Lecture1(Stress Strain MohrCircle FEM)

8/17/2019 Lecture1(Stress Strain MohrCircle FEM)

1/82

Lecture 1

Stress and Small (Infinitesimal) Strain

Finite Element Method

Mohr Circles


2/82

zz zy zx

yz yy yx

xz xy xx

Representation of 9 stress components in a

Cartesian Coordinate

If the body is in equilibrium, then there is only

6 independent stress components

zx xz zx

yz zy yz

xy yx xy

zz yz xz

yz yy xy

xz xy xx

Stress


3/82

If there is no shear stress on a surface, then the traction vector (think of it as a force vector) must

be parallel to the unit normal vector .

T n̂

multiplier scalar aiswherennT T ˆˆ Rearranging this equation

matrixidentityanis I wheren I 0ˆ

From linear algebra, the condition for the existence of a unique solution is 0det I If we expand the determinant of this equation, we find the characteristic equation (an eigenvalue problem)

0322

1

3 I I I

0ˆ

ˆˆ

3

2

1

n

n

n

zz yz xz

yz yy xy

xz xy xx

If we substitute each of to replace , we can solve for the corresponding eigenvectors321 ,, 321 ˆ,ˆ,ˆ nnn

1ˆˆˆ2

3

2

2

2

1 nnn

zx yz xy z y xT ,,,,,


4/82

4

400

065

058

zz yz xz

yz yy xy

xz xy xx

022 xy yy xx p yy xx p

MPa

MPa

MPa

xy yy xx yy xx

4

90.15684

168

2

1

10.125684

168

2

1

5684168

21

4

1

2

1

3

2

1

22

2

2

1

22

1

2

1

22

2,1

2

1

22

2,1


5/82

1 If we substitute

0ˆ

ˆˆ

1

1

1

1

1

1

z

y

x

zz yz xz

yz yy xy

xz xy xx

n

nn

2221

2221

2221

ˆ

ˆ

ˆ

C B A

C n

C B A

Bn

C B A

An

z

y

x

1

1

zz yz

yz yy A

1

zz zx

yz xy B

yz zx

yy xy

C

1


6/82

6


7/82

Y x A General Equation Constituting an Eigenvalue ProblemIf = eigenvalues, then this equation becomes:

x x A This equation is known as eigenvector equation and it can be represented by the

following homogeneous equations:

rseigenvectoasknownarefor solution10

01e.g.matrixidentity

0

x

I

x I A


8/82

thereforezero,notissince x

0

.

....

.

.

21

22221

11211

nnnn

n

n

A A A

A A A

A A A

I A

This equation is known as the characteristic equation:


9/82

81

24 A

0

81

24

03012

01x284

2

or

551.3

449.8

arerootsor seigenvalueThe

2

1


10/82

To determine the eigenvectors, substitute the eigenvalues into the equation:

08124

2

1

2

1

x

x

which gives an eigenvector corresponding to eigenvalue 1 of 8.449:

0449.881

2449.84

2

1

x

x

1

45.0

2

1

x

x

and an eigenvector corresponding to the second eigenvalue 2 of 3.551:

0551.381

2551.34

2

1

x

x

225.0

1

2

1

x

x

The matrix A is said to be positive-definite, as both its roots are positive.

