MR12

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STRUCTURAL AND GEOTECHNICAL ENGINEERING DEPARTMENT

ROCK MECHANICS 2ROCK MECHANICS 2

Giovanni Barla

Politecnico di Torino

LECTURE 12 - OUTLINEThe Finite Element Method2D and 3D Problems

(a) Global equations(b) Global stiffness matrix(c) Boundary conditions(d) Solution of global equations

GLOBAL EQUATIONS - GLOBAL STIFFNESS MATRIXGLOBAL EQUATIONS - GLOBAL STIFFNESS MATRIX

Global Stiffness Matrix - Direct Stiffness MethodGlobal Stiffness Matrix - Direct Stiffness Method

The essence of the direct stiffness method is to assemble theindividual terms of the element stiffness matrix [k]e into theoverall global stiffness matrix [K]G

[K]G [u]G = [R]G (12.1)

To illustrate the stages of the assembly process it is convenientto consider as an example the triangular three noded element CST

(1,2)

(3,4)(5,6)

node number(1,2) degrees of freedom

[k]e=

k11 k12 k16

k61 k62 k66 6x6 34

12

56

1,2,3,,6 define the degrees offreedom for nodes , and (in the local numbering sequence)

a11

a22

a15 a16

a24 a25 a26a23

a34 a35 a36

a44 a45 a46

a56

a66

a55

a12 a14a13K11

a33

K12 K13The matrix for the CST element is a6x6 matrix (the terms in the matrixare called aij and the indices are referredto the degrees of freedom); in order toobtain the global stiffness matrix it isuseful to use the partitioned matrix [kij]3x3, where each term kij representsa 2x2 matrix with indices which arereferred to the nodes

SYMMETRIC


2

As a consequence the stiffness matrix for the CST matrix can be writtenas (with reference to the node numbering):

[k]e =

k11 k12 k13

k22 k23

k33SYMMETRIC

1

2 3

1 2

3

R1R2R3R4R5

=

K11 K12 K13 0 0K22 K23 K24 0

K33 K34 K35K44 K45

K55

u1u2u3u4u5

u1v1

EXAMPLE

SYMMETRIC


GLOBAL STIFFNESS MATRIX - EXAMPLEGLOBAL STIFFNESS MATRIX - EXAMPLE

1 2

1

2

3 5

4 6

Global stiffness matrix for a FEMmesh with two LST elements

ASSEMBLY OF ELEMENT 1

k111 k112 k114 k113k122 k124 k123

k144 k143k133

K111 K112 K113 K114K122 K123 K124

K133 K134K144

SYM SYM


1 2

1

2

3 5

4 6

2

k233 k234 k235 k236k244 k246 k245

k266 k265k255

K111 K112 K113 K114K122 K123 K124

K133+ K233K144+ K244

SYM SYM

K134+ K234 K235 K236K245 K246K255 K256

K266


Global stiffness matrix for a FEMmesh with two LST elements


ASSEMBLY OF ELEMENT

1 2 3

4 5 6

7 8 9

FEM Mesh

1 2 3 4 5 6 7 8 9x x 0 x x 0 0 0 0

x x x x x 0 0 0x 0 x x 0 0 0

x x 0 x x 0x x x x x

xx x

x x 0x 0 x x

12

34567

9

8

BANDWIDTH

symmetric



3

The structure of the assembled stiffness matrixThe structure of the assembled stiffness matrix

1. The global stiffness matrix is symmetric. The non zero terms in this matrix arise only from the connection between degrees of freedom through the elements. Thus for each row in the global stiffness matrix, the lastnon zero term corresponds to the highest degree of freedom to which a particular degree of freedom isconnected.2. The global stiffness matrix is generally sparse (i.e. the matrix has many zero terms) and banded (i.e. the non-zero terms are concentrated along the main diagonal)3. Because of the sparse and banded nature of the stiffness matrix it is not necessary to store the complete stiffness matrix. Reductions in storage can be obtained by taking into account the symmetry of the matrix and only storing the diagonal and upper triangle of terms.

