6. chapter 6 - stress and strain _a4

7/31/2019 6. Chapter 6 - Stress and Strain _a4

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______________________Basics of the Finite Element Method Applied in Civil Engineering

59

STRESS AND STRAIN DISTRIBUTION IN ELASTIC CONTINUA

6.1 BASIC RELATIONSHIPS

The fundamental problem of elastic continua is to evaluate the stress field of

a massive structure subjected to body forces, distributed loads (or boundary

stresses) and external forces.

The stress state over the domain of the structure is described by the stress

vector

[ ] Tzxyzxyzyx = (6.1)

with x ... zx the stress components, while the strain state is described by the

strain vector

[ ] Txzyzxyzyx = (6.2)

with x zx the strain components.

In the differential formulation, A(u) = 0, B(u) =0, the physicalphenomenon is defined by the following equations:

a. The Navier equilibrium, expressed for an infinitesimal element (the

equilibrium of an elementary volume)

=+

+

+

=+

+

+

=+

+

+

0

0

0

z

zyzxz

y

yzyxy

xxzxyx

fyxz

fxxy

fzyx

(6.3)

withfx,fy andfz the elementary forces acting along thex,y andzaxes.


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Chapter 6 Stress and Strain Distribution in Elastic Continua_________________________

60

Fig. 6.1 Stress, shear strain and displacement notations in global coordinates

b. The strain - displacement relationships are

x

ux

= ;

y

vy

= ;

z

wz

= ;

x

v

y

uxy

+

= ;

y

w

z

vyz

+

= ;

zu

xwzx

+

= (6.4)

z

yx

x

y

0 M

N M

N

0

2

1 zx

xz

xy yx

zy

yz

x

z

y

x y

z

M

w

M

u v

d


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61

where u, v, and w the displacement components along the axes of a 3DCartesian coordinate system.

Following the standard notations in The Elasticity Theory (which implies

small deformations), the strain - displacement relationships are written in a

matrix form:

[ ]d =

=

+

+

+

=

=

w

vu

xz

yz

xy

z

y

x

z

u

x

w

y

w

z

v

x

v

y

uz

w

y

vx

u

zx

yz

xy

z

y

x

0

0

0

00

00

00

(6.5)

with [ ]Twvu=d the vector of Cartesian components of displacementand [ ] a matrix of differential operators (first order derivatives).

c. For elastic continua (which implies a linear elastic behavior of the

material), the stress - strain relationship will be linear with the following

form:

( ) 00 E += (6.6)where:

the stress vector;

0 the vector of initial strains, such as those due to temperature

changes, shrinkage or crystal growth (as alkali-aggregate reactions);

0 the vector of initial residual stresses (as the tectonic stresses

inside a rock mass).


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62

In order to explain the matrix notation, lets consider the particular case oftwo-dimensional (2D) plane stress, with the stress vector

=

xy

y

x

(6.7)

The stress - strain relationships (expressed by the Hookes law) are:

yxxxEE

=1

0,

yxyyEE

1

0, = (6.8)

yxxyxyE

)1(20, +=

By replacing and grouping the terms in relationship (6.8), the elasticitymatrix yields:

=

2

100

01

01

1 2

EE (6.9)

Similarly, for a full three-dimensional (3D) problem, the elasticity matrix

yields:

( )( )( )

+

=

a

a

a

bb

bb

bb

E

00000

00000

00000

0001

0001

0001

211

1

E with

)1(221

=a and

= 1

b (6.10)


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63

Resuming briefly, for the differential approach applied to a strain/stressanalysis:

the unknown function u in A(u) = 0 is the displacements vectord =d(x,y,z);

the Navier equations [ ]d = with the Hookes law E = are thepartial differential equations A(u) = 0;

the boundary conditions B(u) = 0 are expressed as applied forces (ordistributed loads) on the boundaries of the structure and/or

prescribed displacements according to the structures supports.

6.2 THE VARIATIONAL APPROACH

The natural variational principle is attached to the problem. The

corresponding functional is the total potential energy of the system

+

+= V V

T

V

TTdVdVdVE 00 E

2

1

WUddVV

c

TTT +=+++ Rpdfd (6.11)

where:

- the first 3 terms (inside the first pair of brackets) represent the strain

energy Ucaused by the induced stresses, initial stresses 0 and initial

strains 0;- the last 3 terms (inside the second pair of brackets) represent the

potential energy of the external loads W due to body forces f and

distributed loads p; a term as cT

R can be added as the energy due

to the concentrated loads Rc, being the displacements of theapplying points of concentrated loads;

- V is the volume of the domain (the structure) and the boundarysubjected to distributed external loads.