For nth order matrix, it would involve polynomial of the nth degree


11/82

First Invariant (independent of coordinate system) of Stress Tensor

Second Invariant of Stress Tensor

Third Invariant of Stress Tensor

z y x I 1

222222222

1

2

1 zx yz xy z y x z y x I

222

3 2 xy z xz y yz x zx yz xy z y x I

Second Invariant of Deviatoric Stress Tensor

22222222222226

1

2

1 zx yz xy z x z y y x zx yz xy z y x sss J

Third Invariant of Deviatoric Stress Tensor

1

222

3

3

1

2

I swhere

ssssss J

ii

xy z xz y yz x zx yz xy z y x


12/82

300

00

001 I

m

m

m

m

Deviator Stress

Hydrostatic Stress

m zz yz xz

yzm yy xy

xz xym xx


13/82

Stress Vector zx yz xy z y x

T ,,,,,

Deviatoric Stress Vector3

1

mI s

Vector Equivalent of the

Kronecker Delta 0,0,0,1,1,1T m

Displacement Vector u

Strain Vector zx yz xy z y x

T ,,,,,

Deviatoric Strain Vector

3volme

Octahedral Normal Stress31 I oct

Octahedral Shear Stress2

3

2 J oct

Volumetric Strain z y xvol


14/82

0

0

0

Z z y x

Y

z y x

X z y x

zz yz zx

yz yy xy

zx xy xx

Differential Equation of Equilibrium in Cartesian Coordinate


15/82

Small Strain

zz zy zx

yz yy yx

xz xy xx

z z z

y y y

x x x

z

u

y

u

x

u

z

u

y

u

x

u

z

u

y

u

x

u

u

Representation of 9 strain components in a

Cartesian Coordinate

For symmetry reasons, then there is only 6

independent strain components

z

x

x

z xz zx

y

z

z

y zy yz

x

y

y

x yx xy

uu

uu

uu

2

1

21

2

1

Shear Strain

Normal Strain

Compatibility of Strain to ensure

that a single valued displacement

can be found from integrating the

strain

z x z x

z y y z

y x x y

xz xx zz

yz zz yy

xy yy xx

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

y z x z z y x

x y z y y x z

z x y x x z y

zx yz xy zz

yz xy zx yy

xy zx yz xx

22

2

22

22

2

22

22

2

22

xy xy 2

1


16/82

Deviator Strain

3

3

3

v zz yz zx

yzv yy xy

zx xyv

xx

dev

zz yy xxv Volumetric Strain


17/82

Stress-Strain Relationship

DElasticity Matrix

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

E

)1(2

)21(00000

0)1(2

)21(0000

00

)1(2

)21(000

0001)1()1(

000)1(

1)1(

000)1()1(

1

)21)(1(

)1(

)1(2

E G

Young’s ModulusShear Modulus

Poisson’s ratio

Lame’s Constant

)21)(1()21(

2

E G

Isotropic Material


18/82


19/82

Transverse Isotropic Material E.g., Shale

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

G

E G

E

E

E

E

2

1

2

1

1

2

122

21

21

1

00000

00000

00)1(2000

0000001

0001

1 isotropyof planethetonormalProperties

isotropyof planetheinProperties

2

1

E

E

zx

yz

xy

zz

yy

xx

zx

yz

xy

zz

yy

xx

nm

nm

nn

nnn

nnnnn

n

E

)21)(1(00000

0)21)(1(0000

00)21(2

1000

000)1(

000)1()1(

000)1()()1(

)21)(1(

2

211

2

211

2

21

2

1

12

22

2

12

22

21

2

2

2

211

2

m E

G

n E

E

2

2

2

1


20/82

Plane Stress Problem: Thin Plate Plane Strain Problem: Tunnel, Retaining Wall

000

0

0

yy yx

xy xx

000

0

0

yy yx

xy xx

yy xx xx xxG

2

1

G

xy

xy2

yy xx yy yyG

2

1

yy xx zz

xy

y

x

xy

yy

xx

G

E E

E E

100

01

01

xy

yy

xx

xy

yy

xx E

2100

01

01

1 2

xy

yy

xx

xy

yy

xx E

2

2100

01

01

211


21/82

Principal Stress Space and Yield Criterion

3

1cos 0 All principal stresses are the same (isotropic stress)

Principal stresses are not the same (general condition)

E F D

nnn ˆ,ˆ,ˆThree orthogonal Unit vector along the stress path plane

1

1

1

3

1 ̂Dn

1

1

0

2

1 ̂E n

1

1

2

6

1ˆF n

planeon planeon Along diagonal


22/82

The components of stresses in the D, E and F directions are given by

the inner products of stresses with the three unit vectors:

p D 33

1321

322

1 E

32126

1 F

Mean stress31 I p

qF E 3

2

3

22

1

133221

2

3

22

21

2

122 By Pythagorean Theorem,

Or the deviatoric stress q is

2

1

2

2

1

2

12

13

2

32

2

21

2

1

133221

2

3

22

21

3

)()()(2

1

I I q

q

q

Lode Angle

23

321

3

2tan

E

F


23/82

Yield Criterion for Metal (Tresca in 1864)

