLdV
By carrying out the integration we obtain:
KNL = EAl2 − L2
2L3
1 0 −1 00 1 0 −1−1 0 1 00 −1 0 1
Institute of Structural Engineering Method of Finite Elements II 24
Bar element
Finally the force vector can be obtained as:
f =∫V
BTL S11dV
Performing the integration:
fint = EA l2 − L2
2L2
−cosθ−sinθcosθsinθ
Institute of Structural Engineering Method of Finite Elements II 25
The Newton-Raphson method
We recall the linearized equilibrium equation:
KT∆un = −Ri = fext − finti
Typically external loads are scaled by a factor λ:
KT∆un = −Ri = λfext − finti
Institute of Structural Engineering Method of Finite Elements II 26
The Newton-Raphson method
For a given value of λ, the above corresponds to a Newton-Raphsoniteration where:
R = Ri + ∂Ri
∂un ∆un = finti − λfext + KT∆un = 0
The iteration is repeated until
|R| <= tol
where tol is the tolerance.
Institute of Structural Engineering Method of Finite Elements II 27
The Newton-Raphson method
Typically:
Different (increasing) values are given to the load factor
Displacements are obtained using the Newton-Raphson method
The response of the structure is obtained for progressivelyincreasing loading
The load-displacement curve of the structure can be obtained
Institute of Structural Engineering Method of Finite Elements II 28
The Newton-Raphson method
The above procedure is called load control and can be illustrated asfollows:
Increment load
Apply load increment
Solve withNewton-Raphson
Repeat for nextincrement
Institute of Structural Engineering Method of Finite Elements II 29
The Newton-Raphson method
The above procedure is called load control and can be illustrated asfollows:
Increment load
Apply load increment
Solve withNewton-Raphson
Repeat for nextincrement
Institute of Structural Engineering Method of Finite Elements II 29
The Newton-Raphson method
The above procedure is called load control and can be illustrated asfollows:
Increment load
Apply load increment
Solve withNewton-Raphson
Repeat for nextincrement
Institute of Structural Engineering Method of Finite Elements II 29
The Newton-Raphson method
The above procedure is called load control and can be illustrated asfollows:
Increment load
Apply load increment
Solve withNewton-Raphson
Repeat for nextincrement
Institute of Structural Engineering Method of Finite Elements II 29
The Newton-Raphson method
The above procedure is called load control and can be illustrated asfollows:
Increment load
Apply load increment
Solve withNewton-Raphson
Repeat for nextincrement
Institute of Structural Engineering Method of Finite Elements II 29
The Newton-Raphson method
The above procedure is called load control and can be illustrated asfollows:
Increment load
Apply load increment
Solve withNewton-Raphson
Repeat for nextincrement
Institute of Structural Engineering Method of Finite Elements II 29
The Newton-Raphson method
In many cases the solution path is not monotonic
Phenomena such as snap-through and snap-back can be present
The Newton-Raphson method will either fail or not provide thefull path in those cases
Such phenomena are often of interest in practice e.g.when theovercritical behavior of a structure needs to be known
Institute of Structural Engineering Method of Finite Elements II 30
Example
Solution path involving snap-through:
Institute of Structural Engineering Method of Finite Elements II 31
Example
Solution path involving snap-through:
Institute of Structural Engineering Method of Finite Elements II 31
Example
Solution path involving snap-through:
Institute of Structural Engineering Method of Finite Elements II 31
Example
Solution path involving snap-through:
Institute of Structural Engineering Method of Finite Elements II 31
Path-following methods
In this category of methods:
It is attempted to follow the full solution path includingsnap-back and snap-through phenomena
The load factor is included as an additional unknown in thenonlinear system of equations
An additional equation (constraint) is added in the system
The choice of this constraint leads to different methods
Institute of Structural Engineering Method of Finite Elements II 32
Path-following methods
After adding ∆λ as an unknown and the constraint as an equationthe new system of equations is obtained:[
R (u, λ)g (u, λ)
]=[
00
]where:
u is the vector of nodal unknownsλ is the load factorR is the residual of the Newton-Raphson methodg is the additional constraint
Institute of Structural Engineering Method of Finite Elements II 33
Path-following methods
The above system can be solved by employing the Newton-Raphsonmethod. The linearization of the system yields: R
(ui, λi
)+∂R
(ui, λi
)∂u ∆u + ∂R
(ui , λi)∂λ
∆λ
g(ui , λi)+ ∂g
(ui , λi)∂u ∆u + ∂g
(ui , λi)∂λ
∆λ
=[
00
]
where:
∆u is the nodal displacement increment∆λ is the load factor increment
Institute of Structural Engineering Method of Finite Elements II 34
Path-following methods
The above system can be solved by employing the Newton-Raphsonmethod. The linearization of the system yields:[
KT −fexthT s
] [∆u∆λ
]=[
Ri
−g i
]
where:
∆u is the nodal displacement increment∆λ is the load factor increment
h Is the gradient of g with respect to u: h = ∂g∂u
s is the derivative of g with respect to λ: s = ∂g∂λ
Institute of Structural Engineering Method of Finite Elements II 34
Path-following methods
The coefficient matrix in this system is not symmetric.
