Industrial Applications of Computational … Applications of Computational Mechanics Shear Walls and...
Transcript of Industrial Applications of Computational … Applications of Computational Mechanics Shear Walls and...
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Industrial Applications of ppComputational MechanicsShear Walls and Fluids
Prof. Dr.-Ing. Casimir KatzSOFiSTiK AG
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FEM ReminderFEM - Reminder
A mathematical methodTh l ( ti ) ld i d The real (continuous) world is mapped on to a discrete (finite) one.
We restrict the space of solutions. We calculate the optimal solution withinWe calculate the optimal solution within
that space on a global minimum principleD ’t t l l i i Don’t expect local precision
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Plates (Slabs and shear walls)Plates (Slabs and shear walls)
Classical plate bending solution (Kirchhoff) w = p
Classical solution for shear walls (Airy stress function F)stress function F)
F = 0 FE / Variational approach for bending plates FE / Variational approach for bending plates
= ½ D dV = Minimum FE / Variational approach for shear walls
= ½ D dV = MinimumKatz_03 / Computational Mechanics3
= ½ D dV = Minimum
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Shear Walls YShear Walls
Unknowns:X
Displacements u=ux and v=uy Rotation z is not defined (see Cosserat)
Strainsx = u/ x ; y = v/ y2 / + / 2 xy = xy = v/ x + u/ y
stresses [ ]x , y , xy [ , z ]
Plane Strain Condition z = 0 Plane Stress Condition = 0
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Plane Stress Condition z = 0
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Plane Stress ConditionPlane Stress Condition
1 01 0
x xE 2 1 0
10 0 (1 ) / 2
y y
xy xy
1 0x xnE t
2 1 01
0 0 (1 ) / 2y y
xy xy
E tnn
( )xy xy
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Plane Strain ConditionPlane Strain Condition
1 0 1 01 0
x x
y yE
0 0 1 2 / 2 01 1 20 1
xy xy
z z
Plane Strainz = 0
Axissymmetric Condition Axissymmetric Conditionz = u/r
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Remarks
1 0
1 01 0
0 0 1 2 / 2 01 1 2
x x
y y
xy xy
E
Shear modulus G = E/(2*(1+)) at Position 3,3
0 1
xy xy
z z
Incompressible limit» For = 0.5 the matrix becomes singular
Extensions for anisotropic behaviour via inverse matrix» E-Modulus in fibre direction x = x/Ex + xy• y/Ey
» E-Modulus transverse to fibre direction (E ) analog» E-Modulus transverse to fibre direction (E90) analog» Poisson ratio for off diagonal term is not uniquely defined» Rotation of axis of Isotropy creates a fully populated matrix
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» Special effects for foams possible
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Membrane elementMembrane element
constant strain triangular elements CST constant strain triangular elements CST» Linear displacements» Constant stress» Constant stress
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Enhanced triangular elementsEnhanced triangular elements
LST Element with 6 nodes LST – Element with 6 nodes» Complete quadratic function space
• Drilling-Degrees of Freedom- The displacements of the mid nodes areThe displacements of the mid nodes are
calculated from the end nodes including the rotation
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Membrane ElementsMembrane Elements
Quadrilateral bilinear elements Quadrilateral bilinear elements» Linear Displacements» Constant stress» Constant stress
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EnhancementsEnhancements
Q d ti Sh f ti Quadratic Shape functions» Lagrange Elements (nine noded)» Serendipity (without central node)
» „Isoparametric“
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DisadvantagesDisadvantages
U if l di t d l l d Uniform loadings creates nodal loads as:1/6 , 2/3 , 1/6
Thus coupling with beam elements is difficult, i.e we need also isoparametric , pbeam elements
Special coupling conditions (friction no Special coupling conditions (friction, no tension etc) also difficult, i.e. we need i t i i t f l tisoparametric interface elements
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Drilling Degrees IDrilling Degrees I
Same principle as with the triangular element Same principle as with the triangular element Further mathematical tricks required to reduce
th f th h f ti d t ththe space of the shape functions compared to the Serendipity-ElementM t d l l d t t d t d / Moments as nodal loads not easy to understand / handle
f f Advantages for folded structures or shells expected
My own benchmarks showed poor quality of results.
