Plane problems in FEM analysis. Advanced Design for Mechanical System - Lec 2008/10/092 from the...

Plane problems in FEM analysis

Advanced Design for Mechanical System - Lec 2008/10/09 2

from the real system, to the mechanical

model, to the mathematical model


F=KD

from the real system, to the mechanical

model, to the mathematical model


Introduction

Necessary preliminaries from solid mechanics theory are reviewed.

Plane elements of several types are discussed

Particular attention it is posed to element displacement fields and what they portend for element behaviour.

Treatment of loads and calculation of stress are discussed

The application of plane element to the frame structural connection is presented


Two-dimensional Elements

By definition, a plane body is flat and of constant thickness.

1. Thin or thick plate elements .2. Two coordinates to define position.3. Elements connected at common nodes and/or along common

edges.4. Nodal compatibility enforced to obtain equilibrium equations5. Two types

- Plane stress - Plane strain


Plane stress

x,y ,xy 0

z ,xz ,yz = 0 Tx

y

T

T

x

y

z 0

Plate with a Hole

Plate with a Fillet


1. A state of stress in which normal stress and shear stresses directed perpendicular to the plane of the body are assumed to be zero.

2. If x-y plane is plane of body then only nonzero stresses are: x,y ,xy

3. Zero stresses: z ,xz ,yz

4. Stresses act in the plane of the plate and can be called membrane stresses. They are constant through the z-direction

5. The deformation field is tridimensional (z0 ), the thickness is free to increase or decrease in response to stress in xy plane.

6. The component of the deformation perpendicular to the plane of the body is due to the transversal contraction (Poisson effect).

As the thickness of a plane body increase, from much less to greater than in-plane conditions of the body, there is a transition of behaviour

from plane stress toward plane strain


Plane strain

x , y , xy 0

z , xz , yz = 0

xz

y

Dam Subjected to Horizontal Load

x

z

y

Pipe subjected to external pressure

on its surface

z 0


1. A state of strain in which normal strain and shear strains directed perpendicular to the plane of the body are assumed to be zero.

2. If x-y plane is plane of body then only nonzero strains are: x , y , xy

3. Zero strains: z , xz , yz

4. Two principal stresses act in plane of the body

5. The third principal stress, perpendicular to the plane of the body, depends on the first two principal stresses and on the Poisson coefficient

6. The value of the third principal stress guarantee the deformation perpendicular to plane of the body is zero and prevent thickness change.

7. Stresses are constant through the z-direction


x

y

x

y

xy

xy

xy

xy

dxdy

x

y

Two-dimensional State of Stress

xy

y

x


yx

xyp

2xy

2

yxyx2

2xy

2

yxyx1

σσ

τ2θ2tan

τ2

σσ

2

σσσ

τ2

σσ

2

σσσ

Principal stresses

2

x1

2

1

P


dxx

uu

dyy

vv

dyy

u

dxx

v

x

v

y

u

dy

dx

u

v

A B

D

x, u

y, v

Two-dimensional State of Strain

Displacements and rotations of lines of an element in the x-y plane

x

v

y

uγ

angle

right in the change

y

vε

x

uε

lenght original the

todivided

lenghtin change

xy

y

x


x

y

z

xy

yz

zx

x

y

z

xy

yz

zx

E

( )( ) (1 )

(1 )

(1 )

1 1 2

1 0 0 0

1 0 0 0

1 0 0 0

0 0 0 0 0

0 0 0 0 0

0 0 0 0 0

22

22

22

Constitutive relation

Let the material linearly elastic and isotropic

E = elastic modulus

= Poisson coefficient


yxz

zx

yz

1

0

0Plane stress

x,y ,xy 0

z ,xz ,yz = 0

Dεσ

2

12

00

01

01

1

E

D


0

0

0

00000

00000

00000

0001

0001

0001

)21)(1(

2)21(

2)21(

2)21(

xy

y

x

xz

yz

xy

z

y

x

E

x , y , xy 0

z , xz , yz = 0Plane strain


xy

y

x

xy

y

xE

2

)21(00

01

01

)21)(1(

Dεσ

2

)21(00

01

01

)21)(1(

E

D


Strain-displacement relations

uε

v

u

xy

y

x

x

v

y

uy

vx

u

xy

y

x

0

0

FE theory makes extensive use of strain-displacement relations to obtain the strain field from a displacement field

The strain definitions suitable if the material has small strains and small rotations are


Ndu

3

3

2

2

1

1

321

321

000

000

),(

),(

v

u

v

u

v

u

NNN

NNN

yxv

yxuu

where N is the shape function matrix

Displacement interpolation

Displacements in a plane FE are interpolated from nodal

displacement u(x,y), and v(x,y) as follow:


BdNdε

According to the previous equation

- u depends only on the ui

- v depends only on the vi

- u and v use the same interpolation polynomials

This is a common arrangement but it is not mandatory

From the strain definition, we obtain

where B is the strain-displacement matrixNB


The strain energy per unit volume of an elastic material in terms of strain and in matrix format is:

Upon integrating over element volume V and substituting from

we obtain:

where the element stiffness matrix is

For a given E, the nature of k depends entirely on B , the behavior of an element is governed by its shape functions.