)(31 metal for failureat stresstensileT

T F E 223 31

Yield Criterion for Metal (Von Mises in 1913)

Equation of a circle

stressYield AxialY

Y

2

13

2

32

2

21

22


24/82

Mohr Coulomb Yield Criterion (1773)

cos2

131

sin2

1

2

13131

cos2sin1sin1 31 c

321 For

planetheat Looking

F D 3

2

3

11

F E D 6

1

2

1

3

12

F E D 6

1

2

1

3

13

Irregular Hexagonal Shape

321 For sin22cos62sin3sin13 DF E c

The size of the yield surface depends upon

pstressmeanand D

Perspective View

Triaxial CompressionTest

Triaxial Extension

Test


25/82

Normally, Mohr-Coulomb Yield Criterion is expressedin 3-D stress space as:

0cos3

sinsincos

3

sin21

c J I F

=Lode Angle

6

2

333sin

2

3

2

3 with

J

J

Lode Angle

23

321

3

2tan

E

F


26/82

Drucker-Prager Yield Criterion (1952)

Difficult representation

in Sharp Corner

1. Modify Von Mises Yield Criterion so that

sin3

cos6

sin3

sin6

c pq

Normally, Drucker-Prager Yield Criterion is

expressed in 3-D stress space as:

012 k I J F

k

relates to frictional component of shearing resistance

relates to cohesion component of shearing resistance


27/82

2. Modify Mohr Coulomb Yield Criterion by Lade and Duncan (1975) for cohesionless soil (c=0)

constantmaterialais3

321 where p

32222)1(2)3(2)(3

DF E F F E D

3

3

2

2

)sin3(

sin16

)sin3(

sin12

1

where

3. Modify Mohr Coulomb Yield Criterion by Matsuoka and Nakai (1974) for cohesionless soil (c=0)

constantmaterialais)( 133221321 where p

0)()1(23)31(226222323 F E D DF E F

32

32

sinsinsin99

)sinsinsin1(3

where

0

3

21

K F

I

I I

Normally, Matsuoka and Nakai Yield Criterion

is expressed in 3-D stress space as:


28/82

d r d r r r 111 )()(

1sideonForceRadial

d r d r r r 333 )()(

3sideonForceRadial

)2

()()2

sin()()(

2sideonForce Normal

2312

d dr

d r r

)

2

()()

2

sin()()(

4sideonForce Normal

4314

d dr

d r r

dr

r 2)(

4sideonForceShear

dr r 4)(2sideonForceShear

Concept of Stress and Strain in

Plasticity


29/82

direction)tangential(involumeunit per componentforce body

direction)radial(involumeunit per componentforce body

S

R

directionradialinforceEquation of Equilibrium

02

)(2

)()()(

42

4231

dr Rrd dr

d dr

d dr d r d r

r r

r r

Divide both sides by drd

0)()(

2

1)()( 4242

31

Rr

d dr

r r r r r r

r

r r

)(

Area gets smaller

to the limit

d

r


30/82

Divide the equation by r

0)()(2

1)()( 4242

31

Rr d dr

r r r r r r

01

Rr r r

r r r

0

Rr d

r

r r r r r

r

r

r

r

r

r

r

r

)()(

r r

r

r

Expand by product rule

Similarly the equation of equilibrium in the tangential direction

021

S r r r

r r


31/82

01 R

r r r r r r

021

S

r r r

r r

Differential equation of equilibrium in Polar Coordinate

0 X

y x xy x

Differential equation of equilibrium in Cartesian Coordinate

0

Y

x y

xy y

Make use of Airy stress function to find the elastic stress distribution of a circular hole in an

infinite medium under isotropic stress condition

o

i

P

P

PressureExternal

PressureInternal

Differential equation of equilibrium in Cartesian Coordinate


32/82

0

X y x

xy x

q q

0

Y x y

xy y

For no horizontal acceleration, X=0 Body force is simply the weight of the body g