To exploit the symmetry and sparsity of KT the followingpartitioning procedure is employed:
∆uI = KT−1fext, ∆uII = KT
−1Ri
∆λ = −g i + hT ∆uII
s + hT ∆uI , ∆u = ∆λ∆uI + ∆uII
Institute of Structural Engineering Method of Finite Elements II 35
Path-following methods
The choice of the additional constraint g is crucial
Different alternatives exist
In some cases problem specific constraints are needed
Some widely used alternatives are demonstrated in the following
Institute of Structural Engineering Method of Finite Elements II 36
Path-following methods: Load control
Setting g = λ− λ results in load control:
g = λ− λ ⇒ h = ∂g∂u = 0, s = ∂g
∂λ= 1
g i = λi − λ
[KT −fext0 1
] [∆u∆λ
]=[
Ri
λ− λi
]
∆λ = λ− λi , ∆u = KT−1(λfext − fint
i)
Institute of Structural Engineering Method of Finite Elements II 37
Path-following methods: Displacement control
Setting g = T · u− u results in displacement control:
g = T · u− u ⇒ h = ∂g∂u = TT , s = ∂g
∂λ= 0
g i = T · ui − u
[KT −fextT 0
] [∆u∆λ
]=[
Ri
u − T · ui
]
In the above T =[
0 0 . . . 1 . . . 0 0]
where the entry 1corresponds to a selected dof
Institute of Structural Engineering Method of Finite Elements II 38
Path-following methods: Displacement control
Displacement control in a pathinvolving snap-through
Displacement control in a pathinvolving snap-back
Institute of Structural Engineering Method of Finite Elements II 39
Path-following methods: Displacement control
Displacement control in a pathinvolving snap-through
Displacement control in a pathinvolving snap-back
Institute of Structural Engineering Method of Finite Elements II 39
Path-following methods: Displacement control
Displacement control in a pathinvolving snap-through
Displacement control in a pathinvolving snap-back
Institute of Structural Engineering Method of Finite Elements II 39
Path-following methods: Arc-length
Setting g =√
(u− u0)T · (u− u0) + (λ− λ0)2 −∆s results in arclength control (Crisfield 1981):
g i =√
(ui − u0)T · (ui − u0) + (λi − λ0)2 −∆s
h = ∂g∂u =
(ui − u0)
g , s = ∂g∂λ
=(λi − λ0)
g
KT −fext(ui − u0)T
g
(λi − λ0)
g
[ ∆u∆λ
]=[
Ri
−g i
]
In the above, superscript 0 refers to the beginning of the step
Institute of Structural Engineering Method of Finite Elements II 40
Path-following methods: Arc-length
We observe that for i = 0 the system becomes:
KT −fext(u0 − u0)T
g
(λ0 − λ0)
g
[ ∆u∆λ
]=[
Ri
−g i
]
Institute of Structural Engineering Method of Finite Elements II 41
Path-following methods: Arc-length
We observe that for i = 0 the system becomes:
[KT −fext0 0
] [∆u∆λ
]=[
Ri
−g i
]
Institute of Structural Engineering Method of Finite Elements II 41
Path-following methods: Arc-length
We observe that for i = 0 the system becomes:
[KT −fext0 0
] [∆u∆λ
]=[
Ri
−g i
]
The coefficient matrix is singular!
To obtain the load factor at the first iteration a predictor step isintroduced.
Institute of Structural Engineering Method of Finite Elements II 41
Path-following methods: Arc-length
For the predictor step the system is solved for the external loads:
∆up = KT−1fext
Then the increment of the load factor is computed as:
∆λp = ± ∆s‖∆up‖
Institute of Structural Engineering Method of Finite Elements II 42
Path-following methods: Arc-length
The sign of the increment is determined by the current stiffnessparameter:
κ = fextT ∆u
∆uT ∆u
Institute of Structural Engineering Method of Finite Elements II 43
Path-following methods: Arc-length
Setting g = (∆up)T (u− u1)+ ∆λp (λ− λ1) results in analternative method for arc length control (Riks 1972):
g i = (∆up)T(ui − u1
)+ ∆λp
(λi − λ1
)
h = ∂g∂u = ∆up, s = ∂g
∂λ= ∆λp
[KT −fext
(∆up)T ∆λp
] [∆u∆λ
]=[
Ri
−g i
]
In the above, superscript 1 refers to the predictor step
Institute of Structural Engineering Method of Finite Elements II 44
Path-following methods: Arc-length
Illustration of arc-length methods:
Crisfield: Riks:
Institute of Structural Engineering Method of Finite Elements II 45