The Finite Element Method for the Analysis ofNon-Linear and Dynamic Systems: Computational
Plasticity Part I
Prof. Dr. Eleni ChatziDr. Giuseppe Abbiati, Dr. Konstantinos Agathos
Lecture 4 - Part A - 28 October, 2020
Institute of Structural Engineering Method of Finite Elements II 1
Learning Goals
To understand a basic lumped plasticity model that consists ona spring-slider system.
To understand the algorithmic procedure of a nonlinear staticfinite element analysis.
To understand how to use the Return Mapping and N-Rmethods in this problem.
References:
René de Borst, Mike A. Crisfield, Joris J. C. Remmers, Clemens V.Verhoosel, Nonlinear Finite Element Analysis of Solids andStructures, 2nd Edition, Wiley, 2012.
Institute of Structural Engineering Method of Finite Elements II 2
https://www.wiley.com/en-us/Nonlinear+Finite+Element+Analysis+of+Solids+and+Structureshttps://www.wiley.com/en-us/Nonlinear+Finite+Element+Analysis+of+Solids+and+Structureshttps://www.wiley.com/en-us/Nonlinear+Finite+Element+Analysis+of+Solids+and+Structures
Material Nonlinearity
Almost all materials are inherently non-linear.Meaning that:
F 6= KU
A typical Force-Displacement curve for such response looks like this:
Institute of Structural Engineering Method of Finite Elements II 3
Material Nonlinearity
Important to note: Not all Nonlinear behavior is Inelastic (e.g.hyperelastic materials)
Source: Dirk Mohr - Lecture 4 (ETHZ)
Institute of Structural Engineering Method of Finite Elements II 4
https://ethz.ch/content/dam/ethz/special-interest/mavt/virtual-manufacturing/ivp-dam/Studium/Vorlesungsunterlagen/Dynamic%20Behavior%20of%20Materials%20and%20Structures/Downloads/Lecture4/Lecture%204%20(Dirk%20Mohr,%20ETH,%20Integration%20Algorithms%20in%20Plasticity).pdf
Lumped Plasticity: a Spring-Slider System
Before we visit the continuum mechanics equations, let us firstexamine a simple (lumped) plasticity model.
if the applied force H is smaller than adhesion, sliding isprevented
if the applied force H is higher than adhesion, sliding starts
The deformation u is decomposed in an elastic and a plastic part as:
u = ue + up
Source: Dirk Mohr - Lecture 4 (ETHZ)
Institute of Structural Engineering Method of Finite Elements II 5
https://ethz.ch/content/dam/ethz/special-interest/mavt/virtual-manufacturing/ivp-dam/Studium/Vorlesungsunterlagen/Dynamic%20Behavior%20of%20Materials%20and%20Structures/Downloads/Lecture4/Lecture%204%20(Dirk%20Mohr,%20ETH,%20Integration%20Algorithms%20in%20Plasticity).pdf
Lumped Plasticity: a Spring-Slider System
Let us slightly refine the model, to better reflect an actual system byexamining a block sliding on a rough surface (De Borst, Crisfield, etal. 2012):
The horizontal displacement u of point A is initially caused bydeformation (elongation) of the spring (ue), since for low forcelevels, the adhesion and the friction between the block and the floorprevent sliding of the block.When the maximum shear force - exerted by adhesion and friction -is exhausted, sliding (up) occurs.
Institute of Structural Engineering Method of Finite Elements II 6
Lumped Plasticity: a Spring-Slider System
The problem is therefore decomposed as follows:
u = ue + up → u̇ = u̇e + u̇p
ue is elastic since, upon removal of the force, the deformation in thespring also disappears.
However, the displacements of the block - once triggered - do notdisappear (permanent). Deformations that cannot be recoveredduring unloading are termed inelastic or plastic.
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Lumped Plasticity: a Spring-Slider System
The surface between the floor and the sliding block is not perfectlysmooth ⇒horizontal sliding (up) will also generate a vertical displacement ofthe block, vp.
These horizontal and vertical displacement components can beassembled in corresponding vectors:
up =
[up
vp
]ue =
[ue
v e
]
For the present example, the vertical elastic force component v e hasno physical meaning and is simply equal to zero.
