MR13

1

STRUCTURAL AND GEOTECHNICAL ENGINEERING DEPARTMENT

ROCK MECHANICS 2ROCK MECHANICS 2

Giovanni Barla

Politecnico di Torino

LECTURE 13 - OUTLINEThe Finite Element Method2D and 3D Problems

(a) The Computer code(b) Structure discretization(c) Examples:

- Beam problem - Circular opening in a plate (Kirsch problem)

Real Problem

FEM Mesh

Approximate displacements [u]

Compute stiffness matrix [k]e

Equilibrium Equations[R]=[K][u]

Boundary Conditions

Flow Diagram for a typical FEM CodeFlow Diagram for a typical FEM Code

IN OUR DESIGN WORKWE USE THE PHASE2 CODE

(Rock Engineering Group - Toronto)

PHASE2 is a 2D FEM code

The code uses 3 different activities as follows:

MODEL: prepares mesh COMPUTE: performs computations INTERPRET: plots results

[u], [], []

STRUCTURE DISCRETIZATIONSTRUCTURE DISCRETIZATION

y

x

1 2 3 4

5 6 7 8

9 10 11 12

As discussed in Lecture 12, thestructure is discretized and nodes and elements are numbered

The FEM mesh is defined element by element as follows (e.g.):Element Nodes i j k l

2 2 3 7 6

2

NOTES FOR PREPARING THE FINITE ELEMENT MESHNOTES FOR PREPARING THE FINITE ELEMENT MESH

The discretization process need be performed with great care, mainly if the problem to be solved contains curved contours such as in the case of tunnels, when it may be appropriate to adopt high order isoparametric elements (e.g. quadrilateral 8 noded elements)

When the problem domain is characterised by the presence of strain gradients zone (as may be expected near holes, in corner zones, etc.) it is essential to use a high number of elements, in relation to the type of element adopted: CST,LST, etc.

When solving problems which contain non homogeneouszones, it is essential that nodes are located along the contoursfrom one zone to the other one

12 cm

48 cm

x

y

E=20000 kN/cm2, =0.25h=thickness=1 cm

Shear stress with parabolicdistributionEquivalent Load P=40 kN

ANALYSES PERFORMEDMESH C-1 128 elements CSTMESH C-2 512 elements CSTMESH L-1 32 elements LSTMESH L-2 128 elements LST

CST LST

BEAM WITH KNOWN SHEAR LOADING CONDITIONSBEAM WITH KNOWN SHEAR LOADING CONDITIONS

RESULTS FOR CST AND LST ELEMENTSRESULTS FOR CST AND LST ELEMENTS

Displacements and stressesElement Type Mesh Total Number of

UnknownsDisplacement

Stressxat x=12 cm

y= 6 cm

CST

LST

C-1C-2

L-1L-2

160576

160576

0.458340.51282

0.532590.53353

TheoryUpper value 0.53374 60.000

51.22557.342

59.14560.024

NOTES: (1) triangular LST elements are shown to be preferable with respect to CST elements(2) as the displacement approximation is improved with higher degree interpolation

functions, the results obtained compare more favourably with analytical solution

FINITE ELEMENT PROGRAM VALIDATION (see LAB WORK)

Phase2CODE

3

FEM MESH

continue

Main Assumptions:

Circular Opening Req = 2,91 m

Uniform state of stress pv = 5 MPa

Plane strains

Stress ratio k0 = 0,5

Inside Pressure pi = 0

Small strains

No gravity loading

MateriaI:

Homogeneous, isotropic

Linearny elastic

1. Validation for a Homogeneous ContinuousIsotropic Linearly Elastic Material

continue

Kirsch Solution:

Stresses

Displacements

where:

= 0 sidewalls = 90 crown

+

=

+

+

+

+=

+

+=

2senrR3

rR2

12

pp

2cosrR3

12

ppr

R1

2pp

2cosrR3

rR4

12

ppr

R1

2pp

4

4eq

2

2eqhv

r

4

4eqhv

2

2eqhv

4

4eq

2

2eqhv

2

2eqhv

r

( ) ( )

( )

+

=

++

=

2senrR

42pprG4

Ru

2cosppr

R44pp

rG4R

u

2

2eq

vh2

2eq

vh2

2eq

vh2

2eq

r

Closed Form Analytical Solutions

continue

FEM MODEL:

Discretization

Material Properties: - Deformation Mdulus E = 800 MPa- Poissons ratio = 0.18

Numerical Solutions - Assumptions

Analysis 4

Analysis 3

Analysis 2

Analysis 1

210180

122930

95720

60110

NODESNumber of Elements on the Opening Contour

continue

4

Mezzo ILE, k0 = 0,5 - Calotta ( = 0)

0

1

2

3

4

5

6

3 5 7 9 11 13 15 17 19distanza x [m]

r, [M

Pa]

Sigma teta - Kirsch Sigma r - Kirsch

continue

Distance (m)

Stress Distribution at the CrownStress Distribution at the CrownMezzo ILE, K0 = 0,5 - Piedritto ( = 0)

0

2

4

6

8

10

12

3 4 5 6 7 8 9 10 11distanza x [m]

r , [M

Pa]

Sigma teta - Kirsch Sigma r - KirschS i 3 S i 4

continue

Stress Distribution at the WallStress Distribution at the Wall

Distance (m)

Mezzo ILE, K0 = 0,5 - Spostamenti radiali

-1

4

9

14

19

24

29

3 8 13 18 23 28 33 38 43 48distanza x [m]

ur [m

m]

Piedritto - Kirsch Calotta - Kirsch

continue

Distance (m)

Displacement Distribution Displacement Distribution

Wall - Crown -

Mezzo ILE Piedritto ( = 0)

0

2

4

6

8

10

12

3 4 5 6 7 8 9 10 11distanza x [m]

r , [M

Pa]

Sigma teta - Kirsch Sigma r - Kirsch Sigma teta - Phases Sigma r - Phases

= 10,7 MPaerr.=14,5%

r = 1,38 MPa

continue

Distance (m)

Stress Distribution for Analysis 1 Stress Distribution for Analysis 1

5


0

2

4

6

8

10

12

3 4 5 6 7 8 9 10 11distanza x [m]

r , [M

Pa]


= 12,1 MPaerr.= 3.2 %

r = 0,38 MPa

Distance (m)

continue

Stress Distribution for Analysis 2Stress Distribution for Analysis 2 Mezzo ILE Piedritto ( = 0)

0

2

4

6

8

10

12

3 4 5 6 7 8 9 10 11distanza x [m]

r , [M

Pa]


= 12,15 MPaerr.= 2,8 %

r = 0,32 MPa

continue

Distance (m)

Stress Distribution for Analysis 3Stress Distribution for Analysis 3


0

2

4

6

8

10

12

3 4 5 6 7 8 9 10 11distanza x [m]

r , [M

Pa]


= 12,33 MPaerr.= 1,4 %

r = 0,11 MPa

continue

Stress Distribution for Analysis 4Stress Distribution for Analysis 4

Distance (m)

N elementi = 10 N elementi = 80Stress Distribution for Analysis 1 Stress Distribution for Analysis 1 Stress Distribution for Analysis 4 Stress Distribution for Analysis 4

continue

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