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EnhancementsEnhancements
Bili f i (Wil ) Bilinear non conforming (Wilson)u = ... + (1-s2) q1 + (1-t2) q2v = ... + (1-s2) q1 + (1-t2) q2
May model constant curvatures exactlyay ode co sta t cu atu es e act y Static Condensation
P h T f lfill d i h i k Patch-Test fulfilled with a trick(Jacobi-Determinant is treated as constant)
Newer approach: Assumed strains
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Newer approach: Assumed strains
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PatchtestPatchtest
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Patchtest ()Patchtest ()
Moment constant linear quadratic
Mesh x=0 x=l/2 x=0 x=l/2 x=0 x=l/2Mesh x 0 x l/2 x 0 x l/2 x 0 x l/2
Reference 1500 1500 1200 600 1200 300
R Q4+2 1500 1500 1051 600 940 337
V Q4+2 1322 1422 1422 701 773 452V Q4+2 1322 1422 1422 701 773 452
R Q4 1072 1072 1072 428 659 240
V Q4 687 578 578 187 393 172
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Patchtest ()Patchtest ()
Moment constant linear quadratic
Mesh x=0 x=l/2 x=0 x=l/2 x=0 x=l/2Mesh x 0 x l/2 x 0 x l/2 x 0 x l/2
Reference 0 0 50 50 100 50
R Q4+2 0 0 50 50 87.5 50
V Q4+2 58 28 65 80 130 73V Q4+2 58 28 65 80 130 73
R Q4 438 0 364 8 376 8
V Q4 502 220 380 294 366 11
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Convergence of displacementsConvergence of displacements
(Q ) (Q )Mesh u (Q4) u (Q4+2)
1 x 8 0.715 1.035
2 x 16 0.939 1.036
4 x 32 1.010 1.038
8 x 80 1.021 1.039
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Drilling Degrees IIDrilling Degrees II
Better approach with a strain field (Hughes/Brezzi) Better approach with a strain field (Hughes/Brezzi)
symm grad v c symm grad v d 2
skew grad v d
Constraint about the rigid body rotations Adding deformation energy makes the element
Adding deformation energy makes the element stifferC bi ti ith f i d i Combination with nonconforming modes is recommended
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LoadsLoads
N d l l d i t l d Nodal loads are no point loads
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LoadsLoads
R l ti f h f l d Resolution of a mesh for loads
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MomentsMoments
Th i d f th t ! There is no degree for that ! Possibilities
» Use more than one node» Kinematic Constraints» Kinematic Constraints» Drilling degrees of freedom
phi=u/y
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Result EvaluationResult-Evaluation
Displacements u Displacements u Support Forces (Residuals)
f = K u pf = K u –p (for all degrees of freedom!)
Stresses in elementsStresses in elements» Centre (Super convergent points)» Gauss-Points» Nodes of elements (Extrapolation!)
Mean values in nodes Error estimates Super-Convergent-Patches
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Post processing forces
Mean values in nodes Smoothing of local singularities Smoothing of local singularities
» May be omitted in general as the FE-solution ill d thi i li itl h t th i dwill do this implicitly, or has not the required
precision for that.
I l f f Integrals of forces» Very good method to obtain better results
especially for design purpose
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EquilibriumEquilibrium
As it is the base for our solution it should be As it is the base for our solution it should be fulfilled for the residual forces even if system and loadings are completely garbageloadings are completely garbage.
It is not fulfilled within the elementsIt i t f lfill d t th d f th l t It is not fulfilled at the edges of the elements
It is not fulfilled within general cuts across the elements
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Nodal stressesNodal stresses
St di ti b t Stresses are discontinuous between elements
• The jump value of the stresses is a measure for the quality (e g error) of the solution forfor the quality (e.g. error) of the solution for that mesh
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Nodal stressesNodal stresses
M l f t i d t bt i Mean value of stresses in nodes to obtain „nicer“ pictures» Discontinuity of Thickness» Discontinuity of E-Modulus» Discontinuity of Geometry
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SingularitiesSingularities
Ri id F ti h lf Rigid Footing on a half space
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SingularitiesSingularities
R E t t C Re-Entrant Corners
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Adaptive Mesh RefinementAdaptive Mesh Refinement
At those locations where we have a large error estimate we At those locations where we have a large error estimate we refine the mesh either geometrically (h-Version) or we increase the Polynomial degree (p_Version) or both (hp-Version).