The general formula for k

dV

dVU

dVU

T

TTT

2

1

2

1

2

1

DBBk

kdddDBBd

DεεT

BdNdε

NB


•Let any element d.o.f. , say the i-th d.o.f., be increased from zero to the value di.

•This is accomplished by applying to the d.o.f. a force that increases from zero to Fi..

•The work is Fidi/2, just as it would be stretching a linear spring as amount di.

•This work is stored as strain energy U.

•The previous equation says that work Fidi/2 is equal to strain energy in the element when the displacement field is that produced by di and the element shape functions.

•Example: if di = u1, we can see that the element displacement field is u(x,y)=N1u1 and v(x,y) =0.


• Stresses are in general functions of the coordinates, so that each stress has a rate of change with respect to x and y.

• In a plane problem the rate of change satisfy the equilibrium equations

where Fx and Fy are body forces per unit volume.

• As for deformation, they are called compatible if displacement boundary conditions are met and material does not crack apart or overlap itself.

• If displacement and stress fields satisfy equilibrium, compatibility, and boundary conditions on stress, then the solution is exact.

Nature of the FE approximation

0yx

and 0yx

yyxy

xxyx FF


• Let be elements based on polynomial displacement fields, as for most elements in common use:

- the compatibility requirement is satisfied exactly within elements

- equilibrium equations and boundary conditions on stress are satisfied in average sense

- at most point within the FEA model the equilibrium equations are not satisfied

- as a mesh is repeatedly refined, pointwise satisfaction is approached more and more closely

NOTE This discussion also applies to three-dimensional elastic problems.

How is the exact solution approached by FEA?


1. Two degrees-of-freedom per node.2. These are x and y displacements.3. ui - x displacement at ith node.4. vi - y displacement at ith node.

Triangular Elements

T T

x

y

x

y

i j

m

Thin Plate in Tension Discretized Plate Using Triangular Elements


3

3

2

2

1

1

3

2

1

v

u

v

u

v

u

d

d

d

d

i

ii v

ud

1 (x1, y1)

2 (x2, y2)

3 (x3, y3)

x

y

3 nodes numbered counterclockwise!6 d.o.f.

Constant strain triangle (CST)

The CST element is the earliest and simplest finite element


yxyxv

yxyxu

654

321

,

,

Funzioni di spostamento lineare

1. Ensures compatibility between elements.

2. Displacements vary linearly along any line.

3. Displacements vary linearly between nodes.

4. Edge displacements are the same for adjacent elements if nodal displacements are equal.

In terms of generalized coordinates i, (equal to the d.o.f. of the element) its displacement field is


Strain field

1. We see that strain do not vary within the element; hence the name “constant strain triangle”.

2. The element also be called “linear triangle”, because its displacement field is linear in x and y.

3. Element sides remain straight as the element deform.

The resulting strain field is:uε

5361

xyyx

xyyx x

v

y

u

y

u

x

u


Were xi and yi are nodal coordinates (i=1,2,3)

xij = xi-xj

yij = yi-yj

2A= x21y31-x31y21 is twice the area of the triangle

The sequence 123 must go counterclockwise around the element if the area of the element (A) is to be positive.

The strain field obtained from the shape function, in the form is:Bdε

3

3

2

2

1

1

122131132332

211332

123123

000

000

2

1

v

u

v

u

v

u

yxyxyx

xxx

yyy

Axy

y

x

B


Were t is the element thickness

(plain stress conditions)

The integration to obtain K is trivial because B and E contain only constants

Stiffness matrix

The stiffness matrix for the element is:

tAk TEBB

2

12

00

01

01

1

E

E


Observations

Stress along the x axis in a beam modeled by CSTs and loaded in pure bending

The CST gives good results in a region of the FE model where there is little strain gradient. Otherwise it does not work well.

This is evident if we ask the CSTR element to model pure bending:

-The x-axis should be stress-free because it is the neutral axis;-The FE model predicts x as a square wave pattern.

The element results enable to represent an x that varies linearly with y


CST also develop a spurious shear stress when bent:

- v2 creates a shear stress that should not be present

Despite defects of CST, correct results are approached as a mesh of CST elements is repeatedly refined


6 nodes 12 d.of.

1

2

3

6

45

v2

v4

u2

u4

u3

v3

u5

v5

u6

v6

u1

v1

Linear stress triangle

4

3

2

2

1

1

v

u

v

u

v

u

d

6

5

4

3

2

1

d

d

d

d

d

d

The LST element has midside nodes in addition to vertex nodes.

The d.o.f. are ui and vi at each node i, i=1,2,…6, for a total of 12 d.o.f.