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64

Fig. 6.2 Finite elements model used in a structural analysis

P

P

P

P

PP P

Pz

xy

ph

ph

ph

ph

4

12

3 56

78

P

PP

P

PP P

P

P

zxy


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65

The displacement field d(x,y,z) which satisfies the structural equilibriumunder the loading system and the prescribed boundary conditions has tominimize or make stationary the functionalE, i.e. 0=E .

6.3 THE FINITE ELEMENT EQUATIONS

The finite element model of the given structure implies that the structure isseparated by discretization lines (or surfaces) into a number of finiteelements (see figure 6.2). The displacement function over the domain

subjected to analysis, ),,( zyxd in the 3D space or ),( yxd in the 2D space,

is expressed in terms of nodal displacement i by means of the shape

functions ),,( zyxN . In the general form:

eii zyxzyx NNd == ),,(),,( (6.12)

An example of a 2D approximation is shown in figure 6.3. For the

quadrilateral element [1, 2, 3, 4], considering only one degree of freedom

(DOF) per node, the displacement field over the area is:

44332211 ),(),(),(),(),( NNNNd yxyxyxyxyx +++= (6.13)

Fig. 6.3 Linear displacement approximation over the elements area

Assuming a linear interpolation over the element, i.e. the shape functions

Ni(x,y) are linear (first degree polynomials):

2

13

4

x

y

),( yxd ),( bxd 1

2

3

4


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66

axx )()0,( 121 d +=

a

xabx

+= )(),( 343 d (6.14)

=

=+=4

1

)]0,(),([)0,(),(i

iib

yxbxxyx Ndddd

By replacing, the displacement field yields:

43211),( d

++

+

+=

ab

xy

b

y

ab

xy

ab

xy

a

x

ab

xy

b

y

a

xyx (6.15)

Note that

=

),,(

),,(

),,(

),,(

zyxw

zyxv

zyxu

zyxd is the vector of unknown displacement

components along the coordinates axes of a certain point inside the finite

element, and

=

i

i

i

i

w

v

u

are the displacements of any node i.

The functions ),,( zyxiN have to be chosen such that the exact nodal

displacements i yield when the co-ordinates of the node i are inserted:

IN =),,( iiii zyx (identity matrix) (6.16)

while for all the other nodes the function Ni (x,y,z) vanishes

0NN === .....),,(),,( kkkijjji zyxzyx (null matrix) (6.17)

The strain field. With known displacements all over the element, the strains

at any point can be determined.

[ ] [ ] ee BNd === (6.18)

For the 2D approximation, the derivative operator leads to the following

expressions of strain:


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67

[ ]

=

xy

y

x

0

0

, the derivative operator (6.19)

x

ux

= ; 44332211 uNuNuNuNu +++= ;

44

33

22

11 u

x

Nu

x

Nu

x

Nu

x

Nx

+

+

+

=

(6.20)

y

vy

= ; 44332211 vNvNvNvNv +++= ;

44

33

22

11 v

y

Nv

y

Nv

y

Nv

y

Ny

+

+

+

=

x

v

y

uxy

+

= ;

44

33

22

11

44

33

22

11 v

x

Nv

x

Nv

x

Nv

x

Nu

y

Nu

y

Nu

y

Nu

y

Nxy

+

+

+

+

+

+

+

=

e

xy

y

x

v

u

v

u

v

u

v

u

x

N

y

N

x

N

y

N

x

N

y

N

x

N

y

Ny

N

y

N

y

N

y

Nx

N

x

N

x

N

x

N

B =

=

=

4

4

3

3

2

2

1

1

44332211

4321

4321

0000

0000

(6.21)


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68

With the shape functions already known, the matrix B can be easilyobtained.

The stresses field. For a linear elastic material, the stress field

eBEE == (6.22)

where E is the elasticity matrix, containing the appropriate material

properties. For an isotropic material only the Hooke modulus E and the

Poisson ratio are necessary.

The potential energy of an element. Substituting the main relationships

into the functional corresponding to the elements region (domain):

( ) += VeTT

eeVe

TT

ee dVdVE 02

1BEBB

ddVdVTT

eVe

TT

eVe

TT

e pNfNEB 0 (6.23)

Note that now the functional depends only on the nodal displacements e.

The integrals are expressed in terms of the known functions N and B and in

terms of the known constant values E, 0, 0, fand p.

Using the notations:= Ve

T dVEBBk (6.24)

=+++= ddVdVdVTT

eVe

TT

eVe

TT

eVe

TT

e pNfNEBBr 00

pf rrrr +++= 00 (6.25)

the new expression:

rkT

ee

T

eeE =2

1(6.26)

becomes the total potential energy of the finite element.


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69

The total potential energy of the structure will be obtained by adding thecontributions of all elements in the mesh and the contribution of the external

concentrated forces.