Strong advantages compared to a uniform refinement Loads are not allowed to be defined for nodes or elements,
but are required in a more general geometric way. For any design purpose we need results for all load cases For any design purpose we need results for all load cases
at the same location. => a mesh for every load case makes life not easier.
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Detail of supportsDetail of supports
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Detail of supportsDetail of supports
El t tSingularity!
Element stress is discontinuous
Nodal stresses as mean valuesas mean values are not correct at this pointat this point
Separate the nodal stresses in groups
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RemarksRemarks
Axissymmetric case Axissymmetric case» The strains are neither constant nor linear nor quadratic» None of the classical elements may describe this exactly» None of the classical elements may describe this exactly» Integral of loads has to include the radius» But not the nodal loads !» But not the nodal loads !
3D case» Most of the plain strain issues are also valid» Most of the plain strain issues are also valid» Two more shear stresses» Elements as Hexahedra or Tetrahedra or something in g
between» Mesh generation is a complex topic !
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Construction StagesConstruction Stages
Classical Approach: Classical Approach:We build and then we switch gravity on
There are many cases especially those including There are many cases, especially those including non linearities where the simulation of the construction process becomes essentialp» Dam construction» Tunnelling» Bridges
Effects to consider:St th» Stress path
» Change of forces due to creep» Adding or removing parts of the structure
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» Adding or removing parts of the structure
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A simple beam exampleA simple beam example
Two single span Two single span beams
• Connected to a continuous beam=> Creep will> Creep will change the forces towards the
ti• Changed to a
single span beam
continuous case
single span beam=> Moment distribution will be
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different
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Removal of central supportRemoval of central support
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How is this done best ?How is this done best ?
Fi t P i i l St i i t ! First Principle: Strain increments! Consider each load case not in total but as
difference to the primary state before New stresses are old stresses + tangentialNew stresses are old stresses + tangential
stiffness times strain increments
new old tE
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Incremental loadsIncremental loads
Th t f th i l d The stresses of the primary load case are in equilibrium with the loadings and support forces of the primary estate
The load vector of the new case is the difference between the total load vector and the residual load vector of the primaryand the residual load vector of the primary stresses.
Ttotal primaryP P B dV
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How it worksHow it works
+
=
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Last not leastLast not least
P ti ll d ti t d t Partially removed or activated systems» Shotcreet hardening» Tunnelling» Loss of strength due to many effectsg y» Icing and deicing a soil
The full set of tools The full set of tools» Factor for Stiffness
F t f i t» Factor for primary stress» Factor for primary loading
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Enlargement of Metro StationMarienplatz Munich
Old TunnelDesigned to withstand the fullDesigned to withstand the full overburden pressure
New TunnelDesigned to support/ mobilize the soil’s load carrying
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the soil s load carrying capacity
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Analytical ModelAnalytical Model
Analysis at a set of representative cross-section slices under plane strain conditionsunder plane strain conditions
• Incorporating 3D stress redistribution effects by stiffness reduction method (-Method)
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Primary Estate Plane system with hole and liningsReduced axial strain of Tunnel („Softening“)
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Stiffness reduction method(-Method)
Disc with relaxed tunnel corePrimary Estate Disc with hole and supporting lining
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Stiffness reduction method(-Method)
Displacements of
Primärspannungs-Zustand
Reduced axial strain of Tunnel („Softening“)
Disc with hole and supporting lining
Displacements of unsupported face
Arching in longitudinal direction
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(„ g )
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Stiffness reduction method(-Method)
Interaction between soil and
Primärspannungs-Zustand
Entspannter Tunnelkern („Aufweichen“)
Final 2D system withhole and linings
Interaction between soil and shotcrete
Loading of the linings
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(„ ) g
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Finite element modelFinite element model
Gravel
Clay
Sand
Ice shield
Pilot tube
Clay
Existing tunnelNew tunnel
Ice shield
Sand
Clay
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Simulation stages:1st phase, primary stress state
Simulating the historic construction process
generating a model loading state that reflects the situation prior to construction activityy
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Simulation stages: 2nd phase, preparatory steps
Drainage of lower ifaquifer
Installation of pilot tunneltunnel
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Simulation stages: 3rd phase, tunnelling process
Soil freezing Excavation base
Installation shotcrete lining
Relaxation of calotte region
lining
Defrostingcalotte region
Drainage turning-off
Excavation calotte g
Installation shotcrete lining
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Simulation stages: 3rd phase, Tunnelling process
Soil freezing Excavation base
Installation shotcrete lining
Relaxation of calotte region
lining
Defrostingcalotte region
Drainage turning-off
Excavation calotte g
Installation shotcrete lining
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Simulation stages: 3rd phase, Tunnelling process
Soil freezing Excavation base
Installation shotcrete lining
Relaxation of calotte region
lining
Defrostingcalotte region
Drainage turning-off
Excavation calotte g
Installation shotcrete lining
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Soil freezingSoil freezing
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DefrostingDefrosting
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Simulating the 3D cross cutting process
There are some more phases in 3D ....