0

y x

xy x 0

g

x y

xy y

Also need the Compatibility Equations in terms of Stresses

x

u x

Components of Strain

y

v y

x

v

y

u xy

Differentiate twice w.r.t. yDifferentiate twice w.r.t. x Differentiate w.r.t. x and then y

y x x y

xy y x

2

2

2

2

2

Arrive the Compatibility Equation in terms of Strain

Use Hook’s Law to transform into Compatibility Equation in terms of Stress


33/82

p y q

)(1 y x x E

)(1 x y y E

xy xy xy

E G )1(21

y x x y

xy x y y x

2

2

2

2

2

)1(2)()(

Substitute into the compatibility equation, we get:

0

y x

xy x

0

g

x y

xy y

Next, to differentiate the equilibrium equation

Differentiate w.r.t. x

Differentiate w.r.t. y and then add the

two equations and it becomes:

2

2

2

22

2 y x y x

y x xy

For Plane Stress Condition:


34/82

The Compatibility Equation in terms of Stresses is obtained by substituting

back into

y x x y

xy x y y x

2

2

2

2

2

)1(2

)()(

2

2

2

22

2 y x y x

y x xy

)1()()(

2

2

2

2

x y

x y y x )(2

2

2

2

y x

y x

)1(2

2

2

2

2

2

2

2

x x y y

x y y x)(

2

2

2

2

y x

y x

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

y x y x x x y y

y x y x x y y x


35/82

02

2

2

2

2

2

2

2

y x x y

y x y x

02

2

2

2

y x y x

The Compatibility Equation in terms of StressesFor Plane Stress Condition:

The Compatibility Equation in terms of Stresses for Plane Stress Condition

including body forces are similarly derived and given as follows:

y

Y

x

X

y x y x 12

2

2

2

F Pl S i C di i


36/82

For Plane Strain Condition:

y x x E 111 2

xy xy xy E G

)1(21

x y y E

111 2

02

2

2

2

y x y x

The Compatibility Equation in terms of StressesFor Plane Stress Condition also the same for

Plane Strain Condition


37/82

The Compatibility Equation in terms of Stresses for any general case with

body forces are similarly derived as follows:

y

Y

x

X

y x y x

1

12

2

2

2

The usual method of solving both the equilibrium and compatibilityequations is to introduce a new stress function (introduced byAiry in 1862, that is why call Airy Stress Function)

0

y x

xy x

0

g x y

xy y

For the equilibrium equations

to be satisfied, it has been shown that the new stress function must also satisfy the following expressions for the stress

components:

gy x

y

2

2

gy y

x

2

2

y x xy

2

S b tit t th i 22 2


38/82

Substitute these expressions

into the Compatibility equation

gy x

y

2

2

gy y

x

2

2

y x xy

2

02

2

2

2

y x y x

The stress function must also satisfy the following expressions:

024

4

22

4

4

4

y y x x

If this stress function can be satisfied, then the equilibriumequation and compatibility equation will also be satisfied .

Thus, the solution of many 2D problems (including body

forces) can be derived by finding a solution which satisfy thestress function incorporating the boundary conditions.

Th t f ti ( ith t b d f ) i l di t


39/82

The stress function (without body force) in polar coordinatesmust also satisfy the following expressions for the stress

components:

2

2

2

11

r r r

r 2

2

r

r r r r r r

111 2

2

The Compatibility Equation in terms of Stresses in polar coordinates is:

01111

2

2

22

2

2

2

22

2

r r r r r r r r

St Di t ib ti S t i l b t A i


40/82

Stress Distribution Symmetrical about an Axis

The Compatibility Equation in terms of Stresses in polar coordinates, when the Stress Function depends on r only, is:

01111

2

2

22

2

2

2

22

2

r r r r r r r r

011211

32

2

23

3

4

4

2

2

2

2

dr

d

r dr

d

r dr

d

r dr

d

dr

d

r dr

d

dr

d

r dr

d

Which is an ordinary differential equation, which can be reduced to a linear differential equation with constant

coefficients by introducing a new variable t such that

The solution has four constants of integration, which must be determined from the boundary conditions.

