FEM Lecture 3
Transcript of FEM Lecture 3
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The Finite Element Method for the Analysis ofNon-Linear and Dynamic Systems
Prof. Dr. Eleni Chatzi
Lecture 3 - 9 October, 2012
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Introduction to Nonlinear Analysis
Conclusion from the previous example:
The basic problem in general Nonlinear analysis is to find a state of
equilibrium between externally applied loads and element nodal forces
tR t F= 0tR=t RB+
t RS+t RC
tF=
m
tVm
tB(m)T t(m) tdV(m)
where RB: body forces, RS: surface forces, RC: nodal forces
We must achieve equilibrium for all time steps whenincrementing the loading
Very general approach
Includes implicitly also dynamic analysis!Institute of Structural Engineering Method of Finite Elements II 2
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Types of Response Diagrams
Basic Types
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Types of Response Diagrams
Complex Types
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Solution Algorithms for NL equations
Root finding for single variable NL problems f(x) = 0
Bisection Method
Assumption: f[a, b] and continuous
If f(a)>0, f(b)
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Solution Algorithms for NL equations
Root finding for single variable NL problems f(x) = 0
Newton (Raphson) Method
Defined by the recurrence relation
xk+1=xk f(xk)f (xk)
terminate when |xk+1 xk| ,
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Newtons method for FE
Assume the general case where Kis a nonlinear function of thedisplacement U:
K(U)U= F
In order to write it in the form f(x) = 0, we define the residual r(U):
r(U) =K(U)U F r(U) = 0
Then, the Newton iteration formula becomes:
Uk+1 = Uk T1(Uk)r(Uk)
The slope of the tangent (the tangent stiffness) is
T(Uk) = dr(Uk)
dU =
dK(U)U F
dU |U=Uk
T(Uk) =K(Uk) +dK(Uk)
dU Uk
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Incremental Analysis
The basic approach in incremental analysis is:Find a state of equilibrium between externally applied loads and
element nodal forces between successive time steps t
t+tR t+tF= 0
Assuming that t+tR is independent of the deformations we have
t+tR= tF + F
We know the solution tFat time t and F is the increment in thenodal point forces corresponding to an increment in the
displacements and stresses from time tto time t+ t. This we canapproximate by
F= tKU
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Incremental Analysis
Newton-Raphson MethodAssume the tangent stiffness matrix:
tK= tFtU
We may now substitute the tangent stiffness matrix into theequilibrium relation
tKU= t+tR tF
which gives us a scheme for the calculation of the displacements:
t+tU= tU + U
The exact displacements at time t+ tcorrespond to the appliedloads at t+ t, however we only determined these approximately aswe used a tangent stiffness matrix thus we may have toiterate to
find the solution.Institute of Structural Engineering Method of Finite Elements II 9
I l A l i
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Incremental Analysis
We may use the Newton-Raphson iteration scheme to find the
equilibrium within each load increment
t+tK(i1)U(i) = t+tR t+tF(i1)
(out of balance load vector)
t+tU(i) = t+tU(i1)
+ U(i)
with Initial Conditions
t+tU(0) = tU; t+tK(0) = tK; t+tF(0) = tF
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M difi d N (R h )M h d
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Modified Newton (Raphson)Method
It may be expensive to calculate the tangent stiffness matrix. In theModified Newton-Raphsoniteration scheme it is only calculated in thebeginning of each new load step
In the quasi-Newtoniteration schemes the secant stiffness matrix is used
instead of the tangent matrixInstitute of Structural Engineering Method of Finite Elements II 11
S i l C id ti
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Special Considerations
Standard Newton-Raphon methods perform poorly for buckingproblems, where the slope at limit points is exactly equal to 0
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S i l C id ti
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Special Considerations
The Arc-Length Method for Nonlinear Post-Buckling
Also called Modified Riks Method.
Control the size of the load step using aparameter .
Solve for both and U in each Newtoniteration.
Assume F = independent of geometry. Then, can be thought of as a normalized loadparameter and the residual is given by
r(U, ) =K(U)U F
The load increment is computed using
=
s2 U2n
where the reference arc length is
s20 = F
nloadstepInstitute of Structural Engineering Method of Finite Elements II 13
Si l B E l R isit d
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Simple Bar Example - Revisited
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Simple Bar Example Revisited
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Simple Bar Example - Revisited
0 0 (1) 1 1 (0) 1 (0)
4(1) 3
7
1 (1) 1 (0) (1) 3
1 (1)1 (1) 4
1 (1)1 (1)
Load step 1: 1:
( )
2 106.6667 10
1 110 ( )
10 5
Iteration 1 : ( 1)
6.6667 10
6.6667 10 < (elastic section!)
1.3333
a b a b
a Y
a
b
b
t
K K u R F F
u
i
u u u
u
L
u
L
=
+ =
= =
+
== + =
= =
= = 3
1 (1) 3 1 (1) 4
0 0 (2) 1 1 (1) 1 (1)
10 < (elastic section!)
6.6667 10 ; 1.3333 10
( ) 0
Y
a b
a b a b
F F
K K u R F F
= =
+ = = 1 3
Convergence in one iteration!
6.6667 10u =
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Simple Bar Example Revisited
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Simple Bar Example - Revisited
1 1 (1) 2 2 (0) 2 (0)
4 3 4(1) 3
7
2 (1) 2 (0) (1) 2
2 (1) 3
2
Load step 2: 2 :( )
(4 10 ) (6.6667 10 ) (1.333 10 )6.6667 10
1 110 ( )
10 5Iteration 1:( 1)
1.3333 10
1.3333 10 < (elastic section!)
a b a b
a Y
t
K K u R F F
u
i
u u u
=
+ =
= =
+
=
= + =
=
(1) 3
1 (1) 4 1 (1) 2 (1) 4
1 1 (2) 2 2 (1) 2 (1) (2) 3
2.6667 10 > (plastic section!)
1.3333 10 ; ( ( ) ) 2.0067 10
( ) 2.2 10
b Y
T
a b b Y Y
a b a b
F F E A
K K u R F F u
=
= = + =
+ = =
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Simple Bar Example Revisited
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Simple Bar Example - Revisited
The procedure is repeated and the results of successive iterations aretabulated in the accompanying table.
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The Continuum Mechanics Incremental Equations
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The Continuum Mechanics Incremental Equations
The basic ProblemEstablish the solution using an incremental formulation. Two mainapproaches exist for establishing equilibrium
Lagrangian Formulation:Track the movement of all particles of the body (located in aCartesian coordinate system), in their motion from the original to thefinal configuration (pathline)
Eulerian Formulation:The motion of the material through a stationary control volume isconsidered (streamlines). Mainly used in fluid mechanics.
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Lagrangian vs Eulerian Formulation - 1D Example
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Lagrangian vs. Eulerian Formulation - 1D Example
Spatial or Eulerian coordinates (x): These coordinates are used to locate apoint 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= (X, t) =X(1 + 2t+t2)
The inverse of the map gives us X in terms of x, i.e.,
X= 1
(x, t) = x
(1 + 2t+ t2)
Then, the displacement of the material point X is
u(X, t) =(X, t) (X, 0) =X(2t+ t2)
The velocity of the material point is (Langrangian Description)
v(X, t) = u
t= 2X(1 +t)
Alternatively we can express the velocity in terms ofx (Eulerian Description)
v(X, t) =v(1(x, t), t) = 1x(1 +t)
(1 + 2t+t2)
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Lagrangian Formulation
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Lagrangian Formulation
In solids we use the Lagrangian approach as the solution process moves from time
t to t+ titeratively following elements of the body in their motion.
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