geotechnical finite element lecture note...
Civil Engineering Analysis and Modelling (CIVL3140)
Dr. Zhenhe (Song) Song
[email protected] GHD Pty Ltd
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Part 1 Geomechanics (Plaxis) Dr. Zhenhe Song
[email protected] Part 2 Hydraulics (Fluent) A/Prof. Tongming Zhou (Unit coordinator)
[email protected] Part 3 Structures (Multiframe) Mr. Philip Christensen
[email protected] 2
Yusuke Suzuki
[email protected]
Wensu Chen
[email protected]
Wen Gao
[email protected]
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All the students to set up PLAXIS Version 9 software before tutorial. If you get your laptop this year, you may have PLAXIS 2010, you need to reinstall Plaxis V9 Please try to run PLAXIS in your laptop and make sure it works well. Please ask help from the IT support if you have any problems to open PLAXIS. IT Support: Keith Russell
[email protected]
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2x2hrs sessions per week First 2hrs: Lecture (Theory) Second 2hrs: Tutorials (Practice) 4 weeks in total 6% Weekly Practice; 14% Assignment 40% Exam (combined) 5
This note has incorporated the note from previous teaching by Prof. Yuxia Hu
The development of tutorial questions by Dr. Long Yu
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Finite element analysis in geotechnical engineering: theory, David M. Potts, Lidija Zdravkovi Finite element analysis in geotechnical engineering: application, David M. Potts, Lidija Zdravkovi Guidelines for the use of advanced numerical analysis, David Potts, Kennet Axelsson, Lars Grande, Helmut Schweiger and Michael Long
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Modelling and FEM in Geotechnical Engineering
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Stability Loading on Structure Movement
Footing; Retaining Wall and Deep Excavation; Piles and Bridge Abutment; Embankment, Dams and Seawalls; Tunnel; Stockpile; Dynamic (Seismic Analysis)
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Soils are neither elastic, nor homogeneous. Soils around the world vary. Same soil with different saturations and consolidations behaves differently. Soil properties are difficult to measure. In situ vs laboratory testing …
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New civil engineer: 14/04/2005
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Geotechnical engineering is complex. It is not because you’re using the FEM that it becomes simpler; The quality of a tool is important, yet the quality of a result (mainly) depends on the user’s understanding of both the problem and the tool; The design process involves considerably more than analysis.
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Traditional methods of analysis often use techniques that based on assumptions that over simplify the problem at hand. These methods lack the ability to account for all of the factors and variables the design engineer faces and may severely limit the accuracy of the solution.
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Equilibrium (stress) Compatibility (strain) Constitutive Relationship (stressstrain) Boundary Condition
Solution of Geotechnical Problems
“Exact” or Closed Form
Finite/ Boundary Element
Limit Equilibrium
Boundary Element
Numerical
Finite Difference
Limit Analysis
Finite Element
Empirical, Based on Experience
Discrete Element
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Method of Analysis
Solution Requirements Stress Equilibrium
Compatibility
Limit equilibrium
(P)
X
Slip-line method
(P)
X
Limit Analysis -Lower Bound -Upper Bound
Constitutive behaviour Rigid plastic Rigid plastic
Design Information Stability
Displacements X X
Perfectly plastic
X
X X
X
Displacement finite element
Any P– partially satisfied 18
Receive Design Prescriptions (from a client) Obtain Soil Properties (Site investigations and lab testing) Model Geotechnical Problem Verification Detailed Design Report 19
Numerical modelling Geotechnical model
Silo
http://www.cofs.uwa.edu.au/Researh/centrifugeprojects.html
Physical modelling
http://www.pbase.com/image/41209293
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Plain strain or axisymmetric
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CL Footing (B/2)
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Discretisation (mesh):Divide the model field (soil and/or structure) into parts (nodes and elements) Displacement Approximation: Over each part (element), displacement is expressed as function of nodal values Element Equation: Use an approximate variational principle (e.g. minimum potential energy) to derive an element equation KUE=PE Global Equation: Then assemble the parts of element equation to form a global equation KU=P Boundary Condition: Formulate boundary conditions and modify global equations. Loads affect P, displacement affect U Solutions: Solve displacement values at nodes and then stress and strain can be evaluated 23
CL Footing (B/2)
Element x x
x
Node x
Gauss point (integration point)
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Element Type
Degree of Freedom per Element
Plane Strain
Axisymmetric
Integration rule Gauss point
Constraints per Element
Ratio Degrees of Freedom Constraints
Suitable
Integration rule Gauss point
Constraints per Element
Ratio of Degrees of Freedom Constraints
Suitable
3-noded constant Strain triangle
1
1-point
1
1
Y
3-point
3
1/3
N
6-noded linear Strain triangle
4
3-point
3
4/3
Y
6-point
6
2/3
N
10-noded quadratic Strain triangle
9
6-point
6
3/2
Y
12-point
10
9/10
N
15-noded cubic Strain triangle
16
12-point
10
8/5
Y
16-point
15
16/15
Y
4-noded quadrilateral
2
2x2
3
2/3
N
3x3
5
2/5
N
8-noded quadrilateral
6
3x3
6
1
Y
3x3
9
2/3
N
12-noded quadrilateral
10
4x4
10
1
Y
4x4
13
10/13
N
17-noded quadrilateral
16
5x5
14
8/7
Y
5x5
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16/19
N 25
Sloan, S. W. and Randolph, M. F. (1982) “Numerical prediction of collapse loads using finite element analysis”, Int. J. Num. Ana. Meth. Geo.