Institute of Structural Engineering Method of Finite Elements II 8
Lumped Plasticity: a Spring-Slider System
The elastic displacement ue is linked to the horizontal force H viaHooke’s law:
H = kue
where k is the spring constant.If H = 0⇒ ue = 0
For the plastic displacement up, this is not the case.When slipping initiates it continues to grow under constant force.
Institute of Structural Engineering Method of Finite Elements II 9
Lumped Plasticity: a Spring-Slider System
Source: Dirk Mohr - Lecture 4 (ETHZ)
Institute of Structural Engineering Method of Finite Elements II 10
https://ethz.ch/content/dam/ethz/special-interest/mavt/virtual-manufacturing/ivp-dam/Studium/Vorlesungsunterlagen/Dynamic%20Behavior%20of%20Materials%20and%20Structures/Downloads/Lecture4/Lecture%204%20(Dirk%20Mohr,%20ETH,%20Integration%20Algorithms%20in%20Plasticity).pdf
Lumped Plasticity: a Spring-Slider System
We will assume that during slipping the rate of the inelasticdeformation u̇p can be determined.For the moment, let us assume that the ratio between the horizontal“plastic” velocity u̇p and the vertical “plastic” velocity v̇p can bedefined from experiments and is defined by the dilatancy angle ψ:
tan(ψ) =v̇p
u̇p
This turns out to be a function of the total accumulated plasticdisplacement/slip (equivalent plastic strain), which is linked to theso-called “hardening law”.For now, let us ignore this.
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Lumped Plasticity: a Spring-Slider System
The overall displacement vector u is therefore decomposed as follows:
u = ue + up → u̇ = u̇e + u̇p
u : total displacement of A [m]
ue : spring elongation (elasticdisplacement) [m]
up : block sliding (plasticdisplacement) [m]
k : spring stiffness[Nm
]ψ : dilatancy angle [rad ]
H : horizontal force [N]
V : vertical force [N]
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Lumped Plasticity: a Spring-Slider System
A mathematical model of the spring-slider system is derived thatexpresses the relationship between displacement and force rates.
u̇ = u̇e + u̇p
u̇e =
[u̇e
v̇ e
] u̇e = Ḣk : horizontal elastic vel. [ms ]v̇ e = 0 : vertical elastic vel.
[ms
]
Institute of Structural Engineering Method of Finite Elements II 13
Lumped Plasticity: a Spring-Slider System
A mathematical model of the spring-slider system is derived thatexpresses the relationship between displacement and force rates.
u̇ = u̇e + u̇p
u̇p = λ̇m
m =
[1
tanψ
] λ̇ : plastic multiplier [m]tanψ = v̇pu̇p : ratio between plastic vert.and horiz. velocities [d .l .]
The value of the plastic multiplier λ̇ can be determined from therequirement that during plastic flow, the stresses remain bounded.
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Lumped Plasticity: a Spring-Slider System
As analogously done for displacements, we define the force responserate of the spring-slider system.
ṙ = Ke u̇e = Ke (u̇− u̇p)with,
ṙ =
[Ḣ
V̇
], Ke =
[k 00 0
] Ke : elastic stiffness matrixḢ : horizontal force rate
[Ns
]V̇ : vertical force rate
[Ns
]Institute of Structural Engineering Method of Finite Elements II 15
Lumped Plasticity: a Spring-Slider System
But when does sliding initiate?
We mentioned that slip occurs when the maximum shear force -exerted by adhesion and friction - is exhausted.
However, a formal criterion needs to be provided that sets theborderline between purely “elastic” displacements and slip (sliding ofthe block), i.e., when “plastic” displacements occur.
For our present system we only have two force components, namely,H and V . The simplest assumption is that sliding starts when theCoulomb friction, defined by the friction angle, φ augmented withsome adhesion, c , is fulfilled.
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Lumped Plasticity: a Spring-Slider System
The following Coulomb yielding function f to define the borderlinebetween purely elastic spring elongation and plastic block sliding.
ϕ : friction angle, c : adhesion coefficient.
f (H,V , ϕ, c) = H + Vtanϕ− c < 0 : elastic spring elongationf (H,V , ϕ, c) = H +Vtanϕ− c = 0 : plastic sliding of the blockf (H,V , ϕ, c) = H + Vtanϕ− c > 0 : physically impossible !!!
Institute of Structural Engineering Method of Finite Elements II 17
Lumped Plasticity: a Spring-Slider System
The following Coulomb yielding function f to define the borderlinebetween purely elastic spring elongation and plastic block sliding.
