MR15

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STRUCTURAL AND GEOTECHNICAL ENGINEERING DEPARTMENT

ROCK MECHANICS 2ROCK MECHANICS 2

Giovanni Barla

Politecnico di Torino

LECTURE 15 - OUTLINEThe Finite Element Method2D and 3D Problems - Nonlinear materials

(a) Solution procedures- Incremental method: elasto-plastic model- Initial stress method: no-tension model

(b) Examples

INCREMENTAL APPROACH (this approach is the one most frequently used, as ingeotechnical engineering the interest is often to simulate excavation andconstruction stages in incremental form)

R

uu1 u2

With the first increment (i=1):

[u]1 = [K]0-1 [R]1 (14.1)

With the second increment (i=2):1) Evaluate the tangent stiffnesselement by element based on the state of stress - strain computed at the end of increment 1 and obtain the global system stiffness [K]1

2) We can compute:

[u]2 = [K]1-1 [R]2 (14.2)

R1

KG2

KG1

KG0

a

b b

c c

d

Tangent stiffness solution

Truesolution

Incremental solution

FEM FOR NONLINEAR MATERIALSFEM FOR NONLINEAR MATERIALS

continue

DIAGRAMMATIC REPRESENTATIONOF THE INCREMENTAL APPROACH

R2

With increment i:

[u]i = [K]i-1-1 [R]i (14.3)

where the stiffness matrix [K]i-1 is computed using the stresses and strainsappropriate to the end of increment i. It is possible to use the tangent parameters so as to obtain the tangent global stiffness matrix.

There are different ways for evaluating [K]i-1, with the main purposeto improve the solution to be obtained for each incremental step

It is clear from the diagrammatic representation of the incremental approach that the accuracy of the solution is highly dependent on the number of increments and the size of each increment. As the loading level is increased, one is moving away from the true solution, so thatcare need be exercised with respect to the constitutive law which isintroduced to analyse the problem

continue


2

FLOW DIAGRAM - INCREMENTAL APPROACH

INPUT GEOMETRY, MATERIALS, INITIAL STRESSES

COMPUTE [k]0 ELEMENT BY ELEMENT AND [K]0 FOR THE FEM MODEL

DIVIDE [R] IN INCREMENTS

APPLY INCREMENT [R]iAND COMPUTE [u]i,[]i, ,[]i

i=m YES

NO

STOP

COMPUTE [k]i AND [K]i

continue

FEM ELASTOPLASTIC SOLUTIONincremental approach - example

FEM ELASTOPLASTIC SOLUTIONincremental approach - example

from Potts e Zdravkovic,1999

Displacement u0 is applied The material behaviour is elasto-plastic ideally plastic with the Mohr-Coulomb yield criterion and the associated flow rule:

F([],[k]) = Q([],[m])

Obtain: F([],[k]) --> F([],[k]) = J - ( tanc + p) g()=0

which is written: F([],[k]) =J

+ p) g()ctan(

-1 = 0

h = v = 50 kN/m2

200

400

600

800

1000

J(kPa)

300 600 900 1200

p (kPa)

displacement Increments of displacements with:a = 3% in the vertical direction

elastic: J = 0.866 (p-50) (14.2) at yielding: J = 0.693 p (14.3)J = 0.693 p

J = 0.866 (p-50)

12

34

5

6

E=10 MPa, =0.2, c=0.0, =30=

INITIAL STRESS Approach: applied in two different cases of interest:a) no-tension material b) elastoplastic behaviour

It is once again an iterative approach!!!

a) no-tension material


b) elastoplastic behaviour

continue

EXAMPLE: Axial loading of a specimenEXAMPLE: Axial loading of a specimen

The specimen is loaded axially. The elements shown(i.e. A) attain a no-tension behaviour; the other elementsfollow a linearly elastic law (ILE)

It is known from a FEM-ILE solution in plane strainconditions that element A will undergo a state of stress with 3 < 0 (tensile stress) and 1 > 0 (compressivestress), as shown in the diagram below

Illustration of the initial stress approach

hh

n

n

13

FEM FOR NONLINEAR MATERIALSa) no-tension material


continue

R

A

u=v=0

3

It is obvious that the state of stress in A is not in line with the no-tensionstress strain law. The unbalanced state of stress in A is to be transferredto the surrounding elements as illustrated in the figure below:

nn

nnhhh h

13

Unbalanced state of stress which is tobe transferred from element A to thesurrounding elements

Following a FEM-ILE analysis, the state of stressin A is such that 3< 0. A stress transfer is totake place so that the new state of stress is asshown in A

AA

continue

The stress transfer of (0, 3) can take place as follows:

x

x

yy

3

1 = 0

x = 3 cos2 + y sin2 - 2xysin2

y = 3 sin2 + ycos2 + 2xysin2

xy = xy (cos2 - sin2) + (3 - y ) sincos

From which:

x = 3 cos2

y = 3 sin2

xy = 3/2 sin2

We can write:

[R] =v [B]T [] dVwhere: []T = [x y xy]

We perform a new FEM-ILE analysis, where the model is loaded with theonly system - [R], where the tensile stresses are assumed to be zero

At the end of this step the FEM model may again be subjected to tensile stresses (in our case the A element), which are to be transferred to thesurrounding elements. The process is iterated up to obtaining 3 ~ 0

hh

A

A 3

3

Iteration nr.

1 2 3 4 5

3g

g gg g

g

g

gg

DIAGRAMMATIC REPRESENTATION OF THE STRESS TRANSFER APPROACH

CASE STUDIES

Underground Cavern

elastic solution

no-tension solution

From Zienkiewicz,1971

CRACKING OF A REINFORCED CONCRETE BEAMDISTRIBUTION OF CONCRETE STRESSES AT VARIOUS SECTIONS



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