2

12112

10987

265

24321

,

,

yxyxyxyxv

yxyxyxyxu

In terms of generalized coordinates i its displacement field is

yxx

yx

yx

xy

y

x

11610583

12119

542

22

2

2

and the resulting strain field is

The strain field can vary linearly with x and y within the element, hence the name “linear strain triangle” (LST).

The element may also be called a “quadratic triangle” because its displacement field is quadratic in x and y


Shape function

Displacement modes associated with nodal d.o.f. v2=1 and v5=1

•LST has all the capability of CST, few they are, and more.

•The strain x can vary linearly with y

•If problem of pure bending is solved, exact results for deflection and stresses are obtained


Comparison of CST and LST Elements

1 Linear Strain Triangle6 Nodes12 D-O-F

4 Constant Strain Triangles6 Nodes12 D-O-F


120 mm

480 mm

Parabolic Load40 kN (Total)

4 x 16 mesh


Test # of Nodes # of g.d.l. # of Elements

4 x 16 Mesh 85 160 128 CST

8 x 32 Mesh 297 576 512 CST

2 x 8 85 160 32 LST

4 x 16 297 576 128 LST


Run g.d.l

Tip

deflection

(mm)

Stress

(MPa)X-location

Y-location

1 160 0.45834 51.225 0 120

2 576 0.51282 57.342 0 120

3 160 0.53259 59.145 0 120

4 576 0.53353 60.024 0 120

Exact Solution 0.53374 60.000 0 120


Bilinear Quadrilater (Q4)

xyyxyxv

xyyxyxu

8765

4321

,

,

8 d.o.f.

The Q4 element has 4 nodes and 8 d.o.f.



yxy

v

x

u

xy

v

yx

u

xy

y

x

yx

xy

8463

87

42

) oft independen in (linear

) oft independen in (linear

yccxcc 4321

The element strain field is

•The name bilinear arise because the form of the expression for u and v is the product of two linear polynomials

where ci are constants

Strain field


Correct deformation mode of a rectangular block in pure bending:

•Plain section remain plane

•Top and bottom edges became arcs of practically the same radius

•Shear strainxy is absent

Q4 element that bent also develop shear strain

Cannot exactly model a state of pure bending, despite its ability to represent an x that varies linearly with y.

Deformation of a bilinear quadrilater under bending load:

•Top and bottom edges remain straight

•Right angles are not preserved under pure moment load

•Shear strain appear everywhere (y0)


• Cannot model a cantilever beam under transverse shear force where the moment and the axial strain x vary linearly with x.

Qualitative variation of axial stress and average transverse shear stress

Q4 element results too stiff in bending because an applied bending moment is resisted by spurious shear stress as well as by the expected flexural

stresses (locking)


Shape functions

Ndu If the generalized coordinates i are expressed in terms of nodal d.o.f.,we obtain the displacement field in the form where

Shape function N2


Strain field

Bdε Ndε

• Equilibrium is not satisfied at every point unless 4 = 8 = 0

(constant strain)• The element converges properly with mesh refinement and in most

problem it works better than the CST element which always satisfied the equilibrium equations

• Non rectangular shape are permitted.

The element strain field is


a

a

Comparative examples

•The test problem chosen here is that of a cantilever beam of unit thickness loaded by a transverse tip force.

•Plane stress conditions prevail.

•Support conditions are consistent with a fixed end but without restraint of y-direction deformation associated with the Poisson effect.


Results

• Simple beam element solves the problem exactly when transverse shear deformation is included in the formulation.

• As expected, CST elements perform poorly.• Q4 element are better but not good.• LST elements give an accurate deflection but a disappointing stress.

Distortion and elongation of elements are seen to reduce accuracy


Elementi quadrilateri a 8 nodi (Q48)

2

162

152

14132

1211109

28

27

265

24321

,

,

xyayxayaxyaxayaxaayxv

xyayxayaxyaxayaxaayxu

b

ya

x

16 g.d.l.

The Q8 element has midside nodes in addition to vertex nodes.

The d.o.f. are ui and vi at each node i, i=1,2,…8, for a total of 16 d.o.f.



Shape functions (Q8)

iiii vNvuNu

b

ya

x

26

222

112

1

114

111

4

111

4

1

N

N

The displacement field in terms of shape function is

As examples, two of the eight shape functions are

The displacements are quadratic in y, which means that the edge deform into a parabola when a single d.o.f on that edge is nonzero.


Strain field

2

161582

7

136125103

162

15141311

287542

2

22

22

22

yxyx

yx

xyxyx

yxyyy

xy

y

x

dB

x

v

y

uy

vx

u

xy

y

x

•Q8 element can represent exactly all states of constant strain, and state of pure bending, if it is rectangular.

•Non rectangular shape are permitted.

(no term in x2)

….

The element strain field is :


a

a

Final comparison

Q8 elements are the best

performers

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