ϕ : friction angle, c : adhesion coefficient.
f (H,V , ϕ, c) = H + Vtanϕ− c < 0 : elastic spring elongationf (H,V , ϕ, c) = H +Vtanϕ− c = 0 : plastic sliding of the blockf (H,V , ϕ, c) = H + Vtanϕ− c > 0 : physically impossible !!!
Institute of Structural Engineering Method of Finite Elements II 17
Lumped Plasticity: a Spring-Slider System
The following Coulomb yielding function f to define the borderlinebetween purely elastic spring elongation and plastic block sliding.
ϕ : friction angle, c : adhesion coefficient.
f (H,V , ϕ, c) = H + Vtanϕ− c < 0 : elastic spring elongationf (H,V , ϕ, c) = H +Vtanϕ− c = 0 : plastic sliding of the blockf (H,V , ϕ, c) = H + Vtanϕ− c > 0 : physically impossible !!!
Institute of Structural Engineering Method of Finite Elements II 17
Lumped Plasticity: a Spring-Slider System
The following Coulomb yielding function f to define the borderlinebetween purely elastic spring elongation and plastic block sliding.
ϕ : friction angle, c : adhesion coefficient.
f (H,V , ϕ, c) = H + Vtanϕ− c < 0 : elastic spring elongationf (H,V , ϕ, c) = H +Vtanϕ− c = 0 : plastic sliding of the blockf (H,V , ϕ, c) = H + Vtanϕ− c > 0 : physically impossible !!!
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Coulomb Yield Function
No plastic strain occurs when the force state stays in the elasticdomain.
ϕ : friction angle, c : adhesion coefficient.
f (H,V , ϕ, c) = H + Vtanϕ− c < 0→ u̇p = 0→ u̇e = u̇
ṙ = Ke u̇
Institute of Structural Engineering Method of Finite Elements II 18
Coulomb Yield Function
Plastic strain occurs when the force state belongs to the yieldingsurface.
ϕ : friction angle, c : adhesion coefficient.
f (H,V , ϕ, c) = H + Vtanϕ− c = 0→ u̇p 6= 0→ u̇e = u̇− u̇p{ṙ = Ke (u̇− u̇p)ḟ = 0
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Coulomb Yield Function
The force states can move either to the elastic domain or within theyielding surface (Prager’s consistency condition).
ϕ : friction angle, c : adhesion coefficient.
ḟ (H,V , ϕ, c) = Ḣ + V̇ tanϕ = nT ṙ = 0
with n =
[1
tanϕ
], ṙ =
[Ḣ
V̇
]Institute of Structural Engineering Method of Finite Elements II 20
Lumped Plasticity Model
As long as we stay on the yielding surface, both of the followingconditions must be fulfilled:
{ṙ = Ke (u̇− u̇p)ḟ = 0
→
{ṙ = Ke
(u̇− λ̇m
)ḟ = 0
→
{ṙ = Ke
(u̇− λ̇m
)nT ṙ = 0
Since m and n are constant, the system is linear and therefore it’sconvenient to recast it in matrix form:
[I Kem
nT 0
] [ṙ
λ̇
]=
[Ke u̇
0
]
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Lumped Plasticity Model
[I Kem
nT 0
] [ṙ
λ̇
]=
[Ke u̇
0
]↓[
ṙ
λ̇
]=
[Ke − KemnT Ke
nT KemKem
nT KemnT Ke
nT Kem−1
nT Kem
][u̇0
]
The inverse of square block matrix A =
[A11 A12A21 A22
]reads,
A−1 =
[A−111 + A
−111 A12B
−1A21A−111 −A
−111 A12B
−1
−B−1A21A−111 B−1]
where,
B = A22 − A21A−111 A12The Matrix Cookbook
Institute of Structural Engineering Method of Finite Elements II 22
https://www.math.uwaterloo.ca/~hwolkowi/matrixcookbook.pdf
Lumped Plasticity Model: Tangent Stiffness
Instantaneous tangent stiffness of the spring-slider system:
ṙ =(
Ke − KemnT KenT Kem
)u̇
λ̇ =(
nT Ke
nT Kem
)u̇
[Ḣ
V̇
]=
[k 00 0]−
[k 00 0
] [1
tanψ
] [1 tanϕ
] [k 00 0
][1 tanϕ
] [k 00 0
] [1
tanψ
][u̇v̇
]
Institute of Structural Engineering Method of Finite Elements II 23
Lumped Plasticity Model: Tangent Stiffness
Instantaneous tangent stiffness of the spring-slider system:
ṙ =(
Ke − KemnT KenT Kem
)u̇
λ̇ =(
nT Ke
nT Kem
)u̇
[Ḣ
V̇
]=
[k 00 0]−
[k 00 0
] [1 tanϕ
tanψ tanψtanϕ
] [k 00 0
]k
[u̇v̇]
Institute of Structural Engineering Method of Finite Elements II 23
Lumped Plasticity Model: Tangent Stiffness
Instantaneous tangent stiffness of the spring-slider system:
ṙ =(
Ke − KemnT KenT Kem
)u̇
λ̇ =(
nT Ke
nT Kem
)u̇
[Ḣ
V̇
]=
[k 00 0]−
[k2 00 0
]k
[u̇v̇]
It is interesting to note that the spring-slider system has no stiffnesswhen the force state belong to the yielding surface.
