Finite Element Method in Geotechnical Engineering
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1 Finite Element Method in Geotechnical Engineering Short Course on + Dynamics Boulder, Colorado January 5-8, 2004 Stein Sture Professor of Civil Engineering University of Colorado at Boulder
2 Contents Steps in the FE Method Introduction to FEM for Deformation Analysis Discretization of a Continuum Elements Strains Stresses, Constitutive Relations Hooke s Law Formulation of Stiffness Matrix Solution of Equations
3 Steps in the FE Method 1. Establishment of stiffness relations for each element. Material properties and equilibrium conditions for each element are used in this establishment. 2. Enforcement of compatibility, i.e. the elements are connected. 3. Enforcement of equilibrium conditions for the whole structure, in the present case for the nodal points. 4. By means of 2. And 3. the system of equations is constructed for the whole structure. This step is called assembling. 5. In order to solve the system of equations for the whole structure, the boundary conditions are enforced. 6. Solution of the system of equations.
4 Introduction to FEM for Deformation Analysis General method to solve boundary value problems in an approximate and discretized way Often (but not only) used for deformation and stress analysis Division of geometry into finite element mesh
5 Introduction to FEM for Deformation Analysis Pre-assumed interpolation of main quantities (displacements) over elements, based on values in points (nodes) Formation of (stiffness) matrix, K, and (force) vector, r Global solution of main quantities in nodes, d d D r R k K K D = R
6 Discretization of a Continuum 2D modeling:
7 Discretization of a Continuum 2D cross section is divided into element: Several element types are possible (triangles and quadrilaterals)
8 Elements Different types of 2D elements:
9 Elements Example: Other way of writing: u x = N 1 u x1 + N 2 u x2 + N 3 u x3 + N 4 u x4 + N 5 u x5 + N 6 u x6 u y = N 1 u y1 + N 2 u y2 + N 3 u y3 + N 4 u y4 + N 5 uy 5 + N 6 u y6 or u x = N u x and u y = N u y (N contains functions of x and y)
10 Strains Strains are the derivatives of displacements. In finite elements they are determined from the derivatives of the interpolation functions: or ε xx = u x x = a 1 + 2a 3 x + a 4 y = N x u x ε yy = u y y = b 2 + 2b 4 x + b 5 y = N y u y γ xy = u x y + u y x = (b + a ) + (a + 2b )x + (2a + b N )y = x u + N x y u y ε = Bd (strains composed in a vector and matrix B contains derivatives of N )
11 Stresses, Constitutive Relations Cartesian stress tensor, usually composed in a vector: Stresses, σ, are related to strains ε: σ = Cε In fact, the above relationship is used in incremental form: C is material stiffness matrix and determining material behavior
12 Hooke s Law For simple linear elastic behavior C is based on Hooke s law: C = 1 ν ν ν ν 1 ν ν E ν ν 1 ν (1 2ν)(1+ ν) ν ν ν 2
13 Hooke s Law Basic parameters in Hooke s law: Young s modulus E Poisson s ratio ν Auxiliary parameters, related to basic parameters: Shear modulus Oedometer modulus E E(1 ν) G = E 2(1+ ν) oed = (1 2ν)(1+ ν) Bulk modulus E K = 3(1 2ν )
14 Hooke s Law Meaning of parameters E = σ 1 σ 2 in axial compression ν = ε 3 ε 1 in axial compression E oed = σ 1 ε 1 in 1D compression axial compression 1D compression
15 Hooke s Law Meaning of parameters K = p ε v in volumetric compression G = σ xy γ xy in shearing note: σ xy τ xy
16 Hooke s Law Summary, Hooke s law: σ xx σ yy σ zz σ xy σ yz σ zx = 1 ν ν ν ν 1 ν ν E ν ν 1 ν (1 2ν)(1+ ν) ν ν ν 2 ε xx ε yy ε zz ε xy ε yz ε zx
17 Hooke s Law Inverse relationship: ε xx ε yy ε zz ε xy ε yz ε zx = 1 E 1 ν ν ν 1 ν ν ν ν ν ν σ xx σ yy σ zz σ xy σ yz σ zx
18 Formulation of Stiffness Matrix Formation of element stiffness matrix K e K T = B CBdV Integration is usually performed numerically: Gauss integration n pdv = α i p i i=1 coefficients α and position of sample points can be chosen such that the integration is exact Formation of global stiffness matrix e Assembling of element stiffness matrices in global matrix (summation over sample points)
19 Formulation of Stiffness Matrix K is often symmetric and has a band-form: # # # # # # # # # # # # # # # # # # # # # # # # # # # # (# are non-zero s)
20 Solution of Equation Global system of equations: KD = R R is force vector and contains loadings as nodal forces Usually in incremental form: Solution: D = K 1 R D = n i=1 D K D = R (i = step number)
21 Solution of Equations From solution of displacement D d Strains: ε i = B u i Stresses: σ i = σ i 1 + C d
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