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DesignDesign
Finite Element Analysis is linear in general Design is based on ultimate loads and Design is based on ultimate loads and
plasticity The real ultimate loading depends on all
elements within the structure. Thus, you may be either
» Not economical» Not economical» Not save
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Example of a BeamExample of a Beam
5 “membrane” elements ClassicalEach obtaining its individualreinforcement
beam element
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M = 250 kNM = 250 kN
Stress Reinforcement Element Theoret. FE Classical FE
1 11 11 11 03 4851 -11.11 -11.03 4852 -5.56 -5.51 106 3 0.00 0.00 0 4 5.55 5.51 694 5 11.11 11.03 1973 1391
Sum 1973 2676Su 973 676Beam design with distribution
from FE-Results
2658
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M=250 kNm N= 500 kNM=250 kNm, N= -500 kN
Stress Reinforcement Element Theoret. FE Classical FE
1 -13.89 -13.81 540 9852 -8.33 -8.29 160 3 -2 79 -2 78 543 2.79 2.78 544 2.79 2.73 344 5 8.33 8.29 1262 1040
S 1802 2583Sum 1802 2583Beam design with
distribution from FE-Results
2244
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M=250 kN N= 1000 kNM=250 kN, N= -1000 kN
Stress Reinforcement Element Theoret. FE Classical FE
1 -16.67 -16.58 1359 14862 -11.11 -11.06 492 3 -5 56 -5 56 1073 5.56 5.56 1074 0.00 0.05 0 5 5.56 5.48 1359 691
S 2718 2776Sum 2718 2776Beam design with distribution
from FE-Results
2260
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Design of Shear WallsDesign of Shear Walls
Direction of principal stresses Direction of reinforcements Direction of cracksDirection of cracks Models available from Baumann / Leonhardt or
Stiglat/WippelStiglat/Wippel Analysis based on minimum of deformation work A bunch of detailed problems be careful e g A bunch of detailed problems - be careful e.g.
inclined compressive reinforcement
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Structured shear wallsStructured shear walls
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Structured Shear WallsStructured Shear Walls
BeamsMod. Stiff. in nodes
Rigid Nodes
FESolutionBeams in nodes Nodes Solution
N (left) 25.6 133.2 133.2 117.8
V (left) 200.1 187.7 195.2 191.8
M (left) -2818 -2216 -2251 -2316
N (right) -25.6 -133.2 -133.2 -117.8
V (right) 137.4 149.8 142.3 145.5
M (right) -2612 -2146 -2112 -2199
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Example of a barrageExample of a barrage
308.3
308.0
305.0
2.00
2279.0
.0
296.0
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ProblemProblem
Stability of a barrage with a height of 32 50 m and Stability of a barrage with a height of 32,50 m anda width of 21 m.
Load cases to be considered: Load cases to be considered:» Seepage» Temperaturep» Ice pressure» Earthquake
On the water side there is an additional brickworkand a so called „Intze-Keil“ to ensure watertightnesstightness
Non linear material for rock and dam
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Friction between Soil and DamFriction between Soil and Dam
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Seepage through damSeepage through dam
308.3
308.0
305.0
302.0300 0 286 0
284.0
282.0
280.0279.0
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300.0
98.0
96.0
288.0
286.0
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Stresses including seepageStresses including seepage
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Transient TemperaturesTransient Temperatures
12.0
15.0
0
1 0
Winter
11.0
9.00
4.00
7.00
S
8.00
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Summer
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EarthquakeEarthquake
There is a fluid structure interaction problem There is a fluid structure interaction problem A simple approach just adds some mass to the
b di t th di t ib ti fbarrage according to the distribution of WestergardY h t dd l d di th You have to add a mass value depending on the distance of the nodes and their depth.