Institute of Structural Engineering Method of Finite Elements II 23
Lumped Plasticity Model: Plastic Multiplier
Instantaneous plastic multiplier of the spring-slider system:
ṙ =(
Ke − KemnT KenT Kem
)u̇
λ̇ =(
nT Ke
nT Kem
)u̇
λ̇ =
[1 tanϕ
] [k 00 0
][1 tanϕ
] [k 00 0
] [1
tanψ
][u̇v̇
]
Institute of Structural Engineering Method of Finite Elements II 24
Lumped Plasticity Model: Plastic Multiplier
Instantaneous plastic multiplier of the spring-slider system:
ṙ =(
Ke − KemnT KenT Kem
)u̇
λ̇ =(
nT Ke
nT Kem
)u̇
λ̇ =
([k 0
]k
)[u̇v̇
]It is interesting to note that only plastic displacement incrementoccurs when the force state belongs to the yielding surface.
Institute of Structural Engineering Method of Finite Elements II 24
Integration of the Force-Displacement Response
Discussion
The rate equations listed in the previous slide are in generalnon-symmetric, since in general φ 6= ψ, which further implies thatm 6= n.
As a consequence, the matrix formed by the outer product mnT willbe non-symmetric, thus rendering the tangential stiffness matrix thatsets the incremental relation between ṙ and u̇ also non-symmetric.
The example we used is a special (simplified) case, but is used hereas an intro to continuum plasticity, which follows in the next lectureand where the off-diagonal terms do not vanish.
Institute of Structural Engineering Method of Finite Elements II 25
Integration of the Force-Displacement Response
Force-displacement response of the spring-slider system.
Let’s imagine to turn this into a computer program:
1: function [rj+1] = elementForce (uj+1)2: ...3: end
Institute of Structural Engineering Method of Finite Elements II 26
Integration of the Force-Displacement Response
Elastic domain:
f (r) < 0
↓ṙ = Ke u̇
↓∆r = Ke∆u
Plastic domain (yielding surface):
f (r) = 0
↓
ṙ =
(Ke − K
emnTKe
nTKem
)u̇
↓
∆r =
(Ke − K
emnTKe
nTKem
)∆u
How to handle the case when we are moving from the elastic to theplastic domain?
Institute of Structural Engineering Method of Finite Elements II 27
Return Mapping Algorithm
Institute of Structural Engineering Method of Finite Elements II 28
Return Mapping Algorithm
Return mapping algorithm Step #1.
rj : initial restoring force (onset of load step j + 1).
Institute of Structural Engineering Method of Finite Elements II 28
Return Mapping Algorithm
Return mapping algorithm Step #2.
rj : initial restoring force (onset of load step j + 1).re = rj + Ke∆uj+1 : elastic predictor of the restoring force (end of loadstep).
Institute of Structural Engineering Method of Finite Elements II 28
Return Mapping Algorithm
Return mapping algorithm Step #3.
rj : initial restoring force (onset of load step j + 1).re = rj + Ke∆uj+1 : elastic predictor of the restoring force (end of loadstep).rj+1 = re −Kem∆λj+1 : exact restoring force (end of load step) thatsatisfies f (rj+1) = 0.