(x3, y3) u 3, v 3
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y x
Function: u(x,y) = a1 + a2x + a3y v(x,y) = b1 + b2x + b3y
v u 1 (x1, y1) u 1, v 1
u=?
2 (x2, y2) u 2, v 2
u1 = u(x1, y1) = a1 + a2x1 + a3y1 u2 = u(x2, y2) = a1 + a2x2 + a3y2 u3 = u(x3, y3) = a1 + a2x3 + a3y3
u1 u2 u3
1 x1 1 x2 1 x3
y1 y2 y3
a1 a2 a3
Solve for a 1, a 2, a 3 26
(x 2 y 3 Function of (x,y) N
N1 N2 N3
(x 3 y1 (x1y 2
x 3 y 2 ) x(y 2 y 3 ) y(x 3 x 2 ) 2A x1y 3 ) x(y 3 y1 ) y(x1 x 3 ) 2A x 2 y1 ) x(y1 y 2 ) y(x 2 x1 ) 2A
u1
Function of (x,y)
U
v1 u2
u
N1
0
N2
0
N3
0
v
0
N1
0
N2
0
N3 v2 u3 v3
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(x6, y6) u 6, v 6 6
x
(x1, y1) u 1, v 1
u1 u2 u3 u4 u5 u6
(x5, y5) 5 u ,v 5 5
v u
1
Function: u(x,y) = a1 + a2x + a3y + a4x2 + a5xy + a6y2 v(x,y) = b1 + b2x + b3y + b4x2 + b5xy + b6y2
y
(x3, y3) u 3, v 3 3
u=?
2 (x2, y2) u 2, v 2
4 (x4, y4) u 4, v 4
x1 x2 x3 x4 x5 x6
1 1 1 1 1 1
x12 x 22 x 32 x 42 x 52 x 62
y1 y2 y3 y4 y5 y6
y12 y 22 y 32 y 42 y 52 y 62
x1 y 1 x2 y2 x3 y 3 x4 y4 x5 y 5 x6 y6
a1 a2 a3 a4 a5 au6
1
v1 u2 v2
U
u v
N1 0
0 N1
N 2 0
0 N 2
N 3 0
0 N N 3 0
4
0 N
4
N 0
5
0 N
5
N 0
6
0 N 6
u3 v3 u4 v4 u5 v5 u6 v6
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Strain within an element: Displacement: u(x,y) = a1 + a2x + a3y + a4x2 + a5xy + a6y2 v(x,y) = b1 + b2x + b3y + b4x2 + b5xy + b6y2 Strain: xx
yy
xy
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y x
(b2
a 3 ) (a5
5 u
1
2 4
u a 2 2a 4 x a5 y x v b3 b5 x 2b6 y y u y
v
2b4 ) x (2a 6
e
b5 ) y
B U
e 29
Constitutive Relation Stress and strain can be written in vector form and then expressed as
D Linear isotropic elasticity
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Body forces and surface tractions applied to the element may be generalized into a set of forces acting at the nodes 6 Based on an appropriate variational principle (e.g. minimum potential energy) to derive element equations:
e
K U
e
where
K
e
T
B DBdv
P
e
P1x
5
1
2 4 P1Y
In order to get [Ke], integration (gaussian integration) must be performed for each element. Basically, the integral of the function is replaced by weighted sum of the function at a number of integration points 31
The stiffness for the complete mesh is evaluated by combining the individual element stiffness matrixes assembly) This produces a square matrix K of dimension equal to the number of degree-of-freedom in the mesh The global vector of nodal forces P is obtained in a similar way by assembling the element nodal force vectors The assembled stiffness matrix and force vector are related by:
K U
P
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1 K11 1 K11
1 K12
1 K13
1 K14
1 K 22
1 K 23 1 K 44
1 K 24 1 K 43
1 K12 1 K 22
1 K13 1 K 23
K141 1 K 24
1 K 33
1 K 34 1 K 44
1 K 33
1 K11
K 332
K 342
K 362
K 352
2 K 44
2 K 46 K 662
K 452 K 652 K 552
K121 1 K 22
1 K13 1 K 23 1 K 33
K 332
1 K14 1 K 24 1 K 34 1 K 44
K 342 K 442
K 352 2 K 45
K 362 K 462
K 552
K 562 K 662 33
CL Footing (B/2)
1
4
10x(B/2)
10x(B/2)
2
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Find symmetrical features, central line can be a roller boundary. (CL) (1) Soil domain needs to be large enough to avoid boundary effect. (10x(B/2), 10x(B/2)) The bottom boundary can be fixed boundary. (2) The side boundary can be roller boundary. (3) Top boundary is normally a free boundary. (4) 34
Element size: the smaller, the more accurate Element type: the higher order, the more accurate Boundary conditions: domain size, realistic Constitutive model: complexity
economy
Soil parameters: realistic, measurable Understanding of the real problem
numerical
model 35
Less elements to reduce computation time Smaller elements to increase accuracy
Optimum Mesh Combination of coarse and fine mesh
How ?
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Footing (B/2)
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Displacement control (prescribed displacement) or load control (prescribed load) ? 2-dimensional or 3-dimensional analysis ? Plain strain or axisymmetric ? Drained, undrained or consolidation analysis? Construction Stages
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Pre-processing Define problem(2D or 3D? Plain strain or Axisymmetric? Soil model? Drained or undrained?); define domain (size?); define boundary condition; generate mesh (element type? mesh density?); input soil/foundation parameters (worked out soil parameter from site investigation).
2) Calculation FEM Calculation Steps
3) Post-processing Process calculation results, such as soil stress/strain distribution; soil deformations, et al.
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