? How ?» Excel sheet (if you know the coordinates)
With bi l d di t ib ti (if h d» With a cubic load distribution (if you have good software)
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Stresses including EarthquakeStresses including Earthquake
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Near Collapse with 7 xpEarthquake forces
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Simple Fluid-ElementsEulerian Approach
Differential equation connecting velocity of nonisco s fl id to press re
pp
viscous fluid to pressure:
up
The density of the fluid is constant
up f
The density of the fluid is constant Velocities are so small that higher order effects
may be neglected (Convection Cavitation etc)may be neglected, (Convection, Cavitation etc). Shear stresses (viscous effects) are neglected
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Fluid Structure Interaction
Flow is described by velocities (Eulerian-A h)Approach)
Structure is described by displacements y p(Lagrangian Approach)
Coupling quite complex Coupling quite complex
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Lagrange Approach for Fluids
Description of Stresses with Compression and Shear modulus:
)21(3
E
EK
1 01 0
0 0 1 2 / 2 01 1 2
x x
y y
xy xy
E
Quite Common in)1(2
EG
0 1
xy xy
z z
4 2 2 Quite Common in Soil Mechanics
Limit State G => 0, => 0.5
4 2 203 3 32 4 20
x x
y y
K G K G K G
K G K G K G
Some viscous effects possible
3 3 30 0 02 2 40
y y
xy xy
z z
G
K G K G K G
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03 3 3
K G K G K G
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Water
Compression modulusK = 2000 N/mm2
dynamic Viscosityh 0 00000000165 N / 2h = 0.00000000165 Ns/mm2
Shear Modulus very smally
GG *
G
GG
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First TestHEAD WATERTANKHEAD WATERTANK
NODE 1 0.00 0.0 ; 10 = 1.905
301 5 08 0 0 ; 310 = 1 905301 5.08 0.0 ; 310 = 1.905
MAT 1 K 2E6 G 2E3 GAM 10.0MAT 1 K 2E6 G 2E3 GAM 10.0
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Katz+Bellmann - Beratende Ingenieure VBIWINGRAF 3397 13.04.99
0.00
1.00
Ein Wassertank M 1 : 19X
Y
Z Verschobene Struktur aus LF 11 Frequenz 70.5 Hertz in 2.00-facherÜberhöhung
1.00 2.00 3.00 4.00 5.00
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Katz+Bellmann - Beratende Ingenieure VBIWINGRAF 3397 13.04.99
0.00
01.00
2.00
Ein Wassertank M 1 : 19X
Y
Z Verschobene Struktur aus LF 12 Frequenz 101.2 Hertz
1.00 2.00 3.00 4.00 5.00
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1st Eigen frequency1st Eigen frequency
G-Modulus PlaneStrain
K+Gplane strain
Plane Stress
200 000 70.61 70.45 69.18
2000 8 19 7 08 6 942000 8.19 7.08 6.94
20 3.55 0.71 0.69
0.2 4.50 0.07 0.069
Reference 0.36 0.36 0.36
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FlawsFlaws
Locking of incompressible media => Three field approximation> Three field approximation
Potential of free surface is missingBOUN 1 TITLE 'Free Surface'
BOUN 1 301 10 CY 10.0
==> Frequencies become bounded
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Katz+Bellmann - Beratende IngenieureVBI
WINGRAF 319731 03 99 VBI31 03 99
0.00
1.00
2.00
Eink
M 1 :19X
Y
Z Verschobene Struktur aus LF 42 Frequenz 0.144
1.00
2.00
3.00
4.00
5.00
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FlawsFlaws
Stiffness of surface now larger then fluid stiffness
R t ti l d f ti h tiff Rotational deformations have no stiffness=> Introduce Penalty function
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Katz+Bellmann - Beratende Ingenieure VBIWINGRAF 3197 31.03.99 Katz+Bellmann Beratende Ingenieure VBIWINGRAF 3197 31.03.99