Institute of Structural Engineering Method of Finite Elements II 28
Return Mapping Algorithm
The Newton-Raphson algorithm is used to build a return mappingalgorithm:
{̂rj+1, ∆λ̂j+1} :
{εr = r̂j+1 − re + Kem∆λ̂j+1 = 0εf = f (̂rj+1) = 0
↓[rk+1j+1
∆λk+1j+1
]=
[rkj+1
∆λkj+1
]−[∂εr∂r
∂εr∂∆λ
∂εf∂r
∂εf∂∆λ
]−1 [εkrεkf
]where:
re = uj + Ke∆uj+1 is the elastic predictor.
r1j+1 = rj and ∆λ1j+1 = 0 is the initialization.
k is the iteration index.
Institute of Structural Engineering Method of Finite Elements II 29
Return Mapping Algorithm
The Newton-Raphson algorithm is used to build a return mappingalgorithm:
{̂rj+1, ∆λ̂j+1} :
{εr = r̂j+1 − re + Kem∆λ̂j+1 = 0εf = f (̂rj+1) = 0
↓[rk+1j+1
∆λk+1j+1
]=
[rkj+1
∆λkj+1
]−[
I KemnT 0
]−1 [εkrεkf
]where:
re = uj + Ke∆uj+1 is the elastic predictor.
r1j+1 = rj and ∆λ1j+1 = 0 is the initialization.
k is the iteration index.
Institute of Structural Engineering Method of Finite Elements II 29
Return Mapping Algorithm
The Newton-Raphson algorithm is used to build a return mappingalgorithm:
{̂rj+1, ∆λ̂j+1} :
{εr = r̂j+1 − re + Kem∆λ̂j+1 = 0εf = f (̂rj+1) = 0
↓[rk+1j+1
∆λk+1j+1
]=
[rkj+1
∆λkj+1
]−
[I− KemnT
nT DemKem
nT KemnT
nT Kem−1
nT Kem
] [εkrεkf
]where:
re = uj + Ke∆uj+1 is the elastic predictor.
r1j+1 = rj and ∆λ1j+1 = 0 is the initialization.
k is the iteration index.
Linear Jacobian → convergence in one iteration !!!Institute of Structural Engineering Method of Finite Elements II 29
Consistent Tangent Stiffness
The Jacobian computed for the last iteration of the Newton-Raphsonalgorithm provides the consistent tangent stiffness matrix:
{̂rj+1, ∆λ̂j+1} :
{εr = r̂j+1 − re + Kem∆λ̂j+1 = 0εf = f (̂rj+1) = 0
↓[rk+1j+1
∆λk+1j+1
]=
[rkj+1
∆λkj+1
]−[∂εr∂r
∂εr∂∆λ
∂εf∂r
∂εf∂∆λ
]−1 [εkrεkf
]↓[
rk+1j+1∆λk+1j+1
]=
[rkj+1
∆λkj+1
]−
[∂r∂εr
∂r∂εf
∂∆λ∂εr
∂∆λ∂εf
][εkrεkf
]
Institute of Structural Engineering Method of Finite Elements II 30
Consistent Tangent Stiffness
The Jacobian computed for the last iteration of the Newton-Raphsonalgorithm provides the consistent tangent stiffness matrix:
{̂rj+1, ∆λ̂j+1} :
{εr = r̂j+1 − re + Kem∆λ̂j+1 = 0εf = f (̂rj+1) = 0
↓
Kj+1 =∂rj+1∂uj+1
=∂rj+1∂∆uj+1
= −∂rj+1∂εr
∂εr∂∆uj+1
with,
∂ (∆uj+1) = ∂ (uj+1 − uj) = ∂uj+1 −���>
constant∂uj = ∂uj+1
Institute of Structural Engineering Method of Finite Elements II 30
Consistent Tangent Stiffness: Spring-Slider System
For the spring-slider system, the consistent tangent stiffness matrixreads:
εr = r̂j+1 − re + Kem∆λ̂j+1 = r̂j+1 −
(r̂j + K
e(
∆uj+1 −m∆λ̂j+1))[
∂r∂εr
∂r∂εf
∂∆λ∂εr
∂∆λ∂εf
]=
[I− KemnT
nT KemKem
nT KemnT
nT Kem−1
nT Kem
]↓
Kj+1 =∂rj+1∂uj+1
=∂rj+1∂∆uj+1
= −∂rj+1∂εr
∂εr∂∆uj+1
=
(I− K
emnT
nTKem
)Ke
with,
∂ (∆uj+1) = ∂ (uj+1 − uj) = ∂uj+1 −���>
constant∂uj = ∂uj+1
Institute of Structural Engineering Method of Finite Elements II 31
Return Mapping Algorithm: Code Template
1: ∆uj+1 ← uj+1 − uj2: re ← rj + Ke∆uj+13: if f (re) ≥ 0 then4: rj+1 ← re5: ∆λj+1 ← 06: εr ← rj+1 − re + Kem∆λj+17: εf ← f (rj+1)8: repeat
9:
[rj+1
∆λj+1
]←
[rj+1
∆λj+1
]−
[∂εr∂r
∂εr∂∆λ
∂εf∂r
∂εf∂∆λ
]−1 [εrεf