[K]G =

BANDWIDTH


Boundary conditionsBoundary conditions

Once the global stiffness matrix has been established, the boundaryconditions must be added. A large system of simultaneous equations is thuswritten which have to be solved to give values for the nodal displacements.

Loading conditions (line loads and surcharge pressures) case (a):- If line or point forces are prescribed, these can be assembled directly into the righthand side vector [R]G- If pressure boundary conditions are defined, these must first be expressed as equivalent nodal forces before being added to [R]G- Body forces also contribute to [R]G, as do the forces from xcavated and constructedelements (to be discussed in the following)

Displacement boundary conditions (the equation correspondingto the prescribed degree of freedom - displacement component - must beeffectively eliminated during the solution process - see below), case (b)

Loading/Displacement conditions, case (c)

NOTE: in Classical Theory of Elasticity the boundary conditionsand define respectively the First and Second Problem of Elasticity;

descibes the so called Mixed Problem

case (a)

case (b)

case (c)

u=0

u=0

u=0

u=0

v=0

v=0

The stresses on theboundaries are known

The displacements on theboundaries are known

On the surface boundary the stresses are known and on other boundaries the displacements are given

Boundary conditions: examplesBoundary conditions: examples

[K]G [u]G = [R]G (12.1)Consider the equilibriumequations of our system:

where: [K]G global stiffness matrix [u]G nodal displacements vector[R]G nodal forces vector

This can be written as:

=

K11 K12 K1j K1n

Kj1 Kj2 Kjj Kjn

...

...Kn1 Kn2 Knj Knn

u1u2

uj

un

R1R2

Rj

Rnwhere reference is made to the node numbering, so that the stiffness

matrices are much simpler and easier to describe(2 degrees of freedom per node in 2D)

(12.2)

Boundary conditionsBoundary conditions

4

Say that the a certain degree of freedom, for example uj, is prescribed:

uj aj (prescribed value) (12.3)

It follows that the jth equation in the system is not required and can bereplaced by the simpler equation (12.3).

Each other equation contains a term Kij aj (for i=1,2,n) which is alsoknown and can therefore be transferred to the right hand side. Equation (12.3) reduces to:

=

K11 K12 0 K1n ...0 0 1 0 ...

Kn1 Kn2 0 Knn

u1u2

uj

un

R1 -K1j aj ...aj...

Rn -Knj aj

Modification due to displacement boundary conditionsModification due to displacement boundary conditions

Line of symmetry

u=0 u=0

u=v=0

The FEM mesh used to analysethe problem consists of 42eight noded isoparametric elements. Only the soil is discretized. The foundation is represented by appropriate boundary conditions.

y

x

In solving the problem, the nodal displacements along AB, BC and CD are known

A

B C

D

Along the vertical side boundaries AB and CD, u=0: nodal reactions (forces) in the horizontaldirection will be generated as a result of the analysis; for all nodes along the bottom boundary of the mesh (BC) u=0 e v=0: nodal reactions (forces) will be generated in both horizontal and vertical directioss for all nodes along the bottom boundary

On the surface, if the footing is a smooth flexible footing, a uniform surcharge pressure isapplied via the vertical nodal forces Fy to the nodes beneath the position of the footing

Boundary conditions - ExampleBoundary conditions - Example

As there is a vertical plane of symmetrythrough the centre of thefooting, the analysis needs to consider only half of the problem

Solution of global equationsSolution of global equations

Once the global stiffness matrix has been established and the boundaryconditions added, a large system of simultaneous equations is obtained(N equations with N unknowns, where N = nl, with n=number ofnodes in the mesh and l=number of unknowns per node). There are several different mathematical techniques for solving large systems of equations: direct methods (e.g.: Gaussian elimination; most finite elementprograms adopt this method) iterative methods (e.g.: the conjugate gradient method which isshown to be more effective for solving 3D problems)

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