MR16

1

STRUCTURAL AND GEOTECHNICAL ENGINEERING DEPARTMENT

ROCK MECHANICS 2ROCK MECHANICS 2

Giovanni Barla

Politecnico di Torino

LECTURE 16 - OUTLINEThe Finite Element Method2D and 3D Problems - Nonlinear materials

(a) Solution procedures- Initial stress method: elasto-plastic model

(b) Case Studies(c) A more mathematical approach(d) Incremental - Iterative (mixed) approach(e) Case Studies

“INITIAL STRESS”/”STRESS TRANSFER” APPROACH

The “initial stress”, also called “stress transfer” approachcan be used to solve elasto-plastic problems

Let us first remind the following:

(a) The constitutive law for an elasto-plastic material is written as follows,in incremental form:

[Δσ] = [Cep] [Δε] where [Cep] is the elasto-plastic matrix which is used instead of the stiffness matrix [C] which applies for a linealrly elastic material, always retaining theincremental form.This equation applies when the state of stress is such that:

F([σ],[k]) = 0

If F([σ],[k]) < 0, the material behaviour is elastic (ILE)

(b) The previous constitutive law holds true for small strains. In stress analyses where significant increments of the applied load occur in conjunction with significant increments in strain, the state of stress maybe such that we go far from the yield surface continue

FEM FOR NONLINEAR MATERIALSb) elasto-plastic material

FEM FOR NONLINEAR MATERIALSb) elasto-plastic material

hσy

(Δσ1)el

(Δε1)el

h

h

σ

ε

σ0

ΔR1

ΔR1

“INITIAL STRESS” APPROACHDiagrammatic representation of uniaxial loading

The specimen undergoes an incrementof load ΔR1, starting with the state of stress σ0, assuming that the appliedstress is greater than the yield stress σy

E1

E1

σ1

(Δσ1)ep

(Δσ)1 = (Δσ)1,el - (Δσ)1,ep

h

continue

2

(a) The analysis starts with the application of the load vector [ΔR]1assuming that the initial condition is given by [σ]0; we can compute:

[Δε]1,el e [Δσ]1,el

(b) A new state of stress [σ]1 is computed as sum of [σ]0 present priorto the application of [ΔR]1 and [Δσ]1,el

(c) Based on [Δε]1,el the increment of stresss [Δσ]1,ep, is computed accordingto the elasto-plastic behaviour:

[Δσ]1,ep = [Cep] [Δε]1,el

(d) Then, the following stress difference can be computed:[Δσ]1 = [Δσ]1,el - [Δσ]1,ep

from which:

[R]1 = ∫Ve [B]T [Δσ]1dV,

with [Δσ]1 representing the initial stress

“INITIAL STRESS” APPROACHFrom uniaxial loading to a general case (e) The displacements, strains and stresses due to [R]1 can be computed

as follows, always for the element under consideration:

[Δu]1 = [K]-1 [R]1

[Δε]2 = [B] [Δu]1

[Δσ]2 = [C] [Δε]2 - [Δσ]1

(f) Go back to (b) and repeat steps (b) to (e) up to obtaining a small valueof [Δσ] which can be considered as acceptable

(g) The computation cycle continues up to the maximum stress levelapplied to the FEM model

A MORE MATHEMATICAL APPROACH

Newton iteration process with tangent stiffness

x

ψ(x)

ψ(xn)

dψdx n

xn+1 xn

dxψ(x)n

ψ(x)=0

If a trial value of xn is foundsufficiently close to the correctone but at which ψ(x) # 0, theimproved, trial solution can beobtained by finding:

xn= xn+1 + Δxn+1where:

Δxn+1 = -ψ(xn)/

1

True Solution

It is worth while to re-examine the whole problem mathematically by applying the Newton method of solving general nonlinear equations of the form:

d

x

ψ(x)

ψ(xn)

dψdx

n

(ψ)oddx

Xn+1 Xn

1

An alternative approach would beto use at all stages a constant valueof :

ddx

ψ(x)o

where:Δxn+1 = -ψ(xn)/

Convergence of such a processwould be slower and is illustratedin the figure on the left

Newton iteration process with constant stiffness

A MORE MATHEMATICAL APPROACH

True Solution

The same ideas can fairly obviouslybe extended to nonlinear equationsof many variables

3

A MORE MATHEMATICAL APPROACHNEWTON-RAPHSON MODIFIED METHOD (MNR)

K0

Δui

Δu1

Δu2

ψ1ψ2

ΔR1

True solution

Displacement

load

[K]j [Δu]j = [ψ]j-1

j=iteration number[K]o[Δu]j = [ψ]j-1

• Tangent stiffness:

• Constant stiffness:

Unbalanced load

hhh

hhh0.5

1.0

σmσy

Elastic solution

Elastoplastic solution

Circular hole in a plate

Eεσy

2σm

From Zienkiewicz,1971

“INITIAL STRESS” APPROACHCASE EXAMPLES

Deep circular tunnel Phase2 Solutions

Rpl

σθ

r

p0

σ1σ1

σ3σ3


From Zienkiewicz,1971

Near surface tunnel with lining


4

INCREMENTAL - ITERATIVE (MIXED) APPROACH

ui-1

Ri-1

True solution

displacement

load

Ri

Ri+1

Ui+1ui

h

h

h

A very efficient solution approach for solving nonlinear problems is the incremental - iterative (mixed) approach, where within eachincremental step the iterative approach is applied up to obtaininga nearly negligible value of the unbalanced force

continue

5 cm

2.5 cm

σ1= 743 kPa

σ3=293 kPa

x

y The model to be used is shown in the figure(it comprises 200 Q4 elements - 4 nodes - and231 nodes). Two plane strain analyses wereperformed by using the incremental - iterativeapproach. The constitutive law is of hyperbolictype (see lecture 3)

v=0

u=0

u=0,v=0

Analysis 1Analysis 1


The initial state of stress is hydrostatic(σ1 = σ3=293 kPa). The specimen is loaded in one step only up to σ1 = 743 kPa

The initial state of stress is hydrostatic(σ1 = σ3=293 kPa). The specimen is loaded in three steps up to σ1 = 743 kPa:step 1, Δσ1 = 200 kPastep 2, Δσ1 = 150 kPastep 3, Δσ1 = 100 kPa

Input Parameters

FEM SOLUTION OF A NONLINEAR ELASTIC PROBLEMincremental - iterative approach



100

200

300

400

500

σ1 - σ3 (kPa)600

500 1000 1500 ε1(%)

(σ1 - σ3)f =515 kPa

Iteration Number = 30Δτmax = 5 kPa

Δτmax (kPa)

50

100

10 20 30Iteration nr.




100

200

300

400

500

σ1 - σ3 (kPa)600

500 1000 1500 ε1(%)

(σ1 - σ3)f = 515 kPa


Δτmax (kPa)

50

100

10 20 30

Increment 1

Increment 2

Increment 3

Δσ1 = 200 kPa

Δσ1 = 150 kPa

Δσ1 = 100 kPa

h

h

h

Iteration Number = 18Δτmax = 1.5 kPa




Iteration nr.

MR16

Documents

Transcript of MR16