]10: εr ← rj+1 − re + Kem∆λj+111: εf ← f (rj+1)12: until ‖ε‖ >= Tol13: Kj+1 ← − ∂r∂εr
∂εr∂uj+1
14: else if f (re) < 0 then15: rj+1 ← re16: Kj+1 ← Ke17: end if
Institute of Structural Engineering Method of Finite Elements II 32
Associated vs. Non-Associated Plastic Flow
Some concluding remark:
nT = [1,tanϕ] : outward normal of the yielding surface (in thestress/force space)
mT = [1,tanψ] : direction of the plastic deformation flow (inthe strain/displacement space)
m = n : the plastic deformation flow and the normal to theyielding surface are co-linear. This is the so called associatedplasticity case that holds, for example, for metals.
m 6= n : the plastic deformation flow and the normal to theyielding surface are not co-linear. This is the so callednon-associated plasticity case that holds, for example, for soils.
Institute of Structural Engineering Method of Finite Elements II 33
Nonlinear Static Analysis (r,u) (Force Control)
So far we focused on calculating the force response of a singleelement given a displacement trial ...
... but we want to solve the static response of a model subjected toan external load history.
Therefore, we need to solve the following balance equation,
r (uj)− fj ,ext = 0
where j is the analysis step index and,
uj : displacement vector
r (uj) : restoring force vector
fj ,ext : imposed load vector
Institute of Structural Engineering Method of Finite Elements II 34
Nonlinear Static Analysis (r,u) (Force Control)
1: for j = 1 to J do2: uj ← uj−13: for i = 1 to I do4: [ri,j ,Ki,j ]← element (Ziuj)5: rj ← rj + ZTi ri,j6: Kj ← Kj + ZTi Ki,jZi7: end for8: res← rj − fj,ext9: repeat
10: jac← Kj11: uj ← uj − jac−1res12: for i = 1 to I do13: [ri,j ,Ki,j ]← element (Ziuj)14: rj ← rj + ZTi ri,j15: Kj ← Kj + ZTi Ki,jZi16: end for17: res← rj − fj,ext18: until ‖res‖ >= Tol19: end for
Institute of Structural Engineering Method of Finite Elements II 35
Nonlinear Static Analysis (r,u) (Displacement Control)
In case the the restoring force is bounded, balance may not besatisfied for high loads ...
... imposing a displacement history is more convenient in this case.
Therefore, we need to solve both balance and compatibilityequations, {
r (uj) + LTλj = 0
Luj − uj ,ext = 0
where j is the analysis step index and,
uj : displacement vector
r (uj) : restoring force vector
uj ,ext : imposed displacementvector
λj : imposed load vector
L : collocation matrix
Institute of Structural Engineering Method of Finite Elements II 36
Nonlinear Static Analysis (r,u) (Displacement Control)
1: for j = 1 to J do2: uj ← uj−13: λj ← λj−14: for i = 1 to I do5:
[ri,j ,Ki,j
]← element
(Ziuj
)6: rj ← rj + ZTi ri,j7: Kj ← Kj + ZTi Ki,jZi8: end for9: res←
[rj + L
TλjLuj − uj,ext
]10: repeat11: jac←
[Kj L
T
L 0
]12: ∆x← −jac−1res13: uj ← uj + ∆x (1)14: λj ← λj + ∆x (2)15: for i = 1 to I do16:
[ri,j ,Ki,j
]← element
(Ziuj
)17: rj ← rj + ZTi ri,j18: Kj ← Kj + ZTi Ki,jZi19: end for20: res←
[rj + L
TλjLuj − uj,ext
]21: until ‖res‖ >= Tol22: end for
Institute of Structural Engineering Method of Finite Elements II 37