= cTeEE R (6.27)

If we designate as the vector of nodal displacements of the discretized

structure (obtained by an augmentation process, as shown in Chapter 4 the

DSM), then:

c

Tm

Tm

TE Rrk

=

112

1*. (6.28)

With the notations

=m

1

kK ; +=m

c

1

RrR (6.29)

the total potential energy of the structure becomes

RKTTE =

2

1(6.30)

Notethat the functional has the simple, quadratic form.

Remark: If the initial stress system is self equilibrating as must be the case

of normal residual stresses, then 00

= r . However, if an excavation ismade in a rock mass where known tectonic forces are present, a material

removal will cause a force imbalance, which results from the previous term.

The structural equations. The stationary condition for the total potential

energy is:

0

,1

=

= nii

E

(6.31)

which yields to:

* In the augmentation process, the local node number (at element level) isre laced b the lobal node number at structure level


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70

K = R (6.32)

Note: Kis now identified with the stiffness matrix of the structure (called

global or master stiffness matrix) and consequently k is the element

stiffness (or elemental stiffness) matrix; R is the nodal forces vector and

consequently r are nodal forces corresponding to body forces rf, initial

stresses r0, distributed forces rp and initial strains r0 acting on each

element.

The finite element approximation reduces to a problem of minimizing the

total potential energy E defined in terms of a finite number of nodal

displacements. This process leads to the formulation of the simultaneous set

of algebraic equations.

Assessing the strain and stress fields. Once the nodal displacements have

been determined by solving the overall structural equations, the

displacements, strains and the stresses at any point of the structure can be

found. Returning to the element level

d = N e

e = B e (6.33)

e = E B - E 0 + 0

The boundary conditions. In structural analyses, the boundary conditions

are expressed as prescribed support displacements.

Without substitution of a minimum number of prescribed displacements in

order to prevent the rigid body movements of the structure, it is impossible

to solve the structural system, because in such a situation the displacements

can not be uniquely determined by the forces. Mathematically this can be

interpreted as matrix Kbeing singular.

For fixed boundaries, when all displacement components of a node i are

zero, the prescribing procedure means the substitution of


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71

0 =

=

=

0

00

i

i

i

i

w

vu

. (6.34)

This is equivalent to reducing the number of equilibrium equations, by

deleting those corresponding to the ith

node from the system. Moreover, the

corresponding columns are also deleted.

For the prescribed not null displacements of a certain number of nodes

( 0n ), in order to substitute the known displacements, the structural

system should be rewritten by partitioning:

=

=

n

u

n

u

nnu

unu

R

R

KK

KKRK (6.35)

The index u stands for unknown and the index n for known, so that u are

the nodal displacements to be determined and n are the prescribed

displacements.

After partitioning, the system can be written as

nnunnuRKK =+

nnnuun RKK =+ (6.36)

and the first group of equations

{ }*}{][}{}{][ RKRK == nunnnn (6.37)

allows the determination of u . The second group of equations can be used

for nodal reaction assessment, corresponding to the prescribed

displacements.

In order to avoid the reorganization of data in the computer memory (forpartitioning) it is often convenient to proceed with a direct solution of the

original equation system, by using an artifice: the diagonal coefficient of the


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72

matrix Kii at the node concerned is multiplied by a very large number andsimultaneously the term at the right-hand side of the equation is replaced

multiplying the prescribed displacement value by the same large number.

This has the effect of replacing the selected equation with ni KK = ,

specifying that the displacement in question i is equal to the prescribed

value n . If K is large enough i.e. several order of magnitude larger then

iiK , the solution of the structural equation will have ni (i.e. a very

large stiffness spring K is added in the node i and a very large

concentrated force nK is applied along the corresponding direction n).

Example: Applying the variational approach to the 2D truss element

For the two-node truss element the internal energy Ue is

=L

e dxEAU02

1

where the strain is related to the nodal displacements through:

[ ] eee

e

e

ji

e

j

i

j

i

u

u

Lu

u

x

N

x

N

xBu

u=

=

=

= 11

1 ; [ ]11

1=

LB

This form is symmetrically expanded by inserting = B ue into the second

andT

= (ue)

TB

Tinto the first :

[ ] [ ] dxu

u

LEA

LuuU

e

j

e

iL

o

e

j

e

i

e

= 11

1

1

11

2

1

The nodal displacements can be written outside the integrals, giving

[ ] eeTee

j

e

iLe

j

e

i

e )(u

udx

L

EAuuU uKu

2

1

11

11

2

1

0 2=

=

in which


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=

=L TLe dxEAdx

LEA

00 2 1111 BBK

is the element stiffness matrix. If the elongation stiffness EA is constantalong the element,

=

== 11

11

11

1120 L

EAL

L

EAdxEA

LTeBBK

which is the same element stiffness matrix derived using the DSMpreviously.

6. chapter 6 - stress and strain _a4

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