0.00
1.00
2.00
Ein Wassertank M 1 : 19X
Y
Z Verschobene Struktur aus LF 43 Frequenz 0.697 Hertz in 0.500-facherÜberhöhung
1.00 2.00 3.00 4.00 5.00
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FlawsFlaws
„spurious modes“
Mass matrix has still modes which are suppressed in the stiffness matrixsuppressed in the stiffness matrix=> Projection by Kim/Yong
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EigenfrequenciesEigenfrequencies
G-Modulus f1 f2 f3
200 000 70.455 101.210
2000 7 101 10 246 13 2542000 7.101 10.246 13.254
20 0.951 1.846 2.730
0.2 0.371 0.582 0.730
0 02 0 361 0 557 0 6890.02 0.361 0.557 0.689
Reference 0.357 0.555
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Katz+Bellmann - Beratende Ingenieure VBIWINGRAF 3397 13.04.99
0 0041 0
0.00260.00280.00300.00320.00340.00360.00380.0041
-1.00
0 00090.00110.00130.00150.00170.00190.00210.0023
0.00
-0.0006-0.0004-0.00020.00000.00020.00040.00060.0009
1.00
-0.0023-0.0021-0.0019-0.0017-0.0015-0.0013-0.0011-0.0009
0 0041-0.0038-0.0036-0.0034-0.0032-0.0030-0.0028-0.0026
2.00
Ein Wassertank M 1 : 25X
Y
Z
-0.0043-0.0041
Verschobene Struktur aus LF 81 Frequenz 0.361 HertzSpannungsmittelwert (1.Invariante) im Knoten, Lastfall 81, von -0.0043bis 0.0043 Stufen 2.1329e-04 MPa
0.00 1.00 2.00 3.00 4.00 5.00
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bis 0.0043 Stufen 2.1329e 04 MPa
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Katz+Bellmann - Beratende Ingenieure VBIWINGRAF 3397 13.04.99
0 0058 .00
0.00370.00400.00430.00460.00490.00520.00550.0058 -1
.
0 00120.00150.00180.00210.00240.00270.00300.0034
0.00
-0.0009-0.0006-0.00030.00000.00030.00060.00090.0012
1.00
-0.0034-0.0030-0.0027-0.0024-0.0021-0.0018-0.0015-0.0012
0
0 0058-0.0055-0.0052-0.0049-0.0046-0.0043-0.0040-0.0037 2.
00
Ein Wassertank M 1 : 25X
Y
Z
-0.0061-0.0058
Verschobene Struktur aus LF 82 Frequenz 0.557 Hertz in 0.500-facherÜberhöhungSpannungsmittelwert (1.Invariante) im Knoten, Lastfall 82, von -0.0061
0.00 1.00 2.00 3.00 4.00 5.00
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Spannungsmittelwert (1.Invariante) im Knoten, Lastfall 82, von 0.0061bis 0.0061 Stufen 3.0490e-04 MPa
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Katz+Bellmann - Beratende Ingenieure VBIWINGRAF 3397 13.04.99
0 0077
-1.0
0
0.00480.00520.00560.00600.00640.00690.00730.0077
0 00160.00200.00240.00280.00320.00360.00400.00440.00 8
0.00
-0.0012-0.0008-0.00040.00000.00040.00080.00120.0016
1.00
0 0044-0.0040-0.0036-0.0032-0.0028-0.0024-0.0020-0.0016
2.00
-0.0072-0.0068-0.0064-0.0060-0.0056-0.0052-0.0048-0.0044 2
Ein Wassertank M 1 : 25X
Y
Z
-0.0080-0.0076
Verschobene Struktur aus LF 85 Frequenz 0.891 Hertz in 0.100-facherÜberhöhungS itt l t (1 I i t ) i K t L tf ll 85 0 0080
0.00 1.00 2.00 3.00 4.00 5.00
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Spannungsmittelwert (1.Invariante) im Knoten, Lastfall 85, von -0.0080bis 0.0081 Stufen 4.0254e-04 MPa
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Three Gorges Dam ChinaThree Gorges Dam China
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Problem of Water in troughProblem of Water in trough
Ship elevator investigated by Ship elevator investigated by Krebs & Kiefer Karlsruhe with SOFiSTiKS i i f l hi t i t h Seismic response of sloshing water in trough
CFD Model with ca 1 Million cells 3D model of water with ca 1000 lagrange
elements
Katz_03 / Computational Mechanics90
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Comparison CFD / FEMComparison CFD / FEM
Katz_03 / Computational Mechanics91