t er

Assume the stress function to be in the form as:


41/82

Assume the stress function to be in the form as:

The corresponding stress function without body force is obtained:

DCr r Br r A 22

loglog

C r Br

A

r r r 2)log21(

12

2

2

r

0 r

2

2

2

11

r r r r

C r B

r

A2)log23(

2

r r r

1

0 BFor The solution becomeswhich may be used to represent the stress

distribution in a hollow cylinder subjected to

uniform pressure on the inner and outer surfaces

C r

Ar 22 C r

A 22

Apply the boundary conditions: pp


42/82

Apply the boundary conditions: obr r iar r p p

C r

Ar 22

ir pC

a

A 2

2

or pC

b

A 2

2

From which,

22

22

ab

p pba A io

22

22

2ab

b pa pC oi

C r

Ar 22 C r

A2

2 and substitute back into the equation


43/82

22

22

222

22 1

ab

a pa p

r ab

p pba oiior

22

22

222

22 1

ab

a pa p

r ab

p pba oiio

To find the radial displacement, use r uor

r u

r E For Plane Stress Condition

112 234 dddd


44/82

0112

32234

dr

d

r dr

d

r dr

d

r dr

d check

DCr r Br r A 22 loglog

DCr r A 2log

Cr r

Adr

d 2

1

C

r A

dr

d 2

122

2

33

3 12

r A

dr

d

44

4 16

r A

dr

d

4

16

r A

3

12

2

r A

r )2

1(

122 C r

Ar

0)21

(1

3 Cr

r A

r

416r

A 414 r A C r r

A 211 24 0121 24 r C

r A

4

16

r

A 01

64

r

A

Derivation of Elasto-Plastic Solution to compute radial crown displacement of a tunnel under plane


45/82

Derivation of Elasto Plastic Solution to compute radial crown displacement of a tunnel under plane

strain, homogeneous, isotropic stress condition for c’ and ’ material

er

i p o pa

o p

2

r hi h i th di l


46/82

r

r p p er oorr e

which is the radial

stress in the elastic

zone

2

1

r

r p

E r

u er o

r

e

which is the elastic radial displacement

For the stresses in the plastic zone, needs to use the Mohr-Coulomb failure criterion

r

r

r N

sin1

sin1

12

)1(

2)1(

r

r

r

r

r

N

N c

N

N c

pa

r r r i

N

rr

which is the radial stresses

in the plastic zone

r r r r i N

rr cc pa

r r

cotcot

)1(

which is the radial stressesin the plastic zone

sin1

i1


47/82

cossin1

sin1

sin1cossin1

c p

c p

or

or

e

e

which is the radial stress at

the elastic-plastic boundary

Plastic radius1

cot

cotcos)sin1(ln

r r r i

r r o

N

c p

cc p

e aer

cos)sin1( c p p oicrit which is the critical internal

support pressure

Concept of Stress and Strain in Finite Element Formulation


48/82

Plasticity is concerned with predicting the maximum loading which can be

applied to a body without causing:

•Excessive Yield

•Flow

•Fracture

Plasticity Rule is based on 3 assumptions:

1. The yield criterion (F), represented by a surface in the stress space,

at which plastic deformation may develop

2. The hardening law (h being a hardening parameter), that governs the

possible changes in shape, size and position of the yield surface with

an increase in plastic strains

3. The plastic flow rule governing the increment of the plastic strains


49/82

For elasto-plastic materials undergoing infinitesimal deformation,

Total strain increment = Elastic strain increment + Plastic strain increment

plel d d d The elastic strain component can be represented using the generalized Hooke's law:

D = Elastic Constitutive Modulus

Plastic strains are irreversible

plel

d d D Dd d

1st assumption in Plasticity Rule:


50/82

The yield criterion (F), represented by a surface in the stress space, at

which plastic deformation may develop

plhF F ,

h is the vector of the hardening parameters governing the changes of the yield surface with

increasing plastic strains

If F < 0 , material is elastic (stress state within the yield surface)

If F = 0, material is in plastic equilibrium (stress state fulfills the yield criterion or stress

state at the yield surface)

F > 0 Not Admissible (stress state cannot be outside the yield surface)

2nd and 3rd assumptions in Plasticity Rule:


51/82

2. The hardening law (h being a hardening parameter), that governs the

possible changes in shape, size and position of the yield surface withan increase in plastic strains

3. The plastic flow rule governing the increment of the plastic strains

Plastic Flow Rule states that during the plastic strain increment along theyield surface, the plastic strain increment is proportional to the

gradient of the Plastic Potential

Plastic Potential plhQQ ,

Plastic Multiplier Increment d

Q

d d pl

This means that the vector representing the plastic strain increment is directed as the

outward normal vector at the point that corresponds to the current stress state.

In the superimposed space of stresses and strains:


52/82

• The flow rule is called Associated Flow Rule if the yield surface is coincided

with the plastic potential

• The flow rule is called Non-Associated Flow Rule if the yield surface is not

coincided with the plastic potential

Associated Flow Rule Non-Associated Flow Rule

Forming of Elasto-Plastic Constitutive Matrix for Finite Element Analysis


53/82

During plastic strain increment, if the stress points remain on the yield surface, then the

following consistency condition is assumed:

0

pl pl

T T

d h

h

F d

F dF

By substituting the equations relating stress increment to total strain increment and the flow rule,

the following equation is obtained:

0

pl pl

T T

d h

h

F d

F dF

plel d d D Dd d

Q

d d pl

0

d Qh

h

F Q D

F Dd

F pl

T T T

Re-arranging this equation so that the plastic multiplier is a function of the total


54/82

strain as follows:

pe

T

H H

Dd F

d

Q D

F H

T

ewhere

Qh

h

F H

pl

T

p

= so called Hardening Modulus

Finally, substitute the equation describing the plastic multiplier into stress versus strain equation,

the elasto-plastic constitutive matrix governing the incremental direct stress versus strain

relationship in the non-linear range is given below:

d Dd ep

T FQ


55/82

pe

ep

H H

DF Q

D D

D

The matrix is symmetrical if F=Q (Associated Flow Rule)

For Strain Hardening Material, H p>0

For Perfectly Plastic Material, H p=0

For Strain Softening Material, H p < 0

Finite Element Method

I th FEM l i d fi i ti i di ti d i t i l


56/82

• In the FEM, a complex region defining a continuum is discretized into simplegeometric shapes called elements.

• The properties and the governing relationships are assumed over these elementsand expressed mathematically in terms of unknown values at specific points in theelements called nodes.

• An assembly process is used to link the individual elements to the given system.When the effects of loads and boundary conditions are considered, a set of linear ornonlinear algebraic equations is usually obtained.

• Solution of these equations gives the approximate behavior of the continuum orsystem.

• The continuum has an infinite number of degrees-of-freedom (DOF), while the

discretized model has a finite number of DOF. This is the origin of the name, finite

element method.

Examples of Perfectly Plastic Material

k i ld C i i


57/82

Drucker-Prager Yield Criterion

012 k I J F

k

relates to frictional component of shearing resistance

relates to cohesion component of shearing resistance

Mohr-Coulomb Yield Criterion

0cos3

sinsincos

3

sin21

c J I F

=Lode Angle

6

2

333sin

2

3

2

3 with

J

J

Von Mises Yield Criterion

Tresca Yield Criterion

For the particular case of a triaxial compression test, 2=3, the

i b


58/82

equation becomes:

cos2sin3131 c

Discontinuities in the yieldsurface gradient in Mohr-

Coulomb Material : Numerical

difficulties

Matsuoka and Nakai Yield Criterion eliminates

this drawback

0

3

21

K F

I I I

Stress-Strain Constitutive Relationship

For Elastic and Isotropic Material expressed Strain in terms


59/82

For Elastic and Isotropic Material expressed Strain in terms

of Stress (plane stress case)

0

0

0

100

01

01

xy

y

x

xy

y

x

xy

y

x

G

E E

E E

12

'

'

0

E

ModulusShear GratiosPoisson

ModulussYoung E

straininitial

0

1 E

Stress-Strain Constitutive Relationship

For Elastic and Isotropic Material expressed Stress in terms of Strains


60/82

For Elastic and Isotropic Material expressed Stress in terms of Strains

(plane stress case)

0

0

0

2

2100

01

01

1 xy

y

x

xy

y

x

xy

y

x E

0 E stressinitial0 For Elastic and Isotropic Material expressed Stress in terms of Strains

(plane strain case)

0

0

0

2

2100

01

01

211 xy

y

x

xy

y

x

xy

y

x E

Strain and Displacement Relationships (for small strains and small rotations)


61/82

x

v

y

u

y

v

x

u xy y x

,,

v

u

x y

y

x

xy

y

x

0

0

Du or

In Matrix Form

Equilibrium Equations (in elastic theory, stresses must satisfy the following

equilibrium equations)


62/82

0

x

xy x f y x

0

y

y xy f y x

forcesbodyare f and f x x

Boundary Conditions


63/82

The boundary S of the body can be divided

into two parts:

Su (displacement condition) and

St (traction condition) and is described as:

uS onvalueknownvvalueknownu vu _ _

,

t y x S onvalueknown yt valueknown xt t t _ _

,

All types of loads (distributed surface loads, body forces, concentrated

forces and moments) are converted to point forces acting at the nodes

General Formulation of the Finite Element Stiffness Matrix


64/82

Displacement (u, v) inside a plane element are interpolated from nodal

displacement (ui, vi) using shape functions Ni as follows:

d

.

.

....00

...00

2

2

1

1

21

21 N uor v

u

v

u

N N

N N

v

u

vector nt displacemenodal

vector nt displacemeu

matrix functionshape N

d

From the strain-displacement relationship, the strain vector is:

BDND dd


65/82

matrixnt displacemestraintheis

DN Bwhere

Bor DN Du

d ,d

The strain energy stored in an element is:

vol

xy xy y y x x

vol

T dV dV U 2

1

2

1

volT

vol

T dV E dV E U 21

21

kd d 2

1

d d 2

1 T

vol

T T

EBdV BU

The Element Stiffness Matrix k is:


66/82

volT

EBdV Bk

Given the material property E, the behavior of k depends on the B

matrix only, which in turn depends on the shape functions N.

Thus the quality of the finite element analysis is also determined by

the choice of the shape functions.

Constant Strain Triangle (CST)


67/82

Using the strain-displacement relationship, the strain within the element is

related to the displacement at the nodes:


68/82


69/82

69

Other types of elements


70/82

Assemble Global Stiffness Matrix


71/82


72/82

matrixnt displacemestraintheis

DN Bwhere

Bor DN Du

d ,d vol

T EBdV Bk

d

.

.

....00

...00

2

2

1

1

21

21 N uor v

u

v

u

N N

N N

v

u

Du

Stress Calculation


73/82

0

0

0

2

2100

0101

211 xy

y

x

xy

y

x

xy

y

x E

d EB E

xy

y

x

xy

y

x

For linear elastic material, solution is direct and converge at each load step

For non-linear elastic material and material behaves as elasto-plastic, iterative

method is required because the stiffness matrix is non-symmetrical


74/82

vol

epT BdV E Bk


75/82


76/82

Mohr-Coulomb Yield Criterion (Elasto-Plastic Model)


77/82

For non-associated flow rule,

replace internal friction angle bydilution angle


78/82

Sandip Shah,

Ph.D. Thesis,

Department of

Civil

Engineering,

University ofToronto

For Hoek- Brown Yield Criterion (Elasto-Plastic Model)

Carranza-Torresa C and Fairhurst C 1999 The elasto-plastic response


79/82

Carranza-Torresa, C. and Fairhurst, C. 1999. The elasto-plastic response

of underground excavations in rock masses that satisfy the Hoek-Brown

failure criterion. International Journal of Rock Mechanics and Mining

Sciences, vol. 36, pp. 777-809.

Mohr Circles


80/82


81/82


82/82

Lecture1(Stress Strain MohrCircle FEM)

Documents

Transcript of Lecture1(Stress Strain MohrCircle FEM)